The Classification of the Finite Simple Groups, Number 9, Part 9: Part V, Chapters 1-8: Theorem C5 and Theorem C6, Stage 1

Table of contents :
Cover
Title page
Preface
Chapter 1. Introduction to Theorem ?₅
1. Statement of Theorem ?₅
2. The Four Stages of the Proof of Theorem ?₅
Chapter 2. General Group-Theoretic Lemmas, and Recognition Theorems
1. Signalizer Functor Theory
2. Signalizers
3. Fusion
4. ?-Groups
5. Uniqueness Subgroups
6. Miscellaneous
7. Components
8. Rigidity, Semirigidity, and Terminality
9. Centralizers of Components
10. Recognition Theorems
Chapter 3. Theorem ?₅: Stage 1
1. Introduction
2. The Strong Balance Lemma
3. Theorem 1: Balance
4. ?₂(?^{?}) Field Triples
5. Corollaries to Theorem 1
6. Theorem 2: Signalizers in ?
7. The Centralizer of a Sylow 2-Subgroup of ?_{?’}(?)
8. Sufficient Conditions for Faithful Action on ?
9. ?₂(?^{?}) Field Triples Do Not Exist
10. A Covering 2-Local Result
11. Theorem 3: Γ_{?,2}^{?}(?)\le?
12. Theorem 4: Γ_{?,2}(?)\le?
13. Theorem 5
Chapter 4. Theorem ?₅: Stage 2
1. Introduction
2. The Principal Subsidiary Theorems
3. Some Generalities
4. Uniqueness Subgroups from ?-Terminal ?-Components
5. Some Sporadic ?-Components
6. Theorem 1: Generalities
7. Theorem 1: ?-Terminal ?-Components
8. Theorem 1: The ?-Terminality of ?₆
9. Theorem 1: Standard 2-Components
10. Theorem 2
11. Corollaries: \BtK^{?}(?), and Components in ?_{?}-??ℯ?(?)
12. Theorem 3: Regular Triples and Mates
13. Theorem 3: ? Must Be 3
14. The Nondegenerate Case: Theorem 4
15. Theorem 5: The Degenerate Case ?>?
16. Theorem 5: The Degenerate Case ?=?
17. Theorem 5: The Degenerate Case \overline?2
8.4. Generation with respect to Non-elementary ?-Groups, ?>2
8.5. Other
9. ?-Structure, ? odd
9.1. Self-centralizing ?_{?²}-Subgroups, ?≥5
9.2. Sylow ?-Subgroups ?, and ?(?) and ?(?)
9.3. Other ?-Subgroups
9.4. Other
10. 2-Structure
10.1. Orthogonal Groups over ?₃
10.2. Sylow 2-Subgroups and Their Overgroups in Quasisimple ?-Groups
10.3. Other Involution Centralizers
10.4. Other
11. ?₂-Groups and ?_{?}-Groups, ? odd
12. {2,?}-Structure
12.1. Wide 2-Components
12.2. Subgroups of Order 2?
12.3. Components for Permutable Subgroups of Orders 2 and ?
12.4. 2-Local ?-Rank
12.5. Mates
12.6. ?-Components in ?₂-Groups
12.7. Other
13. {2,3}-Neighborhoods
14. Semirigidity
14.1. Conditions for Semirigidity
14.2. ?₄-Semirigidity
15. Miscellaneous
16. Preliminary Lemmas for Theorem ?₆*: Stage 1
16.1. ?-Ranks, ?_{?}ⁱ-Groups, and Flat ??_{?}-Groups
16.2. Pumpups and Subcomponents
16.3. 3-Structure
16.4. Other
Bibliography
Index
Back Cover

Citation preview

M ATHEMATICAL

Surveys and Monographs Volume 40, Number 9

The Classification of the Finite Simple Groups, Number 9 Part V, Chapters 1–8: Theorem C5 and Theorem C6 , Stage 1 Inna Capdeboscq Daniel Gorenstein Richard Lyons Ronald Solomon

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10.1090/surv/040.9

The Classification of the Finite Simple Groups, Number 9 Part V, Chapters 1–8: Theorem C5 and Theorem C6 , Stage 1

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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M ATHEMATICAL

Surveys and Monographs Volume 40, Number 9

The Classification of the Finite Simple Groups, Number 9 Part V, Chapters 1–8: Theorem C5 and Theorem C6 , Stage 1 Inna Capdeboscq Daniel Gorenstein Richard Lyons Ronald Solomon

Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

Editorial Board Robert Guralnick Benjamin Sudakov Michael A. Singer, Chair Constantin Teleman Michael I. Weinstein The authors gratefully acknowledge the support provided by grants from the National Security Agency (H98230-07-1-0003 and H98230-13-1-0229), the Simons Foundation (425816), and the Ohio State University Emeritus Academy. 2020 Mathematics Subject Classification. Primary 20D05, 20D06, 20D08; Secondary 20E25, 20E32, 20F05, 20G40. For additional information and updates on this book, visit www.ams.org/bookpages/surv-40.9

The ISBN numbers for this series of books includes ISBN ISBN ISBN ISBN ISBN ISBN ISBN ISBN ISBN

978-1-4704-6437-0 978-1-4704-4189-0 978-0-8218-4069-6 978-0-8218-2777-2 978-0-8218-2776-5 978-0-8218-1379-9 978-0-8218-0391-2 978-0-8218-0390-5 978-0-8218-0334-9

(number (number (number (number (number (number (number (number (number

9) 8) 7) 6) 5) 4) 3) 2) 1)

Library of Congress Cataloging-in-Publication Data The first volume was catalogued as follows: Gorenstein, Daniel. The classification of the finite simple groups / Daniel Gorenstein, Richard Lyons, Ronald Solomon. p. cm. (Mathematical surveys and monographs: v. 40, number 1–) Includes bibliographical references and index. ISBN 0-8218-0334-4 [number 1] 1. Finite simple groups. I. Lyons, Richard, 1945– . II. Solomon, Ronald. III. Title. IV. Series: Mathematical surveys and monographs, no. 40, pt. 1–;. QA177 .G67 1994 512.2-dc20 94-23001 CIP Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to [email protected]. c 2021 by the American Mathematical Society. All rights reserved.  The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America. ∞ The paper used in this book is acid-free and falls within the guidelines 

established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

26 25 24 23 22 21

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Dedicated to the memory of John H. Conway, Bernd Fischer, Simon Norton, Michael O’Nan, and Charles Sims

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Contents Preface

xi

Chapter 1. Introduction to Theorem C5 1. Statement of Theorem C5 2. The Four Stages of the Proof of Theorem C5

1 1 3

Chapter 2. General Group-Theoretic Lemmas, and Recognition Theorems 1. Signalizer Functor Theory 2. Signalizers 3. Fusion 4. p-Groups 5. Uniqueness Subgroups 6. Miscellaneous 7. Components 8. Rigidity, Semirigidity, and Terminality 9. Centralizers of Components 10. Recognition Theorems

11 11 13 15 16 19 20 23 24 27 29

Chapter 3. Theorem C5 : Stage 1 1. Introduction 2. The Strong Balance Lemma 3. Theorem 1: Balance 4. L2 (pp ) Field Triples 5. Corollaries to Theorem 1 6. Theorem 2: Signalizers in M 7. The Centralizer of a Sylow 2-Subgroup of Op (M ) 8. Sufficient Conditions for Faithful Action on T 9. L2 (pp ) Field Triples Do Not Exist 10. A Covering 2-Local Result 11. Theorem 3: ΓoP,2 (G) ≤ M 12. Theorem 4: ΓP,2 (G) ≤ M 13. Theorem 5

33 33 39 39 44 49 53 54 59 60 62 64 74 81

Chapter 4. Theorem C5 : Stage 2 1. Introduction 2. The Principal Subsidiary Theorems 3. Some Generalities 4. Uniqueness Subgroups from p-Terminal p-Components 5. Some Sporadic p-Components 6. Theorem 1: Generalities vii Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

89 89 91 93 95 102 103

viii

CONTENTS

7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

Theorem 1: A-Terminal p-Components Theorem 1: The A-Terminality of A6 Theorem 1: Standard 2-Components Theorem 2 Corollaries: BtKp (G), and Components in Cp − Chev(p) Theorem 3: Regular Triples and Mates Theorem 3: p Must Be 3 The Nondegenerate Case: Theorem 4 Theorem 5: The Degenerate Case K > I Theorem 5: The Degenerate Case K = I Theorem 5: The Degenerate Case I < J Theorem 5: The Degenerate Case I = J

105 115 123 143 144 146 158 160 164 173 174 192

Chapter 5. Theorem C5 : Stage 3 1. Introduction 2. The Principal Subsidiary Theorems 3. Theorem 1 4. Maximal Symplectic Triples 5. Theorem 2: p = 3 and Constrained Neighborhoods 6. Theorem 2: Nonconstrained Neighborhoods of Type F i24 7. Theorem 2: Nondegenerate Constrained Neighborhoods 8. Theorem 2: Constrained Neighborhoods of Sporadic Type

209 209 210 211 219 224 226 227 231

Chapter 6. Theorem C5 : Stage 4 1. Introduction 2. The Case G∗ = Ω7 (3) or Ω± 8 (3) 3. The Case G∗ = F i22 4. The Case G∗ = F i23 5. The Case G∗ = F i24 or F i24 6. The Case G∗ = F2 7. The Case G∗ = Co1 8. The Case G∗ = F1 9. The Case G∗ = 2 D5 (2), 2 E6 (2), or U7 (2)

237 237 237 279 282 283 288 291 291 292

Chapter 7. Theorem C∗6 : Stage 1 1. Theorems C6 and C∗6 2. The Proof of Theorem C7 3. Theorems 1 and 1∗ 4. Theorem 2 5. Theorem 2 Completed: The U3 (8) Case

309 309 314 315 317 325

Chapter 8. Preliminary Properties of K-Groups 1. Small Groups 2. Outer Automorphisms and Covering Groups 3. Pumpups and Subcomponents 3.1. Key Generalities 3.2. Pumpups of Specific Groups 3.3. Pumpups and p-Rank 3.4. Pumpups and Schur Multipliers 3.5. Pumpups of A6

343 343 345 347 347 349 353 356 357

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CONTENTS

4. 5. 6. 7.

8.

9.

10.

11. 12.

13. 14.

15.

3.6. p-Subcomponents and q-Subcomponents, p = q 3.7. Action of CAut(K) (x) on E(CK (x)) 3.8. Pumpups and the Sets Cp , p Prime 3.9. Solvable Components 3.10. Other Computations in Groups of Lie Type Representations Factorizations Signalizers and Balance 7.1. Signalizers 7.2. Balance and Cores Generation 8.1. Generation with respect to Non-elementary p-Groups, p=2 8.2. Generation with respect to Elementary Abelian p-Groups, p=2 8.3. Generation with respect to Elementary Abelian p-Groups, p>2 8.4. Generation with respect to Non-elementary p-Groups, p>2 8.5. Other p-Structure, p odd 9.1. Self-centralizing Ep2 -Subgroups, p ≥ 5 9.2. Sylow p-Subgroups P , and Z(P ) and J(P ) 9.3. Other p-Subgroups 9.4. Other 2-Structure 10.1. Orthogonal Groups over F3 10.2. Sylow 2-Subgroups and Their Overgroups in Quasisimple K-Groups 10.3. Other Involution Centralizers 10.4. Other C2 -Groups and Cp -Groups, p odd {2, p}-Structure 12.1. Wide 2-Components 12.2. Subgroups of Order 2p 12.3. Components for Permutable Subgroups of Orders 2 and p 12.4. 2-Local p-Rank 12.5. Mates 12.6. p-Components in C2 -Groups 12.7. Other {2, 3}-Neighborhoods Semirigidity 14.1. Conditions for Semirigidity 14.2. Z4 -Semirigidity Miscellaneous

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ix

359 361 368 378 379 380 384 393 394 394 397 399 399 400 403 415 422 424 424 426 430 434 435 435 446 456 461 463 469 469 472 477 482 484 485 490 492 495 495 497 497

x

CONTENTS

16. Preliminary Lemmas for Theorem C∗6 : Stage 1 16.1. p-Ranks, Tpi -Groups, and Flat TGp -Groups 16.2. Pumpups and Subcomponents 16.3. 3-Structure 16.4. Other

505 505 506 510 512

Bibliography

515

Index

519

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Preface With this book we begin Part V of our program, the analysis of the special even case. Three theorems belong to that Part: C4 , C5 , and C6 . All three assume that the K-proper simple group G being analyzed is of even type. Theorem C4 , to be treated in a later book, concerns the case e(G) ≤ 3. Thanks to the monumental Quasithin Theorem of M. Aschbacher and S. D. Smith [ASm1] [ASm2], which we now include in our Background Results, it is really the case e(G) = 3. This book contains a complete proof of Theorem C5 , which covers the “bicharacteristic” subcase of the e(G) ≥ 4 problem. The outcome of this theorem is that G is isomorphic to one of the six sporadic groups for which e(G) ≥ 4, or one of six groups of Lie type which exhibit both characteristic 2-like and characteristic 3-like properties. Finally, in Chapter 7, we begin the proof of Theorem C6 and its generalization Theorem C∗6 , which cover the “p-intermediate” case. This case will be shown in the end to be satisfied by no K-proper simple group G. In this case, for some odd prime p and x ∈ G of order p such that mp (CG (x)) ≥ 4, CG (x)/Op (CG (x)) has a component K which is neither in the “generic” set Gp nor in the “characteristic plike” set Cp , and satisfies a condition close to p-terminality. In particular, K might have cyclic Sylow p-subgroups (with a couple of exceptions). In the preceding book in this series, we had promised complete proofs of Theorems C6 and C∗6 in this book, but because of space considerations, we postpone the completion of those theorems to the next volume. This volume has had a long gestation period, a zeroth draft of Theorem C5 having been written by the second and third authors in the early 1990s, suggesting the concept of groups of bicharacteristic type as a unifying approach to a local characterization of many of the large sporadic simple groups. The prehistory of this approach dates to John Thompson’s second paper [Tho70] on simple N -groups, where he characterized the groups P Sp4 (3) and G2 (3) by their 2-local and 3-local structures. Indeed, P Sp4 (3) is the primordial bicharacteristic group, with incarna tions as U4 (2), Ω5 (3) and Ω− 6 (2), as well as W (E6 ) and the group of the 27 lines in a cubic surface; and it is a progenitor of the Fischer groups, which are prominent in this volume. Not long after Thompson’s work, K. Klinger and G. Mason [KMa1] proved the following theorem.

Theorem. Let G be a finite group of characteristic 2-type with F ∗ (G) = E(G). Let p be an odd prime and suppose that G is also of characteristic p-type. Then every 2-local subgroup of G has p-rank at most 2. xi Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

xii

PREFACE

A finite group G is called quasithin if the 2-local p-rank of G is at most 2 for every odd prime p. Thus, an immediate corollary of the Klinger-Mason Theorem is the following statement. Corollary. Let G be a finite group of characteristic 2-type with F ∗ (G) simple. Then either G is quasithin or there exists an odd prime p such that G has 2-local p-rank ≥ 3 and G is not of characteristic p-type. In the late 1960s, Gorenstein and J. H. Walter pioneered a strategy for the classification of finite simple groups which are not of characteristic 2-type. Their strategy was implemented by a large team of group theorists, mostly during the 1970s. Later, Gorenstein and Lyons [GL1] developed a strategy for the classification of finite simple groups (CFSG) which are of characteristic 2-type but for which there exists an odd prime p such that G has 2-local p-rank ≥ 4 and G is not of characteristic p-type. The above corollary shows that, in conjunction with a classification of simple groups of characteristic 2-type that are either quasithin or satisfy e(G) = 3, this would complete the CFSG. And indeed this strategy was implemented in the first proof of CFSG, with the e(G) = 3 theorem of Aschbacher and the Quasithin Theorem of Aschbacher and Smith. From Thompson’s original N -Group argument through Klinger-Mason up to the Trichotomy Theorem of Gorenstein-Lyons [GL1], a significant subcase of the analysis was the so-called “GF (2)-type” case, i.e., the case when a 2-central involution z of G has the property that F ∗ (CG (z)) is a 2-group of symplectic type, in the sense of Philip Hall. As an independent problem (sometimes called the O2 extraspecial problem), this generated a large body of research, starting with the work of Z. Janko and his students and culminating in classification theorems by F. G. Timmesfeld, S. D. Smith and others in the mid-1970s. This entailed characterization theorems for many of the sporadic simple groups, as well as most simple groups of Lie type defined over the field F2 . In the 1980s, with the first proof of CFSG largely complete, Gorenstein and Lyons began a rethinking of the classification strategy. They observed that the endgame for a unifying methodology for the classification of most of the alternating groups and groups of Lie type was to classify simple groups G having an element x of prime order p such that CG (x) has p-rank ≥ 3 and CG (x) has a p-component K with K/Op (K) either an alternating group (with p = 2) or a quasisimple group of Lie type of characteristic r = p. Of course this endgame fails to include the sporadic simple groups. Many of these are quasithin groups, but most of the remaining ones have elements of order p whose centralizers are either p-constrained or have components which are either of Lie type in characteristic p or sporadic. This motivated the replacement of the dichotomy between characteristic 2type and non-characteristic 2-type groups with the dichotomy between groups of even type and those of odd type. While all involution centralizers in groups of characteristic 2-type are 2-constrained, involution centralizers in groups of even type are allowed to have components K, but only with K of Lie type in characteristic 2 (plus or minus a small finite number of groups, mainly sporadic). It is still required that involution centralizers be core-free. A similar notion of p-type emerged for odd primes p, allowing centralizers of elements of order p with p-components K where K/Op (K) is of Lie type in characteristic p (again plus or minus a handful of groups). Here Op (CG (x)) is required only to have odd order for x of order p. From this point of view, the residual case arising in the endgame of their classification

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PREFACE

xiii

strategy was the Bicharacteristic Case, i.e., the case when G is a finite simple group of even type for which there exists an odd prime p such that G has 2-local p-rank ≥ 3 and G is of p-type. A draft treating the Bicharacteristic Case, with some additional simplifying hypotheses, most notably that G has 2-local p-rank at least 4, was completed by Gorenstein and Lyons in 1992, the outcome being the five groups F i23 , F i24 , Co1 , F2 , and F1 . A strengthened version (but still only for 2-local p-rank ≥ 4) is Theorem C5 , which is the principal content of the current volume. In addition to the five sporadic groups just mentioned, F i22 appears, as do six groups of Lie type. Careful definitions and a precise statement appear in Chapter 1. A word about recognition theorems: the six large sporadic groups in the conclusion of Theorem C5 are recognized by well-established characterizations by M. Aschbacher, R. Griess, U. Meierfrankenfeld, and Y. Segev. Our original Background Results included the statement that these groups were characterized by their full centralizer of involution patterns. However, that is an impractical approach, given the above-mentioned characterizations, especially since some such characterization theorems would still eventually be needed to obtain the six groups up to isomorphism. Thus, we change the Background Results as follows. We add two books by Aschbacher [A2], [A19], and two papers, one by Segev on the Baby Monster [Seg1] and one by Griess, Meierfrankenfeld and Segev, on the Monster [GrMeSeg]. (And we remove the result that these six groups are characterized up to isomorphism by their centralizer of involution patterns.) What is needed from these publications are centralizer of involution characterizations of the six large groups, plus a couple of results from Aschbacher’s Sporadic Groups [A2] relating to Co2 , the binary Golay code, and the Todd module for M24 . We have now supplemented the Background Results listed in our initial volume [I1 ] with these four references, the Quasithin Theorem of Aschbacher and Smith [ASm1] [ASm2], and the four references mentioned in the Preface to the previous volume in this series [GLS8]. The six large sporadic groups are in some sense on the border between largerank generic and non-generic groups. When we round them up in Theorem C5 , we also capture six groups of Lie type which can be considered to lie on that 2 2 border as well: Ω7 (3), P Ω± 8 (3), U7 (2), D5 (2), and E6 (2). Characterizing the first three of these – the orthogonal groups over F3 – was one of the last details to be attended to in the original proof of the classification, and was accomplished by Aschbacher [A23]. We have relied heavily on that paper to get from a standard component isomorphic to L± 4 (3) or 2U4 (3) to an endgame configuration. For the final recognition of these and the last three groups of Lie type over F2 , we use tools similar to those used in our previous volume [GLS8], namely, the Curtis-Tits theorem as well as methods of Wong and Finkelstein-Solomon [GLS8, pp. 18–32]. As usual, we continue the notational conventions established in the second volume of this series [IG ]. We refer to the first seven chapters of this book as [V1 ], . . . , [V7 ], and the eighth chapter as [VK ]. As usual, the chapter [VK ] consists of supporting K-group lemmas for the main chapters [V3 ]–[V7 ], and thus logically precedes them. During the preparation of this book, the first author received support from the Leverhulme Trust, and the fourth author received support from a Simons Foundation Grant. We gratefully acknowledge that support. Inna Capdeboscq, Richard Lyons, and Ronald Solomon September, 2020 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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10.1090/surv/040.9/01

CHAPTER 1

Introduction to Theorem C5 1. Statement of Theorem C5 Having completed Part III of the classification proof in volume 8 [GLS8] (modulo Part II, in the case of groups of even type), and Part IV in volume 6 [GLS6], we now take up consideration of Part V, the special even case1 . In this part are three main theorems, C4 , C5 , and C6 . An enormous part of Theorem C4 is contained in the Quasithin Theorem for groups of even type, proved by M. Aschbacher and S. D. Smith, now included in our Background Results [ASm1], [ASm2]. The remaining portion of Theorem C4 , the classification of K-proper simple groups of even type with e(G) = 3, will be treated in a later volume of this series. The main subject of this volume is Theorem C5 , which is of particular interest since it is here that large sporadic groups emerge from the analysis – namely, the Monster F1 , the Baby Monster F2 , the large Conway group Co1 , and the three Fischer groups F i22 , F i23 , and F i24 . (Though the group J4 has huge order, it is quasithin.) We also state Theorem C6 and a companion Theorem C∗6 in Chapter 7, and begin their proofs there. Thus in this introductory section, we confine ourselves to discussing Theorem C5 . We shall state the result first, then review the relevant terminology and notation. Formal definitions will follow in later chapters. Theorem C5 . Assume that G is a K-proper simple group satisfying the following conditions: (a) G is of restricted even type; (b) σ0 (G) = ∅, and p ∈ σ0 (G); and (c) Lop (G) ⊆ Cp . Then p = 3 and G is isomorphic to one of the following twelve groups: − (a) Ω7 (3), P Ω+ 8 (3) or Ω8 (3); − 2 (b) U7 (2), Ω10 (2), or E6 (2); or (c) Co1 , F i22 , F i23 , F i24 , F2 , or F1 .

Recall that in this series, G is always a K-proper simple group, and “group” always means “finite group.” We review some terminology from [I2 ], noting in particular a change in the definition of σ0 (G). This is the first place in our series where this change affects the mathematics. In particular, the primes p ∈ σ0 (G) will, among other properties, be odd primes for which mp (G) ≥ 4, where mp (G) is the p-rank of G. 1 Accordingly,

we refer to the 8 chapters of this volume as V1 , V2 , V3 , V4 , V5 , V6 , V7 , and

VK . 1 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

1. INTRODUCTION TO THEOREM C5

2

Recall that Ip (X) is the set of all elements of order p in the group X. Moreover, (a) I2o (X) = {x ∈ I2 (X) | m2 (CX (x)) ≥ 3}, and   (b) If p is an odd prime, then Ipo (X) = x ∈ Ip (X)  mp (CX (x)) ≥ 4 . Recall that K = ∪r Chev(r) ∪ Alt ∪ Spor = Chev ∪ Alt ∪ Spor is the set of all quasisimple K-groups: the simple groups of Lie type, the alternating groups An , n ≥ 5, the 26 sporadic groups, and all covering groups of these simple groups. For example, Chev(r), defined only for primes r, is the set of all quasisimple groups of Lie type in characteristic r. For any prime p, Kp is the set of all K ∈ K such that Op (K) = 1. Next, Lop (G) is the subset of Kp consisting of those groups K such that for some x ∈ Ipo (G), CG (x)/Op (CG (x)) has a component isomorphic to K. G is of even type if and only if (a) m2 (G) ≥ 3; (b) O2 (CG (x)) = 1 for all x ∈ I2o (G); and (c) Lo2 (G) ⊆ C2 . Three further technical requirements must be added to give the full definition of “restricted even type” [I2 , 8.8]. However, for the purposes of proving Theorem C5 , we shall only use one of these extra requirements, and that one only in Chapters 3 and 4 and rarely at that. This requirement, which can be ignored at a first reading, is expressed as Lo2 (G) ⊆ Co2 . What are the sets C2 and Co2 ? First, the set C2 is a specifically enumerated subset of K2 [I2 , 12.1], [V3 , 1.1], which is a variant of Chev(2) ∩ K2 ; C2 contains almost all groups in Chev(2) ∩ K2 , together with a number of other groups in K2 that have some Chev(2)-like properties, among them the groups L2 (q), q a Fermat or Mersenne prime, for which involution centralizers are 2-groups. The “relaxing” of the set Chev(2) to C2 is what allows some sporadic groups and groups in Chev(3) to be included in Theorem C5 . Second, the set Co2 is a proper subset of C2 , defined by simply removing from C2 the groups L2 (q), q a Fermat or Mersenne prime larger than 17. Returning to our main discussion, analogues Cp ⊆ Kp of C2 have likewise been defined [I2 , 12.1], [V3 , 1.1] for all odd primes p; indeed, for p > 11, Cp =  Chev(p) ∩ Kp ∪ {Ap , A2p , A3p }. On the other hand, for example, C3 includes A9 , eight groups in Chev(2) − Chev(3), and sixteen sporadic groups. In Chapter 3 we shall repeat the full definition of Cp , for every prime p. Next, for any odd prime p, the 2-local p-rank m2,p (G) of G is defined by m2,p (G) = max{mp (N ) | N ≤ G, O2 (N ) = 1}, and σ(G) = {p | p is an odd prime and m2,p (G) ≥ 4}. Remark 1.1. Originally we defined σ(G) by the condition m∗2,p (G) ≥ 4, where is the largest p-rank found in 2-local subgroups of G of odd, or twice odd, index (as opposed to all 2-local subgroups of G). Thus, m∗2,p (G) ≤ m2,p (G), and so the old σ(G) is a subset of the new one. There will be a price to be paid for this change, however, when we deal with the uniqueness theorems of Part II, in later volumes. m∗2,p (G)

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2. THE FOUR STAGES OF THE PROOF OF THEOREM C5

3

Finally, we refer to [I2 , Section 8] and [III1 , 1.2] for the rather technical definition of strong p-uniqueness subgroup of G. The most relevant features of the definition are reviewed in Section 1 of Chapter 3. It is a variant of the notion of strongly p-embedded subgroup. We define σ0 (G) = {p ∈ σ(G) | G has no strong p-uniqueness subgroup}. At several junctures in our proof, we will obtain a contradiction by constructing a strong p-uniqueness subgroup. 2. The Four Stages of the Proof of Theorem C5 The roots of our proof of Theorem C5 lie in J. G. Thompson’s N -group paper [T2], the Bender-Thompson signalizer lemma [IG , 23.2(ii)], K. Klinger and G. Mason’s analysis of groups simultaneously of characteristic 2-type and characteristic p-type [KMa1], and an unpublished manuscript of the second and third authors, which widened the Klinger-Mason work to include large sporadic groups, and which is superseded by this volume. Like the Gorenstein-Walter Alternative in the study of groups of generic type [III3 , (1H)], there is a dichotomy in our analysis for Theorem C5 , ultimately about whether certain signalizer functors are trivial or not. The target groups will emerge in the trivial signalizer functor case. Throughout this section we assume that G satisfies the hypotheses of Theorem C5 . Fix p ∈ σ0 (G). In particular, p ∈ σ(G), so m2,p (G) ≥ 4. The following set is of central importance in all stages of the proof:    Bp∗ (G) = B ∈ Epm (G)  m = m2,p (G), IG (B; 2) = {1} . This is the set of all “witnesses” to the fact that m2,p (G) is as large as it is. (Recall that IG (B; 2) is the set of all B-invariant 2-subgroups of G, Epm (G) is the set of all elementary abelian subgroups of G of order pn , n ≥ m, and Ep (G) = ∪m≥2 Epm (G). For an elementary abelian p-group P we occasionally write E1 (P ) for the set of all subgroups of P of order p.) Thus, (2A)

Bp∗ (G) = ∅.

We also define Sp (G) to be the set of elementary abelian p-subgroups of G that are maximal with respect to inclusion, or equivalently:    Sp (G) = B ∈ Ep (G)  Ip (CG (B)) ⊆ B . If (2B)

Bp∗ (G) ∩ Sp (G) = ∅,

and we choose any B ∈ Bp∗ (G)∩Sp (G), then the Bender-Thompson signalizer lemma [IG , 23.3(ii)] yields that Op (CG (b)) has even order for some b ∈ B # , provided that CG (b) is p-constrained for all b ∈ B # . This raises the possibility that Op might be a (nontrivial) signalizer functor on B, leading eventually to the existence of a strong p-uniqueness subgroup. Of course there are obstacles to such a plan. First, CG (b) may not be p-constrained for all b ∈ B # . Second, G must be balanced for the prime p in order for Op to be a signalizer functor. But these difficulties can be circumvented, unless there are unbalancing or strongly unbalancing components in CG (b)/Op (CG (b)) for certain p-elements b. When b ∈ Ipo (G), however, such components are in Lop (G) ⊆ Cp , and as a result the dreaded unbalancing cannot

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1. INTRODUCTION TO THEOREM C5

4

occur – unless caused by the existence of a single bad component, L2 (33 ) for p = 3. (L2 (33 ) is unbalanced with respect to a field automorphism of order 3.) With some work, however, we are able to argue around the possible occurrence even of this exceptional unbalancing. Moreover, with further work we can move away from elements of Bp∗ (G) and show that G has a strong p-uniqueness subgroup, contradicting the assumption that p ∈ σ0 (G). In this way we prove the first two assertions of the following theorem. Theorem C5 : Stage 1. The following conditions hold: (a) G is balanced with respect to any element of Ep4 (G); (b) Bp∗ (G) ∩ Sp (G) = ∅; (c) For any A ≤ B ∈ Bp∗ (G), Op (CG (A)) has odd order; and (d) Op (CG (x)) has odd order for any x ∈ Ipo (G). The last two assertions are easy consequences of the first two, plus a little signalizer functor theory. Further details may be found in the introductory section of Chapter 3. Following Stage 1, we have a weak symmetry between the primes 2 and p. Namely, (2C)

(1) Lo2 (G) ⊆ C2 and Lop (G) ⊆ Cp ; and (2) O2 (CG (x)) = 1 and |Op (CG (y))| is odd for all x ∈ I2o (G) and y ∈ Ipo (G).

A stronger symmetry was assumed in the unpublished manuscript referenced above, and an even stronger one was assumed in [KMa1], which Stages 2 and 3 generalize. A lot of terminology is needed to state Stage 2. First, we distinguish those B ∈ Bp∗ (G) for which CG (B) has even order (which we prefer) from the others, setting    Bp∗,c (G) = B ∈ Bp∗ (G)  |CG (B)| is even  (2D) Bp∗,c (G) if Bp∗,c (G) = ∅ p B∗,o (G) = otherwise. Bp∗ (G) In the end, it will turn out that Bp∗,c (G) = ∅, but we are far from that conclusion at this point. Since Bp∗ (G) = ∅, clearly (2E)

Bp∗,o (G) = ∅.

The first main goal in Stage 2 is to show that for any B ∈ Bp∗,o (G), any Binvariant 2-subgroup T of G such that CT (B) = 1 is cyclic or of symplectic type2 . The key to this is the first assertion of the following theorem. Theorem 2.1. Let A ≤ B ∈ Bp∗ (G) with |B : A| ≤ p. If |B : A| = p, assume that B ∈ Bp∗,o (G). Then CG (A) is p-constrained. Moreover, if B ∈ Bp∗,o (G) and T ∈ IG (B; 2) with CT (B) = 1, then T is cyclic or of symplectic type. The p-constraint also implies that m2 (CG (B)) ≤ 1 for all B ∈ Bp∗ (G). An analogous statement holds as well for hyperplanes A as above. We define a symplectic pair in G to be a pair (B, T ) such that B ∈ Bp∗ (G), T ∈ I∗G (B; 2), and T is cyclic or of symplectic type. In particular, by definition of 2 In

this volume, “symplectic type” will be understood to exclude “cyclic.”

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2. THE FOUR STAGES OF THE PROOF OF THEOREM C5

5

I∗G (B; 2), T is a B-invariant 2-subgroup of G and is maximal having this property, with respect to inclusion. The above theorem is a source of symplectic pairs, and we shall prove that symplectic pairs must exist in G. We call a symplectic pair (B, T ) faithful if and only if CB (T ) = 1, and trivial if |T | = 2. One of our next goals is to show that every symplectic pair is either faithful or trivial. Unfortunately, there is a single counterexample G to this assertion: the group G = Ω− 8 (3), with p = 3. Here a subgroup H = H+ × H− exists with (3). A Sylow 3-subgroup B of H is isomorphic to E34 and centralizes H± ∼ = Ω± 4 Z(H+ ) ∼ = Z2 . Then (B, O2 (H+ )) is a symplectic pair with O2 (H+ ) ∼ = Q8 ∗ Q8 , and hence it is neither trivial nor faithful. However, we locate the exceptional configuration that is the source of this counterexample and assume it away in much of Stage 2. (We return to this configuration in Stage 4, showing that it leads to G∼ = Ω7 (3) or Ω± 8 (3).) We analyze nonfaithful symplectic pairs in Stage 2. Faithful symplectic pairs will not enter the picture until Stage 3. If (B, T ) is a symplectic pair and we let t be the (unique!) involution in Z(T ), then B ≤ CG (t) and so O2 (CG (t)) ≤ T by the maximality of T . If (B, T ) is not faithful, then it is immediate that F ∗ (CG (t)) = O2 (CG (t)), for otherwise CB (T ) ≤ CB (O2 (CG (t))) = 1. But O2 (CG (t)) = 1 as G is of even type, so CG (t) possesses a component K, and one can argue that such a component K exists of order divisible by p. Such triples (B, t, K) are a focus of our analysis. We make the following definition: BtKp (G) is the set of all triples (B, t, K) such that B ∈ Bp∗ (G), t ∈ I2 (CG (B)), and K is a component of CG (t) of order divisible (2F) by p. We digress briefly to point out two technicalities. The first pinpoints some exceptional configurations, and in particular the configurations leading to the anomalous case G ∼ = Ω− 8 (3) mentioned above. If p = 3, we set (2G)

   BtK3exc (G) := (B, t, K) ∈ BtK3 (G)  K/Z(K) ∼ = L± 4 (3) or G2 (3) .

Triples of this type are not analyzed in Stage 2; instead, as mentioned above, they are worked out at length in Stage 4, the final stage, where they are shown to lead to G ∼ = Ω7 (3) or P Ω± 8 (3). The second technicality arises at one point of our analysis where we need to make a replacement for B. (2H)

Let (B, t, K), (B ∗ , t, K) ∈ BtKp (G). Then (B ∗ , t, K) is a mate of (B, t, K) if and only if CB ∗ (K) = CB (K), I∗CG (t) (B ∗ ; 2) = I∗CG (t) (B; 2), and, if B ∗ = B, then p does not divide | Out(K)|.

Now let us return to the main strand of our analysis of nonfaithful symplectic pairs. Let (B, t, K) ∈ BtKp (G). Our next goal is to find b ∈ B # such that CK (b) has a p-component I, and I ≤ J for some (unique) p-component J of CG (b). While I is easy to find, the existence of J is not automatic; there is no general “L-balance” property to guarantee it, since we are dealing simultaneously with 2-components and p-components. When J does exist, we say that the triple (K, b, I) is regularly embedded, or just regular. Furthermore, we will be interested only in triples (K, b, I) such that either [K, b] = 1 or the following hold: CB (K) = 1, b ∈ K,

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1. INTRODUCTION TO THEOREM C5

and I is terminal (as a p-component) in K. Such triples (K, b, I) we call broad. Terminality here means by definition that I   CG (x) for all x ∈ Ip (CK (I)). Suppose now that (B, t, K) ∈ BtKp (G), b ∈ B # , I is a p-component of CK (b), and I is regular and broad. Let J be the unique p-component of CG (b) containing I. We then call the triple (K; I; J) a nonconstrained {t, b}-neighborhood in G, and say that it derives from (B, t, K) or from t and B. Set CG (b) = CG (b)/Op (CG (b)) and notice that we then have (2I)

(1) Either K = I or K/O2 (K) ↓p I/Op (I); and (2) Either I = J or I/O2 (I) ↑2 J/O2 (J).

As K and I are generically in Chev(2), while J is generically in Chev(p), the two conditions (2I) are rarely compatible with each other, and the groups G that do eventually emerge from this analysis merit the description “bicharacteristic.” (See [V2 , 8.1] for the definitions of ↑2 and ↓p .) Remark 2.2. This concept of neighborhood should not be confused with the concept as defined and used in our treatment of the Generic Case [GLS5], [GLS7], [GLS8]. There, a neighborhood consisted of a set of components in p-local subgroups, for a fixed prime p. Here, a neighborhood defines both some 2-local structure and some p-local structure for some odd prime p. To make the distinction clearer, we could write “{2, p}-neighborhood” everywhere in this volume instead of just “neighborhood,” but we have chosen not to do so, for the sake of brevity. This remark applies also to constrained neighborhoods, to be defined below. We say that the nonconstrained {t, b}-neighborhood (K; I; J) is nondegenerate if and only if K > I and I < J. Otherwise it is degenerate. We shall prove in Stage 2 that whenever there is (B, t, K) ∈ BtKp (G) (and thus, whenever there is a nonfaithful symplectic pair, as noted above), then p = 3 and there exists a nondegenerate nonconstrained {t, b}-neighborhood (K; I; J) for some b. A small catch, insignificant in the end, is that b may not lie in the given B but rather in some other B ∗ ∈ Bp∗ (G). We also make an initial determination of possible isomorphism types for K, I, and J. These are displayed in Table 1. In every case, (B, t) is a trivial symplectic pair. The table also lists the value of m3 (B) = m2,3 (G) in each case. The winnowing of possible nonconstrained neighborhoods is not yet complete at this stage. Indeed, neighborhoods of types (4), (6), and (7) will turn out not actually to occur, nor will Ω− 8 (3) be a true possibility for J. They will be ruled out in Stage 3. However, the reader familiar with sporadic groups will already recognize that neighborhoods of type (1) occur in F1 , type (2) in the nonsimple group F i24 , type (3) in F2 , and type (5) in F i23 and F i24 . (The notation (n)K, where K is a simple group and n ≥ 1, means “either the group K or a covering group of K by a cyclic group of order n.”) The precise statement of Stage 2 is as follows. Theorem C5 : Stage 2. The following conditions hold: (a) Let B ∈ Bp∗ (G) and T ∈ IG (B; 2) with CT (B) = 1. Then T is cyclic or of symplectic type; (b) Symplectic pairs exist in G; (c) If B ∈ Bp∗ (G), then m2 (CG (B)) ≤ 1, while if B ∈ Bp∗,o (G) and A is a hyperplane of B, then m2 (CG (A)) ≤ 2;

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2. THE FOUR STAGES OF THE PROOF OF THEOREM C5

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Table 2.1. Stage 2: Nonconstrained {t, b}-Neighborhoods (K; I; J)

(1) (2) (3) (4) (5) (6) (7)

m2,3 (G) K I J 6 2F2 2F i22 (3)F i24 6 F i23 Ω7 (3) P Ω± 8 (3) 2 5 (2) E6 (2) 2U6 (2), D4 (2) F i22 5 F i22 U4 (3) L4 (9) 5 2F i22 2U4 (3) Ω7 (3), P Ω± 8 (3) 4 Sp8 (2), (2)F4 (2) Sp6 (2) U6 (2), D4 (2) (2)U6 (2), U5 (2) L± 4 (3), 3U4 (3) 4 U4 (2) Sp8 (2), (2)D4 (2) [3 × 3]U4 (3)

(d) If BtKp (G) = ∅, then p = 3; (e) Every symplectic pair is trivial or faithful, unless possibly p = 3 and BtK3exc (G) = ∅; and (f) Suppose that p = 3, (B, t, K) ∈ BtK3 (G), and BtK3exc (G) = ∅. Then there exists a mate (B ∗ , t, K) of (B, t, K), an element b ∈ (B ∗ )# and a nondegenerate nonconstrained {t, b}-neighborhood (K; I; J) deriving from (B ∗ , t, K), such that the following hold: (1) B ≤ K; (2) t ∈ Syl2 (CG (K)) ∩ I∗G (B; 2), and (B, t) is a symplectic pair; (3) K = E(CG (t)); (4) (K; I; J) and m2,3 (G) are as in one of the rows of Table 1 (here J∼ = J/O3 (J)); and (5) If K ∼ = 2 2E6 (2), then for some b1 ∈ (B ∗ )# , there is a nonconstrained {t, b1 }-neighborhood (K; I1 ; J1 ) such that I1 ∼ = 2U6 (2) and J 1 ∼ = F i22 . The purpose of Stage 3 is twofold. First, as mentioned above, we wish to rule out the configurations in Table 1 not corresponding to an actual sporadic group. To do this it is convenient to add an assumption: that all symplectic pairs in G are trivial. With this extra assumption, further local analysis shows that the nondegenerate nonconstrained {t, b}-neighborhood (K; I; J), guaranteed to exist by Stage 2, is given by a row of Table 2.2. The table lists our eventual target group G∗ in each case. In the course of this analysis, we also show that J is actually quasisimple, so the last column is headed by J rather than J. Of course we must eventually confront the possible existence of a faithful symplectic pair (B, T ), and the analysis of such a pair is the second purpose of Stage 3. Our goal in this situation has some similarities to the goal of Stage 2. (In particular, see Remark 2.2.) We again focus on a commuting pair {z, b} consisting of an involution z and an element b of order p. Actually z = Ω1 (Z(T )), and b ∈ B # is chosen to maximize |CT (b)|. We set Cb = CG (b) and are able to prove that Cb

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1. INTRODUCTION TO THEOREM C5

Table 2.2. Stage 3: Nonconstrained {t, b}-Neighborhoods (K; I; J)

G∗ m2,3 (G) K I J F i22 4 2U6 (2) U4 (2) U4 (3) F i23 5 2F i22 2U4 (3) Ω7 (3) F i24 5 2F i22 2U4 (3) P Ω+ 8 (3) F i24 6 F i23 Ω7 (3) P Ω+ 8 (3) 2 F2 5 2 E6 (2) 2U6 (2) F i22 Table 2.3. Sporadic-Type Constrained Neighborhoods (CG (z), b, J)

G∗ m2,3 (G) T = F ∗ (CG (z)) CG (z)/T J 1+8 + Co1 4 2+ Ω8 (2) 3Suz F1 6 21+24 Co 3F i24 1 + has a subgroup J such that the following conditions hold:

(2J)

(1) J is a p-component of E(Cb ), Op (Cb ) = 1 and mp (C(b, J)) = 1; (2) T = F ∗ (CG (z)), and R := CT (b) = O2 (CJ (z)) is extraspecial; and (3) In Cb := Cb /C(b, J), z is 2-central and R = F ∗ (CC b (z)).

(Here C(b, J) equals CCG (b) (J/Op (J)) by definition; in this case, as J is quasisimple, C(b, J) = CG (b J).) When these conditions hold, we say that (CG (z), b, J) is a constrained {z, b}-neighborhood with respect to (B, T ). Moreover, if J ∈ Spor, we say that (CG (z), b, J) is of sporadic type. There are just two sporadic type constrained neighborhoods that we aim for; they occur for p = 3 and are described in Table 2.3. Again in this table, G∗ is our target group when such a neighborhood exists. There is a notion of nondegeneracy for constrained neighborhoods as well. It applies when there exists b ∈ B ∩J −b such that D := b, b  and JD := E(CJ (D)) satisfy the following conditions:

(2K)

(1) JD is quasisimple (and JD is called a core of the neighborhood); (2) For some d ∈ D − b, JD  CG (d); and (3) For every d ∈ D# such that JD  CG (d), the subnormal closure Jd of JD in CG (d) is quasisimple and equals E(CG (d)); moreover, mp (C(d, Jd )) = 1. The groups Jd such that JD  CG (d) are called the neighbors of J (and include J itself).

The non-sporadic type constrained neighborhoods that we aim for again occur for p = 3 only. They are nondegenerate with core U4 (2) or U5 (2), with neighbors restricted to be among the groups D4 (2), U5 (2), U6 (2), and SU6 (2) only. We say that such neighborhoods are of U4 (2)-type; they occur in U7 (2), Ω− 10 (2), and 2 E6 (2), our three target groups in Chev(2).

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2. THE FOUR STAGES OF THE PROOF OF THEOREM C5

9

Now we can state Stage 3. Chiefly, it gives four possible outcomes for the {2, 3}-local structure of G, but it also includes some extra bits of information in a few specific cases. These extra bits will be useful in Stage 4. Theorem C5 : Stage 3. We have p = 3 and one of the following holds: (a) BtK3exc (G) = ∅; (b) All symplectic pairs in G are trivial. There exists a nonconstrained {t, b}neighborhood (K; I; J) for some commuting b ∈ I3 (G) and t ∈ I2 (G), with K, I, and J as in Table 2.2, and with C(b, J) = b. Moreover, in the case labelled G∗ = F i24 in the table, CG (t)/ t ∼ = Aut(F i22 ) and |CG (b) : J| is odd; (c) There exists a constrained {z, b}-neighborhood (CG (z), b, J) of sporadic type for some faithful symplectic pair (B, T ). Furthermore, T = O2 (CG (z)), CG (z)/T , and J as in Table 2.3. Moreover, in the case labelled G∗ = F1 in the table, there is an involution t ∈ CG (z, b) such that E(CG (t)) ∼ = 2F2 ; (d) There exists a constrained {z, b}-neighborhood (CG (z), b, J) of U4 (2)-type, with core Um (2), where m = m2,3 (G) ∈ {4, 5}. Moreover, if J ∼ = D4 (2) or U5 (2), then m2,3 (G) = 4. In Stage 4 we systematically consider the four cases of Theorem C5 : Stage 3, and their various subcases, one by one. This often involves very detailed local analysis, reminiscent of local analysis in the 1960’s and 1970’s that started from a specific centralizer of involution in a proposed simple group. Theorem C5 : Stage 4. G is isomorphic to one of the following groups: − (a) Ω7 (3), P Ω+ 8 (3), or Ω8 (3);  (b) Co1 , F i22 , F i23 , F i24 , F2 , or F1 ; or (c) 2 D5 (2), U7 (2), or 2E6 (2). The groups in (a) result from conclusion (a) of Theorem C5 : Stage 3; the groups in (b), from conclusions (b) and (c) of Theorem C5 : Stage 3; and the groups in (c), from conclusion (d) of Theorem C5 : Stage 3. In Case (a) of Stage 3, BtK3exc (G) = ∅, we follow to a large extent the ideas and methods of Aschbacher in [A23]. The group Ω7 (3) is recognized by constructing a weak Phan system of type B3 (3) generating a subgroup G0 of G. We then quote a Background Result [GHN] stating that that implies G0 /O3 (G0 ) ∼ = Ω7 (3) or Spin7 (3). Our 2-local information Ω implies that O2 2 (G0 ) = 1, so G0 ∼ = 7 (3). As usual, we quote the Bender-Suzuki theorem to finish the proof, after showing that either G = G0 or G0 is strongly embedded in G. − Likewise, the groups G∗ = P Ω+ 8 (3) and Ω8 (3) are recognized by constructing ∗ a covering group G0 of G by the Curtis-Tits theorem [IA , Section 2.9] and the method of Wong and Finkelstein-Solomon [V2 , 10.2], respectively, and then showing that either G = G0 or G0 is strongly embedded in G. Since G is simple, G ∼ = G∗ . In Cases (b) and (c) of Stage 3, every possible case of 2- and p-local data that we have corresponds to one of our six sporadic simple target groups G∗ , or to F i24 . Conversely, every sporadic simple target group corresponds to just one such set of local data. Our original plan was to prove that G and G∗ have the same centralizer of involution pattern. Though feasible, this would be lengthy to carry out and still require Background Results to yield the desired isomorphism G ∼ = G∗ . Instead,

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1. INTRODUCTION TO THEOREM C5

we develop enough information about involution centralizers to quote authoritative recognition theorems for the six sporadic groups in question. For this purpose, and for the Quasithin Theorem, we add the following publications to our Background Results: [A2] M. Aschbacher, Sporadic Groups, Cambridge: Cambridge University Press, 1994. [A19] M. Aschbacher, 3-Transposition Groups, Cambridge: Cambridge University Press, 1997. [ASm1] M. Aschbacher and S. D. Smith, The classification of quasithin groups. I, Mathematical Surveys and Monographs, vol. 111, American Mathematical Society, Providence, RI, 2004, Structure of strongly quasithin K-groups. [ASm2] M. Aschbacher and S. D. Smith, The classification of quasithin groups. II, Mathematical Surveys and Monographs, vol. 112, American Mathematical Society, Providence, RI, 2004, Main theorems: the classification of simple QTKE-groups. [GrMeSeg] R. L. Griess, Jr., U. Meierfrankenfeld, and Y. Segev, A Uniqueness Proof for the Monster, Ann. of Math. (2) 130 (1989), no. 3, 567-602. [Seg1] Y. Segev, On the Uniqueness of Fischer’s Baby Monster, Proc. London Math. Soc. (3) 62 (1991), no. 3, 509-536. In Case (d) of Stage 3, we essentially have a neighborhood in the sense of [GLS8]. By arguments somewhat similar to those in that generic situation, we construct a subgroup G0 ≤ G isomorphic to a covering group of our target group G∗ . The groups G0 ∼ = 2 D5 (2) and (3)2E6 (2) are recognized by the Curtis-Tits ∼ theorem, while G0 = U7 (2) is recognized by the method of Wong and FinkelsteinSolomon [V2 , 10.3]. We then prove that either G0 = G or NG (G0 ) is a strong 3-uniqueness subgroup of G. Since 3 = p ∈ σ0 (G), it follows that G0 = G, giving the desired conclusion G ∼ = G∗ .

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10.1090/surv/040.9/02

CHAPTER 2

General Group-Theoretic Lemmas, and Recognition Theorems 1. Signalizer Functor Theory The reader is referred to [IG , Sections 20–22] for the basics of signalizer functor theory. Here we mention just a few results of particular importance for this volume. For a prime p and a group X of order divisible by p, Ep (X) is the set of noncyclic elementary abelian p-subgroups E of X, and for m ≥ 2, Epm (X) is the subset of Ep (X) defined by the condition mp (E) ≥ m. We can omit the superscript p if X is a p-group. Occasionally we also write E1 (P ), when P is a p-group, for the set of all subgroups of P of order p. Let p be a prime and let A ∈ Ep (G) with mp (CG (A)) ≥ 3. We write  Θ1 (G; A) = Op (CG (a)) | a ∈ A# . and notice the trivial but fundamental property Θ1 (G; Ag ) = Θ1 (G; A)g for all g ∈ G. Theorem 1.1. If A ∈ Ep3 (G) and G is balanced (i.e., 1-balanced) with respect to A, then Θ1 (G; A) is the closure and completion of the 1-balanced signalizer functor on A defined by Θ1 (a) = Op (CG (a)). In that case Θ1 (G; A) is a p -group and hence a proper subgroup of G. Moreover, if π is the union of the sets of prime divisors of |Op (CG (a))| as a varies over A# , then Θ1 (G; A) is a π-group. Let P be a p-subgroup of a group X. We define certain “generated cores” as follows: ΓP,k (X) = NX (D) | D ≤ P, mp (D) ≥ k ΓP,k (X) = CX (D) | D ≤ P, mp (D) ≥ k We also define ΓoP,2 (X) = NX (D) | D ≤ P, mp (D) ≥ 2, mp (DCP (D)) ≥ 3 , if mp (P ) ≥ 3 Γoo P,2 (X) = NX (D) | D ≤ P, mp (D) ≥ 2, mp (DCP (D)) ≥ 4 , if mp (P ) ≥ 4 Typically the last two will be used for X = G and P ∈ Sylp (G). The final one, Γoo P,2 (G), plays a fundamental role in Chapters 3 and 4 when p is odd. We give two results connecting signalizer functors with generated cores. Lemma 1.2. Suppose that A, B ∈ Ep2 (G), [A, B] = 1, and G is balanced with respect to AB. Then Θ1 (G; A) = Θ1 (G; B). Proof. For any a ∈ A# and b ∈ B # , balance yields Op (CG (a)) ∩ CG (b) ≤ Op (CG (b)) ≤ Θ1 (G; B). 11 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

 Then Op (CG (a)) = Op (CG (a)) ∩ CG (b) | b ∈ B # ≤ Θ1 (G; B); so letting a vary  over A# , Θ1 (G; A) ≤ Θ1 (G; B). The result follows by symmetry. Lemma 1.3. Suppose that p is a prime, mp (G) ≥ k ≥ 3, and G is balanced with respect to every element of Epk (G). Then the following conditions hold: (a) Let A ∈ Epk (G) and B ∈ E2 (A). Then Θ1 (G; A) = Θ1 (G; B); (b) Suppose that A0 , A1 , . . . , An is a sequence of elements of Ep2 (G) such that mp (Ai ) ≥ k either for all odd i or for all even i, 0 ≤ i ≤ n. Assume also that for all i = 1, . . . , n, Ai−1 ⊆ Ai or Ai−1 ⊇ Ai . Then Θ1 (G; A0 ) = Θ1 (G; An ); (c) In (b), ΓAi ,2 (G) ≤ NG (Θ1 (G; A0 )) for all i = 0, . . . , n; (d) Suppose that k = 4 and p > 2. Let P ∈ Sylp (G) and let A ∈ E4 (P ). Then Γoo P,2 (G) ≤ NG (Θ1 (G; A)); and (e) Suppose that k = 3 and p > 2. Let P ∈ Sylp (G) and let A ∈ E3 (P ). Then ΓoP,2 (G) ≤ NG (Θ1 (G; A)). Proof. In (a), the proof only uses the noncyclicity of A and B, and is essentially the same as the proof of Lemma 1.2. Then (b) is an immediate consequence. In (c), let A ∈ E2 (Ai ) be arbitrary. By the assumptions of (b), Ai lies in an element of Epk (G), so G is balanced with respect to A and hence by (b) and Lemma 1.2, Θ1 (G; A0 ) = Θ1 (G; Ai ) = Θ1 (G; A). This group is NG (A)-invariant, and as A was arbitrary, (c) follows. Suppose that k = 4. First we claim that Θ1 (G; A) = Θ1 (G; B) for any A, B ∈ E4 (P ). To see this let U  P with U ∼ = Ep2 . Then the sequence A, CA (U ), U CA (U ), U, U CB (U ), CB (U ), B satisfies the conditions of (b), and the claim follows. Moreover, for any p-subgroup Q of G with mp (Q ∩ P ) ≥ 4, NG (Q) permutes E4 (Q) and hence normalizes the group W = Θ1 (G; B) for B ∈ E4 (Q) (which is independent of B). As we can take some B ∈ E4 (Q ∩ P ), W = Θ1 (G; A) and so NG (Q) ≤ NG (Θ1 (G; A)). Now let D ≤ P with mp (D) ≥ 2 and mp (DCP (D)) ≥ 4. We must prove that NG (D) ≤ NG (Θ1 (G; A)). Expand DCP (D) to R ∈ Sylp (DCG (D)). By the previous paragraph NG (R) ≤ NG (Θ1 (G; A)). As NG (D) = DCG (D)NNG (D) (R) by a Frattini argument it is enough to show that DCG (D) ≤ NG (Θ1 (G; A)). Now since mp (D) ≥ 2 and p is odd, there exists V  DCP (D) such that V ∼ = Ep2 and V ≤ D, by [IG , 10.11]. Then V lies in an element E ∈ Ep∗ (DCP (D)) ⊆ E4 (P ). Hence Θ1 (G; A) = Θ1 (G; E) = Θ1 (G; V ), using (a). But DCG (D) normalizes V as V ≤ D, so DCG (D) ≤ NG (Θ1 (G; V )) = NG (Θ1 (G; A)), completing the proof of (d). The proof of (e) is entirely similar.  Lemma 1.4. Suppose that E ∈ Ep4 (G), F ≤ E, and mp (F ) ≥ 3. Suppose that G is balanced with respect to F , and 2-balanced with respect to E. Then Θ1 (G; F ) = Θ2 (G; E). Proof. Let D ∈ E2 (E) and f ∈ F # . Then by 2-balance, ΔG (D) ∩ CG (f ) ≤ Op (CG (f )) ≤ Θ1 (G; F ). Letting f vary over F # we get ΔG (D) ≤ Θ1 (G; F ), and then letting D vary over E2 (E) we get Θ2 (G; E) ≤ Θ1 (G; F ).

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For the reverse inclusion, note that for D ∈ E2 (F ) and f ∈ F # ,



Op (CG (f )) ∩ CG (D) = (Op (CG (f )) ∩ CG (d)) ≤ Op (CG (d)) d∈D #

d∈D #

= ΔG (D) ≤ Θ2 (G; E). Letting D vary over E2 (F ) we get Op (CG (f )) ≤ Θ2 (G; E), and letting f vary over F # gives Θ1 (G; F ) ≤ Θ2 (G; E), completing the proof.  Definition 1.5. Let p be a prime and X a group. An unbalancing triple in X (for the prime p) is a triple (x, y, K) such that x and y are commuting elements of X of order p, K is a p-componenit of CX (x), and Op (CX (y)) ∩ CX (x) does not centralize K/Op (K). A locally unbalancing triple in X is a triple (x, y, K) such that x and y are commuting elements of order p in X, K is a p-component of CX (x), and Op (CX (x, y)) does not centralize K/Op (K). Since Op (CX (y) ∩ CX (x)) ≤ Op (CX (x, y)), every unbalancing triple is also a locally unbalancing triple. The converse, however, does not hold. 2. Signalizers Lemma 2.1. Let B be an elementary abelian p-group acting faithfully on a p group W . Choose b ∈ B # such that |CW (b)| is maximal. Then B/ b acts faithfully on CW (b). Proof. Suppose false, so that [CW (b), b ] = 1 for some b ∈ B − b. Then D := b, b  centralizes CW (b). For any d ∈ D# , CW CW (d) =  (d) ≥ CW (b) so CW (b) by the assumed maximality. Therefore W = CW (d) | d ∈ D# = CW (b), contradicting the faithfulness of the action of B on W . The lemma follows.  Lemma 2.2 (Bender-Thompson). Let p be an odd prime and X a p-constrained group. Let A ∈ Sp (X), i.e., A is an elementary abelian p-subgroup of X containing every element of order p in CX (A). Then any A-invariant p -subgroup of X lies in Op (X). Proof. This is [IG , 23.3(ii)].



Relaxing the p-constraint hypothesis, we have: Lemma 2.3. Let p be an odd prime and X a group with Op (X) = 1. Let A ∈ Sp (X). Let W be an A-invariant p -subgroup of X. Then W centralizes Op (X). If W = 1, then [W, L] = 1 for some component L of X. Proof. Let Y = Op (X)AW , a p-solvable and thus p-constrained group. By Lemma 2.2, W ≤ Op (Y ) so [W, Op (X)] ≤ [Op (Y ), Op (Y )] = 1, proving the first assertion. If also [W, L] = 1 for every component L of X, then W ≤ CX (F ∗ (X)) = Z(Op (X)) by the F ∗ -theorem [IG , 3.6]. As W is a p -group, W = 1. This proves the second statement.  Lemma 2.4. Suppose that p is an odd prime and A ∈ Sp (X). Then CX (A) has a normal p-complement. Proof. Let P ∈ Sylp (CX (A)). Since A ∈ Sp (X), Ω1 (P ) = A ≤ Z(CX (A)).  The result follows from [IG , 16.11].

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Definition 2.5. Suppose that K is a component of X, p is a prime dividing |K|, and A is a p-subgroup of X. Then K is strongly locally balanced with respect to A in X if and only if IAutX (K) (AutA (K); p ) = {1}. Lemma 2.6. Let p be an odd prime and X a K-group with Op (X) = 1. Let A ∈ Sp (X). Suppose that for all subgroups B ≤ A, every component of CX (B) is strongly locally balanced with respect to A in CX (E). Then IX (A; p ) = {1}. Proof. Note that our assumption with B = 1 states that every component K of X is strongly locally balanced with respect to A; in particular, it is 1-balanced with respect to A. In more detail, let a ∈ A# ; we argue that H := Op (CX (a)) = 1. Suppose that H = 1. By [IG , 23.5], [Op (X), H] = 1. Hence there is a p-component K of E(X) such that [K, H] = 1. Set N = NX (K) and NA = N ∩ A. By [IG , 20.6], a ∈ NA and if we set N = N/CN (K), then 1 = H ≤ Op (CKaH (a)). This latter group is invariant under NA , so injects into an element of IAutX (K) (AutA (K)), which, however, is 1 by strong local balance. This is a contradiction, so Op (X) = 1 for all a ∈ A# . Let D be any subgroup of A and put Y = CX (D) and Y = Y /Op (Y ). We claim that the hypotheses apply to Y and A. Obviously A ∈ Sp (Y ) since D ≤ A, so A ∈ Sp (Y ). Let B ≤ A and let L be a p-component of CY (B). Thus L is a p-component of CX (BD), so by assumption L is strongly locally balanced with respect to A. Hence L is strongly locally balanced with respect to A, proving our claim. As in the previous paragraph it follows that L in particular is 1-balanced with respect to A. Now let Z = A ∩ Z(X). We proceed by induction on |A/Z|. If A = Z, then by Lemma 2.4, X = CX (A) has a normal p-complement, so X is a p-group and the result is trivial. Let W ∈ IX (A; p ). Suppose that |A/Z| = p and write A = Z × a. Then CW (A) ∈ ICX (a) (A; p ), so by induction in CX (a)/Op (CX (a)), CW (A) ≤ Op (CX (a)). Therefore CW (A) = 1 by the first paragraph of the proof. To show that W = 1 it remains to show that [W, A] = 1. If A fails to normalize some com ponent K of X, let K ∗ = K A = K a and KA = E(CK ∗ (A)) = E(CK ∗ (a)), a diagonal of K ∗ . Then A centralizes KA . But by [IA , 16.9], there exists x ∈ Ip (KA ) − Z(KA ). Then [x, A] = 1 but [x, KA ] = 1 so x ∈ A. This is a contradiction since A ∈ Sp (X). We have proved that A normalizes every component of X. As all such components are strongly locally balanced with respect to A, [IG , 23.5] implies directly that [W, A] = 1, completing the proof in this case. Finally suppose that |A/Z| ≥ p2 . Write A = Z × D, so that D is noncyclic. For any d ∈ D# , CW (d) ∈ ICX (d) (A; p ) and Z d ≤ Z(CX (d)), so by induction CW (d) ≤ Op (CX (d)) = 1, by the first paragraph. As D is noncyclic and d ∈ D# is arbitrary, W = 1, and the proof is complete.  Lemma 2.7. Let p be an odd prime, X a K-group, and A ∈ Sp (X). Suppose that W ∈ IX (A; p ) and W ≤ Op (X). Then in X := X/Op (X), there exists a component L such that [W , L] = 1, and for some B ≤ A, Lp (CLB  (B)) has a component L0 such that L0 is not strongly locally balanced with respect to A. Proof. Let X be a minimal counterexample; then Op (X) = 1. Let J be the product of all components L of X such that [W, L] = 1. Since W is A-invariant,

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so is J. By minimality X = AW J and J = E(X). Since W = 1, there exists by Lemma 2.6 a subgroup B ≤ A and a component L0 of CX (B) such that L0 is not strongly locally balanced with respect to A. By Lp -balance, L0 is a component of  Lp (CLB  (B)) for some component L of J. The proof is complete. Definition 2.8. Let p be a prime and K ∈ Kp . Then K is strongly locally balanced with respect to p if and only if whenever X is a group in which K is a component, and A ∈ Sp (X) with A normalizing K, then IAutX (K) (AutA (K); p ) = {1}. In [IA , 7.7.11, 7.7.12], some groups are listed that are strongly locally balanced with respect to some odd prime p, notably all groups in Cp , with the exception of L2 (33 ) for p = 3. Lemma 2.9 (Thompson). Let p and q be distinct primes, and let E ∼ = Ep2 act on the q-group Q. Suppose that there exist at most two subgroups E1 , E2 ≤ E of order p such that CQ (Ei ) = 1. Then Q = CQ (E1 ) × CQ (E2 ). Proof. Let Qi = CQ (Ei ), i = 1, 2. Obviously Q1 ∩ Q2 ≤ CQ (E) = 1 and Q = Q1 Q2 . Let Z be a minimal E-invariant normal subgroup of Q. Then Z ≤ Z(Q) and in Q := Q/Z, Q = Q1 × Q2 by induction. Without loss Z ≤ Q1 . Then [Q1 , Q2 ] ≤ Z, so [Q1 , Q2 , E1 ] = 1. As [Q1 , E1 ] = 1, [E1 , Q2 , Q1 ] = 1. But  [E1 , Q2 ] = Q2 so [Q1 , Q2 ] = 1. The proof is complete. 3. Fusion Lemma 3.1. Let x ∈ Ip (X) and Q ∈ Sylp (CX (x)). Suppose that J(Q) is abelian. Then NX (J(Q)) controls X-fusion of x in J(Q). Proof. Clearly x ∈ J(Q). Suppose g ∈ X and xg ∈ J(Q). Expand J(Q) to R ∈ Sylp (CX (xg )). Then J(R) ∼ = J(Q) ≤ R so J(Q) = J(R). As Qg ∈ g gh Sylp (CX (x )), Q = R for some h ∈ CX (xg ). Then xgh = x and J(Q)gh =  J(Qgh ) = J(R) = J(Q), as required. Lemma 3.2. Let T ∈ Syl2 (X) with m2 (T ) = 1 or T ∼ = Q8 ∗ Z2n for some n > 1. Then Ω1 (Z(T )) ≤ Z ∗ (X). Proof. Write z = Ω1 (Z(T )) and suppose that for some g ∈ X and y ∈ I2 (T ) − {z}, y g = z. Then m2 (T ) > 1. Using Sylow’s theorem we may assume that CT (y)g ≤ T . Then Z(T )g ≤ T so z g  ≤ Φ(Z(T ))g ≤ Φ(T ) ≤ Z(T ). Thus, z g = z = y g , so z = y, a contradiction. Hence z is weakly closed in T , and the  Z ∗ -theorem [IG , 15.3] completes the proof. Lemma 3.3. Suppose that p is an odd prime and B ∼ = Epp+1 . Let x ∈ B # , and let B = B0 ×x be a decomposition of B preserved by an automorphism h ∈ Aut(B) of order (pp − 1)/2 acting Frobeniusly on B0 and trivially on x. Let H ≤ Aut(B) with h ∈ H. Then one of the following holds: (a) x is H-invariant; (b) x is H-conjugate into B0 ; (c) |H : NH (x)| is divisible by p.

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2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

Proof. We consider the action of H on E1 (B), specifically the H-orbit Ω containing x, and impose the obvious condition that |Ω|, which divides |H|, must therefore divide | Aut(B)| = |GLp+1 (p)|. As p is odd, h is transitive on E1 (B0 ) and has two equal-size orbits on B0# . Therefore the h-orbits on E1 (B) are {x}, E1 (B0 ), Ω1 and Ω2 , of cardinalities 1, (pp −1)/(p−1), a, and a, respectively, where a = (pp −1)/2. Assuming that (a) and (b) fail, Ω is the union of {x} with one or both of Ω1 and Ω2 . If only one of the Ωi ’s is fused to x, then |Ω| = (pp + 1)/2. But by Zsigmondy’s theorem [IG , 1.1], this has a prime divisor not occurring in | Aut(B)|p = (p − 1)(p2 − 1) · · · (pp+1 − 1).  Thus both Ωi ’s are fused to x and |Ω| = pp , whence (c) holds. 4. p-Groups If K ∈ Kp and p is a prime, we define  − mp (Z(K)),  m  p (K) = mp (K)  is a universal covering group of K. where K Lemma 4.1. Let f : L → K be a covering of quasisimple groups, and p a prime.  p (K), and K contains an elementary abelian p-group E of rank Then m  p (L) = m m  p (K) such that E ∩ Z(K) = 1. Proof. Since m  p (K) depends only on the universal covering group of K,  p (K). If f is universal, then by definition L contains F ∼ m  p (L) = m = Epm such that m = m  p (L) and E ∩ Z(L) = 1. Hence E ∩ ker f = 1 and the result follows.  Lemma 4.2. Suppose that K and L are simple groups and L contains copy of  p (K). K or of a covering group of K. Then for any prime p, m  p (L) ≥ m Proof. We are given that X ≤ L with X quasisimple and X/Z(X) ∼ = K. Let   the last term of the derived series of the L → L be a universal covering of L, and X  By Lemma 4.1, X  contains an elementary abelian p-subgroup preimage of X in L.  = 1. But Z(L)  ∩ E ≤ Z(X)  ∩ E = 1, E such that mp (E) = m  p (K) and E ∩ Z(X)  so m  p (L) ≥ mp (E), as claimed. Lemma 4.3. Let T /Z ∼ = Q8 , with |Z| = 2. Then the involution of T /Z splits over Z. Proof. Otherwise T has a unique involution, so T ∼ = D8 , a = Q16 and T /Z ∼ contradiction.  Lemma 4.4 (MacWilliams). Let T be a 2-group with m2 (T ) ≥ 5. Then T possesses a normal E23 -subgroup and T is connected. Proof. See [III2 , 2.8].



Lemma 4.5. If P is a 3-group and m3 (P ) ≥ 4, then P is connected. Proof. See [IG , 10.22(iii)].



Lemma 4.6. Let P be a p-group. Assume that A ≤ P ≥ B with A ∼ =B∼ = Ep5 . Then A and B are 3-connected. Proof. This follows directly from Lemma 4.4 and [IG , 10.23(i)] for p = 2, and  [IG , 10.23(ii)] for p > 2.

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Lemma 4.7. Let T be a 2-group and x ∈ I2 (T ), and suppose that CT (x) has a normal subgroup T0 = 1 such that CT (x)/T0 ∼ = Z4 or Z2 . Then one of the following holds: (a) x ∈ Z(T ); (b) There exists y ∈ I2 (T0 ) such that CT (y) > CT (x); (c) CT (x) ∼ = Z2 × Z4 or Z2 × Z2 , and m2 (T ) = 2; (d) CT (x) ∼ = Z2 ×Z4 and T ∼ = F (n) for some n ≥ 5 (see [IG , 10.27]). In particular, T has a dihedral maximal subgroup and a cyclic center, m2 (T ) = 3, and T /Ω1 (Z(T )) ∼ = Z2 × D2n−2 ; or (e) T ∼ = Z4  Z2 . Proof. Suppose that x ∈ Z(T ) and set C = CT (x) < T . Choose u ∈ NT (C)− C with u2 ∈ C. If there exists y ∈ Z(C) ∩ T0# with [y, u] = 1, then (b) holds. Note that [C, C] ≤ T0 and u normalizes [C, C] ∩ Z(C), so if [C, C] = 1 then y can be chosen in [C, C] ∩ Z(C), and (b) holds. So assume that [C, C] = 1. The same argument may be made with 2 (C) in place of [C, C], since C/T0 has exponent 2 or 4. So we may assume that C is abelian of exponent 2 or 4. If m2 (C) > 2, then as C/T0 is cyclic, Ω1 (T0 ) ∩ T0u is nontrivial and y may be chosen in it. So assume that m2 (C) = 2. Then C ∼ = Z2 × Z2 , Z2 × Z4 , or Z4 × Z4 . In the first two cases, either (c) or (d) holds by [IG , 10.24, 10.27]. In the third case, u acts freely on both Ω1 (C) and C/Ω1 (C), and so C u ∼ = Z4  Z2 . Then C char C u and C contains precisely two elements of xT , namely x and xu , while the third involution of C lies in Z(T ). It follows that NC (C u) ≤ C u CT (x) = C u, so T = C u and (e) holds. The proof is complete.  Lemma 4.8. Let T , x, and T0 satisfy the hypotheses of Lemma 4.7. Let X be a K-group with T ∈ Syl2 (X), and suppose that O2 (X) = 1 and all components of E(X) are C2 -groups. Suppose also that Z(X) has even order. If m3 (X) ≥ 2, then conclusion (a) or (b) of Lemma 4.7 holds for T and CT (x). Proof. Suppose that the lemma fails, so that T and C := CT (x) satisfy (c), (d), or (e) of Lemma 4.7. If E(X) has a simple component, then Z(T ) is noncyclic. Hence Z(T ) = Ω1 (C) and T = C, which is absurd as E(X) then has cyclic Sylow 2-subgroups. On the other hand, suppose that E(X) has a nonsimple component K ∈ C2 [V3 , 1.1]. In particular m2 (K) > 1, so either m2 (K) = 2 or else T ∩ K/Z(K) ∈ Syl2 (K/Z(K)) is a section of Z2 × D2n for some n ≥ 3. In any case K has 2-rank at most 3 and sectional 2-rank at most 4. Given that Z(K) = 1, [VK , 10.32] yields a contradiction. Therefore, E(X) = 1. Let S be a minimal characteristic subgroup of O2 (X) admitting P faithfully, where P ≤ X with P ∼ = E32 . Then Φ(S) ≤ Ω1 (Z(S)). By the action of P , |S/Φ(S)| ≥ 24 . In particular T ∼  Z4  Z2 or F (n) for any n ≥ 5, nor is T of maximal class; so by = assumption and [IG , 10.24], m2 (T ) = 2 and C ∼ = Z4 × Z2 . Clearly Φ(S) > 1. If |Φ(S)| = 2 then as m2 (S) ≤ 2, S does not admit P faithfully. Hence, |Φ(S)| = 4 and Φ(S) = Z(S) = Ω1 (S) is centralized by P . In particular x ∈ S. Let X = X/Φ(S). As |CxS (x)| ≤ |CxS (x)| ≤ 8 by [IG , 9.16], |S| ≤ 24 . Therefore SP ∼ = A4 × A4 . As P centralizes Z(S) = Ω1 (S), it follows easily that S ∼ = Q8 × Q8 . Modulo central

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2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

elements, x must interchange these factors by the Krull-Schmidt theorem, and it  follows that |CS (x)| = 23 , contradiction. The lemma follows. Lemma 4.9. Suppose that X is a p-constrained group, p an odd prime, and mp (Op p (X)) ≤ 2. Then mp (X) ≤ 3. Proof. Without loss, Op (X) = 1. Let Q = Op (X) and C = CX (Q) ≤ Q. By [IG , 11.16(i)], X/C embeds in GL2 (p), so mp (X/C) ≤ 1. But mp (C) ≤ mp (Q) ≤ 2, so mp (X) ≤ mp (C) + mp (X/C) ≤ 3.  Lemma 4.10. Suppose that F ∗ (X) is a p-group, p odd, in which every characteristic abelian subgroup is cyclic. Let Q = F ∗ (X). Then the following conditions hold: (a) If Q is cyclic, then Q ∈ Sylp (X); and (b) If mp (X) ≥ 4, then Q ∼ = p1+2k ∗ Zpm for some m ≥ 1 and k ≥ 2. Proof. In (a), CX (Q) = Q, so X/Q embeds in the abelian group Aut(Q) and Op (X/Q) = 1. Hence X/Q is a p -group, as desired. In (b), Philip Hall’s theorem [IG , 10.3] and (a) give Q ∼ = p1+2k ∗Zpm for some m and k. If k = 1, then mp (Q) = 2 so mp (X) ≤ 3 by Lemma 4.9, contrary to assumption. Thus k ≥ 2 and the lemma holds.  Lemma 4.11. Let T ∈ Sylp (X) and A = J(T ). Assume that T has no nontrivial abelian direct factor, A = CX (A), and NX (A)/A has one class of elements of order p. Then T is indecomposable. Proof. Suppose for a contradiction that T = T1 × T2 with T1 = 1 = T2 . Then A = J(T1 ) × J(T2 ) = A1 × A2 , where we have put Ai = A ∩ Ti , i = 1, 2. Since T has no abelian direct factor, Ti > Ai , i = 1, 2. Let N = NX (A) ≥ T and N = N/A. Choose xi ∈ Ip (T i ), and set Vi = [A, xi ], i = 1, 2. As A = CX (A), V1 = 1 = V2 . By assumption x1 is N -conjugate to x1 x2 , so V1 ∼ = [A, x1 x2 ] = V1 × V2 . This is a  contradiction as V2 = 1, and the lemma is proved. Lemma 4.12. Let T be a 2-group of symplectic type and X a group of automorphisms of T of odd order. Then [T, X] is extraspecial. Proof. We have T = EF with E extraspecial, [E, F ] = 1, and either F is of maximal class with |F | > 8, or F is cyclic with F = 1. If F is cyclic, then F = Z(T ) and F ≤ CT (X)  T . Hence, [T, CT (X), X] = [CT (X), X, T ] = 1 so [T, X, CT (X)] = 1 by the Three Subgroups lemma. Also [T, T ] = Ω1 (F ) = Ω1 (Z(T )). Now T := T /[T, T ] = [T , X] × CT (X) with [T , X] elementary abelian. As [T, X, CT (X)] = 1, Z([T, X]) ≤ Z(T ) = F so Z([T, X]) ≤ Ω1 (F ). Hence, [T, X] is extraspecial, as claimed. If F is of maximal class, on the other hand, then as |F | > 8, C = Z(CT ([T, T ])) is a cyclic subgroup of T of maximal order and is characteristic in T . Moreover, |T : CT (C)| = 2 and CT (C) is the product of an extraspecial group and C. Then  [T, X] = [CT (C), X], which is extraspecial by the previous case.

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5. UNIQUENESS SUBGROUPS

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5. Uniqueness Subgroups Throughout this section, as throughout this series, G is a K-proper simple group. Also, if (x, K) ∈ ILop (G), then C(x, K) = CCG (x) (K/Op (K)). Lemma 5.1. Let (x, K) ∈ ILo2 (G). Suppose that K is terminal in G, with K ∈ C2 . Suppose that ΓD,1 (K) = K for every D ∈ E22 (Aut(K)). Then m2 (C(x, K)) = 1. Here terminality means that K is quasisimple and K is a component of CG (y) for all y ∈ CG (K) [II3 , p. 133]. Proof. Suppose on the contrary that Q ∈ Syl2 (C(x, K)) and m2 (Q) > 1. By [II3 , Theorem PU4 ], K is standard in G. By [II3 , Corollary PU2 ], there exists E ∈ E22 (C(x, K)) and g ∈ G − NG (K) such that E g ≤ NG (K). By [IG , 18.8], ΓE g ,1 (K) < K. But by assumption K =  ΓE g ,1 (K). This contradiction completes the proof. Lemma 5.2. Let K be a standard component of the centralizer of some involution of G, with O2 (K) = 1. Let M = NG (K) and Q ∈ Syl2 (CG (K)). If Qg ≤ M for some g ∈ G − M , then one of the following holds: (a) Qg ∩ K = 1; or (b) Qg /Z embeds in Out(K) for some Z ≤ Z(Qg ). Proof. Without loss, Qg , Q is a 2-group. Let Z = Qg ∩ KCG (K). Then Qg /Z embeds in Out(K). If Z ≤ Z(Qg ), then (b) holds. Otherwise, choose u ∈ Z − Z(Qg ) and v ∈ Qg such that [u, v] = 1. Write u = uK uQ with uK ∈ K and uQ ∈ Q. g g Then uK ∈ CG (u) ≤ ΓQg ,1 (G) ≤ NG (K g ). Also uK = uu−1 Q ∈ Q , Q so uK , Q  g g g g g g is a 2-group. As Q ≤ CG (K )  NG (K ), Q  uK , Q  = uQ , Q . We have [u, v] = [uK , v][uQ , v]. But [uQ , v] ∈ Q∩Qg = 1. Therefore [u, v] = [uK , v] ∈ Qg ∩K, so (a) holds.  Lemma 5.3. Suppose that x ∈ I2 (G), K is a simple component of CG (x), and K is standard in G. Let M = NG (K) and Q ∈ Syl2 (CG (K)). Let r be the largest rank of an elementary abelian 2-subgroup of Aut(K) disjoint from Inn(K), and assume that m2 (Z(Q)) > r. If there exists g ∈ G−M such that m2 (Qg ∩M ) > r, then there exists h ∈ G−M such that Qh ≤ M . Proof. This is a mild generalization of [IG , 18.11]. Since K is standard, NG (M ) = M . Choose g ∈ G − M such that m := m2 (V ) is maximal, where V = Qg ∩ M . Thus m > r. Replacing Q by an M -conjugate, we may assume that V, Q is a 2-group. Let V0 be the subgroup of V inducing inner automorphisms on K, so that m2 (V0 ) ≥ m − r > 0 and V0 ≤ K × CG (K). As V0 ≤ Qg and K is standard, ΓV0 ,1 (G) ≤ M g . Since V, Q is a 2-group, the projection V1 of V0 on CG (K) lies in Q. If V0 ∩ K = 1, then the conclusion of the lemma holds by [IG , 18.12]. So assume that # V0 ∩ K = 1, whence V0 ∼ = V1 . If V1 ∩ Z(Q) = 1, then choose v ∈ V0 projecting −1 onto an element of Z(Q); we get Q ≤ CG (v) ≤ M g , so Qg ≤ M , as desired. Thus we may assume that V1 ∩ Z(Q) = 1. By assumption there is W ≤ Ω1 (Z(Q)) with −1 −1 m2 (W ) > r. Then W V1 = W × V1 ≤ CG (V0 ) ≤ M g so (W V1 )g ≤ Qg ∩ M . But

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2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

m2 (W V1 ) = m2 (V1 ) + m2 (W ) > m − r + r = m, contradicting our maximal choice of m. The proof is complete.  6. Miscellaneous Lemma 6.1. Suppose that F ∗ (X) = Op (X) is elementary abelian, for some prime p. Let P ∈ Sylp (X). If no element of Ip (X/Op (X)) acts quadratically on Op (X), then J(P ) = Op (X). Proof. This is a straightforward application of the Thompson Replacement theorem [III8 , 1.1].  1 The width w(S) of an extraspecial or symplectic type r-group S is by definition 2 mr (S/Φ(S)) . Lemma 6.2. Let B be an elementary abelian p-group, p odd, acting faithfully on a 2-group S. Let k = ordp (2). Then m2 (S/Φ(S)) ≥ kmp (B). If S is of symplectic type, then w(S) ≥ mp (B). If S is of symplectic type and p > 3, then w(S) ≥ 2mp (B). Proof. Without loss, S = [S, B]. Since B is abelian and acts faithfully, V := S/Φ(S) contains the direct sum of at least mp (B) nontrivial irreducible F2 Bsubmodules V1 , V2 , . . . with pairwise distinct kernels. Each such module has order 2k , the size of the field of pth roots of unity over F2 . This implies the first statement. Now assume that S is of symplectic type. By Lemma 4.12, S is extraspecial. As k ≥ 2, we immediately deduce that w(S) ≥ [k/2]mp (B) ≥ mp (B). The final inequality also follows unless perhaps k = 3. Then the nonsingularity of the commutation form on S/Φ(S) implies that along with every module Vi occurs its dual ∼ ∗ Vi∗ , with the same kernel. 6 When k = 3, Vi = Vi , so we get mp (B) distinct pairs of modules, and w(S) ≥ 2 mp (B). The lemma is proved.  Lemma 6.3. Let p be an odd prime and P = p1+2w an extraspecial p-group of exponent p and width w. If P embeds in Ln (2), then n ≥ pw k, where k = ordp (2). Proof. Let V be the corresponding n-dimensional F2 P -module. As Z(P ) is cyclic, we may assume that V is irreducible. Let A  P with A elementary abelian of order p1+w . Write z = Z(P ). Then CV (z) = 0 and P permutes transitively the set of pw hyperplanes of A disjoint from z. Hence by Clifford’s theorem, V ↓A is the sum of at least pw irreducible nontrivial representations of A with distinct kernels. As each such representation requires k dimensions, the lemma follows.  Lemma 6.4. Suppose that K is a perfect group and m2 (K) ≥ 3. Then m2 (CK (t)) ≥ 3 for all t ∈ I2 (K). Proof. Let S ∈ Syl2 (K). Then by [IG , 10.11], there exists U  S such that U ∼ = E22 . By the Thompson transfer lemma, t has a K-conjugate t ∈ CK (U ). In  view of [IG , 10.20(i)], m2 (CS (t )) ≥ 3, and the lemma follows. In the next three lemmas we consider the following situation: (6A)

(1) X is a K-group, p is an odd prime, and B ∈ Ep (X); and (2) K is a p-component of X not normalized by B.

Lemma 6.5. Assume (6A). If mp (B) ≥ m2,p (X), then p = 3. Proof. See [IG , 8.7(i)].

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6. MISCELLANEOUS

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Lemma 6.6. Assume (6A), with O2 (X) = 1 and K ≤ E(X). If mp (B) ≥ mp (X) − 1, then p = 3, |K B | = 3, and K B is the unique nontrivial orbit of B on  the set of components of O 3 (E(X)). Proof. Without loss we can pass modulo O2 (E(X)) and assume that E(X) = K1 × · · · × Kn with each Ki simple. Without loss X = E(X)B. Let B1 = NB (K1 ) and B ≤ P ∈ Sylp (X), and put Pi = P ∩ Ki ∈ Sylp (Ki ) for each i = 1, . . . , n.  Without loss K1B = K1 K2 · · · Kpr where pr = |B : B1 |. Obviously every element of B # ∩ K1B projects nontrivially on every Ki , 1 ≤ i ≤ pr . Let P i = CPi (B1 ) = 1, r 1 ≤ i ≤ pr . Then B1 centralizes P 0 = P 1 · · · P p −1 and mp (B1 ∩ P 0 ) ≤ mp (P 1 ), 0 p 0 so B1 P ∈ E (X) with mp (B1 P ) ≥ mp (B) − r + d(pr − 1) ≥ mp (X) − r + pr − 2, where d = mp (P 1 ). Therefore pr − r ≤ 2, whence p = 3 and r = 1, with m3 (B1 P 0 ) = m3 (X) and m3 (B) = m3 (X)−1. By a similar argument B1 P 0 normalizes K4 , . . . , Kn . It remains to show that a second nontrivial B-orbit {K4 , K5 , K6 } cannot exist. If it did, then B4 := NB (K4 ) = B1 , so CP4 P5 (B1 ) would similarly contain an E32 subgroup P 00 with B1 ∩ P 0 P 00 = 1. Then m3 (B1 P 0 P 00 ) =  m3 (B1 ) + 4 = m3 (B) + 3 > m3 (X), a contradiction. The lemma follows. Lemma 6.7. If (6A) holds, then |X|2 /|CX (B)|2 > 8. Proof. Let X = X/O2 (X). Then clearly |X : CX (B)|2 ≥ |X : CX (B)|2 so without loss, we may assume that O2 (X) = 1. Let {K = K1 , . . . , Kpr } = K B , B T ∈ Syl2 (CX (B)), and T ≤ S ∈ Syl2 (X). Then the mapping of T ∩ K  Bprojection B B is injective, so |T ∩ K | ≤ |K|2 while |S ∩ K | = to any component of K   r pr |K|2 . Hence |S : T | ≥ |S ∩ K B : T ∩ K B | ≥ |K|2p −1 ≥ 42 . The lemma follows.  Lemma 6.8. Suppose that X is a K-group and B ∈ Ep (X), p an odd prime. Let K be a p-component of X such that K/Op (K) ∈ Cp . If mp (B) > m2,p (X), then B normalizes K. Proof. Without loss, Op (X) = 1. Suppose that B does not normalize K. By [IG , 8.7(i)], p = 3 and (as K ∈ T3 ) K ∼ = L2 (8) ∈ Chev(3); and |B : NB (K)| = |K B | = 3. Let K B = {K = K1 , K2 , K3 }.We may write NB (K) = CB (K)×x×f  where either f induces a nontrivial field automorphism on K or f = 1, and either  x is a diagonal element of I3 ( K B ) or x = 1. If x = 1, write x = x1 x2 x3 with xi ∈ I3 (Ki ), 1 ≤ i ≤ 3; then since [f, x] = 1, [f, xi ] = 1 for each i. If x = 1, choose any mutually commuting xi ∈ I3 (CKi (f )), 1 ≤ i ≤ 3. Then f  normalizes some T ∈ Syl2 (K1 ) and [T, x2 , x3 ] = 1. Hence A := CB (K)×x2 , x3 ×f  normalizes T . Thus m2,p (X) ≥ mp (A) = mp (CB (K)) + 2 + mp (f ) ≥ mp (NB (K)) + 1 = mp (B), a contradiction.  Lemma 6.9. Suppose that p is a prime and K is a p-component of the group X with Op (X) = 1. Let B ∈ Ep (X) and set A = NB (K). Let k be an integer with 0 < k < mp (A). If K = ΓA,∗−k (K), then K ≤ ΓB,∗−k (X). Proof. This is contained in [GL1, I.22-4].



Lemma 6.10. Suppose that p is a prime and K is a p-component of X. Let T ∈ Sylr (Op (X)) for some prime r = p, and suppose that CK (T ) does not cover KOp (X)/Op (X). Then CK (T ) ≤ Op p (K).

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2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

Proof. Let X = X/Op (X), N = NX (T ), and C = CX (T ), so that C  N .  Without loss X = KOp (X). Then K = O p (X). By the Frattini argument   N = X, so C  X = K. If C = K, then O p (C) = O p (C) = O p (K) = K.    But O p (C) ≤ O p (X) = K so O p (C) lies in CK (T ), which therefore covers K in this case. Otherwise, as K is quasisimple, C ≤ Z(K) = Op (K) and the result follows.  Lemma 6.11. Let X ≤ GL2 (p), p prime, with p dividing |X| but Op (X) = 1. Then X contains SL2 (p). Proof. Since Op (X) = 1, X has at least two Sylow p-subgroups P, Q of order  p. Then X ≥ P, Q = SL2 (p). Lemma 6.12. Suppose that p is an odd prime, H ≤ GL3 (p), and H contains a transvection. If |H|p ≤ p2 , then either H is p-closed (i.e., has a normal Sylow  p-subgroup), or O p (H) ∼ = SL2 (p). 

Proof. There is no loss in assuming that O p (H) = H ≤ SL3 (p). Assume that H does not have one of the claimed forms. If H embeds in a parabolic subgroup Y  of SL3 (p), then Y is not p-closed, so O p (Y ) is an extension of SL2 (p) by a natural  module; and H, which has at least two Sylow p-subgroups, covers O p (Y /Op (Y )). As |H|p ≤ p2 and Y is irreducible on Op (Y ), H ∼ = SL2 (p), contrary to assumption. Thus H does not embed in a parabolic subgroup of SL3 (p). By the Borel-Tits theorem Op (H) = 1, and so by the Baer-Suzuki theorem there are transvections t1 , t2 ∈ H such that t1 , t2  is not a p-group. Therefore t1 , t2  ∼ = SL2 (p). On the other hand, [IA , 6.5.3] gives a list of proper subgroups of SL3 (p), into one of which H must embed. Since H contains SL2 (p) but does not embed in a parabolic subgroup, this easily yields a contradiction, proving the lemma.  Lemma 6.13. Suppose p is a prime, X has cyclic Sylow p-subgroups, Op (X) = 1, and Op (X) = 1. Then X is p-closed. Proof. If p = 2, then by a theorem of Burnside [IG , 16.5], X is a 2-group. Suppose that p > 2. Let P ∈ Sylp (X). If F ∗ (X) = Op (X) then P ≤ CX (Op (X)) = Op (X) and X is p-closed. If F ∗ (X) = Op (X) then X has a component K. By [IG , 16.11], there exists an element x ∈ K − Op (K) of order p. Then x, Op (X) is a noncyclic p-subgroup of X, contradicting the fact that P is cyclic.  Lemma 6.14. Let p be a prime. Suppose the group XD acts on the p -group W ,  where D is a noncyclic elementary abelian p-group, [X, D] = 1, and X = O p (X). If W is a K-group, then  [X, W ] = [X, CW (d)] | d ∈ D# . Proof. Obviously it suffices to show that the right side, which we call W0 , contains [X, W ]. We proceed by induction on |X|. If |X| ≤ p, then DX is elementary abelian and the result holds by [IG , 11.25] (which uses the K-group assumption). Otherwise, let X1 = Ip (X) = 1. The group N = [x , W ] | x ∈ Ip (X) is generated by normal subgroups of W , so N  W . But then CX (W/N ) contains Ip (X) = X1 . Now X/X1 acts on W/N , so by induction    N [X, W ]/N = [X, CW/N (d)] | d ∈ D# ≤ N [X, CW (d)] | d ∈ D# N ≤ N W0 /N.

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7. COMPONENTS

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But by induction, the groups generating N all lie in W0 , so N ≤ W0 . Hence  [X, W ] ≤ W0 , completing the proof. Lemma 6.15. Suppose that A = x, z ∼ = Ep2 , A ≤ X, and T is a p -subgroup of X on which x acts nontrivially. If all cyclic subgroups of A except x are NX (T ) ∩ NX (x)-conjugate, then [x, CT (z)] = 1. Proof. If false, then by the assumed conjugacy, [x, CT (y)] = 1 not only for y = z but also for all y ∈ A − x. But then by Lemma 6.14,  [x, T ] = [x, CT (y)] | y ∈ A# = [x, CT (y)] | y ∈ A − x = 1, 

a contradiction. 7. Components

Lemma 7.1. Suppose that X is a K-group with O2 (X) = 1. Let y ∈ I3 (X) and let Y be a 3-component of CX (y). Suppose that [O2 (X), Y ] = 1 and there exists x ∈ I3 (X) such that CX (x) is a 5 -group. Then Y ≤ E(X). Proof. By L3 -balance, Y ≤ L3 (X). Set W = O3 (X). It then suffices to  show that [W, Y ] = 1. Let F = F ∗ (W ); since Y = O 3 (Y ), it suffices by [IG , 3.17(ii)] to show that [F, Y ] = 1. By assumption [O2 (F ), Y ] = 1, so it suffices to show that [E(F ), Y ] = 1. But E(F ) is the central product of Suzuki groups and covering 3 groups of 2B2 (2 2 ), on each of which x induces a possibly trivial field automorphism. If E(F ) = 1, it follows that 5 divides |CE(F ) (x)|, contrary to assumption. Thus E(F ) = 1. The proof is complete.  In the following lemma, x can have any order. Lemma 7.2. Let p be a prime. Suppose that E ≤ G with E ∼ = Ep2 , and  K = O p (K) is a component of CG (e) for all e ∈ E # . Let x ∈ G and suppose that [KE, x] = 1. Then K ≤ E(CG (x)). Proof. Let e ∈ E # . We have K ≤ Lp (CCG (e) (x)) = Lp (CCG (x) (e)) ≤ Lp (CG (x)) by Lp -balance. Let W = Op (CG (x)), a K-invariant group. Then as K is a component of CG (e), [K, CW (e)] = 1. As e is arbitrary in E ∼ = Ep2 ,  [K, W ] = 1. Hence K ≤ E(CG (x)). ∼ Ep2 . Let Lemma 7.3. Let p be a prime. Suppose that E ≤ G with E = p x ∈ CG (E) and suppose that K = O (K) is a component of CG (e, x) for all e ∈ E # . Then K ≤ E(CG (x)). Proof. We have K ≤ Lp (CCG (x) (e)) ≤ Lp (CG (x)) by Lp -balance. Let W = O (CG (x)), a K-invariant group. Then p

[K, CW (e)] ≤ [E(CG (e, x)), Op (CG (e, x))] = 1. As e is arbitrary in E ∼ = Ep2 , [K, W ] = 1. Hence K ≤ E(CG (x)).



Lemma 7.4. Let p be an odd prime. Let X = Lp (X)Op (X) be a K-group and suppose that R ∈ Syl2 (Op (X)) with CX (R) covering X/Op (X). Then L2 (CX (R)) covers X/Op (X).

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2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

Proof. Set W = Op (X) and F = F ∗ (W ). Without loss, O2 (W ) = 1, whence F = O2 (W )E(W ). Set C = CX (F ). We will show that (7A)

C covers X/W.

Since C ∩ W = CW (F ) = Z(O2 (W )) ≤ Z(C), it will follow that C/Z(C) is a product of components, whence [C, C] = L2 (C) = E(C), which also covers X/W . As C  X, [C, R] ≤ C ∩ W ≤ Z(C), so [L2 (C), R] = 1 by the Three Subgroups lemma. Thus L2 (C) ≤ CX (R), and as L2 (C)  X, L2 (C) ≤ L2 (CX (R)), which will imply the result. Let C ∗ = CX (R ∩ F ). Since C ∗ ≥ CX (R), C ∗ covers X/W . Moreover, C  C ∗ and C ∗ /C embeds in CAut(F ) (R ∩ F ), which is solvable (see [IA , 7.1.1b]). Hence (C ∗ )(∞) ≤ C, and as X/W is perfect, (C ∗ )(∞) still covers X/W . Thus (7A) holds and the proof is complete.  8. Rigidity, Semirigidity, and Terminality Definition 8.1. Let p be a prime and K, L ∈ Kp . We say that L ↑p K, or equivalently K ↓p L, if and only if there exists x ∈ Ip (Aut(K)) and a component L0 of CK (x) such that L ∼ = L0 /Op (L0 ). In this case we may say that L ↑p K (or K ↓p L) via x. Definition 8.2. Let J be a quasisimple group, x ∈ I2 (Aut(J)), and K a component of CJ (x). We say that K is Z4 -semirigid in J (with respect to x) if and only if Sylow 2-subgroups of CAut(J) (K) are cyclic of order at most 4. Notice that if |CAut(J) (K)|2 = 2, then K is semirigid in J [IG , 7.1]. To define Z4 -semirigidity of components of involution centralizers in an ambient simple group G of even type, we must set up the situation just as we did in [IG , Sec. 7] for the notion of semirigidity. Thus, let (x, K) ∈ ILo2 (G), let Q ∈ Syl2 (C(x, K)), and expan Q to R ∈ Syl2 (CG (K)). We are interested in involutions t ∈ I2 (R) satisfying one of the following conditions: (8A)

(1) t ∈ Q; or (2) t ∈ R − Q and K has a trivial pumpup in CG (y) for all y ∈ I2 (Q).

In these respective cases we choose an involution z as follows: (8B)

(1) z = x; or (2) z ∈ I2 (Z(R)).

In either case, z ∈ Q and (z, K) ∈ ILo2 (G); moreover, [t, K z] = 1. Now we can define Z4 -semirigidity in G. Definition 8.3. Let G be a group of even type and let (x, K) ∈ ILo2 (G), Q ∈ Syl2 (C(x, K)), and Q ≤ R ∈ Syl2 (CG (K)). We say that (x, K) is Z4 -semirigid in G if and only if whenever t ∈ I2 (R) satisfies (8A), K has a (nontrivial) vertical pumpup J in CG (t), and z is as in (8B), then K is either semirigid or Z4 -semirigid in J with respect to z. Lemma 8.4. Let G be a group of even type and let (x, K) ∈ ILo2 (G) be Z4 semirigid in G. Let Q ∈ Syl2 (C(x, K)). Then one of the following holds: (a) K is terminal in G;

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8. RIGIDITY, SEMIRIGIDITY, AND TERMINALITY

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(b) m2 (Q) ≤ 3, with Q having a maximal subgroup embeddable in a dihedral group if m2 (Q) = 3; (c) There exists t ∈ I2 (Z(Q)) such that the pumpup (t, J) of (x, K) in CG (t) is nontrivial, CQ (J)  Q, x ∈ CQ (J), and Q/CQ (J) is cyclic of order at most 4. Proof. We parrot the first part of the proof of [IG , 7.4], using Lemma 4.7 instead of [IG , 7.3]. Assume that (a) and (b) fail and expand Q to R ∈ Syl2 (CG (K)). Choose t ∈ I2 (R) such that (x, K) has a nontrivial pumpup (t, J) in CG (t), with t ∈ Q if possible. According as t ∈ Q or t ∈ R − Q, set P = Q or R, and refine the choice of t ∈ P so that P1 := CP (t) is as large as possible. If K is semirigid in J, we simply repeat the proof of [IA , 7.3], so assume that Sylow 2-subgroups of CAut(J) (K) are isomorphic to Z4 . If P1 = P , then as Z(R) ≤ Z(Q), t ∈ Z(Q). Then conclusion (c) follows immediately, the last assertion as (x, K) is Z4 -semirigid in G by assumption. We therefore assume that P1 < P and argue to a contradiction. According as P = Q or R, set z = x or choose any z ∈ I2 (Z(R)). Let P2 = CP1 (J). Since P1 centralizes K, z, and t, P1 normalizes JJ z (in case of a diagonal pumpup) or J (in case of a vertical pumpup). In either case, by [IG , 6.19] and our semirigidity hypothesis, P2  P1 with P1 /P2 cyclic of order 2 or 4. Moreover, the pumpup of K is clearly nontrivial in CG (u) for every u ∈ I2 (P2 ). Hence by our choice, |CP (u)| ≤ |P1 | for every u ∈ I2 (P2 ). As t ∈ Z(P ), Lemma 4.7 implies that m2 (P ) ≤ 3, with P ∼ = F (n) for some n in case of equality; the latter condition implies that P has a dihedral maximal subgroup. Whether P = Q or R, Q inherits the properties stipulated in (b). The proof is complete.  Lemma 8.5. Suppose that p is a prime. Assume that in a sequence (x0 , K0 ) < · · · < (xn , Kn ) of pumpups in ILp (G), all pumpups are trivial or diagonal. If p = 2 and K0 /O2 (K0 ) is a proper quotient of a quasisimple group of 2-rank 1, assume that Ki /O2 (Ki ) is simple for all i = 0, . . . , n. If mp (C(xn , Kn )) = 1, then mp (C(x0 , K0 )) = 1. Proof. This is [III8 , 2.4] or [III2 , 1.11].



We recall three definitions from [IG ]. Definition 8.6. Let (x, K) ∈ ILop (G). We say that (x, K) is Gilmanmaximal in G if and only if whenever (x, K) 1. Choose U  Q with U ∼ = Ep2 . For each u ∈ Ipo (G) ∩ Q, let (u, Ku ) be a pumpup of (x, K) in CG (u). By one-step rigidity, either Ku is a trivial pumpup of K, or Ku is one of p p-components Ku = Ku1 , . . . , Kup cycled by x and generating their direct product modulo Op (CG (u)). We say in the latter case that K diagonalizes in CG (u). Define U = {u ∈ Ip (Q) | Ku is a trivial pumpup of K}. It suffices to show that Ipo (G) ∩ Q ⊆ U. For then Gilman-maximality immediately implies the only other condition for Ipo -terminality. Namely, if z ∈ Ip (Z(Q)) ∩ Ipo (G) and Q ≤ Q∗ ∈ Sylp (C(z, Kz )), Kz being the trivial pumpup of K in CG (z), then Q = Q∗ . Notice that mp (Q) ≥ mp (C(u, Ku )) for any u ∈ Ipo (G)−U (and we may assume that such elements exist). But for such an element u, mp (C(u, Ku1 )) ≥ mp (Ku2 · · · Kup ) = (p − 1)mp (K). Therefore by (a) or (b), mp (Q) ≥ 4. We show that U # ⊆ U. As mp (CQ (U )) = mp (Q) ≥ 4 by [IG , 10.20], U # ⊆ Ipo (G). Let u ∈ U # and suppose by way of contradiction that K diagonalizes in CG (u). Then CQ (U ), which contains x and centralizes u Lp (CK (u)), must normalize Ku1 · · · Kup . Consequently Q0 := NCQ (U) (Ku1 ) normalizes each Kui and |CQ (U ) : Q0 | = p. In particular mp (Q0 ) ≥ mp (CQ (U )) − 1 = mp (Q) − 1. But Lp (CK (u)) projects (mod Op (CG (u))) onto each Kui , so Q0 ≤ C(u, Ku1 · · · Kup ). As each Kui contains an element of order p outside Op p (Kui ) by [IG , 16.11], mp (C(u, Ku1 )) ≥ p − 1 + mp (Q0 ) ≥ p − 2 + mp (Q) > mp (Q), contradicting the Gilman-maximality of (x, K). Hence, U # ⊆ U. Next, if u ∈ CQ (U ) − U , and u ∈ Ipo (G), then by the previous paragraph and [IG , 6.20], the pumpup of (x, K) in CG (u) is trivial. Thus CQ (U ) ∩ Ipo (G) ⊆ U. Similarly, if u ∈ Q ∩ Ipo (G) and mp (CCQ (U) (u)) > 1, then u ∈ U by another application of [IG , 6.20]. Thus to complete the proof, we need only assume that there is u ∈ Ip (Q − CQ (U )) ∩ Ipo (G) − U with CCQ (U) (u) =: R being cyclic, and derive a contradiction. Note that x ∈ Z(Q) so x = Ω1 (R). In this situation CQ (u) normalizes Ku1 · · · Kup and in particular R does as well. As x cycles the Kui , R = x. Thus |CQ (u)| = p2 . Hence by [III2 , 7.1], mp (Q) ≤ 2p − 3. If p = 3, then mp (Q) ≤ 3, contradiction. Thus, mp (K) > 1. But then mp (C(u, Ku1 )) ≥ 2(p − 1) > 2p − 3, as shown above. This contradiction completes the proof. 

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9. CENTRALIZERS OF COMPONENTS

27

9. Centralizers of Components We include in this section several specific applications of the p-component uniqueness theorems of [II3 ], and in particular on Lemma 5.1, despite their dependence on K-group properties. The idea in all of them is to bound the 2-structure of C(t, K), and in particular the rank m2 (C(t, K)), whenever (t, K) ∈ ILo2 (G) with K of certain specific isomorphism types. Here, as always, G is a K-proper simple group, and we assume in addition that G has even type. Lemma 9.1. Suppose that G is of even type. Let (t, K) ∈ ILo2 (G) and suppose  that K/O2 (K) ∼ = L± 4 (3), G2 (3), M11 , J3 , F i24 , F3 , or F1 . Then m2 (C(t, K)) = 1. Proof. In every case it follows easily from [IA , 5.3] as well as [IA , 2.2.10, 4.9.6, 5.2.9] and the definition of C2 [I2 , 12.1] that no vertical pumpup of K lies in C2 . Also by [IA , 6.1.4] and [I2 , 12.1], no covering group of K, except 2U4 (3), lies in C2 . Hence by Lemma 8.5 and [IG , 6.10], if m2 (C(t, K)) > 1, then there exists a terminal long pumpup (t0 , K0 ) of (t, K) with K0 one of the hypothesized groups of this lemma, and such that m2 (C(t0 , K0 )) > 1. However, by [VK , 8.5, 8.8], ΓD,1 (K0 ) = K0 for every four-subgroup D ≤ Aut(K0 ). Therefore Lemma 5.1 is contradicted and the lemma is proved.  For the next result we need a preliminary calculation. Lemma 9.2. Let K = F i23 and T ∈ Syl2 (K). Then T is indecomposable. Proof. From [IA , 5.3tud], T ∈ Syl2 (CK (u)) for some u ∈ I2 (K) such that CK (u) ∼ = 2F i22 . Also a Sylow 2-center of F i22 has order 2, so Z(T ) ≤ [T, T ]. Hence T has no nontrivial abelian direct factor. Let A = J(T ); then A = CK (A) ∼ = E211 and NK (A)/A ∼ M has one class of involutions. The lemma then follows by = 23 Lemma 4.11.  Lemma 9.3. Let (t, K) ∈ Lo2 (G) with K terminal in G. (a) If K/Z(K) ∼ = HS or L± 3 (3), then m2 (C(t, K)) = 1; , then C(t, K) = t; (b) If K ∼ F i = 23 or U (2), then m2 (C(t, K)) ≤ 2. (c) If K ∼ A = 6 4 Proof. As usual, by [II3 , Theorem PU4 ], K is standard in G. Let Q ∈ Syl2 (C(t, K)). By standardness, ΓQ,1 (G) ≤ M := NG (K). Part (a) is immediate from Lemma 5.1, in view of the generational properties of K; see [VK , 8.8] and [IA , 7.3.4]. Suppose (b) fails. Then |Q| > 2. Let T ∈ Syl2 (CG (t)) with Q ≤ T . As Out(F i23 ) = 1, T =P ×Q with P ∈ Syl2 (K).By Lemma 9.2, P is indecomposable. Suppose that m2 (Q) = 1. By the Z ∗ -theorem, tg ∈ T − Q for some g ∈ G. Thus g ∈ M . As all involutions in K are 2-central, we may assume that tg ∈ Z(T ). Then by Burnside’s lemma we may assume that g ∈ NG (T ). But if Q is cyclic, then as |Q| ≥ 4 and Z(P ) is a four-group, t = Ω1 (Φ(Z(T )))  NG (T ), contradiction. Similarly if Q is quaternion then t  NG (T ) by the Krull-Schmidt theorem. Therefore m2 (Q) > 1. Choose g ∈ G − M such that V := Qg ∩ M has (1) maximal 2-rank; and (2) maximal order, subject to (1). By [II3 , Corollary PU2 ], m2 (V ) ≥ 2. Without loss V ≤ T . If V ∩ P = 1 then Q ≤ CG (V ∩ P ) ≤ M g and so V = Qg by maximality. If V ∩ P = 1 then V projects isomorphically into Q,

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28

2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

so ΓV,1 (Q) ≤ M g and M g contains a maximal elementary abelian subgroup of Q. In particular it contains a central involution z of Q, so some element of V projects onto z and then V ∼ = Q by maximality. Hence, in any case, V = Qg . By Lemma 5.2 and the fact that Out(K) = 1, either Q is abelian or Qg ∩K = 1. In either case, since all involutions of K are 2-central in K, we may modify g by an element of M and assume that tg1 ∈ Z(T ) − Q for some t1 ∈ I2 (Z(Q)) ⊆ Z(T ). As CG (t1 ) ≤ M , T ∈ Syl2 (CG (t1 )) and so by Burnside’s lemma, tg1 = th1 for some h ∈ NG (T ). Clearly then h ∈ M . Hence Qh ∩ Q = 1. If Q is not abelian, then by the Krull-Schmidt theorem and the fact that Q ∩ Qh = 1, Q has a direct factor isomorphic to P . But then V ∼ = P and so ΓV,1 (K) ≥ ΓP,1 (K) = K as K has no strongly embedded subgroup. But this contradicts [IG , 18.8]. So Q is abelian, and as Φ(Z(P )) = 1, Q is elementary abelian. Now ΓZ(P ),1 (K) = K by [VK , 8.9], so as m2 (Q) > 1, ΓQh ,1 (K) = K, again a contradiction. Thus, (b) holds. In (c), assume that m2 (Q) ≥ 3. By [VK , 8.4], K is outer well-generated for the prime 2, so by [II3 , Corollary PU2 ] and [IG , 18.11], there is g ∈ G − M such that Qg ≤ M . Then as m2 (Q) ≥ 3, K = ΓQg ,1 (K) ≤ M g by [VK , 8.5], contradicting  [IG , 18.8] and completing the proof. Lemma 9.4. Let (u, H) ∈ ILo2 (G) with H ∼ = U4 (2). Suppose that there exists t ∈ I2 (CG (u)) such that L := E(CK (t)) ∼ = A6 and L is a component of CG (t). Then m2 (C(u, H)) ≤ 2. Proof. Suppose false and let Q ∈ Syl2 (C(u, H)) be t-invariant. We argue that (u, H) is terminal in G. Note that by [VK , 14.2b], H is semirigid in G. Hence by [IG , 7.4], if H is not terminal, there exists z ∈ I2 (Z(Q t)) ∩ Q such that the subnormal closure J of H in CG (z) satisfies J > H, and Q = u CQ (J). Then t normalizes J and so L is a component of CJ (t). Hence J cannot be a diagonal pumpup of H, nor can it be isomorphic to L4 (4), by [IA , 4.9.1, 4.9.2] and the BorelTits theorem. Therefore by [VK , 3.10], J/Z(J) ∼ = L± 4 (3). Moreover, as m2 (Q) ≥ 3, m2 (CQ (J)) ≥ 2 and so m2 (C(z, J)) ≥ 2. This contradicts Lemma 9.1 and proves the terminality of H. Now Lemma 9.3c completes the proof.  Corollary 9.5. Suppose that t ∈ I2 (G) and K ∼ = F i23 is a component of CG (t). Then Sylow 2-subgroups of CG (K) have a subgroup of index 2 which is cyclic or quaternion. Proof. Let Q ∈ Syl2 (CG (K)). By Lemma 9.3b we may assume that K is not standard. Hence K is not terminal. Take a chain K = K0 , K1 , . . . , Kn of components of involution centralizers such that Kn is standard and each Ki is a nontrivial pumpup (possibly diagonal) of Ki−1 . Then Kn ∼ = F i24 or K, by [VK , 3.10]. For i < n, therefore, Ki ∼ = K. Moreover, for 1 ≤ i < n, m2 (CG (Ki )) ≥ m2 (K) = 11. Thus in any case m2 (CG (Kn−1 )) ≥ 2. If n > 1 then as Kn−1 is semirigid, m2 (CG (Kn )) ≥ m2 (CG (Kn−1 )) − 1 ≥ 2 by [IG , 7.4]. But this violates Lemma 9.1 or 9.3b. Therefore n = 1. Again by semirigidity and assuming the corollary is false, there is z ∈ I2 (Z(Q)) such that K1 ∼ = F i24 is a component of CG (z). Then Q = z×Q0 , where Q0 = CQ (K1 ). Again by Lemma 9.1, m2 (Q0 ) = 1  so Q0 is cyclic or quaternion. The proof is complete.

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10. RECOGNITION THEOREMS

29

10. Recognition Theorems The twelve groups in the conclusion of Theorem C5 will require a variety of recognition theorems, in addition to the Curtis-Tits theorem [IA , 2.9.3]. First, we recognize Ω7 (3) by a weak Phan system. Although this by itself only determines Spin7 (3) or Ω7 (3) modulo a 3-group, we have enough 2-local information to “kill the core” easily. The recognition theorem is due to Gramlich, Horn, and Nickel [GHN, Lemma 4.4]. For the meaning of “standard Phan-pair,” see [III13 , 1.5] and [III17 , 4.3]. Theorem 10.1. Let {L, I, H} be a weak Phan system of type B3 (3) in G. That is, L, I form a standard Phan-pair in L, I ∼ = U3 (3), I, H form a standard Phanpair in I, H ∼ = B2 (3), and [L, H] = 1. Let X = L, I, H. Then X/O3 (X) is a nontrivial homomorphic image of Spin7 (3). We need two recognition theorems a` la Wong [Wo1] and Finkelstein-Solomon [FinS1], for the target groups C o = Ω− 8 (3) and U7 (2). The theorems are very similar to some of the results in [III13 , Sec. 3]. We set up the situations, state the theorems, and show how the arguments in [III13 , Sec 3] adapt to give the proofs. In both cases we let V be a natural module for C o and let W = F3 v or F4 v be a 1-dimensional subspace with (v, v) = 1. We write C o (V ) for the derived subgroup of the isometry group of V . We assume given a surjective homomorphism f : C o (W ⊥ ) → K

(10A)

where K is a subgroup of G. Note that C o (W ⊥ ) ∼ = Ω7 (3) or SU6 (2). We write V = V0 ⊥ V1 ⊥ · · · ⊥ V7 where V7 = W and Vi , 1 ≤ i ≤ 6, is an isometric copy of W . Moreover, V0 = 0 in the unitary case, while in the orthogonal case, dim V0 = 1 but V0 is not isometric to W . For every nonempty I ⊆ X := {0, 1, 2, . . . , 7} we define

(10B)

I = X − I  VI = Vi and i∈I

CIo = CC o (VI  ) ∼ = Ω(VI ) or SU (VI ) Note that VI  = VI⊥ . o In the Ω− 8 (3) case, the subgroup T of C = Ω(V ) of even permutation matrices fixing V0 is isomorphic to A7 and permutes {V1 , . . . , V7 } naturally. In the U7 (2) case, the subgroup T of all permutation matrices is isomorphic to Σ7 and permutes {V1 , . . . , V7 } naturally. In either case, we assume the existence of a subgroup N ≤ G and a homomorphism λ:N →T

(10C) with the following properties: (10D)

(1) For all t ∈ C o (W ⊥ ) ∩ T , f (t) ∈ N and λ(f (t)) = t; (2) For all I ⊆ X and all u ∈ N such that 7 ∈ I ∪ λ(u)(I), o ). f (CIo )u = f (Cλ(u)(I)

We then can define HD , for any subset D ⊆ X such that D − {0} = {1, . . . , 7} and |D| ≥ 2, as follows: choose t ∈ T and F ⊆ X − {7} with t(F ) = D; then choose

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30

2. GENERAL GROUP-THEORETIC LEMMAS, AND RECOGNITION THEOREMS

nt ∈ N such that λ(nt ) = t; and then set HD = f (CFo )nt . Then as established in [III13 , 3.4], for any such D,

(10E)

(1) HD is independent of the choices of F , t, and nt ; o ) if 7 ∈ D; and (2) HD = f (CD n = (3) For any n ∈ N with λ(n) = t(∈ T ), we have HD Ht(D) .

The theorems are then as follows: Theorem 10.2 (Theorem Ω− 8 (3)). Suppose that G has subgroups K and N , and with the notation just described, there is an isomorphism f : Ω7 (3) ∼ = C o (W ⊥ ) → K as in (10A) and a homomorphism λ as in (10C) such that the conditions (10D) hold. Then there exists a isomorphism g : Ω− 8 (3) → K, N  such that g agrees with o f on a Ω− (3) subgroup of C . 6 Proof. Theorem 3.1 of [III13 ], which characterizes Ω− 8 (q), q odd, q > 3, does (3). However, the proof extends to cover this case with not cover the case C o ∼ = Ω− 8 the changes indicated below. References to results “3.x” are to [III13 ]. The proof of Theorem 3.1 holds without change through Lemma 3.4. The proof of Lemma 3.5 fails if |D| = 4 and o ∼ CD = Ω+ 4 (3) (see [III17 , 9.1d]). Accordingly, in this case, the final two sentences of the proof of Lemma 3.6 must be altered as follows. If D ∪ F = {0, 1, . . . , 7}, then CFo ∼ = Ω− 4 (3), and we may simply reverse the roles of D and F . So, we may assume that D = {1, 2, 3, 4} and F = {5, 6, 7}. In this case, let D∗ = D ∪ {0}, o o o D1∗ = D∗ − {1} and D2∗ = D∗ − {2}. Then CD ∗ = CD ∗ , CD ∗  and [HD ∗ , HF ] = 1 i 1 2 for i = 1, 2 by part (a) of Lemma 3.6. Hence, [HD∗ , HF ] = 1. Since HD ≤ HD∗ , we get [HD , HF ] = 1 in this case as well, completing the proof of Lemma 3.6. The proof resumes with Definition 3.11 of a function h : C7o ∪ C1o → H extending the function f : C7o → H. In the proof of Lemma 3.12, we replace the second paragraph with the following: o . Recall that σ = (2, 1, 6). Since στ −1 = (1, 6, 7) centralizes Now let x ∈ C{0,2,3} o C{0,2,3} , it follows that xσ = xτ . Then f (x)nσ = f (xσ ) = f (xτ ). Conjugating by o o nγ which centralizes both f (C{0,2,3} ) = H{0,2,3} and f ((C{0,2,3} )σ ) = H{0,1,3} , by (3I), we see that f (x)nτ = f (xσ ) = f (xτ ), whence

−1

h0 (x) = (f (xτ ))nτ = f (x) o o o o for all x ∈ C{0,2,3} . Now, C{1,7}  = C{0,3,4,5,6} , C{0,2,3}  by [III17 , 9.1]. Hence, o h0 (x) = f (x) for all x ∈ C{1,7} , as claimed, and the lemma follows. Now the proofs of Lemma 3.13 and Theorem 3.19 hold without change, completing the proof of the theorem. 

Fortunately, although Theorem 3.1 of [III13 ] is stated only for C o ∼ = Um+1 (2) with m ≥ 7, the proof holds without change when m = 6, yielding the following theorem. Theorem 10.3 (Theorem U7 (2)). Suppose that G has subgroups K and N , and with the notation described above, there is a surjective homomorphism f : SU6 (2) ∼ = C o (W ⊥ ) → K as in (10A) and a homomorphism λ as in (10C) such that

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10. RECOGNITION THEOREMS

31

 the conditions (10D) hold. Then there exists an isomorphism g : U7 (2) → K N extending f . The other groups of Lie type in Theorem C5 will be recognized by the CurtisTits theorem. As for the sporadic groups in Theorem C5 , we note first that complete proofs of the following two theorems about Fischer’s groups [Fi2] were published by Aschbacher [A19], making use of Fischer’s unpublished lecture notes (University of Warwick, 1969). Theorem 10.4. Let X be a finite simple group and t ∈ I2 (X). (a) If CX (t) ∼ = 2U6 (2), then X ∼ = F i22 . (b) If CX (t) ∼ = 2F i22 , then X ∼ = F i23 . Proof. By [A19, Theorems 31.1, 32.1] and the Z ∗ -theorem [IG , 15.3], X is determined up to isomorphism in each case. Since F i22 and F i23 exist and have the requisite property, the theorem is true.  Theorem 10.5. Let X be a finite simple group, z ∈ I2 (X), and C = CX (z). Assume that Q := F ∗ (C) ∼ = 21+12 , F ∗ (C/Q) ∼ = 3U4 (3), and |C/Q : F ∗ (C/Q)| = 2. Assume also that z is not weakly closed in Q with respect to X. Then X ∼ = F i24 . Proof. This follows similarly from [A19, Theorem 34.1].



Finally, we quote results of Aschbacher; Segev; and Griess, Meierfrankenfeld, and Segev, for the next two recognition theorems. Theorem 10.6. Let X be a finite simple group, z ∈ I2 (X), and C = CX (z). Assume that Q := F ∗ (C) ∼ = 21+8 and C/Q ∼ = Ω+ 8 (2). Assume also that z is not weakly closed in Q with respect to X. Then X ∼ = Co1 . Proof. This is [A2, Lemma 49.15].



Theorem 10.7. Let X be a finite simple group and t, z ∈ I2 (X). Let K = CX (t), C = CX (z), and Q = F ∗ (C). (a) [Seg1] If Q ∼ = Co2 , F ∗ (K) ∼ = 22E6 (2), and K = F ∗ (K) u = 21+22 , C/Q ∼ ∗ ∼ for some involution u ∈ F (K), then X = F2 ; and (b) [GrMeSeg] If Q ∼ = 21+24 , C/Q ∼ = Co1 , and K ∼ = 2F2 , then X ∼ = F1 .

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10.1090/surv/040.9/03

CHAPTER 3

Theorem C5 : Stage 1 1. Introduction As explained in the opening chapter of this volume, for the proof of Theorem C5 , we assume the following about our K-proper simple group G: (1A) Here

(1) G is of even type; (2) σ0 (G) = ∅; and (3) For some p ∈ σ0 (G), every element of Lop (G) is a Cp group.  σ0 (G) = p | p is an odd prime, m2,p (G) ≥ 4,

 and G has no strong p-uniqueness subgroup .

We review some other relevant definitions. First, G is of even type, by definition, if and only if (1B)

(1) m2 (G) ≥ 3; (2) O2 (CG (t)) = 1 for all t ∈ I2 (G); and (3) Lo2 (G) ⊆ C2 .

Moreover, if G is of restricted even type, as it is throughout this volume, then in addition, Lo2 (G) ⊆ Co2 . Indeed, if G is of restricted even type, then G satisfies two more technical requirements [I2 , 8.8], but they are of no significance in this volume. Because of the importance of the sets C2 , Co2 , and Cp for Theorem C5 , we repeat the definitions of the sets C2 and Cp here for reference [I2 , 12.1], as well as the definition of Co2 . Note that Cp is a subset of Kp , i.e., it consists of certain quasisimple K-groups K such that Op (K) = 1. The elements of Cp are also called Cp -groups, and the elements of Co2 are also called Co2 -groups. Definition 1.1. Let p be a prime (possibly p = 2). A simple K-group K is a Cp -group if and only if K ∈ Chev(p), or K ∼ = Ap , A2p , or A3p , or one of the following holds: (a) p = 2 and either (1) K ∼ = L2 (q), q ∈ FM9 (i.e., q is a Fermat or Mersenne prime or 9); (2) K ∼ = L3 (3), L4 (3), U4 (3), or G2 (3); or (3) K ∼ = M11 , M12 , M22 , M23 , M24 , J2 , J3 , J4 , Co2 , Co1 , HS, Suz, Ru, F i22 , F i23 , F i24 , F3 , F2 , or F1 ; (b) p = 3 and either 1 (1) K ∼ = U5 (2), U6 (2), Sp6 (2), D4 (2), 3D4 (2), F4 (2), 2F4 (2 2 ) , or Sp4 (8); or 33 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

3. THEOREM C5 : STAGE 1

34

∼ M11 , J3 , M c, Ly, Suz, O  N , Co3 , Co2 , Co1 , F i22 , F i23 , F i , (2) K = 24 F5 , F3 , F2 , or F1 ; (c) p = 5 and either 5 1 5 (1) K ∼ = 2B2 (2 2 ), 2F4 (2 2 ) , or 2F4 (2 2 ); or (2) K ∼ = J2 , HS, M c, Ly, Ru, He, Co3 , Co2 , Co1 , F5 , F3 , F2 , or F1 ; (d) p = 7 and K ∼ = He, O  N , Co1 , F i24 , F3 , or F1 ; or (e) p = 11 and K ∼ = J4 . Furthermore, a quasisimple K-group K ∈ Kp is a Cp -group if and only if K/Op (K) is a Cp -group, with the following exceptions: For p = 2: The groups SL2 (q), q ∈ FM9, 2A8 , SL4 (3), SU4 (3), Sp4 (3), and [X]L3 (4) with X of exponent 4, are not included among the C2 -groups. For p = 3: The groups 3A6 and 3O  N are not included among the C3 -groups. Definition 1.2. Co2 = C2 − {K | K ∼ = L2 (q), q > 17, q a Fermat or Mersenne prime}. The definition of strong p-uniqueness subgroup is given in [I2 , Section 8] and [III1 , 1.2]. It is a variation of the notion of strongly p-embedded subgroup, and we shall repeat its relevant features presently. We fix an (odd) prime p satisfying the condition (1A3). Recall that Ep (G) is the set of all noncyclic elementary abelian p-subgroups of G. Also for any p-group B and subgroup X ≤ G, IX (B; 2) is the set of all B-invariant 2-subgroups of X. Moreover, m2,p (G) is the largest p-rank among all 2-local subgroups of G. We define    Bp (G) = B ∈ Ep (G)  IG (B; 2) = {1}    (1C) Bp∗ (G) = B ∈ Bp (G)  mp (B) = m2,p (G) Thus Bp∗ (G) = ∅. The definition of σ0 (G) implies that G possesses a 2-local subgroup of p-rank ≥ 4, so we have mp (B) ≥ 4 for all B ∈ Bp∗ (G). In particular, mp (G) ≥ 4. Recall that Sp (X), for any group X with mp (X) ≥ 2, is the set of all elementary abelian p-subgroups A of X that contain every element of order p in CX (A). Thus Sp (X) consists precisely of the maximal elements of Ep (X) with respect to inclusion. In this chapter, we prove Theorem C5 : Stage 1. Suppose that G satisfies (1A). Then (a) G is balanced with respect to any element of Ep4 (G); (b) Bp∗ (G) ∩ Sp (G) = ∅; (c) For every x ∈ Ipo (G) (i.e., every element x ∈ G of order p such that mp (CG (x)) ≥ 4), Op (CG (x)) has odd order; and (d) For every A ≤ B ∈ Bp∗ (G), Op (CG (A)) has odd order. For the notion of balance and related ideas, see the signalizer functor method sections [IG , Sections 20-22]. Several key arguments in the proof of Theorem C5 : Stage 1 rely on constructing a strong p-uniqueness subgroup of G, thereby contradicting the assumption that p ∈ σ0 (G).

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1. INTRODUCTION

35

For clarity we repeat some of the definitions of various kinds of uniqueness subgroups. A proper subgroup M of G is called a p-uniqueness subgroup (in G) provided ΓP,2 (G) ≤ M for some Sylow p-subgroup P of G. A proper subgroup M of G is strongly p-embedded in G if and only if ΓP,1 (G) ≤ M for some P ∈ Sylp (G). Certain proper subgroups M of G that have very particular structures (the details are technical, and omitted here) and satisfy NG (D) ≤ M for many p-subgroups 1 = D ≤ M are called almost strongly p-embedded in G. Strongly p-embedded subgroups are by definition included in the set of almost strongly p-embedded subgroups. For details see [I2 , Sec. 8] and [III1 , Definition 1.2]. A proper subgroup M of G is a p-component preuniqueness subgroup if and only if M possesses a p-component K such that CG (x) ≤ M for all x ∈ Ip (CM (K/Op (K))). Alternatively, M may be called a K-preuniqueness subgroup. In the course of Stage 1, we prove that G contains a so-called LCp -uniqueness subgroup. This is by definition a proper subgroup M of G such that (a) ΓP,2 (G) ≤ M for some P ∈ Sylp (G); (b) Op (M ) has even order, and for any T ∈ Syl2 (Op (M )), CM (T ) is a p group; and (c) For any A ∈ Sp (M ), every A-invariant p -subgroup of M lies in Op (M ). Finally, a proper subgroup M of G is a strong p-uniqueness subgroup of G if (a) Either M is a p-component preuniqueness subgroup of G that is almost strongly p-embedded in G, or M is an LCp -uniqueness subgroup of G; and (b) Either (1) or (2) holds: (1) Op (M ) = 1, but Op (X) = 1 for all M < X ≤ G; or (2) M = NG (K) for some K ≤ G with K ∈ Chev(2). Moreover, mp (CM (K)) ≤ 1. As explained in [I2 ], if M is a strong p-uniqueness subgroup satisfying condition (b2) of the definition, then in fact it follows that M is almost strongly p-embedded in G. As for the proof of Theorem C5 : Stage 1, the critical conclusions are (a) and (b), and most particularly (b). Conclusions (c) and (d) then follow easily. Thus most of the analysis takes place under the indirect hypothesis (1D)

Bp∗ (G) ∩ Sp (G) = ∅,

that is, some elementary abelian p-subgroup B of rank m2,p (G) and lying in a 2-local subgroup of G contains every element of order p in CG (B). When the definition of strong p-uniqueness subgroup is unraveled, and pcomponent uniqueness theorems [II3 , Theorems PU2 –PU4 ] are taken into account, we see that conclusion (b) of Stage 1 will follow from the following theorem, which we shall prove.

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36

3. THEOREM C5 : STAGE 1

Theorem 1.3. Let B ∈ Bp∗ (G). If B ∈ Sp (G), then G contains a maximal subgroup M such that M contains a Sylow p-subgroup P of G, Op (M ) has even order, and either (a) or (b) holds: (a) M is a p-component preuniqueness subgroup with respect to some p-component L of M and if Q ∈ Sylp (CM (L/Op (L))), then either (1) mp (Q) ≥ 2, mp (L) ≥ 2, and M controls the G-fusion of noncyclic subgroups of Q in P ; or (2) mp (Q) ≤ 1, and M is almost strongly p-embedded in G; or (b) The following conditions hold: (1) ΓP,2 (G) ≤ M ; (2) P acts faithfully on a Sylow 2-subgroup of Op (M ); and (3) For any A ∈ Sp (M ), Op (M ) contains every A-invariant p -subgroup of M . Indeed, if conclusion (b) of Theorem 1.3 holds, then by definition, and by a Frattini argument, M is an LCp -uniqueness subgroup. Moreover as Op (M ) = 1, and as M is maximal, Op (X) = 1 for all M < X ≤ G. Thus M is a strong puniqueness subgroup in this case. On the other hand if conclusion (a1) of Theorem 1.3 holds, then by [II3 , Theorem PU2 ], M is almost strongly p-embedded in G. The same condition holds trivially if (a2) holds, so M satisfies the first part of the definition of a strong p-uniqueness subgroup. As M is a maximal subgroup of G with Op (M ) = 1, the second part of that definition is trivially satisfied whether conclusion (a1) or (a2) of Theorem 1.3 holds. Thus M is a strong p-uniqueness subgroup of G, as desired. As usual, we divide the proof of Theorem C5 : Stage 1 into a number of separate theorems. The first is conclusion (a) of Theorem C5 : Stage 1. Theorem 1.4 (Theorem 1). Assume that (1A) holds and A ∈ Ep4 (G). Then G is balanced with respect to A. If Theorem 1 fails, then we show that G has a strong p-uniqueness subgroup as in Theorem 1.3a. Indeed, because the elements of Lop (G) are Cp -groups, only a single configuration can occur as an obstruction to Theorem 1. It arises as the unique exceptional case in the following strong local balance property of Cp -groups [IA , 7.7.10, 7.7.12]. Keep in mind that throughout this chapter p is a fixed odd prime. Proposition 1.5. Let X be a group with Op (X) = 1 and let K be a component of X which is a Cp -group. Let A ∈ Sp (X) and let W be an A-invariant p -subgroup of X. If AW normalizes K and [W, K] = 1, then p = 3, K ∼ = L2 (27), and the image of AW in Aut(K) is isomorphic to Z3 × A4 , with the Z3 direct factor generated by a nontrivial field automorphism of K. Since AutW (K) is a nontrivial p -group invariant under AutA (K), K is not strongly locally 1-balanced with respect to p, whence [IA , 7.7.12] applies to yield the proposition. In particular, it follows in the present situation that if (x, y, K) is an unbalancing triple, or more generally a locally unbalancing triple, for some x ∈ Ipo (G), then necessarily p = 3 and K/Op (K) ∼ = L2 (27), and y ∈ Ip (CG (x)) normalizes K and induces a nontrivial field automorphism on K/Op (K). (See [V2 , 1.5].)

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1. INTRODUCTION

37

Now, the crux of the verification of balance with respect to elements of Ep4 (G), and hence the crux of the proof of Theorem 1, is essentially to establish the following statement: (1E)

If p = 3 and (x, y, K) is a locally unbalancing triple in G with x ∈ I3o (G), then m3 (C(x, K)) = 1.

It turns out that a very similar result can be proved for arbitrary p with K/Op (K) ∼ = L2 (pp ) where K is a p-component of CG (x), x is of order p, and y ∈ Ip (CG (x)) induces a nontrivial field automorphism on K/Op (K). Furthermore, this more general result is important at later stages of the analysis. We shall therefore refer to such a triple (x, y, K) as an L2 (pp ) field triple. [Thus when p = 3, an L2 (pp ) field triple is in particular a locally unbalancing triple.] Following Theorem 1, we digress briefly to analyze L2 (pp ) field triples. There are three important corollaries to Theorem 1. The first one is that Theorem C5 : Stage 1 hinges on conclusion (b), and hence on Theorem 1.3. Corollary 1.6 (Corollary 1a). If conclusion (b) of Theorem C5 : Stage 1 holds, then so do conclusions (c) and (d) of that theorem. From this point on we assume that Theorem 1.3 fails, so that Bp∗ (G) ∩ Sp (G) = ∅, The next corollary concerns the subgroup Γoo P,2 (G) of G. It is defined for P any p-subgroup of G with mp (P ) ≥ 4 by (1F)

Γoo P,2 (G) = NG (Q) | Q ≤ P , mp (Q) ≥ 2, mp (QCP (Q)) ≥ 4 .

oo Notice that if R ≤ P and mp (R) ≥ 4, then Γoo R,2 (G) ≤ ΓP,2 (G).

Corollary 1.7 (Corollary 1b). For any P ∈ Sylp (G), G contains a proper subgroup M with the following properties: (a) Γoo P,2 (G) ≤ M ; and (b) Op (M ) has even order. The proof of Corollary 1b is based on signalizer functor theory, and follows directly from the balance property established in Theorem 1. Choosing P ∈ Sylp (G) to contain B ∈ Bp∗ (G) ∩ Sp (G), we show that M = NG (Θ1 (G; B)) has the required properties, where Θ1 (G; B) denotes the closure of the balanced B-signalizer functor on G. The signalizer functor theorem [V2 , 1.1] implies that Θ1 (G; B) is a p -group. Also the facts that B is contained in a 2-local subgroup of G and B ∈ Sp (G) allow us to prove that |Θ1 (G; B)| is even. Corollary 1c and Theorems 2–5 concern a fixed (1G)

P ∈ Sylp (G) and a subgroup M as guaranteed by Corollary 1b.

Notice that NG (P ) ≤ Γoo P,2 (G) ≤ M , so in particular P ≤ M . Theorem 1 has the following additional consequence.  = Corollary 1.8 (Corollary 1c). Let x ∈ Ipo (M ), set N = NG (x) and N N/Op (N ). Then Op (N ) ≤ M . Moreover, N ≤ M or one of the following holds: ∼ p1+4 ∗ Zpm for some m ≥ 1; or  ) = Op (N ) = (a) F ∗ (N

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3. THEOREM C5 : STAGE 1

38

  N  such that E  ∼ E = C  (E),  then (b) There is E = Ep3 , and if we put C N p  ∼      CE ≤ Op (N ), Ω1 (CE ) = E, and O (N )/CE = Ep2 SL2 (p); or n ) ∼ (c) E(N = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or 2 G2 (3 2 ), n ≥ 3 (with p = 3)  ))) = 1. and mp (CN (E(N  Note in (b) that x  ∈ E. To sharpen Theorem 1, we need two key preliminary results. The first is: Theorem 1.9 (Theorem 2). If A ∈ Sp (M ), then every A-invariant p -subgroup of M is contained in Op (M ). Now let T be a P -invariant Sylow 2-subgroup of Op (M ). The second key result is a partial result toward proving (1H)

CP (T ) = 1.

The purpose is to be able to use “covering 2-local” methods to study subgroups of G containing an element A ∈ Sp (G) and a 2-subgroup S such that S ≥ [A, S] = 1. We prove (1H) under a variety of additional assumptions which fit our eventual purposes. For example, we prove (1H) when P has a self-centralizing Ep2 subgroup. We do not state the other results here. With the aid of these results, we are able to eliminate field triples entirely. Corollary 1.10 (Corollary 2). G does not contain an L2 (pp ) field triple. Now for our fixed subgroup M and Sylow p-subgroup P of M , we are able to focus on p-uniqueness properties of M . The first step concerns the subgroup defined by   ΓoP,2 (G) = NG (Q)  Q ≤ P, mp (Q) ≥ 2, mp (QCP (Q)) ≥ 3 . We prove Theorem 1.11 (Theorem 3). ΓoP,2 (G) ≤ M . Just as Theorem 1 depends on verification of balance with respect to elements of Ep4 (G), so Theorem 3 depends on the same result for elements of Ep3 (G). However, if the desired conclusion fails for some A ∈ Ep3 (G), then by [IG , 20.6], G contains an unbalancing triple (x, y, K) for some x, y ∈ A# and in view of Corollary 2 (for / Ipo (G)). A major portion of the p = 3), K/Op (K) is not a Cp -group (whence x ∈ proof of Theorem 3 is devoted to using the existence of M to restrict the possible isomorphism types of K/Op (K) to a very limited set. Next, following the argument of [GL1, Proposition 2.1, pp. 650–663], we prove Theorem 1.12 (Theorem 4). ΓP,2 (G) ≤ M . Because mp (P ) ≥ 4 and every element of Lop (G) is a Cp -group, the proof of Theorem 4 is considerably easier than the corresponding result of [GL1]. Finally we show Theorem 1.13 (Theorem 5). CP (T ) = 1, and M is a maximal subgroup of G. Together, Corollary 1b(b) and Theorems 2, 4, and 5 will show that alternative (b) of Theorem 1.3 holds, contradicting our assumption that G has no strong puniqueness subgroup. This will complete the proof of Theorem 1.3, and with it, the proof of Theorem C5 : Stage 1.

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3. THEOREM 1: BALANCE

39

2. The Strong Balance Lemma As noted in the Introduction, many of the exceptional configurations that we must treat have to do with the group L2 (33 ) in the case p = 3. The following lemmas give the precise details; the application of course is to proper subgroups X of G. Lemma 2.1 (Strong Balance Lemma). Let X be a K-group, let A ∈ Sp (X), and  = X/Op (X) and assume that all let W be an A-invariant p -subgroup of X. Set X  lie in Cp . Then one of the following holds: components of E(X)  = 1; (a) W   of X  and an element a ∈ NA (K) (b) p = 3 and there exists a component K 3 ∼  such that K = L2 (3 ) and a induces a nontrivial field automorphism on  K. Proof. Without loss, Op (X) = 1. Assume that W = 1. By [V2 , 2.7], there exists a subgroup B ≤ A and a p-component K of CX (B) such that in CX (B) = CX (B)/Op (CX (B)), K is not strongly locally balanced for the prime p, indeed some A-invariant p -group does not centralize K. We choose B and K satisfying these conditions so that |B| is minimal. We claim that the following conditions hold, which will complete the proof: (2A)

(1) B = 1; (2) p = 3, K ∼ = L2 (33 ), and some a ∈ NA (K) induces a nontrivial field automorphism on K. ∗

Namely, let K ∗ be the subnormal closure of K in X. Then K is a product of components, which are by assumption in Cp . Then by [VK , 3.2], K ∈ Cp . As K is not strongly locally balanced with respect to A, (2A2) follows directly from [IA , 7.7.12]. Thus, CK (a) ∼ = A4 . Furthermore, suppose that B = 1 and choose a hyperplane D of B. Let KD be the subnormal closure of K in CX (D). If KD is a trivial pumpup of K, then CKD /O3 (KD ) (a) ∼ = A4 , so KD /O3 (KD ) is not strongly locally balanced with respect to A. Hence D and KD satisfy the same conditions as B and K, and our minimal choice of B is contradicted. Likewise if KD is a diagonal pumpup of K, then KD = K1 K2 K3 with B/D ∼ = Z3 cycling the Ki ’s and each Ki /O3 (Ki ) ∼ = L2 (33 ). As a normalizes K, a normalizes KD . Replacing a by au for suitable u ∈ B − D, if necessary, we still have the condition CK (a) ∼ = A4 and can arrange that a normalizes K1 . Then CK1 /O3 (K1 ) (a) ∼ = A4 , so D and K1 satisfy the same conditions as B and K, and our minimal choice of B is again contradicted. The only alternative is that KD is a vertical pumpup of K. By [VK , 3.57], however, a then cannot induce a field automorphism of K of order 3. This contradiction shows that (2A1) holds as well.  3. Theorem 1: Balance We prove Theorem 1 in this section: Proposition 3.1. Under hypothesis (1A), for any A ∈ Ep4 (G), G is balanced with respect to A. We actually prove a stronger statement, based on the following definition: Definition 3.2. Let A ∈ Sp (G). Then G is strongly balanced with respect to A if and only if for each a ∈ A# , ICG (a) (A; p ) = Op (CG (a)).

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3. THEOREM C5 : STAGE 1

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We shall prove Proposition 3.3. Under hypothesis (1A), G is strongly balanced with respect to any A ∈ Ep4 (G) ∩ Sp (G). Strong balance with respect to A implies balance with respect to A, since for all a, b ∈ A# , Op (CG (b)) ∩ CG (a) is an A-invariant p -subgroup of CG (a), so lies in Op (CG (a)) by strong balance. Proposition 3.1 is a consequence of Proposition 3.3. Indeed, in the proof of Proposition 3.1, any A ∈ Ep4 (G) may be expanded to an elementary abelian subgroup maximal with respect to inclusion, i.e., an element A∗ ∈ Ep4 (G) ∩ Sp (G). By Proposition 3.3, G is strongly balanced, hence balanced, with respect to A∗ , and so G is balanced with respect to A. Suppose then that Proposition 3.3 fails. Choose a counterexample A ∈ Ep4 (G). By Lemma 2.1 applied to the groups CG (a) (a ∈ A# ), p = 3 and there exist a, b ∈ A# and a p-component L of CG (a) such that b induces a nontrivial field automorphism on L/O3 (L) ∼ = L2 (33 ). Among all quadruples (A, a, b, L) such that A ∈ E34 (G) ∩ S3 (G), a, b ∈ A, and (a, b, L) is a locally unbalancing triple (i.e., L/O3 (L) ∼ = L2 (33 ) and b induces a nontrivial field automorphism on L/O3 (L)), we now choose a quadruple for which A is maximal. Using the maximality, we argue that (3A)

A normalizes L.

Indeed suppose that the A-orbit containing L, on the set of 3-components of CG (a), consists of L = L1 , . . . , Ln , n = 3m , m ≥ 1. Let A0 = NA (L), so that b ∈ A0 . Then A ∩ L2 · · · Ln = 1. However, b, and indeed A0 , normalizes each of L2 , . . . , Ln , and b centralizes a subgroup of L1 isomorphic to A4 . Also A0 centralizes an element bi ∈ Li of order 3 for each i = 1, . . . , n, and we may choose the bi so that A∗ := A0 b1 , b2 , . . . , bn  is a 3-group. It is then elementary abelian, and as b ∈ A0 , (A∗ , a, b, L) is a competing quadruple. But A0 ∩ L2 · · · Ln = 1 so |A∗ /A0 | ≥ | b2 , . . . , bn  | = 3n−1 > n = |A/A0 |. This contradicts the maximality of |A| and proves (3A). Set D = CA (L/O3 (L)). Thus |A : D| ≤ |CAut(L2 (33 )) (f )|3 = 32 , where f is a field automorphism, and so in fact A = D b, c where c ∈ Syl3 (CL (b)). As an immediate consequence: Lemma 3.4. m3 (D) = m(A) − 2 ≥ 2, with strict inequality if m3 (A) > 4. We next prove Lemma 3.5. For any d ∈ D# , the pumpup of L in CG (d) is trivial. We write Ld for the pumpup of L in CG (d). Proof. Assume false, and let C = CG (d) and C = C/O3 (C). Suppose first that the pumpup J is vertical. Since D ≤ A ∈ E34 (G), d ∈ I3o (G). Thus J is a C3 group. But then by [VK , 3.57], no element of A can induce a field automorphism of order 3 on the subcomponent L ∩ J ∼ = L/O3 (L) ∼ = L2 (33 ). As b induces such an automorphism we have a contradiction.

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3. THEOREM 1: BALANCE

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Therefore the pumpup is the product J = J1 J2 J3 of three 3-components cycled by a. Then some hyperplane E of A normalizes each Ji , and E ∩ a, b = b  for some b inducing a nontrivial field automorphism on J1 . But E centralizes ei ∈ Ji of order 3, i = 2, 3, and we may choose the ei so that they generate a 3-group. Set E ∗ = E e2 , e3 . Then E ∗ is elementary abelian and |E ∗ | > 3|E| = |A|. But (d, b , J1 ) is a locally unbalancing triple, so the maximality of |A| is contradicted. The lemma follows.  We next prove Lemma 3.6. For some d ∈ D# , O3 (CG (d)) has even order. Proof. Suppose false. Set L0 = L3 (CL (D)); then L0 covers L/O3 (L). Let a = Ca /O3 (Ca ). Let t ∈ L0 be an involution centralizing b Ca = CG (a) and C 0 = L  we can choose t) is dihedral of order 28; as L and set Ct = CG (t). Then CL (  = O7 (C  ( t) is a a (nontrivial) cyclic 7-group W ≤ L0 so that W  ( L0 t)). As CL  # 3 -group, it follows that W ≤ O3 (CG (D t)). Indeed for any e ∈ D , since the pumpup of L in CG (e) is trivial, L0 covers a component of CG (e)/O3 (CG (e)) and similarly we see that W ≤ O3 (CG (e, t)) = O3 (CCt (e)). Thus W ≤ ΔCt (D).

(3B) Let P ∈ Syl3 (CG (D)) and set (3C)

B = D(P ∩ CL (D)),

so that B is elementary abelian of rank m3 (B) = m3 (D) + m3 (L) = m3 (D) + 3 > m3 (D) + 2 = m3 (A). By our maximal choice of A, G is balanced with respect to B. Let Θ1 be the 1-balanced signalizer functor on B, i.e., Θ1 (b) = O3 (CG (b)), and set X = Θ1 (G; B). Then X = Θ1 (G; E) for every noncyclic subgroup E ≤ B, by [V2 , 1.2]. In particular, X = Θ1 (G; D), which by the signalizer functor theorem [V2 , 1.1] is a 3 -group. We argue that (3D)

[W, O2 (Ct )] = 1.

Set V = O2 (Ct ), so that V is W × D-invariant. As D is noncyclic it suffices to show that Vd := [W, CV (d)] = 1 for each d ∈ D# . But Vd = [W, Vd ] and W ≤ Ld , the (trivial) pumpup of L in CG (d). Thus Vd ≤ Ld . Now Vd = [W, Vd ] is a W invariant 2-group, while CLd /O3 (Ld ) (t) ∼ = D28 . So Vd ≤ O3 (Ld ). But O3 (Ld ) has odd order by our assumption, so Vd = 1. Thus (3D) holds. Likewise, we show that [W, E(O3 (Ct ))] = 1. Indeed let I be a component of  E(O3 (Ct )) with [I, W ] = 1. Let d ∈ D# , I ∗ = I d , and I0 = CI ∗ (d). Then [I0 , W ] has odd order. But I0 has even order. It follows that W normalizes I0 (since [W, d] = 1). Hence W normalizes I. If I ∗ = I, then [W, CI ∗ (d)] odd implies [W, I] odd so [I, W ] = 1, contradiction. If I ∗ = I, then W maps into O 3 (CAut(I) (Id )) = 1 where Id = CI (d).  As G is of even type, O2 (Ct ) = 1. It follows that [W, O 3 (E(Ct ))] = 1. Since W ≤ ΔCt (D), there exists a component I of E(Ct ) which is DW -invariant and such that if we put H = IDW and H = H/CH (I), then 1 = W ≤ ΔH (D) and

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3. THEOREM C5 : STAGE 1

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D∼ = D. (See [IG , 20.6].) Hence I is not locally m3 (D)-balanced with respect to D. Recall that m3 (D) ≥ 2. If m3 (D) > 2, then by [IA , 7.7.5], I ∼ = Am for some m ≥ 29. But then I ∈ C2 , a contradiction. Thus m3 (D) = 2, and by [IA , 7.7.4], I ∼ = Am for some m ≥ 11  7 ∼ or I = L3 (q ) for some q ≡  (mod 3),  = ±1, with a generator of W inducing a nontrivial field automorphism on I. As I ∈ C2 the only possibility is the latter, with q a power of 2, and D induces inner-diagonal automorphisms on I, acting as the image of a nonabelian 3-subgroup of GL3 (q 7 ). For each d ∈ D# , we see from the structure of CLd /O3 (Ld ) (t) ∼ = D28 that [W, O3 (CI (d))] ≤ O3 (Ld ). Hence   (3E) [W, O3 (CI (d))] | d ∈ D# ≤ O3 (Ld ) | d ∈ D# = Θ1 (G; D) = X, a 3 -group. However, by [VK , 8.32], the group on the left side of (3E) is not a  3 -group, a contradiction. This completes the proof of the lemma. This immediately yields Lemma 3.7. m3 (A) < m2,3 (G). Consequently G is balanced with respect to any element of B3∗ (G). Proof. Using the previous lemma for the first inequality below, we have m2,3 (G) ≥ m3 (CG (d)) ≥ m3 (DCL (D)) = m3 (D) + 3 > m3 (D) + 2 = m3 (A). The second assertion of the lemma holds by our maximal choice of A.



As in (3C), expand D to B = D(B ∩ L) ∈ Ep (CG (D)), which we may assume is A-invariant. Let M = NG (Θ1 (G; B)). By our maximal choice of A, G is balanced with respect to B, and Θ1 (G; B) is a 3 -group of even order by the signalizer functor theorem and Lemma 3.6. Thus (3F)

ΓB,2 (G) ≤ M < G.

In particular NG (D) ≤ M . We let L0 = L3 (CL (D)) ≤ NG (D) ≤ M and K = L3 (L0 Θ1 (G; B)) and next prove Lemma 3.8. K is a 3-component of M . Proof. Set I = L3 (M ) and M = M/O3 (M ). Then since L0 is a 3-component of CL (D), L0 ≤ I by L3 -balance, and the normal closure I0 of L0 in I is a product L1 · · · Ln of 3-components of I, permuted transitively by D. Since L0 has a trivial pumpup Ld in CG (d) for each d ∈ D# , it follows with the B3 -property that L0 is a component of CI 0 (d) for each d ∈ D# . As L0 /O3 (L0 ) ∼ = L2 (33 ), it follows by [VK , 3.6] that I 0 = L0 and I0 = L1 . As Θ1 (G; B) ≤ O3 (M ) by construction,  K ≤ L3 (L0 O3 (M )) = I0 . On the other hand as L0 = O 3 (L0 ), [L0 , O3 (M )] is generated by its subgroups Ud := [L0 , CO3 (M ) (d)] as d ranges over D# , by [V2 , 6.14]. Now Ud = [L0 , Ud ] by [IG , 4.3(i)] so Ud ≤ Ld and then Ud ≤ O3 (CG (d)). O (M ) = L0 [L0 , O3 (M )] so I0 = Thus [L0 , O3 (M )] ≤ Θ1 (G; B). But I0 = L0 3 

O 3 (I0 ) ≤ K, proving the lemma.



Now expand D to a (B ∩ K)A-invariant Sylow 3-subgroup Q of CM (K). (We continue the notation M = M/O3 (M ).) Thus [Q, B ∩ K] = 1. Also set KR = L3 (CK (R)) for any 1 = R ≤ Q, so that K Q = K. We write Ku for Ku if u ∈ I3 (Q). Thus KQ ≤ Ku and K u = K as well.

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Lemma 3.9. For any u ∈ I3 (Q), Ku lies in a 3-component Lu of CG (u) and Ku covers Lu /O3 (Lu ). Proof. Note that since m3 (K) = 3, I3 (Q) ⊆ I3o (G). We know that the conclusion holds if u ∈ D# . It suffices then to show that if u, v ∈ I3 (Q) with [u, v] = 1, and if the conclusion holds for u, then it holds for v. But by L3 -balance, the subnormal closure I of Kv in CG (v) is the product of a u-orbit of 3-components of CG (v), and L3 (CKv (u)) is a single 3-component contained in Lu and covering Lu /O3 (Lu ). By L3 -balance and [VK , 3.11d], if the desired conclusion fails, then I/O3 (I) is isomorphic to either L2 (39 ) or the direct product of three copies of L2 (33 ), with the image of Kv a diagonal of this direct product. In either case I/O3 (I) is covered by Kv , ΓB∩K,2 (CG (v)), and hence by I ∩ M . But K v   M , so M ∩ I cannot have the asserted structure. This contradiction completes the proof.  Before advancing we make a simple observation. Lemma 3.10. Let E ∈ E34 (G). Then G is 2-balanced with respect to E. Moreover, if B ∩ K ≤ E, then NG (E) ≤ M . Proof. As the only obstruction to 1-balance in C3 is L2 (33 ), which is locally 2-balanced for the prime 3 by [IA , 7.7.4c], the first assertion follows. Suppose that B ∩ K ≤ E. Since G is balanced with respect to B and B ∩ K is noncyclic, it follows from [V2 , 1.2, 1.4] that Θ1 (G; B) = Θ1 (G; B ∩ K) = Θ2 (G; E). As NG (E) normalizes Θ2 (G; E) and M = NG (Θ1 (G; B)), the lemma follows.



Now we can prove that M is a 3-component preuniqueness subgroup. Lemma 3.11. For any u ∈ I3 (Q), CG (u) ≤ M . Proof. Let u be given and observe that O3 (CG (u)) ≤ ΓB∩K,2 (G) ≤ M . Let Lu be as in Lemma 3.9. Then B ∩ K ∈ Syl3 (Lu ) and Lu ≤ M . Hence Lu NCG (u) (B ∩ K) ≤ M . Thus L3 (CG (u)) ≤ M ; if Lu  M , then  even CG (u) ≤ C (u)

M by a Frattini argument. We may then assume that Lu G is the product L1 · · · Ln (≤ M ) of n ≥ 2 isomorphic 3-components, where L1 = Lu . Expand B ∩ K to E ∈ Syl3 (L1 · · · Ln ), so that m3 (E) ≥ 6. Then NG (E) ≤ M by Lemma 3.10, and a Frattini argument completes the proof.  Lemma 3.12. M is a maximal subgroup of G.

Proof. Suppose false, so that M < X < G for some subgroup X of G. Thus X is a K-group. We have O3 (X) ≤ ΓQ,1 (G) ≤ M . For each u ∈ I3 (Q), Ku is a 3-component of CG (u), hence of CX (u). Thus, using [VK , 3.6], we find that in  = X/O3 (X), K  is a 3-component. Since NX (K ∩ M ) ≤ ΓB,2 (G) ≤ M , we X conclude that L3 (X) ≤ M . Then whether K  X or K  X, we use a Frattini argument, preceded by a 2-balance argument in the latter case, as in the previous lemma, to conclude that X ≤ M , a contradiction.  The following result will show that M , K, and Q satisfy conclusion (a) of Theorem 1.3 (with K in place of L there), and hence as argued after Theorem 1.3,

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M is a strong 3-uniqueness subgroup of G. (Recall that Q contains the noncyclic 3group D.) This contradiction to p ∈ σ0 (G) will complete the proof of Propositions 3.3 and 3.1. Lemma 3.13. If g ∈ G and m3 (Qg ∩ M ) ≥ 2, then g ∈ M . Proof. We suppose false and set V = Qg ∩ M . We choose P ∈ Syl3 (M ) containing QA(B ∩ K). Without loss we may assume that V ≤ P . We also assume, as we may, that g ∈ G − M was chosen so that (3G)

(1) m3 (V ) is maximal; and (2) Subject to (1), |V | is maximal.

By assumption, m3 (V ) ≥ 2. By [VK , 8.25], L2 (33 ) is outer well-generated for the prime 3. For this reason, and since L2 (33 ) is simple, [II3 , 16.13, 16.9] imply that V = Qg and Ω1 (V ) ≤ (B ∩ K) × Q. However, V ∩ Q = 1 by [II3 , 16.9]. It follows that Ω1 (Q) is elementary abelian of rank at most 3. Therefore if we put E = (B ∩ K) × Ω1 (Q), we have E = J(P ). But E ≤ ΓV,1 (G) ≤ M g . It follows that E g = E y for some y ∈ M g . Thus gy −1 ∈ NG (E) ≤ M , the last by Lemma 3.10. −1 So M = (M g )y = M g . As M is maximal in G, g ∈ M , a contradiction, and the lemma is proved.  As noted before Lemma 3.13, the proof of Proposition 3.3 is now complete, and with it the proof of Proposition 3.1. We note the following consequence from signalizer functor theory. Proposition 3.14. Let A, B ∈ Ep4 (G). Then the following conditions hold: (a) Θ1 (G; A) ∼ = Θ1 (G; B), where Θ1 is the 1-balanced signalizer functor; (b) If A, B is a p-group, then Θ1 (G; A) = Θ1 (G; B). Proof. We first prove (b). Let P = A, B and choose (by [IG , 10.11]) U  P with U ∼ = Ep2 . Then in the chain A, CA (U ), U CA (U ), U every adjacent pair consists (in some order) of an element E ∈ Ep4 (G) and a noncyclic subgroup F ≤ E. Thus by Proposition 3.1, G is balanced with respect to E, and so Θ1 (G; E) = Θ1 (G; F ). Consequently Θ1 (G; A) = Θ1 (G; U ). By the same argument Θ1 (G; B) = Θ1 (G; U ), proving (b). In (a) we choose g ∈ G such that −1 Ag , B is a p-group, by Sylow’s Theorem, and deduce Θ1 (G; A) = Θ1 (G; Ag )g ∼ =  Θ1 (G; Ag ) = Θ1 (G; B), by (b). The lemma is proved. We remark that part (a) of the proposition implies that if Op (CG (a)) = 1 for some a ∈ A# , then Op (CG (b)) = 1 for some b ∈ B # . 4. L2 (pp ) Field Triples By an L2 (pp ) field triple for an odd prime p we mean a triple (x, y, K) such that x ∈ Ip (G), K is a p-component of CG (x), and y ∈ Ip (CG (x)) induces a field automorphism of order p on K/Op (K) ∼ = L2 (pp ). Thus much of the previous section relied on an analysis of certain L2 (33 ) field triples. In this section we prove some results that will be useful later about L2 (pp ) field triples for odd p in general. Specifically we shall prove:

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Proposition 4.1. Assume (1A). Let (x, y, K) be an L2 (pp ) field triple in G for the prime p. Then the following conditions hold: (a) Either Op (CG (b)) = 1 for all b ∈ Ipo (G), or mp (C(x, K)) = 1; and (b) If mp (C(x, K)) = 1, then x is not p-central in G. The argument that ruled out L2 (33 ) field triples in Section 3 does not quite apply here, since it may be that mp (C(x, K) ∩ CG (y)) = 1 while mp (C(x, K)) > 1. On the other hand, we have balance to work with. We first prove (a) by contradiction, assuming that Op (CG (b)) = 1 for some b ∈ Ipo (G) and that mp (C(x, K)) > 1 for some L2 (pp ) field triple (x, y, K). Choose (x, y, K) so that if Q is a y-invariant Sylow p-subgroup of C(x, K), then (4A)

(1) mp (Q) is maximal; and (2) Subject to (1), |Q| is maximal.

In particular, x ∈ Z(Q) and (4B)

mp (Q) ≥ 2.

x = Cx /Op (Cx ). Expand Q y to R ∈ Sylp (Cx ). By Set Cx = CG (x) and C Proposition 3.14a and the remark following the proposition, Θ1 (G; A) = Θ1 (G; B) for any A, B ∈ Ep4 (R). Set X = Θ1 (G; A) for any such A, and M = NG (X). Thus, NG (R) ≤ M as NG (R) permutes E4 (R). Similarly, using signalizer functor theory, Proposition 3.14, and the fact that G is balanced with respect to all elements of Ep4 (G), (4C)

ΓB,2 (G) ≤ M for all B ∈ Ep4 (M ).

In particular, Op (Cx ) ≤ ΓB,2 (Cx ) ≤ M . Set E ∗ = R∩K ∼ = Epp , fix any E0 ∈ E∗ (Q) and set E = E0 E ∗ , an elementary abelian group of rank ≥ p + 2 > 4. Then CK (E0 ) ≤ ΓE,2 (G) ≤ M . As E0 centralizes K/Op (K) we conclude that (4D)

K ≤ M.

Since x ∈ R ≤ M , we can consider the pumpup L of K in M . We set M = M/Op (M ) and next prove Lemma 4.2. K covers L. In particular, L ∼ = L2 (pp ). Proof. By Lp -balance, L is either a single x-invariant p-component of M or the product of p p-components cycled by x, and in either case, K is a p-component of CL (x). Also as E leaves K invariant, E leaves L invariant. We can assume that K < L, for otherwise the lemma holds. If L is quasisimple, then CLy (x) = K y is isomorphic to L2 (pp ) extended by a field automorphism of order p. However, by [VK , 3.57], no such K-group L exists. Hence L is the product of p p-components L1 , L2 , . . . , Lp cycled by x. Also E0 = E1 x, where E1 = NE0 (L1 ) leaves each Li invariant, 1 ≤ i ≤ p. Since E1 centralizes K, which projects onto each Li , it follows that E1 centralizes Li , 1 ≤ i ≤ p, and hence centralizes L. Also E1 = 1 as E0 is noncyclic. Let e ∈ E1# and let J be the pumpup of K in CG (e). Since e ∈ Ipo (G), the p-components of J/Op (J) are Cp -groups and K is a p-component of CJ (x). It follows  therefore from [VK , 8.43] that J ≤ K, ΓE ∗ ,2 (G), whence J ≤ M . Since J = K J , J ≤ L. On the other hand, L0 = Lp (CL (e)) covers L, so J is an x-invariant product

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of components of L. The only possibility therefore is that J is product of p pcomponents J1 , J2 , . . . , Jp with J i = Li , 1 ≤ i ≤ p.  some u ∈ x, y−x Finally, as y induces a nontrivial field automorphism on K, leaves L1 invariant and induces a nontrivial field automorphism on L1 and hence also on J1 /Op (J1 ). Thus, (e, u, J1 ) is an L2 (pp ) field triple in G. On the other hand,  2 ), if we let F ∈ Sylp (L2 ) with F centralizing E1 (as shown above, E1 centralizes L then F ∼ = Epp and mp (F E1 ) = p + mp (E1 ) = p − 1 + mp (E0 ). But F E1 centralizes L1 and hence centralizes J1 /Op (J1 ). We conclude that mp (C(e, J1 )) ≥ mp (F E1 ) > mp (E0 ) = mp (Q), contrary to our choice of (x, y, K), and the lemma is proved.  Essentially the same argument yields: Lemma 4.3. The following conditions hold: (a) Q ∈ Sylp (CM (L)); and (b) K has a trivial pumpup in Ct = CG (t) for every t ∈ Ip (Q). Proof. We first verify (b). Let J be the subnormal closure of K in Ct . Then E ∗ x centralizes t and has rank p + 1 ≥ 4. Hence ΓE ∗ x,2 (G) ≤ M . We conclude now as in the preceding lemma with the aid of [VK , 8.43] that J ≤ M and K ≤ J,  ∼ whence J = K. Thus J/Op (J) ∼ =K = L2 (pp ) and so J is a trivial pumpup of K, proving (b). Now, expand Q to a y-invariant Sylow p-subgroup V of CM (L) = CM (K) and let v ∈ Ip (Z(V )) with v centralizing y. Then v centralizes x, so v ∈ Q. Also v  as v centralizes K. Hence by (b), the pumpup I of K in Cv is trivial. centralizes K But y ∈ Cv and y leaves I invariant as it leaves K invariant. Given the action of y  it is immediate that (v, y, I) is an L2 (pp ) field triple in G. On the other hand, on K, V ≤ Cv and V centralizes K, so V leaves I invariant and centralizes I/Op (I). Since Q ≤ V , our choice (4A) of (x, y, K) forces Q = V , so Q ∈ Sylp (CM (L)), proving (a).  We next prove Lemma 4.4. Let X be a subgroup of G such that Lp (X ∩ L) covers L as well as a component of X/Op (X). Then X ≤ M . Proof. Replacing X by an L-conjugate we may assume that E ∗ ≤ X. Since E ∗ Q = E ∗ × Q has rank ≥ p + 2 ≥ 4, ΓE ∗ ,2 (G) ≤ M by (4C). As mp (E ∗ ) ≥ 3, it follows that Op (X) ≤ M . Let J be the p-component of X such that  XL p (X ∩ L)   and excovers J/Op (J). As L ≤ M , J ≤ Op (X)L ≤ M . Let J0 = J pand E ∗ to E0 ∈ Sylp (J0 ). Then E0 ≤ NG (E ∗ ) ≤ M , and either E ∗ = E0 or mp (E0 ) ≥ 2p ≥ 4. In either case ΓE0 ,2 (G) ≤ M , again by (4C). Now by a Frattini argument X = Op (X)J0 NX (E0 ), so it remains to show that J0 ≤ M . But  J0 ≤ Op (J0 )JCJ0 (E ∗ ) ≤ M , proving the lemma. This enables us to prove: Lemma 4.5. M is an L-preuniqueness subgroup of G. Proof. Since Q ∈ Sylp (CM (L)), the definition requires us to prove that CG (t) ≤ M for all t ∈ Ip (Q). Given t, the pumpup Lt of K in CG (t) is trivial by Lemma 4.3. Thus Lp (CL (t)) covers both Lt /Op (Lt ) and L. Hence by Lemma  4.4, CG (t) ≤ M .

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Finally we complete the proof of the proposition: Lemma 4.6. Proposition 4.1a holds. Proof. Assume false and continue the above argument. We emulate Lemmas 3.12 and 3.13. First, we show that M is maximal in G. Let X be such that M ≤ X < G.  = X/Op (X). For any t ∈ Ip (Q), CX (t) = CM (t) by Lemma 4.5, so CX (t) Let X  we see that X t is has a component Xt := Lp (CL (t)). Using the Bp -property in X  As this holds for any t ∈ Ip (Q) and Q is noncyclic, we quasisimple and equals L.    X.  Then Lemma 4.4 yields X ≤ M , so X = M . conclude by [VK , 3.6] that L Thus M is a maximal subgroup of G. Next we show that if g ∈ G and mp (Qg ∩ M ) ≥ 2, then g ∈ M . We assume false and choose g ∈ G − M such that V := Qg ∩ M has rank at least 2, and subject to these conditions, we maximize mp (V ) first, and subject to this, |V |, as in (3G). We choose P ∈ Sylp (M ) containing R, and may assume by an M -conjugation if necessary that V ≤ P . Now L ∼ = L2 (pp ) is simple and outer well-generated for p g [VK , 8.25], so V = Q and Ω1 (V ) ≤ E ∗ × Q by [II3 , 16.13, 16.9]. Since E ∗ is elementary abelian and V ∩ Q = 1 by [II3 , 16.9], Ω1 (V ) is also elementary abelian. It follows that Ω1 (Q) is elementary abelian, as is J(P ) = E ∗ × Ω1 (Q). Then J(P ) ≤ ΓV,1 (G) ≤ M g . By Sylow’s Theorem, J(P )g = J(P )y for some y ∈ M g . As NG (J(P )) ≤ M by (4C), we can argue to the contradiction g ∈ M just as at the end of the proof of Lemma 3.13. The lemma is proved.  Now we turn to Proposition 4.1b. We assume that it fails and argue to a contradiction by showing that G has a normal subgroup of index p, contrary to the simplicity of G. We fix an L2 (pp ) field triple (x, y, K) with x ∈ Z(P ) and x, y ≤ P for some P ∈ Sylp (G), and such that mp (C(x, K)) = 1. Then P ∈ Sylp (CG (x)). We  = N/Op (N ), and Q = CP (K).  By assumption, set C = CG (x), N = NG (x), N mp (Q) = 1, so Q is cyclic as p is odd. Thus K = Lp (C) = Lp (N ). Put E = P ∩K, so that E ∼ = Zp  Zp , and P = QE y as Out(L2 (pp )) ∼ = Z2p [VK , = Epp , E y ∼ 2.5]. It follows that B := J(P ) = E x is the unique element of E∗ (P ). We also set Z = Ω1 (Z(P )), so that Z = x, Z1  ∼ = Ep2 , where Z1 = E ∩Z ∼ = Zp . Finally, we set A = Z y ∼ = Ep3 . Then A ∈ Sp (P ). We fix this notation and first prove Lemma 4.7. N contains a normal subgroup N0 of index p not containing y, and with Sylow p-subgroup QE.   N . Clearly QE ∈ Proof. As noted above, K = Lp (N )  N , so KCN (K)    Sylp (KCN (K)), but y ∈ KCN (K). Since N/KCN (K) embeds in Out(K) ∼ = Z2p , the result follows.  Now Gr¨ un’s theorem (see [IG , 16.9]) yields: Lemma 4.8. x is G-conjugate to some element of P − x. Proof. Indeed, otherwise x would be weakly closed in P with respect to G, and as x ∈ Z(P ), [IG , 16.9] would imply that N controls G-fusion in P , in which case G would have a normal subgroup of index p by the preceding lemma.  We next prove

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Lemma 4.9. x is not G-conjugate to any element of B − x. Proof. B = J(P ) is weakly closed in P with respect to G, so by [IG , 16.9], NG (B) controls G-fusion in B. Hence it suffices to prove that x  NG (B).  has order By [VK , 9.3], NK (B) contains a cyclic p -subgroup H such that H 1 p  centralizing x and acting Frobeniusly on E.  Since p is odd, H is (p − 1) with H 2  transitive on E1 (E). Note that if x is G-conjugate to any element of E1 (E), then it is conjugate to Z1 . As x Z1 ≤ Z(P ), x is then NG (P )-conjugate to Z1 by Burnside’s Lemma. However, this fusion is impossible since Z1 ≤ [Ω1 (P ), Ω1 (P )] (as can be seen from E y ∼ = Zp  Zp ) while x ∈ [Ω1 (P ), Ω1 (P )] (as can be seen from P/E, in which there is a cyclic maximal subgroup, so Ω1 (P/E) is abelian). Thus x is not G-conjugate into E.  Set NG (B) = NG (B)/CG (B). Since Q centralizes B, P = y ∼ = Zp and H ∼ =H  Now as P ∈ Sylp (G), [V2 , 3.3] yields that NG (B) acts on B in the same way as H. stabilizes x, giving the desired assertion.  Now we can prove Lemma 4.10. x is NG (A)-conjugate to some element of A − Z. Proof. By the structure of P , every element of Sp (P ) other than B is P conjugate to A. Since x is G-conjugate to some element of P − x, but to no element of B − x, it follows that xg ∈ A − x for some g ∈ NG (D), where D is as in the Alperin-Goldschmidt theorem [IG , 16.1]. Then Z, xg  ≤ Z(D) so  A = Z, xg  = Ω1 (Z(D)) and g ∈ NG (A). The lemma is proved. Now set S = NP (A). Note that x, y  y Q as p is odd and Q is cyclic. Thus S = QE1 , where E1 = NE (A) ∼ = Ep2 . In particular, Z = Ω1 (Z(S)). Furthermore, [E1 , A] = Z1 and [Q, A] ≤ x. Finally, clearly S ∈ Sylp (NN (A)). But NG (S) leaves x invariant as x is not G-conjugate to any element of Z − x, so in fact S ∈ Sylp (NG (A)). We set NG (A) = NG (A)/CG (A) and next prove  Lemma 4.11. S = E 1 ∼ = Zp and O p (NG (A)) ∼ = SL2 (p).

Proof. S = QE 1 with |Q| ≤ p and E 1 ∼ = Zp or Ep2 . Also = Zp . Thus S ∼ NG (A) ≤ Aut(A) ∼ = GL3 (p). If S  NG (A), then NG (A) = CG (A)NNG (A) (S) by the Frattini argument. But then as Z = Ω1 (Z(S)), it follows that Z  NG (A). Thus x is NG (A)-conjugate only to elements of Z, contrary to the preceding lemma.  Therefore S = Op (NG (A)), and now the lemma follows from [V2 , 6.12].   Let t be a 2-element of NG (A) such that t = Z(O p (NG (A))). Then t centralizes S = E 1 and so by the Frattini argument t can be chosen to normalize S. Finally we prove Lemma 4.12. Z = [A, t]. Proof. Set A1 = [A, t]. Since NG (A) ≤ GL3 (p), t has two eigenvalues equal to −1 on A, and so A1 ∼ = Ep2 . On the other hand, as t normalizes S, t leaves Z invariant. But again x is weakly closed in Z, so t leaves x invariant and hence t ∈ N . Since t is a 2-element, t ∈ N0 (N0 as in Lemma 4.7). Thus t centralizes

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P/QE and hence centralizes A/A ∩ QE = A/Z. Since A1 = [A, t] ∼ = Ep2 and t  leaves Z invariant, this forces A1 = Z. Now we reach an immediate contradiction. Indeed, t ∈ Z(NG (A)), so NG (A) leaves Z = [A, t] invariant. As x ∈ Z, this contradicts Lemma 4.10, and Proposition 4.1 is proved. 5. Corollaries to Theorem 1 We prove Corollaries 1a, 1b, and 1c in this section. We consider Corollary 1a first, thus proving Proposition 5.1. Assume (1A), and suppose that Bp∗ (G) ∩ Sp (G) = ∅. Then Op (CG (x)) has odd order for all x ∈ Ipo (G). Moreover, if A ≤ B ∈ Bp∗ (G), then Op (CG (A)) has odd order. Proof. Let A ≤ B ∈ Bp∗ (G). By hypothesis, B ∈ Sp (G), so B < B ∗ for some elementary abelian B ∗ ≤ G. Then B ∗ normalizes Op (CG (A)). As mp (B ∗ ) > mp (B) = m2,p (G), Op (CG (A)) has odd order, proving the second assertion. For the first assertion fix any B ∈ Bp∗ (G). By the previous paragraph Op (CG (b)) has odd order for all b ∈ B # . In particular by the Signalizer Functor Theorem, Θ1 (G; B) has odd order where Θ1 is, as usual, the 1-balanced signalizer functor on B (which exists by Theorem 1). Now choose any x ∈ Ipo (G) and let E ∈ Ep4 (G) with x ∈ E. Then by Proposition 3.14, Θ1 (G; E) ∼ = Θ1 (G; B), so Θ1 (G; E) has odd  order. Consequently Op (CG (x)) has odd order, and the corollary is proved. Throughout the remainder of this chapter we assume that (1A) holds and Bp∗ (G) ∩ Sp (G) = ∅. Next we consider Corollary 1b, thus proving: Proposition 5.2. Let P ∈ Sylp (G). Then there is a subgroup M < G such that (a) Γoo P,2 (G) ≤ M ; and (b) Op (M ) has even order, as does Op (CG (x)) for some x ∈ Ipo (M ). In particular, Op (M ) = 1; that is why M = G. Proof. The proof actually depends not just on Theorem 1 but also on the somewhat stronger Proposition 3.3. Let B ∈ Bp∗ (G) ∩ Sp (G). Then mp (B) = m2,p (G) ≥ 4. Replacing B by a conjugate, we may assume that B ≤ P . Since B ∈ Bp∗ (G), there is a B-invariant 2-subgroup T = 1 of G. Then T = CT (b) | b ∈ B # , so for some b ∈ B # , CT (b) = 1. Thus CT (b) ∈ ICG (b) (B; p ). But by Proposition 3.3, G is strongly balanced with respect to B, which by definition implies that CT (b) ≤ Op (CG (b)). Hence Op (CG (b)) has even order for some b ∈ B # . Now by Theorem 1, the 1-balanced signalizer functor Θ1 is defined on any E ∈ Ep4 (G). As a result, for any such E and any noncyclic F ≤ E, Θ1 (G; E) = Θ1 (G; F ), a p -group, by [V2 , 1.2, 1.1]. We set W = Θ1 (G; B) and M = NG (W ). Thus W = Op (CG (b)) | b ∈ B # has even order by the previous paragraph, and W ≤ Op (M ), proving (b).

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We argue that P ≤ M . Indeed, let g ∈ NG (P ). Then B, B g  is a p-group, so Θ1 (G; B) = Θ1 (G; B g ) = Θ1 (G; B)g , the first equality by Proposition 3.14b. Hence g ∈ NG (Θ1 (G; B)) = M , so NG (P ) ≤ M . In particular P ≤ M . Similarly, if R ≤ M with mp (R) ≥ 4, then NG (R) ≤ M . Namely, we may assume that R ≤ P . Let g ∈ NG (R). Then for any A ∈ Ep4 (R), A, Ag , B ≤ P so by Proposition 3.14b, Θ1 (G; B) = Θ1 (G; A) = Θ1 (G; Ag ) = Θ1 (G; A)g , forcing g ∈ M. It follows immediately that if R is any p-subgroup of G with mp (R ∩ M ) ≥ 4 then NG (R) ≤ M . One follows a subnormal chain from R ∩ M to R. Now to prove (a) we take any 1 = Q ≤ P such that mp (Q) ≥ 2 and mp (QCP (Q)) ≥ 4; we must show that NG (Q) ≤ M . As mp (Q) ≥ 2 and p is odd, there is U  Q such that U ∼ = Ep2 . Then U  QCP (Q), so by [IG , 10.20], mp (CQCP (Q) (U )) = mp (QCP (Q)) ≥ 4. Thus U ≤ F for some F ∈ Ep4 (P ), so Θ1 (G; U ) = Θ1 (G; F ) = Θ1 (G; B). Hence NG (U ) ≤ NG (Θ1 (G; B)) ≤ M . In particular, CG (Q) ≤ CG (U ) ≤ M . Since QCG (Q)  NG (Q) it suffices by a Frattini argument to choose R ∈ Sylp (QCG (Q)) and show that NG (R) ≤ M . But this follows from the previous paragraph as R can be chosen so that R ∩ M ≥ QCP (Q), of p-rank at least 4. The proof of Proposition 5.2 is complete.  For the remainder of this chapter we fix P ∈ Sylp (G) and a corresponding subgroup M satisfying the conclusions of Proposition 5.2. Next, we prove a sharpened version of Corollary 1c: Proposition 5.3. Let x ∈ Ipo (M ), set N = NG (x), K = Lp (N ), and  = N/Op (N ). Then M contains Op (N ) as well as a Sylow p-subgroup of N . N Moreover, if N ≤ M , then either (a), (b), or (c) holds:  ) = Op (N ) ∼ (a) F ∗ (N = p1+4 ∗ Zpm for some m ≥ 1; moreover, mp (N ) ≤ 6; or ∗   N  such that E ∼ E = (b) F (N ) = Op (N ). There is E = Ep3 , and if we put C  E ≤ Op (N  ), Ω1 (C E ) = E,  and O p (N  )/C E ∼ CN (E), then C = Ep2 SL2 (p). Moreover, mp (N ) = 4; or n  ∼ (c) (1) K = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or 2 G2 (3 2 ), n ≥ 3 (with p = 3 in the last case);  = Op (N  ) is cyclic; (2) F ∗ (CN (K)) (3) N = K(M ∩ N );  0 is a Borel subgroup of K;  and (4) If K0 = K ∩ M , then K (5) M ∩ N is p-solvable. Without loss, R := CP (x) ∈ Sylp (CM (x)). Thus mp (R) ≥ 4, so as remarked before Corollary 1.7, oo Γoo R,2 (G) ≤ ΓP,2 (G) ≤ M by Theorem 1. In particular, if S ≤ R with mp (S) ≥ 4, then NG (S) ≤ M , and so R ∈ Sylp (CG (x)). Also Op (N ) ≤ Γoo R,2 (G) ≤ M . We assume N ≤ M . We immediately obtain Lemma 5.4. Let N1  N and set R1 = R ∩ N1 . If N1 ≤ M , then NG (R0 ) ≤ M for any 1 = R0 char R1 . Proof. Indeed N = N1 NN (R1 ) ≤ N1 NN (R0 ) by the Frattini argument and the fact that R0 char R1 . Since N ≤ M and N1 ≤ M , NG (R0 ) ≤ M , as asserted. 

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5. COROLLARIES TO THEOREM 1

51

 = Op (N  ). We prove Next, set Q = R ∩ Op p (N ), so that Q  ) = Q,  then (a) or (b) of Proposition 5.3 holds. Lemma 5.5. If F ∗ (N Proof. If mp (Q) ≥ 4, we contradict the preceding lemma with N1 = Op p (N ) and R0 = R1 = Q, so mp (Q) ≤ 3. Suppose Q contains a noncyclic elementary abelian characteristic subgroup E. We argue that conclusion (b) of the proposition holds. We have NG (E) ≤ M by the preceding lemma applied with N1 = Op p (N ). There exists U ≤ E with U  R and U ∼ = Ep2 . Now U ≤ A ∈ E∗ (R) for some A, and mp (A) ≥ 4 as x ∈ Ipo (G), and thus NG (U ) ≤ ΓA,2 (G) ≤ M . In particular E = U , so mp (E) = 3. Moreover CN (E) ≤ CG (U ) ≤ M . Hence if we put R1 = CR (E) ∈ Sylp (CN (E)), then NG (R1 ) ≤ M by the preceding lemma, and thus mp (R1 ) = 3, whence Ω1 (R1 ) = E. In particular, x ∈ E. Letting CE = CG (E) we conclude from [V2 , 2.4] that CE E ≤ Op (N  ). Now N  /C E embeds in the has a normal p-complement, and thus C  ∼  )/C E parabolic subgroup H of Aut(E) = GL3 (p) stabilizing x, and so L := O p (N  p ∼ embeds in O (H) = Ep2 SL2 (p). Now L is not p-closed, since then we would have  N = Op (N )NN (R) ≤ M . So L/Op (L) ∼ = SL2 (p). In particular O p (N/N1 ) ∼ = SL2 (p), whence mp (N ) ≤ 1 + mp (Q) ≤ 4, so equality holds. Moreover, to complete the verification of (b) we only have to rule out the possibility L ∼ = SL2 (p). But (p) then |R : C (E)| = p, and since E = Ω (C (E)) we would have if L ∼ SL = 2 R 1 R mp (R) = 3, a contradiction. Thus the proposition holds in this case. Now assume that Q contains no such subgroup E, whence Q ∼ = p1+2k ∗ Zpm for some k ≥ 2 and m ≥ 1, by [V2 , 4.10b]. But k ≤ 2 as mp (Q) ≤ 3, so Q ∼ = p1+4 ∗Zpm , p ∼ as asserted. As O (Out(Q)) = Sp4 (p), which has p-rank 3 by [IA , 3.3.3], mp (N ) ≤ 3 + 3 = 6. The proof is complete.   ), so for the balance of the proof we  = F ∗ (N Thus Proposition 5.3 holds if Q can assume the contrary, whence K := Lp (N ) = 1. We next prove Lemma 5.6. K ≤ M . Proof. Suppose false. Set S = R∩K x. Since K x  N , Lemma 5.4 implies that NG (S1 ) ≤ M for any S1 char S, and in particular mp (S) ≤ 3. Moreover, as Γoo R,2 (G) ≤ M , the definition (1F) implies that (5A)

mp (S1 CR (S1 )) ≤ 3 for any noncyclic S1 char S.

If K has a p-component of p-rank 1, let J be the product of all such pcomponents J1 , J2 , . . . , Jr . Then x ∈ / J as each Ji is simple, so r ≤ 2. In particular, R leaves each Ji invariant and hence Zi = Ω1 (Z(R)) ∩ Ji ∼ = Zp for all i. Thus Z := x, Zi |1 ≤ i ≤ r ≤ Z(R) and Z is noncyclic, so NG (Z) ≤ M . But J x  N and Z char R ∩ J x, so Lemma 5.4 is contradicted. Hence each component of K has p-rank ≥ 2.  is a Cp -group. It On the other hand, as x ∈ Ipo (G), each component of K follows from [VK , 15.24] that R = S1 CR (S1 ) for some noncyclic S1 char S. This contradicts (5A) and completes the proof of the lemma.  Now we conclude n  ∼ Lemma 5.7. K = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or 2 G2 (3 2 ), n ≥ 3 (p = 3).  = 1. Moreover, mp (CN (K))

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52

3. THEOREM C5 : STAGE 1

∼E≤S  If mp (S) ≥ 2, then there is Ep2 = Proof. This time set S = CR (K).  and with E  R. As usual, since mp (R) ≥ 4, NG (E) ≤ M . But NG (E) covers K as Op (N ) ≤ M , it follows that K ≤ M , contrary to the preceding lemma. Thus mp (S) = 1.   Now M covers Γoo R,2 (K) and x centralizes K. Since M does not cover the Cp   not covered by M is isomorphic  group K, and mp (R) ≥ 4, any component K0 of K to one of the listed groups by [VK , 8.36]. Let B ∈ E∗ (R). Then by [IG , 8.7(iii)],  0 . As ΓB,2 (N ) ≤ M , CB (K  0 ) = x. It follows that  0 ∈ Cp , B normalizes K and as K  0 )) + 1 = mp (K  0 ) + 1. If a component K  1 of N  other than K 0 mp (N ) ≤ mp (Aut(K   existed, then K1 would contain a noncentral element of order p, so mp (CN (K0 )) ≥  1   1 cannot exist, so K 0 = K  and the lemma x) > 1, contradiction. Thus, K mp ( K follows.  Now we quickly argue that Proposition 5.3c holds. Indeed, if K0 = K ∩ M ,  0 = Γoo (K)  is a Borel subgroup of K  by [VK , 8.36]. Set Y = CN (K)  and then K R,2  so that S is cyclic and S ∈ Sylp (Y ). Since x  Y , [V2 , 6.13] implies S = CR (K),  In particular, S = Op (N  ). Also if we put T = R ∩ K x, then that F ∗ (Y ) = S.   /K.  Thus mp (T ) ≥ 4 by the structure of K, so NG (T ) ≤ M and hence M covers N  is solvable, so is N  /K  and likewise K  0 is solvable. M = K(M ∩N ). Also as Out(K) Hence M ∩ N is p-solvable and we see that all parts of (c) hold. This completes the proof of Proposition 5.3. Proposition 5.3 has the following direct and useful consequence. Proposition 5.8. If x ∈ Ipo (M ) and CM (x) has a p-component I, then NG (x) ≤ M . In particular, I is a p-component of CG (x) and I/Op (I) is a Cp -group. Proof. If CG (x) ≤ M , then as I is a p-component of CM (x), I is a pcomponent of CG (x). Since x ∈ Ipo (P ), it follows that I/Op (I) is a Cp -group. Thus it suffices to set N = NG (x) and prove that N ≤ M .  = N/Op (N ). Since I ≤ N , M ∩ N is not p-solvable. Hence if N ≤ M , Set N one of Proposition 5.3ab must hold. Let R ∈ Sylp (M ∩ N ). Then mp (R) ≥ 4, so  = F ∗ (N  ). Thus I acts as usual R ∈ Sylp (N ). Set Q = R ∩ Op p (N ), so that Q  On the other hand Q ≤ Op p (N ) ≤ M so Q ≤ Op p (N ∩ M ). As faithfully on Q.  I]  is a p -group. This contradiction completes the proof.  I ≤ Lp (N ∩ M ), [Q, Our analysis also enables us to prove the following additional result that will be needed in later sections. Lemma 5.9. Let (x, y, K) be an L2 (pp ) field triple in G, and H a K-subgroup of G containing x, y. If I = Lp (H ∩ K) covers K/Op (K), then I has a trivial pumpup in H. Proof. Our conditions imply that I is a p-component of CH (x), so the assertion of the lemma is meaningful. Let J be the pumpup of I in H, so that J is either a single x-invariant p-component of H or the product of p p-components J1 , J2 , . . . , Jp of H cycled by x; and in either case I is a p-component of CJ (x). Also as y leaves I invariant, it leaves J invariant.

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6. THEOREM 2: SIGNALIZERS IN M

53

 = H/Op (H). We can assume that I < J,  otherwise the lemma holds. Set H As in Lemma 4.2, [VK , 3.57] yields a contradiction if J is quasisimple, so J = J1 J2 · · · Jp . Again as in Lemma 4.2, some u ∈ x, y − x leaves Ji invariant and induces a nontrivial field automorphism on Ji , 1 ≤ i ≤ p. On the other hand, conjugating (x, y, K) and H by a suitable g ∈ G, we can assume without loss that x, y ≤ P and R = P ∩ J ∈ Sylp (J). Since Γoo R,2 (G) ≤ M and R ∩ Ji ∼ = Epp , 1 ≤ i ≤ p, it follows at once that J ≤ M . Also S = CR (x) ∈ Sylp (K), whence Op (K) ≤ ΓS,2 (G) ≤ M . Since K = Op (K)I, also K ≤ M . If the pumpup L of K in M is trivial, then K = L is a component of M = M/Op (M ).  cycles J1 , J2 , . . . , Jp Hence K = I is a component of J, contrary to the fact that x  Thus we can assume that K does not have a trivial pumpup in with diagonal I. M , and so without loss we can take M as H. Thus Ji is a p-component of M , 1 ≤ i ≤ p. Finally, u centralizes some w ∈ Ip (R ∩ J2 ). Then K1 = Lp (CJ1 (w)) is a p-component of CM (w) covering J1 , and mp (CM (w)) ≥ mp (w K1 ) ≥ 4. Hence by Proposition 5.8, CG (w) = CM (w) has K1 as a p-component. Furthermore, u induces a nontrivial field automorphism on K 1 = J 1 , so (w, u, K1 ) is an L2 (pp ) field triple in G. However, R2 = R ∩ J2 ∼ = Epp centralizes J 1 , so mp (C(u, K1 )) ≥ p. This contradicts Proposition 4.1a. (Note that by Proposition 5.2b and our hypothesis, Op (CG (a)) = 1 for some a ∈ Ipo (G), so Proposition 4.1a indeed applies.) The proof is complete.  6. Theorem 2: Signalizers in M For the balance of this chapter, we set M = M/Op (M ). In this section we establish the following key property of p-signalizers in M . Proposition 6.1. If A ∈ Sp (M ), then Op (M ) is the unique maximal Ainvariant p -subgroup of M . Proof. If M is p-constrained, the proposition follows from the Bender-Thompson lemma [V2 , 2.2], so we can assume that L = Lp (M ) = 1. If each component of L is a Cp -group, the proposition follows from [VK , 2.1], applied to X = M , unless p = 3 and L has a 3-component K with K ∼ = L2 (33 ), and some a ∈ A# induces a nontrivial field automorphism on K. In that case, an element of E∗ (P ) normalizes K by [IG , 8.7(iii)], so since m3 (P ) ≥ 4, Q = CP (K) = 1. Thus, there is x ∈ I3 (Q) centralizing a. Since x centralizes K of 3-rank 3, x ∈ I3o (M ). Also I = L3 (CK (x)) covers K and so is a 3-component of CM (x). Hence CG (x) ≤ M and I is a 3-component of CG (x) by Proposition 5.8. Thus (x, a, I) is an L2 (33 ) field triple in G, whence m3 (C(x, I)) = 1 by Propositions 4.1a and 5.2b. This in turn implies that m3 (Q) = 1, so Q is cyclic and K  M . It follows that x ∈ Z(P ). But now as a induces a nontrivial field automorphism on I/O3 (I), Proposition 4.1b yields a contradiction. We can therefore assume that for some p-component K of L, K is not a Cp group. Again set Q = CP (K). For any x ∈ Ip (Q), we have x ∈ / Ipo (M ), otherwise Proposition 5.8 would again imply that K is a Cp -group. In particular, Op (M ) = 1, otherwise we could choose x ∈ Z(P ) ∩ Q. Therefore, K is simple.

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Suppose Q = 1. Then mp (KQ) ≤ 3 (otherwise Ω1 (Z(Q))# ⊆ Ipo (M )). Also P does not leave K invariant, Q  P and we could again choose x ∈  otherwise Z(P ) ∩ Q. Setting K ∗ = K P , we see that the only possibility is p = 3 and K ∗ = L is the product of three 3-components K = K1 , K2 , K3 of 3-rank 1 cycled ∗ by some element of P with K = F ∗ (M ). However, in this case any B ∈ E∗ (P ) leaves each Ki invariant by [IG , 8.7(iii)]. Thus B ∩ K2 = 1 and so if we take x ∈ B # ∩ K2 , then x ∈ Q and x ∈ Ipo (P ), contrary to what we have shown above. We conclude that Q = 1, whence L = K and L = F ∗ (M ) is simple. Finally, again by Proposition 5.8, for every y ∈ Ipo (M ), every component of CK (y) is a Cp -group. But now as mp (P ) ≥ 4, [VK , 7.3] implies that A leaves invariant no nontrivial p -subgroups of M , so the proposition holds in this case as well.  7. The Centralizer of a Sylow 2-Subgroup of Op (M ) In addition to the subgroups P ∈ Sylp (G) and M that have already been fixed, we henceforth fix a P -invariant Sylow 2-subgroup of Op (M ). By Corollary 1b, T = 1. For the 2-local analysis that we need to complete the proof of Theorem C5 : Stage 1, it is important at places to have the condition (7A)

CP (T ) = 1.

We will eventually establish (7A) in Theorem 5, but unfortunately it is difficult at this point to establish it unconditionally. In this section we shall begin to analyze the consequences for the structure of M of the assumption CP (T ) = 1. Then in the next section, we use these consequences to establish (7A) under various additional technical hypotheses. These results then are sufficient for the needed applications in the proof of Theorems 3 and 4, which will be used in the proof of Theorem 5. We begin with the following connection between CG (T ) and M . Lemma 7.1. COp (M ) (T ) = Op (CM (T )) = Op (CG (T )). Proof. Since M = Op (M )NM (T ), Op (M )Op (CM (T ))  M , so Op (CM (T )) ≤ Op (M ). Hence Op (CM (T )) ≤ COp (M ) (T ). Also Op (CG (T )) is invariant under the Sylow p-subgroup P of NG (T ). Therefore Op (CG (T )) ≤ Γoo P,2 (G) ≤ M so Op (CG (T )) ≤ Op (CM (T )). It remains to show that COp (M ) (T ) ≤ Op (CG (T )).  = C/Op (C). Let ΓC = Γoo (C). Then Let C = CG (T ), C0 = C ∩ Op (M ), and C P,2 ΓC ≤ C ∩ M , and C ∩ M normalizes C0 . We argue first that  = 1. 0 , Op (C)] (7B) [C Indeed choose any B ∈ E∗ (P ), so that mp (B) ≥ 4. Since P ∈ Sylp (G), it follows that B ∈ Sp (G), and in particular B ∈ Sp (N ), where N = NG (T ). Thus by 1 = the Bender-Thompson lemma [V2 , 2.2], applied to the p-constrained group C  C 0 ≤ Op (C  C 0 B, 1 ). This yields (7B). Op (C)

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0 = 1. As Op (C)  = 1, it follows Proceeding by contradiction, assume that C  0 , J].     from (7B) that there is a component J of C such that [C0 , J] = 1. Thus J ≤ [C oo  We fix such a component J. Let A ∈ E3 (B). Then ΓA,2 (C) ≤ ΓP,2 (C) ≤ C ∩ M , so 0 is invariant under ΓA,2 (C). Hence by definition [IA , p. 394], J does not have the C weak signalizer property with depth 3 = mp (A). By [IA , 7.7.17], J is isomorphic to one of the following groups:

(7C)

(1) An , n = kp2 + r, k > 0, where 2 ≤ r < p or (r, p) = (4, 3); (2) D4 (4), F4 (4), E6 (2), or E6± (4), all with p = 3; or (3) L3 (4), with p = 3, or F i22 , with p = 5.

On the other hand (7D)

 C ∩ M does not cover J,

0 , J]  would be a p -group, a contradiction. for if it did, [C Note that since T ∈ Syl2 (Op (M )) is centralized by C ∩ Op (M ) = C0 , C0 = 0 = W  has Z(T ) × W , with W of odd order. As Z(T ) ≤ O2 (Z(C)) ≤ Op (C), C odd order. Also Op (C) ≤ C ∩ M normalizes W , so [W, Op (C)] has odd order. It follows that W centralizes Op (C)/O2 (Op (C)). But 0 , J]  = [W  , J]  J ≤ [C  Hence there is a 2-component J of L2 (C) so CC (Op (C)/O2 (Op (C))) covers J.  mapping onto J.  Choose an involution z ∈ Z(T ). By L2 -balance, J ∗ = J CG (z) is the product of T -conjugate 2-components of CG (z), and J is a 2-component of CJ ∗ (T ). But G is of even type, so J ∗ ≤ E(CG (z)) and J is a component of CJ ∗ (T ). In particular the components of J ∗ are C2 -groups, and hence by [VK , 3.3], J is a C2 -group. This immediately rules out the possibility (7C1). By (7D), J ≤ ΓC . As ΓC ≥ ΓB,2 (C) for some B ∈ E∗ (P ), we have mp (Aut(J)) > mp (B) − 2 ≥ 2. This rules out (7C3). Furthermore, by [VK , 8.38], J ≤ ΓC in the cases (7C2), a final contradiction proving the lemma.  This immediately implies Lemma 7.2. Op (CG (T )) = Z(T ) × W with W of odd order. Proof. Since Op (CG (T )) ≤ Op (M ) and T ∈ Syl2 (Op (M )),

Z(T ) ∈ Syl2 (Op (CG (T ))).

As Z(T ) ≤ Z(CG (T )) the result follows.



Let R = CP (T ). Since P ∈ Sylp (NG (T )), R ∈ Sylp (CG (T )). We choose any B ∈ E∗ (P ), so that mp (B) ≥ 4. Since T = 1 there exists a hyperplane A of B and an involution z ∈ Z(T ) such that [A, z] = 1. We fix A and z. Thus A ∈ E3 (P ). We have RAT ≤ CG (z). Set 

J = O p (E(CG (z))).

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We next prove Lemma 7.3. R centralizes O2 (CG (z)) and E(Op (CG (z))) and acts faithfully on J. Proof. As O2 (CG (z)) = 1 (G is of even type), it suffices by the F ∗ -theorem [IG , 3.6] to show that R centralizes X := O2 (CG (z))E(Op (CG (z))). Notice that X = ΓA,2 (X) ≤ Γoo P,2 (G) ≤ M . Let Y be an RAT -invariant Sylow 2-subgroup of X. We have T CG (T ) ≤ CG (z) as z ∈ Z(T ), so Y0 := CY (T ) ≤ Op (CG (T )) and consequently Y0 ≤ Op (CM (T )). Therefore Y0 ≤ Op (M ) by the preceding lemma. However, Y0 T is a 2-group and T ∈ Syl2 (Op (M )), so Y0 ≤ T . Thus R centralizes Y0 = CY (T ). We conclude from the A×B lemma that R centralizes Y . In particular R centralizes O2 (CG (z)) and normalizes every component of E(Op (CG (z))). But every such component is a p -group and a Co2 -group as G is of restricted even type.  Therefore by [VK , 15.23], [R, E(Op (CG (z)))] = 1, proving the lemma. We next use the weak signalizer property to prove Lemma 7.4. The following conditions hold: (a) [T, J] = 1;  (b) J = Lp (CG (T )) = O p (E(CG (T ))); and (c) P normalizes J.

 Proof. RAT acts on J. Since T ≤ Op (M ), T0 := T CM (z) is a p -group invariant under CM (z). As ΓA,2 (G) ≤ Γoo P,2 (G) ≤ M , T0 is invariant under ΓA,2 (CG (z)). Therefore T0 , and hence T , centralizes every component of J that has the weak signalizer property with depth 3. Suppose that (a) fails and J1 is a component of J and [T, J1 ] = 1. By [IA , 7.7.17], J1 /O2 (J1 ) is as given in (7C). But J1 is a Co2 -group, hence not An for any n ≥ 9, so J1 /O2 (J1 ) is as in (7C2, 3). Notice also that if J1 ≤ M , then [T, J1 ] ≤ Op (M ), so [T, J1 ] is a p -group, whence [T, J1 ] = 1, contradiction. Thus, J1 ≤ M . It follows that (7E)

mp (CM (z)) ≤ 3.

For otherwise, choosing B0 ∈ Ep4 (CM (z)), we would have J1 ≤ ΓB0 ,∗−2 (J) ≤ Γoo P,2 (G) ≤ M (using [VK , 8.38] again), contrary to what we just saw. Thus (7E) holds. In particular, mp (A) = 3. We argue that A normalizes J1 . Suppose false and let {J1 , . . . , Jk } be an A-orbit on the set of all components of J. Let A0 be a hyperplane of A containing NA (J1 ). Then A0 has p orbits on {J1 , . . . , Jk }, and with [IG , 16.9], mp (CJ1 ···Jk (A0 )) ≥ p. But mp (A0 CJ1 ···Jk (A0 )) ≤ mp (A0 CJ1 ···Jr (A0 )) ≤ mp (ΓA,2 (CG (z))) (7F) ≤ mp (CM (z)) ≤ 3. This implies that A0 ≤ J1 · · · Jk , and then p = k = 3. Let Ai be the projection of A0 on Ji , i = 1, 2, 3. Then as A0 lies on a diagonal of J1 J2 J3 , m3 (A1 A2 A3 ) = 3m3 (A0 ) ≥ 6, a contradiction as A1 A2 A3 ≤ CCG (z) (A0 ) ≤ ΓA,2 (CG (z)) ≤ CM (z). We have proved that A leaves J1 invariant. Let Q ∈ Sylp (CM (z)), so that mp (Q) ≤ 3. If J1 is as in (7C2), then since m3 (Q) ≤ 3, mp (CCG (z) (A0 )) = 3 for all A0 ∈ E2 (A), and hence mp (CJ1  (A0 )) ≤ 3.

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7. THE CENTRALIZER OF A SYLOW 2-SUBGROUP OF Op (M )

57

But this contradicts [VK , 9.24]. Therefore (7C3) holds. Then mp (Aut(J1 )) = 2, and it follows that CA (J1 ) = 1. For any a ∈ CA (J1 )# , such a ∈ A# , a ∈ Ipo (M ) since A was chosen inside an element of Ep∗ (M ). But it follows easily from Proposition 5.3 that as CG (a) involves J1 ∼ = L3 (4) or F i22 , CG (a) ≤ M . Therefore J1 ≤ M in this case as well, again a contradiction. This completes the proof of (a).  We have proved that every component of O p (CG (z)) is centralized by T , and so is a component of CG (T ). Conversely, every p-component L of CG (T ) centralizes Z(T ) and hence centralizes Op (CG (T ))/O2 (Op (CG (T ))), so is a 2-component of CG (T ). By L2 -balance, L lies in a product of 2-components of CG (z) of order  divisible by p; all these are centralized by T , so L is a component of O p (CG (z)). This proves (b). As P normalizes T , (c) follows immediately and the lemma is proved.  We next prove Lemma 7.5. Let I be a component of J. Then (a) If I ≤ M , then I is a component of M ; and (b) If I ≤ M , then p = 3, P ≤ I, and one of the following holds: ∗ ∼ (1) I = J ∼ = L± 4 (3) or 2U4 (3), and F (M ) = J(P ) = E34 ; ∗ ∗ ∼ ∼ ∼ (2) I = Co2 and F (I ∩ M ) = F (M ) = U6 (2). Proof. If I ≤ M , then as M = Op (M )NM (T ) and I is a component of CG (T ), I is a component of M , proving (a).  Suppose that I ≤ M . As P ∈ Sylp (G) and P normalizes J, P ∩ I P ∈  P Sylp ( I ). Also by [IG , 8.7(iii)], J(P ) normalizes I. (Note that O2 (I) = 1 as G is of even type.) Now if mp (Aut(I)) ≤ 2, then by [V2 , 6.9], I ≤ ΓB,2 (G) ≤ M for any B ∈ E4 (P ), contradiction. Thus mp (Aut(I)) ≥ 3. Suppose now that I ∈ Chev(2). By [IA , 7.3.6], applied to any B ∈ E4 (P ), I ≤ M unless possibly p = 3 and I has level 2. But even then by [VK , 8.35], I ≤ M , contradiction. Thus I ∈ Chev(2). Likewise I ∼  L2 (q) for any q ∈ FM9 = since mp (Aut(L2 (q))) < 3 for such q. Suppose next that I ∈ Chev(3). As mp (Aut(I)) ≥ 3, p = 3 and I/Z(I) ∼ = ± L4 (3) or G2 (3), as I is a Co2 -group. In the latter case m3 (O3 (H)) ≥ 4 for all proper parabolic subgroups H ≤ I, and I is generated by two such subgroups containing P ∩ I, so I ≤ M , contradiction. If I/Z(I) ∼ = L± 4 (3), then I ∩ M = H, 4 ± a parabolic subgroup of I of the shape 3 H where H + ∼ = SL2 (3) × SL2 (3) and H− ∼ = A6 . Moreover since m3 (O3 (H1 )) = 3 for any maximal parabolic H1 (other than H) containing I ∩ P , we have CP (I) = 1 and then P ≤ I, for otherwise I ≤ Γoo P,2 (G) ≤ M . In particular I  CG (T ). Therefore H = I ∩ M  CM (T ). Let B ∈ E4 (P ). It follows that B  M and, as P ∈ Syl3 (H), that B = F ∗ (M ). This is one of the exceptional configurations in (b). Suppose finally that I ∈ Spor. We mp (Aut(I)) ≥ 3, and if equality holds,  Phave ), it follows from [VK , 8.37] that I ∼ then CP (I) = 1. As I ≤ I0 := Γoo = Co2 P,2 ( I is P -invariant, CP (I) = 1, and I0 ∼ = U6 (2) · 2. Thus (b) holds in this case and the proof is complete.  Now we can attain our objective. Proposition 7.6. CP (T ) acts faithfully on Lp (CG (T )). Moreover, if I is any p-component of Lp (CG (T )), then the following conditions hold:

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58

3. THEOREM C5 : STAGE 1

(a) I is a component of E(CG (T )) and of E(CG (z)) for all z ∈ I2 (Z(T )); (b) Either I ≤ M , or p = 3, CP (I) = 1 and one of the following holds: ∼ (1) I/Z(I) ∼ = L± 4 (3) and O3 ((I ∩ M )/Z(I)) = E34 ; or ∗ ∗ ∼ ∼ and F (I ∩ M ) F (M ) U (2); (2) I ∼ Co = = 6 = 2 (c) If I ≤ M , then I is a component of M and I is one of the following groups: (1) p = 3: I ∈ C3 ∩ Chev(2) = {L2 (8), A6 , U3 (3), U4 (2), U5 (2), U6 (2), 1 Sp6 (2), D4 (2), 3D4 (2), F4 (2), 2F4 (2 2 ) , Sp4 (8)}, or I∼ = U7 (2), Sp8 (2), 2 D5 (2), 2 E6 (2)a , L3 (3), L± 4 (3), G2 (3), M11 , J3 , Co1 , Co2 , Suz, F in , 22 ≤ n ≤ 24, or Fi , 1 ≤ i ≤ 3; 5 1 5 (2) p = 5: I ∼ = A5 , 2B2 (2 2 ), 2F4 (2 2 ) , 2F4 (2 2 ), or I ∈ Spor ∩ Cp ∩ Co2 = {J2 , Co1 , Co2 , HS, Ru, F3 , F2 , F1 }; (3) p = 7: I ∼ = L2 (7), Co1 , F i24 , F3 , or F1 ; (4) p = 11: I ∼ = J4 ; or (5) p = 17: I ∼ = L2 (17). (d) If I ≤ M and P acts transitively on the set I then P normalizes I; and (e) If I ≤ M and Z(P ) is cyclic, then F ∗ (M ) = I.

M

of M -conjugates of I,

Proof. Parts (a) and (b), and the first assertion of (c), are established in the preceding discussion. Suppose I ≤ M . Since I is a component of CG (z), I ∈ Co2 . On the other hand, for any x ∈ Ipo (M ) and any p-component Ix of CI (x), Ix /Op (Ix ) is a Cp -group, by Proposition 5.8. This restricts the possibilities for I to the listed groups in (c), by [VK , 3.100]. Suppose next that I ≤ M and P is transitive on the M -conjugates I = I 1 , . . . , I pn of I, but P does not normalize I. Let Q = CP (I), so that Q ∈ Sylp (CM (I)) by the transitivity of P . We argue that Ip (Q) ⊆ Ipo (M ). Recall that I is quasisimple with O2 (I) = 1, so I is simple. Let x ∈ Ip (Q), and suppose by way of contradiction that mp (CM (x)) ≤ 3. Then x normalizes, say, I 1 , . . . , I k , with k ≡ pn ≡ 0 (mod p). Hence CI i (x) contains an subgroup Z i ≤ Z(P ∩ I i ) of order p for each i = 1, . . . , k. Put Z = Z1 · · · Zk . Then x Z ≤ CM (x), so mp (x Z) ≤ 3. Thus, k = p = 3 and x ∈ Z. So x centralizes P ∩ I 1 · · · I k , whence mp (I i ) = 1 for each i. Let B ∈ E∗ (P ); then B normalizes each I i by [IG , 8.7(iii)], so [B, Z] = 1. Therefore [B, x] = 1 so mp (CG (x)) ≥ mp (B) ≥ 4, a contradiction proving our assertion. By what we just showed and Proposition 5.8, since I is a p-component of CM (x) for every x ∈ Ip (Q), CG (x) ≤ M for every such x, so [II3 , Theorem PU3 ] will imply that M is a strong p-uniqueness subgroup, once we prove that M is a maximal subgroup of G. This will contradict our basic assumption and complete the proof of (d). Let M1 be a maximal subgroup of G containing M . Then Op (M1 ) ≤ Γoo P,2 (G) ≤ M , so Op (M1 ) ≤ Op (M ). For each u ∈ Ip (Q), let Iu = Lp (CIOp (M ) (u)), so that Iu is a p-component of CG (u) and hence of CM1 (u), and Iu /Op (Iu ) is covered by IQ := Lp (CIOp (M ) (Q)). Let H be the pumpup of IQ in M1 . Then by Lp -balance, H ≤ Lp (M1 ) and I Q covers a p-component of Lp (CH (u)) for each u ∈ Ip (Q). Let m = mp (Q) and note that m ≥ mp (I 2 · · · I k ) ≥ mp (I)(p − 1) ≥ 2. Hence by

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8. SUFFICIENT CONDITIONS FOR FAITHFUL ACTION ON T

59

∼ I, Aapm +r , or (for p = 3 [VK , 3.6], H is a single p-component and H/Op (H) = and m = 2 only) Suz or O  N . The last two cases are impossible since 2 = m ≥ (p − 1)mp (I) but mp (I) > 1. Suppose that H/Op (H) ∼ = Aapm +r . Then I ∼ = Ar . As elements of E∗ (P ) normalize I, we have m = mp (Q) = mp (P ) − mp (Ar ) = mp (G) − mp (Ar ). On the other hand, mp (G) ≥ mp (H) ≥ mp (Ar ) + mp (Aapm ) ≥ mp (Ar ) + apm−1 . Therefore apm−1 ≤ mp (G) − mp (Ar ) = m. As clearly m ≥ 2, we have a contradiction. This proves that I ∼ = H/Op (H). The same argument shows that for each i = 1, . . . , pn , there is a component Ji of M1 such that Ii ∩ Ji covers Ji /Op (Ji ) ∼ = I. Set J ∗ = Ii ∩ Ji | i = 1 . . . , pn  . Now a Frattini argument shows that Lp (M1 ) ≤ Op (M1 )J ∗ NG (S), where S = Ω1 (P ∩ J ∗ ). But since elements of E∗ (P ) normalize each J1 , . . . , Jpn , it follows that mp (SCP (S)) ≥ 4. (This is obvious if mp (I) > 1, and if mp (I) = 1, we saw above that S ≤ B for some B ∈ E∗ (P ).) Therefore NG (S) ≤ Γoo P,2 (G) ≤ M , so     Lp (M1 )Op p (M1 ) ≤ M . Letting S1 = Ω1 (P ∩ Lp (M1 )Op p (M1 )), we similarly find that NG (S1 ) ≤ Γoo P,2 (G), so a final Frattini argument shows that M1 = M . Hence M is a maximal subgroup and  (d)is proved. ∗

Finally in (e), let I =

I

P

∗

∗

. Then Z(P ) ∩ I = 1. If F ∗ (M ) > I , then

∗

Ω1 (Z(P )) ≤ I , since I is simple. This contradicts the assumed cyclicity of Z(P ). ∗ ∗ Therefore F ∗ (M ) = I . In particular (d) applies and yields I = I. This completes the proof of the proposition.  The most awkward case of Proposition 7.6b – I/Z(I) ∼ = L± 4 (3) – is restricted because of 2-local structure: Lemma 7.7. Suppose that p = 3 and I/O2 (I) ∼ = L± 4 (3) in Proposition 7.6. Then I is standard in G and m2 (CG (I)) = 1. In particular, m2 (T ) = 1. Proof. By assumptiion, T centralizes I. Let S ∈ Syl2 (CG (I)) with T ≤ S and choose an involution z ∈ Z(T ). By the proposition, I is a component of CG (z). By [V2 , 9.1], m2 (C(z, I)) = 1. Therefore m2 (CS (z)) = 1 so z ∈ Z(S) and m2 (S) = 1.  Hence m2 (T ) = 1 as well, as claimed. 8. Sufficient Conditions for Faithful Action on T In this section we prove three results with p-local hypotheses that yield that CP (T ) = 1. These rely heavily on Proposition 7.6. Proposition 8.1. Suppose that p ≥ 5 and there is D ≤ P such that D = CP (D) ∼ = Ep2 . Then CP (T ) = 1. Proof. Suppose false and let I be as in Proposition 7.6. As p ≥ 5, I is a component of M by that proposition. Since D < P , Z(P ) is cyclic, and Proposition 7.6e implies that I = F ∗ (M ). Thus mp (Aut(I)) ≥ 4. Using this condition and [IA , 4.10.3, Table 5.6.1], we see that the only choice in Proposition 7.6c is p = 5 and I = F1 . Since F1 is complete, d induces an inner automorphism on I, whence d ∈ I. But P ∩ I has a normal subgroup R ∼ = 51+6 , so with [IG , 9.16], |CP (d)| ≥ 2  |CRd (d)| ≥ |CRd/Z(R) (d)| > 5 , a contradiction. The lemma is proved.

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3. THEOREM C5 : STAGE 1

60

Proposition 8.2. Suppose G contains an L2 (pp ) field triple (x, y, K). Let R ∈ Sylp (CG (x)) and assume that R ≤ P . Then the following conditions hold: (a) CP (T ) = 1; ∼ (Zp  Zp ) × Zp and Z(P ) = [R, R] ∩ Z(R) ∼ (b) R = = Zp ; and (c) NP (R) shears x to Z(P ). Proof. Let Q = CR (K/Op (K)). By Propositions 4.1 and 5.2b, Q is cyclic and x ∈ Z(P ). The latter conclusion implies that x is not characteristic in R. Since R/Q ∼ = Zp  Zp by [VK , 9.4a], the second center Z2 (R/Q) has exponent p, and so 1 (Z2 (R)) ≤ Q. As x is not characteristic in R, Z2 (R) has exponent p, so R has no normal Zp2 -subgroup. Thus, Q = x and R ∼ = Zp × (Zp  Zp ). Since R = P , NP (R) fuses the p subgroups of Z(R) other than Z = Z(P ) ∩ [P, P ]. It follows that Z = Z(P ), so (b) and (c) hold. Suppose then that CP (T ) = 1. By Proposition 7.6e, I = F ∗ (M ). Thus P lies between a Sylow p-subgroup of I and one of Aut(I), with Z(P ) cyclic. Further conditions which must be satisfied are mp (P ) ≥ mp (R) = p + 1; |CM (x)|p ≤ |R| = p2+p ; and CM (x), since it embeds in CG (x), must have at most one non-p-solvable composition factor, which, if it exists, can only be isomorphic to L2 (pp ) or (if p > 3) L2 (p). These conditions and [VK , 3.104a] reduce us to the following possibilities:   p = 3 with O 3 (M ) ∼ = U6 (2), I ∼ = D4 (2), or O 3 (M ) ∼ = L± 4 (3). A further condition that must be satisfied is that NK (J(R)) ≤ Γoo P,2 (G) ≤ M , so NK (J(R)) ≤ CM (x). Hence in CM (x) there must exist a section isomorphic to Z3 × F27·13 , where F27·13 is a Frobenius group of the indicated order. But none of the remaining groups have such a section, by [VK , 3.104b]. This contradiction completes the proof.   = C/Op (C). Proposition 8.3. Suppose that x ∈ Ip (G). Set C = CG (x) and C ∗   × Q,  where K is a p-component of C, Q  is cyclic, and Assume that F (C) = K  = (5, HS), (5, Ru), (7, He), or (7, F i24 ). Then CP (T ) = 1. (p, K) Proof. Without loss, Q ≤ R := CP (x) ∈ Sylp (CG (x)). From [IA , 5.3mprv] we see that R ∩ K ∼ = p1+2 in every case. As mp (P ) ≥ 4, x is not p-central in G, so x is not characteristic in R. Let Z = Z(R ∩ K). As Z(R) = Q × Z, it follows that Q = x. Moreover, NP (R) > R and NP (R) normalizes Z = Z(R) ∩ [R, R], so NP (R) shears x to Z. It follows that Z(P ) = Z. Let I be as in Proposition 7.6. Since p ≥ 5, I ≤ M , and then as Z(P ) is cyclic, F ∗ (M ) = I by Proposition 7.6e. In particular mp (Aut(I)) ≥ 4, and as in the proof of Proposition 8.1, the only possibility in Proposition 7.6c is that (p, I) = (5, F1 ).  But then by [IA , 5.3z], |CM (x)|5 ≥ 57 > 54 = |CG (x)|5 , a contradiction. 9. L2 (pp ) Field Triples Do Not Exist In this section we prove the following proposition. Proposition 9.1. G does not contain an L2 (pp ) field triple. Suppose by way of contradiction that (x, y, K) is an L2 (pp ) field triple in G. By Proposition 8.2, (9A)

CP (T ) = 1.

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9. L2 (pp ) FIELD TRIPLES DO NOT EXIST

61

Without loss, R = CP (x) ∈ Sylp (CG (x)) with y ∈ R. Set Z = Z(P ). By Proposition 8.2, R ∼ = (Zp  Zp ) × Zp , Z = [R, R] ∩ Z(R) ≤ K, and some u ∈ NP (R) fuses the p elements of E1 (x Z) − {Z}. Now as P and hence Z u acts faithfully on T by (9A), [V2 , 6.15] yields Lemma 9.2. If we set T0 = CT (x), then [Z, T0 ] = 1. In particular, T0 = 1. Since T0 is R-invariant, [VK , 7.1] implies Lemma 9.3. T0 ≤ Op (CG (x)). Expand T0 to an R-invariant Sylow 2-subgroup W0 of Op (CG (x)), and set N0 = NG (W0 ). Then CG (x) = Op (CG (x))CN0 (x) by the Frattini argument and I0 = Lp (CN0 (x)) covers K/Op (K). We let W be an R-invariant 2-subgroup of G containing W0 and maximal subject to the condition that in N := NG (W ), (9B)

CN (x) contains a p-component I which covers K/Op (K).

We fix W , N , and I and note that y induces a nontrivial field automorphism on I/Op (I). Set E = R ∩ K, so that E ≤ I. We let J be the subnormal closure of I in N and next prove Lemma 9.4. I covers J/Op (J).  = N/Op (N ), so that I ∼ Proof. Suppose false. Set N = L2 (pp ) is a component    [VK , x) and I < J. Since y induces a nontrivial field automorphism on I, of CJ( 3.57] implies that J is not quasisimple, so J is necessarily the product of p pcomponents J1 , J2 , . . . , Jp cycled by x with diagonal I, by Lp -balance. Expand R to V ∈ Sylp (N ). Since mp (R) ≥ 4 and Γoo P,2 (G) ≤ M , it follows oo  that V ≤ M and that ΓV,2 (G) ≤ M . Since mp (CV (Ji )) ≥ 2 for each i, this in turn yields that J ≤ M . In particular, I ≤ M . But also Op (CG (x)) ≤ Γoo R,2 (G) ≤ M , so K ≤ M . But K = I, so K is not a component of M as I is not a component of J. If L denotes the pumpup of K in M , then as with J, [VK , 3.57] implies that L is not quasisimple, so likewise L is the product of p p-components L1 , L2 , . . . , Lp cycled by x. Since y normalizes K, y leaves L invariant, so some u ∈ x, y − x leaves each Li invariant. Since u induces the same action as y on K, u induces a nontrivial field automorphism on each Li , 1 ≤ i ≤ p. Now u centralizes some a ∈ Ip (L2 ). Then a centralizes L1 , so H = Lp (CL1 (a)) covers L1 and a ∈ Ipo (M ). But by Proposition 5.8, CG (a) ≤ M and H is a p-component of CG (a). Now, (a, y, H) is an L2 (pp ) field triple and clearly mp (C(a, H)) ≥ 2, so Proposition 4.1a is contradicted, and the lemma is proved.  Observe next that as Γoo R,2 (G) ≤ M , W ≤ Op (N ) ≤ M . But also as x ∈ R and R ∈ Sylp (CG (x)), R contains an element of Sp (G), whence by Proposition 6.1, Lemma 9.5. W ≤ Op (N ) ≤ Op (M ). We next prove Lemma 9.6. W ∈ / Syl2 (Op (M )).

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62

3. THEOREM C5 : STAGE 1

Proof. Suppose false, so that NM (W ) contains a Sylow p-subgroup of M ,  = N/Op (N ), so which without loss we can assume to be P . Thus P ≤ N . Set N     that I = J is a component of N . Let Q1 = CP (I), so that x ∈ Q1 . If Q1 is noncyclic, then so is Q0 = CQ1 (x). However, I covers K/Op (K), so Q0 ≤ CG (x) centralizes K/Op (K), whence mp (C(x, K)) ≥ 2, contrary to Proposition 4.1a. It follows that Q1 must be cyclic, whence J  N and Q1  P . Thus x = Ω1 (Q) ≤ Z(P ), so Proposition 4.1b is contradicted.  We complete the proof by contradicting the maximality of W . Without loss, W ≤ T , whence W < T by the preceding lemma and consequently NT (W ) > W . Thus if we let S ∈ Syl2 (Op (M ) ∩ N ), then S > W . On the other hand, as Op (N ) ≤ Op (M ), S ∩ Op (N ) ∈ Syl2 (Op (N )).  = N/Op (N ). Since S is E-invariant and E ∈ Sylp (I) with I = J,  Again set N , S ∩ J = 1 by [VK , 9.4b]. But [S, E, E] = [S, E], so as J is a component of N        [S, E] ≤ J, whence [S, E] = 1. Now [VK , 9.4c] yields that S centralizes J. Hence if we let X be the preimage of S in N , then X is RI-invariant and S ∈ Syl2 (X). Thus NIX (S) covers I by the Frattini argument, so I1 = Lp (NIX (S)) does  I0 also as well. But now if we set I0 = Lp (CI1 (x)), then as x centralizes I1 = I,  As I0 ≤ CG (x) and mp (C(x, K)) = 1, clearly I0 covers K/Op (K). covers I. Finally, set H = NG (S), so that RI0 ≤ CH (x). Given the structure of CG (x), it is immediate that I0 is contained in a p-component L of CH (x), whence also L covers K/Op (K). Thus S, H, L satisfy the same conditions as W, N, I. However, S > W , so our maximal choice of W is contradicted. This completes the proof of Proposition 9.1. Proposition 9.1 has the following consequence. Proposition 9.7. Let A ∈ Sp (G) and let x ∈ A# . If every component of CG (x)/Op (CG (x)) is a Cp -group, then every A-invariant p -subgroup of CG (x) is contained in Op (CG (x)). Proof. This is now immediate from Lemma 2.1 applied to X = CG (x), since Proposition 9.1 rules out the exceptional configuration of that lemma.  10. A Covering 2-Local Result In this section we prove a technical result needed for the main argument in the  = (5, HS) next section. The setup is the following; the application will be to (p, K) and (7, He), with L = F2 and F1 , respectively, and to (5, Ru) and (7, F i24 ), with no L.

(10A)

 = C/Op (C); (1) p > 3, x ∈ Ip (G), C = CG (x), and C p   x where K is a p-component of C such (2) O (C) = K ×   = 2; that mp (K) (3) R ∈ Sylp (C), R ≤ P , and R ∩ K ∼ = p1+2 ; p  (4) A ∈ E3 (R) with IG (A; p ) = Op (M );  = ∅ or {L} for some simple L such that ↑p (L) = (5) ↑p (K) ∅;  < mp (Aut(K))  = 2 and m2,p (Aut(L)) < (6) m2,p (Aut(K)) mp (Aut(L)) = 3; and (7) 1 = W0 ∈ IC (A; 2).

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10. A COVERING 2-LOCAL RESULT

63

Note that x ∈ Ω1 (Z(R)) ≤ A. We prove: Proposition 10.1. Assume (10A). Then K ≤ M . We assume that (10A) holds and K ≤ M , and argue to a contradiction in a sequence of lemmas. Lemma 10.2. Z(P ) ∼ = Zp . Proof. We have Z(R) = x × Z(R ∩ K) ∼ = p1+2 . But = Ep2 , as R ∩ K ∼  Z(P ) ≤ Z(R). The result follows.  mp (CG (x)) < 4 ≤ mp (G), so x ∈  K.

Lemma 10.3. Suppose that A ≤ M g for some g ∈ G. Then M g does not cover

∼ Proof. Suppose false and set N = M g and I = Lp (K∩N ), so that I/Op (I) =  K and I is a p-component of CN (x). Let J be the subnormal closure of I in N ,  = N/Op (N ). If J is so that J has the usual form by Lp -balance. Also put N the product of p p-components cycled by x, then M similarly has such a product −1 of components (cycled by xg ). As mp (C) = 3 < p, x cannot normalize all such  As Op (C) ≤ M by (10A4), it follows components of M and so C ∩ M involves K. p  from the structure of O (C) that K ≤ M , contrary to assumption. Thus J is a ∼  or L. single p-component of N , and by our assumption, N =K ∼  Thus M has a component J 0 = K or L. If there are p or more such components,  (or then as we just argued, x cannot normalize them all and so C ∩ M involves K  L, which itself involves K). Therefore again K ≤ M , a contradiction. Therefore P  and L are simple and their automorphism groups normalizes J 0 and CM (J 0 ). As K have rank at most 3 by assumption, while mp (P ) ≥ 4, CP (J 0 ) = 1. Therefore CZ(P ) (J 0 ) = 1 = Z(P ) ∩ J0 , so Z(P ) is noncyclic. This contradicts Lemma 10.2 and completes the proof.   Then for Lemma 10.4. Let A ≤ H < G be such that I := Lp (K ∩H) covers K. ∼  some p-component J of H, I ≤ J and J/Op (J) = K or L. Moreover, H/Op (H) does not contain the product of p p-components isomorphic to J/Op (J). Proof. Since I is a p-component of CH (x), we can consider the subnormal closure J of I in H, so that J has the usual form by Lp -balance. If J is the product of p p-components cycled by x, then mp (J) ≥ p > 3. Hence if S ∈ Sylp (H), g g J ≤ Γoo S,2 (G). As P ∈ Sylp (G), it follows that J ≤ M for some g ∈ G. Thus K ∩M  contradicting Lemma 10.3. Therefore, J is a single pcontains I, which covers K,  component. Set H = H/Op (H). By Lp -balance, I is a component of CJ(x).  or L, proving the first assertion. Hence by assumption, J ∼ =K Let J ∗ be the product of all p-components of H isomorphic modulo Op (H) to  If there are as many as p such components, then as in the previous paragraph, J. g J ∗ ≤ Γoo S,2 (G) ≤ M for some g ∈ G, and Lemma 10.3 is again contradicted. This completes the proof.  Now choose a pair (H, W ) such that

(10B)

(1) (2) (3) (4)

 A ≤ H < G and I := Lp (K ∩ H) covers K; 1 = W ∈ IH (A; 2); Subject to (1) and (2), |W | is maximal; and Subject to (1), (2), and (3), |H|p is maximal.

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By our assumption (10A), the conditions (10B1, 2) are satisfied for H = C and W = W0 . By (10A4), W ≤ Op (M ). Without loss we assume that W ≤ T . Let N = NG (W ) and W ∗ = NT (W ). By maximality of W , W contains a  = H/Op (H). Let J be as in Lemma 10.4, so Sylow 2-subgroup of Op (H). Let H  < mp (J)  = mp (AutA (J)).  that K ∩ J covers K/Op (K). By (10A6), m2,p (Aut(J)) W  ] = 1. Now N ∩ H covers J.  Taking fixed points of x, K ∩ CJ (x) covers Hence [J,  CJ(x) and hence covers K. Therefore N and W ∗ satisfy (10B1, 2), so W = W ∗ , whence W = T . Then |H|p ≥ |N |p = |P | = |G|p . Let S ∈ Sylp (H) ⊆ Sylp (G). ∼ K  or L, and it has no By Lemma 10.4, H has a p-component J with J = more than p − 1 such p-components. Therefore S normalizes J. As Z(S) is cyclic,  = 1, so H  embeds in Aut(J).  But mp (Aut(J))  ≤ 3 by assumption while CH (J) mp (H) = mp (G) ≥ 4. This contradiction completes the proof of Proposition 10.1. 11. Theorem 3: ΓoP,2 (G) ≤ M We continue the above notation for M and P . In this section we prove Proposition 11.1. ΓoP,2 (G) ≤ M . We argue by contradiction in a sequence of lemmas. Ultimately we shall emulate the proof of Proposition 9.1. However, considerable preparatory analysis is required before we are in a position to do so. Since we are proceeding by contradiction, [V2 , 1.3e] implies that there exists A ∈ E3 (P ) such that (11A)

G is not balanced with respect to A.

As a consequence, we obtain Lemma 11.2. There exists A ∈ Ep3 (G) satisfying the following conditions: (a) mp (A) = 3 and A ∈ Sp (G); and (b) For some x ∈ A# , mp (CG (x)) = 3 and CG (x) has a p-component K such that K/Op (K) is not a Cp -group. Proof. Indeed, let A ∈ Ep3 (G) with G not balanced with respect to A. If A ≤ A∗ ∈ Ep4 (G), then G is balanced with respect to A∗ by Proposition 3.1, so G is balanced with respect to A as well, contradiction. Thus mp (A) = 3 and A ∈ Sp (G), proving (a). Furthermore, by [IG , 20.6], G contains an unbalancing triple (x, y, K) with x, y ∈ A# . If K/Op (K) is a Cp -group, then again by [VK , 7.1], p = 3 and K/O3 (K) ∼ = L2 (33 ). In particular, (x, y, K) is an L2 (pp ) field triple. But this contradicts Proposition 9.1. Thus K/Op (K) is not a Cp -group. This in turn implies that x ∈ / Ipo (G), so mp (CG (x)) = 3 and hence (b) also holds. Thus any such A satisfies the stated conditions.  Now choose A ∈ Sp (G) and x ∈ A# such that (11B)

(1) Conditions (a) and (b) of Lemma 11.2 hold, and A ≤ M; (2) Subject to (1), if possible, A contains some U ∼ = Ep2 such that NM (U ) contains a Sylow p-subgroup of M .

Now if U exists as in (11B2), we may replace A and U by M -conjugates and assume that P ∈ Sylp (NM (U )). Then replacing A by a NM (U )-conjugate we may

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assume that in addition, A ≤ P . Hence we have the conditions (11C)

A ≤ P and U ∼ = Ep2 with U  P.

If no U exists as in (11B2), we use an M -conjugation to assume that A ≤ P and choose any normal Ep2 -subgroup U of P . Thus (11C) holds in any event. We define a subgroup V ∼ = Ep2 of A as follows. If A ≥ U for some U as in (11B2), set V = U . Otherwise, set V = CA (U ). In the latter case as A ∈ Sp (G), [A, U ] = 1. Thus in both cases, A ≥ V ∼ = Ep2 and [U, V ] = 1. As A normalizes U , A normalizes U V , which is elementary abelian. Moreover x ∈ V . For if x ∈ V , then [x, U ] = 1. But U # ⊆ Ipo (G) and x ∈ Ipo (G), so x ∈ U . Hence A0 = x U ∈ Sp (G) as mp (CG (x)) = 3. So the condition in (11B2) is achievable (by A0 ), whence U = V and so x ∈ U , a contradiction. Thus, A = x × V. We immediately obtain Lemma 11.3. ΓUV,2 (G) ≤ M . Proof. If V = U , then as mp (CP (U )) = mp (P ) ≥ 4, NG (U ) ≤ Γoo P,2 (G) ≤ M , so we can assume that V = U . Expand V U to E ∈ Sp (G). Then E ≤ NG (U ) ≤ M . Hence if mp (E) ≥ 4, then likewise ΓE,2 (G) ≤ M , whence ΓV U,2 (G) ≤ M , so we can also assume mp (E) = 3. Since U ≤ E, it follows from our choice of A that E does not satisfy the conditions of Lemma 11.2, so G is balanced with respect to E. Since E is connected to B ∈ Ep∗ (G) via the chain B, CB (U ), CB (U )U, U, E, and G is balanced with respect to B, CB (U )U , and E, [V2 , 1.2] yields now that ΓE,2 (G) ≤ M , and the proof is complete.  Our choice of A also enables us to prove: Lemma 11.4. For any v ∈ V # , every component of CG (v)/Op (CG (v)) is a Cp -group. Proof. If v ∈ Ipo (G), the lemma holds as every element of Lop (G) is a Cp group. In particular, this is the case if V = U , so we can assume that V = U and v∈ / Ipo (G). Now as V centralizes U with V = U , the latter condition implies that E =VU ∼ = Ep3 and E ∈ Sp (G). Since U ≤ E and U = V , it follows therefore from our choice of A that for every e ∈ E # , every component of CG (e)/Op (CG (e)) is a Cp -group (this is automatic if mp (CG (e)) ≥ 4). In particular, this is the case for e ∈ V # , and the lemma is proved.  Together with Propositions 9.7 and 6.1, this enables us to prove Lemma 11.5. Op (M ) contains every A-invariant p -subgroup of G. Proof. Let X be an A-invariant p -subgroup of G. Then  X = CX (v) | v ∈ V # , so it suffices to prove that each CX (v) ≤ Op (M ). Hence without loss, X centralizes some v ∈ V # . Proposition 9.7 and the preceding lemma imply that X ≤ Op (CG (v)).

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It will therefore suffice to prove that Y := Op (CG (v)) ≤ Op (M ). But U ≤ CG (v) and so Y = CY (u) | u ∈ U # . Thus we need only show that W = CY (u) ≤ Op (M ) for any u ∈ U # . Expand V U to E ∈ Sp (G). Then W is E-invariant and so again by Proposition 9.7, W ≤ Op (CG (u)). Finally, Q = CP (U ) ≤ CG (u) and mp (Q) = mp (P ) ≥ 4. Since Γoo Q,2 (G) ≤ M , it follows that Op (CG (u)) ≤ M . Also as mp (Q) = mp (P ), Q contains an element of Sp (G), so Op (CG (u)) ≤ Op (M ) by Proposition 6.1. Since W ≤ Op (CG (u)), the lemma is proved.  Recall that x ∈ A# , mp (CG (x)) = 3, and CG (x) has a p-component K with K/Op (K) not a Cp -group. We set (CG (x))= CG (x)/Op (CG (x)), and next prove: Lemma 11.6. The following conditions hold: (a) V leaves K invariant; (b) Lp (CK (v)) ≤ M for every v ∈ V # ; and  (c) V acts faithfully on K.  Proof. If K ∗ = K V > K, then K ∗ is the product of p-components K = K1 , K2 , . . . , Kr with r ≥ 3. By [IG , 16.11], each Ki contains an element of order p not in Z ∗ (Ki ). Since x centralizes each Ki , it follows that mp (CG (x)) ≥ 4, which is not the case. Thus (a) holds.  = Let I be a p-component of CK (v) for v ∈ V # . Set N = NG (v) and N N/Op (N ), and let J be the pumpup of I in N , so that J has the usual form by  is a Cp -group. As I is a Lp -balance. By Lemma 11.4, every component of N   is not a p-component of CJ (x), I is a Cp -group by [VK , 3.2]. In particular as K  This implies (c). Cp -group, I does not cover K. It remains to prove (b), i.e., that I ≤ M. Suppose first that J is the product of p p-components J1 , J2 , . . . , Jp cycled by x with diagonal I. If V = U , then v ∈ Ipo (M ) and as Lp (N ) contains p p-components, Proposition 5.3 implies that N ≤ M , whence I ≤ M . We can therefore assume that V = U . We set E = U V (∼ = Epk , k = 3 or 4). By Lemma 11.3, ΓE,2 (G) ≤ M . Suppose E moves, say, J1 . Then J ∗ = J1E is the product of ≥ p p-components of N transitively permuted by E. Since v ∈ E centralizes J ∗ , [IG , 3.28(ii)] yields that J ∗ ≤ ΓE,2 (G) ≤ M . But it is immediate that mp (J ∗ v) ≥ 4, so v ∈ Ipo (M ). Again I ≤ CG (v) ≤ M by Proposition 5.3. Thus we can also assume that E leaves each Ji invariant, 1 ≤ i ≤ p. Thus by [VK , 9.33], mp (ΓE,2 (J ∗ v)) ≥ 4. Hence mp (M ∩ N ) ≥ 4, so again v ∈ Ipo (M ) and hence I ≤ CG (v) ≤ M by Proposition 5.3 in this case as well, proving (b) if J is not quasisimple. We can now assume that J is quasisimple. Suppose for a contradiction that I ≤ M , whence CG (v) ≤ M . If v ∈ Ipo (M ), then by Proposition 5.3, J ∼ = L2 (pn ), n n 2  = 1. Since n ≥ 3, U3 (p ), n ≥ 2, or G2 (3 2 ), n ≥ 3 (with p = 3), and mp (CN (J))  x) and mp (CG (x)) = 3, it follows from [VK , 3.23] that I is a component of CJ(  However, this J ∼ = L2 (pp ) with x inducing a nontrivial field automorphism on J. contradicts Proposition 9.1. Thus v ∈ / Ipo (M ), whence V = U . We again consider E = U V and again have m3 (E) = 3 and ΓE,2 (G) ≤ M .

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 Again using [IG , 3.28(ii)], Since I ≤ M , J ≤ M , so ΓE,2 (G) does not cover J.    x) and E leaves J invariant. But as J is a Cp -group with I a component of CJ( mp (CG (x)) = 3, we are set up to apply [VK , 8.17]. That lemma has three possible outcomes, each of which we show leads to a contradiction. If [VK , 8.17a] holds, then J ∼ = L2 (pp ). Hence some hyperplane E0 of E induces inner automorphisms on J and as NG (E0 ) ≤ M , it follows that M contains a Sylow p-subgroup of J. As mp (J v) ≥ 4, v ∈ Ipo (M ), contradiction. If [VK , 8.17b] holds, then letting R0 ∈ Sylp (N ) with AE ≤ R0 , we conclude first that v char R0 , whence R0 ∈ Sylp (G). In particular mp (R0 ) ≥ 4. Now [VK , 8.17b] implies that x centralizes a normal Ep2 -subgroup U0 of R0 . Since mp (CG (x)) = 3, x ∈ U0 and A0 := x U0 ∼ = Ep3 with A0 ∈ Sp (G). Then A0 and x satisfy (a) and (b) of Lemma 11.2. Choosing g ∈ G such that R0g = P , it is trivial that Ag0 and xg still satisfy those two conditions, and now Ag0 contains the normal Ep2 -subgroup U0g of P . Hence by our original choice of A and x, U ≤ A and V = U , contradicting what we saw above. Finally if [VK , 8.17c] holds, then  = (5, HS), (5, Ru), (7, He), or (7, F i24 ). (p, J) In this case we apply Proposition 10.1 to get J ≤ M , a final contradiction. To see that the conditions (10A) hold, first let R0 ∈ Sylp (N ) with AU V ≤ R0 . As  From | Out(K)| ≤ 2, R0 = RJ × RQ where RJ = R0 ∩ J and RQ = CR0 (J). 1+2 p ∼ [IA , 5.3mprv], RJ = p . Since A ∈ S (R0 ) with mp (A) = 3, RQ is cyclic. In particular mp (R0 ) = 3, so R0 ∈ Sylp (G), whence v = Ω1 (RQ ) cannot be characteristic in R0 . Therefore 1 (R0 ) = 1, so RQ ∼ = Zp . In particular R0 = AU V ≤ P . Now (10A1, 2, 3) hold with R0 in place of R, v in place of x, and J in place of K. Condition (10A4) holds by Lemma 11.5. In condition (10A5), it is clear from [IA , 4.9.6, 5.2.9] that any pumpup of a sporadic group must be sporadic, and therefore from [IA , 5.3] we get ↑5 (HS) = {F2 }, ↑7 (He) = {F1 }, while ↑5 (Ru) = ↑5 (F2 ) = ↑7 (F i24 ) = ↑7 (F1 ) = ∅, as desired. Then (10A6) follows immediately from [IA , Tables 5.6.1, 5.6.2]. Finally, for (10A7) we show that for Z = Z(RJ ), [Z, CT (v)] = 1, and as T is A ≤ P -invariant, CT (v) is a nontrivial element of ICG (v) (A; 2). Notice that Z(R0 ) = Z v, and as R0 = CP (v) < P , NP (R0 ) fuses E1 (Z(R0 )) − {Z}. As T is P -invariant, the subgroups [Z, CT (y)] are then all NP (R0 )-conjugate. We must prove that they are all nontrivial. But otherwise [Z, T ] = 1, which contradicts Proposition 8.3. This completes the proof of the lemma.  Next, set C = CG (x), and let R ∈ Sylp (C) with A ≤ R and R ∩ M ∈ Sylp (CM (x)). Now we can prove:  is simple. Lemma 11.7. K = Lp (CG (x)) and K  has components besides K  which are Proof. We may assume that if Lp (C)  not Cp -groups, then we have chosen K from the beginning to maximize mp (K)  among all such components of C. Set Z = Ω1 (Z(R)) ≥ x. If Z is cyclic, then x = Z char R, so R ∈ Sylp (G), a contradiction as mp (R) ≤ 3 < mp (P ). Thus Z is noncyclic. Since A ∈ Sp (G), Z ≤ A and hence Z ∩ V = 1.

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 is nonsimple. Then by [VK , 9.5], [Z,  K]  = 1. As Z ∩ V = 1, Suppose that K   is this contradicts the faithful action of V on K (Lemma 11.6c). Therefore K simple.  has Suppose that C has a p-component J = K. Since mp (CG (x)) ≤ 3, C  are both R-invariant. If J is simple, then at most two components, so J and K  hence of CA (K).  Z ∩ J = 1. Hence Z1 := Z ∩ J, x is an Ep2 -subgroup of CZ (K),   This again contradicts the faithful action of V on K. Therefore J is nonsimple. In  > mp (K), particular mp (J) ≥ 2, so mp (K) = 1 as mp (C) ≤ 3. Moreover, as mp (J) our initial choice implies that J is a Cp -group. But there are no such Cp -groups, by [VK , 2.7]. The proof is complete.  We also need the following result.  In Lemma 11.8. If A ≤ M g for some g ∈ G, then M g does not cover K. particular, K ≤ M . Proof. Since A ≤ M , the second assertion is immediate from the first. Suppose the first assertion is false and set N = M g and I = Lp (K ∩ N ), so that  and I is a p-component of CN (x). Let L be the subnormal closure of I/Op (I) ∼ =K  = N/Op (N ). If L I in N , so that L has the usual form by Lp -balance. Also put N  i /Z(L i ) ∼  is the product of p p-components L1 , L2 , . . . , Lp cycled by x, then as L = K,  i is not a Cp -group. If mp (Li ) ≥ 2, then clearly there is y ∈ Ip (L2 ) ∩ Ipo (N ). In L  i is simple, so again by [IG , 8.7(iii)], every element of Ep∗ (N ) the contrary case, L leaves each Li invariant, and consequently such an element y exists in this case as  1 , Proposition 5.8, applied to N , yields that CG (y) ≤ N well. Since y centralizes L  1 is a Cp -group, contradiction. We conclude that L is a single x-invariant and that L p-component of N .  Since A leaves K invariant, Let S ∈ Sylp (N ) with A ≤ S and set Q = CS (L). A leaves I and hence L invariant, so A leaves Q invariant. Suppose that Q = 1, so  it is faithful on I and hence that A ∩ Q = 1 as A ∈ Sp (G). Since V is faithful on K,   = I.  If L is  on L, so A ∩ Q ≤ CA (K) = x. Thus, A ∩ Q = x. In particular, L not S-invariant, it is immediate whether mp (L) ≥ 2 or mp (L) = 1 that x ∈ Ipo (N ), whence mp (CG (x)) ≥ 4, contradiction. Thus L is S-invariant. Hence Q  S, so there is z ∈ Ip (Z(S)) ∩ Q. Then z centralizes A, so as usual this forces z = x. Thus x ∈ Z(S) and as mp (S) = mp (N ) ≥ 4, again mp (CG (x)) ≥ 4, contradiction. We conclude that Q = 1, whence  = F ∗ (N  ) is simple. L x) with I not a Cp -group. Since mp (S) ≥ 4, Finally, I is a component of CL ( [VK , 3.83] implies that there is t ∈ Ipo (N ) such that CN (t) has a p-component J with J not a Cp -group. However, again by Proposition 5.8, CG (t) ≤ N and J is a  Cp -group, contradiction.  Now we can pin down the possibilities for K. Lemma 11.9. The following conditions hold:  ∼  ∼ (a) p = 3 and K = L3 (4), U3 (8), or M12 , or p = 5 and K = F i22 , or p = 13 ∼ F1 ;  = and K

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 Moreover, if p = 3 or 13, then (b) A induces inner automorphisms on K. AutK (A ∩ K) contains O p (SL2 (p)); (c) For any A-invariant p -subgroup X ≤ CG (x), if X0 = X ∩ Op (CG (x)), then |X : X0 | ≤ 2;  ∼ (d) x ∈ Sylp (C(x, K)) unless possibly K = U3 (8); and  ∼ (e) Sylow p-subgroups of CG (x) are of exponent p unless K = U3 (8). Proof. By the preceding lemma K ≤ M . But by Lemmas 11.5 and 11.6b,   is simple, K  is not K0 = Op (CK (v))Lp (CK (v)) | v ∈ V # ≤ M . Furthermore, K a Cp -group, and mp (C) = 3 (recall our shorthand C = CG (x)). Hence (a) and (b)  = Op (C),  follow from [VK , 8.16ac]. Furthermore, if X is as in (c) and we set Q  then X centralizes Q by the Bender-Thompson lemma [V2 , 2.2]. However, by  = K Q  (as Op (C)  = 1). Hence if we set C  = C/C(x, K), Lemma 11.7, F ∗ (C) ∼  ≤ 2,     then C embeds in Aut(K) and X = X. But now [VK , 8.16d] yields that |X| ∼ proving (c). Finally, suppose that K =  U3 (8), let S ∈ Sylp (C(x, K)), and recall that R ∈ Sylp (C). Since mp (C) = 3, S is cyclic. If S > x, or if S = x with R not of exponent p, then x = 1 (R) ∩ Ω1 (Z(R)) char R, whence x is p-central in G. This is, however, impossible as mp (G) ≥ 4. Thus S = x and R has exponent p, proving (d) and (e).  We next extend Lemma 11.8 to more general proper subgroups of G. Lemma 11.10. Let H be a proper subgroup of G containing A such that I =  Then for some p-component J of H, I ≤ J and I covers Lp (K ∩ H) covers K. J/Op (J). Proof. Since I is a p-component of CH (x), we can consider the subnormal  closure J of I in H, so that J has the usual form by Lp -balance. Also I/Op (I) ∼ =K is of p-rank 2 by the preceding lemma. Hence if J is the product of p p-components g cycled by x, then mp (J) ≥ 4, so if S ∈ Sylp (H), Γoo S,2 (G) ≤ M for some g ∈ G; g g g  clearly J ≤ Γoo S,2 (G), so J ≤ M . But then I ≤ M (as I ≤ J), so M covers K, contrary to Lemma 11.8. Thus J is a single x-invariant p-component of H. Also as A = V x leaves K invariant, A leaves I and hence also J invariant.  = H/Op (H). Then A leaves invariant a Sylow p-subgroup Q of Now set H   Thus CH (J). If Q = 1, then as usual x ∈ Q (as A ∈ Sp (G) and V is faithful on K).    x centralizes J and hence I = J, so the lemma holds in this case.  is simple. Since We can therefore assume that Q = 1, whence J = F ∗ (H) ∼    I is component of CJ( x) and I = K, the preceding lemma and [VK , 3.13, 3.8g]  ∼  ∼ force p = 3, K = L3 (64), D4 (4), or 3D4 (4); or K = U3 (8), with = L3 (4), and J ∼ 3 3 2 J ∼ = U3 (8 ), D4 (8), D4 (8), U9 (2) or E6 (2). If J ∼ = L3 (64), U3 (83 ), 3D4 (4), or 3D4 (8), then W := IJ (A; 3 ) is not a 3 group, by [VK , 8.47]; but it follows from Lemma 11.5 that W ≤ O3 (M ), which is absurd. Thus J ∼ = D4 (4), D4 (8), U9 (2), or 2 E6 (2). Let B0 ∈ E34 (J) with A normalizing B0 . In the last two cases we can even take B0 ∈ E35 (J). Then by g [IA , 7.3.6], AJ ≤ ΓB0 ,2 (G) ≤ Γoo P g ,2 (G) ≤ M for some g ∈ G. But this contradicts Lemma 11.8. The proof is complete.  This in turn yields the analogue of Lemma 11.9c.

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Lemma 11.11. Let H be a K-subgroup of G containing A such that I =  and let X be an A-invariant p -subgroup of H. Then X leaves Lp (K ∩ H) covers K Op (H)I invariant and if we set H0 = Op (H)IXA, then |X : X ∩ Op (H0 )| ≤ 2. ∼K  is a component of H  by the  = H/Op (H), so that I = Proof. Again set H preceding lemma.  then clearly X0 ≤ Op (H0 ). If X leaves I invariant, and we set X0 = CX (I),   and V maps faithBut X/X0 is isomorphic to a V -invariant p -subgroup of Aut(I)  With Lemma 11.9ab and [VK , 8.16d], we see that |X/X0 | ≤ 2, fully into Aut(I). so the desired conclusion holds.  In the contrary case, let I ∗ = I X , so that I∗ = I× I0 , where I0 is a nontrivial  each isomorphic to I.  Furthermore, as X and I are product of components of H,  x-invariant, so is I0 and as mp (I) = 2, it follows at once that mp (CI ∗ x (x)) ≥ 4, contrary to mp (CG (x)) = 3. Hence this possibility is excluded and the lemma is proved.  We are almost in a position to follow the proof of Proposition 9.1 to obtain a contradiction. The missing piece is to obtain CV (T ) = 1, which we prove in the next two lemmas. Lemma 11.12. If p = 5, then V acts faithfully on T . Proof. Suppose  false, so that CP (T ) = 1. Let I be as in Proposition 7.6 and set I ∗ = I M , C0 = CM (x), and K0 = L5 (C0 ). Then by [VK , 8.16be] and Lemma 11.6b, we have K0 /O5 (K0 ) ∼ = D4 (2). By L5 -balance, the subnormal closure K1 of K0 in M is a product of 5-components and K 0 is a component of CK 1 (x). Given the possibilities for I in Proposition 7.6c, it follows from [VK , ∗ 3.12], however, that no component of K 1 is isomorphic to I, whence [I , K 1 ] = 1. Then CI ∗ (x) centralizes CK 1 (x) and hence centralizes K 0 . As m5 (CM (x)) = 3 and ∗ m5 (K0 ) = 2, x ∈ Z(P ) ∩ I . But this contradicts m5 (CG (x)) = 3 as m5 (P ) ≥ 4. The lemma follows.  Lemma 11.13. V acts faithfully on T . Proof. Assume false. By Lemmas 11.12 and 11.9, and Proposition 7.6, we  ∼ have p = 3 with K = L3 (4), U3 (8) or M12 . In any case A induces inner automor phisms on K, so A = x × A0 , where A0 = A ∩ K. Since A = x × V , V projects 0 ) contains Q8 . isomorphically onto A0 . Moreover, by Lemma 11.9b, AutK (A We wish to prove that (11D)

V = A0 .

Recall that A ≤ R ∈ Syl3 (C), where C = CG (x). Set Z = Ω1 (Z(R)). Now   ∼ either R = A (only possible in the L3 (4) case) or O 3 (C) = P GL3 (4) × Z3 or ∼  M12 × Z3 or K = U3 (8), in which cases Z = x × (Z ∩ [Ω1 (R), Ω1 (R)]) ∼ = E32 .  of  ∼ (Note that if K = U3 (8), the condition m3 (C) = 3 prevents any element of C   order 3 from inducing a field automorphism on K, so AutΩ1 (R) (K) ≤ P GU3 (8).) We claim that xG ∩ E1 (A) = E1 (A) − E1 (A0 ). ∼ Q8 or SL2 (3) or GL2 (3) is transitive on A# , so its three orbits First, AutK (A0 ) = 0 on E1 (A) are {x}, E1 (A0 ), and E1 (A) − E1 (A0 ) − {x}, of sizes 1, 4, and 8,

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respectively. Now A ∈ S3 (G) contains 3-central elements of G of order 3, while x is itself not 3-central in G. Letting R ≤ S ∈ Syl3 (G) we have NS (R) > R, and xG ∩ E1 (A) then consists of x together with exactly one of the other AutK (A0 )orbits. If A = R, then NG (A) controls G-fusion of x in A, and | xNG (A) | is divisible by 3 as NS (R) > R. So the claim holds in this case. If A < R, then NS (R) normalizes Z(R), so shears x to Z ∩ [Ω1 (R), Ω1 (R)]. Thus the claim holds. As a result, m3 (CG (t)) = 3 for all t ∈ A − A0 . Now we prove (11D). If V = U , then m3 (CG (v)) ≥ 4 for all v ∈ V # ; by the previous paragraph this implies that V ≤ A0 , so (11D) holds in this case. Suppose G then that V = U , but V ≤ K. Choose v ∈ V − A0 . Then v ∈ x , so we could have chosen v and V U instead of x and A, contradicting our choice (11B) of x and A. Thus (11D) holds. As a consequence, V is strongly closed in A with respect to G. We now begin to consider the structure of M . As NG (V ) ≤ ΓUV,2 (G) ≤ M, we have (M ∩ K)∼ = U3 (2), P GU3 (2), or P ΓU3 (2) in the various cases. Since V is strongly closed in A by (11D), NG (A) ≤ NG (V ) ≤ M . Also if Z(P ) is noncyclic, then as Z(P ) contains no conjugate of x, Ω1 (Z(P )) = V . Since the lemma fails, V ∩ CP (T ) = 1, so we can apply Proposition 7.6. Let I be as in that proposition. Suppose first that I ≤ M and I/O2 (I) ∼ = L± 4 (3), with |I|3 = |G|3 , as in Proposition 7.6b and Lemma 7.5. By Lemma 7.7, Ω1 (T ) = y ∼ = Z2 and I  CG (y). Hence M/O3 (M ) is covered by CG (y), so also by NG (I). Now CM (x) involves Z3 × U3 (2); the same condition therefore holds with NG (I) in place of M . But this implies by [VK , 4.13] that m3 (CNG (I) (x)) ≥ 4, a contradiction as m3 (CG (x)) = 3. Thus by Proposition 7.6, either I ≤ M or I ∩ M ∼ = U6 (2) · 2. In any case ∗ F (I ∩ M ) is simple. Since either Z(P ) is cyclic or Ω1 (Z(P )) = V ∼ = E32 admits Q8 in M , it follows that P permutes the components of E(M ) transitively. Then P normalizes F ∗ (I ∩ M ), by Proposition 7.6d and, in the case F ∗ (I ∩ M ) ∼ = U6 (2), because otherwise P would have a normal E3n -subgroup with n ≥ 12, contradicting m3 (CG (x)) = 3. Again because of the structure and embedding of Z(P ) in M , F ∗ (M ) = F ∗ (I ∩ M ). Now we have the following conditions:

(11E)

(1) M ≤ Aut(F ∗ (I ∩ M )); (2) m3 (M ) ≥ 4; (3) m3 (CG (x)) = 3, and CM (x) contains a subgroup J such that J/O3 (J) ∼ = Z3 × U3 (2); (4) F ∗ (I ∩ M ) is simple, as in Proposition 7.6bc.

Conditions (1) and (2) rule out the following isomorphism types for F ∗ (I ∩ M ): 1 L2 (8), A6 , U4 (2), Sp6 (2), 3D4 (2), 2F4 (2 2 ) , Sp4 (8), L± 3 (3), M11 , J3 . Condition (3) rules out U5 (2), U6 (2), F4 (2), U7 (2), Sp8 (2), 2 D5 (2), 2 E6 (2), G2 (3), Co1 , Co2 , Suz, ∼ ± F i22 , F i23 , F i24 , F3 , F2 , and F1 , leaving only L± 4 (3), and D4 (2). If I = L4 (3) we use [VK , 4.13] to get the contradiction m3 (CM (x)) = 4. Thus, I∼ = D4 (2).

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∼ Given the structure of CM (x), x induces a graph automorphism on I with CI (x) = P GU3 (2).  In all three cases for K, a Sylow 3-subgroup W of CI (x) acts faithfully on K, so by [VK , 8.33], W contains an E32 -subgroup W0 such that one of the following holds:   ∼  = O3 (C  (w)) | w ∈ W # ; (1) K = L3 (4) or M12 , and K K or  (11F)  = L3 (C  (w)) | w ∈ W # , with  ∼ (2) K = U3 (8), and K K L3 (CK (w)) ∼ = L2 (8) or 1 for each w ∈ W # .  there is w ∈ W # such that (accordingly) O3 (C  (w)) ≤ As M does not cover K, 0 K  ∼ M or L3 (CK  (w)) is not covered (mod 3 -core) by M . But w ∈ I = D4 (2) and I ≤ M , so m3 (CM (w)) ≥ 4, i.e., w ∈ I3o (M ). Hence by Proposition 5.3, M contains a Sylow 3-subgroup X of CG (w) and as CG (w) ≤ M , CG (w) is 3-constrained or L3 (CG (w)) has a unique 3-component L, and L/O3 (L) ∼ = L2 (3n ), n ≥ 3, U3 (3n ), n 2 n ≥ 2, or G2 (3 2 ), n ≥ 3. In any case, with the help of Lemma 2.1 and Proposition 9.1,  ICG (w) (W0 × x ; 3 ) = O3 (CG (w)), as W0 × x ∈ S3 (G). But W0 × x normalizes a 3 -group H ≤ CG (w) covering O3 (CK  (w)), and hence such that H ≤ M . However, H ≤ O3 (CG (w)) ≤ M as w ∈ I3o (M ) and Γoo P,2 (G) ≤ M . This contradiction rules out (11F1), so (11F2) holds. Then L3 (CG (x, w)) has a 3-component Lxw such that Lxw /O3 (Lxw ) ∼ = L2 (8). By L3 -balance, Lxw ≤ L3 (CG (w)) = L and Lxw is a 3-component of CL (x). As L/O3 (L) ∈ Chev(3) has twisted Lie rank 1, the only possibility is that x induces 3 1 a field automorphism on L/O3 (L) ∼ = 2 G2 (3 2 ). But then CL/O3 (L) (x) ∼ = 2 G2 (3 2 ) has 3-rank 2, and centralizes x, w. Thus m3 (CG (x)) ≥ 4, a final contradiction completing the proof of the lemma.  Now we can follow the proof of Proposition 9.1 to obtain a contradiction. The preceding lemma lets us get started, using the following result. Lemma 11.14. For some v ∈ V # , all subgroups of x, v − v of order p are CP (v)-conjugate. Proof. Suppose first that V = U and set S = CP (U ), so that mp (S) = mp (P ) ≥ 4 and A ≤ S. Set S1 = CS (A). Since A ∈ Sp (G), it follows that A = Ω1 (S1 ). Thus mp (S1 ) = 3, so S1 < S and hence there is t ∈ NS (S1 ) − S1 . Since A char S1 , t leaves A invariant. But t ∈ S centralizes U , so t does not centralize x (as A = U x). Set v = [x, t], so that v ∈ U # (as U  S). Since t centralizes U , it follows that all subgroups of x, v−v of order p are t-conjugate, so the lemma holds in this case. Suppose then that V = U . Since A = x × V with V = CA (U ), x does not centralize U . Since U ∩ Z(P ) = 1, it follows that [U, x] = z with z ∈ Z(P )# . As A ∈ Sp (G), z ∈ A, indeed z ∈ CA (U ) = V . Likewise all subgroups of z, x − z of order p are U -conjugate, so the lemma holds in this case as well.  Fix v as in the preceding lemma. By Lemma 11.13, [v, T ] = 1. As T is P invariant, we have (11G)

[CT (x), v] = 1.

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We use this to prove Lemma 11.15. If X is an A-invariant Sylow 2-subgroup of Op (CG (x)), then CK (X) ≤ Op (CG (x)). Proof. Set T1 = [CT (x), v], so that T1 = 1, T1 is A-invariant and T1 = [T1 , v]. By Lemma 11.9c, |T1 : T1 ∩ Op (CG (x))| ≤ 2, so v centralizes T1 /T1 ∩ Op (CG (x)). Since T1 = [T1 , v], it follows that T1 ≤ Op (CG (x)). Expand T1 to an A-invariant Sylow 2-subgroup X of Op (CG (x)). As all Ainvariant Sylow 2-subgroups of Op (CG (x)) are conjugate in CCG (x) (A), it will suffice to prove that CK (X) ≤ Op (CG (x)). But v induces a nontrivial inner auto by Lemmas 11.6 and 11.9b and v does not centralize X as T1 = 1. morphism on K  is As mp (CG (x)) = 3, x is the unique subgroup of order p in C(x, K). Since K simple, v = v1 xi for some i and some v1 ∈ Ip (K). Therefore [v1 , X] = 1, so CK (X)  Now by [V2 , 6.10], CK (X) ≤ Op (CG (x)), as asserted. does not cover K.   We can therefore consider By the Frattini argument, Lp (NCG (x) (X)) covers K. a proper subgroup H of G with the following properties:

(11H)

 (1) A ≤ H and I = Lp (K ∩ H) covers K; (2) W is a maximal A-invariant 2-subgroup of H, and X ≤ W; (3) Subject to (1) and (2), |W | is maximal; and (4) Subject to (1), (2), and (3), |H|p is maximal.

For such a choice of H and W , we prove Lemma 11.16. W ∈ Syl2 (Op (M )) and |H|p = |P |. Proof. By Lemma 11.5, W ≤ Op (M ). For simplicity of notation, assume W ≤ T . Set W1 = NT (W ) and put H ∗ = NG (W ∗ ), where W ∗ = Ω1 (Z(W )) or J(W ). Then W1 A ≤ H ∗ as W ∗ char W . Also X ≤ W ≤ W1 . We argue that  Our choice of H will then imply first that W1 = W , I := Lp (K ∩ H ∗ ) covers K. whence W = T . But then P ≤ H ∗ and our choice of H will yield that |H|p = |P |. By Lemma 11.11, W leaves Op (H)I invariant and if we set H0 = Op (H)IW A and W0 = W ∩ Op (H0 ), then |W : W0 | ≤ 2. Also the maximality of W implies that W0 ∈ Syl2 (Op (H0 )). Hence by the Frattini argument I0 = Lp (NH0 (W0 )) covers  0 = H0 /Op (H0 ). But x centralizes I0 and x leaves I0 invariant as it leaves I0 in H  But W0 invariant. Hence also I ∗ = Lp (CI0 (x)) covers I0 . Thus I∗ = I0 ∼ = K. ∗ ∗ p ∗ ∗   / A as I is simple. Since K = Lp (CG (x)) A = A ∩ I = 1 as A ∈ S (G), and x ∈  = x , it follows that I ∗ ≤ K, so I ∗ covers K.  Hence to complete the and CA (K) ∗ ∗ proof, we need only show that I ≤ H for at least one of the two choices of W ∗ .  and X ≤ But again |W/W0 | ≤ 2 by Lemma 11.11. Also as I ∗ covers K Op (CG (x)) ∩ W , X ≤ W0 . Lemma 11.15 thus yields that CH0 (W0 ) does not cover I∗ . However, I ∗ normalizes W0 , W leaves I ∗ invariant, and W0 ∩ Op (I ∗ ) ∈ Syl2 (Op (I ∗ )). Since I∗ ∼ = L3 (4) or M12 or U3 (8) (with p = 3) or F i22 (with p = 5) or F1 (with p = 13), it follows from [VK , 6.1] that I ∗ normalizes Ω1 (Z(W )) or J(W ). Correspondingly, taking W ∗ = Ω1 (Z(W )) or J(W ), we conclude that  I ∗ ≤ NG (W ∗ ) = H ∗ , as required. Now we obtain an immediate final contradiction. By the preceding lemma and our notational assumption, W = T . Thus P ≤ H. But by Lemma 11.10, I ≤ J

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 where H  = H/Op (H). We know that for some p-component J of H and I = J,  Set A = V x with V inducing faithful inner automorphisms on I as I covers K.  Q = CP (J). Since mp (CG (x)) = 3, it follows at once that Ω1 (CQ (x)) = x, forcing Q to be cyclic. Since mp (J) = mp (K) = 2, J  H, whence Q  P . Thus x ∈ Z(P ), so mp (CG (x)) = mp (P ) ≥ 4, contradiction. This completes the proof of Proposition 11.1 and hence the proof of Theorem 3. When combined with Proposition 5.3, Theorem 3 immediately yields:  = N/Op (N ), and Proposition 11.17. Let x ∈ Ipo (M ), set N = NG (x), N  K = Lp (N ). If N ≤ M , then the conditions of Proposition 5.3c hold. In particular, n  = E(N ) ∼ K = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or 2 G2 (3 2 ), n ≥ 3 (with p = 3), and NM (x) is p-solvable.  ) = Op (N ) ∼ Proof. If false, then by Proposition 5.3, F ∗ (N = p1+4 ∗ Zpm or    )) = 3. HowF (N ) = Op (N ) has a normal Ep3 -subgroup. In any case mp (Op (N o ever, as x ∈ Ip (M ), M contains a Sylow p-subgroup R of N by Proposition 5.3a.  = Op (N  ) and hence Without loss R ≤ P . Set Q = R ∩ Op p (N ), so that Q o mp (Q) = 3. However, as ΓP,2 (G) ≤ M by Proposition 11.1b, NG (Q) ≤ M . But N = Op (N )NN (Q) by the Frattini argument and Op (N ) ≤ M by Proposition 5.3, so N ≤ M , contrary to assumption.  ∗

12. Theorem 4: ΓP,2 (G) ≤ M In this section we establish Theorem 4. Thus we prove Proposition 12.1. ΓP,2 (G) ≤ M . We assume false and derive a contradiction in a sequence of lemmas. The argument is a considerably shortened adaptation of the proof of [GL1, Proposition 2.1, pp. 650–663]. There is thus Q ≤ P with mp (Q) ≥ 2 such that NG (Q) ≤ M . We consider the set D of all p-subgroups D of G such that (12A)

(1) mp (D) ≥ 2; (2) mp (NM (D)) ≥ 2; and (3) NG (D) ≤  M.

In view of the existence of Q, D = ∅. We define D∗ to be the subset consisting of all D ∈ D satisfying the following conditions:

(12B)

(1) (2) (3) (4)

mp (NM (D)) is maximal; Subject to (1), |NM (D)|p is maximal; Subject to (1) and (2), |Op (NG (D))| is maximal; and Subject to (1), (2), and (3), |D| is maximal.

We fix D ∈ D∗ and write N = NG (D). By (12B4), D = Op (N ). We also let R ∈ Sylp (N ∩ M ), so that mp (R) ≥ 2. The first set of lemmas pins down the structure of N . We first prove Lemma 12.2. R ∈ Sylp (N ). Proof. Expand R to S ∈ Sylp (N ). If R < S, then S1 = NS (R) > R; so NG (S1 ) ≤ M by definition of D and D∗ . But then S1 = R by definition of R,  contradiction. Thus R = S ∈ Sylp (N ), as asserted.

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Without loss, R ≤ P . In particular, D ≤ P . We also immediately obtain Lemma 12.3. mp (D) = 2 = mp (DCG (D)). Proof. If mp (D) ≥ 3, then as ΓoP,2 (G) ≤ M by Theorem 3, N = NG (D) ≤ M , contrary to the definition of D. Thus mp (D) = 2. Suppose there is x ∈ Ip (G) − D with x ∈ DCG (D). Since DCG (D)  N , R ∩ DCG (D) ∈ Sylp (DCG (D)), so we can choose x ∈ R. Thus mp (DCP (D)) ≥ mp (DCR (D)) ≥ 3. Again as ΓoP,2 (G) ≤ M , it follows that N = NG (D) ≤ M , giving the same contradiction, and the lemma is proved.  This in turn yields: Lemma 12.4. N is p-constrained. Proof. Indeed, if false, then K = Lp (N ) contains some x of order p in K − Op p (K), by [IG , 16.11]. But K and hence x centralizes D as D  N , so D x ≤ DCG (D). Therefore, x ∈ D by the preceding lemma. Thus x ∈ Op (N ) and hence  x ∈ Op p (K), which is not the case. Next, set Z = Ω1 (Z(P )) and N0 = NG (Z). We prove Lemma 12.5. Zp ∼ = Z ≤ D and N0 ≤ M . Proof. Since mp (D) = mp (DCG (D)) and Z centralizes D, Z ≤ D, whence Z ≤ Z(D). If mp (Z) > 1, then as mp (D) = 2, it follows that Ep2 ∼ = Z = Ω1 (D). However, as mp (CG (Z)) = mp (P ) ≥ 4, NG (Z) ≤ ΓoP,2 (G) ≤ M . Since Z char D, this yields N ≤ NG (Z) ≤ M , contradiction. Thus Zp ∼ = Z ≤ D. Write Z = z. Suppose next that N0 ≤ M . Since z centralizes P , z ∈ Ipo (M ) and P ≤ N0 . Thus N0 is given by Proposition 11.17. In particular, if K0 = Lp (N0 ), then / Z0 , so Z is noncyclic, K0 /Op (K0 ) is simple. But then Z0 = Z ∩ K0 = 1 and z ∈ contrary to the preceding paragraph.  This immediately yields: Lemma 12.6. Op (N ) ≤ M and D ∈ Sylp (Op p (N )). Proof. Op (N ) centralizes D ≤ Op (N ), so centralizes Z. Since N0 = NG (Z) ≤ M , the first assertion holds. But now if we set S = R ∩ Op p (N ), then as S ∈ Sylp (Op p (N )), N = Op (N )NN (S) by the Frattini argument, so NN (S) ≤ M . Thus NG (S) ≤ M . Since R ≤ NG (S) and D ≤ S, our choice of D in D∗ forces  D = S, so D ∈ Sylp (Op p (N )), and the lemma is proved. We next determine the structure of D. ∼ Ep2 . Lemma 12.7. D = Proof. First, Z is not characteristic in D, otherwise N = NG (D) ≤ NG (Z) = N0 ≤ M , contradiction. In particular, if we set E = Ω1 (Z(D)), then as Z ≤ E, we have Z < E, so E ∼ = Ep2 as mp (D) = 2. Let Ep2 ∼ = U  P . Then Φ(P ) centralizes U . Hence if we set F = E ∩ Φ(D), then F centralizes U and F char D. By the preceding paragraph, F = Z. Hence if F = 1, then E = ZF , in which case E centralizes U . But then mp (ECP (E)) ≥ 3 (whether or not E = U ), so NG (E) ≤ ΓoP,2 (G) ≤ M . However, N ≤ NG (E) as E char D, so N ≤ M , again a contradiction. We conclude that F = 1. Thus E ∩ Φ(D) = 1. Since Φ(D)  D and E = Ω1 (Z(D)), this forces Φ(D) = 1.  Hence D is elementary abelian, whence D = E ∼ = Ep2 , as asserted.

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As a consequence, D ∈ Sp (G).  = N/CN (D) ∼ Now set N = AutG (D). We next prove Lemma 12.8. The following conditions hold:  ) ∼ (a) O p (N = SL2 (p); (b) All elements of E1 (D) are N -conjugate; (c) R ∼ = p1+2 ; and (d) D ∈ Sylp (CG (D)). Proof. As D ∈ Sylp (Op p (N )) and N is p-constrained, CG (D) = CN (D) = D × Op (N ),  ) = 1. But D ∼  ≤ GL2 (p). so (d) holds. For the same reasons, Op (N = Ep2 , so N p  ∼  Also D < R as D < P . Hence O (N ) = SL2 (p) by [V2 , 6.11], whence N acts transitively on the elements of E1 (D), proving (a) and (b). By (a) and (d),  O p (N/Op (N )) is a split extension of DOp (N )/Op (N ) ∼ = Ep2 by SL2 (p). So  R∼ = p1+2 , completing the proof. We can also prove Lemma 12.9. p ≥ 5. Proof. Suppose p = 3. Since mp (P ) ≥ 4, P contains a normal E81 -subgroup A by [IG , 10.17]. Hence P is connected. But E32 ∼ = D ∈ S3 (G), a contradiction.  We have now pinned down the precise structure of N . To eliminate this configuration, we study the permutation action of CG (D) on the set of maximal D-invariant 2-subgroups of G. To do so, we need a few preparatory results. Lemma 12.10. N0 is p-constrained. In particular, Op (N0 ) contains every Dinvariant p -subgroup of N0 . Proof. The second assertion is immediate from the first and [V2 , 2.2]. Suppose the first assertion fails, whence L = Lp (N0 ) = 1. Set Q = P ∩ Op p (N0 ), so that Z ≤ Q and Q  P . 0 = N0 /Op (N0 ), so that Z centralizes L.  Writing D = Z x, we have Set N I := Lp (CL (x)) = Lp (CL (D)) ≤ Lp (N ) = 1. In particular, if L1 , L2 , . . . , Lr denote the p-components of L, D leaves each Li invariant, 1 ≤ i ≤ r.  i is simple. If L0 = 1, Let L0 be the product of those Li , 1 ≤ i ≤ r, such that L  Thus then Z(P ) ∩ L0 = 1, forcing Z ≤ L0 , which is impossible as Z centralizes L.   each Li is nonsimple, 1 ≤ i ≤ r. Set Zi = P ∩ Op p (Li ), 1 ≤ i ≤ r. Then as D leaves Li invariant, Di = CZi (D) = 1. But D ∈ Sylp (CG (D)), so each Di ≤ D.  whence clearly D ∈ If any Di ≤ Z, then D = Di Z centralizes L, / Sylp (CG (D)). It follows that Z = Di for each i, 1 ≤ i ≤ r.  i by [VK , Next, if any Zi is noncyclic, then D induces inner automorphisms of L 2.6], whence Zi = Di = Z, contradiction. We conclude that each Zi , 1 ≤ i ≤ r, is cyclic. Suppose now that r > 1. By [IG , 10.11], P ∩ Li contains a D-invariant Ep2 subgroup Ui with Z ≤ Ui . Since D ∈ Sylp (CG (D)), D ≤ Li for any i, so D = Ui and hence D does not centralize Ui , so if we write D = Z x, then [Ui , x] = Z for each i, 1 ≤ i ≤ r. Also U1 centralizes U2 and U1 = U2 as Z1 and Z2 are cyclic. It follows that x and hence D centralizes some element u ∈ U1 U2 − Z. Thus u ∈ D,

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so D = Z u ≤ U1 U2 . Since U1 U2 is elementary, this in turn forces D = U1 U2 , =L  1 is quasisimple. whence U1 = U2 , contradiction. Thus r = 1 and hence L  Now as D ∈ Sylp (CG (D)) and p ≥ 5, with Lp (CL (D)) = 1, [VK , 1.6] implies ∼ SL (q), q ≡  (mod p),  = ±1; and D/Z induces outer diagonal auto= that L p  On the other hand, by the structure of N  (O p (N ) ∼ morphisms on L. = SL2 (p)), there is y ∈ N such that with Z = z and some choice of x ∈ D − z , z y = z n and  xy = xn , where n, n are primitive roots (mod p) and nn ≡ 1 (mod p). In partic D.  However, by [VK , 2.1], y must induce the ular, y ∈ N0 = NG (Z), so y acts on L   which same power map on Z as it does on the outer-diagonal subgroup of Out(L),  2   is incident to D/Z. Therefore n ≡ n (mod p), so n ≡ 1 (mod p), contrary to the fact that n is a primitive root (mod p) and p ≥ 5. The lemma is proved.  We also need the following result.  = Lemma 12.11. Let H be a proper subgroup of G containing P and set H ∗ p ∗ ∼  = L2 (p ) and D∩F   = (H) H/Op (H). Then Op (H) ≤ M . If H ≤ M , then F (H)  Z. Proof. Set K = Lp (H) and Q = P ∩ Op p (H). Since mp (P ) ≥ 4, [VK , 9.1a] implies that mp (KQ) ≥ 3. Since ΓoP,2 (G) ≤ M , Op (H) ≤ M and the usual Frattini argument yields that H ≤ M if K ≤ M . Hence K ≤ M , so M does not  Again as Γo (G) ≤ M , it follows now from [VK , 9.1b] that K  ∼ cover K. = L2 (pp ) P,2  In with Q = 1 and some d ∈ D inducing a nontrivial field automorphism on K. ∗  = F (H).  Also d is sheared to D ∩ K ∼ particular, K = Zp by elements of P ∩ K. Since P ≤ M and Z = Z(P ), Z = D ∩ K. The lemma follows.  Recall that T is a P -invariant Sylow 2-subgroup of Op (M ). We know by Proposition 8.1 that P acts faithfully on T. In particular, T = 1. We let

W = I∗G (D; 2),

the set of all maximal D-invariant 2-subgroups of G with respect to inclusion. As stated above, we study the permutation action of CG (D) on W. Using the previous lemma, we can prove: Lemma 12.12. T ∈ W. Proof. Since D ∈ Sp (M ), T is a maximal D-invariant 2-subgroup of M by Theorem 2. Hence the lemma holds if NG (T ) ≤ M . We can therefore set H = NG (T ) and assume that H ≤ M and likewise that T < W for some W ∈ W, so that W1 = NW (T ) > T . We have P ≤ H. By the preceding lemma, Op (H) ≤ M , so the maximality of T in M implies that T ∈ Syl2 (Op (H)). Thus if we set  = H/Op (H), then W 1 = 1. But by the preceding lemma, if we put K = Lp (H), H  ∼  = F ∗ (H).  As p ≥ 5 and D ∈ Sp (H), W 1 = 1 by Lemma then K = L2 (pp ) and K 2.1, a contradiction completing the proof.  We next prove Lemma 12.13. CG (D) acts intransitively on W.

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Proof. Suppose false. Now T ∈ W and so for any y ∈ N = NG (D), likewise T y ∈ W (as y leaves D-invariant). Thus T yu = T for some u ∈ CG (D). But then yu ∈ NG (T ), so y ∈ NG (T )CG (D). Since y was arbitrary, it follows that N ≤ NG (T )CG (D). On the other hand, CG (D) ≤ CG (Z) ≤ N0 ≤ M by Lemma 12.5, so as N ≤ M , we conclude that NG (T ) ≤ M .  = H/Op (H). By Lemma 12.11, K  ∼ Set H = NG (T ), K = Lp (H), and H = p ∗   L2 (p ) with K = F (H). Also as shown in that proof Z = D ∩ K, so Z  NH (D). On the other hand, as N ≤ HCG (D) and N acts transitively on the elements of E1 (D) by Lemma 12.8b, so therefore does NH (D). This contradicts Z  NH (D) so the lemma holds.  We choose W ∈ W not CG (D)-conjugate to T in such a way that V := T ∩ W has maximal order. In particular, T ≤ W and W ≤ T , so V < T and V < W . We prove Lemma 12.14. V = 1 and NT (V ), NW (V ) is not a p -group. Proof. Set T1 = NT (V ) and W1 = NW (V ), so that T1 > V and W1 > V . Moreover, T1 and W1 are D-invariant. Our choice of W implies that Y = T1 , W1  is not a p -group. Indeed, suppose false. Then for some g ∈ CY (D), T1 , W1g  is a 2-group, and we expand it to T2 ∈ W. Then T ∩ T2 ≥ T1 > V so by our maximal choice of V , T2h = T for some h ∈ CG (D). Now W1gh ≤ T2h = T , so W1gh ≤ W gh ∩ T . In particular |W gh ∩ T | ≥ |W1 | > |V |, so again by our maximal choice of V , W ghk = T for some k ∈ CG (D). As ghk ∈ CG (D), this contradicts our choice of W . Thus Y is not a p -group, as asserted. It therefore remains to prove that V = 1, so assume false. We claim first that Z does not centralize W . Indeed, otherwise W ≤ N0 ≤ M . But then as D ∈ Sp (M ), W ≤ Op (M ) by Theorem 2(a). Since T ≤ Op (M ), T and W are both D-invariant Sylow 2-subgroups of Op (M ), so they are conjugate in COp (M ) (D), contrary to our choice of W . Therefore, [Z, W ] = 1. We apply [V2 , 6.15]. Since R ∼ = p1+2 acts faithfully on T with Z = Z(R), all elements of E1 (D) − Z are R-conjugate and hence CT (x) = 1 for every x ∈ D − Z. Since Z does not centralize W , CW (x) = 1 for some x ∈ D − Z. Hence we can choose x ∈ D − Z so that both T0 = CT (x) = 1 and W0 = CW (x) = 1. But T0 and W0 are D-invariant, and CG (x) is p-constrained by Lemmas 12.8b and 12.10, whence T0 , W0  ≤ Op (CG (x)) by [V2 , 2.2]. Since T0 > V and W0 > V , we can therefore repeat the argument of the first paragraph with T0 , W0 , Op (CG (x)) in the roles of T1 , W1 , Y , respectively, to conclude that T and W are CG (D)conjugate, contrary to the choice of W . Thus V = 1 and the lemma is proved. 

The lemma shows that NG (V ), which contains T1 D, is a 2-local subgroup of G in which the set of all D-invariant 2-subgroups does not generate a p -group. Let

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79

H be the set of all 2-local subgroups of G satisfying these conditions and choose H ∈ H such that

(12C)

(1) A maximal D-invariant 2-subgroup T2 of H ∩ M containing T1 has maximal order; (2) Subject to (1), mp (H) is maximal; (3) Subject to (1) and (2), |O2 (H)| is maximal; and (4) Subject to (1), (2), and (3), |H| is maximal.

Since T2 ≤ Op (M ) by Theorem 2, we can clearly assume without loss that  = H/Op (H). We T2 ≤ T . We set J = Lp (H), Hd = CH (d) for d ∈ D# , and H first prove Lemma 12.15. The following conditions hold:  and J ∈ Cp ; (a) J is simple, J = F ∗ (H), (b) D ≤ J; (c) The subgroup generated by all D-invariant 2-subgroups of J is not a p group; and  d ) contains every D-invariant p -subgroup of H d. (d) For any d ∈ D# , Op (H Proof. By our choice of H, the subgroup Y = IH (D; 2) is not a p -group. Likewise by Lemmas 12.8b and 12.10, for each d ∈ D# , every D-invariant p -subgroup of CG (d) is contained in Op (CG (d)), so every D-invariant p -subgroup of Hd is contained in Op (Hd ). In particular, (d) holds. Likewise H is not p-constrained, otherwise Y ≤ Op (H) by [V2 , 2.2], contrary to the fact that Y is not a p -group. Thus J = 1. We let J1 , J2 , . . . , Jr be the p-components of J. Expand D to S ∈ Sylp (H), so that S ≤ M g for some g ∈ G. Note that if mp (S) ≥ 3, then ΓoS,2 (G) ≤ M g . In particular, Op (H) ≤ M g . If, furthermore,  then H ≤ M g . However, D ∈ Sp (M g ) (as D ≤ H and D ∈ Sp (G)), M g covers H, in which case Y ≤ Op (M g ) by Theorem 2, again a contradiction. Thus M g and  hence ΓoS,2 (G) does not cover H. Set Q = S ∩ Op p (H). If mp (Q) ≥ 2, then mp (JQ) ≥ 3 by [IG , 16.11], so  and then by the Frattini the previous paragraph applies. Then M g covers JQ g  argument, M covers H, contradiction. Thus mp (Q) ≤ 1. Suppose Q = 1. Then as D ∈ Sp (H), D ∩ Q = d = 1 and as Op (H) d  H, it follows that the subgroup of H generated by all D-invariant 2-subgroups of Hd is not a p -group, contrary to (d). We conclude that Q = 1. Similarly if r ≥ 3 or if r = 2 and mp (Ji ) ≥ 2 for i = 1, 2, then mp (S) ≥ 3 and the previous paragraph again applies. We get that  contradiction. Thus r ≤ 2 and if equality M g covers J and then that M g covers H, holds, then, say, mp (J1 ) = 1. Suppose then that r = 2. In particular, D leaves J1 and J2 invariant. Since D ∈ Sp (H) and J1 and J2 are simple (as Q = 1), it follows that D = D1 × D2 , where Zp ∼ = Di ≤ Ji , i = 1, 2. We argue that Y is a p -group. Set di  = Di ,  d . Using (d), i = 1, 2, and let X be a D-invariant 2-subgroup of H. Then J1 ≤ H 2 we have  d1 ] ≤ J1 ∩ X  ≤ J1 ∩ Op (H  d ) ≤ Op (J1 ) = 1. [X, 2

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 ≤H  d so X  ≤ Op (H  d ). Since X was arbitrary, we conclude that Y ≤ F, Thus X 1 1  so Y and hence also Y is a p -group, which is not the case. Thus r = 1 and as  with J simple. Q = 1, it follows that J = F ∗ (H) Since Y = 1 but p ≥ 5, we conclude from Lemma 2.1 that J ∈ Cp . This completes the proof of (a).  with p ≥ 5 and J = F ∗ (H)   ∈ Sp (H) Next, suppose (b) fails. Since Ep2 ∼ =D  ∼   with J simple, [VK , 9.2] implies that either J = Lp (q), q ≡  (mod p),  = ±1,  or mp (J) = 1 with some d ∈ D# inducing a non-inner diagonal automorphism on J, #  with some d ∈ D inducing a nontrivial field automorphism on J. However, in the first case, mp (S) ≥ 4 and mp (S ∩ J) ≥ 3, and then ΓoS,2 (G) and hence M g covers J  ≥ 3, M g covers H  by the Frattini argument, by [IA , 7.3.6], as p ≥ 5. Since mp (J) contrary to what we have shown above. Thus mp (J) = 1. But now if we let E be a D-invariant Sylow p-subgroup of J, then E is cyclic and D0 = D ∩ E = 1 as D ∈ Sp (H). By [VK , 1.2], E ∼ = Zpn for some n ≥ 2. Hence |NE (D)| ≥ p2 , so a Sylow p-subgroup of N = NG (D) is not of exponent p, contrary to Lemma 12.8c. Thus (b) holds. Now we verify (c). Let Y1 be the subgroup generated by all D-invariant 2subgroups of J. If (c) fails, then Y1 is a p -group and hence so is Y1 . But D ≤ J. It is immediate therefore from the definition of Y that Y ≤ Y1 , CY (D). But again CY (D) ≤ Op (Hd ) for any d ∈ D# , so CY (D) is a p -group. However, CH (D) permutes the set of D-invariant 2-subgroups of J and so leaves invariant their join Y1 . Hence Y1 CY (D) = Y is a p -group, which is not the case. Thus (c) also holds and the proof is complete.  We sharpen the preceding result. Lemma 12.16. The following conditions hold: (a) J is unambiguously of Lie type of characteristic r, r odd, r = p; and (b) mp (H) ≤ 2.  ≥ 3, it follows once Proof. With S and g as in the preceding lemma, if mp (J) g o  so Γ (J)  < J.  again that M does not cover J, S,2  Furthermore, by the preceding lemma, D ≤ J and if Y is the subgroup gener  then Y is not a p -group. But for each ated by all D-invariant 2-subgroups of J, #  contains every D-invariant  Hence D   2-subgroup of CJ(d). d ∈ D , Op (CJ(d)) #    does not centralize some D-invariant 2-subgroup X of J and for every d ∈ D ,  ≤ Op (C (d)).  Since Ep2 ∼  and p ≥ 5, all the conditions of [VK ,  ∈ Sp (J) CX (d) =D J 7.4] are satisfied, so J has the stated form and mp (S) = mp (H) ≤ 2.  We complete the proof of the proposition by arguing to the contrary that mp (H) ≥ 4. We need the following additional properties of H. Lemma 12.17. The following conditions hold: (a) (b) (c) (d)

O2 (H) ∈ Syl2 (Op (H)); O2 (H) ≤ T2 ; H = NG (O2 (H)); and  CH (O2 (H)) does not cover J.

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13. THEOREM 5

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Proof. Let T0 be a T2 D-invariant Sylow 2-subgroup of Op (H). Then J0 = NJ (T0 ) covers J by the Frattini argument, so the subgroup generated by all Dinvariant 2-subgroups of J0 is not a p -group. Hence if H0 = NG (T0 ), the corresponding assertion holds for H0 . Also NG (T0 ) contains a Sylow p-subgroup of H by the Frattini argument, so mp (H0 ) ≥ mp (H). Since T2 D ≤ H0 and O2 (H) ≤ T0 , it follows therefore from our choice of H in (12C) that T0 = O2 (H). In particular, H ≤ H0 , so again by our choice of H, H0 = H. Thus (a) and (c) hold. Next, expand T0 T2 to T ∗ ∈ W, so that T ∩ T ∗ ≥ T1 > V and hence T ∗ = T y for some y ∈ CG (D). Then y ∈ NG (Z) = N0 ≤ M , so T ∗ ≤ M . Thus T0 T2 ≤ M and hence T0 ≤ T2 by definition of T2 in (12C), so (b) also holds.  by [V2 , 7.4], Finally, suppose (d) fails, in which case I = L2 (CH (T0 )) covers J, and D ≤ I. Since H = NG (T0 ), I is a 2-component of CG (T0 ). Hence if we take t ∈ I2 (Z(T0 )), we can consider the pumpup K of I in CG (t). By L2 -balance, K is the product of 2-components K1 , K2 , . . . , Km transitively permuted by T0 and I is a 2-component of CK (T0 ). Since D ≤ I and D ∈ Sylp (CG (D)), we must have m = 1. Thus K = K1 . But G is of restricted even type, so O2 (CG (t)) = 1 and K is a Co2 -group. Since I is a 2-component of CK (T0 ), [VK , 3.3] implies that  so I is unambiguously of Lie type likewise I is a Co2 -group. However, I/Z(I) ∼ = J, of characteristic r, r odd, r = p. Since p ≥ 5, it follows therefore from [VK , 9.9] that mp (I) ≤ 1, contrary to the fact that Ep2 ∼ = D ≤ I. Thus (d) also holds, and the lemma is proved.  Now we derive the desired final contradiction. With T0 = O2 (H), set H0 =  T2 acts faithfully on J.  Also by the preceding O (H)JT2 . Since J = F ∗ (H),  lemma, T0 ≤ T2 , T0 ∈ Syl2 (Op (H)), and CJ (T0 ) ≤ Op (H). It follows that CH0 (T0 ) ≤ O2 (H)T0 . Since J is unambiguously of Lie type of characteristic r, r odd, by Lemma 12.16a, we conclude therefore from [VK , 6.2] that p

 for one of T ∗ = Ω1 (Z(T2 )) or T ∗ = J(T2 ). NH0 (T ∗ ) covers J, Thus H ∗ = NG (T ∗ ) covers J and D ≤ H ∗ , so the subgroup generated by the D-invariant 2-subgroups of H ∗ is not a p -group, inasmuch as the corresponding  But T2 ≤ T and T3 = NT (T2 ) ≤ H ∗ as T ∗ char T2 . It follows assertion holds in J. therefore from our choice of H in (12C) that T3 = T2 , whence T = T2 . But P leaves T invariant, so as T ∗ char T , P ≤ H ∗ . Thus mp (H ∗ ) ≥ 4. However, mp (H) < 4 by Lemma 12.16b, so our choice of H in (12C) is contradicted. This completes the proof of Proposition 12.1 and with it the proof of Theorem 4. 13. Theorem 5 In this section we prove Theorem 5, by proving the following two propositions. Proposition 13.1. Suppose CP (T ) = 1. Then M has a p-component L with L ∈ C2 , M is an L-preuniqueness subgroup of G, and M is almost strongly pembedded in G. Proposition 13.2. M is a maximal subgroup of G. We prove the second proposition first. Suppose that M ≤ H < G. We first prove Lemma 13.3. Op (M ) = Op (H).

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Proof. As P ≤ M ≤ H, clearly Op (H) ≤ ΓP,2 (G) ≤ M ≤ H, so Op (H) ≤  = H/Op (H) and W = Op (M ). We must prove that W  = 1. Op (M ). We set H Set H0 = ΓP,2 (H), so that H0 ≤ M ≤ NH (W ). By [VK , 8.41] we conclude  = 1, completing the proof. that W  We next take B ∈ E∗ (P ) and prove Lemma 13.4. Op (M ) = Θ1 (G; B). Moreover, M = NG (Op (M )). Proof. Since by definition M = NG (Θ1 (G; B)), we have Θ1 (G; B) ≤ Op (M ) and it suffices to prove that Op (M ) ≤ Θ1 (G; B). As  Θ1 (G; B) = Op (CG (b)) | b ∈ B # is the closure of the 1-balanced functor, it suffices to show that COp (M ) (b) ≤ Op (CG (b)) for all b ∈ B # . But since mp (B) ≥ 4 and B ∈ Sp (G), this is immediate from Proposition 3.3.



Proposition 13.2 follows immediately from Lemmas 13.3 and 13.4, as they imply that H ≤ NG (Op (H)) = NG (Op (M )) = M . Now we turn to Proposition 13.1. Assume then that CP (T ) = 1. Then Proposition 7.6 applies, and we let I be as there. Thus I is a component of CG (T ) as well as of CG (z) for all z ∈ I2 (T ). Moreover if I ≤ M , then the structure of I is given in Proposition 7.6c, and I in particular is simple. If I ≤ M , on the other hand, then Proposition 7.6b applies. We first rule out Proposition 7.6b, thus proving Lemma 13.5. I ≤ M . ∼ Proof. Otherwise, by Proposition 7.6b, we have I/O2 (I) ∼ = L± 4 (3) or I = Co2 , with P ≤ I in either case. By [VK , 8.40] and Theorem 4, I ≤ ΓP,2 (G) ≤ M , contrary to assumption.  Now we set L = Lp (IOp (M )), so that L is a p-component of M and L = I. We will show that M is an L-preuniqueness subgroup and is almost strongly p(13A) embedded in G. We prove (13A) by contradiction. Note that if M is strongly p-embedded in G, then it follows immediately from the definitions that M is an L-preuniqueness subgroup. We consider first the case (13B)

I = F ∗ (M ).

Thus mp (Aut(M )) = mp (M ) ≥ 4. From Proposition 7.6 and [VK , 3.101], I is isomorphic to one of the following groups with p = 3: U5 (2), U6 (2), U7 (2), Sp8 (2), D4 (2), 2 D5 (2), F4 (2), 2 E6 (2), L± 4 (3), G2 (3), Co1 , Co2 , Suz, F in , 22 ≤ n ≤ 24, Fi , 1 ≤ i ≤ 3, or else p = 5 and I ∼ = F1 .

(13C)

Lemma 13.6. If I = F ∗ (M ) and x ∈ Ipo (M ) ∩ I, then NG (x) ≤ M .

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13. THEOREM 5

83

 = N/Op (N ) and suppose that N ≤ M . Proof. Let N = NG (x) and N ) = K  × Q,  where Q  is cyclic and By Proposition 11.17, Op (N ) ≤ M and F ∗ (N  K ∈ Chev(p) has twisted rank 1. Without loss R := CP (x) ∈ Sylp (N ), and then  p ) = {1}. Moreover, R contains an element of Sp (G), so by [VK , 7.2], IN (R; Op (N ) ≤ Op (M ) by Theorem 2. On the other hand as I is a component of NG (T ), [T, I] = 1, so T ≤ N as x ∈ I. As T is R-invariant, it follows from the previous paragraph that T ≤ Op (N ) ≤ Op (M ), whence T ∈ Syl2 (Op (N )). A Frattini argument gives N = Op (N )NN (T ), so it suffices to show that NN (T ) ≤ M . But I  NG (T ), so NG (T ) = INNG (T ) (P ∩I) by a Frattini argument. As mp (P ∩I) ≥ 2, certainly NNG (T ) (P ∩I) ≤ ΓP,2 (G) ≤ M , and as I ≤ M we get NG (T ) ≤ M , completing the proof.  Lemma 13.7. If I = F ∗ (M ) and x ∈ Ipo (M ), then NG (x) ≤ M .  = N/Op (N ), and assume without Proof. Again let N = NG (x) and N loss that N ≤ M and R = CP (x) ∈ Sylp (N ). By Lemma 13.6 we may assume that x ∈ I, so x induces a non-inner automorphism on I. From (13C) we see that p = 3 and I ∼ = U6 (2), D4 (2), or 2 E6 (2). Moreover by Proposition 11.17, NM (x) is 3-solvable. This last condition rules out 2 E6 (2), and the condition m3 (CM (x)) ≥ 4 rules out D4 (2), by inspection of [IA , Table 4.7.3A]. Thus I ∼ = U6 (2), and with  [IA , 4.8.2, 4.8.4] we see that the only possibility is that O 3 (M ) = I x ∼ = P GU6 (2) and CI (x) ∼ GU (2) × L (2). In particular, C (x) is a {2, 3}-group, and R ∼ = = 3 2 M ∗    with E32 × (Z3  Z3 ). As m3 (R) = 5, Proposition 11.17 yields F (N ) = K × Q  cyclic and K  ∼  which Q = L2 (34 ) or U3 (32 ). In either case a Borel subgroup of K, is covered by N ∩ M , involves an extension of a 3-group by an element of order 5 acting nontrivially; this contradicts the fact that CM (x) is a {2, 3}-group. The proof is complete.  Since NG (P ) ≤ M , in order to prove that M is strongly p-embedded in G, it suffices to prove that CG (x) ≤ M for all x ∈ Ip (M ), which we shall do (still under the assumption (13B)). In view of Lemma 13.7, we can restrict attention to those x ∈ Ip (M ) − Ipo (M ). From [VK , 3.102] we obtain Lemma 13.8. Assume that F ∗ (M ) = I and let x ∈ Ip (M ) − Ipo (M ). Then ∼ p = 3 and I ∼ = U6 (2), D4 (2), or L± 4 (3). Moreover, in the first two cases, CI (x) = P ΓL2 (8), or G2 (2) or P GU3 (2), respectively.  = C/O3 (C). As To analyze these residual cases, set C = CG (x) and C ΓP,2 (G) ≤ M and m3 (CM (x)) ≥ 2, CM (x) contains a Sylow 3-subgroup of CG (x), which we may assume is R := CP (x). In particular as m3 (R) = 3, O3 (CG (x)) ≤  Moreover, if C is 3-constrained, then ΓR,2 (G) ≤ M , so M does not cover C. R0 := R ∩ O3 3 (C) is noncyclic, so C ≤ O3 (C)NG (R0 ) ≤ ΓR,2 (G) ≤ M . Hence let us assume that C has a component J. If J ≤ M , then m3 (R ∩ J) x ≥ 2, which yields by Frattini arguments that L3 (C) ≤ M and then C ≤ M . Therefore J ≤ M . If J ∗ is a 3-component of C(x, J), then m3 (R ∩ J ∗ ) x ≥ 2, so J ≤ ΓR,2 (C) ≤ M , contradiction. Thus, J = L3 (C).    Since J ≤ M , we have ΓR,2  (J) ≤ J0 < J and m3 (C(x, J)) = 1, where we have ∼ A6 , M11 , L3 (4) or SL3 (4), U3 (3), put J0 = J ∩ M . Therefore by [VK , 8.42ab], J = or L2 (8), and CM (x) is 3-solvable in every case.

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∼ L± (3). Then by Lemma 7.7, Ω1 (T ) =: y ∼ Suppose that I = = Z2 . In par4 ticular [y, R] = 1. This implies first that y ∈ C and then that y ∈ IC (R; 2).  we conclude from [VK , 8.42c] that either y ∈ O3 (C) or Given the structure of J, ∼   J/Z(J) = L3 (4), with y inducing a unitary automorphism on J in the latter case. In either case let W be an R-invariant Sylow 2-subgroup of O3 (C) y. Then since R contains an element of S3 (G) and R ≤ ΓR,2 (G) ≤ M , W ≤ O3 (M ) by Theorem 2. But T ∈ Syl2 (O3 (M )) and Ω1 (T ) = y, so Ω1 (W ) = y. In particular  in the case y ∈ O3 (C), O3 (C) has odd order. Let C0 = O 3 (CJ (y)). There or else J/Z(  J)  ∼ 0 ∼ 0 = J, fore C = L3 (4) and C = U3 (2). But C0   CCG (y) (x), and F ∗ (CG (y)) = IX, where O2 (X) = 1 and m2 (X) = 1, by Lemma 7.7. As Out(I) is a 2-group, x induces an inner automorphism on I; since I has a normal E34 -subgroup, m3 (CI (x)) = 3. But m3 (R) = 3, so CX (I) is a 3 -group and then   J)  ∼ C0 ≤ O 3 (CG (y)) ≤ I ≤ M . Hence C0 ≤ CM (x) is 3-solvable, so J/Z( = L3 (4) 1+2 ∼  and C0 = U3 (2). By [VK , 9.29], therefore, R ∩ I contains Z3 × 3 . But then  so M covers J,  contradiction. J = ΓR∩I,2 (J) Therefore I ∼ U (2) or D = 6 4 (2). But CM (x) is 3-solvable so the only possibility ∼ ∼  is I = D4 (2) and CI (x) = P GU3 (2). As m3 (C(x, J)) = 1, it follows that AutG (J) ∼  contains a subgroup H with H/O3 (H) = P GU3 (2). Given the possibilities for J,  J)  ∼ we must have J/Z( L (4), by [V , 1.9]. But then an element of R of order = 3 K  whence by [VK , 8.22], ΓR,2 (J)  = J,  a 3 induces a non-inner automorphism on J, contradiction. We have proved: Lemma 13.9. If F ∗ (M ) = I, then M is strongly p-embedded in G. Moreover,  M is an L-preuniqueness subgroup of G, where L = O p (IOp (M )). As we have remarked, the second statement is a trivial consequence of the first. Since we are proving (13A) by contradiction, we have Lemma 13.10. F ∗ (M ) = I. We set Q = CP (I) and assume, as we may by replacing P by an M -conjugate, that Q ∈ Sylp (CM (I)). We next consider the case mp (Q) = 1. In this case P normalizes I, so ZQ := Ω1 (Z(P )) ∩ Q = 1 = Ω1 (Z(P )) ∩ I =: ZI . For x ∈ Ip (M )  = C/Op (C). we again set C = CG (x) and C Lemma 13.11. Suppose that mp (Q) = 1 and x ∈ Ipo (M ). Then CG (x) ≤ M . Proof. Again we may assume that R := CP (x) ∈ Sylp (C). Proposition 11.17  = applies, and again since ΓP,2 (G) ≤ M , we have C ≤ M unless possibly F ∗ (C) n n ∼     J × Op (C), with Op (C) cyclic and J = L2 (p ), n ≥ 3, U3 (p ), n ≥ 2, or (for p = 3) n 2 G2 (3 2 ), n ≥ 3. Again setting J0 = J ∩ M , we have that J0 is a Borel subgroup of  In particular by [VK , 4.3], since n ≥ 2 in all cases, no subgroup of J0 of order J. Q  J0 × Op (C).  It follows that Z Q ≤ Ω1 (Op (C))  =  x. p is normal in J0 . But Z Therefore x ∈ Q, whence Lp (CM (x)) covers I and hence C ≤ M by Proposition 5.8. The proof is complete.  Lemma 13.12. Assume that mp (Q) = 1. Let x ∈ Ip (M ) − Ipo (M ). Then CG (x) ≤ M . Proof. Since mp (Q) = 1, mp (AutM (I)) ≥ 3. This condition rules out several 1 possibilities from Proposition 7.6: for p = 3, I ∼ = L2 (8), A6 , 2F4 (2 2 ) , L± 3 (3),

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13. THEOREM 5

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∼ A5 , 2B2 (2 25 ), 2F4 (2 21 ) , J2 , Co2 , HS, Ru, and F3 . In M11 ; and for p = 5, I = particular mp (I) > 1, which implies I  M as mp (Q) = 1. Another consideration is that M controls p-transfer in G since ΓP,2 (G) ≤ M [IG , 16.14, 15.7]. Therefore  Sp4 (8) or 3D4 (2) for M = O p (M ) by the simplicity of G. This implies that I ∼ = 5 2 ∼ 2  F4 (2 ) for p = 5. For in those cases, mp (I) = 2; therefore as p = 3, and I = mp (M ) ≥ 4, some element y ∈ Ip (M ) induces a non-inner automorphism on I; therefore y induces a field automorphism on I (or a quasi-field automorphism in the 3D4 (2) case) and so y ∈ O p (M ), contradiction. The remaining groups are ruled out because by [VK , 3.103], they force mp (I) ≥  3 and mp (x CI (x)) ≥ 3, yielding mp (C ∩ M ) ≥ 4 as Q = 1. By Lemmas 13.11 and 13.12, if mp (CM (I)) = 1, then M is strongly p-embedded in G. As we are proving (13A) by contradiction, we have Lemma 13.13. mp (Q) ≥ 2. This case occupies the rest of this section. When mp (I) ≥ 2, it is direct to prove that M is strongly p-embedded in G: Lemma 13.14. We have mp (I) = 1. Proof. If mp (I) ≥ 2, we show that M is strongly p-embedded in G, a contradiction since we are proving (13A) indirectly. Since ΓP,2 (G) ≤ M , M controls G-fusion in P by [IG , 16.14]. Therefore by [II3 , Theorem PU∗1 ], it suffices to show that CG (x) ≤ M for all x ∈ Ip (Q). But for such x, mp (CM (x)) ≥ 4 since mp (Q) ≥ 2, mp (I) ≥ 2, and I is simple. Moreover, CM (x) has a p-component covering I, so CG (x) ≤ M by Proposition 5.8. The proof is complete.  Now by Lemma 13.14 and Proposition 7.6, (13D)

5

(p, I) = (3, L2 (8)), (5, 2B2 (2 2 )), (5, A5 ), (7, L2 (7)) or (17, L2 (17)).

Fix B ∈ E∗ (P ). Then B leaves I and hence also Q invariant. Thus B centralizes elements z1 ∈ Ip (P ∩ I) and z0 ∈ Ip (Z(Q)). We set Z = z0 , z1  (∼ = Ep2 ). Since B ∈ E∗ (P ), Z ≤ B. Also Q centralizes Z. If P leaves I invariant, then z1 ∈ Z(P ) and we can choose z0 ∈ Z(P ), so in this case On the other hand, if P does not leave I invariant and we set ≤ Z(P ).  PZ ∗ ∗ I = I , then I is the product of p-components I = I 1 , . . . , I r of M for some r = ph , h ≥ 1. Then Q induces a permutation of {I i | 2 ≤ i ≤ r} and as r − 1 is prime to p, Q has at least two orbits on this set of I i . Hence in this case we can, and do, choose z0 to centralize I r . Using Proposition 5.8, we immediately obtain Lemma 13.15. One of the following holds: (a) ΓZ,1 (G) ≤ M ; or (b) P normalizes I, and Z ≤ Z(P ). Proof. If P does not leave I invariant, then Lp (CM (z)) = 1 for any z ∈ Z # as Z centralizes Lr . Since z ∈ Ipo (M ) (as z ∈ B), CG (z) ≤ M by Proposition 5.8. On the other hand, if P leaves I invariant, then Z ≤ Z(P ) by the choice of Z.  Since z0 centralizes I, Proposition 5.8 also yields: Lemma 13.16. CG (z0 ) ≤ M .

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3. THEOREM C5 : STAGE 1

86

Recall that L = Lp (IOp (M )), and L is a p-component of M . Set L0 = Lp (CL (z0 )). Thus L0 = L = I and L0 is a p-component of CG (z0 ) by the previous lemma. We shall verify that M is an L-preuniqueness subgroup of G. Suppose false, in the remaining lemmas of this section. Thus, there exists x ∈ Ip (Q) such that CG (x) ≤ M . Set H = CG (x). Since L0 is a p-component of CG (z0 ) and x centralizes L0 = L, we can consider the subnormal closure J of J0 := Lp (CL0 (x)) in H. Then J 0 = L and by Lp balance, J is either a single z0 -invariant p-component of H or the product of p p-components of H cycled by z0 , with J0 a p-component of CJ (z0 ) in either case. Also z1 ∈ J0 as z1 centralizes z0 , x. We fix this notation and first prove Lemma 13.17. mp (CM (x)) = 3. Proof. First, x centralizes Z ∼ = Ep2 . Also x = z0  by the preceding lemma as H ≤ M . Since Z ∩ Q = z0 , x ∈ Z, so as Z centralizes x, mp (CM (x)) ≥ 3. But if mp (CM (x)) ≥ 4, then as Lp (CM (x)) covers I, Proposition 5.8 implies that H = CG (x) ≤ M , again a contradiction. The lemma follows.  We next prove Lemma 13.18. ΓZ,1 (G) ≤ M . Moreover, Op (H) ≤ M . Proof. Since Op (H) ≤ ΓZ,1 (H), the second assertion is immediate from the first. We can assume that C = CG (z) ≤ M for some z ∈ Z # , for otherwise the lemma holds. Hence Z ≤ Z(P ) by Lemma 13.15, so P ≤ C. Also z ∈ / z0  by Lemma 13.16.  = C/Op (C) and apply Proposition 11.17 to C, setting K = Lp (C). We Set C n  ∼ conclude that K = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or 2 G2 (3 2 ), n ≥ 3 (with p = 3)  0 is a Borel subgroup of K.  Also Op (C) ≤ M , and that if K0 = K ∩ M , then K    But C = K(C ∩ M ), and CC (K) = Op (C) is cyclic. Thus z = Ω1 (Op (C)). mp (CM (x)) = 3 < mp (P ) by Lemma 13.17, so certainly z ∈ x. Hence x acts  nontrivially on K. Again as mp (CM (x)) = 3, it follows from [VK , 4.4] that x induces a nontrivial  ∼ field automorphism on K = L2 (pp ). Thus (z, x, K) is an L2 (pp ) field triple, which contradicts Proposition 9.1. The lemma is proved.  The lemma quickly yields: Lemma 13.19. J ≤ M and J0 covers J/Op (J). Proof. From its definition, J 0 is a p-component of CM (x), so it suffices to  show that J ≤ M . But Op (H) ≤ M , so it suffices to show that M covers J. If J is the product of p p-components J1 , J2 , . . . , Jp of H cycled by z0 with diagonal J0 , then J = ΓZ,1 (J) by [IG , 8.7(iii)]. But ΓZ,1 (H) ≤ M by the preceding lemma, so J ≤ M in this case. Thus by Lp -balance, assume that J is a single z0 -invariant p-component of H and J0 is a p-component of CJ (z0 ). In this case, mp (CM (x)) = 3 and J0 covers a p-component of J ∩ M/Op (J ∩ M ). Thus [VK , 8.46] implies that p = 5, J0 ∼ = A5 ,  ∼     and J = A10 or Co1 with ΓZ,1 (J) ≥ J0 × J1 , where J1 is a component of CM (x) ∼ J1 = ∼ A5 . Now NJ (Z x ∩ J) ≤ ΓP,2 (H) ≤ CM (x). But J0 and such that J1 =

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13. THEOREM 5

87

J1 are conjugate in NJ (Z x ∩ J), by [VK , 8.46], hence in M . Thus since J 0 = I is a component of M , so is J 1 . Let Q0 = CQ (J 1 ) = CP (J 0 J 1 ), so that x ∈ Q0 . Since mp (CM (x)) = 3 it follows that mp (CQ0 (x)) = 1, whence Q0 is cyclic. Hence P normalizes J 0 , J 1 , and Q0 (recall that p = 5); and then x ∈ Z(P ), which is a  contradiction as mp (CM (x)) = 3 < mp (M ). The proof is complete. As a consequence, we have Lemma 13.20. J is not normal in H. Proof. Suppose false. Since J0 covers J/Op (J) and z1 ∈ J0 , z1  = Ω1 (R) for some Sylow p-subgroup R of J. Since J  H, H = JNH (R), whence H =  JNH (z1 ), so H ≤ M by Lemmas 13.18 and 13.19, contradiction. Now we can attain our objective. Lemma 13.21. M is an L-preuniqueness subgroup of G.

 Proof. We assume false and continue the above analysis. Let J ∗ = J H , so that J ∗ is the product of isomorphic p-components J = J1 , J2 , . . . , Jr and r ≥ 2 by the preceding lemma. In particular, z1 centralizes Ji /Op (Ji ) for i > 1. Since NH (z1 ) ≤ M and Op (Ji ) ≤ Op (H) ≤ M , Ji ≤ M for i > 1, and as J ≤ M , it follows that J ∗ ≤ M . Then as mp (J ∗ ) = r ≥ 2 and ΓP,2 (G) ≤ M , a Frattini argument yields H = J ∗ NJ (R) ≤ M , where R ∈ Sylp (J ∗ ). The proof is complete.  Now Lemma 13.21, along with the fusion-control condition ΓP,2 (G) ≤ M (Theorem 4), allows us to quote [II3 , Theorem PU2 ]. The result is that M is almost strongly p-embedded in G, contrary to our assumption that (13A) fails. This contradiction completes the proof of Proposition 13.1 and with it the proof of Theorem 5. Likewise this completes the proof of Theorem C5 : Stage 1.

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10.1090/surv/040.9/04

CHAPTER 4

Theorem C5 : Stage 2 1. Introduction On the basis of the hypotheses of Theorem C5 and the statement of Theorem C5 : Stage 1, our setup is now as follows:

(1A)

(1) G is a K-proper simple group; (2) p is an odd prime such that m2,p (G) ≥ 4; (3) G is of restricted even type, so that O2 (CG (x)) = 1 for all x ∈ I2 (G), m2 (G) ≥ 3, and Lo2 (G) ⊆ Co2 ; (4) Op (CG (x)) is of odd order for all x ∈ Ipo (G), and Lop (G) ⊆ Cp . Moreover, for all A ≤ B ∈ Bp∗ (G), Op (CG (A)) has odd order; (5) G contains no strong p-uniqueness subgroup; (6) Bp∗ (G) ∩ Sp (G) = ∅; and (7) G is balanced with respect to any element of Ep4 (G).

Note that p is fixed throughout. In the next chapter, we will eventually see that p = 3. Recall that for a group X and prime p dividing |X|, Sp (X) = {E ∈ Ep (X) | Ip (CX (E)) ⊆ E}    Bp∗ (X) = B ∈ Ep (X)  IX (B; 2) = {1} and mp (B) = m2,p (X) . As long as m2,p (X) > 1, both these sets are nonempty. The statement of Theorem C5 : Stage 2 requires a considerable amount of terminology. We give formal definitions here for terms introduced in Chapter 1. Definition 1.1.

   Bp∗,c (G) = B ∈ Bp∗ (G)  CG (B) has even order  Bp∗,c (G) if Bp∗,c (G) = ∅ Bp∗,o (G) = otherwise. Bp∗ (G)

Thus, Bp∗,o (G) is nonempty. The notion of symplectic pair is central to our analysis. Recall that I∗G (B; 2) is the set of maximal elements of IG (B; 2) with respect to inclusion. Definition 1.2. A symplectic pair (B, T ) consists of subgroups B ∈ Bp∗ (G) and T ∈ I∗G (B; 2) such that T is cyclic or of symplectic type. The symplectic pair (B, T ) is trivial if T ∼ = Z2 and faithful if CB (T ) = 1. A key part of Theorem C5 : Stage 2 is that symplectic pairs exist, and (apart from a single exceptional situation) every symplectic pair is trivial or faithful. The 89 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

4. THEOREM C5 : STAGE 2

90

non-faithful case is intimately connected with triples (B, t, K) as in the following definition. Definition 1.3. BtKp (G) is the set of all triples (B, t, K) such that B ∈ t is an involution in CG (B), and K is a component of E(CG (t)) of order divisible by p.

Bp∗ (G),

While studying a triple (B, t, K) ∈ BtKp (G) we sometimes need to replace it with another triple, called a mate. Definition 1.4. Let (B, t, K) and (B ∗ , t, K) be elements of BtKp (G). Then (B , t, K) is a mate of (B, t, K) if and only if the following conditions hold: (a) ICG (t) (B ∗ ; 2) = ICG (t) (B; 2); (b) CB ∗ (K) = CB (K); and (c) If B ∗ = B, then p does not divide | Out(K)|. ∗

In order to build a nonconstrained neighborhood starting from a triple (B, t, K), we use an element b ∈ B # such that CK (b) has a p-component I contained in a p-component of CG (b). Hence we make the following definitions. Definition 1.5. KbI p (G) is the set of all triples (K, b, I) such that for some (B, t, K) ∈ BtKp (G), b ∈ B # and I is a p-component of E(CK (b)). The triple (K, b, I) is regular, or regularly embedded, if I lies in a (necessarily unique) p-component J of CG (b). The triple (K, b, I) is broad if either [K, b] = 1 or the following hold: CB (K) = 1, b ∈ K, and I is terminal in K (with respect to p). Definition 1.6. KbIJp (G) is the set of all quadruples (K, b, I, J) such that (K, b, I) ∈ KbI p (G) is regular and broad, and J is the p-component of Lp (CG (b)) such that I ≤ J. Definition 1.7. If (K, b, I, J) ∈ KbIJp (G), we may use the following associated notation and terminology: Cb = CG (b), C b = Cb /Op (Cb ), and J is the image of J in C b . Moreover, there is an involution t and B ∈ Bp∗ (G) such that [B, t] = 1 and K is a component of E(CG (t)); we say that (K, b, I, J) derives from t and B, or from (B, t, K) ∈ BtKp (G). (We are not asserting that t and B are uniquely determined by (K, b, I, J).) Then I is a p-component of CCt (b) and so J is tinvariant and I is a p-component of CJ (t). We call (K; I; J) a nonconstrained {t, b}-neighborhood in G, and say that it, too, derives from t and B, or from (B, t, K). Definition 1.8. Let (K; I; J) be a nonconstrained {t, b}-neighborhood in G for some t ∈ I2 (G) and b ∈ Ip (G). Then (K; I; J) is nondegenerate if and only if K > I and I < J. Otherwise, (K; I; J) is degenerate. Finally, when p = 3, triples (B, t, K) ∈ BtK3 (G) such that K/Z(K) ∼ = L± 4 (3) ± or G2 (3) will be handled differently in Stage 4. There K/Z(K) ∼ = L4 (3) (and K ∈ C2 , since G is of even type) will be shown to lead to G ∼ = Ω7 (3) or P Ω± 8 (3). To exclude these triples from consideration in parts of the analysis of Stages 2 and 3, we define   ∼ L± (3) or G2 (3)}. BtK3 (G) = (B, t, K) ∈ BtK3 (G)  K/Z(K) = exc

4

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2. THE PRINCIPAL SUBSIDIARY THEOREMS

91

Table 1.1. Stage 2: Nonconstrained {t, b}-Neighborhoods (K; I; J)

(1) (2) (3) (4) (5) (6) (7)

m2,3 (G) K I J 6 2F2 2F i22 (3)F i24 6 F i23 Ω7 (3) P Ω± 8 (3) 2 5 (2) E6 (2) 2U6 (2), D4 (2) F i22 5 F i22 U4 (3) L4 (9) 5 2F i22 2U4 (3) Ω7 (3), P Ω± 8 (3) 4 Sp8 (2), (2)F4 (2) Sp6 (2) U6 (2), D4 (2) (2)U6 (2), U5 (2) L± 4 (3), 3U4 (3) 4 U4 (2) Sp8 (2), (2)D4 (2) [3 × 3]U4 (3)

Now we can at last state Theorem C5 : Stage 2. For convenience we reproduce Table 1 of Chapter 1 as Table 1.1. Theorem C5 : Stage 2. The following conditions hold: (a) Let B ∈ Bp∗ (G) and T ∈ IG (B; 2) with CT (B) = 1. Then T is cyclic or of symplectic type; (b) Symplectic pairs exist in G; (c) If B ∈ Bp∗ (G), then m2 (CG (B)) ≤ 1, while if A ≤ B ∈ Bp∗,o (G) with |B : A| = p, then m2 (CG (A)) ≤ 2; (d) If BtKp (G) = ∅, then p = 3; (e) Every symplectic pair in G is trivial or faithful, unless possibly p = 3 and BtK3exc (G) = ∅; (f) Suppose that p = 3 and (B, t, K) ∈ BtK3 (G), and BtK3exc (G) = ∅. Then there exists a mate (B ∗ , t, K) of (B, t, K), an element b ∈ (B ∗ )# and a nondegenerate nonconstrained {t, b}-neighborhood (K; I; J) deriving from (B ∗ , t, K), such that the following conditions hold: (1) B ≤ K; (2) t ∈ Syl2 (CG (K)) ∩ I∗G (B; 2), and (B, t) is a symplectic pair; (3) K = E(CG (t)); (4) (K; I; J) and m2,3 (G) are as in one of the possibilities in Table 1.1; and (5) If K ∼ = 2 2E6 (2), then for some b1 ∈ (B ∗ )# , there is a nonconstrained {t, b1 }-neighborhood (K; I1 , J1 ) with I1 ∼ = 2U6 (2) and J 1 ∼ = F i22 . The reader will note that nonconstrained neighborhoods of types (1), (2), and (3) in Table 1.1 occur in the groups F1 , F i24 , and F2 , respectively, and those of type (5) occur in F i23 and F i24 . 2. The Principal Subsidiary Theorems As usual, we divide the proof into several subsidiary theorems. The first result, while not in the language of symplectic pairs or neighborhoods, is the key to all that follows.

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4. THEOREM C5 : STAGE 2

92

Theorem 1. Let A ≤ B ∈ Bp∗ (G) with |B : A| ≤ p. If |B : A| = p, assume that B ∈ Bp∗,o (G). Then CG (A) is p-constrained. Theorem 2. The following conditions hold: (a) Symplectic pairs exist in G, and Bp∗,c (G) = ∅; (b) Let B ∈ Bp∗ (G) and T ∈ IG (B; 2), and suppose that CT (B) = 1. Then T is cyclic or of symplectic type; (c) If B ∈ Bp∗ (G), then m2 (CG (B)) ≤ 1; and (d) If A ≤ B ∈ Bp∗,o (G) with |B : A| = p, then m2 (CG (A)) ≤ 2. After Theorem 2 is proved (in Section 10), parts (a), (b), and (c) of Theorem C5 : Stage 2 will have been established. Therefore we may, and shall, assume thereafter in this chapter that (2A)

BtK3exc (G) = ∅ (if p = 3).

Under this assusmption, Theorem 3 asserts that any element of BtKp (G) leads to a nonconstrained neighborhood. Theorem 3. Let (B, t, K) ∈ BtKp (G). Then p = 3. Moreover, if BtK3exc (G) = ∅, then there is a mate (B ∗ , t, K) of (B, t, K), an element b ∈ (B ∗ )# , a 3-component I of E(CK (b)), and a 3-component J of CG (b) such that (K, b, I, J) ∈ KbIJ(G). In particular, (K, b, I) is regular and broad, and (K; I; J) is a nonconstrained {t, b}neighborhood in G. Corollary 1. If p > 3, then all symplectic pairs are faithful. Theorem 4. Assume BtK3exc (G) = ∅ and let (B, t, K) ∈ BtK3 (G) and b ∈ B # . Suppose that (K; I; J) is a nondegenerate nonconstrained {t, b}-neighborhood in G. Then (K; I; J) satisfies the conclusions of Theorem C5 : Stage 2 (f ), and (K; I; J) is as in Table 1.1. Theorem 5. Assume BtK3exc (G) = ∅ and let (B, t, K) ∈ BtK3 (G) and b ∈ B . Suppose that (K; I; J) is a nonconstrained {t, b}-neighborhood in G. Then there exists b ∈ B and a nonconstrained {t, b }-neighborhood (K; I  ; J  ) which is nondegenerate. #

We emphasize that in Theorem 5, the second neighborhood has the same subgroup K. Hence we get the following corollary. Corollary 2. Let (B, t, K) ∈ BtK3 (G) and assume that BtK3exc (G) = ∅. Then K has one of the isomorphism types in Table 1.1. In this elaborate situation, we now prove that Theorems 1–5 imply Theorem C5 : Stage 2 (and Corollary 2). Clearly, Theorems 2 and 3 directly imply (a), (b), (c), and (d) of Theorem C5 : Stage 2. Remark 2.1. The proofs of parts (a)–(d) of Theorem C5 : Stage 2 hold even if p = 3 and BtK3exc (G) = ∅. Hence these parts, and part (c) in particular, will be available in Chapter 6 when we consider the case BtK3exc (G) = ∅. For the remaining parts, we may assume that if p = 3, then BtK3exc (G) = ∅. Let (B, T ) be a symplectic pair in G which is not faithful. Write t = Ω1 (Z(T )), so that BT ≤ CG (t). Let B0 = CB (T ). As (B, T ) is not faithful, B0 = 1. Let

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3. SOME GENERALITIES

93

U = Op (CG (t)). As T ∈ I∗G (B; 2) and T ≤ CG (t), T contains a Sylow 2-subgroup U2 of U . Thus B0 centralizes U2 . As F (U ) = O2 (U ) and the components of U are C2 -groups, [B0 , U ] = 1 (see [VK , 15.23]). As O2 (CG (t)) = 1, B0 acts faithfully on   O p (E(CG (t))). In particular there exists a component K of O p (E(CG (t))), and so (B, t, K) ∈ BtKp (G). By Theorem 3, p = 3 (proving Corollary 1) and there exists a mate (B ∗ , t, K) of (B, t, K), an element b ∈ (B ∗ )# and a nonconstrained {t, b}-neighborhood in G. By Theorem 5, there exists b ∈ (B ∗ )# and a nondegenerate nonconstrained {t, b }-neighborhood (K; I  ; J  ). By Theorem 4, applied to (B ∗ , t, K) and (K; I  ; J  ), t ∈ I∗G (B ∗ ; 2). As (B ∗ , t, K) and (B, t, K) are mates, t ∈ I∗CG (t) (B ∗ ; 2) = I∗CG (t) (B; 2) so t ∈ I∗G (B; 2). Thus T = t and (B, T ) is a trivial symplectic pair. This proves (e) of Theorem C5 : Stage 2. Finally, given (B, t, K) ∈ BtK3 (G), we may choose, by Theorem 3, a mate ∗ (B , t, K), an element 1 = b ∈ B ∗ , and a nonconstrained {t, b}-neighborhood (K; I; J). Then by Theorem 5, there is 1 = b ∈ B ∗ and a nondegenerate nonconstrained {t, b }-neighborhood (K; I  ; J  ). By Theorem 4, applied to (B ∗ , t, K) and (K; I  ; J  ), the conditions of part (f) of Theorem C5 : Stage 2 hold for B ∗ , b , and (K; I  ; J  ). It remains to check assertions (f1) and (f2). Since (B, t, K) and (B ∗ , t, K) are mates, condition (f 2) then holds for B, as asserted. Moreover, condition (f 1) holds for B ∗ . Even if B ∗ = B, it still holds for B. For in that case, Out(K) is a p -group, by definition of mate. Thus B ≤ KC(t, K). Since B ∗ ≤ K and O2 (CG (t)) = 1, CB ∗ (K) = 1. By definition of mate, CB (K) = CB ∗ (K) = 1. But B ∈ Ep∗ (CG (t)) so B ≤ K, as desired. Thus Theorem C5 : Stage 2 follows from Theorems 1–5. 3. Some Generalities We shall use the following variant of Γoo P,2 (G). Definition 3.1. Let P be a p-subgroup of G of rank at least 4. Then Γo1 P,2 (G) is defined to be  NG (Q) | Q ≤ P, mp (Q) ≥ 4 or mp (CP (Q0 )) ≥ 4 for all Ep2 ∼ = Q0 ≤ Q . Lemma 3.2. Suppose that A ∈ E p4 (G) and Op (CG (a)) = 1 for some a ∈ A# . Set Θ1 (G; A) = Op (CG (x)) | x ∈ A# . Then Θ1 (G; A) is a nontrivial {2, p} -group normalized by Γo1 P,2 (G) for any p-subgroup P ≤ G such that A ≤ P . Proof. Define Θ1 (G; D) similarly for any elementary abelian subgroup D of P . Then if D and C are elementary abelian of rank at least 4 and mp (D ∩ C) ≥ 2, (1A7) and [V2 , 1.2] imply that Θ1 (G; D) = Θ1 (G; D ∩ C) = Θ1 (G; C). As NG (Q) permutes the elementary abelian subgroups of Q of a given rank it follows that if mp (Q) ≥ 4, then NG (Q) normalizes Θ1 (G; D) for any D ∈ Ep4 (Q), while if mp (CP (Q0 )) ≥ 4 for any Ep2 ∼ = Q0 ≤ Q, then Θ1 (G; Q0 ) = Θ1 (G; D) for any such  Q0 and any D ∈ Ep4 (P ), whence NG (Q) normalizes Θ1 (G; D). Lemma 3.3. Let B ∈ Bp∗ (G) and D ∈ Ep (G) with 1 = A ≤ B ≤ D. Let L be a p-component of CG (A). Then D normalizes L. In particular, B normalizes L. Proof. For any a ∈ A# , L   Lp (CLp (CG (a)) (A) by Lp -balance. But the components of CG (a)/Op (CG (a)) all lie in Cp by (1A4). By repeated use of [III11 , 1.1c], L/Op (L) ∈ Cp .

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By Theorem C5 , Stage 1, B < E for some E ∈ Ep (G). Therefore we may assume that B < D. Then mp (D) > mp (B) = m2,p (G), and the lemma follows directly from [V2 , 6.8].  Definition 3.4. For 1 = C ≤ A ≤ B ∈ Bp∗ (G), a p-component L of CG (A) will be said to be C-terminal in G if and only if for every a ∈ C # , the pumpup of L in CG (a) is trivial, i.e., L covers a component of CG (a)/Op (CG (a)). Lemma 3.5. Let B ∈ Bp∗ (G), 1 = A ≤ D ≤ B, and let L be an A-terminal p-component of CG (A). Assume that mp (A) ≥ 2 and mp (DL) > mp (B). Then the following conditions hold: (a) [A, IG (D; 2)] = 1; and (b) For any involution z ∈ CG (D), [A, Op (CG (z))] = 1. Proof. Let T ∈ IG (D; 2). Choose any a ∈ A# and set Ta = [CT (a), A]. Let La be the pumpup of L in CG (a), so that by assumption, A ≤ C(a, La ). By Lemma 3.3, B normalizes every p-component of CG (a). Hence so do A and [CT (a), A] = Ta , and Ta is D-invariant. As A ≤ C(a, La ), Ta = [Ta , A] ≤ C(a, La ). Therefore, NCG (a) (Ta ) covers DL/Op (L). As mp (DL) > mp (B) = m2,p (G) by assumption, Ta = 1. But a is arbitrary and A is noncyclic by assumption, so [T, A] = 1. Thus (a) holds. In particular, in (b), since D ≤ CG (z), A centralizes some Sylow 2-subgroup of E(Op (CG (z))). As any component of E(CG (z)) is in C2 , it follows from [VK , 15.23] that [A, E(Op (CG (z)))] = 1. But O2 (CG (z)) = 1 as G is of even type. Therefore by (a) and [IG , 3.17(ii)], A centralizes F ∗ (Op (CG (z)))  and then Op (CG (z)), completing the proof of the lemma. Lemma 3.6. Let B ∈ Bp∗ (G), 1 = A ≤ B, and let LA be an A-terminal pcomponent of CG (A). Assume that mp (A) ≥ 3 and mp (ALA ) > mp (B). Let z ∈ I2 (LA ) and set X = O2 (CLA (z)). Assume that X is a p -group and X centralizes every CLA (z)-invariant 2-subgroup of LA . Then the following conditions hold: 

(a) [X, O2 (CG (z))O p (E(CG (z)))] = 1; and (b) If CLA /Op (LA ) (z) is solvable, then X = 1. Proof. Since Op (LA ) has odd order by (1A4), CCG (A) (z) normalizes LA and then X. Set R = [X, O2 (CG (z))], so that R = [R, X]. By Lemma 3.5, with D = A, [A, O2 (CG (z))] = 1, so O2 (CG (z)) ≤ NCG (A) (LA ). As X ≤ LA , it follows that R ≤ LA . Clearly R is CLA (z)-invariant so by assumption [R, X] = 1. Thus, R = 1. Now X is a p -group by hypothesis. It follows that for any a ∈ A# , as the pumpup of LA in CG (a) is trivial, X ≤ Op (CG (a, z)). Therefore X ≤ ΔCG (z) (A).  Let J be any component of O p (E(CG (z))). Then J is 3-balanced with respect to p, since otherwise by [IA , 7.7.5], J/Op (J) ∼ = An for some n > p3 , contradicting the fact that J ∈ C2 as G has even type. Since mp (A) ≥ 3, it follows by [IG , 20.6] that [X, J] = 1. Thus (a) holds. Finally, to prove that X = 1 in (b), it suffices to show that [X, E] = 1, where E = E(Op (CG (z))). For then X is a subgroup of odd order in CCG (z) (F ∗ (CG (z))) ≤ O2 (CG (z)). Assume then that CLA /Op (LA ) (z) is solvable. By Lemma 3.5b, E ≤ CG (A z), so E normalizes LA and maps into CAut(LA /Op (LA )) (z). But this group is solvable by our assumption and the Schreier property, whereas E is perfect. Therefore E must centralize LA /Op (LA ). Let T ∈ Syl2 (E). As Op (LA ) has odd order, CCG (A) (T ) covers ALA /Op (LA ). Hence by our assumption, mp (CCG (A) (T )) =

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mp (ALA ) > mp (B) = m2,p (G). The only possibility is that T = 1, whence E = 1 and [X, E] = 1, as desired. The proof is complete.  4. Uniqueness Subgroups from p-Terminal p-Components In this section we prove several related results that construct subgroups of G with p-uniqueness properties from p-terminal pairs (x, K) satisfying mp (C(x, K)) ≥ 2. The basic such result, Proposition 4.3 below, is our first goal. We assume (1A) throughout. Recall from [II3 , (1A), p. 131] that for any prime q, a maximal subgroup M of G satisfies the q-component preuniqueness hypothesis relative to a qcomponent K ∗ of M if and only if the following conditions hold for some Q and T: (4A)

(1) Q ∈ Sylq (CM (K ∗ /Oq (K ∗ ))) and mq (Q) ≥ 2; (2) For every x ∈ Iq (Q), CG (x) ≤ M ; (3) Q ≤ T ∈ Sylq (M ) and, if q > 2, then mq (T ) ≥ 4.

Moreover, in this situation, if mp (K ∗ ) ≥ 2, certain additional assumptions about M -control of G-fusion of certain subgroups of Q imply that M is strongly pembedded in G [II3 , Theorems PU∗1 and PU∗2 , p. 132]. As one of our ultimate goals is to construct a strong p-uniqueness subgroup of G, it is useful to have situations in which the p-component preuniqueness hypothesis holds. A further result [II3 , Theorem PU∗4 ] strengthens the hypotheses, assuming that K is terminal in G, i.e., K is a component of CG (x) for every x ∈ Ip (CG (K)). This “terminal implies standard” result generalizes theorems applying to p = 2 by Aschbacher, Gilman, and Solomon, which in turn generalized a result of Powell and Thwaites [PoTh1]. We shall use it in the following form. Theorem 4.1. Let G be a K-proper simple group. Let p be any prime and suppose that (x, K) ∈ ILp (G). If p > 2, assume that (x, K) ∈ ILop (G). If K is terminal in G and mp (K) ≥ 2, then either K is standard in G or M has a strongly p-embedded maximal subgroup in which K is subnormal. Proof. This follows directly from [II3 , Theorem PU∗4 ] if p = 2. If p > 2, it will follow once we check that mp (NG (K)) ≥ 4. Since (x, K) ∈ ILop (G) this is a consequence of the following lemma, with X = CG (x).  Lemma 4.2. Suppose that X is a group and p an odd prime such that mp (X) ≥ 4. If K is any p-component of X such that mp (K) ≥ 2, then mp (NX (K)) ≥ 4. Proof. Let S ∈ Sylp (X), so that mp (S) ≥ 4. Hence we are done if S  S  normalizes K. Otherwise K S has at least 3 components so mp ( K ) ≥  Sand mp (K) + 2 ≥ 4 as p is odd (see [IG , 16.11]). As K  K , the proof is complete.  We shall use Theorem 4.1 not only for our fixed odd prime p, but also for the prime 2. Now for our basic result:

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Proposition 4.3. Let (x, K) ∈ ILop (G) and Q ∈ Sylp (C(x, K)), and set K = K/Op (K). Assume that the following conditions hold: (a) (x, K) is p-terminal in G; (b) mp (Q) ≥ 2; (c) mp (K) ≥ 2; (d) K is simple, or (K, p) = (SU6 (2), 3); and (e) (K, p) = (D4 (2), 5) or (A6 , 3). Then there exists a maximal subgroup M of G satisfying the p-component pre uniqueness hypothesis relative to the p-component K1 = O p (KOp (M )) of M and the p-subgroup Q. Furthermore, one of the following holds: (a) mp (Qg ∩ M ) ≥ 2 for some g ∈ G − M ; moreover, K1  M , and for any such g, ΓQg ∩M,1 (K1 ) < K1 ; or (b) M is strongly p-embedded in G, K ∼ = K, Op (M ) = 1, M = NG (K), [K, K g ] = 1 for all g ∈ G − M , K ∈ C2 ∩ Cp if CG (K) has even order, and CG (t) ≤ M for all t ∈ I2 (CG (K)). Moreover for any T ∈ Syl2 (CG (K)), if m2 (T ) ≥ 2 then there is h ∈ G − M such that m2 (T h ∩ M ) ≥ 2. Remark 4.4. Note that in conclusion (b), K is a standard component, considered as a component of CG (u) both for all involutions u ∈ CG (K) and for all elements u ∈ Ip (CG (K)). Remark 4.5. The key consequence of assumption (d) is the following: Suppose that K is a p-component of a group X = KCX (K), with mp (CX (K)) ≥ 2. Let P ∈ Sylp (X). If E ≤ P with E ∼ = Ep2 , then mp (CP (E)) ≥ 4. This property holds trivially if K is simple. In the additional case (K, p) = (SU6 (2), 3), it holds by [VK , 9.25]. We let x, K, and Q be as in Proposition 4.3, and proceed in a sequence of lemmas. Again as (x, K) ∈ ILop (G), Lemma 4.2 gives (4B)

mp (NG (K)) ≥ mp (CG (x)) ≥ 4.

We first prove Lemma 4.6. In conclusion (a) of Proposition 4.3, the condition mp (Qg ∩ M ) ≥ 2, g ∈ G − M , implies that ΓQg ∩M,1 (K1 ) < K1  M . Proof. Since p divides |M g ∩M | and g ∈ G−M , M is not strongly p-embedded in G. Hence K1  M by [II3 , Theorem PU∗3 ]. Suppose that the lemma fails. Since M satisfies the p-component preuniqueness hypothesis with respect to K1 and Q, K1 = ΓQg ∩M,1 (K1 ) ≤ ΓQg ,1 (G) ≤ M g . As K1  M , Qg ∩ M normalizes some Sylow p-subgroup of CM (K1 /Op (K1 )). By [IG , 10.11], there is therefore a Qg ∩ M -invariant Q0 ∼ = Ep2 with Q0 centralizing g K1 /Op (K1 ). Then CQg ∩M (Q0 ) = 1, so Q0 ≤ M . For each u ∈ Q# 0 we set Ku = Lp (CK1 (u)) and note that Ku covers not only K1 /Op (K1 ) but also a component of CG (u)/Op (CG (u)). Hence Ku covers a component of CM g (u)/Op (CM g (u)). It follows from Lp -balance and the (Bp )-property that there is a product L of pcomponents of Lp (M g ) such that L1 := Lp (K1 ∩ L) covers K1 /Op (K1 ); moreover, setting M g = M g /Op (M g ), L1 ∼ = K1 /Op (K1 ) and L1 is a component of CL (u) for  every u ∈ Q# . As (K/O (K), p) = (A6 , 3) by the hypothesis of Proposition 4.3, it p 0

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follows from [VK , 3.6] that L = L1 . Now K1 ≤ M g , so K1 = O p (K1 ) ≤ L. But also Op (M g ) ≤ ΓQ0 ,1 (G) ≤ M , so K1  K1 Op (M g ). Therefore K1 = L   M g . g If L = K1g , then g ∈ NG (K1 ) = M , a contradiction. Hence [K 1 , K 1 ] = 1, and in any case K1g ≤ NG (K1 ) = M , so K1g ≤ CM (K1 /Op (K1 )). We now have symmetry between (K1 , M, M g ) and (K1g , M g , M ). In particular K1g   M . Let R ∈ Sylp (K). Then Rg ≤ CM (K/Op (K)) so ΓRg ,1 (G) ≤ M . This implies that Lp (M g ) ≤ K1g ΓRg ,1 (G) ≤ M and indeed by Lp -balance, Lp (M g ) ≤ Lp (M ). Hence Lp (M g ) = Lp (M ), whence M = NG (Lp (M )) = M g and g ∈ M , a final contradiction.  We next prove Lemma 4.7. Under the assumptions of Proposition 4.3, suppose that K is a component of a subgroup M < G, M satisfies the p-component preuniqueness hypothesis with respect to K, and M is strongly p-embedded in G. Then Op (M ) = 1, K∼ = K, and CG (t) ≤ M for all t ∈ I2 (CM (K)). Proof. If Op (M ) = 1 then M is a strong p-uniqueness subgroup of G, contradicting (1A5). So Op (M ) = 1, whence K ∼ = K. Let C = CG (t) and C0 = C ∩ M . As mp (K) ≥ 2 by assumption, mp (C0 ) ≥ 2, and since M is strongly p-embedded in G, C0 is strongly p-embedded in C. Let R ∈ Sylp (K). As mp (R) ≥ 2, Op (C) ≤ ΓR,1 (G) ≤ M , so X := [K, Op (C)] satisfies X = [X, K]  K as K   M . But also X ≤ Op (C) as K ≤ C, so X ≤ Op (K) = 1. Also Op (C) ≤ O2 (C) = 1 as G has even type, so K centralizes Op (E(C))F (C), which therefore lies in M . Let  L = O p (E(C)). By the F ∗ -theorem [IG , 3.6], [K, L] = 1. If L ≤ M , then C ≤ M by a Frattini argument, as desired. So assume that L ≤ M . Then L ∩ C0 is strongly p-embedded in L. In particular, L is a single component. Now K = K (∞) maps to Inn(L) and so by the F ∗ -Theorem, K ≤ LZ(F (C)), whence K = [K, K] ≤ L. Therefore K is a component of the strongly p-embedded subgroup L ∩ C0 of L. By [IA , 7.6.2], either L ∼ = A2p , p ≥ 5, or L ∼ = F i22 , with p = 5 and K ∼ = D4 (2). As L ∈ C2 only the second case can hold, but that contradicts assumption (e) of the proposition. The proof is complete.  Lemma 4.8. Under the assumptions of Proposition 4.3, if K is standard in G (as a p-component), then the conclusion of Proposition 4.3 holds. Proof. Let N = NG (K), so that CG (t) ≤ N for all t ∈ Ip (Q). We argue that the conclusion of Proposition 4.3 holds with M = N . Expand Q to Q∗ ∈ Sylp (CG (K)). Then CQ∗ (x) = Q. In particular, Z(Q∗ ) contains some z ∈ Ip (Z(Q)). Then Q ∈ Sylp (CG (K z)) by the assumed p-terminality of (x, K), whence Q = Q∗ . Thus, Q ∈ Sylp (CG (K)). Next, we claim that the p-component preuniqueness hypothesis holds for N with respect to K. By Lemma 4.2, mp (N ) ≥ 4. Thus to prove the claim it remains to show that N is a maximal subgroup of G. Suppose false, so that N < M < G for some M . Then Op (M ) ≤ ΓQ,1 (G) ≤ N , so as mp (Q) ≥ 2, we have [Op (M ), K] = 1. Thus by Lp -balance and [VK , 3.6], and again as mp (Q) ≥ 2, the subnormal closure of K in M is isomorphic to Suz or O  N , and p = 3 with K ∼ = A6 . This contradicts assumption (e) of the proposition, so the claim is valid. As Q ∈ Sylp (CG (K)), it remains to argue that if conclusion (a) fails for N , then conclusion (b) holds. Indeed the failure of (a) implies that N is strongly

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4. THEOREM C5 : STAGE 2

p-embedded in G, by [II3 , Theorem PU∗2 ]. By Lemma 4.7, for any involution t ∈ CN (K) = CG (K), CG (t) ≤ M . In particular (t, K) ∈ ILo2 (G) and K is terminal in G as a component of an involution centralizer. Also K ∼ = K by Lemma 4.7, so m2 (K) ≥ 2. As K is standard, it commutes with none of its distinct conjugates. The final sentence of (b) then follows from [II3 , Corollary PU2 ]. This completes the proof.  Thus we now assume that K is not standard. Lemma 4.9. K is not terminal in G with respect to the prime p. Proof. Assume then that K is terminal, though not standard. In particular, K is quasisimple. Then there is a strongly p-embedded subgroup M < G with  K   M , by Theorem 4.1. As K is not standard, K M is the product of two or more mutually commuting conjugates K = K1 , K2 , . . . of K. Since mp (K) = mp (K2 ) ≥ 2, and in view of strong p-embedding, M satisfies the p-component preuniqueness hypothesis with respect to K. Lemma 4.7 then implies that CG (t) ≤ M for all t ∈ I2 (CM (K)), and K ∼ = K. Thus m2 (K) ≥ 2 and m2 (CM (K)) ≥ m2 (K2 ) ≥ 2. Hence M satisfies the 2component preuniqueness hypothesis with respect to K. By [II3 , Theorem PU∗3 ], applied for the prime 2, M is strongly embedded in G. As K   M this contradicts  the Bender-Suzuki Theorem [II2 , Theorem SE]. The proof is complete. Let P ∈ Sylp (KQ) with Q ≤ P . For any u ∈ Ip (Q), let Ku be the (trivial) pumpup of K in CG (u). Lemma 4.10. There exists a subgroup M < G such that: (a) ΓP,2 (G) ≤ M ; (b) ΓQ,1 (G) ≤ M ; and (c) Op (M ) = 1. Proof. By the hypothesis of Proposition 4.3, K is p-terminal in G. As K is not terminal in G, there is y ∈ Ip (Q) such that the pumpup Ky of K in CG (y) is not quasisimple. In particular, Op (CG (y)) = 1. Now by Remark 4.5, for any P0 ≤ P such that P0 ∼ = Ep2 , mp (CP (P0 )) ≥ 4. (G) = Γ (G). Set Thus by Definition 3.1, Γo1 P,2 P,2  Θ1 (G; E) = Op (CG (e)) | e ∈ E # for any E ∈ E2 (P ). Choose any A ∈ E4 (P ) and set M = NG (Θ1 (G; A)). By Lemma 3.2, Θ1 (G; A) is a nontrivial p -subgroup of G, so that M < G with Op (M ) = 1, and moreover ΓP,2 (G) ≤ M , so that (a) and (c) hold. It remains to show that NG (u) ≤ M for every u ∈ Ip (Q), and by a Frattini argument it is enough to show that CG (u) ≤ M . We have K = ΓP,2 (K) ≤ M . Furthermore, any noncyclic p-subgroup R of K is K-conjugate, hence M -conjugate, to a subgroup of P , and so NG (R) ≤ M . Now x ∈ Z(Q). For any u ∈ Ip (Q), the pumpup Ku of K in CG (u) lies in M . Indeed this is trivial if u = x, and otherwise, as (x, K) is p-terminal, Ku ≤ KOp (CG (u)) ≤ KΘ1 (G; x, u) ≤ M . Moreover, Ru := P ∩ K ∈ Sylp (Ku ) and by the previous paragraph, NG (Ru ) ≤ M . By a Frattini argument, NG (Ku ) ≤ M .

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If Ku  CG (u), then CG (u) ≤ M by what we just saw. Otherwise, let Ku = K1 , K2 , . . . be the distinct CG (u)-conjugates of Ku , and set L = K1 K2 · · ·  CG (u). Then L = Ku NG (Ru ) ≤ M . Then there is S ∈ Sylp (L) such that S = Ru (S ∩ Q) ≤ o1 P . Hence it is immediate from the definition that Γo1 S,2 (G) ≤ ΓP,2 (G) ≤ M . By Remark 4.5, mp (CS (E)) ≥ 4 for any E ∈ E2 (S). Hence NG (S) ≤ Γo1 S,2 (G) ≤ M . Thus CG (u) ≤ LNG (S) ≤ M by a Frattini argument, so (b) holds and the lemma is proved.  Now let M ∗ be a maximal subgroup of G containing M , and set 

K ∗ = O p (KOp (M ∗ )). Lemma 4.11. M ∗ and K ∗ satisfy the p-component preuniqueness hypothesis, and Op (M ∗ ) = 1. ∗

Proof. Set M = M ∗ /Op (M ∗ ). We show first that Q ∈ Sylp (CM ∗ (K)). Expand Q to Q∗ ∈ Sylp (CM ∗ (K)). On the one hand, CQ∗ (x) ≤ C(x, K), so CQ∗ (x) = Q. Hence z ∈ Q for some z ∈ Ip (Z(Q∗ )). But then Kz = Lp (CK ∗ (z)) so Q∗ ≤ C(z, Kz ). By definition of p-terminality, Q is a Sylow p-subgroup of C(z, Kz ), whence Q = Q∗ , as desired. In view of Lemma 4.10ab and assumptions (b) and (c) of our proposition, all ∗ that remains to verify for our claim is that K is a component of M . But by p-terminality, Lp -balance, and the (Bp )-property [IA , 7.1.3], K is a component of CM ∗ (y) for all y ∈ Ip (Q). Therefore K y lies in a Q-invariant component I of ∗ M , and as mp (Q) ≥ 2, [VK , 3.6] implies that either our claim holds or (K, p) = (A6 , 3). In view of assumption (e) of the proposition, our claim holds. Finally, as Op (CG (y)) ≤ M ∗ for all y ∈ Ip (Q), and for some such y, Ky is not quasisimple, it  follows that Op (M ∗ ) = 1. Now we quickly complete the proof of Proposition 4.3. As Op (M ∗ ) = 1, but G has no strong p-uniqueness subgroup, M ∗ cannot be almost strongly p-embedded in G. Therefore by Lemma 4.11 and [II3 , Theorem PU∗2 ], and with M ∗ in place of M , conclusion (a) of the proposition holds, completing its proof. Under additional hypotheses, we can rule out conclusion (b) of Proposition 4.3. We begin these variations with the following lemma. Lemma 4.12. Suppose that the assumptions of Proposition 4.3, and its conclusion (b), hold. Suppose also that either mp (CG (K)) > 2 or CK (y) has even order for some y ∈ Ip (K). Then m2 (CM (K)) ≥ 2. Proof. By conclusion (b), M is strongly p-embedded in G, so CG (K) = CM (K). Suppose the lemma fails, so that m2 (CM (K)) < 2. Since Op (M ) = 1, it follows by the Z ∗ -theorem [IG , 15.3] that F := F ∗ (CM (K)) is a p-group, and as mp (Q) ≥ 2, also mp (F ) ≥ 2. Choose u ∈ I2 (K) with mp (CK (u)) > 0, if possible, and set C = CG (u) and C0 = C ∩ M . Thus CM (K) ≤ C0 . By our assumptions and the simplicity of K, mp (C0 ) ≥ 3. Notice that since F ≤ O2 (C0 ) but O2 (C) = 1 as G is of even type, C0 < C. As F ≤ C, C0 is strongly p-embedded in C. In particular Op (C) ≤ C0 . As CM (K) ≤ C, Op (C) ∩ CM (K) ≤ Op (CM (K)) ≤ Op (M ) = 1. Thus Op (C) acts faithfully on K. Since C0 is strongly p-embedded in C, there is a p-component J of C such that  = C/Op (C), J is as in [IA , 7.6.1]. Since 1 = F ≤ Op (C0 ), C = JC0 , and setting C

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we have [F, Op (C)] = 1. But clearly J ≤ [J, F ], so [J, Op (C)] = 1. Hence J is actually a component of C. Therefore J ∈ C2 , as G is of even type. On the other hand, mp (C0 ) ≥ 3 and so it follows from [IA , 7.6.1] that J ∈ Chev(p) is of twisted rank 1 and p-rank at least 3. Therefore J ∈ C2 , a contradiction.  Now we can prove Proposition 4.13. Assume the hypotheses of Proposition 4.3. Assume also that every four-subgroup of Aut(K) contains an involution v such that p divides |CK (v)|. Then conclusion (a) of Proposition 4.3 holds. Proof. If false, then conclusion (b) of Proposition 4.3 holds. Since K ∼ = K is simple, our hypothesis implies that K has elements of order 2p. Hence m2 (CM (K)) ≥ 2 by Lemma 4.12. Let T ∈ Syl2 (CM (K)). By conclusion (b) of Proposition 4.3, K  M , m2 (T g ∩ M ) ≥ 2 for some g ∈ G − M , and Γ := ΓT g ∩M,1 (K) ≤ M g . However, our hypothesis implies that p divides |Γ|. Thus p divides |M g ∩ M |. But M is strongly p-embedded in G, so g ∈ M , a contradiction. The proposition is proved.  Recall that K is said to be outer well-generated for the prime 2 if and only if for every four-subgroup V ≤ Aut(K) such that V ≤ Inn(K), K = ΓV,1 (K). If K satisfies this condition, and the hypotheses of Lemma 4.12, we can again rule out conclusion (b) of Proposition 4.3. Thus we have the following strengthened version of Proposition 4.3. Proposition 4.14. Assume the hypotheses of Proposition 4.3. In addition, assume that (a) Either K contains an element of order 2p, or mp (Q) ≥ 3; and (b) K is outer well-generated for the prime 2. Then conclusion (a) of Proposition 4.3 holds. Proof. Suppose by way of contradiction that conclusion (b) of Proposition 4.3 holds. Again by strong p-embedding of M , CG (K) = CM (K). Let T ∈ Syl2 (CM (K)), so that CG (y) ≤ M for all y ∈ I2 (T ). Thus, K is terminal in G, with respect to the prime 2. Also as Op (K) ≤ Op (M ) = 1, m2 (K) ≥ 2. Hence by [II3 , Theorem PU∗4 ], K is standard in G with respect to the prime 2. By [II3 , 16.13], there is g ∈ G − M such that T g ≤ M and T ≤ M g . Then  ΓT g ,1 (K) < K, and in particular T ∩ T g = 1. Without loss there is S ∈ Syl2 (M ) containing T g and T . Then T T g is a 2-group in which T and T g , as Sylow subgroups of CM (K) and CM (K)g respectively, are normal subgroups. As T ∩ T g = 1, [T, T g ] = 1. Let U = Ω1 (T ). Then by outer well-generation, U g induces inner automorphisms on K so by the assumed simplicity of K, U g ≤ Ω1 (S ∩ KT ) = Ω1 (S ∩ K) × U. If there is u ∈ U such that ug ∈ K, then |CM (ug )| is divisible by p while CG (u) ≤ M , so |M ∩ M g | is divisible by p and g ∈ M , contradiction. Therefore, U g ∩ K = 1 so U g projects onto U . Since [U, U g ] ≤ [T, T g ] = 1, U is then abelian. We argue that (4C)

CM (K u) is a p -group for all u ∈ U # .

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Indeed, let u ∈ U # and choose v ∈ U such that v g projects onto u. If (4C) fails, then v g centralizes a nontrivial p-element of CM (K). As in the previous paragraph this leads to g ∈ M , a contradiction. This proves (4C). We know, since K  M and Op (M ) = 1, that Op (CM (K)) = 1. Since U is noncyclic, (4C) implies that Op (CM (K)) = 1. Thus F := F ∗ (CM (K)) is the direct product of simple p-components. As U = Ω1 (T ) with T ∈ Syl2 (CM (K)), (4C) implies that F is simple. Therefore by [VK , 10.31], F has a strongly embedded n subgroup or F ∼ = L2 (q), q ≡ ±3 (mod 8), 2 G2 (3 2 ) , n ≥ 1, or J1 . o Choose any x0 ∈ Ip (K). Then x0 ∈ Ip (G) and CG (x0 ) ≤ M as M is strongly p-embedded in G. But then F is a component of CG (x0 ) and so F ∈ Cp . Thus n F ∼ = L2 (q), q ≡ ±3 (mod 8), or p = 3 and F ∼ = 2 G2 (3 2 ) , n ≥ 1. Now no element of CM (K), indeed of M , of order p induces an outer automorphism on F . For it would have to be a field automorphism, and then as F  M , M > O p (M ). As M is strongly p-embedded in G, M controls p-transfer in G, so O p (G) < G, contradiction. Therefore, Ω1 (Q) induces inner automorphisms on F . As mp (Q) > 1 we conclude that F ∼ = L2 (pn ) with n odd, n ≥ 3, and p ≡ ±3 n 2 (mod 8), or p = 3 with F ∼ = G2 (3 2 ), n ≥ 3. Finally, for each u ∈ U # , CG (u) = CM (u), so O2 (CF (u)) ≤ O2 (CG (u)) = 1 and E(CF (u))  E(CG (u)). n ∼ But if F = L2 (pn ), then as n ≥ 3, O2 (CF (u)) = 1; and if F ∼ = 2 G2 (3 2 ), then as n ∼ n ≥ 3, E(CF (u)) = L2 (3 ) is a component of E(CG (u)) not in C2 . In either case, we have a contradiction, and this completes the proof of the proposition.  We close this section with some simple consequences of conclusion (a) of Proposition 4.3. Lemma 4.15. Let M , K1 , and Q satisfy the hypotheses and conclusion (a) of Proposition 4.3. Set M = M/Op (M ) and assume that the outer automorphism group of K 1 is a p -group and Z(K 1 ) = 1. Let Q ≤ P ∈ Sylp (M ) and R = P ∩ K1 . Then the following conditions hold: (a) There exists Q1 ≤ R such that Q1 ∼ = Q and ΓQ1 ,1 (K 1 ) < K 1 ; (b) If Ω1 (Q1 ) = Ω1 (R), then K 1 has a strongly p-embedded subgroup; (c) mp (Q) ≤ mp (R); (d) If mp (Q) = mp (R) or Q1 ∩ Z(Rh ) = 1, for some h ∈ K1 , then NG (P ) ≤ M ; and (e) If Z(R) is cyclic, then P ∈ Sylp (G), and either NG (P ) ≤ M or K 1 has a strongly p-embedded subgroup. Proof. Our hypotheses imply that P = R × Q. By [II3 , 16.11], g ∈ G − M may be chosen so that Qg ≤ M and Q ≤ M g . Modifying g by a suitable element of M , we may assume that Qg ≤ P . By [II3 , 16.9], ΓQg ,1 (K 1 ) < K 1 . Hence the projection of Qg on R is injective, and its image Q1 satisfies (a) and (b). Then (c) follows immediately. If mp (Q) = mp (R), then it follows that there is 1 = u ∈ Q1 ∩ Z(R). Suppose, as in (d), that Q1 ∩ Z(Rh ) = 1 for some h ∈ K1 . Then Rh ≤ ΓQg ,1 (G) ≤ M g by the p-component preuniqueness hypothesis. As −1

Q ≤ M g , it follows that M g contains some P1 ∈ Sylp (M ). Then P1g ≤ M and there is m ∈ M such that g −1 m ∈ NG (P1 ). Thus, NG (P1 ) ≤ M , and (d) follows by Sylow’s theorem. Finally, to prove (e), let h ∈ NG (P ) and suppose that Z(R)

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is cyclic. By (c), mp (R) ≥ mp (Q) ≥ 2, so R is nonabelian, hence indecomposable. If Z(Q) is noncyclic, then as Z(P ) = Z(R) × Z(Q), we have Z(Q) ∩ Z(Q)h = 1, and so h ∈ M and NG (P ) ≤ M . Thus in proving (e) we may assume that Z(Q) is cyclic. Hence Q is also indecomposable. By the Krull-Schmidt theorem, h stabilizes {Ω1 (Z(R)), Ω1 (Z(Q))}. If h is a p-element, then h normalizes Z(Q), so h ∈ M . This proves that P ∈ Sylp (G). And if h ∈ M , then h interchanges Ω1 (Z(R)) and Ω1 (Z(Q)), so Q ∼ = R. Thus Q1 = R, and (b) implies that K 1 has a strongly p-embedded subgroup. The proof is complete.  5. Some Sporadic p-Components As an application of the previous section, we rule out the existence in G of certain p-component pairs, mostly of sporadic type. We prove the following two results together. Lemma 5.1. Let (x, K) ∈ ILop (G) be p-terminal. Then (p, K/Op (K)) is not any of the following: (5, J2 ), (5, HS), or (7, He). Lemma 5.2. Let (x, K) ∈ ILop (G). Then (p, K/Op (K)) is not any of the following: (5, M c), (5, Co3 ), (5, Co2 ), (5, He), (5, Ru), (5, F3 ), (7, Co1 ), (7, O  N ), (7, F i24 ), (7, F3 ), or (11, J4 ). Proofs of Lemmas 5.1 and 5.2. For the pairs (p, L) listed in Lemma 5.2, L 5, LA ∼ = Ap , and A = B. Lemma 7.3. Assume (7B). Then for any 1 = C ≤ A, LC /Op (LC ) ∼ = Akp for some 1 ≤ k ≤ 3. Moreover, mp (CA (LC /Op (LC ))) = m − k + 1 ≥ 2. Proof. By Lemma 6.2 and [VK , 3.98], either the first conclusion holds or p = 7 with LC /Op (LC ) ∼ = Co1 or F i24 . However, these two cases are ruled out by Lemma 5.2.  C ). Then mp (C  C = LC /Op (LC ) and C ∗ = CA (L Let L  C ) (LA )) = k −1, so Aut(L ∗ mp (C ) ≥ mp (A)−k+1. On the other hand, m2,p (Akp ) ≥ m2,p (A4 ×Akp−4 ) = k−1,  so mp (A) = m = m2,p (G) ≥ mp (C ∗ ) + k − 1. As m ≥ 4, the lemma follows.

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 c )) is minimal, and set C ∗ = CA (L  c ), so that Now fix c ∈ A# so that mp (CA (L ∗ ∼  mp (C ) ≥ 2 and LC ∗ covers Lc = Akp , 1 ≤ k ≤ 3. By the minimality and Lemma 7.3, the pumpup of Lc in CG (b) is trivial for all 1 = b ∈ C ∗ . As a result, C ∗ ≤ A, LC ∗ is C ∗ -terminal in G and LC ∗ /Op (LC ∗ ) ∼ = Akp , (7C) k = 1, 2, 3, p > 5. Moreover, mp (C ∗ ) = m − k + 1. We shall show that this leads to a contradiction. The argument will at the same time be effective against the following similar configuration: C ∗ is an arbitrary subgroup of A such that LC ∗ is C ∗ -terminal in G and LC ∗ /Op (LC ∗ ) ∼ (7D) = A15 , k = 3, p = 5. Moreover, m5 (C ∗ ) = m − 2. In either of the situations (7C) and (7D), let z ∈ I2 (LA ) map on a root involution of LA ∼ = Ap , and set X = O2 (CLC ∗ (z))L2 (CLC ∗ (z)), so that [X, C ∗ ] = 1. Lemma 7.4. Suppose that (7C) or (7D) holds. Then the following conclusions hold: (a) C ∗ centralizes IG (C ∗ ; 2); (b) C ∗ centralizes Op (CG (z)); and (c) X/X ∩ Op (LC ∗ ) ∼ = Akp−4 . Proof. We have mp (C ∗ ) = mp (A) − k + 1 ≥ 2 and mp (C ∗ LC ∗ ) = mp (C ∗ ) + mp (LC ∗ ) = mp (C ∗ ) + k > mp (A). As LC ∗ is C ∗ -terminal in G, (a) and (b) follow directly from Lemma 3.5, with the role of D there played by C ∗ . Since Op (LC ∗ ) has odd order, the image of X in LC ∗ /Op (LC ∗ ) is isomorphic to L2 (Y )O2 (Y ), where Y = CAkp (u) and u ∈ Akp is a root involution. As p ≥ 7 and p is odd, or kp = 15, this image is isomorphic to Akp−4 , completing the proof.  Lemma 7.5. Suppose (7C) or (7D) holds. Then X is a p -group and C ∗ = A. Proof. By Lemma 7.4b, Op (CG (z)) ≤ CG (C ∗ ). Suppose that p divides |X| and set Y := [X, Op (CG (z))]. Then Y may be computed in CG (C ∗ ), where it is a normal subgroup of Op (CG (z)). Also as z ∈ LC ∗ − Op (LC ∗ ), Op (CG (z)) normalizes LC ∗ and hence normalizes X. Thus Y  X ≤ LC ∗ as well. As O2 (CG (z)) = 1, we have O2 (Y ) = 1. But Op (LC ∗ ) has odd order, so Y ∩ Op (LC ∗ ) ≤ O2 (Y ) = 1. Set CG (C ∗ ) = CG (C ∗ )/Op (CG (C ∗ )). Then Y ∼ = Y = [X, Op (CG (z))]  X. ∼ Akp−4 , we have k = 2 or 3, and Y = X or 1. Since p divides |X| and X = But Y is a p -group, so Y = 1 and hence Y = 1. Note that by Lp -balance, the subnormal closure X ∗ of L2 (X) in CG (z) is a product of p-components of CG (z); but as Y = 1 those p-components are components of E(CG (z)), and hence lie in C2 . On the other hand, since LC ∗ pumps up trivially to La for each 1 = a ∈ C ∗ , L2 (X) is a component of CX ∗ (a) for all 1 = a ∈ C ∗ . As mp (C ∗ ) ≥ 2, it follows by [VK , 3.5] that X ∗ ∼ = Arp2 +kp−4 for some r ≥ 0. But p ≥ 7 and k ≥ 2, or p = 5 and k = 3, so X ∗ ∈ C2 , contradiction. We conclude that X is a p group. It follows immediately that k = 1, and then X/X ∩ Op (LC ∗ ) ∼ = Ap−4 and  mp (C ∗ ) = mp (A) − k + 1 = mp (A), so C ∗ = A.

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Of course in (7D), X/X ∩ Op (LC ∗ ) ∼ = A11 , which is not a 5 -group. As a result of this contradiction: Lemma 7.6. The situation (7D) does not occur. Now we have our first major reduction. Lemma 7.7. The case p > 5, LA ∼ = Ap in Lemma 7.2 does not occur. Proof. Suppose false and continue the above argument. Then X/X∩ Op (LA ) ∼ = Ap−4 is nonabelian simple if p > 7, and of order 3 if p = 7. Now X ≤ Op (CG (z, b)) for all b ∈ A# , since the pumpup of LA in CG (b) is trivial. Therefore X ≤ ΔCG (z) (A). Suppose that X ≤ Op (CG (z)). Then by [IG , 20.6], CG (z) has a p-component I such that I/Op (I) is not locally m = mp (A)-balanced with respect to p. But m ≥ 4, so by [IA , 7.7.5], I/Op (I) ∼ = An for some n > pm . Thus m2,p (G) ≥ mp (CG (z)) ≥ pm−1 > m, contradiction. Consequently, X ≤ Op (CG (z)). But Op (CG (z)) ≤ CG (A z) by Lemma 7.4b, and X   CG (A z). Thus X   Op (CG (z))  CG (z). If p = 7, then |X| is odd, so X ≤ O2 (CG (z)) = 1, a contradiction. Thus p > 7. But then since Op (CG (A)) has odd order, X ∩ Op (LA ) ≤ O2 (X) ≤ O2 (CG (z)) = 1. So X is a component of CG (z), whence Ap−4 ∈ C2 . But p is odd and at least 11, so Ap−4 ∈ C2 , a contradiction. This completes the proof.  For the next four lemmas we consider the case A = B, p > 3, and LA /Op (LA ) ∼ = L2 (pn ), n ≥ 1, from Lemma 7.2a. (This picks up the case p = 5, LA /Op (LA ) ∼ = A5  b | is maximal and that we have not analyzed yet.) We choose b ∈ A# such that |L  b ). Thus LC /Op (LC ) ∼  b and LC is C-terminal in G. set C = CA (L =L Lemma 7.8. Assume (6B) with A = B, and suppose that LA /Op (LA ) ∼ = L2 (pn ), n ≥ 1, and p > 3. Let C be as above. Then n = 1, p ∈ FM, and one of the following holds: (a) C = A; (b) |A : C| = p = 5, and LC /Op (LC ) ∼ = A10 ; or (c) |A : C| = p, and (LC /Op (LC ), p) = (J2 , 5), (HS, 5), or (He, 7). Proof. First let a ∈ A# and consider the possible isomorphism types for the a ∼  a ∈ Cp . One such is L pumpup L = L2 (ppn ). Other than this there are only a few possibilities, given by [VK , 3.97], and only occurring in cases where n = 1 and p ∈ FM. By [VK , 3.97], (LC /Op (LC ), p) = (Co1 , 5), (F2 , 5), or (F1 , 7). For in those cases, mp (Aut(LC /Op (LC ))) = 3, so mp (C) ≥ mp (G) − 3 ≥ 2 and Lemma 5.3 is contradicted. Also Lemma 5.2 rules out several possibilities, leaving (He, 7), (A10 , 5), (A15 , 5), (HS, 5), and (J2 , 5). If LC /O5 (LC ) ∼ = A15 , then LA /O5 (LA ) ∼ = A5 and m2,5 (G) ≥ m5 (C) + m2,5 (A15 ) = m5 (C)+2, so Lemma 7.6 is contradicted. In the other cases, C < A as LA /Op (LA ) ∼ = L2 (p), but mp (CAut(LC ) (LA )) = 1 so mp (A/C) ≤ 1, and |A : C| = r p. Thus the lemma holds unless LC ∼ = L2 (pp n ). Again mp (CAut(LC /Op (LC )) (LA )) = 1. Therefore |A : C| = p with LC ∼ = L2 (ppn ), or A = C with LC ∼ = L2 (pn ). Accordingly, set k = p or 1. We choose z ∈ I2 (LC ) and set X = O2 (CLC (z)). As Op (LC ) has odd order, it follows from [VK , 7.6] that X centralizes every CLC (z)-invariant 2-subgroup of LC . Also |X : X ∩ Op (LC )| is the odd share of pkn − , where  = ±1

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and pkn ≡  (mod 4); so X is a p -group. Finally CLC (z) is solvable, and so Lemma 3.6 applies. We conclude that X = 1. Thus pkn −  = 2a for some a. But p > 3 by hypothesis, so kn = 1 and p ∈ FM. As k = 1, C = A. Thus (a) holds, the lemma does not fail, and the proof is complete.  We will now rule out options (b) and (c) of Lemma 7.8. If Lemma 7.8b holds, we set J = A15 , while in the three cases of Lemma 7.8c, we set J = Co1 , F2 , or F1 , respectively. Thus, mp (J) = 3 in these cases. In either case, we choose c ∈ C # and a shortest pumpup chain (7E)

(c, Lc ) = (a0 , L0 ) < (a1 , L1 ) < · · · < (am−1 , Lm−1 ) < (am , Lm )

such that (am , Lm ) ∈ ILop (G) is Ipo -terminal and p-terminal in G. This is possible by  c or to J, [IG , 6.22, 6.27]; note that by [VK , 3.97], each Li /Op (Li ) is isomorphic to L p hence is centerless of p-rank at least 2. Indeed O (Aut(Li /Op (Li ))) ∼ = Li /Op (Li ), for all i = 0, . . . , m. Lemma 7.9. Lemma 7.8c does not occur.  c , so Proof. Consider the chain (7E). By Lemma 5.1, Lm /Op (Lm ) ∼  L = ∼ Lm /Op (Lm ) = J. By Lemma 5.3, mp (C(ai , Li )) = 1 for any i with Li /Op (Li ) ∼ = J. Thus by the minimality of the chain, Lm−1 /Op (Lm−1 ) ∼ = Lc /Op (Lc ), and of  c , 0 ≤ i < m. course mp (C(am−1 , Lm−1 )) ≥ 2 and Li /Op (Li ) ∼ =L Let Qi ∈ Sylp (C(ai , Li )) and Ri ∈ Sylp (Li ), 0 ≤ i ≤ m, chosen so that Pi := Ri × Qi ∈ Sylp (Ni ), where Ni = NCG (ai ) (Li ). We also may assume that ai ∈ Qi−1 , i = 1, . . . , m. Now Qm is cyclic, so Nm = CG (am ). If |Qm | > p, then am  = Ω1 (Φ(Z(Pm ))) char Pm , so Pm ∈ Sylp (G). Then mp (G) = mp (J) + 1 = 4 < mp (B), contradiction. There  c , and so fore Qm = am . From [IA , 5.3lyz], O p (CLm /Op (Lm ) (am−1 )) ∼ = Zp × L CQm−1 (am ) = am−1 , am . Thus Qm−1 is of maximal class [IG , 10.24]. Suppose that |Qm−1 | > p2 . Then Qm−1 is nonabelian and Z(Qm−1 ) = am−1 . If LC /O5 (LC ) ∼ = J2 , then Rm−1 is elementary abelian, so am−1  = Φ(Pm−1 ) ∩ Z(Pm−1 ) char Pm−1 , whence Pm−1 ∈ Syl5 (G). Hence Z(Pm−1 ) ∼ = E53 . But Z(Pm ) ∼ = E52 , a contradiction. Otherwise, Pm−1 = Rm−1 × Qm−1 is a decomposition into indecomposable groups (with Rm−1 ∼ = p1+2 ). Thus as p is odd, any automorphism of Pm−1 of order p centralizes Z(Pm−1 ). Let Pm−1 ≤ P ∈ Sylp (G) with CP (am−1 ) ∈ Sylp (CG (am−1 )). Then Z(Rm−1 ) centralizes NCP (am−1 ) (Pm−1 ), which lies in Nm−1 . Hence Pm−1 is Sylow in CG (am−1 )). Similarly as NP (Pm−1 ) centralizes am−1 , it follows that Pm−1 = P ∈ Sylp (G). In particular, a Sylow pcenter of G is noncyclic. The same must therefore hold for Pm ∈ Sylp (CG (am )). But J has cyclic Sylow p-center, so am  must be p-central in G. We have already seen, however, that this leads to a contradiction. Therefore Qm−1 = am−1 , am  ∼ = Ep2 . As mp (NCG (c) (Lc )) > mp (A) ≥ 4, m > 1. Since the pumpup (am−2 , Lm−2 ) < (am−1 , Lm−1 ) is trivial, CQm−2 (am−1 ) embeds in and then is isomorphic to Qm−1 . Hence Qm−2 is of maximal class. If |Qm−2 | = p2 , then Qm−2 and Qm−1 are G-conjugate and so the chain can be shortened to omit the m−1 term, contrary to our choice. Thus, Qm−2 is nonabelian of maximal class. As with Qm−1 we find that this leads to Pm−2 ∈ Sylp (G), Sylow

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p-centers of G are noncyclic, and then am  is p-central in G, a contradiction. The proof is complete.  Lemma 7.10. If Lemma 7.8b occurs, then p = 5 and there exists a 5-terminal pair (x, K) ∈ ILo5 (G) such that m5 (C(x, K)) ≥ 2 and K/O5 (K) ∼ = A10 or A15 . c ∼ Proof. Again consider the chain (7E). We have L = A10 . Hence by [VK , ∼ 3.97], Lm /O5 (Lm ) = A10 or A15 (= J). In the first case, the lemma holds with (y, I) = (am , Lm ), by [V2 , 8.5]. So assume that Lm /O5 (Lm ) ∼ = J; again we are done unless Qm is cyclic, which we then assume. As in the proof of Lemma 7.9, let Qi ∈ Syl5 (C(ai , Li )) and Ri ∈ Syl5 (Li ), 0 ≤ i ≤ m, chosen so that Pi := Ri × Qi ∈ Syl5 (Ni ), where Ni = NCG (ai ) (Li ). We also may assume that ai ∈ Qi−1 , i = 1, . . . , m. We repeat the arguments of the third paragraph of the proof of Lemma 7.9, using the fact that the centralizer of a 5-cycle in A15 is isomorphic to Z5 × A10 . We conclude that Qm = am  and Qm−1 is of maximal class, with CQm−1 (am ) = am−1 , am . Suppose that Qm−1 ∼ = E52 . Again since m5 (NCG (c) (Lc )) > 4, m > 1. As (am−2 , Lm−2 ) < (am−1 , Lm−1 ) is a trivial pumpup, CQm−2 (am−1 ) embeds in and then is isomorphic to Qm−1 . Hence Qm−2 is of maximal class. If |Qm−2 | = 52 , then Qm−2 and Qm−1 are G-conjugate and so the chain (7E) can be shortened to omit the m − 1 term, contrary to our choice. Thus, Qm−2 is nonabelian of maximal class. In particular Qm−1 contains p conjugates of am−1  and one of am−2 . Hence am  is conjugate to ak  for one of k = m − 2, m − 1. Therefore m5 (Pk ) = m5 (Pm ) = 4. But Lp (CG (ak )) has p-components of type A10 and of type A15 , so m5 (Pk ) ≥ 5, contradiction. Therefore |Qm−1 | > 52 . Then Z(Pm−1 ) = Rm−1 × am−1 , with Rm−1 ∼ = E52 . g = 1, whence g Hence for any 5-element g ∈ NCG (am−1 ) (Pm−1 ), Rm−1 ∩ Rm−1 normalizes Lm−1 . It follows that Pm−1 ∈ Syl5 (CG (am−1 )) ⊆ Syl5 (G), the last as am−1  = Z(Pm−1 ) ∩ Φ(Pm−1 ) char Pm−1 . As NG (Pm ) controls G-fusion in Z(Pm ) by Burnside’s theorem, am−1  is weakly closed in Z(Pm ). Now am−1 acts on Lm /Op (Lm ) like an element h ∈ Lm ∩ am−1 , am  mapping on a 5-cycle in Lm /Op (Lm ). Then h is conjugate to all elements of Lm mapping on 5-cycles, and in particular to a 5-cycle in Lm−1 ∩ Lm . By the weak closure above, am−1 is conjugate to no such element. Hence am−1  is not conjugate to h. As am am−1  is completely fused in Qm−1 , h is conjugate to am . But again, h is conjugate to an element of Z(Pm−1 ), so am is 5-central in G. This is a contradiction since  mp (CG (am )) = 4 < mp (G), and the lemma follows. Lemma 7.11. Lemma 7.8b does not hold. Proof. Suppose false. Then with Lemma 7.10, Proposition 4.3 applies and  yields a subgroup M and 5-component K1 = O 5 (KO5 (M )) of M . Suppose that conclusion (b) of Proposition 4.3 holds. Then K  M by the proposition. Moreover, since K ∼ = A10 or A15 does not lie in C2 , CG (K) has odd order; and O5 (M ) = 1. As m5 (M ) = m5 (G) ≥ 5, R := O5 (CG (K)) = 1, indeed m5 (R) > 1. Let z ∈ K be a root involution, and set C = CG (z) and C0 = C ∩ M . Then R ≤ O2 (C0 ), but O2 (C) = 1, so C0 < C. Hence C0 is strongly 5-embedded in C and m5 (C0 ) ≥ m5 (R) > 1. Therefore O5 (C) ≤ C0 and F ∗ (C/O5 (C)) = (C/O5 (C))(∞) is simple, with a strongly 5-embedded subgroup

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∼ A6 or A11 as a normal subgroup. But no such containing a copy of E(CK (z)) = simple K-group exists, by [IA , 7.6.1]. This contradiction proves that conclusion (a) of Proposition 4.3 holds. Set M = M/O5 (M ) and let Q ∈ Syl5 (CM (K)). By Proposition 4.3a and [II3 , 16.11], there is g ∈ G − M such that Qg ≤ M and Q ≤ M g , and Γ := ΓQg ,1 (KO5 (M )) lies in M g and satisfies Γ < K. Thus by [VK , 8.11], Γ ∼ = A5 × A5 ∼ ∼ ∼ if K = A10 , while Γ = A5 × A5 × A5 or A10 × A5 if K = A15 . Let Γ0 = [Γ, Qg ] ≤ CM g (K g /O5 (K g )). Then Γ0 contains a Sylow 5-subgroup of K, so m5 (Q) = m5 (Qg ) ≥ m5 (K) = m5 (Aut(K)). As Γ < K, m5 (Q) = m5 (K) and by [IA , 7.6.1], K ∼ = A10 . Thus, m5 (P ) = 4 < m5 (G), so P ∈ Syl5 (G), whence we can take g ∈ NG (P ) − M such that g 5 ∈ P . Furthermore, Qg ∈ Syl5 (Γ0 ) so Qg ≤ K, and then Qg = P ∩ K. A symmetric argument, with M g , g −1 , and Qg −1 playing the roles of M , g, and Q, shows that Q = P ∩ K g . Hence Qg ≤ P ∩ K = Qg . So g 2 ∈ NG (Q). As g 5 ∈ P ≤ NG (Q), g ∈ NG (Q) ≤ NG (M ), a contradiction. The proof is complete.  Lemmas 7.8, 7.9, and 7.11 yield: Lemma 7.12. Suppose that A = B, LA ∼ = L2 (pn ), n ≥ 1, and p > 3. Then C = A, n = 1, and p ∈ FM. The next lemma provides the final step of the proof of Proposition 7.1 when A = B. It is more general, however, and will also be useful when |B : A| = p. Lemma 7.13. Suppose CG (A) has a p-component L with (L/Op (L), p) = 5 (L2 (8), 3) or (2B2 (2 2 ), 5). Then L is A-terminal in G. Proof. Suppose false. Then by [VK , 3.96a], there exists a hyperplane A1 of A 3 such that LA1 /Op (LA1 ) ∼ = Sp4 (8), 3D4 (2), SU6 (2), U6 (2), Co3 , or 2 G2 (3 2 ) (p = 3), 5 or 2F4 (2 2 ) (p = 5), these possibilities being forced by the condition LA1 /Op (LA1 ) ∈ 3 Cp (Lemma 6.2). The 2 G2 (3 2 ), Co3 , and (S)U6 (2) cases are impossible since then m2,3 (LA1 ) − m3 (O3 3 (LA1 )) = 3, forcing m2,3 (G) ≥ m3 (A) − 1 + 3 > m3 (B), a contradiction. 5 Thus LA1 /Op (LA1 ) ∼ = Sp4 (8), 3D4 (2), or 2F4 (2 2 ). Moreover, [VK , 3.96b] implies that La is a trivial pumpup of LA1 for any a ∈ A# 1 . These three cases are quite similar; we shall apply Proposition 4.14. First, we argue that there exists a p-terminal pair (x, K) ∈ ILop (G) such that K/Op (K) ∼ = LA1 /Op (LA1 ) and mp (C(x, K)) > 1. Namely, choose a ∈ A# ; by [I , 6.22], there A 1 exists an Ipo -terminal long pumpup (x, K) of (a, La ); and by [IA , 6.27(ii)], (x, K) is p-terminal. Taking a shortest pumpup chain (a, La ) = (x0 , K0 ) < · · · < (xn , Kn ) = (x, K), we have mp (C(a, La )) ≥ mp (A1 ) ≥ 2 and indeed mp (C(xi , Ki )) ≥ 2 for all 0 ≤ i < n, by minimality. Thus mp (CG (xi )) ≥ 4 and mp (CG (xi+1 )) ≥ 4 (0 ≤ i < n) since mp (Ki ) ≥ 2 and mp (C(xi , Ki )) > 1. Therefore all Ki /Op (Ki ) lie in Cp , so by [VK , 3.96b] they are all isomorphic to LA /Op (LA ). By [V2 , 8.5], mp (C(x, K)) ≥ 2, as asserted. 5 Whether K := K/Op (K) ∼ = Sp4 (8), 3D4 (2) or 2F4 (2 2 ), for a long root involu tion t ∈ K, p divides |CK (t)|. Moreover, O 2 (Out(K)) = 1 if p = 5 or K ∼ = 3D4 (2), while if K ∼ = Sp4 (8), then K is outer well-generated (for the prime 2), by [VK ,

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8.4]. Finally for any E ∈ Ep2 (Aut(K)), K = ΓE,1 (K), by [IA , 7.3.3]. However, Proposition 4.14 contradicts this last assertion.  Lemmas 7.2, 7.7, 7.12, and 7.13, the last applied with A = B, complete the proof of Proposition 7.1 in the case A = B. For the rest of the proof of Proposition 7.1, we continue to assume that (6B) holds for some (B, A), but the conclusions of Proposition 7.1 hold for no such pair (B, A). We thus are in the case |B : A| = p. In view of Lemmas 7.2 and 7.13, we have ∼ L2 (pn ), n > 1, or L± (p), or (LA , p) = (3G2 (3), 3) or LA = 3 (7F) 1 (2F4 (2 2 ) , 5). The most complicated case, LA ∼ = L2 (pn ), we save for last. We next verify A-terminality for the other cases, and rule out L± 3 (p) for p > 3. Lemma 7.14. Suppose that LA ∼ = L± 3 (p). Then LA is A-terminal in G . ± p Proof. Suppose LA ∼ = L± 3 (p). By [VK , 3.11], ↑p (LA ) ∩ Cp = {L3 (p )}, except if (p, LA ) ∼ = (3, U3 (3)), when ↑3 (LA ) ∩ C3 = {U3 (33 ), D4 (2), 3D4 (2)}. Let a ∈ A# . If La is a nontrivial pumpup of LA , then for some C ≤ A with a ∈ C, p 3 ∼ either LC /Op (LC ) ∼ = L± 3 (p ), or p = 3 and LC /O3 (LC ) = D4 (2) or D4 (2). But ± := CA (LC /Op (C)) then A normalizes LC , centralizing an L3 (p) subgroup, so A0  is a hyperplane of A by [IA , 7.1.4c, 4.7.3A]. If m2,p (LC /Op (LC )) ≥ p, then m2,p (G) ≥ m2,p (A0 LC ) ≥ mp (B) − 2 + p > mp (B), a contradiction. This rules p out LC /Op (LC ) ∼ = L± 3 (p ) or D4 (2), in which the centralizers of 2-central invop lutions contain SL2 (p ) and L2 (2) × L2 (2) × L2 (2), respectively. Thus, assume CC := LC /O3 (LC ) ∼ = 3D4 (2). Then some a0 ∈ A# induces a nontrivial graph au tomorphism on LC , and there exists b0 ∈ I3 (CLC (A)) such that E := E(CLC (b0 )) ∼ = L2 (8) and CE (a0 ) has even order. Let B ∗ = A b0  ∼ = B, so that B ∗ ∈ B∗,c (G). Then for some hyperplane A∗ of B ∗ , L3 (CG (A∗ ))/O3 (L3 (CG (A∗ ))) ∼ = L2 (8). So (B ∗ , A∗ ) satisfies (6B), and L3 (CG (A∗ )) is A∗ -terminal by Lemma 7.13. As we are assuming that no such pair exists, this is a contradiction, so LA is A-terminal in G. The proof is complete. 

Lemma 7.15. If LA ∼ = L± 3 (p), then p = 3. ∼ L± (p). Then Proof. Suppose for a contradiction that p ≥ 5 and LA = 3 mp (ALA ) = mp (A) + mp (LA ) = mp (A) + 2 > mp (B). Fix an involution z ∈ LA . By Lemma 3.5, [Op (CG (z)), A] = 1. Hence Op (CG (z)) normalizes LA , and as mp (ALA ) > mp (B) = m, it follows that (7G)

COp (CG (z)) (LA ) has odd order.

Let Lz := Lp (CLA (z)). Since Op (CG (A)) has odd order by (1A4), Lz is a {2, p}-component with Lz /O{2,p} (Lz ) ∼ = SL2 (p). By (7G) and [VK , 4.12], O2 (CG (z)) has a unique involution, and so [Lz , O2 (CG (z))] = 1. Also, Op (CG (z)) normalizes LA and Lz , so [Lz , Op (CG (z))] ≤ Op (CG (z)) ∩ Lz ≤ Op (Lz ) = z O{2,p} (Lz ). It follows that [Lz , E(Op (CG (z)))] = 1. Since G is a group of  even type, it follows that CLz (O p (E(CG (z)))) = CLz (F ∗ (CG (z))) ≤ O2 (Lz ) = z. Notice that for every d ∈ Ip (Lz ), A, d = A × d ∈ Bp∗ (G), centralizing z. Hence,

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by [IG , 8.7(iii)], A, d, and thus d, normalizes every component of O p (E(CG (z))).  Since Lz = O p (Lz ), Lz normalizes every component of E(CG (z)). It follows, with the Schreier property, that Lz ≤ E(CG (z)). But now Lz is a p-component of  CE(CG (z)) (A). As A normalizes every component of O p (E(CG (z))), Lz lies in a single component K of E(CG (z)). By the (Bp )-property and [IA , 6.1.4], Lz ∼ = SL2 (p). Since LA is A-terminal in G and Lz is a component of CLA (z), it follows (with the (Bp )-property) that Lz ∼ = SL2 (p) is a component of CK (a) for every a ∈ A# . Hence by [VK , 3.5], p = 5 and K/Z(K) ∼ = A5m−1 +5 , which is absurd as K ∈ C2 . The lemma follows.  Lemma 7.16. If (p, LA ) = (3, 3G2 (3)), then LA is A-terminal in G. Proof. By [VK , 3.11], ↑3 (3G2 (3)) = ∅, so if LA is not A-terminal in G, then for some b ∈ A# , the pumpup of LA in CG (b) is diagonal, with A permuting the 3-components transitively. However, by Lemma 3.3, B normalizes all 3-components  of CG (b). This is a contradiction as A ≤ B, so the lemma is proved. 1 Lemma 7.17. If p = 5 and LA ∼ = 2F4 (2 2 ) , then there exists a hyperplane A1 of B and an A1 -terminal 5-component LA1 of CG (A1 ) such that LA1 /O5 (LA1 ) ∼ = LA 5 or 2B2 (2 2 ).

Proof. If LA is A-terminal, then we are done with A1 = A. Otherwise, in view of L5 -balance and Lemma 3.3, there is a ∈ A# such that LA pumps up vertically 5 a ∼ in CG (a), to La . By [VK , 3.11], L = 2F4 (2 2 ), with some a0 ∈ A# inducing a  a . Since |B : A| = 5 and Out(LA ) is a 5 -group, nontrivial field automorphism on L # there is b ∈ B ∩ LA . Let A1 = b A0 , where A0 = A ∩ C(a, La )). Then A1 is another hyperplane of B. Set LA1 = L5 (CLa (A1 )). Then LA1 is a 5-component 5 of CG (A1 ) and LA1 /O5 (LA1 ) ∼ = 2B2 (2 2 ) by [IA , 4.8.7]. By Lemma 7.13, LA1 is  A1 -terminal, so the proof is complete. In view of the preceding four lemmas, we turn to the A-terminality question for LA ∼ = L2 (pn ). The next lemma is useful both before and after A-terminality is proved. a ∼ Lemma 7.18. Suppose that for some a ∈ A# , L = L2 (pn ), n ≥ 2, and for all  a ∈ A0 := A ∩ C(a, La ), the pumpup of La in CG (a ) is trivial. Then pn = 32 . 

a ∼ Proof. Note that we are not assuming that L = LA . Suppose that the lemma is false. Let z ∈ I2 (LA ) ⊆ I2 (La ), W0 = O2 (CLa (z)), and W = CW0 (A0 ). Since pn = 32 and n ≥ 2, and Op (CG (a)) has odd order by (1A4), W0 projects  a . In particular W0 = 1. Then for all a0 ∈ A# , nontrivially on a p -subgroup of L 0 W0 ≤ Op (CCG (a0 ) (z)) = Op (CCG (z) (a0 )). That is,

Op (CCG (z) (a0 )) = ΔCG (z) (A0 ). (7H) W0 ≤ a∈A# 0

a . Since LA is a component of CLa (A), A induces field automorphisms on L Therefore A/A0 is cyclic. So mp (A0 ) ≥ mp (A) − 1 = mp (B) − 2 ≥ 2.  a , and so mp (La ) ≥ 2p in that Moreover if A0 < A, then A acts nontrivially on L case. In any case, mp (A0 La ) > mp (B).

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4. THEOREM C5 : STAGE 2

Now mp (A0 LA0 ) > mp (B) and LA0 is A0 -terminal by hypothesis. Also z ∈ LA ≤ LA0 . Hence by Lemma 3.5, A0 centralizes Op (CG (z)), whence A0 acts  faithfully on O p (E(CG (z))). Set Q = O2 (CG (z)). Thus [A0 , Q] = 1, so [W0 , QE(Op (CG (z)))] ≤ [O2 (CCG (A0 ) (z)), O2 (CCG (A0 ) (z))E(CCG (A0 ) (z))] = 1. 

Therefore W0 acts faithfully on O p (E(CG (z))). By (7H) and [IG , 20.6], there  exists a component E of O p (E(CG (z))) such that if we put N = NCG (z) (E), then  := N/CN (E), W 0 is a nontrivial subgroup of Δ  (A 0 ) of odd A0 W0 ≤ N and in N N  is not locally mp (A0 )-balanced. order. In particular, E  ∼ On the other hand, E is a C2 -group. By [VK , 7.10], E = Lp (q), p dividing 0 faithfully inducing field automorphisms (of odd order) on E,  and q − , with W   A0 acting on E like the image of a nonabelian p-subgroup of GLp (q). In particular mp (A0 ) = 2, so m = 4. Therefore some element a1 ∈ A induces a nontrivial field  a . By [VK , 7.9b], this implies that a1 acts nontrivially on automorphism on L  0 in Aut(E)  is W0 . But W0 induces field automorphisms on E, so the image of W   disjoint from [Aut(E), Aut(E)]. Thus, a1 cannot normalize E. Therefore 4 = m =  m2,p (G) ≥ mp (CG (z)) ≥ 2p, a contradiction. The lemma is proved. Lemma 7.19. Suppose that LA ∼ = L2 (pn ), n ≥ 2. Then pn = 32 . Moreover, if LA is not A-terminal in G, then there exists a ∈ A# such that ∼   a ), La = J3 , 3J3 , O  N , Sp6 (2), or F4 (2), and La is A0 -terminal where A0 = CA (L  ∼ ∼  a = Sp6 (2). Finally, |A : A0 | = 3 unless L  a = O N or F4 (2), in unless possibly L which cases |A : A0 | = 9. Proof. Suppose that pn = 32 . By Lemma 7.18, LA is not A-terminal in G. a ∼ Thus, by [VK , 3.11], L = L2 (ppn ) for some a ∈ A# , with some a ∈ A inducing a  a . But then a further nontrivial pumpup La of nontrivial field automorphism on L 1 # La in CG (a1 ) for some a1 ∈ A could not exist; by [VK , 3.11], it would have to be 2 La1 /Op (La1 ) ∼ = L2 (pnp ), and then the image of A in Out(La1 /Op (La1 )) would be a noncyclic p-group, which is absurd as p is odd. Thus La would be A0 -terminal,  a ), contradicting Lemma 7.18. where A0 = CA (L n a ∼  Therefore p = 32 . The argument of the last paragraph also shows that L = 6 # L2 (3 ) for any a ∈ A . Indeed, keeping in mind that 3A6 is not a C3 -group, we see a ∼ from [VK , 3.44abc] that for any a ∈ A# , L = A6 , A9 , Sp4 (8), Suz, 3Suz, Co1 , or one of the groups in the statement of the lemma, with |A : A0 | as claimed and A0 -terminality holding. In the A9 and Sp4 (8) cases, m3 (C(a, La )) ≥ 2 as m3 (CG (A)) > 4 by (6A). But this contradicts Lemmas 5.3 and 5.4. a ∼ If L = Co1 , then |A : A0 | = 9 so m3 (A0 ) = m3 (B) − 3. On the other hand,  a ) + m3 (A0 ) > m3 (B), m2,3 (Co1 ) = 4 by [IA , 5.6.2], so m2,3 (La A0 ) ≥ m2,3 (L contradiction. a ∼ Finally we assume L = Suz or 3Suz and argue to a contradiction. From  A )) = 2. Hence m = m3 (B) = m3 (A) + 1 ≤ [IA , 5.3o] we see that m3 (CAut(La ) (L m3 (A0 ) + 3. Let z ∈ La be a 2-central involution. Then CLa (z)/O2 (CLa (z))  ≤ C  (z) contains P Sp4 (3) (whose Schur multiplier is a 3 -group), so there is E La ∼ E33 and E∩Z( =   a ) = 1. Choose E ≤ CL (A0 z) mapping isomorphically with E L a

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 and set B ∗ = A0 E = A0 × E. As [B ∗ , z] = 1, on E, m = m2,3 (G) ≥ m3 (B ∗ ) = m3 (A0 ) + 3 ≥ m so equality holds throughout and B ∗ ∈ B∗,c (G). Thus by (6B3), B ∈ B∗,o (G) = B∗,c (G), whence CG (B) contains an involution t. As A ∩ La ≤ O3 3 (La ), t ∈  a ), then m3 (A0 ) + 3 = m = m2,3 (G) ≥ m3 (BLa ) ≥ NG (La ). If t ∈ CG (L  m3 (A0 ) + m  3 (Suz) = m3 (A0 ) + 5 by [IA , 5.6.1], contradiction. Thus CAut(La ) (B) has even order. As m3 (A) = m3 (A0 ) + 2, it follows from [VK , 3.70] that B = A, a contradiction completing the proof.  In view of Lemmas 7.14, 7.15, 7.16, 7.17, and 7.19, in order to complete the proof of Proposition 7.1, it suffices to rule out the unwanted possible pumpups in Lemma 7.19. The argument unfortunately is lengthy and we devote the next section to it. 8. Theorem 1: The A-Terminality of A6 We continue the argument of the previous section to complete the proof of Proposition 7.1 by contradiction. Thus, throughout this section, (6B) holds with |B : A| = p = 3, LA ∼ = A6 , and LA is not A-terminal because for some a ∈ A# , a ∼ (8A) L = J3 , 3J3 , O  N, Sp6 (2), or F4 (2), as in Lemma 7.19.  a /Z(L a ) ∼ We shall soon see that the worst culprit is L = J3 , but first we need a few specialized lemmas. x = Cx /O3 (Cx ), Lemma 8.1. Suppose that x ∈ I3o (G) and set Cx = CG (x), C  ∼  and Lx = L3 (Cx ). Let P ∈ Syl3 (Cx ). If Lx = F4 (2) or O N and m3 (C(x, Lx )) = 1, then x is not weakly closed in P with respect to G. Proof. Suppose false. Then clearly Lx  NG (x), NG (P ) ≤ NG (x), and  x )| = 2 by [IA , 2.5.12, 5.3s], so P = R × Q, where P ∈ Syl3 (G). Also, | Out(L  x ) is cyclic, and Q and R are both normal in R = P ∩ Lx ∈ Syl3 (Lx ), Q = CP (L NG (P ). Let E = J(P ). Then by [VK , 9.18], E ∼ = E35 and AutLx (E) ∼ = W (F4 ) 1+4 ∼ (the Weyl group) or 2− D10 according as Lx = F4 (2) or O  N .  For any subgroup H < G such that x ∈ H, we set IH = xH . Then (8B)

H = IH NH (x)

by the weak closure property of x and a Frattini argument. Moreover, [P, P ] = [R, R] does not contain x. Again by weak closure, all Sylow 3-subgroups of G containing x lie in NG (x), so x is not in the commutator subgroup of any of them. Hence, x is not in the commutator subgroup of any Sylow 3-subgroup of IH . Then by [VK , 9.30], (8C)

m3 (IH ) = 1 and IH is either a 3-component or 3-nilpotent.

We shall construct a strong 3-uniqueness subgroup of G; this will contradict (1A5) and complete the proof. Since m3 (E ∩ R) > 3, every maximal elementary abelian subgroup of P (with respect to inclusion) has rank at least 4, and so by (1A7), G is balanced for the prime 3. Let Z = Ω1 (Z(P )) ∼ = E32 and set W = Θ1 (G; Z) and M = NG (W ). By [V2 , 1.2], for any E32 ∼ = E ≤ P , Θ1 (G; E) = Θ1 (G; EZ) = W . Taking E ≤ Z, we have m3 (EZ) ≥ 3, so by [V2 , 1.1], W is

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a 3 -group. For any Q ≤ P with m3 (Q) ≥ 2, Θ1 (G; E) = W for all E ∈ Ep2 (Q). As NG (Q) permutes Ep2 (Q), NG (Q) ≤ NG (W ) = M . Thus ΓP,2 (G) ≤ M . By  x = ΓR,2 (L  x ), so [IA , 7.3.1] and [VK , 8.31], L (8D)

NG (x) ≤ O3 (CG (x))Lx NG (R) ≤ M.

We claim that (8E)

ΓP,1 (G) ≤ M.

Let y ∈ E1 (P ). By [VK , 9.18], y is NG (x)-conjugate into E, hence M -conjugate into E as ΓP,2 (G) ≤ M . Thus, in proving that NG (y) ≤ M , our current goal, we may assume that y ∈ E, and y = x. Let H = NG (y), so that E ≤ H. To show H ≤ M , it is enough by (8B) and (8D) to show that IH ≤ M . But IH  H and in particular IH is E-invariant. Clearly O3 (IH )NIH (x) ≤ ΓE,2 (G) ≤ M so by (8C), we may assume that IH is a 3-component of 3-rank 1. But then m3 (Aut(IH /O3 (IH ))) ≤ 2 by [VK , 1.1], so as m3 (E) ≥ 5, certainly IH ≤ ΓE,2 (G) ≤ M in this case as well. This establishes (8E). If we assume that (8F)

W = 1,

then ΓP,1 (G) ≤ M < G, i.e., M is strongly 3-embedded in G. We next establish a parallel conclusion if (8F) fails; if W = 1, we show that M := NG (I) is strongly 3-embedded in G, where I = ING (y) for all y ∈ E ∩ L# x. Now W = 1 implies by balance that O3 (CG (y)) = 1 for all y ∈ I3 (E). Hence for such an element y, if we set Hy = NG (y), we have that either IHy is a cyclic 3-group containing x, in which case Hy ≤ NG (x), or IHy is quasisimple of 3-rank 1 and contains x. In the latter case, x is inverted in IHy , so yx−1 ∈ (yx)G . But G-fusion in E = J(P ) is controlled in NG (E) by [IG , 16.9], hence in NG (x)∩NG (E) ≤ NG (E ∩Lx ). It follows that y ∈ E ∩Lx , and then similarly that CE (IHy ) ≤ E∩Lx . Moreover, if IHy is a component, no y  ∈ E∩Lx induces an outer automorphism on IHy ; for if it did, it would be a field automorphism, whence the coset y  x would be fused, contradicting the fact that NG (E) normalizes E ∩ Lx . Thus CE (IHy ) = E ∩ Lx . It follows that IHy = IHy for all y, y  ∈ (E ∩ Lx )# , and IHy ≤ NG (x) for all y ∈ E −Lx . Let I = IHy for some (all) y ∈ E ∩L# x , so I is the component of E(CG (y)) containing x. Then E(CLx(y)), which lies in E(CG (y)) by L3 -balance, centralizes I. Therefore I centralizes E(CLx (y)) | y ∈ E ∩ L# x = Lx , by [IA , 7.3.3] and [VK , 8.31]. Clearly NG (P ), which normalizes E ∩ Lx and x, normalizes I. So NG (x) ≤ Lx , NG (P ) ≤ NG (I). It follows that ΓP,1 (G) ≤ ΓE,1 (G), Lx  ≤ M , so M is strongly 3-embedded in G. Whether (8F) holds or not, let M ∗ be a maximal subgroup of G containing M ; then M ∗ also is strongly 3-embedded in G. But M ∗ = NM ∗ (x)IM ∗ , and as with H above, CP (IM ∗ /O3 (IM ∗ )) = 1, so M ∗ = ΓP,1 (M ∗ ) ≤ M . Thus M = M ∗ is maximal in G. Let L be the subnormal closure of Lx in M . Then m3 (L0 ) ≥ m3 (Lx ) = 4 for all 3-components L0 of L, so [L, IM ] ≤ O3 (IM )., Hence CL (x) covers L/O3 (L),  and so L = O 3 (Lx O3 (M )). As CG (x) ≤ M , M is a 3-component preuniqueness subgroup of G. If O3 (M ) = 1, then M is a strong 3-uniqueness subgroup of G by definition [I2 , 8.7], as desired. This is certainly the case if W = 1, by the definition of M . So assume that O3 (M ) = 1. Then Lx = L  M . If Lx ∼ = F4 (2) ∈ Chev(2), then

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by definition [I2 , 8.7], M is a strong 3-uniqueness subgroup as m3 (CM (Lx )) = 1. Finally, if Lx ∼ = O  N , let z ∈ I2 (Lx ), C = CG (z), and C0 = C ∩ M . Then E(C0 ) ∼ = 4L3 (4), and m3 (C0 ) ≥ m3 (E(C0 ) x) = 3. Since 4L3 (4) ∈ C2 , C0 < C and so C0 is strongly 3-embedded in C. But no strongly 3-embedded subgroup in any K-group has such a structure, by [IA , 7.6.1]. The proof is at last complete.  Lemma 8.2. Suppose that (x, K) ∈ ILo3 (G) with K/O3 (K) ∼ = F4 (2). Then m2,3 (G) ≥ 6. Proof. By Lemma 5.4a, m3 (C(x, K)) = 1. Let P ∈ Syl3 (CG (x)), Q = CP (K/O3 (K)), and R = P ∩ K, so that P = R × Q as Out(K/O3 (K)) is a 3 -group. Again let E = J(P ) = J(R) × x ∼ = E35 ; by [VK , 9.18], every element of Ip (P ) is K-conjugate into E. Therefore by Lemma 8.1 and [IG , 16.9], x  NG (E). As a result, x ∈ Syl3 (C(x, K)); for otherwise, x would be characteristic in EQ ∈ Syl3 (CG (E)) and hence normal in NG (E) by a Frattini argument, contradiction. Set N = AutG (E) and N1 = NN (x). By [VK , 5.16], one of the following holds: (1) F ∗ (N ) = E(N )Z(N ) with E(N ) ∼ = Ω5 (3) and E a natural E(N )-module; or (8G) (2) N = T N1 where T  N , T ∼ = E34 , and T permutes regularly the set of complements to E ∩ R in E. Suppose first that (8G1) holds. Identify E with the natural Ω5 (3)-module V . From the structure of NN (x) we see that x is a nonsingular 1-space in ∼ V . Choose, by [IA , 4.7.3A], y ∈ E # ∩ K such that E(CK  (y)) = Sp6 (2), where  K = K/O3 (K). Set D = x, y, so that CN (D) has a normal W (C3 ) ∼ = Z2  Σ3 subgroup. It follows that D is a nondegenerate plane in V . But then there is d ∈ D such that d is nonsingular but not isometric to x. Hence NN (d) has a normal ∼ Ω− 4 (3) = A6 -subgroup. On the other hand, the pumpup of L3 (CK (y)) in CG (d) lies in C3 and so is Sp6 (2) or F4 (2) by [VK , 3.84]. Therefore NN (d) is solvable, a contradiction. Hence, (8G2) holds. We have Q = x and P = R × Q ∈ Syl3 (NG (x)). Set M = NG (E), so that E ∩ K  M , and let S be the stabilizer of the chain E > E ∩ K > 1. Then T is the image of S in N . By a Frattini argument, NM (S) covers N/T . Since g → [g, x] is an isomorphism between S/E and E ∩ K preserving the action of CN (x), N1 has a W (F4 )-subgroup, necessarily acting irreducibly on T . Therefore S/E ∩ K is isomorphic to either E35 or 31+4 . But W (F4 ) does not embed in Aut(31+4 ) ∼ = GSp4 (3) since its commutator subgroup contains Q8 ∗ Q8 and so has 2-rank at least 3, unlike Sp4 (3). Thus, Φ(S) = E ∩ K. Let S0 = [S, z] where z is a central involution of AutK (E). Then |S0 | = 38 and S = S0 x, a semidirect product. Also CS0 (z) = 1 so S0 is abelian. By the action of AutK (E), S0 ∼ = E38 or S0 is homocyclic of exponent 9. y = Suppose that S0 has exponent 9. Again choose y ∈ E # ∩ K with K ∼ E(CK  (y)) = Sp6 (2). Then y ∈ S0 . Let L be the pumpup of L3 (CK (y)) in CG (y). If any 3-element of CG (y) fails to normalize any 3-component of L, then clearly m2,3 (G) ≥ 2m3 (Ky ) = 6, as desired. Otherwise, as L/O3 (L) ∈ C3 , we have L/O3 (L) ∼ = Sp6 (2) or F4 (2), by [VK , 3.84], and S0 normalizes L. Then  y ) is cyclic, so AutS (L) contains Z9 × Z9 × Z9 . But this is Ω1 (CS0 (L)) ≤ CE∩K (K 0 impossible by [VK , 9.27].

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∼ E38 . Let t ∈ N1 be an involution mapping on Thus, we may assume that S0 = a reflection in W (F4 ). Then |CE∩K (t)| = |CS0 /E∩K (t)| = 33 , so CS0 (t) ∼ = E36 and m2,3 (G) ≥ 6. The proof is complete.  Now we can prove a ∼ Lemma 8.3. For any a ∈ A# , L = A6 , O  N , J3 , or 3J3 . a ∼ Proof. Otherwise, by Lemma 7.19, L = Sp6 (2) or F4 (2). First assume that o F4 (2) ∈ L3 (G). Then m = m2,3 (G) ≥ 6 by Lemma 8.2. By (6A), m3 (CG (a)) ≥ a ∼ 7. If L = F4 (2), then m3 (C(a, La )) ≥ 3 and Lemma 5.4 gives a contradiction. a ∼ So L = Sp6 (2), and m3 (C(a, La )) ≥ 4. Choose (x, K) ∈ ILo3 (G) such that K/O3 (K) ∼ = Sp6 (2), and subject to this, so that m3 (C(x, K)) and then |C(x, K)|3 is maximal. Let Q ∈ Syl3 (C(x, K)). Then m3 (Q) ≥ 4 so Q is connected [V2 , 4.5]. If (x, K) is not one-step rigid, then for some y ∈ I3 (Q), (x, K) pumps up to (y, L) with Sp6 (2) ↑3 (L/O3 (L)) ∈ C3 . Hence by [VK , 3.84], L/O3 (L) ∼ = F4 (2) and m3 (C(y, L)) ≥ m3 (CQ (y)) − 1 ≥ 2. Again Lemma 5.4 gives a contradiction. Hence, (x, K) is one-step rigid. By [V2 , 8.10], (x, K) is 3-terminal in G. But this too contradicts Lemma 5.4. Therefore F4 (2) ∈ Lo3 (G). Then m3 (CG (a)) > m3 (B) ≥ 4, whence m3 (C(a, La )) ≥ 2, and Lemma 5.4b gives a contradiction. The proof is complete.  We next prove a ∼ Lemma 8.4. If L = O  N for some a ∈ A# , then m3 (C(a, La )) ≥ 2. Proof. Suppose false and let P ∈ Syl3 (CG (a)), R = P ∩ La , and Q = P ∩ C(a, La ). Thus P = R × Q with R ∼ = E34 and Q cyclic. As [A, LA ] = 1, m3 (A) = 3 and so m = 4. By Lemma 8.1, a is not weakly closed in P with respect to G, so a  NG (P ). In particular, Q = a. Let W = AutG (P ) and Wa = NW (a), so that |Wa | = 2b .5 with 6 ≤ b ≤ 8, by [IA , 5.3s]. If W is transitive on E1 (P ) then |W | = 2b .5.112 and W , like GL5 (3), has cyclic Sylow 11-subgroups. By [III17 , 9.9], W is solvable, and as |E1 (P )| = 112 , W has a self-centralizing normal Sylow 11-subgroup, which is clearly impossible. As Wa is transitive on R# but |E1 (R)| + 1 does not divide | Aut(P )|, it follows that aW = E1 (P ) − E1 (R) and R is W -invariant. Let z be the 2-central involution of AutLa (P ) ≤ W . Then CW (z) normalizes CP (z) = a so Wa = CW (z). By [VK , 9.18c], no involution of Wa acts on R as a reflection. It follows easily by the Z ∗ -theorem that z ∈ Z ∗ (W ). As |W : Wa | = 34 and O2 (Wa ) = 1, T := O2 (W ) has order 34 and is inverted by z. Commutation with a then is an AutLa (P )-isomorphism from T to R. In particular R = CP (T ). Moreover, as P contains elements of order 3 which are 3-central in G, all elements of R are 3-central. Note also that CP (LA ) = A. Let 1 = u ∈ A ∩ La . Let S ∈ Syl3 (NG (P )), so that P ≤ S and S covers T . If Lu ∼ = A6 , then Lu is S-invariant because m2,3 (G) = m = 4, and so [S, S] ∩ P ∩ Lu = 1, a contradiction since P ∩ Lu = P ∩ LA ≤ [a, S]. If Lu ∼ = O  N , then similarly Lu  CG (u) and so a Sylow 3-center of G (lying in CG (u)) has rank at least 5, whence a is 3-central, contradiction. Therefore by Lemma 8.3, Lu ∼ = J3 or 3J3 . Finally, P0 := CP (Lu /O3 (Lu )) = CR (Lu ) is a subgroup of A that is conjugate in La , hence in NLa (P ), to a Sylow 3-subgroup PA of LA . Then R = P0 PA . Since

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NLu (P ) ∩ CG (PA ) shears P/R to PA , centralizing R, it follows that some subgroup H of NG (P )∩CG (R) shears P/R to P0 , centralizing R. But this is impossible, since H, centralizing P0 , normalizes P ∩ Lu . This contradiction completes the proof.  Now we reach the critical configuration. a ∼ Lemma 8.5. For some a ∈ A# , L = J3 or 3J3 . Proof. We shall prove that ∼ J3 or 3J3 for a = If Lb /O3 (Lb ) ∼ = O  N for some b ∈ A# , then L some other a ∈ A# . This statement and Lemma 8.3 then imply the lemma. Suppose that (8H) fails, and let b and Lb be as there. By Lemma 8.3, for every a ∈ A# , a ∼ L = A6 or O  N, and in the latter case, m3 (C(a, La )) ≥ 2 by Lemma 8.4. It follows that for every hyperplane A1 of A, LA1 /O3 (LA1 ) ∼ = A6 , so A centralizes LA1 /O3 (LA1 ). By Lemma 3.3, A normalizes all 3-components of CG (A1 ). Hence for any T ∈ IG (A; 2), NCG (A1 ) ([CT (A1 ), A]) covers ALA1 /O3 (LA1 ), of 3-rank m3 (A) + m3 (LA1 ) = m3 (B) + 1 > m. Thus [CT (A1 ), A] = 1, and as A1 was arbitrary, [T, A] = 1. Choose z ∈ I2 (LA ). Then [A, O2 (CG (z))] = 1 = [A, O3 (CG (z))], just as part (a) implied part (b) in Lemma 3.5. Let J = L3 (CLb (z)), so that J/O{2,3} (J) ∼ = 4L3 (4) ∈ C2 . We prove that J is a component of CG (z), a contradiction as G is of even type. Namely, J = [J, A]  centralizes O3 (CG (z)), so J acts on X = O 3 (E(CG (z))) with kernel contained in O2 (J). We have A = A0 × A2 with A0 = CA (Lb /O3 (Lb )) and A2 = A ∩ J. As m3 (A) = m2,3 (G) − 1, it follows from [V2 , 6.6] that A has at most one nontrivial orbit on the components of X, and that orbit, if it exists, has length 3. As AutJ (A2 ) contains Q8 , A2 must permute the components of X trivially. Hence J = [J, A2 ] does, as well. By L3 -balance, and as [J, O3 (CG (z))] = 1, J lies in a product of components of CG (z) permuted transitively by any a ∈ CA (Lb /O3 (Lb )). As this last group is noncyclic by Lemma 8.4, J lies in a single component K of CG (z). Then J is a 3-component of CK (a ) for any a ∈ CA (Lb /O3 (Lb )). By [VK , 3.5], J = K, as promised. The proof is complete.  (8H)

a ∼ For the rest of this section, fix a ∈ A# with L = J3 or 3J3 . Fix z ∈ I2 (LA )  a ). Thus, we may write A = A0 c where c ∈ I3 (LA ) and and set A0 = CA (L 0 LA = L3 (CLA0 (c)). We also set Cc = CG (c). We first prove Lemma 8.6. The following conditions hold: (a) B∗,c (G) = ∅; (b) O2 (CG (z)) = O2 (CLa (z)) ∼ = Q8 ∗ D8 ; and and L centralizes E(CG (z)). J (c) La ∼ = 3 a Moreover, condition (c) holds with a replaced by any element of A# 0 . Proof. By Lemma 3.3, B normalizes La . Since CG (B) is 3-constrained by  a . By Lemma 6.4b, A is of maximal (6B3), B = A b where b acts nontrivially on L  a ) has 3-rank in CG (ALA ). Thus A ∩ La ≤ O3 3 (La ), and the image of B in Aut(L rank 2. In particular as m  3 (J3 ) = 3 by [IA , 5.6.1], m3 (ALa ) > m3 (B) = m2,3 (G).

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Thus C(a, La )∩CG (A) has odd order. But also no involution of Aut(J3 ) centralizes an E32 -subgroup [IA , 5.3h]. Hence CG (B) has odd order. As B ∈ B∗,o (G), it follows that B∗,c (G) = ∅ by definition of B∗,o (G), so (a) holds. Moreover, m3 (CG (z)) < m, so m3 (CG (z)) = m3 (A). By [IG , 8.7(iii)], A normalizes every 3-component of E(CG (z)). Since ↑3 (3J3 ) = ↑3 (J3 ) = ∅ by [VK , 3.8], LA0 is A0 -terminal. By Lemma 3.5, as m3 (A0 LA0 ) = m − 2 + 3 > m, [A0 , O3 (CG (z))] = 1. In particular, O2 (CG (z)) ≤  a has centralizer of odd order, O2 (CG (z)) acts faithfully on L a . CG (a), and as AL ∼  Since T := O2 (CLa (z)) = O2 (CAut(La ) (z)) = D8 ∗ Q8 and CLa (z) acts irreducibly on T/Φ(T), we have O2 (CG (z)) ≤ O2 (CLa (z)). By the action of CLa (z), O2 (CG (z)) = O2 (CLa (z)) or z. Suppose O2 (CG (z)) = z; we derive a contradiction. Now CG (A0 z) has a  subnormal subgroup H = O 3 (H) ≤ CLA0 (z) with H/O{2,3} (H) ∼ = 21+4 − A5 . Thus H is a 3-component. Let T ∈ Syl2 (O3 (H)). By L3 -balance and the property (B3 ) [IA , 7.1.3], T ≤ O3 (H) ≤ O3 (CG (z)). Now Y := F ∗ (O3 (CG (z))) is the central 3 product of either Z2 or copies of 22B2 (2 2 ) with some number r of Suzuki groups ni 2 B2 (2 2 ), ni > 1, 1 ≤ i ≤ r. (Possibly r = 0, but |Y | > 2 as T ≤ O3 (CG (z))). Then CY (A0 ) has a Sylow 2-subgroup S with Ω1 (S) elementary abelian. But CY (A0 )  CG (A0 z), so Ω1 (S) = z by the structure of LA0 . Hence A0 is fixedpoint-free on a Sylow 2-subgroup of Y / z, which is absurd. So O2 (CG (z)) > z, whence (b) holds. Now take an involution t ∈ O2 (CG (z)), t = z. Then [t, CO2 (La ) (z)] ≤ O2 (La )∩ O2 (CG (z)) = 1. By [IA , 5.3h] and [IG , 16.21], AutLa (t, z) = Aut(t, z) ∼ = Σ3 . It follows that [t, O2 (La )] = 1, so La is quasisimple. By L2 -balance, X := E(CG (z)) ≤ E(CG (t)), and since t ∈ z La , X = E(CG (t)). Similarly, X = E(CG (tz)). Then NG (X) contains Γz,t,1 (G) ≥ Γz,t,1 (La ) = La by [VK , 8.8]. As [z, X] = 1, [La , X] = 1. Then since O2 (CG (z)) ≤ La , O2 (Z(La )) centralizes O2 (CG (z))X = F ∗ (CG (z)). Therefore O2 (Z(La )) = 1, and (c) follows. The final sentence holds since LA0 is A0 -terminal, and a was an arbitrary a ∼ element of A subject to L  = J3 or 3J3 . This completes the proof. 

Since [A0 , O3 (CG (z))] = 1 by Lemma 3.5, O 3 (E(CG (z))) = 1. We fix 

an arbitrary component K   O 3 (E(CG (z))). The next lemma highly restricts the isomorphism type of K. Lemma 8.7. K is a component of Cc . Proof. By Lemma 8.6c, E(CG (z)) ≤ Cc . Let D ∈ Syl2 (LA ) with z ∈ Z(D), so that D ∼ = D8 . For any t ∈ I2 (D), E(CG (z)) ≤ E(CG (t)) by L2 -balance; but then equality holds as t ∈ z LA ⊆ z G . Hence K is a component of CG (t). By [V2 , 7.2], K ≤ E(Cc ), with K a component of CE(Cc ) (t) for all t ∈ I2 (D). Let  L = K E(Cc ) ; as t permutes the components of L transitively for each t ∈ I2 (D), by L2 -balance, L is a single component, and L ∈ C3 . Then as K   CL (t) for all  t ∈ I2 (D), it follows from [VK , 3.5bd] that K = L. The lemma is proved. By Lemma 8.7 and (1A3, 4), (8I)

K ∈ C2 and K/O2 (K) ∈ C3 . Lemma 8.8. Let t ∈ I2 (E(CG (z))). Then m3 (CA0 (t)) ≤ 1.

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Proof. Suppose false, and that A1 ≤ CA0 (t) with A1 ∼ = E32 . For each b ∈ La is a component of CG (b) by Lemma 8.6. Also [t, La ] ≤ [E(CG (z)), La ] = 1. Hence by [V2, 7.2], La ≤ E(CG (t)); moreover, by L3 -balance, for each b ∈ E(CG (t)) is a single b-invariant component for which La is a A# 1 , either L := La component of CL (b), or L is the product of three components permuted transitively by b, in which La is a diagonal subgroup. As A1 is noncyclic, L must be a single component, and by [VK , 3.5], L = La . Since E(CG (z)) is a nontrivial product of C2 -groups, m2 (E(CG (z))) > 1, and so m2 (C(t, La )) > 1. Therefore [V2 , 9.1] is contradicted and the lemma follows.  A# 1 ,

We write I3∗ (K) = {d ∈ I3 (K) | m3 (CK (d)) = m3 (K)}. Lemma 8.9. Suppose that A induces inner automorphisms on K. Then for any d ∈ I3∗ (K), La is a component of CG (d). Proof. Let A1 = CA (K). As O3 (K) = 1, A = A1 ×A2 where A2 = A∩K and c ∈ A1 . Let A∗2 ≤ CK (d) with m3 (A∗2 ) = m3 (K) ≥ m3 (A2 ), and set A∗ = A1 × A∗2 . Now B = A b for some b ∈ LA ≤ La , and we set B ∗ = A∗ b. Then since [K, La ] = 1 and b lies in an A4 -subgroup of LA , m3 (B ∗ ) ≥ m3 (B) and B ∗ normalizes a nontrivial 2-subgroup (of LA ). With Lemma 8.6a, B ∗ ∈ B∗,o (G), and A∗ is a hyperplane of B ∗ . Now A2 , and hence A, is centralized by some 3-central y ∈ K of order 3. As A y centralizes LA , of even order, m3 (A y) < m3 (B) and so y ∈ A. Thus y ∈ A0 and so La   CG (y). But A∗2 contains a K-conjugate of y as m3 (A∗2 ) = m3 (K). Hence La   CG (A∗2 ). It follows that LA   CG (A∗ ). Hence the entire development of this section and the previous section is valid with A∗ in place of A and we conclude by Lemma 8.6, applied to A∗ , that La is a component of CG (a∗ ) for any a∗ ∈ CA∗ (La ), and in particular for any a∗ ∈ A∗ ∩ K = A∗2 . As d ∈ A∗2 , the proof is complete.  Lemma 8.10. We have m3 (K) > 1. Proof. Suppose on the contrary that m3 (K) = 1. Since K/Z(K) ∈ C3 by (8I), K ∼ = L2 (8). As A has maximal rank in CG (z), we have A = CA (K)(A ∩ K) x, where A ∩ K = 1 and x ∈ A induces a (possibly trivial) field automorphism on K. In any event CK (x) has even order. Then, in CG (A ∩ K) we see BLa , and CK (CA (K) x La ) has even order. But m3 (CA (K) x La ) = m3 (BLa ) − 1 =  m3 (B). Thus B∗,c (G) = ∅, contradicting Lemma 8.6a. The lemma follows. Lemma 8.11. If A1 is any hyperplane of A, then CK (A1 ) has odd order. Proof. Otherwise CK (A1 La ) has even order; but m3 (A1 La ) = m3 (B), so we  reach the contradiction B∗,c (G) = ∅ as in the previous lemma. Lemma 8.12. Let t ∈ I2 (K) and assume that A induces inner automorphisms on K. Then there is no D ∈ E32 (CK (t)) such that D# ⊆ I3∗ (K). Proof. If D exists, then La is a component of CG (d) for all d ∈ D# , by Lemma 8.9, and so La is a component of CG (t) by [V2 , 7.2] and [VK , 3.5]. Then CK (t) ≤ C(t, La ), so by [V2 , 9.1], m2 (CK (t)) = 1, whence m2 (K) = 1, which is impossible as K ∈ C2 . The lemma follows.  Now we can prove

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Lemma 8.13. K ∼ = A6 . Proof. As noted above, m3 (A) = m3 (CG (z)), so A normalizes all components  of O 3 (E(CG (z))), by [IG , 8.7(iii)]. By (8I), K ∈ C2 and K/Z(K) ∈ C3 . By Lemmas 8.10 and 8.11, m3 (K) > 1 and every hyperplane of A has an odd order centralizer on K. Also for every involution t ∈ K, m3 (CA0 (t)) ≤ 1, by Lemma 8.8, and, if A maps into Inn(K), CK (t) contains no E32 -subgroup D with D# ⊆ I3∗ (K), by Lemma 8.12. If K ∈ Alt, then K ∼ = A6 as K ∈ C2 ∩ C3 . Suppose that K ∈ Spor. Then contradictions are reached by the following lemmas depending on K. If K ∼ = J3 , [V2 , 9.1] is contradicted (see Lemma 8.6c). If K ∼ = M11 , then Lemma 8.11 is contradicted. If K ∼ = F i22 , F i23 , F i24 , or F1 , then K has a subgroup H = ± E3m3 (K) Ωm3 (K) (3), and any involution t ∈ H with m3 (CO3 (H) (t)) ≥ 2 provides a contradiction to Lemma 8.12. A similar contradiction arises if K ∼ = Suz with H = E35 M11 ; here involutions of M11 invert elements of order 5, so they centralize noncyclic subgroups of O3 (H). If K ∼ = Co2 , F3 , or F2 , then I3∗ (K) = I3 (K), and Z2 × E32 -subgroups are easily found, so Lemma 8.12 is again contradicted. If K ∼ = Σ3 × A9 , an involution t ∈ Sol(H) centralizes = Co1 , then in a subgroup H ∼ D ∼ = E32 with D# consisting of 3-central involutions of K, again contradicting Lemma 8.12. The other sporadic groups do not satisfy (8I) (see [V3 , 1.1]). So we may assume that K ∈ Chev − Alt. Thus K ∈ Chev(3) ∩ C2 or K ∈ Chev(2) ∩ C3 . ∼ G2 (3) or L4 (3), then m3 (K) = 4 by [IA , 3.3.3]; and a Cartan subIf K/Z(K) = group, which is a four group, normalizes an E34 -subgroup of a Borel subgroup, so Lemma 8.12 is violated. If K/Z(K) ∼ = U4 (3), then the root subgroups for two orthogonal long roots centralize a common involution as well as a common short root group; since m3 (K) = 4, Lemma 8.12 is again violated. A similar argument rules out K ∼ = P Sp4 (3) = C2 (3). If K ∼ = L± 3 (3), then there is a hyperplane A1 ≤ A whose image in Aut(K) is that of a long root group, which centralizes an involution of K, violating Lemma 8.12. If K ∈ Chev(2) ∩ Chev(3), then by Lemma 8.10 and [IA , 2.2.10], K ∼ = U3 (3) or P Sp4 (3). Therefore we may assume that K ∈ C3 ∩ Chev(2) − Chev(3). If K ∼ = U5 (2), Sp6 (2), or F4 (2), then using [IA , 4.8.2a, 4.7.3A, 4.10.3a] we see that I3∗ (K) = I3 (K). A long root involution centralizes a parabolic subgroup of type SU3 (2), Sp4 (2), or Sp6 (2); these have 3-rank ≥ 2 so Lemma 8.12 1 is violated. If K ∼ = 2F4 (2 2 ) , then K has a single class of elements of order 3 [IA , 4.7.3A], with centralizers of even order, so Lemma 8.11 is contradicted as m3 (K) = 2. The same lemma is likewise contradicted if K ∼ = Sp4 (8), whether or not some element of A induces a nontrivial outer automorphism on K. Now as [KA, z] = 1, m3 (KA) = m3 (A), whence m3 (AutA (K)) ≥ m3 (K). If K ∼ = D4 (2), this implies that AutA (K) ≤ Inn(K). But I3∗ (K) = I3 (K) and K contains the direct product of three L2 (2)’s, so Lemma 8.12 is contradicted. Similarly, if K ∼ = U6 (2), then A acts diagonalizably on K and so a hyperplane of A centralizes an element of I2 (K), contradicting Lemma 8.11. The only remaining case in [V3 , 1.1] is K ∼ = 3D4 (2). If that occurs and AutA (K) ≤ Inn(K), then Lemma 8.11 is contradicted as every element of I3 (K) centralizes an involution of

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K [IA , 4.7.3A]. Thus AutA (K) Inn(K) = Aut(K), and so AK ∼ = CA (K) × Aut(K). We choose R ∈ Syl3 (LA ), so that R ∼ = E32 . Since [A, LA ] = [K, La ] = 1, we have [AK, R] = 1. But AK contains Z3 × Aut(L2 (8)), so it contains u × D with D ∼ = E32 and u ∈ I2 (K). Then u centralizes B ∗ = D × CA (K)R, of rank 2+m3 (A)−3+m3 (R) = m3 (A)+1 = m. Therefore B ∗ ∈ B∗,c (G). This contradicts Lemma 8.6a and completes the proof.  Lemma 8.14. K is terminal in G. Proof. Let R ∈ Syl2 (CG (K)) with z ∈ R ∩ La ∈ Syl2 (La ). Set RK = {t ∈ I2 (R) | K   CG (t)}; we must prove that RK = I2 (R). Obviously z ∈ RK . By Lemma 8.6c, [K, La ] = 1, and as La has one class of involutions, I2 (R ∩ La ) ⊆ RK . By L2 -balance and [VK , 3.5], if E ∈ E22 (CG (K)) and t ∈ I2 (R) with E # ⊆ g and [E, t] = 1, then t ∈ RK . ∪g∈K RK By Lemma 8.9, La is a component of CG (d) for any d ∈ I3∗ (K), and hence La is a component of CG (K) by Lemma 8.6c. By [IA , 5.3h], every involution of NR (La ) paragraph I2 (NR (La )) ⊆ RK . centralizes a four-subgroup of La , so by theprevious If R normalizes La we are done; otherwise, LR a ≤NR (La ) and every involution of R − NR (La ) surely centralizes a four-subgroup of LR a , so I2 (R − NR (La )) ⊆ RK by the previous paragraph again, This completes the proof of the lemma.  Now we can reach a final contradiction completing the proof of Proposition 7.1. By Lemmas 8.13 and 8.14, and [II3 , Theorem PU∗4 ], CG (z) has a component K ∼ = J3 , and choose = A6 that is standard in G. We have [K, La ] = 1 with La ∼ Q ∈ Syl2 (CG (K)) such that Q ∩ La ∈ Syl2 (La ). By [VK , 8.4], K is outer wellgenerated (for the prime 2), so by [IG , 18.11] there exists g ∈ G − NG (K) such that Qg ≤ NG (K). But |Qg | ≥ |La |2 > 25 = | Aut(K)|2 , so CQg (K) = 1, contradicting [IG , 18.8]. This completes the proof of Proposition 7.1. 9. Theorem 1: Standard 2-Components We complete the proof of Proposition 6.1 in this section. By Proposition 7.1, there exists a pair (B, A) satisfying the basic setup (6B) and such that a p-component LA of CG (A) is A-terminal in G, with the isomorphism type of LA = LA /Op (LA ) and the value of the prime p as in one of the pairs 5

(9A)

(1) (L2 (8), 3) or (2B2 (2 2 ), 5); (2) (if A = B) (L2 (p), p ≥ 5, p ∈ FM); or (3) (if |B : A| = p) (A6 , 3), (L± 3 (3), 3), (3G2 (3), 3), or 1 (2F4 (2 2 ) , 5).

We define B0 ∈ Ep (CG (A)) as follows:

(9B)

(1) If A = B, then B0 = B; 5 (2) If |B : A| = p and LA ∼ = L2 (8) or 2B2 (2 2 ), then B0 = A × b, where b ∈ Ip (CG (A)) induces a nontrivial field automorphism on LA ; 5 2 2 (3) If |B : A| = p and LA ∼ = L± 3 (3), 3G2 (3), or B2 (2 ), B0 = A × b, where b ∈ Ip (LA ) centralizes a 2-central involution z ∈ LA ; (4) If |B : A| = 3 and LA ∼ = A6 , then B0 = A.

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To see that such a subgroup B0 exists, note in (9B2) that Lemma 6.5 implies that Ip (CG (A)) contains an element b as asserted, and in (9B3) that LA contains elements z and b as asserted, by [VK , 12.4]. Note also in (9B2, 3) that CLA (B0 ) = p p CLA (b) has even order and B0 ∼ = B, so B0 ∈ B∗,c (G) = B∗,o (G). In these cases, as |B0 : A| = p, B0 and A satisfy (6B), and we therefore can replace B by B0 and assume that B = B0 . The following properties are then immediate in all cases (see Lemma 3.3):

(9C)

(1) B = B0 = A × b ∈ Bp∗,c (G) = Bp∗,o (G) unless possibly p = 3 and LA ∼ = A6 , in which case B0 = A < B; (2) B0 normalizes LA and La for all a ∈ A# ; (3) mp (B0 LA ) > m; and (4) B0 centralizes an involution z that is 2-central in LA .

Indeed, in cases (9B1, 4), B0 = A so (9C4) holds trivially. We fix z as in (9C4) and study the relationship between CG (z) and LA , eventually showing that LA is a (quasisimple) standard component of the centralizer of some involution of G, which leads fairly quickly to a final contradiction. We make one further stipulation about z. Suppose that LA ∼ = A6 but B ∈ B∗,c (G), i.e., CG (B) has even order, and LA  CG (A). In this case we fix an involution t ∈ CG (B). Then t normalizes LA , and so we may (9D)

choose z so that [t, z] = 1.

Lemma 9.1. mp (CG (z)) = mp (B0 ), i.e., B0 ∈ Ep∗ (CG (z)). Moreover, B0 normalizes all components of E(CG (z)). Proof. In the cases where B0 = B, the first assertion is obvious from (9C4) as mp (B) = m = m2,p (G). So assume that p = 3, LA ∼ = A6 , so that B0 = A. If m3 (CG (z)) > m3 (B0 ) = m − 1, then there is B1 ∈ E3∗ (CG (z)) with m3 (B1 ) = m. If possible choose B1 to contain A, but in any event choose B1 so that Q := B1 , A is a 3-group. If B1 ≥ A, then A and B1 both lie in S3 (Q) as [Q, z] = 1, so A ∩ B1 ≥ Ω1 (Z(Q)). Hence, in any event, there is a ∈ A ∩ B1# . Then B1 ≤ CG (a, z) so B1 normalizes La . Indeed for any involution y ∈ A6 , CAut(A6 ) (y) is a 2-group;  a . But this contradicts (6B3). Thus the first assertion holds, so B1 centralizes L and it implies the second by [IG , 8.7(iii)].  In view of (9C3), the next lemma is immediate from Lemma 3.5, with the role of D in that lemma being played by B0 . Lemma 9.2. The following conditions hold: (a) A centralizes every element of IG (B0 ; 2); and  (b) A centralizes Op (CG (z)) and acts faithfully on O p (E(CG (z))). This quickly implies: Lemma 9.3. O2 (CG (z)) embeds in O2 (CAutG (LA ) (z)). Proof. By Lemma 9.2, O2 (CG (z)) centralizes A and so acts on LA . As Op (LA ) has odd order, CAutG (LA ) (z) is covered by CNG (LA ) (z). So O2 (CG (z)) maps into O2 (CAutG (LA ) (z)), and it remains to show that W := CO2 (CG (z)) (LA ) = 1. Suppose false. Then Lemma 9.2 gives W ≤ C := CG (ALA ). By Lemma 6.4b, A ∈ Sp (C). As A ≤ Z(C) and p is odd, C has a normal p-complement, so

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W ≤ Op (C). In particular, Op (C) has even order. This implies, as C is B0 LA invariant, that m2,p (G) ≥ mp (B0 LA ) > m, which is a contradiction. The proof is complete.  Lemma 9.4. We have that NCG (z) (LA )(∞) centralizes LA . Proof. In all cases of (9A), CAut(LA ) (z) is solvable, whence the result.



Lemma 9.5. Let K be a component of CG (z). Then A∩K = 1 and Ip (CK (A)) ⊆ A. Finally, if A induces inner automorphisms on K, then A = (A ∩ K) × CA (K) and A ∩ K ∈ Sp (K). Proof. First we show that X := E(Op (CG (z))) = 1. Suppose that X = 1. Since [B0 , z] = 1, B0 permutes the components of X, and by [VK , 4.1], CX (B0 ) has even order. Now X centralizes A by Lemma 9.2, and z ∈ LA , so X ≤ NCG (z) (LA ). Therefore by Lemma 9.4, [X, LA ] = 1. Set F = XOp (LA ), so that [F, LA ] ≤ Op (LA ). Then let F0 = CX (B0 )Op (LA ); we have [F0 , B0 LA ] ≤ Op (LA ). Let R ∈ Ep∗ (B0 LA ), so that mp (R) > m. But R normalizes the p -group F0 Op (LA ) of even order, so m2,p (G) ≥ mp (R) > m, contradiction. Therefore, E(Op (CG (z))) = 1 and so p divides |K|. Suppose that there is d ∈ Ip (CK (A)) − A; we argue to a contradiction. By Lemma 9.1, A = B0 , so |B : A| = p and LA ∼  A6 . Let A∗ = A, d, so = ∗ that mp (A ) = m. Then (6B3) implies that d normalizes LA , and even that Lp (CLA (d)) = 1. In particular, d acts nontrivially on LA . By [VK , 4.8], there exists 1 = T ≤ O2 (CLA (z)) such that [T , d] = T . There is an A∗ z-invariant 2-group T ≤ CLA (z) mapping isomorphically on T . For any a ∈ A# , La = LA Op (La ) and so T ≤ Op (CLa (z)) ≤ Op (CCG (a) (z)). Thus, T ≤ Op (CCG (z) (a)), and so T ≤ ΔCG (z) (A). As p divides |K| it follows first that T normalizes K, then that T = [T, d] ≤ ΔK (A), and finally that K is not locally mp (A)-balanced with respect to p [IG , 20.6]. Since mp (A) ≥ 3, K/Z(K) ∼ = An for some n > p3 by [IA , 7.7.5], which is absurd as K is a C2 -group. Thus, Ip (CK (A)) ⊆ A ∩ K. As Op (K) ≤ O2 (K) = 1, this immediately implies the other two assertions of the lemma.  Now we choose an important elementary abelian 2-subgroup D of LA . Definition 9.6. D is an elementary abelian 2-subgroup of LA chosen to contain z and to satisfy the following further conditions: 1  3G2 (3) or 2F4 (2 2 ) , then D is a maximal elementary abelian (a) If LA ∼ = subgroup of LA ; (b) If LA ∼ = 3G2 (3), then |D| = 23 and AutLA (D) = Aut(D); and 1 (c) If LA ∼ = 2F4 (2 2 ) , then |D| = 22 and D is normal in a parabolic subgroup 1 of Aut(LA ) ∼ = 2F4 (2 2 ) of type A1 (2). By [VK , 4.9], such a D exists, and we fix one. In case LA ∼ = A6 or L2 (p), p > 5, we also choose a second D∗ with the same properties, such that D, D∗  ∼ = D8 ; any results proved for D will also hold for D∗ . Set r = m2 (D). The pairs (LA , r) 5 are then (L2 (8), 3), (2B2 (2 2 ), 5), (L2 (p), 2), (A6 , 2), (L± 3 (3), 2), (3G2 (3), 3), and 1 (2F4 (2 2 ) , 2). Moreover, in all the cases there exists y ∈ NLA (D) of prime power

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∼ order (relatively prime to p unless LA ∼ = A6 or L± 3 (3)) such that y / y∩Op (LA ) = Z2r −1 permutes the involutions of D regularly. We set ND = D y . Using D and y we can prove the following basic result connecting 2-local and p-local structure (cf. Lemma 8.7): Lemma 9.7. Let K be a component of CG (z). Let a ∈ A# and let Ka be a component of CK (a) of order divisible by p. Then Ka is a component of CG (a), and [Ka , LA ] = 1. a = Ca /Op (Ca ). Let Proof. As usual, we write Ca = CG (a) and C Ta = [Op p (Ca ) ∩ CG (z), Ka ]. Since z ∈ LA ≤ Lp (Ca ) by Lp -balance, with Op (Ca ) of odd order, z centralizes a ) and so Ta = [Op (C a ), K  a ]. Moreover Ta = [Ta , Ka ] by [IG , 4.3(i)]. On the Op (C other hand, Ka ≤ K  E(CG (z))  CG (z), and Ta ≤ CG (z), whence Ta ≤ K. Thus Ta ≤ CK (a). As Ka is a component of CK (a), Ta = [Ta , Ka ] ≤ Op p (Ca ) ∩ Ka ≤ O{2,p} p (Ka ) ≤ O2 (Z(Ka )). Therefore Ta = 1. a ), K  a ] = 1. But Op (C a ) = F ∗ (O2 (C a )), and K  a is genIn particular [Op (C   erated by 2-elements, so by [IG , 3.17(iii)], [O2 (Ca ), Ka ] = 1. By L2 -balance, the  a of C  (     component K Ca z ) lies in L2 (Ca ), so Ka ≤ E(Ca ).  a is a component of C  (  A , and L  A is a component of C a , Now K z ) with z ∈ L Ca

 a is a component of C a distinct from L A . z ) solvable. Therefore K with CAut(LA ) ( To complete the proof it remains to show that Ka is a component of Ca . Returning to Ta above, we see that as Ta = 1,

(9E)

Ka centralizes Op (Ca ) ∩ CG (z).

 ≤ L  A , [D, Ka ] is a p -group. But D centralizes z and a, and Ka is Since D a component of CG (a, z) of order divisible by p, so [D, Ka ] = 1 and Ka is a  A , [y, Ka ] is a p -group; as component of CG (a D). In a similar way since y ∈ L y centralizes a and normalizes D, [y, Ka ] = 1. Now conjugating (9E) by y, we get [Op (Ca ) ∩ CG (d), Ka ] = 1 for all d ∈ D# . As Op (Ca ) has odd order and D is noncyclic, [Op (Ca ), Ka ] = 1. Therefore Ka is a component of Ca and the lemma is proved.  We also need an analogue of Lemma 9.7 for solvable components. For the purposes of the next lemma, we make the following definition. Definition 9.8. A complete (S)U3 (2)-component of a group H is a subgroup L   H such that O 3 (L) ∼ = SU3 (2) or U3 (2), |L : O 3 (L)| = 3, and ∼ L/O3 (L) = SL2 (3). Furthermore,  3 Lcs 3 (H) = O (L) | L is a normal complete (S)U3 (2)-component of H . For example, L ∼ = GU3 (2) or P GU3 (2) satisfy these conditions, as does C3D4 (2) (x), 3 s where x ∈ I3 ( D4 (2)) is 3-central. Note that Lcs 3 (H)  L3 (H) (see [IG , Def. 13.6]). Lemma 9.9. Suppose that p = 3. Let K be a normal component of CG (z). Let a ∈ A# and let Ka be a normal complete (S)U3 (2)-component of CK (a). Then O3 (Ka ) ≤ O3 (CG (a)), and [O 3 (Ka ), La ] = 1.

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Proof. We have Ka   CG (z, a). Therefore, Ka ≤ O{2,3} (CG (z, a)), and Ka normalizes the 3-component La of CG (a) as it centralizes z. Let N =  = N/C(a, La ). Then K  a embeds in C z ). Let R = NCG (a) (La ) and N  a ) ( Aut(L   O3 (Ka ). If R = 1, then N contains U3 (2) or SU3 (2). But C z ) does not  ( Aut(La )

 = 1. Thus, [R,  L  a ] = 1. involve U3 (2), by [VK , 10.61]. This is a contradiction, so R Moreover as R ≤ O3 (CG (z, a)), (9F)

[R, CO3 (CG (a)) (z)] ≤ [O3 (CG (z, a)), O3 (CG (z, a))] = 1.

 L  a ] = 1 so R centralizes a 2-subgroup D1 ≤ La and element  N D ] ≤ [R, Now [R,  1 = D,  and D  1  D . Conjugating (9F) y1 ∈ NLa (D1 ) such that z ∈ D1 , D y1  = N by powers of y1 , we get that R centralizes ΓD1 ,1 (O3 (CG (a))) = O3 (CG (a)). Hence [R, La ] = 1 and in particular [R, LA ] = 1.  ≤ O3 (C  ( Next, as R ≤ O3 (CG (a, z)) and O3 (CG (a)) has odd order, R Ca z )).       z ))∩ Since z ∈ La and [R, La ] = 1, it follows that [R, E(Ca )] = 1. Then R ≤ O3 (C  ( Ca

a )) ≤ O3 (C  (E(C a ))) ≤ O3 (C a ). As [R, O3 (Ca )] = 1, R ≤ O3 (Ca ), CCa (E(C Ca proving the first assertion of the lemma. Now define Ca = Ca /O3 (Ca ). Let S ∈ Syl2 (Ka ), so that SL2 (3) ∼ = Ka   CC (z), and S = O2 (Ka ). Hence Ka is a solvable component of CC (z). By solvable a

a

L2 -balance [IG , 13.8], either S ≤ La or [S, La ] ≤ O2 (La ). In the first case, Ka a ∼ embeds in CAut(La ) ( z ). The only possibilities, as p = 3, are L = L± 3 (3) or 3G2 (3),   =  with Z(S) z . Since S ≤ La this forces z = Z(S), which is a contradiction as [z, R] = 1 = [Z(S), R]. Therefore, [S, La ] ≤ O2 (La ), so [S, La ] ≤ O2 (La ) O{2,3} (La ). Now [S, CO3 (La ) (z)] ≤ [O{2,3} (CG (a, z)), O{2,3} (CG (a, z))] = An argument like that in the first paragraph, using ND , yields [S, O3 (La )] = Hence [S, La ] = 1, completing the proof.

= 1. 1. 

Now we can analyze the layer of CG (z), which we write as the product of its components: E(CG (z)) = K1 · · · Kk , where k ≥ 1 by Lemma 9.2. When k = 1 we may write K1 = K. Note that by Lemma 9.5, A ∩ Ki = 1 for each i; in particular p divides each |Ki |. Lemma 9.10. The following conditions hold: (a) If k = 1, then one of the following holds: (1) K ∈ Chev(2), mp (K) ≥ 2, and mp (AutG (K)) ≥ mp (A) ≥ 3; moreover, if p does not divide | Out(K)|, then A ∈ Ep∗ (K); (2) p = 3, A ∈ E3∗ (K), and K/Z(K) ∼ = L± 4 (3), G2 (3), Co2 , Co1 , Suz,  F i22 , F i23 , F i24 , F3 , F2 , or F1 ; (3) p = 3 = |B : A|, A ∈ E3∗ (K), and K ∼ = J3 ; (4) p = 5, A ∈ E5∗ (K), and K ∼ = F1 ; or (5) p = 5 = |B : A|, A ∈ E5∗ (K), and K/Z(K) ∼ = Co1 or F2 ; and (b) If k ≥ 2, then for any i = j between 1 and k, and any 1 = a ∈ A ∩ Kj , Ki is a component of CG (a). Moreover, for each i, Ki is simple and Ki ∈ C2 ∩ Cp . Proof. If k = 1, then K = F ∗ (KA) by Lemma 9.2. Then as mp (A) ≥ m − 1 ≥ 3, (a) follows by [VK , 12.46a] (if |B : A| = p), [VK , 12.47a] (if A = B),

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and Lemma 9.5, except for the assertions in (a) that A ∈ Ep∗ (K). In proving this we p have that Out(K) is a p -group. Now if A = B or LA ∼ = A6 , then A ∈ E∗ (CG (z)) as mp (B) = m and by Lemma 9.1, implying the desired result. Otherwise, note that by Lemma 9.5, A ∈ Sp (K). Now as [B, z] = 1 and k = 1, B normalizes K. Hence CB (K) = 1. But then if A ∈ Ep∗ (K) and we choose A∗ ∈ Ep∗ (K), we have A∗ × CB (K) ≤ CG (z), so m2,p (G) ≥ mp (A∗ ) + 1 > mp (A) + 1 = m, a contradiction. This completes the proof of (a). Suppose finally that k ≥ 2. Let 1 ≤ i ≤ k and choose j in the same range with i = j. By Lemma 9.5 we may choose a ∈ A# ∩ Kj . Then by Lemma 9.7, Ki is a component not only of CG (z) but also of CG (a). Therefore O2 (Ki ) ≤ Op (Ki ) = 1. By our hypotheses (1A), Ki ∈ C2 , so Ki is simple, and Ki ∈ Cp . The proof is complete.  Now we can prove Lemma 9.11. The following conditions hold: (a) E(CG (z)) = E(CG (d)) for all d ∈ D# ; (b) [E(CG (z)), ND ] = 1; and (c) Under any of the following conditions, [E(CG (z)), LA ] = 1 and all Ki are simple: (1) k ≥ 2; (2) k = 1 and K ∈ Chev(2); or 1 2 ∼ 2  (3) LA ∼ = A 6 , L± 3 (3), 3G2 (3), or F4 (2 ) (and |B : A| = p), or LA = L2 (p), p > 5, p ∈ FM (and A = B). Proof. We first prove that if (c1) or (c2) holds, then all assertions of the lemma hold. Indeed under these conditions, for any i = 1, . . . , k we have either    (1) Ki = O p (E(CKi (a))) | a ∈ A# ; or    (9G) # , with O 3 (E(CKi (a)))Lsc (2) Ki = 3 (CKi (a)) | a ∈ A p = 3. For if k ≥ 2, a single a ∈ Kj , j = i, proves (9G1); while if k = 1, then K ∈ Chev(2) by assumption, and [VK , 12.46b] and [VK , 12.47b] yield our claim.  But for any a ∈ A# , O p (E(CKi (a))) and (for p = 3) Lsc 3 (CKi (a)) centralize LA by Lemmas 9.7 and 9.9, so [Ki , LA ] = 1. In particular, it follows that E(CG (z)) is B0 LA -invariant. But mp (B0 LA ) > mp (B) = m2,p (G) and so O2 (E(CG (z))) = 1, whence Ki is simple, as claimed in (c). Moreover, since ND ≤ LA and ND is transitive on the involutions of D, the conclusions (a) and (b) follow if (c1) or (c2) holds. It remains to consider the case k = 1 and K ∈ Chev(2), whence K is as given in Lemma 9.10a, and in particular A ∈ Ep∗ (K). Moreover by [VK , 12.46c] and [VK , 12.47c], (9H)

CAut(K) (A) is a p-group.

On the other hand, every d ∈ D# centralizes A and z, so normalizes K. It follows from (9H) that [D, K] = 1. Then by L2 -balance, K ≤ E(CG (d)), so equality holds as z and d are conjugate, proving (a). It follows that ND and ΓD,1 (G) normalize K, and if LA ∼ = A6 or L2 (p), p > 5, then ΓD,D∗ ,1 (G) normalizes K.

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∼ L2 (8), 2B2 (2 52 ), or A5 , then ND is a p -group, and as [ND , A] = 1, (9H) If LA = implies that [ND , K] = 1. Thus in these cases (b) holds. Hence for the rest of the proof we may assume that (c3) holds (and still (c1) 1 2 2  and (c2) fail). We have LA = ΓD,1 (LA ) if LA ∼ = L± 3 (3), 3G2 (3), or F4 (2 ) , and LA = ΓD,D∗ ,1 (LA ) if LA ∼ = A6 or L2 (p), p > 5, by [IA , 7.3.4, 7.6.1] and [VK , 8.7]. Thus LA normalizes K, and since z ∈ CLA (K), [K, LA ] = 1. Finally, mp (KLA ) = m + 1 > m2,p (G) so O2 (K) = 1, whence Z(K) = 1. Thus (c) holds, which implies (b) in these cases. The proof is complete.  Lemma 9.12. For every 1 = C ≤ A, LC = LA is simple. Proof. Let W = Op p (LC ), so that W has odd order. By Lemma 9.11a, E(CG (z)) is normalized by ΓD,1 (W ) = W . Then for any d ∈ D# , E(CG (z)) = E(CG (d)) is centralized by [W, d]. Also O2 (CG (d)) centralizes C and d and therefore normalizes LC , so [O2 (CG (d)), CW (d)] ≤ O2 (CG (d)) ∩ W = 1. Therefore [CW (d), z] centralizes O2 (CG (d))E(CG (d)) = F ∗ (CG (d)). Hence [CW (d), z] = 1. As d is arbitrary, it follows that [W, z] = 1. This implies that LC is quasisimple, and then LC = W LA = LA as LA is A-terminal. Finally, if LA is not simple, then as |W | is odd, the only possibility by [IA , 6.1.4] is that LA ∼ = 3G2 (3). But then with Lemma 9.11c, [F ∗ (CG (d)), CW (d)] = 1, so CW (d) = 1 and then W = 1, contradiction.  In particular, Lemmas 9.12 and 9.11 immediately imply: Lemma 9.13. LA ∼ = 3G2 (3). The next result is preliminary to considering the terminality of components of CG (z). Lemma 9.14. Let t ∈ I2 (NG (LA )). Let K ∗ be a nontrivial product of components of CG (z) centralized by t. Then there is g ∈ LA such that (a) [K ∗ , g] = 1; and (b) Either [t, Dg ] = 1 or LA t ∼ = P GL2 (9) or P GL2 (p). Proof. If t centralizes LA , the result holds with g = 1. So suppose that [LA , t] = 1. 5 First assume that LA ∼ = L2 (8) or 2B2 (2 2 ). Then D = Ω1 (T ) = Z(T ) for some T ∈ Syl2 (LA ), T ∈ Syl2 (Aut(LA )), and ND is a maximal subgroup of LA . If t normalizes D, it follows that t centralizes D, and the conclusion holds with g = 1. In any case [t, Dg ] = 1 for some g ∈ LA , by Sylow’s theorem. Suppose that t does not normalize D. Then CLA (K ∗ ) ≥ D, Dt  = LA , so g ∈ CLA (K ∗ ), as desired. If LA ∼ = L2 (5), then either conclusion (b) holds or t induces an inner automorphism on LA . In the latter case, the proof may be completed as in the previous paragraph. In the remaining cases, by Lemma 9.11c, [E(CG (z)), LA ] = 1 so it suffices to show that some LA -conjugate of t centralizes D. If LA ∼ = A 6 , L± 3 (3), or L2 (p), this holds, with the exception of the P GL2 (9), P GL2 (p) configurations in (b), by [VK , 1 4.10]. If LA ∼ = 2F4 (2 2 ) , it holds by the Thompson transfer lemma and the fact 1 1 [IA , 3.3.2d] that every involution of Aut(2F4 (2 2 ) ) ∼ = 2F4 (2 2 ) is inner. In view of Lemma 9.13, this completes the proof.  We are now ready to consider the terminality of the components Ki of CG (z). Our objective for the moment is to show that k = 1 and K1 is standard in G. By

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[II3 , Theorem PU∗4 ], for standardness it suffices to show that K1 is terminal in G. We first prove Lemma 9.15. Let 1 ≤ i ≤ k. Then (z, Ki ) has no diagonal pumpups in G. Proof. Suppose false, and let t ∈ I2 (CG (Ki z)) be such that the subnormal closure of Ki in CG (t) is the product L1 L2 , where L1 and L2 are components of CG (t) interchanged by z, and Ki = E(CL1 L2 (z)). Take any a ∈ A ∩ Ki# . Then there are aj ∈ Ip (Lj ), j = 1, 2, such that a ∈ a1 , a2  ∼ = Ep2 , az1 = a2 and Ca1 ,a2  (z) = a. Set a0 = a−1 1 a2 , so that z inverts a0 . Now LA = La   CG (a) by Lemma 9.12. As z ∈ LA inverts a0 ∈ CG (a), a0 ∈ LA . Note that [t, a, a0 ] = 1, so t normalizes LA and maps into CAut(LA ) (a0 ). From [VK , 4.11], either t centralizes LA or one of the following holds: (1) LA ∼ = A6 , and the image of t in Aut(LA ) lies in the Σ6 subgroup of Aut(LA ); (2) LA ∼ = L± 3 (3), and t induces a non-trivial automorphism (9I) on LA ; or 1 (3) LA ∼ = 2F4 (2 2 ) , and t induces an inner automorphism on LA corresponding to a 2-central involution. By Lemma 9.14, applied with K ∗ = Ki , we may in each case choose g ∈ CLA (Ki ) such that [t, Dg ] = 1. By Lemma 9.11, Ki is a component of CG (d) for all d ∈ Dg . But as Dg is noncyclic, Lemma 6.19 of [IG ] implies that there exists an involution d ∈ Dg centralizing L1 L2 , so Ki   L1 L2 , which is absurd. The lemma follows.  The cases (9J)

k = 1, p = 3, K ∼ = 3D4 (2) or L3 (q), q ≡  = ±1 (mod 3), L3 (CK (a)) = 1 for all a ∈ A ∩ K # ,

are badly behaved and will require a change of (B, A) to a new pair still satisfying the basic setup (6B). ∗ ∗ Lemma 9.16. Suppose that (9J) holds. Then there exists a pair (B , A ) sat∗ # isfying (6B) such that LA∗ = LA and K = L3 (CK (a)) | a ∈ A ∩ K .

Proof. As k = 1, CA (K) = 1, so m3 (A) ≤ m3 (Aut(K)) ≤ 3 in these cases. Thus m3 (A) = 3 = |B : A| and some a0 ∈ A# induces a graph automorphism on K (in the 3D4 (2) case) or a field automorphism (in the other cases). 3 By [V  theorem, there exists A0 ∈ E2 (CK (a0 )) such that K , 8.34a] and Sylow’s # K = L3 (CK (a)) | a ∈ A0 and A, A0  is a 3-group. We set A∗ = A0 a. Thus [LA , A∗ ] = 1 by Lemma 9.11c. As A, A0  is a 3-group, it follows by L3 -balance that LA ≤ L3 (CG (A∗ )). Consider the structure of C := CG (A). We have LA   C, B ≤ C, and if B ∈ B∗,c (G), then CC (B) has even order. Since m3 (B) = m2,3 (G), it follows, with [IG , 8.7] and by the structure of LA , that B normalizes LA . By (6B), we may write B = A × b with [LA , b] = 1. Moreover, if B normalizes K then we may suppose that b = CB (K). Suppose first that LA ∼ = L2 (8). Then by our choice of B and z (see (9B) and (9C)), b induces a field automorphism on LA and centralizes z. Then b normalizes

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K so [A∗ , b] ≤ [K, b][a0 , b] = 1. We set B ∗ = A∗ b in this case, so that [B ∗ , z] = 1 and B ∗ ∈ B∗,c (G), as required. Now suppose that LA ∼ = L2 (8), so that LA ∼ = L± 3 (3) or A6 . As m = 4,  LA  C; since Out(LA ) is a 3 -group and m2,3 (ALA ) = 3 + m2,3 (LA ) = 4, in fact A ∈ E3∗ (CC (LA )). Therefore B = A × (B ∩ LA ). In particular as [K, LA ] = 1 by Lemma 9.11, A∗ = (A∗ ∩K) a0  = (A∗ ∩K)(A∗ ∩A) centralizes LA . If LA ∼ = L± 3 (3), ∗ ∗ ∗ we take B = A b where b ∈ Syl3 (CLA (z)), and have B ∈ B∗,c (G), as required. Assume then that LA ∼ = A6 . We take B ∗ = A∗ (B ∩ LA ). Since B ∩ LA normalizes a ∗ four-group in LA , B ∈ B∗ (G). It remains to show that if B ∈ B∗,c (G), then B ∗ ∈ B∗,c (G). But in this case there exists an involution t ∈ CC (B). Also LA  CG (A), so we have chosen t and z so that [t, z] = 1 (9D). Hence, t normalizes K, centralizing A. By [VK , 8.34b], and because of the isomorphism type of K, t centralizes a Sylow 3-subgroup of K invariant under A, and we could have chosen A∗ to centralize t,  whence t centralizes A∗ (B ∩ LA ) = B ∗ . The proof is complete. We assume through Lemma 9.23 below that (9K)

1 ≤ i ≤ k and Ki is not terminal in G.

Thus i is fixed. We expand D ≤ Q0 ≤ Q ≤ R, where Q0 ∈ Syl2 (NLA (D) ∩ CG (z)), Q ∈ Syl2 (CG (Ki z)), and R ∈ Syl2 (CG (Ki )). The expansion Q0 ≤ Q is possible because by Lemma 9.11bc, either LA ≤ CG (Ki ) or 5 k = i = 1 with LA ∼ = L2 (5), L2 (8), or 2B2 (2 2 ), in which cases [NLA (D), Ki ] = 1. Note also that Q = CR (z). Recall the notion of semirigidity [IG , 7.1, 7.2]. Using [IG , 7.4], we obtain Lemma 9.17. Assume (9K). If (z, Ki ) is semirigid in G, then Q is dihedral or semidihedral.  Proof. Otherwise, by [IG , 7.4], there is a nontrivial pumpup (z, Ki ) < ( z , K) for some involution z ∈ Z(Q). Then by semirigidity, Q = z × Q∗ for some subgroup Q∗ such that Ki is not a component of CK (u) for any involution u ∈ Q∗ . But as |Q : Q∗ | = 2, we have D ∩ Q∗ = 1, and Ki is a component of CG (u) for any  u ∈ I2 (D). This is a contradiction, and the lemma follows. The following configuration is the most difficult to eliminate, and we shall first reduce to it. (9L) p = 5, Ki ∼ = A5 , and A = B. Lemma 9.18. Assume that (9K) holds but (9L) fails. If k ≥ 2, then there exists (v, J) ∈ ILo2 (G) such that v ∈ R, Ki < J, and one of the following holds: (a) (p, Ki ) = (3, U3 (3)), J/Z(J) ∼ = U4 (3), and m2 (CAut(J) (Ki )) = 1; (b) (p, Ki ) = (3, A6 ), J ∼ = L4 (3), and m2 (CAut(J) (Ki )) = 2; or (c) (p, Ki ) = (5, A5 ), J/Z(J) ∼ = M12 or J2 , m2 (CAut(J) (Ki )) = 2, |B : A| = 1 5 2  2 ∼ 2 5, LA = F4 (2 ) or B2 (2 2 ), and m2 (LA ) > 2. Proof. Since k ≥ 2, CG (Ki ) ≥ Kj × LA for any j = i, by Lemma 9.11c. In particular as [Kj , D] = 1, Q is not dihedral or semidihedral, so by Lemma 9.17, (z, Ki ) is not semirigid in G. By Lemmas 9.10 and 9.11c (for the simplicity of Ki ∈ Cp ), and as G is of even type, it follows from [VK , 14.5] that (p, Ki ), J, and

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132

4. THEOREM C5 : STAGE 2

m2 (CAut(J) (Ki )) are as asserted. In conclusion (c), the fact that (9L) fails implies that |B : A| = 5. Then the rest of (c) follows from (9A) and [IA , 3.3.3]. The proof is complete.  Still assuming that (9K) holds but (9L) fails, choose (v, J) as in Lemma 9.18, if possible with v ∈ Q. For any a ∈ A∩Ki# , La = LA , and LA is one of the possibilities listed in Proposition 7.1, but not 3G2 (3) (Lemma 9.13). Then v ∈ CG (Ki ) so v ∈ CG (a). We next show that v cannot normalize LA = La . Lemma 9.19. Assume that (9K) holds but (9L) fails, and that k ≥ 2. Choose (v, J) as above. Then LvA = LA . Proof. Suppose first that there exists g ∈ CLA (Ki ) such that [v, Dg ] = 1. If m2 (D) > m2 (CAut(J) (Ki )), then J is contained in a component of CG (d) for some involution d ∈ Dg . This is an obvious contradiction as Ki is a component of CG (d) by Lemma 9.11a. Therefore m2 (D) ≤ m2 (CAut(J) (Ki )) ≤ 2, whence m2 (D) = 2, and either (b) or (c) of Lemma 9.18 holds. Accordingly, −1 1 g −1 ∼ 2 2  LA ∼ and J1 = J g . Thus = A6 or L± 3 (3), or LA = F4 (2 ) . Let v1 = v [v1 , D] = 1, [v1 , Ki ] = 1, and Ki ≤ J1 , inasmuch as v and g centralize Ki . Since  mp (Ki ) ≤ 2, and CA (O p (E(CG (z)))) = 1, there is a nontrivial product M of components of E(CG (z)) other than Ki , and an involution u ∈ CM (v1 ). Note that Z(M ) = 1 and [M, LA ] = 1 by Lemma 9.11c. Set E = D u ∼ = E23 . Then  [E, v1 , z] = 1. Also E ≤ CCG (v1 ) (Ki ) so CE (J1 ) = 1. Choose v ∈  CE (J1 ) and 

let J1∗ be the subnormal closure of Ki in CG (v  ). Then J1 = KiJ1 ≤ J1∗ , J1 is a component of CJ1∗ (v1 ), and Ki is a component of CJ1∗ (z). By [VK , 3.67, 3.18], J1∗ = J1 . If p = 5, then we may assume by an LA -conjugation that v  ∈ z, u, as u ∈ M centralizes LA . Let S ∈ Syl2 (LA ) with D  S and Z(S) = z. Then S centralizes v  and Ki , so S maps to CAut(J1 ) (Ki ) ∼ = E22 . As z ∈ [S, S], we have [J1 , z] = 1, a contradiction. Therefore p = 3 and so J1 ∼ = J ∼ = L4 (3). Now D # embeds in CAut(J1 ) (Ki ) so by [VK , 3.5], there is d ∈ D such that Ki is not a component of CJ1 (d), contradicting Lemma 9.11. Therefore no g exists as in the previous paragraph. By Lemma 9.14, p = 3, LA ∼ = A6 , and J ∼ = L4 (3), with LA v ∼ = P GL2 (9). Assume next that m2 (CM (v)) ≥ 2, where M = CE(CG (z)) (Ki ). Then by [VK , 3.5], there is x ∈ I2 (E(CG (z))) such that [v, x] = 1 and Ki < I for some component I of CJ (x). We see from [IA , 4.5.1] that I ∼ = J or P Sp4 (3), and in any case CAut(I) (Ki ) is solvable. Since [x, LA ] = 1, it follows that [I, LA ] = 1, a contradiction as z ∈ LA . Therefore m2 (CM (v)) = 1, whence k = 2. We have m2 (CKj (v)) = 1, where we have put j = 3−i. Since Kj ∈ C2 ∩C3 , it follows from [VK , 11.13] that Kj ∼ = A6 and Kj v ∼ = P GL2 (9). In particular, as CA (Ki Kj ) = 1 by Lemma 9.2, m ≤ m3 (K1 ) + m3 (K2 ) + 1 = 5. On the other hand, consider CG (v), with a component J ∼ = L4 (3). Also W := O2 (CKj (v)) has order 5 and lies in O2 (CG (z, v)). Therefore by [IA , 20.6], W acts faithfully on a locally unbalanced component Iv ∈ C2 of CG (v). It follows by [VK , 7.14] that Iv ∼ = A6 . Hence m ≥ m3 (CG (v)) ≥ m3 (JIv ) = 6, a contradiction. This completes the proof of the lemma.  Lemma 9.20. Assume that (9K) holds but (9L) fails. Then k = 1.

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Proof. Suppose that k ≥ 2 and choose v and a as specified before Lemma 9.19, so that by that lemma, LvA is a component of CG (a) distinct from LA . As z ∈ LA , [v, z] = 1, so v ∈ CR (z) = Q. We set H = L2 (CLA LvA (v)), a diagonal of LA LvA , with H ≤ LA , v ≤ CG (Ki ). Let U = DDv and D1 = CU (v) ∼ = D. Then U # normalizes LA . By our choice of v, for any u ∈ D1 , the pumpup of Ki in CG (u) is not isomorphic to L4 (3) or a covering group of U4 (3) (if p = 3) or to M12 or J2 (if p = 5). Thus by [VK , 3.67, 3.18], [J, u] = 1 for any u ∈ D1# . Therefore D1 embeds in CAut(J) (Ki ), so m2 (D) = m2 (D1 ) = m2 (CAut(J) (Ki )) = 2. Hence m2 (LA ) = 2, whence by Lemma 9.18, J ∼ = L4 (3) and p = 3. Again by [VK , 3.5], Ki < I for some component I of CJ (u), for some u ∈ D1# . Then as [D, u] ≤ [D, D1 ] = 1, the  action of D on I contradicts [VK , 3.5]. The lemma follows. We write K = K1 . Lemma 9.21. Assume that (9K) holds but (9L) fails. Then the following conditions hold: (a) Q = R is dihedral or semidihedral; (b) LA ∼ = A6 or L3 (3) (if p = 3), or L2 (p) (if p ≥ 5); and (c) K is simple. Proof. By Lemma 9.2, CA (K) = 1, so as mp (A) ≥ 3, it follows immediately by [VK , 14.4] that (z, K) is semirigid in G. By Lemma 9.17, Q is dihedral or semidihedral. As D ≤ Q, m2 (D) = 2, which implies that LA ∼ = A6 or L± 3 (3) 1 2  (p = 3), F4 (2 2 ) (p = 5), or L2 (p) (p ≥ 5). In particular, unless LA ∼ = L2 (5), [K, LA ] = 1 and K is simple by Lemma 9.11c, and so Q contains a Sylow 2subgroup of LA , as z was chosen 2-central in LA . Thus (b) holds. Moreover, (c) holds unless possibly p = 5 and LA ∼ = A5 . But then A = B so m5 (Aut(K)) ≥ 4. From the definition of C2 and [IA , 5.6.1, 4.10.3] this yields K ∈ Chev(2) or K ∼ = F1 , so K is simple, thanks to Lemma 9.11c. It remains to show that Q = R. If |Q| > 4, then Z(Q) = z = Z(R) so Q = CR (z) = R. If Q is a four-group but Q = R, then Q = D = Z(R) × z and −1 LA ∼ = A5 . Hence Z(R) = z g  for some g ∈ ND ≤ NG (K). Then Rg ≤ C(z, K),  so |Q| = |C(z, K)|2 ≥ |R| > |Q|, contradiction. The proof is complete. Now we can attain our objective. Lemma 9.22. Suppose Ki is not terminal in G for some i, 1 ≤ i ≤ k. Then Ki ∼ = A5 , p = 5, and A = B. Proof. Suppose false and continue the above argument. As Q = R, there  ∈ ILo2 (G). By Lemma exists z ∈ I2 (Q) and a nontrivial pumpup (z, K) < ( z , K) 9.15, it is a vertical pumpup. Now CA (K) = 1 and mp (CG (LA )) = mp (A) ≥ m−1 ≥ 3. Choose any a ∈ A∩K # . By Lemmas 9.7 and 9.9, every component of CK (a) is a component of CG (a) and hence of CK (a); and for every normal complete (S)U3 (2)component Ka of CK (a), O3 (Ka ) ≤ O3 (CK  (a)). Now [VK , 12.48] yields that p = 3 and one of the following holds:

(9M)

(1) (9J) holds;   ∼ (2) K ∼ K) = U4 (2) and K/Z( = L± 4 (3); or ∗ and there is A ≤ K and 1 = a∗ ∈ A∗ , such (3) K ∼ F i = 23 ∗  is a 3-group, and CK (a∗ ) has A ∩ K, A, A that A∗ ∼ = a component L which is not a component of CK (a∗ ).

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If (9J) holds, we choose A∗ and B ∗ as in Lemma 9.16. Then L3 (CG (A∗ )) has a 3-component LA∗ containing z. The previous development applies to A∗ , and as L3 (CK (a)) = 1 for some a ∈ (A∗ )# , it yields a contradiction.   ∼  = 1 by K) z , K)) Suppose that K ∼ = L± = U4 (2) and K/Z( 4 (3). Then m2 (C( [V2 , 9.1]. Note that as A ≤ K, z ∈ CG (A) so z normalizes LA and CLA (z).  = 1 as |C z ) contains a four-group V , then CV (K) If CLA (  (K)| = 2. But Aut(K) LA has one class of involutions, so K pumps up in CG (z), which is not the case. Thus m2 (CLA ( z )) = 1. Therefore, LA ∼ z ∼ = A6 and LA  = P GL2 (9), so that z )) = 1 and z inverts W . Then W maps into CAut(K) W := O5 (CLA (  (K) so  = 1. But C(  ≤ CG (A) normalizes LA , so W = O5 (CL (  [W, K] z , K) z ))  C( z , K). A z )) = 1, a contradiction. It follows that W ≤ O2 (CG ( Thus K ∼ = F i23 . Hence NG (K) = K × CG (K) ≤ CG (LA )CG (A) ≤ NG (LA ) ≤ LA CG (z) ≤ NG (K). Hence equality holds and B normalizes K. We construct B ∗ ≥ A∗ such that (B ∗ , A∗ ) satisfies the basic setup (6B). If LA ∼ = L3 (3), then B = A×b with [b, z] = 1 (see (9C)) and we take B ∗ = A∗ ×b ∈ B∗,c (G). Suppose then that LA ∼ = A6 . Given the structure of NG (K), we again have B = A × b with b ∈ CG (K), and take B ∗ = A∗ × b. By [IA , 5.6.2], m2,3 (K) < 6, so   W := O 2 (CG (B)) = O 2 (CCG (K) (b)). Therefore if B ∈ B∗,c (G), then W = 1 and so B ∗ ∈ B∗,c (G). In any event, B ∗ normalizes LA , so B ∗ normalizes a foursubgroup of LA , and B ∗ ∈ B∗ (G). Also by L3 -balance, LA ≤ L3 (CG (A∗ )), so (6B) is satisfied by (B ∗ , A∗ ). Therefore the development to this point holds for A∗ as well as A, and the existence of L in (9M3) yields a contradiction. The proof is complete.  The next lemma gives more detail about non-terminal components. Lemma 9.23. Suppose that 1 ≤ i ≤ k and Ki is not terminal in G, so that p = 5, A = B, and Ki ∼ = A5 . Then the following conditions hold: (a) For some t ∈ I2 (CG (Ki )), Ki ≤ J where J/Z(J) ∼ = M12 or J2 , and J is a component of CG (t); and  (b) Set C = CG (Ki ), C = C/Op (C), and F = j =i Kj × LA , a direct product. Then F ∗ (C) = F and Op (C) has odd order. Proof. We have k ≥ 2 since CA (E(CG (z))) = 1. Thus by Lemma 9.17, (z, Ki ) is not semirigid in G, and as p = 5, (a) follows by [VK , 14.5]. For (b), note that as k ≥ 2, F ≤ C, with the help of Lemma 9.11c. Fix b ∈ A# ∩ Ki . Thus C ≤ CG (b). By Lemma 9.7, each Kj , j = i, is a simple component of CG (b), and is A-invariant by Lemma 9.5. Moreover LA = Lb   CG (b) by Lemma 9.12. It follows that F is a direct product as asserted, and F  F ∗ (C). Set H = CC (F ), so that H is A-invariant. As A centralizes LA and mp (A) = mp (B) = m2,p (G), we have mp (A) = mp (AH). But A ∩ H ≤ CA (K1 · · · Kk ) = 1 by Lemma 9.2. Hence H is a p -group, whence F = F ∗ (C). Finally, mp (NG (C)) ≥ mp (ALA ) > mp (A), so Op (C) has odd order.  We next prove Lemma 9.24. Suppose that Ki is terminal in G for some i = 1, . . . , k. Then k = 1. Proof. Without loss we take i = 1 and assume for a contradiction that k ≥ 2. We may further assume that for some  ≥ 1, K1 , . . . , K are all the Ki ’s that

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are terminal, and m2 (K1 ) ≤ m2 (K2 ) ≤ · · · ≤ m2 (K ). By [II3 , Theorem PU∗4 ], K1 , . . . , K are standard. We set C1 = CG (K1 ) and C 1 = C1 /Op (C1 ) and argue that  = 1; suppose false. Then for any a ∈ A# ∩ K1 , C1 ≤ CG (a), and K2 , . . . , Kk are components of CG (a) by Lemma 9.5, hence components of C1 . Since K2 is standard, K2  C1 . But also LA   CG (a), so K2 < E(C1 ). Therefore as K2 is simple, a Sylow 2subgroup Q of C1 has noncyclic center and m2 (Q) ≥ 4. If there is no four-subgroup of Aut(K1 ) disjoint from Inn(K1 ), then by [V2 , 5.3], Qg ≤ NG (K1 ) for some g ∈ G − NG (K1 ). As K1 is standard, Qg ∩ C1 = 1 so Qg embeds in Aut(K1 ). On the other hand, m2 (Q) ≥ m2 (K2 LA ) = m2 (K2 ) + m2 (LA ) ≥ m2 (K1 ) + 2. Therefore there is a four-subgroup of Aut(K1 ) disjoint from Inn(K1 ), contrary to our assumption. We may therefore assume that such a four-subgroup of Aut(K1 ) exists. Since K1 ∈ C2 ∩ Cp , it follows by [VK , 11.16] that K1 ∼ = L± 4 (3), and hence  for any four-group E ≤ Aut(K1 ), K1 = ΓE,1 (K1 ) by [IA , 7.3.4]. By [V2 , 5.1], m2 (Q) = 1, a contradiction. Hence,  = 1. Since k ≥ 2 and K2 , . . . , Kk are not terminal, A = B, p = 5, and K2 ∼ = ··· ∼ = o A , by Lemma 9.22. Fix (t, J) ∈ IL (G), in accordance with Lemma 9.23a, Kk ∼ = 5 2 such that [t, Kk ] = 1, Kk ≤ J, and J/Z(J) ∼ = M12 or J2 . Set Ck = CG (Kk ), ∗  ∗  C k Ck = Ck /Op (Ck ), and K = K2 × · · · × Kk−1 × LA . Then K1  Ck , so K ∗  by Lemma 9.23b. In particular K1 and K are t-invariant. Also |Op (Ck )| is odd by Lemma 9.23. We argue that m2 (CK ∗ (t)) ≤ 2.

(9N)

If not, choose E ≤ K Op (Ck ) with CG (t) ≥ E ∼ = E23 . Then E centralizes Kk so E normalizes J. But m2 (CAut(J) (Kk )) = 2 and so there is an involution x ∈ E that centralizes J. Now [x, t] = 1, so by L2 -balance, J ≤ E(CG (x)). Now the terminal component K1 is a component of CG (x) and [K1 , Kk ] = 1, so [K1 , J] = 1. Take a nontrivial element b ∈ A ∩ K1 , so that J ≤ CG (b). But Kk is a component of CG (b) by Lemma 9.7 and Kk < J, a contradiction. This proves (9N). As a consequence, k ≤ 3. Let us now restrict the possibilities for K1 . Let C1 = CG (K1 ) and C 1 = C1 /Op (C1 ). If k = 3, then by Lemma 9.23, C 1 has exactly three components, so a Sylow 2-subgroup Q of C1 has noncyclic center. Using [V2 , 5.3] as in the second paragraph of the proof, we find that Qg acts faithfully on K1 for some g ∈ G. Consequently m2 (Aut(K1 )) ≥ m2 (K2 K3 LA ) ≥ 6. On the other hand if k = 2, then mp (AutA (K1 )) ≥ mp (A) − 1 ≥ 3. In either case we conclude from [VK , 11.15] that m2 (CK1 (t)) ≥ 3. As m2 (CAut(J) (Kk )) = 2, there is an involution y ∈ K1 ∩ CG (J). Consider the subnormal closure Jy of Kk (or equivalently of J) in CG (y). Then Kk ≤ J ≤ Jy . Since [y, z] = [y, Kk ] = 1, Kk is a component of CJy (z). This implies that Jy cannot be a diagonal pumpup of Kk , as J > Kk . As J/Z(J) ∼ = M12 or J2 , it also implies by [VK , 3.67] that J = Jy . But LA centralizes y ∈ K1 and Kk , so LA normalizes J and maps into the solvable group CAut(J) (Kk ). Hence [LA , J] = 1. This is a contradiction, since z ∈ LA and Kk is a component of CG (z), hence of CJ (z). This completes the proof.  ∗

At last we can prove

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Lemma 9.25. E(CG (z)) = K1 is quasisimple and standard in G. Proof. Suppose false. By Lemmas 9.22 and 9.24, and [II3 , Theorem PU∗4 ], A = B, p = 5, and K1 , . . . , Kk are all non-terminal and isomorphic to A5 . As CA (K1 · · · Kk ) = 1, k ≥ 4. Let F , C, and C be as in Lemma 9.23b, relative to i = k, and set F0 = F Op (CG (Kk )), so that F 0 = F = F ∗ (C). Choose t and J as in Lemma 9.23a. We may assume, replacing t by a CG (Kk )-conjugate if necessary, that t, z is a 2-group. Subject to that condition, we choose t if possible so that t ∈ CF0 (z). If in fact t ∈ CF0 (z), then t induces inner automorphisms on K1 · · · Kk , and so m2 (CK1 K2 (t, z)) ≥ 4. As [K1 K2 , Kk ] = 1 and m2 (CAut(J) (Kk )) = 2, there exists x ∈ I2 (K1 K2 ) centralizing t, z, and J. Then the subnormal closure J ∗ of Kk in CG (x) contains J, and Kk is a component of CJ ∗ (z), so by [VK , 3.67], J = J ∗ . But then as k ≥ 4, K3 ≤ CCG (x) (Kk ) maps into the solvable group CAut(J) (Kk ), and so [K3 , J] = 1. Choosing b ∈ B ∩ K3# , we get J ≤ CG (b), which is absurd as Kk is a component of CG (b) by Lemma 9.7. Therefore, t ∈ CF0 (z). By our choice of t, CF0 (J, z) has odd order. For otherwise any involution t∗ ∈ CF0 (J, z) could have been chosen instead of t. Indeed then the subnormal closure J ∗ of J in CG (t∗ ) would again be a vertical pumpup of (z, Kk ) and then J ∗ = J by [VK , 3.67] again, justifying the choice of t∗ . (t, z) maps injectively into CAut(J) (Kk ), which has 2-rank It follows  that   CF 0  ≤ 2. 2. Thus m2 CF t, z On the other hand, F has at least k − 1 ≥ 3 components isomorphic to A5 , and t must permute these. But since t, z is a 2-group, there is a t-invariant Sylow 2-subgroup R of F containing z. If LA ∼ = A5 , then Ω1 (R) = Z(R) so [R, z] = 1. It follows easily that m2 (CR (t)) > 2, contradicting the previous paragraph. Thus, LA ∼ = A5 , so t normalizes K 2 ·· · K k , which is centralized by z. As k ≥ 4, we again get the contradiction m2 (CF ( t, z )) > 2, completing the proof.  We now write K = K1 = E(CG (z)). The possibilities for K are given in Lemma 9.10a. We set M = NG (K) and prove Lemma 9.26. M is a maximal subgroup of G. Proof. Let M1 be a maximal subgroup of G containing M . As K is standard and [D, K] = 1 by Lemma 9.11a, we have ΓD,1 (G) ≤ M ≤ M1 . For any d ∈ D# , it follows that CO2 (M1 ) (d) ≤ O2 (CG (d)) = 1, as G is of even type. As D is noncyclic, O2 (M1 ) = 1. Then by L2 -balance, K ≤ E(M1 ). If O2 (M1 ) = 1, then by standardness of K, K  NG (Ω1 (Z(O  2 (M1 )))) = M1 , so M1 = M . Thus we may assume that O2 (M1 ) = 1. Set K ∗ = K E(M1 ) . Then by L2 -balance, K ∗ is the product of components of M1 permuted transitively by each involution in D. But D is noncyclic, so K ∗ is a single component. We have K  CK ∗ (d) for all d ∈ D# . Thus if K < K ∗ , [VK , 3.5] implies that K is not one of the groups specified in Lemma 9.10a (using [IA , 4.10.3] for p-ranks), contradiction. So K = K ∗  M1 . The lemma follows. 

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Let us eliminate some possibilities for K. ∼ Lemma 9.27. K/Z(K) =  L± 4 (3), G2 (3), J3 , or F1 . Proof. Since m2 (CG (K)) > 1, this follows directly from [V2 , 9.1].



Lemma 9.28. One of the following holds: (a) p = 3 and K/Z(K) ∼ = Suz, Co2 , Co1 , F i22 , F i23 , F i24 , F3 , or F2 ; (b) p = 3 and K ∼ = U5 (2), U6 (2), D4 (2), or F4 (2); (c) p = 3 = |B : A| and K ∼ = U4 (2), 3D4 (2), Sp4 (8), or Sp6 (2); or 5 (d) p = 5 = |B : A| and K ∼ = Co1 , F2 , or 2F4 (2 2 ). Proof. If K ∈ Spor, then by Lemmas 9.25, 9.10a, and 9.27, the appropriate conclusions hold. Otherwise, these lemmas imply that K ∈ Chev(2). In particular (9O)

[LA , K] = 1 and K is simple,

by Lemma 9.11c. First suppose that p ≥ 7. Then LA ∼ = L2 (p) and A = B. By [VK , 3.20], and as mp (A) ≥ 4, there exists a ∈ A# such that CK (a) has a component I such that mp (I) > 1. By Lemma 9.7, I ∈ Cp . But by [IA , 4.9.6], I ∈ Chev(2). As mp (I) > 1, this is impossible by [IA , 2.2.10] and the definition of Cp [V3 , 1.1]. Thus, p ∈ {3, 5}. By [VK , 12.5], there is c ∈ CLA (B) of order p such that NLA (c) contains an  = C/Op (C). Since involution z0 . In particular, [c, K] = 1. Set C = CG (c) and C [B, c] = 1, mp (C) > mp (B) = m so Op (C) has odd order. We claim that we may assume that (9P) Suppose first that (9Q)

K ≤ Lp (C).  K = Lp (CK (a)) | a ∈ A# .

By Lemma 9.7, Lp (CK (a)) is a product of components of Lp (CG (a)) and is centralized by c, so (9P) follows from (9Q) and Lp -balance. Suppose then that (9Q) fails. By [VK , 12.46b], p = 3 and K ∼ = U4 (2), 3D4 (2), or U3 (8). Then m3 (Aut(K)) = 3 and so |B : A| = 3 and m = 4. The first two of these three groups are allowed conclusions of the lemma, so assume that K ∼ = U3 (8). By [VK , 15.26], K central Since m = 4, L3 (C)  has at most five 3-components, and they must be izes O3 (C). normalized by K. Then (9P) follows from the Schreier property. Notice that z0 normalizes c and K, and K  CG (z0 ) as K is standard. Hence   C  (z0 ). By L2 -balance, K  either lies in a single component of C  or is the K C diagonal of two such components interchanged by z0 . On the other hand for any a ∈ A# , by Lp -balance, any component of Lp (CK (a)) either lies in a single pcomponent of C or is a diagonal of p p-components cycled by a. The latter case is  lies in a single p-component of C.  If thus impossible. Hence if (9Q) holds, then K ∼ (9Q) fails, then as above we may assume that p = 3 and K = U3 (8) ∈ C3 . As all  lie in C3 , K  must lie in a single component in any case. Say K  components of C  lies in the component Lc .  c ∈ Chev(p). Then by [IA , 4.9.6], the component K  of C  (z0 ) Suppose that L Lc  lies in Chev(p), hence in Chev(p) ∩ Chev(2). Also, mp (Aut(K)) ≥ mp (A) ≥ 3. Thus by [IA , 2.2.10, 4.10.3a], K ∼ = U4 (2) and conclusion (c) holds. Therefore we may  c ∈ Cp − Chev(p) by our main hypothesis (1A). Now [VK , 11.14] assume that L yields the lemma. 

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138

4. THEOREM C5 : STAGE 2

In particular, Lemma 9.28 implies that p = 3 when A = B, and so Lemma 9.29. LA ∼  L2 (p), p ≥ 5, p ∈ FM. = Lemma 9.30. The following conditions hold: (a) CG (ALA ) has odd order; (b) LA = Lp (CG (A)); (c) K is simple; (d) If p = 3 and LA ∼ = L2 (8), then D ∈ Syl2 (CG (A)); and 5 (e) If p = 5 and LA ∼ = 2B2 (2 2 ), then for any D0 ∈ Syl2 (LA ) with D ≤ D0 , D0 ∈ Syl2 (CG (A)). Proof. By Lemmas 6.4 and 9.12, A ∈ Ep∗ (CG (ALA )), so CG (ALA ) has a normal p-complement, which is B0 LA -invariant. But mp (B0 LA ) > m by (9C3), so CG (ALA ) has odd order. In particular LA = Lp (CG (A)), proving (a) and (b). 5 By Lemmas 9.11c and 9.29, K is simple unless possibly LA ∼ = L2 (8) or 2B2 (2 2 ), with p = 3 or 5, respectively. Assume that one of these two cases holds. Let D0 ∈ Syl2 (CG (A) ∩ NG (D)). Then D0 normalizes LA , and by the previous paragraph D0 acts faithfully on LA . But | Out(LA )| is odd, and so D0 ∈ Syl2 (LA ). In particular, if p = 3, then D0 = D, and we have (d) and (e). By Lemma 9.28, if (c) fails, then K ∈ Spor and so | Out(K)| ≤ 2 [IA , 5.3]. Thus A ≤ K, so CG (K) ≤ CG (A). Consequently, by (d), (e), and Lemma 9.11b, ND ≤ CG (K) and a Sylow 2-subgroup of CG (K) is contained in D0 . Thus Ω1 (O2 (K)) ≤ D. But ND acts without fixed points on D, and centralizes K, so O2 (K) = 1. The lemma is proved.  5 Lemma 9.31. If LA ∼ = L2 (8) or 2B2 (2 2 ), then NK (LA ) = CK (LA ).

Proof. This is because [ND , K] = 1 (Lemma 9.11) and CAut(LA ) (ND ) = 1.  Lemma 9.32. Suppose that LA ∼ = L2 (8) or 2B2 (2 2 ). Let x, y, v ∈ Ep (K) with # ∼ x, y = Ep2 . If LA is a component of CG (w) for all w ∈ x, y , then LA is a component of CG (v) and is centralized by CK (v). 5

Proof. First, v ∈ K centralizes ND and lies in CK (x), in which LA is a component. Hence v normalizes LA , and indeed centralizes it by Lemma 9.31. Now by Lp -balance, [V2 , 7.2], and [VK , 3.5], LA is a component of CG (v). Hence, as NK (v) centralives ND , NK (v) normalizes and then centralizes LA , by Lemma 9.31 again.  Lemma 9.33. If A ≤ K, then [K, LA ] = 1. Proof. Suppose [K, LA ] = 1. By Lemmas 9.29 and 9.11, and (9A), LA ∼ = 5 L2 (8) or 2B2 (2 2 ), with p = 3 or 5, respectively. By Lemmas 9.11 and 9.28, K ∈ Spor, and either p = 3 with m3 (K) ≥ 4 (see [IA , 5.6.1]) or p = 5 with K ∼ = Co1 or F2 . Let R ∈ Sylp (K) with A ≤ R. By Lemma 9.12, CK (a) normalizes and then centralizes LA for all a ∈ A# . If p = 3, then R is connected by [IG , 10.17]. Hence by Lemma 9.32, LA is centralized by CK (z) | z ∈ I3 (R), which equals K since K has no strongly 3-embedded subgroup [IA , 7.6.1]. Thus the lemma holds in this case.

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9. THEOREM 1: STANDARD 2-COMPONENTS

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∼ Co1 or F2 . Let R ∈ Syl (K) and choose, by Suppose that p = 5, so K = 5 [VK , 8.23], C ≤ R with C ∼ = E52 and C containing no 5-central element of K of order 5. Let C ∗ = CZ(R) ∼ = E53 . Let U  R with U ∼ = E52 . Then C ∗ and A are both connected to U , and hence connected to each other. Therefore K = ΓC,1 (K) ≤ NK (LA ) = CK (LA ), by [VK , 8.24] and Lemma 9.32. The proof is complete.  Recall that M = NG (K). Now we can prove Lemma 9.34. LA  M and [K, LA ] = 1. Proof. By Lemma 9.30ab, |CG (ALA )| is odd and LA = Lp (CG (A)). Now if A ≤ K, then CG (K) ≤ CG (A) ≤ NG (LA ) so LA  CG (K) by Lemma 9.33. As CG (ALA ) has odd order, LA = Lp (CG (K))  M , as desired. We may therefore assume that A ≤ K, so that p divides | Out(K)|. By Lemma 9.28 and [IA , 2.5.12, 4.7.3A, 5.3], it follows that p = 3 with K ∼ = U6 (2), 3D4 (2),  5 2 D4 (2), or Sp4 (8), or p = 5 with K ∼ = F4 (2 2 ). Moreover, KA ∼ = O p (Aut(K)). By Lemma 9.11c, [K, LA ] = 1. If K ∼ = D4 (2), then any a ∈ A − K induces a graph automorphism on K. Hence as CA (K) = 1, m3 (A) ≤ 3, by [IA , 4.7.3A]. Thus m3 (B) = 4. Therefore LA ∼ = L2 (8), for otherwise m2,3 (G) ≥ m3 (K) + m2,3 (LA ) = 4 + 1 > m3 (B), contradiction. In particular B0 = B, and as A ≤ B0 it follows that m3 (KB0 ) > 4. But [KB0 , z] = 1 by (9C4), so m2,3 (G) > 4 = m3 (B), again a contradiction. Thus, K ∼ = D4 (2). Set C = A ∩ K, a hyperplane of A. Also set CC = CG (C). By Lemma 9.12, LA is a component of CC . Hence LA is a component of CG (K). Suppose that LA  NG (K) and choose g ∈ NG (K) with LgA = LA . Unless LA ∼ = L2 (8) or 5 2 2 B2 (2 ), we have mp (LA ) ≥ 2 and m2,p (LA ) ≥ 1. Hence, m2,p (CC ) ≥ m2,p (LA ) + mp (LgA ) + mp (C) ≥ 3 + mp (C) = mp (A) + 2 > m, ∼ L2 (8) or 2B2 (2 2 ). As A ∈ Ep∗ (CG (ALA )), we have a contradiction. Therefore, LA = p A ∈ S (CCC (LA )), and it follows that A1 := A ∩ LgA = 1, and A = C × A1 . As 5 g LA ∼ = L2 (8) or 2B2 (2 2 ), A1 lies in a subgroup A∗1 ∼ = Zp2 of LA , and [A∗1 , A z] ≤ [A∗1 , CLA A1 ] = 1. Set A∗ = C × A∗1 ≤ CG (z), so that A = Ω1 (A∗ ). Then A∗ , like  A, acts faithfully on K. However, Y := O p (Aut(K)) ∼ = P GU6 (2), Aut(3D4 (2)), 5 2 Sp4 (8) f , or F4 (2 2 ) f , with f a field automorphism of order p. Thus by [VK , 9.23], Y contains no copy of A∗ . This contradiction completes the proof of the lemma.  5

Lemma 9.35. We have A ∈ Ep∗ (AK). Proof. Suppose that the lemma is false. Then as AK ≤ CG (LA ), there is A∗ ∈ Ep (CG (LA )) such that mp (A∗ ) = mp (A) + 1. As LA has even order we must have |B : A| = p. By Sylow’s theorem we may take A∗ so that A, A∗  is a p-group. Then LA is a component of CG (A, A∗ ). Working down a subnormal chain from A, A∗  to A∗ , we see by Lp -balance that LA ≤ Lp (CG (A∗ )), which contradicts  (6B3) as mp (A∗ ) = m. The lemma follows. We now wish to prove that LA is a component of CG (x) for some involutions x ∈ K.

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4. THEOREM C5 : STAGE 2

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We first define A := {a ∈ Ip (CG (LA )) | LA is a component of CG (a)}. Observe that A# ⊆ A, and A is closed under CG (LA )-conjugacy. The next lemma will be applied to elements x of order p as well as to involutions x. Lemma 9.36. Let x ∈ CM (LA )# be of order 2 or p. Suppose that there is F ∈ Ep (CM (x)) such that mp (F ) > 1 and F # ⊆ A. Then LA   CG (x), or else the following conditions hold: (a) p = 3, F ∼ = A6 ; = E32 , and LA ∼ (b) If x is an involution, then K ∼ = Suz, and x is neither 2-central nor half 2-central in M ; and (c) If x has order 3, then there is no involution t ∈ CG (F ) such that LA t ∼ = Σ6 . In particular, if |F | ≥ 25, then LA   CG (x). Proof. Since LA is a component of CG (a) for all a∈ F # , we may apply [V2 , E(C (x)) is the product 7.2] to conclude that LA ≤ E(CG (x)). By Lp -balance, LA G of components in a single a-orbit for every a ∈ F # . Hence LA lies in a single F invariant component J of CG (x), and LA is a component of CJ (a) for all a ∈ F # . If LA = J, then by [VK , 3.5], and the facts that Lo2 (G) ⊆ C2 and Lop (G) ⊆ Cp (to rule out alternating groups), we have p = 3, mp (F ) = 2, LA ∼ = Suz = A6 , and J/Z(J) ∼  or O N . Thus (a) holds. Moreover, since z ∈ LA   CJ (a) for any a ∈ F # , we see from [IA , 5.3os] that if we put I = E(CJ (z)), then I/Z(I) ∼ = L3 (4). Suppose that x is an involution. By L2 -balance and Lemma 9.25, I is a component of CK (x). Hence by [VK , 3.14] and Lemma 9.28, K ∼ = Suz. As E(CK (x)) = 1, it is clear from [IA , 5.3o] that x is neither 2-central nor half 2central in M = NG (K). Thus (b) holds. Finally suppose that x has order 3 and that (c) fails, so that there is an involution t such that [F, t] = 1 and LA t ∼ = Σ6 . Thus CLA (t) contains elements of order 3. But the image of t in Aut(J) is an involution centralizing F . By [VK , 3.70], since J/Z(J) ∼ = Suz or O  N , this implies that CLA (t) is a 3 -group, a contradiction. The proof is complete.  As a result, A extends from A over appropriately connected subsets of Ip (CG (LA )). Lemma 9.37. Let A0 ∈ Ep (AK) with mp (A0 ) ≥ 3 or 2 according as p = 3 or 5. If A0 has an AK-conjugate which is 3-connected or 2-connected to A (according as p = 3 or 5), then A# 0 ⊆ A. Proof. We have A# ⊆ A. Clearly, A is closed under AK-conjugation, since [AK, LA ] = 1. Therefore it suffices to show that if A1 ∈ E33 (AK) or E52 (AK) with A# 1 ⊆ A, and x ∈ Ip (CAK (A1 )), then x ∈ A. But this is an immediate consequence of Lemma 9.36.  Lemma 9.38. Let y, w ∈ I2 (CM (LA )) with [y, w] = 1, and suppose that there exists a ∈ A such that [y, a] = 1. If LA   CG (w), then LA   CG (y). Proof. Let Ly be the subnormal closure of LA in CG (y). Then Ly ≤ E(CG (y)) as G is of even type, and Ly has the usual structure, by L2 -balance. Moreover,

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since a centralizes y and LA , it normalizes Ly and LA   CLy (a) as a ∈ A. If Ly is the product of two w-conjugate components, each component is then a-invariant and then centralized by a, as LA projects onto both components. Hence, LA   Ly since a ∈ A, which is absurd. Therefore Ly is a single component. Suppose that LA < Ly ; then Ly ∈ C2 , LA   CLy (w), and LA   CLy (a). This contradicts [VK , 3.50], however, so the lemma follows.  Lemma 9.39. There exists a four-subgroup V ≤ K such that LA   CG (v) for all v ∈ V # . ∼ F i22 , we shall find x ∈ I2 (K) for which we Proof. Except in the case K = can check that LA   CG (x) by application of Lemma 9.36. We shall then find a four-group V ≤ K and element a ∈ A such that [V, a] = 1 and [V, xg ] = 1 for some g ∈ K. (The latter condition is automatic if x was chosen to be 2-central in K.) Since [V, xg ] = 1, LA   CG (v) for all v ∈ V # by Lemma 9.38. Suppose first that K ∼ = U4 (2), U5 (2), Sp6 (2), F4 (2), or Co2 , with p = 3, p or K ∼ = Co1 with p = 5. Then p does not divide | Out(K)| so A ∈ E∗ (K) by Lemma 9.10. But in each of these cases, by [VK , 9.14], J(U ) is elementary abelian for U ∈ Sylp (K), so A ⊇ {a ∈ Ip (K) | mp (CG (a)) = mp (K)}. It follows from [IA , 4.8.2a, 4.10.3a, 4.7.3A, 5.3] that A ⊇ Ip (K). In the four classical Chev(2) cases, m3 (CK (x)) ≥ 2 for the high root involution x of a Borel subgroup; in the F4 (2) case, the same inequality holds for any x ∈ I2 (K) which is 2-central in Aut(K); and in the two sporadic cases, the same inequality holds for x 2-central by [IA , 5.3]. Hence LA   CK (x). Since I3 (K) ⊆ A in these cases, it suffices by Lemma 9.38 to find a four-group V ≤ K such that |CK (V )| is divisible by 3. Such a V is provided by [VK , 12.6], so the lemma holds in these cases. A similar argument applies if K ∼ = D4 (2); one must first apply Lemma 9.35 to obtain A ∈ E3∗ (AK). This implies that A induces inner automorphisms on K. Again by [IA , 4.8.2a], A ⊇ I3 (K), and we may take x to be a high root element of a Borel subgroup, and V is provided by [VK , 12.6]. Likewise if K ∼ = U6 (2), then by Lemma 9.35, m3 (A) ≥ m3 (K) = 4, implying by [VK , 9.17] that A acts on K like a diagonalizable subgroup of GU6 (2). This again leads to I3 (K) ⊆ A, and again we may take x to be a high root element and V from [VK , 12.6]. Suppose next that p = 3 and K ∼ = Aut(K), m3 (A) = 3, = 3D4 (2). Then KA ∼ m = 4, |B : A| = 3, and KA  KBLA , so KBLA = KA × L, where L = LA unless LA ∼ = L2 (8), in which case L = LA b. Let x be a high root involution in K; then CKA (x)/O2 (CKA (x)) ∼ = P ΣL2 (8) so m3 (CKALA (x)) ≥ 2 + m3 (L) = 4. Hence Bp∗,c (G) = ∅, so B ∈ Bp∗,c (G). Therefore there is t ∈ I2 (CG (B)), and so if LA ∼ = A6 , we must have LA t ∼ = Σ6 . Consequently, by Lemma 9.36c, any y ∈ I3 (KA) with CA (y) noncyclic lies in A. In particular, for any U ∼ = E32 such that U  P for some P ∈ Syl3 (AK), U # ⊆ A. Using [IA , 4.7.3A] we conclude that I3 (K) ⊆ A. Indeed [KB, t] = 1, so another application of Lemma 9.36 yields that I3 (KA) ⊆ A. And yet another application then yields LA   CG (x). Finally V is provided by [VK , 12.6]. 5 Next, if K ∼ = Sp4 (8) or 2F4 (2 2 ), then A = Ω1 (P ) for some P ∈ Sylp (KA), so Ip (KA) ⊆ A, and we may take x to be a high root element, which is half 2-central or 2-central in Aut(K), respectively. Then mp (CKA (x)) ≥ 2 so Lemma 9.36 yields LA   CG (x), as desired. Again as Ip (KA) ⊆ A, [VK , 12.6] completes the proof in these cases.

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4. THEOREM C5 : STAGE 2

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The remaining possibilities for K are all sporadic, so | Out(K)| ≤ 2 and then A ∈ Ep∗ (K) by Lemma 9.35. Let P ∈ Sylp (K), and let x be a 2-central involution of K. If p = 3, then K ∼ = F2 and p = 5, whence by [VK , 9.19], P is connected. By Lemma 9.37, I5 (K) ⊆ A, and by [IA , 5.3], m5 (CK (x)) = 2. Hence by Lemma 9.36, LA   CG (x), and [VK , 12.6] completes the proof as usual. We may now assume that p = 3. We use the fact [V2 , 4.6] that (9R)

any two E35 -subgroups of a 3-group are 3-connected.

We freely use the information in [IA , 5.3]. If there is x ∈ I2 (K) such that CK (x) contains A1 ∼ = E35 , and there is a four-group V ≤ CK (x) centralized by an element of A# , then the lemma holds. Indeed m3 (A) = m3 (K) ≥ 5, so A1 is 3-connected to 1 a conjugate of A. Thus A# 1 ⊆ A by Lemma 9.37, whence LA   CG (x) by Lemma 9.36, and Lemma 9.38 yields the desired statement about V # . This argument handles the cases K ∼ = F i23 , F i24 , and F2 , with E(CK (x)) ∼ = 2F i22 , 2F i22 , and 2 2 E6 (2), respectively. If K ∼ = F3 , then m3 (CK (u)) ≥ 5 for all u ∈ I3 (K), so I3 (K) ⊆ A by Lemma 9.37, and any involution x and Z3 × E22 subgroup satisfy the required conditions. If K ∼ = Suz, then m3 (CK (u)) = 5 for all u ∈ I3 (K) that are not of class 3C. But for v ∈ I3 (K) of class 3C, CK (v) has a single class of involutions t, and m3 (CK (v, t)) = 2. As m3 (CK (x)) = 3 for a 2-central involution x, all u ∈ I3 (CG (x)) satisfy m3 (CG (u)) ≥ 5, so I3 (CG (x)) ⊆ A and Lemmas 9.36 and 9.38 again give the desired conclusion, with a Z3 × E22 subgroup lying in CK (x). If K ∼ = Co1 , then K contains A1 X with A1 = CA1 X (A1 ) ∼ = E36 normalized by X ∼ = 2M12 and inverted by the involution of Z(X). As X is perfect, it contains involutions x such that m3 (CA1 (x)) = 4, and these must be 2-central. Again A# 1 ⊆ A. As m2 (CK (a)) ≥ 2 for every a ∈ I3 (K), Lemmas 9.36 and 9.38 again give the desired conclusion. The last case is K ∼ = F i22 . As K permutes transitively the class 2A of 3transpositions with 2-point stabilizers 22+8 U4 (2) and U4 (3) · 2, K contains R × E with R ∼ = E34 . We take x ∈ I2 (R). Again Lemmas 9.37 and 9.36 give = Σ3 and E ∼ I3 (RE) ⊆ A and LA   CG (x). We choose V within E(CK (O3 (R))) ∼ = U4 (3), and the proof is complete.  Lemma 9.40. Let H = CG (LA ). Then for any e ∈ I2 (H), m2 (CH (e)) ≥ 3. Proof. Let Q ∈ Syl2 (H). We have Q ≤ CG (z) ≤ M , so Q normalizes K. The possibilities for K are given in Lemma 9.28, and with [IA , 3.3.3, 5.6.1], m2 (K) ≥ 4. Hence by [IG , 10.11], there is U  Q with E22 ∼ = U ≤ K. By [IG , 10.20], m2 (CG (U )) ≥ 4. Let e ∈ I2 (H). If e ∈ KCH (K), then write e = bc with b ∈ K, c ∈ CH (K), and b2 = c−2 . Applying the Thompson transfer lemma in K = K/Z(K) to b, we see that b has a K-conjugate in CH (U ), so e does as well. Hence whether e ∈ U or not, m2 (CH (e)) ≥ 3. Thus, we may assume that e induces a noninner automorphism on K, and m2 (CK (e)) = 1, and derive a contradiction. In particular as Out(Sp6 (2)) = 1, K ∼ = Sp6 (2). If m2 (K) ≥ 5, then by MacWilliams’ theorem [V2 , 4.4], m2 (CK (e)) > 1. So assume that m2 (K) ≤ 4. Using [IA , 3.3.3, 5.6.1] and Lemma 9.11c, we are reduced to the cases K ∼ = U5 (2), F4 (2), or U4 (2). In the first two cases, by [IA , 4.9.1,

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143

1 ∼ U4 (2), then e 4.9.2], CK (e) contains Sp4 (2) or 2F4 (2 2 ) , so m2 (CK (e)) > 1. If K =  normalizes J(Q ∩ K) ∼ = E24 , so m2 (CK (e)) > 1. The proof is complete.

Lemma 9.41. LA is standard in G (as a component of the centralizer of an involution). Proof. It suffices by [II3 , Theorem PU∗4 ] to show that LA is terminal in G. Let H = CG (LA ) and Q ∈ Syl2 (H), and let V be as in Lemma 9.39. Suppose U ≤ Q with U ∼ = E22 , and y ∈ I2 (CG (U )). Then if LA   CG (u) for all u ∈ U # , it follows from [VK , 3.5] and the fact that Lo2 (G) ⊆ C2 (G is of even type) that LA   CG (y). Consequently, LA   CG (y) for all y ∈ I2 (H) such that y ∈ U ≤ H with U ∼ = E22 connected to V or an H-conjugate of V . By the previous lemma, every y ∈ I2 (H) is so connected, via a normal E22 of a Sylow 2-subgroup of H  containing one of CH (y). The proof is complete. Now Proposition 6.1 follows quickly. Let Q ∈ Syl2 (CG (LA )), so that Q ≤ CG (z) ≤ M , Q ∩ K ∈ Syl2 (K), and ΓQ,1 (G) ≤ M , as LA is standard and M = NG (LA ). By [II3 , Corollary PU2 ], there is g ∈ G − M such that m2 (Qg ∩ M ) ≥ 2, and then ΓQg ∩M,1 (LA ) < LA . This rules out LA ∼ = L± 3 (3), by [IA , 7.3.4]. In the remaining four cases for LA , LA is outer well-generated for p = 2, by [VK , 8.4] for LA ∼ = A6 , and because I2 (Aut(LA )) ⊆ Inn(LA ) for the other possibilities 1 5 LA ∼ = L2 (8), 2F4 (2 2 ) , 2B2 (2 2 ). Then by [II3 , 16.13], g can be chosen so that Qg ≤ M . Now Qg acts faithfully on LA by [IG , 18.6], so |K|2 ≤ |Q| = |Qg | ≤ | Aut(LA )|2 ≤ 25 , 23 , 212 , 210 , respectively, with p = 3 in the first two cases and p = 5 in the last two. But these inequalities are false since K and p are as in Lemma 9.28, a final contradiction. This completes the proof of Proposition 6.1. 10. Theorem 2 Proposition 6.1, combined with the Thompson dihedral lemma, yields the following proposition. Notice that parts (a) and (b) are parts (c) and (d) of Theorem 2. Proposition 10.1. (a) If B ∈ Bp∗ (G), then m2 (CG (B)) ≤ 1. p (b) If A ≤ B ∈ B∗,o (G) with |B : A| = p, then m2 (CG (A)) ≤ 2. (c) If t ∈ I2 (G) and B ∈ Bp∗ (G) with B ≤ CG (t), then t is the largest elementary abelian normal 2-subgroup of CG (t). Proof. Let A and B be as in (b) and suppose that U ∈ E2 (CG (A)) with m2 (U ) = 3. By Proposition 6.1 and (1A4), CG (A) is p-constrained with Op (CG (A)) of odd order. Therefore there is a U -invariant Sylow p-subgroup Q of Op p (CG (A)), and CU (Q) = 1. By the Thompson dihedral lemma, U Q contains the direct product of A and three copies of D2p . Hence m2,p (G) ≥ m2,p (U Q) ≥ mp (A) + 2 > m2,p (G), a contradiction. Thus, (b) holds. The proof of (a) is similar. In (c), suppose that t < V  CG (t) and V is elementary abelian. Since t exists, B ∈ Bp∗,o (G). By (a), t = CV (B). Hence for some hyperplane A ≤ B, m2 (CV (A)) ≥ 3, contradicting (b). Hence (c) holds.  Recall that a symplectic pair in G is an ordered pair (B, T ) such that B ∈ Bp∗ (G), T ∈ I∗G (B; 2), and T is cyclic or of symplectic type. Moreover, a symplectic pair (B, T ) is called trivial if T ∼ = Z2 and faithful if CB (T ) = 1. Note that in a symplectic pair (B, T ), T = 1 as B ∈ Bp∗ (G).

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The following proposition yields parts (a) and (b) of Theorem 2, and hence, with Proposition 10.1, completes the proof of Theorem 2. Proposition 10.2. The following conditions hold: (a) Symplectic pairs exist in G, and Bp∗,c (G) = ∅; and (b) Let B ∈ Bp∗ (G) and T ∈ IG (B; 2), and suppose that CT (B) = 1. Then T is cyclic or of symplectic type. Proof. To prove (b), let B ∈ Bp∗ (G) and T ∈ IG (B; 2) with CT (B) = 1, but suppose that T is not cyclic or of symplectic type. Set C = CT (B). Since C = 1, B ∈ Bp∗,o (G). By P. Hall’s theorem [IG , 10.3] there is a noncyclic characteristic abelian subgroup U of T . By Proposition 10.1, [B, U ] = 1. Therefore there is a hyperplane A of B such that [B, CU (A)] = 1. Then m2 (CU (A)) > 1, but by Proposition 10.1, (10A)

m2 (CU (A)C) ≤ 2.

In particular m2 (CU (A)) = 2, so p = 3 and CU (A) = [CU (A), B]. As [C, B] = 1 and C = 1, this implies that C ∩ CU (A) = 1 and then that m2 (CU (A)C) > m2 (CU (A)) = 2, a contradiction. Hence (b) holds. To prove (a), choose any B ∈ B∗,o (G) and T ∈ I∗G (B; 2). We may assume that T is not cyclic or of symplectic type, and hence CT (B) = 1, by (b). Let U be a noncyclic elementary abelian characteristic subgroup of T . Then U = [U, B] so for some hyperplane A of B, [CU (A), B] = 1. By Proposition 10.1, mp (CU (A)) ≤ 2, so equality holds and p = 3. As usual, CG (A) is p-constrained. Another application of the Thompson dihedral lemma, to the group CU (A)Q, with Q a CU (A)-invariant Sylow p-subgroup of Op p (CG (A)), shows that G contains A × H1 × H2 , with H1 ∼ = H2 ∼ = D2p . Let B1 = AOp (H1 ), t ∈ I2 (H2 ), and T1 ∈ I∗G (B1 ; 2) with t ∈ T1 . By (b), (B1 , T1 ) is a symplectic pair, and the proposition follows.  Now that Theorem 2 is proved, we may assume for the remainder of this chapter, as noted in Section 2, that if p = 3, then (10B)

BtK3exc (G) = ∅. 11. Corollaries: BtKp (G), and Components in Cp − Chev(p)

Lemma 11.1. Let (B, T ) be a symplectic pair in G and t = Ω1 (Z(T )). If  CG (t) is 2-constrained, or more generally if O p (E(CG (t))) = 1, then (B, T ) is a faithful symplectic pair. Proof. Let C = CG (t) and assume then that F ∗ (C) = O2 (C)E(C) is a p group. Since T ∈ I∗G (B; 2), T ∩ F ∗ (C) ∈ Syl2 (F ∗ (C)). Then by [VK , 15.23], since every component of E(C) is a C2 -group, CB (T ∩ F ∗ (C)) ≤ CB (F ∗ (C)) ≤  B ∩ O2 (C) = 1. Consequently CB (T ) = 1, so (B, T ) is faithful. In order to understand non-faithful symplectic pairs, therefore, we need to investigate triples (B, t, K) ∈ BtKp (G). We begin this investigation in Section 12. Corollary 11.2. Suppose p = 3 and B ≤ X ≤ G for some B ∈ B3∗ (G). Then  E(X) = O 3 (E(X)).

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11. COROLLARIES: BtKp (G), AND COMPONENTS IN Cp − Chev(p)

145

Proof. Otherwise E(O3 (X)) is a nontrivial commuting product of covering groups of Suzuki groups. Let T be a B-invariant Sylow 2-subgroup of O3 (E(X)). By [VK , 10.48], CT (B) = 1, so by Proposition 10.2, T is of symplectic type.  However, Ω1 (T ) is noncyclic abelian by [VK , 10.48], a contradiction. Corollary 11.3. If (B, t, K) ∈ BtKp (G), then Ω1 (O2 (E(CG (t)))) = t or 1. Proof. Let T = t Ω1 (O2 (E(CG (t)))), a nontrivial B-invariant elementary abelian 2-group witth [B, t] = 1. By Proposition 10.2, T = t, implying the result.  In the rest of this section we use Proposition 6.1 to rule out certain isomorphism types of groups in Lop (G) − Chev(p). Definition 11.4. Let p be an odd prime and let H ∈ Cp be simple. Suppose that the Schur multiplier of H is a p -group and that one of the following holds: (a) | Out(H)| is a p -group; or (b) For any P ∈ Sylp (Aut(H)), Ω1 (P ) is abelian. We then say that H is barely in Cp if and only if there exists E ∈ Ep∗ (Aut(H)) and subgroups E2 ≤ E1 ≤ E such that |E : E1 | ≤ p, |E1 : E2 | ≤ p, CH (E1 ) has even order, and Lp (CH (E2 )) = 1. For example, L2 (8) and Sp4 (8) are barely in C3 . Define Cop to be the set of all H ∈ Cp that are barely in Cp , and define C∗p = Cp − Cop . Corollary 11.5. Let B ∈ Bp∗ (G) and b ∈ B # . Let H be a p-component of CG (b). Then H/Op (H) ∈ C∗p , or p = 3 and H/O3 (H) ∼ = L2 (8). Proof. Suppose by way of contradiction that H := H/Op (H) ∈ Cop . By (1A6), mp (CG (b)) > mp (B) = m2,p (G). Let D ∈ Ep∗ (CG (b)). Then by [IG , 8.7(iii)], D normalizes H. Suppose first that p does not divide | Out(H)|. Then D ∩ H ∈ Ep∗ (H) and D = CD (H) × (D ∩ H). Since H ∈ Cop , there are E0 ∈ Ep∗ (H) and subgroups E2 ≤ E1 ≤ E0 such that |E0 : Ei | ≤ pi , CH (E1 ) has even order, and Lp (CH (E2 )) = 1. Set Di = CD (H) × Ei , i = 0, 1, 2. Then mp (D0 ) = mp (D) so D0 ∈ Ep∗ (CG (b)), |D0 : Di | ≤ pi , i = 0, 1, 2, CH (D1 ) has even order, and Lp (CH (D2 )) = 1. But mp (D0 ) = mp (CG (b)) > m2,p (G) so mp (D1 ) ≥ m2,p (G), whence equality holds and D1 ∈ Bp∗,o (G). Hence by Proposition 6.1, Lp (CG (D2 )) = 1. Hence by Lp -balance, Lp (CG (D2 b)) = 1, so Lp (CH (D2 )) = 1, a contradiction. Thus the lemma holds in this case. Now suppose that Ω1 (P ) is abelian for any P ∈ Sylp (Aut(H)). Again let D ∈ Ep∗ (CG (b)) and write D = CD (H) × (D ∩ H) × Dout , where Dout is an arbitrary complement to CD (H) × (D ∩ H). Now H ∈ Cop , and we let E2 ≤ E1 ≤ E ≤ Aut(H) be as in the definition of this term. Since Ω1 (P ) is abelian, we may assume that AutD (H) ≤ E by Sylow’s theorem. Then let Di be the preimage in D of Ei ∩ AutD (H), i = 1, 2. We conclude that |D : D1 | and |D1 : D2 | are at most p, CH (D1 ) has even order, and Lp (CH (D2 )) = 1, using Lp -balance to deduce the last condition from Lp (CH (E2 )) = 1. Then as in the previous paragraph, mp (D1 ) ≥ m2,p (G) so D1 ∈ Bp∗,o (G), and Lp (CG (D2 )) ≥ Lp (CG (D2 b)) ≥ Lp (CH (D2 )) = 1, contradicting Proposition 6.1. The proof is complete. 

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146

4. THEOREM C5 : STAGE 2

Corollary 11.6. Let 1 = b ∈ B ∈ Bp∗ (G) and let H be a p-component of CG (b). Then either H/Op (H) ∈ Chev(p) or one of the following holds: (a) p = 3 and H/O3 3 (H) ∼ = A9 , U5 (2), U6 (2), D4 (2), or 3D4 (2); (b) p = 3, mp (H) ≥ 3, and H ∈ Spor; viz., H/O3 3 (H) ∼ = J3 , Co1 , Co2 , Co3 , M c, Ly, Suz, O  N , F i22 , F i23 , F i24 , F5 , F3 , F2 , or F1 ; or (c) p = 5 and H/O5 (H) ∼ = Ly. Proof. By Corollary 11.5, we may assume that H/Op (H) ∈ C∗p − Chev(p). Then by [VK , 3.81], p and H are as claimed.  12. Theorem 3: Regular Triples and Mates We recall that (10B) is now in force. Recall also from Definition 1.5 that KbI p (G) is the set of all triples (K, b, I) such that for some (B, t, K) ∈ BtKp (G), b ∈ B # and I is a p-component of E(CK (b)). The triple (K, b, I) is regular, or regularly embedded, if I lies in a (necessarily unique) p-component J of CG (b). The triple (K, b, I) is broad if either [K, b] = 1 or the following hold: CB (K) = 1, b ∈ K, and I is terminal in K (as a p-component). Lemma 12.1. Let (B, t, K) ∈ BtKp (G) and b ∈ B # . Let I be a p-component of CK (b). Then B normalizes I. Proof. Suppose false. As G is of even type, O2 (K)= 1 and so B normalizes K by [IG , 8.7(iii)], which also implies that p = 3 and I B = I1 ∗ I2 ∗ I3 , with Z( I B ) ∼ = Z3 and I1 ∼ = I2 ∼ = I3 ∼ = 3A6 , 3A7 , 3M22 , or SL3 (q), q ≡ 4 or 7 (mod 9),  = ±1. From [IA , 5.2.8, 5.3] it follows that K ∈ Alt ∪ Spor. Likewise by [IA , 4.8.2, 4.8.4, 4.9.1, 4.9.2], b induces an inner-diagonal automorphism on K and K is not a classical group. (Note that if K ∼ = L9 (q), then Z(I1 ) = Z(I2 ).)  ∼ Then from [IA , 4.7.3A], we see that K = E6 (q)a and b acts as an element of  Z(P ) for some P ∈ Syl3 (K). The congruence on q implies that O 3 (Aut(K)) = Inndiag(K). Then m3 (CAut(K) (B)) ≥ m3 (K) = 5 as K is simple [IA , 4.10.3a]. Let  x ∈ B − NB (I1 ). Then O 3 (CAut(K) (B)) ≤ CInndiag(K) (b, x) =: Y and F ∗ (Y ) ∼ = L3 (q) × E32 . Therefore 5 ≤ m3 (CAut(K) (B)) ≤ 2 + m3 (Aut(L3 (q))) = 4, again by the congruence on q. This contradiction completes the proof.  Next we prove an elementary sufficient condition for a triple (K, b, I) ∈ KbI p (G) to be regularly embedded. It is stated in terms of the following notion. Definition 12.2. Let (B, t, K) ∈ BtKp (G). Let F ≤ B and let H be a component of CK (F ). We say that H is B-wide if and only if for any normal subgroup D of O2 (Z(H)) t such that t ∈ D and O2 (Z(H)) t /D is cyclic, we have (12A)

m2 (H t /D) ≥ mp (B/CB (H)) + 1, with strict inequality if D = 1.

If H is not B-wide, we say that it is B-narrow. We remark that in the simplest cases O2 (Z(H)) = 1, and then the requirement is m2 (H t) ≥ mp (AutB (H)) + 2. Proposition 12.3. Let (B, t, K) ∈ BtKp (G) and (K, b, I) ∈ KbI p (G) with b ∈ B. Let Cb = CG (b) and C b = Cb /Op (Cb ). Then the following conditions hold: (a) [COp (C b ) (t), I] = 1; and

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12. THEOREM 3: REGULAR TRIPLES AND MATES

147

(b) Either (K, b, I) is regularly embedded or the following conditions hold: (1) [Op (C b ), t, I] = 1; (2) I is B-narrow; and (3) For any subgroup F ≤ B and component H of CI (F ), H is B-narrow. Note in (b) that I is a component of CK (b) and H is a component of CK (b F ), so the term “B-narrow” makes sense. Since Op (Cb ) has odd order by (1A4), Proof. Set P = Op p (Cb ).  CP (t) = CP (t) ≤ Op p (CG (b, t)). But I ≤ O p (E(CG (b, t))), and therefore [Op p (CG (b, t)), I] = 1. This yields (a). Next, suppose that (K, b, I) is not regularly embedded but (b1) fails, i.e., [P , t, I] = 1. With (a) it follows that [P , I] = 1. Set L = Lp (Cb ); we claim that I ≤ L. Indeed, I is a component of CCb (t), so I ≤ L2 (Cb ) by L2 -balance. As Op (Cb ) has odd order by (1A4), I ≤ L2 (C b ). As F ∗ (O2 (C b )) = Op (C b ) and I = O p (I), it follows by [IG , 3.17(ii)] that [O2 (C b ), I] = 1. Hence I ≤  CL2 (C b ) (O2 (C b ))(∞) = E(C b ) = L, so I = O p (I) ≤ L, as claimed.  L Let J = I . If J is a p-component of L, then (K, b, I) is regularly embedded, contrary to assumption. Hence, as I is a component of CJ (t), J is the product of two p-components J1 , J2 interchanged by t, with I on the diagonal. Thus I/Op (I) is a homomorphic image of J1 /Op (J1 ) ∈ Cp , so I/Op (I) ∈ Cp . As Op p (Cb ) has odd order, so does Z(I). Set E = CB (I). Since [I, E] = 1 and E has odd order, [J, E] = 1. Also set Et = [Op p (J), t]. Since [E, t] ≤ [B, t] = 1, EEt = E × E t . Choose Ai ∈ Ep (Ji ) of maximal rank subject to the following conditions: Ai ∩ Op p (Ji ) = 1 and IJ2 (A2 ; 2) = {1}. Thus by [VK , 12.37], applied with the roles of J, J/Z, and K there played by Ji /Op (Ji ), I/Op (I), and K, we have mp (A1 ) ≥ mp (I/Z(I)) − 1, and if m2,p (I/Z(I)) > 0, then mp (A2 ) > 0. Then mp (B) = m2,p (G) ≥ m2,p (EEt J) (12B)

≥ mp (EEt A1 A2 ) ≥ mp (EEt A1 ) ≥ mp (EEt ) + mp (J1 /Op p (J1 )) − 1 = mp (E) + mp (Et ) + mp (I/Z(I)) − 1.

If E t = 1, and B maps into Inn(I) ∼ = I/Z(I), then equality holds throughout, whence mp (A2 ) = 0 and m2,p (I/Z(I)) = 0. By [VK , 12.37b], mp (I) = 1 so |B : E| = p. Then choosing b0 ∈ Ip (CJ1 (E) − Op p (J1 )) and setting B1 = E b0 , we have B1 ∼ = B and m2 (CG (B1 )) ≥ m2 (J2 ) ≥ 2, contradicting Proposition 10.1a. Suppose next that B does not map into Inn(I). Hence some element b0 ∈ B induces a non-inner automorphism on I. By [VK , 3.58], K ∈ Chev(2), so also I ∈ Chev(2). But also I/Z(I) ∈ Cp . As p divides | Out(I)| we conclude that I/Z(I) is one of the following groups: L2 (8), Sp4 (8), U6 (2), D4 (2), or 3D4 (2), with p = 3, 5 5 or 2B2 (2 2 ) or 2F4 (2 2 ), with p = 5. But also B ∈ Ep∗ (BI), which rules out the case I∼ = D4 (2), in which b0 induces a graph automorphism on I. In every other case b0 centralizes elements of Ep∗ (Ji ), i = 1, 2, and m  p (I) = mp (I). Hence we may choose  p (J1 /Op p (J1 )). Also choose A2 ∈ Ep (CJ2 (b0 )) A1 ∈ Ep (CJ1 (b0 )) with mp (A1 ) = m such that A2 ∩Op p (J2 ) = 1, IJ2 (A2 b0  ; 2) = {1} and, subject to these conditions,

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4. THEOREM C5 : STAGE 2

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to maximize mp (A2 ). This time mp (B) = m2,p (G) ≥ m2,p (EJ b0 ) ≥ mp (EA1 b0  A2 ) ≥ mp (EA1 b0 ) = mp (E) + mp (I/Z(I)) + 1 ≥ mp (B), since mp (Out(I)) ≤ 1 in every case. Hence equality holds throughout, whence 5 A2 = 1. This forces I ∼ = L2 (8) or 2B2 (2 2 ). In both cases EA1 is a hyperplane of EA1 b0  ∼ = B centralizing J2 /Op (J2 ) and CJ2 (EA1 b0 ) has even order. Hence m2 (CG (EA1 )) ≤ 2 by Proposition 10.1. But m2 (J2 ) ≥ 3, contradiction. To complete the proof of (b1), we return to (12B) and consider the case that E t = 1 and B maps into Inn(I). Then t centralizes Z(J) = Op (J), whence I = E(CJ (t)) ∼ = J 1 as p is odd. Hence if I = K, then Op (J 1 ) = 1 as O2 (K) = 1; as a result, mp (A1 ) = mp (J1 ), so (12B) now reads (12C)

mp (B) ≥ m2,p (EJ) ≥ mp (EA1 A2 ) ≥ mp (EA1 ) = mp (E) + mp (I/Z(I)),

and we argue to a contradiction as in the paragraph following (12B). Thus, I/Op (I) ↑p K. As K ∈ C2 it follows easily that I/O3 (I) ∼ = 3G2 (3) (with p = 3). Thus by [VK , 12.37], mp (A1 ) = mp (I/Z(I)), (12C) holds, and we argue to a contradiction the same way. This completes the proof of (b1). It remains to prove (b3), since (b2) is the special case F = 1 of (b3). Let F ≤ B and let H be a component of CI (F ). Without loss we may expand F if necessary and assume that F = CB (H). Since I normalizes but does not centralize P , by (b1), the same holds for H. Therefore there is a subgroup D of Z(H t) = Z(H) t such that Z(H) t /D is cyclic and H does not centralize P 0 := CP (D). Choose such a subgroup D of maximal order. By (a), t ∈ D. By the maximality of D, H t /D acts faithfully on P 0 . Hence if we set R = CP 0 (F ), then it follows from the A × B lemma that H t /D acts faithfully on R. Let R be the preimage of R in Cb . Set k = m2 (H t /D). By the Thompson dihedral lemma, there are E2k ∼ = V /D ≤ H t /D and Epk ∼ = X ≤ R such that V X/D is the direct product of k copies of D2p . As CX (V ) = 1 and [V, F ] = 1, XF = X × F has p-rank k + mp (F ). If |D| is even, then since [D, XF ] = 1, we get mp (B) = m2,p (G) ≥ mp (XF ) = mp (F ) + k. If |D| is odd, on the other hand, then there is a 2-element v ∈ V − D centralizing a hyperplane of X and hence a hyperplane of XF , by the structure of V X/D. Hence mp (B) ≥ mp (XF ) − 1 = mp (F ) + k − 1 in this case. We conclude that k ≤ mp (B/F ) + 1, with strict inequality if |D| is even. By definition, this means that H is B-narrow, so (b3) holds and the proposition is proved.  Corollary 12.4. Let (B, t, K) ∈ BtKp (G) and (K, b, I) ∈ KbI p (G) with b ∈ B. If I ≤ Lp (CG (b)), then (K, b, I) is regularly embedded. Proof. If not, our assumptions yield [Op p (CG (b)), I] ≤ Op (CG (b)), which contradicts Proposition 12.3b1.  Recall also that two elements (B, t, K) and (B ∗ , t, K) of BtKp (G) are mates of each other if and only if the following conditions hold: (a) ICG (t) (B ∗ ; 2) = ICG (t) (B; 2); (b) CB ∗ (K) = CB (K); and

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(c) If B ∗ = B, then p does not divide | Out(K)|. Our goal in this section is to prove the following proposition, which yields part of Theorem 3. Proposition 12.5. Let (B, t, K) ∈ BtKp (G). If p = 3 assume that BtK3exc (G) = ∅. Then there is a mate (B ∗ , t, K) ∈ BtKp (G) of (B, t, K) and a triple (K, b, I) ∈ KbI(G) with b ∈ (B ∗ )# such that (K, b, I) is regularly embedded and broad. Moreover, the following conditions hold: (a) If CB (K) = 1, then (K, b, K) is regularly embedded for all b ∈ CB (K);  (b) If CB (K) = 1, then K = O p (E(CG (t))) and b ∈ K. Before proving Proposition 12.5, we recall some notation and terminology. If (K, b, I) ∈ KbI p (G) is regular and broad, there exists a p-component J of CG (b) such that I ≤ J. The set KbIJp (G) is defined as the set of all such quadruples (K, b, I, J) arising this way. Also KIJp (G) is the set of all triples (K; I; J) such that (K, b, I, J) ∈ KbIJp (G) for some b. The elements of KIJp (G) are called nonconstrained neighborhoods. More specifically, if (K, b, I) arose from (B, t, K) ∈ BtKp (G), we call (K; I; J) a nonconstrained {t, b}-neighborhood. Thus the proposition has an immediate corollary. Corollary 12.6. Let (B, t, K) ∈ BtKp (G). If p = 3, assume that BtK3exc (G) = ∅. Then there is a mate (B ∗ , t, K) of (B, t, K), a triple (K, b, I) with b ∈ (B ∗ )# , and a nonconstrained {t, b}-neighborhood (K; I; J). Once Corollary 12.6 is proved, and it is shown that p = 3 (which we shall do in the next section), Theorem 3 will have been proved. To begin the proof of 12.5, let (B, t, K) ∈ BtKp (G) and assume that the conclusion of the proposition fails. By (10B), Lemma 12.7. If p = 3, then K/Z(K) ∼  L± = 4 (3) or G2 (3). We first prove 

Lemma 12.8. If CB (K) = 1, then K = O p (E(CG (t))). 

Proof. Indeed, if H were a second component of O p (E(CG (t))), then H would be B-invariant by [IG , 8.7(iii)], and as O2 (CG (t)) = 1, CH (B) would contain an element of order p. But since B ∈ Bp∗ (G), mp (B) = mp (CG (t)) and so CB (K) = 1, contradiction.  Therefore, when CB (K) = 1, what must be proved is the existence of a suitable regularly embedded triple (K, b, I) ∈ KbI p (G). By [VK , 12.45], there is a mate (B ∗ , t, K) ∈ BtKp (G) of (B, t, K), and a triple (K, b, I) ∈ KbI p (G) such that b ∈ B ∗ ∩ K and I is terminal in K. When CB (K) = 1, we let B ∗ = B, choose an arbitrary b ∈ CB (K)# , and set I = K, so that (K, b, I) ∈ KbI p (G) in this case as well. We fix this notation. By our choices, Lemma 12.9. (K, b, I) is broad. Moreover, I = K or I is terminal in K. Since we are proceeding by contradiction, we may assume that (12D)

(K, b, I) is not regularly embedded.

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150

Set A = CB (K), E = CB (I), Cb = CG (b), C b = Cb /Op (Cb ), and P = Op (C b ), and fix this notation. By Proposition 12.3, [P , t, I] = 1.

(12E)

The analysis splits into two major subcases according as A = 1 or not. Through Lemma 12.19, we assume that A = 1,

(12F)

except that Lemmas 12.11b and 12.13, as to be noted, are proved without this assumption, for later application. Lemma 12.10. Either p = 5 and (K, mp (AutB (K))) = (2J2 , 2), or p = 3 and (K, mp (AutB (K))) = (A6 , 2), (L± 3 (3), 2), (M11 , 2), (F3 , 5), (2Sp6 (2), 3), or (2Suz, 5). Proof. As observed above, we may choose b so that b ∈ A and I = K. By (12D) and Proposition 12.3, for every B ∗ ∈ Ep∗ (CG (t)) with b ∈ B ∗ , every F ≤ B ∗ , and every component H of CK (F ), H is B ∗ -narrow. Therefore [VK , 12.2] applies and yields the lemma, except for the additional possibility of p = 3 and K ∼ = 2J2 . However, in this case, m3 (K) = m2,3 (K) = 2 with m2,3 (K) achieved in a subgroup K ≥ H ∼ = SL2 (3)×A5 . Hence an element of B∗ (G) normalizes a Q8 ×E22 subgroup. As this is not of symplectic type, Proposition 10.2b is contradicted. This completes the proof.  Before considering the remaining cases of Lemma 12.10, we make a reduction. Part (b) of the next lemma will be proved even if A = 1. Lemma 12.11. The following conditions hold: (a) (K, a, K) is not regularly embedded for any a ∈ A# ; and (b) Suppose that p = 3 and K ∼ = 2Suz. If a ∈ B # is such that CK (a) has a ∼ component I1 = 6U4 (3), then (K, a, I1 ) is not regularly embedded. Proof. In (b), if A = 1, whence B ≤ K ∼ = 2Suz, then a ∈ bK (see [IA , 5.3o]), so (K, a, I1 ) is not regularly embedded by (12D). Therefore we may assume that A = 1 and that b is chosen in A. For uniformity, we put I1 = K in (a). Assume by way of contradiction that (K, a, I1 ) is regularly embedded. We argue that (K, b, K) a = Ca /Op (Ca ). is regularly embedded, contrary to (12D). Let Ca = CG (a) and C By (1A4), Op (Ca ) has odd order. We have I1 ≤ J1 for some p-component J1 of Ca . As I1 ≤ K, b centralizes I1 ,  so b normalizes J1 . If I1 ≤ Lp (CJ1 (b)), then I1 = O p (I1 ) ≤ Lp (CJ1 (b)) ≤ Lp (Cb ) by Lp -balance. Hence K ≤ Lp (Cb ). But then (K, b, K) is regularly embedded by Corollary 12.4, a contradiction. Thus, I1 ≤ Lp (CJ1 (b)), so we have [J1 , b] = 1, I1  E(CJ1 (t)) and [I1 , b] = 1. In particular, I1 < J1 and t, b ∼ = Z2p embeds in CAut(J1 ) (I1 ). Suppose that (b) holds. Since I1 ∼ = 6U4 (3), the only possibility, by [VK , 3.36], is that J1 ∼ = 3Ω7 (3);  but then C  (I1 ) is a 2-group, a contradiction. Thus we may assume henceforth Aut(J1 )

that I1 = K. We consider the possible K’s in Lemma 12.10, one by one.

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∼ 2Sp6 (2), then similarly, by [VK , 3.31], and as Op (Cb ) has odd order, If I1 = ∼  J1 = Co3 and F ∗ (CJ1 (t)) = I1 , a contradiction. If I1 ∼ = 2J2 , F3 , or 2Suz, then I1 ↑2 J1 /O2 (J1 ) ∈ Spor ∩ Cp by [III11 , 1.1], but this is impossible by inspection of [IA , 5.3]. If I1 ∼ = M11 , then from [IA , 5.3] we see that J1 ∼ = M c, which has no element of order 33, a contradiction.  Suppose that I1 ∼ = L± 3 (3). Then as J1 ∈ C3 , it follows from [VK , 3.79a] that ± J1 /Z(J1 ) ∼ = Ln (3), n ≥ 4, L3 (9), or P Sp6 (3). In the last two cases, and also if ∼    J1 /Z(J1 ) = L± 4 (3), CAut(J1 ) (I1 ) is a 2-group, contradiction, by [IG , 7.1.4c, 4.5.1]. Thus J1 /Z(J1 ) ∼ = L± n (3), n ≥ 5. Choose x ∈ I2 (J1 t) such that [x, t] = 1,   := L3 (C  (x)) ∼ [I1 , x] = 1, and L = SL± (3). Let L = L3 (CJ (x)). Then J1



n−1

1

I1 = O 3 (I1 ) ≤ L. Since I1 ≤ L2 (CG (t, x)), L2 -balance implies that L ≤  L2 (CG (x)) = E(CG (x)), as G is of even type. Let L∗ = LE(CG (x)) , so that L is a component of CL∗ (a) and the components of L∗ lie in C2 . As I1   CL∗ (t), [VK , 3.54] implies that [L∗ , a] = 1. Hence L = L∗ ∈ C2 , a contradiction. Therefore I1 ∼ = A6 and p = 3. Using [VK , 3.79b] and the fact that J1 ∈ C3 , we obtain that J1 must be a projective orthogonal group of dimension at least 6 over F3 or F32 , or U4 (2), U5 (2), ±  L2 (34 ), L± 3 (9), P Sp4 (9), or L4 (3). In the last six cases, no element of I3 (Aut(J1 )) centralizes t and a component of CJ1 (t) isomorphic to I1 , by [IA , 4.8.2, 4.9.2] in the first two of these cases, and the Borel-Tits theorem in the last four cases. ± Next, suppose that J1 ∼ (9), m ≥ 3. From [IG , 4.5.1], t = Bm (9), m ≥ 3, or Dm  acts on J1 like an involutory isometry of the natural orthogonal J1 -module V with an eigenspace Vt of dimension 3. Let x ∈ I2 (J1 t) with [t, x] = 1, acting on V with an eigenspace Vx of dimension 5 and containing Vt . Thus [I1 , x] = 1 and there  of C  (x) containing I1 and isomorphic to B2 (9). Let L be the is a component L J1  in L3 (CJ (x)). Then I1 = O 3 (I1 ) ≤ L. As I1 ≤ L2 (CG (x)) by full preimage of L 1  L2 -balance, it follows that L ≤ L2 (CG (x)) = E(CG (x)). Let L∗ = LE(CG (x)) , a product of one or three components of CG (x). Then CL∗ (a) has the component L∼ = B2 (9). It follows easily that a component L1 of L∗ lies in Chev(3). But as a ± component of E(CG (x)), L1 ∈ C2 . Therefore L1 ∼ = L± 4 (3), L3 (3), B2 (3), or G2 (3), none of which involves B2 (9), a contradiction. ± Hence, J1 ∼ (3), m ≥ 3, and again t acts on J1 like an involutory = Bm (3) or Dm+1 isometry of the natural orthogonal J1 -module V , this time with an eigenspace Vt of dimension 4 and − type. Let x ∈ I2 (J1 t) with [t, x] = 1, acting on V with an eigenspace Vx containing Vt , and of dimension 2m and − type. As in the previous paragraph, there is a product L∗ of one or three components of CG (x), permuted transitively by a, and a component L of CL∗ (a) with L/O2 (L) ∼ = Ω(Vx ) ∼ = − ∗ ∗ Ω2m (3). Moreover since Vx ⊇ Vt , I1 ≤ L ≤ L . If L is not quasisimple then as its ∼ components lie in C2 , L∗ is the product of three Ω− 6 (3) = U4 (3) components, which ∗ violates [V2 , 9.1]. Thus, L is quasisimple. We have L   CL∗ (a) and I1 = K   CL∗ (t). By [VK , 3.45], it follows that m = 3 and L∗ = L ∼ = Ω− 6 (3). Note that B maps into CAut(J1 ) (t), of 3-rank at most 4, so m3 (CB (J1 )) ≥ m − 4. Then as x ∈ J1 and m3 (C  (x)) ≥ m3 (U4 (3)) = 4, m3 (CG (x)) ≥ m3 (C  (x)) + m3 (CB (J1 )) ≥ m. J1

J1

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152

4. THEOREM C5 : STAGE 2

Choosing any B ∗ ∈ E3∗ (CG (x)), we then have (B ∗ , x, L) ∈ BtK3exc (G), contradicting (10B). The proof is complete.  Remark 12.12. The lemma shows, in view of (a), that when A = 1, any element of A# may be taken as b. We next eliminate the case K ∼ = 2Suz. As with Lemma 12.11b, our proof is valid when A = 1 as well. Lemma 12.13. K ∼  2Suz. = Proof. Suppose false. As p = 3, and by [IA , 5.3o], there is a ∈ B # ∩ K such that I1 := E(CK (a)) ∼ = 6U4 (3). By Lemma 12.11b, (K, a, I1 ) is not regularly embedded. Hence we may assume that b = a and I = I1 . By [IA , 5.3o], E := CB (I) = b A. By Corollary 11.3, Z(K) = t. We make a preliminary observation. Suppose that u ∈ I2 (CG (t)) and H := E(CK (u)) = 1, so that H ∼ = 2Suz, 2L3 (4), 2M12 , or 2J2 with t ∈ H, by [IA , 5.3o]. Let L be the pumpup of H in CG (u), so that L is quasisimple as t ∈ H, and H is a component of CL (u) by L2 -balance. Then by [VK , 3.38a], t ∈ Z(L), forcing H = L. In particular, t ∈ Z(O2 (CG (u))). We divide the analysis into two cases, in both of which we use the preceding observation. Case 1: K = F ∗ (CG (t)). Let S ∈ Syl2 (CG (t)). By the Z ∗ -theorem, t is Gconjugate to some u ∈ S − t. As t = O2 (CG (t)), also u = O2 (CG (u)). But then if E(CK (u)) = 1, it follows from our observation that t ∈ O2 (CG (u)), whence t = u, contradiction. Hence E(CK (u)) = 1, so if we set C = CCG (t) (u), it follows − from [IA , 5.3o] and [VK , 10.52] that C/ t is an extension of Q38 by Ω− 6 (2) or O6 (2), and that we can choose u so that t, u = Z(S). By Burnside’s lemma, tx = u for some x ∈ NG (S), so x normalizes C = CG (Z(S)). However, by [VK , 10.52], [O2 (C), O2 (C)] ∼ = Z2 . It follows that S ∈ Syl2 (G) and hence that x can be taken to be a 2-element interchanging t and u. Moreover, by [VK , 12.32], we can choose u ∈ I, whence b ∈ C. As OutG (C/O2 (C)) is covered by NG (b), we may take x to normalize b, by a Frattini argument. Now u ∈ I, so [u, CP (t)] = 1 by Proposition 12.3a. Since x normalizes b, it normalizes Op p (Cb ), so conjugating by x we get [t, CP (u)] = 1. It follows that t, u centralizes CP (t), CP (u) and hence [tu, P ] = 1. But then I = [I, tu] centralizes P , contradicting Proposition 12.3b. Case 2: K < F ∗ (CG (t)). Note that t ∈ K and by [VK , 3.38a], the pumpup of K in CG (y) is trivial for any y ∈ I2 (CG (K)). Hence K is terminal in G, and so it is standard, by [II3 , Theorem PU∗4 ]. We claim that there is an involution u ∈ CG (t) such that u inverts b, E(CK (u)) ∼ = 2L3 (4), and I0 := E(CI (u)) ∼ = A6 . Indeed by [VK , 12.32], an element w ∈ K exists with all these properties except that w is not an involution, but rather w2 = t. Hence we need only prove that CG (K) contains an element v of order 4 such that v 2 = t, and then take u = vw. If m2 (CG (K)) = 1, this is clear unless t ∈ Syl2 (CG (K)), which would imply the contradiction F ∗ (CG (t)) = K. If m2 (CG (K)) > 1, then by [II3 , Corollary PU2 ], there exists g ∈ G − NG (K) and a four-group U ≤ CG (K) such that U g ≤ NG (K) and (by [IG , 18.8]) ΓU g ,1 (K) < K. Thus by [VK , 8.6] any involution x ∈ U g induces an inner automorphism on K corresponding to an element x0 ∈ K of order 4 with x20 = t. We take v = xx0 , proving the claim.

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Now as I centralizes CP (t) but not P , the same holds for I0 . As [I0 , u] = 1, we may assume, replacing u by ut, if necessary, that I0 does not centralize P 0 := CP (u). Now H := E(CK (u)) ∼ = 2L3 (4) with t ∈ H. We apply our preliminary observation to u and H to conclude that H is a component of CG (u). On the other hand, I0 ≤ CK (u)(∞) = H and P 1 := [P 0 , I0 ] = 1. Since H   CG (u), it follows that H ∩Op p (CG (b)) covers P 1 , a nontrivial I0 -invariant 3-group. But |H|3 = 32 = |I0 |3 , so no such 3-group exists. This contradiction completes the proof of the lemma.  We next prove Lemma 12.14. K ∼ = M11 or F3 . Proof. In either case, ↑2 (K) ∩ C2 = ∅ and the Schur multiplier of K is trivial, by [VK , 3.10] and [IA , 6.1.4]. Hence if m2 (CG (K)) ≥ 2, there exists a terminal pair (t1 , K1 ) with K1 ∼ = K and m2 (C(t1 , K1 )) ≥ 2, by [V2 , 8.5] and [IG , 6.10]. Therefore m2 (CG (K)) = 1 by [V2 , 9.1]. As A = 1 and no element of C2 has 2-rank  1, O2 (CG (t)) ∼ = Q8 , and O 3 (CG (t)) ∼ = SL2 (3) × K. Since m3 (CG (t)) = m ≥ 4 but m3 (M11 ) = 2, K ∼ ∈ Syl = F3 . Now b 3 (CG (t K))  and [CP (t), K] = 1 by Proposition 12.3a, so t inverts P / b . In particular, as [ b , t] = 1, Ω1 (P ) has exponent 3, and is abelian or the product of an abelian group and an extraspecial group.  Since K contains a faithful extension of E25 by L5 (2), m3 (P / b ) ≥ 31. Take a four-group t,  x ≤ CG (t). Replacing x by tx if necessary, we may assume that m3 (CP (x)/ b ) ≥ 16. Hence certainly m3 (CG (x)) > 6 = m3 (CG (t)) = m2,3 (G) by  [IA , 5.6.1], a contradiction. The lemma follows. Lemma 12.15. Suppose that either p = 5 and K ∼ = 2J2 , or p = 3 and K ∼ = 2Sp6 (2). Then the following conditions hold: (a) For any a ∈ A# , t is the unique involution of CCG (t) (K) normalizing a; and (b) p = 3, K = E(CG (t)) ∼ = 2Sp6 (2) is terminal in G, and m ≤ 5. Proof. Let u ∈ I2 (C(t, K)) and let L be the pumpup of K in CG (u). Since t ∈ K, L is a single component, and by [VK , 3.38a], K = L. Thus, K is terminal in G. Suppose that u can be chosen so that in addition, u = t and u normalizes a for some a ∈ A# . By Remark 12.12, we can take b = a. Replacing u by tu if necessary, we may assume that P 1 := [CP (u), K] = 1. However, K   CG (u) by the previous paragraph. As P 1 = 1 it follows that p divides |Op p (K)|, contradicting the simplicity of K/O2 (K). Hence no such u exists, proving (a). Suppose now that (b) fails. Suppose first that K = E(CG (t)). Then A acts faithfully on O2 (CG (t)), which is of symplectic type by Proposition 10.2. If mp (A) = 1, then as mp (B) ≥ 4 and m5 (Aut(J2 )) = 2, we have p = 3, m3 (B) = m = m3 (K) + 1 = 4, so the assertions of (b) hold, contradiction. If m3 (A) ≥ 2, then by (a), m2 (CO2 (CG (t)) (a)) = 1 for all a ∈ A# . Hence p = 3. But if m3 (A) ≥ 3, then some a ∈ A# centralizes a Q8 ∗ Q8 subgroup of O2 (CG (t)), contradiction. Thus m ≤ 5, and (b) holds again, contradiction. Finally, suppose that E(CG (t)) has components K, K1 , . . . , Kn , n ≥ 1. Then B normalizes every Ki by [IG , 8.7(iii)], so A ∩ Ki = 1, i ≥ 1. If n > 1 then choosing a ∈ A ∩ K1# , we have m2 (CCG (tK) (a)) ≥ m2 (K2 ) > 1 as K2 ∈ C2 . This is impossible by (a). So n = 1. Similarly, if p = 5 and m5 (Aut(K1 )) = 1, then as

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m5 (K) = 2, A must act nontrivially on O2 (CG (z)), whence m2 (O2 (CG (z))) ≥ 2, again contradicting (a). Hence if p = 5, then m5 (Aut(K1 )) > 1. Next, by (a), (12G)

I2 (NK1 (a)) ⊆ {t} for all a ∈ A# .

Unless K1 is unambiguously in Chev(2), however, this condition fails by [VK , 12.8] for any a ∈ A# ∩ K1 which is p-central in Aut(K1 ). As B ∈ Ep∗ (CG (t)), B must contain such a p-central element a, and a ∈ A# as a ∈ K1 . Therefore, K1 is unambiguously in Chev(2). Similarly, K1 is simple, because of [VK , 12.10]. Thus, I2 (NK1 (a)) = ∅ for all a ∈ A# . Suppose that some a ∈ A# induces an outer automorphism on K1 . Then by [IA , 4.9.1, 4.7.3A] and (12G), it induces an outer diagonal automorphism. But then (12G) is contradicted by [VK , 12.7]. Hence, A induces inner automorphisms on K1 , and so as O2 (CG (t)) = 1, A = (A ∩ K1 ) × CA (K1 ), with A ∩ K1 ∈ Ep∗ (K1 ). Suppose that A ∩ K1 is noncyclic. Then 

Γ2A∩K1 ,1 (K1 ) = 1. Hence by [IA , Theorem 7.3.3], and as mp (A ∩ K1 ) = mp (K1 ), K1 ∼ = L3 (2n ), 2n ≡ 1 2   = ±1 (mod 3) if p = 3, and F4 (2 2 ) if p = 5. In the first case, every element of I3 (K1 ) is strongly real; in the second case, for any x ∈ Ip (K1 ), CK1 (x) has even order, by [IA , 4.8.7]. This contradicts (12G). Therefore, A ∩ K1 is cyclic. But then by Burnside’s transfer theorem, A ∩ K1 is inverted in K1 , and the proof is complete.  We now eliminate the 2Sp6 (2) case. We split the analysis into two parts, first proving Lemma 12.16. If K ∼ = 2Sp6 (2), then m2 (O2 (CG (t))) = 1. Proof. By Lemma 12.15, we have that K = E(CG (t)) ∼ = 2Sp6 (2) is terminal in G, and we put R = O2 (CG (t)). Suppose that m2 (R) ≥ 2, choose u ∈ I2 (R)−{t}, # and set V = t, u ∼ = E22 . By the terminality, t ∈ O2 (CG (v)) for every v ∈ V , so  t centralizes ICG (v) (V ; 2 ) . As v is arbitrary, t centralizes IG (V ; 2 ). Choose, by [VK , 3.34], a ∈ B ∩ K with I1 := E(CK (a)) ∼ = SL2 (9) and t ∈ I1 . Then V ≤ CG (a), so [O3 3 (CG (a)), t] = 1 by the previous paragraph and (1A4). Hence (K, a, I1 ) is regularly embedded; otherwise Proposition 12.3, applied to a and I1 in place of b and I, would give [O3 3 (CG (a)), t] ≤ O3 (CG (a)), contradiction. Let J1 be the 3-component of CG (a) containing I1 . Set J1 = J1 /O3 (J1 ), so that J1 ∈ C3 . Then I1 ∼ = SL2 (9) is a component of CJ1 (t) with t ∈ I1 . Notice that since a ∈ K and Out(K) = 1, CCG (a) (t) = CCG (t) (a) = CK (a)C(t, K), and since K = E(CG (t)), C(t, K) is 2-constrained. Thus, E(CCG (a) (t)) = E(CK (a)) = I1 . It  follows therefore by [VK , 3.66] that J1 ∼ = L± 3 (9) and that CAut(J1 ) (I1 ) is a {2, 5}group. As in Lemma 12.11 this is impossible: b centralizes I1 , so b normalizes J1 and 3 3  (J1 ), implying [b, O (J1 )] = 1; by L3 -balance, O (J1 ) ≤ L3 (CG (b)), [J1 , b] ≤ O  3K so K = I1 ≤ L3 (CG (b)) and (K, b, K) is regularly embedded, contradiction.  Thus m2 (R) = 1, as claimed. Lemma 12.17. K ∼ = 2Sp6 (2).

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Proof. Suppose false and continue the above analysis, so that A = b. In this final case, m = m3 (B) = 4, and since O3 3 (CG (b)) has odd order, CO3 3 (CG (b)) (t) ≤ O2 (CCG (t) (b)) = b. By [VK , 10.51], K contains a subgroup W ∼ = D8 ∗ D8 ∗ D8 with Z(W ) = t. Then t inverts P/ b for some P ∈ Syl3 (O3 3 (CG (b))), and if we let b ≤ X ≤ P with X/ b elementary and irreducible under W , then m3 (X/ b) = 8. Since W centralizes b, X is not extraspecial, by [VK , 15.25]. Hence X ∼ = E39 , and then m3 (CX (w)) = 5 for any w ∈ I2 (W ) − {t}, contrary to m = 4. The lemma is proved.  By Lemmas 12.10, 12.13, 12.14, 12.15, and 12.17, K∼ = A6 or L± 3 (3), with p = 3; and m3 (A) ≥ 2. Lemma 12.18. If for some a ∈ A# there exists an involution u ∈ C(t, K) − t normalizing a, then [u, a] = 1, m = 4, and u may be chosen so that CG (u) has a component L ≥ K such that L ∼ = L± 4 (3) or 2U4 (3). Proof. Let a and u satisfy the hypothesis of the lemma. We take b = a, so that I = K. Then t, u acts on P and [P , t, K] = 1 by (12E). Replacing u by ut if necessary, we may assume that W := [CP (u), t, K] = 1. Then W = [W , K] by [IG , 4.3(iv)]. Therefore there is W ≤ O3 3 (CG (a)) ∩ CG (u) mapping onto W and such that W = [W, K]. But then L2 -balance implies that K has a vertical pumpup L in CG (u), with W ≤ L. Since L ∈ C2 , it follows from [VK , 3.110] that L ∼ = L4 (3) if K ∼ = L3 (3); L/Z(L) ∼ = A6 ; and K ∼ = U3 (3). = U4 (3) if K ∼ If a ∈ CG (u), then as [K, a] = 1, it follows that [L, a] = 1. Since W ≤ O3 3 (CG (a)) it follows that [L, W ] ≤ O3 (L). But W = [W, K] and K ≤ L so W = 1, a contradiction. Thus u inverts a, as asserted. It follows that for any a ∈ A# , the only involution of C(t, K) centralizing a is t. It remains to show that m = 4, or equivalently m3 (A) = 2. Let us look more closely at CG (t). If K = E(CG (t)), then A acts faithfully on R := O2 (CG (t)), which is B-invariant and hence of symplectic type by Proposition 10.2. By the previous paragraph t is the only involution of CR (a) for any a ∈ A# , which forces m3 (A) = 2, as desired. So we may assume that E(CG (t)) has components K, K1 , . . . , Kn , all in C2 , with n ≥ 1. Again using the previous paragraph we obtain that n = 1, A acts faithfully on K1 , and m2 (R) = 1. By the previous paragraph and [VK , 12.11, 12.9],  either m3 (A) ≤ 2 or K1 is simple and K1 ∈ Chev(2), and Γ2A,1 (K1 ) = 1 < K1 . By [IA , Theorem 7.3.3], m3 (A) ≤ 2. Thus m3 (A) ≤ 2 in any case, so m = m3 (A) + 2 = 4. The proof is complete.  Lemma 12.19. Suppose that for all a ∈ A# , there are no involutions in CG (t) − {t} normalizing a. Then m = 4 and there is u ∈ I2 (CG (t)) such that  O 3 (E(CG (u))) ∼ = L± 4 (3) or 2U4 (3). Proof. We have m3 (A) ≥ 2. Let K, K1 , . . . , Kn be the components of CG (t). As in the previous lemma, if n ≥ 1, then n = 1 and A acts faithfully on K1 . By [VK , 12.8, 12.10], K1 is simple and unambiguously in  Chev(2). Again Γ2A,1 (K1 ) = 1, yielding K1 ∼ = L3 (2c ),  = ±1, 2c ≡  (mod 3), or

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F4 (2 2 ) . Again a 3-central element of K1 in A is real in the first case, and centralized by an involution in the second case [IA , 4.7.3A], contradicting our hypothesis. Therefore K = E(CG (t)) and A acts faithfully on R, which, being B-invariant, is of symplectic type by Proposition 10.2. Because of our hypothesis, m2 (CR (a)) ≤ 1 for all a ∈ A# . It follows that m3 (A) = 2 and R has width at least 2. In particular, m = 4 and moreover, if we expand R to Q ∈ Syl2 (C(t, K)), then Ω1 (Z(T )) = t. Note that if K were terminal in G, then since K is outer well-generated by [VK , 8.4], we would have Qg ≤ NG (K) for some g ∈ G − NG (K) by [II3 , 16.13], and ΓQg ,1 (K) < K; but m2 (Qg ) ≥ m2 (R) = 3, so this violates [VK , 8.5]. So K is not terminal in G. Since Z(Q) is cyclic it follows from [IG , 7.4] that there is u ∈ I2 (Q) such that the pumpup L of K in CG (u) is vertical and K is not semirigid in L. By [VK , 14.4], since L ∈ C2 , we have L ∼  = L± 4 (3) or 2U4 (3), and the lemma follows.

2

1

The preceding two lemmas lead to the contradiction BtK3exc (G) = ∅, so they imply that Proposition 12.5a holds. Hence since we are proceeding by contradiction, we conclude: Lemma 12.20. A = 1. Then Propositions 12.3 and 10.1a, (10B), and [VK , 12.1] combine to show that (a) and (b) of Proposition 12.5 hold, or p = 3 and K is one of the following groups. Note that K ∼  2Suz by Lemma 12.13. = (12H)

Co1 , 2Co1 , Suz, F i22 , 2F i22 , F i23 , F i24 , F3

As we are assuming that Proposition 12.5 fails, we have Lemma 12.21. K is isomorphic to one of the groups in (12H). The argument eliminating most cases is one that we have used before. We find B ∗ ∈ E3∗ (CG (t)) and b ∈ (B ∗ )# such that I = E(CK (b)) is quasisimple, b is normalized by an involution u ∈ K t − t, I0 := E(CI (u)) = 1 and I0 ≤ H = E(CK (u)) with H quasisimple. These exist in most cases by [VK , 12.33]. Let L be the pumpup of H in CG (u). As usual, replacing u by tu if necessary, we may arrange that P 1 := [CP (u), I0 ] = 1. Then L ∩ CG (b) covers P 1 as L   CG (u) and  P 1 = [P 1 , I0 ]. Let P1 = O p (P0 ), where P0 is the full inverse image in CG (b) of P 1 . Then P1 = [P1 , I0 ]. However, given the possibilities for I0 and L, in most cases I0 leaves invariant no p-nilpotent subgroup of L of odd order with nontrivial Sylow p-subgroup, thus yielding a contradiction. In one case, we must use I0 × t instead of I0 to obtain the desired contradiction. With this argument, and referring freely to [IA , 5.3], we can reduce to the following cases. Lemma 12.22. CG (t) ∼ = 2Co1 , or K ∼ = F i or F3 . 24

Proof. Refer to the general argument above. If K ∼ = (2)F i22 , take b with I ∼ = (2)U4 (3) and B ∗ ∈ E3∗ (CK (b)) ⊆ E3∗ (K) [IA , 5.6.1]. Then there is an involution u inverting b and centralizing I = I0 , and with u ∈ H ∼ = (2)2U6 (2). Thus, u ∈ L, so Z(L) = 1. By L2 -balance and [VK , 12.34a], either L = H (type “a”), L is the product of two t-conjugate components with diagonal H (type “b”), or L ∼ = (2)F i22 . In each case the projection of I0 on a component of L acts nontrivially on R/Op (R), where R is the corresponding projection of P1 . We therefore have a contradiction, by [VK , 12.34b].

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If K ∼ = F i23 , take b with I ∼ = Ω7 (3), u inverting b and centralizing I = I0 , and then H ∼ = F i24 since m3 (CG (u)) ≤ m = m3 (K) = 6. = 2F i22 . Note that L ∼ Thus either L is of type a or b, or L ∼ = F i23 by [VK , 12.34a]. Again we have a contradiction by [VK , 12.34b].  Next, suppose that K ∼ = Suz. We take b with I = O 3 (CK (b)) ∼ = 3U4 (3), u inverting b, H ∼ = A6 . Since m3 (B) = m3 (CG (t)) and A = 1, = L3 (4), and I0 ∼ B ≤ K and C(t, K) is a 3 -group. Hence b ∈ Syl3 (O3 3 (CCG (t) (b))), whence CP (t) ≤ P 1 . Thus, I0 × t acts faithfully on P 1 . Now L is either of type a or b, or L ∼ = L3 (16) or Suz, by [VK , 12.34a]. Note also that if L ∼ = Suz, then as H is a component of CL (t), t induces an inner automorphism on L. But then by [VK , 12.34c], I0 t ∼ = A6 × t acts on no 3-nilpotent subgroup R of L of odd order and faithfully on R/O3 (R). This is a contradiction. If K ∼ = J2 and = Co1 , take b with CK (b) ∼ = 3Suz, u inverting b with I0 ∼ ∼ H = G2 (4). Then either L is of type a or b or else L ∼ = G2 (16) or Co1 , by [VK , 12.34a]. Again [VK , 12.34b] yields a contradiction. Finally, suppose that K ∼ = 2Co1 and CG (t) > K. As Out(K) = 1, F ∗ (CG (t)) > K. As above, B ≤ K and so a Sylow 2-subgroup of CF ∗ (CG (t)) (K) is B-invariant, hence cyclic or of symplectic type by Proposition 10.2. Thus CG (K) contains an element of order 4 with square t. Since b is inverted by an element of K mapping on an involution of K/ t ∼ = Co1 , there is thus an involution u ∈ CG (t) inverting b, and we reach the same contradiction as in the Co1 case. Now the lemma follows by Lemma 12.21.  We eliminate these last three cases in succession, using [IA , 5.3lwx]. Lemma 12.23. K ∼  F i24 . = Proof. Suppose false and take b with I ∼ = P Ω+ 8 (3). By [VK , 10.23], a Sylow 2-subgroup S of I has cyclic center and has a normal four-subgroup V whose involutions are cycled by a 3-element of NI (V ). Thus AutI (V ) ∼ = Σ3 . Since I acts faithfully on P , we have [CP (v), w] = 1 for any two distinct v, w ∈ V # . Without loss, take v ∈ Z(S) and set Y = CI (v). Then Y has a normal subgroup of index 8 which is the central product of four copies of SL2 (3). Also, Y /O2 (Y ) acts irreducibly on O2 (Y )/ v. Now V ≤ O2 (Y ) as V  S. Since w acts nontrivially on R := CP (v), Y / v acts faithfully on R. But O2 (Y )/ v ∼ = E28 , so by the Thompson dihedral lemma, m3 (P ) ≥ 8. Thus m3 (CG (v)) ≥ 8, so m ≥ 8. However, as A = 1, m = 7 by [IA , 5.6.1].  Lemma 12.24. CG (t) ∼  2Co1 . = Proof. Originally due to Finkelstein and Solomon [Fin3], this lemma is greatly eased by our K-proper hypothesis. Suppose that CG (t) ∼ = 2Co1 . There is a subgroup N ≤ CG (t) of odd index such that W := O2 (N ) = F ∗ (N ) ∼ = E212 , ∼ N = NCG (t) (W ), and N/W = M24 , with W the binary Golay code under the action of N/W . Moreover, W = J(S), where S ∈ Syl2 (N ) ⊆ Syl2 (CG (t)). Besides the orbit {t} of length 1, the N -orbits on W # have lengths 759, 759, and 2576. 1 = N1 /W . Then N = CN (t). We argue that Let N1 = NG (W ) and N 1 (12I)

N = N1 .

 | is a subsum of 1 + 759 + 759 + 2576, containing the 1 term 1 : N If false, then |N and at least one other term. The only possibilities, since |N1 : N | divides |L12 (2)|,

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are |N1 : N | = 72 31 and 212 − 1. In particular, |N1 : N | is odd, so O2 (N1 ) = 1.  ) fixes CW (N ) = t so C  (N  ) ≤ Z(N  ) = 1. For any prime r > 2, Also CN1 (N N1 2 1 )| ≤ r , which forces [N  , O2 (N 1 )] = 1 = Or (N 1 ). Therefore F ∗ (N 1 ) contains |Or (N    N and is a simple K-group. By [VK , 15.22], N1 = N . This proves (12I). Finally since W = J(S), N controls the G-fusion of the conjugates of t that it contains, by [IG , 16.9]. Hence t is weakly closed in W with respect to G. But every involution of CG (t) ∼ = 2Co1 has a conjugate in W [IA , 5.3l], so t is weakly closed in CG (t) with respect to G. This contradicts the Z ∗ -theorem [IG , 15.3], so the lemma follows.  Lemma 12.25. K ∼ = F3 . Proof. Suppose false and refer to [IA , 5.3x]. Choose b ∈ I3 (K) such that I = E(CK (b)) ∼ = G2 (3), and let u be an involution inverting b, so that I0 := E(CI (u)) ∼ = L2 (8). Then I0 acts faithfully on [P , I]. Now b ∈ Syl3 (O3 3 (CG (b, t))) so t inverts [P , I] b / b. Thus, Ω1 ([P , I]) is the product of an abelian group and (possibly) an extraspecial group. Let du = dim(C[P ,I]b/b (u)) and define dtu similarly. Without loss, du ≥ dtu . By [VK , 15.21], m3 (C[P ,I]b (u)) ≥ 7, a  contradiction since m2,3 (G) = m = m3 (K) = 5 by [IA , 5.6.1]. This at last completes the proof of Proposition 12.5. 13. Theorem 3: p Must Be 3 With Proposition 12.5 and Corollary 11.6, we can show that if p ≥ 5, involutions centralized by an element of Bp∗ (G) have no p-components in the layer of their centralizers. Together with Proposition 12.5, this will complete the proof of Theorem 3. Proposition 13.1. Assume that p > 3. Then BtKp (G) = ∅. In particular if B ∈ Bp∗ (G) and t ∈ I2 (CG (B)), then E(CG (t)) is a p -group and B acts faithfully on a Sylow 2-subgroup of F ∗ (CG (t)). We will need part of the following lemma, which will be useful in other contexts. Lemma 13.2. Let (B, t, K) ∈ BtKp (G), p an odd prime. Then K ∼  F1 . = Proof. Suppose that K ∼ = F1 . Let Q ∈ Syl2 (C(t, K)). By [V2 , 9.1], m2 (Q) = 1. As Out(K) = 1 [IA , 5.3z], Ct = K × W , with F ∗ (W ) =: Q0 ≤ Q. We argue that |Q| = 2. By the Z ∗ -theorem, tg ∈ Ct − t for some g ∈ G. Then tg ∈ t K ≤ CG (Q) so Qh ≤ Ct , where h = g −1. As Qh is a Sylow   subgroup  h t = of C(th , K h )  CG (th ), it follows that Qh ∈ Syl2 (Qh )CG (t,t ) . Let C  h ∈ Syl2 (N  ) for some N   C  (th ). As m2 (Qh ) = 1, it is Ct /C(t, K). Then Q K h clear from the structure of CK (t ) [IA , 5.3z] that N ∼ = Z2 . We have proved that Ct = K × t. Since B ≤ Ct , we have mp (K) ≥ 4, and hence p = 3 or 5 [IA , 5.6.1]. Fix x ∈ Ip (K) such that L := E(CK (x)) = 1 and mp (L) is maximal subject to this condition. Set C = CG (x) and C = C/Op (C). By L2 -balance, L ≤ L2 (C). We claim that (13A)

L ≤ Lp (C).

Indeed, if p = 3, the maximality guarantees that L ∼ = 3F i24 , while if p = 5, then L ∼ = F5 ; and mp (L x) = mp (K) [IA , 5.3, 5.6.1]. We have m2 (L) = 11 or 6

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according as p = 3 or 5, and m2,p (G) = mp (K) < m2 (L) − 1 [IA , 5.6.1], so L must centralize Op (C), by the Thompson dihedral lemma. Consequently L centralizes O2 (C)/O{2,p} (C) and so (13A) follows. If L is the diagonal of the product L∗ of two t-conjugate p-components of C, then using [IA , Theorems 5.6.1, 5.6.2] we find m2,p (L∗ ) ≥ m2,p (L) + mp (L) > mp (K) = m2,p (G), a contradiction. Thus L ≤ L0 for some p-component L0 of C, with L  CL0 (t). Since ↑2 (F i24 ) = ↑2 (F5 ) = ∅ by [VK , 3.10], L covers L0 /Op (L0 ). Note that CG (L t, x) = t × CK (L x) with CK (L x) = x [IA , 5.3], so C(x, L0 ) = W t with W of odd order and W/ x inverted by t. In particular, L0 = Lp (C). Moreover, since Op (C) has odd order by (1A4), t must invert an element y ∈ W of order p. Then CL0 (t, y) covers  LOp (L0 )/Op (L0 ), whence y centralizes O p (CL0 (t)) = L. Now choose w ∈ Ip (L) such that E(CK (x, w)) ∼ = D4 (3) or U3 (5), according as p = 3 or p = 5. Let u ∈ x, w − x. By Lp -balance and the (Bp ) property, E(CK (x, w)) ≤ E(CK (u)). This forces E(CK (u)) ∼ = L, whence u ∈ xK [IA , 5.3]. Now t, y centralizes L x, so t, y centralizes u and acts on E(CG (u)) ∼ = L. But as u ∈ xK and t centralizes E(C), t also centralizes E(CG (u)). As t inverts y, therefore, ycentralizes E(CG (u)) = E(CK (u)). Thus y centralizes  # E(CK (u)) | u ∈ x, w = K, the last by [VK , 8.30]. Now we have NG (K) ≥ y, t × K ∼ = D2p × F1 . Let z ∈ I2 (K) with CK (z) ∼ = 2F2 , and let Lz be the subnormal closure of CK (z) in Cz . If Lz is a diagonal pumpup of CK (z), then m2,p (G) ≥ 2mp (CK (z)) > mp (K) = m2,p (G) [IA , 5.6.1], contradiction. Therefore Lz is a single component, and as z ∈ Lz , Lz = CK (z) by [VK , 3.38b]. Then as CCt (CK (z)) = t, z, CG (Lz ) has a dihedral or semidihedral Sylow 2-subgroup R with z ∈ Z(R), by [IG , 10.25]. If Cz has a component J = Lz , it follows that z ∈ J and so R ∩ J ∈ Syl2 (J) with R ∩ J a quaternion group. This, however, is impossible as J ∈ C2 . Hence, Lz = E(Cz ). Now t, y ≤ C(z, Lz ) so [y, O2 (Cz )] = 1. As O2 (Cz ) ≤ R, the only possibility is that p = 3, O2 (Cz ) ∼ = Q8 , and CG (Lz ) ∼ = GL2 (3). Now let x∗ ∈ I3 (Lz ) with Lx∗ ,z := E(CLz (x∗ )) ∼ = 2F i22 [IA , 5.3y]. As ∗ (∞) ∗ |CK (x ) |2 ≥ |2F i22 |2 , the only possibility is that x ∈ xK [IA , 5.3z]. Since L0 is the unique 3-component of C and C(x, L0 ) is solvable, we must have Lx∗ ,z ≤ Lx∗ := L3 (CG (x∗ )). As z = Z(R), it follows that CG (Lz ) acts faithfully on Lx∗ := Lx∗ /O3 (Lx∗ ) ∼ = 3F i24 , centralizing the image of Lx∗ ,z , which is isomorphic to 2F i22 . But by [IA , 5.3v], the centralizer of that image in Aut(Lx∗ ) is a 2-group,  so cannot contain GL2 (3). This contradiction completes the proof. Now we can prove the main result of this section. Proof of Proposition 13.1. It suffices to prove that BtKp (G) = ∅. For  then, if B ∈ B0∗ (G) and t ∈ I2 (CG (B)), a component K  O p (E(CG (t))) would p yield a triple (B, t, K) ∈ BtK (G), contradiction; hence E(CG (t)) is a p -group. Then for a B-invariant S ∈ Syl2 (F ∗ (CG (t))), CB (S) ≤ CB (E(CG (t))O2 (CG (t))) = CB (F ∗ (CG (t))) = 1, using [VK , 15.23]. Suppose for a contradiction that (B, t, K) ∈ BtKp (G). By Proposition 12.5, (B, t, K) may be chosen so that there is (K, b, I) ∈ KbI p (G) such that b ∈ B and (K, b, I) is regularly embedded, and either I = K or I is terminal in K, with b ∈ K and CB (K) = 1 in the latter case.

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Suppose first that I = K. Then I ∈ C2 and CB (K) = 1. Let J be the pcomponent of CG (b) with I ≤ J, so that I is a component of CJ (t). Note that Op (J) has odd order by (1A4). Set J0 = J/[J, Op (J)] and let I0 be the image of I in J0 . By Corollary 11.5, J0 /Op (J0 ) ∈ C∗p . We quote [VK , 3.82] with J0 and I0 playing the roles of J and K there, and τ there induced by the involution t. Assumption (b) of that lemma holds by Proposition 10.1. We conclude that p = 3, a contradiction. In particular, I < K, I is terminal in K with respect to the prime p. By [VK , 9.31] and Lemma 13.2, K ∈ Chev(2), so I ∈ Chev(2) by [IA , 4.9.6]. Thus by terminality and [VK , 3.52], mp (CB (I)) = 1, so mp (AutB (I)) = m − 1 ≥ 3. Again since (K, b, I) is regularly embedded, and by Corollary 11.5, there is a pcomponent J of CG (b) with I ≤ J and J/Op (J) ∈ C∗p , and I is a component of CJ (t). Now [VK , 3.82] applies with I in the role of K there, and we again reach the contradiction p = 3. The proof is complete.  As discussed following the statement of Proposition 12.5, this completes the proof of Theorem 3. 14. The Nondegenerate Case: Theorem 4 In this section, we carry out the proof of Theorem 4. We let (B, t, K) ∈ BtK3 (G) and assume that BtK3exc (G) = ∅. We let (K, b, I) ∈ KbI(G) with b ∈ (B ∗ )# for some mate (B ∗ , t, K) of (B, t, K), chosen if possible so that I = K, and we suppose that (K, b, I) is regular and broad. Thus there exists a nonconstrained {t, b}-neighborhood (K; I; J). In particular K ≥ I ≤ J. We set Cb = CG (b), C b = Cb /O3 (Cb ), and Ct = CG (t), and call (K; I; J) nondegenerate if and only if K > I and I < J. Note that I   CK (b) and I   CJ (t). We shall establish Theorem 4, by proving the following proposition. Proposition 14.1. With the above notation, assume that (K; I; J) is nondegenerate. Then the following conditions hold: (a) B ≤ K; (b) t ∈ I∗G (B; 2) and t ∈ Syl2 (CG (K)); (c) K = E(Ct ); (d) Up to isomorphism, (K; I; J) and m = m2,3 (G) = m3 (K) are one of the following: (1) (2F2 ; 2F i22 ; (3)F i24 ), with m = 6; (2) (F i23 ; Ω7 (3); P Ω± 8 (3)), with m = 6; (3) ((2) 2E6 (2); (2)U6 (2) or D4 (2); (3)F i22 ), with m = 5; (4) (F i22 ; U4 (3); L4 (9)), with m = 5; (5) (2F i22 ; 2U4 (3); Ω7 (3) or P Ω± 8 (3)), with m = 5; (6) (Sp8 (2) or (2)F4 (2); Sp6 (2); U6 (2) or D4 (2)), with m = 4; or (7) ((2)U6 (2), U5 (2), Sp8 (2), or (2)D4 (2); U4 (2); L4 (3), U4 (3), 3U4 (3), or [3 × 3]U4 (3)), with m = 4; and

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(e) If K ∼ = 2 2E6 (2) in (d3), then for some b1 ∈ (B ∗ )# , there is a nonconstrained {t, b1 }-neighborhood (K, I1 , J1 ) satisfying (d3) such that I1 ∼ = 2U6 (2) and J 1 ∼ = F i22 . The stringent restrictions on (K; I; J) derive mainly from the tension between the conditions K ∈ C2 and J ∈ C3 . Still, there are many further cases to be ruled out, even in the nondegenerate case, where they are least common. We note that in our setup, the element b lies in B ∗ but not necessarily in B. However, it suffices to prove the result for B ∗ in place of B. For, the only conclusions involving B are (a) and (b), and we argue that if they hold for B ∗ , then they hold for B. Since (B, t, K) and (B ∗ , t, K) are mates, ICt (B; 2) = ICt (B ∗ ; 2) by definition. Thus, condition (b) of Proposition 14.1 holds for B if and only if it holds for B ∗ . Assume that (a) holds for B ∗ and that B = B ∗ . Then by Definition 1.4c, p = 3  does not divide | Out(K)|, so B ≤ O 3 (Ct ) ≤ KC(t, K); but B ∗ ≤ K, so C(t, K) is  a 3 -group as m = m3 (B ∗ ) = m3 (Ct ) and O2 (Ct ) = 1; thus K = O 3 (Ct ) ≥ B, as required. As a result, we may assume that (14A)

B ∗ = B,

and in particular b ∈ B. The first lemma holds for any nonconstrained neighborhood, degenerate or not. Lemma 14.2. K > I if and only if C(t, K) is a 3 -group. Proof. Since B ∈ B3∗ (G), m3 (B) = m3 (Ct ). Hence B normalizes K by [IG , 8.7]. Moreover, C(t, K) is not a 3 -group if and only if CB (K) = 1. In that case, we have chosen b ∈ CB (K) so K = I. In the contrary case, b ∈ CB (K) so K > I, as asserted.  Now we begin the proof of Proposition 14.1. In particular, we assume that (K; I; J) is nondegenerate. Lemma 14.3. We have B ≤ K, m = m3 (K), K = E(Ct ), and C(t, K) is a 2-group of 2-rank 1. Proof. Assume that B ≤ K. As K > I, C(t, K) is a 3 -group. Hence some a ∈ B induces an outer automorphism on K, and in particular 3 divides | Out(K)|. As K ∈ C2 , it follows that K ∈ Chev(2). Recall that b ∈ K. Now [VK , 11.12] reduces us to one of two specific configurations: (A) K ∼ = U4 (2), whence m = m3 (B) = m3 (P GU6 (2)) = 5, and a = U6 (2) and I ∼ may be chosen so that H := E(CK (a)) ∼ = U5 (2). Moreover, since I is a component of CJ (t) and [t, b] = 1, with m2,3 (J t, b) ≤ 5, J/O3 3 (J) ∼ = P Sp4 (9), L± 4 (3), or Ω7 (3), by [VK , 3.28]. (B) K ∼ = D4 (2), and m = m3 (B) = m3 (Inndiag(2E6 (2))) = 6, and = 2E6 (2)a , I ∼ a may be chosen so that H = E(CK (a)) ∼ = 2 D5 (2); moreover, J/O3 (J) ∼ = F i22 by [VK , 11.12]. In both cases, we may assume, moreover, that [I, a] = 1, so that CB (I) = a, b ∼ = E32 .  Now, a normalizes J and maps into O 3 (CAut(J) (I)) = 1, by [VK , 3.65] in (A) and [IA , 5.3] in (B). Let Ja = L3 (CJ (a)), so that J a = J and I ≤ Ja . Hence the subnormal closure H ∗ of I in CG (a) contains Ja . By L3 -balance, H ∗ is a 3component of CG (a) or the product of three such 3-components cycled by b, and

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in any case Ja is a 3-component of CH ∗ (b). On the other hand, I is a component of CH ∗ (t), so any component H0 of H ∗ /O3 (H ∗ ) is isomorphic to either I or L5 (4) (in case (A)) or I or D5 (4) (in case (B)), by [VK , 3.10]. Hence L5 (4) involves L± 4 (3) or P Sp4 (9), or D5 (4) involves F i22 , both of which are impossible by Sylow 3-structure. This contradiction completes the proof that B ≤ K. The remaining assertions follow directly. Since m3 (B) = m2,3 (G), B ∈ E3∗ (Ct ) and m3 (B) = m3 (K). Now O2 (Ct ) = 1 as G has even type, and so C(t, K) is a 3 -group. By Proposition 10.1, m2 (C(t, K)) = 1. Thus C(t, K) has a normal 2-complement by Frobenius’s theorem, so C(t, K) is a 2-group. In particular, K =  E(Ct ). Thus conclusions (a) and (c) of Proposition 14.1 hold. We turn to (e). By our setup and Proposition 10.1, the hypotheses of [VK , 13.3] hold, with the roles  of X and J there played here by J t and J in C b = Cb /O3 (Cb ). Recall that O3 (Cb ) has odd order by (1A4). Therefore if [VK , 13.3a] holds, i.e., m2,3 (J t) − m3 (Z(J t)) ≥ m, then as [J t , b] = 1, we have m2,3 (J t, b) > m = m2,3 (G), a contradiction. Likewise [VK , 13.3b] does not hold, by our assumption. If conclusion (e), (f), (g), (k), or (l) of [VK , 13.3] holds, then conclusion (d4), (d5), (d2), (d6) or (d3) of Proposition 14.1 holds. We consider the remaining cases of [VK , 13.3] in succession. Case (c). Thus K ∼ = Co2 , I ∼ = U4 (2), and J/O3 (Z(J)) ∼ = L± 4 (3), with m = 4. If C(b, J) had even order, then m2,3 (Cb ) ≥ m3 (J b) > 4 = m, contradiction. So |Cb |2 ≤ | Aut(J)|2 ≤ 210 . Let T ∈ Syl2 (Ct ). As Out(K) = 1, T = R × Q where R = T ∩ K ∈ Syl2 (K) and Q = CT (K). By Lemma 14.3, m2 (Q) = 1. Let A = J(T ) = A0 × t where by [VK , 10.41], A0 = CK (A0 ) = J(R) ∼ = Aut(M22 ). By the = E210 and NK (A0 )/A0 ∼ same lemma, A0 contains representatives of every class of involutions in K. Let N = NG (A). By the Z ∗ -Theorem and the above remark, tG ∩ A = {t}. But by [V2 , 3.1], tG ∩ A = tN . By a Frattini argument, tN = tNG (CT (A)) . Hence t is not characteristic in CT (A), whence Q = t. Note that by [VK , 10.41], I2 (K) = z K ∪ y K ∪ xK , where z, y, x are extremal in T ; z = Z(T ) ≤ [T, T ], y ∈ [CT (y), CT (y)]; and in the N N N decomposition A# 0 = z ∪ y ∪ x the cardinalities are 1023 = 77 + 330 + 616. The N -orbits in A are represented by t, z, y, x, zt, yt, and xt. As z ∈ [CT (z), CT (z)] and similarly for y, y and z are not in tN . Suppose that T ∈ Syl2 (G). Since z ∈ tN , zt ∈ tN by Burnside’s lemma. Hence tN is the union of {t} with one or more of the following: (yt)N , xN , and (xt)N , these being of cardinalities 330, 616, and 616, respectively. Of course |tN | divides | Aut(N )| = |GL11 (2)|, but no appropriate sum of orbit cardinalities has this property. Therefore T ∈ Syl2 (G). As Z(T ) = z, t with z ∈ [T, T ], we must have zt ∈ tN . The possibilities for |tN | are then those of the previous case increased by |(zt)N | = 77, viz, 408, 694, 1024, 1310, and 1640, as well as 78. Of these, only 408 = 17.24 and 1024 = 210 divide |GL11 (2)|. So |N/A| = |CN/A (t)||tN | = |N ∩K/A0 |.408 or |N ∩K/A0 |.210 . In the first case, 2 O (N ∩ K/A0 ) ∼ = M22 centralizes F (N/A), whose order divides 17.24. It follows easily that E(N/A) contains M22 as a non-normal subgroup with index dividing 17.48. In particular E(N/A) is quasisimple, but no quasisimple K-group has order

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as required. Therefore |tN | = 210 . Let S ∈ Syl2 (N ) with T ≤ S; z ∈ Z(S)  then and N/A = (S/A).(N ∩ K)A/A. It follows that z N ⊆ A0 , and so z N = A0 , as 11 divides |N ∩ K|. Thus A0  N and tN = A − A0 . Suppose that W/A := O2 (N/A) = 1. Then since N ∩ K normalizes CA (W ) = CA0 (W ) and is irreducible on A0 , [W, A0 ] = 1 and W/A ∼ = A0 as N -module. Therefore |W : CW (t)| = |A0 | = 210 so N = CN (t)W . Let v ∈ I3 (NK (A)). Then as NK (A) has a single class of elements of order 3, and has Sylow 2-subgroups isomorphic to E32 , |CA0 (v)| = |CW/A (v)| ≥ 24 . Also |CAut(M22 ) (v  )|2 = 23 for v  ∈ I3 (M22 ), so |CN (v)|2 ≥ 211 . This contradicts the inequality established in the first paragraph of this case, however. Therefore O2 (N/A) = 1. It follows that L := F ∗ (N/A) = E(N/A) is simple, containing Aut(M ), where M ∼ = M22 has index dividing 210 . But by [VK , 15.12] there is no such simple K-group L with 210 | Aut(M22 )| dividing | Aut(L)|. This contradiction rules out Case (c) of [VK , 13.3]. Case (d). Here (K; I; J) = (2Suz, 6U4 (3), 3Ω7 (3)) and m = 5. In this case t induces an inner automorphism on J and there is u ∈ tJ ⊆ Cb such that [t, u] = 1 and L := L3 (CI (u)) satisfies L ∼ = Ω5 (3). But in Ct , u acts on I = E(CK (b)), and the structure of L is therefore impossible, by [VK , 12.31]. This is a contradiction, so Case (d) of [VK , 13.3] does not occur. Case (h). Here (K; I; J) = (F3 , G2 (3), G2 (9)). In this case there is an involution u ∈ NK (b)) inducing an (involutory) graph-field automorphism on I, by [IA , 5.3x]. Then u normalizes J. But by [IA , 2.5.12], Out(J) ∼ = Z4 , with t mapping to an involution of Out(J) as it induces a field automorphism on J. Hence as [K, t] = 1, u ∈ CJt (t) = t × I, so u induces an inner automorphism on I, contradiction. So Case (h) of [VK , 13.3] does not occur. Case (i). Here we may assume that (K; I; J) = (2F2 , 2F i22 , F i23 ), for otherwise (d1) of the proposition holds. There is a 2-element u ∈ CK (b) inducing an outer automorphism on I, by [IA , 5.3y]. Then u normalizes J. But | Aut(J) : I| is odd and so every 2-element in NAut(J) (I) maps into Inn(I), contradiction. So Case (i) of [VK , 13.3] does not occur. Case (j). Suppose next that [VK , 13.3j] holds but Proposition 14.1d7 does not, so that (K; I; J/O3 (Z(J))) = (D5 (2), U4 (2), L± 4 (3)) with m = 4. Let b1 ∈ B # have minimal (2-dimensional) support on the natural K-module, and H = E(CK (b1 )) ∼ = 2 D4 (2). Then because m2 (H) = 6 and m3 (AutB (H)) = 3, H is B-wide by definition. Hence by Proposition 12.3, (K, b1 , H) is regularly embedded, so that H ≤ L for some 3-component L of CG (b1 ), and H is a component of CL (t). Therefore L/O3 (L) ∈ C3 and H ∼ = 2 D4 (2) ↑2 L/O3 3 (L). These last two conditions are incompatible, however, by [VK , 11.10]. Hence if [VK , 13.3j] holds, then Proposition 14.1d7 holds. Case (m). Here (K; I; J) = (2 D5 (2), D4 (2), (3)F i22 ), with m = 5. By [IA , 5.3t], CJ (t) ∼ = Aut(D4 (2)). However, CJ (t) covers CJ (t) as O3 (J) has odd order. Therefore OutCJ (t) (I) = Out(I) ∼ = Σ3 , so OutCt (I) = Out(I). However, as K  Ct , OutCt (I) ≤ OutAut(K) (I) ∼ = Z2 . Thus [VK , 13.3m] does not occur. This completes the proof of Proposition 14.1d. We next prove Proposition 14.1b:

Lemma 14.4. t ∈ Syl2 (CG (K)) and t ∈ I∗G (B; 2).

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Proof. Recall that t ∈ Q ∈ Syl2 (C(t, K)), with m2 (Q) = 1 by Lemma 14.3 and Proposition 10.1. We first show that Q = t. As b ∈ K, [Q, b] = 1, and so Q ≤  = N/CN (J). As [J, t] = 1, Q ∼  embeds in C := C N := NG (J). Set N =Q Aut(J) (I). If C has elementary abelian Sylow 2-subgroups, then Q = t, as desired. Otherwise, ∼ by Proposition 14.1d and [VK , 3.55], (K; I; J) = (2F i22 , 2U4 (3), P Ω+ 8 (3)) and C = + ∼  J ∼ D8 , so we may assume that Q = Z4 . Therefore Q = P SO8 (3). By [VK , 10.22], there is an involution u ∈ QJ such that [u, t] = 1, and H := L3 (CJ (u)) and H0 := L3 (CJ (t, u)) satisfy H ∼ = P Sp4 (3). As H0 × b ≤ CCt (u) = L4 (3) and H 0 ∼ and Ct /Q embeds in Aut(K), it follows using [IA , 5.3t] H0 ≤ CCt (u)(∞) ≤  that E(CK (u)). By L2 -balance, H0 ≤ E(CG (u)), so H = H0H ≤ E(CG (u)). Let M be a component of E(CG (u)) on which H projects nontrivially. Then m3 (M b) ≥ m3 (H) + 1 = 5. But m = 5, so it follows that M  CG (u) and H ≤ M . Consequently H is a component of CM (b). Since M ∈ C2 , we must have M = H, by [VK , 3.93]. But as m3 (CG (u)) = m, this contradicts our assumption  that BtK3exc (G) = ∅. The proof is complete. Finally we prove Lemma 14.5. Proposition 14.1e holds. Proof. Assume that (K; I; J) satisfies K ∼ = 2 2E6 (2). Then m = 5, and by # [IA , 4.7.3A], there is b1 ∈ B such that I1 := E(CK (b1 )) ∼ = (2)U6 (2) is terminal in K. As m3 (B) = 5 and m2 (I1 ) ≥ 9 by [VK , 10.38], I1 is B-wide, so (K, b1 , I1 ) is regular by Proposition 12.3b. Also (K, b1 , I1 ) is broad as b1 ∈ B ≤ K. Hence there is a component J1 of CG (b1 ) such that I1 is a component of CJ1 (t), and (K; I1 ; J1 ) is a nonconstrained {t, b1 }-neighborhood. It therefore remains only to show that J 1 := J1 /O3 (J1 ) ∼ = F i22 . Note that as O3 (CG (b1 )) has odd order, J 1 is a nontrivial pumpup of I 1 ∼ = (2)U6 (2). Hence J 1 ∼ = 2F i22 by [VK , 3.17]. The proof is complete.  This completes the proof of Proposition 14.1 and with it, the proof of Theorem 4. 15. Theorem 5: The Degenerate Case K > I We continue the notation of the opening two paragraphs of the previous section, including B = B ∗ (see (14A)), b, t, K, I, J, Cb , C b , and Ct . We prove Theorem 5 in this section and the next three. Thus we prove: Proposition 15.1. There exists b ∈ B and a nonconstrained {t, b }-neighborhood (K; I  ; J  ) which is nondegenerate. Suppose that the proposition fails. We assume given a nonconstrained {t, b}neighborhood (K; I; J). Then (K; I; J) is itself degenerate, and we have three cases: (15A)

(1) K > I and I = J; (2) K = I and I < J; and (3) K = I and I = J.

If (15A1) holds, we will generally argue to a contradiction, but in some cases we will produce b ∈ B and a nondegenerate nonconstrained {t, b }-neighborhood (K; I  ; J  ). That is the content of this section. Note that in this case, we may assume that b ∈ K, by Proposition 12.5b. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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In cases (15A2, 3) we will argue to a contradiction, in the next three sections. Lemma 15.2. Suppose that K > I and I = J. Then O2 (I) = 1, and K is isomorphic to one of the following groups: (a) Co1 , Suz, F i22 , F i23 , F i24 , or F2 , with I ∼ = 3Suz, 3U4 (3), U4 (3), Ω7 (3), P Ω+ 8 (3), or F i22 , respectively; (b) Co2 or F3 , with I ∼ = U4 (2) or G2 (3), respectively; (c) D7 (2), D6 (2), or Sp12 (2) (with I ∼ = SU6 (2)); D6− (2) or Sp10 (2) (with ∼ ∼ I = U5 (2)); or D5 (2) (with I = U4 (2)); (d) (2)F4 (2) or Sp8 (2), with I ∼ = Sp6 (2); or Sp8 (2), with I ∼ = U4 (2); or ∼ Sp6 (8), with I = Sp4 (8); (e) U7 (2), with I ∼ = SU6 (2); 2E6 (2), with I ∼ = U6 (2) or D4 (2); or 2 D5 (2), with ∼ I = D4 (2) or U5 (2); or (f) K/Z(K) ∼ = U4 (2). = U6 (2), U5 (2), or D4 (2), with I ∼ Proof. Since I = J and O3 (Cb ) has odd order, O2 (I) = 1. Also CB (K) = 1 by Lemma 14.2, so m3 (AutCt (K)) = m3 (B) ≥ 4. If K ∈ Spor, then | Out(K)| = 1 or 2 so m3 (K) ≥ 4. Thus (a) or (b) holds by [IA , 5.6.1, 5.3] and Lemma 13.2; I is uniquely determined up to conjugacy by the conditions that it be terminal in K and m3 (CK (b)) = m3 (B) = m3 (K). If K ∈ Chev(3), then as K ∈ C2 , | Out(K)| is not divisible by 3. But 1 = I  E(CK (b)) by our setup, whereas E(CK (b)) = 1 by the Borel-Tits theorem, contradiction. As m3 (Aut(L2 (q))) < 4 for q ∈ FM9, while K ∈ C2 , we may assume that K ∈ Chev(2). Hence I ∈ Chev(2), by [IA , 4.9.6]. Now I ∼ = J ∈ C3 . In particular as O3 (Cb ) has odd order, Z(I) has odd order. If I/Z(I) ∈ Chev(2) ∩ Chev(3), then I/Z(I) ∼ = L2 (8), Sp4 (2) , G2 (2) , or U4 (2), by [IA , 2.2.10]. Otherwise, by [V3 , 1.1], I/Z(I) ∼ = U5 (2), U6 (2), Sp6 (2), D4 (2), F4 (2), Sp4 (8), 3D4 (2), or 2F4 (2 2 ) . 1

As b ∈ K ∈ Chev(2), the structure of centralizers of inner automorphisms of K 1 [IA , 4.8.2, 4.8.4, 4.7.3A] rules out I ∼ = 3D4 (2) and I ∼ = 2F4 (2 2 ) . If I ∼ = L2 (8), then similarly we find that K ∼ = Sp4 (8), 3D4 (2), U3 (8), L4 (8), L2 (83 ), or G2 (8), whence m3 (Aut(K)) = 3 [IA , 4.10.3a, 2.5.12]. But as CB (K) = 1, m3 (Aut(K)) ≥ m3 (B) ≥ 4, contradiction. Likewise if I ∼ = Sp4 (8), then as I is terminal in K, the only possibility is K ∼ = Sp6 (8), as in conclusion (d). So we may assume that q(I/Z(I)) = 2. Since b ∈ K, q(K) ≤ 2 by [IA , 4.2.2], so q(K) = 2. [We use the notation q(K) for the “level” of K ∈ Chev. That is, if K/Z(K) ∼ = [L, L] and L = d Σ(q) ∈ Lie, then q(K) = q. Equivalently but more technically, if (L, σ) is a σ-setup for L, then q(K) = q(L, σ). See [IA , 2.1.13].] If 3 divides | Outdiag(K)|, then K ∼ = U3n (2), n ≥ 2, or 2E6 (2). In view of conclusions (e) and (f), we may assume that K ∼ = U3n (2), n ≥ 3. Then since q(I) = 2 and I is terminal, with CCt (b) containing B ∈ E3∗ (Ct ), I ∼ = U3n−2 (2), which is not the case. Thus, we may assume that 3 does not divide | Outdiag(K)|. As D4 (2) is allowed in (f), we may assume that Out(K) is a 3 -group, so in particular B/ b acts faithfully on I, and hence m3 (Aut(I)) ≥ 3. Hence it remains to consider the cases I/Z(I) ∼ = Un (2), 4 ≤ n ≤ 6, Sp6 (2), D4 (2), and F4 (2).

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∼ Sp6 (2), then the By [IA , 4.8.2, 4.8.4, 4.7.3A], F4 (2) does not occur. If I = Dynkin diagram of K has a double bond, and as I is terminal in K, K/Z(K) ∼ = Sp8 (2) or F4 (2). If I ∼ = D4 (2), then either K ∼ = Dn± (2) for some n, or K ∼ = 2E6 (2), by [IA , 4.8.2, 4.7.3A]. The only possible orthogonal group, as m3 (K) = m3 (B) = m3 (b I) = 5, is K ∼ = D5− (2), as in (e). Thus, we are reduced to the case I/Z(I) ∼ = Un (2), 4 ≤ n ≤ 6. If K ∼ = Um (2), then as B ≤ K and I is terminal in K, m ≤ 7, as in (e) and (f). If K is any other classical group, then with [IA , 4.8.2], K is as in (c), or K ∼ = Sp8 (2), 2 D5 (2), or D4 (2), as in (d–f). Finally if K is of exceptional type, then from [IA , 4.7.3A] we see that K/Z(K) ∼ = 2E6 (2). Note that if K ∼ = [X]2E6 (2) with X ∼ = [X]U6 (2), so X = 1. The proof is complete.  = 1, Z2 or Z2 ×Z2 , then I ∼ Lemma 15.3. Suppose that K > I and I = J. Let Q0 be a nontrivial Binvariant 2-subgroup of C(t, K). Then m2 (Q0 ) = 1 or Q0 ∼ = Q8 ∗ Z for some cyclic Z. Proof. We have t ∈ Q0 , B normalizes Q0 , and B ∩ K centralizes Q0 . As K is one of the groups in Lemma 15.2, | Out(K)|3 ≤ 3. So |B : B ∩ K| ≤ 3. Then by Proposition 10.1, m2 (Q0 ) ≤ 2, with strict inequality if B ≤ K. We may therefore assume that |B : B ∩ K| = 3, m2 (Q0 ) = 2, and m2 (CQ0 (B)) = 1. Now Q0 ∈ IG (B; 2) is of symplectic type by Proposition 10.2. Since m2 (Q0 ) = 2, and Q0 = [Q0 , B] ∗ CQ0 (B), [Q0 , B] ∼ = Q8 , and CQ0 (B) is cyclic, completing the proof.  To deal with the possibility that O3 (Cb ) might be nontrivial, we use our conditions (1A6, 7). Thus, there exists B ∗ ∈ E3 (G) such that B < B ∗ , and G is balanced ∗ as any element of E34 (G). Let W = Θ1 (G; B ∗ ) = with respect to B ∗as#well O3 (CG (x)) | x ∈ (B ) and M = NG (W ). Expand B ∗ to P ∈ Syl3 (G) such that CP (b) ∈ Syl3 (Cb ). Then we have Lemma 15.4. Γoo P,2 (G) ≤ M and K ≤ M . Either O3 (CG (x)) = 1 for all x ∈ I3 (P ) with m3 (CP (x)) ≥ 4, or M < G. Proof. By [V2 , 1.3d], Γoo P,2 (G) ≤ M . Let B ≤ R ∈ Syl3 (Ct ). By the same (G) ≤ M . If K is as in Lemma 15.2ab, then K = Γoo result, Γoo R,2 R,2 (K) ≤ M by [VK , 8.39]. If, on the other hand, K is as in Lemma 15.2cdef, then by [VK , 8.28], there exists B0 ∈ E4 (R) and a hyperplane B1 ≤ B0 such that IK (B1 ; 2) = {1}. But then K = ΓB1 ,2 (K) ≤ Γoo P,2 (G) ≤ M , the first containment by [IA , 7.3.1] and the Borel-Tits theorem. The lemma follows.  Lemma 15.5. Suppose that K > I and I = J. Then J  Cb . Proof. Suppose false and let g ∈ Cb with J = J g . As K > I, CB (K) = 1, so m3 (Aut(K)) ≥ m3 (B) = m2,3 (G) ≥ m2,3 (II g b) ≥ m2,3 (I) + m3 (I) − m3 (Z(I)), with strict inequality at the end if O3 (I) = 1. However, the outer inequality is false by [VK , 12.44], except for the case (K, I) = (Sp6 (8), Sp4 (8)), in which case m2,3 (I) = 1, m3 (I) = 2, and m3 (Aut(K)) = 4. However, in this case, there is b0 ∈ B # inducing a field automorphism of order 3 on I and J. If b0 normalizes J g , then m2,3 (JJ g b, b0 ) ≥ 5 > m2,3 (G). If b0 does not normalize J g , then Cb has at least three 3-components isomorphic to J, and clearly then m2,3 (Cb ) > 4, again a contradiction. This proves the lemma. 

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15. THEOREM 5: THE DEGENERATE CASE K > I

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Lemma 15.6. Suppose that K > I and I = J. Assume that the following conditions hold: (a) CAut(K) (b I) has odd order; and (b) K = I g | g ∈ K, bg ∈ B ∩ K. Then there exists y ∈ I3 (CG (B)) ∩ O2 (CG (K)) such that y t = y −1 . Proof. Set C = C(b, J), so C  Cb by Lemma 15.5. Thus O3 (C) ≤ O3 (Cb ), and since O3 (Cb ) has odd order, so does O3 (C). As I = J, t ∈ C, and hence t ∈ C − O3 (C).

(15B)

Now CG (t I b) maps into CAut(K) (I b), which has odd order by our hy  pothesis. Therefore O 2 (CC (t)) ≤ C(t, K), so O 2 (CC (t)) is a 3 -group by Lemma  14.2. Let Q0 be a B-invariant Sylow 2-subgroup of O 2 (CC (t)). Then by Lemma 15.3, m2 (Q0 ) = 1 or Q0 ∼ = Q8 ∗ Z for some cyclic Z. In either case, t char Q0 so Q0 ∈ Syl2 (C). Moreover by [V2 , 3.2], t ∈ Z ∗ (C). In view of (15B), t inverts an element y ∈ I3 (O3 3 (C)), indeed we may assume that [B, y] = 1 by the A × B-lemma. A similar argument proves that t ∈ Z ∗ (CG (K)). It remains to show that [K, y] = 1. For then, y = [y, t] ∈ O2 (CG (K)). Let b ∈ bK ∩ B, say b = bg , g ∈ K. Then b centralizes t, y. We have t ∈ C(b, J)  Cb , so conjugating by g, t ∈ C(b , J g )  CG (b ). In particular y = [y, t] ∈ C(b , J g ). Also J g = I g O3 (J g ) ≤ I g W . (By Lemma 15.4, K ≤ M .) Since [B, y] = 1, y ∈ M . Letting g vary and using hypothesis (b), we see that y normalizes I g W | g ∈ K, bg ∈ B ∩ K = I g | g ∈ K, bg ∈ B ∩ K W = KW and centralizes KW/W . Let L = L3 (CKW (y)), so that LW = KW and L is t-invariant. By Lemma 15.3 and Proposition 10.1, a B-invariant Sylow 2-subgroup S of the 3 -group C(t, K) has rank at most 2 and Ω1 (CS (B)) = t. Hence by [VK , 15.11], (∞) C(t, K) is solvable and so K = Ct . But W has odd order so K = CKW (t)(∞) ≤ L. In particular, [K, y] = 1, and the proof is complete.  We now assume through Lemma 15.11 that (15C)

K and I are as in Lemma 15.2a

and aim for a contradiction. Lemma 15.7. The conclusions of Lemma 15.6 hold. Moreover, m2 (Q) = 1. Proof. For hypothesis (a) of Lemma 15.6a, the tables [IA , 5.3] show that CInn(K) (b I) = b. Then, in the cases where Out(K) = 1 (K ∼ = Suz, F i22 , or F i24 ), the desired statement can only fail if there is an involution in Aut(K) − Inn(K) centralizing b I. But the tables again show that | b I|3 > |CK (v)|3 for every outer involution v ∈ Aut(K), contradiction. Thus hypothesis (a) holds. Since | Out(K)| ≤ 2, B ≤ K, so B ∈ E3∗ (K). In particular [B, Q] = 1, so m2 (Q) = 1 by Proposition 10.1. Also by [VK , 8.27], there is g ∈ K such that bg ∈ B and I, I g  = K. This implies hypothesis (b) of Lemma 15.6. The proof is complete.  As a result, we can set X = O2 (CG (K)) and fix y ∈ I3 (X) with y t = y −1 .

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Next, by [IA , 5.3], we can fix u ∈ I2 (K) with H := E(CK (u)) ∼ = L3 (4), G2 (4), 2U6 (2), 2F i22 , 2F i22 , 22E6 (2), or 2F2 , according to the respective isomorphism type of K. Lemma 15.8. We have H  E(CG (u)). Proof. Let X = O2 (CG (K)) = 1 and D = t CG (X). Because of y, t ∈ CG (X). Also B ≤ K ≤ CG (X) ≤ D. Let R ∈ Syl2 (O3 (CG (X))) be B tinvariant. Then R t is cyclic or of symplectic type, and t = Ω1 (CRt (B)). But CRt (B) = t CR (B) has rank 1 by Proposition 10.1, so CR (B) = 1 and thus R = 1. Hence O3 (CG (X)) has odd order.  Let D = D/O3 (CG (X)). Since I covers J, I ≤ L3 (D) = E(D). Hence K =

I

K

≤ E(D), and the subnormal closure L

of K in D satisfies K   CL (t) and I   CL (b). Hence L is a single component, and by [VK , 11.11], K = L. Let N be the pumpup of H in CG (u). We argue that N ≤ D. Namely, H u ≤ K ≤ CD (t X), so N = H E(CG (u)) is t X-invariant. Whatever type of pumpup N is of H, CAut(N ) (H) is 2-closed by [VK , 10.74c]. On the other hand CX (t) ≤ O2 (Ct ) = 1,  so X = [X, t] and then [X, N ] = 1. That is, N ≤ D, as asserted. Since N = H N and H ≤ K   D, N ≤ K. This implies that H = N , and the lemma follows.  Note that as H u ≤ K, C(u, H) ≥ CG (K)  y. Hence as t inverts y, (15D)

t ∈ O2 (C(u, H)).

Now let T ∈ Syl2 (C(u, H) ∩ CG (t)) and expand T to S ∈ Syl2 (C(u, H)). Thus by [VK , 10.74a], T /Q is elementary abelian and |T /Q| ≤ 4, with equality only if K∼ = Suz, Co1 , or F i24 . We next prove Lemma 15.9. We have T < S. Proof. Suppose false, so that T = S. If H = E(CG (u)), then C(u, K) is 2constrained and O2 (C(u, K)) ≤ O2 (CG (u)) = 1, so as t ∈ Z(T ), t ∈ O2 (C(u, H)), contradicting (15D). Thus, CG (u) has a component H0 = H. Now u is a direct factor of T , so u ∈ H0 . If |T /Q| = 2, so that T = Q u, this implies that Q contains a Sylow 2-subgroup of H0 , which is impossible as H0 ∈ C2 and m2 (Q) = 1. Thus, unless K ∼ = Q × E22 , and similarly, a Sylow 2-subgroup R0 of = F i24 , T ∼ H0 embeds in Z2 × Q. Moreover, R0 ∩ K =: v ∼ = Z2 and (since K ∼ = Suz or Co1 ), v ∈ uK . Since H0 ∈ C2 , m2 (R0 ) > 1, so t, v = Ω1 (R0 ). But since E(Ct ) ∼ = E(CG (u)) ∼ = E(CG (v)), t ∈ v G . Hence by Burnside’s lemma and transfer theorem, H0 is not simple, a contradiction. We are therefore reduced to the case K ∼ = F i24 , with |T /Q| = 4, so T = Q u, v, where v 2 ∈ Q and CK (v) ∼ = F i23 . If v cannot be chosen so that v 2 = 1, then m2 (Q v) = 1 and a Sylow 2-subgroup of H0 embeds in Q v, again giving a contradiction. So we may assume that v 2 = 1. Let Hv = E(CK (v)) ∼ = F i23 . Now R0 embeds in Q v. Since H0 ∈ C2 , Z(H0 ) = 1 and so v may be chosen in tH0 . We may also assume that u was chosen so that b ∈ H ∼ = = F i22 and E(CH (b)) ∼ U4 (3), whence E(CHv (b)) ∼ = Ω7 (3). Then t centralizes J, while v acts on J like a reflection. That is, L3 (CG (v, b)) has a 3-component M such that M/O3 (M ) ∼ = Ω7 (3). Since v ∈ tG , there is b ∈ I3 (Ct ) such that L3 (CCt (b )) has a 3-component

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15. THEOREM 5: THE DEGENERATE CASE K > I

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Mb ∼ = K, so Mb is a component of CK (b ). However, = M . Then Mb ≤ Ct no such component of a 3-centralizer in Aut(K) exists [IA , 5.3]. This contradiction proves the lemma.  (∞)

As a consequence, Lemma 15.10. We have Q = t. Moreover, |CS (t)| = |T | = 2|T /Q| ≤ 8 and T is abelian. Proof. Since CS (t) = T < S, t is not characteristic in T . But Φ(T ) ≤ Q, so t = Ω1 (Q) ≤ Φ(T ). As Φ(Q) ≤ Φ(T ), Φ(Q) = 1, whence Q = t. The second assertion follows immediately as t ∈ K and |T /Q| ≤ 4.  By Lemma 15.10 and [VK , 10.74d], C(u, K) has no component isomorphic to H, so H  CG (u). Next we show Lemma 15.11. u ∈ H0 for some component H0 = H of CG (u). Proof. Again by (15D), t ∈ O2 (C(u, H)) so [t, H0 ] = 1 for some component H0 = H of CG (u). We assume for a contradiction that u ∈ H0 . If H0t = H0 , then by the structure of T = CS (t), |T | = 8 and |T /T ∩ O2 (H0 H0t )| = 4, so Z(H0 ) = 1 and T acts faithfully on H0 H0t , a contradiction as u ∈ T . Thus, H0t = H0 . Let S0 = S ∩ H0 . If Ω1 (T ) = t, u, choose s ∈ I2 (CS0 (T )); thus s = t or tu. Either way, CS0 (t) = CS0 (tu) contains a four-group, which then lies in Ω1 (CS (t)) = Ω1 (T ) = t, u, so u ∈ H0 , contradiction. Thus, we may assume that T = t, u, v ∼ = Suz, Co1 , or F i24 , and by [VK , 10.74b], there is = E23 . Hence K ∼ an involution g ∈ CK (u) ≤ CG (t, u) such that [v, g] = u. If T ∩ H0 ≤ t, u, then u = [T, g] ≤ H0 H0g . Since u ∈ H0 and t normalizes H0 , with T = CG (H0 t), H0 = H0g and |CH0g (t)| ≤ 4. But this implies that H0g ∩ S has maximal class, and since Z(H0g ) = 1, H0g ∈ C2 , contradiction. Therefore T ∩ H0 ≤ t, u. Again as u ∈ H0 , CS0 (t) = T ∩ H0 = t  where t = t or tu. But this is absurd as u ∈ Z(S)  and |H0 |2 > 2. The lemma is proved. Now we can prove Lemma 15.12. If K > I and I = J, then Lemma 15.2a does not occur. Proof. Assume false and continue the above argument. By [VK , 10.69], there is no group H0 t with H0 ∈ C2 such that Z(H0 ) = 1, as Lemma 15.11 requires, but also with t2 = 1 and CS1 (t) abelian of order at most 8, as Lemma 15.10 requires,  where S1 ∈ Syl2 (H0 t). This contradiction proves the lemma. Lemma 15.13. If K > I and I = J, then K ∼  F3 . = ∼ F3 and CK (b) = b × I Proof. Suppose by way of contradiction that K = with I ∼ = G2 (3). By [IA , 5.3x], there exists E ∈ E25 (K) such that N := NK (E) is a non-split extension of E ∼ = E23 , so = E25 by L5 (2). Let b ∈ I3 (N ) be such that CE (b ) ∼  ∼ that CN (b ) = Z3 × E23 L3 (2). As 7 divides |CK (b )| it follows from [IA , 5.3x] that b ∈ bK , so we can assume without loss that b = b ∈ N . As Out(K) = 1 and B ≤ K, it follows that [B, Q] = 1, m2 (Q) = 1 by Proposition 10.1, and T ∈ Syl2 (Ct ) has the structure T = R × Q, with R ∈ Syl2 (K). As K

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4. THEOREM C5 : STAGE 2

has one class of involutions and Z(R) =: z ∼ = Z2 , the Z ∗ -theorem and Burnside’s NG (T ) lemma imply that t ∩ Z(T ) = {t}. As R is not a quaternion group, z ∈ tNG (T ) so tNG (T ) = {t, tz}. In particular t is not characteristic in T , so Z(Q) = t and by the Krull-Schmidt theorem, Q = t. Thus as CK (R) = Z(R) by [IA , 5.3x], CG (T ) = Z(T ) and |NG (T ) : T | = 2. Let R0 = O2 (CK (z)) ∼ = 21+8 + . By [VK , 15.13], NG (T ) − T contains a 2-element g ∈ CG (R0 ). By [VK , 10.44], E ≤ R0 , so g ∈ CG (E). Set F = t × E = Z(T )E. Then tg ∈ F − E − {t}, as E # ⊆ z G and t and z are not conjugate. In particular g ∈ NG (F ). Moreover, as N is transitive on F − E − {t}, tG ∩ F = tNG (F ) = F − E and z G ∩ F = E # . Thus, E  NG (F ). We have NG (F ) = CNG (F ) (E)NK (F ) with |NG (F ) : NCG (t) (F )| = |F − E| = 25 . It follows that x → [x, t] is an isomorphism from CNG (F ) (E)/CG (F ) to E preserving the action of AutN (E). In particular, CNG (F ) (b) has a perfect subgroup M of order 2a 3 · |L3 (2)|, a = 6 or 7, and M ≤ CN (b)M ≤ J. But |J|2 = 26 , so this is a contradiction, and the lemma is proved.  Lemma 15.14. If K > I and I = J, then Lemma 15.2b does not hold. In particular, K ∈ Chev(2). ∼ U4 (2). Now Out(K) = 1, so ∼ Co2 and I = Proof. Suppose false, so that K = T = R×Q where T ∈ Syl2 (Ct ) and R ∈ Syl2 (K). We repeat paragraphs 2, 3, 4, 5, 6, and 8 of the nondegenerate case K ∼ = Co2 . (This is Case (c), following Lemma 14.3.) Thus Q = t, A = J(T ) = A0 × t where A0 = J(R) ∼ = E210 , and N := NG (A) = N0 W where N0 = NK (A0 ) ∼ = E210 Aut(M22 ) and W = O2 (N ) = CN (A0 ). Here W/A ∼ = A0 as N0 -modules. Moreover, if we let P ∈ Syl3 (N0 ), then |CA0 (x)| = 24 and |[CA0 (x), y]| = 22 for all x, y ∈ P such that P = x, y. Furthermore, the fact that m2 (CA0 (x)) = 4 means that x ∈ bK [IA , 5.3k], so without loss we may take b ∈ P and y ∈ I, so y ∈ J. Let W0 := [CW (b), y] = [W0 , y] ≤ J. By the Borel-Tits theorem, W0 y embeds in a parabolic subgroup M of J ∼ = I ∼ = U4 (2) with W0 embedding in O2 (M ). However, as |W0 : Z(W0 )| = 1 or 4, no such embedding is possible, a contradiction. The lemma is proved.  Lemma 15.15. If K > I and I = J, then Lemma 15.2c does not hold. Proof. Suppose false, so that K ∼ = D7 (2), D6± (2), D5 (2), Sp12 (2) or Sp10 (2). Let V be the natural K-module. In every case the support of b on V has maximal dimension among all elements of B # . Now choose, however, b ∈ B # having minimal such dimension, and set I  = E(CK (b )) ∼ = D6− (2), D5∓ (2), D4− (2),  Sp10 (2), or Sp8 (2), respectively. Accordingly, m2 (I ) = 15, 10, 6, 15, or 10, and so m2 (I  ) ≥ m3 (K) + 2 = m3 (B) + 2. Hence, I  is B-wide and so by Proposition 12.3, (K, b , I  ) is regular. Let J  be the 3-component of CG (b ) containing I  , so that (K; I  ; J  ) is a nonconstrained {t, b }-neighborhood. As (K, I  ) does not appear in Table 1.1, (K; I  ; J  ) must be degenerate, i.e., I  /O3 (I  ) ∼ = J  /O3 (J  ). But  then K and I must appear in Lemma 15.2, which they do not. This contradiction completes the proof.  Lemma 15.16. If K > I and I = J and Lemma 15.2d holds, then K ∼ = Sp8 (2) and there exists a nondegenerate nonconstrained {t, b }-neighborhood (K; I  ; J  ) for some b ∈ B # . Proof. If K ∼ = Sp6 (2). Since = Sp8 (2), fix b ∈ B # such that I  = E(CK (b )) ∼  m2 (I ) = 6 by [IA , 3.3.3], while m3 (K) = 4 by [IA , 4.10.3a], (K, b , I  ) is regular

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by Proposition 12.3. Therefore there exists a nonconstrained {t, b }-neighborhood (K; I  ; J  ). If (K; I  ; J  ) is nondegenerate, we are done, so we may assume that (K; I  ; J  ) is degenerate, whence J  /O3 (J  ) ∼ = I  . Hence we can ignore the case K ∼ = U4 (2), and it suffices to derive a contradiction in the cases = Sp8 (2), I ∼ I = J when K ∼ = (2)F4 (2), Sp8 (2), or Sp6 (8), with I ∼ = Sp6 (2), Sp6 (2), or Sp4 (8), respectively. Since no proper parabolic subgroup of K contains b × I in these cases, it follows from the Borel-Tits theorem that CK (b I) has odd order. Indeed as every 1 outer involution of Aut(F4 (2)) fixes only a subgroup isomorphic to 2F4 (2 2 ), of 3rank 2, even CAut(K) (b I) has odd order. Moreover, condition (b) of Lemma 15.6 holds by [VK , 8.26]. We conclude from Lemma 15.6 that there is y ∈ I3 (CG (KB)). Let u be a long root element of K. Thus, CB (u) contains some A ∼ = E33 . Then by the Borel-Tits theorem there is a B-invariant parabolic subgroup P ≤ K with u ∈ O2 (P ). Hence O2 (P ) is of symplectic type, by Proposition 10.2. But Sylow 2-subgroups of K have noncyclic centers (containing a high short root subgroup),  so O2 (P ) has a noncyclic center, contradiction. Lemma 15.17. If K > I and I = J, then Lemma 15.2e does not hold. Proof. Suppose false. If K ∼ = 2 D5 (2) or 2E6 (2), choose b ∈ B # such that  I := E(CK (b )) ∼ = D4 (2) or U6 (2), respectively. Using [IA , 3.3.3, 4.10.3] we see that I  is B-wide, so (K, b , I  ) is regular by Proposition 12.5. If the resulting nonconstrained {t, b }-neighborhood (K; I  ; J  ) is nondegenerate, it must appear in Table 1.1, but it does not. Thus it is degenerate, and we switch our attention to it and derive a contradiction. Thus in effect, in these cases we may assume that b = b and I = I  ∼ = D4 (2) or U6 (2). Returning to the general case, the hypotheses of Lemma 15.6 hold by [VK , 3.71, 8.26], so we fix y ∈ I3 (CG (K)) with y t = y −1 . In NG (y) =: Ny we conclude that K ≤ L2 (CNy (t)) ≤ L2 (Ny ). Let L be the subnormal closure of K in Ny (or equivalently in Cy ). Thus L is the product  of one or two 2-components of Ny permuted transitively by t. Since K = I K , L is the subnormal closure of I in Ny . But [y, b] = 1 and [I, y] = 1 = [J, y], so L is the pumpup of J in Cy (i.e. the pumpup of J with respect to the prime 3). Hence, L is a single 3-component of Ny and K is a component of CL (t). Hence by [VK , 11.9], L/O3 (L) ∈ C3 . However, m3 (Ny ) ≥ m3 (K y) ≥ 4, so L/O3 (L) ∈ C3 by our basic assumption. This contradiction completes the proof.  

Lemma 15.18. If K > I and I = J, then Lemma 15.2f does not hold. ∼ Proof. Suppose false, so that K/Z(K) ∼ = U6 (2), U5 (2), or D4 (2), and I = U4 (2). Again by Lemma 15.6, with the help of [VK , 3.71, 8.26], we fix y ∈ I3 (CG (K)) with y t = y −1 . We repeat the third paragraph of the proof of Lemma 15.17, giving the t-invariant 3-component L of CG (y) such that K  E(CL (t)), L/O3 (L) ∈ C3 , and L is the pumpup of J in CG (y). Note that O3 (CG (y)) has odd order since B y ≤ CG (y). If L/O3 (L) ∼ = K, then K ↑2 L/O3 3 (L) and I ↑3 L/O3 (L). But there is no such L/O3 (L) ∈ C3 , by [VK , 11.8]. Hence, K∼ = L/O3 (L).

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There is a hyperplane B0 of B centralizing a 2-central involution z ∈ K. Let C = CG (z) and R = O2 (C ∩ K) ∼ = 21+k , k = 6 or 8. Then B1 := B0 × y ≤ C with B1 ∼ = B, so B1 ∈ B∗,c (G). Let S = RO2 (C). With Proposition 10.2, S and CS (R) are B1 t-invariant and cyclic or of symplectic type. It follows, as NK (R) is irreducible on R/ z, that S = RCS (R). As t inverts y ∈ B1 , t ∈ CS (R). But by the structure of Ct , CCS (R) (t) = z×Q1 for some 2-group Q1 ≤ C(t, K). Without loss, Q1 ≤ Q. Now Q normalizes R and S so [Q, Q1 ] ≤ CS (R). As t ∈ CS (R), [Q, Q1 ] = 1. Hence Q1 ≤ Z(Q) with t ∈ Q1 , so Q1 = 1. Hence CS (R) is cyclic or of maximal class. But also C(y, L) has odd order, otherwise Proposition 10.1 is contradicted in CG (y) by B1 or B1 ∩ L y. As L/O3 (L) ∼ = K, it follows by [VK , 10.49] that z ∈ Syl2 (CCG (y) (R)). So CCS (R) (y) = z, whence CS (R) = z or CS (R) ∼ = Q8 . Suppose that R = O2 (C). Then [E(C), y] = 1 as [R, y] = 1 = O2 (C). Now E(C) z is Q-invariant, so it has a Q-invariant Sylow 2-subgroup U . But [U, R] ≤ [E(C), R] = 1, so U maps into the image of z in Aut(K). So U  z × Q, whence U = z × Q1 with Q1  Q. In particular t ∈ Q1 . If U = z, t, then Sylow 2-subgroups of E(C) z are of maximal class; thus, E(C) has a component M of 2-rank 1, so M ∈ T2 , contradicting the fact that G is of even type. Therefore Q1 > t. Since C(t, K) is solvable, t ∈ Z ∗ (E(C) z). Obviously z ∈ tG . If m2 (Q1 ) = 1, then by Burnside’s lemma t is weakly closed in t, z with respect to E(C) z, so t ∈ Z ∗ (E(C) z), contradiction. Therefore m2 (Q1 ) > 1, so by Lemma 15.3, Q = Q0 ∗ Z with Q0 = [Q0 , B] ∼ = Q8 and Z cyclic of order ≥ 4. As B0 normalizes E(C) z, Q0 ≤ Q1 and so Ω2 (Q) ≤ Q1 . Again Z(U ) = t, z and t is weakly closed in t, z with respect to E(C) z. As t ∈ Z ∗ (E(C) z), and by the Z ∗ -theorem [IG , 15.3], there is g ∈ E(C) z and u ∈ U − t, z such that ug = t. By Sylow’s theorem we may assume that CU (u)g ≤ U . But t is a square in CU (u) so tg ∈ Φ(U ). Hence tg = t, a contradiction. Therefore, R = O2 (C). Suppose next that R ∩ O2 (C) = z , or equivalently R ≤ O2 (C). Then CO2 (C)t (t) = z × Q1 with Q1 a B0 -invariant normal subgroup of Q. Now either Q1 = t or t = Ω1 (Φ(z×Q1 )) char z Q1 (because of B0 -invariance when m2 (Q1 ) = 2). Accordingly, either O2 (C) has maximal class or O2 (C) ≤ z × Q. In either case, CR (O2 (C)) > z and so [O2 (C), R] = 1 by the irreducible action of NK (R) on R/Φ(R). On the other hand, R ≤ O3 (CG (z, y)) = O3 (CC (y)), so again by the irreducible action of NK (R) on R/ z and by [VK , 7.8], either [E(C), R] = 1 or m2,3 (Ct ) ≥ m3 (E(C)R/ z) ≥ m2 (R/ z). However, the latter conclusion fails as m3 (E(C)R) ≤ 5 or 6, and correspondingly m2 (R/ z) ≥ 6 or 8. Hence, [F ∗ (C), R] = 1, which is absurd as R is nonabelian. Therefore R ≤ O2 (C), so R < O2 (C). By what we saw above, O2 (C) = S = R ∗ Q0 with Q0 ∼ = Q8 and [Q0 , y] = [O2 (C), y] = Q0 . Also CQ0 (t) = z. Now b may be taken to centralize z, with CR (b) ∼ = Q8 ∗ Q8 . In particular, b normalizes R, O2 (C), and hence Q0 . Since [t, b] = 1 and t acts nontrivially on Q0 /Φ(Q0 ), we have Q0 ≤ Cb , and then Q0 embeds in Aut(J). But

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16. THEOREM 5: THE DEGENERATE CASE K = I

173

O2 (CI (b)) ≤ R and I = J, so |Q0 ∩ J| = 2, which is impossible by the structure of Aut(J) (J ∼  = U4 (2)). This contradiction completes the proof of the lemma. Now Lemmas 15.12, 15.14, 15.15, 15.16, 15.17, and 15.18, together with Lemma 15.2, show that: Lemma 15.19. We have K = I. That is, either (15A2) or (15A3) holds. 16. Theorem 5: The Degenerate Case K = I As discussed in the introduction to the previous section, we analyze the cases (15A2, 3) in this section and the next two, thereby completing the proof of Proposition 15.1. In this short section we mention some initial results holding in both cases (15A2, 3). We again let Ct = CG (t), Cb = CG (b), and C b = Cb /O3 (Cb ), and have K = I  E(Ct ),  = and J the 3-component of Cb containing I. We also set N = NCb (J) and N ∼ AutN (J) = N/CN (J). The two present cases of (15A) do not mix, in the following sense. Lemma 16.1. If J = I, then for all b0 ∈ CB (K)# , if we let Jb0 be the subnormal closure of I in CG (b0 ), I covers Jb0 /O3 (Jb0 ). Remark 16.2. Equivalently, if J > I, then I does not cover Jb0 /O3 (Jb0 ) for any b0 ∈ CB (K)# . This will allow us at certain points in the analysis to replace b by another element of CB (K)# . Proof. Suppose on the contrary that J > I, but I covers Jb0 := Jb0 /O3 (Jb0 ) for some b0 ∈ CB (K)# , where Jb0 is the subnormal closure of I in CG (b0 ). Consider the action of t, b0  ∼ = Z6 on J. Since IO3 (Jb0 )   CG (b0 ), and by the B3 -property, I is a component of CJ (b0 ) as well as a component of CJ (t). As I < J ∈ C3 , this  is impossible by [VK , 13.2], however. The lemma is proved. Lemma 16.3. No component of Ct is isomorphic to M11 . Proof. If K1   Ct and K1 ∼ = M11 , then m2 (C(t, K1 )) = 1 by [V2 , 9.1]. Since all components of Ct lie in C2 , F ∗ (C(t, K1 )) = O2 (C(t, K1 )) is cyclic or quaternion, whence m3 (C(t, K1 )) ≤ 1. But then m3 (B) ≤ 1 + m3 (Aut(K1 )) = 3, a contradiction.  Lemma 16.4. K is isomorphic to one of the following groups: L2 (17), A6 , U4 (2), L2 (8), Sp4 (8), U5 (2), U6 (2), 2U6 (2), Sp6 (2), 2Sp6 (2), D4 (2), 3D4 (2), 1 2 F4 (2 2 ) , G2 (4), L3 (4), F4 (2), 22E6 (2), L± 3 (3), M12 , J2 , J3 , Co2 , Co1 , 2HS, Suz, F i22 , 2F i22 , F i23 , F i24 , F3 , F2 , or 2F2 . Proof. Now K ∈ C2 , J ∈ C3 . Since Z(K) = O2 (K) and O3 (J) has odd order, either K ∼ = J with K simple or K is isomorphic to a component H of CJ (τ ), where τ ∈ I2 (Aut(J)) is induced by t. Suppose that J ∈ Spor. We use the definition of C3 [I2 , 12.1] and the tables [IA , 5.3] freely. If J ∼ = M11 or (3)M c, then K ∼ = M11 , contradicting Lemma 16.3. We get the following possible pairs (J, K): ((3)J3 , L2 (17)), (J3 , J3 ), (Co1 , Co1 or G2 (4)), (Co2 , Co2 ), (Co3 , 2Sp6 (2) or M12 ), ((3)Suz, Suz or L3 (4) or M12 or J2 ), (O  N, A6 ), ((3)F i22 , F i22 or 2U6 (2) or D4 (2) or Sp6 (2)), (F i23 , F i23 or 2F i22 or [2 × 2]U6 (2)),

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174

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((3)F i24 , 2F i22 or F i23 or F i24 or [2 × 2]U6 (2)), (F5 , 2HS), (F3 , F3 ), (F2 , F2 or 22E6 (2) or F4 (2)), (F1 , F1 or 2F2 ). By Corollary 11.3, K ∼ = [2 × 2]U6 (2), and by Lemma 13.2, K ∼ = F1 . Moreover, K ∼ = A5 , then CB (K) violates Proposition 10.1. In particular the = A5 , for if K ∼ lemma holds if J ∈ Spor. Now if J ∈ Alt, then J ∼ = K, or J ∼ = A6 ∼ = A9 ; but in the latter case as K ∈ C2 , we have K ∼ = A5 , contradiction. Next suppose that J ∈ Chev(2). By the Borel-Tits theorem, τ must be a field, graph, or graph-field automorphism. From [IA , 2.2.10, 4.9.1, 4.9.2] and [V3 , 1.1], the only possibilities, if K ∼  J , are J ∼ = U4 (2) or U5 (2) with K ∼ = A6 , J ∼ = U6 (2) = 1 2  ∼ ∼ ∼ ∼ 2 or D4 (2) with K = Sp6 (2), J = F4 (2) with K = F4 (2 ) , and J = Sp4 (8) with 3 K ∼ = 2B2 (2 2 ). The last of these again violates Proposition 10.1, however. And if K ∼ = J , then K ∼ = L2 (8), A6 , U4 (2), U3 (3), U5 (2), U6 (2), Sp6 (2), D4 (2), 3D4 (2), 1 2  F4 (2), F4 (2 2 ) , or Sp4 (8). We are now reduced to the case J ∈ Chev(3) − Chev(2), as J ∈ C3 . Hence K ∈ Chev(3) ∩ C2 . By hypothesis, (B, t, K) ∈ BtK3exc (G). Hence by definition of  C2 and [IA , 2.2.10], K ∼ = L3 (3). This completes the proof. Lemma 16.5. Suppose that O3 (CG (x)) = 1 for some x ∈ I3o (G). Let P ∈ Syl3 (G) with B ≤ CP (t) ∈ Syl3 (Ct ). Choose any A ∈ E4 (P ), and set Θ = Θ1 (G; A) = O3 (CG (a)) | a ∈ A# and MΘ = NG (Θ). Then Θ is a nontrivial 3 -group and Γoo P,2 (G) ≤ MΘ . Proof. By (1A6), G is balanced with respect to any element of E34 (G). In particular by [V2 , 1.1, 1.3d], Θ is a 3 -group independent of the choice of A ≤ P , and Γoo P,2 (G) ≤ M . By our assumption and Sylow’s theorem, O3 (CG (x)) = 1 for some x ∈ I3 (P ) with m3 (CP (x)) ≥ 4. Hence Θ = 1, completing the proof.  17. Theorem 5: The Degenerate Case I < J In this section we rule out the case (15A2). Thus we prove: Proposition 17.1. It is not the case that K = I and I < J. We assume that K = I and I < J and eventually derive a contradiction. Note that by Lemma 16.1, this assumption remains valid if we replace b by any other element b of CB (K)# , and J by the subnormal closure of I in CG (b ).    t )). Lemma 17.2. m2,3 (J) ≤ m3 (Z(J)) + m3 (Aut(K)) + m3 (CAut(J) (K Proof. First, m2,3 (G) = m3 (B) ≤ m3 (CB (K)) + m3 (Aut(K)). Also, CB (K) ≤    t )). Finally, m2,3 (J) − CCb (K t) so m3 (CB (K)) ≤ m3 (C(b, J)) + m3 (CAut(J) (K m3 (O3 (J)) + m3 (C(b, J)) ≤ m2,3 (G). Combining these inequalities yields the lemma.  Now since K ∈ C2 is a component of CJ (t) with J = J/O3 (J) ∈ C3 , and in view of Lemma 17.2, the possibilities for K and J are given in [VK , 11.2]. We first treat alternative conclusion (a) of [VK , 11.2]. Thus we assume through Lemma 17.23 that ± J ∈ Chev(3), and K ∼ = A6 , L2 (8), U4 (2), L± 3 (3), L4 (3), 2U4 (3), (17A) or G2 (3). Lemma 17.3. Assume (17A). Then K ∼ = A6 , L2 (8), U4 (2), or L± 3 (3).

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Proof. Otherwise (B, t, K) lies in BtK3exc (G), which is empty by (10B).



We can limit the possible J ∈ C3 . Since O3 (Cb ) has odd order by (1A4), K∼ = I ↑2 J/O3 (J). Lemma 17.4. Assume (17A). Then the following conditions hold: (a) One of the following holds: (1) K ∼ = L2 (8) and J ∼ = G2 (3); ± ∼ (2) K ∼ A and J/Z(J) = 6 = L2 (34 ), U4 (2), L± 4 (3), or L3 (9); (3) K ∼ = L± = U4 (2) and J/Z(J) ∼ 4 (3); or ± ± ∼ ∼ J/Z(J) L (4) K = L3 (3) and = 4 (3), correspondingly, or L3 (9); and (b) C := CAut(J) (I t ) is a 2-group, which is cyclic except if J/Z(J) ∼ = L± 4 (3) ∼ and K = A6 , in which case C embeds in D8 . Proof. We have K ∼ = I, and by (17A), J ∈ Chev(3). 1 If K ∼ = 2 G2 (3 2 ) , then the only possibility, by [IA , 4.5.1, 4.9.1] is that = L2 (8) ∼ 1 := CJ (t) ∼ J ∼ t induces a graph-field automorphism on = G2 (3), and H = 2 G2 (3 2 )   with I = F ∗ (H). Therefore CAut(J) (I t ) = CAut(J) (H t ) = t by [IA , 4.9.1]. In particular, (a1) and (b) hold in this case. In the remaining cases we apply [VK , 11.3]. Note that as [t, B] = 1, m3 (B) − m3 (CB (J)) ≤ m3 (CAut(J) (t)). (For the one case in which Z(J) = 1, namely  J ∼ = 3U4 (3) or [3 × 3]U4 (3), we use [IA , 6.4.4].) On the other hand m2,3 (J t ) + m3 (CB (J)) ≤ m2,3 (G) = m3 (B). Therefore [VK , 11.3] applies with AutJt (J) and J/Z(J) in the roles of X and J there. In particular, either J is as in (a2), (a3), or (a4), or K ∼ = A6 with J ∼ = D4± (3) ± or D5 (3). In those cases, t acts on J like an involutory isometry of some natural orthogonal module V with −1-eigenspace Vt of dimension 4 and type −. Let u ∈ I2 (CJ (t)) be the image of an involution of Ω(V ) with −1-eigenspace Vu on V such that Vt ⊆ Vu , Vu has codimension 2 in V , and Vu is of − type or + type according as dim V = 8 or 10. Since Vu ⊇ Vt , [K, u] = 1. Also clearly CJ (u) has a 3-component + Ju containing K and with J u ∼ = Ω− 6 (3) or Ω8 (3) according to the dimension of V . As remarked at the end of the fourth paragraph of the proof of [VK , 11.3], m3 (CCb (u)) = m3 (B). Let B1 ∈ E3∗ (CCb (u)); then B1 ∈ B3∗,o (G). Let H be the  pumpup of K in Cu := CG (u). As Ju = K Ju ≤ Cu , Ju ≤ H, indeed, Ju is a 3-component of CH (b). Then K ↑2 H as well. It then follows from [VK , 13.1] (note that K ∼  SL2 (q) for any odd q) that H = Ju with Ju /O3 (Z(Ju )) ∼ = = U4 (3) (as H ∈ C2 ). But then (B1 , u, H) ∈ BtK3exc (G), which set is empty by (10B). This contradiction completes the proof of (a). Part (b) then follows from [VK , 3.53]. The proof is complete.  We wish to prove that E(C(t, K)) = 1. To that end we let Q ∈ Syl2 (CG (K)) and R = O2 (Ct ) with R ≤ Q. Lemma 17.5. Assume (17A) and that E(C(t, K)) = 1. Then the following conditions hold: (a) R is of symplectic type with Ω1 (Z(R)) = Ω1 (Z(Q)) = t and Q ∈ Syl2 (C(t, K)); (b) Either R contains Q8 ∗ Q8 , or K ∼ = U4 (2) with m3 (C(t, K)) = 1; and (c) K is standard in G.

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Proof. Since R ∈ ICt (B; 2), R is of symplectic type by Proposition 10.2. Since E(C(t, K)) = 1, R = F ∗ (C(t, K)). As t ∈ R, Z(Q) ≤ Z(R), and (a) follows. In particular, Q ≤ C(t, K). If m3 (C(t, K)) ≥ 2, then it follows that R contains Q8 ∗ Q8 . Otherwise m3 (Aut(K)) ≥ m3 (B) − 1 = 3, and as K is as in Lemma 17.3, K∼ = A6 and m3 (B) ≥ 5, then R contains = U4 (2). Hence (b) holds. Similarly, if K ∼ Q8 ∗ Q8 ∗ Q8 . It remains to prove (c). Suppose that K is not standard in G. By [II3 , Theorem PU4 ], K is not terminal in G. If R contains Q8 ∗ Q8 , then m2 (R) ≥ 3 and R has no maximal subgroup embeddable in a dihedral group; hence neither does Q. Therefore if K is Z4 semirigid in G, there exists by [V2 , 8.4] an involution t0 ∈ I2 (Z(Q)) such that K has a nontrivial pumpup in CG (t0 ). But Ω1 (Z(Q)) = t, so this is absurd. Therefore K is not Z4 -semirigid in G. Note that since Ω1 (Z(Q)) = t, Q ∈ Syl2 (CG (K)). By Lemma 17.3 and [VK , 14.6], K ∼ = A6 and (t, K) < (u, H) ∈ ILo2 (G) for some ∼ u ∈ Q and H = L4 (3); moreover, CQ (u)/CQ (H) embeds in D8 . If m3 (B) = 4, then for some B1 ∈ E34 (H), (B1 , u, H) ∈ BtK3exc (G), contradicting (10B). Thus, m3 (B) ≥ 5 and so R contains Q8 ∗ Q8 ∗ Q8 , as shown in the first paragraph. Then |CR (u)| ≥ |CR/t (u)| ≥ 8. If strict inequality holds, it follows that CR (H) = 1; but of course t ∈ CR (H). Choosing v ∈ I2 (CR (H)), we have CR (v) containing 21+4 ± , while CR (v)/CR (H) embeds in D8 ; thus t ∈ CR (H), contradiction. Hence, CR (H) = 1 and |CR (u)| = 8. So CR (u) ∼ = D8 ; but then u must have an additional fixed point on CR (CR (u)), a contradiction. Thus R does not contain Q8 ∗ Q8 . In view of (b), we have m3 (B) = 4 with K ∼ = U4 (2). Then since BtK3exc (G) = ∅ by (10B), the only possibility for J is J ∼ = L4 (4), by [VK , 3.10]. Let a ∈ B∩K with CK (a) ∼ = GU3 (2), and set Ja = E(CJ (a)) ∼ = SL3 (4). Then Ja ≤ E(CCG (a) (u)) ≤ L2 (CG (a)) by L2 -balance. Set Ca = CG (a) and a = Ca /O3 (Ca ). Then either Ja ≤ E(C a ) or Ja acts nontrivially on O3 (C a ), C    u acts centralizing CO3 (Ca ) (u). Suppose the latter and let S ∈ Syl2 (Ja ); then S   faithfully on O3 (Ca ), centralizing a. The Thompson dihedral lemma then yields a ). As the components of E(C a ) are m2,3 (Ca ) ≥ 5, a contradiction. So Ja ≤ E(C ∼     C3 -groups, Ja lies in a component L   Ca with L = 3Suz, by [VK , 3.91]. But CG (t, a) has a SU3 (2)-solvable component contained in Ja . Therefore CL (t) has ∼ such a solvable component. But this is absurd as L = 3Suz (see [IA , 5.3o]). The proof is complete.  Now let Q ≤ T ∈ Syl2 (Ct ). We next prove: Lemma 17.6. Assume (17A). If E(C(t, K)) = 1, then the following conditions hold: ∼ U4 (2); (a) K = (b) m2 (Q) ≤ 2; (c) R = Q0 or Q0 ∗ Z, where Q0 ∼ = Q8 and Z is cyclic of order at least 4; (d) t  NG (T ); and (e) t is 2-central in G. Proof. Since m2 (CR (B)) = 1 by Proposition 10.1, and since CB (K) = 1, R is the central product of Q8 -subgroups possibly together with one cyclic or quaternion group. Hence if (c) fails, then m2 (R) ≥ 3 and so m2 (Q) ≥ 3.

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Suppose that m2 (Q) > 1. Let M be a maximal subgroup of G containing in G, K ≤ L2 (M ) by L2 -balance. NG (K). As M is a K-group and K is standard  Moreover, for any involution u ∈ Q, K  K, CO2 (M ) (u) , so [K, CO2 (M ) (u)] = 1 and hence, as m2(Q) > 1, K ≤ E(M ). Suppose K  E(M ). Then again as m2 (Q) > 1, L := K M ∼ = A10 , by [VK , 3.5] and standardness. But B ≤ NG (K) ≤ M and as m3 (B) = 4 > m3 (Aut(L)), CB (L) = 1. Let b1 ∈ CB (L)# ; then the subnormal closure of K in CG (b1 ) contains L. But this subnormal closure is given in Lemma 17.4 and so cannot contain A10 , by order considerations. This contradiction proves that M = NG (K). By [II3 , Corollary PU2 ], m2 (Qg ∩ M ) ≥ 2 for some g ∈ G − M . Moreover, by Lemma 17.3 and [VK , 8.4], K is outer well-generated for the prime 2. Hence by [IG , 18.11, 18.8], Qg ≤ M for some g ∈ G − M , and ΓQg ,1 (K) < K. It follows by [IA , 7.3.3] that m2 (Q) = m2 (Qg ) = 2 or else K ∼ = L2 (8). By the first paragraph of the proof, (c) holds. In particular m3 (C(t, K)) = 1, so m3 (AutB (K)) ≥ 3. In view of Lemma 17.3, this implies (a), and (b) follows as well. Since (d) implies (e), it remains to prove (d). Suppose it fails and choose g ∈ NG (T ) with tg = t. Since K is standard and t = Ω1 (Z(Q)) with Q ∈ Syl2 (CG (K)), T ∈ Syl2 (NG (K)) and Q ∩ Qg = 1. Let z = Z(T ∩ K). As Qg  T , either Qg ∩ K = 1 or z ∈ Qg . If Qg ∩ K = 1, then as Qg ∩ Q = 1, Qg embeds in g g CAut(K) (T ∩ K) ∼ = z, which is absurd.  Thus, Q ∩ K = 1, and so z ∈ Q ∩ K. −1

−1

Let S = (T ∩ K)g . Then Z(S) = z g ≤ Q, so S ≤ M and S ∩ K = 1. Hence a subgroup of S of index at most | Out(K)| = 2 embeds in Q. As m2 (Q) ≤ 2 and  m2 (K) = 4, this is a contradiction, so the proof is complete. Recall that b ∈ CB (K)# and K < J   Cb . By [VK , 11.5, 11.7], J/O3 (J) ∼ = But if J/O3 (J) ∼ = U4 (3), then t ∼J tz by [VK , 10.1f]. As T ∈ Syl2 (Ct ) and tz ∈ Z(T ), Burnside’s argument implies that t ∼NG (T ) tz, contradicting Lemma 17.6d. We have proved the first part of the next lemma.

L± 4 (3).

Lemma 17.7. Assume (17A). If E(C(t, K)) = 1, then J ∼ = L4 (3). Moreover, t is not a square in NG (b). In particular, R ∼ = Q8 . Proof. As R = F ∗ (C(t, K)) = Q0 or Q0 ∗Z as in Lemma 17.6c, m3 (C(t, K)) ≤ 1 and m2,3 (G) = m3 (Ct ) = 4. It follows that J  NG (b). On the other hand, by the structure of J and CJ (t), t induces a graph automorphism on J, which is not a square in Aut(J) [IA , 4.5.1, 2.5.12]. Hence t is not a square in NG (b). Now if R = Q0 ∗ Z with Z ∼ = Z2n , n ≥ 2, then Z ≤ Cb . This is a contradiction as t ∈ Φ(Z).  Hence, R = Q0 , as asserted. Lemma 17.8. Assume (17A). Suppose that E(C(t, K)) = 1. Then R = Q. Moreover, Q b ∼ = SL2 (3) is a direct factor of Ct . ∼ L4 (3) by Lemma 17.7, C (t) = ∼ Aut(K) by Proof. Note first that as J = J [VK , 10.6]. It suffices to prove that b is not inverted by any 2-element y ∈ Ct . If such a y exists, then we may alter y by an element of CJ (t, y) and assume that [y, K] = 1, whence y 2 ∈ CRb (b) = t. Hence y 2 = 1 by Lemma 17.7. Then y maps into CAut(J) (K), which is generated by the image of t, so replacing y by yt if necessary, we may assume that y centralizes J. Note that CO3 (Cb ) (t) ≤ O{2,3} (Cb ∩ Ct ) = 1, so O3 (Cb ) is inverted by t and centralized by J = [J, t]. Thus J ∼ = L4 (3). By L2 -balance, K lies in E(CG (y)).

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 Since J = K J , J ≤ M   CG (y) and M ∈ C2 , with K ↑2 M . By [VK , 3.88], M = J. But m2,3 (G) = m3 (Ct ) = 4 = m3 (J). Thus choosing B ∗ ∈ E3∗ (J), we have (B ∗ , y, M ) ∈ BtK3exc (G). But this set is empty by (10B), contradiction. Hence the lemma is proved.  Now we attain our objective. Lemma 17.9. Assume (17A). Then E(C(t, K)) = 1. Proof. By Lemma 17.8, Ct ∼ = SL2 (3) × Aut(U4 (2)). We have T ∈ Syl2 (Ct ) ⊆ Syl2 (G) by Lemma 17.6e. By the Z ∗ -theorem, tg ∈ T − {t} for some g ∈ G. Then −1 −1 [Q b , tg ] = 1 so (Q b)g ≤ Ct . But by [VK , 10.27], since tg lies in an SL2 (3) −1 subgroup of Ct , tg is 2-central in Ct . Altering g in Ct , we may assume that −1 tg ∈ Z(T ). By Burnside’s lemma t ∈ Z(NG (T )), contradicting Lemma 17.6d. The proof is complete.  Now let K1 , . . . Kn , n ≥ 1, be the components of Ct other than K. Lemma 17.10. Assume (17A). Suppose b1 ∈ CB (K)# , u ∈ I2 (C(t, K)), u = t, and [b1 , u] = 1. Assume either that u ∈ I2 (Ki t) ∪ I2 (O2 (Ct )) for some i = 1, . . . , n, or that m2 (CC(t,K) (u)) ≥ 3. Let J1 be the subnormal closure of K in CG (b1 ). Then J1 is quasisimple. Moreover, K ∼ = A6 and J1 /Z(J1 ) ∼ = L± 4 (3) or U4 (2). Finally, m2 (CC(t,K) (u)) ≤ 3. Proof. Since t, u centralizes K b1 , it normalizes J1 ; and to show that J1 is quasisimple it suffices to set W = O3 (J1 ) = O{2,3} (J1 ) and show that Wv := [CW (v), K] = 1 for all v ∈ t, u# . By [IG , 4.3(iv)], Wv = [Wv , K]. Let Kv be the pumpup of K in CG (v). As K ≤ Kv   CG (v), Wv ≤ Kv . If Kv is a trivial or diagonal pumpup of K, it is obvious that Wv ≤ O2 (Kv ) = 1. If Kv is a vertical pumpup of K, then given that Kv ∈ C2 and K is as in Lemma 17.3, we conclude from [VK , 7.7] that Wv = 1. Hence J1 is quasisimple.  Hence without loss we may assume b1 = b and By Lemma 17.4, J1 > K. J1 = J. Note also that by our assumptions and [VK , 2.8, 15.9], since K1 ∈ C2 , we may fix E ∈ E2∗ (CC(t,K) (u)), and m2 (E) ≥ 3. Suppose that the desired conclusion fails. By Lemma 17.4b, the image of ∼ t, u in Aut(J) is cyclic, or J ∼ = L± 4 (3), K = A6 , and CAut(J) (K) embeds in D8 . Replacing u by a suitable element of E, we may assume that [J, u] = 1. Let H be the subnormal closure ofK in CG (u), and J0 = L2 (CJ (u)), so that J 0 = J. By L2 -balance and as J0 = K J0 ≤ H, H is a vertical pumpup of K, i.e., K ↑2 H ∈ C2 via t. Also J0 is a 3-component of CH (b), hence a component, by the B3 -property, and [t, b] = 1. ± ± In case (K, J/Z(J)) = (L2 (8), G2 (3)), (U4 (2), L± 4 (3)), (L3 (3), L3 (9)), (L3 (3), ± ± 4 ∼ L4 (3)), or (A6 , L2 (3 ) or L3 (9) or U4 (2)), then by [VK , 3.49], H = J, and H/Z(H) ∼ = L± 4 (3) or G2 (3). Suppose that H/Z(H) ∼ = L± 4 (3) or G2 (3). Then m2 (C(u, H)) = 1 by [V2 , 9.1]. Also m2 (CAut(H) (K)) ≤ 2, so m2 (CC(t,K) (u)) ≤ 3. Thus m2 (E) = m2 (CC(t,K) (u)) = 3 so m2 (CAut(H) (K)) = 2. Therefore K ∼ = A6 , so the lemma holds in this case. Otherwise the previous paragraph and Lemma 17.4 show that H ∼ = J ∼ = U4 (2)

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and m2 (C(u, H)) ≥ m2 (CC(t,K) (u)) − 1. By [V2 , 9.4], m2 (C(u, H)) ≤ 2, so m2 (CC(t,K) (u)) ≤ 3 and the proof is complete.  Lemma 17.11. Assume (17A). Suppose b1 ∈ CB (K)# , u ∈ I2 (K1 t), u = t, and u inverts b1 . Let J1 be the subnormal closure of K in CG (b1 ). Then J1 is quasisimple. Moreover, K ∼ = A6 and J1 /Z(J1 ) ∼ = L± 4 (3) or U4 (2). Either J1 is a ∼ component of CG (u) or else J1 = U4 (2) lies in a U5 (2)-component of CG (u). Except for this U5 (2) case, m2 (CC(t,K) (u)) ≤ 3. Proof. The proof of Lemma 17.10 remains valid with the following changes. (0) J is changed to J1 . (1) u normalizes b1  and centralizes K, so u normalizes J1 , centralizing K. (2) In the argument that H = J1 , [VK , 3.49] now allows H ∼ = U5 (2),  J1 ≤ H. Lemma 17.12. Assume (17A). Then K ∼ = A6 . Moreover there exists b1 ∈ CB (K)# and u ∈ I2 (NK1 t (b1 )) satisfying the hypotheses, and hence conclusions, of Lemma 17.10 or 17.11, with [J1 , u] = 1. In particular J1 ∼ = U4 (2) or J1 /Z(J1 ) ∼ = ± L4 (3). Moreover, either J1 is a component of CG (u) and m2 (CC(t,K) (u)) ≤ 3, or J1 ≤ H   CG (u) with J1 ∼ = U5 (2). = U4 (2) and H ∼ Proof. Since K1   Ct , and 3 divides |K| and hence |C(t, K1 )|, the possible isomorphism types of K1 are given in Lemma 16.4. Then the existence of b1 and u in K1 normalizing b1  follows from [VK , 12.12]. Then Lemmas 17.10 and 17.11 complete the proof.  Lemma 17.13. Assume (17A) and that B ∩ C(t, KK1 ) = 1, and choose b1 ∈ B # ∩ C(t, KK1 ). Let J1 be the subnormal closure of K in CG (b1 ). Then the conclusions of Lemma 17.10 hold, and [J1 , K1 ] = 1. Moreover, J1 ∼ = U4 (2), K ∼ = A6 , t ∈ K1 , and m2 (K1 ) = 2. Remark 17.14. Recall that E(O3 (CG (t))) = 1 by Corollary 11.2. So if 3 divides |C(t, KK1 )|, then the hypothesis of Lemma 17.13 holds. Proof. For any u ∈ I2 (K1 ), Lemma 17.10 applies. Furthermore, as [K1 , b1 ] = 1 = [K1 , K], K1 normalizes J1 and maps into CAut(J1 ) (K), which is a 2-group by Lemma 17.4. Therefore [J1 , K1 ] = 1. Since [J1 , t] = 1, t∈ K1 . For any u ∈ I2 (K1 ), the pumpup H of K in CG (u) then contains J1 = K J1 , and indeed J1 is a component of CH (b1 ). But if J1 /Z(J1 ) ∼ = L± 4 (3), then by [VK , 3.89], J1 = H. As K1 ∈ C2 , m2 (K1 ) ≥ 2, so m2 (C(u, H)) ≥ 2. This contradicts [V2 , 9.1], however. ∼ ∼ Hence, J1 /Z(J1 ) ∼  L± = 4 (3), so by Lemma 17.10, J1 = U4 (2) and K = A6 . It remains to show that m2 (K1 ) = 2. Suppose false and fix Q1 ∈ Syl2 (K1 ) and u ∈ I2 (Z(Q1 )), so that J1 is a component of CG (u) and Q1 ≤ R1 ∈ Syl2 (C(u, J1 )) for some R1 . Then m2 (R1 ) ≥ m2 (Q1 ) ≥ 3. But by [VK , 14.2], J1 is semirigid in G, so by [IG , 7.4], either J1 is terminal in G or there exists a pumpup (v, Jv ) of (u, J1 ) such that v ∈ Z(R1 ), [v, t] = 1, and R1 = u (R1 ∩ C(v, Jv )). In the first case, as J1 is outer well-generated by [VK , 8.4], there exists g ∈ G such that R1g normalizes J1 and ΓR1g ,1 (J1 ) < J1 . But since m2 (R1 ) ≥ 3, [VK , 8.3] is contradicted. In the second case, if Jv /Z(Jv ) ∼ = L± 4 (3), [V2 , 9.1] is contradicted as m2 (C(v, Jv )) ≥ 2. The only other possibility, by [VK , 3.10k], is that Jv ∼ = L4 (4). But as K   CJv (t),  this is impossible by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2].

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Lemma 17.15. Assume (17A). Then we have n = 1. Moreover, E(Ct ) = KK1 , and F ∗ (CCt (KK1 )) = O2 (Ct ) is cyclic or of symplectic type with central involution t. Proof. Suppose that n > 1. By [IG , 8.7] and Corollary 11.2, B ∩ K2 = 1. Choose b1 ∈ B ∩ K2# and u ∈ I2 (K1 ). By Lemmas 17.10 and 17.13 m2 (CC(t,K) (u)) ≤ 3 and t ∈ K1 . Choose v ∈ I2 (K2 ) − Z(K2 ); then K1 ∩ t, v = 1 and t, v ≤ CC(t,K) (u). It follows that m2 (CK1 (u)) ≤ 1, so m2 (K1 ) = 1, a contradiction as K1 ∈ C2 . Therefore, n = 1. The remaining statements follow from Proposition 10.2 and the fact that G is of even type.  Lemma 17.16. Assume (17A). Then C(t, KK1 ) is a 3 -group. Proof. If false, then as KK1  Ct , there exists b1 ∈ B ∩ C(t, KK1 )# . As O2 (Ct ) = 1, b1 acts nontrivially on R := O2 (Ct ), which is of symplectic type with t ∈ [R, R] by Lemma 17.15. Moreover, [R, b1 ] contains a subgroup R0 ∼ = Q8 . Let u ∈ I2 (K1 ) − {t}, so that [b1 , u] = 1. Let J1 be the subnormal closure of K in CG (b1 ). By Lemma 17.10, K ∼ = A6 and J1 /Z(J1 ) ∼ = U4 (2) or L± 4 (3). # Let b2 ∈ B ∩ K1 . Then as [b2 , b1 ] = 1 = [b2 , K], b2 normalizes J1 and maps into CAut(J1 ) (K). The last group is a cyclic 2-group or D8 by Lemma 17.4b, so [b2 , J1 ] = 1. Let J2 be the subnormal closure of J1 in CG (b2 ). By [IG , 8.7(iii)], B normalizes all 3-components of J2 , so b1 does and J2 is a single component. But t acts on J2 = J2 /O3 (J2 ) and K ≤ J1 ≤ J2 , so by [VK , 13.1], J2 is a trivial pumpup of J1 . Then by [VK , 3.39], the image of t in Aut(J2 ) does not lie  Since [R, K b2 ] = 1, however, R0 embeds in in a Q8 subgroup of CAut(J2 ) (K).  C   (K), a contradiction. The lemma is proved. Aut(J2 )

Lemma 17.17. Assume (17A). Then K1 ∈ C3 . Thus, K1 ∼ = A6 , U4 (2), L2 (8), 1 Sp4 (8), U5 (2), U6 (2), Sp6 (2), D4 (2), 3D4 (2), F4 (2), 2F4 (2 2 ) , L± 3 (3), J3 , Co2 , Co1 , Suz, F i22 , F i23 , F i24 , F3 , F2 , or F1 . Proof. Assume K1 ∈ C3 . In particular K1 ∼  A6 . Let b0 ∈ B ∩ K # , and J0 = the subnormal closure of K1 in CG (b0 ). By Proposition 12.5, J0 is a 3-component  J0 /O3 (J0 ), and of CG (b0 ). Since K1 ∈ C3 , but m3 (CG (b0 )) ≥ m3 (B) ≥ 4, K1 ∼ = the entire prior development applies to K1 , K, b0 , J0 in place of K, K1 , b, and J, yielding K1 ∼ = A6 , contradiction. Thus K1 ∈ C3 , and the final list is immediate  from Lemma 16.4 and [V3 , 1.1]. The lemma is proved. Lemma 17.18. Assume (17A). Then K1 ∼ = A6 or L± 3 (3), and m2,3 (G) = 4. Proof. Suppose that K1 ∼ = L2 (8). As m3 (Ct ) ≥ 4 and CB (KK1 ) = 1 by Lemma 17.16, there is b1 ∈ I3 (C(t, K)) such that K1 b1  ∼ = Aut(L2 (8)). Then b1 induces a field automorphism on K1 , so CK1 (b1 ) contains an involution u, and m2 (CK1 t (u)) = 4, contradicting Lemma 17.10. In all other cases except for the desired K1 ∼ = A6 and K1 ∼ = L± 3 (3), viz., 1 3 ∼ K1 = U4 (2), Sp4 (8), U5 (2), U6 (2), Sp6 (2), D4 (2), D4 (2), F4 (2), 2F4 (2 2 ) , J3 , Co2 , Co1 , Suz, F i22 , F i23 , F i24 , F3 , F2 , F1 , [VK , 12.18] implies that there exists b1 ∈ I3 (B ∩ K1 ) and u ∈ I2 (K1 ) such that [b1 , u] = 1 and m2 (CK1 t (u)) ≥ 4, contradicting Lemma 17.10. This completes the proof that K1 ∼ = A6 or L± 3 (3). Then since m3 (Aut(K1 )) = 2 and by Lemma 17.16, m2,3 (G) = m3 (B) = m3 (Ct ) = 4, as asserted. 

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∼ L± (3), choose b1 so that b1 ∈ L1 ∼ Now fix b1 ∈ I3 (K1 ). If K1 = = SL2 (3), 3 L1 ≤ K1 . Thus in any case b1  is 3-central in K1 and is normalized by an involution u ∈ K1 . Let J1 be as in Lemma 17.12. Lemma 17.19. Assume (17A). Then J1 ∼ = U4 (2). Proof. Otherwise by Lemma 17.12, J1 /Z(J1 ) ∼ = L± 4 (3) is a component of CG (u). As m2,3 (G) = 4, this contradicts (10B), whence the lemma.  By Lemma 17.19 and [IA , 4.9.2], CJ1 (t) = K v ∼ = Σ6 , where v 2 = 1 and CK (v) ∼ = Σ4 . By [VK , 10.64], v may be chosen to be 2-central in J1 . Lemma 17.20. Assume (17A). Then [K1 , v] = 1. Proof. Since [v, t] = 1 and v normalizes K, v normalizes E(C(t, K)) = K1 . Let D ∈ Syl3 (K1 ) with b1 ∈ Z(D). As [D, K b1 , t] = 1, D maps into CAut(J1 ) (K t), which is a 2-group by Lemma 17.4b. Thus [D, J1 ] = 1 so [D, v] = 1. As CAut(K1 ) (D) is a 3-group, [K1 , v] = 1.  Lemma 17.21. Assume (17A). Then [K1 , J1 ] = 1, and K1 ∼ = A6 is a component of CG (v). Proof. We have (B, t, K1 ) ∈ BtK3 (G). First let b0 ∈ I3 (B ∩ K). As K v ∼ = Σ6 , after replacing v by a suitable involution in Kv, there is an involution u0 ∈ CKv (b0 ) − K. By Proposition 12.5, K1 lies in a product J0 of 3-components of CG (b0 ). If K1 does not cover J0 /O3 (J0 ), then K1 and J0 satisfy (17A). Applying Lemma 17.10 to K1 , b0 , u0 , and J0 instead of K, b1 , u, and J1 , we conclude that m2 (CC(t,K1 ) (u0 )) ≤ 3. But CK (t, u0 ) contains Σ4 , so m2 (CC(t,K1 ) (u0 )) ≥ 4, contradiction. Therefore K1 covers J0 /O3 (J0 ). We also conclude that K1 ∼ = A6 . Moreover, the first paragraph of the proof of Lemma 17.10 may be repeated with K1 , b0 , u0 , and J0 instead of K, b1 , u, and J1 , and we conclude that J0 is quasisimple. As K1 ≤ J0 , we have K1 = J0   CG (b0 ). Note that K1 has at most 2-conjugates in CG (b0 ), since m2,3 (G) = 4. Since J1 ∼ = U4 (2), b0 lies in a subgroup A ∼ = E33 of J1 . Then A normalizes K1 ; but K v is maximal in J1 , so J1 = K, v, A normalizes K1 . As [K, K1 ] = 1, it follows that [J1 , K1 ] = 1. Now the subnormal closure L1 of K1 in CG (v) is a product of components of CG (v) by L2 -balance, and K1 is a component of CL1 (t). But because K1 is a component of CG (b0 ), K1 is also a component of CL1 (b0 ). As [t, b0 ] = 1, it follows from [VK , 13.2] that L1 = K1 , so K1 is a component of CG (v). The proof is complete.  Lemma 17.22. Assume (17A). Then m2 (C(v, K1 )) ≥ 3 and K1 is terminal in G. Proof. By our choice, v is 2-central in J1 . Let S = O2 (CJ1 (v)) ∼ = Q8 ∗ Q8 , so that v ∈ Φ(S). By Lemma 17.21, [S, K1 ] = 1, and by Lemma 17.18, m2,3 (G) = 4. Hence by (10B), no involution centralizer in G has an L± 4 (3) or 2U4 (3) component. Since K1 ∼ = A6 , it follows that K1 is semirigid in G by [VK , 14.4]. As v ∈ Φ(S) and S is not of maximal class, [IG , 7.4] implies that K1 is terminal in G. Since  m2 (K v) = 3, the lemma follows. Lemma 17.23. (17A) does not hold.

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Proof. Suppose false and continue the above argument. Let R1 ∈ Syl2 (CG (K1 )). By Lemma 17.22 and [II3 , Theorem PU4 ], K1 is standard in G, and m2 (R1 ) ≥ 3. This contradicts [V2 , 9.3], so the lemma follows.  Next we consider the following configurations, from [VK , 11.2b]: (17B)

K=I∼ = U6 (2), SU6 (2), or D4 (2). = Sp6 (2), and J ∼

Lemma 17.24. Assume (17B). Then m3 (C(b, J)) = 1, and C(b, J) has odd order. Proof. Suppose that m3 (C(b, J)) > 1. Note m3 (J) − m3 (O3 3 (J)) = 4. By [IG , 6.22, 6.27], choose a 3-terminal long pumpup (a, H) of (b, J). By [VK , 3.92], every pumpup in a chain from (b, J) to (a, H) is trivial or diagonal, so m3 (C(a, H)) ≥ 2 by [V2 , 8.5], and H/O3 (H) ∼ = D4 (2), U6 (2), or SU6 (2). Now Proposition 4.3 is applicable and we argue that conclusion (b) of that proposition holds. Indeed otherwise, as in conclusion (a), there exists E ∈ E32 (Aut(H/O3 (H))) such that ΓE,1 (H) < H, contradicting [IA , 7.3.3]. Therefore if we set M = NG (H), O3 (H) = 1 and M is strongly 3-embedded in G. Working back along a pumpup chain from (a, H) to (b, J) we see that J = H. Furthermore, [J, J g ] = 1 for all g ∈ G, and since CG (u) ≤ M for all u ∈ I2 (CM (J)), by Proposition 4.3, M contains some T ∈ Syl2 (G). Note also that for P ∈ Syl3 (Ct ), P ∩ K = 1, which quickly yields Ct = ΓP,1 (Ct ) ≤ M . We may assume that CT (t) ∈ Syl2 (Ct ). As t induces a graph automorphism of order 2 on J, T = T0 t where T0 = T ∩ JCM (J). Let T1 = CJ∩T (t) ∈ Syl2 (K). By the Thompson transfer lemma, there exists g ∈ G such that tg ∈ T0 and T1g ≤ T . In particular g ∈ M . It follows by [VK , 12.14], as |T1g ∩T0 | ≥ |T1 |/2 = 28 , that for some y ∈ I2 (T1 ), CJ (y g ) contains an element x of I3 (J). Let C = CG (y g ). If C ≤ M g , then x ∈ M ∩ M g so g ∈ M , contradiction. Therefore C ≤ M g . Notice however that if we let P1 ∈ Syl3 (M ), then CG (y) contains P0 := CP1 (J), and m3 (P0 ) ≥ 2 by our assumption. Hence C contains x ∈ M as well as P0g ≤ M g . It follows that C0 := C ∩ M is strongly 3-embedded in C, and contains x as well as a G-conjugate R of P0g . Expand R to S1 ∈ Syl3 (C ∩ CM (J)) and choose S2 ∈ Syl3 (C ∩ J) with x ∈ S2 . As M is strongly 3-embedded in G, M controls 3-fusion in G. Therefore S1 and S2 are commuting nontrivial disjoint stongly closed subgroups of a Sylow 3-subgroup of C. However, since C0 is strongly 3-embedded in C, no such pair of disjoint subgroups exists, by [VK , 8.45]. This contradiction completes the proof that m3 (C(b, J)) = 1. Consequently, C(b, J) has a normal 3-complement by Burnside’s theorem. As  O3 (CG (b)) has odd order by (1A4), the lemma follows. Still assuming (17B), let B < A ∈ S3 (G). Then A centralizes B ∩ J so A normalizes J. Let z ∈ I2 (J) be a 2-central (high root) involution of J. We may choose z to be a high root involution in K ∼ = C3 (2)orB3 (2) as well, according as K∼ = U6 (2) or D4 (2). Let A1 ∈ Syl3 (CK (z)). Expand A1 to A0 ∈ E3∗ (CCb (z)). Then m3 (A0 ) = m3 (B), so A0 ∈ B3∗,c (G). Let RK = O2 (CK (z)). Then RK ≤ O2 (CJ (z)) by [VK , 10.65a], and we expand RK to an A0 -invariant Sylow 2-subgroup R of and R = [R, A0 ∩ K], by [VK , 10.65b]. O{2,3} 2 (CJ (z)). Then R ∼ = 21+8 + Finally, set Cz = CG (z) and S = O2 (Cz ).

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Lemma 17.25. Assume (17B). Then R ≤ S and S is of symplectic type. Moreover, E(O3 (Cz )) = 1. Proof. By Proposition 10.2 and Corollary 11.2, S is of symplectic type and E(O3 (Cz )) = 1. Suppose that R ≤ S. Now NJ (R) acts irreducibly on R/ z by [VK , 10.50], so R ∩ S = z. Consider the action of t × (A0 ∩ K) on S. We have [CS (t), A0 ∩ K] ≤ S ∩ K ≤ S ∩ RK ≤ S ∩ R = z, so [CS (t), A0 ∩ K] = 1. By the A × B lemma, [S, A0 ∩ K] = 1. But as noted before this lemma, R = [R, A0 ∩ K]. By the Three Subgroups lemma, [S, R] = 1. On the other hand, R ≤ O3 (CG (z, b)) = O3 (CCz (b)). By the irreducible action of NJ (R) on R and [VK , 7.8], [E(Cz ), R] = 1. (Note that m3 (Cz ) ≤ m2,3 (G) ≤ 6 while m2 (R/ z) = 8.) Therefore R ≤ CCz (F ∗ (Cz )) = Z(F ∗ (Cz )), which is absurd as R is not abelian. This completes the proof.  Lemma 17.26. Assume (17B). Then m3 (C(t, K)) = 1 and m3 (B) = m2,3 (G) = 4. Proof. Let b ∈ P ∈ Syl3 (C(t, K)) and set Pb = CP (b). Then since  [K t, b , Pb ] = 1, Pb maps into O 3 (CAut(J) (K t )) = 1. Hence Pb ≤ C(b, J) so m3 (Pb ) = 1 by Lemma 17.24. This proves the first statement, and the second follows directly.  As usual, let Q ∈ Syl2 (C(t, K)). Lemma 17.27. Assume (17B). Then m2 (Q) = 2 and m2 (CQ (b)) = 1. Proof. We have [B ∩ K, Q] = 1 with |B : B ∩ K| = 3 by Lemma 17.26. Hence m2 (Q) ≤ 2 by Proposition 10.1b. Likewise as B = (B ∩ K) b, CQ (b) ≤ CG (B) and Proposition 10.1 implies that m2 (CQ (b)) = 1. It remains to show that m2 (Q) > 1. Note that if Q is cyclic, then as b ∈ C(t, K), we get the contradiction b ∈ O2 (C(t, K)) ≤ O2 (Ct ) = 1, by Burnside’s theorem and the fact that G is of even type. Hence we may assume that Q is quaternion. Let T ∈ Syl2 (Ct ). As Out(K) = 1, T = R × Q where R = T ∩ K. Now R is indecomposable but not quaternion, by [VK , 10.35]. Hence t char T by the Krull-Schmidt theorem. Therefore, T ∈ Syl2 (G). Notice also that as components of Ct lie in C2 , C(t, K) must be 2-constrained. As b ∈ C(t, K) and O2 (Ct ) = 1, the only possibility is that O2 (C(t, K)) ∼ = Q8 and |Q| ≤ 24 . Thus |G|2 = |T | ≤ 213 and Z(T ) is noncyclic. However, |J t |2 ≥ 213 and Sylow 2-centers of J t are cyclic, a contradiction. The proof is complete.  Lemma 17.28. Assume (17B). Then E(Cz ) = 1. Proof. Suppose false, so that b acts nontrivially on S = F ∗ (Cz ). By Lemma 17.25, R ≤ S so R ≤ CS (b). But S is of symplectic type, so [S, b] and CS (b) commute and are of symplectic type. Now t ∈ Cz and [t, b] = 1, so [S, b] is tinvariant. Let S0 = C[S,b] (t). Then C[S,b] (b) = z, so S0 = [S0 , b] z. As R ≤ CS (b), S0 centralizes R. In particular S0 centralizes RK = R ∩ K = O2 (CK (z)). But CAut(K) (RK ) is the image of Z(RK ). Hence S0 , as a subgroup of Ct , lies in Z(RK ) × C(t, K). Thus [S0 , b] ≤ C(t, K). If [S0 , b] is elementary abelian, then t ∈ S0 and m2 (Q) > 2, contradicting Lemma 17.27. So Φ([S0 , b]) = 1. But then z = Ω1 (Φ([S0 , b])) ≤ C(t, K), a final contradiction.  Lemma 17.29. Assume (17B). Then E(Cz ) = 1.

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Proof. Suppose on the contrary that L is a component of E(Cz ). By Lemma 17.25, [L, R] = 1. Since CAut(J) (R) is a 2-group and m3 (C(b, J)) = 1, it follows that m3 (CG (R)) = 1, and any element of I3 (CG (R)) is CG (R)-conjugate to b±1 . In particular, m3 (L) ≤ 1, and equality holds, by Lemma 17.25. Thus, L  Cz , so L is CG (R)-invariant, whence b ∈ L. On the other hand, set C0 = CL (t). Then C0 ≤ CG (t R) ≤ CG (t RK ) ≤ Z(RK ) × C(t, K). Indeed CG (t R) ∩ Z(RK ) ≤ CZ(RK ) (R) = z. Therefore C0 z / z embeds in C(t, K). Hence by Lemma 17.27, (17C)

m2 (CL (t) z / z) ≤ 2.

Moreover, if equality holds, then since [Z(RK ) t , b] = 1, C0 z / z contains an involution commuting with the (nontrivial) image of b. If equality holds in (17C), then m3 (L) = 1 and L ∈ C2 , with L containing an element of order 6. By [VK , 15.10], no such group L exists. Therefore m2 (CL (t) z / z) = 1. By [VK , 15.10], L ∼ = L2 (q), q ∈ FM. But then A0 ∈ B3∗,o (G) normalizes a four-subgroup of L, centralizing z, contradicting Proposition 10.2b. The proof is complete.  Lemmas 17.28 and 17.29 immediately imply: Lemma 17.30. (17B) does not hold. Now we turn to the final case of (15A2) and [VK , 11.2], assuming through Lemma 17.42 that (K, J/O3 (J)) ∼ = (2Sp6 (2), Co3 ), (2U6 (2) or D4 (2), F i22 ), (17D)

(22E6 (2), F2 ), (2F i22 , F i23 or F i24 ), (F i23 , F i24 ), or (2F2 , F1 ).

(It is impossible that Z(K) be noncyclic, by Proposition 10.1c. Also, K ∼  M11 , by = Lemma 16.3.) In every case Out(K) is a 3 -group, so as O2 (K) = 1, B = (B ∩ K) × CB (K). For any D ∈ E3∗ (K), the group B ∗ := D ×CB (K) is isomorphic to B and centralizes t, so B ∗ ∈ B3∗,c (G). Moreover, (B ∗ , t, K) ∈ BtK3 (G). In arguing that the case ∗

K = I, I < J is impossible, there is therefore no loss in replacing B by B . In particular, by such a replacement, we may assume by [VK , 3.46] that B ∩K contains some w such that E(CK (w)) ∼ = 2A6 , U4 (2), U4 (2), (17E) 2U6 (2), 2U4 (3), Ω7 (3), or 2F i22 , in the respective cases of (17D).

Lemma 17.31. One of the following holds: (a) E(Ct ) > K, and O2 (Ct ) is cyclic; or (b) K ∼ = 2Sp6 (2) and Ct = KC(t, K) with O 2 (C(t, K)) ∼ = SL2 (3) and O2 (Ct ) ∼ = Q8 . Proof. Let R = O2 (Ct ). If R is cyclic, then (a) holds. For then CB (K), which is nontrivial as K = I, centralizes KR but not E(Ct )R = F ∗ (Ct ); hence, E(Ct ) > K, as claimed. So we may assume that R is noncyclic. Suppose first that K ∼ = 2Sp6 (2) and J ∼ = Co3 .

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17. THEOREM 5: THE DEGENERATE CASE I < J

185

Suppose also for the moment that there exists b1 ∈ CB (K)# such that Rb := CR (b1 ) > t. Then by Lemma 16.1 and (17D), the subnormal closure J1 of K in CG (b1 ) satisfies J1 := J1 /O3 (J1 ) ∼ = Co3 . As Rb centralizes b1 and K, it normalizes  = N/O3 (CG (b1 )). Then as J1 is complete by J1 . Let N = NCG (b1 ) (J1 ) and N   . Therefore   It t is a direct factor of CN (K). [IA , 5.3j], J1 is a direct factor of N follows that t is a direct factor of Rb . But by Proposition 10.2, Rb is of symplectic type with t = Ω1 (Z(Rb )). Hence, Rb = t, contradiction. Thus CR (b) = t for all b ∈ CB (K)# . If CB (K) is noncyclic, then R = t and CB (K) acts nontrivially on E(C(t, K)), so (a) holds. Otherwise CB (K) is cyclic, and O2 (Ct ) is the central product of Q8 ’s. As m3 (K) > 1, K  Ct . Then R centralizes the hyperplane B ∩ K of B, so m2 (R) < 3 by Proposition 10.1. Hence, R∼ = Q8 , and (b) follows in this case. Therefore we may assume that K ∼  2Sp6 (2). By (17D), = K∼ = 2U6 (2), D4 (2), 22E6 (2), 2F i22 , F i23 , or 2F2 .  SL2 (q) Let w ∈ B ∩ K # be as in (17E) and set Kw = E(CK (w)). Note that Kw ∼ = for any odd q. Also J ∼ = F i22 , F i22 , F2 , F i23 or F i24 , F i24 , or F1 , respectively. By [VK , 3.46], if we set Jw = L3 (CJ (w)), then J w = E(CJ (w)) ∼ = U4 (3), U4 (3), +  F i22 , Ω7 (3) or P Ω+ (3), P Ω (3), or 3F i , respectively. Clearly, J w is t-invariant 24 8 8 and Kw is a component of CJw (t). Let H be the subnormal closure of Kw in CG (w). L2 -balance, H is a trivial,  J By w ≤ H, and J w > K w , H is diagonal, or vertical pumpup of Kw . As Jw = Kw a single 2-component. But also H is the subnormal closure of Jw in CG (w), so H  := H/O{2,3} (H) is quasisimple, is also a 3-component, by L3 -balance. Thus H    H ∈ C3 , Jw is a component of CH (b), and Kw is a component of CH (t). As [t, b] = 1,  = Jw .  SL2 (q)) that H it follows from [VK , 13.1] (and our remark that Kw ∼ = Since [R, w] = 1 and t is the unique minimal normal subgroup of R, R then  w ), whose Sylow 2-subgroups in turn embed in D8 by [VK , embeds in CAut(Jw ) (K 3.46]. As [B, t] = 1, it follows that [R, B] = 1. By Proposition 10.2, m2 (R) = 1, so R is cyclic. Since B ∩ C(t, K) = 1, we conclude that E(C(t, K)) = 1, whence (a) holds and the proof is complete.  Lemma 17.32. E(Ct ) > K, and O2 (Ct ) is cyclic. Proof. Suppose false and consider the configuration of Lemma 17.31b. Let T ∈ Syl2 (Ct ) and S = T ∩ K. As Out(K) = 1, T = S ∗ P , where P ∈ Syl2 (C(t, K)) and P ∼ = Q8 , Q16 , or SD16 . By [VK , 11.6], Z(S) = t, whence Z(T ) = t and so T ∈ Syl2 (G). Also Ct = K ∗ (P b), so similarly S ∈ Syl2 (Cb ). Since S is not of maximal class, there is U  S with U ∼ = E22 ; then U  T . Now S ≤ J with J ∼ = Co3 , and it is clear from [IA , 5.3j] that all the involutions in U # are Jconjugate to t. Hence by [IG , 16.21], applied in J, there is a 3-element v ∈ NJ (U ) cycling U # . Let T0 = CT (U ) = S0 ∗ P , where we put S0 = CS (U ). By a Frattini argument we may assume that v normalizes T0 . In particular T0 / t ∼ = T0 / u for any u ∈ U − t. Similarly, arguing in J, we see that S0 / t ∼ = S0 / u for any such u. But T0 / t ∼ = (S0 / t) × P/ t, while T0 / u ∼ = (S0 / u) ∗ P . Let Z = Z(S0 / t). Therefore Z(T0 / t) ∼ = Z × Z(P/ t), while Z(T0 / u) ∼ = Z, a  contradiction as T0 / t ∼ = T0 / u. The proof is complete. Lemma 17.33. O2 (Ct ) = t.

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186

4. THEOREM C5 : STAGE 2

Proof. Since O2 (Ct ) is cyclic, it is centralized by b, so O2 (Ct ) embeds in CAut(J) (K), which is elementary abelian by [VK , 11.4b]. The result follows.  Let K1 be a component of Ct distinct from K. Fix w ∈ B ∩ K as in (17E). Lemma 17.34. One of the following holds: (a) K1 is isomorphic to one of the groups listed as “K” in (17D); moreover, there is a 3-component J1 of CG (w) such that (K1 ; K1 ; J1 ) ∈ KIJ3 (G); we have symmetry between K, J, b and K1 , J1 , w; or (b) K1 is terminal in G and is isomorphic to one of the groups listed in (18A) below. Moreover, K1 is a normal 3-component of CG (w). Proof. We have (B, t, K1 ) ∈ BtK3 (G), and let w ∈ B ∩ K ≤ CB (K1 ) be as in (17E). Let I1 = CK1 (w) = K1 , and let J1 be the subnormal closure of I1 in CG (w). Then by Proposition 12.5, J1 is a 3-component of CG (w), and (K1 ; I1 ; J1 ) ∈ KIJ(G). Suppose that I 1 < J 1 in C 1 = CG (w)/O3 (CG (w)). Then the results of this section so far apply to (K1 ; I1 ; J1 ) as well as (K; I; J). Then (a) follows from Lemmas 17.23 and 17.30. We may therefore assume that I 1 = J 1 , so that K1 covers a component  2Sp6 (2). Let Qw ∈ of CG (w)/O3 (CG (w)). We first consider the case K ∼ = Syl2 (CK (w)). For any u ∈ I2 (Qw ), we claim that K1   CG (u). Let Ku be the subnormal closure of K1 in CG (u). Then K1 is a component of CKu (t) by L2 balance. But also Ku is w-invariant, and K1 ≤ Ju   CKu (w), where we have put Ju = Ku ∩ J1 . Since K1 covers J1 /O3 (J1 ), it covers Ju /O3 (Ju ), and hence, with   the B3 -property, O 3 (Ju ) is a component of CKu (w). As K1 ≤ O 3 (J1 ), equality must hold, i.e., K1   CKu (w). By [VK , 13.2], therefore, Ku = K1 , proving our claim. Thus, K1 ∈ C2 ∩ C3 . We will encounter this situation in general in the next section, but for now we just note that this condition, combined with Lemma 16.4, forces K1 to be isomorphic to one of the groups in (18A) below. As a result, by [VK , 14.6], one of the following holds. Here Q1 ∈ Syl2 (C(t, K1 )) and Q1 ≤ R1 ∈ Syl2 (CG (K1 )). (1) K1 is Z4 -semirigid in G; or ∼ A6 and for some z ∈ I2 (CG (K1 )), (z, K1 ) has (2) K1 = an L4 (3) pumpup (v, H) with Q2 /CQ2 (H) embedding in D8 , where Q2 ≤ CC(z,K1 ) (v). Moreover, z and Q2 (17F) satisfy one of the following: (1) z = t and Q2 = CQ1 (v); or (2) K1 has a trivial pumpup in CG (y) for all y ∈ I2 (Q1 ), CQ2 (H) ∩ Q1 = 1, and Q2 = CR1 (v). ∼ Still assuming that K =  2Sp6 (2), suppose that (17F1) holds. By [VK , 11.4], Q1 ∈ Syl2 (C(t, K1 )) has rank at least 4. By [IG , 7.4] and [V2 , 8.4], if K1 is not terminal in G, there exists z ∈ I2 (Z(Q1 )) such that K1 has a nontrivial pumpup Kz in CG (z), and indeed in CG (v) for every v ∈ I2 (Q∗1 ), for some Q∗1  Q1 with Q1 /Q∗1 cyclic. As Qw is noncyclic, some C(t, K1 )-conjugate of Q∗1 meets Qw nontrivially, contradicting the fact that K1 has a trivial pumpup in CG (u) for all u ∈ I2 (Qw ), as we saw above. Thus, K1 is terminal in G in this case.

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∼ Suppose next that K =  2Sp6 (2) and (17F2a) holds. We claim that there still exists z ∈ I2 (Z(Q1 )) such that K1 has a nontrivial pumpup in CG (z). Namely, since K ∼ = 2Sp6 (2), it follows from [VK , 11.6] that Z(Q1 ) is noncyclic. Let H, v, and Q2 be as in (17F2). Now Q2 = CQ1 (v) ≥ Z(Q1 ) so Z(Q1 ) ≤ Z(Q2 ). As Q2 /CQ2 (H) embeds in D8 , and Z(Q1 ) is noncyclic, our claim holds unless possibly Q2 = CQ2 (H)×Z(Q1 ). But even then, E(CH (z)) ∼ = P Sp4 (3) for some z ∈ Z(Q1 )# , so the claim holds. Hence Q1 normalizes H and Q1 /CQ1 (H) embeds in D8 . Since K1 has trivial pumpup in CG (u) for all u ∈ I2 (Qw ), Qw meets every C(t, K1 )conjugate of CQ1 (H) trivially. Therefore, |Qw | ≤ 23 . On the other hand it is clear from [VK , 3.46] that |Qw | > 23 , contradiction. Still assuming K ∼ = 2Sp6 (2), suppose that (17F2b) holds. Since Q1 ∩ K ∈ Syl2 (K), we see from the possibilities for K that m2 (R1 ) ≥ m2 (Q1 ) ≥ 5. By [V2 , 4.4], R1 is connected. But also if E22 ∼ = E ≤ R1 and K1 is a component of CG (e) for all e ∈ E # , then by [VK , 3.5], K1 is a component of CG (a) for all a ∈ I2 (CR1 (E)). The connectedness then implies that K1 is terminal in this case. (The same argument applies even if K ∼ = 2Sp6 (2), as long as R1 is connected.) It remains to consider the case K∼ = 2Sp6 (2). Since m2 (CG (B)) = 1 by Proposition 10.1, t = Z(K) in this case. Assume that K1 is not terminal. If (17F1) holds, then as Q1 has sectional 2-rank 6, it follows by [V2 , 8.4] that K1 has a nontrivial pumpup in CG (v) for every v ∈ I2 (Q∗1 ), for some Q∗1  Q1 with Q1 /Q∗1 abelian. But t ∈ [Q1 ∩ K, Q1 ∩ K] and K1   CG (t), a contradiction. Still assuming K ∼ = 2Sp6 (2), suppose (17F2a) holds. Let P ≤ K contain Q1 ∩K and be the preimage in K of a parabolic subgroup of K/ t with P/ t = (R/ t)L, R/ t ∼ = L3 (2). By [VK , 10.51], R ∼ = D8 ∗ D8 ∗ D8 . Let v and H be as in = E26 , L ∼ (17F2). Then Q2 normalizes P . Let QH = CQ2 (H), so that Q2 /QH ∼ = D8 and t maps onto Z(Q2 /QH ). If some involution x ∈ QH induces an inner automorphism with t = Z(R0 ), so R0 embeds in Q2 /QH , on R, then CR (x) contains R0 ∼ = 21+4 ± which is absurd. Therefore QH ∩ R = 1, whence CR (v) embeds in Q2 /QH . On the other hand, |CR (v)| ≥ |CR/t (v)| ≥ 23 = |Q2 /QH |, so CR (v) ∼ = Q2 /QH ∼ = D8 . But then CR (v) must also contain an element of CR (CR (v)) − t, a contradiction. To complete the proof that K1 is terminal, we finally consider the case (17F2b), As remarked above, we may assume that R1 is not connected, so m2 (R1 ) ≤ 4 by [V2 , 4.4]. Let C1 = CG (K1 ) and  C 1 = C1 /O2 (C1 ). Then K is a component of C1  = L is a single 2-component which, as a pumpup CC1 (t), so by L2 -balance, K of 2Sp6 (2), is isomorphic to K itself or to Co3 . In any case C 1 = LCC 1 (L) and R1 = (R1 ∩ L)CR1 (L). As m2 (R1 ) ≤ 4, every involution u ∈ R1 has the form u = u1 u2 , where 1 = u1 ∈ L, u2 ∈ CR1 (L), and u21 = u22 has order 1 or 2. By the Thompson Transfer lemma, u has an L-conjugate centralizing a fixed U  R1 with U ∼ = E22 . As [L, K1 ] = 1 and K1 is a component of CG (x) for all x ∈ U # , K1 is a component of CG (u) by [VK , 3.5]. This completes the proof that K1 is terminal, hence standard in G [II3 , Theorem PU4 ]. Finally, we show that K1  CG (w) by showing that K1 = J1 , i.e., that J1 is quasisimple. For this we use [VK , 3.46e] to obtain an involution u ∈ K −t normalizing w. Then as O3 (J1 ) has odd order, it suffices to show that [CO3 (J1 ) (v), K1 ] = 1 for all v ∈ t, u# . But this follows from the terminality of K1 . 

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Lemma 17.35. Suppose Lemma 17.34a holds. Then one of the following holds: (a) K ∼ = D4 (2) or F i23 ; or (b) K1 ∼ = 2U6 (2), D4 (2), or F i23 . Proof. Suppose that the lemma fails. Since (a) fails, Z(K) ∼ = Z2 , so Z(K) = t by Proposition 10.1, as K is B-invariant. Since (b) fails, Z(K1 ) = 1 so Z(K1 ) = t by symmetry. In particular t ∈ J ∩ J1 . Let Y1 = CK1 (b). Because (b) fails and by [VK , 12.30], b could have been chosen so that t ∈ W1 := [Y1 , Y1 ]. As Y1 ≤ K1 ≤ C(t, K), Y1 normalizes the 3-component J of CG (b), and [K, Y1 ] = 1. Thus Y1 maps into CAut(J) (K), which is abelian by [VK , 11.4]. Hence W1 centralizes J,  and in particular [J, t] = 1, which is absurd. The proof is complete. Since we have symmetry between K and K1 , Lemma 17.35 immediately implies that if Lemma 17.34a holds, we may assume that (17G)

Either K ∼ = F i23 or D4 (2), or K ∼ = 2U6 (2). = K1 ∼

Lemma 17.36. If Lemma 17.34a holds, then {K, K1 } ⊆ {F i23 , D4 (2)}. Proof. Suppose false, so that by the preceding discussion, we may assume that ∼ 2U6 (2). We claim that K is 2-terminal in G. Namely, let Q ∈ Syl (C(t, K)). K1 = 2 By [VK , 14.2], K is semirigid in G, so since Q ≥ Q∩K1 is obviously not of maximal class, if K is not terminal in G, then there is a maximal subgroup Q1 ≤ Q such that K has a nontrivial pumpup in CG (u) for all u ∈ I2 (Q1 ). But t ∈ Φ(Q ∩ K1 ) ≤ Φ(Q) ≤ Q1 and K is a component of CG (t), contradiction. Thus, the claim holds. Now we could have chosen b ∈ B ∩ K1# to be 3-central, in which case [b, u] = 1 for some u ∈ I2 (K1 ) − {t} by [VK , 12.13]. But then the four-group t, u maps into CAut(J) (K) ∼ = Z2 , so replacing u by tu, we may assume that [J, u] = 1. By the previous paragraph, K   CG (u). But this implies that K   CJ (u) = J, a contradiction. The lemma follows.  Now we can prove Lemma 17.37. Lemma 17.34b holds. Proof. Suppose false and continue the above argument. We have K ∼ = F i23 or D4 (2), with J ∼ = F i24 or F i22 , respectively, and CAut(J) (K) ∼ = Z2 in either case. As in the previous proof we could have chosen b ∈ B ∩ K1 so that [b, u] = 1 for some u ∈ I2 (K1 t) − {t}, by [VK , 12.15]; and u could be chosen so that [J, u] = 1. As J = K J , it follows that K has a vertical pumpup Ku in CG (u). Let Q ∈ Syl2 (C(t, K)) with u ∈ Q, and note that by [VK , 14.6], K is semirigid in G. Let V be the set of involutions v ∈ CG (K) such that the pumpup of K in CG (v) is isomorphic to J. By [IG , 7.4] and [VK , 3.90], Q has a maximal subgroup Q1 such that I2 (Q1 ) ⊆ V, indeed such that the subnormal closure K ∗ of K in CG (Q1 ), as well as in CG (v) for any v ∈ I2 (Q1 ), is isomorphic to J. By the Thompson transfer lemma, the subnormal closure Kv of K in CG (v) is isomorphic to J, for any v ∈ I2 (Q∩E(Ct )). Now Q∩K1 has normal 2-rank (far) exceeding 3, so Q1 ∩K1 has an E23 subgroup A  Q∩K1 . Thus any u ∈ I2 (Q∩E(Ct)) centralizes some four-subgroup V ≤ Q1 ∩ K1 . As V maps into CAut(Ku ) (K) ∼ = Z2 , it follows that Ku = Kv for some (hence all) v ∈ V # . Hence Ku = Kv whenever u and v are commuting involutions in Q ∩ E(Ct ). By Sylow’s theorem, Ku = Kv

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whenever u and v are commuting involutions in E(C(t, K)). Now K1 was an arbitrary component of C(t, K), so no such component has a strongly embedded subgroup, whence the commuting graph of I2 (C(t, K)) is connected. Hence, letting K ∗ = Kv for any v ∈ I2 (Q ∩ E(Ct )), E(C(t, K)) maps to CAut(K ∗ ) (K) ∼ = Z2 , whence E(C(t, K)) centralizes K ∗ and K ∗ is a component of C := CG (E(C(t, K))). Then CC (K ∗ t) ≤ CC (K t) = CG (F ∗ (Ct )) = t by Lemma 17.33. It follows that K ∗ = E(C) is C(t, K)-invariant, and then centralized by O 2 (C(t, K)). We assume that notation has been chosen so that the components of Ct are K, K1 , . . . , Kk , . . . , Kn , where 1 ≤ k ≤ n, and for 1 ≤ i ≤ k, Ki satisfies Lemma 17.34a, while for k < i ≤ n, Ki satisfies Lemma 17.34b. Then for each component K1 , . . . , Kk the pumpup Ki∗ of Ki in CG (t0 ) is independent of t0 ∈ I2 (K), and is isomorphic to F i24 or F i22 according as Ki ∼ = F i23 or D4 (2). Moreover, Ki∗ is centralized by O 2 (C(t, Ki )) and in particular by K. As [K ∗ , Ki ] = 1, CK ∗ (t0 ) normalizes Ki∗ for all t0 ∈ I2 (K). But by [VK , 8.48], K ∗ = K, CK ∗ (t0 ) | t0 ∈ I2 (K) ≤ NG (Ki∗ ). As [K ∗ , Ki ] = 1, it follows as usual that [K ∗ , Ki∗ ] = 1. By symmetry, [Ki∗ , Kj∗ ] = 1 for all 1 ≤ i < j ≤ k. Also, [K ∗ , Kj ] = 1 for all k < j ≤ n, so by symmetry, [Ki∗ , Kj ] = 1 for all 1 ≤ i ≤ k < j ≤ n. Set K0 = K, K0∗ = K ∗ , and L := K0∗ K1∗ · · · Kk∗ Kk+1 · · · Kn , a t-invariant commuting product of components isomorphic to F i24 or F i22 , or F i23 or D4 (2). From [IA , 4.10.3a, 5.6.1, 5.6.2] we have m3 (Ki∗ ) = m2,3 (Ki∗ ) + 1 and m3 (Ki∗ ) = m3 (Ki ) + 1 for each i = 0, . . . , k. Also B normalizes each Ki and induces inner automorphisms on the simple group Ki for each i = 0, . . . , k, as B ∈ E3∗ (Ct ). Finally, CAut(Ki∗ ) (Ki ) is a 2-group for all i = 0, . . . , k, so CB (K0 K1 · · · Kk ) = CB (K0∗ K1∗ · · · Kk∗ ). Therefore m2,3 (L) ≥ m2,3 (K ∗ ) +

k 

m3 (Ki∗ ) + m3 (CB (K0∗ K1∗ · · · Kk∗ ))

i=1

= −1 +

k 

m3 (Ki∗ ) + m3 (CB (K0∗ K1∗ · · · Kk∗ ))

i=0

=k+

k 

m3 (Ki ) + m3 (CB (K0 K1 · · · Kk ))

i=0

= k + m3 (B) = k + m2,3 (G). As k ≥ 1, this is a contradiction. The proof is complete.



Write E(C(t, K)) = K1 . . . Kn , n ≥ 1. As K1 was arbitrarily chosen, Lemma 17.37 applies to every Ki . Lemma 17.38. CJ (E(C(t, K))) covers J.  ∼ Proof. If K ∼ = 2Sp6 (2), let L = O 3 (CJ (w)) = 31+4 SL2 (9); otherwise let L = E(CJ (w)). In either case L = E(CK (w))L . Then K, L covers J, by [VK , 3.46] and [IA , 5.3j]. By Lemma 17.34b, L normalizes K1 . As [K, K1 ] = 1, it follows that [L, K1 ] = 1. Hence [K, L , K1 ] = 1. As K1 was an arbitrary component of C(t, K), the lemma follows. 

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∼ 2Sp6 (2). Thus J/O3 (J) ∼ Lemma 17.39. K = = F i22 , F i23 , F i24 , F2 , or F1 , and m2 (J) = m2 (J) ≥ 10. Proof. It suffices by (17D) and [IA , 5.6.1] to prove the first assertion. Suppose that K ∼ = 2Sp6 (2). Let u ∈ I2 (K1 ) and Ju = L3 (CJ (u)). By Lemma 17.38, Ju covers J/O3 (J). Clearly K ≤ Ju and so Ju = K Ju lies in the pumpup Ku of K in CG (u). In particular Ku is a vertical pumpup of K, and so by [VK , 3.31], Ku ∼ = [2 × 2]D4 (2) or Co3 . As Ku involves J u ∼ = Co3 , Ku ∼ = Co3 . But as G is of  even type, this is a contradiction since Co3 ∈ C2 . Now let Q1 ∈ Syl2 (CG (K1 )) with Q1 containing a Sylow 2-subgroup of J t as well as one of K2 . . . Kn . Such a subgroup Q1 exists by Lemma 17.38 since O3 (J) has odd order. Lemma 17.40. K1 ∼ = U5 (2), U6 (2), D4 (2), F4 (2), Co2 , Co1 , F i22 , or F2 . ∼ A6 , U4 (2), L2 (8), Sp6 (2), 3D4 (2), 2F4 (2 2 ) , Proof. Suppose false. Then K1 =  Sp4 (8), J3 , Suz, F i23 , F i24 , F3 , F1 , L3 (3), or U3 (3). In every case, by [VK , 8.4], K1 is outer well-generated for the prime 2. Hence as K1 is terminal in G and m2 (CG (K1 )) ≥ 2, it follows from [II3 , Theorem PU4 ] and [IG , 18.11] that if we put M = NG (K1 ), there exists g ∈ G − M such that Qg1 ≤ M , CQg1 (K1 ) = 1, and Γ := ΓQg1 ,1 (K1 ) < K1 . By Lemma 17.39, in particular, m2 (Aut(K1 )) ≥ m2 (Q1 ) ≥ 10. Hence by [IA , 5.6.1, 3.3.3, 2.5.12], K1 ∼ = F i23 , F i24 or F1 . Then [V2 , 9.3b, 9.1] and Lemma 13.2 give a contradiction.  1

As a consequence of Lemma 17.40 and [IA , 4.10.3, 5.6.1], m3 (K1 ) ≥ 4.

(17H)

As [K1 , K] = 1 and m3 (CJ (K1 )) = m3 (J), we deduce from (17D) and [IA , 4.10.3, 5.6.1] that (17I)

m3 (K) ≥ 4 and m3 (CG (K1 )) ≥ m3 (J) ≥ 5.

Our next goal is to prove Lemma 17.41. NG (K1 ) is a strongly 3-embedded maximal subgroup of G. Proof. Suppose first that (17J)

O3 (CG (x)) = 1 for some 1 = x ∈ A ∈ E34 (G).

We then let Θ and MΘ be as in Lemma 16.5, so that 1 = Θ ≤ O3 (MΘ ) and Γoo P,2 (G) ≤ MΘ , for any P ∈ Syl3 (G) containing A. Replacing P by a conjugate we may assume that P contains P1 ∈ Syl3 (NG (K1 )). Using (17H) and (17I) we conclude easily that K1 ≤ MΘ , CG (K1 ) ≤ MΘ , and then by a Frattini argument, NG (K1 ) ≤ MΘ . In this case (17J) we let M be a maximal subgroup of G containing MΘ . On the other hand, if (17J) fails, i.e., O3 (CG (x)) = 1 for all x ∈ I3o (G), we let M be a maximal subgroup of G containing NG (K1 ). Thus, NG (K1 ) ≤ M in either case. In both cases we claim that M = NG (K1 ). Namely, K1 is standard in G and m2 (CG (K1 )) ≥ m2 (J) > 1. Hence taking E ∈ E22 (CG (K1 )) we see that O2 (M ) ≤ ΓE,1 (O2 (M )) ≤ NG (K1 ), whence O2 (M ) ≤ CG (K1 ). By this fact, L2 -balance, and the noncyclicity of E, K1 lies in a component K0 of M . As K1  ΓE,1 (K0 ), it

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follows from [VK , 3.5] and Lemma 17.40 that K0 = K1 . By standardness, K1  M , proving our claim. Next, we verify that M satisfies the 3-Component Preuniqueness Hypothesis. The only part of that hypothesis that needs checking is that (17K)

CG (v) ≤ M

for all v ∈ I3 (CG (K1 )). First, for v ∈ B ∩ K # , K1 has a trivial pumpup in CG (v) since K1 = I1 and I 1 = J 1 , and in view of Lemma 16.1. But m3 (B ∩ K) ≥ 4 since B ∈ E3∗ (Ct ). Therefore if O3 (CG (v)) = 1, so that (17J) holds, O3 (CG (v)) ≤ ΓB∩K,3 (G) ≤ Γoo P,2 (G) ≤ M , so K1   CG (v). If O3 (CG (v)) = 1, then it is obvious that K1   CG (v). As K1 is standard in G, K1  CG (v), and (17K) holds for v ∈ B ∩ K # . Since m3 (CP (K1 )) ≥ 4, CP (K1 ) is connected by Konvisser’s theorem [IG , 10.17]. Hence to verify (17K) for all v ∈ I3 (CG (K1 )) it suffices to show that if Y ∈ E32 (CG (K1 )) and (17K) holds for all v ∈ Y # , then (17K) holds for any v ∈ I3 (CG (K1 Y )). But this is a consequence of [VK , 3.5] and Lemma 17.40. Therefore, M satisfies the 3-Component Preuniqueness Hypothesis. Now by [II3 , Theorem PU∗2 ], either the conclusion of the lemma holds or there exist g ∈ G − M and a noncyclic 3-group Y ≤ CG (K1 ) such that Y g ≤ M and ΓY g ,1 (K1 ) < K1 . Suppose the latter occurs. Then the four possibilities in Lemma 17.40 with K1 ∈ Chev(2) are ruled out by [IA , 7.3.3]. The other four possibilities are sporadic, so Z(K1 ) and Out(K1 ) are 3 -groups, whence we may take Y ∈ Syl3 (CG (K1 )), by [II3 , 16.11]. Thus |Y | ≥ |J|3 ≥ 39 , the last by Lemma 17.39. Therefore, ΓY g ,1 (K1 ) = K1 by [VK , 8.44], a contradiction completing the proof of the lemma.   = M/O3 (M ). Then J   M . Lemma 17.42. Let M = NG (K1 ) and M Proof. First, J is a 3-component of CG (b), where b ∈ CB (K)# . For any b1 ∈ B ∩ K1# , b1 then normalizes J and centralizes J0 := L3 (CJ (K1 )), which covers J by Lemma 17.38. By L3 -balance, (b, J) < (b1 , J1∗ ) where J1∗ is a 3-component of J ∗ , which in turn is the subnormal closure of K in CG (b1 ). Then K is a component of CJ ∗ (t). Hence the pair (K, J ∗ /Z(J ∗ )) is also given by (17D). Therefore since  F i24 , 3F i24 ∈ ↑3 (F i23 ) [IA , 5.3v], J ∗ is a trivial pumpup of J, so that J∗ = J. ∼    Now B ∩ K1 is noncyclic and J =  A6 , so by [VK , 3.5], J   M , as claimed.  Now we can complete the proof of Proposition 17.1. As K1 is standard, there exists by [II3 , Theorem PU∗1 ] an involution u ∈ CG (K1 ) and an element g ∈ G − M such that ug ∈ M − CG (K1 ). By Lemma 17.41, M ∩ M g is a 3 -group, so CM (ug )  and by (17D) and [IA , 5.3], in every is a 3 -group. But then ug normalizes J, g case for J, |CJ (u )| is divisible by 3, a contradiction. This completes the proof of Proposition 17.1. The proposition has the following direct corollary. Corollary 17.43. Let (B, t, K) ∈ BtK3 (G) and suppose that CB (K) = 1. Then O2 (K) = 1. Proof. By Proposition 12.5, there exists b ∈ CB (K)# such that K ≤ J for some 3-component J of CG (b). If K does not cover J/O3 (J), then K, I = K, and J contradict Proposition 17.1. Thus K covers J/O3 (J). As O3 (J) has odd order,  O2 (K) = 1, as asserted.

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18. Theorem 5: The Degenerate Case I = J In this section we complete the proof of Theorem 5 and thereby the proof of Theorem C5 : Stage 2. The remaining degenerate case to be ruled out is (15A3). Thus we prove: Proposition 18.1. It is not the case that K = I and I = J. Throughout this section we assume that K = I and I = J, and we aim derive a contradiction. Since O3 (CG (b)) has odd order, O2 (K) = O2 (I) = 1. Thus, K is simple, whence K∼ = J ∈ C2 ∩ C3 . Indeed we have: Lemma 18.2. For some n ≥ 0, E(Ct ) = K × K1 × · · · × Kn . Here each Ki ∈ C2 ∩ C3 , and for any b0 ∈ B ∩ K # , if we let Ji be the subnormal closure of Ki in CG (b0 ), then Ki ∼ = Ji /O3 (Ji ). Proof. Let K, K1 , . . . , Kn be the components of E(Ct ). Then for each i = 1, . . . , n, (B, t, Ki ) ∈ BtK3 (G), by Corollary 11.2. All these components Ki are simple by Corollary 17.43. Also by Theorem 3, there exists a nonconstrained {t, b0 }neighborhood (Ki , Ii , Ji ) for some b0 . Since [K, Ki ] = 1, [Ki , b0 ] = 1, and then by Lemma 16.1 we may take b0 to be arbitrary in B ∩ K, and the lemma holds.  It follows directly from Lemmas 16.4 and 13.2 that K, and each Ki , lying in C2 ∩ C3 , is isomorphic to one of the following groups: (18A)

A6 , U4 (2), L2 (8), U5 (2), U6 (2), Sp6 (2), D4 (2), 3D4 (2), F4 (2), 1 2 F4 (2 2 ) , Sp4 (8), J3 , Suz, Co2 , Co1 , F i22 , F i23 , F i24 , F3 , F2 , L3 (3), or U3 (3).

However, no two of K, K1 , . . . , Kn need be isomorphic. A variety of arguments seems to be needed to rule out all these possibilities for K. Throughout, we fix Q ∈ Syl2 (C(t, K). ∼ We choose K = A6 if possible. Thus, (18B)

If K ∼  A6 , then Ki ∼  A6 for any i = 1, . . . , n. = =

Lemma 18.3. One of the following holds: (a) n ≥ 1; (b) K is standard in G; or (c) C(t, K) ∼ = GL2 (3) and m3 (Aut(K)) ≥ 3; or (d) K ∼ = A6 and m2,3 (G) > 4. Proof. Suppose that (a) and (b) fail, so that n = 0 and K is not terminal in G. Then F := F ∗ (C(t, K)) = O2 (C(t, K)). By Proposition 10.2, F is of symplectic type. If m3 (Aut(K)) ≤ 2, F contains Q8 ∗Q8 as m3 (CB (K)) ≥ m3 (B)−m3 (Aut(K)) ≥ 2. Thus m2 (Q) ≥ 3 and no maximal subgroup of Q is embeddable in a dihedral group. But, unless (d) holds, K is Z4 -semirigid by [VK , 14.6] and the fact that BtK3exc (G) = ∅, so by [V2 , 8.4] there is z ∈ I2 (Z(Q)) such that the subnormal closure Kz of K in CG (z) satisfies Kz > K. But then Q/CQ (Kz ) is cyclic and t ∈ CQ (Kz ). As t ∈ [F, F ] ≤ [Q, Q], this is a contradiction.

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Therefore m3 (Aut(K)) ≥ 3. Since K = I, m3 (C(t, K)) > 0. Thus F contains Q8 , so t ∈ [F, F ]. By [VK , 14.6], K is semirigid in G. Suppose that there is z ∈ I2 (Z(Q)) such that the subnormal closure Kz of K in CG (z) satisfies Kz > K. Then Q/CQ (Kz ) is cyclic so [Kz , t] = 1, a contradiction. Therefore no such z exists, so by [IG , 7.4], Q is dihedral or semidihedral. As m3 (C(t, K)) > 0 and F is of symplectic type, C(t, K) ∼ = GL2 (3). Thus (c) holds and the proof is complete.  Lemma 18.4. K ∼  J3 , F i24 , or F3 . = Proof. Suppose on the contrary that K has one of these isomorphism types. Then by [IA , 6.1.4] and [VK , 3.10, 10.39], if we let T ∈ Syl2 (Ct ), the following conditions hold: (1) For any X ∈ C2 with X/O2 (X) ∼ = K, ↑2 (X) ∩ C2 = ∅; (2) I2 (CAut(K) (T )) ≤ Inn(K); and (18C) (3) T normalizes no subgroup R ∼ = Q8 of Aut(K). Now as G is of even type, we may choose a sequence (t, K) = (t0 , K0 ) < · · · < (tn , Kn ) of pumpups with Kn terminal and n as small as possible. Then by (18C1), every pumpup in the chain is trivial or diagonal. If n > 0 then by minimality of n, the last pumpup is diagonal. But Kn is terminal, hence standard by [II3 , Theorem PU4 ]. Thus Kn commutes with none of its distinct conjugates, a contradiction. Therefore n = 0, i.e., K is already standard in G. Let Q ∈ Syl2 (CG (K)). By [V2 , 9.1], m2 (Q) = 1, whence t ∈ Z ∗ (NG (K)). Since all components of Ct lie in C2 , which contains no component of 2-rank 1, F ∗ (C(t, K)) = O2 (Ct ) ≤ Q. On the other hand, since K = I, B0 := CB (K) = 1. The only possibility is that O2,3 (Ct ) = B0 Q0 ∼ = SL2 (3), with B0 ∼ = Z3 , Q0 ∼ = Q8 , and either Q = Q0 or Q ∼ = Q16 . Expand Q ≤ T ∈ Syl2 (Ct ). If there is x ∈ NG (T ) − Ct , then Qx0  T with x Q0 ∩ CG (K) = 1. Then T ∩ K normalizes Qx0 , which embeds in Aut(K), contrary to (18C3). Therefore NG (T ) ≤ Ct , and in particular T ∈ Syl2 (G). Set C0 = CG (B0 ). As I = J , K covers a component of C 0 := C0 /O3 (C0 ). By  (1A4), O3 (C0 ) has odd order. Thus CC 0 (K t ) is covered by CC0 (K t), which equals B0 t. It follows that CC 0 (K) is solvable, and as O3 (C 0 ) = 1, t must invert some y ∈ I3 (C0 ) with y ∈ O3 (C 0 ). Now y centralizes K = J. Let J0 = CJ (y). Then J0 covers J and is tinvariant, whence K ≤ J0 . In particular y ∈ CG (K). As Q ∈ Syl2 (CG (K)), CG (K) = O2 (CG (K))C(t, K) [IG , 15.3]. Hence y = [t, y 2 ] ∈ O2 (CG (K)). As a result B0 Q0 acts faithfully on a 3-subgroup R = [R, t] of O2 (CG (K)). Let z ∈ I2 (K ∩ Z(T )) and set Cz = CG (z). As T ∈ Syl2 (G), also T ∈ Syl2 (Cz ). Since [t, O2 (Cz )] = 1, we must have [y, O2 (Cz )] = 1. Therefore there is a component L of Cz such that [L, y] = 1. Notice that T may be chosen to be permutable with B0 , and in that case |T : CT (B0 )| ≤ 8. It follows by [V2 , 6.7] that B0 normalizes all components of Cz . Hence, so does Q0 = [Q0 , B0 ]. But L ∈ C2 , so Out(L) does not involve A4 , by [VK , 2.3]. Therefore Q0 induces inner automorphisms on L, and by [III11 , 13.1], Q0 ≤ L. In particular R = [R, t] ≤ L. Since Q0 B0  CLB0 (t) and L ∈ C2 , it

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∼ M11 or L ∈ Chev(3) with q(L) = 3. Since follows from [III11 , 13.1] that L = Q0  CL (t) and L ∈ C2 we actually have L ∼ = M11 , U3 (3), G2 (3), or L3 (3). As B0 R maps faithfully into Aut(L), U3 (3) and M11 are impossible. Suppose that L∼ = SL2 (3) and Z(M ) = t. Since = G2 (3). Then CL (t) = M ∗ Q0 B0 with M ∼ | Out(K)| ≤ 2, O2 (M ) induces inner automorphisms on K with CO2 (M ) (K) = t. This forces M ≤ K, whence t ∈ O2 (K) = 1, a contradiction. Hence, the only choice is L ∼ = L3 (3). Then T ∩ L = Q0 u is semidihedral, for some involution u. Suppose that u ∈ KQ = K × Q. Then the projection of u on Q is an involution so [Q0 , u] = 1, contradiction. Thus u induces a noninner automorphism on K. But as t ∈ T ∩ L, T normalizes L and hence T ∩ L  T . Therefore u Q/Q  T /Q so u centralizes a Sylow 2-subgroup of K. This contradicts (18C2), and so the proof is complete.  Lemma 18.5. K ∼  L2 (8) or Sp4 (8). = Proof. Suppose that K has one of these isomorphism types. Choose B ∗ ∈ E (G) such that B < B ∗ , and choose b ∈ CB (K)# . B ∗ exists by (1A6). Then CG (b) has a 3-component J such that K = J   C b . Since B ∩ K = 1, B ∗ normalizes J. Let J0 = J if K ∼ = L2 (8); if K ∼ = Sp4 (8), choose b0 ∈ B ∩ K # and 2 ∼ let J0 = O (CJ (b0 )). In either case, J 0 = L2 (8) and B ∗ normalizes J0 . Hence B ∗ has a hyperplane B0 either centralizing J 0 or inducing only field automorphisms on J 0 . In any case, B0 centralizes an involution of J0 and therefore normalizes a Sylow 2-subgroup T1 of J0 containing it. But |B0 | = |B ∗ |/3 ≥ |B|, so 3

B0 ∈ B∗,o (G) and T1 ∈ IG (B0 ; 2) with CT1 (B0 ) = 1. Hence by Proposition 10.2, T1 is of symplectic type, a contradiction as T1 ∼ = E23 . This completes the proof.  1 Lemma 18.6. K ∼ = 2F4 (2 2 ) , 3D4 (2), Sp6 (2), or F4 (2).

Proof. Suppose false. Let B < B ∗ ∈ S3 (G); again B ∗ exists by (1A6). Then B centralizes b and B ∩ K, so B ∗ normalizes J. Set A = CB ∗ (J) and write B ∗ = A∗ × A where A∗ ≥ B ∩ J = B ∩ K and F ∗ (JA∗ ) = J. Thus A∗ = B ∩ K unless possibly K ∼ = 3D4 (2) and JA∗ ∼ = Aut(K). By [VK , 12.17], there exists A1 ∈ E3 (JA∗ ) with the following properties: ∗

(18D)

(1) m3 (A1 ) = m3 (A∗ ) − 1; and (2) CJ (A1 ) contains an involution u such that u = Ω1 (Z(S)) for any S ∈ Syl2 (J).

Set B1 = A1 × A. Then by (18D1), m3 (B1 ) = m3 (A∗ A) − 1 ≥ m3 (B), and [B 1 , u] = 1. As O3 (J) has odd order, B1 ∈ B3∗,o (G). Since J ∈ Chev(2), the Borel-Tits theorem implies that there is a B1 -invariant parabolic subgroup P of J with u ∈ R := O2 (P ). Again as O3 (J) has odd order, Proposition 10.2 implies that R is cyclic or of symplectic type, and u = Ω1 (Z(R)). Let S ∈ Syl2 (J) with S ≤ P ; as F ∗ (P ) = R it follows that u = Ω1 (Z(S)). This contradicts (18D2), so the proof is complete.  Lemma 18.7. K ∼ = L± 3 (3). ∼ L± (3), then m3 (C(t, K)) ≥ m2,3 (G) − m3 (Aut(K)) ≥ 4 − Proof. If K = 3 2 = 2, so as O2 (C(t, K)) = 1 and all components of E(C(t, K)) are C2 -groups,

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m2 (C(t, K)) ≥ 2. By [IA , 7.3.4], K = ΓD,1 (K) for every D ∈ E22 (Aut(K)), so by [V2 , 5.1], K is not terminal in G. By Lemma 18.3, n ≥ 1. As K1 is given in (18A), m2 (K1 t) ≥ 3, with strict inequality unless K1 ∼ = L± 3 (3) or A6 . The last does not occur by our choice of K (18B), and in the first two cases, K1 ∩ Q ∼ = SD16 or Z4  Z2 , so K1 t ∩ Q = (K1 ∩ Q) × t has no maximal subgroup embeddable in a dihedral group. Thus again [V2 , 8.4] yields z ∈ I2 (Z(Q)) such that Kz > K, and Q/CQ (Kz ) is cyclic with t ∈ CQ (Kz ). Since K1 ∼ = A6 and K1 is one of the groups in (18A), it follows from [VK , 12.16] applied to X = C(t, K) that there is b ∈ CB (K)# and u ∈ I2 (Q) with u = t and [b, u] = 1. Replacing u by tu if necessary we may assume that u ∈ CQ (Kz ). Thus the subnormal closure Ku of K in CG (u) satisfies Ku ≥ Kz > K. Now [b, K u] = 1 so b normalizes Ku . Thus b, t ∼ = Z6 acts on K u ∈ C2 with K   CKu (t). By [VK , 3.54], we have [Ku , b] = 1. Hence Ku = K Ku lies in the subnormal closure Jb of K in Cb . But by Lemma 16.1, the only composition factor of Jb of order divisible by 3 is isomorphic to K, a contradiction. This completes the proof of the lemma.  Lemma 18.8. K ∼ = F i22 . Proof. Suppose false. Let A = CB (K), so that m3 (A) = m2,3 (G) − m3 (K). Set CA = CG (A) and C A = CA /O3 (CA ). Then K covers J 0 for some 3-component J0 of CA , and O3 (CA ) has odd order. Set C0 = CCA (J 0 ). Then t ∈ O3 (C0 ). So whether t acts nontrivially on O3 (C 0 ) or E(C 0 ), there exists v ∈ I3 (C0 ) − A [IG , 16.9]. Let A∗ = A v. By [IA , 5.6.1, 5.3t], there exists D ∈ E3 (J0 ) such that m3 (D) = m3 (J0 )−1 and CJ0 (D) contains an involution u, indeed CJ 0 (u) ∼ = 2U6 (2). ∗ := ∗ D × A ∈ B∗,o (G). Then B Let Ku = E(CK (u)) ∼ = 2U6 (2). The subnormal closure Lu of Ku in CG (u) cannot be a vertical pumpup, since Ku would be an anchored subcomponent of CLu (t) and no such K-groups Lu exist, by [VK , 3.33]. Suppose that Lu = Ku . Then (B ∗ , u, Lu ) ∈ BtK3 (G) and CB ∗ (Lu ) ≥ A = 1. Hence by Corollary 17.43, O2 (Lu ) = 1, a contradiction. Therefore, Lu is a diagonal pumpup of Ku , say Lu = Lu1 Lu2 as a product of components. But A∗ centralizes u and Ku , so it normalizes and then centralizes Lu . As m3 (Lu ) = 8, it follows that m2,3 (G) ≥ m3 (CG (u)) ≥ 8 + m3 (A∗ ) > 4 + m3 (A∗ ) = m3 (B), a contradiction. The lemma is proved.  Lemma 18.9. K ∼  A6 . = ∼ A6 . Since m3 (C(t, K)) ≥ 4 − m3 (Aut(K)) = 2, Proof. Suppose that K = and no C2 -groups have 2-rank 1, certainly m2 (Q) ≥ 2. We claim that K is not standard in G. If it is standard, then as K is outer well-generated by [VK , 8.4], Qg ≤ NG (K) for some g ∈ G − NG (K), with ΓQg ,1 (K) < K, by [IG , 18.11]. By [VK , 8.1], Q ∼ = E22 . Since t ∈ Z(C(t, K)), this implies that C(t, K) has a normal 2-complement, whence |C(t, K)| = 4. As 3 divides |C(t, K)| this is a contradiction, so our claim holds. Therefore, K is not terminal in G [II3 , Theorem PU4 ]. We next prove that n ≥ 1.

(18E) ∗

Suppose false, so that F := F (C(t, K)) = O2 (Ct ) admits CB (K) faithfully, and is of symplectic type by Proposition 10.2. Also m2,3 (G) > 4 by Lemma 18.3.

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196

4. THEOREM C5 : STAGE 2

We argue that K is semirigid in G. Otherwise, (t, K) < (u, H) for some (u, H) ∈ ILo2 (G) with H ∼ = L4 (3), by [VK , 14.3]. Since m2,3 (G) > 4, we have m3 (CB (K)) ≥ 3 and F contains Q8 ∗ Q8 ∗ Q8 . Now Fu := CF (u) acts on H, with t acting nontrivially and AutFu (H) ≤ CAut(H) (K) ∼ = D8 . Then |Fu | ≥ |CF/Φ(F ) (u)| ≥ 23 . But Fu cannot be isomorphic to D8 ; if it were, u would have an additional fixed point on CF (Fu ). Therefore, CFu (H) contains an involution v ∈ F − t (recall that [H, t] = 1). The pumpup Hv of K in CG (v) is then equal to the pumpup of H in CG (v), which by [VK , 3.10] is either trivial or diagonal. But now Fv := CF (v) contains Z2 × F0 with F0 ∼ = Q8 ∗ D8 . In the trivial pumpup case, as t = Z(F0 ), F0 embeds in CAut(H) (K) ∼ = D8 , a contradiction. In the diagonal pumpup case, F0 has a subgroup F 0 of index 2 acting on H and centralizing K; then F 0 ∼ = Q8 × Z2 or Q8 ∗ Z4 , and in either case, Q8 embeds in CAut(H) (K) ∼ = D8 , an impossibility. This proves semirigidity of K. Hence by [IG , 7.4], and because K is not terminal, there exists u ∈ I2 (Z(Q)) such that K pumps up nontrivially in CG (u). As n = 0, F ∗ (C(t, K)) = O2 (Ct ), so u ≤ Z(O2 (Ct )) = t by Proposition 10.1. This contradicts the fact that K   Ct , so we have proved (18E). As a consequence, C(t, K) ≥ E(C(t, K))×t so Z(Q) is noncyclic and m2 (Q) ≥ 3. Letting Kv be the subnormal closure of K in CG (v), for any v ∈ I2 (C(t, K)), we can deduce that (18F)

Kv > K for some v ∈ I2 (Z(Q)).

Namely, if K is semirigid in G, this holds by [IG , 7.4]. Otherwise, again (t, K) < (u, H) for some H ∼ = L4 (3), by [VK , 14.3]. Letting Qu = CQ (u) and QH = CQu (H), we need only show that Z(Q) ∩ QH = 1. But Qu /QH embeds in CAut(H) (K) ∼ = D8 , so as Z(Q) is noncyclic, (18F) holds unless possibly Z(Q) maps isomorphically on Qu /QH ∼ = E22 . But in this case too, (18F) holds; otherwise, by [IA , 7.3.3], K  ΓQu /QH ,1 (H) = H, a contradiction. Let v be as in (18F) and set Qv = CQ (Kv ). Then Q/Qv embeds in CAut(Kv ) (K) and hence, by [VK , 3.39], in D8 . Also K < Ku for all u ∈ Qv . In particular we let U be the set of all 2-subgroups U ≤ C(t, K) such that either U is not embeddable in D8 or some C(t, K)-conjugate of U contains Ω1 (Z(Q)), and deduce that any U ∈ U contains an involution u such that Ku > K. Suppose that b1 ∈ CB (K)# and [b1 , U ] = 1 for some U ∈ U, and choose u ∈ I2 (U ) such that Ku > K. Then b1 normalizes Ku and centralizes K. Let J1 be the subnormal closure of K in CG (b1 ). By Lemma 16.1, K covers J1 /O3 (J1 ). It follows that K covers a component of CKu (b1 )/O3 (CKu (b1 )), so by the B3 -property, K   CKu (b1 ). If Ku is a diagonal pumpup of K, then b1 must normalize, and then centralize, each component of Ku , contradicting K   CKu (b1 ). Thus, Ku is a vertical pumpup of K relative to the prime 3, which contradicts [VK , 13.2] as Ku ∈ C2 . Thus no such U and b1 can exist. This imposes strong conditions on C(t, K). If n ≥ 2 we could take b1 ∈ B ∩ K2 and any U ∈ Syl2 (K1 t), thereby contradicting the previous paragraph. Thus n = 1. Moreover, as K1 is one of the groups in (18A), and 3-central elements of I3 (K1 ) lie in a conjugate of B, and even J-3-central elements do if K1 ∼ = J3 , we can find U = U1 × t ∈ U with U1 ≤ K, and b1 ∈ K1 , satisfying the conditions of the previous paragraph unless K1 ∼ = A6 , by [VK , 12.15]. (We call an element of K J-p central if it lies in Z(J(P )) for some P ∈ Syl3 (K).) Finally, if m2 (O2 (Ct )) ≥ 3 or

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O2 (Ct ) contains Q8 , then O2 (Ct ) ∈ U and we can take any b1 ∈ B ∩ K1 to reach the same contradiction. It follows that O2 (Ct ) admits no automorphism of order 3. Now we have F ∗ (Ct ) ∼ = A6 × A6 × O2 (Ct ), with Aut(O2 (Ct )) a 2-group. Hence m2,3 (G) = m3 (B) = 4. Also, there is symmetry between K and K1 . Since m2,3 (G) = 4 and by (10B), there are no pairs (w, H) ∈ ILo2 (G) such that H/Z(H) ∼ = L4 (3). Hence K is semirigid in Kv , by [VK , 14.3]. Thus by [IG , Definition 7.1], there exists R ≤ Q such that Q = R t and K is not a component of CG (u) for any u ∈ I2 (R). If O2 (Ct ) > t, we choose an involution u ∈ R ∩ O2 (Ct ) and any b1 ∈ I3 (K1 ), and argue as above that K is a component of both CKu (t) and CKu (b1 ), contradicting [VK , 13.2]. Thus, O2 (Ct ) = t. Moreover, R ∩ K1 = 1, so Ku > K for all u ∈ I2 (K1 ). We next show that (18G)

For any b0 ∈ I3 (K1 ), K   CG (b0 ).

By Sylow’s theorem and Lemma 16.1, we may assume that b0 = b, so K ≤  J   CG (b) with J = O 3 (J) = KO3 (J). Set X = O3 (J); it suffices to show that [X, K] = 1. There is u ∈ I2 (K1 ) such that bu = b−1 , and J is u, t-invariant. As X has odd order it suffices to show Ww := [CX (w), K] = 1 for all w ∈ u, t# . By [IG , 4.3(iv)], Ww = [Ww , K], a K-invariant {2, 3} -subgroup of Kw , the subnormal closure of K in CG (w). If Kw is a trivial or diagonal pumpup of K, then Ww projects trivially on each of its components, so is trivial. If Kw is a vertical pumpup of K, the triviality of Ww follows from [VK , 7.7] inasmuch as A6 ↑2 Kv ∈ C2 . Thus (18G) holds. Let u ∈ I2 (K1 ). By (10B) and since m2,3 (G) = 4, Ku /O2 (Ku ) ∼ = L± 4 (3). Thus, by [VK , 3.10], one of the following holds: (1) Ku ∼ = A6 × A6 ; (2) Ku ∼ = 2HS, HS, L4 (2), L5 (2), or Sp4 (4); or (18H) (3) Ku ∼ = U4 (2) or U5 (2). Suppose that (18H1) holds. Let Bu ∈ Syl3 (Ku ) and let Ku1 and Ku2 be the components of Ku . Then (Bu , u, Ku1 ) ∈ BtK3 (G). Let bu ∈ (Bu ∩ Ku2 )# . By Propositions 12.5 and 17.1, (Ku1 , Ku1 , Ju1 ) ∈ KIJ(G), where Ju1 is the subnormal closure of Ku1 in CG (bu2 ). Hence our analysis of (K; I; J) applies equally well to (Ku1 , Ku1 , Ju1 ). In particular we have F ∗ (CG (u)) = Ku × u with Ku ∼ = A6 × A6 . However, by [IG , 7.4], if we let R = Q ∩ K1 t ∼ = D8 × Z2 , then R = t × R0 where R0 ∼ = D8 centralizes Ku u. This contradicts the F ∗ -theorem. Hence (18H1) does not hold. Suppose that (18H2) holds. Then B ∩ K ∈ Syl3 (Ku ) contains some a of order  = 3 such that La := E(CKu (a)) ∼ = A5 [IA , 4.8.2, 5.3m]. Set C = CG (a) and C C/O3 (C). Then O3 (C) ≤ O2 (C) and La ≤ L2 (C) by L2 -balance. We argue  a , O3 (C)]  = 1. From the structure of Ct , that La ≤ L3 (C), by showing that [L O3 3 (CC (t)) = t (B ∩ K). Also B ∩ K contains a Sylow 3-subgroup of La . Hence    a, and so t inverts O3 (C)/ a. As [La , t] = La , La centralizes O3 (C), CO3 (C)  (t) =  as asserted.  a is a component of C  (u). Hence,  lie in C3 , and L The components of E(C) E(C)  a lies in a component of C  isomorphic to A9 . In addition, by by [VK , 13.4], L  has a component isomorphic to (18G) and the symmetry between K and K1 , C

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198

4. THEOREM C5 : STAGE 2

A6 . Therefore, m2,3 (C) ≥ m2,3 (Z3 × A9 × A6 ) = 5 > m2,3 (G), a contradiction. Thus (18H2) does not hold. Finally, suppose that Ku ∼ = Um (2), m = 4 or 5. Then B ∩ Ku ≤ A ∼ = E3m−1 for some A ≤ Ku . As A centralizes B ∩ Ku and m2,3 (G) = 4, A normalizes the component K1 of CG (B ∩Ku ), by (18G) and the symmetry between K and K1 . Set H = E(A, K) ≤ Ku , so that by [VK , 8.2], H is simple, K ≤ H, and m3 (H) ≥ 3. But [K, K1 ] = 1, so H normalizes and then centralizes K1 . In particular H ≤ CG (b). Since K   CG (b) by (18G), this is a contradiction, and the proof of the lemma is complete.  Having ruled out K ∼ = A6 , we now have symmetry between K and each Ki , 0 < i ≤ n. Summarizing Lemmas 18.4, 18.5, 18.6, 18.7, 18.8, and 18.9, and using this symmetry, we have: Lemma 18.10. The isomorphism types of K and each Ki , i > 0, are one of the following: Un (2), 4 ≤ n ≤ 6, D4 (2), Suz, Co2 , Co1 , F i23 , or F2 . Lemma 18.11. K ∼  U4 (2). = ∼ U4 (2). First assume that K is not standard in G. Proof. Suppose that K = Then there exists u ∈ I2 (Q) such that the subnormal closure Ku of K in CG (u) satisfies Ku > K. As K is semirigid in G by [VK , 14.2], it follows from [IG , 7.4] that we may take u ∈ Z(Q), and then Q = Q0 t for some Q0 such that K has a nontrivial pumpup in CG (v) for all v ∈ I2 (Q0 ). We claim that n ≥ 1. If the claim is false, then by Lemma 18.3, C(t, K) ∼ = GL2 (3). In particular m2,3 (G) = 4. As m2,3 (G) = 4 but m2,3 (U4 (2) × U4 (2)) = 5, Ku is a vertical pumpup. But also by (10B), Ku /Z(Ku ) ∼ = L± 4 (3). Hence by [VK , # 3.10], Ku ∼ = L4 (4). Choose a ∈ (B ∩ Ku ) such that La := E(CKu (a)) ∼ = SL3 (4)    by and set C = CG (a) and C = C/O3 (C). Then O3 (C) ≤ O2 (C) and La ≤ L2 (C)  L2 -balance. Let W = O3 (C). Since La is a component of CG (u, a), it acts trivially    u acts faithfully on W . on CW  (u). Hence if La acts nontrivially on W , then La   u But this implies m2,3 (G) ≥ 5, by the Thompson dihedral lemma and as La   ] = 1, whence L  a ≤ E(C).  As a , W centralizes  a. This contradiction shows that [L  a is a component of C  (u), it follows  are in C3 , and L all components of E(C) E(C)  has a component M  ∼ a ≤ M . But from [VK , 3.91] that C = 3Suz such that L CG (t, a) has a SU3 (2)-solvable component contained in La . Therefore CM (t) has ∼ 3Suz (see [IA , 5.3o]). This = such a solvable component. But this is absurd as M proves our claim that n ≥ 1. Continuing to assume that K is not standard in G, we note that by the Thompson transfer lemma, every involution in K1 is K1 -conjugate into Q0 ∩ K1 , and so K has a nontrivial pumpup in the centralizer of any involution of K1 . Now by Lemma 18.10 and [VK , 12.15], there are a ∈ E ∈ E3∗ (K1 ) and v ∈ I2 (K1 ) such that [a, v] = 1. Without loss, a ∈ B, by Proposition 12.5. Let Kv be the subnormal closure of K in CG (v). Then K is a component of CKv (t). As K covers a component of CG (a)/O3 (CG (a)) by Lemma 18.4, it follows from the B3 -property that K is also a component of CKv (a). But this violates [VK , 13.2]. We have proved that K is standard in G. As K is outer well-generated by [VK , 8.3], there exists by [IG , 18.11] an element g ∈ G − NG (K) such that Qg ≤ NG (K) and ΓQg ,1 (K) < K. By [VK , 8.3],

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199

this implies that m2 (Qg ) ≤ 2, so m2 (Q) ≤ 2. If K1 exists, it is simple and so m2 (Q) ≥ m2 (K1 t) ≥ 3. Therefore K1 does not exist, so F := F ∗ (C(t, K)) = O2 (C(t, K)). As CB (K) = 1 and F is of symplectic type, F contains a copy of Q8 . Hence Q does as well, whence by [VK , 8.3], m2 (Q) = 1. In particular CB (K) = b is cyclic, and m2,3 (G) = 4. Moreover, since all components of Ct are C2 -groups, C(t, K) must be solvable, indeed a {2, 3}-group. Let X = CG (b). Then CC(b,J) (t) = CC(t,K) (b) = b, t so C(b, J) = Y t where Y has odd order and t inverts Y / b. As t ∈ O3 (X), O3 (Y ) > b. We claim that t is 2-central in G. Let T ∈ Syl2 (Ct ) with Q ≤ T and write z = Z(T ∩ K). By [III17 , 8.8d], Z(T ) = z, t. If T ∈ Syl2 (G), choose g ∈ NG (T ) − T with g 2 ∈ T . Then as Q  T , Qg  T and so the image of Qg in Aut(K) is normalized by InnT (K), contrary to [VK , 10.39]. This proves our claim. Finally, let Cz = CG (z). We have O 2 (CK (z)) = L1 ∗L2 with L1 ∼ = L2 ∼ = SL2 (3) and z ∈ L1 ∩ L2 . Thus L1 and L2 are solvable components of CCz (t). As all components of Cz are C2 -groups it follows from [VK , 3.105] that O2 (L1 L2 ) ≤ O2 (Cz ). Consequently if L is any component of Cz , then L ∩ T ∈ Syl2 (L) and L ∩ T ≤ CT (O2 (L1 L2 )) = z × Q. Hence m2 (L z / z) = 1, which is impossible as L ∈ C2 . Thus, Cz is 2-constrained. Then t ∈ Z(T ) ≤ O2 (Cz ) and so 1 = [Y, t] ≤  O2 (Cz ) = 1, a contradiction. The lemma is proved. By Lemmas 18.10, 18.11, and 13.2, we see that to complete the proof of Proposition 18.1, and with it the proofs of Theorem 5 and Theorem C5 : Stage 2, we need show that the following cases are impossible: (18I) K∼ = Suz, Co1 , F i23 , F2 , D4 (2), Co2 , U5 (2), U6 (2) We assume henceforth that K is one of the above groups, and in fact choose the isomorphism type of K to be as early as possible in the above list. Thus for example if K ∼ = D4 (2), then there are no instances of K ∼ = Suz. By [IA , 4.10.3, 5.6.1], (18J)

m3 (B ∩ K) ≥ 4, and hence a Sylow 3-subgroup of K is connected

by [IG , 10.17]. Lemma 18.12. K ∼  Suz, Co1 , F i23 , F2 , or D4 (2). = Proof. The main work in the proof is to construct a 3-component uniqueness subgroup M . We first consider the case Θ1 (G; B) = 1.

(18K)

Because G is balanced with respect to every element of E34 (G) (1A7), (18L)

|Θ1 (G; B)| is an odd 3 -number,

and (18K) is equivalent to (18M)

O3 (CG (x)) = 1 for some x ∈ I3o (G),

and it implies that for any E ∈ E34 (G), O3 (CG (x)) = 1 for some x ∈ E # . We set (18N)

 = M/O3 (M ), M = NG (Θ1 (G; B)) < G, and M

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4. THEOREM C5 : STAGE 2

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and conclude that for any Sylow 3-subgroup P of G containing B, Γoo P,2 (G) ≤ M.

(18O)

In particular P ≤ M . We fix P to contain a Sylow 3-subgroup of CG (t). By (18L), Θ1 (G; B) ≤ O{2,3} (M ).

(18P) We first show that

K ≤ M.

(18Q)

It suffices to show that K ≤ Γoo P,2 (G). As P ∩K is connected, and 3 divides |C(t, K)| (as K = I), and as K is simple, ΓP ∩K,1 (K) ≤ Γoo P,2 (G). Thus, if (18Q) failed, K would have a strongly 3-embedded subgroup. Given the possible isomorphism types of K, no such subgroup exists, by [IA , 7.6.1], so (18Q) holds. As a consequence, O3 (J) ≤ O3 (CG (b)) ≤ Θ1 (G; B) ≤ O{2,3} (M ), so  J = O3 (J) ≤ M and J = K. Let K∗ be the subnormal closure of K in M . As K = I and I = J, L3 balance implies that K∗ is the product of one or three 3-components of M permuted transitively by b, and J is a component of CK∗ (b). Likewise by L2 -balance, K∗ is the product of one or two 2-components permuted transitively by t, and K is  = J is a component of both C  (t) and C  (b). a component of CK∗ (t). Now K K∗ K∗ Given the possible isomorphism types of K, we see from [VK , 3.85] that (18R) K∗ /O2 (K∗ ) ∼ =K∼ = K∗ /O3 (K∗ ). Recall from (1A6) that there is A ∈ S3 (G) such that B < A. Then A ≤ M , and  ∗ ) > CB (K  ∗) = since B ∩ K is of maximal 3-rank in both K and Aut(K), CA (K CB (K) > 1. In particular, (18S)

 ∗ )) ≥ 2. m3 (CM (K

Our next goal is to prove that (18T)

 ∗ )), CG (v) has a unique 3-component Kv For all v ∈ I3 (CM (K such that Kv ≤ K∗ and Kv /O3 (Kv ) ∼ = K.

First, choose u ∈ I3 (K) such that CJ (u) has a 3-component L with (18U) L∼ = 3U4 (3), 3Suz, Ω7 (3), F i22 , or U4 (2), according to the isomorphism type of K. Clearly m3 (CG (u)) ≥ 4. Let Lu be the subnormal closure of L in CG (u). By L3 -balance, Lu is the product of 3components permuted transitively by b, and L is a 3-component of CLu (b). But also t acts on Lu , and E(CLu (t)) is the component of CK (u) contained in L and  It follows that Lu /O3 (Lu ) ∈ C3 and then, by [VK , 13.2], that L covers covering L. Lu /O3 (Lu ). In particular as O3 (CG (u)) ≤ M and Lu is perfect, Lu ≤ K∗ .  ∗ ))−B. We prove (18T) first for 3-central v ∈ I3 (CM (K ∗ )); Now let v ∈ I3 (CM (K  ∗ )), then (18T) holds for all thus, we assume that if v is not 3-central (in CM (K  ∗ )) 3-central v. In any case, therefore, we may assume that there is c ∈ I3 (CM (K such that c, v ∼ = E32 , [c, v , u] = 1, and CG (c) has a 3-component Kc ≤ K∗ such that Kc /O3 (Kc ) ∼ = K. (Namely, take c = b if v is 3-central, and otherwise take c to be 3-central.) By L3 -balance, the subnormal closure Kv of L3 (CK∗ (c, v))

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18. THEOREM 5: THE DEGENERATE CASE I = J

201

in CG (v) is a pumpup of Kc . Note that by [VK , 3.85], ↑3 (K) ∩ C3 = ∅, so Kv is a trivial or diagonal pumpup of Kc . But also u acts on Kv and L3 (CKc (u)) has a 3-component contained in Lu and covering Lu /O3 (Lu ). This rules out the diagonal pumpup case, so Kv ≤ K∗ and (18T) is proved. Continuing with v as in (18T), let Cv = CG (v). Let Pv ∈ Syl3 (Cv ) with Pv ∩ M ∈ Syl3 (CM (v)). Since m3 (Kv ) ≥ 4 and Kv ≤ K∗ , m3 (Pv ) ≥ 4, so Γoo Pv ,2 (G) ≤ M .   But now Γoo (G) contains O (C ), then K N (P ∩K ), then L (C 3 v v Cv v v 3 v ) and finally Pv ,2 Cv , the last two by Frattini arguments. We have proved: (18V)

 ∗ )), CG (v) ≤ M. For all v ∈ I3 (CM (K

We are almost ready to quote [II3 , Theorems PU2 , PU3 ], but we must first show that M is a maximal subgroup of G. Let N be a maximal subgroup of G such that M ≤ N . Choose E ∈ E32 (CM (K ∗ )). Then for any v ∈ E # , L3 (CK∗ (v))   CM (v) = CG (v) = CN (v) by (18V), so L3 (CK∗ (v)) ≤ L3 (N ) by L3 -balance.  ∗ is a component of C  (  = N/O3 (N ). By the B3 -property, K Let N N v ). As E is   noncyclic, it follows from [VK , 3.5] that K∗ is a component of N . As m3 (K∗ ) ≥ 4, oo it follows easily that N = Γoo P,2 (N ) ≤ ΓP,2 (G) ≤ M , so M = N , proving maximality. Quoting the above-mentioned theorems, we conclude that one of the following holds: (1) M is strongly 3-embedded in G; or (2) K∗  M , and there exists E ∈ E32 (CM (K ∗ )) and g ∈ G − M such that E g ≤ M .  ∗ ) and P ∩K∗ are disjoint and noncyclic. Suppose that (18W2) holds. Now CP (K Hence each contains a normal E32 -subgroup of P , which in turn is centralized by an element of E g . Hence by (17K), M ∩ M g ≥ ΓE g ,1 (M ), which contains an g E34 -subgroup. But ΓP,4 (G) ≤ Γoo P,2 (G) ≤ M , and so M ∩ M contains a Sylow g 3-subgroup P0 of G. Then M and M are conjugate in NG (P0 ) ≤ M , so M = M g and g ∈ M , contradiction. Therefore M is strongly 3-embedded in G. Next, let u ∈ I3 (K) be as above (18U), and choose a four-group F ≤ CK (u). Let y ∈ F # and set C = CG (y). We claim that CO3 (M ) (y) = 1. Since O3 (M ) has odd order, this is obvious if C ≤ M , as O2 (CG (y)) = 1. Suppose then that C ≤ M . Then C0 := C ∩ M is strongly 3-embedded in C. Since C ≥ CP (K) × u  and CP (K) is noncyclic, m3 (C0 ) ≥ 3. Thus, F ∗ (C/O3 (C)) = O 3 (C/O3 (C)) is a single simple component isomorphic to a twisted rank 1 group in Chev(3), with O3 (C0 ) = O3 (C) in every case [IA , 7.6.1, 7.6.2]. Let R ∈ Syl3 (C0 ); then NC0 (R) acts irreducibly on Z(R). However, as [C0 , y] = 1, C0 normalizes K∗ and  ∗ ), and R ∩ K∗ = 1 = CR (K∗ ). Hence Z(R) ∩ K∗ = 1 = CZ(R) (K  ∗ ) and CM ( K Z(R) ∩ K∗  NC0 (R), contradicting the irreducibility on Z(R). This proves that CO3 (M ) (y) = 1. As y ∈ F # was arbitrary, O3 (M ) = 1, a contradiction. We have proved that (18K) fails, that is, (18W)

(18X)

O3 (CG (x)) = 1 for all x ∈ I3o (G).

In particular, J = K. Choose A ∈ S3 (G) again with B < A. Since m3 (A) > m2,3 (G), A normalizes K, by [IG , 8.7(i)] applied in CG (b). Again as m3 (B ∩ K) = m3 (Aut(K)), m3 (CA (K)) > m3 (CB (K)) ≥ 1. For any v ∈ CA (K)# ⊆ I3o (G), the pumpup of K in CG (v) is not diagonal, by [IG , 8.7(i)] again, and it is not vertical

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4. THEOREM C5 : STAGE 2

since ↑3 (K) ∩ C3 = ∅ [VK , 3.85]; hence, K   CG (v). We argue that (18Y)

K is terminal in G (with respect to p = 3).

For this we use the element u ∈ I3 (K), the component L of CK (u), and the subnormal closure Lu of L in CG (u) (see (18U)). Now u, L, and Lu are CA (K)invariant. As K   CG (a) for all a ∈ CA (K)# , it follows that L   CLu (a) for all such a. Since CA (K) is noncyclic, it follows from [VK , 3.5] and (18X) that Lu = L. This prevents K from having a diagonal pumpup in CG (v) for any v ∈ I3 (CG (K)). Again [VK , 3.85] implies that K has no vertical pumpup in CG (v) for any such v. These facts, and (18X), imply (18Y). Now define [K] to be the product of K and all its G-conjugates that lie in CG (K) [IG , 18.2, 18.3]. We argue that (18Z)

M := NG ([K]) is a strongly 3-embedded maximal subgroup of G.

By [II3 , Theorem PU4 ], in proving (18Z), we may assume that K is standard in G (with respect to the prime 3), so that [K] = K and M = NG (K). Let M ≤ N < G. Let P ∈ Syl3 (M ). Then ΓCP (K),1 (G) ≤ M . In particular O3 (N ) ≤ M . By [VK , 3.5], and again as CP (K) is noncyclic, K covers a component of N/O3 (N ), and then K   N . Hence K  N so N = M , proving the maximality of M . To prove the strong 3-embedding in (18Z), we use [II3 , Theorem PU2 ]. Accordingly, we let Q ∈ Syl3 (CG (K)), assume that m3 (Qg ∩ M ) ≥ 2 for some g ∈ G − M , and that (18AA)

Γ := ΓQg ∩M,1 (K) < K,

and aim to derive a contradiction. Let P ∈ Syl3 (M ) with Q ≤ P , and set T = P ∩K. If K ∼ = D4 (2), then Γ = K by [IA , 7.3.3], a contradiction. So we ignore this case. Hence K ∈ Spor, so Out(K) is a 3 -group. In particular by [II3 , 16.11], we may assume that Qg ≤ M . Without loss, Qg ≤ P . We see from [IA , 5.3] that P = T × Q with Z(T ) ∼ = E32 or Z3 according as ∼ K = Suz or not. We argue that (18BB)

NG (P ) ≤ M.

If false, then we could have taken g ∈ NG (P ) − M . By standardness, Q ∩ Qg = 1, so in particular Z(Q) ∩ Z(Q)g = 1. Unless K ∼ = Suz, this implies that |Z(Q)| = |Z(T )| = 3. If Z(Q)g = Z(T ), g −1 then T projects injectively to Q, so |Q| ≥ |T |. But then Qg projects onto T ,  so ΓT,1 (K) = Γ < K and K has a strongly 3-embedded subgroup, contrary to [IA , 7.6.1]. Thus Z(Q)g = Z(T ), so Qg ∩ T = 1. As Qg  P , [Qg , T ] = 1, so Qg ≤ Z(T ) × Q and so Qg ∩ Q = 1, a contradiction. Hence, (18BB) holds if K∼  Suz. = Suppose that K ∼ = Suz. If Z(Q) is noncyclic, then E := Ω1 (Z(Q))g projects onto Z(T ) and so Γ ≥ ΓE,1 (K) = ΓZ(T ),1 (K) = K by [VK , 8.21], contradiction. So Z(Q) is cyclic, and as m3 (Q) > 1, Z(Q) ∩ [Q, Q] = 1. By [VK , 9.10], Z(T ) ≤ [T, T ] and T is indecomposable. Hence, by the Krull-Schmidt theorem, g normalizes Ω1 (Z(Q)), contradicting Q ∩ Qg = 1. This completes the proof of (18BB). As a consequence, P ∈ Syl3 (G) but P ≤ M g (since otherwise, M and M g would then be conjugate in NG (P ) ≤ M ). Since P = T × Q, T ≤ M g , and consequently no element of Qg acts on K like a 3-central element of K.

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∼ F2 or Co1 , then Γ = K by [VK , 8.21], so we may ignore Therefore, if K = those cases. Now suppose that K ∼ = F i23 . If K were terminal with respect to p = 2, then by [V2 , 9.3b], |C(t, K)| = 2, contradicting the fact that b ∈ I3 (C(t, K)). Note also that by [V2 , 9.5], Sylow 2-subgroups of C(t, K) have a quaternion or cyclic subgroup of index 2. Hence by [VK , 12.43], m3 (CB (K)) = m3 (C(t, K)) = 1. Thus m2,3 (G) = m3 (B) = m3 (K)+1 = 7 [IA , 5.6.1]. As a result, G contains no subgroup isomorphic to K ×K [IA , 5.6.1, 5.6.2]. As K is not terminal with respect to p = 2, it follows from [VK , 3.10] that (t, K) < (u, K1 ) for some involution u and component K1 ∼ = F i24 . In particular, P contains a copy S of a Sylow 3-subgroup of F i24 . Now |Z(S)| = 3 so S must project injectively into either T or Q. We saw in the last paragraph that |Q| < |T |, so |T | ≥ |S|. As |T | = 313 and |S| = 316 , we have a contradiction. To complete the proof of (18Z) we must rule out the case K∼ = Suz. We still have (18AA) and (18BB). We have P = T ×Q ∈ Syl3 (G) with T ∈ Syl3 (K) and Q ∈ Syl3 (CG (K)). We also have g ∈ G − M with Qg ≤ T and Qg ∩ Q = 1 and ΓQg ,1 (K) < K. Since NG (P ) ≤ M and NG (P ) controls G-fusion in Z(P ) by Burnside’s lemma, while K and CG (K) are normal in M , Z(Q) and Z(T ) are strongly closed in Z(P ) with respect to G. Now E := J(T ) ∼ = E32 , and CE (x) = Z(T ) ∼ = E32 = E35 , T /E ∼ for all x ∈ T − E, by [VK , 8.21]. We claim that if 1 = N  X ≤ T and X is not abelian, then N ∩ Z(T ) = 1. Namely, there exists x ∈ X − E. Choose z ∈ Z(X)# ∩N . If z ∈ E, then z ∈ CE (x) = Z(T ), as desired. If Z(X)∩N ∩E = 1, on the other hand, then N ∩ E = 1 so N is abelian. Then as X is not abelian, X ≤ N Z(T ). If X ∩E > Z(T ), then for y ∈ X ∩E −Z(T ), [y, z] = 1, contradiction. So X ∩ E = Z(T ), whence |X| ≤ 34 . But N ∩ Z(X) = 1, so |Z(X)| ≥ 33 and hence X is abelian, contradiction. This proves our claim. Suppose now that Q is nonabelian. If Qg ∩ T = 1, then the previous paragraph, with X = Qg and N = Qg ∩ T , shows that Qg ∩ Z(T ) = 1. Since projection from Qg to T is injective, there is x ∈ Z(Q) such that xg ∈ Z(T ), contradicting the strong closure proved above. Thus, Qg ∩ T = 1. The projection H of Qg on T must satisfy 1 < H ∩ E < H since Qg is not abelian. Then 1 = [H, H] ∩ Z(H) ≤ Z(T ). Then choosing 1 = y ∈ Z(Q) ∩ [Q, Q], we have y g ∈ ([H, H] ∩ Z(H))Z(Q) ≤ Z(P ), and so by the strong closure proved above, y g ∈ Q, a contradiction. It follows that Q is abelian. If H k ∩ Z(T ) = 1 for some k ∈ K, then for some y ∈ Q# , y gk ∈ (H ∩ Z(T ))Q ≤ Z(P ), again contradicting strong closure. Thus no element of H lies in ∪k∈K Z(T )k , which contains E. Hence Q ∼ = E32 . Moreover, each x ∈ I3 (Qg ) acts on K with Kx := E(CK (x)) ∼ = A6 , and with CK (x) having Sylow Px = Qx Z(T ) = CPx (Kx )×Z(T ), where Qx is the projection of Qg on T , and Z(T ) ≤ Kx . Let H = Qx Kx | x ∈ I3 (Qg ). By L3 -balance, and since O3 (CK (x)) = 1 for all x ∈ I3 (Qg ), Kx ≤ E(H). Since m3 (K) = 5 < 6, Kx lies in a single component of E(H), namely Z(T )E(H) . Hence E(H) is quasisimple. If CQx Kx (E(H)) = 1 for some x ∈ I3 (Qg ), then E(H) = Kx = Ky for all y ∈ I3 (Qg ). Otherwise, CQx Kx (E(H)) = 1 and by [VK , 3.5], E(H) = K. We have shown that either H = K, as desired, or E(H) = Kx for all x ∈ I3 (Qg ). We assume the latter

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204

4. THEOREM C5 : STAGE 2

and derive a contradiction. We have QQg × E(H) ≤ CG (Qg ) ≤ M g , which implies that E(H) ≤ K g and Z(T ) is a Sylow 3-center of K g . Therefore there is h ∈ M g h such that Z(T ) = Z(T h ) = Z(T ) . But Z(T ) = Z(P ) ∩ [P, P ] as Q is abelian. h ≤ CG (Z(T )), we may take h ∈ CG (Z(T )). Let Since NG (P ) ≤ M , and P, P X = {x ∈ Z(T ) | E(CK (x)) ∼ = 3U4 (3)}. As ΓQ,1 (G) ≤ M , we see, using [VK , 3.5], that E(CK (x)) is a component of CG (x) for all x ∈ X. As |G|3 = 32 |Suz|3 , E(CK (x))  CG (x) for all x ∈ X. Therefore h normalizes E(CK (x)) | x ∈ X , which is equal to K by [VK , 8.21]. Thus, h ∈ M , a contradiction. This completes the proof of (18Z). Now, observe from [IA , 5.3louy] that if K ∼ = Suz, Co1 , F i23 , or F2 , then for every involution u ∈ M , m3 (CM (u)) ≥ 2 and CM (u) is not 3-solvable. But if CM (u) < CG (u), then CM (u) is strongly 3-embedded in CG (u), and we get a contradiction from [IA , 7.6.1, 7.6.2], according to which CM (u) is 3-solvable. Therefore, M is strongly embedded in G, an impossibility by [II2 , Theorem SE], unless K∼ = D4 (2). In this final case, let S ∈ Syl2 (K) and choose U  S with U ∼ = E22 . By [VK , 10.47], every u ∈ U # is a root involution, and hence m3 (CK (u)) ≥ 3, indeed m3 (CM (u)) ≥ 5. Since CM (u) is strongly 3-embedded in CG (u), and CK (u) involves Σ3 ×Σ3 ×Σ3 , it follows easily (with [IA , 7.6.1]) that CG (u) ≤ M . Factoring O2 (M ) under U and using that G has even type, we conclude that O2 (M ) = 1. Similarly, factoring O2 (M ) under B and using (1A4), we get O2 (M ) = 1. Therefore, as CB (K) = 1, F ∗ (M ) = E(M ) is a direct product of components K × K1 × · · · × Kn , n ≥ 1. Each Ki is a component of CG (u) for any u ∈ U # , and also of CG (a) for any a ∈ B ∩ K. Moreover, there is B < A ∈ S3 (M ), and by [VK , 10.47], m2,3 (G) ≥ m3 (CM (u)) ≥ m3 (A) − 1 ≥ m3 (B). Replacing u by a K-conjugate if necessary, we have (CA (u), u, Ki ) ∈ BtK3 (G) with CA (u) ∩ K = 1. Hence each Ki is isomorphic to one of D4 (2), Co2 , U6 (2), or U5 (2). We argue next that each Ki is terminal in G. We already know that Ki is a component of CG (u) for u ∈ U # . Now Ki is semirigid in G by [VK , 14.2] and m2 (K) > 2. Hence by [IG , 7.4], either Ki is terminal in G or S has a maximal subgroup S0 such that Ki is not a component of CG (v) for any v ∈ I2 (S0 ). But the latter possibility does not occur since S0 ∩ U = 1. Thus Ki is terminal, as claimed. ∼ Likewise K is terminal in G. Otherwise  L we would have K < L with L = K ×K, D4 (4), or F i22 , by [VK , 3.10], and L = K ≤ M . Hence L∩M would be strongly 3-embedded in L and of 3-rank at least m3 (K) = 4, a contradiction in view of [IA , 7.6.1]. Now if some Ki ∼ = U5 (2), then we let Qi ∈ Syl2 (CG (Ki )) with Qi ∩K ∈ Syl2 (K). As U5 (2) is outer well-generated, there exists, by [II3 , Corollary PU2 ] and [IG , 18.8], an element g ∈ G − M such that Qgi ≤ M and CQgi (Ki ) = 1. This is impossible as |Qgi | ≥ |K|2 = 212 > 211 = | Aut(Ki )|2 . If some Ki ∼ = U6 (2), we take Q ∈ Syl2 (CG (K)) and set Si := Q ∩ Ki ∈ Syl2 (Ki ). Let m be the maximum value of m2 (Qg ∩ M ) as g varies over G − M . By [II3 , Corollary PU2 ], m ≥ 2. Then as J(Si ) ∼ = E29 , m ≥ m2 (Q ∩ M g ) ≥ 5 for some g ∈ G − M . Now, NAut(Ki ) (J(Si )) = Si H with H an extension of L3 (4) by a

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18. THEOREM 5: THE DEGENERATE CASE I = J

205

field automorphism, so m2 (H) = 4. As m ≥ 5, J(Si ) ∩ Qg = 1 for some g ∈ G − M . Thus m ≥ 9. But m2 (Aut(K)) < 9 [IA , 3.3.3, 2.5.12] so for some g ∈ G − M , Qg does not act faithfully on K, contradicting ΓQg ,1 (K) < K. Suppose that some Ki ∼ = Co2 . Choose u ∈ I2 (K) ∩ Qi . Then CM (u) ≥ Ki , so by [IA , 7.6.1, 7.6.2], CM (u) is not strongly 3-embedded in CG (u). Therefore, CG (u) ≤ M . On the other hand, Out(Ki ) = 1, and so Qgi ≤ M for some g ∈ G−M , by [IG , 18.8] and [II3 , Corollary PU2 ]. But CKi (ug ) has an element v of order 3, by inspection of [IA , 5.3k]. Then v ∈ M ∩ M g so g ∈ M by strong 3-embedding, a contradiction. Therefore all Ki ∼ = D4 (2). Let Q ∈ Syl2 (CG (K)) and S1 = Q ∩ K1 ∈ Syl2 (K1 ), and let m be the maximum value of m2 (Qg ∩ M ) as g ranges over G − M . Fix g such that that maximum is achieved. As G − M is closed under inversion, m is also the maximum value of m2 (Q∩M g ). By [II3 , Corollary PU2 ], m ≥ 2. Hence as all Ki  M by standardness, and | Out(K)|2 = 2, some involution x ∈ Qg ∩ M induces an inner automorphism on K1 . As ΓQg ,1 (G) ≤ M g , m ≥ m2 (CK1 (x)) ≥ 5 by [VK , 10.47]. Then some x ∈ Qg ∩ M lies in O2 (N ) where N is a Qg ∩ M -invariant parabolic subgroup of K1 of type L4 (2). As O2 (N ) ∼ = E26 , m ≥ 6. Considering the action of Qg ∩ M g on K we see that there is a Q ∩ M -invariant parabolic subgroup N1 of K of type L4 (2) such that some involution y ∈ Qg ∩ M acts on K like an element of O2 (N1 ). By [VK , 10.47], 3 divides |CM (y)|. But y ∈ Qg so CG (y) ≤ M g . Hence M ∩ M g contains an element of order 3, so M = M g and g ∈ M by strong 3-embedding. This contradiction completes the proof of Lemma 18.12.  Lemma 18.13. K ∼  Ur (2), r = 5 or 6, or Co2 . = Proof. Suppose to the contrary that K ∼ = Ur (2), r = 5 or 6, or Co2 . If possible choose such a counterexample so that t ∈ [Q, Q], where Q ∈ Syl2 (C(t, K)), and subject to this, if possible so that K is standard. Let z be a 2-central involution of K, and set Ct = CG (t) and Cz = CG (z). Also set P = CK (z). By [VK , 12.42], z may be chosen to centralize a hyperplane B0 of B. Note that in every case, |B ∩ K| = 34 . Let A ∈ S3 (G) with B < A, by (1A6). Then A normalizes both b and B ∩ K = B ∩ J, so A normalizes J. Set R = O2 (P ) ∼ = 21+2(r−2) or 21+8 , and A0 = CA (z). Then by [VK , 12.42], |A : A0 | = 3. As A > B, it follows that |A : B| = |A0 : B0 | = 3 and A0 ∈ B3∗,c (G). We first assume that E(Cz ) = 1, and L is a component of E(Cz ). Since R = F (CK (z)) = [R, B0 ] and CK (z)/R ∼ = GU3 (2) or U4 (2) or Sp6 (2) ∼ , there is b ∈ B0 ∩ K = B0 ∩ P such that according as r = 5 or 6 or K Co = 2 0  . As B ≤ A , B normalizes all components of Cz of order divisible by 3, P = bP 0 0 0 0  normalizes L. We also set P and in particular P = bP 1 = O2,3 0 ) or P according  (P P ∼ as K = U5 (2) or not, and choose b1 ∈ B0 ∩ P1 such that P1 = b1 . We argue that (18CC)

∗

(18DD)

[L, b1 ] = 1.

This will have the immediate important consequence (18EE)

[L, P1 ] = 1.

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∼ We use that fact that P1  P  CG (t, z). Suppose, for example, that K = U5 (2) and [L, b1 ] = 1. Since L ∈ C2 , Out(L) is 2-nilpotent by [VK , 2.3]. But P/P1 ∼ = SL2 (3), and it follows that O 3 (P ) induces inner automorphisms on L. Note that we may assume that t normalizes L, since otherwise E(CLLt (t)) is a component of CG (t, z), while P ≤ Sol(CG (t, z)), so by [IG , 6.19], O 2 (P ) centralizes L, as desired. It follows that CL (t) contains a 2-element u whose projection on L/Z(L) is also the projection on L/Z(L) of some element v ∈ O 3 (P ). We may choose v to invert the projection b1 (of order 3) of b1 on L, and b1 is also centralized by    t. Then v ∈ P  CG (t, z), b1 ∈ P , whence b1 ∈ P1 . Then b1 = b1 ∈ L, so  P as P1 = b1 ≤ L. Thus L is a proper covering group of L/ z, and CL/z (t) has a normal subgroup isomorphic to P1 / z = EH with E ∼ = E26 and H ∼ = 31+2 acting irreducibly on E. A similar but easier analysis in the other two cases shows that if [L, b1 ] = 1, then CL/z (t) likewise has a normal subgroup E1 H1 with E1 ∼ = E28 and H1 ∼ = U4 (2) or Sp6 (2) acting naturally or via a spin module on E1 . The preimages in L of E1 and E in these cases are extraspecial groups. Also, t centralizes B0 ∩ L, with |A0 : B0 | = 3 and A0 ∈ E3∗ (Cz ). There are no such C2 -groups L, by [VK , 10.70]. This contradiction establishes (18DD) and (18EE). Now (18EE) implies that CB0 (L) = 1, so (A0 , z, L) ∈ BtK3 (G) with CA0 (L) = 1. Hence L ∼ = U5 (2) or U6 (2) or Co2 . As O2 (P1 ) is extraspecial, z ∈ [O2 (P1 ), O2 (P1 )]. Thus by our choice, t ∈ [Q, Q]. Now L is semirigid in G by [VK , 14.2, 3.10], and m2 (O2 (P1 )) > 2, so it follows by a standard argument that L is terminal and standard in G. Indeed otherwise by [IG , 7.4], letting O2 (P1 ) ≤ QL ∈ Syl2 (C(z, L)), there would exist a maximal subgroup QL,0 of QL such that L is not a component of CG (x) for any x ∈ I2 (QL,0 ); but z ∈ [QL , QL ] ≤ QL,0 , contradiction. By our choice, then, K is terminal and standard in G. Also, as with K and J, for every w ∈ CA0 (L)# , the subnormal closure of L in CG (w) is a 3-component Lw such that L covers Lw /O3 (Lw ). Our next goal is to show that (18FF)

[K, L t] = 1.

Let W = {w ∈ I3 (B0 ∩ K) | E(CK (w)) ∼ = U4 (2)}. By [VK , 8.19], K = E(CK (w)) | w ∈ W, and by (18EE), CK (L) = 1, so it suffices to fix w ∈ W, set N = E(CK (w)), and show that N normalizes L. We argue that Lw = L, i.e., Lw is quasisimple. Choose a four-group V ≤ NKt (w, z) centralizing L, and set W = O3 (Lw ). Then W has odd order so it suffices to show that for each v ∈ V # , [CW (v), L] = 1. But [v, z] = 1 so the subnormal closure of L in CG (v) is L, by terminality. Therefore Lw = L, as claimed. In particular, L is a component of CG (w). Next, B0 ∩ K ≤ A 0 so B0 ∩ K normalizes all components of CG (w), whence so  does (B0 ∩ K)CK (w) , which contains N . Thus (18FF) holds. Let M be a maximal subgroup of G containing NG (K). Then by L2 -balance, K ≤ L2 (M ). Factoring O2 (M ) under a four-group of L t, and using terminality of K, we see that [K, O2 (M )] = 1 so K ≤ E(M ). Again using terminality, this time together with [VK , 3.5], we have that K  E(M ). By standardness, K  M . So M = NG (K). A similar analysis shows that L, which is also terminal, is normal in CG (K); hence M = NG (L). We interchange (t, K) and (z, L) if necessary to ensure that |K| ≤ |L|. Let S ∈ Syl2 (CG (K)) and SL = S ∩ L ∈ Syl2 (L).

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18. THEOREM 5: THE DEGENERATE CASE I = J

207

If K ∼ = U5 (2), then L ∼ = U5 (2) or |L|2 > | Aut(K)|2 . Also, K is outer wellgenerated by [VK , 8.4]. Hence by [II3 , Corollary PU2 ] and [IG , 18.8, 18.11], there is g ∈ G − M such that S g ≤ M and ΓS g ,1 (K) < K. In particular |S| ≤ | Aut(K)|2 , so L ∼ = U5 (2). Again by outer well-generation, Ω1 (S g ) induces inner automorphisms on K, and so has the same image in Aut(K) as Ω1 (T ), where T ∈ Syl2 (K). Thus ΓT,1 (K) < K, so K has a strongly embedded subgroup, a contradiction. Therefore both K and L are in {U6 (2), Co2 }. Choose g ∈ G − M such that m := m2 (S g ∩ M ) is maximal. By [II3 , Corollary PU2 ], m ≥ 2. Then ΓS g ,1 (L) ≤ −1

L ∩ M g . Since L ∩ M g is conjugate to Lg ∩ M , and as SL ≤ S, m ≥ m2 (CL (x)) for all x ∈ I2 (S g ). Suppose for the moment that L ∼ = K ∼ = U6 (2). We have J(SL ) ∼ = E29 by [VK , 2.9], so m ≥ 5. Now S g ∩ M normalizes some L-conjugate of the parabolic subgroup NL (J(SL )), which is an extension of E29 by an extension of L3 (4) by a field automorphism. Hence some involution in S g ∩ M lies in an L-conjugate of J(SL ). Therefore m ≥ 9 = m2 (Aut(K)). Hence AutS g ∩M (K) contains the image of J(T ) for some T ∈ Syl2 (K), whence ΓS g ∩M,1 (K) ≥ ΓJ(T ),1 (K) = K, a contradiction. Finally suppose that L ∼ = Co2 . The argument is similar. This time J(SL ) ∼ = E210 and NL (J(SL )) is an extension of J(SL ) by Aut(M22 ), the complement having 2-rank 5 by [IA , 5.6.1]. Since Out(L) = 1, m ≥ 6, and as in the previous case this leads to ΓS g ∩M,1 (K) ≥ ΓJ(T ),1 (K) = K by [VK , 8.10], a contradiction. We have proved that (18CC) is false, i.e., (18GG)

E(Cz ) = 1.

In particular, (18HH)

z ∈ tG .

Moreover, by L2 -balance and the fact that G is of even type, (18II)

E(CG (z, t)) = 1, whence E(Ct ) = K.

This in turn implies that F ∗ (C(t, K)) = O2 (Ct ). Since O2 (Ct ) is B-invariant, it is of symplectic type (note B ∩ C(t, K) = 1), and so (18JJ)

t = Ω1 (Z(Q)).

Let X = O3 (C(b, J)). Then A0 normalizes X, so A0 normalizes R × R1 for some Sylow 2-subgroup R1 of X. As [z, A0 ] = 1, R × R1 is of symplectic type by Proposition 10.1, whence R1 = 1 and X has odd order. In particular, t ∈ C(b, J) − X. Let C0 = CG (B0 ) and note that B0 is a hyperplane of B and of A0 . Hence by Proposition 6.1, C0 is 3-constrained. We claim that (18KK)

t ∈ O3 (C0 ).

Suppose that t ∈ O3 (C0 ). Let T ∈ Syl2 (O3 (C0 )) with t ∈ T . We may choose T to be invariant both under B and under Ax0 for some x ∈ C0 such that Ax0 , B is a 3-group. Now by Proposition 10.1, t is the unique involution in CG (B), up to CG (B)-conjugacy. Likewise z x is the unique involution in CG (Ax0 ), up to CG (Ax0 )conjugacy. By Proposition 10.2, T is cyclic or of symplectic type with Ω1 (Z(T )) = t invariant under Ax0 . Hence z x is conjugate to t in CG (Ax0 ), contradicting (18HH).

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This proves (18KK). As C0 is 3-constrained, t must invert some y ∈ O3 3 (C0 ) of order 3. Indeed by the A × B lemma, we may fix such an element y that centralizes B. Notice that b ∈ B0 , so C0 normalizes J. Of course t normalizes J and centralizes J, so y = [y , t] centralizes J. Thus CJ (y) is t-invariant and covers J, whence K ≤ CJ (y), i.e., (18LL)

y ∈ CG (KB).

Let H be the subnormal closure of L3 (CJ (y)) in CG (y). Then H is t-invariant and K is a component of CH (t), with K ≤ CJ (y). It follows easily that H is a single 3-component. It is also a 2-component since O3 (CG (y)) has odd order, owing to the fact that B y is elementary abelian and y ∈ B. Thus if H/O3 (H) ∼ = K, then K ↑2 H/O3 3 (H) and K ↑3 H/O3 (H), via t and b, respectively. But there is no such H/O3 (H) ∈ C3 , by [VK , 13.2]. Hence, K ∼ = H/O3 (H). Set B1 := B0 × y ≤ Cz , with B1 ∼ = B, so B1 ∈ B∗,c (G). Let S = RO2 (Cz ). With Proposition 10.2, S and CS (R) are B1 t-invariant, and cyclic or of symplectic type. It follows, as NK (R) is irreducible on R/ z, that S = RCS (R). Since t inverts y ∈ B1 , t ∈ CS (R). But since CAut(K) (R) is just the image of z in Aut(K) [IA , 3.1.4, 5.3], CCS (R) (t) = z × Q1 for some 2-group Q1 ≤ C(t, K) not containing t. As Q centralizes z, t it normalizes Q1 . Hence if Q1 = 1, then t ∈ Q1 , by (18JJ), contradiction. Therefore Q1 = 1 and CCS (R) (t) = z. So CS (R) is cyclic or of maximal class. On the other hand m3 (CB1 (R)) ≥ m3 (b, y) = 2, and so CB1 (S) = 1. Therefore CB1 (O2 (Cz )) = 1, so E(Cz ) = 1. This contradicts (18GG) and completes the proof of Lemma 18.13.  In view of (18I), Lemmas 18.12 and 18.13 complete the proof of Proposition 18.1, and with it, the proofs of Theorem 5 and Theorem C5 : Stage 2.

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10.1090/surv/040.9/05

CHAPTER 5

Theorem C5 : Stage 3 1. Introduction In the previous stage, we have analyzed nonconstrained neighborhoods and the trivial symplectic pairs that accompany them. In this chapter, we will tighten the conclusion of Stage 2 under the assumption that all symplectic pairs are trivial. This will narrow that conclusion to configurations actually existing in certain sporadic groups. The other purpose of this chapter is to carry out a corresponding analysis of faithful symplectic pairs and their accompanying 2- and p-local configurations, which we call “constrained neighborhoods.” Again, throughout this chapter, p is the odd prime fixed in [V4 , (1A)]. We shall soon see that constrained neighborhoods do not exist if p > 3, but the existence of faithful symplectic pairs implies the existence of constrained neighborhoods. Since all symplectic pairs are faithful if p > 3, by Theorem C5 : Stage 2, one corollary of our results is thus that p = 3. Definition 1.1. Let (B, T ) be a faithful symplectic pair in G. Let Ω1 (Z(T )) = z and b ∈ B # . We say that (CG (z), b, J) is a constrained {z, b}-neighborhood with respect to (B, T ), or, for brevity, just a constrained {z, b}-neighborhood, if and only if the following conditions hold for Cb = CG (b): (a) J is a p-component of E(Cb ), Op (Cb ) = 1, and mp (C(b, J)) = 1; (b) T = F ∗ (CG (z)), and R := CT (b) = O2 (CJ (z)) is extraspecial; and b := Cb /C(b, J), z is 2-central and R  = F ∗ (C  ( (c) In C Cb z )). In addition, we have the following notion of nondegeneracy for constrained neighborhoods. Definition 1.2. Let (CG (z), b, J) be a constrained {z, b}-neighborhood with respect to the symplectic pair (B, T ), and let L ∈ K. Then (CG (z), b, J) is nondegenerate with core L if and only if there exists b ∈ B ∩ J − b such that if we set D = b, b , the following conditions hold: (a) JD := E(CJ (D)) ∼ = L; (b) There exists d ∈ D − b such that JD  CG (d); and (c) For every d ∈ D# such that JD  CG (d), the subnormal closure Jd of JD in CG (d) is quasisimple and equals E(CG (d)); moreover, mp (C(d, Jd )) = 1. We distinguish two types of constrained neighborhoods; one will lead to G ∈ Spor and the other to G ∈ Chev(2). Definition 1.3. A constrained neighborhood (CG (z), b, J) is of sporadic type if and only if J ∈ Spor. Definition 1.4. Let (CG (z), b, J) be a constrained {z, b}-neighborhood with respect to (B, T ). Then (CG (z), b, J) is of U4 (2)-type if and only if p = 3, 209 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Table 1.1. Nonconstrained {t, b}-Neighborhoods (K; I; J)

G∗ m2,3 (G) K I J F i22 4 2U6 (2) U4 (2) U4 (3) F i23 5 2F i22 2U4 (3) Ω7 (3) F i24 5 2F i22 2U4 (3) P Ω+ 8 (3) F i24 6 F i23 Ω7 (3) P Ω+ 8 (3) 2 F2 5 2 E6 (2) 2U6 (2) F i22 Table 1.2. Sporadic-Type Constrained Neighborhoods (CG (z), b, J)

G∗ m2,3 (G) T = F ∗ (CG (z)) CG (z)/T J 1+8 + Co1 4 2+ Ω8 (2) 3Suz F1 6 21+24 Co 3F i24 1 + (CG (z), b, J) is nondegenerate with core U4 (2) or U5 (2), and for every d ∈ D# such that JD  CG (d), the normal closure Jd of JD in CG (d) is isomorphic to D4 (2), U5 (2), U6 (2), or SU6 (2). For convenience, we restate Theorem C5 : Stage 3 and reproduce Tables 2.2 and 2.3 of Chapter 1 as Tables 1.1 and 1.2. Each row in each table is labelled by a group G∗ which will be the eventual target group in Stage 4. Theorem 1.5 (Theorem C5 : Stage 3). We have p = 3. Moreover, one of the following holds: (a) BtK3exc (G) = ∅; (b) All symplectic pairs in G are trivial. There exists a nonconstrained {t, b}neighborhood (K; I; J) for some commuting b ∈ I3 (G) and t ∈ I2 (G), with K, I, and J as in Table 1.1 and C(b, J) = b. Moreover, in the case labelled G∗ = F i24 in the table, CG (t)/ t ∼ = Aut(F i22 ) and |CG (b) : J| is odd; (c) There exists a constrained {z, b}-neighborhood (CG (z), b, J) of sporadic type for some faithful symplectic pair (B, T ), and T = O2 (CG (z)), CG (z)/T , and J are as in Table 1.2. Moreover, in the case labelled G∗ = F1 in the table, there is an involution t ∈ CG (z, b) such that E(CG (t)) ∼ = 2F2 ; (d) There exists a constrained {z, b}-neighborhood (CG (z), b, J) of U4 (2)-type, with core Um (2), where m = m2,3 (G) ∈ {4, 5}. Moreover, if J ∼ = D4 (2) or U5 (2), then m2,3 (G) = 4. 2. The Principal Subsidiary Theorems We divide the proof of Theorem C5 : Stage 3 in two, as follows. Theorem 1. Suppose that if p = 3, then BtK3exc (G) = ∅. Assume also that every symplectic pair in G is trivial. Then p = 3 and there is a nonconstrained {t, b}-neighborhood (K; I; J) for some commuting b ∈ I3 (G) and t ∈ I2 (G), with K, I, and J as in one of the rows of Table 1.1, but not the row labelled G∗ = F i24 . In particular, J is quasisimple. Moreover, b ∈ Syl3 (C(b, J)).

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Theorem 2. Assume that some symplectic pair in G is faithful. Then p = 3. Moreover, if BtK3exc (G) = ∅, then one of the following holds: (a) There is a nonconstrained {t, b}-neighborhood (K; I; J) for some commuting t ∈ I2 (G) and b ∈ I3 (G) satisfying the row of Table 1.1 labelled G∗ = F i24 . Moreover, CG (t)/ t ∼ = Aut(F i22 ), C(b, J) = b, and |CG (b) : J| is odd; or (b) There is a constrained {z, b}-neighborhood (CG (z), b, J) with respect to some faithful symplectic pair (B, T ), satisfying conclusion (c) or (d) of Theorem C5 : Stage 3. Thus, we assume throughout this chapter that If p = 3, then BtK3exc (G) = ∅. 3. Theorem 1 In this section we analyze the consequences of Theorem C5 : Stage 2 under the additional assumption that (3A)

Every symplectic pair is trivial.

By Theorem C5 : Stage 2 and (3A), there exists a symplectic pair (B, t) such that t2 = 1 = t. Set Ct = CG (t). As t ∈ I∗G (B; 2) by definition of symplectic pair, t ∈ Syl2 (Op (Ct )). Since G is of even type, F ∗ (Ct ) = E(Ct ) t and every component K of E(Ct ) has order divisible by p. Thus (B, t, K) ∈ BtKp (G). By Theorem C5 : Stage 2 and (3A), for any (B, t, K) ∈ BtKp (G), the following hold:

(3B)

(1) (2) (3) (4)

p = 3; B ≤ K; t ∈ Syl2 (CG (K)); and There exist B ∗ ∈ B3∗,o (G) centralizing t, and an element b ∈ (B ∗ )# , such that the structure of a {t, b}neighborhood (K; I; J) is one of the possibilities in [V4 , Table 1.1], but not (2 2E6 (2); D4 (2); F i22 ).

We establish Theorem 1 by proving: Proposition 3.1. Assume (3A). Then p = 3 and there exists (B, t, K) ∈ BtKp (G), b ∈ B # , and a nonconstrained {t, b}-neighborhood (K; I; J) satisfying one of the rows of Table 1.1 (but not the row labelled G∗ = F i24 ), and so that b ∈ Syl3 (C(b, J)). Remark 3.2. The group G = F i24 possesses a {t, b}-neighborhood (2F i22 ; 2U4 (3); P Ω+ 8 (3)), but does not satisfy (3A). Remark 3.3. Our proof of Theorem 1 will show that any (B, t, K) ∈ BtK3 (G) will do (for suitable b) unless K ∼ = D4 (2), in which case we shift to another (B1 , t1 , K1 ) ∈ BtK3 (G) for which K1 ∼ = 2U6 (2). We take an arbitrary (B, t, K) ∈ BtK3 (G) and an element b ∈ I3o (G) satisfying (3B). We set Cb = CG (b), C b = Cb /O3 (Cb ), and Ct = CG (t). Lemma 3.4. No elementary abelian 3-subgroup of G of rank m2,3 (G) normalizes an extraspecial 2-subgroup of G.

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5. THEOREM C5 : STAGE 3

212

Proof. Otherwise, expanding the 2-group to a maximal 2-signalizer, we would  obtain a nontrivial symplectic pair by [V4 , 10.2], contradicting (3A). Lemma 3.5. Assume that b contains a Sylow 3-subgroup of O3 3 (J). Then b ∈ Syl3 (C(b, J)) under either of the following conditions: (a) m2,3 (J b) = m2,3 (G); or (b) Some elementary abelian 3-subgroup of J b of rank m2,3 (G) − 1 normalizes an extraspecial 2-subgroup of J. Proof. Suppose that m3 (C(b, J)) > 1 and let E be an E32 -subgroup of C(b, J). Let R be a nontrivial 2-subgroup of J and D an elementary abelian 3subgroup of J b normalizing R with m3 (D) = m2,3 (G) (if (a) holds) or m3 (D) = m2,3 (G) − 1 and R extraspecial (if (b) holds). By Sylow’s theorem in C(b, J), we may assume that [D, E] = 1. Since b contains a Sylow 3-subgroup of O3 3 (J), m3 (ED) > m3 (D). But DE normalizes R so m2,3 (CG (b)) > m3 (D). Thus if (a) holds, we have an immediate contradiction. If (b) holds, then DE ∈ B3∗ (G), and Lemma 3.4 is contradicted as R is extraspecial. Therefore m3 (C(b, J)) = 1. By Burnside’s theorem, C(b, J) has a normal 3-complement, which has odd order by [V4 , (1A4)]. Thus, C(b, J) has odd order, so there is a t-invariant P ∈ Syl3 (C(b, J)). By the previous paragraph, P is cyclic; as [t, b] = 1, P ≤ Ct . On the other hand, by [VK , 3.51] and (3B4), b ∈ Syl3 (CCt (I b)), and so P = b, completing the proof of the lemma.  With this and Theorem C5 : Stage 2, we conclude: Lemma 3.6. If Proposition 3.1 fails, then (K; I; J) is one of the following: (a) (b) (c) (d)

(F i23 ; Ω7 (3); P Ω− 8 (3)), with m2,3 (G) = 6; (F i22 ; U4 (3); L4 (9)), with m2,3 (G) = 5; (Sp8 (2) or (2)F4 (2); Sp6 (2); U6 (2) or D4 (2)), with m2,3 (G) = 4; or ((2)U6 (2) or U5 (2) or Sp8 (2) or (2)D4 (2); U4 (2); L± 4 (3) or 3U4 (3)), but not (2U6 (2); U4 (2); U4 (3)).

Proof. If (K; I; J) is none of the claimed examples, nor one of the possibilities given in Proposition 3.1, then by Theorem C5 : Stage 2, J is P Ω± 8 (3) with m2,3 (G) = 5, or (3)F i24 with m2,3 (G) = 6, or [3 × 3]U4 (3) with m2,3 (G) = 4. But in these three cases, by [VK , 12.40], J b contains a 3-group of rank m2,3 (G) normalizing an extraspecial 2-subgroup of G, contradicting Lemma 3.4. Therefore, (K; I; J) is as claimed in Proposition 3.1, and we argue that W := O3 (J) = 1. Namely, CW (t) ≤ CO3 (Cb ) (t) ≤ O3 (CCb (t)) = O3 (CCt (b)) = t. The last equation holds since C(t, K) = t (as O2 (Ct ) = 1) and K is locally balanced with respect to b [IA , 7.7.1b, 7.7.8]. Then as O3 (Cb ) has odd order, CW (t) = 1 and t inverts W elementwise. But J = [J, t], so W ≤ Z(J). As W is a {2, 3} -group, it follows by [IA , 6.1.4] that W = 1. Hence (K; I; J) is as claimed in Proposition 3.1, contrary to assumption. Therefore (K; I; J) is as asserted in the lemma. By [VK , 12.41], in these remaining cases, J b contains a 3-group of rank m2,3 (G) − 1 normalizing an extraspecial 2-subgroup of G, so b ∈ Syl3 (C(b, J)) by Lemma 3.5. The proof is complete.  Lemma 3.7. Lemma 3.6a does not occur.

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3. THEOREM 1

213

∼ P Ω− (3) and I ∼ Proof. Suppose false. Then J = = Ω7 (3), so by [VK , 3.60], 8 there is u ∈ I2 (J) such that H := L3 (CJ (u)) satisfies H ∼ = L4 (3) and [t, u H] = 1. and H = [H, H] ≤ C (u) = C (u), the only possibility [IA , 5.3u], As K ∼ F i = 23 K [Ct ,Ct ] by Lagrange’s theorem, is CK (u) ∼ 2F i . = 22 Let L = CK (u) = E(CK (u)) and let N be the subnormal closure of L in CG (u). Then N has the usual form by L2 -balance, and H ≤ N . Now H = L3 (CJ (u)) is a 3-component of CG (u, b), hence one of E(CN (b)), by the B3 -property. If N is a vertical pumpup of L, then L4 (3) ↑3 N/O3 (N ) and 2F i22 ↑2 N , which is impossible by [VK , 3.10] and [IA , 5.3]. So N is the product of two 2F i22 -components and hence L4 (3) ↑3 F i22 , a contradiction by [IA , 5.3t]. This completes the proof.  Lemma 3.8. Lemma 3.6b does not occur. Proof. The proof is similar to the preceding. This time we take u ∈ I2 (CJ (t)) such that if we set H := L3 (CJ (u)), then H ∼ = L3 (9), and H0 := E(CH (t)) ∼ = U3 (3). We can do this by [VK , 3.59]. Then H0 ≤ CK (u). Set L = E(CK (u)). As K ∼ = F i22 and H0 ∼ (2), or Sp = U3 (3), we conclude [IA , 5.3t] that H0 ≤ L ∼ = 2U6 (2), D 6 (2). 4 Again let N be the subnormal closure of L in CG (u). Then H = H0H ≤ N . This time we find that a component N1 of N is a 3-pumpup of, or covering group of, L3 (9). Hence by [VK , 3.11], N1 ∈ Chev(3). But as N1 is a 2-pumpup of, or covering group of, 2U6 (2), D4 (2), or Sp6 (2), this is impossible, by [IA , 4.9.6, 2.2.10]. The lemma is proved.  Lemma 3.9. Lemma 3.6c does not occur. Nor does Lemma 3.6d with K ∼ = Sp8 (2). ∼ U6 (2) or D4 (2), there Proof. Suppose that Lemma 3.6c occurs. Whether J =  exists a parabolic subgroup P of J such that O2 (P ) is extraspecial and m3 (P b ) = 4 = m2,3 (G). Hence Lemma 3.4 is violated, and the first statement follows. Suppose then that Lemma 3.6d occurs with K ∼ = Sp8 (2). Then m2,3 (G) = 4, and by [IA , 4.8.2], there is b1 ∈ B # such that I1 := E(CK (b1 )) ∼ = Sp6 (2) is terminal in K. As m3 (B) = 4 and m2 (I1 ) = 6 by [IA , 3.3.3], I1 is B-wide, so (K, b1 , I1 ) is regular by [V4 , Proposition 12.5]. Also (K, b1 , I1 ) is broad as b1 ∈ B ≤ K. Hence there is a component J1 of CG (b1 ) such that I1 is a component of CJ1 (t). b = Cb /O3 (Cb ), and C = C(b1 , J1 ). Since m2,3 (G) = 4, Set Cb1 = CG (b1 ), C 1 1 1 it follows easily that J1  Cb1 . Hence C  Cb1 , whence O3 (C) ≤ O3 (Cb1 ), with both of odd order by [V4 , (1A4)]. Suppose that I1 = J1 . Then t ∈ C − O3 (C). But CC (t)/ t embeds in CAut(K) (I1 b1 ), which has odd order, so t ∈ Syl2 (C). Therefore t inverts some x ∈ I3 (C). Now J1 b1 , x contains an E34 subgroup y )) for some 2-central involution y ∈ J1 , contradicting Lemma normalizing O2 (CJ1 ( 3.4 or [V4 , 10.2]. Thus, I1 < J1 , and so Sp6 (2) ↑2 J1 . By [VK , 13.3], J1 /Z(J1 ) ∼ = U6 (2) or D4 (2), leading to a contradiction as in the first paragraph. The proof is complete.  Lemma 3.10. If Lemma 3.6d occurs, then the following conditions hold: (a) J/Z(J) ∼ = U4 (3); and (b) C(b, J) = b × W where W is a {2, 3} -group.  Proof. We have m2,3 (G) = 4 and J ∼ = L± 4 (3) or 3U4 (3). Then J b contains Z3 × (SL2 (3) ∗ SL2 (3)), so b ∈ Syl3 (C(b, J)) by Lemma 3.5. As O3 (CG (b)) has odd order by [V4 , (1A4)], (b) follows. Suppose that (a) fails, so that J ∼ = L4 (3).

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5. THEOREM C5 : STAGE 3

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∼ Aut(U4 (2)), so there is an involution u ∈ CJ (t) − I such By [VK , 10.6], CJ (t) = that CI (u) ∼ = Σ6 [IA , 4.9.2]. Set H = E(CI (u)) ∼ = A6 . Then by [IA , 4.5.1] and [VK , 3.60], H = E(CJ (u)) and Y := CJ (H) ∼ = Z4 . As CJ (t) = I u, t inverts Y . Now u centralizes t and b, and induces a non-inner automorphism on I = E(CK (b)), with CI (u) ∼ = Σ6 . Given the possibilities for K we conclude from [VK , 12.27] that K ∼ = (2)U6 (2) or (2)D4 (2), u induces a non-inner automorphism on K, and N := E(CK (u)) = CK (u) ∼ = Sp6 (2). We claim that (3C)

N   CG (u).

Let M be the subnormal closure of N in CG (u), so that M has the usual structure by L2 -balance. As m2,3 (G) = 4 and m3 (N ) = 3, M must be a single component, and N is a component of CM (t). Now H ≤ I ≤ K, so H ≤ CK (u) = N ≤ M . As H   CC b (u), CM (b) has a 3-component H0 with H0 /O3 (H0 ) ∼ = A6 . The only possibility, by [VK , 12.26] with M playing the role of K there, is that M = N , proving (3C). We claim next that CG (N )/O{2,3} (CG (N )) ∼ = D8 . Namely, u maps onto a Sylow 2-subgroup of Out(K), which is isomorphic to Z2 or Σ3 . Hence as C(t, K) = t, O 2 (CG (t, u)) = CK (u) = N , and in particular b ∈ N . Therefore CG (N ) ≤ CG (H b). But C(b, J)/O{2,3} (CG (b)) is covered by b, and CAut(J) (H) ∼ = D8 , by [VK , 3.60]. So CG (Hb)/O{2,3} (CG (Hb)) is the direct product of the image of b with a subgroup of D8 . As b ∈ N and t inverts Y , our claim follows. so CG (u) = N × CG (N u). Moreover, in Finally,  Out(N ) = 1 [IA , 2.5.12], C b, Y t ∼ = D8 and u ∈ Z(Y t ), so the projection of Y t on CG (N u) contains D8 . Let S ∈ Syl2 (CG (u)). Thus S = S1 × S2 , where S1 ∈ Syl2 (N ), |S1 | = 29 , S2 ∼ = D8 , and u = Z(S2 ). By [VK , 10.35], S1 is indecomposable, so u char S by the Krull-Schmidt theorem. Hence, S ∈ Syl2 (G), so |G|2 ≤ 212 . On the other hand since u ∈ KCG (K), |CG (t)|2 > |K t |2 ≥ 213 . This contradiction completes the proof.  Lemma 3.11. If Lemma 3.6d holds, then K/Z(K) ∼ = U6 (2) or D4 (2). Proof. Suppose false, so that by Lemmas 3.6d and 3.9, K ∼ = U5 (2) and m2,3 (G) = 4. Let B < A ∈ S3 (G) [V4 , (1A6)]. Then A ≤ CG (b), and A normalizes J by [IG , 8.7(i)]. Now J contains SL2 (3) ∗ SL2 (3), so by Lemma 3.4, m3 (C(b, J)) = 1. In particular m3 (CA (J)) = 1, so A ∼ = E35 . By [VK , 9.16b] and [IA , 6.4.4], if we let A ≤ P ∈ Syl3 (CG (b)), then A = J(P ). The group AutJ (A) ∼ = A6 is irreducible on A/ b. Let N = NG (B) and C = CG (B), so that N = CNN (A) by a Frattini argument. By (c) of Theorem C5 : Stage 2, m2 (C) = 1, so t ∈ Z ∗ (C) by [IG , 15.3]. By [V4 , (1A4)], O3 (C) has odd order, so t inverts an element a ∈ O2 (C) of order 3. Without loss we may replace A by B a and assume that a ∈ A, so that t ∈ NN (A). Let X = AutG (A); then AutX (B) = AutG (B) contains AutK (B) ∼ = Σ5 . As b < B ≤ A, X is irreducible on A. Also the image  t of t in X is a reflection t) ∼ on A with CX ( = Z2 × Σ5 or E22 × Σ5 . But by [VK , 5.13], there are no such subgroups X of GL(A) ∼ = GL5 (3). This contradiction completes the proof of the lemma.  Lemma 3.12. If Lemma 3.6d holds and K/Z(K) ∼ = D4 (2), then Z(K) = 1.

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∼ 2D4 (2). Then by [VK , 9.13], b is 3-central in K, Proof. Suppose that K = and we argue that b is 3-central in G. Let P ∈ Syl3 (CG (b)) with P ∩ K ∈ Syl3 (K). Then by [VK , 9.7], Z(P )/ b = Z(P/ b) and A = Z(P ∩ I). If b is not 3-central in G, then b has 3 NG (P )-conjugates in E1 (Z(P )). By [VK , 3.35], E(CK (b1 )) ∼ = Sp4 (3) for at least one such conjugate b1 . Hence, CG (b) contains Sp4 (3). On the other hand, it follows from Lemma 3.5 that b ∈ Syl3 (C(b, J)), whence |C(b, J)| has odd order. Hence by [VK , 1.7], CG (b) does not contain Sp4 (3), a contradiction. Thus, P ∈ Syl3 (G). In particular |G|3 = 37 . Again by [VK , 3.35], there is b1 ∈ Z(P )# such that b = Cb /O3 (Cb ). Then Ib1 := E(CK (b1 )) ∼ = Sp4 (3). Let Cb1 = CG (b1 ) and C 1 1 1 3  |Cb1 : Ib1 |3 = 3 , and hence Ib1 centralizes O3 (Cb1 ). Therefore Ib1 ≤ L3 (Cb1 ), so (K, b1 , Ib1 ) is regularly embedded; it is also obviously broad in K. Thus, by [V4 , Prop. 12.5], I1 ≤ J1 for some 3-component J1 of L3 (Cb1 ). As t ∈ O2 (I1 ) − O3 (J1 ) by [V4 , (1A4)], I1 < J1 . But then (K; I1 ; J1 ) must be as in [VK , 13.3], a contradiction.  Lemma 3.13. Suppose that Lemma 3.6d holds with Z(K) = 1. Let z ∈ K be z = Cz / z, and a 2-central involution and set R = O2 (CK (z)), Cz = CG (z), C Rz = O2 (Cz ). Then the following are true: (a) t ∈ Syl2 (CG (K)); (b) t ∈ [Cz , Cz ]; and (c) One of the following holds: with t ∈ Rz ; (1) R ≤ Rz ∼ = 21+10 + (2) t, z  Cz ; or (3) There exists (B ∗ , t∗ , K ∗ ) ∈ BtK3 (G) with K ∗ ∼ = 2U6 (2). Proof. By Lemma 3.11, K ∼ = D4 (2) or U6 (2). Part (a) holds by (3B3). Let T ∈ Syl2 (Ct ). Since K ∼ = D4 (2) or U6 (2), Z(T ∩ K) = z, Z(T ) = z, t, 1+8 ∼  Note that | Out(K)|2 = 2, so t ∈ [T, T ], R = 2+ , and CK (z) acts irreducibly on R. G while z ∈ [T, T ]. Thus, z ∈ t . If z ∼G tz, then t is weakly closed in Z(T ) and so T ∈ Syl2 (G); but then by Burnside’s lemma, z ∼NG (T ) tz, which is impossible. We have shown that z is weakly closed in Z(T ) with respect to G. As t ∈ [G, G], (b) follows, by [III8 , 6.3b]. Note that z, t is the unique maximal normal abelian 2-subgroup of CG (z, t). Suppose that there is a noncyclic abelian subgroup U char Rz . Then CU (t) ≤ z, t. If |U | > 4, then CU (t) is a noncyclic normal abelian subgroup of CG (z, t), so it equals z, t, whence t ∈ U and |CU (t)| > 4, contradiction. So |U | = 4. By (b), [U, t] = 1, so U = z, t. Hence, (c2) holds in this case. Therefore we may assume that every characteristic abelian subgroup of Rz is cyclic, so Rz is of symplectic type or cyclic, by P. Hall’s theorem. Consequently either Rz is extraspecial or Rz = z, or Rz has a characteristic cyclic subgroup (Φ(Rz ) or Z(Rz )) on which t acts nontrivially. As t ∈ [Cz , Cz ], Rz is extraspecial or Rz = z .  Suppose that R ≤ Rz . By the irreducibility of CK (z) on R, R ∩ Rz = z . Hence CRz t (t) = t, z, so Rz t is of maximal class. Thus if Rz > z, some cyclic maximal subgroup of Rz is normal in Cz , or Rz ∼ = Q8 . In any case, t ∈ [Cz , Cz ],

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216

5. THEOREM C5 : STAGE 3

contradicting (b). Therefore, Rz = z = CCz (E(Cz )). Let T ≤ S ∈ Syl2 (Cz ) and S0 = S ∩ F ∗ (Cz ). Let L be any component of E(Cz ). Since L ∈ C2 , we cannot have S ∩ L of maximal class and z ∈ L, so CS0 (t) ≤ z, t. By the irreducibility  it follows that R ≤ S0 . Since Rz = z, L/L ∩ z is simple. of CK (z) on R,  implies that if E1 is the product of the Indeed the irreducibility of CK (z) on R simple components of E(Cz ), then the projection of R on E1 , being a normal 2-subgroup of CG (t, z), must be trivial. Hence R lies in the product Ez of all those components of E(Cz ) containing z. We may assume that L was chosen in  := L/ z. Notice that Ez so that R projects nontrivially on the simple group L E(CG (z, t)) = E(CCt (z)) = 1 and in particular t normalizes L. Then F ∗ (CL (t))  of index at most has an elementary abelian subgroup (the projection of R on L) 2, and CL (t) either is a {2, 3}-group or has a unique non-{2, 3}-group composition  ≤ 4 and factor, isomorphic to U4 (2), since K ∼ = D4 (2) or U6 (2). Moreover, m3 (L) 3  has a 2-fold covering group in C2 . Furthermore, L  ∼ L U (3) as BtK = 4 exc (G) = ∅; ∗ 3 ∗ also L ∼ 2D (2) or 2F (2), since then (B , z, L) ∈ BtK (G) for any B ∈ E3∗ (L), = 4 4 contradicting Lemma 3.9 or 3.12. Likewise, we may assume that L ∼ = 2U6 (2) since otherwise, (c3) holds with t∗ = z, K ∗ = L, and B ∗ ≤ L. Thus by [VK , 10.68], L ∼ = 2G2 (4), then Z(S) is noncyclic by = 2G2 (4) or 2Sp6 (2). Note that if L ∼ [IA , 6.4.1a, 6.4.2a], so t ∈ Z(S); but then CL (t) has an A5 -composition factor, a contradiction. Finally if L ∼ = 2Sp6 (2), then L  Cz as m3 (Cz ) ≤ 4. This implies   a contradiction as m2 (Sp6 (2)) = 6 [IA , 3.3.3]. that R ≤ L, so 8 = m2 (R) ≤ m2 (L), We have proved that R ≤ Rz . Therefore Rz is extraspecial. Recall that R ≤ T ≤ S ∈ Syl2 (Cz ). If [t, L] = 1 for some component L of Cz , then we have CLt ∩S (t) = t, z, so z ∈ L = Lt and S ∩ L has maximal class, a contradiction as L ∈ C2 . Therefore [t, E(Cz )] = 1 so E(Cz ) = 1. Then t normalizes both R and C = CRz (R) with CC (t) = z or t, z. z as a transvection, In any case, C t is of maximal class. If t ∈ Rz , then t acts on R ∼   so the image of t in Out(Rz ) = O(Rz ) is outside Ω(Rz ), contradicting t ∈ [Cz , Cz ]. , and (c1) holds. The proof is complete.  Therefore t ∈ Rz ∼ = 21+10 +

Lemma 3.14. Suppose that Lemma 3.6d holds with K ∼ = D4 (2). Then there exists (B ∗ , t∗ , K ∗ ) ∈ BtK3 (G) such that K ∗ /Z(K ∗ ) ∼ = U6 (2). Proof. We may assume that (c1) or (c2) of Lemma 3.13 holds. Let R, Rz , Cz , z be as in that lemma. We have O2 (CG (z, t)) = R × t, and as K ∼ and C = D4 (2), H := O2,3 (CG (z, t)) = R t A, where A ∈ E33 (K) and A  AE ≤ CK (z) with ∼ AE = L1 ×L2 ×L3 , and all Li ∼ = E28 is the tensor product of = L2 (2). Moreover, R the natural modules for L1 , L2 , and L3 . In particular the groups Ai = Li ∩ A ∼ = Z3 ,  i = 1, 2, 3, are the only subgroups of A of order 3 which are fixed-point-free on R. Suppose that t, z  Cz , as in Lemma 3.13c2. Then |Cz : CG (t, z)| ≤ 2, and in particular H  Cz . Thus R = O2 (H) ∩ [H, H]  Cz . Also O3 (Cz /R) = AR/R so AR  Cz , and then by the previous paragraph, NCz (A) permutes the set {A1 , A2 , A3 }. We argue that (3D)

|Cz /R t |2 = 23 or 24 .

The lower bound comes from AE. Note that because of the irreducible action of  ≤ Z(R z ), and hence, as R is extraspecial, Rz = RC, where C := CR (R). AE, R z But then CC (t) = z, t, so C is of maximal class. If |Cz /R t |2 > 24 , then since

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3. THEOREM 1

217

 > R t, whence |C| ≥ 8. But then NCz (A) permutes {A1 , A2 , A3 }, Rz = CCz (R) z, t is not characteristic in Rz , contradiction. Thus (3D) holds. Still assuming t, z  Cz , we have tR ∈ Z(Cz /R). Since t ∈ [Cz , Cz ], it follows that t ∈ [S, S]R where t ∈ S ∈ Syl2 (Cz ). If t ∈ Z(S), then t ∈ [S, S]R by the structure of Ct , so t ∈ Z(S). It follows that CG (z, t) = HE, and Cz = CG (z, t) g where tg = tz and g 2 ∈ CG (z, t). By a Frattini argument we may choose g ∈ NS (A). We write F := z, t, s1 , s2 , s3  ∼ = E25 , where si  = Li ∩ S, i = 1, 2, 3. Thus by (3D), NS (A) = F g with g 2 ∈ F . Then [NS (A), NS (A)] ≤ CF (g). Since tg = tz, t, z ∩ [NS (A), NS (A)] = z. Now, t ∈ [Cz , Cz ] = H[NS (A), NS (A)], so t ∈ [Cz , Cz ] = H s, where [NS (A), NS (A)] = z, s and [A, s] = 1. But then t ∈ CHs (A) = A × z, a contradiction. Therefore Lemma 3.13c2 fails, so we may assume that Lemma 3.13c1 holds. Consequently CRz (b) ∼ = Q8 ∗ Q8 ∗ D8 , forcing C b to contain P O6− (3), by [VK , 10.7]. In particular, AutC b (I) = Aut(I). It follows by [VK , 12.25] that in Ct , NCt (b) contains an involution t∗ such that [t∗ , I] = 1 and t∗ inverts b. By [VK , 10.63], CK (t∗ ) ∼ = Sp6 (2).  Also t∗ maps into CAut(J) ( t I) ∼ = Z2 . So replacing t∗ by t∗ t if necessary, we ∗ have [t , J] = 1. Now the pumpup K ∗ of CK (t∗ ) in CG (t∗ ) contains I and covers J, and so K ∗ /Z(K ∗ ) ∼ = U6 (2), by [VK , 3.26]. Choosing any B ∗ ∈ E3∗ (K ∗ ), we have ∗ ∗ ∗ 3  (B , t , K ) ∈ BtK (G), completing the proof. Lemma 3.15. Suppose that Lemma 3.6d holds with K ∼ = U6 (2), but t, z  Cz and Lemma 3.13c1 holds. Then there exists no (B ∗ , t∗ , K ∗ ) ∈ BtK3 (G) such that K/Z(K) ∼ = D4 (2). , so Rz = R∗D with t ∈ D = CRz (R) ∼ Proof. By assumption Rz ∼ = 21+10 = D8 . + z = Cz / z, and set C z = Cz /Rz and H0 = CK (z)(∞) , so that We have C 0 ∼  and D.  Indeed R  is a natural U4 (2)-module for H = U4 (2), normalizing R     H0 , and as H0 is perfect, [H0 , D] = 1. Let H = AutCz (Rz / z), a subgroup of + AutAut(Rz ) (Rz / z) ∼ (2). As Rz = O2 (Cz ), O2 (H) = 1. Also, by Lemma = O10 3.4, m3 (H) ≤ 3, and equality holds as m3 (H0 ) = 3. Let F ∈ E33 (H0 ). We are going to quote [VK , 5.19], and so we must check that if b1 ∈ I3 (H0 ) with |[Rz / z , b1 ]| = 24 , then CH (b1 ) normalizes R. Since F contains representatives of all H0 -conjugacy classes of elements of order 3, we may assume that b1 ∈ F . Take preimages F0 ∼ = F of F and b0 ∈ I3 (Cz ) of b1 in CK (z); then b0 ∈ bK .  Hence [F0 , CRz (b0 )] = [F0 , CR (b0 )] = CR (b0 ) ∼ = Q8 ∗ Q8 . As F0 ≤ O 3 (CG (b0 )) = b0  J0 where J0 := L3 (CG (b0 )) ∼ = J, it follows that CR (b0 ) = R ∩ J0 . Hence R = [R, b0 ](R ∩ J0 ), so R is invariant under CH (b1 ), as required by the hypotheses of [VK , 5.19]. We conclude by that lemma that H normalizes R and D, i.e., D  Cz and R  Cz . By assumption t, z  Cz , so all four involutions in D are Cz -conjugate to t. As a result, |Cz : CG (t, z)| = 4 and so H0 contains a Sylow 3-subgroup (Z3  Z3 ) of Cz ; thus, all E33 -subgroups of Cz are conjugate (to F0 ). Furthermore, as O2 (H) = 1, F ∗ (H) = E(H) = H0 . Thus F is self-centralizing in H. It follows that D ∈ Syl2 (CCz (F0 )), and then as z char D, D ∈ Syl2 (CG (F0 )). Suppose finally that K ∗ ∼ = D4 (2) for some (B ∗ , t∗ , K ∗ ) ∈]BtK3 (G). Let z ∗ ∈ ∗ ∗ I2 (Z(T )) for some T ∈ Syl2 (K ∗ ). Then z ∗ is 2-central in CG (t∗ ). As Cz ∼ = CG (t∗ ), t∗ ∈ z G . It follows easily that t shears to z ∗ in NG (T ∗ ), and so z ∗ ∈ z G . Replacing (B ∗ , t∗ , K ∗ ) by a conjugate triple we may assume that z ∗ = z. Let

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5. THEOREM C5 : STAGE 3

∼ 21+8 . Then CK ∗ (z) contains F ∗ ∼ R∗ = O2 (CK ∗ (z)) = = F . As [F ∗ , t∗ ] = 1, the + previous two paragraphs imply that t∗ ∈ tG . But this is impossible as E(Ct ) ∼ =  U6 (2) ∼ = D4 (2) ∼ = E(CG (t∗ )). The proof is complete. Lemma 3.16. Suppose that Lemma 3.6d holds with K/Z(K) ∼ = U6 (2). Then there exists (B ∗ , t∗ , K ∗ ) ∈ BtK3 (G) such that K ∗ ∼ = 2U6 (2). Proof. Suppose false, so that K ∼ = U6 (2). Without loss we assume that [z, I] = 1 and z inverts b. We apply Lemma 3.13. If part (c3) holds, we are done. Suppose that t, z  Cz . Then |Cz : CG (t, z)| ≤ 2. If t ∈ Z(Cz ), then t ∈ [Cz , Cz ], contradiction. So suppose |Cz : CG (t, z)| = 2. Then C0 := CG (t, z)(∞) = (∞) Cz  Cz . But |CG (t, z)/C0 | = 2 or 4. If it is 2, then CG (t, z) = t × C0 and t ∈ [Cz , Cz ], contradiction. Therefore |CG (t, z)/C0 | = 4. Hence CG (t)/ t ∼ = Aut(U6 (2)). If CG (t, z)/C0 were cyclic then there would exist x ∈ CG (t, b) such that x2 ∈ tCK (b). But CK (b) = b×I ≤ b J, so modulo J, x2 ≡ t, which is impossible as t maps to a noncentral involution of Out(J) ∼ = E22 . = D8 . Thus CG (t, z)/C0 ∼ ∼ Thus, Ct = Aut(U6 (2)) × Z2 . Consequently, CG (z, t) = RH × t, where H ∼ = Aut(U4 (2)) and [RH, RH] = C0 . Since t ∈ [Cz , Cz ], Cz /C0 ∼ = D8 . Now H is  ×  z := Cz /R t = H v , with | v | = 2. Therefore there is w ∈ Cz complete, so C  such that w  is an involution, w  ∈ H, and the image of w in Cz /C0 has order 4. As  centralizes or a result, w2 ∈ R t − C0 = Rt. Moreover, we may assume that w 2  of order 3 such that C  ( inverts some x ∈H x ) = 1. Therefore w ∈ CRt ( x) = z t. R But tw = tz and z w = z, so [w, w2 ] = 1, which is absurd. Therefore t, z  Cz and we may assume that Lemma 3.13c1 holds, i.e., R ≤ and t ∈ Rz . Consequently CRz (b) ∼ Rz ∼ = Q8 ∗ Q8 ∗ D8 , forcing C b to = 21+10 + contain P GO6− (3), by [VK , 10.7]. In particular, some involution of C b centralizes t and induces an outer automorphism on I. It follows by [VK , 12.19] that in Ct , CCt (b) contains an involution t∗ such that I t∗  ∼ = Aut(I), t∗ induces a graph ∗ ∼ Replacing t∗ by tt∗ if necessary, we may automorphism on K, and CK (t ) = Sp6 (2). ∗ ∗ ∗ ∼ assume that I := E(CJ (t )) = I and I t ∼ = Aut(I). Let K ∗ be the subnormal ∗ ∗ closure of CK (t ) in CG (t ). As m2,3 (G) = 4, K ∗ is a single component. As I ∗ ∩K ∼ = Σ6 , I ∗ is a component of CK ∗ (b) isomorphic to I. Thus U4 (2) ↑3 K ∗ /O3 (K ∗ ) and Sp6 (2) ↑2 K ∗ ∈ C2 . So K ∗ /Z(K ∗ ) ∼ = K, and we may assume that Z(K ∗ ) = 1. Then the whole analysis we have done for Ct applies as well to CG (t∗ ). We have CK (z, t∗ ) = F M where F ∼ = E25 and M ∼ = Σ6 . Hence F M ≤ ∗ (∞) CG (z, t, t ) ≤ CK ∗ (z, t) and so z is 2-central in K ∗ as well as in K. Then just  so |CR (t∗ )| ≥ 28 . as t ∈ O2 (Cz ), we have t∗ ∈ O2 (Cz ). Hence t∗ centralizes R, ∗ ∗ ∼ But CR (t ) = O2 (CK (t )) = E25 . This contradiction completes the proof of the lemma.  Lemma 3.17. Suppose Lemma 3.6d holds with K ∼ = U4 (3). = 2U6 (2). Then J ∼ Proof. By [VK , 3.78] and [IA , 6.4.1], NK (b) = I × t × b, z, with z a 2central involution of K inverting b. Since CAut(J) (I) ∼ = Z2 [IA , 4.5.1], Ct,z (J) = 1. ∼  Therefore m3 (J) ≤ m2,3 (G) = 4, so J = U4 (3) by [IA , 6.4.4], as claimed. Now Lemmas 3.6–3.12, 3.14, 3.16, and 3.17 together prove Proposition 3.1 and Theorem 1.

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4. Maximal Symplectic Triples Having analyzed the situation in which every symplectic pair of G is trivial, we shall now consider the contrary case, in the remainder of this chapter. In view of Theorem C5 : Stage 2, we assume throughout these sections that (4A)

G contains a faithful symplectic pair.

In this case it will again turn out that p = 3, but we do not know that yet. For any faithful symplectic pair (B, T ), T is of course of symplectic type, and we set z = Ω1 (Z(T )) and T0 = O2 (CG (z)). By [V4 , Proposition 10.2], T0 is also of symplectic type. As T ∈ I∗G (B; 2), T0 ≤ T , and we will eventually see that T = T0 . For now, however, we prove: Lemma 4.1. Let (B, T ) be a faithful symplectic pair, z = Ω1 (Z(T )), and T0 = O2 (CG (z)). Then |T : T0 | ≤ 2 and T0 = F ∗ (CG (z)). Moreover, Sylow 2-centers of G are cyclic and contain conjugates of z. Proof. Set C = CG (z) and C = C/ z. We argue first that E(C) ≤ Op (C). Otherwise, there exists (B, z, K) ∈ BtKp (G), whence p = 3 by Theorem C5 : Stage 2 and BtK3exc (G) = ∅ by our assumption. By Theorem C5 : Stage 2, z ∈ I∗G (B; 2), a contradiction as z ∈ T and (B, T ) is faithful. Hence, E(C) ≤ Op (C), as asserted. In particular, E(C)T is a B-invariant p -group. Since T ∈ I∗G (B; 2), T ∈ Syl2 (E(C)T ). Set S = T ∩ E(C) ∈ Syl2 (E(C)). Then S is B-invariant, hence of symplectic type or trivial. In particular if E(C) = 1, then 1 = S  T , so z ∈ S, and indeed as Ω1 (Z(O2 (C))) = z, z = Ω1 (Z(K1 )) for every component K1 of E(C). Thus Sylow 2-subgroups of the simple group K1 /Z(K1 ) embed in the direct product of an elementary abelian group and a dihedral group, by the structure of T . However, as K1 ∈ C2 , this is impossible by [VK , 10.33]. Thus, E(C) = 1, so T0 = F ∗ (C). It remains to show that |T : T0 | ≤ 2. Assume that T0 < T . Thus T 0 < T , T 0  T , and CT (T 0 ) ≤ T 0 . As T is of symplectic type, T = T1 T2 , where T1 is extraspecial, T2 is embeddable in a semidihedral group, [T1 , T2 ] = 1, and T1 ∩ T2 = z. Hence, T = T 1 × T 2 , with T 1 elementary abelian and T 2 embeddable in a dihedral group. As T 0 ≥ CT (T 0 ), we have T 0 = T 1 × T 0,2 , where T 0,2 is a proper normal subgroup of T 2 containing its centralizer in T 2 . It follows that T 2 is dihedral and |T 2 : T 0,2 | = 2 (with T 0,2 cyclic or dihedral). Thus |T : T0 | = 2 and the proof is complete.  This has the following interesting corollary. Corollary 4.2. Let (B, T ) be a faithful symplectic pair and write z = Ω1 (Z(T )). Then for any b ∈ B # , z inverts Op (CG (b)). Proof. G is balanced with respect to B and Op (CG (b)) has odd order for all b ∈ B # , by [V4 , (1A4,7)]. Hence by [V2 , 1.1, 1.3b], Θ := Θ1 (G; B) has odd order and is invariant under ΓB,2 (G). In particular, T = ΓB,3 (T ) normalizes Θ. Let W = CΘ (z). Then W normalizes F ∗ (CG (z)) = T0 by Lemma 4.1. Hence [W, T0 ] ≤ Θ∩T0 = 1, so W ≤ O 2 (CG (T0 )) = 1. Thus z inverts Θ, and as Op (CG (b)) ≤ Θ by definition, the corollary follows. 

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Definition 4.3. A maximal symplectic triple is a triple (B, T, b) such that (B, T ) is a faithful symplectic pair, b ∈ B # , and |CT0 (b)| is maximized among all such pairs (B, T ) and elements b ∈ B # . Here, as above, T0 = O2 (CG (Ω1 (Z(T )))). Obviously, maximal symplectic triples exist. Moreover, for any such triple, [V2 , 2.1] yields that (4B)

B/ b acts faithfully on CT0 (b).

In the main result of this section, we closely follow [KMa1, Theorem D], for the most part. Proposition 4.4. Let (B, T, b) be a maximal symplectic triple. Set Cb = CG (b) and C b = Cb /Op (Cb ). Then [CT0 (b), Op (C b )] = 1. We assume that the proposition is false and derive a contradiction in a sequence of lemmas. Set m = m2,p (G) and R = CT0 (b), and let P ∈ Sylp (Op p (Cb )) with B, P  a p-group and P invariant under R. Let D be a critical subgroup of P of class at most 2 and exponent p [IG , 11.11]. Then D b shares these properties so we may assume that b ∈ D. By criticality, R acts nontrivially on D. By maximality and [V4 , 10.2], R is of symplectic type. We first prove Lemma 4.5. D is extraspecial, and Z(D) = CD (z) = b. Proof. Observe first that P0 := CP (z) satisfies [P0 , T0 ] ≤ T0 . But then [P0 , R] ≤ P ∩ T0 = 1. Suppose that mp (P0 ) ≥ 2, in which case P0 contains a B-invariant Ep2 -subgroup U . Set B ∗ := U CB (U ). Then mp (B ∗ ) ≥ mp (B) and [B ∗ , z] = 1, so B ∗ ∈ Bp∗,c (G) and B ∗ acts on T0 = F ∗ (CG (z)). Thus U does not centralize Su := CT0 (u) for some u ∈ U # . But [U, R] ≤ [P0 , R] = 1, so |RSu | > |R| and hence |CT0 (u)| > |CT0 (b)|. Expand T0 to T ∗ ∈ I∗G (B ∗ ; 2). Then (B ∗ , T ∗ ) is a (faithful) symplectic pair by [V4 , 10.2] as B ∗ centralizes z ∈ T ∗ . But z = Ω1 (Z(T ∗ )), so our maximal choice of (B, T, b) is violated. Thus, mp (P0 ) = 1, whence b = Ω1 (P0 ). Next, set V = Z(D), so that b ∈ V and b = CV (z) by the preceding paragraph (as D has exponent p). Suppose now that V > b, so that V = b × W , where z inverts W and W = 1. Since R is of symplectic type, R therefore acts faithfully on W . But B/ b acts faithfully on R by (4B), so the width w(R) is at least mp (B) − 1 = m − 1 by [V2 , 6.2]. Hence mp (W ) ≥ 2m−1 . As z inverts W , it follows that for some involution x ∈ R, 1 mp (CW (x)) ≥ (2m−1 ) = 2m−2 . 2 But x also centralizes b ∈ W , so (4C)

mp (CG (x)) ≥ mp (CV (x)) > 2m−2 .

As m ≥ 4, (4C) implies that mp (CG (x)) > m, contrary to m = m2,p (G). Therefore V = b. Since D has exponent p and class at most 2, D is extraspecial with center b. But also as shown above, P0 is cyclic. Since D has exponent p, it follows that  b = P0 ∩ D = CD (z), and the proof is complete. As an immediate corollary we obtain Lemma 4.6. m ≤ 5 and p = 3.

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 := D/ b. Let w be the width Proof. By the preceding lemma, z inverts D of R. Since B/ b acts faithfully on R (4B), w ≥ kmp (B/ b) = k(m − 1), where  ≥ 2w and for some x ∈ I2 (R − z), p = 3 if k = 1 [V2 , 6.2]. Again we have mp (D) w−1 mp (CD (x)) ≥ 2 . Let E be the full preimage of CD (x) in D. As x centralizes b, E = CD (x), so E is extraspecial, and mp (E) ≥ 2w−2 . Now if w ≥ 2(m − 1), then as m ≥ 4 and [x, E] = 1, m ≥ mp (E) ≥ 22(m−1)−2 > m, contradiction. Thus, k = 1, so p = 3 and mp (E) ≥ 2m−3 . As mp (E) ≤ m, we get m ≤ 5 and the lemma is proved.  The Klinger-Mason argument easily eliminates the case m = 5. Thus we prove Lemma 4.7. m = 4. Proof. Suppose that m = 5. Then as B/ b acts faithfully on R, R contains a subgroup S ∼ (see [V2 , 6.2]). Let x ∈ I2 (S) − z; then CS (x) contains a = 21+8 + subgroup S1 ∼ . As CD (z) = b, S1 acts faithfully on D1 := CD (x), and so = 21+6 + mp (D1 /Φ(D1 )) ≥ 23 . Also D1 is extraspecial, and so the width w(D1 ) ≥ 4. But as m = 5 and D1 centralizes x, mp (D1 ) ≤ 5, so in fact w(D1 ) = 4. Finally, there is y ∈ I2 (S1 − z) and S2 ≤ S such that D8 ∗ D8 ∼ = S2 ≤ CS1 (y) with S2 acting faithfully on D2 := CD1 (y). Thus D2 is extraspecial of width 2. But S2 ≤ R centralizes b = Z(D2 ), so S2 embeds into CAut(D2 ) (b) ∼ = Sp4 (3). However, m2 (S2 ) = 3, while m2 (Sp4 (3)) = 2 by [IA , 4.10.5d], contradiction. The proof is complete.  Because of the possible existence of elements of orders 2 or 3 with nonconstrained centralizers, elimination of this remaining case is more delicate. This time, there is Q8 ∗ Q8 ∗ Q8 ∼ = S ≤ R. We fix x ∈ I2 (S − z), with x 2-central in G if possible. As usual, there is Q8 ∗ D8 ∼ = S1 ≤ CS (x) with S1 acting faithfully on D1 := CD (x) and centralizing b. Hence, w(D1 ) ≥ 2. Since m3 (CG (x)) ≤ m = 4, it follows that w(D1 ) ≤ 3. Thus (4D)

w(D1 ) = 2 or 3.

The same argument shows that w(CD (xz)) = 2 or 3, so (4E)

w(D) = 4, 5, or 6.

In particular, m3 (P ) ≥ m3 (D) ≥ 5. As m = 4, it follows that CG (P ) has odd order. This yields (4F)

P = F ∗ (C b ).

First, as in [KMa1], we prove Lemma 4.8. w(T0 ) ≤ 6. Proof. Since w(D) ≤ 6, m3 (D/ b) ≤ 12 < 24 , whence w(R) ≤ 3. Maximality of |CT0 (b)| implies therefore that w(CT0 (a)) ≤ 3 for all a ∈ B # . Suppose now that w(T0 ) ≥ 7. Since T0 = CT0 (B0 )  |B : B0 | = 3 , clearly B contains some E ∼ = E32 that centralizes a symplectic type subgroup T1 of T0 of width 2. Then T2 := CT0 (T1 ) is E-invariant of symplectic type of width at least 5. Since E has only four Z3 -subgroups, it follows that for some e ∈ E # , CT2 (e) is of symplectic type of width at least 2, whence CT0 (e) = T1 CT2 (e) has width at least 4, contradicting the previous paragraph. The lemma follows. 

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We next prove Lemma 4.9. The following conditions hold: (a) m2 (CG (D1 )) = 1; (b) w(D1 ) = 2; and (c) m3 (CG (x)) = 3. Proof. First, (b) follows from (c) and (4D), as m3 (D1 ) = 4 if w(D1 ) = 3. Moreover, if (c) fails, then m3 (CG (x)) ≥ 4 as m3 (D1 ) ≥ 3 and [D1 , x] = 1. We claim that if (a) fails, then there is u ∈ I2 (CG (D1 )) such that CG (u) contains a 31+6 subgroup. Indeed, there is then E22 ∼ = U ≤ CG (D1 ). But b = Z(D1 ), so U ≤ Cb , whence U acts on D. For each u ∈ U # , CD (u) is extraspecial. As D1 < D, there is u ∈ U # such that D1 < CD (u). Then w(CD (u)) ≥ 3 and CD (u) contains a 31+6 subgroup, as claimed. We see then that if the lemma fails, there is some t ∈ I2 (CG (D1 )) such that m3 (CG (t)) ≥ 4; moreover either CG (t) contains a 31+6 subgroup, or t = x and CG (t) ≥ D1 S1 . We derive a contradiction from the existence of t. Since m = 4, m3 (CG (t)) = 4 and CG (t) contains some B ∗ ∈ B3∗ (G). Expand t to T ∗ ∈ IG (B ∗ ; 2). By Theorem C5 : Stage 2 and Lemma 4.1, either T ∗ = t, or |T ∗ : F ∗ (CG (t))| ≤ 2 with T ∗ of symplectic type.  Suppose that T ∗ = t. Then CG (t) contains a component K = O 3 (K) [V4 , 11.2]. By Theorem C5 : Stage 2, since m = 4, (1) K = E(CG (t)) ∼ = U5 (2), (2)U6 (2), Sp8 (2), (2)D4 (2), or (2)F (2); and (4G) 4 (2) t ∈ Syl2 (CG (K)). It follows by [VK , 9.22] that Aut(K) (and hence CG (t)) contains no 31+6 subgroup. As observed above, this implies that t = x and CG (t) ≥ D1 S1 . Now as F ∗ (Cz ) = T0 is of symplectic type, z is a Sylow 2-center of G. But F ∗ (CG (x)) ≥ K so x ∈ z G , and so x is not 2-central in G. By our choice of x, no involution of (S1 × x) − Z(S1 ) is 2-central in G. This, however, leads to a contradiction as follows. By (4G), CD1 S1 t (K) = t = x, so D1 S1 maps injectively into Aut(K). Moreover, S1 ≤ O2 (CG (Z(S1 ))). If K/Z(K) ∼ = F4 (2), then by [IA , 4.7.3A], Sylow 2-subgroups of CK/Z(K) (b) are isomorphic to Q8 × Q8 . But S1 = Ω1 (S1 ) ∼ = Q8 ∗ D8 . Hence some involution u ∈ S1 induces an outer automorphism on K, whence u ∈  F4 (2). Then, by (4G) and [VK , O2 (CG (Z(S1 ))), contradiction. Hence, K/Z(K) ∼ = 4.6], K/Z(K) ∼ = U6 (2), and some noncentral involution of S1 acts on K like a 2central involution y of K. Thus y ∈ S1 t − Z(S1 ), so by what we just saw, no involution of y, t is 2-central in G. By [VK , 2.10a], for some S2 ∈ Syl2 (CG (t)), y, t ≥ Z(S2 ). Hence no involution of Z(S2 ) is 2-central in G. But as S2 ∈ Syl2 (CG (t)), Z(S2 ) contains a Sylow 2-center of G, the desired contradiction. We conclude that |T ∗ : T0∗ | ≤ 2 where T0∗ = F ∗ (CG (t)). Since also F ∗ (CG (z)) = T0 is of symplectic type, both t and z generate Sylow 2-centers of G, and thus t ∈ z G . In particular T0 and T0∗ are G-conjugate. By Lemma 4.8, w(T0∗ ) ≤ 6. But D1 ≤ Aut(T0∗ ) as T0∗ = F ∗ (CG (t)), so by [V2 , 6.3], m2 (T0∗ /Φ(T0∗ )) ≥ 18. This  contradicts w(T0∗ ) ≤ 6 and completes the proof of the lemma. Next, modifying an argument from [KMa1], we prove: Lemma 4.10. There exists an elementary abelian 2-group V ≤ CG (x) such that V is D1 -invariant and noncyclic, with CV (b) = x.

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Proof. Suppose that CG (x) contains a D1 -invariant noncyclic elementary abelian 2-subgroup W containing x. Let W1 = CW (b) ≥ x, so that [W 1 , D1 ] ≤ D. Since [W1 , D1 ] ≤ W , it follows that [W 1 , D1 ] = 1. But m2 (CG (D1 )) = 1 by Lemma 4.9a, so W1 = x. Hence the lemma holds with W = V , so it suffices to prove the existence of such a subgroup W . Let H = O3 (E(CG (x))), so that H/Z(H) is a direct product of Suzuki groups. Then D1 normalizes some Y ∈ Syl2 (H) and by [VK , 10.48], if H = 1, then Ω1 (Y ) is elementary abelian and noncyclic. Hence we can take W = Ω1 (Y ) x in this case, so we may assume that H = 1. Set X = O2 (CG (x)) and suppose next that X = F ∗ (CG (x)). If Z(X) is cyclic, then x = Ω1 (Z(X)). However, it then follows as in the preceding lemma that x ∈ z G , whence m3 (CG (x)) = m3 (Cz ) = 4, contrary to Lemma 4.9c. Hence, Z(X) is noncyclic and we can take W = Ω1 (Z(X)).  Thus we can assume that F ∗ (Cx ) > X, whence L = E(CG (x)) = O 3 (L) = 1 as H = 1. Let L1 , . . . , Lr be the components of L. Since O2 (CG (x)) = 1 and m3 (CG (x)) = 3, r ≤ 3. Suppose that r = 3, whence m3 (Li ) = 1, 1 ≤ i ≤ 3. Let Q ∈ Syl3 (CG (x)). As m3 (Q) = 3, CQ (L) = 1. Moreover, if Q1 = NQ (L1 ), then |Q : Q1 | ≤ 3. Also by [VK , 1.1], a Sylow 3-subgroup of Aut(Li ) splits over Inn(Li ). Since m3 (CG (x)) = 3, this forces Q1 ≤ L, whence Q1 is abelian. However, D1 ≤ CG (x) and D1 ∼ = 31+4 contains no abelian subgroup of index 3, a contradiction. Hence, r ≤ 2, and in particular, D1 normalizes L1 . If [L1 , D1 ] = 1, then as m3 (D1 ) = 3 and there exists x1 ∈ I3 (L1 ) − Z(L1 ), we have m3 (CG (x)) ≥ 4, contradiction. Hence [L1 , D1 ] = L1 . If b ∈ CD1 (L1 ), then L1 ≤ Cb , whence L = [L, D1 ] ≤ P , a contradiction as L has no normal 3complement. Thus b ∈ CD1 (L1 ), so D1 acts faithfully on L1 . As D1 = [D1 , S1 ] ∼ = 31+4 , L1 ∈ C2 , and m3 (L1 ) ≤ 3, this contradicts [VK , 9.21]. The proof is complete.  Now we complete the proof of Proposition 4.4 by deriving a final contradiction. With V as in the preceding lemma, we have V1 := [V, b] = 1, so D1 acts faithfully on V1 and CV1 (b) = 1. Let y ∈ D1 − b. Then CD1 (y) = y × E1 with E1 ∼ = 31+2 and Z(E1 ) = b. As all subgroups of order 3 in y, b other than b are D1 -conjugate, CV1 (y) = 1, and we let W be a subgroup of CV1 (y) on which E1 acts irreducibly and faithfully. Thus, m2 (W ) = 6. y = Cy /O3 (Cy ). Since m3 (Cy ) ≥ m3 (CD (y)) ≥ 5 Consider Cy := CG (y) and set C y lies in C3 . Choose an by (4E), O3 (Cy ) has odd order and every component of C E1 W -invariant Sylow 3-subgroup Q of O3 3 (Cy ). If [W, Q] = 1, then CW (Q) = 1 by the irreducible action of E1 , so m2,3 (W Q) ≥ 5 by the Thompson dihedral lemma,  acts faithfully on against m = 4. Therefore [W, Q] = 1. Let L = L3 (Cy ). Thus W    L = E(Cy ), as does E1 . Observe next that as w(D) ≥ 4, CD (y) contains a subgroup E ∼ = 31+6 with  b = Z(E), and since W = [W, b], E likewise acts faithfully on L. Let M be the product of those 3-components M1 , . . . , Mr of L such that m3 (Mi )−m3 (O3 3 (Mi )) ≥  not in M  is isomorphic to L2 (8) by [VK , 1.3], and so is 2. Any component of L )). It follows that E1 W centralized by b by [VK , 1.1]. Thus W centralizes E(CL (M . Note that r ≤ 2, since otherwise an involution of M1 would acts faithfully on M

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224

5. THEOREM C5 : STAGE 3

centralize an E35 -subgroup of M y, contrary to m = 4. Thus E normalizes each Mi , as does W = [W, b]. Since E1 acts irreducibly on W , it follows that E1 W acts faithfully on some  1 . Thus Aut(M 1 ) contains a copy of E1 W ∼ Mi , let us say on M = E26 31+2 as well 1+6 ∼  1  y ) ≥ 5, as one of E = 3 . As M1 ∈ C3 , [VK , 12.39] now gives that m2,3 (M whence m2,3 (Cy ) ≥ 5, contradicting m = 4. This at last completes the proof of Proposition 4.4. 5. Theorem 2: p = 3 and Constrained Neighborhoods We continue to assume (4A), and we fix a maximal symplectic triple (B, T, b). As in the previous section we let z = Ω1 (Z(T )), T0 = F ∗ (CG (z)), Cb = CG (b), C b = Cb /Op (Cb ), R = CT0 (b), Cz = CG (z), and m = m2,p (G). We also set J = Lp (Cb ). Just as Theorem C5 : Stage 2 made a preliminary reduction to a small number of configurations when BtKp (G) = ∅, the following proposition makes an analogous reduction when a faithful symplectic pair exists. It also settles that p = 3 and proves that T = T0 . Proposition 5.1. Assume (4A). With the above notation, (CG (z), b, J) is a constrained {z, b}-neighborhood with respect to (B, T ). Moreover, the following conditions hold: (a) p = 3;  (b) J ∼ = P Ω+ 8 (3), U5 (2), (3)U6 (2), D4 (2), (3)Suz, Co2 , Co1 , (3)F i24 , F3 , F2 , or F1 ; (c) B = (B ∩ J) b and m = m3 (CJb (z)), unless J/Z(J) ∼ = U6 (2) and OutB (J) = 1; (d) O3 (Cb ) = 1 and C(b, J) is a cyclic 3-group; (e) T = F ∗ (Cz ) = T0 , and R = O2 (CJ (z)) is extraspecial; and b = Cb /C(b, J), R  = F ∗ (C  ( (f) In C z  is a Sylow 2-center in J Cb z )), and  b . and C The first assertion follows directly from the various parts and Definition 1.1. As before, R is of symplectic type with z = Ω1 (Z(R)), and B/ b acts faithfully on R. By Proposition 4.4, [R, Op (C b )] = 1, so [J, z] = 1. We fix a p-component J1 of Cb such that [J 1 , z] = 1 and argue in a sequence of lemmas that J = J1 and the conclusions of Proposition 5.1 hold. We first prove Lemma 5.2. B normalizes J1 , z ∈ J1 , and Op (Cb ) = 1. Proof. Since b ∈ B and mp (B) ≥ 4, every component of J lies in C3 . Then since B centralizes z ∈ Cb , B normalizes all components of J by [IG , 8.7(ii)]. Set S = [R, B]; then S normalizes all components of J, and is a 2-group of symplectic type by [V4 , 10.2]. By [VK , 2.2], z induces an inner automorphism on J, and by our choice, z acts nontrivially on J 1 . Now Op (Cb ) has odd order by [V4 , (1A4)], so if z ∈ J1 , then z has nontrivial projections x, y on J 1 and C(b, J1 ), respectively. But B centralizes z and normalizes J1 , so B centralizes x, y ∼ = E22 . However, m2 (CG (B)) = 1 by (c) of Theorem C5 : Stage 2, a contradiction. Therefore z ∈ J1 . Finally, by Corollary 4.2, z inverts Op (Cb ). Hence J1 = [z, J1 ] centralizes  Op (Cb ). But z ∈ J1 and Op (Cb ) has odd order. Hence, Op (Cb ) = 1.

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This in turn yields Lemma 5.3. J = J1 and C(b, J) is a cyclic p-group. Proof. Set C = C(b, J1 ), and suppose that mp (C) ≥ 2. Since B normalizes J1 , it normalizes C and hence normalizes some Ep2 -subgroup U of C. Set B ∗ = U CB (U ), so that mp (B ∗ ) ≥ mp (B) = m. Then U centralizes z ∈ J1 , whence B ∗ centralizes z and hence B ∗ ∈ Bp∗,c (G). On the other hand, R normalizes J1 as z ∈ J1 and Op (Cb ) = 1. Therefore CR (J1 ) = 1. However, [R, U ] ≤ CT0 (b) = R as U ≤ Cz and T0 = F ∗ (CG (z)). But [R, U ] ≤ C as R normalizes C and U ≤ C. Thus [R, U ] ≤ R ∩ C = 1, the last as R acts faithfully on J1 . Now U acts on T0 , so Su := [CT0 (u), U ] = 1 for some u ∈ U # . Then RSu ≤ CT0 (u) and as U centralizes R by the preceding paragraph, RSu > R. As u ∈ B ∗ ∈ Bp∗ (G) with [B ∗ , z] = 1, as usual this contradicts the fact that (B, T, b) is a maximal symplectic triple. We conclude that mp (C) = 1. Since b ∈ Z(C), this implies that C is a cyclic p-group, whence also J1 = J, and the lemma is proved.  Next, set R1 = CT (b). Since |T : T0 | ≤ 2, B centralizes T /T0 , so T = R1 T0 and (5A)

|R1 : R| = |T : T0 |.

Also R1 is of symplectic type by [V4 , 10.2], and hence z = Ω1 (Z(R1 )). We prove: Lemma 5.4. R1 ∈ I∗Cb (B; 2). Proof. Expand R1 ≤ R∗ ∈ I∗Cb (B; 2) and R∗ ≤ T ∗ ∈ I∗G (B; 2). As usual, (B, T ∗ ) is a faithful symplectic pair. Since R1 ≤ T ∗ and z = Ω1 (Z(R1 )), likewise z = Ω1 (Z(T ∗ )). Thus T ∗ ≤ Cz , so T ∗ T0 is B-invariant. As T ∗ is maximal, T0 ≤ T ∗ T0 = T ∗ . Therefore T = R1 T0 ≤ T ∗ . But T ∈ I∗G (B; 2), so T = T ∗ . Hence  R∗ ≤ CT (b), so R∗ = R1 , as required. b = Cb /C(b, J) and let φ : Cb → C b be the natural mapping. We Now we set C shall apply [VK , 10.62] to obtain most of the conclusions of Proposition 5.1. The  J,  z, R,  R 1 , B,  and r in that lemma will be played by C b , J,  z, R,  R 1 , roles of X,  and m = m2,p (G) in the current analysis. Likewise the roles of X, J, A, φ, B, B, and z there will be played by Cb , J, C(b, J), φ, B, and z here. The hypotheses of [VK , 10.62] are easily verified with the help of (4B) and Lemmas 5.3 and 5.4. Conclusions (a) and (b) of Proposition 5.1 will follow from [VK , 10.62a], once we rule out the possibility that p = 5, J ∼ = F1 , and m = 4. Conclusion (c) holds + ∼ if J P Ω (3) because R  C (z), forcing z to be in class t2 in [IA , 4.5.1]; hence = J 8  3 E∗ (CAut(J) (z)) ≤ Inn(J), so B maps into Inn(J). Conclusion (c) also holds if J = 3Suz and 3F i24 because all the preimages in J of elements of J/Z(J) of order 3 are themselves of order 3 [IA , 5.3]. If J = D4 (2), then as B is of maximal rank in CCb (z), B induces inner automorphisms on J and conclusion (c) holds. Indeed, conclusion (c) holds in all other cases as 3 does not divide | Out(J)|. Conclusion (d) holds by Lemma 5.3. Conclusion (e) is a consequence of (5A) and [VK , 10.62c], and conclusion (f) is direct from [VK , 10.62bc]. The following lemma therefore will complete the proof of Proposition 5.1. Lemma 5.5. p = 3.

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∼ F1 . Proof. If false, then by the above discussion, p = 5, m2,5 (G) = 4, and J = By [VK , 12.28], there is a ∈ J ∩ B # such that H := E(CJ (a)) ∼ F and R ∩ H∼ = 5 = ∼ 2 . Set E = a, b E . Then C (T ) ≤ C (T ) = 1 so there exist independent 21+8 = 5 E B + a1 , a2 ∈ E # such that Ti := [CT (ai ), E] = 1, i = 1, 2. Let Hi be the subnormal closure of H in CG (ai ), i = 1, 2. Since m2,5 (G) = 4, Hi is a single 5-component and E normalizes all 5-components of CG (ai ). If Hi is a trivial pumpup of H for some i, then E ≤ C(ai , Hi ) and so Ti = [CT (ai ), E] ≤ C(ai , Hi ). Therefore m2,5 (G) ≥ m5 (Hi E) ≥ m5 (H) + m5 (E) = 3 + 2 = 5, contradiction. So Hi is a vertical pumpup of H for i = 1, 2, whence Hi /O5 (Hi ) ∼ = F1 by [VK , 3.8], with [Hi /O5 (Hi ), a3−i ] = 1. Let C = CG (z). As R ∩ H is extraspecial, z is 2-central in H and then 2-central in each Hi , by [VK , 12.29]. Hence in C := C/O5 (C), CC (ai ) has a component I i ∼ = Co1 , with [I i , a3−i ] = 1. But = CHi (z)/O5 (CHi (z)) ∼ m5 (C) ≤ m2,5 (G) = 4, so by L5 -balance, C has a component L with Co1 ↑5 L. But there are no such K-groups L, by [VK , 3.8]. This contradiction completes the proof of the lemma and with it the proof of Proposition 5.1.  6. Theorem 2: Nonconstrained Neighborhoods of Type F i24 We begin the analysis of the cases of Proposition 5.1 with the single case of a maximal symplectic triple (B, T, b) for which J = E(Cb ) ∼ = P Ω+ 8 (3). In this case we prove that G has a nonconstrained neighborhood (K; I; J) ∼ = (2F i22 ; 2U4 (3); P Ω+ 8 (3)). More precisely: Proposition 6.1. Let (B, T, b) be a maximal symplectic triple with J := E(Cb ) ∼ = P Ω+ 8 (3). Then m = 5 and there exists an involution t ∈ J such that the following conditions hold: ∼ 2U4 (3); (a) I := E(CJ (t)) = (b) CG (t) has a component K ∼ = 2F i22 with I = E(CK (b)); (c) CG (t)/ t ∼ = Aut(F i22 ); and (d) C(b, J) = b and |Cb : J| is odd. Remark 6.2. Since m3 (F i22 ) = 5 = m, there is E35 ∼ = B ∗ ≤ K with b ∈ B ∗ ∈ Thus (a) and (b) will imply that (K; I; J) is, in fact, a nonconstrained {t, b}-neighborhood. In particular, it will then follow from Theorem C5 : Stage 2 that C(t, K) = t. B3∗,c (G).

Proof. By Proposition 5.1, B = (B ∩ J) × b, C(b, J) =: P is a cyclic 3group, and R ≤ J. Since B/ b acts faithfully on R = CT (b), 1 = [R, B] ≤ J. By Proposition 5.1 and [VK , 10.25], m = m2,3 (J b) = 5. Now R is a maximal B-invariant 2-subgroup of Cb , so by [VK , 10.24], R ∼ and (B ∩ J)R is the = 21+8 + ∼ central product L1 L2 L3 L4 with Li = SL2 (3) for all i = 1, 2, 3, 4. Set S = CT (R), so that S = [T, b] is extraspecial by [V2 , 4.12]. Also let t ∈ I2 (R) − {z} with t ∈ L1 L2 , and set Ct = CG (t). Then by [VK , 10.24], I := E(CJ (t)) ∼ = 2U4 (3) with t, z ≤ I. In particular t = Z(I), I is a component of CCt (b) and t = z. t = Ct /X, so We determine the structure of Ct . Set X = O3 (Ct ) and C ∼    that I = U4 (3) is a component of CCt (b). Let K be the subnormal closure of t . Since m3 (I) = 4 and m3 (Ct ) ≤ m = 5, K  is a single component and I in C

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 ≤ 1. In particular, K  C t . Suppose for a contradiction that K  = I.  m3 (CCt (K))    We have S ≤ Ct and z ∈ S = [S, b]. Then as b centralizes I, so does S. Hence I normalizes the 3 -group XS, and consequently I ∩ XS ≤ O3 (I) = t. However,  > I.  z ∈ I ∩ S, so z = t, which is not the case. We conclude that K ∼     Thus, I ↑3 K. As m = 5, K = F i22 , by [VK , 3.27].   Then since m3 (K) = ∗  K  = F (C t ) and b ∈ K.  Furthermore, b ∈ Syl3 (C  5 = m3 (C),  (I)) by Aut(K)

[IA , 5.3t], which in turn implies  that P = b. 3   Next, O (CK (b)) = I × b by [IA , 5.3t]. Hence by order considerations, b  in Ct . Let Q ∈ Syl3 (I). By [VK , 9.7], is not 3-central in K, the preimage of K ∼ Y := Z(Q) = Z3 , whence Z(Q b) = Y × b and Y = Z(Q b) ∩ Φ(Q b). As b is not 3-central in K, it follows that NK (Q b) − Q b contains a 3-element v and bv = by for some y ∈ Y # . Thus if we set U = Y b, all subgroups of U − Y are NK (U )-conjugate. On the other hand, CAut(J) (I) embeds in D8 . But O3 (CCb (t)) = CX (b) and clearly this group centralizes I as I  CCb (t), so CX (b) embeds in D8 . Hence U , which centralizes t ∈ CX (b), centralizes all of CX (b). It follows therefore from the preceding paragraph that [CX (u), U ] = 1 for all u ∈ U − Y . Hence [X, Y ] = 1.   = O 3 (C t ). Therefore, K is But then CCt (X) contains Y Ct , which covers K quasisimple. Since t = Z(I), this forces K ∼ = 2F i22 , and so (a) and (b) hold. As remarked above, it now follows that (K; I; J) is a nonconstrained neighborhood and that C(t, K) = t. Let Ct,b = CCt (b). In particular Ct,b / t embeds 10  in CAut(K)  (b), which by [IA , 5.3t] yields |Ct,b |2 ≤ 2 . However, CJ (t) ≤ Ct,b 10 and |CJ (t)|2 = 2 by [VK , 10.24], so |CCb (t) : CJ (t)| is odd and |Ct,b |2 = 210 .  which is t > K,  so C t ∼ Since |CK (b)|2 = 28 by [IA , 5.3], it follows that C = Aut(K), (c). Finally, the above arguments are valid for any involution t ∈ Li Lj −{z} (i = j). But if |Cb : J| were even, there would exist, by [VK , 10.26], a choice of such an element t for which |CCb (t) : CJ (t)| would be even, contradiction. Therefore, |Cb : J| is odd, so (d) holds. The proof is complete.  7. Theorem 2: Nondegenerate Constrained Neighborhoods We continue to assume (4A), and continue the notation introduced at the beginning of Section 5. Proposition 5.1 gives us that p = 3 and (CG (z), b, J) is a constrained neighborhood with respect to (B, T ). It also provides a list of possible structures for J; because of Proposition 6.1 we shall assume that J ∼  D4 (3). = We set up the situation for further analysis of these possibilities. Recall that R = CT0 (b) = CT (b) (by Proposition 5.1). Note that Z(J) ≤ b, since by Proposition 5.1, C(b, J) is a cyclic 3-group containing b. Also by Proposition 5.1, O3 (Cb ) = 1 and m = m3 (B) = m3 (CCb (z)), with z a 2-central involution of J. Table 7.1 lists the possibilities for J, for m = m2,3 (G) = m3 (B) = m3 (CCb (z)), and for a component JD = E(CJ (b )) for a suitable b ∈ B − b such that z ∈ JD [VK , 12.50]. We set D = b, b  .

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Table 7.1. The Subgroup JD

J m = m2,3 (G) JD = E(CJ (D)) = E(CJ (b )) U5 (2) 4 U4 (2) SU6 (2) 4 or 5 Um (2) U6 (2) 4 or 5 Um (2) D4 (2) 4 U4 (2) Co2 4 U4 (2) Co1 5 3Suz Suz 4 3U4 (3) 3Suz 4 [3 × 3]U4 (3) F i24 6 P Ω+ 8 (3)  3F i24 6 P Ω+ 8 (3) F3 4 G2 (3) F2 5 F i22 F1 7 3F i24 By [VK , 12.50] and the fact that z ∈ CR (b ), (1) CR (b ) acts faithfully on JD ; (2) B/D acts faithfully on CR (b ); and (3) F ∗ (CJb  (b )) = JD b  Z(J).

(7A)

For any d ∈ D# , set Cd = CG (d) and let Jd be the subnormal closure of JD in Cd . Thus Jb = J. Notice that as z ∈ JD ≤ Jd and z inverts O3 (Cd ) (Corollary 4.2), which has odd order, O3 (Cd ) = 1. As d ∈ B, the components of Cd are C3 -groups, and by [IG , 8.7(ii)], B normalizes them all. In particular, (7B)

For any d ∈ D# , Jd is a single component and O3 (Cd ) = 1.

We also set Des = {d ∈ D# | Jd > JD }, the set of all “essential” elements of D. Note that b ∈ Des . Lemma 7.1. Suppose that d ∈ D# and [CT (d), D] = 1. Then d ∈ Des . Proof. Set S = [CT (d), D] = 1 and suppose by way of contradiction that Jd = JD . Now in every case of Table 7.1, m3 (JD ) − m3 (Z(JD )) ≥ m − 1. (This is trivial for the first five rows. It holds by [IA , 5.6.1] if JD ∈ Spor; and by [IA , 3.3.3, 6.4.4a] in the remaining cases.) Therefore m3 (JD D) > m. As B normalizes all components of Cd , S = [S, D] normalizes Jd , so S, like D, centralizes Jd = JD , and so is JD D-invariant. But then m = m2,3 (G) ≥ m3 (JD D) > m, contradiction. The lemma follows.  We will use the structure of Jd for various d ∈ D# to determine the structure of CT (d), and thereby the structure of T = F ∗ (Cz ). Lemma 7.2. Let d ∈ Des . Then one of the following holds: (a) Jd ∼ = J; (b) Jd /Z(Jd ) ∼ = J/Z(J) ∼ = F i24 or U6 (2); or

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(c) JD ∼ = U4 (2) and both J and Jd are found in the first five rows of Table 7.1, in the first column. Proof. We have JD ↑3 Jd ∈ C3 , and m2,3 (Jd ) − m3 (Z(Jd )) + 1 ≤ m2,3 (Jd d) ≤ m. ∼ U4 (2) or U5 (2). Then by [VK , 3.9], either (a) or (b) holds, Suppose JD = or (m; JD ; Jd ) is one of the following: (4; 3U4 (3); 3F i22 ), (4; [3 × 3]U4 (3); Co1 ), (6; D4 (3); D4 (33 )), or (4; G2 (3); (3)F i24 , D4 (3), 3D4 (3), or G2 (33 )). In all these m2,3 (Jd ) − m3 (Z(Jd )) + 1 > m [IA , 5.6.2, 4.5.1]. This contradicts the previous paragraph, however. We may therefore assume that JD ∼ = U4 (2) or U5 (2). Since Jd ∈ C3 , either Jd is isomorphic to one of the groups J in the first five rows of Table 7.1, or Jd ∼ = P Sp4 (33 ), by [VK , 3.11]. The last case is impossible as then m2,3 (Jd ) = 6 > m. Hence if JD ∼ = U4 (2), then (c) holds. Finally, if JD ∼ = U5 (2), then Jd /Z(Jd ) ∼ = U6 (2)  by [VK , 3.11], and the proof is complete. Lemma 7.3. Let d ∈ Des . Then C(d, Jd ) is a cyclic 3-group. Proof. Let Ed ∈ Syl3 (C(d, Jd )) be B-invariant. It suffices by (7B) to show that Ed is cyclic. If Ed is noncyclic, then it contains a B-invariant E32 -subgroup U containing d, and B ∗ := U CB (U ) ∈ B3∗,c (G) as [z, B ∗ ] = 1. Since U is noncyclic and acts faithfully on T = F ∗ (Cz ), there is u ∈ U # such that [CT (u), U ] = 1. But m3 (Jd U ) > m2,3 (G) by the previous lemma and as U is noncyclic. Hence the subnormal closure H of Jd in CG (u) must be a nontrivial pumpup. As d ∈ B ∗ , H is a single 3-component by [IG , 8.7(ii)]. Also, m2,3 (H) − m3 (Z(H)) + 1 ≤ m, By [VK , 3.29, 3.84], given the possible isomorphism types of Jd and the fact that H ∈ C3 , we must have Jd ∼ = U5 (2). But then m2,3 (G) ≥ m2,3 (Jd × U ) = 5, so by Table 7.1, JD ∼ = U5 (2). Since Jd > JD by assumption, this is a contradiction, and the lemma is proved.  Lemma 7.4. The following conditions hold: (a) b ∈ Des ; (b) Des  = D; and (c) Unless possibly J/Z(J) ∼ = U5 (2), we have C(b, J) = b = U6 (2) and JD ∼ and C(d, Jd ) = d for each d ∈ Des . Proof. Since J > JD , (a) holds. Next, CD (T ) = 1 as T = F ∗ (CG (z)). In particular [T, b] = 1, so [CT (d), D] = 1 for some d ∈ D − b. By Lemma 7.1, d ∈ Des , and (b) follows. Finally, to prove (c), we may assume that JD ∼ = U5 (2), so m = 4 if J/Z(J) ∼ = U6 (2). This implies that D induces inner automorphisms on J. Let E = C(b, J); then E is a cyclic 3-group by Proposition 5.1, and [D, E] = 1. Hence E embeds in CAut(Jd ) (JD ), which has Sylow 3-subgroups of exponent 3 by Lemma 7.2 and [VK , 3.56]. Therefore (c) holds for b. Similarly, for any d ∈ Des − b, let Ed = C(d, Jd ), so that b ∈ Ed and Ed is a cyclic 3-group by Lemma 7.3. Then CEd (b) acts faithfully on J, so it embeds into CAut(J) (JD ). Thus CEd (b) has exponent 3 by Lemma 7.2 and [VK , 3.56]. Hence we are done unless [Ed , b] = 1, in which case b induces a non-inner automorphism on Jd . As JD is a component of CJd (b), we must have Jd /Z(Jd ) ∼ = U6 (2). Since JD ∼ = U5 (2), we have JD ∼ = U4 (2) and m2,3 (G) = 4 by Table 7.1. But by [VK ,

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4.5], there exists b0 ∈ I3 (bJd ) such that E(CJd (b0 )) b0 , d ∼ = U5 (2) × E32 . Hence  m2,3 (G) ≥ 5, a contradiction. This completes the proof. By Lemma 7.4b, we may assume that b ∈ Des . We now briefly focus on the cases in which JD ∼ = U4 (2), to wit, (7C)

{Jd | d ∈ Des } ⊆ {U5 (2), (3)U6 (2), D4 (2), Co2 }

by Lemma 7.2. Lemma 7.5. Assume (7C). Then b  NCz (B). Proof. Suppose b  NCz (B) and choose d ∈ Des − b (Lemma 7.4b). Then d := Nd / d Td ∼ letting Td = CT (d) and Nd = NJd d (Td ), we have B ≤ Nd and N = GU3 (2), U4 (2), Σ3 × Σ3 × Σ3 , or Sp6 (2), according to the type of Jd in (7C). Since Nd ≤ Cz , b  NNd (B). As b = 1,the  only possibilities are that  ∼  d ∼  N (2) with b ∈ Z( N ), or J (2) with GU D b in one of the three Σ3 -direct = 3 d d = 4 d . In these situations, however, CT (b) = z. As CT (b) ≥ CT (D) = factors of N d d d ∼ CT (D) = 21+4 , this is a contradiction. This completes the proof.  Lemma 7.6. Assume (7C). Then T is the central product of k copies of Q8 , where 4 ≤ k ≤ 2 + |Des | ≤ 10. Proof. For any d ∈ Des , CT (d) = O2 (CJd (z)) is the central product of 3 or 4 copies of Q8 , of which two generate CT (D). For any d ∈ D# −Des , CT (d) = CT (D) by Lemma 7.1. Hence k ≤ 2 + |Des |, and since Des  = D by Lemma 7.4b, k ≥ 4. The lemma follows.  Lemma 7.7. J ∼  Co2 , and Jd ∼ = = Co2 for any d ∈ D# . Proof. Suppose J ∼ = Co2 . Let  C z = Cz /T . By Lemma 7.4c and as Out(Co2 ) = 1 [IA , 5.3k], we have CC z (b) = b × L with L ∼ = (Cz ∩ J)/O2 (Cz ∩ J) ∼ = Sp6 (2). By L3 -balance and as m = 4, the subnormal closure M of L in C z is then a single,  := M /O3 (M ) is a vertical pumpup of L. normal, 3-component. By Lemma 7.5, M ∼ | so k ≥ 9; by Lemma  Suppose that M = Sp6 (8). Then 218 − 1 divides |M  # 7.6, this forces k = 10, i.e., Des = D# . Let d ∈ D ∩ M ; then d, b ≤ A ≤  M b for some nonabelian A of order 33 and exponent 9, and d ∈ [A, A]. Hence there is a preimage A ≤ CG (d) of A with A ∼ = A and d ∈ [A, A]. However, d ∈ Syl3 (C(d, Jd )) by Lemma 7.4c, and | Out(Jd )|3 = 3, so A/A ∩ Jd is abelian. Thus, d ∈ Jd , so Jd = F ∗ (CG (d)) ∼ = SU6 (2). But then by [VK , 9.28], any abelian subgroup of CG (d) of order 36 has 3-rank at least 5. As d lies in a Z9 × Z9 × Z9 subgroup of M , this is a contradiction. ∼ Hence by [VK , 3.25], M = Sp8 (2) or F4 (2), and in either case NC z (B) contains a Σ4 -subgroup N 0 permuting a basis {b0 , b1 , b2 , b3 } of B naturally, where b0 = b  and for i = 1, 2, 3, bi ∈ I3 (M ) and O 2 (CL (bi )) ∼ = Sp4 (2). Let Tbi = CT (bi ) for each i = 0, 1, 2, 3. By [VK , 10.42], CTb (bi ) = z for all i = 1, 2, 3, and hence 0

CTb (bj ) = z for all i = j, i, j ∈ {0, 1, 2, 3}. It follows that the subgroups Tbi / z i of T / z are independent, and so |T | ≥ 21+32 , contradicting Lemma 7.6. Thus, J ∼ = Co2 for some d ∈ D# , the same arguments with d and Jd in = Co2 . If Jd ∼ place of b and J reach a similar contradiction. This completes the proof. 

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8. THEOREM 2: CONSTRAINED NEIGHBORHOODS OF SPORADIC TYPE

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Proposition 7.8. Suppose that (CG (z), b, J) is a constrained {z, b}-neighborhood with respect to (B, T ), and J is as in (7C). Then (CG (z), b, J) is of U4 (2)-type and (d) of Theorem C5 : Stage 3 holds. Proof. By (7B) and Lemmas 7.3 and 7.4b, (CG (z), b, J) is nondegenerate. By Lemma 7.7 and Table 7.1, it is of U4 (2)-type, and the remainder of part (d) of Theorem C5 : Stage 3 holds, completing the proof.  8. Theorem 2: Constrained Neighborhoods of Sporadic Type To complete the analysis of constrained neighborhoods, we consider the cases in which J ∈ Spor. Thus, in view of Table 7.1 and Lemma 7.7, we assume throughout this section that ∼ Co1 , (3)Suz, (3)F i , F3 , F2 , or F1 . (8A) J= 24

We shall eliminate all possibilities for J except those – J ∼ = 3Suz and 3F i24 – occurring in target simple groups Co1 and F1 , respectively. We also make an initial determination of the structure of Cz for those two cases. Specifically, we prove: Proposition 8.1. Assume that (CG (z), b, J) is a constrained {z, b}-neighborhood and J is as in (8A). Then up to isomorphism, (J; T ; Cz /T ) is either + 1+24  (3Suz; 21+8 ; Co1 ). Moreover, (c) of Theorem C5 : Stage 3 + ; Ω8 (2)) or (3F i24 ; 2+ holds. Set C = Cz , C = C/O3 (C), M = O 2 (CJ (z)), and M0 = [M, M ]. Thus T = 1 as T = O2 (C). Also by Proposition 5.1, R = O2 (CJ (z)). In the respective cases of (8A), we have (8B)

1+6 − 1+12 , [(3) × 3]U4 (3)), (R, M ) =(21+8 + , D4 (2)), (2− , (3) × Ω6 (2)), (2+ 1+22 (21+8 , Co2 ), or (21+24 , Co1 ). + , A9 ), (2+ +

Then R ≤ M and so by Proposition 5.1, (8C)

F ∗ (M ) = R or R b ,

the latter only if b ∈ J. Now | Out(J)| ≤ 2, so B ≤ JC(b, J)) = J b, by Lemma 7.4c, and thus B ≤ O 2 (CJ (z)) b = M b. In particular, b ∈ M b. Moreover, b acts nontrivially on R with CR (b ) = 1, whereas every element of Z(M ) acts either  trivially or without fixed points on R, by [VK , 10.43]. Thus, b ∈ M b − Z(M b). Since M  CCb (z) = CC (b), M  CC (b), and so (8D)

M 0 is a component of CC (b).

Let d ∈ Des . By (8A), either (a) or (b) of Lemma 7.2 holds. Therefore if we set Md = O 2 (CJd (z)), M0,d = [Md , Md ], Rd = CT (d), and choose d ∈ D ∩ Jd with D = d, d , then Rd ∼ = R, so (B, T, d) is a maximal symplectic triple, and hence the above conclusions hold similarly, with d, b, Jd , J, Md , M0,d , Rd , and d in place of b, d, J, Jd , M, M0 , R, and b , respectively. We fix all this notation. Let (8E)

N = the subnormal closure of M 0 in C.

By (8D), N has the usual L3 -balance structure. Lemma 8.2. N is quasisimple and [N , b] = 1.

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Proof. Of course, m3 (N ) ≤ m2,3 (G) = m. In every case, for any covering group K of M 0 , we have m3 (K)−m3 (Z(K)) ≥ m−2. (This follows from [IA , 4.10.3] for M 0 ∼ = U4 (2) or D4 (2); from [IA , 3.3.3, 6.4.4a] for M 0 /Z(M 0 ) ∼ = U4 (3); and from [IA , 5.6.1] for M 0 ∈ Spor.) Hence N is quasisimple, for otherwise N has 3 such components K and so m2,3 (G) ≥ m3 (N ) ≥ 3(m−2) > m, a contradiction. Similarly the subnormal closure N d of M 0,d in C is quasisimple for any d ∈ Des . Fix such a d such that d = b.   We have B = (B ∩ M 0 ) b = (B ∩ M 0,d ) d , and so |B : B ∩ M 0 ∩ M 0,d | ≤ 32 . In particular B ∩ M 0 ∩ M 0,d does not centralize M 0 or M 0,d . Thus [M 0 , M 0,d ] = 1, whence N = N d . As [M 0,d , b] = 1 and M 0,d ≤ N d = N , we conclude that  [N , b] = 1, as claimed. Lemma 8.3. The following conditions hold: (a) (J; N ) = ((3)Suz; U5 (2), U6 (2), D4 (2), D5 (2), Sp8 (2), U4 (8), or Co2 ), (Co1 ; 2 D5 (2), 2E6 (2), or D4 (8)), (F i24 ; 3F i22 or Suz), (3F i24 ; 3Suz or Co1 ), or (F3 ; A12 or F5 ); (b) Either N = F ∗ (C) or (J; N ) = (F i24 ; Suz) and |CC (N )|3 = 3; and (c) J ∼ = F2 or F1 . Proof. We use the facts that M 0 ↑3 N ∈ K3 via b (Lemma 8.2) with m3 (N ) ≤ m and M as in (8B). Then (a) follows directly from [VK , 3.21], and in particular (c) holds.   To establish (b), first assume that M 0 is simple. Then CC ( b M 0 ) has b as Sylow 3-subgroup. This can be seen in Cb in view of Lemma 7.4c and  [VK , 3.22]. Now if CC (N ) = 1, then CC (N ) has a 3-element centralized by N b , and so by  b M 0 , so b ∈ CC (N ), contradicting Lemma 8.2. Thus (b) holds if M 0 is simple. Suppose then that M 0 is not simple. Then J ∼ = (3)F i24 . Using Lemma 7.4c 2 and (8B), we see that |CCb (z)|3 = 3 |U4 (3)|3 = 38 , so |CC (b)|3 = 38 . If also |CN (b)|3 = 38 , then CN (b) contains a Sylow 3-subgroup of CC (b). A b-invariant Sylow 3-subgroup P of CC (N ) thus satisfies CP (b) ≤ Z(N ). But b induces an inner automorphism on N (| Out(N )| ≤ 2) and hence on P . As |Z(N )| ≤ 3, we get P = CP (b) ≤ N . Thus CC (N ) = Z(N ) and F ∗ (C) = N in this case. It remains to consider the case |CN (b)|3 < 38 . As M 0 ≤ CN (b), the only possibility is J ∼ = F i24 , N ∼ = Suz, and O 2 (CN (b)) ∼ = 3U4 (3) with |CN (b)|3 = 37 . Again letting P ∈ Syl3 (CC (N )) be b-invariant, we have CN P (b) = CN (b)×CP (b) as N is simple. As |CN (b)|3 = 37 , CP (b) ∼ = Z3 . Since b induces an inner automorphism on N , it follows that |P | = 3, so again (b) holds, and the lemma is proved.  A closer look at N yields more precise information. Lemma 8.4. J ∼  F3 . = ∼ F3 and first consider the case N ∼ Proof. Suppose that J = = A12 . As M ∼ = A9 , b is represented by a 3-cycle in the standard representation of A12 . Similarly, considering M d , d is represented by a 3-cycle. As D = b, d, CN (D) contains an A6 component, and in particular CCz (D) is nonsolvable. However, JD ∼ = G2 (3) has only one class of involutions and CJD (z) is solvable by [IA , 4.5.1]. As JD = E(CJ (d)) is the unique nonsolvable composition factor of CJ (D) and hence of CG (D), CCz (D) must be solvable, a contradiction.

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∼ F5 . By (8B), R ∼ Therefore by Lemma 8.3, N = = 21+8 + , and similarly Rf = CT (f ) ∼ = R for any f ∈ Des . Now for f ∈ D − b, f ∈ Des if and only if 1 = [CT (f ), D] = [CT (f ), b], if and only if CT (f ) ≤ CT (b) (see Lemma 7.1). As a result, since there are just four subgroups of order 3 in D, T = CT (f ) | f ∈ Des  . In particular by [V2 , 6.3], Aut(T ) does not involve 51+4 . But embeds in 21+32 + N∼  = F5 has a 51+4 subgroup, contradiction. The lemma follows. Lemma 8.5. J ∼  Co1 . = Proof. Suppose that J ∼ = 3Suz and we let b0  = Z(JD ). = Co1 . Then JD ∼ In particular b0 is inverted by a 2-element u ∈ J as well as by a 2-element u1 ∈ Jd . Set U = u, u1 . Then U ≤ NG (D) ∩ NG (b0 ) with CD (u) = b and CD (u1 ) = d = b. Thus AutU (D) ∼ = Σ3 . Since Suz has one class of 2-central involutions [IA , 5.3o], we can choose u and u1 to centralize z, whence  U ≤ C. By the preceding paragraph, all subgroups of D of order 3 other than b0 are U -conjugate. Hence for any x ∈ E1 (D)−{b0 }, CN (x)  2 D5 (2), so has a component M 0,x ∼ = D4 (2). By [IA , 4.8.2], this implies that N ∼ = 2 ∼ N = D4 (8) or E6 (2) by Lemma 8.3. In particular, N has a cyclic subgroup of order 19. As Aut(T ) involves N , it follows that m2 (T / z) ≥ 18. On the other hand, R = CT (b) ∼ by (8B). = 21+8 +  ∼ Similarly Rx = CT (x) = R for all x ∈ D − b . Letting b0 , b, x1 , x2 be elements of the four subgroups of D of order 3, we have T = Rb0 Rb Rx1 Rx2 . Now if b0 ∈ Des , then Jb0 ∼ = Co1 , a contradiction as b0 ∈ JD ≤ Jb0 and Co1 is simple. Therefore b0 ∈ Des , so [CT (b0 ), b] = 1. Hence Rb0 = CT (b0 ) ≤ R, so T = Rb Rx1 Rx2 . However, as JD ∼ = 3Suz, it follows from [IA , 5.3o] that S := CT (D) ∼ = 21+6 − . Thus, m2 (T / z) ≤ 12, contrary to what we have shown above. The lemma follows.  A similar argument yields: Lemma 8.6. J ∼  Suz. = Proof. Suppose that J ∼ = 3U4 (3) with b0 ∈ JD ≤ Jd . By = Suz. Then JD ∼ [IA , 5.3o], b0 is inverted by involutions u ∈ J and u1 ∈ Jd centralizing z, and as in the preceding lemma, U = u, u1  induces Σ3 on D = b, b0 . Now U normalizes JD = E(CJ (b0 )) = E(CJd (b0 )), D centralizes JD , and U contains a 3-element acting nontrivially on D. As Out(JD ) is a 3 -group, D is normal, but not central, in some 3-group Q1 ≤ CG (JD ). By [VK , 9.16b], CJD (z) contains a subgroup V ∼ = E32 with b0 ∈ V . Thus V × Q1 ≤ C. Given the possible isomorphism types of N from Lemma 8.3, and the fact that m3 (N ) = m = 4, it follows from [VK , 9.20] that N ∼ = U4 (8); otherwise C (≤ Aut(N )) does not contain a subgroup isomorphic to V × Q1 . Therefore again 19 divides |N |, so m2 (T / z) ≥ 18. This time Rd ∼ for = 21+6 − 1+4 ∼ all d ∈ Des , and CR (D) = R ∩ JD = 2+ . As in the preceding lemma, the facts that Suz is simple and b0 ∈ Z(JD ) force CT (b0 ) ≤ R, and so m2 (T / z) ≤ 10, a contradiction. The proof is complete.  ∼ + Lemma 8.7. If J ∼ = 3Suz, then Cb = J, T ∼ = 21+8 + , and Cz /T = Ω8 (2). ∼ Z2 inverts Z(J) = b Proof. By Lemma 7.4c, F ∗ (Cb ) = J. As Out(J) = [IA , 5.3o], Cb = J. Now JD ∼ = [3 × 3]U4 (3) and since Cb = J, D = Z(JD ). Thus JD has four homomorphic images by a subgroup of D of order 3. By [IA , 5.1.5b], and as

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5. THEOREM C5 : STAGE 3

Out(JD ) ∼ = D8 acts faithfully on Z(JD ) [IA , 6.3.1], these fall into two isomorphism classes, each arising from two subgroups of D. Precisely one of these classes has the isomorphism type of the centralizer of an element of order 3 in Suz. Therefore if x ∈ D − b − b , Jx = JD , whence CT (x) ≤ R = CT (b). It follows that T = RCT (b ), with CT (b ) ∼ = R ∼ = 21+6 − . Moreover, m2,3 (CJ (z)/R) = 2 < m3 (B/CB (J)), so ∗ ICb,z (B; 2) = {R}. Therefore O2 (CJD (z)) ≤ R, whence CT (D) = CR (D) = ∼ 1+8 O2 (CJD (z)) ∼ = 21+4 + . We conclude that T = 2+ . In particular, C/T embeds in Out(T ) ∼ = O8+ (2). Thus N embeds in Out(T )(∞) ∼ = Ω+ 8 (2). Given the possibilities for N in Lemma 8.3, + + we conclude by order considerations that N ∼ = Ω+ 8 (2). Since |O8 (2) : Ω8 (2)| = 2, it + + ∼ follows that O3 (C) = T and C/T = Ω8 (2) or O8 (2). But as R = CT (b) ∼ = 21+6 − , we 6 see with the help of [IA , 4.8.2] that |CC (b)|2 = |R|2 c, where c = |C|/|T ||Ω+ (2)|. As 8 (2), completing the proof.  Cb = J ∼ = 3Suz, |Cb | = 213 , so c = 1 and C/T ∼ = Ω+ 8 The following lemma will complete the analysis of the cases in (8A). We also get useful information for the next chapter about an involution t ∈ z G . Lemma 8.8. If J ∼ = (3)F i24 , then the following conditions hold: ∼ 3F i ; (a) Cb = J = 24 1+24 ∼ Co1 ; and ∼2 and C/T = (b) T = + (c) There is t ∈ I2 (CJ (b )) such that I := E(CJ (t)) ∼ = 2F i22 , CG (t) ∼ = 2F2 , and z is 2-central in both I and CG (t). Proof. Note first that if J ∼ = 3F i24 , then Out(J) ∼ = Z2 inverts Z(J) [IA , 5.3v], so by Lemma 7.4c, Cb = J and (a) holds. ∼ ∼ 1+12 and m = 6. Also In any case, JD ∼ = P Ω+ 8 (3), M = [(3) × 3]U4 (3), R = 2+ N ∼ = 3F i22 or Suz if J ∼ = F i24 , while N ∼ = 3Suz or Co1 if J ∼ = 3F i24 , by Lemma 8.3. By [VK , 12.20], there is t ∈ I2 (CJ (b )) such that (1) I := E(CJ (t)) ∼ = 2F i22 ; (2) t ∈ I0 := E(CI (b )) ∼ = 2U4 (3); and (8F) (3) z is 2-central in I. Let d ∈ Des −b. We have I0 ≤ CJ (b )(∞) = JD ≤ Jd . Let I1 be the subnormal closure of I0 in CCd (t) and hence in CJd (t). Since Jd ∼ = (3)F i24 , but no involution ∼ u ∈ Aut(Jd ) satisfies E(CJd (u)) = 2U4 (3) [IA , 5.3v], b acts nontrivially on I1 . t = Ct /O3 (Ct ). Then I ∼ Let Ct = CG (t) and C = F i22 is a component of   = 1. As I1 ≤ Ct CCt (b). Since m = 6 and m3 (F i22 ) = 5 [IA , 5.6.1], m3 (CCt (I)) t . Since m3 (Ct ) ≤ 6, the and b acts nontrivially on I1 , I is not a component of C  and subnormal closure K of I in Ct is therefore a single 3-component with I < K ∼    I   CK (b). By [VK , 3.8], we must have K = F2 , and then as m3 (K) = 6 = m t = K.   = 1 by [IA , 5.6.1, 5.3y], C and Out(K) Next, set X = O3 (Ct ) and let S ∈ Syl2 (X) be D-invariant. Since [I, S] ≤ X, a 3 -group, it follows from [VK , 7.5] that [CS (b), I] = 1 and |CS (b)/ t | ≤ 2. Indeed, by the same reasoning, for any x ∈ D# such that Jx ∼ = (3)F i24 , [CS (x), I] = 1 and |CS (x)/ t | ≤ 2. We claim that [D, S] = 1. Otherwise, S0 := [CS (y), b] = 1 for some y ∈ D# (as [CS (b), D] ≤ [CS (b), I] = 1), so JD is a component of CG (y). Since B normalizes

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every component of CG (y) by [IG , 8.7], and [S0 , b] = S0 with b centralizing JD , it follows that [S0 , JD ] = 1. Thus JD b normalizes S0 , so m ≥ m3 (JD b) = 7 [IA , 3.3.3], a contradiction, proving our claim. Thus S = CS (b), so |S/ t | ≤ 2, whence X has a normal 2-complement. But O2 (Ct ) = 1, so X = S and Ct /S ∼ = F2 . Furthermore, t ∈ I0 ≤ [Ct , Ct ], so E(Ct ) ∼ = 2F2 , and Ct = E(Ct )S. By (8F3), z is 2-central in I, hence in Ct , by [VK , 12.29]. By [IA , 5.3], CE(Ct ) (z) by Co2 . In particular, C involves Co2 . But by order is an extension of 21+22 + considerations, neither 3F i22 nor Suz involves Co2 , and so N ∼ = Co1 , which in turn implies that J ∼ = 3F i24 and M ∼ = [3 × 3]U4 (3). Therefore (a) holds. By Lemma 8.3b, F ∗ (C) = N , so as Out(Co1 ) = 1, C ∼ = Co1 , which is the second assertion of (b). Also as Cb ∼ = 3F i24 , Z(CCb (t)) ∼ = Z6 [IA , 5.3v]. Since S ≤ Z(CCb (t)), S = t and so Ct ∼ = 2F2 , proving (c). Thus to complete the proof, it remains only to verify that O3 (C) = T ∼ = 21+24 . +

z = Cz /T . By [VK , 5.10], m2 (T / z) ≥ As T ∈ I∗G (B; 2), T ∈ Syl2 (O3 (Cz )). Set C 24. Moreover, if equality holds, then O3 (Cz ) = T ∼ = 21+24 , as desired. For otherwise, as m2 (Co1 ) = 11 and the Schur multiplier of Co1 is a 2-group [IA , 6.1.4], z ))|, ms (O3 (C z )) ≥ 11 by the Thompson for some prime divisor s ≥ 5 of |F (O3 (C  dihedral lemma, which is impossible since Cz acts faithfully on T / z. Thus it remains to show that m2 (T / z) ≤ 24 and T is of + type. Now, , and similarly Rx ∼ R∼ = 21+12 = R for every x ∈ Des , while Rx ≤ R for every x ∈ + D − Des . Since J ∼ = 3F i24 , we have, as in the preceding lemma, m2,3 (CJ (z)/R) = 4 < m3 (B/CB (J)), leading to CT (D) = CR (D) ∼ = 21+8 + , by [VK , 10.24]. It follows  that m2 (T / z) ≤ 8 + 4 · 4 = 24, and the lemma is proved. Lemmas 8.4–8.8 complete the proof of Proposition 8.1. Then Propositions 5.1, 6.1, 7.8, and 8.1 complete the proof of Theorem 2, and with it the proof of Theorem C5 : Stage 3.

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10.1090/surv/040.9/06

CHAPTER 6

Theorem C5 : Stage 4 1. Introduction In this chapter we complete the proof of Theorem C5 . The various alternative conclusions of Stage 3 are used, one by one, to prove: Theorem C5 : Stage 4. G is isomorphic to one of the following groups: − (a) Ω7 (3), P Ω+ 8 (3), or Ω8 (3);  (b) Co1 , F i22 , F i23 , F i24 , F2 , or F1 ; or (c) 2 D5 (2), U7 (2), or 2E6 (2). If conclusion (b) or (c) of Theorem C5 : Stage 3 holds, we consider the (sporadic) target groups G∗ one at a time. If conclusion (a) or (d) of Theorem C5 : Stage 3 holds, then there are three target groups in each case, which we initially consider together, before eventually separating the analysis. 2. The Case G∗ = Ω7 (3) or Ω± 8 (3) In Stages 2 and 3 of Theorem C5 we made the technical assumption BtK3exc (G) = ∅, assuming away certain components of involution centralizers isomorphic to G2 (3), L± 4 (3), and 2U4 (3). We analyze these exceptional cases in this section, ruling out the first one but obtaining the three possibilities G ∼ = Ω7 (3) and P Ω± 8 (3) as our reward. Our treatment of the last two cases owes a great deal to Aschbacher’s work on a similar but broader question [A23], [A9]. We first prove  G2 (3). Proposition 2.1. Let (B, x, K) ∈ BtK3 (G). Then K ∼ = We assume false, and let (2A)

(B, x, K) ∈ BtK3 (G), K ∼ = G2 (3), Q ∈ Syl2 (C(x, K)), Q ≤ T ∈ Syl2 (CG (x)).

We write R = T ∩ K, Z(R) = z and CK (z) = (K1 ∗ K2 ) u , ∼ where K1 = K2 ∼ = K2 u ∼ = GL2 (3) [IA , 4.5.1]. Set = SL2 (3) and K1 u ∼ (2B)

(2C)

Cz = CG (z), Cx = CG (x),

Ri = O2 (Ki ) = T ∩ Ki , i = 1, 2, and R0 = R1 R2 . ∼ Thus R0 = Q8 ∗ Q8 . We first prove:

Lemma 2.2. Assume (2A) and (2B). Then the following conditions hold: (a) Q = x; (b) x ∈ z G ; 237 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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238

(c) (d) (e) (f) (g)

x is not 2-central in G; xz ∈ xCz ; x ∈ [Cz , Cz ]; Cz has no normal Z4 -subgroup; and m2,3 (G) = 4.

Proof. By [V2 , 9.1], m2 (Q) = 1. Note that | Out(K)| = 2 [IA , 2.5.12]. Hence T /R has the rank 1 subgroup QR/R ∼ = Q of index 2, so T /R has sectional rank at most 3. In particular, as x ∈ R, x lies in the center of no Q8 ∗ Q8 subgroup of T , and hence of G. On the other hand, z ∈ R0 ∼ = Q8 ∗ Q8 , so (b) follows. Suppose that T ∈ Syl2 (G). By (b), NG (T ) centralizes Ω1 (Z(T )) = x, z, so by Burnside’s theorem, x, z, and xz are mutually non-G-conjugate. By the Z ∗ theorem, x ∈ uG for some x = u ∈ I2 (T ). As K has one class of involutions by [IA , 4.5.1], it follows by [IA , 2.5.12, 4.9.1] that CG (x) = (K × Q) u with 1 CK (u) ∼ = 2 G2 (3 2 ), and K1u = K2 . We may assume that z ∈ U := CT (u) ∈ Syl2 (CG (u, x)). Now ug = x for some g ∈ G, and we may take g so that U g ≤ T . Since m2 (T /R) = m2 (Q u) = 2, (U ∩ R)g ∩ R = 1. Adjusting g by an element of K we may assume that z g = z. Thus u ∈ xCz . In particular, u, like x, is 2-central in Cz . However, as u interchanges K1 and K2 , it follows that O2 (K1 K2 ) ≤ O2 (Cz ). Hence by [III11 , 13.1], Ki lies in a component Ji of Cz , i = 1, 2, and Ji ∼ = 2An u for some n, Ji ∼ or J ∈ Chev(3) with q(J ) = 3. Clearly J = J M = 11 i i 2 , but 1 then as u is 2-central in Cz , J1 = J2 . As 2An ∈ C2 and |M11 |2 < |Q8 ∗ Q8 |, J1 ∈ Chev(3). By [III11 , 10.2], K1 ≤ J1 , so K1 is a solvable 2-component of CJ1 (x), with Z(K1 ) = z ≤ Z(J1 ). As J1 ∈ C2 , this contradicts [VK , 3.109]. Hence (c) holds. Let S ∈ Syl2 (NG (T )), so that S > T by (c), and choose h ∈ S − T . As Z(T ) = x, z, Qh  T with Ω1 (Qh ) = xz. Therefore Qh ∩ R = Qh ∩ Q = 1, so Qh ≤ CT (RQ) ≤ z Z(Q) with Φ(Qh )∩Q = 1. But Φ(Qh ) ≤ Z(Q). Thus, Φ(Q) = 1 and (a) follows. Since (B, x, K) ∈ BtK3 (G) and m3 (Aut(K)) = m3 (K) = 4, this implies (g). Next, z = Z(T ) ∩ [T, T ]  S, so xS = {x, xz} and (d) holds. As z is then weakly closed in Z(T ) = z, x, (e) is immediate from [III8 , 6.3]. Finally, CG (z, x) has no normal Z4 subgroup, so if A is a normal Z4 -subgroup of Cz , then [A, x] = 1, whence x ∈ [Cz , Cz ], contradicting (e). Thus (f) holds and the proof is complete.  The next lemma will turn out to be the key. Lemma 2.3. I2 (R0 x) ⊆ xCz . Proof. By Lemma 2.2d, I2 (R0 x) ⊆ xG as K has one class of involutions. So choose any g ∈ G and u ∈ I2 (R0 x) with ug = x; we wish to replace g by an element of gCx to achieve g ∈ Cz . We have CR0 (u)g ≤ Cx . Now u = yx with y ∈ R0 and y 2 = 1, so CR0 (u) contains a D8 -subgroup D with z ∈ [D, D]. Thus z g ∈ [Cx , Cx ] ≤ K. As K has one class of involutions, z gh = z for some h ∈ K, as required.  Lemma 2.4. R0 ≤ O2 (Cz ). Proof. We must show that Ri ≤ O2 (Cz ), i = 1, 2. Suppose that this fails, say R1 ≤ O2 (Cz ).

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By [III11 , 13.1], and as G is of even type, R1 ≤ J for some component J ∈ C2 of Cz , and (as z ∈ Z(R1 ) ≤ Z(J)) J ∈ Chev(3) with q(J) = 3, or J ∈ Chev(2). In the first case J1 ≤ J by [III11 , 10.2], contradicting [VK , 3.109]. So J ∈ Chev(2), with z ∈ Z(J). Again as J ∈ C2 , J ∼ = 2U4 (2) ∼ = Sp4 (3) or 2L4 (2) ∼ = 2A8 . Note that C := CG (x, z) = CCG (x) (z) has order 2a 32 with a = 7 or 8. Also m2 (C) = 4 or 5, and O2 (C) = R0 × x is of 2-rank 4. Moreover, for W ∈ Syl3 (C), W ∼ = E32 and AutC (W ) ∼ = Z2 or E22 according as a = 7 or 8. As W normalizes R1 , it normalizes J. Finally, x, z is a maximal W -invariant abelian subgroup of C, and R0 x ∈ IC (W ; 2). Suppose first that J/Z(J) ∼ = L3 (4), so that Z(J) is elementary abelian since J ∈ C2 . If x maps into Inn(J), then CJ (x) is a 2-group of 2-rank at least 5 by [IA , 6.4.1], so CJ (x) ≤ O2 (C), contradicting the previous paragraph. Since CJ (x) is solvable, the only possibility is CJ/Z(J) (x) ∼ = U3 (2). But then AutC (W ) contains Q8 , contradicting the previous paragraph. Thus, J/Z(J) ∼  L3 (4). Suppose next = that J ∼ = 2G2 (4). Since | Out(J)| = 2, W induces inner automorphisms on J. CAut(K) (x) is a 3 As C is solvable, CK (x) is a 3 -group by [VK , 10.73].  C Hence z , W ] = 1. Hence R0 = group, and it follows [J, W ] = 1. Similarly [ J  Cthat z [R0 , W ] centralizes J and in particular J. This contradicts R1 ≤ J, however, 3 so J/Z(J) ∼  G2 (4). Next, if J/Z(J) ∼ = 2B2 (2 2 ), then CJ (x) has a characteristic = elementary abelian subgroup of rank 4 or 5, which is then W -invariant, again contradicting what we saw above. Therefore by [IA , 6.1.4], J/Z(J) ∼ = 2 E6 (2), F4 (2), D4 (2), U6 (2), or Sp6 (2). As m2,3 (G) = 4 by Lemma 2.2g, while m3 (2E6 (2)) > 4, 22E6 (2) is impossible. Let x ∈ S ∈ Syl2 (Cz ) with CS (x) ∈ Syl2 (C), and set Cz = Cz / z. If J/Z(J) ∼ = U6 (2) then S1 := J(S ∩J) ∼ = E2b with b = 10 or 11. So m2 (C) ≥ m2 (CS1 x (x)) ≥ 6, using [IG , 9.16], a contradiction. If J/Z(J) ∼ = F4 (2), then since C is solvable, x induces an inner automorphism on J [IA , 4.9.1]. Let P be a parabolic subgroup of J of type Sp6 (2), and S0 = O2 (P ) and L a Levi subgroup of P . Then [L, S 0 ] = S 0 and Φ(S 0 ) ∼ = Z2 by [VK , 10.46]. Thus S 0 /Φ(S 0 ) ∼ = E214 . Let Sx = CS0 x (x). It follows that |Sx | ≥ 28 . Hence equality holds, Sx ∈ Syl2 (CG (x)), and x ∈ S0 . Therefore Sx = x × (Sx ∩ S0 ) and Φ(Sx ) = Φ(Sx ∩ S0 ) ≤ Φ(S0 ), so |Φ(Sx )| ≤ 22 . However, |[R, R]| ≥ 23 , a contradiction. Hence, J ∼  F4 (2). =  Next, suppose that J/Z(J) ∼ = D4 (2). Let J ∗ = J Cz . If J ∗ > J, then S ∩ J ∗ has a x-invariant section which is the direct product of at least two copies of 21+8 + , and hence |CS∩J ∗ (x)| ≥ 29 > |CG (z, x)|2 , contradiction. Therefore J  Cz . Since m2,3 (G) = 4 = m3 (J), Z(J) is cyclic by (c) of Theorem C5 : Stage 2. Therefore, Z(J) ∼ = Z2 . As x ∈ [Cz , Cz ], x ∈ JCG (J). Since z, x is a maximal W -invariant abelian 2-subgroup of C, x ∈ J. Using [VK , 10.36], we conclude that a Sylow E ∼ 2-subgroup = E25 . Then E ∩ R0 > z, so  W of CJ (x) has a normal subgroup 3 R0 ≤ E ≤ CJ (x). Since |Z(J)| = 2, O (AutCz (J)) = Inn(J). As m3 (C) = 2, it follows that W ≤ J, so R0 W ≤ CJ (x). By the Borel-Tits theorem, CJ (x) ≤ P for some maximal parabolic subgroup P of J. Suppose that P is of type A31 . Then and P/U ∼ U := O2 (P ) satisfies U/ z ∼ = 21+8 = Σ3 ×Σ3 ×Σ3 . Then R0 = [R0 , W ] ≤ + U . If x ∈ U , then |CU (x)| ≥ |U/Φ(U )| = 28 ; but CU (x) ≤ O2 (CJ (x)), which has

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order 26 , contradiction. Thus x ∈ U . As [W, x] = 1, CP (x) ≤ P2 ≤ P1 , where P1 and P2 are parabolic subgroups of types A3 and A21 . Thus, whether or not P is of type A31 , CJ (x) lies in a parabolic subgroup P1 of type A3 . Let U1 = O2 (P1 ), so 2 that U1 / z is a natural module for P1 /U1 ∼ = Ω+ 6 (3). Then x ∈ O (CP1 (W )) ≤ U1 , and U1 is either elementary abelian or extraspecial. In the first case m2 (CJ (x)) ≥ 7, contrary to m2 (C) ≤ 5. So U1 ∼ = 21+6 + . As [W, x] = 1, the involution x maps to a 2-central involution of P1 / z, and |CP1 (x)|2 ≥ |P1 / z |2 = 212 , a contradiction. Hence, J/Z(J) ∼  D4 (2). = As z ∈ J, the only other possibility is that J ∼ = 2Sp6 (2). As C is solvable and Out(J) = 1 [IA , 2.5.12], x ∈ NCz (J) = JCG (J). Using [IA , Table 5.3j], we see that S ∩ J has a normal E24 -subgroup, whence m2 (CJ (x)) ≥ 2, with strict inequality unless possibly x ∈ J. If R2 ≤ J, then [R2 , J] = 1 and so CJ (x) ≤ CG (x × R2 ) = x × K1 ∼ = Z2 × SL2 (3). Whether x ∈ J or not, this contradicts our inequality on m2 (CJ (x)). Thus, R0 ≤ J. Write x = x1 x2 with x1 ∈ CJ (R0 ), x2 ∈ CG (J), and x21 = x22 ∈ z. Then x1 , x2  ≤ CG (x × R0 ) = z, x, whence we may assume x = x1 ∈ J and then CG (J) = z and Cz = J. By the Borel-Tits theorem, there is a parabolic subgroup P of J with C ≤ P and x × R0 ≤ O2 (P ). As |P |3 ≥ 32 , P is of type Sp4 (2) or L2 (2) × L2 (2). Accordingly, by [VK , 10.45], x ∈ CP (W ) is a long or short root involution. In either case, x is 2-central in J, so |C|2 = |CJ (x)|2 ≥ 29 , a final contradiction completing the proof of the lemma.  We continue to set Cz = Cz / z, let S ∈ Syl2 (Cz ), and set S0 = O2 (Cz ) and  Cz = Cz /S0 . Thus R0 ≤ S0 . We continue to let W ∈ Syl3 (K1 K2 ). We can now prove Lemma 2.5. S0 is of symplectic type. Proof. Otherwise, by P. Hall’s theorem [IG , 10.3], S0 has a noncyclic elementary abelian characteristic subgroup A. Then CA (x) is a normal elementary abelian subgroup of CG (z, x) containing z, so CA (x) = z or z, x. In the latter case, A = z, x. By Lemma 2.3, I2 (R0 x) ⊆ xCz ⊆ z, x, which is absurd. Therefore CA (x) = z, whence A is a four-group and x ∈ [Cz , Cz ], contradicting Lemma 2.2e.  Lemma 2.6. S0 is extraspecial and contains R0 . Proof. Let W1 = W ∩ K1 . By the A × B-lemma, [W1 , CS0 (x)] = 1. But [W1 , CS0 (x)] ≤ [W1 , O2 (CG (z, x))] = R1 , so equality holds and R1 ≤ S0 . Similarly R2 ≤ S0 , so R0 ≤ S0 . Finally if S0 were not extraspecial, it would have a characteristic subgroup A ∼ = Z4 , contradicting Lemma 2.2f. The lemma follows.  Lemma 2.7. The following conditions hold: (a) x ∈ S0 ∼ = D8 ∗ D8 ∗ D8 ; and z ) ∼ (b) F ∗ (C = A6 . Proof. Since S0 is extraspecial, R0  S0 and S0 = R0 ∗ S1 , where S1 = CS0 (R0 ) is W x-invariant, and either extraspecial or of order 2. Suppose that SE := S ∩ E(Cz ) = 1. Then z = Z(S) ≤ SE . Also CS1 SE (x) ≤ CCx (R0 ) = z, x, so S1 E(Cz ) x has Sylow 2-subgroups of maximal class. As z ∈ SE , SE must be quaternion or semidihedral, contradicting the fact that all components of E(Cz )

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lie in C2 [VK , 10.30]. Thus E(Cz ) = 1, whence F ∗ (Cz ) = S0 . Then since x ∈ CG (R0 ) − R0 , S1 is extraspecial. Indeed as S1 x is of maximal class, S1 ∼ = D8  or Q8 . Now if x ∈ S1 , then x induces a transvection on S 0 , and since Cz embeds in Out(S0 ) ∼ = O6± (2), x ∈ [Cz , Cz ], contradicting Lemma 2.2e. Therefore x ∈ S1 , ∼ whence S1 = D8 and (a) holds. z ) = 1. As W ∈ Syl3 (Cz ) normalz embeds in Out(S0 ) ∼ Now C = Σ8 , and O2 (C z ) = 1. So F ∗ (C z ) = E(C z )O3 (C z ). izes no nontrivial {5, 7}-subgroup of Σ8 , O3 (C Since all involutions in R0 x are Cz -conjugate by Lemma 2.3, and x ∈ CS0 (W ) ≤ z )), O3 (C z ) centralizes R0 x and hence is trivial. Thus F ∗ (C z ) = E(C z ), CS0 (O3 (C ∗  ∼ which, as a subgroup of A8 , is readily seen to be simple. As W ≤ Cz , F (Cz ) = An , 6 ≤ n ≤ 8. However, if n = 7 or 8, then as elements of order 5 or 7 in Aut(S0 ) fix no noncentral involutions, all 70 noncentral involutions in S0 are Cz -conjugate. However, z G ∩ R0 ⊆ z and so t ∈ z G , a contradiction. Therefore n = 6, proving (b).  We can now quickly reach a contradiction completing the proof of Proposition 2.1. We have S0 = R0 ∗ S1 and let A be the Z4 -subgroup of S1 ∼ = D8 . Since x ∈ S1 , S1 = CS0 (W ), and therefore A is the unique Z4 -subgroup of CS0 (W ). But then z ) normalizes A. Hence A  Cz , which contradicts Lemma with [VK , 5.15], E(C 2.2f. This completes the proof of Proposition 2.1. Next, at much greater length, we consider the following situation. (1) (t, K) ∈ ILo2 (G) with K ∼ = L± 4 (3) or 2U4 (3) and m2,3 (G) = m3 (CG (t)); (2) Ct = CG (t), T ∈ Syl2 (Ct ), Q = CT (K) ∈ Syl2 (C(t, K)), and R = T ∩ K ∈ Syl2 (K); (3) Z = Z(R) = z × Z(K) with | z | = 2, O 2 (CK (z)) =: (2D) K0 = K1 ∗ K2 , K1 ∼ = K2 ∼ = SL2 (3), z ∈ K1 ∩ K2 , and Cz = CG (z); and (4) Ri = O2 (Ki ) = R ∩ Ki , i = 1, 2, R0 = R1 R2 , W ∈ Syl3 (K1 K2 ), Wi = W ∩ Ki , i = 1, 2. Note that (2D1) is equivalent, given Proposition 2.1, to the statement that (B, t, K) ∈ BtK3exc (G) for some B ∈ B3∗ (G). Moreover, given (2D1), most of (2D2 − 4) is notation; and the rest follows from [VK , 10.2, 10.1]. We shall prove: Proposition 2.8. If (2D) holds, then G ∼ = Ω7 (3) or P Ω± (3). 8

Through Lemma 2.89, we assume that (2D) holds. We first prove some general lemmas, then deal with the local structure of G, in the case T ∈ Syl2 (G) from Proposition 2.12 through Lemma 2.21, and then in the case T ∈ Syl2 (G) through Lemma 2.77. Then we construct G0 ≤ G of the desired isomorphism type and prove that G0 = G, ending with Lemma 2.89. Lemma 2.9. We have (a) m2 (Q) = 1; (b) z ∈ tG ; (c) For any 3-subgroup P ∗ ≤ Ct , Φ(P ∗ ) lies in an element of B3∗ (G); and (d) m2,3 (G) ≤ 5. Proof. Part (a) holds by [V2 , 9.1].

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For (b), suppose that t = z h for some h ∈ G. Note that Out(K) is a 2group, so (K1 K2 )h ≤ O 2 (Ct ) ≤ KC(t, K). If (K1 K2 )h ∩ K ≤ t, it follows that (K1 K2 )h / t ∼ = A4 × A4 is involved in C(t, K), which is absurd in view of (a). Thus, (K1 K2 )h ∩ K > t. Hence K contains a subgroup H ∼ = Q8 with t ∈ H. In particular t ∈ K so K ∼ = U4 (3) has a single class of = 2U4 (3). Then K/ t ∼ involutions and a non-quaternion Sylow 2-subgroup, so all involutions of K/ t split over t. This contradicts the existence of H, and (b) follows. Finally, a Sylow 3-subgroup P0 of Ct has the form (P0 ∩ K) × CP0 (K t), and |CP0 (K t)| ≤ 3 since Q = F ∗ (CG (K t)) has 2-rank 1. (Note that CG (K t) has no components; any such components would have to be in C2 and hence would have 2-rank at least 2.) Now, there is E34 ∼ = P1  P0 ∩ K such that P0 /P1 is elementary abelian. As m2,3 (G) = m3 (CG (t)) by (2D1), (c) follows, as does (d),  since m3 (Aut(K)) = 4. The following set will figure prominently in our analysis. Definition 2.10. V1 = {v ∈ I2 (CT (R0 )) | E(CK (v)) ∼ = P Sp4 (3)}. Then V1 has the following properties.

(2E)

(1) For any v ∈ V1 , v induces a graph automorphism on K; (2) I2 (CT (R0 )) ⊆ V1 ∪ z, t# , m2 (CT (R0 )) ≤ 3, and CAut(K) (R0 ) ∼ = D8 or E22 according as K/Z(K) ∼ = U4 (3) or L4 (3); (3) For any v ∈ V1 , CAut(K) (v) ∼ = Z2 × Aut(P Sp4 (3)) maps onto the subgroup P O6± (3)/ Inn(K) of Out(K) of index 2; (4) For any v ∈ V1 , v ∼CK (z) vz if and only if K ∼ = U4 (3); moreover, if K ∼ = 2U4 (3), then v ∼CK (z) vtz; (5) If K ∼ = L4 (3), and v ∈ V1 , then AutCt (K) = Aut(K) if and only if v is T -conjugate to vz; and  (6) If w ∈ I2 (Ct ) and O 3 (CK (w)) ∼ = A4 × A4 , then wK ∩ V1 = ∅.

Assertions (2E1, 3, 6) follow from [IA , 4.5.1]; (2E2) from [VK , 10.2b, 10.1b] and the fact that m2 (Q) = 1; (2E4) from [VK , 10.1f, 10.2c]; and (2E5) from [VK , 10.2c]. Lemma 2.11. The following conditions hold: (a) z is 2-central in G; (b) If v ∈ V1 or v = 1, then z is weakly closed in z, t, v with respect to G; (c) t ∈ [Cz , Cz ]; and (d) V1 ⊆ [Cz , Cz ]. Proof. By [VK , 10.2b, 10.1a], and Lemma 2.9a, Ω1 (Z(T )) = z, t, v where ∼ L4 (3)). Therefore (b) implies v = 1 or v ∈ V1 (and the latter can occur only if K = that z is weakly closed in Z(T ) with respect to G. Hence (b) immediately implies (a), and (c) as well in view of [III8 , 6.3]. Let us prove (b). Suppose first that z g = y ∈ Z(T ) − z, t for some g ∈ G. By (2E2), y ∈ V1 , so E(CG (y, t)) ∼ = P Sp4 (3). By L2 -balance and as G is of even type, g −1 if we put u = t , then CE(Cz ) (u) has a component L ∼ = P Sp4 (3). Let   L∗ = LE(Cz ) ,

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a product of components permuted transitively by u. As components of E(Cz ) lie in C2 , L∗ ∼ = P Sp4 (3), L± 4 (3), 2U4 (3), or L4 (4), by [VK , 3.10]. (Note that ∗ ∼ L = P Sp4 (3) × P Sp4 (3) is impossible as m2,3 (G) ≤ 5 by Lemma 2.9d.) Also by solvable L2 -balance, either R0 ≤ O2 (Cz ), or R1 ≤ J1 for some component J1 of Cz . If R1 ≤ J1 , then by [III11 , 13.1] and the facts that J1 ∈ C2 and Z(R0 ) = z ≤ Z(Cz ), z ∈ Z(J1 ) and J1 ∈ Chev(2) ∪ Chev(3), with q(J1 ) = 3 if J1 ∈ Chev(3). In the latter case by [III11 , 10.2], K1 ≤ J1 . But as J1 ∈ C2 , this is impossible by [VK , 3.109]. Thus J1 ∈ Chev(2). Hence, either R1 ≤ O2 (Cz ) or R1 ≤ J1 ∈ Chev(2). Similarly, R2 ≤ J2   Cz with z ∈ J2 ∈ Chev(2), or else R2 ≤ O2 (Cz ). Suppose that [L∗ , R0 ] = 1. In this case if L∗ /Z(L∗ ) ∼ = L± 4 (3), then W must ∗ project faithfully on C(z, L ) and so m2,3 (G) ≥ 6, contradicting Lemma 2.9d. Thus L∗ ∼ = P Sp4 (3) or L4 (4). Hence by [VK , 10.72], m2 (t CL∗ (t)) ≥ 4. However, t CL∗ (t) ≤ CCt (R0 ), so m2 (CCt (R0 )) ≥ 4, contradicting (2E2). Hence, [L∗ , R0 ] = 1. In particular R0 ≤ O2 (Cz ), and so L∗ = J1 above. Therefore L∗ = J1 ∈ Chev(2) ∩ C2 with z ∈ Z(L∗ ), a contradiction. We have proved that z G ∩ Z(T ) ⊆ z, t. Now if z is not weakly closed in Z(T ), then by Lemma 2.9b, z G ∩ Z(T ) = {z, zt}. If T ∈ Syl2 (G), then by Burnside’s lemma, z NG (T ) = {z, zt}, so |NG (T ) : T | is even, contradiction. Thus T ∈ Syl2 (G). Let T < T ∗ ∈ Syl2 (G) and let g ∈ NT ∗ (T ) − T with z g = zt. Then tg ∈ Z(T ) − z, t. Hence (zt)g ∈ z G with (zt)g = z g tg ∈ Z(T ) − z, t, contradicting the previous paragraph. This completes the proof of (b). Finally to prove (d), let v ∈ V1 and expand CT (v) ≤ Tv ∈ Syl2 (CG (v)). Then as R0 ≤ CT (v) and m2 (CT (R0 )) ≤ 3, Ω1 (Z(Tv )) ≤ Ω1 (Z(CT (v))) = z, t, v. By (b), z is weakly closed in Z(CT (v)), and then also in Z(Tv ). Now (d) follows from [III8 , 6.3].  Now, through Lemma 2.21, we assume that T ∈ Syl2 (G), and we shall prove Proposition 2.12. Assume that (t, K) satisfies (2D1) with t 2-central in G. Assume that (t, K) has been chosen, subject to this condition, if possible, so that K/Z(K) ∼ = U4 (3). Then the following conditions hold: (a) F ∗ (Cz ) ≤ O 2 (Cz ) = L × K0 with K0 = O 2 (CK (z)) ∼ = SL2 (3) ∗ SL2 (3); (b) One of the following holds: (1) L ∼ = A4 , with |Q| ≤ 4; or (2) L ∼ = A6 ; (c) K ∼ = 2U4 (3); and (d) tG ∩ V1 = ∅. To prove the proposition, we let S0 = O2 (Cz ) and first prove: Lemma 2.13. The following conditions hold: (a) R0 ≤ S0 ; (b) z G ∩ V1 = ∅; and

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(c) If t ∈ S0 and V1 = ∅, then z G ∩ Ct = z K , and moreover, for any z1 ∈ z G and t1 ∈ tG such that [z1 , t1 ] = 1, the three involutions of z1 , t1  are pairwise non-G-conjugate, and z1 t1 ∈ (zt)G . Proof. Using the fact that t is 2-central in G and hence in Cz , and using  solvable L2 -balance [IG , 13.8], we see that the subgroups Ri = O 3 (Ki ), i = 1, 2, of CG (t, z), which contain z, lie in S0 , thanks to [VK , 10.66]. Thus (a) holds. Now CT (S0 ) ∈ Syl2 (CCz (S0 )) and m2 (CT (S0 )) ≤ m2 (CKQ (S0 )) + m2 (Out(K)) = 2 + m2 (Out(K)) = 4, so E(Cz ) has no product of components containing a copy of P Sp4 (3), which has 2-rank 4. Then with L2 -balance, (b) follows. Suppose that V1 = ∅ and t ∈ S0 . Then by (2E2), Ω1 (Z(T )) = z, t. As t ∈ z G , the three involutions of z, t are pairwise non-G-conjugate, by Burnside’s lemma. It therefore suffices to prove that z G ∩ Ct = z K . For then if z1 and t1 are as in (c), we may assume that t1 = t, whence z1 ∈ z K and z1 , t1  is conjugate to z, t, yielding the second statement of (c). Note that Ω1 (Z(S0 )) = z, t by (a) and (2E2), so Cz = CG (z, t) = CCt (z). In particular, Cz is solvable.  Finally let z1 = z g ∈ z G ∩ Ct , g ∈ G, and let I = O 3 (CK (z1 )). The desired conclusion is that I ∼ = SL2 (3) ∗ SL2 (3), for then z1 ∈ z K . As CG (z1 ) ∼ = Cz is solvable, the only alternative, by [IA , 4.5.1], is that I ∼ = A4 × A4 ; but in that case  V1 = ∅ by (2E6), contrary to assumption. The proof is complete. Lemma 2.14. Suppose that t ∈ S0 . Then V1 = ∅. Proof. Suppose false, so that Lemma 2.13 applies. By the Z ∗ -theorem [IG , 15.3], choose t1 ∈ tG ∩ T − {t}, so that [K, t1 ] = 1 by Lemma 2.9a. Let I =  O 3 (CK (t1 )). Then t1 lies in no K-conjugate of z, t, by Lemma 2.13c; also, as V1 = ∅, we see again by (2E6) that I ∼  A4 × A4 . Thus by [IA , 4.5.1], I ∼ = = L± 3 (3) or ± K A6 . In the L3 (3) case, we see easily that I contains some z1 ∈ z , and moreover t1 is conjugate in K to t1 z1 , contradicting Lemma 2.13c. If I ∼ = A6 and t1 induces a graph automorphism on K, then again t1 ∼K t1 z1 for some z1 ∈ I2 (CK (t1 )), and z1 ∈ z K , giving the same contradiction (see [VK , 10.76]). Therefore, I ∼ = A6 and t1 maps into Inndiag(K). If K ∼ = L4 (3), then t1 maps into Inn(K), and t1 = u or tu, where u ∈ I2 (K) is a non-2-central involution. Let y ∈ I2 (I); then y ∈ uK and yu ∈ z K . Then t1 , yu satisfies the conditions of Lemma 2.13c, so t1 yu ∈ (zt)G ∩ Ct = z K t. But if t1 = u, then t1 yu = y ∈ uK = tK , contradiction; if t1 = tu, then t1 yu = ty so y ∈ z K , contradiction. Therefore, K ∼ = U4 (3) or 2U4 (3), whence an involution By y ∈ I has the form y = z h or y = tz h (h ∈ K), respectively, by [V  K , 10.1de]. [VK , 10.8], however, t1 is K-conjugate to t1 z h in either case, and t1 , z h does not satisfy the conclusion of Lemma 2.13c. This contradiction completes the proof of the lemma.  Lemma 2.15. Suppose that t ∈ S0 . Then (a), (b1), and (d) of Proposition 2.12 hold. Moreover, either K ∼ = 2U4 (3). = L4 (3), or K ∼ Proof. By Lemma 2.14, we may choose v ∈ V1 . If t  Cz , then Cz ≤ Ct . But v ∈ [Ct , Ct ] by (2E1), so v ∈ [Cz , Cz ], contradicting Lemma 2.11d. Therefore, t  Cz .

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But t ∈ Z(T ) ∩ S0 ≤ Z(S0 ). Hence there exists g ∈ Cz of odd order such that tg ∈ Ω1 (Z(S0 )) − z, t. In particular as R0 ≤ S0 by Lemma 2.13a, [tg , R0 ] = 1. By (2E2), tg ∈ V1 , so we may assume that v = tg , proving (d). As z g = z, and m2 (Ω1 (Z(S0 ))) ≤ m2 (CT (R0 )) = 3, g acts as a 3-element on Ω1 (Z(S0 )) = v, z, t. Therefore |Cz : Cz ∩ Ct | = 3, so Cz is a {2, 3}-group. In particular, Cz is solvable. Now Φ(T )∩v, z, t ≤ z, t as v induces a graph automorphism on K; so Φ(S0 )∩ v, z, t ≤ z, t. By the action of g, Φ(S0 ) ∩ v, z, t = z. Hence Ω1 (Z(S0 )) = z × Y where Y = [Y, g] ∼ = E22 . We have Ω1 (Z(T )) = z, t or z, t, v, and in either case, by Burnside’s lemma, t is not G-fused to tz. Hence AutCz (z, t, v) is 3-closed, whence Y  Cz . If C(t, K) has a (normal) SL2 (3)-subgroup J, then by solvable L2 -balance, O2 (J) ≤ S0 and so t = Ω1 (Q) ≤ Φ(S0 ), contradiction. Therefore C(t, K) = Q. In particular W ∈ Syl3 (Cz ∩ Ct ), so we may assume g was chosen to normalize W . Then g normalizes [S0 , W ] = R0 and CS0 (W ). As | Aut(R0 )|3 = |W |, we may replace g by an element of CW g (R0 )# and assume that W g is elementary abelian and [R0 , g] = 1. Now v ∈ Φ(S0 ), so t ∈ Φ(S0 ) and so Q ∩ S0 = t. Therefore |CS0 (W )| ≤ 2|CAut(K) (R0 )|2 ≤ 24 by [VK , 10.1b,  10.2b]. As z, t, v ≤ CS0 (W ), it follows that [CS0 (W ), g] ≤ z, t, v. Hence O 3 (Cz ) satisfies (a) of Proposition 2.12 with L ∼ = A4 . Let C0 = CG (z, t, v) = CCz (Y )  Cz and T0 = T ∩ C0 . Then |T : T0 | ≤ 2 so Φ(Φ(T )) ≤ Φ(T0 ). As v ∈ Φ(T0 ), the action of g (and a Frattini argument) shows that t ∈ Φ(T0 ). Therefore t ∈ Φ(Φ(T )), forcing |Q| ≤ 4. Finally, if K ∼ = U4 (3), then v shears to z in CK (z) by (2E4). Hence t shears to tz in Cz , contradicting Burnside’s lemma. Thus K ∼ = L4 (3) or 2U4 (3). The proof is complete.  Recall that R = T ∩ K ∈ Syl2 (K). Lemma 2.16. Suppose that t ∈ S0 and K ∼ = L4 (3). Then Ct = t × K v for some v ∈ V1 . Moreover, T = t, v × R. Proof. By Lemma 2.15, F ∗ (Cz ) ≤ O 2 (Cz ) = K0 ×L with L ∼ = A4 and L  Cz . Now [Q, K0 ] = Q ∩ K0 = 1. It follows that CQ (L) ≤ Q ∩ CK0 L (L) = Q ∩ K0 = 1. By (2E2), there exists v ∈ L ∩ V1 − t. Suppose that |Q| > 2. Then QL ∼ = Σ4 and t = Ω1 (Q) = [Q, O2 (L)] ≤ O2 (L). Let u ∈ Q be of order 4. Then vu is an involution and since O2 (L)  Cz , vu ∈ [Cz , Cz ]. However, as [Q, K0 ] = 1, vu ∈ CT (R0 ), and so (2E2) implies that vu ∈ V1 . Thus, Lemma 2.11d is contradicted. We conclude that |Q| = 2. Consequently, O2 (L) = v, t  where t = t or tz. Also, Ct has a subgroup C1 = K v × t of index at most 2. If C1 < Ct , then AutCt (K) = Aut(K), so v ∼Cz vz by (2E5). Hence t ∼Cz t z. In any case, we have t ∼Cz tz, and AutG (z, t) has even order. This contradicts the facts that T ∈ Syl2 (G) and  z, t ≤ Z(T ), and so C1 = Ct . The proof is complete. = L4 (3). Lemma 2.17. If t ∈ S0 , then K ∼ Proof. Suppose false and continue the above argument. We follow an argument of Aschbacher [A23]. By [VK , 10.3], there is E ∼ = E24 with E = CR (E)  R, AutK (E) ∼ = Σ5 , and NK (E) centralizing v, t. Moreover, E∗ (R) = {E, E0 , E1 , E2 }, where |z G ∩ E| = |z G ∩ E0 | = 5 < 9 = |z G ∩ Ei |, i = 1, 2, and E0 shares the same

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properties as E. Also E1 and E2 are R-conjugate. Set E ∗ := EZ(T ) = E × t, v ∈ E∗ (T ) and Ei∗ = Ei Z(T ), i = 0, 1, 2. As E1∗ and E2∗ are T -conjugate, either all four of these E24 ’s are G-conjugate, or E ∗ and E0∗ are G-conjugate, or one of E ∗ and E0∗ is weakly closed. Now E ∗  T and E0∗  T , so E ∗ and E0∗ are each in an NG (T )-orbit of odd length. Using Burnside’s lemma we deduce that E ∗ or E0∗ is weakly closed in T . Without loss it is E ∗ . Thus E ∗  NG (T ). In particular, E ∗ is normalized by a subgroup X of NCz (T ) of order 3, and [E ∗ , X] = O2 (L) ∼ = E22 (notation as in Proposition 2.12). Let A = AutG (E ∗ ) and let A0 ∼ = Σ5 be the image of AutK (E) in A. As T ∈ Syl2 (G), A0 contains a Sylow 2-subgroup of A. The only possibilities, by [VK , 15.20], are E(A) ∼ = A7 or E(A) = [A0 , A0 ] ∼ = A5 . Notice that for y ∈ I2 (A0 ), m2 ([E ∗ , y]) = 2 or 1 according as y is in or is not in [A0 , A0 ]. But if E(A) ∼ = A7 , then all involutions of A are A-conjugate, contradiction. Thus, E(A) ∼ = A5 . Then [A0 , A0 ] acts absolutely irreducibly on E = [E ∗ , [A0 , A0 ]], so [E, F (A)] = 1. As X does not centralize t, v, X ≤ A0 . It follows that F (A) ∼ = Z3 and then that F (A) = Z(A) = X. Hence [E ∗ , X] = CE ∗ (A0 ) = t, v. Moreover, [R/E, X] = 1 and so [R, X] = 1. We may write X = g, with tg = v and v g = tv. By [VK , 10.2e], R contains an involution u ∈ z K such that v is K-conjugate to vu. Conjugating by g −1 , we see that t ∼G tu. Conjugating in K, we get tuK ⊆ tK . Choose u ∈ uK ∩ O 2 (CK (z)) ∩ E; u exists as u ∈ z K . Then t ∼G tu . Since E ∗ is weakly closed in T and abelian, tu ∈ tA ⊆ t, v, a contradiction. The proof is complete.  From now through Lemma 2.21, we consider the case t ∈ S0 , still assuming that T ∈ Syl2 (G). Lemma 2.18. Suppose that t ∈ S0 . Then [E(Cz ), t] = 1 and E(Cz ) ∼ = L2 (q), q ∈ FM9. Moreover, V1 ∩ E(Cz ) = ∅. Proof. Since t ∈ S0 but t ∈ Z(T ), t acts nontrivially on some component of E(Cz ). Let J be any component of E(Cz ), so that T ∩ J ≤ CT (R0 ). As m2 (Q) = 1 it follows from (2E2) that T ∩ J has rank at most 3 and sectional rank at most 4. If Z(J) = 1, these conditions yield a contradiction, by [VK , 10.32] and the fact that J ∈ C2 . Hence, Z(J) = 1, whence z ∈ J, and so T ∩ J has rank 2 and sectional rank at most 3. As U3 (4) has sectional 2-rank 4, we conclude by [IA , 5.6.3] and the fact that J ∈ C2 [V3 , 1.1] that J ∼ = L2 (q), q ∈ FM9, as desired, unless J ∼ = M11 ± or L3 (3). In any case J is simple of 2-rank 2. Since m2 (CT (R0 )) ≤ 3, J = E(Cz ). Moreover, (2E2) implies that V1 ∩ J = ∅. Let P = CT (J) ∈ Syl2 (C(z, J)). Note that R0 ≤ S0 ≤ P , whence m2 (P ) ≥ 3 and z ∈ [P, P ]. It remains to assume that J ∼ = M11 or L± 3 (3) and obtain a contradiction. Notice that such a component J cannot be terminal in G. For, ΓD,1 (J) = J for every D ∈ E22 (Aut(J)), by [VK , 8.5]; and hence terminality would imply m2 (C(z, J)) = 1 by [V2 , 5.1], contradiction. By [VK , 14.6], J is Z4 -semirigid in G. Hence by [V2 , 8.4], since J is not terminal in G, and Q8 ∗ Q8 ∼ = R0 ≤ P , there exists a subgroup P0  P such that P/P0 is cyclic and J is not a component of CG (u) for any u ∈ I2 (P0 ). But J is a component of Cz and z ∈ [P, P ] ≤ P0 , a contradiction. This completes the proof.  We fix v ∈ V1 ∩ E(Cz ) and next prove:

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Lemma 2.19. Suppose that t ∈ S0 . Then CQ (v) = t and E(Cz ) ∼ = A6 . Moreover, either v, t or v, tz lies in E(Cz ). Proof. Again let J = E(Cz ) and P = CT (J) ∈ Syl2 (C(z, J)). Let U = v Z(T ∩ J) ≤ J. Since v ∈ Φ(T ), v ∈ Φ(T ∩ J) and so U ∼ = E22 . Also U ≤ CT (R0 ) ∩ CT (v) ≤ z, v Q by (2E2), so U ≤ Ω1 (Cz,vQ (v)) = z, v Ω1 (CQ (v)) = z, v, t ≤ v Z(T ). Since z, v, t is of maximal rank in CT (R0 ) and v ∈ J, z, v, t ∩ J = v, t or v, tz, proving the final statement. In particular Cv,t (J) = 1, so v CQ (v) embeds in CAut(J) (v). But as J ∼ = L2 (q), q ∈ FM9, CAut(J) (v) has no Z4 -subgroup disjoint from the image of v. Therefore CQ (v) = t. Now suppose that q = 9, so that q ∈ FM. Then CAut(J) (U ) = AutU (J), so CT (U ) = U × P and P ∼ = CT (U )/U = CT (v)/U . As CQ (v) = t, P is embeddable in CInndiag(K) (v) ∼ = Aut(P Sp4 (3)) and contains R0 . Let S1 = S ∩ E(CK (v)) ∈ Syl2 (CK (v)). Then S1 centralizes z, t, v and acts on J, so S1 ≤ P × U with S1 ∩ U = 1 (note that S1 ∩ z, t, v = z). Let P1 be the projection of S1 on P . Then P1 = R0 s, where s2 = 1 and s interchanges the two Q8 subgroups of R0 . Moreover as P embeds in Aut(P Sp4 (3)), |P : P1 | ≤ 2. We use [VK , 3.95] to argue that (z, J) is terminal in G. Let I = {y ∈ I2 (P ) | J   CG (y)}. Thus z ∈ I. For any y ∈ I2 (R0 ) − I, it follows from [VK , 3.95] that I2 ([CR0 (y), CR0 (y)]) ∩ I = ∅. But this is a contradiction as z ∈ [CR0 (y), CR0 (y)]. Thus I2 (R0 ) ⊆ I. Likewise for any y ∈ I2 (P ) interchanging the two normal Q8 subgroups of R0 , CR0 (y) ∼ = E23 . Hence again by [VK , 3.95], if J pumps up nontrivially in CG (y), then I2 (CR0 (y)) ⊆ I, a contradiction. It remains to assume that y ∈ I2 (P ) − R0 − I and y normalizes both Q8 subgroups of R0 , and derive a contradiction. By the previous paragraph, P = P1 y with I2 (P1 ) ⊆ I. Since CR0 (y) contains E22 or Q8 , and I2 (R0 ) ⊆ I, the only possibility by [VK , 3.5] is that y acts freely on R0 / z, q = 5, and J pumps up in CG (y) to J ∗ with J ∗ /Z(J ∗ ) ∼ = M12 or J2 . Hence |CAut(J ∗ ) (J)|2 = 4, whence |CP1 (y)| ≤ 4. However, y acts freely on R0 /Z(R0 ) ∼ = E24 and so |CP1 (y)| ≥ |CP1 /Z(R0 ) (y)| > |CR0 /Z(R0 ) (y)| = 4, a contradiction. This proves the claimed terminality. As J is outer well-generated by [VK , 8.4], it follows by [IG , 18.11, 18.8] that P embeds in a (dihedral) Sylow 2-subgroup of Aut(J), an absurdity. Therefore q = 9  so J ∼ = A6 . The lemma is proved. Write J ∩ z, t, v = v, t , where t = t or tz. Thus NJ (v, t ) is transitive on # v, t  . Lemma 2.20. Assume t ∈ S0 . Then K ∼ = 2U4 (3) and tG ∩ V1 = ∅. Proof. Suppose first that K ∼ = U4 (3). Then by (2E4), v ∼Cz vz. By the transitivity just noted, t ∼Cz t z. But t z ≤ Z(T ) and T ∈ Syl2 (G), so AutG (t , z) has odd order, a contradiction. It is harder to rule out K ∼ = L4 (3). In that case, by [VK , 10.2b], CAut(K) (R0 ) is a four-group, so T ∩J ≤ CT (R0 ) = z, v Q. Therefore, Ω1 (Φ(T ∩J)) = Ω1 (Φ(Q)) = t and so t = t. So v ∈ tG . # Consider the neighborhood of (z, t , K1 ) in G. For each u ∈ z, t , let Lu be the subnormal closure of L := K1 in CG (u). We have Lt = K ∼ = L4 (3), and Lz = L. Now U := v, t acts on Ltz . Fix g ∈ I3 (NJ (U )) with tg = v, v g = tv. Then z g = z.

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 ∼ P Sp4 (3). We have CG (v, tz) = CG (t, tvz)g = CG (t, vz)g so L ≤ O 3 (CG (v, tz)) = −1  g 3 ∼ Likewise CG (tv, tz) = CG (t, vz) so O (CG (tv, tz)) = P Sp4 (3). Indeed there is an involution w ∈ NJ (U ) centralizing t and interchanging v and tv. Hence (tz)w = tz and CG (tv, tz)w = CG (v, tz). It follows that E(CLtz (tv))w = E(CLtz (v)) ∼ = P Sp4 (3). As a consequence of the full neighborhood structure, Ltz ∼ = U4 (3) or 2U4 (3), by [VK , 3.19]. Hence K/Z(K) ∼ = U4 (3) by our choice of (t, K) in the statement of Proposition 2.12, a contradiction. This proves the first assertion of the lemma. The second assertion is trivial if t = t. Otherwise t = tz ∼J v so t ∼J vz. By  (2E2), vz ∈ V1 and the proof is complete.

Lemma 2.21. If t ∈ S0 , then the conclusion of Proposition 2.12 holds, with (b2) holding. Proof. In view of Lemmas 2.19 and 2.20 and the facts that Out(R0 ) ∼ = Σ3  Z2 and Out(A6 ) is a 2-group, it suffices to show that S0 = R0 . As v ∈ J, S0 ≤ CT (v). By Lemma 2.19, CQ (v) = t, so CT (v)/ t embeds in CAut(K) (v) ∼ = Z2 × Aut(P Sp4 (3)). But we see from [IA , 4.5.1] that for a 2-central involution z  ∈ Aut(P Sp4 (3)), we have O2 (CP Sp4 (3) (z  )) = O2 (CAut(P Sp4 (3)) (z  )) ∼ =  R0 . It follows that S0 = R0 , as desired. Lemmas 2.15, 2.17, and 2.21 complete the proof of Proposition 2.12. Next, up to but not including Corollary 2.62, we assume that (2D) holds, and whenever it holds,T ∈ Syl2 (G). In this case we choose (t, K) ∈ ILo2 (G) satisfying (2D) so that (1) K ∼ = 2U4 (3), if possible; (2) Otherwise, K ∼ = L4 (3), if possible; and (2F) (3) Otherwise, K ∼ = U4 (3). We choose g ∈ NG (T ) − T with g 2 ∈ T. Before considering the different possibilities for K, we prove three preliminary lemmas. Lemma 2.22. If T ∈ Syl2 (G), then the following conditions hold: (a) Q = t, unless possibly K ∼ = Z2 or Z4 ; = 2U4 (3), in which case Q ∼ ∼ (b) If K/Z(K) = U4 (3), then tz = tg and Z(T ) = t, z; and (c) m2,3 (G) = 4. Proof. By Lemma 2.9a, m2 (Q) = 1. We have Qg  T with Q ∩ Qg = 1. By Lemma 2.9b, z ∈ Qg , so Qg ∩ R0 = 1. Indeed, unless K ∼ = 2U4 (3), z = Z(R) and so Qg ∩R = 1. Therefore Qg embeds in CAut(K) (R0 ), and even in CAut(K) (R) unless K∼ = 2U4 (3). But CAut(K) (R) is elementary abelian by [VK , 10.2b and 10.1a], and CAut(K) (R0 ) ∼ = D8 in the 2U4 (3) case by [VK , 10.1b], so (a) holds. In (b), Z(T ) ≤ z Q, but because of g, t is not characteristic in Z(T ), whence Z(T ) = z, t. As noted above, z ∈ Qg , so tg = tz, as claimed. Finally, m2,3 (G) = m3 (Ct ) = 4 by (2D1) and (a), completing the proof.  Lemma 2.23. Suppose that T ∈  Syl2 (G). Let h ∈ G with z h ∈ Ct . Then h G E(CK (z )) = 1. Moreover, z ∩ K = z K .

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Proof. Suppose that h ∈ G, z h ∈ Ct , and J := E(CK (z h )) = 1. Then from [IA , 4.5.1] we see that J ∼ = A 6 , L± 3 (3), or P Sp4 (3). Let I be the subnormal closure h of J in CG (z ), so that J is a component of CI (t). −1 Suppose that [I, R0h ] = 1. Let I1 be a component of I h ≤ CCz (R0 ). Thus I1 is a component of Cz . We have S1 := t, z CI1 I1t (t) ≤ CG (R0 t) = CCt (R0 ), so by [VK , 10.2b, 10.1b], S1 /S1 ∩ Q embeds in D8 . In particular I1 = I1t . Now I1 is a −1 pumpup, possibly trivial, of J h ∼ = A 6 , L± 3 (3), or P Sp4 (3). Thus by [VK , 10.77], I1 ∼ = A6 and I1 contains t or tz. In particular, C(z, I1 ) ≤ CG (z, t), a solvable group, so I1 = E(Cz )  Cz . It follows that t or tz is 2-central in Cz . Hence t is 2-central in Cz . But z is 2-central in G, so t is as well. This contradicts our assumption that T ∈ Syl2 (G), however. Therefore, [I, R0h ] = 1. As R0h = O2 (K0h ) and K0h = K1h K2h is the product of two solvable components of CCG (zh ) (th ), solvable L2 -balance implies that, say, R1h ≤ I and then K1h normalizes I. Since I ∈ C2 and CZ(I) (K1h ) contains the involution z h , K1h ≤ ICCG (zh ) (I), by [VK , 3.106]. Hence there exists a solvable SL2 (3)-component L1 of CI (th ) such that O2 (L1 ) = R1h . If also R2h ≤ I, then we similarly find SL2 (3) = L2   CI (th ) with R2h ≤ L2 , and [VK , 3.108] gives an immediate contradiction. Hence, [I, R2h ] = 1. There exists y ∈ CG (z h , th ) such that K1hy = K2h . Let I ∗ = I y , so that [I, I ∗ ] = 1 and Ly1 ≤ I ∗ . Then by [VK , 3.107], I ∼ = 2G2 (4), whence It := E(CI (t)) ∼ = U3 (3). Then S2 := CCt (It ) is an = G2 (2) ∼ extension of Q ≤ Z4 by a subgroup of CAut(K) (It ) ∼ = Z4 . In particular |S2 | divides 16. It follows that t normalizes I ∗ and |CI ∗ t (t)| divides 16. However, t normalizes a parabolic subgroup P of I ∗ with m2 (O2 (P )/Φ(O2 (P ))) ≥ 8, so using [IG , 9.16] we get |CI ∗ t (t)| ≥ 32, a contradiction. This completes the proof of the first assertion, and the second is an immediate consequence, given Lemma 2.22b.  Lemma 2.24. Suppose (2D) holds with T ∈ Syl2 (G). Then one of the following holds: (a) t ∼NG (T ) tz; or (b) OutT (K) = φ where φ is the image of some involution v ∈ tNG (T ) ∩ Z(T ) ∩ V1 . Proof. By Lemma 2.22b, we may assume that K ∼ = L4 (3). By [VK , 10.2b], every involution in T mapping into I2 (Z(T /Q)) lies in z, t ∪ V1 . Suppose that t is NG (T )-conjugate to no involution in Z(T ) ∩ V1 . Then tg = tz and so (a) holds. If (a) fails, then tg =: v ∈ Z(T ) ∩ V1 , but v ∼T vz. As v ∈ Φ(T ) and g ∈ NG (T ), t is a direct factor of T , with a complement containing v, z. It then follows from (2E5) that the image of T in Out(K) is as asserted in (b).  Now we begin consideration of the case K ∼ = L4 (3). Lemma 2.25. Suppose that T ∈ Syl2 (G) and K ∼ = L4 (3). Let v ∈ V1 and let J be the subnormal closure of E(CK (v)) in CG (v). Then J = E(CK (v)) or J ∼ = K. Proof. By Lemma 2.22c, m3 (CG (v)) ≤ m2,3 (G) = 4, so J is a single compo K, then J ∼ nent. By [VK , 3.10] and our choice of K, if J ∼ = = L4 (4) or U4 (3). Suppose that J ∼ = U4 (3). We may assume that T0 := T ∩ E(CK (v)) ∈ Syl2 (E(CK (v))), and let D ≤ T0 with D ∼ = E24 . Then as D ≤ J, all involutions of D are J-conjugate, hence G-conjugate to z. But as K ∼ = L4 (3), E(CK (y)) = 1 for some y ∈ D# , by [VK , 10.3de], contradicting Lemma 2.23.

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∼ L4 (4). Let x ∈ I3 (E(CK (v))) be 3-central in J, so Suppose then that J = that CJ (x) ∼ = GL3 (4) and H := E(CJ (x)) ∼ = SL3 (4). Set Cx = CG (x) and C x = Cx /O3 (Cx ). By [VK , 9.32], CK (x) contains A ∼ = E34 , and IAut(K) (A; 2) = {1}. Hence t ∈ I∗Ct (A; 2), whence t ∈ I∗G (A; 2). It follows that O3 (Cx ) ≤ O2 (Cx ) t. By L2 -balance, H ≤ L2 (C x ). If [H, O3 (C x )] = 1, then [H, O2 (C x )] = 1 so H ≤ E(C x ). In this case the subnormal closure I of H in C x is a product of (at most 2) components in C3 , and H is a component of CI (v). Moreover, CCx (t) = CCt (x) is solvable. In particular, by a Frattini argument, t ∈ O3 (Cx ). This, however, contradicts [VK , 3.64] applied to C x . Therefore [H, O3 (C x )] = 1. It follows by [VK , 15.19] that m2,3 (Cx ) > 4 = m2,3 (G), a contradiction. The proof is complete.  We continue to assume that K ∼ = L4 (3) and T ∈ Syl2 (G), and aim for a contradiction (Proposition 2.30 below). The arguments are due to Aschbacher [A23]. We use the following notation. By [VK , 10.3, 10.4], R(= T ∩ K) has exactly two normal E24 -subgroups F0 and F1 . Set Ei = CT (Fi ) = CCt (Fi ) = Fi vi , t for i = 0, 1, where vi2 = 1. If vi = 1 for i = 0 or 1, then CK (vi ) ∼ = Aut(P Sp4 (3)). Also, Ai := AutK (Fi vi , t) ∼ = Σ5 , [Ai , vi , t] = 1, Ai permutes the 5-element set z G ∩ Fi faithfully, and v1 = v0 z. Lemma 2.26. If K ∼ = L4 (3) and T ∈ Syl2 (G), then tz ∼NG (T ) t. Proof. If not, then Lemma 2.24b holds and v0 = 1 = v1 . Without loss, v0 = tg . Since v0 ∈ Z(T ), T = t, v0  R = t, v1  R. Then E0  T ∈ Syl2 (CG (v0 )), so AutK g (E0 ) ∼ = Σ5 . As AutK (E0 ) centralizes v0 , it lies in AutCG (v0 ) (E0 ) = AutK g (E0 ). Therefore AutK g (E0 ) = AutK (E0 ) centralizes t. As E0 = E1 , [AutK g (E1 ), t] = 1, indeed CE1 (AutK g (E1 )) = tz, v0  by [VK , 10.4]. On the other hand, CE1 (AutK (E1 )) = v1 , t = v0 z, t. Let r = v0 zt. Then r ∈ V1 , and E(CK g (r)) ∼ = P Sp4 (3), with z ∈ E(CK (r)) ∩ E(CK g (r)). By Lemma 2.25, the pumpup H of E(CK (r)) in CG (r) is isomorphic to P Sp4 (3) or L4 (3). Hence AutH (E1 ) ∼ = Σ5 , whence AutH (E1 ) = AutK (E1 ). But H is also the pumpup of E(CK g (r)) in CG (r), so similarly AutH (E1 ) = AutK g (E1 ). Therefore AutK g (E1 ) =  AutK (E1 ) and so v0 z, t = v0 , tz, a contradiction. The proof is complete. Now choose h ∈ NG (T ) with th = tz. Since z is weakly closed in Z(T ) with respect to G, z h = z, so h2 ∈ T . We may therefore assume that our element g ∈ NG (T ) − T satisfies tg = tz. Recall from [IA , 1.15.10, 2.5.10] that ΓK is a group of graph automorphisms of K isomorphic to the group of symmetries of the Dynkin diagram of K in the case of an untwisted group with one root length, and trivial in thge case that K is a Steinberg group. Thus in the next lemma, ΓK ∼ = Z2 is the group of graph automorphisms of K ∼ = L4 (3) Lemma 2.27. Assume that K ∼ = L4 (3) and T ∈ Syl2 (G). Let E = E0 and F = F0 . Then the following conditions hold: (a) v0 = 1 = v1 ; (b) E = CG (E) ∼ = E25 ;

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(c) AutG (K) ∩ ΓK = 1; and (d) |z G ∩ F | = |z G ∩ E| = 5. Proof. We have R0 = O2 (CK (z)) = O2 (O 2 (CG (t, z))) = O2 (O 2 (CG (tz, z))) = O2 (CK g (z)). G By Lemma 2.23 and [VK , 10.2g], every  Kelement of z ∩ T maps into the image of := ΓK in Out(K). Then as T ∩ K = z ∩ T R0 by [VK , 10.2h], if we set T0 G t, tz, z ∩ T R0 , then t R ≤ T0 and the image A0 of T0 in Out(K) lies in the image of ΓK . In particular, Ei ∩ T0  T0 , i = 0, 1. Note that T0 = T0g , so tz (T ∩K g ) ≤ T0 and the image of T0 in Out(K g ) is isomorphic to A0 and lies in the image of ΓK g . Now Ei ∩K g = Ei ∩(T0 ∩K g )  T0 ∩K g = T ∩K g . Therefore Ei ∩K g , i = 0, 1, are the normal E24 -subgroups of T ∩K g . In particular AutK g (E∩K g ) ∼ = Σ5 and thus AutK g (E ∩ K g ) does not centralize t, whose projection on K g is z. Suppose that v0 = 1, so that |E| = 26 . Let X = O 2 (CAutK (E) (t, z)) ∼ = A4 . Then in Ct we see that CE (X) ∼ = E23 contains CE (AutK (E)) = v  , t for some v  ∈ V1 ∩E. Without loss we may replace v0 by v  and assume that CE (AutK (E)) = v0 , t. A similar calculation in CG (tz) shows that CE (AutK g (E)) is a hyperplane of CE (X), not containing t. Hence replacing v0 by v0 t if necessary, we may assume that v0 ∈ CE (AutK g (E)). Let H be the pumpup of E(CK (v0 )) in CG (v0 ). As z ∈ E(CK (v0 )) ∩ E(CK g (v0 )), H is also the pumpup of E(CK g (v0 )) in CG (v0 ). Using Lemma 2.25 as at the end of the proof of Lemma 2.26, we conclude that AutK (E) = AutH (E) = AutK g (E), a contradiction as AutK (E) centralizes t but AutK g (E) does not. Therefore v0 = 1, so (a) holds. By [VK , 10.4], CG (E)/ t, which embeds in CAut(K) (F ), is an elementary abelian 2-group. Similarly, CG (E)/ tz is elementary abelian. Hence CG (E) is an elementary abelian 2-group. Then (a) implies (b), and by [VK , 10.2g], (b) implies (c).   Finally, F = z NK (F ) ≤ z G ∩ E with |z NK (F ) | = 5. Also E(CK (y)) = E(CK (ty)) = 1 for all y ∈ F # − z NK (F ) , so by Lemma 2.23 and the fact that tz ∈ tG = z G , z G ∩ E = z NK (F ) . Part (d) follows, and the proof is complete. 

Now let N = NG (E) and N = N/E ∼ = AutG (E) (Lemma 2.27b). Lemma 2.28. Assume that T ∈ Syl2 (G) and K ∼ = L4 (3). Then the following conditions hold: (a) O2 (N ) = CN (F ) = U0 t, where U0  N , U0 is homocyclic abelian of order 28 and exponent 2 or 4, and F = [U0 , t]; and (b) There is y ∈ I2 (F ) − z G such that V := y, z is normal in a Sylow 2subgroup S of G containing T , and yz ∈ y N . In particular, |S : CS (y)| = 2. Proof. We have N ∩ Ct ∼ = Σ5 , both groups permuting z G ∩ F = N ∩ CG (tz) ∼ naturally. Hence N = U (N ∩ Ct ) where U = CN (F ) > 1. Then U stabilizes the chain E > F > 1, so U = O2 (N ). Moreover the mapping u → [u, t] is an N ∩ Ct isomorphism from U to F . In particular the coset tF is completely fused by N . Let U be the full preimage of U in N . Then N is irreducible on U/E. Moreover CU (t ) = E for all t ∈ tF . It follows that U/F has exponent 2. Then by [VK , 5.18a], U/F = U0 /F ×E/F where U0 = [U, N ]; and U0 /F and F are isomorphic N modules. Hence by [VK , 5.18b], U0 is homocyclic abelian of order 28 and exponent 2 or 4, proving (a).

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Hence, |G|2 ≥ |N |2 ≥ |U |2 |Σ5 |2 = 212 , while |Ct |2 ≤ 2| Aut(K)|2 ≤ 210 . Hence t is not half 2-central in G. On the other hand, let T ≤ S ∈ Syl2 (G). By [IG , 10.11], there exists V ∼ = E22 with V ≤ T and V  S. Now Z(S) ≤ Z(T ) = z, t by Lemma 2.27a, so Z(S) = z. By [VK , 10.3cd] and Lemma 2.27c, V ≤ E, and if V ≤ F , then V # ⊆ z G . Since elements of V # are half 2-central in G but elements of tF ⊆ tG are not, indeed V ≤ F . We fix y ∈ V # − z G . Thus, V = y, z and yz ∈ y T . The lemma is proved.  We have U0 ≤ CG (y). Set H = E(CK (y)). Then H ∼ = L2 (9) and yz ∈ H, by [VK , 10.2f]. Let J be the subnormal closure of H in CG (y). Lemma 2.29. If T ∈ Syl2 (G) and K ∼ = L4 (3), then J ∼ = HS or 2HS. Proof. Suppose first that J = J1 J1t is a diagonal pumpup of H. Since m2,3 (G) = 4 it follows from (c) of Theorem C5 : Stage 2 that m2 (C(y, J)) = 1. Hence CU0 (J) is cyclic, whence |U0 /CU0 (J)| ≥ 26 , with strict inequality if U0 is elementary abelian. Now the largest abelian 2-subgroups of Aut(J1 ) are of order 8, so U0 /CU0 (J) must embed in Aut(J1 ) × Aut(J1 ) and have order 26 . Therefore U0 /CU0 (J) cannot be homocyclic of exponent 4, a contradiction. Therefore, J is a single component. Let Qy,t = CG (y, t H), which, by Lemma 2.27c and [IA , 4.5.1], is isomorphic to E22 or Z4 × Z2 . Then Qy,t normalizes J and we let Qy be a Qy,t -invariant Sylow 2-subgroup of C(y, J). Then CQy (t) ≤ Qy,t so by [IG , 10.24, 10.26, 10.27], either m2 (t Qy ) = 2 or t Qy ∼ = F (n) for some n ≥ 5. In any case m2 (t Qy ) ≤ 3. Suppose next that J = H. If J  CG (y), say J = J a for some a ∈ CG (y), then by (c) of Theorem C5 : Stage 2, m2 (C(y, JJ a )) = 1 and we reach a contradiction as in the paragraph before last. So J  CG (y), and U0 /CU0 (J) embeds in Aut(L2 (9)), hence has order at most 23 . Thus |CU0 (J)| ≥ 25 and CU0 (J) has exponent at most 4. As m2 (Qy ) ≤ 3, the only possibility is that t Qy ∼ = F (n) and CU0 (J) ∼ = Z2 ×Z4 ×Z4 embeds in t Qy . But as F (n) is an extension of Z4 by D2n−2 , t Qy has no such subgroup, a contradiction. Therefore, J is a vertical pumpup of H. Consequently by [VK , 3.10], J ∼ = L4 (3), U4 (3), 2U4 (3), U5 (2), P Sp4 (3), Ln (2), n = 4 or 5, Sp4 (4), HS, or 2HS. In all cases m2 (J) > 3 ≥ m2 (Qy ), so J  CG (y). In the first four cases, by (c) of Theorem C5 : Stage 2, m2 (Qy ) = 1, whence the abelian group U0 /CU0 (J) of order ≥ 26 embeds in Aut(J), contrary to [VK , 10.71a]. In the next three cases, we use the fact that yz ∈ H to get a contradiction. Namely, we saw above that yz is 2central in CG (y), hence in J; but in these three cases, involutions of H are not 2-central in J, by [VK , 10.71b], contradiction. Next suppose that J ∼ = Sp4 (4). Let Sy = CS (y) ∈ Syl2 (CG (y)), a maximal subgroup of S, and choose h ∈ S − Sy , so that h interchanges y and yz ∈ J. Clearly J(Sy ) induces inner automorphisms on J, so J(Sy ) = J(Sy ∩ J) × J(Qy ). As Sy ∩ J is indecomposable [VK , 10.34] with Sy ∩ J = J(Sy ∩ J) and yz ∈ [Sy ∩ H, Sy ∩ H] ∼ = E24 , while m2 (Qy ) ≤ 3, the KrullSchmidt theorem implies that y = (yz)h ∈ [Sy ∩ H, Sy ∩ H] ≤ J, contradiction. The proof is complete.  ∼ HS and Let us continue the argument to rule out the remaining cases J = J∼ = 2HS (assuming T ∈ Syl2 (G) and K ∼ = L4 (3)). There is t0 ∈ J of order 2 or 4, respectively, such that t acts on J ∼ = HS or 2HS like t0 , and t20 ∈ Z(J). Moreover, by the structure of Qy,t , if Z(J) = 1, then t = t0 t1 where Z4 ∼ = t1  = CQy (t1 ) ≤ Qy .

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On the other hand if Z(J) = 1, then t = t0 t1 for some t1 ∈ Qy with t21 = 1, Qy embeds in Z4 and, being abelian, is centralized by t. Thus if Z(J) = 1, then [U0 , t] ≤ [Sy , t] ≤ J. It follows in that case, as t ∈ U0 , that CU0 (J) = 1. Hence, Aut(J) has an abelian subgroup of order 28 , which is not the case by [VK , 10.37]. Therefore J ∼ = 2HS. Now [t0 , Qy ] = 1 while Z(F (n)) has order 2, so m2 (t0  Qy ) = 2 by [IG , 10.27]. As m2 (Aut(J)) = 5 [IA , 5.6.1], U0 has exponent 4. Moreover, CQy (t) = t1 , so in CG (y) := CG (y)/C(y, J), U 0 contains U1 ∼ = Z4 × Z4 × Z4 , with [U 1 , t] = CU 1 (t) = Ω1 (U 1 ). But then by [VK , 10.75], t induces an outer automorphism on J, a contradiction. We have proved: Proposition 2.30. Suppose that (2D) holds and T ∈ Syl2 (G). Then K/Z(K) ∼  = L4 (3). We next follow Aschbacher [A23] and Finkelstein [Fin2] to prove: Proposition 2.31. Suppose that (2D) holds, T ∈ Syl2 (G), and (2F) holds. Then K ∼ = 2U4 (3). ∼ The argument is lengthy. We suppose that K =  2U4 (3). Hence by Propositions 2.30 and 2.12, K∼ = U4 (3) and T ∈ Syl2 (G). We eventually reach a contradiction after Lemma 2.61. By Lemma 2.22a, Q = t , and hence m2,3 (G) = m3 (K) = 4. We first have to deal with the following situation, which we assume through Lemma 2.50:

(2G)

(1) y ∈ I2 (Ct ) with J := E(CK (y)) = 1; (2) m2 (CG (y)) ≥ 7; and (3) L is a component of the subnormal closure of J in CG (y), and QL ∈ Syl2 (CG (L)).

Note that m2 (Ct ) ≤ 1 + m2 (Aut(K)) = 1 + 5, so in (2G), y ∈ tG . We also define H := {(u, H) ∈ ILo2 (G) | m2 (CG (u)) ≥ 7} Thus, in (2G), (y, L) ∈ H. Lemma 2.32. Let (u, H) ∈ H. Then H/Z(H) ∼  L± = 4 (3). Therefore Proof. If H/Z(H) ∼ = L± 4 (3), then (2D) applies to (u, H). m2 (C(u, H)) = 1 by Lemma 2.9. But then m2 (CG (u)) ≤ 1 + m2 (Aut(H)) = 1 + 5, contradicting the definition of H.  Lemma 2.33. Let (u, H) ∈ H with H ∼ = L4 (4) or (2)G2 (4). Suppose that u ∈ I2 (CG (u)) and H  := E(CH (u )) ∼ = U4 (2) or U3 (3), respectively. Suppose also that either u = t or H  is a component of CG (u ). Let x ∈ I3 (H) be 3-central in H. Then x lies in no element of B3∗ (G). 

Proof. Suppose false and let x ∈ B ∈ B3∗ (G). Since B ∈ S3 (G), m3 (CG (B)) > 4.

Now H  and H share a Sylow 3-subgroup PH . We may assume that x ∈ I3 (Z(PH )) and set I = E(CH (x)) ∼ = SL3 (4) and I0 = O3 (CH  (x)) ∼ = 31+2 . Thus,  I0  CI (u ).

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Set Cx = CG (x) and C x = Cx /O3 (Cx ). As m3 (CG (x)) ≥ m3 (CG (B)) > 4, O3 (Cx ) has odd order. Hence u, u ∈ I2 (C x ) with I ∼ = I   CC x (u). If H  is a component of CG (u ), then I0 ∼ = I 0 ≤ I ∩ O3 (CC x (u )). If, on the other hand,   ∗ t = u , then F (CC x (u )) is the direct product of u  and a nontrivial 3-group, by inspection of Ct and the Borel-Tits theorem. Suppose that [I, O3 (C x )] = 1. Then [I, O2 (C x )] = 1 by [IG , 3.17] as F ∗ (O2 (C x )) = O3 (C x ).   E(C x ) But I ≤ L2 (C x ), so I ≤ E(C x ). Then M := I is SL3 (4) or the product of two copies of SL3 (4), or else L3 (4) ↑2 M /O2 (M ) via u. But as x ∈ I3o (G), components of E(C x ) lie in C3 . Hence by [VK , 3.80a], M ∼ = 3Suz. Since m2,3 (G) = 4 < 6 = m3 (M ), M = E(C x ). As Suz is balanced with respect to the prime 2 [IA , 7.7.1], the conclusions of the previous paragraph about O3 (CC x (u )) or F ∗ (CC x (u )) are then contradicted. Therefore [I, O3 (C x )] = 1. In particular as H   CG (u), I × u acts faithfully on O3 (C x ), centralizing x. Using the Thompson dihedral lemma and the fact that m2 (I) = 4, we conclude that m2,3 (Cx ) ≥ 5 > m2,3 (G), a contradiction. The proof is complete.  Lemma 2.34. Let (u , H  ) ∈ H. Suppose that H  ∼ = P Sp4 (3) ∼ = U4 (2). Let   x ∈ I3 (H ) be 3-central in H . Then x lies in no element of B∗ (G). Proof. Suppose false and choose a counterexample with x ∈ B ∈ B∗ (G). Then m3 (CG (x)) ≥ m3 (CG (B)) > 4 as B∗ (G) ∩ S3 (G) = ∅. Let QH  ∈ Syl2 (CG (H  )). Then m2 (QH  ) ≥ 7 − m2 (Aut(H  )) = 3, by [IA , 3.3.3]. Hence by [V2 , 9.3c], H  is not terminal in G. But by [VK , 14.2], H  is semirigid in G, so there exists u ∈ I2 (Z(QH  )) and a nontrivial pumpup (u , H  ) < (u, H) ∈ ILo2 (G). Then m2 (CG (u)) ≥ m2 (H  ) − m2 (Z(H  )) + m2 (QH  ) ≥ 4 + 3, so (u, H) ∈ H. Moreover if H is a diagonal pumpup, then CG (u) contains E35 ×Z2 , contradicting m2,3 (G) = 4. Thus, H is a vertical pumpup of H  , whence by Lemma 2.32 and [VK , 3.10k], H ∼ = L4 (4). As H  is a component of CG (u ), Lemma 2.33 applies and completes the proof.  Lemma 2.35. In (2G), J ∼ = U4 (2). Proof. Suppose that J ∼ = U4 (2). Then by [VK , 3.10k], L ∼ = J, L± 4 (3), 2U4 (3), ∼ or L4 (4). As (y, L) ∈ H, Lemmas 2.32 and 2.33 force L = J. (Note that a 3central element of J lies in an element of B3∗ (G) by Lemma 2.9c.) If L = J, then Lemma 2.34 gives a contradiction. Finally if L is a diagonal pumpup of J, then as  m3 (J) ≥ 3, m3 (CG (y)) ≥ 6 > m2,3 (G), a contradiction. Lemma 2.36. In (2G), J ∼  U3 (3). = Proof. Suppose false. By Lemma 2.32, L/Z(L) ∼ = U4 (3). By Lemma 2.9c, a 3central element of J lies in an element of B3∗ (G). If L ∼ = (2)G2 (4), then Lemma 2.33 gives a contradiction. Therefore L ∼ = U3 (3) by [VK , 3.10o]. Then m2 (C(y, L)) ≥ 4. As ΓD,1 (L) = L for all D ∈ E22 (Aut(L)), the usual argument shows that L is not terminal in G. If L pumps up to L∗ ∼ = (2)U4 (3) in CG (u) for some u ∈ I2 (C(y, L)), then using [V2 , 4.7] we may assume that u is 2-central in C(y, L), and so m2 (C(u, L∗ )) > m2 (C(y, L)) − m2 (CAut(U4 (3)) (U3 (3))) ≥ 4 − 1, and Lemma

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2.32 gives a contradiction. The only other possibility, by [VK , 3.10o], is that L pumps up to L∗ ∼ = (2)G2 (4). Then Lemma 2.33 gives a contradiction, completing the proof.  Lemma 2.37. In (2G), J ∼ = A6 and, replacing y by ty if necessary, we may arrange that J = L. Proof. By Lemmas 2.35 and 2.36, and [IA , 4.5.1], J ∼ = A6 . Suppose that J = L. Then m2 (QL ) ≥ 7 − m2 (Aut(A6 )) = 4. Since J < K, J is not terminal in G. But L is semirigid in G by [VK , 14.3] (note that no L4 (3) component exists in the centralizer of any involution of G by our choice of K (2F)). So by [IG , 7.4] we may choose u ∈ I2 (Z(QL )) such that L < L∗ (L∗ )y for some component L∗ of CG (u). Indeed by [IG , 8.5], L is invariant under every E ∈ E2∗ (CG (y)), and by [IG , 7.4] we may assume that u lies in such a group E, so that m2 (CG (u)) ≥ m2 (CG (y)) ≥ 7. In particular u = t. If we may choose u = ty, then the lemma holds, so we may assume that u ∈ t, y. But u ∈ CCt (J), whence by [VK , 3.62], E(CK (u)) ∼ = U4 (2). Hence, (2G) holds with u in place of y, so Lemma 2.35 is then contradicted, completing the proof.  Lemma 2.38. In (2G), L ∼  U4 (2) or U5 (2). = Proof. Suppose false. In either case m2 (Aut(L)) = 4 by [IA , 3.3.3, 4.9.2], so m2 (QL ) ≥ 3. If L ∼ = U5 (2), then m3 (L) = 4 and so CG (y) contains B × E23 with 3 (G), contradicting (c) of Theorem C5 : Stage 2. B ∈ B E34 ∼ = ∗ Hence, L ∼ U = S ∈ Syl3 (L) and write Z(S) = x, so = 4 (2). Let Z3  Z3 ∼ that by Lemma 2.34, x lies in no element of B3∗ (G). This condition implies that no pumpup of L can be diagonal, or isomorphic mod center to L± 4 (3). Therefore either L is standard or there is u ∈ I2 (QL ) such that L < L∗ ∼ = L4 (4) for some component L∗ of CG (u). In the standard case, since m2 (QL ) ≥ 3, [V2 , 9.3c] is contradicted. Thus L∗ exists. But L is semirigid in G by [VK , 14.2], and t ∈ CG (y) normalizes L. Hence by [IG , 7.4] we may assume u chosen so that [u, t] = 1 and u ∈ Z(QL ). Then t ∈ CG (u) with E(CL∗ (t)) ≥ E(CL∗ (t, y)) = J. By [IA , 4.9.1, 4.9.2] and the Borel-Tits theorem, E(CL∗ (t)) ∼ = U4 (2), L4 (2), or Sp4 (4). As E(CL∗ (t)) ≤ K it must be that E(CK (u)) = E(CL∗ (t)) ∼ = U4 (2). However, as u ∈ Z(QL ), m2 (CG (u)) ≥ m2 (L) + m2 (QL ) ≥ 4 + 3 = 7, so (2G) is satisfied with u in place of y. This contradicts Lemma 2.35, and the proof is complete.  Now, if (2G) holds, Lemmas 2.32, 2.37, and 2.38, and [VK , 3.10l], force either LLt ∼ = A6 × A6 or (2H)

(2G) holds and J < L ∈ {Sp4 (4), L4 (2), L5 (2), HS, 2HS}.

In each case of (2H), J contains a Sylow 3-subgroup of L, and we fix (see [IA , 4.8.2]) x ∈ I3 (J) such that I := E(CL (x)) ∼ = A5 , and set Cx = (2I) CG (x), C x = Cx /O3 (Cx ), and P = O3 (Cx ). ∼ J or We let K1 be the subnormal closure of J in Cty := CG (ty). Thus K1 = J × J, or K1 ∈ ↑2 (A6 ) ∩ C2 , which set is enumerated in [VK , 3.10]. Lemma 2.39. Assume (2H). Then the following conditions hold: (a) I t ∼ = Σ5 ; (b) O3 (Cx ) = 1; and (c) P = Ω1 (P ), P/ x ∼ = E3a , a ≤ 6, and y inverts P/ x.

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256

6. THEOREM C5 : STAGE 4

Proof. As J ≤ L, t centralizes a Sylow 3-subgroup of CL (x). But in each  case x × I = O 3 (CL (x)) [IA , 4.8.2, 5.3m], so t centralizes a Sylow 3-subgroup of I. As CL (t, x) ≤ CCt (x), a solvable group, [I, t] = 1. Hence, I t ∼ = Σ5 , proving (a). Note that since C(t, K) = t, (Lemma 2.22), F ∗ (CCt (x))/ t is a 3-group by the Borel-Tits theorem, and so O3 (CCt (x)) = t. As t ∈ O3 (I t), it follows that O3 (Cx ) has odd order. Let Y = O3 (Cx ). Since x ∈ J ≤ K1 , C(ty, K1 ) ≤ Cx , so CY (ty K1 ) ≤ O2 (C(ty, K1 )) ≤ O2 (CG (ty)) = 1. As O{2,3} (CAutG (tyK1 ) (x)) = 1 by [VK , 7.13], CY (ty) = 1. Similar arguments show that CY (t) = CY (y) = 1, and (b) follows. Recall that P = O3 (Cx ). We similarly find that CP (ty K1 ) ≤ O3 (C(ty, K1 )) ≤ O3 (Cty ) = 1, and as Out(K1 ) is a 3 -group, CP (ty) embeds in O3 (CK1 (x)). Hence by [VK , 12.51], CP (ty) ∼ = 31+2 , Z3 × 31+2 , or E3n for some n ≤ 4. Similarly, we have the same possibilities for CP (t). On the other hand, given the choice of x ∈ J ≤ L,  CP (y) = x. Since P = CP (y)CP (t)CP (ty), (c) follows. Lemma 2.40. Assume (2H). Then [P, I] = 1. Proof. Suppose false. Then [I, E(Cx )] = 1. Since CCt (x) = CCx (t) is solvable, t normalizes all components of E(Cx ). As I = [I, t] = I (∞) , I ≤ E(Cx ) and the normal closure of I in E(Cx ) has the form M M y for some component M of Cx . Of course, I is a component of CM M y (y). Since [I, O3 (Cx )] = 1, M is the image of a component M of Cx . If M ∈ C3 , then by [VK , 3.80b], the only possibility is M ∼ = A9 , with y acting as a root involution. But then t acts on M as the product of three disjoint transpositions, and F ∗ (CCx (t))/ t is not a 3-group, contradiction. Therefore, M ∈ C3 . By our general hypothesis, this implies that x ∈ I3o (G), so that m3 (Cx ) ≤ 3. Suppose that I < M . We have I   CM (y), m3 (M x) ≤ 3, and F ∗ (CM t (t))t is a 3-group. Hence by [VK , 3.94], M ∼ = SL3 (4) or L3 (4), with CM (t) ∼ = SU3 (2) or ∼ V ∈ Syl U3 (2), respectively. Let Q8 ∼ (C (t)) and v = Z(V ). As Out(K) = = D8 M 2 and v = t, v induces a nontrivial inner automorphism on K. But x ∈ CK (v) and so by [VK , 9.16a], m3 (CK (x)) = 4, contradiction. Hence, M is a diagonal or trivial 2-pumpup of I. Let UK ∈ Syl3 (CK (x)) and UK ≤ U ∈ Syl3 (CG (x)). Since x ∈ J and m3 (UK ) = 3, it follows from [VK , 9.26] that UK ≥ V = x × V0 with V0 ∼ = 31+2 . As m3 (U ) = 3, the components y of M M are normalized by U . But M M y x contains a Sylow 3-subgroup of J, which therefore lies in Z(UK ) but is disjoint from Φ(U ) and hence from Φ(UK ). Hence m3 (UK ) ≥ m3 (Z(UK )V0 ) ≥ 4, contradiction. This completes the proof.  By [VK , 12.35], there exists an involution s ∈ L t such that [I t , s] = 1 and s inverts x. Then s, y ∼ = E22 . Lemma 2.41. Assume (2H). Let F be an I t, y, s-invariant subgroup of P that is minimal subject to [F, I] = F . Then F ∼ = E34 , and F is isomorphic to the core of the natural permutation F3 I-module. Moreover, F ≤ Z(P ).

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Proof. Set P = P/ x. By [VK , 5.11], the nontrivial irreducible F3 A5 modules are of dimensions 4 and 6. Neither one supports an invariant symplectic form, in the first case since A5 does not embed in Sp4 (3) (which has 2-rank 2), and in the second case by [VK , 5.11]. Suppose F is not elementary abelian. Then X := HomF3 A5 (F ∧ F , x) = 0. If F is irreducible, this contradicts what we just saw. Otherwise F is indecomposable with V := F /CF (I) of dimension 4. But then F ∧ F has a filtration with quotients isomorphic to V ∧ V and V , and hence X = 0, contradiction. Thus F is elementary abelian, and as |F x / x | ≥ 34 , we have 35 ≤ |F | ≤ 37 . As F is indecomposable, some involution of s, y must centralize F . Thus m3 (F ) ≤ m2,3 (G) = 4. Therefore x ∈ F and F is the core of the natural permutation F3 I-module. Since |P : F | ≤ 33 by Lemma 2.39c, P/F is a trivial I-module. As F is abelian, commutation determines a morphism of I-modules from (P/F ) ⊗ F to x. As P/F is a trivial module and F is irreducible, that morphism can only be trivial. Hence F ≤ Z(P ).  Set F ∗ = F × x. For some t1 ∈ {t, ty}, t1 is a reflection on F and so E := CF ∗ (t1 ) ∼ = E34 with x ∈ E. We fix this notation. Lemma 2.42. Assume (2H). Suppose that P is elementary abelian, and x ∈ E1 ≤ P with either E1 = E ≤ F ∗ or P = E1 F ∗ . Then P = J(O3 (CG (E1 ))) and NG (E1 ) ≤ NG (P ). This applies for E1 = CP (τ ) for any τ ∈ {t, ty}. In particular, NG (E) ≤ NG (P ). Proof. As x ∈ E1 , we have CG (E1 ) ≤ Cx . Since P = O3 (Cx ) ∼ = E3m with 5 ≤ m ≤ 7, and m2,3 (G) = 4 and O3 (Cx ) = 1, we have P = F ∗ (Cx ). As y inverts P/ x, Cx = P CG (x, y). Let Uy = C(y, L). Then as x ∈ L, O 2 (CG (x, y)) = O 2 (Uy ) × I. Since [Uy , I] = 1, [F ∗ , Uy ] = 1. Note then that if P = E1 F ∗ , then CUy (E1 ) = CUy (P ) = 1. So in any event, our hypotheses imply that Either Uy ≤ CG (E1 ) or Uy ∩ CG (E1 ) = 1. As O3 (Uy ) = 1, it follows in either case that O3 (CG (E1 )) ≤ P O3 (CI (E1 )). Any element of I3 (I) has a Jordan block of size 3 on P , whence in any case P = J(O3 (CG (E1 ))) and so NG (E1 ) ≤ NG (P ). Now, CP (t1 ) = E. If τ = t1 , then as both t1 and y invert P/F ∗ , P = CP (τ )F ∗ . But x ∈ CP (τ ) so we may take  E1 = CP (τ ) to complete the proof. We set N = NG (P ), N = N/CG (P ), and NE = NK (E). Then I is a component of CN (y). Since |P | ≤ 37 , |N |5 = 5, and so I lies in a 2-component H of N , by L2 -balance. Lemma 2.43. Assume (2H) and suppose that t1 = t. Then P is elementary abelian and A6 ∼ = N E ≤ H. Moreover, O2 (CN (t)) = 1. Proof. Suppose that t1 = t. Then E ≤ O 2 (Ct ) = K and so by [VK , 9.16b], E = CK (E) and NE /E ∼ = A6 . One consequence is that as |N |5 = 5, N E ≤ H. Another is that N E does not normalize x. As Φ(P ) ≤ x, it follows that Φ(P ) = 1. Finally, CN (t) normalizes CP (t) = E, and as E = O2 (NCt (E)), we  have O2 (CN (t)) = 1. The lemma is proved.

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6. THEOREM C5 : STAGE 4

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Lemma 2.44. Assume (2H). Then Φ(P ) = O2 (CN (t)) = 1. Moreover, if t1 = ty, then the following conditions hold: (a) K1 ∼ = U4 (3), U5 (2), or A6 × A6 ; (b) AutN (E) contains A6 , Σ5 , or Z4 × Z4 , respectively. Proof. By Lemma 2.43, we may assume t1 = ty. So m3 (CG (ty)) ≥ 4. From the structure of Ct , we have CG (J t, y) = t, y, v, an elementary abelian group, with either v = 1 or E(CK (v  )) ∼ = U4 (2) for all v  ∈ t, y, v − t, y. Suppose that v = 1. Then unless L/Z(L) ∼ = HS, m2 (CAutL (J)) = 1, and replacing v by a suitable element of t, y, v − t, y if necessary, we may assume that [L, v] = 1. Let Kv = E(CK (v)) ∼ = U4 (2) and let H0 be the subnormal closure of J (and Kv and L) in CG (v). Then as Kv ↑2 H0 , H0 ∼ = L± 4 (3), 2U4 (3), or L4 (4), by [VK , 3.10]. But also L ↑2 H0 . The only possibility is H0 ∼ = L4 (4), with t and y respectively inducing graph-field and field automorphisms on H0 . If L/Z(L) ∼ = E22 , and each u ∈ I2 (CAut(L) (J)) satisfies = HS, then CAut(L) (J) ∼ ∼ E(CL (u)) = J or A8 according as u ∈ Inn(L) or not. Arguing as in the previous case, we deduce that v may be chosen so that for some component H0 of CG (v), Kv ∼ = U4 (2) is a component of CH0 (t) and Lv ∼ = A8 or L is a component of CH0 (y). The only possibility again is H0 ∼ = L4 (4), with t and y acting as in the previous paragraph, and Lv ∼ = A8 . Since m2 (H0 ) = 8, (v, H0 ) ∈ H, and Lemma 2.35 is contradicted, with v and Kv in the roles of y and J. We conclude that v = 1. Let S be a t-invariant Sylow 2-subgroup of C(ty, K1 ). Then t CS (t) ≤ t, y, v. If CS (t) = ty, then S t is of maximal class, so either t ∼S t(ty) or S = ty.  CG (y), so S = ty. As E ≤ CG (ty), m3 (K1 ) ≥ 4. Hence by [VK , But CG (t) ∼ = 3.10] and L2 -balance, K1 ∼ = L± 4 (3), 2U4 (3), U5 (2), or A6 × A6 . By our choice of K, only U4 (3), U5 (2), and A6 × A6 are possible. Hence E ≤ K1 , with AutN (E) containing A6 , Σ5 , or Z4 × Z4 , respectively. As in the previous lemma, x is not AutN (E)-invariant, so P is elementary abelian. Also again, O2 (NCG (ty) (E)) = E and so O2 (CN (ty)) = 1.  

Lemma 2.45. Assume (2H). Then I < H := O 5 (N )  E(N ), and H is quasisimple. Moreover, t1 = ty, K1 ∼ = U5 (2) or A6 × A6 , and m3 (P ) ≤ 6. Proof. By Lemma 2.44, CO2 (N ) (t1 ) ≤ O2 (CN (t1 )) = 1. It follows that O (N ) is inverted by t1 and then centralized by [I, t1 ] = I. Now CN (y) normalizes CP (y) = x, so CN (y) = CNG (x) (y) and hence I  CN (y). By the previous paragraph and L2 -balance, I ≤ E(N ). As |N |5 = |I|5 = 5, 2



E(N )

H = O 5 (N ) = I  is quasisimple. Suppose that either t1 = t (whence CP (t) = E), or t1 = ty and K1 ∼ = U4 (3). Then O 2 (CN (t1 )) = E(CH (t1 )) ∼ = A6 . But E(CH (y)) = I ∼ = A5 , and so H is a 2-pumpup of both A5 and A6 , contrary to [III11 , 1.6ab]. Therefore t1 = ty and K1 ∼ = U5 (2) or A6 × A6 . If m3 (P ) = 7, then as both ty and y invert P/CP (ty) and CP (ty) ≤ F ∗ , we have that P = CP (t)F ∗ and m3 (CP (t)) = 4. Then CP (t) ≤ K and NK (CP (t))/CP (t) ∼ = A6 . Also, NG (CP (t)) ≤ NG (P ) by Lemma 2.42, so NK (CP (t)) maps onto E(CH (t)), whence again H is a 2-pumpup of both A5 and A6 , again a contradiction. The proof is complete.  Lemma 2.46. Assume (2G). Then LLt ∼ = A6 × A6 .

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2. THE CASE G∗ = Ω7 (3) OR Ω± 8 (3)

259 ∗

Proof. Assume false and continue the above argument. Let N = N ±1 ≤ ∗ GL(P ). Now apply [VK , 5.1] to N ; the reflection in question is y. It follows that N ∼ = Σ7 or Σ7 × Z2 , and |P | = 36 . But if K1 ∼ = U5 (2), then t1 centralizes an A5 subgroup of N , and so acts as a transposition or its negative on P . As CP (t1 ) ∼ = E34 , this is a contradiction. Therefore, K1 ∼ = A6 × A6 . If I   N , then I  N  by the 5-structure. Hence  N [ x , I] = 1 so m3 ( xN ) ≤ 3. On the other hand xN ∩K1 has rank 4, contradiction. Therefore I is proper in a component of N . By [VK , 5.1], N ∼ = Σ7 or Σ7 × Z2 . Thus N has no Z4 × Z4 subgroup, a contradiction.  Lemma 2.47. Assume (2G). Then m2 (C(y, LLt )) = 1. Moreover, there exists a subgroup of CG (y) containing LLt and isomorphic to Σ6 × Σ6 . Proof. Since m3 (LLt ) = 4, m2 (C(y, LLt )) = 1 by (c) of Theorem C5 : Stage 2. Then since the only subgroups of Aut(A6 ) of 2-rank 3 contain Σ6 , the second assertion follows from our assumption that m2 (CG (y)) ≥ 7.  ∼ Ei ≤ L with x = E1 ≤ E2 , and set Ci = CG (Ei ), i = 1, 2. Choose E3i = Then Ci ∩ CG (y) contains some B ∈ B3∗ (G). By our basic setup B ∈ S3 (G), so m3 (Ci ) ≥ m3 (CG (B)) > 4. As m2,3 (G) = 4, O3 (Ci ) ≤ O2 (Ci ), i = 1, 2. 2 = C2 /O3 (C2 ). Then Lt ≤ L2 (CC (y)) ≤ L2 (Ci ), Set C 1 = C1 /O3 (C1 ) and C i i = 1, 2, by L2 -balance. , with M /O3 (M ) ∼ t < M Lemma 2.48. Assume (2G). Then L = P Sp4 (3) or ± t    L4 (3) for some component M of E(C2 ), and L is a component of CM (y).  t ≤ E(C 2 ). For otherwise, using the facts Proof. Clearly Lt ≤ C2 . Indeed, L that m2 (CG (y)) ≥ 7 and m2 (C(y, LLt )) = 1, we have that C2 contains a copy 2 ), and the Thompson dihedral lemma yields of Σ6 × Z2 acting faithfully on O3 (C t ≤ M M y for some component m2,3 (G) ≥ 3 + m3 (E2 ) = 5, contradiction. Thus, L  of E(C 2 ), and L  t is a component of C  y (y). M MM t = M . Using (c) of Theorem C5 : Stage 2, we see that M  has no Suppose that L 3    more than two C2 -conjugates, so M  O (C2 ). Let P ∈ Syl3 (C2 ). It follows that P ∩M ∼ = E32 is a direct factor of P . By symmetry, E2 is a direct factor of a Sylow 3-subgroup of CG (P ∩ M ), and so E2 is a direct factor of P . As m3 (C2 ) ≥ 5 and in ) = E2 × Y = P ∩ O3 3 (C2 ), where Y is view of (c) of Theorem C5 : Stage 2, CP (M nontrivial and cyclic; moreover, P is y-invariant and we may assume that Y = [P, y]. D/   y ∼ However, there exists D ∼ = Σ6 . Then = E24 such that y ∈ D ≤ C2 and M D normalizes Y and a hyperplane D0 of D centralizes E2 × Y , contradicting (c) of t < M M y . Theorem C5 : Stage 2. So L y    2 × M ×M y with M ∼ But if M = M , then C2 contains E = A6 , and part (c) y =M  . of Theorem C5 : Stage 2 is contradicted. Therefore M  t ) is y. As M  ∈ C3 , the only possibilities Also the only involution in CM (L  y ± /O3 (M ) ∼ by [VK , 3.42] are M = P Sp4 (3), U5 (2), L4 (3), L2 (34 ), or L± 3 (9). An E26 t ∼   subgroup of CG (y) normalizes E2 and L , hence acts on M . If M = U5 (2), L2 (34 ),  or L± 3 (9), then m2 (Aut(M )) ≤ 4, so NG (E2 ) contains E34 × E22 , contradicting part  (c) of Theorem C5 : Stage 2. This completes the proof.

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6. THEOREM C5 : STAGE 4

260

We let M1 be the pumpup of M in C1 . Then we have m3 (M )−m3 (O3 3 (M )) ≥ 3 and m2,3 (M1 ) − m3 (O3 3 (M1 )) ≥ 2, and so as m2,3 (G) = 4, m3 (CC1 (M 1 )) ≤ 2.

(2J)

Also m3 (E1 M1 ) ≥ 4, so by part (c) of Theorem C5 : Stage 2, m2 (CC1 (M 1 )) ≤ 1.

(2K)

Now return to the possible pumpups of L. We have m2 (C(y, L)) ≥ m2 (Z2 × Σ6 ) = 4, by Lemma 2.47. We shall construct a (finite) pumpup chain (y, L) = (y0 , L0 ) < (y1 , L1 ) < · · · < (yn , Ln ), all terms of which are y-invariant, and study the final term of the chain. At each stage we will choose Qi ∈ Syl2 (C(yi , Li )), and we will have yi+1 ∈ Qi . Since m2 (C(y, L)) ≥ 4, L is not terminal in G by [V2 , 9.3c]. If L0 is not a component of CG (y1 ) for some y1 ∈ I2 (Q0 ), set j = 1 and let L1 be the pumpup of L0 in CG (y1 ). Note that L0 is semirigid in G by [VK , 14.3] (recall our choice (2F)). Hence, we may choose y1 ∈ Z(Q0 ) by [IG , 7.4], and then have Q0 = y (Q0 ∩ Q1 ). Then y centralizes y1 and L0 . If L0 is a component of CG (y1 ) for every y1 ∈ I2 (Q0 ), set j = 2 and L1 = L0 and fix y1 to be central in Q1 ∈ Syl2 (CG (L0 )). Then, again using [VK , 14.3] and [IG , 7.4], choose y2 ∈ I2 (Z(Q1 )) such that L1 pumps up to L2   CG (y2 ) and Q1 = y1  (Q1 ∩ Q2 ) for some Q2 ∈ Syl2 (C(y2 , L2 )). Now y ∈ Q0 ≤ Q1 so y2 and L1 are y-invariant. In neither case can the pumpup (yj−1 , Lj−1 ) < (yj , Lj ) be a diagonal pumpup. For if it were, then m2 (Qj ) ≥ m2 (Qj−1 ) − 1 ≥ m2 (Q0 ) − 1 ≥ 3; hence CG (yj ) would contain E34 × E23 , contradicting part (c) of Theorem C5 : Stage 2. So Lj−1 < Lj , and Lj is y-invariant. We may then assume as well that Qj is y-invariant. If Lj is terminal, we set n = j and have our chain. Otherwise we proceed as in the next paragraph and set n = k, completing our chain. Namely, when Lj is not terminal, a similar construction, starting with (yj , Lj ) instead of (y0 , L0 ), produces a vertical pumpup (yk−1 , Lk−1 ) < (yk , Lk ). Here k = j + 1 or j + 2, and in the latter case Lj = Lj+1 and yj+1 ∈ Z(Qj+1 ) with Qj+1 ∈ Syl2 (CG (Lj )). We have m2 (Qk ) ≥ m2 (Qj ) − 1 ≥ 2. Moreover, since (yj , Lj ) is y-invariant, and in view of [IG , 7.4c(iii)], we may assume that (yk , Lk ) is y-invariant. Lemma 2.49. Assuming (2G) and with the above notation, Lj is terminal in G. Moreover, Lj ∼ = L4 (2), L5 (2), or Sp4 (4). Proof. Suppose Lj is not terminal in G, and consider (yk , Lk ). As G is of even type, L0 ≤ L1 ≤ · · · ≤ Lk , and so L0 = L ∼ = A6 is a component of E(CLk (y)). Also Lj is a component of E(CLk (yj )) and L0 is a component of E(CLj (y)). Since m2 (Qk ) ≥ 2, m3 (Lk ) < 4 by part (c) of Theorem C5 : Stage 2. It follows from [VK , 3.41] that Lk ∼ = HS or 2HS. Since HS contains E32 × Z2 , Lk has no diagonal pumpups; if it did, part (c) of Theorem C5 : Stage 2 would be contradicted. As ↑2 (Lk ) ∩ C2 = ∅ by [VK , 3.10], Lk is terminal in G. But this contradicts [V2 , 9.3a]. This proves terminality. We have m2 (C(yj , Lj )) ≥ 3. Hence by [V2 , 9.3, 9.1], Lj ∈ Chev(3) ∩ C2 and  HS. The second assertion then holds by [VK , 3.43].  Lj /Z(Lj ) ∼ = Now we finally can prove:

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Lemma 2.50. (2G) does not hold. Proof. Suppose false and continue the above argument. Now E1 ≤ Lj and we may suppose E1 to have been chosen so that H := E(CLj (E1 )) ∼ = A5 . By L2 balance, H either acts nontrivially on O3 (C 1 ) or lies in L3 (C 1 ). In the former case, because of the action of the 5-element, m3 (O3 (C 1 )) ≥ 3, contradicting (2J). In the latter case, as m2 (H) > 1, H must lie in M 1 by (2K). Then M 1 is simultaneously a , with M /Z(M ) ∼ 2-pumpup of A5 , and a 3-pumpup (possibly trivial) of M = P Sp4 (3) ± or L4 (3), by Lemma 2.48. However, there are no such groups M 1 , by [VK , 3.48], a final contradiction.  Now, assuming K ∼ = U4 (3) (recall (2F)), we divide the argument into two cases: (1) V1 = ∅; and (2L) (2) V1 = ∅. We shall reach a contradiction in each case. Recall from Lemma 2.22 that g ∈ NG (T ) satisfies g 2 ∈ T and tg = tz. Since I2 (K) = z K , I2 (Kt) ⊆ tG . If (2L2) holds, we fix v ∈ V1 , while if (2L1) holds, we set v = 1. In either case we choose F ∈ E24 (CK (v)) and set E = F, t, v. By [VK , 10.10], F is the core of a natural AutK (F ) ∼ = A6 -permutation module if (2L2) holds, and it can be so chosen if (2L1) holds. Subject to this, if possible, we make the choice of F so that F ≤ K g . We then set N = NG (F ) ∩ NG (E) ∩ CG (E/F ). As AutK (F ) is transitive on F # , F # ⊆ z G . Lemma 2.51. The following conditions hold: (a) F ≤ K g ; (b) N permutes F t 2-transitively; and (c) CG (E) = CG (F t) = E. Proof. Clearly F ≤ CG (tz) = Ctg . Suppose that F ≤ K g and choose z1 ∈ F − K g , so that z1 ∈ z G . By Lemma 2.23, E(CK g (z1 )) = 1. Since z1 ∈ K g and  I2 (K g tz) ⊆ tG , it follows from [IA , 4.5.1] that O 3 (CK g (z1 )) ∼ = A4 × A4 , and then from (2E6) that (2L2) holds. Now Q∗ := O2 (O 2 (CG (t, z))) ∼ = Q8 ∗ Q8 is g-invariant by Lemma 2.22b, so Q∗ ≤ K g . Hence by [VK , 10.10], z1 interchanges the two Q8 central factors of Q∗ = O 2 (CK g (z)). It then follows with [VK , 10.14] that   O 3 (CK g (tz1 )) = O 3 (CK g (tz.zz1 )) ∼ = P Sp4 (3). As tz1 ∈ (tz)K ⊆ tG , we conclude that there is s ∈ V1 and h ∈ G such that s = th . Choose F  ∈ E24 (E(CK (s))) and set E  = F  , t, s. Note that F  ≤ E(CK (s)) = E(CG (s, t)) = E(CK h (t)). Thus, E(CK h (t)) ∼ = P Sp4 (3). Let M be the normalizer in G of the chain t F  > F  > 1. It follows that M ∩ K h does not centralize t. But as F  ≤ K, M ∩ K is transitive on tF  − {t}. Hence, M is transitive on tF  . In particular there is g  ∈ M such that    tg = tz. Then F  = (F  )g ≤ K g = E(CG (tz)) = K g . This completes the proof of (a). Whichever case of (2L) holds, N0 := NK (E), NK g (E) lies in N since E ∩ K = F = E ∩ K g . In particular, N0 permutes tF = tzF , with NK (E) and NK g (E) transitive on tF − {t} and tF − {tz}, respectively. This implies (b). Set C = CG (F t) ≥ CG (E). Then C = CCt (F ) maps into CAut(K) (F ), which by [VK , 10.10b] is elementary abelian, generated by the image of E and an automorphism φ such that E(CK (φ)) ∼ = P Sp4 (3). Hence, (c) holds in case (2L2),

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and if it fails in case (2L1), then there is w ∈ C such that E(CK (w)) ∼ = P Sp4 (3) and w2 = t. But then t = Φ(C) is normalized by N , contradicting (b). Hence (c) holds and the lemma is proved.  Now set

N = N/E ∼ = AutN (E). Then by [VK , 10.10], N t := CN (t) ∼ = A6 or Σ6 . By Lemma 2.51b, |N : N t | = 24 . Lemma 2.52. O2 (N ) = CN (O2 (N )) ∼ = A6 or Σ6 . = E24 , and N /O2 (N ) ∼ = Nt ∼ ∼ Aut(F ) ∼ Proof. Let P = CAut(E) (E/F ), so that P/O2 (P ) = = L4 (2) and O2 (P ) = CP (F ) is the direct sum of one or two natural modules for P/O2 (P ). We regard N ≤ P . The natural P/O2 (P )-modules are irreducible for the image of N t in P/O2 (P ). It follows that |N ∩ O2 (P )| = 24 or 1. Since 24 = |N : N t | does not divide |L4 (2) : A6 |, N ∩ O2 (P ) ∼  = E24 and the lemma follows. Let S1 be the preimage of O2 (N ) in N , so that S1 /E ∼ = E24 . Lemma 2.53. Z(S1 ) = F . Proof. Clearly F ≤ Z(S1 ) and Z(S1 ) ≤ CG (E) = E. Also S1 is transitive on F t, so if the lemma fails, then there is v ∈ Z(S1 ) − F t, whence (2L2) holds and by [VK , 10.10], E(CK (v)) ∼ = P Sp4 (3) with CK∩N (v)/F ∼ = A5 . Let L be the subnormal closure of E(CK (v)) in CG (v). Since m2,3 (P Sp4 (3)) = 2 < m3 (P Sp4 (3)) while m2,3 (G) = 4, L is not a diagonal pumpup. If L ∼ = L4 (4), then m2 (L) = 8 by [IA , 3.3.3], so Lemma 2.50 is contradicted. Hence, by [VK , 3.10] and our choice (2F), L ∼ = P Sp4 (3) or U4 (3), and again as m2,3 (G) = 4, L  CK (v). But CK∩N (v) ≤ L, so S0 := [S1 , CK∩N (v)] ≤ L. Therefore |L|2 ≥ |S0 | ≥ 28 , which is not the case. The lemma is proved.  Lemma 2.54. Let s ∈ E − F . Then there is Ks  CG (s) with Ks ∼ = K, F = E ∩ Ks , and (N ∩ Ks )/F ∼ = A6 . In particular, we have symmetry between s and t. Proof. The result is obvious for s = t, and, since N is transitive on F t, it follows for any s ∈ F t. Thus we may assume s ∈ E − F t, whence (2L2) holds. Replacing s by a suitable element of F s, we may assume by [VK , 10.10] that E(CK (s)) ∼ = P Sp4 (3) and CK∩N (s) covers an A5 -subgroup N 1 /S 1 of N /S 1 , and N 1 is irreducible on S 1 . Hence either s ∈ Z(S1 ) or F s is completely fused by S1 . By Lemma 2.53, the latter holds. In particular by a Frattini argument, CN (s) covers N /S 1 . Because of our choice (2F), the pumpup Ls of E(CK (s)) in CG (s) is isomorphic to U4 (3), and as in the previous lemma, Ls  CG (s). Then F = [F, CN (s)(∞) ] ≤ Ls , and the lemma follows for s. As F s is totally fused by N , we are done.  The following corollary will be important when case (2L2) holds. Lemma 2.55. Suppose that case (2L2) holds. Let s ∈ I2 (Ct ) be such that  O 3 (CK (s)) ∼ = A4 × A4 . Then E(CG (s)) ∼ = U4 (3) and we have symmetry between s and t. Proof. Let T1 ∈ Syl2 (Aut(K)) with T1 containing the image of R s (where R = T ∩ K). By [VK , 10.17ab], F has exactly two conjugates under T1 , just one of which centralizes an involution of Rs. Without loss we may assume this conjugate

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is F itself. By [VK , 10.17bc], there is v ∈ V1 ∩ Rs centralizing F , and there exists  s1 ∈ sK such that s1 ∈ v, t F . The result then follows from Lemma 2.54. In the next lemma, Ja (S1 ) is the subgroup of S1 generated by all abelian subgroups of S1 of maximal order [IG , p. 69]. Lemma 2.56. Let D = Ja (S1 )  N . Then D ∼ = E28 , CD (t) = F = [D, t], and N = DCN (t). Proof. First suppose that (2L1) holds, so that |S1 | = 29 . By Lemma 2.53,  = N/F . In this case, we have F t = CS (t). Hence t is F = Z(S1 ), and we put N 1 not a square in S1 , nor is any element of tN = F t. Therefore, Φ(S1 ) ≤ F , so S1 is elementary abelian. The mapping x → [x, t] is then an N/S1 isomorphism between S 1 and F . Let B ∈ Syl3 (N ), so that B = B1 × B2 with B1 ∼ = B2 ∼ = Z3 and CS 1 (Bi ) = ∼ [CS 1 (Bi ), B3−i ] = E22 , i = 1, 2. Therefore Ui := [CS1 (Bi ), B3−i ] has order 16 and class at most 2 with CUi (B3−i ) = 1, whence Ui is abelian. Set D0 = [S1 , B] = U1 U2 , of order 28 . By [V2 , 2.9], D0 is abelian. Hence, D0 = Ja (S1 ), for otherwise, S1 = D0 D1 for some D1 with |D1 | = 28 , and so D0 ∩ D1 ≤ Z(S1 ) = F , which is absurd as |F | = 24 < 27 = |D0 ∩ D1 |. Therefore D0 = D  N and D is homocyclic of exponent 2 or 4, with S1 = D t and [D, t] = CD (t) = F . As S1 CN (t) = N , N = DCN (t). Next suppose that (2L2) holds. Again set D0 = [S1 , B]. Again x → [x, t] is an N/S1 -isomorphism from S 1 to F , so t is not a square in S1 , nor is any element of tN = F t. By the symmetry in Lemma 2.54, the same statement holds with any s ∈ E − F in place of t. Hence again Φ(S1 ) ≤ F , so S1 is elementary abelian. Again using [V2 , 2.9] we see that D0 is abelian. We argue that D0 = Ja (S1 )  N . As F = Φ(S1 ) ≤ D0 , D0  S1 . Suppose D1 ≤ S1 is abelian, D1 ≤ D0 , and |D1 | ≥ |D0 |. Then s ∈ D0 D1 for some s ∈ E − F . Hence |CS1 (s)| = 26 , so |Z(D0 D1 )| ≤ 25 . But D0 ∩ D1 ≤ Z(D0 D1 ) with |D0 ∩ D1 | ≥ |D0 ||D1 |/|S1 | = 26 , contradiction. Thus, D0 = D. Again [D, t] = CD (t) = F and S1 = DE = DCS1 (t) so N = DCN (t); and D is homocyclic of exponent 4 or 2. It remains to show in either case that D is elementary abelian. Suppose false, so that D∼ = Z4 × Z4 × Z4 × Z4 . ∼ BY ≤ NK (F ) ≤ N acting on D, with There exists a Frobenius subgroup F9.4 = |Y | = 4. Then Y cycles a basis of D/F and therefore it cycles a basis of D over Z4 . Hence det does not vanish on Y , contradicting the fact that NK (F ) is perfect. The lemma is proved.  Lemma 2.57. AutNG (K) (F ) = AutK (F ) ∼ = A6 . Proof. Suppose false. Since t ∈ Syl2 (CG (K)), Σ6 ∼ = AutNG (K) (F ) = AutCt (F ) > AutK (F ) by a Frattini argument and by [VK , 10.10]. Also, by [VK , 10.15], there is x with the following properties:

(2M)

(1) (2) (3) (4) (5)

x2 ∈ t; x ∈ CN (t, z) induces transvections on E and F ; x normalizes O2 (CK (z)) and [E, x] = z; 3 divides CK∩N (x); and E(CK (x)) = 1.

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264

6. THEOREM C5 : STAGE 4

In particular, since d → [d, t] is a CN (t)-isomorphism from D/F to F , x induces a transvection on D/F by (2M2). Suppose that x2 = t. Let Dx be defined by Dx /F = CD/F (x), and choose d ∈ Dx − [D, x]F/F . Then [d, x] ∈ F and [F, x]  [d, x, x] = [d, x2 ][d, x]2 = [d, t] ∈ [F, x], a contradiction. Therefore by (2M1), x is an involution. Let y = x or xt, and write F f = [D/F, y] with f ∈ [D, y] ≤ CD (y). Using (2M4), let W ≤ CK∩N (x) with |W | = 3. Then CD/F (y) = [D/F, W ] × F f  and |[D, W ]| = 24 . Suppose that [D, W, y] = 1. Then C[D,W ] (y) = [F, W ], so CD (y)F/F = [D/F, y]. Let Y = O2 (CK (z)). Then AutY (F ) is a four-group acting freely on F , by [VK , 10.16]. In particular Y ≤ N , so F is a direct summand of D as Y -module, and Z := CD (Y ) is a four-group with CZ (t) = z. But x normalizes Y , as does t, so x, t normalizes Z. As [Z, t] = 1, either y = x or y = tx centralizes Z. Hence [D/F, y] = ZF/F . Taking the commutator with t, we get [F, y] = [Z, t] = z. As [F, t] = 1, [F, x] = z, contradicting (2M3). Therefore [D, W, y] = 1. It follows that CD (y) covers CD/F (y), and so |CD (y)| = 6 2 . As y ∈ D, m2 (CG (y)) ≥ 7. As E(CK (y)) = 1 by (2M5), (2G) holds for some L, contradicting Lemma 2.50. The proof is complete.  Lemma 2.58. S1 ∈ Syl2 (CG (F )) and D = CG (D). Proof. Suppose that u is a 2-element of CG (F ) normalizing S1 . Then u normalizes J(S1 ) = D and has a fixed point on S1 /D ∼ = E/F . By the symmetry in Lemma 2.54, we may assume that u normalizes D t. So u ∈ DCDt,u (t) ≤ DE ≤  S1 . This proves the first statement, and the second follows immediately. Now set X = NG (D) and X ∗ = X/D ∼ = AutG (D). For every element or subset U of X, write U ∗ for its image in X ∗ . Lemma 2.59. Case (2L1) holds, i.e., V1 = ∅. ∼ E22 . Lemma 2.54 and a Frattini arProof. Suppose false, so that E/F = gument show that CX ∗ (s∗ ) = (CX (s))∗ = E ∗ × M , E ∗ ∼ = A6 , for all = E22 , M ∼ s ∈ E − F , i.e., 1 = s∗ ∈ E ∗ . By [VK , 3.5] and L2 -balance, M  X ∗ . But also E ∗ ∈ Syl2 (CX ∗ (M )) so some s∗ is 2-central in X ∗ . Therefore E ∗ M contains U ∗ for some U ∈ Syl2 (X), and it follows easily that D = J(U ), whence U ∈ Syl2 (G). In particular by [IG , 16.9], X controls G-fusion in D. Also as U ∈ Syl2 (X) normalizes F , F  X. In particular z G ∩ D = F # . On the other hand, choose u ∈ I2 (NK (F )) − F . Then tu ∈ (tz)G = tG , and CD (tu) ≤ F . Write t = (tu)h , h ∈ G. Let z1 ∈ CD (tu) − F . We argue by a process of elimination that z1 ∈ z G , contradicting the previous paragraph. Note that m2 (CG (z1 )) ≥ m2 (D) = 8. Let K1 = E(CG (tu)) ∼ = K. If E(CK1 (z1 )) = 1, then z1h satisfies (2G), con tradicting Lemma 2.50. So E(CK1 (z1 )) = 1. If O 3 (CK1 (z1 )) ∼ = A4 × A4 , then by Lemma 2.55, m2 (CG (z1 )) < 8, contradiction. The only alternative, by [IA , 4.5.1],  is that z1 ∈ z G , and the proof is complete. Lemma 2.60. X ∗ ∼ = A6  Z2 with t∗ ∈ X ∗ − E(X ∗ ). Proof. Using Lemma 2.54 and a Frattini argument as in the proof of Lemma 2.59, we see that CX ∗ (t∗ ) = t∗  × M where M = (N ∩ K)∗ ∼ = A6 .

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Let Y ∗ = O2 (X ∗ ), so that CY ∗ (t∗ ) ≤ O2 (CX ∗ (t∗ )) = 1 and Y ∗ is abelian. Since Y ∗ embeds in Aut(D) ∼ = L8 (2), mr (Y ∗ ) ≤ 2 for all primes r > 3, and so as ∗ M contains A4 , [Or (Y ), M ] = 1. Likewise |O3 (Y ∗ )| ≤ |L8 (2)|3 /|M |3 = 33 , and ∗ M does not embed in L3 (3), so [Y ∗ , M ] = 1. Thus, by L2 -balance, M ≤ M1 M1t ∗ ∗ for some component M1 of E(X ). By the structure of CX ∗ (t ) and [VK , 3.40], M1 ∼ = M . As Out(M1 ) is a 2-group, it follows easily that X ∗ ∼ = A6  Z2 , as desired (note that any E34 -subgroup of X ∗ is self-centralizing), or t∗ ∈ Z ∗ (X ∗ ). Suppose then, for a contradiction, that X ∗ = t∗  Y ∗ × M . As M is absolutely irreducible on F and D/F , t∗  Y ∗ embeds in GL2 (2), so |Y ∗ | = 1 or 3. Let U ∈ Syl2 (X). Every involution of M acts freely on F and D/F , so every involution of X ∗ acts freely on D, whence D = J(U ) and then U ∈ Syl2 (G). In particular by [IG , 16.9], X controls G-fusion in D. Let u∗ ∈ I2 (M ). Then there is a preimage u ∈ I2 (CX (t)) of u∗ . We claim that CD (tu)# ⊆ z G . Indeed let y ∈ CD (tu)# . We have tu ∼Ct tz ∼G t and m2 (CG (y)) = 8. Write t = (tu)h and set y1 = y h . Then y1 ∈ Ct , and E(CK (y1 )) = 1 by Lemma  2.50. Hence E(CK (y1 )) = 1. By Lemma 2.59 and (2E6), O 3 (CK (y1 )) ∼ = A4 × A4 . Hence by [IA , 4.5.1], y1 ∈ z G , and our claim is proved. Hence, CD (tu)# ⊆ z X . But CD (tu) ≤ F , so F  X. This implies that Y ∗ = 1. 2 Let w ∈ I3 (X) with w∗  = Y ∗ . Then CD (tu)# ⊆ z X ⊆ F ∪ F w ∪ F w . But 2 2 |CF (tu)| = 4 and tu interchanges F w and F w . Hence |F w ∩F w | ≥ |F |−|CF (tu)| = 2  12, so F w = F w , which is absurd as F = F w . The proof is complete. Lemma 2.61. X ∼ = NK (F )  Z2 . Proof. By Lemma 2.60 there exist subgroups H1 , H2 of X such that D ≤ H1 ∩ H2 , H2 = H1t , and (H1 H2 )∗ = H1∗ × H2∗ ∼ = A6 × A6 . There exist fi ∈ I5 (Hi ), ∗ . Using [IA , 4.8.2] i = 1, 2, such that [f1 , f2 ] = 1. Then Ci := CH3−i (fi ) covers H3−i ∼ we see that this forces D3−i := CD (fi ) = E24 and Di = [Di , Hi ] for i = 1, 2. As ∗ -invariant, [Di , H3−i ] = 1. Now [Hi , Hi ] ∩ D = Di , and Di = [Di , fi ] is Ci∗ = H3−i so H1 H2 = [H1 , H1 ] × [H2 , H2 ] and X ∼ = [H1 , H1 ]  Z2 . Then [H1 , H1 ] ∼ = CH1 H2 (t) ∼ =  NK (F ), and the lemma is proved. Now we can finally complete the proof of Proposition 2.31. We continue the above argument to a final contradiction. Write X = (X1 × X2 ) t with Xi ∼ = NK (F ) and X2 = X1t . Let Yi ∈ Syl3 (Xi ), Y = CY /O3 (CY ). Notice that Y1 Y2 ∈ B3∗ (G), CYi = CG (Yi ), i = 1, 2 and C 1 1 1 3 whence Y1 Y2 ∈ S (G) by Theorem C5 : Stage 1, and so m3 (CY1 ) > 4. Hence 2 ∼ Y )) = 1. Then by O3 (CY1 ) has odd order and so X = X2 . Suppose CX2 (O3 (C 1 the Thompson dihedral lemma, CY1 contains the direct product of Y1 ∼ = E32 and four copies of Σ3 , so m2,3 (CY1 ) ≥ 5 > m2,3 (G), contradiction. As O2 (X2 ) is the 2 ), O3 (C Y )] = 1, unique minimal normal subgroup of X2 , it follows that [O2 (X 1     and X2 acts faithfully on E(CY1 ) = L3 (CY1 ). As m2,3 (G) = 4, CY1 has at most 5 components, which must be X2 -invariant as X2 /O2 (X2 ) ∼ = A6 . By the Schreier Y ) and so there is a component L  1 of C Y on which X2 faithfully 2 ≤ E(C property, X 1 1 induces inner automorphisms. Then as m3 (Y1 Y2 ) = 4, elements of Y1# lie in I3o (G), and so with [III11 , 1.1c],  1 ∈ C3 . Moreover, L  1 /Z(L  1 ) contains an isomorphic copy of X2 , and m2,3 (L 1 ) − L  m3 (Z(L1 )) ≤ 2 as m2,3 (CY1 ) ≤ m2,3 (G) = 4. It follows then from [VK , 15.16] that  1 /Z(L 1 ) ∼ L = U4 (3) or M c. In particular m3 (L1 ) ≥ 4.

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6. THEOREM C5 : STAGE 4

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There exists y ∈ Y1# and a four-group V ≤ D ∩ X1 such that [y, V ] = 1 and  3 (L1 /O3 (L1 )) ≥ 4, Y1 V ∼ = Z3 × A4 . Let Ly be the pumpup of L1 in CG (y). As m it follows from part (c) of Theorem C5 : Stage 2 that Ly is not a diagonal or trivial pumpup of L1 , so as X2 ≤ L1 , X2 ≤ CLy (V ). Now Ly /O3 (Ly ) ∈ C3 and X2 ∩ O3 (Ly ) = 1 as O3 (Ly ) has odd order. Also m3 (CLy /O3 (Ly ) (V )) ≤ 3 by part (c) of Theorem C5 : Stage 2, so that [Ly /O3 (Ly ), V ] = 1; and of course m2,3 (y Ly ) ≤ 4. However, these conditions on Ly are incompatible by [VK , 15.14]. Hence, Proposition 2.31 is proved. Propositions 2.31 and 2.12 immediately imply: ∼ 2U4 (3). Corollary 2.62. Assume (2D) holds. Then (2D) holds for some K = We continue to assume (2D), and now assume that K∼ = 2U4 (3). We again follow Aschbacher [A23]. Our goal is to prove that G ∼ = Ω7 (3) or Ω± 8 (3). We first prove: Lemma 2.63. tG ∩ V1 = ∅. Proof. By Proposition 2.12, we may assume that T ∈ Syl2 (G). Fixing g ∈ NG (T ) − T with g 2 ∈ T , we have tg = tz by Lemma 2.22. Let F ∈ E2∗ (K), so that F = CK (F ) ∼ = E25 and F is an indecomposable module for NK (F )/F ∼ = A6 , with socle t [VK , 10.10]. Moreover AutCt (F ) ∼ = A6 or Σ6 is transitive on z G ∩ F and on (tz)K ∩ F , both of which have cardinality 15. Suppose that F ≤ K g . Then F is an indecomposable AutK g (F )-module with socle tz, so F is irreducible for A := AutK (F ), AutK g (F ) ≤ Aut(F ). In particular O2 (A) = 1. As z G = tG by Lemma 2.9b, A is transitive on tG ∩F , of order 16, so |A| = 2a |A6 | with a = 4 or 5. As A contains A6 , O2 (A) = 1 and so by [VK , 15.17], A embeds in Aut(A0 ) for A0 = A5 , A6 , or P Sp4 (3). By order considerations, a = 4 and A embeds in P Sp4 (3). But then |P Sp4 (3)|/|A| = 9, whereas P Sp4 (3) does not embed in A9 (it would have index 7). This is a contradiction.  Therefore, F ≤ K g . By indecomposability, F = z K ∩ F , so there is z  ∈ z K ∩ F − K g . By Lemma 2.23, E(CK g (z  )) = 1. By (2E6) (applied to tz instead of t) and [IA , 4.5.1], every involution in K g z  either is conjugate to z  and z, or is conjugate to an element of V1 . It therefore suffices to prove that tG ∩ K g z  = ∅. But tz  ∈ (tz)K ⊆ tG . Also t and tz are interchanged by g, and tz ∈ K, so t ∈ K g .  Hence tz  ∈ K g z  and the lemma follows. Definition 2.64. E(t, z) is the set of all subgroups E ∈ E26 (CG (t, z)) such that tG ∩ E ∩ V1 = ∅ and |E ∩ K| = 25 . Lemma 2.65. E(t, z) = ∅. Proof. By Lemma 2.63, there exist y, v ∈ G such that v = ty ∈ V1 and such that CT (v) ∈ Syl2 (CCt (v)). Let L = E(CK (v)) ∼ = P Sp4 (3), F = J(T ∩ L) ∼ = E24 ,  and E = F v, t ∼ = E26 . Then E ∈ E(t, z). Now fix an arbitrary E ∈ E(t, z), and fix y, v ∈ G such that v = ty ∈ V1 ∩ E and such that CT (v) ∈ Syl2 (CCt (v)). Let L = E(CK (v)) ∼ = P Sp4 (3), F = J(T ∩ L) ∼ = E24 . Then E = F v, t. Let H be the subnormal closure in AutG (E) of AutK (E), AutK y (E) . ∼ A7 or A8 according as T ∈ Syl (G) or not. We shall show that H = 2 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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By [VK , 10.10], AutK (E) ∼ = A6 and AutL (E) ∼ = A5 . By L2 -balance, L ≤ E(CG (v)), and so L ≤ K y . Thus F = [F, AutL (E)] ≤ K y . It follows by [VK , 10.9d] that M := E(CK y (t)) ∼ = P Sp4 (3). Thus we have symmetry between t and v, and in particular AutK y (E) ∼ = A6 . Moreover, E is a standard permutation module for each of AutK (E) and AutK y (E), by [VK , 10.9c]. The orbit decomposition of A := AutK (E) ∼ = A6 on E # is, for some z  ∈ E # , (2N)

E # = {t} ∪ v A ∪ (tv)A ∪ z A ∪ (tz)A ∪ (z  )A .

We will also refer to orbits by their respective cardinalities: 63 = 1 + 6 + 6 + 15 + 15 + 20. Here E # ∩ K = 1 + 15 + 15 , and by [IA , 4.5.1], O 3 (CK (z  )) ∼ = A4 × A4 . A similar decomposition under A := AutK y (E) holds, with t and v interchanged, and z  replaced by a different representative. 

Lemma 2.66. The following conditions hold: (a) H is irreducible on E; (b) E ∩ K = E ∩ K y ; (c) z H = z A ∪ (z  )A has cardinality 35; and (d) E = CG (E). Proof. As z ∈ L ≤ K y , we have vz ∈ K y − K, which implies (b). As the only A-invariant (respectively A -invariant) submodules of E are t and E ∩ K (respectively v and E ∩ K y ), (b) implies (a). By [VK , 10.10b], CAut(K) (E) lies in the image of E, so CG (E) ≤ EQ. If CG (E) = E, then CG (E) = EQ with Q∼ = Z4 ; but then Φ(EQ) = t is H-invariant, contradicting (a). Thus CG (E) = E, H A  A so holds. By Lemmas 2.11b and 2.23 and (2N), z ⊆ z ∪ (z ) . Since  A(d)  z ≤ E ∩ K < E, (c) follows from (a). Lemma 2.67. We have H = F ∗ (AutG (E)). Moreover, one of the following holds: (a) T ∈ Syl2 (G), H ∼ = A7 , and tG ∩ E = tH = {t} ∪ v A ; or (b) T ∈ Syl2 (G), H ∼ = A8 , and E # = (tG ∩ E) ∪ z H . In any case, E is isomorphic to the core of the natural permutation AutG (E)module. Proof. Since tG = z G , tG ∩ E # ⊆ 1 + 6 + 15 + 6 . As |tAutG (E) | divides |L6 (2)| and v ∈ tG , the only possibilities are tAutG (E) = 1 + 6 and 1 + 6 + 15 + 6 . If tAutG (E) = 1 + 6, then as CAutG (E) (t) ∼ = A6 or Σ6 by [VK , 10.10], we must have that E is the trace-0 submodule of a natural module for AutG (E) ∼ = A7 or Σ7 , as asserted. Then E(CAutG (E) (tv)) ∼ = A5 so tv ∈ tG . Hence either tAutG (E) = tG ∩E or else tG ∩ E = 1 + 6 + 15 . But A7 and Σ7 have no orbit of length 15 on E. Hence tG ∩ E = 1 + 6. In particular, tz ∈ tG , so T ∈ Syl2 (G) by Lemma 2.22b. Thus (a) holds in this case. Suppose then that tAutG (E) = 1+6+15 +6 . By Lemma 2.66c, E # = tAutG (E) ∪ AutG (E) . Note that this is the case if G = P Ω+ z 8 (3), with E preserving an ordered orthonormal basis. It follows that if we set q(t) = 1 and q(z) = q(1) = 0, and extend q to E by making it AutG (E)-invariant, then q is a nondegenerate quadratic form of + type on E. Hence AutG (E) ≤ O(E, q) ∼ = Σ8 with index at most 2 (index 4 is

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6. THEOREM C5 : STAGE 4

impossible), and since Σ8 = Aut(AutG (E)), the isomorphism type of E is the same 12 as in P Ω+ ≥ |T |. Indeed 8 (3), as asserted. Furthermore, |NG (E)|2 ≥ |E||A8 |2 = 2 if equality holds, then AutG (K) = Aut(K) and |Q| = 4, so NCt (E)/E ∼ = Σ6 × Z2 and NG (E)/E ∼ = Σ8 . Thus in any case |NG (E)|2 > |T |2 , whence T ∈ Syl2 (G) and (b) holds. The proof is complete.  Our next goal is to prove: Proposition 2.68. Ki   Cz , i = 1, 2. This is immediate from Proposition 2.12 if T ∈ Syl2 (G), so we shall proceed under the assumption that T ∈ Syl2 (G). Some of the results we prove will be of use beyond the proof of Proposition 2.68. Lemma 2.69. Suppose that T ∈ Syl2 (G). Then one of the following holds: (a) AutG (E) ∼ = A8 , Q ∼ = Z2 , and AutCt (E) ∼ = Σ6 ; or ∼ (b) AutG (E) = Σ8 , Q ∼ = Z4 , and AutCt (E) ∼ = Z2 × Σ6 . The image of Q in AutG (E) is Z(AutCt (E)), and is generated by a transvection with axis E ∩ K and center t. Proof. Let q be the quadratic form on E defined in the proof of Lemma 2.66. We have AutG (E) = Ω(E, q) ∼ = A8 or O(E, q) ∼ = Σ8 . As t is a nonsingular vector, the structure of AutCt (E) ∼ = CAutG (E) (t) follows immediately. Since AutQ (E) is a normal subgroup of AutCt (E) of order |Q/ t | = |Q|/2, and AutCt (E)/ AutQ (E)  embeds in Σ6 ∼ = AutAut(K) (E/ t), the result follows directly. Set M = AutG (E) and Mz = O 2 (CM (z)). As z ∈ E is a singular vector, CM (z) is a parabolic subgroup of type A1 ×A1 and so Mz = M1 ×M2 with M1 ∼ = M2 ∼ = A4 . Lemma 2.70. Suppose that T ∈ Syl2 (G). Then |K1 K2 : NK1 K2 (E)| = 3, and AutK1 K2 (E) = M1 or M2 . Proof. CAutK (E) (z) fixes z and t, which are, respectively, a singular vector and nonsingular vector orthogonal to each other in E (note that tz ∼ t is nonsingular). Hence any w ∈ I3 (CAutK (E) (z)) centralizes a 4-dimensional subspace of E. Thus w centralizes an element of O(E) of order 5, so w corresponds to a 3-cycle in A8 ∼ = Ω(E). As AutK (E) ∼ = A6 is transitive on (E ∩ K/ t)# , 2 2 O (CAutK (E) (z)) ∼ = A4 , and so O (CAutK (E) (z)) = M1 or M2 . Now K1 K2 = O 2 (CG (t, z)), so O 2 (CAutK (E) (z)) = AutK1 K2 (E), proving the second statement. As E = CG (E) and E ∩ K1 K2 ∼  = E23 , the first statement follows as well. For specificity, let us say that AutK1 K2 (E) = M1 . Next, choose t1 ∈ K1 K2 ∩ E − z. Then t1 ∈ tG , tt1 ∈ z G , t1 z ∈ tG , and tz ∈ tG . Thus z, t, t1 are mutually orthogonal vectors in E. Using Witt’s lemma we find h ∈ [M, M ] ∼ = Ω(E) such h t and t . Set K = K and K = K2h , so that that z h = z, and h interchanges 1 3 4 1 2 h K3 K4 = O (CG ( t , z )). The same arguments as for Lemma 2.70 show that |K3 K4 : NK3 K4 (E)| = 3 and AutK3 K4 (E) = M1 or M2 . But th is centralized by AutK3 K4 (E) and not by M1 , so AutK3 K4 (E) = M2 . For any u ∈ tG with z ∈ E(CG (u)), define K(u) = O 2 (CG (z, u)) ∼ = SL2 (3) ∗ SL2 (3). Thus K(t) = K1 K2 , K(th ) = K3 K4 , and for any x ∈ Cz , K(tx ) = K(t)x . Let Qt = O2 (K(t)) and Qth = O2 (K(th )). Lemma 2.71. Suppose that T ∈ Syl2 (G). [Qt , Qth ] = 1.

Then with the above notation,

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Proof. Let Ni , i = 1, 2, be the preimage of Mi in NG (E), and let ai ∈ I3 (Ni ). We may choose a1 ∈ K(t) and a2 ∈ K(th ) to commute with each other. Since there are just two nontrivial a1 -chief factors in both N1 N2 and Qt , [a1 , N1 N2 ] = [a1 , Qt ] = Qt , and similarly [a2 , N1 N2 ] = [a2 , Qth ] = Qth . Hence, Qt and Qth normalize each other. But M1 ∩M2 = 1, and [a1 , E]∩[a2 , E] = 1, so Qt ∩Qth ≤ z.  The action of a1 , a2  then implies that [Qt , Qth ] = 1, as claimed. Now let Δ(t) = I2 (Qth ) − {z}. By construction, t ∈ Δ(t). Lemma 2.72. Assume that T ∈ Syl2 (G). If x ∈ Cz and tx ∈ Δ(t), then K(t) = K(tx ). Proof. First assume that [t, tx ] = 1. Then tx ∈ CG (z, t) so tx acts on K(t). But tx ∈ Qth so [tx , Qt ] = 1. This forces [tx , K(t)] = 1. Hence K(t) ≤ O 2 (CG (z, tx )) = K(tx ) = K(t)x , so x normalizes K(t). Now for the general case, any element of Δ(t) has the form tx for some x ∈ Cz . By the structure of K3 K4 , there is a chain t = t, tx1 , . . . , tx1 ···xn = tx with adjacent terms commuting, and with x1 , . . . , xn ∈ Cz . We conclude that x1 , . . . , xn normalize  K(t), so K(tx ) = K(t), as asserted. We quickly conclude: Lemma 2.73. Assume that T ∈ Syl2 (G). Then K1 , K2 , K3 , K4  = K1 ∗ K2 ∗ K3 ∗ K 4 . Proof. Note that K(th ) = K3 K4 ≤ Cz and K3 K4 normalizes Δ(t). Hence by Lemma 2.72, K(th ) normalizes K(t) = K1 K2 . Conjugating by h, K(t) normalizes K(th ). With Lemma 2.71, [K(t), K(th )] ≤ z and then [K(t), K(th )] = 1, completing the proof.  We have O2 (Ki ) = Ri for i = 1, 2, and we extend this notation to i = 3, 4. Let S = R1 R2 R3 R4 ,

N = NG (S),

Cz = Cz / z ,

and let Δ be the set of all involutions in S projecting nontrivially on exactly two of the Ri ’s. Thus Δ(t) ⊆ Δ. For each u ∈ Δ, let K(u) be the product of the two Ri ’s that centralize u. We check that this agrees with the previous definition of K(u). Lemma 2.74. If T ∈ Syl2 (G), then for any u ∈ Δ we have symmetry between t and u. That is, E(CG (u)) ∼ =K∼ = 2U4 (3), and K(u) = O 2 (CG (z, u)). Proof. If u ∈ tCz , this is clear. We first claim that for all u ∈ Δ, E(CG (u)) ∼ = 2U4 (3) or U4 (3). Suppose for example that u ∈ R1 R4 . Replacing u by a K1 K4 -conjugate, and noting that K1 K4 ≤ Cz , we may assume that [u, t] = [u, th ] = 1. Then K2 ≤ CCt (u) but K1 ≤ CCt (u). Hence by [IA , 4.5.1], K2 ≤ E(CCG (u) (t)) = E(CK (u)) ∼ = U3 (3). h ∼ Similarly, K3 ≤ E(CCG (u) (t )) = U3 (3). Let I be the subnormal closure of K2 in CG (u). Then z ∈ I − Z(I) and z ∈ K3 , so I is also the subnormal closure of K3 in CG (u). Since [K2 , K3 ] = 1, I is not a diagonal pumpup. Hence by [VK , 3.10m], I := I/Z(I) ∼ = U4 (3) or G2 (4). In the latter case, by [VK , 9.35], and as K2 ∗ K3 ≤ I, we cannot have 3-central elements of I in both K2 and K3 . Hence we may assume that x2 ∈ I3 (K2 ) is  ∼ not 3-central in I, whence K3 ≤ H = E(CI (x2 )) with H = L2 (4). As L2 (4) has elementary abelian Sylow 2-subgroups, I ∼ = 2G2 (4) and z ∈ Z(R2 ) = Z(I). But

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then U3 (3) ≤ CI (t) ≤ CG (z, t), a contradiction as CG (z, t) is solvable. Therefore I ∼ = U4 (3). Our claim follows. Finally, suppose that I ∼ = U4 (3). Then since K2 K3 ≤ CI (z), and I has one class of involutions, z is I-conjugate to an element of Δ supported on R2 and R3 . By our claim, E(Cz ) has a component J ∼ = U4 (3) or 2U4 (3). In either case, by [VK , 10.1], a Sylow 2-center of Cz , and hence of G, is noncyclic. This is a contradiction as z, t = Z(T ) and t is not 2-central in G. The proof is complete.  Lemma 2.75. If T ∈ Syl2 (G), then the following conditions hold: (a) (b) (c) (d) (e)

CG (S) = S; CG (R1 R2 ) = SK3 K4 Q; If Q = x ∼ = Z4 , then x acts freely on R3 R4 ; N permutes {K1 , K2 , K3 , K4 }, transitively; and For any t ∈ Δ, O 2 (CN (t )) ∼ = SL2 (3) ∗ SL2 (3).

Proof. By [VK , 10.1b], |CCt (R1 R2 )| divides 23 |Q|, but CR3 R4 Q (t) ≤ CCt (R1 R2 ) with |CR3 R4 Q (t)| = 23 |Q|. (Note that Q centralizes th and z so Q normalizes R3 R4 with Q∩R3 R4 = t.) Hence CR3 R4 Q (t) = CCt (R1 R2 ). Intersecting with CG (R3 R4 ) yields CG (S) ≤ Q. As S is extraspecial, CG (S) = SCG (S), so (c) implies (a). Similarly, R3 R4  CN (R1 R2 ) and K3 K4 is transitive on I2 (R3 R4 ), so CN (R1 R2 ) ≤ K3 K4 CN ∩C(t) (R1 R2 ) ≤ K3 K4 R3 R4 Q = K3 K4 Q. Therefore CN (R1 R2 ) = R1 R2 CN (R1 R2 ) ≤ SK3 K4 Q, which implies (b). In (c), we know by Lemma 2.69b that the image ξ of x in M is a transvection with axis E ∩ K and center t. As [ξ, AutK (E)] = 1 with AutK (E) ∼ = A6 , ξ must correspond to a transposition in Σ8 ∼ = AutG (E). In the notation introduced before Lemma 2.70, [M1 , ξ] = 1 and so M2 ξ ∼ = Σ4 . In particular ξ inverts an element of M2 of order 3. But x normalizes K3 K4 ∩ NG (E), which has a Sylow 3-subgroup of order 3 mapping onto M2 and fixed-point-free on R3 R4 . It follows that (K3 K4 ∩ NG (E)) x /R3 R4 ∼ = Σ3 , and the desired freeness of the action of x follows. In (d), let Δ be the set of all involutions u ∈ S such that u projects nontrivially on all Ri , 1 ≤ i ≤ 4. Then K1 K2 K3 K4 permutes Δ transitively, and I2 (S) − Δ = Δ. Note that t ∈ R3 R4 and th ∈ R1 R2 , so tth ∈ Δ . As tth ∈ z G , Δ ⊆ z G . But Δ ∩ z G = ∅ by Lemma 2.74, so N acts on Δ. Again by Lemma 2.74, N permutes the groups K(u), u ∈ Δ, and hence it permutes {K1 , K2 , K3 , K4 }, as asserted. As K(t) = K(u) for all u ∈ Δ(t), NCG (th ) (K3 K4 ) has an element interchanging K3 and K4 and normalizing K(t). Together with the element h, which interchanges K1 K2 and K3 K4 , this proves the desired transitivity. Finally, (e) is obvious for t = t, and holds for arbitrary t ∈ Δ by Lemma 2.74. The lemma is proved.   = N/S ∼ Now set N = N /S, and let S1 ∈ Syl2 (N ) with S ≤ S1 . Lemma 2.76. We have S = J(S 1 ) and S1 ∈ Syl2 (Cz ). Proof. Since N = NCz (S), the first statement implies the second. Suppose S = J(S 1 ). Then by [IG , 25.2] there exists W ≤ S 1 such that (2O)

, W  ] = 1, and |C (W  )||W  | ≥ |S|. W ∼ = S, [S, W S

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0 = ∩4 N (Ri ), and S 0 = C (W 0 ). Note that S 0 ∩ Ri = 1 for all 1 ≤ i ≤ 4. Let W i=1 W S     Now since [S, W , W ] = 1, W /W0 is trivial, generated by a transposition, or generated by a pair of disjoint transpositions (as a permutation group on {R1 , . . . , R4 }). 0 ) : C (W  )| ≥ |W  /W 0 |. In any of these cases, since S 0 ∩ Ri = 1 for all i, |CS (W S   Combined with (2O), this yields |CS (W0 )||W0 | ≥ |S|. But by Lemma 2.75bc, no 0 normalizes each Ri , and 0 induces a transvection on S. As W involution in W   | AutW 0 (Ri )| ≤ 2, 1 ≤ i ≤ 4, it follows easily that |CS (W0 )||W0 | < |S|, a contradiction completing the proof.  Lemma 2.77. Suppose that T ∈ Syl2 (G). Then K1 K2 K3 K4  Cz , i.e., Cz = N . Moreover, E(Cz ) = 1 and F ∗ (Cz ) = S. Proof. It suffices to show that S  Cz . For then E(Cz ) ≤ CCz (S) ≤ CG (t, z), and as CG (t, z) is solvable, E(Cz ) = 1. Furthermore, the E34 image of K1 K2 K3 K4 in Out(S) ∼ = D4 (2) is self-centralizing, so then O2 (Cz ) = CG (S) = S. Moreover, by Lemma 2.75d, K1 K2 K3 K4  Cz , as desired. If S ≤ O2 (Cz ), then by Lemma 2.76, S  Cz . Therefore we may assume that S ≤ O2 (Cz ), so by Lemma 2.75d, R1 ≤ O2 (Cz ). Now K1 is a solvable component of CG (z, t), so by solvable L2 -balance, R1 ≤ J for some component J of Cz . As z ∈ R1 , z ∈ Z(J). Now J ∈ C2 , so by [III11 , 13.1], J∼ = 2U4 (3) or J ∈ Chev(2). The first possibility is ruled out by [VK , 3.109]. Thus we may assume that J ∈ Chev(2) and argue to a contradiction. If S ≤ J, then by the transitive action from Lemma 2.75d, [J, Ri Rj ] = 1 for some 1 ≤ i < j ≤ 4, so [J, t ] = 1 for some t ∈ Δ. By the symmetry between elements of Δ and t (Lemma 2.74), CG (z, t ) is solvable, a contradiction as J ≤ CG (z, t ). Hence, S ≤ J. Let S ≤ SJ ∈ Syl2 (J). Then S  SJ . Now K1 K2 K3 K4 SJ permutes Δ, in orbits whose sizes have the form 18a, a ≤ 6. Thus |SJ : CSJ (t)| ≤ 8, whence |J|2 = |SJ | ≤ 8|Ct |2 ≤ 215 . In particular J ∼ = 2 E6 (2), F4 (2), or U6 (2). Also m2 (J) ≥ m2 (S) = 8 so with [IA , 3.3.3], J ∼ = D4 (2), 3 Sp6 (2), G2 (4), L3 (4), or 2B2 (2 2 ). But this exhausts the possible isomorphism types of J since Z(J) = 1 and J ∈ Chev(2) ∩ C2 [IA , 6.1.4]. The proof is complete.  In particular, Proposition 2.68 is a direct consequence of Lemma 2.77. Now we have reached the endgame. It has three branches, one for each of the target groups Ω7 (3) and Ω± 8 (3). By Corollary 2.62, K ∼ = 2U4 (3). By Lemma 2.63, there is v = tg ∈ V1 for some g ∈ G. Since CG (v) ∼ = Ct and v centralizes K1 K2 , we clearly can modify g by an element of CG (v) to arrange that z g = z. Therefore (tz)g = vz and g normalizes ∼ K1 K2 . But by [VK , 10.1f], K1 K2 < E(CK (vz)) = E(CG (t, vz))  =−1 P Sp 4 (3). −1 g Conjugating by g , K1 K2 < J ≤ E(CG (tz)), where J = E(CG ( t , tz )) ∼ =  E(C (tz)) G . Of course K1 K2 ≤ Ktz . P Sp4 (3). Set Ktz = J Lemma 2.78. We have Ktz = CG (tz)(∞) , and one of the following holds: (a) Ktz ∼ = A7 ; = P Sp4 (3), T ∈ Syl2 (G), and F ∗ (AutG (E)) ∼ (b) Ktz ∼ = L4 (3), T ∈ Syl2 (G), Q = t, and F ∗ (AutG (E)) ∼ = A7 ; or (c) Ktz ∼ = A8 . = 2U4 (3), T ∈ Syl2 (G), and F ∗ (AutG (E)) ∼ Proof. If Ktz is not a single component, then by L2 -balance, and as G is of even type and Sp4 (3) ∈ C2 , we must have Ktz ∼ = P Sp4 (3) × P Sp4 (3). But then

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m2,3 (G) ≥ m2,3 (J1 ) = 5 > 4, contradiction. Hence, Ktz ∈ C2 ∩ ↑2 (P Sp4 (3)), so by [VK , 3.10], either Ktz is as in the lemma or Ktz ∼ = U4 (3) or L4 (4). If Ktz ∼ = U4 (3), then all involutions of K1 K2 are conjugate and hence lie in z G . But in K, involutions of K1 K2 − z lie in (tz)K , so tz ∈ z G . This is absurd as Cz has no component isomorphic to Ktz , by Proposition 2.12. Thus we may assume that Ktz ∼ = L4 (4). In particular tz ∼G t so by Lemma 2.22, T ∈ Syl2 (G). But then m2 (T ) ≥ m2 (CG (tz)) ≥ 9 [IA , 3.3.3], while m2 (T ) ≤ m2 (Ct ) ≤ 1+m2 (Aut(K)) < 9 [VK , 10.18], contradiction. Hence, Ktz is as claimed. We claim that t ∼G tz if and only if Ktz ∼ = K. Indeed, one direction is obvious. If Ktz ∼ = K then involutions of K1 K2 − z are conjugate in K to tz, but in Ktz to tz.z = t, proving the claim. Furthermore, t ∼G tz if and only if T ∈ Syl2 (G). One direction holds since by Lemma 2.22, if T ∈ Syl2 (G), then Z(T ) = z, t and so tg = tz for g ∈ NG (T ) − T with g 2 ∈ T . For the other, suppose that T ∈ Syl2 (G) and t ∼G tz. Then t ∼NG (T ) (tz) by Burnside’s lemma, so |NG (T ) : T | is even, contradiction. We conclude that Ktz ∼ = K if and only if T ∈ Syl2 (G), as desired. Now Lemma 2.67 completes the proof of (a), (b), or (c), except for the assertion about Q in (b). But if Ktz ∼ = L4 (3), then as CQ (Ktz ) = 1, Q embeds in Ktz /C(tz, Ktz ) ≤ Aut(Ktz ) as a normal cyclic subgroup of a Sylow 2-subgroup, and hence |Q| = 2 by [VK , 10.5], as claimed. Finally, z ∈ K1 ≤ Ktz so z is 2-central in Ktz , and z Aut(Ktz ) = z Ktz . However, CCG (tz) (z) = CCt (z) is solvable. It follows that Ktz  CG (tz) and then CG (tz) = Ktz CCG (tz) (z), whence Ktz = CG (tz)(∞) . The lemma is proved.  Following [III14 ], we set G0 = K, Ktz  . First we treat case (c) of Lemma 2.78. Our argument is inspired by Aschbacher’s identification of D4 (3) in the final section of [A9]. Proposition 2.79. Suppose that Ktz ∼ = 2U4 (3). Then G0 ∼ = Ω+ 8 (3). Proof. By the preceding results, we have that F ∗ (Cz ) ≤ K1 ∗K2 ∗K3 ∗K4  Cz , Cz is a {2, 3}-group, and for each u ∈ I2 (Ki Kj ) − {z}, O 2 (CG (u)) = E(CG (u)) := Lu ∼ = 2U4 (3). Choose Ai ∈ Syl3 (Ki ), 1 ≤ i ≤ 4. Let A = A1 and set H = CG (A). As m3 (H) ≥ 4, O3 (H) has odd order by [V4 , (1A4)]. Notice that K2 ∗K3 ∗K4  CH (z) and CH (z)/A z acts faithfully on K2 K3 K4 . In particular, O2 (CH (z)) = A. Let u ∈ I2 (K3 K4 ) with u = z, and let U = u, z. As K1 ∗ K2 ≤ Lu ∩ Luz , we see that O 2 (CG (v, A)) = CLv (A) = QAv K2 with QAv ∼ = 31+4 and with CQAv (z) = A, for each v ∈ {u, uz}. It follows immediately that O3 (H) = 1. Suppose that E := E(H) = 1. As z acts faithfully on each component E1 of E, and E1 ∈ C3 , it follows that K2 ∗ K3 ∗ K4 is normal in CAutH (E1 ) (z). By [VK , 10.60], we reach a contradiction, and so E(H) = 1 and F ∗ (H) =: Q = O3 (H). As A = CQ (z) and K2 ∗ K3 ∗ K4 acts faithfully on Q, we conclude that Q = QAu ∗ QAuz ∼ = 31+8 .   Moreover, setting Q = Q/A, we see that Q as F3 Ki -module is a direct sum of four 2-dimensional irreducible F3 Ki -modules, i = 2, 3, 4. In particular, Ai acts  Hence, [Q, Ai ] ≤ CQ (Ai ) and CQ (Ai ) ∼ quadratically on Q. = E35 for i = 2, 3, 4.  H = QCQ (z), whence Q(K2 ∗K3 ∗K4 )  H, and H/Q permutes Also, as z inverts Q, the images of K2 , K3 , and K4 . The same argument applies to Hj := CG (Aj ) for all j.

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Next set M = A2 CQ (A2 ) ∼ = E36 and let I = NG (M ). As AA2 ≤ M , we have CG (M ) = M and K3 ∗ K4  CI (z). As CG (z) transitively permutes the Ki ’s, we may choose y ∈ CG (z) with Ay = A2 . Then Qy (K1 ∗ K3 ∗ K4 )  CG (A2 ). We have CQ (A2 ) = A × (Q ∩ Qy ) and so M = AA2 (Q ∩ Qy ) = ACQy (A) = A2 CQ (A2 ). As argued in the previous paragraph for Q and Ai , we see that [Qy , A] ≤ CQy (A). Hence, Qy normalizes M . We set Q1 = Qy . Now, let I = I/M. ∼ We argue that O 2 (I) ∼ = Ω+ 6 (3). First, I is a subgroup of GL(M ) = GL6 (3). Also, CM (z) = AA2 and K3 ∗ K4 ≤ CI (z) with K3 K4 acting faithfully on [M, z]. By the previous paragraph, Q1 ≤ I and, as A ≤ Q1 , we have |Q1 ∩ M | ≤ 35 , whence, as A2 ≤ Q1 ∩ M , we see that Q1 is elementary abelian of order at least 34 . Hence, I ≥ P 1 := Q1 (K 3 ∗ K 4 ). We also have M = ACQ1 (A), and so Q ≤ I with |Q ∩ M | ≤ 35 and |Q| ≥ 34 . Hence, I ≥ P := Q(K 3 ∗ K 4 ). Now, M is a uniserial P -module with composition factors of dimension 1, 4, 1, and CM (P ) = A, and M is a uniserial P 1 -module with composition factors of dimension 1, 4, 1,A and CM (P 1 ) = A2 . It follows that M is an irreducible I-module, whence I is not contained in a parabolic subgroup of GL(M ), and then it follows immediately that F ∗ (I) is simple. As O 2 (CI (z)) = K 3 ∗ K 4 , we conclude with [III11 , 13.1] and [IA , 4.5.1] that O 2 (I) =: L ∼ = Ω+ 6 (3), as claimed. (The structure of CL (A) rules U (3) and G (3).) We may choose a K3 ∗ K4 -invariant complement B2 to out L ∼ = 4 2 CQ (A2 ) in Q. Then B2 (A3 ∗ A4 ) is a complement to M in a Sylow 3-subgroup of I, whence we may assume that L is a complement to M in O 2 (I) with K3 ∗ K4 ≤ L and L ∼ = Ω+ 6 (3). Next, we repeat the analysis with I3 := NG (M3 ), where Q3 = O3 (CG (A3 )) and M3 = ACQ3 (A). Again we get O 2 (I3 ) = M3 L3 with 3 K2 ∗ K 4 ≤ L 3 ∼ = Ω+ 6 (3). Next we claim that |M ∩ M3 | = 3 . Suppose first that |M ∩ M3 | ≥ 35 . Then, as M3 is normalized by K2 ∗ K4 , we have A2 ≤ M3 and M = (M ∩M3 )A2 = CM (A3 ). But |CM (A3 )| = 34 , a contradiction. Hence |M ∩M3 | ≤ 34 and |M M3 /M | ≥ 32 . Now, M A3 /M is a K4 -invariant subgroup of M M3 /M and M M3 /M is the sum of M A3 /M and irreducible 2-dimensional K4 -modules, as K4 module. As M M3 /M is an elementary abelian subgroup of I/M , it follows that |M ∩ M3 | = 33 = |M M3 /M | and we may assume that K4 < L0 < NL (M ∩ M3 ) with L3 ∼ = L3 (3) a complement to M M3 in O 2 (NI (M ∩ M3 )). Now we may choose K0 so that (K0 , K4 ) is a weak Curtis-Tits system in L0 , and then we may extend this to a weak Curtis-Tits system (J0 , K0 , K4 ) for L. As [J0 , K4 ] = 1, we must have J0 ≤ O 2 (CL (K4 )) = K3 . As O 2 (NG (M M3 )) = M M3 L0 = O 2 (NI3 (M M3 )) and as all complements to M M3 in M M3 L0 are M M3 conjugate to L0 by [VK , 5.17], we conclude that L0 is contained in some complement J3 to M3 in O 3 (I3 ). As K4 ≤ L0 ≤ J3 and O 2 (CI3 (K4 )) = K2 , we must have O 2 (CJ3 (K4 )) = K2 . Hence we may extend (K0 , K4 ) to a weak Curtis-Tits system (K2 , K0 , K4 ) for J3 .

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Finally, as [Ki , Kj ] = 1 for all 1 ≤ i < j ≤ 4, (K0 , K2 , K3 , K4 ) is a weak Curtis-Tits system for G1 := K0 , K2 , K3 , K4  ∼ = D4 (3).  3  3 Now K1 ∗ K2 ∗ K3 ∗ K4 = E4 (CG (z)) = E4 (CG1 (z)) . Hence, K1 ≤ G1 . Then t, tz ≤ G1 and E(CG1 (t)) ∼ = K, whence K ≤ G1 . Likewise Ktz ≤ G1 . Finally, G1 = K, Ktz  by [IA , 7.3.2]. So, G1 = G0 ∼  = Ω+ 8 (3), as claimed. In the remaining two cases we first construct an Ω7 (3)-subgroup X ≤ G0 . Thus, for the next two lemmas we assume that either case (a) or case (b) of Lemma 2.78 holds. We essentially repeat the proof of part of [III14 , 12.17], which was modeled after Aschbacher [A9], to obtain: h such that Lemma 2.80. There exist h ∈ K and H ≤ I ≤ Ktz  h (a) K1 and K1 form a standard Phan-pair in K1 , K1h ∼ = U3 (3); (b) [K1 K2 , H] = 1; (c) I = K1h , H ∼ = Ω5 (3), and K1h and H form a standard Phan-pair in I; in particular H ∼ = L2 (3); and (d) H = O 2 (CI (z)).

Proof. We include the proof for completeness. Recall that standard-Phan pairs in certain groups of Lie type of Lie rank 2 are twisted analogues of pairs of fundamental SL2 -subgroups. In particular, a standard Phan-pair in U3 (3) is a pair of noncommuting SL2 (3)-subgroups whose unique involutions commute with each other [III13 , 1.5]. We also remind the reader that a standard Phan-pair in Ω5 (3) arises from a decomposition V = V1 ⊥ V2 ⊥ V3 of the natural F3 Ω5 (3)-module, with V1 and V2 being 2-dimensional of − type, and dim V3 = 1. The pair consists of one of the SL2 (3)-solvable components of Ω(V1 ⊥ V2 ) (i.e. of the pointwise fixer of V3 ), and Ω(V2 ⊥ V3 ) (i.e., the pointwise fixer of V1 ) [III17 , 4.3]. Fix an involution t1 ∈ K1 K2 − z and set t0 = tt1 . Then t0 ∈ z K , whence t1 = tt0 ∈ (tz)K . Then by [III17 , 4.8], there is h ∈ K such that conclusion (a) 2 holds, K1h = K1 , z h = t0 , and th0 = z. h ∼ Let I1 be the subnormal closure of K1h in CG (t1 ). As tz = th1 , I1 = Ktz = Ktz . Moreover, let V be the natural (orthogonal) I1 -module. Now w := zt1 = (t0 tz)h = h (t1 z)h ∈ (K1 K2 − z)h = K1h K2h − t0 , and the actions of w and z on I1 = Ktz coincide. Thus, z acts on I1 like an involution whose support [V, z] on V is 2dimensional of − type. Also, z h = t0 acts like an involution whose support is 4-dimensional of + type. Let W be a nondegenerate 5-dimensional subspace of V containing [V, t0 ], and set I = CI1 (W ⊥ ) ∼ = Ω5 (3) and H = O 2 (CI (z)) = O 2 (CI (w)). h ∼ (Note that if Ktz = Ω5 (3), then V = W and I = I1 .) Thus, H ∼ = Ω3 (3) and (c) holds, as follows from [III17 , 4.3] or our description above of standard Phan-pairs. It remains to verify that [K1 K2 , H] = 1. As T ∈ Syl2 (G), the structure of O 2 (Cz ) is given in Proposition 2.12. As H ≤ O 2 (Cz ) by construction, we conclude that H normalizes K1 K2 and induces inner automorphisms on it, by Proposition 2.12a. But by construction, [H, t1 ] = 1  and t1 ∈ I2 (K1 K2 − z). Hence CK1 K2 (t1 ) is a 3 -group. As H = O 3 (H),  [H, K1 K2 ] = 1. The proof is complete.

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Remark 2.81. In the above argument, once the choice of h is made, there h -module is the further choice of the 5-dimensional subspace W of the natural Ktz containing the support of t0 . The definitions of H and I depend on this choice. In situations where other choices of W might be made (but h remains the same), we shall indicate the dependence of H and I on W by writing H(W ) and I(W ). And in the next lemma, we can also write X(W ), A(W ), D(W ) and A∗ (W ) to show the dependence of X, A, D, and A∗ on W . In the case Ktz ∼ = Ω5 (q), there is of course only one choice of W . Recall that R = T ∩ K.

 Lemma 2.82. Let h and H be as in Lemma 2.80. Set X = K1 , K1h , H . Then the following conditions hold: (a) K, I = X ∼ = Ω7 (3); and (b) E25 (R) = {A, A∗ }, where A and A∗ have the following properties. There exists D ∈ E26 (Ct ) such that A < D ≤ X and F ∗ (AutX (D)) ∼ = A7 ; on the other hand, CX (A∗ ) = A∗ . Proof. By Lemma 2.80, K1 , K1h , and H form a weak Phan system of type B3 (3). Though this does not immediately imply that they generate a homomorphic image of Spin7 (3), it does imply by [V2 , 10.1] that X/O2 (X) ∼ = Ω7 (3) or Spin7 (3). As z ∈ K1   CX (z), z acts on X/O2 (X) like an involution with 4-dimensional support of + type on the natural F3 X/O2 2 (X)-module. As T ∈ Syl2 (G), CX (z) has only two solvable SL2 (3)-components, by Proposition 2.12ab. This rules out X/O2 (X) ∼ = Spin7 (3), so X/O2 (X) ∼ = Ω7 (3). Again as K1   CX (z), it follows from [IA , 4.5.1] that O 2 (CX/O  (X) (z)) ∼ = K1 ∗ K2 × A, 2

with A ∼ = A4 . Then, by Proposition 2.12ab, O2 (CX (z)) = 1. As X has a four-group U with U # ⊆ z X , O2 (X) = 1. Therefore, X ∼ = Ω7 (3), as claimed. Let V be the natural F X-module. Now t1 ∈ K1 K2 and t1 centralizes I = 3  h K1 , H by construction. Therefore, t1 normalizes X and acts on it like an involution of O(V ) with 2-dimensional support disjoint from the support of t0 = z h , which lies in I. Hence t = t0 t1 acts on X like an involution of O(V ) with 6-dimensional (∞) ∼ ∼ ± support.  So E(C = Ω− 6 (3), K = E(CX (t)) ≤ X. X (t)) = Ω6 (3). As K = Ct h As I = K1 , H ≤ X, K, I ≤ X. But X = K1 , I ≤ K, I, so X = K, I. Therefore, (a) holds.  Part (b) arises from applying [VK , 10.21] to the inclusion K ≤ X. We quickly deduce: Proposition 2.83. If Ktz ∼ = Ω7 (3). Moreover, Propo= Ω5 (3), then G0 = X ∼ sition 2.12b1 holds. Proof. This case of Lemma 2.78 is distinguished from the other two by the h h = K, I = X, by . Since h ∈ K ≤ G0 , G0 = Gh0 = K, Ktz fact that I = Ktz Lemma 2.82. 2 We have O 2 (CX (z)) ∼ = Ω+ 4 (3) × Ω3 (3). Suppose that O (Cz ) ≤ X. Then by (∞) ∼ Proposition 2.12ab, L := Cz = A6 and [K1 K2 , L] = 1. Therefore L ≤ CG (t1 ), and by L2 -balance, L ≤ L2 (CG (t1 )). But t1 = (tz)h , with h ∈ K as above. Thus h is a component of CG (t1 ). As Ktz ∼ Ktz = P Sp4 (3) and m2,3 (G) = 4, L2 -balance

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h implies that L ≤ Ktz = I ≤ X. Thus L ≤ CX (z), a solvable group, which is 2  absurd. Therefore O (Cz ) ≤ X, and the proof is complete.

Next we prove Proposition 2.84. If Ktz ∼ = L4 (3), then G0 ∼ = Ω− 8 (3) and Proposition 2.12b2 holds. Proof. Let h be as in Lemma 2.80 and let V be the natural (orthogonal) h 6-dimensional Ktz -module over F3 . As in the proof of Lemma 2.80, z h = t0 acts on h Ktz as an involution in Ω(V ) with 4-dimensional support Vt0 of + type. Let V1 and = CV (t0 ). As Ktz ∼ V2 form an orthogonal frame for Vt⊥ = L4 (3), V1 and V2 are not 0 = V ⊥ V and X = X(W ), i = 1, 2. (See Remark 2.81.) Thus, isometric. Let W i t i i i 0  Xi = K1 , K1h , Hi ∼ = Ω7 .(3) in the language of Lemma 2.82, and K ≤ X1 ∩ X2 . ⊥ Here Hi = O 2 (Cz ∩ CKtz h (W i )) ≤ CG (K1 ), i = 1, 2, and [K2 , Hi ] = 1, as in Lemma 2.80 If X1 = X2 , then H1 , H2  ≤ O 2 (CX1 (K1 )) ∼ = SL2 (3) × A4 . Hence O2 (Hi ) lies in the A4 direct factor, i = 1, 2, and so O2 (H1 ) = O2 (H2 ). However, by [VK , 10.28], applied in O 2 (CKtz h (z)), O2 (H1 ) = O2 (H2 ). This contradiction shows that X1 = X2 . Let Ai = A(Wi ), Di = D(Wi ), and A∗i = A∗ (Wi ), i = 1, 2. By Lemma 2.82b, {A1 , A∗1 } = E25 (R) = {A2 , A∗2 }. As Ktz ∼ = L4 (3), Q = t by Lemma 2.78. Hence by [VK , 10.9b], CG (Ai ) = CCt (Ai ) = Di ∈ E(t, z), so F ∗(AutG (Di )) ∼ = A7 and in particular F ∗ (AutG (Di )) = ∗ NG (Di ) F (AutXi (Di )). Hence Xi = K . If A1 = A2 , then D1 = D2 and it follows that X1 = X2 , contradicting what we showed above. Therefore A1 = A∗2 . It follows that we may decompose natural F3 X1 -, F3 X2 -, and F3 K-modules Y1 , Y2 , and Y as Y = V1 ⊥ · · · ⊥ V6 Y1 = Y ⊥ V7 Y2 = V0 ⊥ Y where V1 , . . . , V7 are isometric nonsingular 1-spaces, and V0 is a nonsingular 1-space not isometric to V1 . Moreover, D1 stabilizes each Vi , i > 0 and AutX1 (D1 ) permutes V1 , . . . V7 faithfully. Now, let V = V0 ⊥ V1 ⊥ · · · ⊥ V7 Vi⊥

and interpret to mean the orthogonal complement in V . By [III17 , 2.40], there exists N < X1 such that N ∼ = A7 permutes the set {V1 , . .. , V7 } naturally and N ∩ K ∼ = A6 centralizes V7 . Now if N ≤ X2 , then X1 = K N ≤ X2 and so X1 = X2 , contradiction. Therefore, N ∩ X2 = N ∩ K. Fix an injection f : Ω(V7⊥ ) → G1 := X2 , N  with image X2 , and an isomorphism f1 : Ω(V0⊥ ) → X1 . Let λ = f1−1 ↓N : N → T , where T := f1−1 (N ). Then λ is an isomorphism, and all the required compatibility conditions of [V2 , Theorem 10.2] hold, with X2 and G1 in the roles of K and H there. By that theorem, there is a homomorphism g : Ω(V ) → G1 and a 6-dimensional subspace U ⊆ V7⊥ such ∼ − that Ω(U ) ∼ = Ω− 6 (3) and g(Ω(U )) = f (Ω(U )) ≤ X2 . Let L1 = f (Ω(U )) = Ω6 (3) ∗ ∗ X2 ∗ and t  = Z(L1 ). Then t ∈ t , and we may change notation so that t = t and L1 = K. Let G2 := im(g), so that K = L1 ≤ G2 . As t = C(t, K), we have G2 ∼ = Ω(V ) ∼ = Ω− 8 (3), and K is the centralizer of a plane in the natural G2 -module. 2 (∞) = Ktz by Lemma 2.78, we have Ktz ≤ G2 . As O (CK (z)) ∼ = Ω+ 4 (3) and CG (tz)

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Thus G0 = K, Ktz  ≤ G2 . Using [III17 , 9.1], we see that G2 = K, Ktz  = G0 , completing the proof.  By Lemma 2.78 and Propositions 2.79, 2.83, and 2.84, G0 ∼ = Ω7 (3) or Ω± (3). 8

Set M = NG (G0 ). We next prove Lemma 2.85. Γt,z,1 (G) ≤ M . (∞)

Proof. We have K = Ct = E(Ct ). Similarly with Lemma 2.78, E(CG (tz)) = Ktz . Also by Lemma 2.77 and Proposition 2.12, E(Cz ) = 1 unless G0 ∼ = Ω− 8 (3), in (∞) ∼ (∞) ∼ which case E(Cz ) = Cz = A6 = CG0 (z) . Therefore   G0 = E(CG (u)) | u ∈ t, z# , so NG (t, z) ≤ M . Now CAut(K) (z) covers Out(K) by [VK , 10.19], so Ct = KCG (z, t) ≤ M . Similarly, z ∈ K1 ≤ Ktz and by [VK , 10.19], CAut(Ktz ) (z) covers Out(Ktz ), so CG (tz) ≤ M . It remains to show that Cz ≤ M . Since K1 ≤ K = E(CG0 (t)), z acts on a natural G0 -module V as an involution of Ω(V ) inverting a 4-dimensional subspace Vz of + type. Thus O 2 (CG0 (z)) ∼ = ⊥ ⊥ Ω+ 4 (Vz ) × Ω(Vz ), and in each case t, which inverts a 2-dimensional subspace of Vz of − type, has the form z(tz), with z ∈ Ω(Vz ) and tz ∈ Ω(Vz⊥ ). If T ∈ Syl2 (G), then we see from Proposition 2.12ab that O 2 (Cz ) and O 2 (CG0 (z)) are isomorphic, hence equal. Moreover Ω(Vz⊥ ) is normal in Cz and has one class of involutions. Hence in that case, Cz = O 2 (CG0 (z))CG (z, tz) ≤ G0 . Finally assume that T ∈ Syl2 (G), so that K1 K2 K3 K4  Cz . Let F be the set of all F ∼ = E22 such that F ≤ Ki Kj for some 1 ≤ i < j ≤ 4, with F # ⊆ tG ; we claim that for any F ∈ F,  (2P) G0 = E(CG (f )) | f ∈ F # . Namely, we use a natural module V such that dim(CV (K1 K2 )) = 4. Suppose first that F ≤ K1 K2 ≤ K. Then for each f ∈ F # , preimages of f in Ω(V ) have an eigenspace of dimension 6 on V , and so E(CG0 (f )) ∼ = 2U4 (3) ∼ = CG (f )(∞) , as G f ∈ t . Hence E(CG0 (f )) = E(CG (f )). Then by [III17 , 9.1], (2P) holds. The same argument proves (2P) if {i, j} = {3, 4}. In the other four cases for {i, j}, we use [VK , 10.24] to conclude that for each f ∈ F # , E(CG0 (f )) ∼ = 2U4 (3). The same argument then establishes (2P), using an appropriate triality twist of V rather than V , in order to apply [III17 , 9.1]. Since Cz permutes {K1 , K2 , K3 , K4 } by Lemma 2.75d, Cz permutes F, and  hence Cz normalizes G0 by (2P). The proof is complete. Now we aim to prove that G = G0 . Recall that U(G; M ; 2) is the set of all 2-subgroups B = 1 of M such that for all g ∈ G, B g ≤ M if and only if g ∈ M . Lemma 2.86. The following conditions hold: (a) F ∗ (M ) = G0 ;

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(b) t, z ∈ U(G; M ; 2); (c) If B ∈ E32 (M ) and ΓB,1 (G) ≤ M , then B ∈ U(G; M ; 2); and (d) M contains a Sylow 2-subgroup of G. Proof. Since Γt,z,1 (G) ≤ M and O2 (CG (u)) = 1 for all u ∈ t, z# , it follows that O2 (M ) = 1. Also t ∈ CCt (K) = Q ∼ = Z4 or Z2 , and [t, G0 ] = 1, so CCt (G0 ) = 1. Hence O2 (M ) = 1, and (a) follows. Next, suppose that t, z ≤ M g , g ∈ G. Then by [IA , 7.3.3], Gg0 = Γt,z,1 (Gg0 ) ≤ M . Hence, Gg0 ≤ M (∞) = G0 , so g ∈ M , proving (b). The proof of (c) is the same. Finally, (d) holds by (b) and [III8 , 6.1c].  Lemma 2.87. If G0 ∼ = Ω7 (3), then G0 = G. Proof. If G0 ∼ = Ω7 (3), then t, z acts as scalars on the −1-eigenspace of z on the natural module. It follows easily that every involution of M ≤ Aut(G0 ) ∼ = SO7 (3) has an M -conjugate in CG (t, z). Hence by [III8 , 6.1d] and Lemma 2.86b, M controls the G-fusion of its involutions. By the Thompson transfer lemma, therefore, I2 (M ) ⊆ I2 (G0 ) = z G0 ∪ tG0 ∪ (tz)G0 . Hence CG (u) ≤ M for all involutions u ∈ M . But then M lies in a strongly embedded subgroup M ∗ of G, so M ∗ and hence M are solvable by [II2 , Theorem SE]. This is absurd, so the lemma is proved.  Lemma 2.88. If G0 ∼ = Ω− 8 (3), then G = G0 . #

Proof. We first claim that E1 (t, z) ⊆ U(G; M ; 2). Let u ∈ t, z . Then CG (u) ≤ M by Lemma 2.85 and so O 2 (CG (u)) ∼ = L± 4 (3) or SL2 (3) ∗ SL2 (3) × A6 , g using Lemma 2.86a. Suppose that g ∈ G and u ∈ M ; we must show that g ∈ M . Note that |L2 (34 )| does not divide |O 2 (CG (u))|, so it follows from [IA , 4.5.1] that ug acts on G0 like an involution of O8− (3). In particular, since t, z has a 4-dimensional eigenspace on the natural module, there is x ∈ M such that [ugx , t, z] = 1. Then (gx)−1 ≤ CG (u) ≤ M , so since t, z ∈ U(G; M ; 2), gx ∈ M and then g ∈ M . t, z This proves our claim. Now suppose that g ∈ M and t = tg ∈ CM (t). Then since t has a 2-dimensional eigenspace of + type on the natural module, ttg does as well, or else ttg ∼G0 z. In any case, by the previous paragraph, ttg  ∈ U(G; M ; 2). Hence if M < G, then by  [III8 , Corollary 6.2], M is solvable, a contradiction. The lemma follows. Lemma 2.89. Assume G0 ∼ = P Ω+ 8 (3). Then G = G0 . Proof. We first claim that z ∈ U(G; M ; 2) and M controls G-fusion of its involutions. Of course Cz ≤ Γt,z,1 (G) ≤ M . Note that z is 2-central in M . We can now repeat the proof of [III16 , Lemma 4.10] almost verbatim, with z here in place of t there, n = 8, and  = +1. This establishes the claim. It suffices now to show that CG (u) ≤ M for all u ∈ I2 (G0 ). For then u ∈ U(G; M ; 2) by the control of fusion. Moreover, if g ∈ M and z g ∈ Cz − {z}, then z, z g  ≤ G0 , so zz g  ∈ U(G; M ; 2). Hence [III8 , Corollary 6.2] will complete the proof. As Cz ≤ M , we may assume by [IA , 4.5.1] that I := E(CG0 (u)) ∼ = 2U4 (3). Also without loss, [u, z] = 1 and z is 2-central in CM (u); also, as z ∈ U(G; M ; 2), CM (u) contains some S ∈ Syl2 (CG (u)). Thus by [IG , 5.30], I ≤ J for some component J of CG (u). We have |J|2 ≤ |M |2 ≤ | Aut(D4 (3))|2 = 215 and S ∩ J normalizes I, with J ∈ C2 . Also CJ (z) ≤ CM (z) is solvable. Hence by [VK , 15.18], J = I.

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As Cz = CM (z) is solvable, I  CG (u). Then as z Aut(J) = z J by [VK , 10.19],  CG (u) = ICCG (u) (z) ≤ M . The proof is complete. Proposition 2.8 follows from Lemma 2.78, Propositions 2.79, 2.83, and 2.84, and Lemmas 2.87, 2.88, and 2.89. 3. The Case G∗ = F i22 In this section and the next five, we shall prove: Proposition 3.1. Suppose that conclusion (b) or (c) of Theorem C5 : Stage 3 holds, and conclusion (a) fails. Then G ∼ = F i22 , F i23 , F i24 , F2 , Co1 , or F1 . We treat each row of the tables [V5 , 1.1, 1.2] in its own section, except that we consider G∗ = F i24 and G∗ = F i24 in the same section. Our goal is thus to prove G∼ = G∗ , except that we aim for a contradiction in the case G∗ = F i24 . Throughout these sections, we assume, by hypothesis, that (3A)

BtK3exc (G) = ∅.

In this section, G∗ = F i22 , so by Theorem C5 : Stages 2 and 3, our setup is as follows:

(3B)

m2,3 (G) = 4, t ∈ I2 (G), b ∈ I3 (CG (t)), and (K; I; J) is a nonconstrained {t, b}-neighborhood such that K ∼ = 2U6 (2), I ∼ = U4 (2), (3), satisfying the conditions of Theorem C and J ∼ U = 4 5 : Stage 2. Moreover, every symplectic pair in G is trivial.

We use the notation of [V4 , Definition 1.7]. In particular, Cb = CG (b), C b = Cb /O3 (Cb ), K is a component of Ct := CG (t), J is a 3-component of Cb , and I is a component of both CK (b) and CJ (t). Also C(t, K) = t and there is B ∈ B∗ (G) with t ∈ I∗G (B; 2); but B will play little role in this section. Lemma 3.2. CG (t) = K γ, where γ = 1 or γ is an involution such that CK (γ) ∼ = Z2 × Sp6 (2). Proof. Since C(t, K) = t and | Out(K)| = 2, and by [IA , 4.9.2], we have CG (t) = K γ where either γ = 1 or Kγ := CK (γ) ∼ = Z2 × Sp6 (2) [VK , 3.30]. Moreover, in the latter case, γ 2 ∈ Z(K) = t. It therefore remains to show that in the latter case, γ 2 = 1. There exists c ∈ I3 (Kγ ) such that CKγ (c) contains A6 . Hence, using [IA , 4.8.2], we see that E(CK (c)) ∼ = U4 (2), whence c ∈ bK . Conjugating in K we may assume that b = c ∈ Kγ . On the other hand since E(CJ (t)) = I, t lies outside a subgroup of Cb of index 2 [IA , 4.5.1]. Hence γ 2 = t, whence γ 2 = 1. The lemma is proved.  Now by [V2 , Theorem 10.4a], we immediately deduce: Lemma 3.3. If γ = 1, then G ∼ = F i22 . Thus, for the rest of this section we assume that (3C)

γ = 1

and we aim for a contradiction. Let T ∈ Syl2 (Ct ) with I ∩ T ∈ Syl2 (I). By [VK , 2.10], Z(T ) = t, z for some involution z ∈ Z(T ∩ I). We fix z and set Cz = CG (z). Lemma 3.4. z is 2-central in G, and there exists T ∗ ≤ G such that |T ∗ : T | = 2, z ∈ Z(T ∗ ), and T ∗ /T interchanges t and tz.

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∼ SL2 (3) ∗ SL2 (3) is centralized by t, whose action on Proof. O 2 (CJ (z)) = J ∼ (3) is that of a reflection fixing the support of z on the natural F3 J= P Ω− 6 module. Consequently there exists an involution g ∈ J interchanging t and tz. Hence |NG (Z(T )) : CG (Z(T ))| is even, so T ∈ Syl2 (G) and z is 2-central in G. The lemma follows.  Now set A = J(T ), NA = NG (A), and NA,t = CNA (t). Lemma 3.5. The following conditions hold: (a) A = CG (A) ∼ = E210 ; (b) NA controls G-fusion in tG ∩ A; (c) NA /A ∼ = Aut(M22 ); (d) G has exactly three classes of involutions, represented by t, z, and, say, u. The orbits of NA on A# are tG ∩ A, z G ∩ A, and uG ∩ A, of respective cardinalities 22, 231, and 770; and (e) T ∗ ∈ Syl2 (G) and |T ∗ | = 218 . Proof. Since CG (A) = CCt (A), part (a) follows from [VK , 2.10], and (b) from [IG , 16.9]. We have A char T  T ∗ , so |tNA | = |NA : NA,t | is even. By [VK , 5.9], |tNA | is a subsum of 1023 = 1 + 21 + 21 + 210 + 210 + 560 which contains 1, and divides | Aut(A)|. The only choices are |tNA | = 22 and 792. A,t ∼ A = NA /A ∼ Let N = AutG (A), so that as γ = 1, N = L3 (4) φ, φ being a field NA  A,0 = O 2 (N A,t ) ∼  automorphism of order 2, and |NA | = |NA,t | · |t |. Let N = L3 (4), which acts absolutely irreducibly on Q := A/ t. A | = 210 .34 .5.7.11, and as N A,0 ∼ Suppose that |tNA | = 792. Then |N = L3 (4), ev A )] = 1.  ery primary component of F (NA ) has order at most 11, and thus [NA,0 , F (N   Hence F (NA ) normalizes t, and by absolute irreducibility, F (NA ) is either trivial or isomorphic to the dual Q∗ , which is absurd by order considerations. Therefore, A ) = 1. Again by order considerations, N A does not have as many as 5 isoF (N A,0 ≤ E(N A ). As |N A,0 | morphic components, whence by the Schreier property, N    is divisible by a prime to the first power, NA,0 ≤ L  NA for some simple compo divides 24 .32 .11, so C  (L)  is solvable by [III17 , 9.9]. As  Then |C  (L)| nent L. NA NA ∗     F (NA ) = 1, L = F (NA ). Given the order of NA , [VK , 15.8] now yields a contraA ) is simple, diction. Therefore, |tNA | = 22. The same arguments show that F ∗ (N ∗  ∼ and so by [VK , 15.8], F (NA ) = M22 . As φ = 1, (c) holds, and so |NA |2 = 218 . By [VK , 5.8], A = J(R) for any R ∈ Syl2 (NA ), so R ∈ Syl2 (G). As |R : T | = 2 = |T ∗ : T |, T ∗ ∈ Syl2 (G), proving (e). By [VK , 5.8] again, A is uniquely determined as NA /A-module, and NA has 3 orbits on A# , of cardinalities 22, 231, and 770. As A = J(R), these orbits are the intersections with A of three distinct conjugacy classes of G, by [IG , 16.9]. To complete the proof it remains to show that any involution x ∈ R has a conjugate in A. By the Thompson transfer lemma, x has a conjugate in [NA , NA ]. In turn, since [NA , NA ] is perfect and |[NA , NA ]|2 = 2|K|2 , x has a conjugate in K. Finally, it follows from [VK , 2.9] that x then has a conjugate in A. Thus, (d) holds and the proof is complete.  Lemma 3.6. We have O2 (CG (t, z)) ≤ O2 (Cz ) ∼ . = 21+10 + Proof. Let Czo = CCz (t). Since z is weakly closed in t, z, Czo / t = ∼ CCt /t (z t / t) = [QM γ]/ t with Q/ t ∼ = 21+8 + , and M γ / t = Aut(U4 (2))

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3. THE CASE G∗ = F i22

281

∼ E29 . We have T ∗ ∈ acting irreducibly on Q/ t, z. By [VK , 2.10], Q/ z = o ∗ Syl2 (Cz ) and T ∈ Syl2 (Cz ), with |T : T | = 2. Since Z(T ) = t, z  T ∗ , Z(T ∗ ) = z and t ∈ Z2 (T ∗ ). Let L = E(Cz ); we argue that L = 1. Supposing that L = 1, we have z ∈ L. z = Cz / z. Then C  ( Let C t), and the latter group has a subgroup of z ( L t)  CC index 2 isomorphic to an extension of E29 by Aut(U4 (2)). In particular, with the  2 ≤ 217 . As Z(L) = 1, however, there are no irreducibility noted above, 214 ≤ |L|  by [VK , 15.6]. Thus, L = 1 so such groups L, F ∗ (Cz ) = O2 (Cz ). Let Rz = O2 (Cz ) and Rzo = O2 (Czo ). Since t z ∈ Z(T ∗ / z), t ∈ Rz . Again by the irreducibility, Rzo ≤ Rz , as claimed. As T ∗ ∈ Syl2 (G) with |T ∗ | = 218 , we have |Rz | = 211 . Write |CRz (b)| = 2a ; thus, a is odd. If Rz has a noncyclic characteristic elementary abelian subgroup E, then (∞) m2 (E) ≤ 3 by the action of M and as |Rz | = 211 . Thus, [E, Cz ] = 1. In particular [E, b] = 1 so from the structure of CG (b, z), E = t, z and t, z  Cz , whence (3D)

|Rz : Rzo | = 2 = |CRz (b) : CRzo (b)|.

Consider Q1 := O2 (CCb (z)) ≥ CRz (b). Since m3 (J b) > 4 = m2,3 (G), C(b, J) has odd order, so Q1 embeds in Aut(J). As t, z  Cz , Q1 ∼ or 21+6 = Z2 ×21+4 + + . But 1+4 in Ct we see that CRzo (b) contains Z2 ×2+ . Thus, (3D) implies that Q1 ∼ = 21+6 + , and 1+4 o o ∼ ∼ [Rz , b] = [Rz , b] = 2+ . Now Rz / z is an elementary abelian maximal subgroup of Rz / z, and Q1 / z is elementary abelian. Therefore [Rz , Rz ] ≤ [Rz , b]. The irreducible action of M then forces Rz / z to be abelian, and then Rz = CRz (b) ∗ , as desired. [Rz , b] ∼ = 21+10 + We may therefore assume that no such E exists, so Rz is of symplectic type by , which it then equals P. Hall’s theorem. In particular Rz contains a copy of 21+10 + by orders. The proof is complete.  Lemma 3.7. The following conditions hold: (a) Cz /O2 (Cz ) ∼ = Aut(U4 (2)); and (b) b ∈ Syl3 (C(b, J)). Proof. To prove (b), recall that in this case every symplectic pair is trivial. Now there is H0 ≤ J with z ∈ Z(H0 ) and H0 ∼ = SL2 (3) ∗ SL2 (3). As m2,3 (G) = 4, C(b, J) cannot contain E32 , for if it did, there would be a nontrivial symplectic pair. Thus, a Sylow 3-subgroup P of C(b, J) is cyclic. Then if (b) fails, P H0 contains Z9 × E32 . But this is impossible if (a) holds. z = Cz /Rz , where Rz = O2 (Cz ). Set Thus, it suffices to prove (a). Set C  ∼ and H H = CCt (z) = CCz (t). As in the previous lemma, Rz ∼ = 21+10 = Aut(U4 (2)). + ∗   In particular t ∈ Z(Rz ) so T = T Rz and we have that |Cz : H| is odd. Again z ) = 3 = m3 (H).  z embeds in as all symplectic pairs are trivial, m3 (C And C +   O (2). It then follows by [V , 15.7] that C = H. This completes the Out(Rz ) ∼ = 10 K z proof.  Lemma 3.8. γ = 1. ∼ Proof. Suppose not, and continue the above  argument. Since O2 (Cz ) = ∗ ∼  , O (C (b)) D ∗Q ∗Q . As F (C ) = b ×J by Lemma 3.7, and O (C 21+10 = 2 Cz 8 8 8 b 3 b) +

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6. THEOREM C5 : STAGE 4

282

has odd order, this implies by [VK , 4.7] that there are involutions u1 , u2 ∈ CG (b) such that  O 3 ((CG (b, ui ))/O{2,3} (CG (b, ui ))) ∼ = Z3 × Li ∼ ∼ with L1 = U3 (3) and L2 = A6 . However, neither C = Cz nor C = Ct possesses an  element b of order 3 such that O 3 ((CC (b ))/O{2,3} (CC (b ))) ∼ = Z3 × Li for i = 1 or 2. As G has just 3 classes of involutions, u1 and u2 are conjugate. Let C1 = CG (u1 ). Then there are b1 , b2 ∈ I3 (C1 ) such that  O 3 ((CC1 (bi ))/O{2,3} (CC1 (bi ))) ∼ = Z3 × Li , i = 1, 2.

1 = C1 /O3 (C1 ). Let Ji be the subnormal closure of L3 (CC1 (bi )) in C1 . Let C   Then Ji has the usual structure by L3 -balance. If J1 is not quasisimple, then by the structure of J1 , m3 (C1 ) > 4 = m2,3 (G), a contradiction. Thus J1 is quasisimple. We have J1 ∼ = U3 (3), U3 (33 ), G2 (8), D4 (2), or 3D4 (2), by [VK , 3.16]. Likewise J1  is b2 -invariant. Then O 3 ((CJ1 (b2 ))/O3 (CJ1 (b2 ))) is involved in Z3 × A6 . By [VK , 3.16], J1 ∼ = U3 (3). Since A6 is not involved in U3 (3), m3 (C  (J1 )) > 1 and so C1

m3 (CC1 (b1 )) > 3, a contradiction. The proof is complete.



By Lemmas 3.8 and 3.3, Lemma 3.9. G ∼ = F i22 . 4. The Case G∗ = F i23 In this section our setup is as follows, by Theorem C5 : Stages 2 and 3: (4A)

m2,3 (G) = 5, t ∈ I2 (G), b ∈ I3 (CG (t)), and (K; I; J) is a nonconstrained {t, b}-neighborhood such that K ∼ = 2U4 (3), = 2F i22 , I ∼ and J ∼ = Ω7 (3), satisfying the conditions of Theorem C5 : Stage 2. Moreover, every symplectic pair in G is trivial.

We continue the notation of [V4 , Definition 1.7], in particular Cb , C b , Ct , as discussed at the beginning of Section 3. Again C(t, K) = t. Lemma 4.1. Ct = K. Proof. We first note that m3 (C(b, J)) = 1. For otherwise, as J contains A4 × (SL2 (3) ∗ SL2 (3)), Cb would contain an elementary abelian 3-subgroup of rank 5 = m2,3 (G) acting nontrivially on an extraspecial 2-group, contradicting the fact that in this case, every symplectic pair is trivial. Therefore, C(b, J) is a cyclic 3-group. In particular C(b, J) has odd order. Now suppose that AutCb (J) > Inn(J). Then AutCb (J) ∼ = SO7 (3), so by [VK , 10.22], there is an involution u ∈ CCb (t) such that H := E(CJ (u)) ∼ = L4 (3) and H0 := E(CH (t)) ∼ = Ω5 (3). Since |Ct : K| ≤ | Out(K)| = 2, b ∈ K, and by the Bp -property for 2 and 3, there is a component H0  E(CK (u, b)) mapping isomorphically onto H0 . As K ∼ = 2F i22 it is clear from [IA , 5.3t] that H0 ≤ L := E(CK (u)) and L is a single component. Let N be the subnormal closure of Lin C G (u), so that H0 ≤ N and N has the usual form by L2 -balance. Since H = H0H , CN (b) has a component H ∼ = L4 (3). As H ∈ C2 , the only possibility by [VK , 3.87] is that H = N . Since m3 (CG (u)) ≥ m3 (b N ) = 5 = m2,3 (G), if we choose any A ∈ E35 (CG (u)), then (A, u, N ) ∈ BtK3exc (G) = ∅, a contradiction. Thus, AutCb (J) = Inn(J).

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5. THE CASE G∗ = F i24 OR F i24

283

As bK = bAut(K) [IA , 5.3t], Ct = KCt,b , where Ct,b = CG (t, b). As | Out(K)| = 2 and O3 (Cb ) has odd order, |Ct : K| = |Ct,b : CK (b)|2 = |CJ (t)|2 /|CK (b)|2 . But |CJ (t)| = 2|I| = |CK (b)|, so |Ct : K| = 1, proving the lemma.



We immediately deduce from Lemma 4.1 and [V2 , Theorem 10.4b]: Lemma 4.2. G ∼ = F i23 . 5. The Case G∗ = F i24 or F i24 We treat these two cases simultaneously. Thus our hypothesis for this section is the following:

(5A)

t ∈ I2 (G), b ∈ I3 (CG (t)), and (K; I; J) is a nonconstrained {t, b}-neighborhood satisfying the conditions of Theorem C5 : Stage 2, of type (K; I; J) = (2F i22 ; 2U4 (3); P Ω+ 8 (3)) or (3)). Moreover, m (G) = 5 or 6, respectively. (F i23 ; Ω7 (3); P Ω+ 2,3 8

For simplicity, we refer to these cases, which lead to F i24 and F i24 , respectively, as Case 1 and Case 2. In Case 2, we shall reach a contradiction before having to identify F i24 . We continue the notation of [V4 , Definition 1.7], in particular Cb , C b , Ct , as discussed at the beginning of Section 3. Again C(t, K) = t, and we have BtK3exc (G) = ∅ by (3A). Recall from [V5 , Section 6] that Case 1 arises only from the existence of a maximal symplectic triple with J ∼ = P Ω+ 8 (3). Hence [V5 , Proposition 6.1] applies, whence in Case 1, ∼ Aut(K) = ∼ Aut(F i22 ); and (1) Ct / t = (5B) (2) C(b, J) = b, and |Cb : J| is odd. In Case 2, on the other hand, [V5 , Proposition 3.1] applies, and as Out(K) ∼ = Out(F i23 ) = 1, (1) Ct = K × t ∼ = F i23 × Z2 ; and (5C) (2) b ∈ Syl3 (C(b, J)). We first prove Lemma 5.1. Assume that Case 2 holds and let t1 ∈ I2 (I) be such that E(CI (t1 )) ∼ = 2U4 (3). Then E(CG (t1 )) = O {2,3} (CG (t1 )) ∼ = 2F i22 . Moreover, O 2 (CG (t1 )) ∼ = 2F i22 or 2F i22 ∗ SL2 (3). Proof. Set K1 = E(CK (t1 )). By [VK , 12.21], K1 = CK (t1 ) ∼ = 2F i22 and in particular t1 ∈ K1 . Since m3 (F i22 ) = 5 [IA , 5.6.1] and m2,3 (G) = 6, the pumpup of K1 in Ct1 is not a diagonal pumpup. By [VK , 3.32], K1 is a component of CG (t1 ). Set X = C(t1 , K1 ). Since Ct = K × t, CX (t) = CCt (K1 ) = t, t1 , so a Sylow 2-subgroup of X is dihedral or semidihedral and any component of X contains t1 . As every component of X is in C2 , and O2 (X) = 1, X is either a 2-group or  isomorphic to GL2 (3). Since | Out(K1 )| = 2, the lemma follows. Now we can determine the 5-structure of G. ∼ E52 ∈ Syl (G) and all subgroups Lemma 5.2. Let W ∈ Syl5 (K). Then W = 5 of W of order 5 are G-conjugate. Moreover, if w ∈ W # , then CG (w) = w × Xw , where Xw ∼ = A9 in Case 1 and Xw ∼ = Σ9 in Case 2.

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6. THEOREM C5 : STAGE 4

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Proof. By [IA , 5.3tu], W ∼ = E52 and all elements of E1 (W ) are K-conjugate. We can choose W and w so that w ∈ I. By the fusion just noted, to complete the proof we need only verify the structure of CG (w) for this particular w. Since w ∈ I, [w, t] = [w, b] = 1. Set H = L3 (CG (w, b)) = L3 (CJ (w)). By the structure of Ct and Cb , (1) H ∼ = A6 ; (2) CG (w, b) = w × b × H in Case 1, and in any case, b ∈ Syl3 (CG (w, b H)); (5D) (3) In Case 1, CG (w, t) ∼ = Z5 × (E × A5 )Z2 , w, t = Z(CG (w, t)) and |E| = 4; and (4) In Case 2, CG (w, t) ∼ = w, t × Σ7 . We shall prove that (5E)

O2 (CG (w)) = w ;

for this, it is sufficient to find a four-subgroup V ≤ CG (w) such that (5F)

O2 (CG (v, w)) = w

for all v ∈ V # . Now (5F) clearly holds for v = t. In Case 1, (5D1, 2) imply that t∈H ∼ = A6 , so we can find our V as a subgroup of H. On the other hand, in Case 2, I = E(CJ (t)) ∼ = Ω7 (3), so H t ∼ = Σ6 by [VK , 12.23a]. This time we take V = t, t0 , where t0 ∈ CHt (t) ∩ tH , so that t1 := tt0 ∈ H. By [VK , 12.23b], t1 ∈ I and E(CI (t1 )) ∼ = 2U4 (3). Hence by Lemma 5.1, K1 := E(CG (t1 )) = O {2,3} (CG (t1 )) ∼ = 2F i22 . In particular, w ∈ K1 . Since O2 (CG (t1 )) = 1 and O 2 (CG (K1 )) ∼ = K1 or K1 ∗ SL2 (3), by Lemma 5.1 again, O2 (CG (w, t1 )) = O2 (CK1 (w)) = w by [IA , 5.3t]. This is the desired condition, and (5E) is proved. As observed above, t normalizes, but does not centralize, H ∼ = A6 . In particular, t ∈ O2 (CG (w)). But by (5D3, 4), t is the unique minimal normal 2-subgroup of CG (w, t). Therefore O2 (CG (w)) = 1. With (5E) we conclude that F ∗ (CG (w)) = N w, where N = E(CG (w)) and Z(N ) ≤ w. By L2 -balance and (5D3, 4),  M := E(CG (w, t)) = O 3 (CG (w, t)) is a component of CN (t), and M ∼ = A5 or A7 . In particular, b ∈ M ≤ N . Similarly, by (5D3, 4), O3 (CG (w, t)) has order 5 · 2a with a ≤ 2. This implies that E(O3 (CG (w, t))) = 1, so O3 (CG (w)) = w. By L3 -balance, H = L3 (CJ (w)) ≤ N ; indeed as b ∈ N , H ≤ N1 for some component N1 of CG (w). As H = [H, t], also N1 = [N1 , t]. But E(CN (t)) is a single component containing b, so either b ∈ N1 or [N1 , b] = 1. In the latter case N1 = H, so CH (t)   CG (w, t). But H t ∼ = A6 or Σ6 , as noted above, so correspondingly CH (t) ∼ = D8 or Σ4 , which is inconsistent with the structure of CG (w, t) given in (5D3, 4). Thus, b ∈ N1 . In particular, M ≤ N1 . Now E(CN1 (t)) = M ∼ = A5 or A7 , and E(CN1 (b)) ∼ = A6 , ∼ so by [VK , 12.22], N1 = A9 and in Case 2, N1 t ∼ = Σ9 . Again in view of (5D3, 4), CG (w, t) ∼ = CN1 w (t) in Case 1 and CG (w, t) ∼ = CN1 w,t (t) in Case 2. Therefore CG (w, t) ≤ N w, t. As t leaves N1 invariant and F (CG (w)) = w, this clearly forces N = N1 , whence CG (w) = N1 w or N1 w, t in the respective cases. The lemma follows.  Lemma 5.3. Cb = JD × b, where F ∗ (JD) = J, D ∩ J = 1, and D ∼ = Z3 and Σ3 in Cases 1 and 2, respectively.

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5. THE CASE G∗ = F i24 OR F i24

285

Proof. This time let W ∈ Syl5 (J), so that |W | = 52 and so W ∈ Syl5 (G), by Lemma 5.2. Consequently, O3 (CG (b)) is a 5 -group. Then as [W, b] = 1 and O3 (CG (w)) = w for each w ∈ W # , O3 (CG (b)) ∩ CG (w) = 1 and thus O3 (CG (b)) = 1. By Lemma 5.2, CG (W ) = W × E, where E ∼ = A4 or Σ4 according as Case 1 or Case 2 holds. In particular, b ∈ Syl3 (CG (W )) so NG (b) covers NG (W )/CG (W ). But by the structure of K, and as W ∈ Syl5 (G), NG (W ) contains an element of order 3 outside CG (W ). On the other hand, AutJ (W ) is a 2-group by [VK , 9.12a]. Hence there is y ∈ I3 (Cb ) inducing a triality on J. Furthermore, our argument shows that Sylow 3-subgroups of NG (W ) are isomorphic to E32 . Finally, by [VK , 9.12b], J y has a single class of elements of order 5, so Cb = J y CG (w, b) for any w ∈ W # . By Lemma 5.2, however, CG (w, b) = w × CXw (b) with CXw (b) ∼ = Z3 × A6 or Z3 × Σ6 in Case 1 or 2, respectively. As CJ (w) ∼  = w × A6 , the proof is complete. Now we let z be an involution of I ∼ = Ω− 6 (3) or Ω7 (3) with four eigenvalues equal to −1 on the natural F3 I-module. Thus by [VK , 10.20], z is 2-central in I and lies in the commutator subgroup of F ∗ (CI (z)) = O2 (CI (z)). We first prove Lemma 5.4. The following conditions hold: (a) z = Z(T ) for some T ∈ Syl2 (G); (b) In Cb , we have (1) z is 2-central in J; × Z3 ; and (2) F ∗ (CCz (b)) ∼ = 21+8 + (3) CCz (b) = AE, where A  CCz (b), A ∩ E = 1, A is the central product of four copies of SL2 (3) permuted transitively by E ∼ = Z3 × Z2 × E0 , A or Σ according as Case 1 or Case 2 holds; where E0 ∼ = 4 4 (c) CCz (t) is an extension of U = O2 (CCz (t)) by V . In Case 1, U/ t ∼ = 21+10 + and V ∼ and V is a crown product = Aut(U4 (2)). In Case 2, U ∼ = E23 ×21+8 + of Σ3 and Aut(U4 (2)); and (d) In both Case 1 and Case 2, (1) Q := F ∗ (CK (z)) ∼ = E22 × 21+8 + ; ∼ (2) Q/Z(Q) = E28 is a natural module for V (∞) ∼ = U4 (2); and (3) [Z(Q), b] = 1 and [Q, b] ∼ . = 21+4 + Proof. Since CCz (b) = CCb (z), part (b) follows directly from Lemma 5.3 and [VK , 10.58]. Likewise as I = E(CG (b, t)) = E(CK (b)), parts (c) and (d) follow from [VK , 10.59]. Hence, it suffices to prove (a). Suppose first that Case 1 holds. By order considerations and [VK , 10.55], z, t = Z(R) for some R ∈ Syl2 (Ct ). Since Z(I) = t in this case, t and tz both have 2-dimensional supports of − type on the natural J-module, so they are CJ (z)-conjugate by Witt’s lemma. Thus, NG (z, t) has a 2-element u with tu = tz. As R ∈ Syl2 (CG (z, t)), we can choose u ∈ NG (R) − R with u2 ∈ R. Then Z(R u) = z. But Z(T ) ≤ R for any T ∈ Syl2 (G) containing R u, so z = Z(T ), proving (a) in this case. Suppose then that we are in Case 2. This time, using [VK , 10.55] again, for some R ∈ Syl2 (Ct ), Z(R) = t, z, y ∼ = E23 with z, y ≤ I. Here z and y have four and two eigenvalues −1 in the natural representation of I, respectively. Moreover, with Witt’s Lemma again, yzt and t are conjugate by a 2-element of CJ (y, z).

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6. THEOREM C5 : STAGE 4

286

Expand R to R∗ ∈ Syl2 (CG (y)). By the conjugacy just noted, we may assume that R∗ is chosen so that z, y ≥ Z(R∗ ). However, with t1 as in Lemma 5.1, y and t1 are J-conjugate, so E(CG (y)) ∼ = 2F i22 by Lemma 5.1. Hence y, z = Z(R∗ ), and now we can repeat the argument of Case 1 to conclude that y ∼NJ (y,z) yz, whence z = Z(T ) for suitable T ∈ Syl2 (G) containing R∗ , Thus (a) again holds, completing the proof of the lemma.  Lemma 5.5. The following conditions hold: (a) |Cz |5 = 5; and (b) F ∗ (Cz ) = O2 (Cz ). Proof. By Lemma 5.4c, 5 divides |CCz (t)|. Hence in view of Lemma 5.2, to prove (a) we need only show that for W ∈ Syl5 (G), CG (W ) contains no conjugate of z. We know from Lemma 5.2 that O 5 (CG (W )) ∼ = A4 or Σ4 . Since 52 divides |Ct |, CG (W ) contains a conjugate of t, and similarly in Case 2, CG (W ) contains a conjugate of t1 as well, by Lemma 5.1. As Ct ∼ = CG (t1 ), t and t1 are not conjugate in G. Thus every involution of CG (W ) lies in tG in Case 1, and in tG ∪ tG 1 in Case 2. On the other hand, E(Ct ) and E(CG (t1 )) are isomorphic to 2F i22 or F i23 , so by [IA , 5.3u], their Sylow 2-subgroups have noncyclic centers, while a Sylow 2-subgroup of Cz has center z by Lemma 5.4a. Thus CG (W ) contains no element of z G , and (a) is proved. Now take W ∈ Syl5 (G) containing w ∈ Syl5 (Cz ). By Lemma 5.2, z ∈ CG (w) = w × X, with X ∼ = A9 or Σ9 . As CX (z) is a 5 -group, z is the product of three or four disjoint transpositions in X. Hence |CCz (w)/ w | divides 27 32 . In particular, CCz (w) is solvable. We use these facts in verifying (b). Suppose that (b) fails, so that E(Cz ) = 1 as G is of even type. Let L ∈ C2 be a component of Cz . By Lemma 5.4a, Cz has no simple components, so Z(L) = 1. It follows by [VK , 2.4] that w ∈ LCCz (L). Since CCz (w) is solvable, [L, w] = 1, whence w ∈ L and L = E(Cz ). Thus, z ∈ L. As L ∈ C2 , |L|5 = 5, and z = Z(T ), we conclude from [VK , 12.52] that | Aut(L)|2 ≤ 212 . However, |CCz (L)|2 ≤ |CCz (w)|2 ≤ 27 by the preceding paragraph. Thus |Cz |2 ≤ 219 . Now, |Ct |2 = 219 , so by Lemma 5.4a, z and t both generate Sylow 2-centers in G. As |Cz |5 = 5 and |Ct |5 = 52 , z and t are not conjugate, a contradiction. The proof is complete.  Now set

R = O2 (Cz ) = F ∗ (Cz ) and again take w ∈ I5 (CCz (t)). Lemma 5.6. The following conditions hold: ; (a) R ∼ = 21+12 + (notation as in Lemma 5.4b); (b) CR (b) = O2 (A) ∼ = 21+8 + (c) z is 2-central in CG (w); and (d) In Case 1, z is not weakly closed in R with respect to G. Proof. Suppose that F is any nontrivial elementary abelian characteristic subgroup of R. Then CF (t)  CCz (t), so by Lemma 5.4d, [CF (t), b] = 1. By the A × B lemma, [F, b] = 1. Thus F  CCz (b), so F = z by Lemma 5.4b. Hence by P. Hall’s theorem, R is of symplectic type. By [V2 , 4.12], [R, w] is extraspecial. By the A×B lemma, [CR (t), w] = 1. But [CR (t), w] ≤ Q (notation as in Lemma 5.4d), so [CR (t), w] ≤ [Q, w]. By the irreducibility of V on [Q, w] ∼ = 21+8 + , we have

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5. THE CASE G∗ = F i24 OR F i24

287

[Q, w] ≤ R, so [CR (t), w] = [Q, w] ∼ = 21+8 + . Hence [R, w] is the central product of [CR (t), w] and its centralizer R1 in [R, w], both of which are t-invariant. But then R1 = z, for otherwise 1 = [CR1 (t), w] ≤ R1 ∩ [CR (t), w] = z, a contradiction. Therefore [R, w] = [CR (t), w] ∼ = 21+8 + . Likewise, by Lemma 5.4d, [CR (t), b] ∼ = 21+4 + , and we conclude similarly that 1+4 ∼ [R, b] = [CR (t), b] = 2+ . However, as CR (b)  CCz (b) = CCb (z), Lemma 5.4b implies that either CR (b) = O2 (A) ∼ or CR (b) = z. The latter is impossible = 21+8 + since R = [R, b]CR (b) and R contains 21+8 , so CR (b) ∼ = 21+8 + + , and (a) and (b) follow. 1+4 ∼ In addition it follows that CR (w) = 2+ . Finally, as seen in the preceding lemma, CCz (w) = w × CX (z) with z ∈ ∼ X∼ = CR (w) ≤ O2 (CX (z)). Clearly = A9 or Σ9 . By the preceding paragraph 21+4 + this forces z to be the product of four disjoint transpositions in X, and in Case 1, CR (w) = O2 (CX (z)). In particular, (c) holds. Finally, E1 ≤ CX (z) for some E1 ∼ = E23 with E1# ⊆ z X (a regular E23 on 8 points). In Case 1, |X|2 = 2|O2 (CX (z))| by inspection of A9 , so z X ∩ O2 (CX (z)) = {z}, and (d) follows. The lemma is proved.  Since O3 (Cb ) = 1 by Lemma 5.3, we can now use the notation C z = Cz /R. We set H z = F ∗ (C z ) and next prove Lemma 5.7. The following conditions hold: (a) H z ∼ = 3U4 (3); (b) O2 (C z ) = Z(H z ) ∼ = Z3 is inverted in C z ; ∼ (c) C z /H z = Z2 or E22 according as Case 1 or 2 holds; and (d) H z is irreducible on R/ z, and CR (Z(H z )) = z. Proof. Continuing the previous proof, we have CR (w) ∼ = 21+4 + , and according as we are in Case 1 or Case 2, (5G) C (w) ∼ = Z5 × Σ3 or C (w) ∼ = Z5 × Σ3 × Z2 . Cz

Cz

∼ Z6 × A4 or Z6 × Σ4 , as the case Write CCz (b) = AE as in Lemma 5.4b. Thus E = may be. Since b J = b × J, clearly A b /O2 (A) ∼ = E35 . We saw in Lemma 5.6 that CR (b) = O2 (A) ∼ , whence = 21+8 + 5 E (5H) C (b) ∼ = 3 (Z2 × A4 ) or C (b) ∼ = E35 (Z2 × Σ4 ), respectively. Cz

Cz

Now set Y = O2 (C z ). We argue that |Y | ≤ 3.

(5I)

This follows from (5G) if [Y , w] = 1. In the contrary case, since |C z |5 = 5 by Lemma 5.5a, and C z involves U4 (2) by Lemma 5.4c, AutC z (Y ) involves U4 (2). + (2). But this contradicts [VK , Hence so does the automizer of Y in Out(R) ∼ = O12 15.4], so (5I) holds. It follows therefore that H z = E(C z )Y . Furthermore, CC z (w) is solvable by (5G), so w normalizes every component of H z . In addition CC z (t) involves U4 (2), of order divisible by 5, and w ∈ Syl5 (C z ). It follows that H z is a single component + (2), and (5G) and (5H) hold, [VK , 15.3] yields containing w. As C z embeds in O12 ∼ E(C z ) = 3U4 (3) (whence H z = E(C z ) ≥ Y ∼ = Z3 ), as well as the remaining assertions of the lemma.  Lemma 5.8. Case 1 occurs.

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6. THEOREM C5 : STAGE 4

288

Proof. Suppose that Case 2 occurs. We use [III2 , 6.1], with z and t in the respective roles of y and x there, to show that t ∈ [G, G], contradicting the simplicity of G. Let Tt ∈ Syl2 (Ct ), chosen so that z ∈ Z(Tt ), and expand Tt to T ∈ Syl2 (Cz ) ⊆ Syl2 (G). As Ct = t × K, Z(Tt ) = t × Z(T ∩ K), with Z(T ∩ K) ∼ = E22 containing three non-K-conjugate involutions. Now, U4 (3), the unique nonsolvable composition factor of Cz , does not involve U6 (2) or F i22 , so by [IA , 5.3u], z is − uniquely determined in Z(T ∩ K), and CK (z) ∼ = (E22 × 21+8 + )(Σ3 Y O6 (2)). For the same reason, z G ∩ Z(Tt ) ⊆ {z, zt}. If z G ∩ Z(Tt ) = {z, zt}, then NT (Tt ) normalizes z, zt = z, t and moves t, so t ∈ z G , a contradiction. Therefore z G ∩ Z(Tt ) = {z}, the key assumption of [III2 , 6.1]. ∼ Finally, O 2 (CCz (t)) is an extension of a 2-group by Z3 ×Ω− 6 (2) = Z3 ×P Sp4 (3). It follows that t centralizes a 3-element v ∈ O2,3 (Cz ) and induces a graph automorphism on E(CCz (v)) z / z ∼ = 3U4 (3). Since Cz = O2 (Cz )NCz (v), we conclude that t ∈ [Cz , Cz ]. Hence by [III2 , 6.1], t ∈ [G, G], the desired contradiction. The proof is complete.  Lemma 5.9. G ∼ = F i24 . Proof. Since Case 1 holds, and by Lemmas 5.6ad and 5.7ac, we may apply [V2 , Theorem 10.5] to obtain G ∼  = F i24 .

6. The Case G∗ = F2 In this section our assumptions are as follows. t ∈ I2 (G), b ∈ I3 (CG (t)), and (K; I; J) is a nonconstrained {t, b}neighborhood satisfying the conditions of Theorem C5 : Stage 2, (6A) of type (K; I; J) = (22E6 (2); 2U6 (2); F i22 ), and with m2,3 (G) = 5. We let T ∈ Syl2 (Ct ) with S := T ∩ I ∈ Syl2 (I) and set Z = Z(T ). By [VK , 10.53], S and T may be chosen so that Z = Z(S) ∼ = E22 . However, a Sylow 2subgroup of J has center of order 2 by [VK , 10.55], and I = CJ (t) by [IA , 5.3t]. Hence there is a 2-element y ∈ NJ (Z) such that ty ∈ Z − t. We set z = tty and immediately deduce that

(6B)

(1) (2) (3) (4)

T ∈ Syl2 (G); z = tty is 2-central in I and G; F ∗ (CI (z)) = O2 (CI (z)); and z = [O2 (CI (z)), O2 (CI (z))].

The third condition holds by the Borel-Tits theorem, while the fourth holds because y normalizes CI (z) = CI (Z) = CJ (Z) and by [VK , 2.11], [O2 (CI (z)), O2 (CI (z))] has order 2 and thus equals CZ (y) = z. Also, (6C)

CK (Z)/Z is a split extension of E220 by U6 (2),

as CK (Z) is the maximal parabolic subgroup of K of type U6 (2).

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6. THE CASE G∗ = F2

289

As in the previous section, we first study elements of order 5 in K and their centralizers in G. By [VK , 12.24], (1) I5 (K) is a single conjugacy class; (2) If w ∈ I5 (K), then CK (w) = w × t × Xw , Xw ∼ = A8 , and 2-central involutions of Xw are 2-central in K; and (6D) (3) If Ct > K (whence |Ct : K| = 2), then CCt (w)/ t, w ∼ = Σ8 . Now y normalizes CG (Z) ≤ Ct , and |CG (Z)|5 = 5 by (6C). Hence for any fixed w ∈ I5 (CG (Z)), we can assume by a Frattini argument that y ∈ CG (w). Thus t is not isolated in a Sylow 2-subgroup of CG (w). Set Cw = CG (w) and w = Cw /O2 (Cw ). By (6D2, 3), O2 (CC (w)) = O2 (CK (w)) = w, so C t (6E) We let

CO2 (Cw ) (t) = w .  = F ∗ (C w ) H

and first prove Lemma 6.1. The following conditions hold: (a) O2 (Cw ) = w; ∼ (b) H = HS;    (c) Cw = H t ∼ = Aut(HS);  and (d) z is 2-central in H; (e) C(b, J) = b. Proof. Since m2,3 (G) = 5 = 1+m2,3 (J) [IA , 5.6.1], C(b, J) has a cyclic Sylow 3-subgroup and a normal 3-complement. Now CO3 (Cb ) (t) ≤ O{2,3} (CCt (b)) = 1 [IA , 4.7.3A], so t inverts O3 (Cb ); but then J = [J, t] centralizes O3 (Cb ), so O3 (Cb ) = 1 as t ∈ J. Next, CG (b, w) = CCb (w) is the direct product of a cyclic 3-group with Z5 × Σ5 × V , and |V | ≤ 2. Since I ∼ = 2U6 (2), I contains some w1 ∈ wK by (6D1). ∼ By [IA , 5.3t], CJ (w1 ) = Z5 × Σ5 . Hence Cw contains a K-conjugate b1 of b and  := E(C  (b1 )) ∼ E(CCw (b1 )) = CCw (b1 )(∞) ∼ = A5 . = A5 . Then L Cw     On the other hand, let H0 be the subnormal closure of Xw in Cw . We have  w , we  t ∈ Syl2 (CCw (Xw )) by (6D). As  t is not isolated in a Sylow 2-subgroup of C  0 , whence H  0 = F ∗ (C w ) = H,  O2 (H)  = 1, and X w is a component w < H have X K   is not a t). Now b1 ∈ b ⊆ CG (t), so the structure of L implies that H of CH (  and so H  is simple. diagonal pumpup of X,  = [H,  t], E(Cw ) has Also by (6E), t inverts O2 (Cw )/ w elementwise, so as H  a (unique) component H covering H. Given the structure of CCw (b) and CCw (t), we see from [VK , 3.47] that H ∼ = HS, and t induces an outer automorphism on H. Thus, (b) and (c) hold. Moreover, CH (b1 ) ∼ = Z3 × Σ5 . As H = O 2 (Aut(H)), it follows that b ∈ Syl3 (C(b, J)), so (e) holds. Likewise, |CG (b1 , w) : w CH (b1 )| ≤ 2. As b1 induces a nontrivial inner automorphism on H, C(w, H) is a 3 -group.   Therefore our conditions imply But also |C(w, H)| is odd as  t ∈ Syl2 (CCw (H)).  that C(w, H) = w = O2 (Cw ), proving (a). By [VK , 12.24] and [VK , 3.68], z is 2-central in H, so (d) holds and the proof is complete.  As a corollary we obtain

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6. THEOREM C5 : STAGE 4

290

Lemma 6.2. Cb = J v × b, where v is an involution and J v ∼ = Aut(F i22 ). Proof. By the structure of Cw , and in the notation of the preceding proof, ∼ Z3 × Σ5 × Z5 × Z2 , CCw (b1 ) = as H t ∼ = Z5 × Σ5 by [IA , 5.3t], so = Aut(HS). However, as J ∼ = F i22 , CJ (w1 ) ∼ Cb contains an involution v such that J v ∼ = Aut(F i22 ). The lemma follows then from Lemma 6.1e.  Similarly, we prove: Lemma 6.3. The isomorphism type of Ct is uniquely determined. In particular Ct = K s for some involution s ∈ Ct − K such that CK (s) ∼ = F4 (2) × Z2 . Proof. Set Y = CG (t, b). By Lemma 6.2, the structure of Y is uniquely determined: Y = b × Y0 , where Y0 = CJv (t), with J v ∼ = Aut(F i22 ) and CJ (t) = I ∼ = 2U6 (2). By [IA , 5.3t], Y0 = I g with g 2 = 1 and CI (g) ∼ = Sp6 (2). As CK (b) ∼ = Z3 × 2U6 (2), Y ≤ K. But | Out(22E6 (2))| = 2 by [IA , 2.5.12, 6.3.1], so |Ct : K| = 2 and Ct = KY = K g. Now O 2 (Y ) ∼ = Z3 × 2U6 (2) ∼ = CK (b), so 2 2 O (Y ) = CK (b). In particular, CAut(K) (O (Y )) = b by [IA , 4.7.3A]. Thus the action of g on K is completely determined by its action on O 2 (Y ). As there exists a graph automorphism γ ∈ Aut(K) such that γ 2 = 1, E(CK (γ)) ∼ = F4 (2), [γ, b] = 1  and CCK (b) (γ) ∼ = Z3 × Sp6 (2), g must act as γ on K, and the lemma follows. Now we turn to Cz . Set M = CK (z) and N = CJv (z). By [VK , 10.54] and [IA , 5.3t], and M/O2 (M ) ∼ (1) O2 (M ) = F ∗ (M ) ∼ = Z2 ×21+20 = U6 (2); + and (6F) and N/O2 (N ) ∼ (2) O2 (N ) = F ∗ (N ) ∼ = 21+10 = Aut(U4 (2)). + Set C z = Cz /O3 (Cz ) and N0 = [N, N ]. We next prove: Lemma 6.4. C z ∼ = Co2 . Proof. Set L = E(C z ). Then N 0 ∼ = (2)U4 (2) is a 3-component of CC z (b), so  N 0 ≤ L by L3 -balance. Also b ∈ Syl3 (CCb (N0 /O2 (N0 ))), so b ∈ Syl3 (CC z (N 0 )).   L Set L1 = N 0 ; then m3 (CC z (L1 )) ≤ 1. Consequently, Aut(L1 ) involves M /O2 (M ) ∼ = (2)U6 (2). In particular, L1 must be a single component, and L1 > N 0 . Thus [L1 , b] = 1, so CC z (L1 ) is a 3 -group. Therefore L = L1 = F ∗ (C z ) is simple. We have N 0 /Z(N 0 ) ∼ = U4 (2) ↑3 L via b with m3 (CL (N 0 )) ≤ 1 and m3 (L) ≤ m2,3 (G) = 5. Moreover, L involves M /O2 (M ) ∼ = U6 (2). It follows from [VK , 3.24] that L ∼ = Co2 or U6 (2). We show that m5 (C z ) > 1. As | Aut(U6 (2))|5 = 5 and Out(Co2 ) = 1, this will force L ∼  U6 (2) and complete the proof of the lemma. Now, we have taken =  with w ∈ CG (Z), so 1 = w ∈ M . But also CCw ( z ) contains a perfect subgroup D ∼ A5 , by Lemma 6.1cd and [IA , 5.3m], so Cz contains w × D with D  2 (D)  = D/O ∼ A5 . Since C z = Cz /O3 (Cz ), D = 1. Hence as w = 1, perfect and D/O2 (D) = m5 (C z ) ≥ m5 (w D) > 1. The proof is complete.  Lemma 6.5. O3 (Cz ) = O2 (Cz ) = F ∗ (Cz ) ∼ . = 21+22 +

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8. THE CASE G∗ = F1

291

∼ E32 , Proof. By Lemma 6.4 and [VK , 3.72], there is E ≤ Cz with b ∈ E = E # ⊆ bCz , and a generator of E ∩ N having eigenvalues ω, ω −1 , 1, 1 (ω 3 = 1) in the natural representation of N 0 ∼ , = U4 (2). Set D = O3 (Cz ). By (6F2), CD (b) ∼ = 21+10 + 1+10 # ∼ for all e ∈ E . But E = (E ∩J)×b with E ∩J ≤ N , and we see so CD (e) = 2+ 1+10 within J that CD (E) ∼ -subgroups and = 21+6 + . Thus, D is the product of four 2 22 |D/ z | = 2 . By [VK , 5.4], D/ z is elementary abelian and C z acts irreducibly on it. Therefore CD (e)  D for each e ∈ E # , so D is the central product of CD (E)  with four 21+4 + -subgroups. The lemma follows. ∼ 22E6 (2), Lemmas 6.3, 6.4, and 6.5 allow us to quote [V2 , Theorem Now as K = 10.7a]. We conclude: Lemma 6.6. G ∼ = F2 . 7. The Case G∗ = Co1 In this section, following (b) of Theorem C5 : Stage 3, our assumptions are as follows. We write Cb , C b , and Cz for CG (b), Cb /O3 (Cb ), and CG (z), respectively.

(7A)

(1) (B, T ) is a faithful symplectic pair with B ∈ B3∗ (G), m3 (B) = 4, and T ∼ = 21+8 + ; (2) b ∈ B # , O3 (Cb ) = 1, z ∈ J := E(Cb ) ∼ = 3Suz, and C(b, J) is a cyclic 3-group; (3) T = F ∗ (Cz ) and Cz /T ∼ = Ω+ 8 (2); and 1+6 ∼ := (4) R CT (b) = 2− and R = F ∗ (CJ (z)).

Lemma 7.1. z is not weakly closed in T with respect to G. Proof. It suffices to show that z is not weakly closed in R with respect to J. But by [IA , 5.3o], any involution u ∈ J − z J satisfies |CJ (u)|2 = 29 < 12 |J|2 . Hence any half 2-central involution of J lies in z J ⊆ z G . But R contains a four-group U normal in a Sylow 2-subgroup of J by [IG , 10.11], so U # ⊆ z G . The lemma is proved.  By Lemma 7.1 and (7A1, 3), we may quote [V2 , Theorem 10.6] and deduce: Lemma 7.2. G ∼ = Co1 . 8. The Case G∗ = F1 In this section we continue the notation Cb , C b , and Cz , and we assume the following (see Theorem C5 : Stage 3):

(8A)

(1) (B, T ) is a faithful symplectic pair with B ∈ B3∗ (G), ; m3 (B) = 6, and T ∼ = 21+24 + # (2) b ∈ B , O3 (Cb ) = 1, z ∈ J := E(Cb ) ∼ = 3F i24 , and C(b, J) is cyclic; (3) T = F ∗ (Cz ) and Cz /T ∼ = Co1 ; (4) z is 2-central in Cb ; (5) There is t ∈ I2 (CG (b, z)) such that K := E(CG (t)) ∼ = 2F2 ; and and R = F ∗ (CJ (z)). (6) R := CT (b) ∼ = 21+12 +

Lemma 8.1. Ct = K.

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6. THEOREM C5 : STAGE 4

292

Proof. As Out(J) inverts b = Z(J), Cb = JC(b, J) so CCz (b) = CCb (z) = CJ (z)C(b, J). Using [IA , 5.3lv] we see that |CCz (b)| = |CJ (z)|, so C(b, J) = Z(J) = b and Cb = J. Similarly, Ct = KC(t, K) and using CCt (b) = CCb (t) = CJ (t) and  [IA , 5.3vy], we get C(t, K) = Z(K) and Ct = K. In view of Lemma 8.1 and (8A1, 3, 5), [V2 , Theorem 10.7b] implies: Lemma 8.2. G ∼ = F1 . Now Proposition 3.1 follows from Lemmas 3.9, 4.2, 5.9, 6.6, 7.2, and 8.2. 9. The Case G∗ = 2 D5 (2), 2 E6 (2), or U7 (2) In the preceding sections we have treated outcomes (a), (b), and (c) of Theorem C5 : Stage 3. We now turn to outcome (d), the last alternative. We assume that G has a constrained neighborhood (CG (z), b, J) of U4 (2)-type, coming from a faithful symplectic pair (B, T ) and a maximal symplectic triple (B, T, b). Thus J is a component of Cb := CG (b), b ∈ B ∈ B3∗ (G), m3 (C(b, J)) = 1, T ∈ I∗G (B; 2) is extraspecial, and T = F ∗ (CG (z)) where z = Z(T ). Moreover, J is isomorphic to one of the groups (9A)

D4 (2), U5 (2), U6 (2), SU6 (2),

and z is a 2-central involution of J and of G. There exists D = b, a ∈ E2 (B) such that JD := E(CJ (D)) ∼ = Um (2) where m = m2,3 (G) = m3 (B) = 4 or 5. For each d ∈ D# we let Jd be the subnormal closure of JD in Cd := CG (d), and we have

(9B)

(1) O3 (Cd ) = 1; (2) Jd  Cd , and if Jd > JD , then C(d, Jd ) is a cyclic 3group; (3) Either Jd = JD or Jd is isomorphic to one of the groups in (9A).

Finally, for some d ∈ D − b, Jd > JD . We fix d and set J  = Jd . Our goal is the following theorem. Theorem 9.1. Under the above hypotheses, G ∼ = 2 D5 (2), U7 (2) or 2 E6 (2). By [VK , 12.38], we have B < B ∗ ∼ = E3m+1 and m3 (CG (b)) = m + 1. We set ∗ ∼ AutG (B ∗ ) W = NG (B )/CG (B ∗ ) = and abuse notation by writing W ∩ H for AutH (B ∗ ), whenever H is a B ∗ -invariant subgroup of G. Our first step will be to pin down the possibilities for W . We will often write Cx for CG (x) and C(d, Jd ) for CCd (Jd ). We set  G 0 = Jd | d ∈ D # . Our next step will be the proof of the following theorem. Theorem 9.2. G0 ∼ = 2 D5 (2), U6 (2), U7 (2) or 2 E6 (2). Finally, we simplicity of G. contradiction in We proceed

will prove that G = NG (G0 ), and so G = G0 , in view of the In the case when G0 = U6 (2), we will have B ∗ ≤ G0 , yielding a this case and thereby proving Theorem 9.1. in a sequence of lemmas.

Lemma 9.3. O3 (CG (x)) = 1 for all x ∈ (B ∗ )# .

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9. THE CASE G∗ = 2 D5 (2),

2

E6 (2), OR U7 (2)

293

∗ respect to B  , by Theorem C5 : Stage 1, O3 (Cx ) =  Proof. As G is balanced,with # #  O3 (Cx ) ∩ Cd | d ∈ D ≤ O3 (Cd ) | d ∈ D = 1 by (9B1).

Lemma 9.4. There exists b ∈ B such that JE := E(CJ (b )) JE ≤ JD . Set E = D b . For each e ∈ E # , let Je be the subnormal in Ce := CG (e). Then for every e ∈ E # , the following hold: (a) Je  Ce ; and (b) Either Je = JE or Je is isomorphic to one of the groups in Moreover, if m2,3 (G) = 4, then E = D.

∼ = U4 (2) and closure of JE

(9A).

Proof. If m2,3 (G) = 4, then JD ∼ = U4 (2), and we take b = 1 and E = D; there is nothing to prove in this case.  Assume then that m2,3 (G) = 5, whence JD = O 3 (CJ (D)) ∼ = U5 (2). Then z ∈ JD is 2-central in JD . Also B ∩ J normalizes JD , so B = b × (B ∩ JD ). As [B ∩ JD , z] = 1, there is b ∈ B ∩ JD such that if we set E = D b  and JE = E(CJ (E)), then JE ∼ = U4 (2) and JE ≤ JD . We let Je be the subnormal closure of JE in Ce := CG (e) for each e ∈ E # , and verify the desired properties, using the inclusion D ≤ E. Now, B normalizes every component of Je by L3 -balance and [IG , 8.7(ii)], so Je is a single component. If Je > JE , then by [VK , 3.11], either Je is as in (9A), or Je ∼ = P Sp4 (33 ) or Co2 . Since m2,3 (G) = 5 < m2,3 (P Sp4 (33 )), we need only rule out Je ∼ has 3-rank 1 = Co2 . But D acts on Je and maps into CAut(Je ) (JE ), which   Cd = Jd . by [IA , 5.3k]. Hence there is 1 = d ∈ CD (Je ). Then as JE ≤ JD , Je ≤ JD As Jd is one of the groups in (9A), it is impossible that Je ∼ = Co2 . Therefore Je is as in (9A). If Je  Ce for some e ∈ E # , then as JE ≤ Je , Ce contains a copy of JE ×JE ×e,  so m2,3 (G) > 5, contradiction. This completes the proof.

We fix E as in Lemma 9.4. Lemma 9.5. We have d = C(d, Jd ) for all d ∈ D# with JD < Jd . In particular, CG (B ∗ ) = B ∗ . Proof. Let d ∈ D# with JD < Jd and let Rd ∈ Syl3 (C(d, Jd )). Then Rd is cyclic by our setup. Hence, Rd acts faithfully on J1 centralizing JD , where we take J1 = J if d ∈ b, while J1 = J  if d ∈ b. It follows that Rd = d by [VK , 3.74]. As O3 (Cd ) = 1 by our setup, we have d = C(d, Jd ), as claimed. Then, [VK , 3.74]  yields that CG (B ∗ ) = B ∗ , completing the proof. Lemma 9.6. O{2,3} (W ) = 1. Proof. Suppose Or (W ) =: Xr = 1 for some prime r > 3. Note that W embeds in GL(B ∗ ) ∼ = GL(n, 3) where n := m3 (B ∗ ) ∈ {5, 6}. If r = 13 or n = 6, it follows that Xr is cyclic. If X13 ∼ = Z13 ×Z13 , then J/Z(J) ∼ = U6 (2) and W ∩J ∼ = Σ6 . Hence, in any case, [W ∩ J, W ∩ J] centralizes Xr . As CB ∗ ([W ∩ J, W ∩ J]) = b, it follows that [Xr , b] = 1 and so Xr  CW (b). But Or (W ∩ Cb ) = 1, so Xr = 1, a  contradiction. As O{2,3} (W ) is a solvable group, the lemma follows. Lemma 9.7. O2 (W ) = 1. Proof. Suppose not. Then by Lemma 9.6, X := O3 (W ) =  1. Let BX = ∼ SU6 (2), CB ∗ (X) = 1. If b ∈ BX , then X  CW (b), a contradiction. If J =

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294

6. THEOREM C5 : STAGE 4

then W ∩ J acts on B ∗ with unique minimal submodule b. But then as BX is a nontrivial submodule of B ∗ , b ∈ BX , a contradiction. Hence, Z(J) = 1 and b⊕(B ∗ ∩J) is the socle of B ∗ as (W ∩J)-module, with B1 := B ∗ ∩J an irreducible (W ∩ J)-module in every case, by (9A). Thus, B1 ≤ BX . Also, b = CB ∗ (W ∩ J). Let E = D b  with b ∈ B ∩ J and JE = E(CJ (b )) as in Lemma 9.4. We claim that [X, W ∩ JE ] = 1. Suppose W ∩ JE ≤ CW (X). If J ∼ = D4 (2), then s := Z(W ∩ J) ≤ [W ∩ JE , O2 (W ∩ J)] ≤ CW (X). Since b = CB ∗ (s), it follows that [X, b] = 1, whence X  CW (b), a contradiction. If J ∼ = D4 (2), then [W ∩ J, W ∩ J] ≤ CW (X). But then X fixes b = CB ∗ ([W ∩ J, W ∩ J]), whence X  CW (b), again a contradiction. Hence, [X, W ∩ JE ] = 1, as claimed. As b ∈ B1 , [X, b ] = 1. Let Y be the preimage of X in NG (B ∗ ) ∩ Cb . Then by the previous paragraph, [Y, NJE (B ∗ )] ≤ CG (B ∗ ) = B ∗ . As JE ≤ Jb , [Y, NJb (B ∗ )] ≤ B ∗ . But [Y, NJb (B ∗ )] ≤ O3 (NJb (B ∗ )). In view of the possibilities for Jb , we have that O3 (NJb (B ∗ )) = B ∗ ∩ Jb in all cases, whence  [Y, NJb (B ∗ )] ≤ B ∗ , a contradiction completing the proof. Lemma 9.8. Let W ∗ be a subgroup of W containing both W ∩ J and W ∩ Jd for some d ∈ D − b with JD < Jd . Then W ∗ acts indecomposably on B ∗ with absolutely irreducible socle containing B ∗ ∩ J. Proof. As noted in the previous proof, the socle of B ∗ as (W ∩ J)-module is either b or b ⊕ (B ∗ ∩ J) = B ∗ . Moreover, the socle of B ∗ / b as (W ∩ J)-module is b, B ∗ ∩ J / b, which is absolutely irreducible. Also, b is not W ∩Jd -invariant. Hence the socle of B ∗ as W ∗ -module is irreducible and contains B ∗ ∩ J, completing the proof.  Lemma 9.9. Suppose m3 (B ∗ ) = 6. Then for all d ∈ E # , Jd /Z(Jd ) ∼ = Un(d) (2) for some n(d) ∈ {4, 5, 6}. Proof. Suppose that m3 (B ∗ ) = 6 and Jd ∼ = D4 (2) for some d ∈ E # . As ∗ m3 (C(b, J)) = 1 and B ≤ Cb , we must have J/Z(J) ∼ = U6 (2). As m = 6, CB ∗ (Jd ) =: D0 ∼ = E32 and D0 ≤ E. As no proper pumpup of Jd is a C3 -group by [VK , 3.84], we have Jv = Jd for all v ∈ D0# . In particular, b ∈ D0 and so # J E = E(CJd (b)) =E(CJ (v)) for all v ∈ D0 by [VK , 3.73]. But then JE = E(CJ (v)) | v ∈ D0# = J by [IA , 7.3.2], a contradiction proving the lemma.



∼ D4 (2) for some d ∈ E # . Let W1 be the Lemma 9.10. Suppose that Jd = subgroup of W generated by all its reflections. Then W1 contains a subgroup W0 with W ∩ Jd < W0 and W0 ≡B ∗ W (D5 ). Moreover, one of the following holds: (a) W1 ≡B ∗ W (D5 ) or W (C5 ); or (b) W1 ∼ = SO5 (3) or O5 (3). ∼ D4 (2), we may and shall assume that b ∈ J ∩ B ∗ ∩ bW1 . In any case, if J = In (a), the notation means that W1 ∼ = W (D5 ) or W (C5 ) in such a way that B ∗ corresponds to the root module taken modulo 3 [III17 , 1.6, 1.5]. The notation W0 ≡B ∗ W (D5 ) has a similar meaning.

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9. THE CASE G∗ = 2 D5 (2),

2

E6 (2), OR U7 (2)

295

Proof. By Lemma 9.9, m3 (B ∗ ) = 5, D = E and C(d, Jd ) = d. Let WJd = W ∩ Jd ≤ W1 , and let zd  = Z(WJd ). Then d = CB ∗ (zd ) and so CW (zd ) ≤ NG (d). In particular, WJd  CW1 (zd ). Also, W1 contains both W ∩ J and W ∩ Jd and hence acts indecomposably on B ∗ by Lemma 9.8. Suppose that O2 (W1 ) = Z(W1 ). Since B ∗ is indecomposable as W -module and m + 1 = 5 is an odd prime, O2 (W1 ) is abelian but noncyclic, whence by [III17 , 1.4], W1 ≡B ∗ W (D5 ) or W (C5 ), as claimed. It follows that we may assume |O2 (W1 )| ≤ 2 and O2 (W1 ) acts as scalars on B ∗ . As O2 (W1 ) = 1 by Lemma 9.7, E1 := E(W1 ) = 1 and F ∗ (W1 ) = E1 × Z(W1 ) with |Z(W1 )| ≤ 2. Since GL2 (3) is solvable, and m3 (B ∗ ) = 5, E1 is simple, in view of Lemma 9.7. Let Qd = O2 (WJd ) ∼ = Q8 ∗ Q8 . Recall that zd  = Z(Qd ) and CB ∗ (zd ) = d, whence W ∩ Jd and Qd are normal in CW (zd ). As Qd is absolutely irreducible on B ∗ ∩ Jd , it follows that Qd CW (Qd ) = Z × Qd , where Z = Z(W1 ) and |Z| ≤ 2. Moreover, Z ∩[W1 , W1 ] = 1, whence zd  = Z(S) for S ∈ Syl2 (E1 ) and 25 ≤ |E1 |2 ≤ 28 . And of course, |E1 | divides |L5 (3)|. We now use these conditions to consider the possibilities for E1 . If E1 ∈ Alt ∪ Spor, then E1 /Z(E1 ) ∼ = M12 . But then we get a contradiction from the existence of a D10 subgroup of Z2 × Σ5 containing a conjugate of zd . So E1 ∈ Chev(r) − Alt for some prime r. If r is odd, then using the structure of CW (zd ) and [IA ; 4.5.1] we see that E1 ∼ = G2 (3), Ω5 (3), L4 (3) or U4 (3). Then by [III17 , 16.10], E1 ∼ = Ω5 (3). Finally, if E1 ∈ Chev(2), then as m2 (S) ≤ 4 and |S| ≤ 28 , we easily see, using [IA , Table 3.3.1], that E1 ∼ = U4 (2) ∼ = Ω5 (3). As E1 acts faithfully on B ∗ , W ≤ O5 (3) by [III17 , 1.15]. Suppose that W1 = Z × E1 with Z ∼ = Z2 . Then −zd ∈ W1 is a reflection inverting d and centralizing B ∗ ∩ Jd . We may choose the W -invariant form on B ∗ so that (d, d) = 1. There exists M ∼ = E25 : A5 with M < W and Z, zd  < O2 (M ). Then |(−zd )M | = 5 yielding 5 M -conjugates of d: d = d0 , d1 , d2 , d3 , d4 , with di having 2-dimensional support on a suitable natural Jd -module. On the other hand, W ∩ Jd contains a reflection r, which may be chosen to interchange d1 and d2 . Then d∗ := d1 d−1 is 2 the center of the reflection r, and (d∗ , d∗ ) = −1. So, d∗ is not W -conjugate to d, whence r ∈ W1 , a contradiction. It follows that W1 ∼ = SO5 (3) or O5 (3), as claimed. Now W1 contains a subgroup W0 with W0 ≡B ∗ W (D5 ). The final statement of the lemma follows from the action of W0 on B ∗ , completing the proof.  Lemma 9.11. Suppose that J ∼ = D4 (2). Then G1 := J, J   ∼ = 2 D5 (2). Moreover # ∼ if W1 = W (D5 ) or W (C5 ), then G1 = G0 := Jd | d ∈ D . Proof. Let W0 be as in Lemma 9.10. We write B ∗ = b0 , b1 , b2 , b3 , b4  with b = b0 and B ∗ ∩ J = b1 , b2 , b3 , b4 , such that W0 acts as a monomial group with respect to the frame {b0  , . . . , b4 }. By Lemma 9.10, we may assume that b = b1 W and D = E = b0 , b1 . Set Ji = E(CG (bi )), 1 ≤ i ≤ 4. As bi  ∈ b 0 , Ji ∼ = D4 (2) for all i. For i = j, let tij ∈ O2 (W0 ) with tij inverting bi and bj but centralizing the other three bk ’s. Also let σij ∈ W0 be the involution interchanging bi and bj and centralizing the other three bk ’s. Now JD = E(CG (D)) ∼ = Ω− 6 (2). Choose a Curtis-Tits system (L3 , L4 ) for −1 JD with b2 b3 ∈ L3 ∼ = SL2 (2) and b4 ∈ L4 ∼ = Ω− 4 (2). Then L4 centralizes a 2-dimensional nondegenerate subspace of + type in a natural orthogonal F2 JD module VD . In particular, dimF2 [VD , b4 ] = 2.

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6. THEOREM C5 : STAGE 4

296

We study C := CG (L4 ). There exists an involution t4 ∈ L4 inverting b4 . Then Q := O3 (CJD (b4 )) ∼ = E33 and NJD (Q)/Q ∼ = Σ4 . As t4 ∈ L4 , [Q, t4 ] > b4 . It follows that [J4 , t4 ] = 1. But C ≤ CG (b4 , t4 ) = CJ4 (t4 ). In particular, by [VK , 10.67], |C|2 divides |Sp6 (2)|2 . Next, observe that L04 := CJ (L4 ) ∼ = L4 with b1  ≤ L04 . If some ρ ∈ CG (b0 ) acts as a triality automorphism on J, then ρ does not centralize L4 . It follows that either CC (b0 ) = b0  × L04 or CC (b0 ) = b0  × L04 τ0  with L04 τ0  ∼ = Σ5 . Similarly, conjugating C(b0 ) by σ01 , we get L14 := CJ1 (L4 ) ∼ = L4 with b0  ≤ L14 , and b1  × L14 is a subgroup of index at most 2 in CC (b1 ). If W ∼ = W (D5 ) or W (C5 ), then JD = E(CG (β))  CG (β) for all β ∈ D −  ∪ b ). (b 0 1 As CG (JD ) < CG (b4 ), we easily see that CG (b0 b1 ) contains b0 b1  ×  −1 b1 , σ01 . Similarly, b0 b1 , σ01 ×JD as a subgroup of index at most 2. Set J+ = b−1 0  −1 CG (b−1 0 b1 ) contains b0 b1 × J− × JD as a subgroup of index at most 2, where J− = b0 b1 , σ01 t01 . Note that J+ and J− are root SL2 (2)-subgroups of J4 . As CJD (L4 ) = 1, we conclude that CC (b0 b1 )contains b0 b1  × J+ as a subgroup of −1 index at most 2, and CC (b−1 b ) contains b b 1 1 × J− as a subgroup of index at 0 0 most 2 ∼ ∼ If [W, W ] ∼ = CW (b0 b−1 = Ω5 (3), then CW (b0 b1 ) ∼ 1 ) = Σ6 , so Jv /Z(Jv ) = U6 (2) −1 for v ∈ {b0 b1 , b0 b1 }. Also v, Jv  has index at most 2 in CG (v). As L4 < JD < Jv , b0 b1  × J+ as a subgroup of index we may compute in Jv that CC (b0 b1 ) contains −1 × J b ) contains b b at most 2, and CC (b−1 1 1 − as a subgroup of index at most 2. 0 0 Hence, these statements hold in all cases. Let EC = E(C). It follows that D ∈ Syl3 (C) and F ∗ (C) = EC . Moreover, either EC = L04 × L14 or EC is simple with CEC (bi ) ∼ = Z3 × A5 or Z3 × Σ5 for i = 0, 1. Furthermore, D  CC (v) with index 2 or 4 for v ∈ {b0 b1 , b−1 0 b1 }. But, as noted above, |C|2 divides |Sp6 (2)|2 , so |C|2 divides 32 · 5 · 7. Therefore by [VK , 15.1], EC ∼ = A8 . As EC ≤ CJ4 (t4 ) it follows from [VK , 10.67] that t4 induces an outer automorphism of J4 , and we have EC ≤ H := CJ4 (t4 ) ∼ = Ω7 (2). By [VK , 5.5], there is a unique conjugacy class of subgroups of H isomorphic to EC , and so EC ∼ = Ω+ 6 (2) centralizes a nondegenerate plane of + type in a suitable 8-dimensional orthogonal space for J4 . We have J± ≤ C ≤ J4 and claim that J± ≤ EC . Namely, the involutions in J± are root involutions in J4 ∼ = D4 (2) and hence neither centralize a subgroup of J4 of order 5 nor invert a subgroup of J4 of order 7. This implies our claim. Now we extend (J− , J+ ) to a Curtis-Tits system (J− , L∗ , J+ ) for EC . From the paragraph before last, we see that EC is a Levi complement in a maximal parabolic subgroup of J4 . It follows that we may extend (J− , L∗ , J+ ) to a Curtis-Tits system for J4 := E(C(b4 )). Indeed, there is a unique pair L13 , L23 of fundamental SL2 (2) subgroups of J4 completing (J− , L∗ , J+ ) to an extended Curtis-Tits system for J4 with central node L∗ . Then, note that L13 × L23 are contained in CG (J− × J+ ) = JD or JD r ∼ = O6+ (2). Indeed, L1 × L2 ≤ CG (J− × J+ × b4 ) ∼ = Ω+ (3) or O + (3). 3

3

4

4

In either case, the pair {O3 (L13 ), O3 (L23 )} is uniquely determined by the condition that these are the two subgroups of O3 (L13 × L23 ) conjugate to b0 b1 . Hence, so is

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9. THE CASE G∗ = 2 D5 (2),

2

E6 (2), OR U7 (2)

297

the pair {L13 , L23 }. Since b−1 2 b3 ∈ L3 , it follows that L3 is a member of this pair, where (L3 , L4 ) is the original Curtis-Tits system for JD . Hence, (J− , L∗ , J+ , L3 ) is a Curtis-Tits system for J4 , and so (J+ , L∗ , L3 , L4 ) is a Curtis-Tits system for 2 D5 (2). Let G2 := J+ , J ∗ , L∗3 , L4 , so G2 ∼ = 2 D5 (2) by the Curtis-Tits theorem [IA , 2.9.3]. Now b4 ∈ G2 with E(CG2 (b4 )) ∼ = D4 (2). So J4 ≤ G2 with b0 , b1 ∈ J4 . Then E(CG (bi )) = E(CG2 (bi )) for i = 0, 1. Hence G1 = J, J   = G2 , the last equality by [IA , 7.3.2]. Finally, if W ∼ = W (D5 ) or W (C5 ), then W ∩ JD  CW (b0 b1 ),  = ±1, and it  follows that JD = E(CG (b0 b1 )), whence G0 = G1 , completing the proof. Lemma 9.12. Suppose that J ∼ = D4 (2) and [W, W ] ∼ = 2 E6 (2). = Ω5 (3). Then G0 ∼ Proof. By Lemma 9.11, if J  = E(CG (d )) with d ∈ D ∩ J # , we have G1 := J, J   ∼ = 2 D5 (2). We choose a Curtis-Tits system (L1 , L2 , L3 , L4 ) for G1 ∼ ∼ with Li = SL2 (2) for i = 1, 2, 3, and L4 ∼ = Ω− 4 (2) = SL2 (4). We may assume that L1 was chosen so that c1  := O3 (L1 ) ≤ D, and that L3 , L4  = JD . If V1 is a natural module for G1 , then dim([V1 , c1 ]) = 4, whence we may assume  W that c1  = b−1 d . As W ∩ J ∼ = W ∩ J ∼ = W (D4 ) and c1  ∈ b , we have ∼ CW (c1 ) = Z2 × Σ6 . It then follows that if J1 = E(CG (c1 )), then J1 /Z(J1 ) ∼ = U6 (2). ⊥ Now, C[W,W ] (c1 ) ∼ = P O4− (3) and B ∗ = c1  ⊕ c1  as C[W,W ] (c1 )-module. If J1 ∼ = SU6 (2), then W ∩ J1 ∼ = Σ6 and B ∗ ≤ J1 with B ∗ an indecomposable [W ∩ J1 , W ∩ J1 ]-module, a contradiction. Hence, J1 ∼ = U6 (2). Write L1 = c1 , t1  with t21 = 1. Then t1 centralizes c1  × JD ≤ CG1 (c1 ) with c1 ∈ D and JD ∼ = Ω− 6 (2). ∗  Then JD ≤ J1 and, as t1 normalizes B ∩ J1 , also c1 ∈ J1 . It follows that [L1 , J1 ] = 1. Choose a Curtis-Tits system (L5 , L4 , L3 ) for J1 extending the system (L4 , L3 ) for JD , with L5 ∼ = L4 . Let ∼ L = L4 , L5  = L3 (4). We consider C := CG (L). We have that L1 ≤ C. Now, C(c1 , J1 ) has a cyclic Sylow 3-subgroup C1 which acts faithfully on J centralizing JD . Hence C1 = c1  by [VK , 3.74]. By [VK , 15.2], CAut(J1 ) (L) = 1, whence CC (c1 ) = O3 (CC (c1 )) c1 . In particular, |C|3 = 3. Let b4  ∈ Syl3 (L4 ). As [V1 , L4 ] is a 4-dimensional orthogonal space of − type, it follows that dim([V1 , b4 ]) = 2, and so b4  ∈ bG1 , whence E(CG (b4 )) = E(CG1 (b4 )) =: J4 ∼ = D4 (2) and L1 is a root SL2 (2) subgroup of J4 . Hence, O2 (CJ4 (t1 )) =: Q4 ∼ and dim(CV1 (t1 )) = 8. Moreover, t1 ∈ z G , whence = 21+8 + ∗ F (CG (t1 )) =: Q1 is a 2-group of symplectic type by hypothesis. Let b4 ∈ F ∈ Syl3 (L). If F ≤ J4 CG (J4 ), then F ≤ B ∗∗ ∈ (B ∗ )G . But all elements of F # are L-conjugate, whereas no E32 subgroup of B ∗ has the property that all non-identity elements x are conjugate and have E(CG (x)) ∼ = D4 (2), a contradiction. Hence, F/ b4  acts by graph automorphisms on J4 and so CJ4 (F ) ∼ = P GU3 (2) ∼ = E32 : SL2 (3) or CJ4 (F ) ∼ = G2 (2). In the former case, t1 lies in a Q8 with a free summand on V1 , against dim(CV1 (t1 )) = 8. Hence, CJ4 (F ) ∼ = G2 (2) and # . Since all elements of F are C (t )-conjugate to b4 , it O2 (CG (t1 , F )) ∼ = 21+4 G 1 +

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6. THEOREM C5 : STAGE 4

298

follows that |Q1 | = 221 . As the minimum dimension of a faithful F2 [J1 ]-module is . Let W1 := Q1 / t1 . By [VK , 5.3], as 20 by [VK , 5.3], it follows that Q1 ∼ = 21+20 + J1 -module, W1 is an F2 -form of ∧3 U , where U is the standard module for SU6 (2), and W1 is absolutely irreducible. The restriction of U to L3 (4) is U1 ⊕U2 with Ui a 3dimensional module for L. It follows that ∧3 U contains ∧3 (U1 )⊕∧3 (U2 ) ∼ = F4 ⊕F4 . Hence, |CQ1 (L)| ≥ 23 . We continue to consider C = CG (L). We have that L1 ≤ C ≤ CJ4 (F ) ∼ = G2 (2). Also CC (t1 ) is a 2-group of size at least 23 . If O3 (C) = 1, then O3 (C) = c1  and C = L1 , a contradiction. Hence, O3 (C) = 1. Suppose that O2 (C) = 1. As involution centralizers in G2 (2) are {2, 3}-groups, O2 (C) = F ∗ (C). As t1 ∈ L1 ≤ C, t1 ∈ O2 (C). It follows that t1 is not 2-central in C. Recalling that C ≤ CJ4 (F ) ∼ = G2 (2), we see that CC (t1 ) ∼ = E23 and |O2 (C)| ≥ 8. As CO2 (C) (c1 ) = 1, it follows that O2 (C) ∼ = Z42 . But then CO2 (C) (t1 ) ∼ = Z4 , a contradiction. Hence, O2 (C) = 1. It follows that C1 := F ∗ (C) is a simple group of order 2a · 3 · 7. Hence, C1 ∼ = L3 (2). Note that C1 ≤ (CG (L4 ))(∞) ≤ (CG (b4 ))(∞) = (CG (b4 ))(∞) ∼ = Ω+ (2). 1

8

Hence,

C1 ≤ (CG (L4 ))(∞) = (CG1 (L4 ))(∞) ∼ = Ω+ 6 (2). + C1 ≤ E(CG (L5 )) =: E5 ∼ As L5 ∈ = Ω6 (2). Now, there exists a Curtis-Tits system (L1 , L∗2 ) for C1 which can be extended to a Curtis-Tits system (L1 , L∗2 , L∗3 ) for E5 , since the involutions in C1 are C1 -conjugate to involutions in L1 , and hence are root involutions in E5 . But L3 ≤ CE5 (L1 ), whence equality holds and so L∗3 = L3 . As L∗2 ≤ C1 ≤ CG (L4 , L5 ), we see that (L∗2 , L3 , L4 , L5 ) is a weak Curtis-Tits system for G2 := L∗2 , L3 , L4 , L5 . Thus G2 ∼ = 2 E6 (2) by the Curtis-Tits theorem [IA , 2.9.3]. Now, E(CG2 (b4 )) ∼ = D4 (2), whence J4 ≤ G2 . Then c1 ∈ G2 and J1 ≤ G2 . As each 3-element v of D is W -conjugate to either c1 or b4 , it follows that E(CG (v)) ≤  G2 for all v ∈ D# . As G2 = E(CG2 (v)) | v ∈ D# by [IA , 7.3.3] it follows that  G2 = G0 , completing the proof. LL 4,

We have now reached the following situation: Jv /Z(Jv ) ∼ = Un(v) (2), 4 ≤ n(v) ≤ 6, for all v ∈ D# . We may assume that n(b) is maximal among all n(v) and set n(b) = n ∈ {5, 6}. First we determine the structure of W in the case when n = 5 using arguments from [III15 , Lemma 10.11]. Lemma 9.13. Suppose that n = 5, i.e., J ∼ = U5 (2). Let W1 be the subgroup of W generated by all its reflections. Then W1 ∼ = Σ6 , with B ∗ the quotient of a standard permutation module by the trivial module. Proof. By Lemma 9.7, O2 (W ) = 1. Suppose O2 (W ) = Z(W ). Since B ∗ is indecomposable as W -module and dim(B ∗ ) = 5 is an odd prime, O2 (W ) is abelian but noncyclic, whence by [III17 , 1.4], W ∼ = W (D5 ) or W (C5 ) acting as a monomial group. But then for some d ∈ D# , Jd ∼ = D4 (2), a contradiction. Hence, E(W ) = 1. Let E1 be a component of W . Since GL2 (3) is solvable and |B ∗ | = 35 , F ∗ (W ) = E1 Z where |Z| ≤ 2 and Z ≤ Z(W ). If E1 ∈ Chev(r) − Alt for some prime r > 3, then E1 ∼ = L2 (11). But then Σ5 ≤ P GL2 (11), which is

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9. THE CASE G∗ = 2 D5 (2),

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E6 (2), OR U7 (2)

299

∼ A6 by [VK , 5.2]. Finally, if not the case. If E1 ∈ Chev(2) ∪ Chev(3), then L = E1 ∈ Alt ∪ Spor − Chev(2) − Chev(3), then as E1 embeds in GL5 (3), we see that E1 ∼ = M11 . Then E1 acts irreducibly on B ∗ and W = E1 or E1 × Z with Z ∼ = Z2 . But W ∩J ∼ = Σ5 and E1 does not contain a subgroup isomorphic to Σ5 by [IA , 5.3a]. Hence, E1 ∼ = Σ5 , we have that W1 = (W ∩J)E1 ∼ = A6 with E1  W . As W ∩J ∼ = Σ6 . It now follows that the lemma holds.  The next two lemmas concern the case that n = 6, i.e., J/Z(J) ∼ = U6 (2). We repeat arguments from [III15 , Lemmas 10.8–10.10] to establish the first assertion of the following lemma. We set d = m + 1 = m3 (B ∗ ). As B ∗ / b embeds in P GU6 (2) as the image of a diagonalizable subgroup of GU6 (2), we may choose V to be an 6-dimensional vector space over F = F4 affording a natural representation for GU6 (2) and we may choose an orthonormal basis {e1 , . . . , e6 } for V such that B ∗ ∩ J acts diagonally with respect to this basis. (Note: if J∼ = U6 (2), the action of elements of B ∗ ∩J is only determined up to scalars.) We let W0 = W ∩ J. Then W0 permutes B := {Fei | 1 ≤ i ≤ 6}. If d = 6, then JB ∗ / b ∼ = P GU6 (2) and we may choose y ∈ B ∗ with JD < L := E(CJ (y)) ∼ = U5 (2). If d = 5, set L = JD acting on e1 , e2 , e3 , e4 . We may assume that L stabilizes e6 and the span of {e1 , . . . , e5 }. Let H be the SL2 (2)-subgroup of L stabilizing the span of {e1 , e2 } as well as each ei , 3 ≤ i ≤ 6. Let t ∈ NH (B ∗ ) be the involution interchanging e1 and e2 , and let τ denote its image in W0 . Then τ acts as a reflection on B ∗ .  Set β = [B ∗ , τ ]. Then dim([V, β]) = 2 and β, t = H ≤ L. Let Lβ = O 2 (CL (β)).  We have Lβ < Jβ := E(CJ (β)) ∼ = U4 (2). Moreover, Jβ = O 2 (CJ (H)). We fix this choice of t, τ, β, Lβ and Jβ , and use this information to study CW (τ ). Lemma 9.14. Suppose that n = 6. Then CW (τ ) has a subnormal subgroup Στ ∼ = Σk for some k ∈ {5, 6}, whose derived group is a component Lτ of CW (τ ) isomorphic to Ak . Moreover, the unique nontrivial Lτ -composition factor in B ∗ is isomorphic to the core of the standard F3 -module for Lτ . Proof. Let Nβ = NG (β). As [B ∗ , τ ] = β, CW (τ ) = CW ∩Nβ (τ ). We claim that Nβ has a component Kβ such that Kβ /Z(Kβ ) ∼ = Uk (2) for some k ∈ {5, 6}, and [Kβ , t] = 1, from which the first statement of the lemma will follow immediately with Στ = AutKβ (B ∗ ). Moreover, we may identify Lτ with a section of Kβ . Then [B ∗ , Lτ ] ≤ B ∗ ∩ Kβ and the second statement follows as well. Suppose that d = 6 and L = E(CG (v)) for all v ∈ D1 − b, where D1 = b, y. As JD < J  , we have J  /Z(J  ) ∼ = U5 (2) or U6 (2) with E(CJ  (b)) = JD . Then D1 acts on J  and JD = E(CJ  (v)) for all v ∈ D1# , contrary to [IA , 7.3.3]. If d = 5, set D1 = D. Then, in all cases, there exists v ∈ D1 − b with Jv a vertical pumpup of L. We fix this choice of v and Jv . Now we consider the subnormal closure Kβ of Jβ in CG (β). Suppose that Kβ = Jβ . As β ∈ H < L < Jv with L = E(CJv (b)), we have H canonically embedded   in Jv . But then Jβ ∩ O 2 (CJv (β)) = Lβ is subnormal in O 2 (CJv (β)). If JD < L,  then L ∼ = SU3 (2) and O 2 (CJv (β)) ∼ = U6 (2). So Lβ ∼ = U4 (2), = U5 (2) and Jv /Z(Jv ) ∼ a contradiction. Hence, JD = L ∼ = U4 (2) and Lβ ∼ = U5 (2) = U2 (2), while Jv /Z(Jv ) ∼  or U6 (2) and O 2 (CJv (β)) ∼ = SU3 (2) or U4 (2), again a contradiction. It follows that Jβ < Kβ . As Jβ ∼ = U4 (2) and β is conjugate into D, we must  have Kβ /Z(Kβ ) ∼ = U5 (2) or U6 (2), as desired.

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300

6. THEOREM C5 : STAGE 4

We now complete our analysis of W , and we identify G0 in the case when m = 5, i.e., d = 6. Lemma 9.15. We have that Σm+2  W and J/Z(J) ∼ = Um+1 (2). Moreover, if m = 5, then G0 ∼ = U7 (2). Proof. By Lemma 9.13, we may assume that J/Z(J) ∼ = U6 (2). First observe that as Lτ ∼ = Ak is a component of CW (τ ) and O2 (W ) = 1 and |W |5 = 5 and W acts indecomposably on B ∗ , we have that Lτ is a component of CE(W ) (τ ), where E(W ) is quasisimple, and CW (E(W )) = Z(W ) acts as scalars on B ∗ . In particular, |Z(E(W ))| ≤ 2. We invoke [VK , 5.1]. Observe that for any u ∈ D# , Ju ∼ = Uj (2) for some j ∈ {4, 5, 6}, so CW (u) has no normal subgroup isomorphic to W (D4 ) or W (D5 ). Then [VK , 5.1] yields that W ∼ = Σ7 or Σ7 × Z2 , proving the first statement of the lemma. Now, if m = 5, we set C o = SU7 (2) acting on V := V1 ⊥ · · · ⊥ V7 as in [V2 , Section 10] and let f : C7o → J be a surjective homomorphism and λ : NG (B ∗ ) → T be a surjective homomorphism onto the subgroup T of C o of all permutation matrices permuting {V1 , . . . , V7 } naturally. Such homomorphisms exist satisfying the conditions [V2 , (10D)]. Let N = NG (B ∗ ). We conclude, by [V2 , 10.3], that G1 := J N  ∼ = U7 (2). As J  is N -conjugate to J, we see that G0 ≤ G1 ,  and then by [IA , 7.3.2], G0 = G1 ∼ = U7 (2), completing the proof. ∼ U5 (2) and Σ6  W . This final case It remains to treat the case m = 4, i.e., J = will occupy the remaining two lemmas. Choose T ∈ Syl3 (N ). Then T = B ∗ A, with A ∼ = E32 , A ∩ B ∗ = 1, and A ∗ acting on B the way a Sylow 3-subgroup of A6 acts on the quotient of a natural permutation module V for W1 ∼ = Σ6 over F3 , modulo its 1-dimensional submodule V0 . One checks then that B ∗ = J(T ), so T ∈ Syl3 (G). We assume, as we may, that T was chosen so that there exists a basis {v1 , v2 , . . . , v6 } for V permuted naturally by W1 , and such that the image of T in W is (v1 , v2 , v3 ), (v4 , v5 , v6 ), and such that D is the image of v1 , v2  modulo V0 . Lemma 9.16. Suppose that J ∼ = U5 (2). Let B ∗ ≤ T ∈ Syl3 (G) and let β = Z(T ). Then H1 ∗ H2  CG (β) with H1 ∼ = H2 ∼ = SU3 (2) and Z(H1 ) = Z(H2 ) = β. ∼ Moreover, if T1 ≤ T and T1 = Z3  Z3 , then Z(T1 ) = β. Proof. By the previous paragraph, B ∗ = J(T ) and we may assume that T has the properties chosen above. Let T = T / β. Then T ∼ = 31+2 × 31+2 . Hence if Z3  Z3 ∼ = T1 ≤ T and Z1 = Z(T1 ), then Z1 = [T1 , T1 , T1 ] = 1 and Z1 = β, as claimed. Note that β is also the image of (v4 v5 v6 )−1 in B0 , and so β ∈ JD = E(CG (D)) ∼ = U4 (2), where D is the image of v1 , v2 , with H2  CG (D, β), β = Z(H2 ) and H2 ∼ = SU3 (2). Moreover, H2  CG (d, β) for all d ∈ D# . Also, the image b45 of v4−1 v5 is in R2 = O3 (H2 ) and there is H1  CG (b45 , β) with H1 ∼ = H2 . Set R1 = O3 (H1 ). Then β ∈ Z(R1 ). Looking in NG (B ∗ ), we see that we may choose notation so that R1 = b12 , ρ and R2 = b45 , σ, where the images of ρ and σ are commuting elements of order 3 in

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9. THE CASE G∗ = 2 D5 (2),

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W . We may assume that R2 = R1a , where a ∈ N maps on (v1 , v4 )(v2 , v5 )(v3 , v6 ) ∈ W . Then by Lemma 9.3 and solvable L3 -balance, either R1 ∗ R2 = O3 (CG (β)) or Ri ≤ Li for i = 1, 2 with Li a 3-component of CG (β), and Li ∈ C3 by our general hypothesis. In the former case, C β := CG (β)/R1 R2 embeds in Sp4 (3) with |C β |3 = 32 . If E(C β ) = 1, then by [III17 , 9.9], O 2 (C β ) ∼ = A6 . But CJD (β) ∼ = GU3 (2), a contradiction. Hence, E(C β ) = 1 and |O3 (C β )| ≤ 3, whence O2 (C β ) = 1. By the 2-local structure of Sp4 (3), we conclude that O 2 (C β ) ∼ = Sp2 (3) × Sp2 (3). Now H1 centralizes b45 and Z(H 1 ) inverts R1 /Z(R1 ), and a symmetrical statement holds for H2 . We conclude that H 1 H 2 = H 1 × H 2 ∼ = Q8 × Q8 , and so [H1 , H2 ] = 1, and the lemma holds. In the latter case, if L1 = L2 , we reach a contradiction, since b12 ∈ L1 , but CG (b12 ) is solvable. So, L1 = L2 =: L. Now, H2 is a solvable 3-component of CL (d) for all d ∈ D# , and Z(H2 ) = Z(L). Also, H1 ≤ L and R1 ∗ R2 ≤ T ∩ L. If m3 (L) ≤ 4 and |L|3 ≤ 36 , then we are done by [III17 , 10.23]. So we may assume m3 (L) = 5 and |L|3 = 37 . Also H2  ΓD,1 (L), whence L ∈ Chev(r) for any r = 3 by [IA , 7.3.3]. By order considerations and [IA , 6.1.4], and as β ∈ Z(L), we have L/Z(L) ∼ = Σ3  Z2 . = M c, U4 (3), or G2 (3). But also NL (B ∗ )/B ∗ embeds in CW (β) ∼ If L/Z(L) ∼ = M c or U4 (3), then NL (B ∗ )/B ∗ contains A6 , a contradiction. Thus, L := L/ β ∼ = E34 . Now, = G2 (3). Then R1 R2 ∼ O 2 (CG (D β)) = D × H2 (B ∗ ∩ JD ) ∼ = E32 × GU3 (2), whence CL (D) ≥ D × H 2 (B ∗ ∩ JD ) ∼ = E32 × P GU3 (2). Let z = Z(H 2 ). Then CL (z) contains a subgroup isomorphic to E32 × SL2 (3), whereas |CL (z)|3 = 32 , a final contradiction, completing the proof.  Lemma 9.17. If J ∼ = U5 (2), then G0 ∼ = U6 (2). Proof. We follow the argument in [III13 , Section 3] for the case (3D2), but with m there taking the value 6, and V0 there equal to 0. Thus, we set C o = SU6 (2) acting on V := V1 ⊥ · · · ⊥ V6 , dim Vi = 1, and let f : C6o → J be an isomorphism and λ : N → T ∼ = Σ6 be a surjective homomorphism. Such homomorphisms exist satisfying the desired compatibility conditions [III13 , 3.1abcd]. Setting H = G1 , we obtain well-defined subgroups HD of H for all proper subsets D ⊆ {1, 2, 3, 4, 5, 6} with |D| ≥ 2. The one result which does not follow as in [III13 ] is Lemma 3.6 for the case D ∪ F ∈ P and |D| = |F | = 3. It is therefore enough to show: (9C)

For D = {1, 2, 3} and F = {4, 5, 6}, [HD , HF ] = 1.

∼ Z3  Z3 with But HD∪{4} contains GU3 (2), which in turn contains some T1 = Z(T1 ) = Z(HD ). Hence, by the preceding lemma, we may assume that Z(HD ) = β. Also, CG (β) ≥ HD ∗ H2 with H2 ∼ = HD , and indeed by inspection in W , we find an element c ∈ W interchanging HD and H2 in NG (β). Hence, H2 = HE for some subset E of {1, 2, 3, 4, 5, 6} with |E| = 3. If D ∩ E = ∅, then D ∪ E ≤ {n} for some n ∈ {4, 5, 6}. But then HD ∗ HE ≤ CJ1 (β) for some J1 ∈ J G , contrary to the structure of U5 (2). Hence, E = {4, 5, 6}, and the desired conclusion holds.

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6. THEOREM C5 : STAGE 4

302

Table 9.1. The Groups G0 , JD , and Jd

G0 m2,3 (G) J D5 (2) 4 D4 (2) U6 (2) 4 U5 (2) U7 (2) 5 SU6 (2) 2 E6 (2) 4 D4 (2)

2

JD Je > JD D3 (2) D4 (2) U4 (2) U5 (2) U5 (2) SU6 (2) U4 (2) D4 (2), U6 (2)

2

It remains to verify Theorem 3.15 of [III13 ] extended to the case C o = SU6 (2). The same argument shows that, given a weak Phan system for C o , say (U1 , U2 , U3 , U4 , U5 ), and taking Li = h(Ui ) ≤ G1 := J, N  for all i, we must verify that L1 , L2  commutes with L4 , L5 , as required to verify that G1 is a homomorphic image of SU6 (2), and hence G1 ∼ = U6 (2). But L1 = H{1,2,3} and L2 = H{4,5,6} , and so we are done by (9C). In view of the structure of CW (d) for d ∈ D# , we see that Jd ≤ G1 for all d ∈ D# , and then, with [IA , 7.3.3], we conclude that G0 = G1 , completing the proof.  Lemmas 9.11, 9.12, 9.15 and 9.17 complete the proof of Theorem 9.2. We now begin the proof of Theorem 9.1 by supposing that it fails. By Theorem 9.2, (9D)

M := NG (G0 ) < G.

Since 3 = p ∈ σ0 (G) by the hypothesis of Theorem C5 , we deduce: Lemma 9.18. M is not strongly 3-embedded in G. Moreover, CG (G0 ) is a 3 -group. Proof. Note that by (9B2), m3 (CG (J)) = 1, and then as [J, b] = 1 = [G0 , b], with J ≤ G0 , CG (G0 ) must be a 3 -group. Suppose that M is strongly 3-embedded in G. Then M is a 3-component preuniqueness subgroup of G with respect to G0 [I2 , Def. 6.1] and therefore also a strong 3-uniqueness subgroup of G [I2 , Def. 8.7]. But since 3 ∈ σ0 (G), G has no strong 3-uniqueness subgroup by definition. This contradiction proves the lemma.  Our goal in all but one case is to prove that M is a strongly 3-embedded subgroup of G, contradicting Lemma 9.18. However, in the case G0 ∼ = U6 (2), we SF] to reach a contradiction. use Holt’s theorem [II2 , Theorem We have G0 = Jd | d ∈ D#  M , where D = b, b  ≤ B ∗ . Let P ∈ Syl3 (M ) with B ∗ < P. We have four cases to consider, tabulated in Table 9.1. In all cases, the 2-central involution z of J = E(CG0 (b)) is also 2-central in G0 . We have D ≤ CG (z) with Uz := F ∗ (CG (z)) a 2-group of symplectic type. Lemma 9.19. The following conclusions hold: (a) NG (D) ≤ M ; and (b) For all d ∈ D# with Jd = JD , D ∈ Syl3 (C(d, Jd )).

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9. THE CASE G∗ = 2 D5 (2),

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E6 (2), OR U7 (2)

303

Proof. By Theorem C5 : Stage 3, Jd = E(CG (d)) for all d ∈ D# such that Jd > JD . As NG (D) permutes {d ∈ D# | Jd > JD }, NG (D) normalizes   E(CG (d)) | d ∈ D# and Jd > JD = Jd | d ∈ D# = G0 . So, NG (D) ≤ M , as claimed in (a). Now, let d ∈ D# with Jd = JD . Let Q ∈ Syl3 (C(d, Jd )) with D ≤ Q. Then NQ (D) ≤ M . As CG (G0 ) is a 3 -group, NQ (D) embeds in a Sylow 3-subgroup of CAut(G0 ) (JD ) ∼ = D by [VK , 12.49]. Hence, D = NQ (D) = Q, as claimed in (b).  Lemma 9.20. Suppose that J ∼ = D4 (2). Then ΓD,1 (G) ≤ M . ∼ 2 D5 (2), clearly JD contains Proof. As G0 ≥ G1 ≥ J with G1 := J, J   = g g a G1 -conjugate D of D. Also, NG (D ) = NG (D)g ≤ M . Let d ∈ D# and Nd = NG (d). As Dg < JD ≤ Jd  Nd , we have Jd CNd (Jd ) ≤ M . If Jd = JD , then by Lemma 9.19ba and a Frattini argument, Nd ≤ CNd (Jd )NG (D) ≤ M. So assume that Jd > JD . If Jd ∼ = U6 (2), then Jd controls the Nd -fusion of D, whence Nd ≤ M . If Jd ∼ = D4 (2) and G0 ∼ = 2E6 (2), then by [IA , 4.7.3A], AutG0 (Jd ) contains a triality so again Jd controls the Nd -fusion of D and Nd ≤ M . In both cases we have used [VK , 3.76]. Suppose finally that Jd ∼ = D4 (2) and G0 ∼ = 2 D5 (2). Then Nd ∩ G0 contains an involution u inverting d, normalizing D, and such that Jd u ∼ = O8 (2). We claim that AutNd (B ∗ ) = AutJd u (B ∗ ). Indeed, W1  W and W1 ∼ = W (D5 ) or W (C5 ), by Lemmas 9.10 and 9.12. This implies that W preserves the unique frame in B ∗ preserved by W1 , and so W = W1 . If AutNd (B ∗ ) > AutJd u (B ∗ ), then as Out(Jd ) ∼ = Σ3 , W would contain an element w of order 3 normalizing d and acting as a triality on CW1 (d) ∼ = W (D4 ). But W ∼ = W (D5 ) or W (C5 ), so no such element w can exist. This contradiction proves our claim. As a result, again Jd controls the Nd -fusion of D, whence Nd ≤ M . Hence, we have Nd ≤ M for all d ∈ D# in all cases, completing the proof.  Lemma 9.21. Suppose that J ∼ = U5 (2) or SU6 (2). Then ΓD,1 (G) ≤ M . Proof. We choose d ∈ D with d = diag(ω, ω −1 , 1, . . . , 1) ∈ G0 , with respect to an orthonormal basis of the natural F4 J-module. Here ω 3 = 1 = ω. (Of course, if G0 ∼ = U6 (2), then d is really only defined up to a scalar multiple.) Then Jd = JD . We fix a permutation matrix t ∈ G0 such that w := dt ∈ JD and we set Jw = Jdt . Then Jv , Jw  = JD , Jw  = G0 by [IA , 7.3.2] for all v ∈ D# . Now let v ∈ D# . As w ∈ JD ≤ Jv , we have CG (Jv ) ≤ CG (w) ≤ NG (Jw ), whence CG (Jv ) normalizes Jv , Jw  = G0 , and so CG (Jv ) ≤ M . In particular, C(d, Jd ) ≤ M . Since Jd  NG (d), Lemma 9.19ba and a Frattini argument give NG (d) ≤ C(d, Jd )NG (D) ≤ M . As w ∈ dM , we have NG (w) ≤ M. N (v) = Again, let v ∈ D# . As w ∈ JD ≤ Jv and Jv  NG (v), we have w G NG0 (v) g h by [VK , 3.75]. Hence, if g ∈ NG (v), then w = w for some w h ∈ G0 . Hence, gh−1 ∈ NG (w) ≤ M and so g ∈ M , completing the proof of the lemma. 

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304

6. THEOREM C5 : STAGE 4

Lemma 9.22. The following conclusions hold: (a) D ∈ U(G; M ; 3); (b) NG (B ∗ ) ≤ M and NG (P ) ≤ M ; (c) P ∈ Syl3 (G); (d) NG (B ∗ ) controls G-fusion in B ∗ , and B ∗ = J(P ); and (e) Uz = O2 (CG0 (z)) and CG (G0 ) = 1; and (f) For some x ∈ I3 (M ), CG (x) ≤ M . Proof. Suppose that Dg ≤ M . By [IA , 7.3.4], G0 = ΓDg ,1 (G0 ) ≤ M g by Lemmas 9.20 and 9.21, whence G0 = Gg0 and g ∈ M , proving (a), Now, (b) and (c) follow directly from (a) and (c) of [III8 , 6.1]. By [IA , 4.10.3c], B ∗ is the unique elementary abelian subgroup of P of maximum rank, whence B ∗ = J(P ) and (d) holds by [IG , 16.9]. As Uz = ΓD,1 (Uz ), we have Uz ≤ O2 (CG0 (z)). But then as CG0 (z) acts irreducibly on O2 (CG0 (z))/ z, equality holds. In particular CG (G0 ) ≤ CG (Uz ) = CG (F ∗ (CG (z))) = z , so (e) follows as [G0 , z] = 1. Finally, (f) follows from (b) and [IG , 17.2].



Lemma 9.23. Suppose v ∈ I3 (M ) with either v ∈ G0 or m3 (CM (v)) = m3 (B ∗ ). Then v m ∈ B ∗ for some m ∈ M . Suppose further that NG (v) ≤ M . Then v ∈ U(G; M ; 3). In particular, d ∈ U(G; M ; 3) for all d ∈ D# . Proof. If v ∈ G0 , then m3 (CM (v)) = m3 (B ∗ ) by [VK , 9.34]. As B ∗ = J(P ) by Lemma 9.22d, it follows that v m ∈ B ∗ for some m ∈ M , as claimed. Now suppose further that NG (v) ≤ M . Suppose that vg = w ≤ M . If m3 (CM (w)) < m3 (M ), then by [VK , 9.34], G0 ∼ = U6 (2) and E(CM (w)) ∼ = L2 (8). However, CG (v) = CM (v) and |CM (v)| is not divisible by 7, by [VK , 9.34]. Hence, m3 (CM (w)) = m3 (B ∗ ), and so wm ∈ B ∗ for some m ∈ M by the previous paragraph. Then wm m1 = v for some m1 ∈ NG (B ∗ ) ≤ M by Lemma 9.22bd. So vgmm1 = v. As NG (v) ≤ M , we have gmm1 ∈ M , whence g ∈ M and v ∈ U(G; M ; 3), as claimed. As NG (d) ≤ M for all d ∈ D# by Lemmas 9.20 and 9.21, the final claim follows.  = 2 E6 (2). Lemma 9.24. G0 ∼ Proof. Suppose G0 ∼ = 2E6 (2). By Lemma 9.18, M is not a strongly 3embedded subgroup of G. Let β = Z(P ). If P ≤ G0 , then by [IA , 4.7.3A], there is x ∈ I3 (P ) with E(CG0 (x)) =: Jx ∼ = 2 D5 (2). Hence m2,3 (G) ≥ 1+m2,3 (2 D5 (2)) = 5, a contradiction. Hence P ≤ G0 and b1 ∈ ∪m∈M Dm for all b1 ∈ I3 (M ) − β M . Thus by Lemma 9.22f, NG (β) ≤ M . Let C = CM (β). Then by [IA , 4.7.3A], C has a normal subgroup L0 = L1 ∗ L2 ∗ L3 with β ∈ Li ∼ = SU3 (2) for each i and with |C : L0 |2 ≤ | Out(G0 )|2 = 2. Moreover there is u ∈ C ∩ P of order 3 cycling the Li ’s. Set Pi = O3 (Li ) and let Qi ∈ Syl2 (Li ) for all i. We claim first that b1 ∈ β M for all b1 ∈ CO3 (C) (L1 ) − β. For if not, then C1 := CM (b1 ) has a normal subgroup K0 ∼ = L0 with |C1 : K0 |2 ≤ 2. It follows that |Q1 ∩ K0 | ≥ 4. Hence, P1 = [P1 , Q1 ∩ K0 ] ≤ O3 (C1 ). But then β ∈ [P1 , P1 ] ≤ [O3 (C1 ), O3 (C1 )] = b1 , a contradiction. Now set C ∗ = CG (β). Then O3 (C ∗ ) = 1 by Lemma 9.3, and O3 (C ∗ ) ≤ O3 (C). As C acts irreducibly on O3 (C)/ β, either O3 (C ∗ ) = β or O3 (C ∗ ) = O3 (C). In the latter case, b1 ∈ O3 (C ∗ ) and b1  ∈ U(G; M ; 3), whence C ∗ = C, a

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9. THE CASE G∗ = 2 D5 (2),

2

E6 (2), OR U7 (2)

305

contradiction. Hence, O3 (C ∗ ) = β and F ∗ (C ∗ ) = E(C ∗ ). By solvable L3 -balance [IG , 13.8], P1 is contained in a 3-component K1 of C ∗ . If K1 is not u-invariant, then E(C ∗ ) ≥ K1 ∗ K2 ∗ K3 . But then b1 ∈ K2 ∗ K3 and K1 ≤ CG (b1 ) ≤ M , a contradiction. Hence, O3 (L0 ) ≤ K1 with β ∈ Z(K1 ). Moreover, P normalizes K1 with |P | = 39 , and [III17 , 9.18], applied to X = K1 P , gives a contradiction completing the proof.  = 2 D5 (2). Lemma 9.25. G0 ∼ Proof. Suppose G0 ∼ = 2 D5 (2). By Lemma 9.22f, there exists w ∈ I3 (M ) such that NG (w) ≤ M . By Lemma 9.23, w ∈ ∪y∈M Dy . By [IA , 4.8.2a], m3 (CM (w)) = 5 = m3 (B ∗ ), so by Lemma 9.23, we may assume that w ∈ B ∗ . By Lemma 9.3, O3 (CG (w)) = 1. Assume first that Jw := E(CM (w)) = 1. As w ∈ ∪y∈M Dy , Jw ∼ = Um (2) for some m ∈ {4, 5}. Replacing w by a NG0 (B ∗ )-conjugate, we may assume that  O 2 (CJw (D)) ∼ = SU3 (2) and E(CJw (b)) ∼ = U4 (2). Then, D acts on  Jw = E(CG (d)) ∩ Jw | d ∈ D# ≤ E(CG (w)), the last by L3 -balance. As m3 (G) = 5 and m3 (Jw ) ≥ 3, we see that Jw projects into a unique 3-component Lw of CG (w). Then ∼ U4 (2). E(CL (b)) = E(CJ (b)) = w

w

∼ P Sp4 (27), then |Lw |3 = 312 . If Lw ∼ So, Lw is a pumpup of U4 (2). If Lw = = 7 Co2 , then |CG (w)|3 ≥ 3 . However, |G|3 = |G0 |3 = 36 , a contradiction in both cases. Hence, Lw ∈ Chev(2). As ΓD,1 (Lw ) ≤ M , Jw  ΓD,1 (Lw ). Thus by [IA , 7.3.3], Lw = Jw . Moreover, ∪m∈M Dm ∩ Lw = ∅. Let d1  ≤ Lw with d1 ∈ ∪m∈M Dm . Then d1  ∈ U(G; M ; 3) and d1  ≤ Lw  NG (w) with Lw ≤ M . Hence, NG (w) ≤ M , contradiction. Therefore there is v ∈ I3 (B ∗ ) with v 3-central in P , CG (v) ≤ M , and 2 O (CM (v)) ∼ = GU3 (2) × Z3 × Z3 . Let Xv = O3 (CG (v)). Suppose E(CG (v)) = 1. Again O3 (CG (v)) = 1, so Xv = F ∗ (CG (v)). Then as Xv ≤ O3 (CM (v)) and CM (v) is irreducible on O3 (CM (v))/Z(P ), we must have Xv = O3 (CM (v)). Without loss, we may assume that O 2 (CM (v)) = Sv × D with Sv ∼ = GU3 (2). But then D ≤ Xv  CG (v), a contradiction as before. Hence, E(CG (v)) =: Lv = 1. If v ∈ Lv , then Sv ∩ Lv = 1. By solvable L3 -balance, O3 (Sv ) ≤ O3 (CG (v)). If Pv ∈ Syl3 (Sv ), then [Pv , Pv ] contains a subgroup d2  conjugate to a subgroup of D. But then d2  < O3 (CG (v))  CG (v), and as before, this implies that CG (v) ≤ M . Finally, suppose that v ∈ Z(Lv ). As Lv ∈ C3 , while (Sv × D)/ v embeds in Aut(Lv ) and |Lv |3 ≤ 36 , and ΓD,1 (Lv ) < Lv , we easily see by [V3 , Def. 1.1], inspection of [IA , Tables 5.3], and [IA , 7.3.4] that any component Jv of Lv lies in Chev(3). We may assume that v ∈ Jv and 33 ≤ |AutG (Jv )|3 l ≤ 35 , contradicting the possibilities  given in [IA , 6.1.4]. Hence, CG (v) ≤ M , a final contradiction. Lemma 9.26. G0 ∼  U7 (2). = Proof. Suppose that G0 ∼ = U7 (2). As in the previous proof, there is w ∈ I3 (B ∗ ) with CG (w) ≤ M and O3 (CG (w)) = 1. Moreover, w ∈ ∪y∈M Dy . Choose any w0 ∈ I3 (B ∗ ) with E(CM (w0 )) = 1 and suppose first that w = w0 . Set Jw := E(CM (w)) = 1. As w ∈ ∪m∈M Dm , CM (w) embeds in Sw × Jw with Sw ∼ = GU3 (2) and Jw ∼ = U4 (2). We may assume that [D, Jw ] = 1 and so Jw = E(CG (w, d)) for all d ∈ D# . Let Lw be the pumpup of Jw in CG (w).

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6. THEOREM C5 : STAGE 4

306

Then Jw  CLw (d) for all d ∈ D# , whence Lw = Jw by [VK , 3.5]. Now, let d1 = diag(ω, ω −1 , 1, 1, 1, 1, 1) ∈ D. Then d1  ∈ U(G; M ; 3) and some G0 -conjugate d2 of d1 is in Jw . Hence, d2  ≤ Lw  CG (w) with Lw < M . Hence, CG (w) ≤ M , and so w ∈ U(G; M ; 3) by Lemma 9.23. Therefore w = w0 , but we shall use the fact that w0  ∈ U(G, M, 3) later. Let w1 ∈ (B ∗ )# with CM (w1 ) =: G1 × G2 with G1 ∼ = G2 ∼ = GU3 (2) and with Z(Gi ) = zi , where E(CM (zi )) = 1. By the previous paragraph, zi  ∈ U(G, M, 3), i = 1, 2. Suppose that w = w1 . Then zi ∈ O3 (CG (w)). It then follows immediately, with solvable L3 -balance, that O3 (CG (w)) = w = Z(L1 ), with L1 quasisimple and O3 (G1 × G2 ) ≤ L1 . As D acts on L1 with ΓD,1 (L1 ) solvable, it follows from [IA , 7.3.4] that L2 ∈ Chev(3) ∪ Alt ∪ Spor. As w ∈ Z(L1 ) and | Aut(L1 )|3 ≥ 37 ≥ | Inn(L1 )|3 , and L1 ∈ C3 , it follows from [IA , Tables 5.3 and 6.1.3] that L1 / w ∼ = Suz with P ∈ Syl3 (L1 ). But |Z(P/ w)| = 3, whereas |Z(P0 )| = 32 for P0 ∈ Syl3 (Suz), a contradiction. Hence, CG (w) ≤ M , contradiction. Thus, w = w1 , and w1  ∈ U(G, M, ). The only remaining possibility is that w = diag(ω, ω, ω −1 , ω −1 , 1, 1, 1). Then ∼ E33 × GU3 (2), O 2 (CM (w)) = F2 × S2 = and we may assume that w0  = [Xw , Xw ], where Xw = O3 (CM (w)). As w0  ∈ U(G; M ; 3), w0 ∈ O3 (CG (w)) and M contains a Sylow 3-subgroup of CG (w). Now, O 3 (S2 ) is a solvable 3-component of CG (w0 , w), whence by solvable L3 -balance, O3 (S2 ) ≤ L2 for some 3-component L2 of CG (w). Now, there is A = E0 × w0  ∼ = E33 with E0 ≤ F2 , A in a Sylow 3-center in CG (w), A ≤ NC(w) (L2 ), and a ∈ wM for any a ∈ A# . Hence CL2 (a) ≤ CM (w) for all a ∈ A# . Also, (9E)

a Sylow 3-subgroup R of AutCG (w) (L2 ) satisfies |R| = 36 and |Ω1 (Z(R))| = 33 .

We have also L2 ∈ C3 as m3 (CG (w)) ≥ 4. Hence, L2 ∈ Alt, and by (9E) and inspection of [IA , 5.3], L2 ∈ Spor. If L2 ∈ Chev(2), then by definition of C3 [V3 , Def. 1.1] and (9E), L2 ∼ = U6 (2), D4 (2), or F4 (2). But then L2 = ΓA,1 (L2 ) ≤ M , contradiction. Therefore L2 ∈ Chev(3). But then (9E) forces q(L2 ) = 33 and so is impossible. This contradiction completes the proof.  Lemma 9.27. We have G0 ∼ = U6 (2) and CG (z) ≤ M . Proof. By the last three lemmas and Table 9.1, G0 ∼ = U6 (2). We know that 2 ∼ F (CG (z)) = Uz ∼ and O (C (z)/U ) U (2) × Z . As CG (z)/Uz embeds = = 21+8 M z 4 3 + in Out(Uz ) ∼ = O8+ (2), and D ≤ CM (z) with ΓD,1 (CG (z)) normalizing O 2 (CM (z)), it follows from [VK , 8.18] that O 2 (CM (z))  CG (z). As B ∩ D = 1, B ∩ D ∈ U(G, M, 3), so NG (B) ≤ M . Moreover, B = J(Q) for some Q ∈ Syl3 (CG (z)). Hence by a Frattini argument, CG (z) ≤ O 2 (CM (z))NG (B) ≤ M . The lemma is proved.  ∗

 U6 (2). Lemma 9.28. We have G0 ∼ = ∼ U6 (2). We will argue that z ∈ U(G; M ; 2). Then Proof. Suppose that G0 = it will follow immediately from Holt’s Theorem [II2 , Theorem SF] that G = M . As G0 = G, this is a contradiction. Let t ∈ I2 (z G ∩ M ). We claim that CG (t) contains an element d of order 3 with d ∈ U(G; M ; 3). Let V be the natural F4 [SU6 (2)]-module. If t ∈ M − G0 , then by [IA , 4.9.2], CG0 (t) ∼ = Sp6 (2) or the centralizer of a transvection in Sp6 (2). In

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9. THE CASE G∗ = 2 D5 (2),

2

E6 (2), OR U7 (2)

307

either case, CG0 (t) contains an element d1 of order 3 with dim([V, d1 ]) = 2, whence t be d1 is conjugate to an element of D, and so d1  ∈ U(G; M ; 3). If t ∈ G0 , let   0  the involution in SU6 (2) lifting t. Then by [III17 , 6.34], CSU6 (2) (t) ≥ L0 × M  0 = SU ([V, L  0 ]) ∼ 0 ∼ with L = SUr (2), where r = dimF4 [V, t], = SUn−2r (2) and M    0 ] is a natural [V, M0 ] is the direct sum of two natural F4 [M0 ]-modules, and [V, L  0 ]-module. Let L0 and M0 be the images of L  0 and M 0 in G0 . If r < 3, F4 [L  0 ]) ≥ 2 and L0 contains a conjugate d1 of an element of D# . Again, then dim([V, L we have d1  ∈ U(G; M ; 3). Finally, if r = 3, then M0 ∼ = U3 (2). But M0 is := embeddable in C CG (z) = CM (z) (Lemma 9.27), and hence in C := C/O2 (C). Now C/O3 (C) embeds in Aut(U4 (2)) ∼ = O6− (2). As |O3 (C)| = 3, M0 embeds in − O6 (2). But by Clifford’s theorem, M0 ∼ = U3 (2) has no characteristic 2 faithful representation in dimension less than 8. This contradiction proves our claim. Now, let g ∈ G with z g ∈ M . By the previous paragraph, z g centralizes some d ∈ I3 (M ) with d ∈ U(G; M ; 3). By [III8 , 6.1f], it follows that g ∈ M , whence z ∈ U(G; M ; 2), contrary to assumption. The proof is complete.  As the preceding two lemmas are in conflict, Theorem 9.1 is proved. Hence the proof of Theorem C5 : Stage 4 is complete, as is the proof of Theorem C5 itself.

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10.1090/surv/040.9/07

CHAPTER 7

Theorem C∗6 : Stage 1 1. Theorems C6 and C∗6 In this chapter we begin the proof of Theorem C6 , which concerns the case in which our K-proper simple group G is of LTp -type for some odd prime p. Thus by assumption, G and p satisfy the following conditions:

(1A)

(1) G is of even type (indeed of restricted even type); (2) Every element of Lop (G) is either a Cp -group or a Tp group; and (3) Some element of Lop (G) is a Tp -group;

and (1B)

p ∈ σ(G).

Recall from the first chapter of this volume that σ(G) = {p | p is an odd prime and m2,p (G) ≥ 4}, i.e., the odd prime p lies in σ(G) if and only if some 2-local subgroup H of G satisfies mp (H) ≥ 4. Here mp (H) is the p-rank of H. Moreover, p ∈ σ0 (G) if and only p ∈ σ(G) but G has no strong p-uniqueness subgroup. Furthermore, for odd p,

(1C)

Ipo (G) = {x ∈ Ip (G) | mp (CG (x)) ≥ 4} and Lop (G) is the set of all components of CG (x)/Op (CG (x)) as x ranges over the elements of Ipo (G). The set of (quasisimple) Cp -groups is defined in [V3 , Sec. 1] and also [I2 , 12.1]. It consists roughly of the groups in Kp which lie in Chev(p) together with Akp , k ≤ 3 (except A3 , of course). For p ≤ 11, a few groups are added and a couple are subtracted from Cp . The (quasisimple) Tp -groups are similarly described in [I2 , §13]. Because of their importance in this chapter, we repeat the definition of the set Tp . For an odd prime p, a quasisimple K-group K with Op (K) = 1 is called a Tp -group, provided that one of the following holds:

(1D)

(1) mp (K) = 1 and K is not a Cp -group (in particular,  L2 (8) if p = 3, and K ∼  L2 (p) or Ap for p > 3, K ∼ = = 5 2 2 K ∼ B (2 ) if p = 5); = 2 (2) p = 3 and K/Z(K) ∼ = L3 (q) (q ≡  (mod 3),  = ±1), A7 , M12 , M22 , J2 , or G2 (8); or K ∼ = 3A6 ; or (3) p = 5 and K ∼ F i . = 22 309

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7. THEOREM C∗ 6 : STAGE 1

310

In particular, mp (K) ≤ 2 in all cases [VK , 16.1]. Also, by [IA , 6.1.4], If p is odd and K ∈ Tp is nonsimple, then p = 3 and K ∼ = SL3 (q), (1E) q ≡  (mod 3),  = ±1, 3A6 , 3A7 , or 3M22 . We shall begin the proof of: Theorem C6 . If G is of LTp -type, p odd, i.e., if (1A) and (1B) hold, then G contains a strong p-uniqueness subgroup. A trivial corollary of the theorem is: Corollary C6 . Let p ∈ σ0 (G). Then G is not of LTp -type. The proof of Theorem C6 is based on the construction of a p-component preuniqueness subgroup M that satisfies the conditions of the component uniqueness theorem [II3 , Theorem PU2 ] (apart from one exceptional subcase). The proof involves an ordering of the elements of Lop (G) that are Tp -groups. To describe it, we divide the set of Tp -groups into five disjoint subsets Tpi , whose elements are called Tpi -groups, 1 ≤ i ≤ 5. This partition is nontrivial only if p = 3 or p = 5. In the following definition,  can take the values 1 and −1. Definition 1.1. T31 = {SL3 (q) | q ≡  (mod 3)} ∪ {3A6 , 3A7 , 3M22 , J2 , G2 (8)}; and Tp1 = ∅ if p > 3; T32 = {L3 (q) | q ≡  (mod 9)}; and Tp2 = ∅ if p > 3; T33 = {L3 (q) | q ≡ 4 or 7 (mod 9), q > 4} ∪ {A7 , M12 , M22 }, and Tp3 = ∅ if p > 3; T34 = {L3 (4)} and T54 = {F i22 }, and Tp4 = ∅ if p > 5; Tp5 = {L ∈ Tp | mp (L) = 1}. It is immediate from the definition and (1D) that every Tp -group is a Tpi -group for exactly one value of i, 1 ≤ i ≤ 5. The separation of the groups L3 (q), q ≡  (mod 3),  = ±1 according to the congruence on q (mod 9) is due to the fact that when q ≡  (mod 9), there is y ∈ Ip (L) such that CL (y) has a component isomorphic to SL2 (q), whereas if q ≡ 4 or 7 (mod 9), then |L|3 = 32 and CL (y) is solvable for every y ∈ I3 (L) (see [IA , 4.8.2, 4.8.4]). Also the groups L3 (4) for p = 3 and F i22 for p = 5 are exceptional because of their poor balance and generational properties. Furthermore, by [VK , 16.3, 3.2], if L is a Cp -group or a Tpi -group and J is a component of CL (y) for some element y of order p acting on L, or J is a quotient of L, then J/Op (J) is a Cp -group or a Tpj -group for some i ≤ j; and if L is a Cp -group then so is J/Op (J). We define (1F)

d = dTp = min{i | Lop (G) ∩ Tpi = ∅}.

It follows immediately, in the presence of (1A2), that the set of elements of Lop (G) that are Tpd -groups is closed under pumping up within the set of all elements of Lop (G). In other words, if (x, K), (y, L) ∈ ILop (G) with (x, K) < (y, L) and

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1. THEOREMS C6 AND C∗ 6

311

K/Op (K) ∈ Tpd , then L/Op (L) ∈ Tpd . We refer to the integer dTp as the Tp depth of G. (Cf. the Gp -depth, analogously defined in [III3 , Def. 5.1].) In this chapter the prime p will have been fixed, and we always write d for dTp . To state the first step in the proof of Theorem C6 , we introduce the following additional terminology. If x ∈ Ipo (G) and K is a p-component of CG (x) with (x, K) Ipo -terminal in G and K/Op (K) a Tp -group, we call (x, K) an Ipo -terminal Tp pair. Moreover, if in addition K/Op (K) is a Tpi -group, and we wish to specify the integer i, we refer to (x, K) as an Ipo -terminal Tpi -pair. We also recall the notion of Ipo -rigidity for a pair (x, K) ∈ ILop (G) [IG , 6.11]. It stipulates that for any long pumpup (x, K) 8, q ≡ 

(mod 3),  = ±1.

First, by [IA , 4.9.1], every u ∈ U −x, z induces a nontrivial field automorphism 1 on K. For each such u, we set Iu = L3 (CK (u)), so that I u ∼ = L3 (q 3 ). Note that 1 q 3 > 3, by (4F). We denote the pumpup of Iu in CG (u) by Ku and we set (CG (u))= CG (u)/O3 (CG (u)). We fix this notation. We immediately obtain Lemma 4.11. For any u ∈ U − x, z , one of the following holds:  u is the direct product of three components isomorphic to Iu cycled by (a) K x; u ∼  u ; or (b) K = L3 (q) with x inducing a nontrivial field automorphism on K  u = Iu . (c) K Proof. We know by L3 -balance that Ku is either the product of three 3components of CG (u) cycled by x or is a single x-invariant 3-component and in / Lo3 (G) by Lemma 4.10, either case Iu is a 3-component of CKu (x). Since SL3 (q) ∈ (a) therefore holds in the first case. Likewise (b) or (c) holds in the second case by  Lemma 4.10 and [VK , 16.6].  u = Iu (for some u ∈ U − x, z) that we need the It is to eliminate the case K element g. However, we first eliminate the other two possibilities of the lemma for every u ∈ U − x, z without recourse to g. We first dispose of the cycling case on the basis of our maximal choice of (x, K). Lemma 4.12. Ku is a single 3-component of CG (u) for every u ∈ U − x, z. Proof. Suppose false, so that Ku = J1 J2 J3 , where each Ji is a 3-component of Ku with Ji ∼ = Iu ∼ = L3 (q 1/3 ), 1 ≤ i ≤ 3. In particular, as m3 (Ji ) = 2 for each i and Ji is simple, it follows that m3 (C(u, J1 )) ≥ 5. Choose w ∈ Lo3 (G) and a 3-component J of CG (w) with J/O3 (J) ∼ = Iu and (w, J) Gilman-maximal. In particular, if W ∈ Syl3 (C(w, J)), then m3 (W ) ≥ 5. Since (x, K) ∈ TJ∗3 (G) and m3 (Q) = 1, J is not I3o -terminal in G. Hence by [V2 , 8.10], (w, J) is not one-step I3o -rigid. Again by Lemma 4.10 and [VK , 16.6], we conclude that there is w∗ ∈ I3o (W ) such that if J ∗ denotes the pumpup of J0 = L(CJ (w∗ )) in Cw∗ , then J ∗ /O3 (J ∗ ) ∼ = L3 (q) (∼ = K). Set W1 = CW (w∗ ). Since m3 (W ) ≥ 4, we have m3 (W1 ) ≥ 3, as a consequence of [IG , 10.17]. Also W1 centralizes J0 /O3 (J0 ), so W1 leaves J ∗ invariant. But J0 is a 3-component of CJ ∗ (w) and w ∈ W1 . Hence if we set W0 = CW1 (J ∗ /O3 (J ∗ )), it follows

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4. THEOREM 2

323

from [IA , 7.1.4c] that W1 = W0 w , whence m3 (W0 ) ≥ 2. We conclude that m3 (C(w∗ , J ∗ )) ≥ 2. But now as J ∗ /O3 (J ∗ ) ∼ = K, it follows once again from Lemma 4.10, [VK , 16.6], and [V2 , 8.5] that G possesses an I3o -terminal pair (x∗ , K ∗ ) ∈ ILo3 (G) with h K ∗ /O3 (K ∗ ) ∼ = L3 (q ∗ ), q ∗ = q 3 for some h ≥ 0, such that either (4G)

(1) h ≥ 1; or (2) m3 (C(x∗ , K ∗ )) ≥ 2.

However, as (x, K) ∈ TJ∗3 (G) and m3 (Q) = 1, this is a contradiction in either case of (4G).  u ∼ By Lemmas 4.11 and 4.12, for any u ∈ U − x, z, either K = L3 (q) (with    x inducing a nontrivial field automorphism on Ku ) or Ku = Iu . The same argument as in the preceding lemma, in fact, establishes the following slightly sharper conclusions. We omit the proof. Lemma 4.13. For each u ∈ U − x, z, one of the following holds: u ∼ (a) K = L3 (q) and m3 (C(u, Ku )) = 1; or  (b) Ku = Iu and m3 (C(u, Ku )) ≤ 3. Note that if (a) holds, then also (u, Ku ) ∈ TJ∗3 (G) and the situation is entirely symmetric with (u, Ku ) in place of (x, K). Next, by [VK , 16.10] and [IA , 4.2.3], there is b ∈ I3 (T0 ) (whence b centralizes U ) such that if we set H = L3 (CK (b)), then H ∼ = SL2 (q) and every u ∈ U − x, z induces a nontrivial field automorphism on H (inasmuch as t ∈ U induces a nontrivial field automorphism on K). For each such u, we put Hu = L3 (CH (u)). Then H u ∼ = SL2 (q 1/3 ). Moreover, Hu ≤ Iu . We let J be the subnormal closure of H in CG (b), so that J has the usual form by L3 -balance. We also put (CG (b))= CG (b)/O3 (CG (b)) and fix this notation as well. For future reference, we remark that the properties of b are shared by any element of the coset b x. Using the preceding lemma, we prove 1 −η 2 Lemma 4.14. We have that J ∼ = L2 (q), L3 (q), P Sp4 (q), or L− 4 (q), or Ln (q ), n = 4, 5 (the last only if q = q02 ,  = 1, q0 ≡ η (mod 3), η = ±1). In every case, x induces an inner automorphism on J. Proof. Suppose false. Since H is a 3-component of CJ (x), this time we conclude from Lemma 4.10 and [VK , 16.11] that one of the following holds:

(4H)

(1) J ∼ = L2 (q 3 ) with x inducing a nontrivial field automor or phism on J; (2) J is the product of three 3-component J1 , J2 , J3 cycled by x with Ji ∼ = H/Z(H), i = 1, 2, 3.

As t ∈ U centralizes b, t leaves H invariant and hence leaves J invariant.  Inndiag(J)  has p-rank 1, some v ∈ t, x − x Suppose (4H1) holds. Since Aut(J)/ ∼  maps into Inndiag(J), and hence into CInndiag(J)  (x) = Inndiag(H) [IA , 4.9.1]. Thus v induces an inner-diagonal automorphism on H, contradiction. Hence, (4H2) holds. Since x cycles J1 , J2 , J3 and t, x ∼ = E32 , some t ∈ t, x − x leaves each Ji invariant, 1 ≤ i ≤ 3. But as x ∈ Z(T ), CT (t ) ∈ Syl3 (CCx (t )), so without loss we can assume that t = t.

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7. THEOREM C∗ 6 : STAGE 1

324

∼ L2 (q 1/3 ), 1 ≤ i ≤ 3, H1 , H2 , H3 are cycled i = Set Hi = L3 (CJi (t)). Then H  t is a diagonal of H 2H 3. 1H by x, and H Since x leaves Kt invariant, it follows by L3 -balance that either each Hi ≤ Kt  t , 1 ≤ i ≤ 3. The first possibility is excluded as then or each Hi centralizes K t ∼ m3 (Kt ) ≥ 3, contrary to the fact [IA , 4.10.3a] that K = L3 (q) or L3 (q 1/3 )) has  t , so H  t does as well, contrary to the fact 3-rank 2. Hence H1 H2 H3 centralizes K  that Ht ≤ It ≤ Kt . The proof is complete. Now we can prove  u = Iu . Lemma 4.15. For each u ∈ U − x, z , we have K ∼ L (q) with x inducing u = Proof. Suppose false for some such u, so that K 3  u . We have Hu   CK (b, x). It follows a nontrivial field automorphism on K u that there is b ∈ b x inducing an inner-diagonal automorphism on Ku . By [IA ,  4.2.3], x induces a nontrivial field automorphism on L3 (CK  u (b )), which is therefore  isomorphic to SL2 (q). Replacing b by b , as we may by the remark before Lemma ∗ ∼ 4.14, and setting H ∗ = L3 (CKu (b)), we have that H = SL2 (q) with x inducing a ∗  . nontrivial field automorphism on H Let J ∗ be the pumpup of H ∗ in CG (b), so that J ∗ has the usual form by L3 balance. As Hu ≤ H ∗ ≤ J ∗ , we have J ∗ = J. But as already noted in this case, the situation is symmetric in (u, Ku ) and (x, K). Hence Lemma 4.14 applies to  By [VK , 16.11], x induces an u as well as x, whence x, u maps into Inn(J). ∗  inner automorphism on L3 (CJ(u)) = H . This is a contradiction as x induces a  nontrivial field automorphism on H ∗ /O3 (H ∗ ), and the lemma follows. Now we bring the element g into the analysis. Again y = xg and L = K g . We set CG (y)∗ = CG (y)/O3 (CG (y)) and use the “star convention”: for X ≤ CG (y), we write X ∗ for XO3 (CG (y))/O3 (CG (y)). Using the preceding lemma, we obtain Lemma 4.16. The following conditions hold: (a) y ∈ x, z; (b) y ∈ xNG (T ) ; (c) y = z; and (d) x induces an inner automorphism on L∗ . Proof. If (a) fails, then y ∈ U − x, z, in which case y induces a nontrivial field automorphism on K, so Iy and Ky are defined. Then Iy ≤ CG (y) with Iy∗ ∼ = L3 (q 1/3 ), and as Qg ∈ Syl3 (C(y, L∗ )) with Q cyclic, we must have Iy ≤ L. Thus L is the pumpup of Iy in CG (y), so by definition Ky = L. But this is impossible since then L∗ = Iy∗ ∼ = L3 (q 1/3 ) by the preceding lemma, contrary to the ∗ ∼  ∼ fact that L = K = L3 (q). Thus (a) holds. By Burnside’s Lemma, x and y are NG (T )-conjugate, so we may take g ∈ NG (T ). But x, z = Ω1 (Z(T )), so g ∈ NG (x, z). Since x, z maps into Inn(K), it therefore maps into Inn(L∗ ). Thus (b) and (d) also hold. Suppose by way of contradiction that y = z. Recall our notation that P = T ∩K ∈ Syl3 (K). Now Ω1 (T )/P has a normal cyclic subgroup (Ω1 (T )∩Q)P/P whose quotient embeds in Ω1 (OutT (K)) and therefore is elementary abelian of order −1 at most 9. On the other hand P ∩ Z(T ) = z. Therefore if we put S = P g , then S ∩ Z(T ) = x and so P ∩ S = 1. It follows that Ω1 (T )/P contains a copy

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∼ P , whence P has a normal cyclic subgroup of index at most 9. However, of S = by [VK , 16.22], P has no normal Z9 -subgroup, and |P | ≥ 34 as q is a perfect cube. This is impossible, and the lemma follows.  Lemma 4.17. Some u ∈ t x, z induces a nontrivial field automorphism on L∗ . Proof. Consider the images E of U and S of T in CG (y)/C(y, L), which in turn embeds in Aut(L∗ ). Suppose first that Ω1 (CS (E)) is nonabelian. Then by [VK , 16.23], every element of U induces a field automorphism or an inner automorphism on L∗ , and the subgroup U0 inducing inner automorphisms on L∗ has order 9. As t x, z generates U , we must have u ∈ U − U0 for some u ∈ t x, z, in which case u induces a nontrivial field automorphism on L∗ , as asserted. We may therefore assume that Ω1 (CS (E)) is abelian. Consequently the commutator subgroup Γ of Ω1 (CT (U )) = Ω1 (CT (t)) lies in C(y, L). In particular, x ∈ Γ. Looking in Cx , we see that z ∈ Γ. It follows that z = y, contradicting Lemma 4.16 and completing the proof.  Now we can prove: Lemma 4.18. We have q = 8. Proof. Suppose false and continue the above argument. First, the eigenvalues of z ∈ K are the distinct cube roots of unity, so z is real G in K, and hence xz ∼ xz −1 . As y = z, it follows that x ∩ E1 (x, z) = E1 (x, z) − {z}, and so we can take a 3-element g ∈ NG (T ) so that y = xg = xz and z g = z. In particular, z ∈ L. Now let u be as in Lemma 4.17, so that u induces a nontrivial field automorphism on both K and L∗ . Moreover, u ∈ U ≤ Z(T0 ) with T0 ∈ Syl3 (CCx (u)). We have Iu = L3 (CK (u))  CCx (u), so as z = Z(T0 ) ∩ K, we must have z ∈ Iu . Thus z ∈ Iu , a U -invariant component of (CG (u)). Set L0 = L3 (CL (u)) and let L1 be the subnormal closure of L0 in CG (u). If  1 = Iu . But L  0 is a component of C  (y) by L3 -balance, and  1 ] = 1 then L [ z, L L1 0 ∼ 0 = L  1 is centralized by x and y, hence L = L∗0 ∼ = L3 (q 1/3 ) ∼ = Iu . Therefore L  1 ] = 1. Consequently, [  1 ] = 1 and in particular, z by z. This contradicts [ z, L z, L centralizes L∗0 . Since z ∈ L we have z ∗ ∈ CL∗ (L∗0 ), whence CL∗ (L∗0 ) = 1. But this  contradicts [IA , 7.1.4c], so the proof is complete. 5. Theorem 2 Completed: The U3 (8) Case In this section we complete the proof of Theorem 2 by considering the case (5A)

TJ3 (G) = ∅ and (x, K) ∈ TJ∗3 (G) with K/O3 (K) ∼ = U3 (8)

and reasoning to a contradiction. We assume (5A) throughout this section. Lemma 5.1. There is no pair (y, L) ∈ I3o (G) such that L/O3 (L) ∼ = U3 (8) and m3 (C(y, L)) > 1. Proof. Suppose false. By [IG , 6.22], there exists a chain (y, L) = (y0 , L0 ) < (y1 , L1 ) < · · · < (yn , Ln ) of elements of ILo3 (G) such that (yn , Ln ) is I3o -terminal in G. Then every Li /O3 (Li ) is in C3 or T3 −T31 , or is a TG3 -group. Since L/O3 (L) is a T32 -group, each Li /O3 (Li )

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is in fact a T32 -group or a flat TG3 -group, by [VK , 16.3]. Therefore there exist nonnegative integers k ≤ n and n1 ≤ · · · ≤ nk such that Li /O3 (Li ) ∼ = U3 (8ni ) for all i ≤ k, and if k < n, then Lk+1 /O3 (Lk+1 ) is a flat TG3 -group. Suppose that k < n. Then as Lk /O3 (Lk ) ∈ Alt ∪ Chev(r) for any r > 2, and by inspection of sporadic possibilities for Lk+1 /O3 (Lk+1 ) in [IA , 5.3], we see that the flat TG3 -group Lk+1 /O3 (Lk+1 ) lies in Chev(2). From the list of flat TG3 -groups (1J) and using [IA , 4.8.2, 4.8.4], we see, however, that the pumpup U3 (8nk ) ↑3 Lk+1 /O3 (Lk+1 ) is impossible. Therefore, k = n. In particular, (yn , Ln ) ∈ TJ3 (G). Since by assumption, (x, K) ∈ TJ∗3 (G), the definition of TJ∗3 (G) implies that nk = 1 (Definition 1.6). Then by [V2 , 8.5], m3 (C(yn , Ln )) ≥ 2, so (yn , Ln ) ∈ TJ3 (G). However, by (5A),  TJ3 (G) = ∅. This contradiction proves the lemma. We recall the notation Cx = CG (x), C x = Cx /O3 (Cx ), T ∈ Syl3 (Cx ), (5B)

P = T ∩ K ∈ Syl3 (K), z = Ω1 (Z(P )),

and t ∈ I3 (T ) inducing a nontrivial field automorphism on K. Also, Q = CT (K) is cyclic by Lemma 5.1. By [VK , 16.24], CT (P ) = Q×z, whence Ω1 (Z(T )) = x, z. We have P = P0 w where, with respect to a suitable orthogonal frame in the natural K-module, P0 ∼ = P0 ∼ = Z3 × Z9 , P 0 is diagonal, w3 = 1, and w permutes the frame in a 3-cycle. We may assume that CP (t) = Ω1 (P0 ) w ∼ = 31+2 . We write Ω1 (P0 ) = b1 , z with bw 1 = b1 z, and set H = L3 (CK (b1 )), so that H ∼ = L2 (8) and t induces a field automorphism on H. Thus CH (t) ∼ = L2 (2). Set (5D) F = x, b1  and E = F z ∼ = E33 . (5C)

Recall that U = x, z, t. Also, as Q is cyclic, x, z ≤ Z(T ) ≤ Q × z. Lemma 5.2. x is not weakly closed in U with respect to G. Proof. This is Lemma 4.2.



Lemma 5.3. We have x ∈ z . If CG (z) has a 3-component J, then one of the following holds: (a) Z(J/O3 (J)) = 1, and z ∈ Z(J ∗ ), where J ∗ is the normal closure of J in CG (z), or (b) T ∈ Syl3 (G), J = L3 (CG (z)), and m3 (J) = 1 with x ∈ J. G

Proof. Suppose that J exists and let J ∗ be its normal closure in CG (z). Let T ≤ T ∗ ∈ Syl3 (CG (z)). Also, let J = J/O3 (J). Then 1 = T ∗ ∩ J ∗  T ∗ , whence Z(T ∗ ) ∩ J ∗ = 1. If T < T ∗ , then Z(T ∗ ) = z, whence z ∈ Z(J ∗ ) and Z(J) = 1, so (a) holds. Also, if Z(J ∗ /O3 (J ∗ )) = 1, then its preimage Z3∗ (J ∗ ) is normal in CG (z) of order divisible by 3, whence Z3∗ (J ∗ ) ∩ Ω1 (Z(T ∗ )) = 1. Now, Ω1 (Z(T ∗ )) ≤ x, z and so, if z ∈ J ∗ , then L3 (CJ ∗ (x)) = 1. But L3 (CG (x, z)) = 1, a contradiction. Hence, if |Z3∗ (J ∗ )| is a multiple of 3, then z ∈ Z(J ∗ ) and Z(J) = 1. Therefore if (a) fails, then T ∈ Syl3 (CG (z)) and Z(J) = 1. We claim that T ∈ Syl3 (G) in  this case. Otherwise, NG (T ) contains a 3-element centralizing either xz or xz −1 , and cycling the three other subgroups of x, z of order 3.

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But z is real in K, so all four subgroups of x, z are conjugate, hence all are 3central, contradiction. This proves the claim. Let J + be the normal closure of J in L3 (CG (z))T . Then z ∈ J + , whence x1 ∈ J + for some x1 ∈ {x, xz, xz −1 }. If J + = J, then J + /O3 (J + ) is the product of at least three simple components of 3-rank ≥ 1, whence T contains a normal E33 -subgroup disjoint from z, which is impossible by the structure of Cx . Thus J + = J with x1 ∈ J. As L3 (CG (x, z)) = 1, it follows that L3 (CG (z)) = J. Suppose that T ∩ J contains some elementary abelian V  T . Then z ∈ V so [P, V ] = 1 and V maps into CAut(K) (P ), which is the image of z. But then V ≤ Q × z so V = x1 . In particular, m3 (J) = 1. Since z is inverted in Cx , xz and xz −1 are NG (z)-conjugate. As J = L3 (CG (z)), J cannot contain xz or xz −1 , so x ∈ J, proving (b).  It follows that CG (z) ∼ = Cx , so x ∈ z G , and the proof is complete. For the next several lemmas, we introduce the following notation. First, b will be an arbitrary element of E − x, z, and Hb = L3 (CK (b)), a 3-component of CG (b, x). Thus b is K-conjugate to an element of F − x = b1 , x − x, and Hb ∼ = L2 (8). We set and let

b = Cb /O3 (Cb ), Cb = CG (b) and C Hb∗

be the pumpup of Hb in Cb . Note that [b, t] = 1.

3 ∗ ∼ Lemma 5.4. H b = L2 (8 ), Sp4 (8) or SL4 (8). 3 ∗ ∼ Proof. Suppose first that H b = L2 (8 ). Since t ∈ CG (b) normalizing Hb , t  ∗ centralizing H  b , it follows normalizes Hb∗ . As x induces a field automorphism on H b that t3 ∈ xHb∗ C(b, Hb∗ ), contrary to the hypothesis that t3 = 1. ∗ ∼ ∗ Now suppose that H b = Sp4 (8) or SL4 (8). As z ∈ Hb = L3 (CHb (x)), we  ∗ such that if V is a suitconclude that x induces an inner automorphism on H b  b is supported on the 2-dimensional subspace  ∗ , then H able standard module for H b CV (x). Then, as z ∈ Hb , if Hb,z := L3 (CHb∗ (z)), then Hb,z /O3 (Hb,z ) ∼ = L2 (8). Let    in C (z) and let J = J/O (J). If Z( J) = 1, then by J be the pumpup of Hb,z G 3 Lemma 5.3, T ∈ Syl3 (G) and J = L3 (CG (z)) with m3 (J) = 1 and x ∈ J. Thus, ∗  = 1. By x ∈ Hb,z , whence x ∈ z Hb , contradicting Lemma 5.3. Therefore Z(J) [VK , 16.12], J ∼ = SU6 (2). (Note that [z, t] = 1 and t induces a field automorphism on Hb,z .) Since m3 (Cx ) = 4 but m3 (J) = 5, x normalizes J and by [IA , 4.8.2, 4.8.4], E(CJ(x)) ∼ = L2 (8). But CG (x, z) = CCx (z) is solvable, a contradiction. This completes the proof. 

Lemma 5.5. The following conclusions hold: (a) T ∈ Syl3 (G);  (b) x is NG (T )-conjugate to both xz and xz −1 ; and (c) If T < T ∗ ∈ Syl3 (CG (z)), then Z(T ∗ ) = z and T ∗ ∈ Syl3 (G). Proof. As x is not G-conjugate to z by Lemma 5.3 and as xz −1 ∈ (xz)K , either x is weakly closed in Z(T ) with respect to G and T ∈ Syl3 (G), or  N (T ) x G = {x , xz , xz −1 }. In the latter case, T ∈ Syl3 (G) and, as Z(T ) ≤ Q × z, all parts of the lemma hold.

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Suppose the former case holds. It follows from Lemma 5.2 that there is some y ∈ U − x, z such that y = xg . Then x induces a field automorphism on K g . It follows that CQ (y) = x and x, y × CP (y) ∈ Syl3 (CG (x, y)) with [CP (y), CP (y)] = z. Thus z is 3-central in K g and there exists h ∈ K g with h3 = z = [h, x]. Hence, g  xz −1 ∈ xK , contrary to assumption, completing the proof. We fix T ∗ ∈ Syl3 (CG (z)) such that T < T ∗ , so that T ∗ ∈ Syl3 (G). Lemma 5.6. The following conditions hold: (a) If L3 (CG (z)) = 1 and J is the product of all 3-components of CG (z) in a T ∗ -orbit, then z ∈ Z(J), and for every 3-component L of CG (z), L/O3 (L) is isomorphic to one of the following: 3U4 (3), (3 × 3)U4 (3), 3Ω7 (3), 3G2 (3), or 3J3 ; (b) Hb∗ is a nontrivial pumpup of Hb , and Hb∗ ≤ ΓE,2 (G). Moreover, if Hb∗ is  ∗ is isomorphic to one of the following: a vertical pumpup of Hb , then H b 3 U3 (8), 3D4 (2), U6 (2), 2 G2 (3 2 ), or Co3 ; and ∗ ∼ (c) Either Q = x, or H b = U3 (8) and |Q| = 9. Proof. Let J be as in (a). As T < T ∗ , Lemma 5.3 says that z ∈ Z(J) and z := for each 3-component L of J, Z3∗ (L) has order divisible by 3. Then, in C   Cz /O3 (Cz ), L ∈ T3 ∪ TG3 , whence L ∈ C3 . We observe that CG (x, z)/O3 (CG (x, z)) has 3-rank 4 and has a normal 3-subgroup of index at most 2. By inspection of  is sporadic, then L ∼  ∼ [IA , Tables 5.3], it follows that if L = 3J3 . Likewise L = SU6 (2), for otherwise m3 (CL (x)) > 4 or CL (x) is nonsolvable, by [IA , 4.8.2, 4.8.4]. Hence,  is as listed in (a). L  0∗ follow from [VK , In (b), as H b ∼ = L2 (8), the possibilities for the structure of H G 16.12] and Lemma 5.4; in particular, by (a), e ∈ z for any e ∈ E − z. However, to complete the proof of (b), we need to rule out the case ∗ ∼ (5E) H = SU6 (2). b

  ∗Q  If (5E) occurs, let Qb = so that H b b has index at most 2 in Cb . Let w 3 D = b, z and choose w ∈ NK (D) such that b = bz and w = 1. There are solvable 3-components S1 , S2 of CHb∗ (D) such that S1 S2 = S1 × S2 ∼ = SU3 (2) × SU3 (2), w ∗    S1 = S2 , and C  (D) = Qb S1 S2  y  with y ∈ I3 (H ) and Si  y ∼ = GU3 (2), C(b, Hb∗ ),

b

Cb

i = 1, 2. Let S0 = S2w , a third solvable 3-component of CG (D). Then S0 maps into b . CAut(H ∗ ) (S1 S2 ) ∼ = Z3 , so S0 = O 3 (S0 ) ≤ Q b c ∈ Z(S2 ), and let Jc = L3 (CHb∗ (c)), so that Now choose c ∈ I3 (S2 ) with  w ∼ U4 (2). Then c ∈ S0 ≤ C(b, H ∗ ); let Lcw be the pumpup of Lc in CG (cw ). Jc = b

−1

As b ∈ Hb∗ , and by [VK , 16.9a], Lcw is a trivial pumpup of Hb∗ . Set L = (Lcw )w , a 3-component of CG (c) covering S1 × S2 . Since Jc covers S1 , Lc covers Jc in  := CG (c)/O3 (CG (c)). Now CL (b) covers S0 × Jc . But this is impossible by C c [VK , 16.9b]. Thus (5E) cannot occur, and (b) is proved. We now argue that Hb∗ is a nontrivial pumpup of Hb . By K-conjugation, we may assume that b ∈ F − x. Let g ∈ NG (T ) with xg = xz and z g = z. As F ∩ K ≤ [T, T ], it follows that b ∈ K g C(xz, K g ). Let b0 ∈ xz, b ∩ K g . Then by the above, b0 ∈ z G , whence L3 (CK g (b0 )) = L3 (CG (xz, b)) =: Hb0 ∼ = Hb

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with z ∈ Hb ∩ Hb0 . Thus Hb0 ≤ Hb∗ . If Hb∗ is a trivial pumpup of Hb , then [Hb∗ , x] ≤ O3 (Hb∗ ). But [Hb0 , x] covers Hb0 /O3 (Hb0 ), a contradiction. Hence Hb∗ is a nontrivial pumpup of Hb , as claimed. Next, we have E = b×x, z. By [IA , 7.3.4, 7.3.8], [IG , 3.28], and [VK , 16.27], and as [Hb∗ , b] = 1, Hb∗ = Γx,z,1 (Hb∗ ) ≤ ΓE,2 (G). Thus, (b) holds. Finally, since  b , (c) follows.  ∗ centralizing H  Q acts faithfully on H b

Lemma 5.7. G is balanced with respect to E, i.e., O3 is an E-signalizer functor. Moreover, one of the following holds: (a) O3 (CG (a)) = 1 for all a ∈ E # ; or (b) There is a maximal subgroup M of G with ΓE,2 (G) ≤ M . Moreover,  = M/O3 (M ), we have K  < E(M ) = F ∗ (M ) = M 1 × M 2 × M 3 , setting M ∼   where Mi = U3 (8) and the Mi are transitively permuted by x. Moreover, x ∈ Syl3 (C(x, K)), and, after replacing t by a suitable element of t x i if necessary, t ∈ M normalizes and induces field automorphisms on M ∗ ∗  ∼ for all i. If b ∈ b1 , z − z, then H b = L2 (8) × L2 (8) × L2 (8) and Hb =  ∗ ∼  L3 (CG (b)) ≤ M ; while if b ∈ x b1 z with  = ±1, then H b = U3 (8), and ∗  again Hb = L3 (CG (b)) ≤ M . Proof. Suppose that G is not balanced with respect to E. Then by [IG , 20.6], there exist a, a1 ∈ E # such that Xa := O3 (CG (a)) ∩ CG (a1 ) normalizes and acts nontrivially on L1 /O3 (L1 ), for some 3-component L1 of CG (a1 ). Then in X := AutL1 Xa (L1 /O3 (L1 )), the image of Xa is nontrivial and lies in O3 (CX (a)). In particular, L1 /O3 (L1 ) is not locally balanced with respect to the image of a in Aut(L1 /O3 (L1 )). By [VK , 7.12], K is locally balanced with respect to any subgroup of Inn(K) of order 3. As E ≤ K x, it follows that a1 = x. As xz = xg for some g ∈ NG (T ) and as b, z ≤ [T, T ], we see that E ≤ xzK g ≤ CG (xz), and again using [VK , 7.12], a1 = xz. Similarly, a1 = xz −1 . By Lemma 5.6a, if J is a 3-component of CG (z), then J/O3 (J) ∈ C3 − {L2 (33 )}, so J/O3 (J) is balanced for p = 3 by [VK , 7.1]. Thus, a1 = z. Hence a1 ∈ E − x, z. Setting a1 = b as in Lemma 5.6b and Hb = L3 (CK (b)), we have the possibilities for the subnormal closure Hb∗ of Hb in Cb listed in that lemma. If L1 is a 3-component of Hb∗ , then by [VK , 7.11, 7.12], L1 is balanced with respect to E, unless possibly a induces a non-inner automorphism on L1 /O3 (L1 ) ∼ = U3 (8). But in that case Hb = L3 (CL1 (x)), Hb /O3 (Hb ) ∼ = L2 (8), and z ∈ Hb . Hence E = b, x, z induces inner automorphisms on L1 /O3 (L1 ), contradiction. Thus L1 is not a 3-component of Hb∗ . Since z ∈ Hb∗ , L3 (CL1 (z)) =: Lz covers L1 /O3 (L1 ). Let L∗z be the pumpup of Lz in CG (z). In view of the possibilities for L∗z /O3 (L∗z ) listed in Lemma 5.6a, we see that Lz /O3 (Lz ), and hence also L1 /O3 (L1 ), is isomorphic to a central quotient of one of the groups listed in Lemma 5.6a. In any case, L1 /O3 (L1 ) ∈ Chev(3) or L1 /O3 3 (L1 ) ∼ = J3 , and thus L1 /O3 (L1 ) is locally balanced by [IA , 7.7.1]. As this exhausts all the possibilities for the pair (a1 , L1 ), we reach a contradiction, whence G is balanced with respect to E. Let  WE = Θ1 (G; E) = O3 (CG (e)) | e ∈ E # , the completion of the O3 signalizer functor on E. Then WE is a 3 -group by [V2 , 1.1]. If WE = 1, then conclusion (a) holds. So, assume WE = 1. Then

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ΓE,2 (G), which normalizes WE by [V2 , 1.3c], is contained in a maximal subgroup M of G. As K = CK (b, x), CK (bz, x) by [IA , 7.3.2], it follows that K ≤ M and  = M/O3 (M ), so K  is a component of C  ( x) and likewise K g ≤ M . Set M E(M ) 2 ∼   [E(M ), x ] = 1. Then by L3 -balance and [VK , 3.13], E(M ) = U9 (2), E6 (2), D4 (8), 3 ) = M 1 ∗ M 2 ∗ M 3 , with M i /Z(M i ) ∼ D4 (8), or U3 (83 ), or E(M = U3 (8), i = 1, 2, 3. ) = F ∗ (M ) and Z(E(M )) = 1. In every case, since m3 (C(x, K)) = 1, E(M Now for every b ∈ E − x, z, we see by Lemma 5.6b that Hb∗ ≤ ΓE,2 (G) ≤ M.  ∗ is a component of C  (b). On the other hand, Lemma 5.6b In particular, H b E(M )  ∗ /O3 (H  ∗ ). gives the possible isomorphism types of H b b ∗ ∼ ) ∼ If E(M = U9 (2), the only possibility is E(CM (b)) = Hb = U3 (8). This must ∼ ∼ hold for all b ∈ b1 , x − x. But also E(CM (x)) = U3 (8). Thus E(CM (u)) = U3 (8) # ) ∼ for all u ∈ b1 , x , contradicting [VK , 16.13]. So E(M = U9 (2). 2 ∼  Suppose that E(M ) = E6 (2). Then with [IA , 4.7.3A], x induces an outer ). In this case we fix b ∈ (E∩K)−z, so that b ∈ I3 (M ) and automorphism on E(M 3 ∗ ∼  H (2). By [V , 16.14], there exists b, y ∈ E (L (M ) x) such that if we U K 3 2 b = 6 2 ∼ ∼    put My = L3 (CM (y)), then b ∈ My , y ∈ E(M ), My = D5 (2), and E(CM y (b)) = 

U5 (2). Let Mby = L3 (CMy (b)). Since Hb∗ = L3 (CM (b)) = O 3 (CM (b))(∞) by [IA , 4.7.3A], Mby = L3 (CHb∗ (y))  CG (b, y). Let I be the subnormal closure of Mby   M in CG (y). Then I has the usual structure by L3 -balance, and My = Mby y ≤ I. So I is a single 3-component, U5 (2) ↑3 I/O3 (I), and I/O3 (I) involves 2 D5 (2). By ) ∼ [VK , 16.15], I/O3 (I) ∈ C3 ∪ T3 ∪ TG3 , a contradiction. Thus, E(M = 2E6 (2). ). But then t ) ∼ If E(M = U3 (83 ), then x acts as a field automorphism on E(M  acts as a field automorphism of order 9 on E(M ), contrary to the fact that t3 = 1. ) ∼ Thus, E(M = U3 (83 ). ) ∼ , and there If E(M = 3D4 (8), then x induces a graph automorphism on M 3 ∗  exists b ∈ (E − z, x) ∩ L3 (M ). Then Hb ↑3 D4 (8) via b, so by [IA , 4.7.3A], 3 ∗ ∼  ∼3 H b = SU3 (8) or L2 (8 ), contradicting Lemma 5.6. Thus, E(M ) = D4 (8). ∼  If E(M ) = D4 (8), then we reach a similar contradiction if (E − x, z) ∩ L3 (M ) = 1. So E ∩ L3 (M ) ≤ x, z. As x induces a graph automorphism on  in this case, E ∩ L3 (M ) = z. Therefore some b ∈ E − x, z induces a field M , and hence on E(C (x)) = K  by [IA , 4.2.3]. But E induces automorphism on M M = ∼ D4 (8). only inner automorphisms on K, contradiction. Hence, M     Hence E(M ) = M1 × M2 × M3 with the 3-components M1 , M2 , M3 permuted transitively by x. Now, b1 , z is diagonally embedded in M1 M2 M3 , and so if b ∈ b1 , z−z, then ±1     ∼ b1 z, then b permutes the E(CM (b)) = H1 H2 H3 with Hi = L2 (8). Also, if b ∈ x ∼ Mi and then E(CM (b)) = U3 (8). By consideration of the possibilities for L3 (CG (b)) listed in Lemma 5.6b, we see that Hb∗ = L3 (CG (b)) ≤ M , and so conclusion (b) holds, after replacing t by a suitable element of t, x, completing the proof.  Now we can prove Lemma 5.8. O3 (CG (a)) = 1 for all a ∈ E # .

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Proof. Suppose not. Then by Lemma 5.7, there is a maximal subgroup M  = M/O3 (M ), F ∗ (M ) = M 1 × M 2 × M 3 of G containing ΓE,2 (G) such that in M ∼  with Mi = U3 (8) for all i, and with the 3-components M1 , M2 , M3 of M permuted transitively by x. Replacing t by a suitable element of t, x − x, we may assume i . that t normalizes each Mi and induces a field automorphism on M Let zi , di  ∈ E32 (CMi (t)) with zi 3-central in Mi , and L3 (CMi (di )) =: Hi , i ∼ where H = L2 (8). We may choose these so that z1 z2 z3 = z and d1 d2 d3 = b1 ∈ K,  i = E(C  (b1 )). Recall that H = and zi , di | i = 1, 2, 3 is a 3-group. Thus H Mi 2H  3 . By Lemma 5.6,  is a diagonal of H 1H L3 (CK (b1 )), so H ≤ K ≤ M and H ∗     with b = b1 , Hb1 lies in M and its image in M is H1 H2 H3 , which is also the image of L := L3 (CHb∗ (d1 ))   L3 (CG (d1 , b1 )), by L3 -balance. Write L = L1 L2 L3 , 1 ∼ L2 (8), and Li ≤ Hi . Let M13 be the i = where Li is a 3-component of L, L subnormal closure of L3 in CG (d1 ). Thus by L3 -balance, M13 is a b1 -invariant product of 3-components of CG (d1 ) and L3 is a 3-component of L3 (CM13 (b1 )).   M3   Since M3 = L3 , M13 is a single 3-component covering M3 /O3 (M3 ). By [VK , 16.12] and Lagrange’s theorem, M13 /O3 (M13 ) ∼ = U3 (8). But then C(d1 , M13 ) contains d1 , z1 , d2 , z2  of 3-rank 4, contrary to Lemma 5.1. This contradiction completes the proof of the lemma.  Recall that b ∈ E −x, z is arbitrary and now Hb = E(CK (b)) and Hb∗ ≤ E(Cb ) is the subnormal closure of Hb in Cb . Corollary 5.9. We have Hb∗ = L3 (Cb ) = E(Cb ). Proof. Suppose that L3 (Cb ) = H2 Hb∗ with H2 a nontrivial product of components of CG (b). As z ∈ Hb ≤ Hb∗ , we have H2 ≤ L3 (CG (z)). Indeed, as the 3-components of CG (z) are as listed in Lemma 5.6a, each component of H2 is either a component L of CG (z) not containing z or a diagonal L of a product L∗ = L1 ∗ L2 ∗ L3 of components of CG (z) cycled by b. In the former case, if T normalizes L, then Z(L) ∩ Ω1 (Z(T )) = 1. But Ω1 (Z(T )) = x, z, and so [L, x] = 1, a contradiction. Hence, in either case, CG (z) contains a product L∗ of three 3components. Now, x normalizes each component of L∗ and, by consideration of the possible isomorphism types for these components, x induces an inner automor ∗ = L∗ /O3 (L∗ ). Then |C  (x)|3 ≥ 33 for each component phism on each. Let L L0 L0 by [VK , 16.25], whence |CL∗ (x)|3 ≥ 39 and so |Cx |3 ≥ 39 . But |Cx |3 ≤ 38 , a contradiction completing the proof.  Corollary 5.10. Let s ∈ I3 (CG (b, z)). Suppose that either Hb∗ ∼ = U3 (8) or s ∈ tE. Then O3 (CG (s)) = 1. In particular, if s ∈ t, E, then O3 (CG (s)) = 1. Proof. Suppose on the contrary that Xs := O3 (CG (s)) = 1. As b, z normalizes Xs and b, z − z ⊆ E − x, z, we may assume that either Xsb := CXs (b) = 1 or Xsz := CXs (z) = 1. In the former case, as O3 (CG (b)) = 1, [IA , 20.6] implies that Hb∗ is not locally balanced with respect to s. By Lemma 5.8 and Lemma 5.6b, and [VK , 7.11], Hb∗ ∼ = U3 (8). But then s ∈ tE by assumption. (Recall that the coset b z is totally fused in K.) Then s induces a field automorphism on H, and hence also on Hb∗ . But then O3 (CAut(H) (s)) = 1, by [IA , 4.9.1], again yielding a contradiction. Hence, we conclude that s, Xsz  normalizes some component J1

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7. THEOREM C∗ 6 : STAGE 1

of CG (z) with [J1 , Xsz ] = 1. But this is impossible in view of Lemma 5.6a and  [IA , 7.7.1bc], yielding a final contradiction. From now on, we restrict b by assuming b ∈ F − x = b1 , x − x . We will rule out all possibilities for Hb∗ , via a sequence of lemmas, thereby achieving a final contradiction. Three cases present particular difficulties. When Hb∗ ∼ = L2 (8) × L2 (8) × L2 (8), we are in the vicinity of G0 = U3 (8)  Σ3 ; and when Hb∗ ∼ = U6 (2) or Hb∗ ∼ = 3D4 (2), we are in the vicinity of G0 = Aut(2 E6 (2)). Of course G0 is not a simple group in either case, but this is not easy to see from the 3-local data that we have.  U6 (2). Lemma 5.11. We have Hb∗ ∼ = Proof. Suppose false. We make a few preliminary observations. First recall that by Lemma 5.6c, Q = x. Next, observe that CHb∗ (x) = Hb s with s inducing a field automorphism on Hb and s3 = 1 = s. Then s induces a field automorphism on K. Suppose that Cx /x covers O 2 (Aut(K)). Then there exists h1 ∈ T of order 9 with b1 = [h1 , s] and with [h1 , H] = 1. As CHb∗ x (H) = x, we must have h1 xi ∈ CG (Hb∗ ) for some i. But then, as s ∈ Hb∗ , [h1 , s] = 1, a contradiction. Changing notation we may assume that s = t and Cx contains x × Kt with index 1 or 2. Now let Qb = CCb (Hb∗ ). Then CQb (x) = b. Let Pb ∈ Syl3 (Hb∗ ) be such that Tb := (Qb × Pb )x ∈ Syl3 (Cb ) with z, t ≤ Tb and with z ∈ Z(Tb ). Recall that bz ∈ bK . Hence, we may choose w ∈ NG (Tb ) with bw = bz, z w = z, and w w2 w3 ∈ Tb . Then Qb ∩ Qw = bz 2 ∈ Qb , contrary to fact. As b  Tb . If b ∈ Qb , then b (x) = 1, whence Qb ∩ Qw CQb (x) = b, it follows that CQb ∩Qw b = 1. Suppose that b w Y := Qb ∩ Pb = 1. Now, CPb (x) = h, t is extraspecial of order 33 with center z. As Y  Tb , CY (x)  CPb (x) and so z ∈ CY (x). But z ∈ Qw b , whence Y = 1. Hence, w w [Qw , P ] = 1 and so Q ≤ Q × z. As Q ∩ Q = 1, we conclude that Qw b b b b b b b = bz and Qb = b. As O3 (Cb ) = 1, we conclude that Cb contains b × Hb∗ x ∼ = Z3 × P GU6 (2) as a subgroup of index at most 2. Again as z is 3-central in Hb∗ and hence in Cb , we may expand CT (b) to Tb ∈ Syl3 (CG (b, z)) ⊆ Syl3 (Cb ). Let B = J(Tb ), so that B∼ = E36 . Then CG (B) = B and ∼ Σ6 . Wb := NBH ∗ (B)/B = b

Let t0 ∈ NHb∗ (B) with τ := t0 B a reflection in Wb . Then [B, τ ] = c where CHb∗ (c) = c × Lb,c and Lb,c ∼ = U4 (2). We may choose t0 so that z ∈ Lb,c with CLb,c (z) ∼ = GU3 (2). We also choose d ∈ B − (B ∩ Hb∗ ) with Lb,c < Lb,d := CHb∗ (d) ∼ = U5 (2). Note that t, x ≤ CT (b) ≤ Tb ≤ BNHb∗ (B). Now, Ω1 (CT (b)) = x, b, z, t. As m3 (B ∩ Hb∗ ) = 4, we have m3 (CB∩Hb∗ (x)) ≥ 2. But m3 (CHb∗ (x)) = 2, so CB∩Hb∗ (x) = z, s for some s ∈ CT (b) inducing a field automorphism on K. Replacing t by s, we may assume that CB∩Hb∗ (x) = z, t. First, recall that bz = bw0 for some w0 ∈ CK (t, z). Since b, t, z ≤ B = J(Tb ) and Tb ∈ Syl3 (Cb ), it follows that bz = bw1 for some w1 ∈ CNG (B) (t, z) as well. ∗ Then O 2 (CG (bz)) = O 2 (Cb )w1 = bz × Hbz x. We wish to determine the pumpup Lc of Lb,c in CG (c). First of all CG (b, z, d) = b, d × S × J with S ∼ = GU3 (2). It follows that = SU (2, 2) ∼ = Σ3 and J = CLb,c (z) ∼

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CG (b, d, [J, J]) = b, d, S, z, and, replacing c by c−1 if necessary, with O3 (S) = cz. Note that, by Corollary 5.10, O3 (CG (v)) = 1 for all v ∈ B # . Let Ld be the pumpup of Lb,d in CG (d). Any allowable 3-component Kd of Ld must satisfy Kd ∼ = U5 (2) or U6 (2) or SU6 (2) (see [VK , 3.11]). Hence, Ld ∼ = U5 (2)×U5 (2)×U5 (2), U5 (2), U6 (2) or SU6 (2). Suppose that Ld = L1 × L2 × L3 with Li ∼ = U5 (2) and with the Li ’s cycled by b. We have t ∈ B ∩ Hb∗ = B ∩ Lb,d ≤ Ld . Now, bLd ⊇ b(B ∩ Ld ), and so bt ∈ bLd . Let Lbt = E(CG (bt)) ∼ = U6 (2) with = Hb∗ . Then x acts on Lbt ∼ CLbt (x) isomorphic to a subgroup of CG (x, bt), which contains x, bt × Jbt with Jbt ∼ = P GU3 (2) as a subgroup of index 1 or 2. However, U6 (2) does not admit any such automorphism [IA , 4.8.2, 4.8.4], yielding a contradiction. Hence, either Ld = Lb,d or Ld /Z(Ld ) ∼ = U6 (2). Suppose that Ld = Lb,d . Since B = CG (B) has 3-rank 6, Ld  CG (d). We ∗ x. We consider the embedding of c, d in the group O 2 (CG (bz)) = bz × Hbz ∗ (x) = H1 t with H1 ∼ have CHbz = L2 (8). Now, CG (bz, d) = CG (b, z, d), as Lb,d = Ld is a normal simple component of CG (d) containing z and centralized by b. Suppose then that CG (bz, d) = CG (b, z, d). Thus, S × J  CG (bz, d). It follows from consideration of the centralizers of 3-elements of P GU6 (2) that ∗ . In particular, as cz is inverted in CG (b, z, d), we must have cz ≤ d ∈ Hbz 3 ∗ ∗ ∗ . As z ∈ Hbz , it follows that c ∈ Hbz . By construction, O (CG (bz, d)) ≤ Hbz t0 ∈ CG (b, z) ≤ CG (bz) and t0 acts as a reflection on B inverting c. Therefore ∗ (c) = c × Lbz,c with Lbz,c ∼ CHbz = U4 (2) and with [J, J] = O 3 (Lbz,c ∩ Lb,c ). Let Lc be the pumpup of Lb,c in CG (c). Then Lc is a nontrivial pumpup of Lb,c . On the other hand, if Ld /Z(Ld ) ∼ = U6 (2) and CG (b, z, d) < CG (bz, d), then ∗ ∗ (d) =: Lbz,d ∼ with CHbz d induces an outer automorphism on Hbz = U5 (2) and with ∗ (c) = c × Lbz,c with Lbz,c ∼ S × [J, J] < Lbz,d . Then CLbz,d (c) = CHbz = U4 (2) and [J, J] < Lbz,c . As z ∈ [J, J], b does not centralize Lbz,c . As before, we conclude that the pumpup Lc of Lb,c in CG (c) is nontrivial. Thus, in any case, Lc is a nontrivial pumpup of Lb,c , and Lc is either a diagonal pumpup or Lc ∈ C3 ∪ TG3 . It follows in the latter case, by [VK , 16.7], that Lc /Z(Lc ) ∼ = U5 (2), D4 (2), D5 (2), Sp(8, 2), Co2 , or P Sp4 (33 ). = U6 (2) or that Lc ∼ Suppose that Lc = J1 × J2 × J3 with Ji ∼ = U4 (2). As b, d normalizes Lc and centralizes Lb,c , some d1 ∈ b, d# centralizes Lc . Then d1 ∈ b and we must have E(CG (d1 )) ∼ = U5 (2) × U5 (2) × U5 (2). But this yields a contradiction as in the fifth paragraph before this one. Hence, Lc is a vertical pumpup and, again by [VK , 16.7], some d1 ∈ b, d − b centralizes Lc , or Lc /Z(Lc ) ∼ = U6 (2). In the first case, Lc ≤ Ld1 , where Ld1 /Z(Ld1 ) ∼ = U5 (2) or U6 (2). Given the listed possibilities, the only ones occurring as subgroups of (S)U6 (2) are U5 (2) and (S)U6 (2). Hence in any case, Lc /Z(Lc ) ∼ = U5 (2) or U6 (2). As t0 inverts c and centralizes b Lb,c , we conclude that t0 centralizes Lc . Set W = AutG (B) ∼ = NG (B)/B. Then, CW (τ ) has a subnormal subgroup Σ isomorphic to Σ5 or Σ6 , and Wb  CW (b). Let B1 be the socle of B as W -module. As bz ∈ bW , CB (W ) = 1. As the socle of B as Wb -module is b⊕(B ∩Hb∗ ), it follows that B ∩Hb∗ ≤ B1 . Suppose that B ∩Hb∗ is W -invariant. Then Σ acts faithfully on CB∩Hb∗ (τ ). But m3 (CB∩Hb∗ (τ )) = 3, a

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334

7. THEOREM C∗ 6 : STAGE 1

contradiction. Hence B1 is irreducible with dim(B1 ) ≥ 5. Moreover, as B has no 1-dimensional W -invariant subspace, B is indecomposable as W -module. Now, CW ([Wb , Wb ]) < NW (b), whence CW ([Wb , Wb ]) is a 2-group. As W embeds in GL6 (3), it follows immediately that O2 (W ) =: X is a 3-group and CB (X) contains B1 . Suppose X = 1. Then CB (X) = B1 < B. As dim(B1 ) = 5, b, d∩B1 = 1. Let d0 ∈ b, d∩B1# . Then X ≤ O3 (W ∩CG (d0 )), so d0 ∈ b. As we argued above for d, we have either U5 (2) ∼ = E(CHb∗ (d))  CG (d0 ) or Ld0  CG (d0 ) ∼ with Ld0 /Z(Ld0 ) = U6 (2) and with b inducing an outer diagonal automorphism on Ld0 . Let Σ0 = W ∩ Ld0  CW (d0 ). Then Σ0 ∼ = Σ5 or Σ6 and [X, Σ0 ] ≤ X ∩ Σ0 = 1. Thus |X| = 3. Hence, |O2 (W )| ≤ 3. Letting A = [Σ, Σ], it follows by L2 -balance that A ≤ E(W ), and either A is diagonal in a product L ∗ Lτ of components of W with L/Z(L) ∼ = A, or A ≤ E for some component E of W normalized by τ with A a component of CE (τ ). As W embeds in GL6 (3) and |GL6 (3)|5 = 5, the diagonal case is impossible. Moreover, as Wb ∼ = Σ6 and CWb (τ ) ∼ = Σ4 × Z2 , it follows that A is not subnormal in W , whence E is a vertical pumpup of A. We invoke [VK , 5.1]. In the hypothesis (c) of that lemma, we take D = b, d. # As noted before, for any u ∈ b, d , E(CG (u)) is a central extension of U5 (2) or U6 (2), so CW (u) has no normal subgroup isomorphic to W (D4 ) or W (D5 ). Now [VK , 5.1] yields that NG (B)/B ∼ = Σ7 or Σ7 × Z2 . In particular, Tb ∈ Syl3 (NG (B)) and, as B = J(Tb ), it follows that Tb ∈ Syl3 (G) and NG (B) controls G-fusion in B. But, bz ∈ bW and CΣ (z) ∼ = Σ3 Z2 . So |bW | ≥ 11, whence |W | ≥ 11 · 6! > 7!, a contradiction, completing the proof of the lemma.  We now rule out the possibility that H pumps up diagonally in CG (b) for some b ∈ F #. Lemma 5.12. The pumpup Hb∗ of H in CG (b) is vertical for all b ∈ F − x, and hence for all b ∈ E − x, z. Proof. Suppose false for some b ∈ F − x. Then by Lemma 5.6b, Hb∗ = J1 × J2 × J3 with Ji ∼ = L2 (8), and with the Ji transitively permuted by x. Take zi ∈ Ji with z = z1 z2 z3 . Let V = b, z1 , z2 , z3 . We have CJi (zi ) =: hi  with h3i = zi . Also, as t induces a field automorphism on H, we see that replacing t by a suitable element of t x, we may assume that t induces field automorphisms on Ji , [hi , t] = zi , and [zi , t] = 1, for all i. Now, we have w ∈ CK (z, t) with bw = bz. Let Hbz = E(CK (bz)) = H w . Then z ∈ H ∩ Hbz and |H ∩ Hbz | ≤ 32 . Also, the pumpup of Hbz in CG (bz) is ∗ := J01 × J02 × J03 with J0i ∼ Hbz = L2 (8) and Hbz diagonally embedded in the product. We consider the embedding of V in CG (bz). As [w, t] = 1, we see that ∗ t induces a field automorphism on each J0i . Moreover, as hi , x normalizes Hbz , ni ni there is some integer ni such that hi x normalizes each J0j . Hence, zi = [hi x , t] ∈ ∗ ∗ ) × Hbz for all i. Now, CG (x, bz, Hbz ) = h0 , x, with either [CG (bz), t] ≤ CG (Hbz 3 ∗ bz) (x) ≤ h0 . Now, h0 = bz or h0 = bz. Hence, h0 may be chosen so that CCG (Hbz ∗ [V, x] = z1−1 z2 , z2−1 z3  and [V, x, x] = z < Hbz . In particular, if e ∈ [V, x] − z ∗ ∗ ) and e0 ∈ Hbz , then z ±1 = [e, x] = [c, x][e0 , x], and e = ce0 with c ∈ CG (Hbz 3 whence [c, x] = 1. Thus, c ∈ h0 . Since c = 1, we have c ∈ bz. Hence, as [V, x] is elementary abelian and contains z, [V, x] ≤ bz, z01 , z02 , z03  with z0i ∈ J0i and z = z01 z02 z03 . Thus, [V, x] ∩ bz, z01 , z02  = 1. As z ∈ bz, z01 , z02 , we see that zi−1 zj ∈ bz, z01 , z02  for some i = j. Renumbering the J0i ’s, we may assume that

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z1−1 z2 ∈ bz, z01 , z02 . Hence by L3 -balance, E(CG (z1−1 z2 )) ≥ J3 , J03 . (Notice that if v ∈ b, z1 , z2 , z3  ∪ bz, z01 , z02 , z03 , then [v, b, z] = 1, and we have O3 (CG (v)) = 1 by Corollary 5.10.) We now argue that it will suffice to show that, for some v ∈ b, zi  − b or v ∈ bz, z0i  − bz, the pumpup of some Jj or J0j , j = i, in CG (v) is not trivial. Without loss, we assume that this holds for v ∈ b, z3  − b and for J1 . Let Ki be b b2 the pumpup of Ji in CG (v) for i = 1, 2. If K1 = J11 ×J11 ×J11 , then as J2 commutes with J1 and normalizes K1 , J2 commutes with K1 , whence K2 = J2K2  commutes with K1 . If, on the other hand, K1 is a vertical pumpup of J1 and [K1 , J2 ] = 1, then J1 ×J2  E(CK1 (b)). However, noting that [b, t] = 1 and J1 t ∼ = P ΣL2 (8), we see that E(CK1 (b)) = J1 , by [VK , 16.12]. Again, we conclude that [K1 , K2 ] = 1. Hence, K1 ∗ K2  E(CG (v)). Conjugating by x, we get K2 ∗ K3  E(CG (v x )) 2 and K3 ∗ K1  E(CG (v x )). It follows that NG (K2 ) ≥ K1 ∗ K2 ∗ K3 , and this group is normalized by x, with K1x = K2 , K2x = K3 and K3x = K1 . Hence, CK1 K2 K3 (x) ≥ K0 with K0 isomorphic to a quasisimple central quotient of K1 containing H. Then H ≤ K0 ≤ K, and so either K0 = H or K0 = K. If K1 is a non-trivial pumpup of J1 , then K0 = K and K1 ∼ = U3 (8) with K1 a component of CG (v) and m3 (CG (v, K1 )) ≥ 2, a contradiction to Lemma 5.1. We now return to our consideration of E12 := E(CG (z1−1 z2 )) ≥ J3 , J03 . Suppose first that J3∗ is a non-trivial pumpup of J3 in E12 . As b, z1  centralizes J3 , b, z1  normalizes J3∗ . By [VK , 16.12], CAut(J3∗ ) (J3 ) has cyclic Sylow 3-subgroups, so some v ∈ b, z1  − b centralizes J3∗ , contrary to the previous paragraph. Hence, J3 is a component of E12 . If [J3 , J03 ] = 1, then since J3 = J03 , the (diagonal) pumpup of J03 in CG (z1−1 z2 ) is J3 × J4 × J5 with bz permuting these three components. But bz normalizes J3 , a contradiction. Hence [J3 , J03 ] = 1. In particular, [J3 , z03 ] = 1. Now consider C03 := CG (z03 ). By the previous argument with v = z03 , both J01 and J02 are components of C03 . But also, as [b, z03 ] = 1 = [J3 , J03 ], J3 ≤ E(C03 ), and so, J3 normalizes both J01 and J02 . As z ∈ z03  × J01 × J02 and [z, J3 ] = 1, we have [J3 , J01 × J02 ] = 1. Without loss, we may assume that [J3 , J01 ] = 1, and b b2 so, the pumpup of J3 in C03 is J01 × J01 × J01 . But, as before, from CG (bz), we see that b normalizes J01 , yielding a final contradiction, and completing the proof.  Lemma 5.13. Suppose that b ∈ F − x and u ∈ I3 (CG (F )) with Hb∗ ∼ = 3D4 (2) ∼ and J := E(CHb∗ (u)) = U3 (3) and u inducing a nontrivial field automorphism on K. Let Y be the subnormal closure of J in CG (u). Then Y ∼ = 3D4 (2). Proof. Now, x induces an inner automorphism on Hb∗ with CHb∗ (x) = b2 ×H for some b2 ∈ xb. Set I = O 3 (CK (u)) ∼ = P SU3 (2). Note that CG (x, u) is solvable and F = b, x. Now, as Q = x by Lemma 5.1, CG (x, b) contains F × H u with index dividing 6, and CG (F, u) = F, u × z, s with z, s ∼ = L2 (2). In particular, as J = O 2 (CHb∗ (u)) and b2 ∈ CHb∗ (x, u), we have b2 , z ≤ CJ (x). Now consider Y . If Y = J, then as z ∈ J and z is contained in every non-trivial normal subgroup of I, I acts faithfully on Y . But Aut(U3 (3)) has no subgroup isomorphic to I, a contradiction. Suppose Y is a diagonal pumpup of J. As CG (x, u)

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is solvable, it follows that x normalizes each component of Y , inducing inner automorphisms, whence m3 (CG (x, u)) ≥ 7. But m3 (Cx ) = 4, a contradiction. Hence, Y is a vertical pumpup of J containing z and normalized by F . In particular, Y ∼ = G2 (8), U3 (33 ), D4 (2) or 3D4 (2) by [VK , 3.11]. As CG (x, u) is solvable, Y ∼ = U3 (33 ), then again as CG (x, u) is = G2 (8) [IA , 4.7.3A, 4.9.1]. If Y ∼ solvable, x induces an inner automorphism on Y . But then m3 (CG (x, u)) > 4, a contradiction as before. Thus Y ∼ = D4 (2) or 3D4 (2). We have b2 ∈ J < Y . By the structure of J and I, both x and b induce outer automorphisms on Y . As b2 ∈ Y , we see that I ≤ [CG (x, u), b2 ] ≤ Y with CI (b2 ) = z, s ∼ = Σ3 . Thus, Ib2  ≤ Y and CY (x) = Ib2 . Suppose finally that Y ∼ = D4 (2). Let P0 = O3 (CY (x)) b2  ∈ Syl3 (CY (x)) and let P0 ≤ R0 , for R0 an x-invariant Sylow 3-subgroup of Y . As z = [P0 , P0 ] ≤ [CR0 (z), CR0 (z)], we see that dim(CV (z)) = 2 for V a standard module for Y . Now, CJ (z) =: P1 Z1 with Z1 ∼ = Z4 and P1 = [P1 , Z1 ] ∼ = 31+2 . In particular, as b2 ∈ CJ (z), b2 ∈ O 3 (CJ (z)) ≤ O 3 (CY (z)) = P1 Q1 ∼ = O 3 (CΩ([V,z]) (z)), ∼ SU3 (2). In particular, b2 acts trivially on CV (z). As b2 ∈ [R1 , R1 ] with P1 Q1 = for a suitable Sylow 3-subgroup R1 of Ω([V, z]), we get dim([V, b2 ]) = 4. It follows that CY (b2 ) ∼ = L2 (2) × L2 (2) × L2 (2) × Z3 . As CG (u, b2 , x) = u, b2 , x × z, s, we see that CY x (b2 ) = Y2 × b2  with Y2 ∼ = SL2 (2)  Z3 and with O3 (Y2 ) the unique minimal normal subgroup of Y2 . By Lemmas 5.6b, 5.11, and 5.12, and Corollary 5.9, the possibilities for Hb∗2 = 3 E(CG (b2 )) are: U3 (8), 3D4 (2), 2 G2 (3 2 ), or Co3 . As Y2 acts faithfully on Hb∗2 , the 3 cases U3 (8), 3D4 (2) and 2 G2 (3 2 ) are immediately ruled out. If Hb∗2 ∼ = Co3 , then 3 ∼ ∗ CHb (u) has no normal subgroup isomorphic to O (Y2 ) = Σ3 × Σ3 × Σ3 [IA , 5.3j], 2  a contradiction. Thus Y ∼  D4 (2) and tthe lemma is proved. = = 3D4 (2). Lemma 5.14. We have Hb∗ ∼ Proof. Suppose that Hb∗ ∼ = 3D4 (2). As t induces a field automorphism on H, t induces an outer automorphism on Hb∗ . As in the previous lemma, x induces an inner automorphism on Hb∗ with CHb∗ (x) = b2  × H for some b2 ∈ xb. By [VK , 3.69], we may replace t and assume that CHb∗ (t) ∼ = G2 (2). The previous lemma then applies, with t in the role of u. Set J = O 2 (CHb∗ (t)) ∼ = U3 (3), I = O 3 (CK (t)) ∼ = P SU3 (2), and Y = the pumpup of J in CG (t), so that Y ∼ = 3D4 (2). Again note that CG (x, t) is solvable. We claim that C2 := CY (b2 ) = b2  × J0 with J0 ∼ = L2 (8). Suppose not. Then C2 / b2  ∼ = P GU3 (2), and F ∗ (C2 ) ∼ = 31+2 . As J ≤ Y and O3 (CJ (z)) is nonabelian, it follows that b2 ∈ z Y . But then by Lemma 5.6a, Hb∗2 ∼ = 3U4 (3), (3 × 3)U4 (3), 3Ω7 (3), 3G2 (3), or 3J3 , contradicting Lemma 5.6b. Hence C2 = b2  × J0 , as that b2 = x. By [VK , 16.18], b2 is claimed. As CG (x, t) is solvable,  it follows are Hb∗ -conjugate, and one of these is x. inverted in Hb∗ , whence bb2  and bb−1 2 Thus each subgroup of F of order 3 is conjugate to b, x or b2 . We consider the possibilities for Hb∗2 = L3 (CG (b2 )). Once again, the possibilities for Hb∗2 are: 3 U3 (8), 3D4 (2), 2 G2 (3 2 ), or Co3 .

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∼ 2 G2 (3 32 ). Let v ∈ I2 (H ∗ ) with NH ∗ (b2 ) = b2 , v × Suppose first that Hb∗2 = b b ∗ H, where bv2 = b−1 2 . As [v, b × H] = 1, it follows that [v, Hb2 ] = 1, by [IA , 7.1.4]. Now, b induces an outer automorphism on Hb∗2 . Write CHb∗ (b) = H t1 , with 2

⊆ b2  Hb∗2 . As t1 ∈ b2 , b, t. Choose t0 ∈ b2 , t1  ∩ b, t# . Then t0 ∈ b2  t±1 1 ±1 t0 ∈ b t , J = L3 (CG (b, t0 )). Let Y0 be the pumpup of J in CG (t0 ). Then Lemma 5.13 applies with t0 in the role of u, and yields Y0 ∼ = 3D4 (2). On the other hand, recall that b2 , z ≤ J. Now by [VK , 16.26], O 2 (Cb2 ×Hb∗

2

b (t0 ))

≥ b2  × (A1 × A2 ) b ,

where A1 ∼ = A2 b ∼ = Z3  Z3 . = A2 ∼ = E33 with t0 ∈ A1 and z ∈ A2 and A1 b ∼ Since z ∈ J ≤ Y0 , it follows that A2 b acts faithfully on Y0 . But this contradicts the structure of Aut(Y0 ) ∼ = Aut(3D4 (2)). Next, suppose that Hb∗2 ∼ = U3 (8) or 3D4 (2). Then x is inner on Hb∗2 with ∗ CHb∗ (x) = (F ∩ Hb2 ) × H. But then as t induces a field automorphism on H, we 2 see that t is outer on Hb∗2 . But J0 is a component of E(CHb∗ (t)) with J0 ∼ = L2 (8), 2 a contradiction. Finally, suppose that Hb∗2 ∼ = Co3 . Let b3  = b, x ∩ Hb∗2 . Then b3 is inverted in ∗ Hb2 , whence b2 b3 is G-conjugate to b2 b−1 3 . It follows that b3 = b. In Cb , b2 is inverted by an involution s2 ∈ CHb∗ (H). Therefore s2 normalizes Hb∗2 and centralizes b×H, so [Hb∗2 , s2 ] = 1. Also t induces a field automorphism on H and centralizes b2 , so there is t1 ∈ I3 (CHb∗ (b)) inducing a field automorphism on H such that t ∈ t1 b2 . 2 Let Q2 ∈ Syl3 (C(b2 , Hb∗2 )). As x ∈ b2 , b, Q2 ≤ CG (x H). So Q2 is cyclic of order at most 9 and Q2 = CQ2 (t1 ) = CQ2 (t) = b2 , whence NG (b2 ) = Hb∗2 × b2 , s2 . Choose A ∼ = E36 with A ≤ CG (b2 ) and z ∈ A. Now, CG (A) = A. Let W = AutG (A) ∼ = NG (A)/A. We have NW (b2 ) ∼ = Z2 × Z2 × M11 . For any A ≤ S ∈ Syl3 (CG (b2 )), we have J(S) = A, so W controls G-fusion of b2 in A, by [V2 , 3.1]. We now calculate | b2 W |. By [VK , 16.28], NW (b2 ) has two orbits on E1 (A ∩ Hb∗2 ), one having representative z and size 55, and the other having size 66. Call a representative of the second orbit d. As b2 ∼ b−1 in CG (Hb∗2 ), we 2 −1 −1 have b2 z ∼ b2 z and b2 d ∼ b2 d. Also, b2 ∼K b2 z. Hence there are at most four W -orbits on E1 (A): (a) (b) (c) (d)

dW ∩L , of size 66; z W ∩L , of size 55; b2 dNW (b2 ) , of size 132; and {b2 } ∪ b2 z NW (b2 ) , of size 111.

Now b2 is not G-conjugate to z, and so | b2 W | = 111 or 177 or 243 or 309. As ∗ ∗ | b2 W | divides |GL6 (3)|, | b2 W | = 35 and bW 2 = A − Hb2 , and A1 := A ∩ Hb2  NG (A). Let s2 be the image of s2 in W , and write W0 = AutHb∗ (A) = s0  × M , 2 where s0 ∈ I2 (Z ∗ (NHb∗ (A))) and M ∼ = M11 . Then CW (s0 ) = s2  × W0 so M ≤ 2 L2 (W ) by L2 -balance. Since |W : W0 | = 2 · 35 it follows by [VK , 3.8] that M covers E(W/O2 (W )). As M ∼ = Aut(M ), we must have O2 (W ) = O3 (W ) ∼ = E35 centralizing A1 and shearing b2 to b2 A1 . Let R0 be the preimage of O3 (W ) in NG (A), and R = [R0 , s0 ]. As s0 centralizes b2 , |R| = 310 and s0 inverts R. By the irreducibility of M on R/A1 and A1 , R ∼ = E310 or Z95 . In the first case, m3 (CRx (x)) ≥ 5, contradicting m3 (Cx ) = 4. In the second case, since s2 inverts

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b2 and centralizes A1 , s2 must centralize R. But then b2 = [b2 , s2 ] centralizes R, a contradiction. This completes the proof of the lemma.  We have now reached the point where the only possibilities for Hb∗ are 2 G2 (3 2 ), Co3 , and U3 (8). We can now rule out the first two of these cases. 3

3 = 2 G2 (3 2 ) or Co3 . Lemma 5.15. We have that Hb∗ ∼

Proof. Suppose not and pick an involution s ∈ CHb∗ (x) with CHb∗ (s) = s×Xs and Xs ∼ = L2 (27) or M12 [IA , 4.5.1, 5.3j]. As Cx /x embeds in Aut(U3 (8)), it follows that if U = O2 (CG (s)) and Ux = O2 (CG (x, s)), then Ux is a Sylow 2subgroup of K with Zx := Z(Ux ) ∈ Syl2 (H). Thus Zx ≤ H ∗ , and we see that Zx = s×(Zx ∩Xs ). Hence, U ∩Zx = s, and so s = Ω1 (CU (x)) and |CU (x)| ≤ 8. A similar argument applies for any v ∈ b, x# such that L3 (CG (v)) ∼ = U3 (8). 3 Moreover, if v ∈ x, b with L3 (CG (v)) ∼ = 2 G2 (3 2 ) or Co3 , then CU (v) = s. In particular, CU (b) = s. By Wielandt’s formula, at worst, |U | ≤ |CU (b)||CU (x)|3 /23 ≤ 210 /23 = 27 . Since a non-trivial F2 [Xs ]-module has dimension at least 10, it follows that [Xs , U ] = 1. As G is of even type and Cx is a 5 -group, it follows from [V2 , 7.1] that Xs ≤ E(CG (s)), whence the subnormal closure Ys of Xs in E(CG (s)) is a 3-pumpup of Xs which is a C2 -group. Therefore, by [VK , 3.8], either Ys = Xs ∼ = M12 or Ys ∼ = M12 × M12 × M12 with the components permuted transitively by b. Since x normalizes Xs , it normalizes Ys . If Ys has three components then either m3 (CYs (x)) ≥ 6 or CYs (x) contains M12 , both of which are impossible by the structure of Cx . Therefore Ys = Xs . Then Ux ≤ O3 (CCG (s) (x)) normalizes Xs by L∗3 -balance, and, as Ux = [Ux , b] and [Xs , b] = 1, we have [Ux , Xs ] = 1. But |Zx ∩ Xs | = 4 and [Zx ∩ Xs , Xs ] = 1, a final contradiction.  We rule out the final possibility in the next lemma. This case was treated by Aschbacher in [A24], under the additional hypothesis that G is of characteristic 2-type, by showing that the subgroup Z (see below) is a weakly closed abelian TIsubgroup of G, and then invoking results of S. D. Smith, Stroth and Timmesfeld. His argument inspired us to try to prove that the subgroup Q (see below) is a 2-group. Lemma 5.16. Let {xi  : 1 ≤ i ≤ 4} be the subgroups of F = x, b of order 3, and let Ki be the pumpup of H in CG (xi ) for i = 1, 2, 3, 4. Then for some i, Ki ∼  U3 (8). = ∼ Σ3 . As H ≤ Ki for all i and t Proof. Suppose not. Let CH (t) = z, s = induces a nontrivial field automorphism on H, t induces a nontrivial field automorphism on Ki for all i. Then for some j = i, CKi (t) = wi , zQi xj  ∼ = P GU3 (2) with wi , z ∼ = Q8 , and s ∈ Z(Qi ). Of course, CKi (t)  CCG (t) (xi ) for all = E32 , Qi ∼ i. Indeed, F ≤ xi  × Ki , and Ci := C(xi , Ki ) is cyclic of order 3 or 9 for all i, by Lemma 5.6c and symmetry among the xi , 1 ≤ i ≤ 4. Hence, CG (xi , t) contains CCi (t) × t × CKi (t) with index 1 or 2. In particular, CCG (t) (xi ) is solvable and O3 (CCG (t) (xi )) is abelian for all i. Let A = x, b, t and let Q0 := Qi | 1 ≤ i ≤ 4. Then Q0 is an A-invariant subgroup of CG (s) ∩ CG (t). Let Ri = O2 (CKi (s)). Then Ri is special of order 23+6 with Z(Ri ) = [Ri , Ri ] = Ω1 (Ri ) =: Z = CH (s). Thus Z  CG (xi , s) for all i.

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5. THEOREM 2 COMPLETED: THE U3 (8) CASE

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For each i, Ci ∼ = Z3 or Z9 acts faithfully on Kj for j = i, centralizing H. Then Ci (s) := CG (xi , s) has a 2-closed normal subgroup Ri Ti of index at most 2, where Ti is a 3-group of order at most 35 . Recall that F = x, b. Then F ≤ Z(Ti ) and so we may assume that Ti = Tj =: T0 ∈ Syl3 (Ci (s)) for all i. We have A = Ω1 (T0 ). First we shall study the embedding of Q0 A in Cs := CG (s). Let U = O2 (Cs ) and Cs = Cs /U . Our goal is to prove that Q0 ≤ U.

(5F)

By the structure of Ci (s), Ci (s) ∩ U ≤ Ri . If Z ∩ U = s, then Ω1 (Ri ∩ U ) = s, whence |Ri ∩ U | ≤ 23 and |U/s| ≤ 28 . Otherwise, as t ∈ CG (s), Z ≤ U and |U/Z| ≤ 224 . As Ci (s) is a {2, 3}-group for all i, we see that Op (C s ) = 1 for all p = 3 and C s n has no 2B2 (2 2 ) component for any n. Therefore O3 (C s ) = 1 so O3 (Cs ) = U . If [E(C s ), Q0 ] = 1, then Q0 lies in the 3-constrained group CC s (E(C s ))A. Then by the Thompson-Bender Lemma, as A contains every element of order 3 in CCs (A), Qi ≤ O3 (Cs ) = U for all i, whence Q0 ≤ U , as desired. Hence, in proving (5F), we may assume [E(C s ), Q0 ] = 1. Let L be a minimal Q0 A-invariant product of isomorphic components of Cs , such that [L, Q0 ] = 1. Since CCs (xi ) is a {2, 3}-group for all i, F normalizes and acts faithfully on L0 for all components L0 of L. As Q0 = [Q0 , F ], F Q0 normalizes each L0 . If Z(L0 ) = 1 for some L0 , then   A A∩Z L0 = 1, and so L0 is Q0 A-invariant and L = L0 . Hence if L = L0 , L0 is simple and m3 (CL (F ))) ≥ 3. But then A ≤ L so L = L0 . Hence, L is a single component of Cs . Moreover, F acts faithfully on L and for at least one i, CQi (L) = s. In particular, Qi ∩ U = s for such a choice of i. Indeed, Qi ≤ O3 (CCs (xi )), so L is not locally balanced with respect to xi . Fix this i through the proof of (5F). Suppose that L ∈ Chev(3). 3 ∼ By [IA , 7.7.1], L = L2 (3 ) and xi induces a field automorphism on L. But then for some j = i, xj is inner on L, whence CLxi  (xj ) is a non-abelian 3-group generated by elements of order 3, contradicting A = Ω1 (T0 ). Hence L ∈ Chev(3). Suppose that L ∈ Chev(r), r = 3. 

Suppose first that L = Γ := ΓrF ,1 (L). Then for some xj , CL (xj ) has a Lie component Lj in Chev(r). As CG (xj , s) is solvable, we must have that r = 2. On the other hand, if L = Γ, then the possibilities for L are enumerated in [IA , 7.3.3]. Thus in any case, either r = 2 or L0 ∼ = L3 (q) with q ≡  (mod 3). As Qi ∼ = Qi /s and Qi acts faithfully on L, we have that O2 (CL (xi )) = 1, whence L ∈ Chev(2), by [IA , 4.2.2]. Hence, if L ∈ Chev(r), then L ∼ = L3 (q) with q ≡  (mod 6). As L ∈ C2 , L must act non-trivially on U . Now Sylow r-subgroups of L are special of type q 1+2 and act faithfully on U . It follows that |U/s| ≥ 2rdn , where d is minimal such that r divides 2d − 1, and q = r n . Since |U/s| ≤ 226 and r ≥ 5 and d ≥ 3,

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340

7. THEOREM C∗ 6 : STAGE 1

we conclude that r ∈ {5, 7} and n = 1. But then as A = Ω1 (T0 ) is abelian, F is inner on L and so O2 (CL (xj )) = Qj for all j. Thus, |Rj ∩ U | ≤ 27 for all j, and so |U/s| ≤ 219 , whereas rd ≥ 20, a contradiction. So, L ∈ Alt ∪ Spor. Suppose that L ∈ Alt. ∼ As m3 (CCs (xi )) = 3, L = An for some n ≤ 11. As Qi ∼ = Qi /s acts faithfully on L, xi fixes exactly 4 points in a standard permutation action of L. So n = 3m + 4 ∈ {7, 10}. Suppose that L ∼ = A10 . Then O 2 (C s ) = L × O 2 (CC s (L)) with A ≤ L. Hence, CC s (L) = 1 and A = T0 . By the structure of centralizers of elements ∼ of I3 (A10 ), xj ∈ xL i for all j. But A10 contains no E32 -subgroup F = E32 with every nonidentity element having exactly four fixed points, contradiction. Thus, L ∼ = A7 . Again, O 2 (C s ) = L × O 2 (CC s (L)). As F projects isomorphically into L, it follows that T0 = A and F ∗ (CC s (L)) = t0  for some t0 ∈ A − F . Without loss, we may assume that Qj = O2 (CL (xj )) for j = 1, 2, while Qk = 1 for k = 3, 4. Then L = Q1 , Q2  ≤ Q0 . Now consider Qt0 := CU (t0 ) = CQt0 (xi ) : 1 ≤ i ≤ 4. Then CQt0 (xi ) = s for i = 1, 2, while CQt0 (xk ) = Qk for k = 3, 4. It follows by Wielandt’s formula that |Qt0 | = 25 . Also, if V is an A-invariant abelian subgroup of Qt0 , then CV (xi ) = s for all i, whence V = s. Hence Qt0 is extraspecial of order 25 . Now, Qt0  CCs (t0 ) and CCs (t0 ) covers L. As Aut(Qt0 ) does not involve A7 , CCs (t0 , Qt0 ) covers L. But F acts faithfully on Qt0 , a contradiction. Finally, suppose that L ∈ Spor. ∼ By [IA , 7.7.1] L = M12 , J2 , or M22 . In the first two cases, some xi induces a 3central element on L, and so CG (xi , s) contains an extraspecial subgroup of order 33 and exponent 3, impossible as A = Ω1 (T0 ). Hence, L ∼ = M22 . As in the A7 case, we conclude that O 2 (C s ) = t0  × L with t0 ∈ A − F . Thus T0 = A. As A ∩ L is inverted in L, we see that F ≤ L and, for each xj , CL (xj ) = xj ×Qj xk  ∼ = Z3 ×A4 , j = k. Hence, Q0 = L ∼ = M22 . As t0 centralizes each Qi , CQ0 (t0 )/CQ0 ∩U (t0 ) ∼ = M22 . Again, setting Qt0 = CU (t0 ), we see this time that CQt0 (xj ) = s for all j, whence Qt0 = s. Hence, the pullback of L in CCs (t0 ) is L ∼ = 2M22 . But in 2M22 [IA , 5.3c], the centralizer of a 3-element is isomorphic to Z3 × Z2 × A4 , not Z3 × SL2 (3), a contradiction. This final contradiction proves (5F), which implies that Q0 = CU (t). The Wielandt formula shows that |Q0 | = 29 . Moreover, if V is a characteristic abelian subgroup of Q0 containing s, then CV (xj ) = s for all j, whence V = s and Q0 is extraspecial. It follows that Q0 = Q1 ∗ Q2 ∗ Q3 ∗ Q4 ∼ = 21+8 . +

We now study CG (t). Our goal is to prove that Q0 acts faithfully on Ut := O3 (CG (t)). For then Ut must have a chief CG (t)-factor V of dimension ≥ 16 on which Q0 acts faithfully. But then |CV (x)| ≥ 36 , so |CUt (x)| ≥ 36 by [IG , 9.16], whereas |CUt (x)| ≤ |O3 (CCx (t))| ≤ 35 , a final contradiction. We therefore assume that [Ut , s] = 1 and argue to a contradiction. Since O3 (CG (t)) = 1 by Corollary 5.10, F ∗ (CG (t)) = Ut Et , with Et = E(CG (t)). As [Ut , s] = 1, s acts non-trivially on Et . Since CG (xi , t) is solvable for all i, F

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5. THEOREM 2 COMPLETED: THE U3 (8) CASE

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is in the kernel of the permutation action of CG (t) on the components of Et . As Q0 = [Q0 , F ], Q0 F normalizes each component of Et . Hence, there is a component J of Et on which Q0 F acts faithfully. Let I be any component of Et . If I ∼ = D4 (q) for some odd q, then some xi induces an inner or field automorphism on I, whence m3 (CIxi  (xi )) ≥ 5, contradicting m3 (CG (xi )) = 4. Hence I does not have this form, and in particular Out(I) is 2-nilpotent, by [III11 , 2.1a]. As Q0 = [Q0 , F ], Q0 induces inner automorphisms  0 of Q0 on J/Z(J) is isomorphic to on Et . As O3 (CG (t)) = 1, the projection Q 1+8 ∼ Q0 = 2+ and invariant under CJ/Z(J) (s). If J ∈ Chev(r), r odd, it follows from [IA , 4.5.1] that J ∼ = D4 (r n ), contradicting  the previous paragraph. If J ∈ Alt ∪ Spor, then as Q0  CJ/Z(J) (s), we see from [IA , 5.2.8, 5.3] that J/Z(J) ∼ = Co1 , Co2 , F i22 , F i23 , F i24 , F5 , F3 , F2 , or F1 . On the other hand |CJ (xi )| divides 24 35 , which rules all the groups out by inspection of [IA , Tables 5.3]. Hence, J ∈ Chev(2). Suppose J has a cyclic Sylow 3-subgroup. Since F acts faithfully on J, some xi induces a field automorphism on J. As CJ (xi ) is a {2, 3}group, J ∼ = L2 (8). But Q0 acts faithfully on J, a contradiction. Hence, if J ∈ T3 , then J ∼ = L3 (2n ), n > 1, 2n ≡  (mod 3). If n > 3, then by [VK , 16.16], CJ (x) is not a solvable {2, 3}-group, contradiction. And if n = 2, then Q0 cannot act faithfully on J, contradiction. As J ∈ C3 ∪ T3 ∪ TG3 , it follows that J is isomorphic to one of the following groups: U3 (8), U5 (2), U6 (2), SU6 (2), Sp6 (2), D4 (2), 3D4 (2), n n F4 (2), 2 F4 (2) , Sp8 (2), 2 D4 (2), D5 (2), Sp4 (2n ), or L− m (2 ), m ∈ {4, 5}, 2 ≡  (mod 3). Suppose first that CJ (xi ) has a normal subgroup isomorphic to P SU3 (2) for some i. Then J ∼ = U3 (8), D4 (2), or 3D4 (2), by [VK , 16.17]. Moreover, xj induces an inner automorphism on J for some j = i. As Qj acts faithfully on J, CJ (xj ) has a normal subgroup isomorphic either to P SU3 (2) or to Q8 . However, this is not the case for any element of J of order 3 [IA , 4.8.2, 4.8.4, 4.7.3A], a contradiction. Hence for all i, CJ (xi ) does not have a normal subgroup isomorphic to P SU3 (2). As Qi acts faithfully on J, it must be that O3 (CKi (t)) is in the kernel of the action on J for all i. Then s  CJ (xi ) for all i. Hence s ∈ Z(Γ), where Γ = ΓF,1 (J). As O2 (J) = 1, Γ < J. Given the above possibilities for J, it follows by [IA , 7.3.4] that 1 J ∼ = O6− (2), J ∼ = Sp8 (2) with Γ ∼ = O8+ (2), or J ∼ = 2F4 (2 2 ) with = Sp6 (2) with Γ ∼ ∼ Γ = Aut(L3 (3)). But s ∈ Z(Γ), a contradiction in each case. Hence, [Q0 , Et ] = 1, whence Q0 F acts faithfully on Ut , yielding a final contradiction as discussed above.  Now the proof of Proposition 4.1 is complete. To recapitulate, if the proposition fails, then K ∼ = U3 (8) by Lemmas 4.10 and 4.18. With the notation b1 , H, F , and E from (5C) and (5D), Lemma 5.6b describes the possible structures for the nontrivial  ∗ = H ∗ /O3 (H ∗ ) of H in CG (b), for any b ∈ E − x, z and in particular pumpup H b b b for any b ∈ F − x. By Lemma 5.8, O3 (Hb∗ ) = 1 for any such b. Then Lemmas 5.11, 5.12, 5.14, 5.15 rule out all possibilities except that Hb∗ ∼ = U3 (8) for all b ∈ F # , ∼ including b = x since K = U3 (8). Finally, Lemma 5.16 gives a contradiction. This completes the proof of Theorem C∗6 : Stage 1.

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10.1090/surv/040.9/08

CHAPTER 8

Preliminary Properties of K-Groups 1. Small Groups Lemma 1.1. Let K ∈ Kp for some odd prime p, with mp (K) = 1. Let P ∈ Sylp (Aut(K)). Then P contains a cyclic complement to P ∩ Inn(K). Moreover, Ω1 (P ) is abelian and mp (P ) ≤ 2. 

Proof. See [III11 , 7.2].

Lemma 1.2. Let p be an odd prime and let K ∈ Kp with p dividing |K|. If p divides | Out(K)|, then p2 divides |K|. 

Proof. See [IIK , 5.1d]. Lemma 1.3. Let K ∈ C3 with m  3 (K) = 1. Then K ∼ = L2 (8).

Proof. This follows directly from the definition of C3 [V3 , 1.1], and [IA , 3.3.3, 1  4.7.3, 5.6.1]. Note that L2 (8) ∼ = 2 G2 (3 2 ) . Lemma 1.4. Let K ∈ Cp , with p an odd prime. Suppose that (a) m2,p (K) − mp (Z(K)) ≤ 1; (b) mp (Aut(K)) > 1; and (c) Lp (CK (x)) = 1 for all x ∈ Ip (K) − Z(K). Then (p, K) is one of the following pairs: (p, L2 (pn )), n ≥ 2, (p, L± 3 (p)), (3,3G2 (3)), 5 1 1 (3,L2 (8)), (5,2B2 (2 2 )), (3,2F4 (2 2 ) ), (3,M11 ), (5,2F4 (2 2 ) ), (5, M c), (5, F3 ), (7, O  N ), (11, J4 ). Proof. Suppose that K ∈ Chev(p) and K is simple. Then mp (CK (t)) ≤ 1 for all involutions t ∈ K. Using [IA , 4.5.1] we see that K ∼ = L2 (pn ) (n ≥ 2 by (b)), 1 ± 2  L3 (p), or G2 (3 2 ) ∼ = L2 (8) with p = 3, as claimed. If K ∈ Chev(p) and Z(K) = 1, as K ∈ C3 . Thus K := K/Z(K) ∼ then K ∼ 3A = U4 (3), Ω7 (3), or G2 (3). In the = 6 U4 (3) case the pair of long root subgroups in a Borel subgroup of K centralize an involution, and lie in an E34 -subgroup, the unipotent radical of a parabolic subgroup of type A1 (32 ). This radical has elementary abelian preimage in K by [IA , 6.4.4], so m2,3 (K) − m3 (Z(K)) ≥ 2, contradicting (a). As U4 (3) ∼ = Ω− 6 (3) ≤ Ω7 (3), it is (3). Thus, the lemma holds if K ∈ Chev(p). also impossible that K ∼ Ω = 7 Suppose next that K ∈ Cp ∩ Chev − Chev(p). If p = 5 then by definition of 5 C5 [V3 , 1.1], either the lemma holds or K ∼ = 2F4 (2 2 ). But in the latter case, by  5 [IA , 4.8.7], O 2 (CK (x)) ∼ = 2B2 (2 2 ) for some x ∈ I5 (K), contradicting (c). Thus, by definition of Cp , we may assume that p = 3 and K ∈ Chev(2) is as in [V3 , 1.1]. Using [IA , 4.8.2, 4.7.3A] we see that (c) is contradicted in every case except 1 K∼ = 2F4 (2 2 ) , as claimed. 343 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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∼ Akp , k ≤ 3. If k = 1 then (b) is contradicted, If K ∈ Alt, then as K ∈ Cp , K = while if k ≥ 2, then (c) is contradicted, except for the single case K ∼ = L2 (32 ), = A6 ∼ p = 3. Therefore we may assume that K ∈ Spor. Using the tables [IA , 5.3], we see that hypothesis (c) narrows down the groups listed in [V3 , 1.1] to (p, K) = (3, (3)M c) and (5, Ly), other than the groups in the conclusion of this lemma. As these two groups contain 2A8 and 2A11 , respectively, they do not satisfy hypothesis (a). The proof is complete.  Lemma 1.5. m  3 (G2 (3)) = 3. Proof. The universal covering group of G2 (3) is K = 3G2 (3), by [IA , 6.1.4]. Let K = K/Z(K). Every root element of K splits over Z(K), since it lies in a GL2 (3)-subgroup by the Chevalley relations [IA , 2.4]. If α and β are fundamental roots with α long, then there exists h ∈ H, the Cartan subgroup, such that h inverts the root group Xα but centralizes Xβ . Then as [X2α+3β , Xα+2β ] = 1, h both inverts and centralizes the commutators of preimages of these root groups, so the two root groups commute elementwise. Similarly, so do the preimages of Xα+3β and Xα+2β , as well as Xα+3β and X2α+3β . The lemma follows.  Lemma 1.6. Suppose that p is a prime, p ≥ 5, and X = KD where K  X, K ∈ Kp with Z(K) = 1, D ∼ = Ep2 , and D ∈ Sylp (CX (D)). Then K ∼ = SLp (q),  = ±1, q ≡  (mod p), and some d ∈ D# induces a diagonal non-inner automorphism on K. Proof. By [IA , 6.1.4], since Z(K) = 1 and p ≥ 5, K/Z(K) ∼ = Lkp (q),  = ±1, q ≡  (mod p). We have D = z, d, where z = Ω1 (Op (Z(K))) and |CKd (d)|p = p2 . Therefore |CKd/Z(K) (d)|p ≤ p2 , whence equality must hold. Let V be the linear or unitary vector space underlying K. If d is induced by an isometry of V of order p, then that isometry is diagonalizable and it follows easily that mp (CKd/Z(K) (d)) ≥ p − 2 > 2, contradiction. Hence d is not so induced, whence by [IA , 4.8.4], if k > 1, then CK (d) has an Lk (q p ) composition factor, which has order divisible by p. As D ∈ Sylp (CKD (d)), this is impossible. Therefore, k = 1, and by [IA , 4.8.4], d maps nontrivially into Outdiag(K). The lemma is proved.  Lemma 1.7. Aut(L± 4 (3)) does not contain a copy of Sp4 (3). 

Proof. Let K = L4 (3). If the lemma were false, O 3 (CK (z)) would contain Sp4 (3) for some z ∈ I2 (K). But this is impossible by [IA , 4.5.1] and order considerations.  n Lemma 1.8. Let K be a quasisimple group with K/Z(K) ∼ = 2B2 (2 2 ) for some odd n > 1. Let T ∈ Syl2 (K). Then Ω1 (T ) is elementary abelian.

Proof. By the Chevalley relations [IA , Table 2.4], [IA , 6.1.4], and [IA , 6.4.1],  Ω1 (T ) is elementary abelian. Lemma 1.9. The automorphism groups of the following groups do not have a subgroup H such that H/O3 (H) ∼ = P GU3 (2): A6 , 3A6 , M11 , U3 (3), L2 (8). Proof. Sylow 3-subgroups S of P GU3 (2) are isomorphic to 31+2 , and no subgroup of order 3 is weakly closed in S with respect to P GU3 (2). Of the listed groups

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2. OUTER AUTOMORPHISMS AND COVERING GROUPS

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the only one involving 31+2 in its automorphism group is U3 (3), which has Sylow 3-subgroups isomorphic to S but with a weakly closed subgroup of order 3.  2. Outer Automorphisms and Covering Groups Lemma 2.1. Let K ∈ Lie be the universal version. Then Outdiag(K) and Z(K) are isomorphic as modules for Out(K). Proof. See [IA , 6.3.1a].



Lemma 2.2. Let K ∈ K. Then Out(K) contains no copy of SL2 (3). Proof. See [III11 , 2.1].



Lemma 2.3. If K ∈ C2 , then Out(K) is 2-nilpotent. Proof. See [III11 , 2.1]; this also follows easily from [IA , 2.5.12].



Lemma 2.4. Let K ∈ C2 . If Z(K) = 1, then 5 does not divide | Out(K)|. Proof. This is an easy consequence of [IA , 6.1.4, 2.5.12].



Lemma 2.5. If q = pn where p is an odd prime, then Out(L2 (q)) ∼ = Z2 × Zn . Proof. This follows directly from [IA , 2.5.12].



Lemma 2.6. Suppose p is an odd prime, K ∈ Kp , and Z(K) is noncyclic. Then  p = 3, K ∼ = 32 U4 (3), and O 3 (Aut(K)) = Inn(K). Proof. The structure of K follows immediately from [IA , 6.1.4]. Out(K) ∼ = D8 by [IA , 2.5.12], and the lemma follows.

Then 

Lemma 2.7. For p odd there is no Cp -group K such that mp (K) ≤ 2 and Z(K) = 1. Proof. This can be checked using the definition of Cp -group [V3 , 1.1], Schur multiplier information [IA , 6.4.1, 6.4.4], rank information [IA , 4.10.3a, 5.6.1], and Lemma 1.5.  Lemma 2.8. Let K ∈ C2 . Then there is a four-group U ≤ K such that U ∩ Z(K) = 1. Proof. We may assume that Z(K) = 1. If K ∈ Spor, then by [IA , 6.1.4, 5.6.1], Z(K) is cyclic and m2 (K) ≥ 3, which implies the desired conclusion. If K ∈ Chev(3)−Chev(2), then by [IA , 6.1.4], K ∼ = 2U4 (3) ∼ = Ω− 6 (3), again with 2-rank bigger than 2 and cyclic center. Therefore we are reduced to the case K ∈ Chev(2). By [IA , 6.4.4], the result holds if K/Z(K) ∼ = Sp6 (2) or L3 (4); as Sp6 (2) embeds in D4 (2), F4 (2), 2E6 (2), and U6 (2) [IA , 4.9.2], and SL3 (4) in G2 (4) [IA , 4.7.3A], the result holds for those groups K/Z(K) as well. Finally, Lemma 1.8 implies the 3 result if K/Z(K) ∼ = 2B2 (2 2 ). By [IA , 6.1.4] and [V3 , 1.1], we have covered all the necessary cases, so the lemma follows.  Lemma 2.9. Let K = U6 (2) and T ∈ Syl2 (Aut(K)). Set T0 = T ∩ K. Then J(T ) = J(T0 ) = F ∗ (NAut(K) (J(T ))) ∼ = E29 and NAut(K) (J(T ))/J(T ) is an extension of P GL3 (4) by a field automorphism. Moreover, every involution of K has a conjugate in J(T0 ).

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346

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Proof. Let P be the parabolic subgroup of Inndiag(K) containing T0 and of type P GL3 (4), with Levi decomposition P = QL. A graph automorphism of Inndiag(K) derives from an automorphism φ of F4 , so NAut(K) (Q)/Q is as claimed in the lemma. P is the stabilizer of a maximal isotropic subspace of the natural GU6 (2)-module, and a matrix calculation shows that Q ∼ = E29 is an F2 -form of V ⊗F4 V φ as P GL3 (4)-module, where V is the natural GL3 (4)-module. From this one calculates that an involution z ∈ Z(T0 ∩ L) satisfies |CQ (z)| = 25 . The same then holds for all involutions in L. Moreover, for any A ≤ T0 ∩ L with A ∼ = E23 , one calculates [CQ (z), A] = 1. As m2 (L) = 4, this implies that Q = J(T0 ) and that any B ≤ T with B ∼ = Q = B must satisfy B = (B∩Q)×b, where b ∈ I2 (T −T0 ) induces a transvection on Q. However, all involutions of T − T0 are P -conjugate modulo Q, by [IA , 4.9.1], and one of them inverts an element v ∈ I3 (L), so if J(T ) = Q, then [Q, v] ∼ = E26 , contradiction. Thus = E22 . However, one calculates that [Q, v] ∼ J(T ) = Q. It remains to prove the final statement. It suffices to show that any involution t ∈ SU6 (2) centralizes a totally isotropic subspace W of the natural SU6 (2)-module V . First we show that CV (t) = [V, t]⊥ . Let ( , ) be the hermitian form on V . Both spaces have dimension 6 − dim[V, t] so it suffices to take v ∈ CV (t) and w ∈ V and observe that (v, [w, t]) = (v, w) + (v, wt ) = (v, w) + (v t , wt ) = 0. As [V, t] ⊆ CV (t), [V, t] is totally isotropic and CV (t) = [V, t] ⊥ X with X nondegenerate of dimension 6−2 dim[V, t]. Then X has a totally isotropic subspace X0 of dimension 3−dim[V, t], so [V, t] ⊥ X0 is the required 3-dimensional totally isotropic subspace of CV (t). The proof is complete.  Lemma 2.10. Let K ∈ K with K/O2 (K) ∼ = U6 (2). Let T ∈ Syl2 (K). Then the following conditions hold: (a) Z(T )/Z(K) = Z(T /Z(K)) ∼ = Z2 ; (b) If Z(K) = 1 and v ∈ I3 (Aut(K)) with E(CK (v)) ∼ = U4 (2), then E(CK (v)) contains a conjugate of Z(T ); (c) J(T ) = CK (J(T )) ∼ = E29 × Z(K), and AutK (J(T )) ∼ = L3 (4); and (d) Let z ∈ I2 (K) be such that 1 = zZ(K) ∈ Z(T /Z(K)) and let R = O2 (CK (z)). Then |Φ(R)| = 2 and R ∼ = Z(K) × 21+8 + . Proof. By [IA , 3.2.2], |Z(T /Z(K))| = 2, and then (a) follows from [IA , 6.4.1,  = SU6 (2) and let V be the natural F4 K-module.  6.4.2]. Let K Then by [IA ,  v ) has dimension 4 and 4.8.2], v has a preimage v ∈ I3 (K) such that V+ := CV ( E(CK (v)) ∼ = SU (V+ ). In particular E(CK (v)) contains the image of a transvection  As this image is conjugate to z, (b) holds. By Lemma 2.9, (c) holds if in K. Z(K) = 1. In general, the preimage of J(T /Z(K)) in K is elementary abelian by [IA , 6.4.4a], so J(T )/Z(K) = J(T /Z(K)) and (c) follows. It clearly suffices to prove (d) when Z(K) is as large as possible, namely, Z(K) ∼ = E22 . In that case by (a) and a Frattini argument, CAut(K) (zZ(K)) (and hence NAut(K) (R)) covers Outdiag(K) ∼ = Z3 , which acts nontrivially on Z(K) by [IA , 6.3.1]. As R/Z(K) is extraspecial, either |Φ(R)| = 2 or Φ(R) = Z(T ) ∼ = E23 . The same holds for [R, R]. It suffices to prove that |Φ(R)| = 2, for then Z(K) is a direct factor of R, as desired. Now, V := R/Z(T ) ∼ = U4 (2), = E28 is a natural module for H := CK (z)/R ∼ hence irreducible over F2 . We have F4 ⊗F2 V ∼ = W ⊕ W ∗ , where W is a natural F4 H-module and is absolutely irreducible. Therefore Hom(V ⊗F2 V, F2 ) ∼ = E22 .

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On the other hand via the commutator mapping, the dual group [R, R]∗ embeds in Hom(V ⊗F2 V, F2 ), so [R, R] ∼ = Z2 . Now the irreducibility of H on V forces the abelian group R/[R, R] to be elementary abelian, whence Φ(R) = [R, R] ∼ = Z2 . The lemma is proved.  Lemma 2.11. Let K = 2U6 (2) and T ∈ Syl2 (K). Then Z(T ) = z Z(K) for some involution z ∈ T − Z(K) such that z = [O2 (CK (z)), O2 (CK (z))]. Proof. This is immediate from Lemma 2.10.



3. Pumpups and Subcomponents In this section, p can be any prime, including 2. Recall that for K, J ∈ Kp , we say that K ↑p J if and only if there is x ∈ Ip (Aut(J)) and a p-component K1 of CJ (x) such that K1 /Op (K1 ) ∼ = K. Also for K ∈ Kp , ↑p (K) is the set of all J ∈ Kp such that K ↑p J. By definition, J ↓p K if and only if K ↑p J, and ↓p (J) is the set of all K ∈ Kp such that J ↓p K. The questions addressed in this section are generally of the following sort: If K, J ∈ Kp and K ↑p J, how do properties of K (in particular, its isomorphism type) influence properties of J (in particular, its possible isomorphism types)? On the other side of the same coin, if we are given the isomorphism type of J, considerable information about the isomorphism type of K is available from [IA , 4.2.2, 4.5.1, 4.5.2, 4.7.3A, 4.8.2, 4.8.4, 4.9.1, 4.9.2] and the Borel-Tits theorem in the case J ∈ Chev, and from [IA , 5.2.8, 5.2.9, 5.3] in the cases J ∈ Alt ∪ Spor. Many of the results in this section follow simply by looking at the other side of the coin. When we say that some K ∈ K belongs to a subset S of K unambiguously, that means that K belongs to exactly one of the following sets: Chev(r) for some prime r, Alt, or Spor, and that one is S. Most groups K ∈ K belong to one of these sets unambiguously; the possible ambiguities are given in detail in [IA , 2.2.10]. 3.1. Key Generalities. Lemma 3.1. Let K ∈ Kp . If K ∈ Chev(r), then ↓p (K) ⊆ Chev(r). If K ∈ Alt, then ↓p (K) ⊆ Alt. If K ∈ Spor, then ↑p (K) ⊆ Spor. Proof. The first two statements hold by [IA , 4.9.6, 5.2.8], or [III11 , 1.1a], and they imply the final statement.  Lemma 3.2. Let K ∈ Kp . If K ∈ Cp , then ↓p (K) ⊆ Cp and any quotient of K lies in Cp . If K ∈ Tp , then ↓p (K) ⊆ Cp ∪ Tp . If K ∈ Gp , then ↑p (K) ⊆ Gp . Proof. The first two sentences hold by [III11 , 1.1cd], and they imply the third.  Lemma 3.3. Suppose that K ∈ C2 or Co2 , and T is a 2-subgroup of Aut(K). Let L be a component of CK (T ). Then L/O2 (L) is a C2 -group or Co2 -group, respectively. Proof. Using L2 -balance we may assume inductively that |T | = 2. The assertion for C2 is then found in [IIIK , 5.1]. If L ∈ C2 − Co2 , then by definition L ∼ = L2 (q) for some q ∈ FM9, q > 17. But then from [IA , 5.2.9, 2.2.10, 5.3], L has no pumpups outside Chev(q), so K ∈ C2 , contradiction. This proves the lemma. 

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Lemma 3.4. Let K ∈ Chev(r). Let p be a prime, p = r, and x ∈ Ip (Inn(K)) with I a p-component of CK (x). Assume that I is terminal in K. Then mp (CInn(K) (I)) = 1. Proof. Suppose false and let D ∈ Ep2 (CInn(K) (I)). Then CK (D) has order  divisible by r, so K = ΓrD,1 (K) by [IA , 7.3.1]. For every d ∈ D# , every Lie com ponent of CK (d) normalizes I since I is a component of CK (d). Hence O r (CK (d))  normalizes I. As d is arbitrary, I is normalized by ΓrD,1 (K) = K, which is absurd. The lemma follows.  ∼ Ep2 , and I ≤ K is Lemma 3.5. Suppose that K ∈ Kp , E ≤ Aut(K) with E = a p-component of CK (e) for all e ∈ E # . Then (a) − (d) hold: (a) E ∩ Inn(K) = 1; (b) One of the following holds: (1) K/Z(K) ∼ = An+kp2 , k > 0, n ≥ 5, I ∼ = An , and E acts on K/Z(K) like a subgroup of Σn+kp2 with k regular orbits and n fixed points; moreover, CK (E) < CK (e) for all e ∈ E # ; or (2) The triple (p, K/Z(K), IZ(K)/Z(K)) is one of the following: (2, M12 or J2 , A5 ), (2, Co1 , G2 (4)), (2, Suz, L3 (4)), 3 (2, He, 22 L3 (4)), (2, Ru, 2B2 (2 2 )), (3, Suz, A6 ), or (3, O  N, A6 ); (c) If (b1) holds with kp2 = 4, then E  CAut(K) (I) ∼ = Σ4 ; and (d) If (b2) holds, then E contains all elements of CAut(K) (I) of order p; in particular mp (CAut(K) (I)) = 2. Proof. See [III11 , 1.16]. The extra assertion in (b1) us easily verified.



Lemma 3.6. Suppose that X is a K-group, p is a prime, Op (X) = 1, D ∈ Ep2 (X), and D permutes transitively the components of E(X). Suppose that L ≤ X and L is a component of CX (d) for all d ∈ D# . Then E(X) is quasisimple. If E(X) = L and p is odd, then one of the following holds, with m = mp (D) and a > 0: (a) K ∼ = Aapm +r , L ∼ = Ar , and K ∈ Cp ; (b) p = 3 and K/Z(K) ∼ = Suz or O  N . Proof. If d ∈ D permutes the components of E(X) nontrivially, then D must act faithfully on E(X). Then for any d ∈ D − d, d does not centralize E(CE(X) (d)), contrary to assumption. Therefore E(X) is quasisimple. The remaining assertions are contained in Lemma 3.5, with the help of the definition of  Cp [V3 , 1.1]. Lemma 3.7. Let K ∈ Kp ∩ Chev(r) for some primes p > 2 and r. Let x ∈ Ip (Aut(K)) and suppose that CK (x) has a p-component L. If CAut(K) (L) contains a subgroup E ∼ = Ep2 with x ∈ E, then there is y ∈ E # such that L is properly contained in a p-component of CK (y). Proof. Let e ∈ E # . By [III11 , 1.13b, 1.14], since p is odd, L ≤ Le for some p-component Le of CK (e). Hence we may assume by way of contradiction that L = Le   CK (e) for all e ∈ E # . But then as K ∈ Chev, Lemma 3.5b yields a contradiction. 

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3.2. Pumpups of Specific Groups. Lemma 3.8. The following conditions hold: (a) ↑2 (M11 ) = {M c}; (b) ↑2 (M12 ) = ↑2 (J2 ) = {Suz}; (c) ↑3 (K) = ∅ if K ∈ {M11 , M12 , J3 , 3J3 , Co2 , Co1 , Suz, F i24 , F2 }; (d) ↑3 (F i22 ) = {F2 }; (e) ↑5 (F5 ) = {F1 }; (f) ↑5 (Co1 ) = ∅; and (g) ↑p (F1 ) = ∅ for all primes p | |F1 |. Proof. By Lemma 3.1, each left side is a subset of Spor, so the lemma can be verified by checking through the tables [IA , 5.3].  Lemma 3.9. The following conditions hold: (a) ↑3 (3Suz) = {Co1 }; (b) ↑3 (U4 (3)) = {F i22 , U4 (33 )}; (c) ↑3 (3U4 (3)) = {Suz, 3F i22 }; (d) ↑3 ([3 × 3]U4 (3)) = {3Suz, Co1 }; (e) ↑3 (Ω7 (3)) = {F i23 , Ω7 (33 )};   3 (f) ↑3 (P Ω+ 8 (3)) = {F i24 , 3F i24 , D4 (3 )}; (g) ↑3 (G2 (3)) = {F i24 , 3F i24 , F3 , D4 (3), 3D4 (3), G2 (33 )}; (h) ↑3 (F i22 ) = {F2 }; (i) ↑3 (3F i24 ) = {F1 }; and (j) ↑3 (F3 ) = {F1 }. Proof. Write each equation as ↑3 (K) = S and let L ∈ S. If L ∈ Spor, then L is one of the claimed groups by inspection of [IA , 5.3]. Otherwise K ∈ Spor by Lemma 3.1, so K ∈ Chev(3) unambiguously, and hence L ∈ Chev(3). By the BorelTits theorem and the fact that Outdiag(L) is a 3 -group, K ↑3 L via a graph, field, or graph-field automorphism x ∈ Aut(L). Using [IA , 4.9.1, 4.9.2], we see that L is one of the claimed groups. The reverse inclusions are then trivially checked.  Lemma 3.10. The following conditions hold: (a) ↑2 (2Sp6 (2)) ∩ C2 = {[2 × 2]D4 (2)}. (b) ↑2 (K) ∩ C2 = ∅ for any K ∈ {L± 4 (3), 2U4 (3), G2 (3), M11 , HS, 2HS}. (c) ↑2 (K) = ∅ for any K ∈ {J3 , Co2 , F i24 , F5 , F3 , F1 }; (d) ↑2 (U5 (2)) = {L5 (4)}; (e) ↑2 (U6 (2)) = {L6 (4)}; (f) ↑2 (2F i22 ) = {F i23 , F i24 }; (g) ↑2 (F i23 ) = {F i24 }; (h) ↑2 (D4 (2)) = {D4 (4), F i22 }; (i) ↑2 (2 D5 (2)) = {D5 (4)}; (j) ↑2 (2U6 (2)) = {F i22 }; (k) ↑2 (22E6 (2)) = {F2 }; (l) ↑2 (2F2 ) = {F1 }; (m) ↑2 (P Sp4 (3)) ∩ C2 = {L± 4 (3), 2U4 (3), L4 (4)}; (3), 2U4 (3), U5 (2), P Sp4 (3), L4 (2), L5 (2), Sp4 (4), (n) ↑2 (A6 ) ∩ C2 = {L± 4 HS, 2HS}; (o) ↑2 (U3 (3)) ∩ C2 = {U4 (3), 2U4 (3), G2 (4), 2G2 (4)}; and (p) ↑2 (L3 (3)) ∩ C2 = {L4 (3)}.

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Proof. Write the left side as ↑2 (K) or ↑2 (K) ∩ C2 , as the case may be. Using [IA , 6.2.2] for (a), as well as [IA , 4.9.1, 5.3, 4.5.1], we see easily that every left side includes the corresponding right side. We must verify the reverse inclusion. So suppose that K ↑2 L ∈ K2 , via x ∈ I2 (Aut(L)). Consider first cases (c–l), so that K ∈ Spor or K is unambiguously in Chev(2). By Lemma 3.1, L ∈ Spor ∪ Chev(2). If L ∈ Spor, then as usual, inspection of the tables [IA , 5.3] shows that for each equation, L must be one of the groups on the right hand side. If L ∈ Chev(2), then K ∈ Chev(2). By [IA , 4.9.2] and the Borel-Tits theorem, x ∈ Aut0 (L), given the possible isomorphism types of K, so x is a graph-field or field automorphism. Then K and L must be as in (d), (e), (h) or (i). (In (k), note that the Schur multipler of E6 (4) has order 3 [IA , 6.1.4].) Now consider the remaining cases (a–b,m–p), in all of which we may assume that L ∈ C2 . Again if L ∈ Spor, we use [IA , 5.3] to verify that of the listed K’s, only the appropriate ones occur in ↓2 (L). (Note in (b) that F5 ∈ C2 .) If L∼ = L2 (q), q ∈ FM9, then clearly ↓2 (L) = ∅. For L ∈ C2 ∩(Alt∪Chev(3)), we have ↓2 (L4 (3)) = {A6 , L3 (3), P Sp4 (3)}, ↓2 (U4 (3)) = ↓(2U4 (3)) = {A6 , U3 (3), P Sp4 (3)}, ↓2 (G2 (3)) = {L2 (8)}, ↓(P Sp4 (3)) = {A6 } = ↓(A8 ), and ↓2 (L) = ∅ otherwise. So we may assume that L ∈ Chev(2) − Alt − Chev(3). Then (a) holds by [IA , 6.3.3b], so we may assume that K ∼  2Sp6 (2). = If x is a graph automorphism, then by [IA , 4.9.2], K/Z(K) ∼ = Sp2n (q), n ≥ 1 or F4 (q), so K ∼ (2) or L± = A6 , as in (n). Then [IA , 4.9.2] yields L ∼ = L± 4 5 (2). Suppose finally that x is a field or graph-field automorphism. Then K ∼ = A6 or U3 (3) ∼  = G2 (2) , so L ∼ = Sp4 (4) or (2)G2 (4). The proof is complete. Lemma 3.11. The following conditions hold: p

(a) ↑p (2F4 (2 2 ) ) = {2F4 (2 2 )} if p > 2; ± p (b) ↑p (L± 3 (p)) = {L3 (p )} if p > 2, except that ↑3 (U3 (3)) = {U3 (33 ), G2 (8), 3D4 (2), D4 (2)}; (c) ↑3 (L3 (9)) ⊆ Chev(3); (d) ↑p (L2 (pn )) = {L2 (ppn )} if n ≥ 2 and pn = 32 ; (e) ↑3 (3G2 (3)) = ∅; (f) ↑3 (U4 (2)) ∩ C3 = {U5 (2), U6 (2), SU6 (2), D4 (2), Co2 , P Sp4 (3)}; and (g) ↑3 (U5 (2)) ∩ C3 = {U6 (2), SU6 (2)}. 1

Proof. For each claimed equation it is trivial to check that the left side includes the right side, by [IA , 4.9.1, 4.9.2, 4.7.3A, 4.8.2]. Let K be a group on the left side, in each case. By inspection of [IA , 5.3], K ∈ Spor. Moreover, in (b), (c), (d), and (e), except for K = U3 (3) ∼ = G2 (2) in (b), K is unambiguously in Chev(p) so by Lemma 3.1, all groups on the right must be in Chev(p). By the Borel-Tits theorem and the fact that Outdiag(K) is a p -group, the only pumpups that can occur in Chev(p) are via graph, field, or graph-field automorphisms. This proves (c) and (d). In (e) we have to note that although G2 (3) ↑3 D4 (3) and 3D4 (3), [IA , 6.1.4] implies that D4 (3) and 3D4 (3) have no exceptional Schur multiplier, so (e) holds as well. For (b), as well as (a), (f), and (g), it remains to consider the case of pumpups L ∈ Chev(2) of K. In (f) and (g), we are limited to L ∈ C3 [V3 , 1.1], and easily verify the assertions by using [IA , 4.8.2, 4.8.4, 4.7.3A], as required. In (a) and (b), since the Dynkin diagrams of type F4 and G2 are not proper subdiagrams of any other diagram or extended diagram, the pumpup must be via a graph, field,

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or graph-field automorphism. This easily yields that L is one of the groups on the right side, completing the proof of the lemma.  Lemma 3.12. If K ∈ K5 and D4 (2) ↑5 K, then K ∈ Chev(2). Moreover, either m5 (K) > 2 or K ∼ = D4 (25 ). Proof. If K ∈ Alt ∪ Chev(r), then D4 (2) ∈ Alt ∪ Chev(r) respectively, so the only possibility is K ∈ Chev(2). From the tables [IA , 5.3] we see that K ∈ Spor. This proves the first statement. Let x ∈ I5 (Aut(K)) with CK (x) having a D4 (2) 5component. It is clear from [IA , 4.8.2] that K is neither a linear nor a unitary group, so by [IA , 6.1.4, 2.5.12], K is simple with 5 not dividing | Outdiag(K)|. If x induces an inner automorphism on K, then m5 (K) ≥ m5 (CK (x)) ≥ m5 (x × D4 (2)) = 3. Otherwise x induces a field automorphism of order 5, whence the lemma.  Lemma 3.13. The following conditions hold: (a) ↑3 (L3 (4)) = {L3 (64), D4 (4), 3D4 (4)}; (b) ↑3 (U3 (8)) = {U3 (83 ), D4 (8), 3D4 (8), U9 (2), 2 E6 (2)}; (c) ↑3 (M12 ) = ∅; (d) ↑5 (F i22 ) = ∅. Proof. Let K ∈ Kp , x ∈ Ip (Aut(K)), and L be a component of CK (x) with L/Op (L) ∼ = L3 (4), U3 (8), M12 , or F i22 , and p = 3, 3, 3, or 5, respectively. In the two sporadic cases K ∈ Spor by [IA , 4.9.6, 5.2.9], but examination of the tables [IA , 5.3] shows that no such K exists. Next assume that L/O3 (L) ∼ = L3 (4) or U3 (8). If x induces a field automorphism on K, then K ∼ = L3 (43 ) or U3 (83 ); if x induces a graph automorphism on K, then K ∼ = D4 (4), D4 (8), 3D4 (4), or 3D4 (8), by [IA , 4.7.3A]. So assume that x induces an inner-diagonal automorphism on K. Using [IA , 4.8.2, 4.8.4, 4.7.3A] we see that in the only cases in which a component of CK (x) has L3 (4) as a composition factor, the component in fact is SL3 (4). On the other hand, the same references show U3 (8) as a component of CK (x) for K ∼ = U9 (2)  and 2 E6 (2). This completes the proof. Lemma 3.14. Suppose that I ∈ K2 is a covering group of L3 (4). Then ↑2 (I) ⊆ {L3 (16), Suz, 2Suz, He, O  N }. Proof. Let K ∈ K2 with I ↑2 K. Since L3 (4) ∈ Alt ∪r>2 Chev(r), K ∈ Chev(2) ∪ Spor. If K ∈ Chev(2), then K ∼ = L3 (16) by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2]. If K ∈ Spor, then by inspection of [IA , 5.3], K is one of the four asserted groups.  Lemma 3.15. Suppose K, L ∈ K3 and K ↑3 L via x ∈ I3 (Aut(L)). If K ∼ = G2 (8), then x ∈ Inn(L). Proof. Using [IA , 2.2.10, 5.3] and Lemma 3.1, we see that L ∈ Chev(2). If x ∈ Inn(L), then by [IA , 4.2.2], the extended untwisted Dynkin diagram of L  G2 (8) by contains a G2 -subdiagram. So L ∼ = G2 (q) for some q, and then K ∼ =  [IA , 4.7.3A], a contradiction. The proof is complete. Lemma 3.16. If U3 (3) ↑3 K ∈ K3 , then K ∼ = U3 (33 ), G2 (8), D4 (2), or 3D4 (2). 3 Moreover, for any b ∈ I3 (Aut(K)), O (CK (b)) is not involved in Z3 × A6 . ∼ G2 (2) ↑ K, K ∈ Chev(r), r = 2 or 3, by [III11 , 1.1] Proof. Since U3 (3) = 3 and [IA , 2.2.10] and inspection of the sporadic tables [IA , 5.3]. If r = 3, then by

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∼ U3 (33 ). the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], the only possibility is K = Suppose that r = 2 and U3 (3) ↑3 K via x ∈ I3 (Aut(K)). Since G2 is a maximal Dynkin diagram, x ∈ Inndiag(K) by [IA , 4.2.2]. Then [IA , 4.9.1, 4.9.2] yield the three stated possibilities for K. Let P ∈ Syl3 (Aut(K)) contain a Sylow 3-subgroup of CK (b). If b ∈ Inn(K), then in every case CP (b) contains 31+2 , see [IA , 4.9.1, 4.9.2]. If b ∈ Inn(K), then m3 (CP (b)) ≥ 4 (for K ∼ = U3 (33 ) and D4 (2)) or CP (b) has exponent 9 (for K ∼ = G2 (8) 3 or D4 (2)), see [IA , 3.2.2, 4.8.2a, 4.7.3A]. In no case is CP (b) embeddable in E33 , so the lemma follows.  Lemma 3.17. Let K ∈ K2 with 2U6 (2) ↑2 K or U6 (2) ↑2 K. Then K ∼ = F i22 or K/Z(K) ∼ = L6 (4). If K ∈ C3 , then K ∼ = F i22 . Proof. By [IA , 5.2.9, 4.9.6, 2.2.10], K ∈ Chev(2) ∪ Spor. Either 2U6 (2) ↑2 K/Z(K) or U6 (2) ↑2 K/Z(K). From [IA , 5.3] we see that if K ∈ Spor, then K ∼ = F i22 . If K ∈ Chev(2), we must have U6 (2) ↑2 K/Z(K), by the Borel-Tits theorem. By [IA , 4.9.1, 4.9.2], the only possibility is K/Z(K) ∼ = L6 (4) ∈ C3 . The proof is complete.  Lemma 3.18. Suppose that K ∈ C2 , t ∈ I2 (Aut(K)), and CK (t) has a compo∼ ± nent I ∼ = U3 (3) or A6 . If K involves L± 4 (3), then K/Z(K) = L4 (3). ± 6 Proof. Suppose K/Z(K) ∼  L± = 4 (3); we show that |K/Z(K)|3 < 3 = |L4 (3)|3 , contradiction. Using I ↑2 K ∈ C2 and [IA , 5.3] for K ∈ Spor, Lemma 3.39 for K ∈ Alt, [IA , 4.9.1, 4.9.2] for K ∈ Chev(2), and [IA , 4.5.1] for K ∈ Chev(3), we find that K/Z(K) ∼ = HS, L± n (2), n ∈ {4, 5}, Sp4 (4), or G2 (4). The desired inequality holds in all cases, completing the proof. 

Lemma 3.19. Let X be a K-group with O2 (K) = 1 such that all components of  X lie in C2 . Let t, v be a four-group acting on X, and let Xu = O 3 (CX (u)) for  # each u ∈ t, v . Suppose that Xv ∼ = Xtv ∼ = P Sp4 (3), while Xt = O 3 (Xv ∩ Xtv ) ∼ = SL2 (3)∗SL2 (3). Suppose also that there is an involution w ∈ Aut(X) interchanging Xv and Xtv by conjugation and centralizing Xt . Then the subnormal closure of Xt in X is isomorphic to U4 (3) or 2U4 (3).  Proof. Without loss X = XtX , and then by L2 -balance, X = E(X). If X is a diagonal pumpup of Xv , then X = X0 X0v = X0 X0tv ∼ = Xv × Xv (recall that  Sp4 (3) ∈ C2 ). In that case t normalizes X0 and X0v . But then Xt = O 3 (CX (t)) ≥  O 3 (CX0 (t)) = 1, so Xt ∩ X0 = 1, contradicting the assumption that Xt ≤ Xv . Hence, X is quasisimple and X ∈ C2 ∩ ↑2 (P Sp4 (3)) = {L4 (4), L± 4 (3), 2U4 (3)}, by Lemma 3.10. If X ∼ = L4 (4), then Out(X) is a four-group, and v and tv both induce graph-field automorphisms on X. Hence, t induces an inner automorphism on X, corresponding to some x ∈ I2 (X). According as x has one or two 2 × 2 Jordan blocks, we  compute that Xt = O 3 (CX (x)) satisfies |O2 (Xt )| = 45 or |Xt |3 = 3, contradicting in either case the assumed structure of Xt . Thus, X ∼ = L4 (4). It remains to assume that X ∼ = L4 (3) and derive a contradiction. We regard X as Ω(V ) for some 6-dimensional orthogonal space V over F3 of type +, so that Aut(V ) is the projective group of all similarities of V . Thus we can consider t, v and w to be acting on V . In particular, v and tv act as reflections with orthogonal

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centers Vv and Vtv , respectively. From the structure of Xt we see that Vt := (Vv ⊥ Vtv )⊥ is 4-dimensional of + type. Hence Vv ⊥ Vtv is of + type, whence Vv and Vtv are not isometric. Also, Xt is supported on Vt . As [w, Xt ] = 1, w acts as a scalar on Vt and hence w preserves the quadratic form on V . But Vvw = Vtv so Vv and Vtv are isometric, contradiction. This completes the proof.  3.3. Pumpups and p-Rank. Lemma 3.20. Let p be a prime, p ≥ 7. Let K ∈ Chev(2) and A ∈ Ep4 (Aut(K)). Then there is a ∈ A# and a component I of CK (a) such that mp (I) > 1. Proof. Without loss, K is the adjoint version. Since p ≥ 7, for any Lie component L, L(∞) is quasisimple of p -index in L. If some element a ∈ A# induces a field automorphism on K, then I = E(CK (a)) satisfies the requirement, because CK (a) embeds in Inndiag(I), mp (CAut(K) (I)) = 1, and mp (Outdiag(I)) ≤ 1 [IA , 4.9.1, 7.1.4c]. So we may assume that A ≤ Inndiag(K). Among all pairs (a, L) consisting of an a ∈ A# and a component L of CK (a), choose one in which |L| is maximal. For any b ∈ CA (L), Lp -balance and [III11 , 1.13b], and the maximality of |L|, combine to imply that L   CK (b). By Lemma 3.4, mp (CA (L)) = 1. In particular, A normalizes L. If mp (L) ≤ 1, then mp (Aut(L)) ≤ 2 by [III11 , 7.2], so mp (A) ≤ mp (CA (L)) + mp (Aut(L)) ≤ 3,  contradiction. Therefore mp (L) > 1, completing the proof. Lemma 3.21. Suppose M0 ∈ K3 , M0 ↑3 N ∈ K3 , and m3 (N ) ≤ m. Then the following hold: (a) If (M0 , m) = (U4 (2), 4), then N ∼ = U5 (2), U6 (2), D4 (2), D5 (2), Sp8 (2), U4 (8), or Co2 ; (b) If (M0 , m) = (D4 (2), 5), then N ∼ = 2 D5 (2), 2E6 (2), or D4 (8); (c) If (M0 , m) = (3U4 (3), 6), then N ∼ = 3F i22 or Suz; (d) If (M0 , m) = ([3 × 3]U4 (3), 6), then N ∼ = 3Suz or Co1 ; (e) If (M0 , m) = (A9 , 4), then N ∼ = A12 or F5 ; and  Co2 or Co1 . (f) M0 ∼ = Proof. Say M0 ↑3 N via x ∈ I3 (Aut(N )). In part (a), N ∈ Chev(2) ∪ Chev(3) ∪ Spor by Lemma 3.1, while in (b), N ∈ Chev(2) ∪ Spor. We then see from [IA , 5.3] that N ∼ = Co2 (in (a)) is the only sporadic possibility in either case. Likewise in (a) if N ∈ Chev(3), then x is a field automorphism by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2]; but then m3 (N ) = m3 (B2 (33 )) = 9, contrary to hypothesis. So in (a) and (b), we may assume that N ∈ Chev(2). If x ∈ Inndiag(N ) then by [IA , 4.9.1, 4.9.2] x is a field automorphism, so N ∼ = U4 (8) or D4 (8), as desired. Therefore we may assume that x ∈ Inndiag(N ), whence q(N ) = 2 by [IA , 4.2.2]. By [IA , 4.7.3A], if N is of exceptional type, then (b) holds with N ∼ = 2E6 (2), the adjoint version as m = 5 [IA , 4.10.3a]. So now assume in (a) and (b) that N is a classical group. In neither case can N be a linear group, since M0 is not a linear group. In case (b) N must be of type D, by [IA , 4.2.2]. Using the rank condition and [IA , 4.10.3a], and using [IA , 4.8.2, 4.8.4] to calculate subcomponents, we verify (a) and (b). Parts (c) and (d) follow from Lemma 3.9. In parts (e) and (f), Lemma 3.1 implies that N ∈ Alt ∪ Spor and N ∈ Spor, respectively, and the assertions follow  from the information in [IA , 5.2.6, 5.3]; note also in (e) that m3 (Co1 ) = 6.

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Lemma 3.22. Let M0 be as in Lemma 3.21abe. Accordingly, let J = Suz, Co1 , or F3 . Let z ∈ I2 (J) be 2-central and set C = CAut(J) (z), C = C/O2 (CInn(J) (z)), and F = E(C) ∼ = M0 . Then CC (F ) = 1. ∼ − Proof. If J = Suz, then F ∼ = Ω− 6 (2) and C = O6 (2). In the other two cases, F = C is simple. The result follows.  Lemma 3.23. Suppose that p is an odd prime, K  X, Op (Z(X)) = 1, and n K∼ = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or (for p = 3 only) 2 G2 (3 2 ), n ≥ 3. Let x ∈ Ip (X) and suppose that mp (CX (x)) ≤ 3 and Lp (CK (x)) = 1. Then K ∼ = L2 (pp ) and x induces a nontrivial field automorphism on K. Proof. Since Lp (CK (x)) = 1, x induces a non-inner automorphism on K, by the Borel-Tits theorem. As K admits no graph automorphism of odd order by [IA , 2.5.12], x induces a field automorphism on K, so CK (x) ∼ = L2 (pn/p ), U3 (pn/p ), n/p or 2 G2 (3 2 ). Then x Ω1 (Op (Z(X))) is noncyclic and disjoint from CK (x), so 3 ≥ mp (CX (x)) ≥ 2 + mp (CK (x)). Consequently mp (CK (x)) = 1. As mp (L2 (pn/p )) = n/p n/p, while mp (U3 (pn/p )) ≥ 2 ≤ m3 (2 G2 (3 2 )) [IA , 3.3.3], we must have K ∼ = p L2 (p ). The lemma is proved.  Lemma 3.24. Suppose that K ∈ K3 , U4 (2) ↑3 K, m3 (K) ≤ 5, and for some M ≤ K, M/O2 (M ) ∼ = U6 (2). Then K ∼ = Co2 or U6 (2). Proof. Say U4 (2) ↑3 K via x ∈ I3 (Aut(K)). Since M exists, m3 (K) ≥ 4. From [IA , 5.3], ↑3 (U4 (2)) ∩ Spor = {Co2 }. Hence with Lemma 3.1, we may assume that K ∈ Chev(r), r = 2 or 3. By the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], if r = 3 then K ∼ = B2 (33 ), so by [IA , 3.3.3], m3 (K) = 9, contrary to assumption. Likewise if r = 2 and x ∈ Inndiag(K), K ∼ = U4 (8) by [IA , 4.9.1, 4.9.2], and m3 (K) = 3 by [IA , 4.10.3a], contradiction. So we may assume that x ∈ Inndiag(K), whence q(K) = 2 by [IA , 4.2.2]. Since U4 (2) is not a linear group, neither is K. If K/Z(K) ∼ = Un (2), then n ≥ 6 because of M , but n ≤ 6 as m3 (K) ≤ 5. By assumption Z(K) is a 3-group, so K ∼ = U6 (2). = U6 (2) or SU6 (2). The latter case is impossible since M/O2 (M ) ∼ Thus the lemma holds if K is a unitary group. As U4 (2) ↑3 K, the table [IA , 4.7.3A] shows that K is not of exceptional type. Hence, given the restrictions on m3 (K), we are reduced to considering K ∼ = C4 (2), C5 (2), D4 (2), D5± (2), or D6− (2). As 11 divides |U6 (2)|, we are further reduced to K ∼ = C5 (2), D5− (2) or D6− (2). Let P ∈ Syl3 (K). With [IA , 4.8.2a], we see that P contains a normal subgroup A ∼ = E35 such that A = CK (A) and NK (A)/A contains a copy of A5 permuting a frame of A naturally. We deduce that P ∼ = (Z3  Z3 ) × E32 . Thus P has an elementary abelian maximal subgroup. However, a Sylow 3-subgroup of M has no such subgroup since m3 (U6 (2)) = 4 and |U6 (2)|3 = 36 . This completes the proof.  Lemma 3.25. Suppose that L ↑3 K via x ∈ I3 (Aut(K)) and m3 (K) ≤ 4. If L ∼ =  Sp6 (8), then K ∼ Sp6 (2) but K ∼ = = Sp8 (2) or F4 (2). Moreover x acts on K like some  x0 ∈ I3 (K), and there are x1 , x2 , x3 ∈ I3 (K) such that O 2 (CK (xi , xj )) ∼ = Σ6 for all 0 ≤ i < j ≤ 3, and x0 , x1 , x2 , x3  ∼ = E34 is normalized by some Σ4 ∼ =Σ≤K permuting {x0 , x1 , x2 , x3 } naturally.

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Proof. Using [IA , 4.9.1, 4.9.2, 4.8.2, 5.3], and [IA , 4.10.3a] for ranks, we find that K ∼ = Sp8 (2) or F4 (2). By [IA , 2.5.12], Out(K) is a 3 -group, so as K is simple, x acts like some x0 ∈ I3 (K). Suppose first that K ∼ = Sp8 (2). Let V be the natural F2 K-module and let B be a maximal elementary abelian subgroup of K containing x0 . Then by [IA , 4.8.2], dim[V, x0 ] = 2 and B is the direct product of four conjugates xi  of x0 , 0 ≤ i ≤ 3.  Then dim[V, xi , xj ] = 4 for i = j so O 2 (CK (xi , xj )) ∼ = Sp4 (2) as claimed. Moreover, there is a basis of V for which the matrices of xi coincide when restricted to their supports. Then the desired Σ can be found as a group of permutation matrices, so the lemma holds in this case. Now suppose that K ∼ = F4 (2). There exist two conjugacy classes of subsystem subgroups M isomorphic to Sp8 (2), coming from subsystems of types C4 and B4 , and interchanged by a graph-field automorphism of K. Let M1 and M2 represent  these conjugacy classes. Now there is yi ∈ I3 (Mi ) such that Li := O 2 (CMi (yi )) ∼ = Sp6 (2), i = 1, 2. Moreover, we can take y1 and y2 to be interchanged by a graph field automorphism of K. Then it is clear from [IA , 4.7.3A] that Li = O 2 (CK (yi )), i = 1, 2, and y1 and y2 represent the only classes of elements of I3 (K) with such a component in their centralizers. So x0 is K-conjugate to either y1 or y2 . The rest of the proof may be carried out in this case within M1 or M2 , as in the previous paragraph. The lemma follows.  Lemma 3.26. Suppose that Sp6 (2) ↑2 K ∈ C2 , m3 (K) ≤ 4, and |K|3 ≥ 36 . Then K/Z(K) ∼ = U6 (2). Proof. We have K ∈ Chev(2) by [IA , 5.2.9, 5.3, 4.9.6]. By the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], K/Z(K) ∼ = Sp6 (4) or L± n (2), n = 6 or 7. Of these, 6 all but Un (2) have |K|3 | < 3 , and m3 (U7 (2)) = 6, yielding the lemma.  Lemma 3.27. Suppose that U4 (3) ↑3 K and m3 (K) ≤ 5. Then K ∼ = F i22 . Proof. By [IA , 4.9.6, 5.2.9, 2.2.10], K ∈ Spor ∪ Chev(3). If K ∈ Chev(3) then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], K ∼ = U4 (33 ), so m3 (K) = 12 by [IA , 3.3.3]. Therefore K ∈ Spor, and the result follows from the tables [IA , 5.3].  Lemma 3.28. Suppose that E(X) = K ∈ C3 , O3 (X) = 1, Z(X) = 1, and t ∈ I2 (X) is such that [K, t] = 1 and E(CK (t)) has a component isomorphic to U4 (2). If m2,3 (X) ≤ 5, then X/Z(X) ∼ = L± 4 (3), Ω7 (3), or P Sp4 (9). Proof. Since K/O3 (K) ↓2 U4 (2), K ∈ Chev by [IA , 5.2.9, 5.3]. Similarly, using [IA , 4.9.1, 4.9.2, 4.9.6] and the definition of C3 [V3 , 1.1], K ∈ Chev(3). If t ∈ Aut0 (K), then K ∼ = P Sp4 (9). So assume that t ∈ Aut0 (K) and use [IA , 4.5.1]. The result is that K/Z(K) ∼ = P Ω± k (3) for some k ≥ 6. If k is even and k ≥ 8, then K t contains a reflection r, and 5 ≥ m2,3 (X) ≥ m3 (CK (t)Z(X)) ≥ 1+m3 (Ωk−1 (3)) ≥ 6, contradiction. If k ≥ 9, then K/Z(K) contains a copy of Ω+ 8 (3), and we again reach the contradiction m2,3 (X) ≥ 6, with [IA , 3.3.3]. The lemma is proved.  Lemma 3.29. Suppose that J is one of the sporadic groups in [V5 , Table 7.1], and J ↑3 H ∈ C3 . Then m2,3 (H) − m3 (Z(H)) ≥ m, where m is as in that table. Proof. By Lemmas 3.8 and 3.9, (J, H) = (3Suz, Co1 ), (F3 , F1 ), or (3F i24 , F1 ). Then Z(H) = 1 and m2,3 (H) = m = 4 or 6, by [IA , 5.6.2], proving the lemma. 

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3.4. Pumpups and Schur Multipliers. Lemma 3.30. Let K = 2U6 (2) and γ ∈ I2 (Aut(K)) with CK/Z(K) (γ) ∼ = Sp6 (2). Then CK (γ) ∼ = Z2 × Sp6 (2). Proof. As Sp6 (2) = O 2 (Sp6 (2)), the only alternative is that CK (γ) ∼ = 2Sp6 (2), in which case m2 (CK (γ)) = 4 by [IA , 6.4.4c]. Let T ∈ Syl2 (K) be γ-invariant. By Lemma 2.10c, J(T ) ∼ = E210 . Hence m2 (CT (γ)) ≥ 5, contradiction. The proof is complete.  Lemma 3.31. ↑2 (2Sp6 (2)) = {[2 × 2]D4 (2), Co3 }. Proof. Let 2Sp6 (2) ↑2 K ∈ K2 via t ∈ I2 (Aut(K)). By [IA , 4.9.6, 5.2.9] and the Borel-Tits theorem, either K ∈ Chev(2) with Z(K) = 1, or K ∈ Spor. In the latter case K ∼ = [2 × 2]D4 (2) by = Co3 by [IA , 5.3]. In the former case, K ∼  [IA , 6.3.3b]. Lemma 3.32. Let K ∈ K2 and t ∈ I2 (Aut(K)). Let K1 ∼ = 2F i22 be a component of CK (t). Then Z(K1 ) ≤ Z(K). Proof. By Lemma 3.1, K ∈ Spor. The result follows by inspection of the  tables [IA , 5.3]. Lemma 3.33. Let K ∈ K and t ∈ I2 (Aut(K)) with CK (t) having a component J∼ = 2U6 (2). Then Z(J) ∩ Z(K) = 1. Proof. Without loss, O2 (K) = 1. If K ∈ Chev(2), the assertion holds by [IA , 6.3.3b]. Otherwise, by [IA , 4.9.6, 5.2.9], K ∈ Spor and if the lemma fails, then U6 (2) ↑2 K/Z(K). But this is impossible by inspection of [IA , 5.3]. The lemma follows.  Lemma 3.34. Let K = 2Sp6 (2) and x ∈ I3 (K) with E(CK/Z(K) (x)) ∼ = A6 . Then E := E(CK (x)) ∼ = SL2 (9). Proof. We use the embedding K → L ∼ = Spin7 (3) [IA , 6.2.2]. Let L = L/Z(L), so K ∼ = Sp6 (2). By the Borel-Tits theorem, E ≤ P for some parabolic (∞) ∼ Ω5 (3) ∼ subgroup P ≤ L. The only choice, by orders, is P = = P Sp4 (3), and then the full inverse image P of P satisfies Z(L) ≤ P (∞) and P (∞) /O3 (P ) ∼ = Spin5 (3) (see [IA , 6.2.1]). Hence, m2 (Z(L)E) ≤ m2 (Spin5 (3)) = 2, and the result follows.  Lemma 3.35. Let K = 2D4 (2) and set K = K/Z(K). Let x1 , x2 , x3 ∈ I3 (K) be such that x1 , x2 , x3 represent the three conjugacy classes in I3 (K) such that   K i := O 2 (CK (xi )) ∼ = U4 (2). Then, of the three inverse images Ki = O 2 (CK (xi )), one is isomorphic to U4 (2), and two are isomorphic to Sp4 (3).  = [2×2]D4 (2) and let τ ∈ Aut(K) be a graph automorphism of Proof. Let K order 3. Then τ cycles the conjugacy classes of x1 , x2 , and x3 [IA , 4.7.3A]. Also τ  and cycles the three involutions of Z(K)  [IA , 6.3.1]. It therefore suffices acts on K  of each K i contains Sp4 (3). If this is false, then  i in K to show that the preimage K i ∼  contains an E24 subgroup E disjoint from Z(K).  K = U4 (2) × Z(K)  However, K embeds in L := Spin8 (3) and so we may regard E ≤ L with E ∩ Z(L) = 1. Let z ∈ Z(L)# and set L = L/ z ∼ = Ω+ 8 (3). By [IA , 6.2.1], all involutions in E ∼ = E24 have trace 0 on the natural L-module. But this is impossible,

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as can be seen by lifting the character on E to characteristic 0 and computing the inner product with the principal character: the result is 1/2. This completes the proof.  Lemma 3.36. Suppose that K ∈ K, Z(K) has odd order, t ∈ I2 (Aut(K)), and CK (t) has a component I ∼ = 6U4 (3). Then K ∼ = 3Ω7 (3). Proof. Clearly K ∈ Chev(3) ∪ Spor, by the structure of CK (t). Since Z(K) has odd order, 2U4 (3) ↑2 (K/Z(K)). But this condition fails if K ∈ Spor [IA , 5.3], so K ∈ Chev(3). If K ∈ Lie(3), then I ∈ Lie(3) as well by [IA , 4.2.2]. But Z(I) has an element of order 3, and therefore Z(K) does too. Hence by [IA , 6.1.4], K/Z(K) ∼ = 3Ω7 (3). Only the last group properly involves = U4 (3) or G2 (3), or K ∼  U4 (3), by order considerations. This completes the proof. Lemma 3.37. Suppose that K ∈ Chev(2) ∩ C2 , b ∈ I3 (Aut(K)), and I is a component of CK (b). Then I ∼ = 2Sp6 (2) or 22E6 (2). Proof. Suppose false. By [IA , 4.2.2], IZ(K)/Z(K) ∈ Lie(2), so Z(I) ≤ Z(K). In particular Z(K) = 1. Using [IA , 6.1.4] to give the possibilities for K, and then [IA , 4.2.2] for the possibilities for I, we see that I ∼ = 2Sp6 (2) and K ∼ = 2F4 (2), and Z(I) = Z(K) =: Z. By [IA , 6.4.1] and the remark following it, every 2-central involution of K splits over Z but not every 2-central involution of I splits over Z. However, a Sylow 2-center of I is generated by root groups for an orthogonal shortlong root pair, and so it is also a Sylow 2-center for K. This is a contradiction, and the lemma follows.  Lemma 3.38. Let K ∈ C2 , t ∈ Aut(K) with t2 = 1, and I  E(CK (t)). Then the following hold. (a) If I ∼ = 2Suz, 2M12 , 2J2 , 2L3 (4), or 2Sp6 (2), then Z(I) ≤ Z(K); and (b) If I ∼ = 2F2 with Z(I) ≤ Z(K), then I = K. Proof. If K ∈ Chev, then I ∼ = 2L3 (4) or 2Sp6 (2), so I is unambiguously in Chev(2), whence K ∈ Chev(2). Therefore Z(I) ≤ Z(K) by the Borel-Tits theorem. Obviously K ∈ Alt as I ∈ Alt, so we may assume that K ∈ Spor. The result then  follows by examination of the tables[IA , 5.3]. 3.5. Pumpups of A6 . Lemma 3.39. Suppose that A6 ∼ = L ↑2 K ∈ K2 . Then one of the following holds: ∼ (a) K = A2n , n ≥ 5; (b) K ∼ = HS or 2HS; (c) K ∼ = L± n (2), n ∈ {4, 5}, or Sp4 (4); (d) K ∼ L = 2 (34 ) or L± 3 (9); ± (e) K is a quotient of Ω± n (3), n ≥ 6, or Ωm (9), m ≥ 5. Moreover, if K ∈ C2 , then CAut(K) (L) embeds in D8 . ∼ L2 (9) ∼ ∼ Proof. We have A6 ∼ = Ω− = [B2 (2), B2 (2)] = 4 (3) = Ω3 (9). If K ∈ Alt ∪ Spor − Chev(2), then (a) or (b) holds by [IA , 5.2.8, 5.3]. If K ∈ Chev(2), then (c) holds by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2]. If K ∈ Chev(r) for some r > 2, then r = 3 by [IA , 4.9.6] and inspection of [IA , 4.9.1, 4.5.1, 4.5.2] completes  the proof. (Note that A8 and Ω5 (3) ∼ = U4 (2) appear in (c).)

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Lemma 3.40. Let X = K x with x ∈ I2 (X), K ∈ K2 , and CX (x) = x × L with L ∼ = A6 . Suppose also that K embeds in GL8 (2). Then K = L. Proof. Note that if Z(K) = 1, then x ∈ Z(K) and the result is trivial. Suppose that Z(K) = 1 and K > L. We refer to Lemma 3.39. If K ∼ = A6+2k , k ≥ 1, then x is the product of k disjoint transpositions and CK (x) contains Σ6 , contrary to assumption. If K ∼ = HS, then CK (x) contains a copy of Aut(L), contradiction. If Lemma 3.39c holds, then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], CK (x) ∼ = Sp4 (2) ∼ = Σ6 , contradiction. In Lemma 3.39d, CK (x) contains P GL2 (9), contradiction. In Lemma 3.39e, |K|3 ≥ 36 , contradicting the fact that  K embeds in GL8 (2). The proof is complete. Lemma 3.41. Suppose that A6 ↑2 L1 ↑2 L2 and A6 ↑2 L2 , with L1 , L2 ∈ C2 . Also  2 ) < 4 for some covering group L  2 of L2 . Then L2 ∼ assume that m3 (L = HS or 2HS. Proof. By assumption, {A6 , L1 } ⊆ ↓2 (L2 ) with L1 ∼  A6 . If L2 satisfies = Lemma 3.39a, then L2 ∈ C2 , contradiction. If L2 satisfies Lemma 3.39c, then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], ↓2 (L2 ) = {A6 }, a contradiction. If L2 satisfies Lemma 3.39de, then m3 (L2 ) ≥ 4 by [IA , 3.3.3]. Moreover, the only  2 of L2 by a 3-group are 3U4 (3), 32 U4 (3), and 3Ω7 (3), nontrivial covering groups L  2 ) ≥ 4 by [IA , 6.4.4]. Hence these groups cannot occur, by [IA , 6.1.4], and so m3 (L HS or 2HS by Lemma 3.39, as claimed.  so L2 ∼ = Lemma 3.42. Suppose K ∈ C3 , K  K y with y 2 = 1, andK > J := E(CK (y)) ∼ = A6 , and m2 (CKy (J)) = 1. Then K/Z(K) ∼ = P Sp4 (3), U5 (2), L± 4 (3), L2 (34 ), or L± 3 (9). Proof. By Lemma 3.39, the only possible alternative is that K is a quotient ± of Ω± n (3), n ≥ 7, or Ωm (9), m ≥ 5. In either case y must act as an involution ± ± (9) of On (3) with a 4-dimensional eigenspace of − type or as an involution of Om with a 3-dimensional eigenspace. In both cases there clearly exists a noncentral ± involution of Ω± n (3) or Ωm (9) centralizing y but acting on K distinctly from y, so m2 (CKy (J)) > 1, contradiction.  Lemma 3.43. Suppose that A6 ↑2 K ∈ C2 − Chev(3) and m3 (K) < 4. Then K∼ = L4 (2), L5 (2), Sp4 (4), HS, or 2HS. Proof. With Lemma 3.39, the condition A6 ↑2 K ∈ C2 − Chev(3) implies that K is as claimed, or else K ∼ = U5 (2). By [IA , 4.10.3], m3 (U5 (2)) = 4, so the lemma follows.  Lemma 3.44. Let L = A6 and K ∈ C3 . Then the following conditions hold: (a) Suppose that L ↑3 K via x. Then K ∼ = L2 (36 ), A9 , J3 , 3J3 , Suz, 3Suz,  O N , Sp4 (8), or Sp6 (2); (b) In (a), m3 (CAut(K) (L x)) = 1 or 2, the latter occurring only if K ∼ = Suz, 3Suz, or O  N ; and (c) If, instead, L |J|2 , so L is a single 2-component. In particular, K ≤ L.  |K/Z(K)|  In every case  J 2 E(CJ (w)) ≤ K 0 = L, so [L, t] = 1 by the previous paragraph. Hence, H = Kw K ↑2 L. Using Lemma 3.10 and the fact that |J0 | divides |J|, we get with one exception that L ∼ = J, whence J0 = J, as desired. The exception occurs for K ∼ = F i24 , with L ∼ = F i23 . However, we also have CL (w)   = 2F i22 and J ∼ CJ0 (w) ≥ E(CJ (w)), whence in this case, CL (w) has a D4 (3) component. But it is false that D4 (3) ↑3 F i23 . This completes the proof of (c). Part (d), for the cases in which K is not a Fischer group, follows from information in [IA , 4.5.1, 5.3]. The remaining cases follow from the observation that 3 ∼ − if L = Ω+ 8 (3), u ∈ I2 (Aut(L)), and M := O (CL (u)) = Ω6 (3), then CAut(L) (M ) embeds in D8 . This holds because for a suitable natural module V for L, u acts on L like an involution of O(L) with a 2-dimensional eigenspace Vu on V ; and then CAut(L) (M ) embeds in O(Vu ) ∼ = O2− (3) ∼ = D8 . Finally, it is clear that CK (w) has sectional 2-rank > 2, which implies (e).  Lemma 3.47. Let K ∈ K2 be simple. Suppose that t ∈ I2 (Aut(K)) with L := E(CK (t)) ∼ = A8 and |CK (L)|2 ≤ 2. Suppose also that b ∈ I3 (Aut(K)) with M := E(CK (b)) ∼ = A5 and m3 (CK (M b)) ≤ 1. Then K ∼ = HS.  HS, then K ∼ Proof. By [III11 , 1.6], if K ∼ = = An for some even n > 8 or K ∼ = L4 (4). In the latter case, however, b ∈ K and m3 (CK (b)) = 3 so m3 (CK (M b)) = 2, contrary to assumption. Taking into account that A5 ↑3 K, we have K ∼ = A8+6k ,  k ≥ 1. But then |CK (L)|2 > 2, contradiction. The lemma is proved. Lemma 3.48. Let M ∈ K3 be a 3-pumpup (possibly trivial) of L = P Sp4 (3), L4 (3), or a covering group of U4 (3). Then it is not the case that M/O3 (M ) ↓2 A5 . Proof. Suppose that M/O3 (M ) ↓2 A5 . If M ∈ Chev(3), then by [IA , 4.9.6], A5 ∈ Chev(3), which is not the case, contradiction. If M ∈ Chev(2), then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], M/O3 (M ) ∼ = L2 (16) or L± 3 (4), none of which involves U4 (2) by Lagrange’s theorem, contradiction. Since L ∈ Chev(r)∪Alt for any r > 3, M ∈ Chev(r) ∪ Alt. Hence M ∈ Spor. The A5 condition forces

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M/Z(M ) ∼ = M12 , J1 or J2 . But none of these three involves L by Lagrange’s theorem, contradiction. The proof is complete.  Lemma 3.49. Suppose that K ↑2 H ∈ C2 via t ∈ I2 (Aut(H)), b ∈ Aut(H) with b3 = 1, and J is a component of CH (b) with J ∈ C3 . Then the following conditions hold: (a) If (K, J/Z(J)) is one of the following, (L2 (8), G2 (3)), (A6 , L2 (34 )), ± ± ± ± (A6 , L± 3 (9)), (U4 (2), L4 (3)), (L3 (3), L3 (9)), or (L3 (3), L4 (3)), then ± ∼ ∼ H/Z(H) = J/Z(J) = L4 (3) or G2 (3); and (b) If (K, J) = (A6 , U4 (2)) and [b, u] = 1, then H ∼ = J. Proof. Note that J/Z(J) ∈ C3 . If H/Z(H) ∼ = J/Z(J), then J/Z(J) ∈ C2 , so ∼ J/Z(J) ∼ (3) in (a), and H J in (b), as Sp4 (3) ∈ C2 . So we may = G2 (3) or L± = 4 assume that J ↑3 H/Z(H). If H ∈ Chev(3), then since H ∈ C2 , Out(H) is a 3 -group. Therefore, ↓3 (H/Z(H)) = ∅ by the Borel-Tits theorem, a contradiction. We reach the same contradiction if H ∼ = L2 (q), q ∈ FM9. If H ∈ Spor, then checking the tables [IA , 5.3], we see that the conditions H ∈ C2 , K ↑2 H and J ∈ ↓3 (H/Z(H)) ∩ C3 are inconsistent. Hence we may assume that H ∈ Chev(2), whence J ∈ Chev(2). Thus (K, J) = (A6 , U4 (2)). By Lemma 3.39, H ∼ = L± n (2), n = 4, 5, or Sp4 (4), and as H ↓3 U4 (2), the only possibility is H ∼ = U5 (2). We must derive a contradiction from the assumption [b, u] = 1. By the Borel-Tits theorem and [IA , 2.5.12], u is a graph automorphism, induced, say, by an automorphism of F4 . The eigenvalues of b are ω, ω, ω, ω, ω −1 , where ω 3 = 1, so b is not real in K; but bu is conjugate to b−1 . A  fortiori, bu = b. Lemma 3.50. Let K ∈ C2 and let a and y be commuting automorphisms of K of orders p and 2, where p is an odd prime. Let L be a component of CK (y) and assume that L is also a component of CK (a). Then (L, p) is not any of the 1 5 2 2 2  2 following: (A6 , 3), (L± 3 (3), 3), (L2 (8), 3), ( F4 (2 ) , 5), ( B2 (2 ), 5). Proof. Our hypotheses imply that mp (Aut(K)) ≥ 2, so K ∼  L2 (p), p ∈ FM. = As K ∈ C2 , K ∈ Chev(3), by the Borel-Tits theorem if p = 3 and since L ∈ Chev(3) if p = 5. If K ∈ Spor, then since L ↑2 K, the only possibility by [IA , 5.3] is K/Z(K) ∼ = HS with L ∼ = A6 ; but HS is not a 3-pumpup of A6 , contradiction. Therefore, by definition of C2 [V3 , 1.1], K ∈ Chev(2) − Chev(3). By the Borel-Tits theorem, y induces a graph, field, or graph-field automorphism on K, with fixed points given in [IA , 4.9.1, 4.9.2]. If L ∼ = A6 , then it follows that K ∼ = Sp4 (4) or L± n (2), n = 4 or 5, so m3 (Aut(K)) = 2 and so Aut(K) cannot 1 5 2 2 2  2 contain Z3 × A6 . If L ∼ = L± 3 (3), F4 (2 ) , or B2 (2 ), the only possibilities are K ∼ = G2 (4), F4 (2), and B2 (25 ), respectively, and we reach similar contradictions. Finally, suppose that L ∼ = L2 (8). Then K ∼ = L2 (82 ) or L3 (8), giving the usual ∼ contradiction, or K/Z(K) = U3 (8). In this final case, any involution y such that CK (y) has an L2 (8) component satisfies CAut(K) (y) ∼ = Z2 × Aut(L2 (8)), so the condition [a, y] = 1 cannot be satisfied. The proof is complete.  3.7. Action of CAut(K) (x) on E(CK (x)). Lemma 3.51. Let K = F ∗ (X) ∈ K. Let b ∈ I3 (X) and set I = E(CX (b)) and m = m3 (X). If K and I are as in [V4 , Table 1.1], and m has the value of m2,3 (G) in that table, then b ∈ Syl3 (CX (I)).

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Proof. It is equivalent to show that b ∈ Syl3 (CX (I b)). If | Out(K)| is not divisible by 3, then the assertion is that (3B)

b ∈ Syl3 (CK (I b)).

This holds, by [IA , 4.8.2] in the classical cases, [IA , 4.7.3A] in the nonclassical Lie type cases, and the sporadic tables [IA , 5.3]. If K/Z(K) ∼ = 2E6 (2) or U6 (2), then m3 (Inndiag(K/Z(K))) = 1 + m3 (K/Z(K)) = 6 or 5, respectively. 

As | Outdiag(K)| = 3 = | Out(K)|3 , we must have O 3 (X) = K, so again it is (3B) that must be proved. Again it follows from the information in [IA , 4.7.3A, 4.8.2].  Lemma 3.52. In Lemma 3.4, suppose that r = 2, p > 3, and there exists J ∈ Cp and t ∈ Aut(J) such that t2 = 1 and CJ (t) has a component I0 such that I0 /Z(I0 ) ∼ = I/Z(I). Suppose also that mp (CAut(K) (x)) ≥ 4. Then mp (CAut(K) (I)) = 1. Proof. We have mp (Inndiag(K)) ≥ 4 − 1 = 3, so K is not a Suzuki or Ree group. Also by [IA , 4.2.2], I ∼ = G2 (q) for any q. Suppose that the lemma is false. If K admits a field automorphism of order p, then q(I) ≥ q(K) ≥ 2p , by [IA , 4.2.2]. On the other hand if K admits no such field automorphism, then since the lemma fails, mp (Outdiag(K)) = 1, whence q(I) ≥ q(K) > 2 and mp (Aut(I)) ≥ mp (Inn(K)) − 1 ≥ 2. It follows in either case by [IA , 2.2.10] that I ∈ Alt ∪ Chev(r) for any r > 2, and hence J ∈ Chev(p) ∪ Alt. Also if J ∈ Chev(2), then as J ∈ Cp , J ∼ = L2 (p) for p = 5 or 7 by [IA , 2.2.10], whence I = J and mp (Aut(I)) = 1, contradiction. Therefore J ∈ Spor. But then we see by inspection of [IA , 5.3] that it is false that q(I) ≥ 2p . Thus q(I) > 2 and mp (Aut(I)) ≥ 2, and the only possibility is p = 5, J ∼ = Co1 , and I ∼ = G2 (4), which we have seen cannot occur. This contradiction completes the proof.  Lemma 3.53. Suppose that J ∈ Chev(3), t ∈ I2 (Aut(J)), and K is a component of CJ (t); and specifically that (J/Z(J), K) is one of the pairs (G2 (3), L2 (8)), ± ± ± ± (L2 (34 ) or U4 (2) or L± 4 (3) or L3 (9), A6 ), (L4 (3), U4 (2)), (L4 (3) or L3 (9), L3 (3)). Set C = CAut(J) (K t). Then C is a 2-group. Either C is cyclic or (J/Z(J), K) = (L± 4 (3), A6 ), in which cases C at any rate embeds in D8 . Proof. In the cases J ∼ = G2 (3), L2 (34 ), and L3 (9), t ∈ Aut0 (J) and so C = t ± by [IA , 7.1.4c]. If J ∼ L (9) or K ∼ = 3 = U4 (2), then t is a graph automorphism and so U (2), then C = t by [IA , 4.9.2]. If J/Z(J) ∼ C = t by [IA , 4.5.1]. If J ∼ = 4 = L4 (3)  (3),  = ±1, then again from [I , 4.5.1], C is cyclic and |C| = 3 − . and K ∼ L = 3 A  ∼ (3), K A ,  = ±1. For each group J It remains to consider J/Z(J) ∼ L = 6 = 4 there are two possible classes of automorphisms t. One is a graph automorphism t, and we quote [IA , 4.5.1] to get C = t in that case. The other is an inner-diagonal automorphism t of J/Z(J) ∼ = P Ω6 (3) induced by an involution in O6 (3) with a −1-eigenspace W of dimension 2 and type −. Then C embeds in O(W ), which  embeds in D8 . Lemma 3.54. Suppose K ∈ C2 , t ∈ I2 (Aut(K)), and CK (t) has a component L∼ = L± 3 (3). Then CAut(K) (L t) is a 2-group.

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∼ L± (3) and the Proof. If K ∈ Chev − Chev(2), then as K ∈ C2 , K/Z(K) = 4 assertion holds; see [IA , 4.5.1]. If K ∈ Chev(2), then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2, 2.2.10], K ∼ = G2 (4) and [IA , 7.1.4c] yields the result. Finally, by [IA , 5.2.9, 5.3], there are no such groups K ∈ Alt ∪ Spor. The lemma is proved.  Lemma 3.55. Suppose that J ∈ C3 , t ∈ I2 (Aut(J)) and I = E(CJ (t)). Let Q ∈ Syl2 (CAut(J) (I)). If J ∼ = D4 (3) and I ∼ = 2U4 (3), then Q ∼ = D8 . In the following cases, on the other hand, Q is elementary abelian: (J/Z(J), I) = (F i24 , 2F i22 ), (D4± (3), Ω7 (3)), (F i22 , 2U6 (2) or D4 (2)), (L4 (9), U4 (3)), (Ω7 (3), 2U4 (3)), (Ω− 8 (3), 2U4 (3)), (U6 (2) or D4 (2), Sp6 (2)), (3), U (2)). and (L± 4 4 Proof. Since Aut(J) embeds in Aut(J/Z(J)), we may assume that Z(J) = 1. + ∼ If J ∼ = P Ω+ 8 (3) and I = 2U4 (3), then on some natural module V for O8 (3), t acts like an involution with eigenspaces of − type and dimensions 6 and 2. Then ∼ ∼ + ∼ Q∼ = D8 . Similarly if J ∼ = Ω− = O2− (3) ∼ 8 (3) and I = 2U4 (3), Q = O2 (3) = E22 . In ∼ the remaining cases in which I = Ω7 (3), 2U4 (3), or U4 (2), considering J/Z(J) as a projective orthogonal group, we have that t acts like a reflection and so generates CAut(J) (I). The cases I ∼ = U4 (3) and Sp6 (2) give |Q| = 2 by [IA , 4.9.1, 4.9.2], respectively. If J ∼ = F i22 , then |Q| = 2 by [IA , 5.3t]. If J ∼ = F i24 , then CInn(J) (I) = ∼ t = Z2 , and some involution u ∈ Aut(J) − Inn(J) centralizes F i23 , which in turn  contains a conjugate of I, so CAut(J) (I) ∼ = E22 . Lemma 3.56. Let J, JD ∈ C3 be as in [V5 , Table 7.1]. Then Sylow 3-subgroups of CAut(J) (JD ) are of exponent 3. Proof. If J/Z(J) ∼ = U6 (2) and JD ∼ = U4 (2), then   O 3 (CAut(J) (JD )) = O 3 (CP GU6 (2) (JD )) ∼ = E32 .  ∼ Z3 , as follows from [IA , 4.8.2] when J is a In all other cases, O 3 (CAut(J) (JD )) = classical group, and [IA , 5.3klovxyz] when J ∈ Spor. 

Lemma 3.57. Suppose K ∈ Cp , p odd, K  X, and x and y are commuting elements of Ip (X) such that [x, K] = 1 and E(CK (x)) ∼ = L2 (pp ). Then y does not induce a nontrivial field automorphism on E(CK (x)). Proof. By Lemma 3.11d, K ∼ = L2 (pp ) with x inducing a field automorphism of order p on K. But then the subgroup of Aut(K) generated by elements of order p is the image of K x. Hence y acts on E(CK (x)) like an element of K x and hence like an element of K. But AutK (E(CK (x))) ∼ = P GL2 (pp ) by [IA , 6.5.1], which contains no field automorphism of order p, proving the lemma.  2

Lemma 3.58. Let K ∈ C2 and K  KB with B ∈ Ep∗ (KB) for some odd prime  p. Suppose that I is a component of O p (E(CK (b))) for some b ∈ B # . Suppose that some element of B induces a non-inner automorphism on I. Then K ∈ Chev(2). Proof. Our hypotheses imply that mp (Aut(K)) ≥ 2. Assuming K ∈ Chev(2), it then follows from the Borel-Tits theorem, [IA , 4.10.3a], and the definition of C2 [V3 , 1.1], that K ∈ Spor. Surveying [IA , 5.3] we see that the only case in which some x ∈ Ip (K) induces an outer automorphism on I is K ∼ = F i24 , with p = 3, I ∼ = D4 (3), and x inducing a graph automorphism on I. But then CK (b, x) ∼ = E32 × CI (x). As b, x ≤ B = (B ∩ K) × CB (K) with m3 (B ∩ K) = m3 (K) = 7 by [IA , 5.6.1],

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we have m3 (AutB (CI (x))) ≥ 5. However, by [IA , 4.9.2beg], m3 (Aut(CI (x))) =  m3 (Aut(G2 (3))) = 4, a contradiction. The proof is complete. Lemma 3.59. Let K = L4 (9) and X = K t where t ∈ I2 (Aut(K)) is a graphfield automorphism. Then there is u ∈ I2 (CK (t)) such that H := E(CK (u)) ∼ = L3 (9) and E(CH (t)) ∼ = U3 (3). Proof. By [IA , 4.9.1], t is determined up to Aut(K)-conjugacy, so we may assume that K is given by 4 × 4 matrices and At = ((A(3) )T )−1 for all A ∈ K, where the map A → A(3) cubes all entries of A. Then u = diag(1, 1, 1, −1) satisfies all the requirements.  Lemma 3.60. Let K = Ω2k (3), k ≥ 3,  = ±1. Suppose that X = K t where t acts on K like a reflection on the natural K-module V . Then there is u ∈ I2 (K) such that H := E(CK (u)) ∼ = Ω− 2k−2 (3) and [t, u H] = 1. Moreover, unless possibly K ∼ = D8 , and if, further,  = (−1)k+1 , = D4 (3), we have CAut(K) (H) ∼ ∼ then CInn(K) (H) = Z4 . Proof. Let v ∈ V generate the center of t. Choose w ∈ v ⊥ that is isometric to v, set W = v, w ⊆ V , and let u ∈ K invert W elementwise and centralize W ⊥ . Since v and w are isometric, W is of − type, so H = E(CK (u)) ∼ = Ω(W ⊥ ) ∼ = − Ω2k−2 (3). It is obvious that [t, u H] = 1. Finally, unless possibly K ∼ = D4 (3), CAut(K) (H) ∼ = O(W ) ∼ = O2− (3) ∼ = D8 ; and if  = (−1)k+1 , then K ∼ = P SO(V ), which implies the final assertion. This completes the proof.  Lemma 3.61. Let K = U4 (3) and X = Aut(K), and let v ∈ I2 (X) with L := E(CX (v)) ∼ = P Sp4 (3). Let z be a 2-central involution of L. Then E(CX (vz)) ∼ = P Sp4 (3). Proof. This follows from Lemma 10.1f below. One can also argue directly that  := 2U4 (3) = Ω(V ), where V is a 6-dimensional v is the image of a reflection v ∈ K  covers K. Then V = [V, v] ⊥ CV (v), and orthogonal space of − type and K ∼   for L := CK ([V, v]) = Ω(CV (v)) covers L. Now z is the image of −1CV (v) u ∈ L  Hence −1V u some reflection u  ∈ L. v is a reflection on V , and maps on vz, which implies the result.  Lemma 3.62. Let K = U4 (3) and let u, v be a four-subgroup of Aut(K). Suppose that E(CK (u, v)) = E(CK (v)) ∼ = A6 . Then E(CK (u)) ∼ = P Sp4 (3). Proof. Let the orthogonal space V =⊥6i=1 Vi be the orthogonal sum of six lines Vi = F3 wi with (wi , wi ) = 1 for i = 1, 6, and (wi , wi ) = −1 for 1 < i < 6. Set H = O(V ) ∼ = O6− (3), so that K ∼ = [H, H]/Z([H, H]). By [IA , 4.5.1], the conditions L := E(CK (v)) ∼ = A6 ∼ = Ω− 4 (3) and m2 (CAut(K) (L)) > 1 determine v up to Aut(K)conjugacy. We may therefore assume that v acts on K like the involution y of H whose −1-eigenspace is V1 ⊥ V2 . Then L is naturally identified with Ω((V1 + V2 )⊥ ), and so u and uv act as scalars (±1) on (V1 + V2 )⊥ . The only possibility is that  u ↓V1 +V2 is a reflection, and the result follows directly. Lemma 3.63. Let K = F i23 and let z, t ∈ I2 (K) be such that CK (z) ∼ = 2F i22 and E(CK (t)) ∼ = 22 U6 (2). Then Z(E(CK (t))) contains a conjugate of z.

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Proof. Assume false. We use [IA , 5.3u]. Let Kt = E(CK (t)) and Z = Z(Kt ). All involutions u ∈ Z satisfy E(CK (u)) = 1 by L2 -balance, so under our hypothesis they must all be conjugate to t. Hence they are all conjugate in NK (Kt ), under a 3-element w. It follows that w can be chosen so that CKt (w) ∼ = U5 (2). However, 11 divides |U5 (2)| but does not divide the order of the centralizer of any 3-element of K, a contradiction proving the lemma.  Lemma 3.64. Let K ∈ C3 be a component of X, let v ∈ I2 (X), and let H be a component of CKK v (v). Also let t ∈ I2 (CX (v)) and suppose that CX (t) is solvable.  SL3 (4). Then H/O3 (H) ∼ = Proof. Suppose that H/O3 (H) ∼ = SL3 (4). Since no covering group of L3 (4) lies in C3 , K v = K. Likewise K t = K as CX (t) is solvable. We then have K/O3 (K) ↓2 L with K/O3 (K) simple in C3 and L a covering group of L3 (4) by a (central) 2-group. If K ∈ Spor ∪ Alt, then by the definition of C3 and the information in [IA , 5.3, 5.2.8], K ∼ = Suz or O  N ; but then CK (t) is nonsolvable, contradiction. As L ∈ Chev(r) only for r = 2, we must have K ∈ Chev(2) − Chev(r) for all r = 2. Then by the Borel-Tits theorem there must exist an involution z ∈ Aut(K) − Inndiag(K) with CK (z) having a component isomorphic to L. Using [IA , 4.9.1, 4.9.2] we see that the only possibility is that K/Z(K) ∼ = L3 (42 ). But then by [V3 , 1.1], K ∈ C3 , contrary to hypothesis. The lemma is proved.  Lemma 3.65. Let K = L± 4 (3), Ω7 (3), or P Sp4 (9), and let t ∈ I2 (Aut(K)) be such that CK (t) has a component I ∼ = U4 (2). Then CAut(K) (I) is a 3 -group. Proof. Suppose false. In all cases Out(K) is a 2-group by [IA , 2.5.12], so CInn(K) (I) is not a 3 -group. By the Borel-Tits theorem, b × I ≤ P for some parabolic subgroup P and some b ∈ I3 (O3 (P )). Since I ≤ P , the only possibility  is K = Ω7 (3) and O 3 (P/O3 (P )) ∼ = I. But in that case O3 (P ) ∼ = E35 is a natural ∼ module for I = Ω5 (3), whence b ∈ CO3 (P ) (I) = 1, a contradiction. The lemma is proved.  Lemma 3.66. Let K ∈ C3 and suppose that t ∈ I2 (K) − Z(K) with t ∈ E(CK (t)) ∼ = SL2 (9). Then K ∼ = L± 3 (9) and CAut(K) (E(CK (t))) is a {2, 5}-group. Proof. We may pass to K/O3 (K) and assume that K is simple. Since E(CK (t)) ∈ C2 , K ∈ C2 . As K is simple, K ∈ Chev(2). As A6 ∈ Chev(r) for r > 3, K ∈ Chev − Chev(3). Likewise as K is simple, K ∈ Alt, and K ∈ Spor by examination of [IA , 5.3]. Thus K ∈ Chev(3). As t ∈ K by assumption, K∼ = L± 3 (9) by examination of [IA , 4.5.1]. As Out(K) is a 2-group by [IA , 2.5.12], ∼  and CK (E(CK (t))) ∼ = GL± 1 (9) = Z9∓1 , the lemma follows. Lemma 3.67. Suppose that I ∈ C2 and t ∈ I2 (Aut(I)), and CI (t) has a component J with J/Z(J) ∼ = M12 or J2 . Then there is no involution u ∈ Aut(I) such that CI (u) has a component isomorphic to A5 . Proof. Clearly I ∈ Spor and indeed by [IA , 5.3], I/Z(I) ∼ = Suz. But again  by [IA , 5.3], Aut(Suz) has no involution centralizer with an A5 component. Lemma 3.68. Let K = HS and t ∈ I2 (Aut(K)) with J := E(CK (t)) ∼ = A8 . Let z be a 2-central involution of J. Then z is 2-central in K.

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Proof. Otherwise, by [IA , 5.3m], z is a direct factor of CK (z). But z ∈ Z(O2 (CJ (z))) with O2 (CJ (z)) extraspecial, a contradiction. The lemma follows.  Lemma 3.69. Let K = 3D4 (2), X = Aut(K) and b ∈ I3 (K) with E(CK (b)) ∼ = L2 (8). Then there is t ∈ I3 (X) − K such that [t, b] = 1 and CK (t) ∼ = G2 (2). Proof. We use [IA , 4.7.3A]. Choose t ∈ I3 (X) − K with I := CK (t) ∼ = G2 (2). Let P ∈ Syl3 (K) contain Q ∈ Syl3 (I), so that Q ∼ = 31+2 . Now |P | = 34 and P contains a conjugate R ∼ = Z3 × Z9 of a Sylow 3-subgroup of CK (b). Then  Q ∩ R = Ω1 (R) contains a conjugate of b, and the lemma follows. Lemma 3.70. Let K = Suz or O  N . Let t ∈ I2 (Aut(K)) be such that CK (t) ≥ F ∼ = A6 for all f ∈ F # . Then CE(CK (F )) (t) = E32 , with E(CK (F )) = E(CK (f )) ∼ is a 3 -group.  Proof. If K = Suz, then O 3 (CK (F )) ∼ = E32 × A6 . As A6 has no element of order 6, the desired conclusion holds if t ∈ Inn(K). Also if t ∈ Out(K), it holds as well since m3 (CK (t)) = m3 (J2 ) = m3 (M12 ) = 2 < 3. If K = O  N , then for t ∈ Inn(K), m3 (CK (t)) = m3 (J1 ) = 1 < 3, while for  t ∈ Inn(K), m3 (CK (t)) = m3 (L3 (4)) = 2 < 3. This completes the proof. Lemma 3.71. Let K = U7 (2), 2E6 (2), or 2 D5 (2). Let b ∈ I3 (K) with I := E(CK (b)) ∼ = SU6 (2), U6 (2), or D4 (2). Then CAut(K) (b I) has odd order. The same conclusion holds if K ∼ = U6 (2), U5 (2), or D4 (2), with I ∼ = U4 (2). Proof. For K = 2E6 (2), this follows from the information in [IA , 4.7.3A]. In the other cases, by [IA , 4.2.2, 4.8.2], I is the only Lie component of CK (b), whence CInndiag(K) (b) induces only inner-diagonal automorphisms of odd order on I. In every case CAut(K) (b) = CInndiag(K) (b) γ, where γ 2 = 1 and γ is a graph automorphism of K of order 2. In the orthogonal group cases, we may take γ to be a transvection on the support of I (and trivial on its orthogonal complement), whence γ induces a (nontrivial) graph automorphism on I and centralizes b, completing the proof in those cases. In the K = U5 (2) and U7 (2) cases, b is not real in Inndiag(K) = K, but the graph automorphism induced by a field automorphism of F4 inverts b, so γ = 1, completing the proof in those cases. Finally suppose that K = U6 (2). Then CInndiag(K) (I) = x × b, t ∼ = Z3 × Σ3 , and b is inverted by a graph automorphism γ  (say, induced again by a field automorphism), and we may take γ = tγ  . Thus γ again induces a graph automorphism on I, and the proof is complete.  Lemma 3.72. K = Co2 possesses a subgroup E ∼ = E32 such that for all e ∈ E # , ∼ ∼ CK (e) = Z3 × Aut(Ke ) with U4 (2) = Ke  CK (e), and the projection of E on Ke is generated by an element of order 3 with two eigenvalues equal to 1 on the natural F4 Ke -module. ∼ HS, in which there is one class of Proof. By [IA , 5.3km], K contains L = elements e of order 3 and a Sylow 3-subgroup isomorphic to E32 . Moreover, CL (e) contains M ∼ = Z3 × Aut(U4 (2)). For if e belonged = Σ5 . This implies that CK (e) ∼ to the other class of elements of Co2 of order 3, CK (e)/O3 (CK (e)) would be a split extension of D8 ∗ Q8 by Σ5 , contrary to [IA , 5.3k]. Now write E = e × e  with e ∈ Ke := E(CK (e)). Let A ∈ E34 (CK (e)) with E ≤ A. Since M c embeds in K, and using [IA , 5.3n], we see that A = J(P ) for

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some P ∈ Syl3 (K) and AutK (A) contains A6 . By [IG , 16.9], N := NK (A) controls K-fusion in A, so NK (A) has just two orbits eN and f N on E1 (A), where f is 3central in K. Both orbits have cardinality divisible by 5, and 3 divides | eN |. Since 52 does not divide | Aut(A)|, the only possibility is | eN | = 30 and | f N | = 10. Let e0 , e1 , e2 , represent the N ∩ Ke -orbits on E1 (A ∩ Ke ), with ei having i eigenvalues equal to 1 on the natural F4 U4 (2)-module. Then e1 ∈ [CK (e1 ), CK (e1 )] N so e1 is 3-central in K; hence e  ∼ e1 . Let Ne = N ∩ Ke . Then | e1  e | = 4 N N while e1  = f  has cardinality 10. A short calculation shows that the only  N N N N possibilities for f  − e1  e are e2  e and ee0  e . Since f N = E ≤ Ke , f N = e1 Ne ∪ ee0 Ne . In particular, as ee0 ∼ f , e ∼ e0 . Therefore e ∼ e2 , and the lemma is proved.  Lemma 3.73. Let K = U6 (2) and let x ∈ I3 (Aut(K)) with U4 (2) ∼ = J  CK (x). Then J = E(CK (x)). Proof. This follows quickly from [IA , 4.8.2, 4.8.4].



Lemma 3.74. Let H be isomorphic to Ω+ 8 (2) or Un (2) with n ∈ {5, 6}. Let v be a 3-element in Aut(H) with Jv := E(CH (v)) ∼ = U4 (2). Let E ∈ E3 (Inndiag(H)) with E ∈ E3∗ (HE). Then the following conclusions hold: (a) The 3-exponent of CAut(H) (Jv ) is 3; and (b) CHE (E) = E. Proof. If H ∼ = D4 (2), then v belongs to one of three H-conjugacy classes Ci , i = 1, 2, 3. Also using [IA , 4.8.2a], we see that E ∩ Ci = ∅, i = 1, 2, 3. If x ∈ Aut(H) − Inn(H) is a 3-element, then by [IA , 4.7.3A], x cycles C1 , C2 , and C3 . Thus x does not centralize v or E, and as v ∈ Syl3 (CH (Jv )), x does not centralize Jv , and (a) holds for D4 (2). Hence, we may calculate in D4 (2), GU5 (2), or GU6 (2). Then (b) follows easily. If H ∼ = GU1 (2) × GU1 (2) ∼ = E32 . Finally, if H ∼ = U6 (2), = U5 (2), then CGU5 (2) (Jv ) ∼ ∼ ∼ then CGU6 (2) (Jv ) = GU1 (2) × GU2 (2) = E32 × Σ3 , and we are done in all cases.  Lemma 3.75. Let x ∈ Ip (X) for some prime p, and let  = ±1. Under either Aut(X) X = x : of the following conditions, x  (a) X = SLn (q) and x is diagonalizable (in X); or (b) Ln (q) ≤ X ≤ P GLn (q) and x is diagonalizable (in X). By “diagonalizable”, in the case  = −1, we mean with respect to some orthonormal basis of the underlying module. Proof. For  = +1 this is just [III17 , 14.1]. For  = −1 the argument is essentially the same. We may assume that x ∈ H, where H is the diagonal subgroup of X with respect to some orthonormal basis. For any automorphism φ ∈ Aut(Fq2 ), the corresponding automorphism of X induces a power mapping on H; and such automorphisms cover Aut(X)/ Inndiag(X). Moreover, CAut(X) (H) covers Outdiag(X). The result follows directly.  Lemma 3.76. Let K = U6 (2) or D4 (2) and let d0 ∈ I3 (K) with  J0 := O 2 (CK (d0 )) ∼ = U4 (2). Let M = NAut(K) (d0 ). Then M covers Outdiag(K) and |M : M ∩ Inndiag(K)| = 2. If K = U6 (2), then M covers Out(K). If K = D4 (2) and τ ∈ Aut(K) is a triality, then M τ  covers Out(K). Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

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Proof. If K = U6 (2), the assertions hold by Lemma 3.75 and [IA , 2.5.12]. If K = D4 (2), then K = Ω(V ) for some 8-dimensional orthogonal F2 K-module V such that dim CV (d0 ) = 6. Then d0 is centralized by some transvection in O(V ), so identifying O(V ) with a subgroup of Aut(K), we have O(V ) ≤ M . As | Aut(K) : O(V )| = 3, the result follows.  Lemma 3.77. Let p be an odd prime and K = SLp (q p ),  = ±1, q ≡  (mod p). Let φ be a field automorphism of K of order p. Then CK (φ) ∼ = SLp (q)  ∼ and CInndiag(K) (φ) = CInn(K) (φ) = P GLp (q). Proof. There exist σ-setups (K, σ) and (K, τ ) for K and L := CK (φ) ∼ =  SLp (q), respectively, such that τ p  = σ and τ induces φ on K. Here K = SLp (F), where F is the algebraic closure of Fq . The first assertion follows directly as K is universal. Let z = Z(K) ∼ = Zp . Then z = z τ ∈ L. By Lang’s theorem [IG , 2.1.1], p τ there is u ∈ K such that u = uz. Hence uτ = uz p = u, so u ∈ CK (σ) = K. Let  = K/ z. Then K   L u ≤ CK (φ) ≤ CInndiag(K) (φ).   = | Outdiag(L)| = p, by [IA , 4.9.1b]. But |CInndiag(K) (φ)|/|L| = | Inndiag(L)|/|L| Since u ∈ L, equality holds throughout. The lemma follows.  Lemma 3.78. Let K = U6 (2) and b ∈ I3 (K) with I := E(CK (b)) ∼ = U4 (2). Then NK (b) = I × b, z with b, z ∼ = Σ3 and z a 2-central involution of K. Proof. The element b is the image of an element of I3 (SU6 (2)) with a 4dimensional eigenspace Vb on the natural SU6 (2)-module. Thus NK (b) preserves the decomposition Vb ⊥ Vb⊥ , and z ∈ SU (Vb⊥ ) is a transvection, hence 2-central in K.  3.8. Pumpups and the Sets Cp , p Prime. Lemma 3.79. The following conditions hold: ∼ ± (a) If K ∈ ↑2 (L± 3 (3)) ∩ C3 , then K/Z(K) = Ln (3), n ≥ 4, L3 (9), or P Sp6 (3); (b) If K ∈ ↑2 (A6 ) ∩ C3 with Z(K) of odd order, then K is a projective orthogonal group of dimension at least 6 over F3 or F32 , or K ∼ = U4 (2), U5 (2), L2 (34 ), L± (9), or P Sp (9). 4 3 Proof. In (a), K ∈ Alt ∪ Spor by [IA , 5.3, 5.2.9]. If K ∈ Chev − Chev(3), then K ↓2 U3 (3) ∼ = G2 (2) so K ∼ = G2 (4) by [IA , 2.2.10, 4.9.1, 4.9.2] and the BorelTits theorem. But G2 (4) ∈ C3 . So K ∈ Chev(3). If K ↓2 L± 3 (3) via a graph-field or field automorphism, then K ∼ = L3 (9). Otherwise we see from [IA , 4.5.1] that K/Z(K) ∼ = L± n (3) or P Sp6 (3), as asserted. Part (b) is immediate from Lemma 3.39 and the definition of C3 [V3 , 1.1]. The lemma is proved.  Lemma 3.80. Suppose that K ∈ C3 , J ∈ C2 and J ↑2 K/Z(K). (a) If J ∼ = Suz; = L3 (4), then K/Z(K) ∼ ∼ (b) If J = A5 , then K ∼ = A9 ; (c) J ∼  F i22 , F i24 , Suz, HS, Sp4 (4), L4 (2), or L5 (2); = (d) If J ∼ = 2HS, then K ∼ = F5 ; (e) If J/Z(J) ∼ = P Ω± n (3), n ≥ 6, then K ∈ Chev(3); ± ∼ (f) If J ∼ = U4 (2), then K ∼ = P Sp4 (9) or L± 5 (3), or K/O3 (K) = P Ωn (3), n ≥ 6.

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Proof. Say J ↑2 K/Z(K) via x ∈ I2 (Aut(K/Z(K))). First suppose that K ∈ Chev(3). Then (e) is trivial, and J ∈ Chev(3) by Lemma 3.1, so we may assume J ∼ = C2 (3)a = B2 (3)a ∼ = U4 (2) and consider it as J ∼ in (f). If x is a field or graph-field automorphism, then it is a field automorphism and K ∼ = P Sp4 (9). Otherwise by = B2 (9). As K ∈ C3 , Z(K) = O3 (K), so K ∼ [IA , 4.2.2], q(K) = 3, and we can use [IA , Table 4.5.1], which implies that K is not of exceptional type or of type Cn ; if x is a graph automorphism, then K ∼ = L± 5 (3) or ± K/Z(K) is a projective orthogonal group as stated (note that L4 (3) ∼ = P Ω± 6 (3)); and if x is an inner-diagonal automorphism, then K/Z(K) ∼ = Bn (3) for some n > 2. Thus the lemma holds if K ∈ Chev(3). If K ∈ Alt − Chev(3), then K ∼ = A5 as J ∈ C2 , and the lemma = A9 , so J ∼ holds. If K ∈ Spor, then K is restricted by the hypothesis K ∈ C3 [V3 , 1.1], and examination of the tables [IA , 5.3] shows that J cannot be as in (c), but J and K must be as in (a) or (d). So the lemma holds in these cases. Thus K is one of the eight groups in C3 ∩ Chev(2) − Chev(3). In particular, q(K) ≤ 2 or q(K) = 8, so x is not a field automorphism. By the Borel-Tits theorem, 1 x is a graph or graph-field automorphism. In the latter case, J ∼ = 2F4 (2 2 ) or 3 2 B2 (2 2 ), which are not mentioned in the lemma. In the former case, by [IA , 4.9.2] and [V3 , 1.1], J ∼ = Sp4 (2) or Sp6 (2), which again are not mentioned in the lemma. This completes the proof.  Lemma 3.81. Let p be an odd prime and K ∈ C∗p . Then K ∈ Chev(p) or one of the following holds: (a) p = 3 and K/Z(K) ∼ = A9 , U5 (2), U6 (2), D4 (2), 3D4 (2); (b) p = 3, K ∈ Spor and m3 (K) ≥ 3; viz., K/Z(K) ∼ = J3 , Co1 , Co2 , Co3 , M c, Ly, Suz, O  N , F i22 , F i23 , F i24 , F5 , F3 , F2 , or F1 ; or (c) p = 5 and K ∼ = Ly. Proof. We must show that any K ∈ Cp − Chev(p) not listed in the lemma lies in Cop . Suppose that K ∼ = Akp , k = 1, 2, or 3. The case p = 3 is trivial, so assume that p ≥ 5. Conditions (a) and (b) in Definition [V4 , 11.4] are easily checked. If k = 1, take E1 = E2 = 1 in that definition; if k = 2, take E1 to be generated by a p-cycle and E2 = 1; if k = 3, take E1 to be generated by two disjoint p-cycles and E2 generated by one of these p-cycles. Then Lp (CK (E2 )) = 1 and |CK (E1 )| is even, as required. Now if K satisfies (3C)

mp (Aut(K)) = 2 and for some x ∈ Ip (Aut(K)), |CK (x)| is even,

then E1 = x and E2 = 1 satisfy the demands of Definition [V4 , 11.4]. The only pairs (p, K), K ∈ Cp , which are not listed in this lemma and do not satisfy (3C) are the following: (1) p = 3, K ∼ = Sp6 (2), F4 (2) or P Sp4 (8); 5 (2) p = 5, K ∼ (3D) = 2F4 (2 2 ), Co1 , F5 , F2 , or F1 ; or (3) p = 7, K ∼ = F1 . This can be verified using [IA , 6.1.4, 4.10.3a, 5.6.1] plus centralizer structures found in [IA , 4.8, 5.3], and, of course, [V3 , 1.1]. 5 Notice that P Sp4 (8) and 2F4 (2 2 ) satisfy condition (b) in Definition [V4 , 11.4], while the rest of the groups in (3D) satisfy condition (a) there.

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We give quadruples (K, mp (E2 ), L, L∗ ) such that L = E(CK (E2 )) is simple of p-rank 2, E2 has index p2 in an element of Ep∗ (Aut(K)), and L∗ is a subgroup of CK (E2 ) having a subgroup x of order 2p disjoint from E2 . These quadru ples show that K satisfies Definition [V4 , 11.4], with E1 = E2 x2 . A dot indicates that L∗ = L. The quadruples are: (Sp6 (2), 1, A6 , Σ6 ), (F4 (2), 2, A6 , Σ6 ) 5 1 and (P Sp4 (8), 1, A6 , Σ6 ) for p = 3; (2F4 (2 2 ), 1, 2F4 (2 2 ) , ·), (F5 , 1, U3 (5), SL2 (5)), (F2 , 1, HS, ·), and (F1 , 2, U3 (5), SL2 (5)) for p = 5, and (F1 , 1, He, ·) for p = 7. This completes the proof.  Lemma 3.82. Let X = KB t where K ∈ C2 , K  X, Z2 ∼ = t ≤ Z(X), and B ∈ Ep∗ (X) for some odd prime p. Assume the following conditions: (a) There are a quasisimple group J with J/Op (J) ∈ C∗p and Op (J) of odd order, an element τ ∈ Aut(J) with τ 2 = 1, and a component of CJ (τ ) isomorphic to K; and (b) For every B ∗ ≤ X with B ∗ ∼ = B and every hyperplane A ≤ B ∗ , ∗ m2 (CX (A)) ≤ 2, and m2 (CX (B )) ≤ 1. Then p = 3. Proof. Suppose that p ≥ 5. If J/Op (J) ∈ Chev(p), then by Lemma 3.81, p = 5 and J/O5 (J) ∼ = Ly. But then K  E(CJ (τ )) ∼ = Ly or 2A11 , a contradiction as K ∈ C2 . Therefore, J/Op (J) ∈ Chev(p). Consequently by [IA , 4.9.6], K ∈ Chev(p). As K ∈ C2 , the only possibilities by [V3 , 1.1] and [IA , 2.2.10] are K ∼ = L2 (p), p = 5 or 7. In particular A := CB (K) contains a hyperplane of B and m2 (CX (A)) ≥  m2 (K t) = 3, contrary to hypothesis (b). The proof is complete. Lemma 3.83. Suppose that F ∗ (X) = K ∈ Kp for some odd prime p, and that mp (X) ≥ 4. Let X be the set of all x ∈ Ip (X) such that CX (x) has a p-component L with L/Op (L) ∈ Cp . If X = ∅, then there exists x ∈ X such that mp (CX (x)) ≥ 4. Proof. Let y ∈ X . We may assume that mp (CX (y)) ≤ 3. Since L/Op (L) ∈ Cp it follows from Lemma 3.2 that K ∈ Cp . If K ∈ Alt then K ∼ = An for some n ≥ 4p as mp (X) ≥ 4; but then it is immediate that mp (CK (y)) ≥ 4, contradiction. Suppose K ∈ Spor. Then as | Out(K)| ≤ 2, mp (K) ≥ 4. But then by [IA , Table 5.6.1] and [V3 , 1.1], K is a Cp -group, contradiction. Therefore K ∈ Chev − Chev(p) (note that the exceptional Chev(3) group that is not a C3 group, namely 3A6 , does not satisfy our rank condition on K). Say that K has level q. If some element g ∈ Ip (X) induces a graph or graph-field automorphism on K, then p = 3 and K ∼ = D4 (q) or 3D4 (q). But the latter is impossible as then m3 (K) = 2 and Out(K) is cyclic, contradicting m3 (X) ≥ 4. Since K ∈ C3 , q > 2 and by [IA , 4.8.2], there is x ∈ I3 (K) such that CK (x) has a component L isomorphic mod center to L4 (q), q ≡  (mod 3). But then L ∈ C3 , as required. So we may assume that X has no such element g. Suppose next that Z(K) = 1. One possibility is K ∼ = E6 (q)u , with p = 3 and q ≡  (mod 3),  = ±1. From [IA , 4.7.3A], whose data we use freely, we see that there is g ∈ I3 (K) such that CK (g) has a component L ∼ = SL6 (q). We can take x = g unless q = 2. In that case, however, all components of centralizers of inner automorphisms of K ∼ = 2 E6 (2) of order 3 lie in C3 , so y has a nonidentity

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image in Outdiag(K). But this implies (see the table) that CK (y) contains a simple component L with m3 (L) ≥ 2. Then m3 (LZ(K) y) ≥ 4 and we can take x = y. The other nonsimple case is K/Z(K) ∼ = Ln (q),  = ±1, q −  ≡ n ≡ 0 (mod p). Hence n ≥ 5, and if q = 2 then n ≥ 6 as y exists. Let g = diag(ω, ω −1 , 1, . . . , 1) (with respect to an orthonormal basis of the underlying space in the unitary case), where ω is a primitive pth root of unity in Fq or Fq2 , according as  = 1 or −1. Then CK (g) has a component L ∼ = SLn−2 (q). Unless K ∼ = SU6 (2), L is not a Cp group and we take x = g. Notice that mp (L Z(K), x) = mp (L)+2 = n−3+2 ≥ 4, as needed, since n − 2 ≥ 3 and p does not divide n − 2. As SU6 (2) is a C3 -group, the exceptional case cannot occur. We may therefore assume that Z(K) = 1. Suppose that p divides | Outdiag(K)|. If K ∼ = E6 (q) with p = 3 and q ≡  (mod 3), then we can use x = g as in the previous paragraph, again unless q = 2. If q = 2, then we see again that y must induce a non-inner automorphism, and now [IA , 4.7.3A] shows that m3 (CK (y)) ≥ 3, so again we can take x = y. On the other hand if K ∼ = Ln (q), q −  ≡ n ≡ 0 (mod p), (but not the C3 -group U6 (2)) we can again try g = diag(ω, ω −1 , 1, . . . , 1), giving a subcomponent L ∼ = SLn−2 (q) of p-rank n − 3, so m3 (x L) = n − 3 + 1 and we can take x = g unless n = 5, a case that requires more attention. If q ≡  (mod 52 ), then we can take x = diag(α, α, α, α, α−4 ) where α5 = ω. So assume q ≡  (mod 52 ). Then |K|5 = 54 , m5 (K) = 3, and K admits no field automorphism  of order 5. As m5 (X) ≥ 4, we must have O 5 (X) ∼ = P GL5 (q), and we can take x = diag(ω, 1, 1, 1, 1). Now we may assume that p does not divide | Outdiag(K)|, so without loss X = K f  where f = 1 or f induces a field automorphism of order p on K. In particular mp (K) ≥ 3. Suppose that f = 1. Then we may take x = f unless p = 3 and K ∼ = U4 (8), U5 (8), Sp6 (8), D4 (8), or F4 (8). In all cases there is x ∈ I3 (K) such that CK (x) has a component L ∼ = SU3 (8) ∈ C3 annd m3 (CK (x)) ≥ 3. So we can assume that f = 1, i.e., X = K. If p is a good prime for K, then by [IA , 4.10.3], mp (CK (y)) = mp (K) = mp (X) ≥ 4 and we can take x = y. So suppose that p is a bad prime for K (but note that we have handled the case K ∼ = E6 (q), q ≡  (mod 3) above). Since mp (K) ≥ 3, K ∼ = G2 (q). Using [IA , 4.7.3AB] we find a suitable x in every case except K ∼ = F4 (2), which is ruled out as it is a C3 -group. The proof is complete.  Lemma 3.84. If L = Sp6 (2), D4 (2), 2 D5 (2), U5 (2), U6 (2), SU6 (2), Sp4 (8), or F4 (2), and L ↑3 K ∈ C3 , then L = Sp6 (2) and K ∼ = F4 (2), or L = U5 (2) and K/Z(K) ∼ = U6 (2). Proof. Suppose that K ∈ C3 and L ↑3 K. By Lemma 3.1, K ∈ Chev(2) or K ∈ Spor. Checking the limited possibilities for K [V3 , 1.1], using [IA , 4.8.2, 4.7.3A, 5.3], yields the result.  Lemma 3.85. Suppose that K ∼ = Suz, Co1 , F i23 , F2 , or D4 (2). Then ↑2 (K) ⊆ C2 and ↑3 (K) ∩ C3 = ∅. Proof. Let K ↑2 L and K ↑3 M ∈ C3 . For the sporadic K’s, L and M lie in Spor by [IA , 4.9.6, 5.2.9]; the result follows directly from [IA , 5.3] and [V3 , 1.1]. For K = D4 (2), the same lemmas give L ∈ Chev(2) ∪ Spor, and then L ∈ Chev(2) ∪ {F i22 , 2F i22 } and L ∈ C2 . On the other hand, as D4 (2) ∈ Chev(3),

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we also have M ∈ Chev(2) ∪ Spor − Chev(3). But no M ∈ Spor has the requisite subcomponent, nor does any C3 -group in Chev(2) − Chev(3), by use of [V3 , 1.1] and [IA , 4.8.2, 4.8.4, 4.7.3A]. The proof is complete.  Lemma 3.86. Suppose that L2 (8) ↑2 K and K ∈ C2 . Then K ∼ = L2 (64), L± 3 (8), or G2 (3). Proof. K ∈ Alt ∪ Spor by [IA , 5.2.9, 5.3], so K ∈ Chev(r), where r = 2 or 3 as K ∈ C2 . If r = 3 then by [IA , 4.5.1, 4.9.1] the only possibility is K ∼ = G2 (3). So assume that K ∈ Chev(2) and CK (x) has a component isomorphic to L2 (8) for some x ∈ I2 (Aut(K)). By the Borel-Tits Theorem and [IA , 4.9.1, 4.9.2bf], K∼  = L2 (82 ) or L± 3 (8), as desired. Lemma 3.87. If K ∈ C2 , then it is not the case that L4 (3) ↑3 K/O2 (K). Proof. Suppose false. Since L4 (3) ∈ Chev(2), K ∈ Chev(2). As m2 (L4 (3)) >  L± 2, K ∼ = = L2 (pn ) for any odd p. Also K/Z(K) ∼ n (3) or G2 (3), n ≤ 4, by the Borel-Tits theorem. Hence K ∈ Spor, but then examination of the tables [IA , 5.3]  shows that it is false that L4 (3) ↑3 K/O2 (K). The proof is complete. Lemma 3.88. Suppose that U4 (2) ↑2 H ∈ C2 and H contains a copy of L4 (3). Then H ∼ = L4 (3). Proof. This follows from Lemma 3.10k and Lagrange’s theorem. Lemma 3.89. There is no H ∈ C2 with A6 ↑2 H and

L± 4 (3) ↑3



H/O2 (H).

Proof. Suppose that H is such a group. Since L± 4 (3) ↑3 H/O2 (H), H ∈ Chev(3) ∪ Spor, by [IA , 4.9.6, 5.2.9]. If H ∈ Chev(3), then by the Borel-Tits 3 theorem and [IA , 4.9.1, 4.9.2], H/O2 (H) ∼ = L± 4 (3 ), so H ∈ C2 , contradiction. Thus, H ∈ Spor. Using [IA , 5.3] we see that the condition L± 4 (3) ↑3 H/O2 (H) forces H/O2 (H) ∼ = F i22 ; but ↑2 (A6 ) contains no covering group of F i22 , a final contradiction.  Lemma 3.90. Let K = F i23 or D4 (2), and L = F i24 or F i22 , respectively. Suppose that M ∈ C2 , M contains L, and K ↑2 M . Then M = L. Proof. By Lemma 3.10, M ∼ = L or M ∼ = D4 (4) and K = D4 (2). As |F i22 |  does not divide |D4 (4)|, the result follows. Lemma 3.91. Suppose that L3 (4) ↑2 K ∈ K2 with K/Z(K) ∈ C3 . Then K ∼ = Suz. Proof. This follows easily from Lemma 3.14, [IA , 5.3], and the fact that L3 (16) ∈ C3 .  Lemma 3.92. If K ∼ = U6 (2), SU6 (2), or D4 (2), then ↑3 (K) ∩ C3 = ∅. Proof. Suppose K ↑3 L ∈ C3 . By [IA , 4.9.6, 5.2.9, 2.2.10], L ∈ Spor ∪ Chev(2) − Chev(3). The sporadic groups L are ruled out by inspection of the tables [IA , 5.3]. The groups L ∈ Chev(2) − Chev(3) listed in [V3 , 1.1] are ruled out by the  usual centralizer information [IA , 4.8.2, 4.8.4, 4.7.3A]. The lemma follows. Lemma 3.93. Let K ∈ C2 and suppose that b ∈ I3 (Aut(K)). Then CK (b) has no component L such that L/Z(L) ∼ = L4 (3).

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Proof. Suppose false. As Z(L) is a 3 -group by [IA , 6.1.4], m3 (L) = m3 (L4 (3)) = 4 [IA , 3.3.3]. Thus, K ∼ = L2 (q) for q ∈ FM9. As L ∈ Chev(2), K ∈ Chev(2) by [IA , 4.9.6]. As L ↑2 K, K ∈ Spor by inspection of the tables [IA , 5.3]. Therefore as K ∈ C2 , K/Z(K) ∈ Chev(3) with | Out(K)| not divisible by  3. So L3 (CK (b)) = 1 by the Borel-Tits theorem, a final contradiction. Lemma 3.94. Let H = M t, y with M  H, O3 (H) of odd order, and M/O3 (M ) ∈ K3 − C3 . Let X = O3 (CM (t)) and suppose that the following hold: (a) t ∈ I2 (H) and F ∗ (CM t (t)) = X × t; and (b) y ∈ I2 (H) and I is a component of CM (y) with I ∼ = A5 and I < M . ∼ ∼ Then M = SL3 (4) or L3 (4), with CM (t) = SU3 (2) or U3 (2), respectively. Proof. Suppose Y := O3 (M ) = 1. As F := F ∗ (CM (t)) ≤ X × t, we see that t inverts y. But then F covers M/Y , a contradiction. Hence, Y = 1 and M ∈ K3 − C3 . If X ≤ Z(M ), then CM (t) ≤ X × t and |CM/X (t)| ≤ 2, a contradiction. Hence X ≤ Z(M ). Then by [IA , 7.7.1], using F ≤ X × t, we have M/Z(M ) ∼ = L3 (4), = L3 (4) or M ∈ Chev(r) with r > 3, as M ∈ C3 . If M/Z(M ) ∼ then using [IA , 6.1.4], the conclusion of the lemma holds, so we may assume that M ∈ Chev(r), r > 3. By the structure of F , it follows from [IA , 4.5.1, 4.9.1] that M ∼ = L2 (5) = L2 (q) for some q = r a . As I is a component of CM (y) with I ∼ and I < M , it follows that M ∼ = Z12 or Z13 , a final = L2 (52 ). But then F ∼ contradiction.  Lemma 3.95. Let K ∈ C2 , z ∈ I2 (Aut(K)), and J  E(CK (z)) with J ∼ = L2 (q), q ∈ FM. Then CAut(K) (J) has abelian Sylow 2-subgroups of rank at most 2. Proof. If K ∈ Spor, then we see from the information in [IA , 5.3] and the definition of C2 that (K/Z(K), q) = (M12 , 5), (J2 , 5), (J2 , 7), or (J3 , 17), and the desired assertion holds. If K ∼ = L2 (q0 ), q0 ∈ FM9, no such z and J exist. As J ∈ Chev(3), K ∈ Chev(3) by [IA , 4.9.6]. Since C2 ∩ Alt ⊆ Chev(2) we may assume then that K ∈ Chev(2), whence J ∈ Chev(2). Let T ∈ Syl2 (CAut(K) (J)). If T ∩ Inn(K) = 1, then CT ∩Inn(K) (z) contains an involution y; but then J ≤ L2 (CK (y)) by L2 -balance, contradicting the Borel-Tits theorem. Hence T embeds in Out(K), and hence in Out(K)/ Outdiag(K). Thus T /T0 embeds in the cyclic group ΦK , where T0 is a subgroup of order at most 2 mapping into the graph automorphism subgroup of Out(K). The result follows.  5

Lemma 3.96. Suppose that (p, L) = (3, L2 (8)) or (5, 2B2 (2 2 )). If K, J ∈ Cp and L 2, contradiction. The proof is complete.  Lemma 4.5. Let X = K b with K = E(X) ∼ = (S)U6 (2) and b ∈ I3 (X) inducing a non-inner automorphism on K such that E(CK (b)) ∼ = U4 (2) or U5 (2). Then there is b0 ∈ I3 (Kb) such that E(CK (b0 )) ∼ = U5 (2). Proof. We may assume that E(CK (b)) ∼ = U4 (2), whence with respect to a suitable orthonormal basis of the natural SU6 (2)-module, b acts on K like diag(ω, ω, 1, 1, 1, 1), ω 3 = 1. Then b0 = diag(ω, ω, ω, ω, ω, 1) has the desired properties.  Lemma 4.6. Let K = U5 (2), (2)U6 (2), Sp8 (2), or (2)D4 (2). Let D1 S1 ≤ Aut(K) ∼ with D1 = F ∗ (D1 S1 ) ∼ = 21+4 = 31+4 and S1 ∼ − . Then K/Z(K) = U6 (2). Moreover, some involution of S1 − Z(S1 ) acts on K like a 2-central involution of K/Z(K). Proof. Since S1 is irreducible on D1 /Z(D1 ), it follows from [IA , 2.5.12] that D1 ≤ Inn(K). Unless K/Z(K) ∼ = U6 (2), we see from [IA , 4.8.2a, 4.10.3a] that |K|3 = 35 (so D1 ∈ Syl3 (K)) but m3 (K) = 4, a contradiction. Hence, K/Z(K) ∼ = U6 (2). We may assume that Z(K) = 1 and identify Inn(K) with K. Let y ∈ I2 (S1 ) − Z(S1 ) and D0 = CD1 (y) ∼ = 31+2 . Then CD1 S1 (y) contains a copy C of SU3 (2). By the Borel-Tits theorem there exists a parabolic subgroup P of K such that y ∈ O2 (P ) and C ≤ CK (y) ≤ P . The latter condition forces O2 (P ) ∼ = 21+8 + ∼ ∼ and P/O2 (P ) = U4 (2). Then y ∈ CO2 (P ) (Z(D1 )) = Q8 , so y = Z(O2 (P )), which is generated by a transvection. The lemma follows.  Lemma 4.7. Suppose K = F ∗ (X) ∼ = U4 (3) and z ∈ I2 (X). Suppose that D ∗ Q ∗ Q . Then there are involutions u1 , u2 ∈ X such that O2 (CX (z)) ∼ = 8 8 8 3 3 ∼ ∼ O (CX (u1 )) = U3 (3) and O (CX (u2 )) = A6 . Proof. Without loss, X = KQ where Q = O2 (CX (z)) ∼ = D8 ∗ Q8 ∗ Q8 . As ∼ ∼ ∼ 2 Q ∗ Q , X/K E . But Out(K) D and Outdiag(K) O2 (CK (z)) ∼ = 8 = 2 = 8 = Z4 , so 8 every involution in Inndiag(K) is induced by some involution of X. By [IA , 4.5.1], inner-diagonal involutions u1 , u2 exist with the claimed centralizer structures, so the lemma is proved. 

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2 2 2 2  Lemma 4.8. Let K = L2 (8), L± 3 (3), 3G2 (3), B2 (2 ), or F4 (2 ) and let ∗ Inn(K) ≤ H ≤ Aut(K). If z is a 2-central involution of H, then F (CAut(K) (z)) is a 2-group. 5

1



Proof. If K = L2 (8), 2B2 (2 2 ), or 2F4 (2 2 ) , then O 2 (Aut(H)) ∈ Lie(2) and the assertion holds by [IA , 3.1.4]. So assume that K ∼ = L± 3 (3) or 3G2 (3). Then as z is 2-central in H, z ∈ Inndiag(K) = Inn(K) and H/ Inn(K) is a 2-group; we use [IA , 4.5.1, 4.9.1, 2.5.12]. It therefore suffices to show that F ∗ (CK (z)) is a 2-group,  which is visible from [IA , 4.5.1]. 5

1

Lemma 4.9. Let K = G2 (3). Then there is E ≤ K such that E = CK (E) ∼ = E23 and AutK (E) = Aut(E). Moreover, every involution of Aut(K) centralizes a Kconjugate of E. Proof. By [IA , 2.5.12, 4.9.1], all non-inner involutory automorphisms of K are conjugate graph-field automorphisms. Since K has one class of involutions it suffices to show that any non-inner automorphism t of K of order 2 centralizes some subgroup E ∼ = Aut(E). Now = E23 such that E = CK (E) and NK (E)/E ∼ 1 2 ∼ C := CK (t) = G2 (3 2 ) ∼ = Aut(L2 (8)). Let E ∈ Syl2 (C); then NC (E) = EF with F ∼ = F7.3 . Also, it is immediate from the involution centralizer structure of K [IA , 4.5.1] that CG (E) = E. As E ∈ Syl2 (K) and F embeds in Aut(E) as a maximal subgroup, NK (E)/E ∼  = Aut(E), as claimed. Lemma 4.10. Let K = A6 , L2 (p) (p ≥ 5, p prime), or L± 3 (3), and let V ≤ K with V ∼ = E22 . Let t ∈ I2 (Aut(K)). Then either (a) [t, V g ] = 1 for some g ∈ K; or (b) K = A6 or L2 (p), and K t ∼ = P GL2 (9) or P GL2 (p), respectively. ∼ L± (3), then any four-subgroup of K is diagonalizable on the Proof. If K = 3 natural module, and so there is a unique K-conjugacy class of four-subgroups of K (by Witt’s lemma in the unitary case). By [IA , 2.5.12, 4.5.1], Aut(K) = Aut0 (K) in these cases and m2 (CK (t)) = 2, implying the desired statement. On the other hand if K ∼ = A6 or L2 (p), then K has one class of involutions so the result holds for t ∈ Inn(K). We may therefore assume, by [IA , 2.5.12, 4.9.1], that K t ∼ = P GL2 (9) or Σ6 , or P GL2 (p), in the respective cases. In the Σ6 case, there are two classes of four-groups, and the result is easily checked.  Lemma 4.11. Let (p, L) = (3, A6 ), (3, L2 (8)), (5, 2F4 (2 2 ) ), (5, 2B2 (2 2 )), or (p ≥ 5, L2 (p)). Let t ∈ I2 (Aut(L)) with p dividing |CL (t)|. Then one of the following holds: (a) L ∼ = A6 and Inn(L) t ∼ = Σ6 ; or 1 (b) L ∼ = 2F4 (2 2 ) and t is conjugation by a 2-central involution of L. 1

5

Proof. Let x ∈ Ip (CL (t)). If L ∼ = L2 (p) or L2 (9), then CInndiag(L) (x) is a p-group, so the only possibility is that L ∼ = L2 (9) and t is a field automorphism, 5 which is (a). If L ∼ = L2 (8) or 2B2 (2 2 ), then Aut(L) has a unique class of involutions, 1 whose centralizers in L are 2-groups, so no x exists. Finally if L ∼ = 2F4 (2 2 ) , then L  1 1 (x)) ∼ has a unique class of elements of order 5, and O 2 (C2 = 2B2 (2 2 ) ∼ = F5.4 , 2 F4 (2 )

1

by [IA , 4.8.7]. As the centralizer of a 2-central involution in 2F4 (2 2 ) is a parabolic 1  subgroup of type 2B2 (2 2 ), (b) follows in this case. The proof is complete.

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Lemma 4.12. Let K = L± 3 (p), p > 3 a prime, and let Inn(K) ≤ H ≤ Aut(K). Let z ∈ I2 (Inn(K)). Then z is the unique involution of O2 (CH (z)). Proof. This is immediate from [III11 , 6.4e].



Lemma 4.13. Suppose that F ∗ (X) = K ∼ = L± 4 (3). Let x ∈ I3 (K) and suppose that CX (x) contains a subgroup J with J/O3 (J) ∼ = Z3 ×U3 (2). Then m3 (CK (x)) = 4. Proof. Suppose false. By assumption, J has a subgroup x × ES, where E∼ = Q8 . Let = E32 and S is a 2-subgroup of NJ (E) ∩ CJ (x) such that S/CS (E) ∼ A = x × E. Then CS (E) ≤ CS (A). Since X is of characteristic 3 type and A ∈ S3 (X) (as m3 (CX (x)) = 3), CS (A) = 1, so S ∼ = Q8 acts faithfully on E. Let y be the involution of S. Without loss we take J = AS. By the Borel-Tits theorem there is a J-invariant parabolic subgroup H of K, and so J embeds in NAut(K) (O3 (H)), an extension of H by outer automorphisms of  K. In particular A ≤ O 3 (Aut(K)) = K, so A ≤ H. Let P ∈ Syl3 (H) with A ≤ P . Suppose first that K ∼ = U4 (3). Let B = J(P ) ∼ = E34 and H0 = NK (J(P )). Then x ∈ B, so x projects nontrivially on H 0 = H0 /B ∼ = A6 . If H ≤ H0 , then as Out(K) ∼ = D8 , y ∈ H0 . Then x and y are commuting elements of H 0 of orders 2 and 3, contradicting the structure of A6 . Therefore H ≤ H0 . It follows that H∼ = 31+4 (SL2 (3) ∗ Z4 ). Again as Q8 is not involved in Out(K), y ∈ H. Since y is a square in S ≤ NAut(K) (Z(O3 (H))), y centralizes Z(O3 (H)). But if y ∈ Z ∗ (H), then y inverts Z(O3 (H)), contradiction; therefore y ∈ Z ∗ (H). Then CP (y) is the product of two orthogonal long root subgroups of P and so CP (y) ≤ B. Hence x ∈ B, contradiction. Hence, K ∼ = L4 (3). Expanding H to a parabolic subgroup maximal with respect to being J-invariant we may assume that either H has rank 2, or H is of rank 1, corresponding to the middle node of the Dynkin diagram. Note that  NAut(K) (P )/P ∼ = Z2 × D8 , and hence NAut(K) (H)/O 3 (H) is a quotient of Z2 × D8 ,  by a Frattini argument. Therefore y ∈ O 3 (H).  Suppose that O 3 (H)/O3 (H) ∼ = L3 (3) or SL2 (3). Then it follows that y lies  in a root SL2 (3)-subgroup of K. Consequently O 3 (CK (y)) ∼ = SL2 (3) ∗ SL2 (3), a Sylow 3-subgroup of which is conjugate to a subgroup of B = J(P ). So x is K-conjugate into B and m3 (CK (x)) = 4, contradiction. Finally suppose that  O 3 (H)/O3 (H) ∼ = SL2 (3) ∗ SL2 (3), whence O3 (H) = B and so x has a nontrivial  image in H = H/O3 (H). Since y ∈ O 3 (H) and [y, x] = 1, the only possibility is  that y ∈ Z ∗ (H). But then again O 3 (CK (y)) ∼ = SL2 (3) ∗ SL2 (3) and we reach a contradiction as in the previous case. The proof is complete.  Lemma 4.14. Let L = L3 (9) and let τ ∈ Aut(L) be an involution inducing a graph automorphism on L. Then τ inverts an element of L of order 5. Proof. By [IA , Table 4.5.1], there is a unique class of involutory graph automorphisms of L, and E(CL (τ )) ∼ = L2 (9). There is an involution t ∈ E(CL (τ )) inverting an element x ∈ I5 (E(CL (τ ))). Hence, tτ is an involution in Lτ inverting x. But tτ is L-conjugate to τ , proving the claim. 

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5. Representations Lemma 5.1. Let B ∼ = E3n with n ∈ {5, 6}. Suppose that B  N with CN (B) = B. Set W := N/B and suppose that the following conditions hold: (a) τ ∈ W is a reflection on B and there is a subnormal subgroup Σ of CW (τ ) with Σ ∼ = Σk for some k ∈ {5, 6}; and (b) A := [Σ, Σ] < E for some component E of W . Then one of the following holds: (a) n = 6, E ∼ = Σ7 or Z2 × Σ7 ; or = A7 , and W ∼ (b) k = 6 and E ∼ = Ω5 (3). Moreover, in the latter case, suppose D ≤ B, D ∼ = E32 , and D = CB (S0 ) with S0 ∼ = Σn−1 generated by reflections and contained in a Σ6 generated by reflections. Then there is u ∈ D# such that CW (u) has a normal subgroup isomorphic to W (Dn−1 ). Proof. Suppose first that E/Z(E) ∼ = Am for some m ≥ 8. Then E contains a Frobenius group of order 8 · 7, whence n ≥ 7, a contradiction. Hence, if E/Z(E) ∼ = Am , then m ≤ 7, and, as A is a component of CE (τ ), we have n = 5, m = 7, and Z(E) = 1. Since 7 does not divide |GL5 (3)|, and A contains a Frobenius group of order 5 · 4, A is absolutely irreducible on B, and conclusion (a) holds. Henceforth, suppose that E/Z(E) ∼  Am for any m. = Suppose next that k = 6, i.e., Σ ∼ = Σ6 . Then by Lemma 3.39, either E ∼ = Sp4 (4) ± ± ∼ or L5 (2) or L3 (9), or E/Z(E) = HS, or E/Z(E) is a projective simple orthogonal group over F3 or F9 of dimension at least 4. As W embeds in GL6 (3), order considerations eliminate Sp4 (4), HS and U3 (9). If E ∼ = L± 5 (2), then E contains an extraspecial group of order 27 , whence E has no faithful representation over F3 of degree less than 8, a contradiction. Suppose that E ∼ = L3 (9). By the structure of A and [IA , 4.5.1, 4.9.1], τ induces a graph automorphism on E, and its action on E is uniquely determined up to E-conjugacy. But then τ inverts some element g ∈ E of order 5 by Lemma 4.14. Since τ is a reflection on B, g = τ τ g  therefore embeds in GL2 (3), which is absurd. Still assuming k = 6, we have that E/Z(E) is a projective simple orthogonal group over F3 or F9 of dimension at least 4. By order considerations, we quickly reduce to the case E/Z(E) ∼ = P Ω± r (3) with r ≥ 5. By [III17 , 6.18], if r ≥ 6, then τ inverts an element of E of order 5, yielding a contradiction as before. Hence, E∼ = Ω5 (3), so (b) holds. Furthermore, the final assertion of the lemma results from [III17 , 1.14, 1.17, 3.10, 3.12]. Hence, we may assume that Σ ∼ = Σ5 . It follows from [III11 , 1.6] that either E ∈ Chev(5) or E/Z(E) ∼ = M12 , J1 , J2 , L2 (16), U3 (4) or L3 (4). Order considerations rule out J1 , J2 , and U3 (4). As L2 (16) contains a Frobenius group of order 16 · 15, it is also ruled out. Suppose that E/Z(E) ∼ = L3 (4). Since CW (τ ) has a subnormal subgroup isomorphic to Σ5 , 4 divides |NW (E)/ECW (E)|. Thus there is a 2-element s ∈ NW (E) inducing an involutory automorphism on E, and such that CE (s) contains a rational element x of order 7. Then, as x is rational on the eigenspaces of s and s ∈ Z(W ), it follows that n ≥ 7, a contradiction. Next suppose that E/Z(E) ∼ = M12 . We freely use the information in [IA , 5.3b]. If AutW (E) ∼ = Aut(M12 ), then W contains a rational element of order 11, contrary to n ≤ 6. Hence W = ECW (E). If Z(E) = 1, then n = 6 as Z(E) ≤ [E, E],

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and E is absolutely irreducible as it contains a Frobenius group F11.5 . Therefore W = E, contradicting det(τ ) = −1. Thus W = E × CW (E). Let τ = τE τC be the corresponding decomposition of τ . Let τE ∈ T ∈ Syl2 (E), s = Z(T ), and R = O2 (CE (s)) ∼ = Q8 ∗ Q8 . Using the Thompson transfer lemma we may replace τ by a conjugate and assume that τE ∈ R − s. Then for some g ∈ R, τ τ g = s. As τ is a reflection and s has four eigenvalues equal to −1, this is impossible. Finally, we have E ∈ Chev(5). As |GL6 (3)|5 = 5, E/Z(E) ∼ = L2 (5). But then E = A, contrary to assumption, providing a final contradiction and completing the proof.  Lemma 5.2. Let L ∈ Chev(2) ∪ Chev(3) with m2 (L) ≤ 4. Suppose that L embeds in SL(V ) with V ∼ = A5 = F53 . Suppose that, for some v ∈ V # , L > CL (v) ∼ or Σ5 . Also suppose that either L acts irreducibly on V or L acts indecomposably on V with dim(Soc(V )) = 4 and v ∈ Soc(V ). Then L ∼ = A6 . Proof. Suppose that L ∈ Chev(2) − Chev(3). As m2 (L) ≤ 4 and |L| divides |SL5 (3)|, we immediately see, using [IA , 3.3.3], that m L/Z(L) ∼ = L± n (2 ) for some m and n. Also, L has no subgroup isomorphic to L3 (2) or L2 (2m ) for m ≥ 3, the latter since L2 (2m ) contains a Frobenius group of order 2m (2m − 1) with no representation of dimension less than 2m − 1 over F3 . It follows that L ∼ = L2 (4), using our conditions on m2 (L) and |L|. But then L = CL (v), a contradiction. Suppose then that L ∈ Chev(3) but L ∼  L2 (9) ∼ = A6 . As |L| divides |SL(V )|, = |L| is not divisible by 3r − 1 for any r > 5 or by 3r + 1 for any r > 2. Also, |L|5 = 5. It follows that L/Z(L) ∼ = SL4 (3), then L leaves = L4 (3), L5 (3), or Ω5 (3). If L ∼ invariant the eigenspaces of Z(L) on V of dimensions 1 and 4, contradiction. If L∼ = A4 . As Hi has no faithful = Ω3 (3) ∼ = L4 (3), then L > H = H1 × H2 with Hi ∼ F3 -representation of dimension less than 3, H has no faithful F3 -representation of dimension less than 6, and hence L ≤ SL(V ), a contradiction. If L ∼ = SL5 (3), then O3 (CL (v)) = 1, a contradiction. Likewise if L ∼ = Ω5 (3), then V is a natural F3 L-module, by [III17 , 1.15]. Hence CL (v) ∼ = Ω± 4 (3) or O3 (CL (v)) = 1. In any  case, CL (v) ∼ = A5 or Σ5 , a final contradiction. Lemma 5.3. Let H = U6 (2) and let V be a faithful F2 H-module of minimum dimension. Then the following conclusions hold: (a) dimF2 (V ) = 20; (b) V is absolutely irreducible; and (c) V is an F2 -form of ∧3 U , where U is the natural module for SU6 (2). Proof. By [IA , 2.8.2], V is the restriction to H of an F2 H-module V whose high weight is in the basic range. Here H is the universal overlying algebraic group and by assumption, Z(H) ∼ = Z3 acts trivially on V . We take ai − ai+1 , 1 ≤ i ≤ 5, as our fundamental system. The fundamental weights are then βi = a1 + · · · + ai , 1 ≤ i ≤ 5. Let β=

5 

δ i βi ,

i=1

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 δi = 0 or 1, be the high weight of V . As Z(H) acts trivially, i iδi ≡ 0 (mod 3). Now the Weyl group W ∼ = Σ6 permutes the ai naturally and dim(V ) ≥ |W : Wβ |. If more than one δi equals 1, then Wβ is the stabilizer of a partition of {a1 , . . . , a6 } into at least 3 parts, so |W : Wβ | > 20. If just one δi = 1, the only choice is β = β3 , and |W : Wβ | = 20. To complete the proof it remains to show that for β = β3 , the corresponding module V β is ∧3 U , which is irreducible of dimension 20. Here U is either one of the natural modules for H. But it is trivial that the high weight for ∧3 U is a1 + a2 + a3 = β3 and that the dimension is 20, and together these facts imply irreducibility. The proof is complete.  Lemma 5.4. The minimal degree of a nontrivial F2 Co2 -representation is at least 22. Proof. Suppose by way of contradiction that K = Co2 and V is a nontrivial 21-dimensional F2 K-module. Let P ∈ Syl3 (K) and Z = Z(P ), so that by [IA , 5.3k],  C := CK (Z) satisfies O3 (C) ∼ = 31+4 and O 3 (C/O3 (C)) is an extension of D8 ∗ Q8  by A5 . In particular, O 3 (C) has no nontrivial mapping into L3 (2). On the other hand, let V0 = CV (Z). Since O3 (P ) is faithfully represented on V , dim[V, Z] = 18 and so dim V0 = 3. Therefore [V0 , P ] ≤ [V0 , C] = 0. Let Y be a K-conjugate of Z lying in P , with Y = Z. (Such a Y exists by examination of the centralizer of a non-3-central element of order 3.) Then V0 ≤ CV (Y ) so V0 = CV (Y ). Set D = Y, Z ∼ = E32 . Lift the character of D on V to a complex character χ. Then χ(z) = −6. Let n = |z K ∩ D|. The 8 − n elements u ∈ D# − z K are all K-conjugate as K has only two classes of elements of order 3. Write χ(u) = m ∈ Z. As dim V0 = 3 we have 1 (21 − 6n + (8 − n)m) 9 whence n = 8 and m = (6 + 6n)/(8 − n). But n ≥ 4 and n is even. As m ∈ Z, n = 4. Thus, n = 6 and m = 21, so CD (V ) = 1, a contradiction. This completes the proof.  3 = (χ, 1D ) =

Lemma 5.5. Any two A8 -subgroups of Sp6 (2) are conjugate. Proof. For any group X let Y(X) be the set of all ordered triples (H1 , H2 , t) such that Z3 × A5 ∼ = Hi ≤ X, i = 1, 2, t ∈ I2 (X), and the following conditions hold: (1) H1 , H2  ∼ = A8 ; (2) O3 (Hi ) ≤ E(H3−i ), i = 1, 2; (5A) (3) t ∈ NH1 (O3 (H2 )); and (4) t ∈ NX (H2 ), and E(H2 ) t ∼ = Σ5 . Note that Y(A8 ) = ∅; we can take h1 = (123), h2 = (678), t = (45)(67), and Hi = CA8 (hi ), i = 1, 2. In view of (5A1), it therefore suffices to let X = Sp6 (2) and show that the conjugation action of X on Y(X) is transitive. Let V be the natural X-module. Let (H1 , H2 , t) ∈ Y(X) and set Wi = O3 (Hi ). Since CX (Wi ) contains A5 , dim[V, Wi ] = 2 and CX (Wi ) = Wi × Ji ∼ = Z3 × Σ6 with Ji = CX ([V, Wi ]) ∼ = Sp(CV (Wi )). Hence Wi is determined up to X-conjugacy. Then J1 and J2 are determined up to X-conjugacy. Now Ji has two conjugacy classes of A5 -subgroups, but nonconjugate A5 ’s contain nonconjugate Sylow 3subgroups. As the conjugacy class of Wi is determined, and Wi ≤ J3−i by definition

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of Y(X), E(Hi ) is determined up to conjugacy in Ji . In particular the pair (W1 , H1 ) is determined up to X-conjugacy. Then since t ∈ NH1 (W2 ), W2 , t ∼ = Σ3 is a Sylow 3-normalizer in E(H1 ), so (W1 , H1 , W2 , t) is determined up to X-conjugacy. Given H1 , there seem to be three different choices for E(H2 ) within J2 , since W1 ≤ H2 . Conditions (5A3, 4) imply, however, that H2 is uniquely determined by H1 and t. Namely, we can think of Wi as generated by a 3-cycle in J3−i ∼ = Σ6 ; W 1 , for example, thus has a fixed point set Ω1 of cardinality 3 in a natural permutation representation of J2 ; and E(H2 ) is the derived subgroup of the stabilizer of a point in Ω1 . Now t normalizes W2 , hence t normalizes J2 , and t acts on J2 like an element τ ∈ J2 . As t centralizes W1 , τ is a transposition supported in Ω1 . Since E(H2 ) t ∼ = Σ5 , the support of τ must be contained in that of E(H2 ), and hence the support of E(H2 ) is determined as the union of the supports of W1 and τ . This completes the proof.  Lemma 5.6. Let K = F i22 and N ≤ K with A := F ∗ (N ) ∼ = E210 and N/A ∼ = M22 . Then A = J(T ) for any T ∈ Syl2 (N ) ⊆ Syl2 (K). Moreover, N has 3 orbits on A# , of lengths 22, 231, and 770. Remark 5.7. Such a subgroup N exists by [IA , 5.3t]. Proof. We use [IA , 5.3t] without comment. Fix T and let z ∈ I2 (Z(T )). Let Cz = CK (z), Q = O2 (Cz ) = t × Q0 ∼ = Z2 × 21+8 + , and C z = Cz /Q. Since there  exists a non-2-central involution t with CK (t ) ∼ = 2U6 (2), Z(T ) = z, by Lemma 2.10a. Thus CC z (t) ∼ = U4 (2), of index 2 in C z ∼ = Aut(U4 (2)). If t ∈ z K then by the (∞)

structure of CK (t, z), CK (t)(∞) = Cz , whence z and t are conjugate in NK (Q), which is absurd. Thus, t ∈ z G , so since CK (t) is nonsolvable, CK (t) ∼ = 2U6 (2). Then CK (t, z)/ t is a maximal parabolic subgroup in CK (t)/ t of type U4 (2), (∞) so Q/Z(Q) is a natural F4 -module for C z ∼ = U4 (2). Now we show that (5B)

A = J(T ).

Let A ∼ = B ≤ T . Clearly m2 (Q) = 6. If m2 (B ∩ Q) = 6, then t ∈ B so B ≤ C z . (∞) Let T0 = T ∩ Ct . Then J(T 0 ) ∼ = E24 is the centralizer of a maximal totally isotropic subspace V0 of Q/Z(Q), and CQ/Z(Q) (J(T 0 )) = V0 . Hence B ∩ Q/Z(Q) = V0 and CT0 (B ∩ Q) = B. Thus if m2 (A ∩ Q) = m2 (B ∩ Q) = 6, then A = B as desired. It therefore suffices to prove that m2 (B∩Q) = 6. Otherwise, m2 (B∩Q) = 5 and B = J(T 0 ) u with u ∈ T 0 . The last condition implies that t ∈ B, so it is still true that B ∩ Q/Z(Q) = V0 . But u induces a field automorphism on V0 , so [B ∩ Q/Z(Q), u] = 1, a contradiction. Thus, (5B) holds. Let c = |NK (A)|/|N |, so that c is odd. The above analysis also shows that A = CK (A) and CNK (A) (z) = NC z (A) = T L, with L ∼ = A5 . Thus |z N | = 17 c|N |/2 .3.5 = 231c. Also, CNK (A) (t)/ t = NCK (t) (A)/ t is a parabolic subgroup of CK (t)/ t U6 (2) with Levi factor L3 (4), by Lemma 2.10c, so |tN | = c|N |/216 32 .5.7 = 22c. Hence |A# | ≥ (22 + 231)c, forcing c = 1 or 3. If c = 3, then |NK (A)/A| = 3|M22 |. By Sylow’s theorem, in this case NK (A) has an element of order 33, which K does not, contradiction. Hence c = 1, and N = NK (A) controls K-fusion in A by [IG , 16.9]. As K has just three classes of involutions, the remaining 210 − 1 − 22 − 231 = 770 involutions of A form a single N -orbit. The proof is complete.  (∞)

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Lemma 5.8. Let K = M22 or Aut(M22 ) and let M be a F2 K-module with a generating set {mi }22 i=1 of distinct elements permuted naturally by K. Suppose that dimF2 M = 10. Then the following conditions hold: (a) As F2 K-module, M is uniquely determined up to isomorphism and is irreducible; (b) The K-orbits on M are of lengths 22, 231, and 770; (c) In the semidirect product KM , M = J(T ) for any T ∈ Syl2 (KM ); and (d) There are no relations mi1 + · · · + mir = 0 with r ≤ 5 and i1 , . . . , ir distinct. Proof. For (a) and (b), it suffices to consider the case K = M22 . Let I := {1, . . . , 22}. For i ∈ I, the stabilizer of mi is Ki ∼ = M21 ∼ = L3 (4). A sum mi + mj , i = j ∈ I, is stabilized by the subgroup K{i,j} , which is maximal in K [IA , 5.3c], so K{i,j} is the stabilizer of mi + mj , which has 231 K-conjugates. There are no relations mi1 + · · · + min = 0 with i1 , . . . , in distinct and n ≤ 5; if there were one, then by triple transitivity, every sum of three mi ’s could be replaced by a shorter sum, so every m ∈ M would be one of the 22 + 231 = 210 − 1 elements already accounted for, a contradiction. m of mi + mj + mk in K contains For distinct i, j, k ∈ I, the stabilizer Kijk the stabilizer K{i,j,k} , which is a self-normalizing subgroup E24 E32 4 of K with 770 distinct K-conjugates. From the list of maximal subgroups of K [IA , 5.3c] we can deduce that the unique maximal subgroup of K containing K{i,j,k} is K[Δ] = NK (O2 (K{i,j,k} )) ∼ = E24 A6 , the setwise stabilizer in K of the unique hexad Δ of the M22 Steiner system containing {i, j, k}. Note that K{i,j,k} = KΔ−{i,j,k} . Since m = K{i,j,k} . By the transitivity of M22 K[Δ] induces A6 on Δ, it follows that Kijk on hexads, either the 1540 elements mi + mj + mk are distinct, or they form 770 pairs of equals, and in the latter case  mi = 0 i∈J

for every hexad J ⊆ I. As 2 − 1 = 22 + 231 + 770, the latter case holds, all elements of M are accounted for, and so the uniqueness of the module is proved. Irreducibility follows from the orbit decomposition. The Steiner system on I puts an F4 -projective plane structure on I − {1}, and [K, K]1 ∼ = L3 (4) induces a collineation group on it. Hence every involution in [K, K]1 , and hence every involution t ∈ [K, K], has exactly 1+5 = 6 fixed points on I. This implies that dim[M, t] ≥ 4, as [M, t] contains at least 8 different elements mi + mj . Another consequence is that for any four-group U ≤ [K, K], U must have an orbit of length 4 on I, and hence [M, U, U ] = 1 as there are no relators on the mi of length 4. Now suppose that T ∈ Syl2 (KM ), J(T ) = M , and B ∈ E210 (T ) with B = M and |B ∩ M | maximal. It follows from the Thompson Replacement theorem that [M, B, B] = 1, and so m2 (BM ∩ [K, K]) ≤ 1. Thus |M ∩ B| ≥ 28 , whence BM ∩ [K, K] = 1 and |M ∩ B| = 29 . Let u ∈ B − M . Then for some g ∈ KM , t = uug is an involution in [K, K]M − M , and dim[M, t] ≤ 2, contrary to what we saw above. This completes the proof of (c) and the lemma.  10

Lemma 5.9. Let K = 2U6 (2) and let P = AL be a parabolic subgroup of K with A = O2 (P ) ∼ = E210 and L ∼ = L3 (4). Then the lengths of the L-orbits on A# are subsums of 1 + 21 + 21 + 210 + 210 + 560.

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∼ F i22 . Let T ∈ Syl (P ) and Proof. By [IA , 5.3t], K = CH (t) in a group H = 2 T ≤ T ∗ ∈ Syl2 (H). By Lemma 5.6, A = J(T ∗ ), and NH (A)/A ∼ = M22 has an orbit Ψ of length 22 on A# . As P/A ∼ = L3 (4), |NH (A) : P | = 22 and so the involution in Z(K) lies in Ψ. Usng the notation of the previous proof, we have Ψ = {mi }22 i=1 and without loss, m1 ∈ A generates Z(K), and L is the stabilizer of m1 in NH (A)/A. Continuing the analysis of the previous proof, we see some L-orbits on A# of length 1 ({m1 }); 21 ({mi |2 ≤ i ≤ 22}); 21 ({m1 + mi | 2 ≤ i ≤ 22}); 210 ({mi + mj | 2 ≤ i < j ≤ 22}); and 210 ({m1 + mi + mj | 2 ≤ i < j ≤ 22}). The remaining 560 elements of A# are those sums (equal in pairs, as before) mi +mj +mk such that the hexad containing {i, j, k} does not contain 1. We need only show that for a triple {i, j, k} of the last sort, m := m1 + mi + mj + mk is another such triple. Note that as there are no relations of length 5 on the mi ’s, m = m and m = m1 + m + mn for any , n. Likewise as {1, i, j, k} does not lie in a hexad, m = m + mn for any , n. Hence, m = mi + mj  + mk for some i , j  , k ∈ {2, . . . , 22}, and it remains to show that {1, i , j  , k } does not lie in a hexad Δ. But if such a Δ existed, m would equal m1 + m + mn where Δ = {1, i , j  , k ,  , n }, contrary to what we have shown. The lemma is proved.  Lemma 5.10. The minimum dimension of a nontrivial F2 -representation of Co1 is 24. Proof. This follows by Clifford’s theorem applied to the restriction of the representation to a 36 · (2M12 ) subgroup [IA , 5.3l]. Note that the minimum degree  of a nontrivial permutation representation of M12 is 12 [IA , 5.3b]. Lemma 5.11. Let M and N be irreducible F3 -modules for A5 and Σ5 , respectively, of dimension at least 2. Then (a) Either M is the (4-dimensional) core of the natural permutation module, or dimF3 (M ) = 6 and M is a projective module; (b) If dimF3 (M ) = 6, then HomF3 A5 (M ∧ M, F3 ) = 0; and (c) N ↓A5 is irreducible. Proof. For any F3 -space V , write V = V ⊗ F9 . By [Is1, Appendix, Table 3], the complex irreducible A5 modules have dimension 1, 3, 3, 4, 5. As A5 has four 3-regular classes, there are four irreducible 3-modular characters. The 5-dimensional F3 [A5 ] permutation module splits as 1 + 4, while the 6-dimensional irreducible F3 [L2 (5)] permutation module is indecomposable with composition factors 1, 4, 1. By [Is1, 15.29] and [F1, Chapter IV, Lemma 4.19], the characters of degree 3 are in distinct blocks of defect 0 and the associated modules are projective. Hence the principal 3-block consists of characters of degree 1 and 4, the latter being the core of the natural permutation module, and both realized over F3 . If dimF3 (M ) = n, then 5 divides |GL(n, 3)|, whence n ≥ 4. Hence, there is one further irreducible F3 [A5 ]-module M of dimension 6 with M ∼ = M1 ⊕M2 , the sum of the two projective irreducible modules of dimension 3. Hence, the possibilities for M are as described in (a). As the Brauer characters of M1 and M2 are real, M1 and M2 are self-dual. We have ∧2 M ∼ = P1 ⊕P2 ⊕P3 , where P1 = ∧2 M1 , P2 = ∧2 M2 and P3 = M1 ⊗M2 . Then HomF3 A5 (Pi , F3 ) = 0 for i = 1 and 2 since Mi is irreducible and odd-dimensional, and for i = 3 as M1 and M2 are distinct and self-dual. Hence (b) holds.

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By Clifford’s Theorem [Is1, 6.5], if N0 := N ↓A5 is not irreducible, then N0 = M1 ⊕ M2 , where the Mi are distinct irreducible F3 [A5 ]-modules which are Σ5 -conjugate. As the irreducible F3 A5 -modules have distinct dimensions, this is impossible, proving (c).  Lemma 5.12. Let M be a nontrivial irreducible F3 [A6 ]-module such that for some involution y ∈ A6 , dimF3 ([M, y]) ≤ 3. Then M is isomorphic to the unique 4dimensional F3 [A6 ]-module, which is the core of either of the natural 6-dimensional permutation modules for A6 over F3 . Moreover, dimF3 ([M, y]) = 2. Proof. Since A6 is perfect, detM (y) = 1 so the −1-eigenspace of y on M has dimension 2. Let F be the algebraic closure of F3 , σ ∈ Aut(F) the Frobenius, and M1 , M2 , and M3 the basic irreducible FSL2 (9)-modules, with dim Mi = i. Thus Z(SL2 (9)) acts trivially on M1 and M3 but inverts M2 . Using the Steinberg tensor product theorem, we see that there are four nontrivial irreducible FA6 -modules: M3 and M3σ , both writable over F9 but not over F3 ; M2 ⊗ M2σ and M3 ⊗ M3σ , both writable over F3 . Now TrM3 (y) = −1 so −1 has multiplicity 4 as an eigenvalue of y on M3 ⊗ M3σ . Since M3 is not writable over F3 , the irreducible F3 A6 -module that involves it (after tensoring with F) is M3 ⊕ M3σ , and again −1 has multiplicity 4. Therefore M ∼ = M2 ⊗ M2σ , the unique irreducible 4-dimensional irreducible F3 A6 module. The uniqueness implies that M is the core of either 6-dimensional natural  permutation module for A6 , and the proof is complete. Lemma 5.13. Suppose that X is an irreducible K-subgroup of GL5 (3) containing a reflection t. Then CX (t) is not isomorphic to Z2 × Σ5 or E22 × Σ5 . Proof. Suppose false and let C = CX (t), f ∈ I5 (E(C)), and v ∈ I3 (E(C)). Consider W := O2 (X). By irreducibility O3 (X) = 1, and as mp (X) ≤ mp (GL5 (3)) ≤ 1 for all primes p > 3, E(C) centralizes F (W ) and then W itself. By L2 -balance, E(C) ≤ E for some component E of X, yielding a contradiction by Lemma 5.1. The lemma is proved.  Lemma 5.14. Let M be the core of a natural permutation module for A6 over F3 . Then Hom(M ∧ M, M ) = 0. Proof. Let N = M ∧ M . If the lemma is false, then N has a submodule P such that N/P ∼ = M . Since dim P = 2, A6 must act trivially on P . There are therefore two dimensions of skew-symmetric bilinear forms on M preserved by A6 (note that M is self-dual). Since M is absolutely irreducible, this is a contradiction. The lemma is proved.  Lemma 5.15. Let K = Ω(V ) where V is an F2 -orthogonal space of dimension 6 and + type. Let L ≤ K with L ∼ = A6 . Then L fixes a nonsingular vector in V . ∼ Proof. Note first that K = A8 has a unique conjugacy class of subgroups isomorphic to A6 , since A6 has a unique nontrivial permutation representation of degree 8 (up to conjugation by Aut(A6 )). It therefore suffices to find an A6 subgroup fixing a nonsingular vector. Let U be the  core of the standard 2 Σ8 -permutation F 8 8 module. Thus U consists of all sums u = i=1 ci ui , ci ∈ F2 , i=1 ci = 0, subject  to the relation 8i=1 ui = 0. Define q(u) = 0 or 1 according as the number of ci is equivalent to 0 or 2 (mod 4). Then q is a nondegenerate Σ8 -invariant quadratic form on U , so we may identify K with Ω(U, q). Let L be the commutator subgroup of the stabilizer in K of u1 + u2 . Then L ∼ = A6 and q(u1 + u2 ) = 1, as desired. 

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Lemma 5.16. Let V = V1 ⊕ V2 be a vector space of order 35 , with |V1 | = 3. Let N be a K-group acting on V with N1 := NN (V1 ) stabilizing V2 and having a normal subgroup W ∼ = W (F4 ) acting naturally on V2 . Then one of the following holds: (a) N = N1 ; (b) N = QN1 , Q ∩ N1 = 1, and Q = O3 (N ) ∼ = E34 permutes regularly the set of complements to V2 in V ; or (c) F ∗ (N ) = E(N )Z(N ) with E(N ) ∼ = Ω5 (3), and V is a natural E(N )module. Proof. Let N + = N ∩ SL(V ) and for any X ≤ N , write X + = X ∩ N + . Of course, NN (X) ≤ NN (X + ). Let z = Z(W ). Then V1 = CV (z) and V2 = [V, z], so N1 = CN (z). Note that since O2 (W ) is absolutely irreducible on V2 and normal in N1 , N1+ /O2 (W ) embeds in Out(O2 (W )) ∼ = Σ3  Z2 , so |N1+ |2 ≤ 28 and O2 (N1+ ) = O2 (W ). Suppose that Q := O3 (N ) = 1. Since O3 (N1 ) = 1, CQ (z) = 1, and the action of O2 (W ) on Q forces |Q| ≥ 34 . Then, CV (Q), which is N1 -invariant, equals V2 , and (b) holds. Thus, we may assume that O3 (N ) = 1. Let R be any nontrivial 2-subgroup of N + permutable with O2 (W ). Then RO2 (W ) has irreducible constituents on N of dimensions 4 and 1, and these must be the constituents of O2 (W ), i.e., RO2 (W ) ≤ N1+ . Applying this to a Sylow 2-subgroup R of N + , we get T ≤ N1+ , so |N + |2 ≤ 28 . Note also that if R is W -invariant, then as W contains a Sylow 3-subgroup of N1+ , R ≤ O2 (W ) and Z(R) = z. In particular if O2 (N + ) = 1, then Z(O2 (N + )) = z and so N = N1 . We may therefore assume that O2 (N + ) = 1. By the last two paragraphs, F (N + ) is a {2, 3} -group. Let P ∈ Syl3 (W ), so that P ≤ [W, W ]. As |N |{2,3} divides |GL(V )|{2,3} = 5.112 .13, and |SL2 (11)|3 = 3, CP (F (N + )) = 1. Therefore K := E(N + ) = 1, whence z = Z(T + ) ≤ K. By the last paragraph, O2 (K) = 1. Now z is a maximal elementary abelian normal subgroup of CN + (z). It follows that z lies in a single component of K, and then K = F ∗ (N + ) is simple as CN (z) is solvable and O2 (N + ) = O3 (N + ) = 1. It remains to show that K ∼ = Ω5 (3). Since O 2 (CN + (z)) ∼ = SL2 (3) ∗ SL2 (3) and |N + | divides |SL5 (3)|, with 2-part at most 28 , K ∈ Alt ∪ Spor [IA , 5.2.8, 5.3]. If K ∈ Chev − Chev(3), then q(K) = 2 and there is a proper parabolic subgroup of 3-rank 2, by the structure of CK (z) and the Borel-Tits theorem. But this forces K to have untwisted rank at least 3, and so K ∼ = L4 (2), contradicting the structure of CN (z). Therefore, K ∈ Chev(3). By  the structure of O 3 (CK (z)), K ∼ = L± 4 (3), G2 (3), or Ω5 (3). By order considerations, ∼ we need only rule out K = L4 (3), which contradicts |K| ≤ |N + | ≤ |N1+ ||E1 (V )| ≤  28 .32 .112 . The proof is complete. Lemma 5.17. Let V be an irreducible 3-dimensional F3 L-module for L ∼ = L3 (3). Then H 1 (L, V ) = 0. Proof. Suppose false and let f : L → V be a cohomologically nontrivial 1-cocycle. Then W := V ⊕ F3 w becomes an indecomposable 4-dimensional F3 Lmodule by the definition gw = w + f (g), for each g ∈ L. Let H ∈ Syl13 (L). Then CW (H) = 0 and we may assume that w ∈ CW (H). The orbits of H on V + w are of cardinalities 1, 13, and 13. It follows that |wL | = 14 or 27, and the latter must hold as 7 does not divide |L|. Hence |L : CL (w)| = 27. But L does not contain a  subgroup of order 24 · 13, a contradiction.

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Lemma 5.18. Let H = Σ5 and K = [H, H] = A5 , and let V be the core of the natural F2 H-permutation module. Then the following conditions hold: (a) H 1 (H, V ) = H 1 (K, V |K ) = 0; and (b) HomH (V ∧ V, V ) = HomK (V |K ∧ V |K , V |K ) = 0. Proof. In (a), a Sylow 2-subgroup of K acts freely on V . Therefore H 1 (K, V |K ) = 0, so any exact sequence 0 → V → W → F2 → 0 of H-modules splits relative to K. An H-complement to V in W is then unique as a P -complement for any P ∈ Syl5 (K), hence invariant under KNH (P ) = H. Thus (a) holds. In (b), let g ∈ K have order 3. The eigenvalues of g on V are 1, 1, and ω ±1 , where ω is a primitive cube root of unity. Hence the eigenvalues of g on V ∧ V are 1 and ω ±1 , each twice. As the only irreducible F2 K-module besides V |K and the trivial module is the natural F4 L2 (4) module, on which g has no fixed points, the last module must be a composition factor of V |K ∧ V |K , along with two trivial composition factors. Therefore HomK (V |K ∧ V |K , V |K ) = 0, which implies (b).  + (2) with O2 (H) = 1. Let Lemma 5.19. Let U4 (2) ∼ = H0 ≤ H ≤ A ∼ = O10 V be a natural F2 A-module and suppose that V = V0 ⊥ V1 with V1 = CV (H0 ) of dimension 2 and V0 = [V, H0 ] the natural U4 (2)-module. Suppose also that m3 (H) = 3. Suppose further that if b1 ∈ I3 (H0 ) with dimF2 [V, b1 ] = 4, then CH (b1 ) normalizes V0 . Then H normalizes V0 and V1 , and |H : CH (V1 )| ≤ 2.

Proof. Let X = O3 (H). Then E(X) is a commuting product of Suzuki ni 3 groups 2B2 (2 2 ) and covering groups of 2B2 (2 2 ). A primitive prime divisor of 22ni + 1 then does not divide |GL10 (2)|, so E(X) = 1. For any prime p > 3, mp (D5 (2)) ≤ 2, so as O2 (H) = 1, [H0 , F (X)] = 1, forcing [H0 , X] = 1 as H0 =  O 3 (H0 ). Since m3 (H) = 3 = m3 (H0 ) and H0 is simple, O3 3 (H) = O3 (H), and H0 then lies in E(H) and indeed E(H) is a single component. If E(H) normalizes V0 , then it normalizes and then centralizes V1 since H0 does. Hence V0 = [V, E(H)] is H-invariant, whence V1 = V0⊥ is as well. So we may assume that H is quasisimple. There exists E ≤ H0 such that E ∼ = E32 , and for every e ∈ E # , either [V, e] = V0 or dimF2 CV0 (e) = 4. In either case, and by our special  assumption, CH (e) normalizes V0 . Therefore ΓE,1 (H) normalizes V0 . If H = H0 , ΓE,1 (H) , then H normalizes V0 and we are done. So we may assume that  ΓE,1 (H) ≤ H0 , ΓE,1 (H) < H. We consider the various possibilities for H to argue to a contradiction. Note that O2 (H) = O3 (H) = 1, by hypothesis and as shown above. All Sylow p-subgroups of D5 (2) for p > 3 are abelian, and so H is simple. If H ∈ Spor, then since m3 (H) = 3, H ∼ = J3 . But then for P ∈ Syl3 (H), Ω1 (P ) ∼ = E33 , whereas H0 contains Z3  Z3 , contradiction. If H ∈ Alt, then since m3 (H) = 3, H ∼ = An , n = 9, 10, or 11. Then H0 has a faithful permutation representation of degree n, which is impossible. So H ∈ Chev(r) for some r. If r = 3, then since m3 (H) = 3, H ∼ = L2 (33 ) by [IA , 3.3.3]. But then |H0 | does not divide |H|,contradiction. Therefore, r = 3. Since m3 (H) = 3 and H0 , ΓE,1 (H) < H, it follows from [IA , 7.3.3] that H ∼ = Sp6 (2) and no element of E # has 4-dimensional fixed point space on the natural F2 H-module W . Let P ∈ Syl3 (H0 ) with E ≤ J(P ). Then P ∈ Syl3 (H). The forbidden conjugacy class of subgroups of H of order 3 has three representatives in J(P ). Hence in H0 ∼ = U4 (2), the forbidden class consists of

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subgroups fixed-point-free on V . As E contains such subgroups (two of them, in fact), we have a contradiction. The proof is complete. 

6. Factorizations Lemma 6.1. Suppose (p, K) = (3, L3 (4)), (3, M12 ), (3, U3 (8)), (5, F i22 ), or  = X/Op (X), with K ∼  A∼ (13, F1 ). Let X = Op (X)IW A and X = I  X, = Ep3 , ∗ p ∼  A ∈ S (X), and W ∈ IX (A; 2) with |W | ≤ 2. If K = L3 (4) or U3 (8), assume that  Suppose also that W0 := W ∩ Op (X)  X A induces inner automorphisms on I.  and CX (W0 ) does not cover I. Then for one of W1 = Ω1 (Z(W )) or J(W ), NX (W1 )  covers I. Proof. By our assumptions, W0 = O2 (X). Set Z = Ω1 (Z(W0 )). Since  (if W  = 1). CX (W0 ) does not cover I and W0  X, CX (W0 ) does not cover W Thus Ω1 (Z(W )) ≤ W0 and then Ω1 (Z(W )) ≤ Z. It therefore suffices to prove  Suppose false. Set C = CX (Z), so that that either CX (Z) or NX (J(W )) covers I.   [C, I] = 1, and put X = X/C. Since J(W )  X, J(W ) = J(W0 ), whence there is  | = 2, so equality holds and E ∈ E∗ (W ) such that E ≤ W0 . Then |E : E ∩ W | ≤ |W  =W  . Also (E ∩ W )Z is elementary abelian, so |E|/|(E ∩ W )Z| ≤ 1, yielding E |Z : E ∩ Z| ≤ 2, i.e., the nonidentity element w ∈ W acts on Z as a transvection. Suppose first that K = L3 (4). Our assumptions imply that ∼ , X = Z3 × IW ) ∼  in X and let T ∈  := C (W with H = U3 (2). Let H be the full preimage of H I 4 Syl2 (HW ) with w ∈ T . Then |T | = 2 < |X|2 so NX (T ) contains a 2-element g such that wg = wz, where z is the unique involution of Z(T ∩ I). Now w z , like z, inverts ∼  O3 (H) = E32 . It follows that there are three conjugates of (the transvection)  ∼ wz generating a group Q such that Q = E32 . In particular, |Z : CZ (Q)| ≤ 23 and a Sylow 3-subgroup Q3 of Q embeds in Aut(Z/CZ (Q)) ≤ GL3 (2). This is a contradiction as |Q3 | ≥ 32 , however.  = C (W ) ∼ If I ∼ = Aut(D4 (2)) and = F i22 , we argue similarly. This time H I g  w = wz for some g ∈ I, where z is a root involution of H (∞) ∼ = D4 (2). We  = I:  namely, four find 5 conjugates of wz that generate a group Q such that Q fundamental root involutions and the lowest root involution. (By the commutator formula, the group so generated contains the positive Sylow 2-subgroup, hence is a parabolic subgroup by Tits’s Lemma [IA , 2.6.7]; but no proper parabolic subgroup containing that Sylow group also contains the lowest root subgroup.) Now a Sylow 3-subgroup of Q has order 35 but embeds in GL5 (2), a contradiction.  = 1 and so W = W0  X, and the If I ∼ = F1 , M12 or U3 (8), then actually W conclusion is trivial. For F1 , we have m2,13 (G) ≤ 1 by [IA , 5.6.2]. For M12 , it  2-subgroup follows from the information in [IA , 5.3] that any nontrivial A-invariant  ∩ I ∼    is a four-group. For U3 (8), A ( A ∩ I; 2) is trivial; of Aut(I) = E32 and IAut(I)  ∼  = 1 and for any non-inner involution w  C (w) note that m2,3 (I)  ∈ Aut(I), I  = L2 (8) has 3-rank 1. 

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Lemma 6.2. Let X = Op (X)JT , where J = Lp (J), J/Op (J) ∈ Chev(r) unambiguously, r and p are distinct odd primes, and T is a 2-group. Let T0 = T ∩ Op (X) and assume that T0  X and T0 ∈ Syl2 (Op (X)), and CX (T0 ) does not cover  where X  = X/Op (X). Then J is covered by either CX (Ω1 (Z(T ))) or J = F ∗ (X), NX (J(T )).  = Proof. As in Lemma 6.1, let Z = Ω1 (Z(T0 ))  X, C = CX (Z), X X/Op (X), and X = X/C = O2 (X)JT . We may assume that X is minimal with respect to containing T0 and covering JT, so by the Frattini argument, O2 (X) is nilpotent. Also J is perfect.  there exists A ∈ E∗ (T ) such Assuming C and NX (J(T )) each fail to cover J,  that A = 1. Therefore A = 1. If [A, O2 (X)] = 1, then [J, O2 (X)] = 1 and by minimality, F ∗ (X) = J. In that case the result follows from a theorem of Aschbacher [GL1, I.28–6]. theorem [IG , 26.10(iv)], So assume that [A, O2 (X)] = 1. Then by Glauberman’s      J covers  aJ and some element a ∈ A acts as a transvection on Z. But a  so by McLaughlin’s theorem [McL3], J ∈ Chev(2) ∪ Alt, a final hence covers J, contradiction.  7. Signalizers and Balance 7.1. Signalizers. Lemma 7.1. Suppose K ∈ Cp , K  X, A ∈ Sp (X), where p is an odd prime. Let W ∈ IX (A; p ). Then [O 2 (W ), K] = 1. Indeed, either [W, K] = 1 or p = 3, K ∼ = L2 (33 ), and some a ∈ A# induces a nontrivial field automorphism on K. Moreover, if P ∈ Sylp (X) and F ∗ (X) = KOp (X), then IX (P ; p ) = {1}. Proof. If [W, K] = 1, then K is not strongly locally balanced with respect to p, so by [IA , 7.7.12], p = 3 and K ∼ = L2 (33 ). If no a ∈ A# acts as a field automorphism on K, then A ≤ (P ∩ K) × CP (K), so A = (P ∩ K) × CA (K) as A ∈ S3 (X). But NAut(K) (P ∩ K) is the unique subgroup of Aut(K) containing P ∩ K but not K. Hence W maps into CAut(K) (P ∩ K) = P ∩ K, so [W, K] = 1. This argument shows in fact that IAut(K) (P ∩ K; p ) = {1}, which together with [V2 , 2.7] implies the final statement.  Lemma 7.2. Suppose X is a K-group, P ∈ Sylp (X) for some odd prime p, and all components of X/Op (X) are Cp -groups. Then IX (P ; p ) = Op (X). Proof. Without loss, Op (X) = 1. Let W ∈ IX (P ; p ). Since P contains some A ∈ Sp (G), it follows from Lemma 7.1 that [W, E(X)] = 1. Also by [V2 , 2.2], W ≤ Op (AW Op (X)), whence [W, Op (X)] = 1. Therefore W ≤ CX (F ∗ (X)) so W ≤ Op (Z(F ∗ (X))) = 1, as required.  Lemma 7.3. Suppose that F ∗ (X) = K is a simple K-group, p is an odd prime, mp (X) ≥ 4, and for every x ∈ Ipo (X), every component of CK (x) is a Cp -group. Let A ∈ Sp (X). Then IX (A; p ) = {1}.

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Proof. Let W ∈ IX (A; p ). Suppose that K ∈ Chev(p). Since mp (X) ≥ 4, it is not the case that p = 3 and K∼ = L2 (33 ). Hence the result follows in that case from [IA , 7.7.11c]. Suppose that K ∈ Alt, so that K ∼ = An , with n ≥ 4p as mp (X) ≥ 4. Let x ∈ K be a p-cycle; then x ∈ Ipo (X) and so An−p ∼ = E(CK (x)) ∈ Cp . Thus n = 4p. Hence the result follows in this case from [IA , 7.7.11a]. Suppose that K ∈ Spor. If p > 3 then p = 5 with K ∼ = F1 . From [IA , 5.3], mp (CK (x)) ≥ 4 for all x ∈ Ip (K). Hence for every a ∈ A# , every component of CK (a) is a Cp -group. As p = 3 it follows from [IA , 7.7.12] that CW (a) ≤ Op (CK (a)). But Op (CK (a)) = 1 in this case so CW (a) = 1 for all a ∈ A# , whence W = 1 as A is clearly noncyclic. Suppose then that K ∈ Spor with p = 3. Note that K is not a pumpup of L2 (33 ), so by the argument of the last paragraph, for any a ∈ A ∩ Ipo (K), CW (a) ≤ Op (CX (a)). Hence as long as (a) Ip (K) = Ipo (K) and (b) K is locally balanced with respect to p, the desired result holds. The balance condition holds by [IA , 7.7.1, Table 5.6.1] as mp (K) ≥ 4. Condition (a) holds except possibly for K∼ = Co3 , F i22 , or F i23 , with at most one conjugacy class in Ip (K) − Ipo (K). But for an element a of one of these classes it is still true by inspection that every component of CK (a) is a Cp -group. Therefore the same argument finishes the proof in this case. Now suppose that K ∈ Chev(r) − Chev(p), r = p. Suppose first that the Schur multiplier of K is a p -group and p is a good prime for K. Also exclude D4 (r n ) if p = 3. Then A induces field and inner automorphisms on K. Moreover, using [IA , 4.10.3] we see that as A ∈ Sp (X), A = a0  × (A ∩ K), where ap0 = 1, a0 induces a (possibly trivial) field automorphism on K, and A ∩ K belongs to the unique conjugacy class of maximal elements of Ep (K). In particular A# ⊆ Ipo (X), and moreover mp (A) = mp (X) ≥ 4. Then by the usual argument CW (a) ≤ Op (CX (a)) for all a ∈ A# . If W = 1, then CW (B) ≤ ΔX (B) for any hyperplane B of A, and we can choose B such that CW (B) = 1. Hence K is not 3-balanced with respect to A, which contradicts [IA , 7.7.7a]. The same argument applies to K = D4 (r n ) with p = 3 unless possibly some a ∈ A# induces a graph automorphism or graph-field automorphism on K. But there exists x ∈ I3o (K) such that CK (x) has a component isomorphic to L4 (r n ), where r n ≡  (mod 3) and  = ±1. As this group is by assumption a C3 -group, r n = 2. In particular K is locally balanced with respect to the prime 3, by [IA , 7.7.8a]. Moreover for every a ∈ A# , every component of CK (a) is (by inspection now) a C3 -group. So the usual argument yields a contradiction. Still assuming that p is a good prime for K, suppose that p divides the order of the Schur multiplier of K, so that K ∼ = Lpk (r n ), r n ≡  (mod p),  = ±1. As mp (X) ≥ 4, pk ≥ 5. Then there is x ∈ Ipo (K) such that CK (x) has a component isomorphic to SLpk−2 (r n ). As this lies in Cp by assumption, the only possibility is p = 3, k = 2, r n = 2, so K ∼ = U6 (2) ∈ Cp . Hence the usual argument yields W = 1. Now assume that p is a bad prime for K. The centralizers of elements of order p in Inndiag(K) can be found in [IA , 4.7.3AB]. In the various cases we find that the following subcomponents L of centralizers of elements of Ipo (K) are in

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Cp by assumption: (K, L) = (F4 (r n ), Sp6 (r n )), (E6 (r n ), L6 (r n )), (E7 (r n ), D6 (r n )), (E8 (r n ), A8 (r n )), (E8 (r n ), D6− (r n )), or (E8 (r n ), SU5 (r 2n )), with p = 3 in the first four cases and p = 5 in the last two. This yields K ∼ = F4 (2) or 2 E6 (2). In both cases we have balance, by [IA , 7.7.8a]. In the F4 (2) case I3 (X) = I3o (X), so the usual argument prevails. In the 2 E6 (2)  case since 2 D5 (2) is not a C3 -group, O 3 (X) = K. But then for every a ∈ A# , every component of CK (a) is by inspection a C3 -group. Hence the usual argument finishes the proof.  Lemma 7.4. Let K ∈ Kp be simple, where p is a prime and p ≥ 5. Suppose that D ≤ K, D ∼ E 2 and D ∈ Sylp (CK (D)). Assume that for each d ∈ D# , = p  ICK (d) (D; 2) ≤ Op (CK (d)), but that IK (D; 2) is not a p -group. Assume also that D does not centralize some element of IK (D; 2). Assume that F ∗ (X) = K. Assume also that if mp (X) ≥ 3, then K ≤ ΓoS,2 (X), where S ∈ Sylp (X). Then K is unambiguously of odd characteristic r = p, and mp (X) ≤ 2. Proof. If K ∈ Chev(p), then as p ≥ 5, [IA , 7.7.11c] implies that IK (D; 2) = {1}, contradiction. Suppose K ∈ Chev(2). If mp (X) ≥ 3, choose E ∈ Ep∗ (S). As ΓE,2 (X) lies in o ΓS,2 (X), K ≤ ΓE,2 (X). Also by [IG , 8.7(ii)], E normalizes K. Then by [IA , 7.3.5], 5 1 5 p = 5 and K ∼ = 2B2 (2 2 ), 2F4 (2 2 ) , or 2F4 (2 2 ). In the first two cases m2,5 (Aut(K)) = 5 1, so IK (D; 2) = 1. So K ∼ = 2F4 (2 2 ). Similarly, m2,5 (K) = 1 in that case, so some element of D induces a nontrivial field automorphism on K. But then D centralizes an E52 -subgroup of K, so m5 (DCK (D)) = 3, contradicting D ∈ Syl5 (CX (D)). Thus, mp (X) = 2. This has strong consequences. As p ≥ 5, we may assume that K ∈ Lie(2). Moreover, as p ≥ 5, p is a good prime for K and p does not divide the order of the Schur multiplier of K. Since IK (D; 2) = {1}, D lies in a parabolic subgroup H of K. Then CD (O2 (H)) = 1 so there is d ∈ D# such that [D, CO2 (H) (d)] =  1. In particular [D, O 2 (CK (d))] = 1. Since p does not divide the order of the Schur multiplier of K, it follows from [IA , 4.2.2] that D induces inner-diagonal automorphisms on every Lie component of CK (d), and in particular on some Lie component L such that IL (D; 2) = 1. Hence D = (D ∩ L) d, and D ∩ L lies in two parabolic subgroups H+ and H− which are opposites. Then IK (CK (d); 2) ≥ O2 (H+ ), O2 (H− ) = L, which is not a p -group, contrary to assumption. We have proved that K ∈ Chev(p) ∪ Chev(2). Notice that now if K ∈ Chev with mp (X) ≥ 3, then with [IA , 7.3.5], K ≤ ΓoS,2 (X), contrary to assumption. So all parts of the lemma hold if K ∈ Chev, and it remains to rule out K ∈ Alt ∪ Spor. If K ∈ Alt, then D is generated by p-cycles as otherwise mp (K) ≥ p ≥ 5 and K = ΓoS,2 (K), contradiction. So K = A2p+r , 0 ≤ r < p. But then as p ≥ 5, Op (CK (d)) centralizes D for every d ∈ D# , so D centralizes every D-invariant 2-group, contrary to assumption. Thus, K ∈ Alt. Suppose finally that K ∈ Spor. Again our assumptions imply that [D, T ] = T = 1 for some d ∈ D# and some 2-subgroup T ≤ CK (d). By [IA , 7.7.1b], T ≤ Op (CK (d)). Hence by [V2 , 2.3] and the Bp -property, T acts nontrivially on some component L of E(CK (d)). From [IA , 5.3] we see that D normalizes all

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components of CK (d), and hence so does T = [D, T ]. Thus L is not strongly locally balanced for p, and in particular L ∈ Cp . Surveying the tables [IA , 5.3], using the conditions L ∈ Cp and D ∈ Sylp (CK (D)), we see that (p, L, K) must be one of the following: (5, A6 , Suz), (5, A7 , F i23 ), (5, A9 , F i24 ), (7, 2L3 (4), F2 ), (11, M12 , F1 ), or (13, L3 (3), F1 ). But in all these cases, a Sylow p-subgroup of L centralizes every 2-subgroup of Aut(L) that it normalizes, so [T, L] = [D, T, L] = 1, a contradiction completing the proof of the lemma.  Lemma 7.5. Let K = F i24 and t ∈ I2 (K) with I := E(CK (t)) ∼ = 2F i22 . Let W ≤ K be an I-invariant 3 -group. Then [W, I] = 1 and if W is a 2-group, then |W/ t | ≤ 2 and W ∩ [K, K] = t. Proof. The order |W | divides |K|3 /|F i22 |3 = 25 .72 .17.23.29. Let P ∈ Syl5 (I). Then P ∼ = E52 and NI (P ) is transitive on E1 (P ). It follows that P acts nonfaithfully, and then trivially, on every Sylow subgroup of F (W ). Hence [W, P ] = 1 by [IG , 3.17], so [W, I] = 1. Finally W ≤ CK (I) = CCK (t) (I) ∼ = E22 , with C[K,K] (I) = t, by [IA , 5.3v]. The proof is complete.  Lemma 7.6. Let K = L2 (pn ), p a prime, p ≥ 5. Let z ∈ I2 (K), and C = CK (z). Then O2 (C) centralizes every C-invariant 2-subgroup of K. Moreover, O2 (C) = 1 if and only if n = 1 and p ∈ FM. Proof. K has one class of involutions, so C contains some Sylow 2-subgroup of K. Hence any C-invariant 2-subgroup of K lies in O2 (C), whence the first statement. As C is dihedral of order pn − , where  = ±1 is such that pn ≡  (mod 4), O2 (C) = 1 if and only if pn −  is a power of 2, which proves the second assertion.  Lemma 7.7. Suppose that K ∈ C2 and t ∈ I2 (Aut(K)), and L ∼ = A6 , L2 (8), U4 (2), or L± (3) is a component of C (t). If W = [W, L] ≤ K and W is a {2, 3} K 3 group, then W = 1. Proof. Suppose not. Then a Sylow 2-subgroup of L acts faithfully on F (W ) by [IG , 3.17], so L acts faithfully and we may assume W is a p-group of exponent p ≥ 5 and class at most 2. As L is not a subgroup of SL2 (p), mp (W/Φ(W )) ≥ 3 and then mp (K) ≥ mp (W ) ≥ 3. Without loss, K is simple. If K ∈ Spor, then by [IA , 5.6.1] and [V3 , 1.1], K ∼ = Co1 , F2 , or F1 , none of which have a suitable subcomponent L. The condition mp (K) ≥ 3 rules out all K ∈ Chev(3) ∩ C2 [IA , 4.10.3], as well as K ∼ = L2 (q), q ∈ FM9. So K ∈ Chev(2). By the Borel-Tits ± 2 theorem and [IA , 4.9.1, 4.9.2], K ∼ = Sp4 (4), L± n (2), n = 4 or 5, L2 (8 ), L3 (8),  L4 (4), or G2 (4). Again mp (K) ≥ 3 is contradicted. The proof is complete. 7.2. Balance and Cores. Lemma 7.8. Let X = E(X)S x with every component of E(X) a C2 -group of order divisible by 3, x3 = 1, and S a 2-subgroup of O3 (CX (x)) such that CS (E(X)) = 1. Then m2 (S) ≤ m3 (X). Proof. Let K1 , . . . , Kn be the components of E(X) normalized by x. By of E(X). [IG , 20.6], S normalizes each Ki and centralizes every other component  Set Si = CS (Ki ), so that ∩ni=1 Si = CS (E(X)) = 1. Thenm2 (S) ≤ ni=1 m2 (S/Si ). Also as each Ki ∈ C2 , O2 (Ki ) = 1, so m3 (E(X)) ≥ ni=1 m3 (Ki ). It therefore

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suffices to assume that n = 1 and show that m2 (S) ≤ m3 (K1 ). Thus in A := AutK1 Sx (K1 ), S embeds in Δ := O3 (CA (x)). As CS (K1 ) = 1 we may assume that O2 (K1 ) = 1, so K1 is simple. We quote [IA , 7.7.1] with p = 3, and check that m2 (Δ) ≤ m3 (K1 ), keeping in mind that K1 ∈ C2 . The only relevant cases for K1 ∼ = An are n = 5 and 8; for K1 ∈ Spor we have m2 (Δ) ≤ 2 ≤ m3 (K1 ); there are no cases for K1 ∈ Chev(3); and for K1 ∼ = L2 (q), q ∈ FM9, m2 (Δ) = 1. This leaves the case K1 ∈ Chev(2),  x ∈ Inndiag(K1 ). Then Δ ∩ Inndiag(K1 ) ≤ D := O3 (O 2 (CInn(K1 ) (x))). But by [IA , 4.2.2], D has odd order unless possibly CK1 (x) has a component isomorphic to m 2 B2 (2 2 ) for some m ≥ 3. Again by [IA , 4.2.2], such a component can only occur if k K1 ∼ = 2F4 (2 2 ) for some k, and it actually does not occur even then, by [IA , 4.7.3A]. So S ∩ Inndiag(K1 ) = 1. It follows from [IA , 2.5.12] that m2 (S) ≤ 2; moreover, if m3 (K1 ) = 1 (so that K1 ∼ = L2 (2n ), L3 (22n+1 ), or U3 (22n )), then m2 (S) ≤ 1. The proof is complete.  Lemma 7.9. Let K = L2 (pn ), p odd, n ≥ 2. Let z ∈ I2 (K). Then the following conditions hold: (a) O2 (CK (z)) = 1 unless pn = 32 ; and (b) If f ∈ Aut(K) is a field automorphism of order p centralizing z, then [f, O2 (CK (z))] = 1. Proof. Let W = O2 (CK (z)). Then |W | is the odd share of pn − , where  = ±1 and pn ≡  (mod 4). As n ≥ 2, part (a) follows. In (b), write n = kp. Then |[f, W ]| = |W |/|CW (f )| is the odd share of (pkp − )/(pk − ). As p is odd,  this share is nontrivial by Zsigmondy’s theorem [IG , 1.1]. Lemma 7.10. Let K ∈ C2 , let E ∈ Ep2 (Aut(K)) for some odd prime p, let Inn(K)E ≤ H ≤ Aut(K), and suppose that ΔH (E) := ∩e∈E # Op (CH (e)) is not a 2-group. Then K ∼ = Lp (2a ) where  = ±1 and 2a ≡  (mod p), E ≤ Inndiag(K) and E is the image of a nonabelian p-subgroup of GLp (2a ), and Δ ∩ Aut0 (K) = 1. Proof. This is immediate from the definition of C2 and [IA , 7.7.4]. Note that the possibility there that p = 5 and K/Z(K) ∼ = F i22 is ruled out by our 2-group hypothesis.  3

Lemma 7.11. Let K = 3D4 (2), U6 (2), 2 G2 (3 2 ), or Co3 . Then K is locally balanced with respect to p = 3. 

Proof. This follows from [IA , 7.7.8b, 7.7.1bc]. Lemma 7.12. Suppose that F ∗ (X) ∼ = U3 (8). O3 (CX (a)) = 1.

If a ∈ I3 (F ∗ (X)), then

 = SU3 (8), a covering group of K. Also Proof. Let K = F ∗ (X) and let K ∼  let Z = Z(GU3 (8)) = Z9 , so that Z(K) = Φ(Z). Notice that [IA , 4.8.4] does not apply to a, since a induces an inner automorphism on K. Let  a be a preimage of  Then   = Φ(Z). Hence there is z ∈ Z such that (z a in K. a3 ∈ Z(K) a)3 = 1. Thus a) is given by [IA , 4.8.2]. It is either a Sylow 3-subgroup CP GU3 (8) (a) = CP GU3 (8) (z of P GU3 (8) or isomorphic to L2 (8) × Z9 . Therefore O3 (CP GU3 (8) (a)) = 1. Thus in any case | a CK (a)|3 ≥ 33 . As Aut(K)/ Inndiag(K) ∼ = Z6 , the only way the lemma could fail would be if O3 (CX (K)) = g ∼ = Z2 induced a graph automorphism on

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K. Then g ∈ Z(CX (a)) so |CK (g)|3 ≥ 33 . But then CK (g) ∼ = L2 (8) by [IA , 4.9.2],  so |CK (g)|3 = 32 , contradiction. The proof is complete. Lemma 7.13. Let K = F ∗ (X) ∈ ↑2 (A6 ) ∩ C2 with K simple. Let x ∈ I3 (X). Then O{2,3} (CX (x)) = 1. Proof. The possibilities for K are given in Lemma 3.10. Using [IA , 7.7.1bc, 7.7.8b], we verify the desired conclusion unless perhaps K ∼ = L4 (2) or Sp4 (4). In both cases Out(K) is a 2-group [IA , 2.5.12], so it suffices to show that O3 (CK (x)) = 1 for all x ∈ I3 (K). But using [IA , 4.8.2], we find that CK (x) ∼ = Z3 × Σ3 or Z3 × A5 in these cases (not respectively). The proof is complete.  Lemma 7.14. Let K ∈ Co2 and Inn(K) ≤ H ≤ Aut(K). Suppose that z ∈ I2 (H) and 5 divides |O2 (CH (z))|. Then K ∼ = A6 . Proof. By assumption, K is not locally balanced with respect to p = 2. If the result fails, then by [IA , 7.7.1abc], K ∈ Chev(r) − Chev(2) for some odd r. As K ∈ Co2 , K/Z(K) ∼ = L± 4 (3) or G2 (3), or L2 (p), p ∈ {5, 7, 17}. But then the result is  easily checked (see [IA , 7.7.8] for L± 4 (3) and G2 (3)). 8. Generation 8.1. Generation with respect to Non-elementary p-Groups, p = 2. Lemma 8.1. Let K = A6 and let Q be a 2-subgroup of Aut(K) such that ΓQ,1 (K) < K. If m2 (Q) > 1, then Q ≤ Inn(K) and Q ∼ = E22 . Proof. Let Q0 = Ω1 (Q) and Q ≤ P ∈ Syl2 (Aut(K)). By [IA , 7.3.4], and as K has no strongly embedded subgroup, E22 ∼ = Q0 ≤ Inn(K). Then Q ≤ NP (Q0 ) = P ∩ Σ6 ∼ = D8 × Z2 , but Q0 ≤ Z(NP (Q0 )). If Q > Q0 , then the projection of Q on the D8 direct factor is Z4 , and so Q0 = Ω1 (Q) ≤ Z(NP (Q0 )), contradiction.  Therefore Q = Q0 , proving the lemma. Lemma 8.2. Suppose that K ∼ = Um (2), m ∈ {4, 5}, and L = E(CK (g)) ∼ = A6 for some graph automorphism g of K. Suppose that L ≤ H ≤ K. Then the following conditions hold: (a) If F (H) = 1, then m = 5, E(H) ∼ = A6 or U4 (2), and |F (H)| = 3; and (b) If m3 (H) ≥ m − 1, then L ≤ E(H), E(H) is simple, and m3 (E(H)) ≥ 3. Proof. L is involved in no parabolic subgroup of K – and hence in no 2-local subgroup of K, by the Borel-Tits theorem. If L acts nontrivially on a subgroup W ≤ K of odd prime power order, then clearly |W | = 3n and n ≥ 4. But then |W L|3 ≥ 36 > |K|3 , contradiction. Consequently [L, F (H)] = 1. As mp (K) ≤ 4 for all odd p [IA , 4.8.2a], L normalizes all components of E(H) and then L = [L, L] ≤ E(H) by the Schreier property. Suppose that F (H) = 1. As [L, F (H)] = 1, we can use [IA , 4.8.2] to conclude the following. F (H) is a cyclic 3-group, for otherwise E(H) ≤ CK (F (H))(∞) = 1, contradiction. The only possibility is that E := E(CK (F (H))) ∼ = U4 (2) and m = 5, in which case E(H) = E or L. Thus (a) holds. Moreover, if m3 (H) = 4, then E(H) = E and the conclusion of (b) holds. Thus we may assume in proving (b) that F (H) = 1.

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By the classification of groups in K with three distinct prime divisors [III17 , 9.9], E(H) is simple. Thus m3 (AutH (E(H))) = m−1. If m = 4, this yields E(H) = K and we are done. If m = 5 and m3 (E(H)) = 2, then m3 (Out(E(H))) ≥ 2, so E(H) ∈ Chev admits a field automorphism and an outer-diagonal automorphism of order 3. But then E(H) ∼ = L± 3 (q) and so m3 (Aut(E(H))) = 3, a final contradiction.  Lemma 8.3. Suppose K ∼ = U4 (2) and T is a 2-subgroup of Aut(K) with m2 (T ) ≥ 2. Under any of the following conditions, K = ΓT,1 (K): (a) m2 (T ) > 2; (b) T contains a quaternion subgroup; or (c) Ω1 (T ) ≤ Inn(K). Proof. Suppose that K = Γ := ΓT,1 (K). Let E ∈ E2 (T ) be of maximal rank subject to the condition that if possible, E ≤ Inn(K). Subject to that, choose E  T if possible. Clearly ΓE,1 (K) < K. By [IA , 7.3.3], E ≤ K and E is a fourgroup all of whose involutions have preimage of order 4 in Sp4 (3) [IA , 7.3.3]. This implies that (c) and (a) fail. Suppose that T ≥ Q ∼ = Q8 . Let y ∈ I2 (Q). Since Q∼ = Q8 , the preimages of y in Sp4 (3) must be involutions. As m2 (T ) ≥ 2, there is a four-group x, y ≤ T . Then Γ ≥ Γx,y,1 (K) = K by [IA , 7.3.3], and the lemma is proved.  8.2. Generation with respect to Elementary Abelian p-Groups, p = 2. Lemma 8.4. Let K be a K-group. Then K is outer well-generated for the prime 2 if K is one of the following: A6 , L± 3 (3), P Sp4 (3), Sp4 (8), He, L2 (q), q 1 odd, q > 3, L2 (8), Sp6 (2), 3D4 (2), 2F4 (2 2 ) , J3 , Suz, F i23 , F i24 , F3 , F1 , U5 (2). Proof. For the first three groups, and for L2 (q), q odd, this is immediate from [IA , 7.3.4, 7.3.3b]. For Sp4 (8), since | Out(Sp4 (8))| = 6 and outer involutions are graph-field automorphisms, the assertion follows from [IA , 7.3.8]. For L2 (8), Sp6 (2), 3D4 (2), F i23 , F3 , and F1 , the assertion is vacuously true as | Out(K)| is odd. 1 Likewise for 2F4 (2 2 ) , there are no involutions in Aut(K) − Inn(K), by [IA , 3.3.2d], so the assertion holds. Suppose that E ∈ E22 (Aut(K)) with E ≤ Inn(K), and let Γ = ΓE,1 (K), Γ = Γ/O2 (Γ), and H = L2 (Γ). Let e1 and e2 be distinct elements of E − Inn(K), and Li = CK (ei ), i = 1, 2. Set g = e1 e2 and L = CK (g). In the sporadic cases we use [IA , 5.3] freely. For K = He, we have CO2 (H) (e) ≤ O2 (CK (e)) for all e ∈ E # , so |O2 (H)| divides 32 . Then for each i = 1, 2, if we put Ki = E(CK (ei )) ∼ = 3A7 , Ki ≤ H, by L2 -balance, and as |Ki |3 = |K|3 , K1 , K2  ≤ J for some 2-component J of H. Then [J, O2 (H)] = 1 so J is quasisimple, and as [K1 , E] = K1 , A7 ↑(J/O2 (J)). 2

Since E(CL (e1 )) ∼ = 3A7 , the only possibility, by [III11 , 1.6c], is that J = K, as desired. For K = U5 (2), | Out(K)| = 2, and Li ∼ = Σ6 for the two elements e1 , e2 ∈ E − Inn(K). We argue that |Γ|2 ≥ |CK (g)|2 ≥ 29 . Namely, g ∈ L1 . The graph automorphism e1 is induced, we may assume, by an automorphism of F4 . Hence, it normalizes a subgroup K1 ∼ = U4 (2) containing L1 . As g ∈ K1 , g is K1 -conjugate

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into J(S) for S ∈ Syl2 (K1 ). But P := NK (J(S)) is a parabolic subgroup of K of type Z3 × A5 and with J(S) ≤ Z(O2 (P )); in particular |CP (g)| is divisible by 29 , as asserted. By Lemma 8.2a, if Γ < K and F (Γ) = 1, then F ∗ (Γ) ∼ = Z3 × A6 or Z3 × U4 (2), both of which are inconsistent with L2 ≤ Γ. Hence F ∗ (Γ) = E(Γ), a simple group as 5-local and 11-local subgroups of K are solvable. Using Lemma 3.39 we have E(Γ) ∼ = A6 or U4 (2); but then |Γ|2 < 29 , a contradiction. If K ∼ = Suz, then by L2 -balance, E(Li ) ≤ H so H contains a pumpup of M12 or J2 . As Suz is the only vertical pumpup of either of these, and L1 contains either a Sylow 5-subgroup or a Sylow 11-subgroup of K, the only possibility is that E(L1 ) = E(L2 )  H. Hence CK (E) covers E(L1 ), so [E, L1 ] = 1. But by [IA , 5.3o], CK (L1 ) = 1, a contradiction. A similar argument rules out K ∼ = J3 , for which Li ∼ = L2 (17), and also rules out K ∼ = F i24 if L1 or L2 is isomorphic to F i23 . Suppose finally that K∼ = F i24 and L1 ∼ = L2 ∼ = [2 × 2] Aut(U6 (2)). If g ∈ O2 (L1 ) then by L2 -balance, E(L1 ) ≤ L2 (CK (g)) so CK (g) ∼ = 2F i22 2. Also as L1 is transitive on O2 (L1 )# , Γ ≥ M := ΓO2 (L1 ),1 (K). As O2 (CK (g)) = 1, we find as usual that O2 (M ) = 1 and 2F i22 ↑2 M0 for some component M0 of M . Hence, M0 ∼ = K or F i23 . But the latter case is impossible as CM (g) = CK (g) maps onto Aut(L). Thus, Γ = K if g ∈ O2 (L1 ), and similarly if g ∈ O2 (L2 ). Assume that g ∈ O2 (L1 ). Then we find that O2 (Γ) = 1 and [2×2]U6 (2) ↑2 J for some component J of Γ. (A diagonal pumpup is impossible by order considerations.) This again implies that J ∼ = F i23 or K. In particular, J does not embed in CK (g). But g is the unique minimal normal subgroup of CK (g), so CK (g) acts faithfully on J. Again it is impossible that J ∼ = F i23 , whose automorphism group has no subgroup isomorphic to CK (g). Thus J = K and the proof is complete.  Lemma 8.5. Let K ∈ K2 be a K-group. Let D ∈ E22 (Aut(K)) with ΓD,1 (K) < ± ∼ K. Then K/Z(K) ∼ = M11 , F3 , L± 4 (3), L3 (3), G2 (3), F1 . Moreover if K = A6 or P Sp4 (3), then m2 (D) = 2. Proof. Let H = ΓD,1 (K). If K ∼ = A6 , then m2 (Inn(K)) = 2. Hence if |D| > 4, then D ≤ Inn(K) and Lemma 8.4 gives the contradiction H = K. If K ∼ = P Sp4 (3) and |D| > 4, then by [IA , 7.3.3b], D acts on K like a subgroup D0 ≤ K each involution of which is not 2-central. But then these involutions do not split over Z(Sp4 (3)), so the preimage of D0 in Sp4 (3) has a unique involution. Hence ± m2 (D) ≤ 2, contradiction. If K/Z(K) ∼ = L± 4 (3), L3 (3), or G2 (3), then H = K by [IA , 7.3.4], contradiction. If K ∼ = M11 , then for any x ∈ D# , CK (x) ∼ = GL2 (3) has a single class of non-central involutions, so H is strongly embedded in K, contradicting the Bender-Suzuki theorem. For K = F3 and F1 , see the remark following [IA , 7.5.6]; F ∗ (H) is a simple K-group with CH (z) = CK (z) for any  involution z ∈ D ≤ H, so F ∗ (H) = K. The proof is complete. Lemma 8.6. Suppose K ∼ = Suz is a K-group, D ≤ Aut(K), D ∼ = E22 , and < K. Then every d ∈ D# acts on K as an involution of K of class 2B, i.e., E(CK (d)) ∼ = L3 (4).

ΓD,1 (K)

Proof. Let H = ΓD,1 (K) and assume H < K. O2 (CK (d)) = 1 for all d ∈ D# [IA , 5.3], O2 (H) = 1.

Then as CO2 (H) (d) ≤

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∼ M12 or J2 . By If D ≤ Inn(K) then for some d ∈ D# , Hd := E(CH (d)) = L2 -balance, Hd ≤ E(H). As m11 (K) = m7 (K) = 1, Hd lies in a single component H0 of E(H), and either Hd = H0  H or Hd ↑2 H0 . In the latter case H0 = K = H by Lemma 3.8, so Hd = H0  H. But D contains some d ∈ K, and then CH0 (d )  CK (d ). Comparing the structures of CK (d ) and CH0 (d ) we see that this is impossible, however. Therefore D ≤ Inn(K) and so by [IA , 5.3], we may assume that for some d ∈  by Ω− D# , CK (d) = CH (d) is an extension of 21+6 − 6 (2). Obviously CK (d ) ≤ CK (d)  ∗ for any d ∈ D − d (see [IA , 5.3]) so by [IA , 7.5.6], H1 := F (H) is simple (and nonabelian). Then CH (d) = [CH (d), CH (d)] = CH1 (d). By [IA , 4.5.1, 5.2.8, 5.3], the structure of CH1 (d) forces H1 ∈ Chev(2). But |H1 |2 = 213 , so H1 ∈ Chev(2), contradiction. This completes the proof.  1 Lemma 8.7. Let K = 2F4 (2 2 ) , T ∈ Syl2 (Aut(K)), and E22 ∼ = U  P , where P 1 is a parabolic subgroup of Aut(K) ∼ = 2F4 (2 2 ) of type A1 (2). Then K = ΓU,1 (K). 1

Proof. We identify Aut(K) with 2F4 (2 2 ). From the Chevalley relations we see that Z(T ) ∼ = Z2 , and P1 := CAut(K) (Z(T )) and P are the two maximal parabolic subgroups of Aut(K) containing T . Now T normalizes ΓU,1 (K) so it suffices to show that T ΓU,1 (K)  Aut(K). But T ΓU,1 (K) contains P1 and is normalized by P , hence is normalized by P, P1  = Aut(K). The lemma is proved.  Lemma 8.8. Let K = J3 , F i24 , F3 , HS, or 2HS, and let E be a four-subgroup of Aut(K). Assume that K is a K-group. Then K = ΓE,1 (K). Proof. For the first three groups K, see the remark following [IA , 7.5.6]. For the last two groups, we may assume that K = 2HS. Set K = K/Z(K) and Γ = ΓE,1 (K). We use [IA , 5.3m] freely. For one thing, O2 (CK (e)) = 1 for all e ∈ I2 (Aut(K)), so O2 (Γ) ≤ CO2 (Γ) (e) | e ∈ E #  ≤ O2 (CK (e)) | e ∈ E #  = 1. As | Out(K)| = 2, E ∩ Inn(K) contains some z ∈ E # . Choose z if possible to be 2-central in Inn(K), and assume first that this is possible. Then z is stable in K by [IA , 6.4.2], so Γ ≥ C z := CK (z). Let e ∈ E − z. By Sylow’s theorem in Γ we may assume that z is 2-central in O 2 (CK (e)), which is covered by CK (e). Now z ∈ Z ∗ (CK (e)), so z ∈ Z ∗ (Γ). Now O2 (Γ)  CK (z), whose structure implies that if O2 (Γ) = 1, then z = O2 (Γ) or Φ(O2 (Γ)), so z ∈ Z ∗ (Γ), contradiction. Therefore, L := F ∗ (Γ) = E(Γ), and L is simple. again by the structure of CK (z). Comparing CK (z) with centralizers of involutions in alternating and sporadic groups and their automorphism groups [IA , 5.2.8, 5.3], we see that if L ∈ Alt∪Spor, then L = K = Γ, as claimed. So we may assume that L ∈ Chev(r) for some r. If r > 2, then either   O r (CL (z)) = O r (CK (z)) would be a central product of groups in Chev(r), or L∼ = L2 (q) would have dihedral Sylow 2-subgroups. Neither of these is the case, so L ∈ Chev(2). Since CL (z) ≥ CK (z)(∞) by the Schreier property, CL (z) has cyclic center, and hence q(L) = 2 or 21/2 . But |L|2 = 2a , 7 ≤ a ≤ 9, so the only possibility is L ∼ = Sp6 (2), whose order, however, does not divide |K|. This contradiction shows that E contains no 2-central involution of Aut(K). Fix e ∈ E # ∩ Inn(K). Then Ke := E(CK (e)) ∼ = A6 , with 3 not dividing |K : K e | and CAut(K) (Ke ) = e, f , where f ∈ I2 (Aut(K) − Inn(K)) and Kf := E(CK (f )) ∼ = A8 . Therefore by L2 -balance, L := E(Γ) is a single component containing K e . If L = K e , then Γ centralizes e = CInn(K) (L); but for any

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e ∈ E−e, O 2 (CInn(K) (e )) does not centralize the involution e = e , contradiction. Therefore Ke ∼ = A6 ↑ L. 2

As a result, the condition |L|3 = 3 and [IA , 4.9.6] show that L ∈ Chev(r) for any odd r. Examining [IA , 5.3] we see likewise that if L ∈ Spor, then L = K, as desired. And if L ∈ Alt ∪ Chev(2), then by [IA , 4.9.1, 4.9.2], L ∼ = A8 ∼ = L4 (2), L5 (2), or Sp4 (4). The last two possibilities violate Lagrange’s theorem, so L ∼ = A8 . It follows that CK (L) = 1, so L = F ∗ (Γ). Now e acts on L as a transposition. If E ≤ Inn(K) then another e ∈ E acts as a transposition, and then ee acts as a conjugate of an involution in K e . But then ee is a 2-central involution of Inn(K), contradicting what we saw above. Hence, E = e, f  with f ∈ Inn(K). The only possibility is CK (f ) ∼ = CK (ef ) ∼ = A8 , which is absurd as e acts nontrivially on L = F ∗ (Γ). This completes the proof.  2

Lemma 8.9. Let K = F i23 and assume that K is a K-group. Let P ∈ Syl2 (K). Then K = ΓZ(P ),1 (K). Proof. Let Γ = ΓZ(P ),1 (K). We use [IA , 5.3u] freely. Every involution of K is 2-central. In particular for some z ∈ Z(P )# , 2F i22 ∼ = CK (z) = CΓ (z). By L2 balance, and the fact that |Γ| divides |K|, CK (z) = CL (z) for some 2-component L of Γ. If z ∈ Z ∗ (L), then as L/O2 (L) is a pumpup of 2F i22 of order dividing |K|, L∼ = K so Γ = K, as required. Thus we may assume that z ∈ Z ∗ (Γ). It follows that for every involution u ∈ Z(P ), z ∈ Z ∗ (CK (u)). But this is clearly false if we take u in class 2C, so the lemma follows.  Lemma 8.10. Let K = U6 (2) and T ∈ Syl2 (K). Then K = ΓJ(T ),1 (K). Proof. By Lemma 2.9, NK (J(T )) is a maximal parabolic subgroup of K of type L3 (4). On the other hand, Z(T ) is generated by a root involution, so CK (Z(T )) contains a copy of U4 (2) and so does not lie in NK (J(T )). Hence, ΓJ(T ),1 (K) is normalized by NK (J(T )), CK (Z(T )) = K, and the lemma follows by the simplicity of K.  8.3. Generation with respect to Elementary Abelian p-Groups, p > 2. Lemma 8.11. Let p be a prime, p ≥ 5. Let K = Akp , k = 2 or 3. Let Q be a noncyclic p-group acting on K, and suppose that Γ := ΓQ,1 (K) < K. Then Q ∼ = Ep2 . Moreover, if k = 2, then F ∗ (Γ) ∼ = Ap × Ap . If k = 3, then ∗ ∼ F (Γ) = Ap × Ap × Ap or A2p × Ap . Proof. Since Γ < K, CQ (K) = 1, and then since p ≥ 5, Q is elementary abelian. If |Q| = p3 then k = 3 and Q = x1 , x2 , x3  where the xi are disjoint p-cycles, so Γ ≥ CK (x1 ), CK (x2 ) = K, contradiction. Therefore |Q| = p2 . Let Ω be a set on which K acts naturally, and let Ωx be the fixed point set of x on Ω for any element or subset x of Q. If Φ := ΩQ is nonempty, then Q has two orbits Ωi of length p, i = 1, 2, and Γ ≥ AΦ∪Ω1 , AΦ∪Ω2  = K, contradiction. Thus every orbit of Q has length p, Γ contains AΩx for every x ∈ Q# , and the result follows easily.  Lemma 8.12. Let K  X with K ∈ Chev and q(K) = 2a , a > 1. Let p be an odd  prime and A ∈ Ep∗ (X). Let E ∈ E2 (A). If p > 3 or mp (E) > 2, then K = Γ2E,1 (K).

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∼ L (q), Proof. If the result is false, then as a > 1, [IA , 7.3.3] implies that K = p q ≡  (mod p),  = ±1, |E| = p2 , and the image of E in Aut(K) is the image of a nonabelian p-subgroup of GLp (q). In particular A = E × f  × CA (K) where either f = 1 or f induces a field automorphism of order p on K. But then CK (f ) contains a subgroup F ∼ = Epp−2 disjoint from Z(K). Hence A∗ := F f  CA (K) is elementary abelian and mp (A∗ ) − mp (A) = mp (F ) − mp (E) = p − 4 > 0, contradicting the  assumption that A ∈ Ep∗ (X). The proof is complete. Lemma 8.13. Suppose that K ∈ Chev(r) is simple, r = p, where p is an odd prime. If mp (K) = 2 and Sylow p-subgroups of K are nonabelian, then p = 3 and n K∼ = L3 (q),  = ±1, q ≡  (mod 9), 3D4 (q), G2 (q), or 2F4 (2 2 ) . Proof. If p divides the order of the Schur multiplier of K, then K ∼ = Ln (q)  (with p dividing n) or E6 (q) (with p = 3),  = ±1, q ≡  (mod p). Accordingly by [IA , 4.10.3a], mp (K) ≥ n − 2 or 5. Therefore K ∼ = L3 (q) as asserted. Otherwise, mp (K) = nm0 = 2 in the notation of [IA , 4.10.3a]. But then since a Sylow p-subgroup P of K is nonabelian, P contains more than one Ep2 -subgroup, by [IA , 4.10.2c]. Then the lemma follows immediately from [IA , 4.10.3c].  Lemma 8.14. Suppose K ∈ Spor − Cp , K is a K-group, and mp (K) = 2 for some odd prime p. Let E ∈ E2 (K) and suppose that Γ := Γc E,1 (K) < K. Then one of the following holds: (a) p = 3, K ∼ = M12 , Γ = 1, AutK (E) contains a copy of SL2 (3), and IAut(K) (E; 3 ) = {1}; ∼ (b) p = 5, K ∼ = F i22 , Γ ∼ =D 4 (2), NK (Γ) = Aut(Γ) is a maximal subgroup of  ∼ K, and IAut(K) (Γ, 5 ) = Z2 ; or (c) p = 13, K ∼ = F1 , Γ = 1, AutK (E) contains a copy of SL2 (13), and IAut(K) (E; 13 ) = {1}. Proof. By definition of Cp [V3 , 1.1] and using [IA , 5.6.1] for p-ranks, we see that the following pairs (p, K) satisfy our conditions: (3, Mn ), n = 12, 22, 23, 24; (3, J2 ); (3, J4 ); (3, HS); (3, He); (3, Ru); (5, Suz); (5, F in ), n = 22, 23, 24; (7, F2 ); (11, F1 ); and (13, F1 ). Except for (3, M12 ), (5, F i22 ), and (13, F1 ), K = ΓE,1 (K) by [IA , 7.5.5]. Let N = NK (E). As N normalizes Γ, it is enough to show that ΓN ≥ ΓE,1 (K), and so it is enough for each e ∈ E # to show that CK (e) ≤ O3 (CK (e))L3 (CK (e))(N ∩ CK (e)). This holds, by inspection of the tables [IA , 5.3], except for (3, Ru). In this case it follows easily from the structure of CK (e) for e ∈ E # that Op (N Γ) = 1. Then by Lp -balance and the fact that mp (K) = 2, N Γ has a component L containing ΓE,1 (K), and we may assume that 3A6 ↑3 L < K. Thus, L ∼ = M24 , J2 , or 3O  N , by [III11 , 13.26a]. But J2 is impossible as L contains E(CK (e)) ∼ = 3A6 for all e ∈ E # ,  and M24 and 3O N are impossible by Lagrange’s Theorem. Therefore (p, K) = (3, M12 ), (5, F i22 ), or (13, F1 ), as asserted. If K = F i22 , then K has a strongly 5-embedded subgroup M ∼ = Aut(D4 (2)), which we may take to contain E. Then EO5 (CK (e))E(CK (e)) ∼ = Z5 × A5 ∼ = CM (∞) (e) for all e ∈ E # , (∞)  and so by [IA , 7.3.3], M = Γ. Clearly ICK (e) (E; 5 ) = {1} for each e ∈ E # , so  IK (E; 5 ) = {1}. But of Aut(K)  by [IA , 5.3], CAut(K) (M ) is a noninner subgroup of order 2. Hence IAut(K) (Γ; 5 ) ∼ = Z2 . Finally, if M < M ∗ ≤ K, then M is

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strongly 5-embedded in M ∗ , so by [IA , 7.6.1], M ∗ = K, whence M is maximal in K. Hence (b) holds. Suppose next that K = M12 . From [IA , 5.3], no maximal subgroup of K contains E and three different E-invariant four-groups, so E must contain only 3-central elements of order 3. Then N ∩ CK (e) contains an element of order 3 for each e ∈ E # , whence N/E contains SL2 (3), hence is isomorphic to GL2 (3) from the list of maximal subgroups in [IA , 5.3]. As F ∗ (CAut(K) (e)) = O3 (CAut(K) (e)) for all e ∈ E # , Γ = 1 and IAut(K) (E; 3 ) = 1. Hence (a) holds. Hence we may assume that K = F1 . If some e ∈ E # is not 13-central, then by L13 -balance and the fact that m13 (K) = 2, N Γ has a component L with L3 (3) ↑13 L. As L3 (3) is unambiguously in Chev(3), L ∼ = L3 (313 ) (which is impossible by Lagrange’s theorem) or L ∈ Chev(3) of level 3. In this last case, since a Sylow 13-subgroup P of LN is not abelian, Lemma 8.13 is contradicted. Therefore, every e ∈ E # is 13-central in K. The proof of (c) is then completed as in the  M12 -case. Lemma 8.15. Suppose that K < L, where K ∼ = L3 (q), q ≡  (mod 3), and η n n ∼ L = L3 (q ), q ≡ η (mod 3), where , η = ±1 and q is a prime power, with n > 1. If (q n − η)/(q − ) is a power of 3, then q = 2, n = 3, and η =  = −1. Proof. Since K < L, we have n = 1; also if η = −1, then n is odd and  = −1. If η =  = 1 then by Zsigmondy’s theorem [IG , 1.1], either q is a Mersenne prime and n = 2, or q n = 64. But in either case, the ratio in question is not a power of 3, in the latter case because q ≡  (mod 3). If η = 1 = −, then n is even and (q 2 − 1)/(q + 1) = q − 1 is a power of 3, but q + 1 = q −  ≡ 0 (mod 3), contradiction. Finally, suppose that η = −1 = , so that n is odd. Then again Zsigmondy’s theorem gives a contradiction unless q 2n − 1 = 64, whence q n = 8 and the lemma follows.  Lemma 8.16. Let p be an odd prime, and suppose that X is a K-group with Op (X) = 1 and Z(X) = 1 containing a simple component K ∈ Cp . Suppose that mp (X) ≤ 3. Let V ≤ X with V ∼ = Ep2 . If Γc V,1 (K) = K, then the following conditions hold: (a) (p, K) = (3, L3 (4)), (3, U3 (8)), (3, M12 ), (5, F i22 ), or (13, F1 ); ∼ (b) Γc V,1 (K) = 1, 1, 1, D4 (2), or 1, respectively; (c) V induces inner automorphisms on K. Moreover, if p = 3 or 13, then V acts on K like a subgroup V0 ≤ K with V0 ∼ = Q8 = V and AutK (V0 ) ∼ (p = 3 only), or group containing SL (p); 2  (d) IAut(K) (V0 ; p ) has order 2, 1, 1, 2, or 1, respectively; and (e) In (b), with p = 5, NK (Γc V,1 (K)) is a maximal subgroup of K. Proof. Obviously V acts faithfully on K. Set X = X/CX (K) and Γ = Γc V,1 (K). Let Vf be the set of all v ∈ V # inducing a field automorphism on K. If Vf = ∅,  then for each v ∈ V # we set Kv = O r (CK (v)) and conclude (see [GL1, I.23– 5 16]) that K = Kv | v ∈ Vf . (The cases (p, K) = (3, L2 (8)) and (5, 2B2 (2 2 )) do not occur, thanks to our assumption K ∈ Cp .) Moreover as mp (X) ≤ 3, mp (CK (v)) ≤ 1  for all v ∈ Vf . It follows that mp (K) = 1, and O p (Kv ) = Lp (CK (v)) for all v ∈ Vf .

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   K0 = O p (Kv ) | v ∈ Vf .

Then K0 ≤ Γ. Moreover, K 0 is invariant under N := NK (V ). By a Frattini   argument, K v ≤ O p (K v )N for each v ∈ Vf , so K 0 (N ∩ K) = K v | v ∈ Vf = K, whence K 0 = K by simplicity. Thus K0 = Γ = K, contrary to hypothesis. We may therefore assume that Vf = ∅. If some v ∈ V induces a graph or graph-field automorphism on K, then p = 3 ∼ G2 (q) or P GL (q),q ≡  (mod 3),  = ±1. But then m3 (CK (v)) > 1 and CK (v) = 3 so m3 (X) ≥ m3 (v CK (v)Z(X)) ≥ 4, contradiction. Next suppose that some v ∈ V induces a non-inner but diagonal automorphism on K. Since mp (K) ≤ mp (X) − mp (Z(X)) ≤ 2, p = 3 and K ∼ = L3 (q), q ≡  (mod 3),  = ±1, q > 2. Then V = v (V ∩ K) contains three Kconjugates of v. Let s be a prime divisor of q 3 − 1 or q 6 − 1 (according as  = 1 or −1) not dividing r n − 1 for any r n < q 3 or q 6 , respectively, where ∼ q is a power of r. Then either E(CK (v)) = SL2 (q) or Os (CK (v)) = 1. Set K0 = Os (CK (v1 ))E(CK (v1 )) | v1 ∈ V # . Then K0 ≤ Γ and Γ contains either three distinct Sylow s-subgroups (each centralized by some element of V ), or three copies of SL2 (q). Using [IA , 6.5.3, 7.3.3] we see that K0 = K, so Γ = K, contradiction. Thus, V induces inner automorphisms on K. There is then no loss in replacing V by its image in K ∼ = K, a simple group of p-rank 2. If K ∼ = An , then n = 2p + k, 0 < k < p, with k > 0 as K ∈ Cp . Then by [IA , 7.5.1a], K = Γ, contradiction. If K ∈ Spor, then Lemma 8.14 gives the same contradiction, with three exceptions: p = 3, K ∼ = M12 ; p = 5, K ∼ = F i22 ; p = 13, K ∼ = F1 . The same lemma gives the additional data in the other parts of our conclusion. Thus by definition of Cp , we may assume that K ∈ Chev(r) − Chev(p) for some r = p. Suppose next that p divides the order of the Schur multiplier of K, whence (as mp (K) = 2) p = 3 and K ∼ = L3 (q),  = ±1, q = r a ≡  (mod 3). If q = 4, then CK (e) is a 3-group for all e ∈ E # , so Γ = 1 and IK (E; 3 ) = {1}. Thus  IAut(K) (E; 3 ) = O3 (CAut(K) (E)), which quickly yields the assertion about Esignalizers. We may therefore assume that q = 4. Similarly if q = 8 then by [IA , 7.3.3], Γ = K forces E to be the image of a nonabelian subgroup of SU3 (8). Hence again CK (e) is a 3-group for all e ∈ E # and the various conclusions follow as they did for q = 4. Consequently we may assume that q = 4 or 8, whence q −  = 3b n, where 3 does not divide n and n > 1. As E is the image of a 31+2 subgroup of SL3 (q), Γ = ΓcE,1 (G) is generated by four distinct homocyclic abelian groups of rank 2 and exponent n. By Lemma 8.15, for any e ∈ E # , O3 (CK (e)) is 1/m not contained in any subgroup of type L± ), m > 1, unless q = 4 or 8, contrary 3 (q to our choice. It then follows easily from [IA , 6.5.3] that Γ = K, contradiction. Now we may assume that p does not divide the order of the Schur multiplier  of K. By [IA , 7.3.3], if we let Ke = O r (CK (e)) for each e ∈ E # , then  K = Ke | e ∈ E # .

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(The only exceptions in that lemma, given that mp (K) = 2, are (p, K) = (3, A6 ), 1 1 (3, U3 (3)), (3, 2F4 (2 2 ) ), and (5, 2F4 (2 2 ) ), with K ∈ Cp in every case, contrary to assumption.) Each Ke is a commuting product of Lie components [IA , 4.9.3]. Suppose next that Sylow p-subgroups of K are not abelian. By Lemma 8.13 n and as K ∈ Cp , p = 3 and K ∼ = G2 (q) or 3D4 (q), q > 2, or 2F4 (2 2 ), n > 1. It is then clear from [IA , 4.7.3A] that every Lie component of every Ke , e ∈ E # , is a 3-component. Thus Ke = L3 (CK (e)) and K = Γ. Thus we may assume that E = Ω1 (P ) where P is an abelian Sylow p-subgroup of K. Now Ke is the product of Lie components L, which are of four types: (a) pcomponents L; (b) groups L ∈ Lieexc such that [L, L] is a p-component and p does not divide |L/[L, L]|; (c) p -groups L; and (d) solvable Lie components. In every case we argue that L ≤ Lp (CK (e))NL (E) ≤ ΓNK (E). Namely, (a): this is trivial if L is a p-component. (b): If [L, L] is a p-component, then E = e (E ∩ [L, L]) and E ∩ [L, L] is the exponent p subgroup of a Sylow p-subgroup of [L, L] so the desired factorization holds by a Frattini argument. (c): If L is a p -group, then by [IA , 4.9.7a, 4.2.2], [E, L] = 1 as p does not divide the order of the Schur multiplier of K. (d): If L is solvable and of order divisible by p, then either L ∼ = A1 (2) with 1  2 A2 (2) since mp (K) = 2.) In both cases it p = 3 or L ∼ = = 2B2 (2 2 ) with p = 5. (L ∼ is clear that L ≤ NK (E). Since Ke ≤ ΓNK (E) for all e ∈ E # , we conclude that K ≤ ΓNK (E). But obviously NK (E) normalizes Γ so K = Γ by the simplicity of K. The proof is complete.  Lemma 8.17. Suppose that p is an odd prime, K  X, v ∈ Ip (Op (Z(X))), and K ∈ Cp . Let x ∈ Ip (X)−v and suppose that mp (CX (x)) ≤ 3 and Lp (CK (x)) = 1. Suppose also that ΓE,2 (X) ≥ K for some E ∈ Ep3 (X) with Op (Z(X)) ≤ E. Suppose that there is R ∈ Sylp (X) with x E ≤ R, and x ∈ E. Then one of the following holds: (a) K ∼ = L2 (pp ) and x induces a nontrivial field automorphism on K; (b) v char R and if mp (R) ≥ 4, then every u ∈ Ip (R) centralizes a normal Ep2 -subgroup of R; (c) (p, K) = (5, HS), (5, Ru), (7, He), or (7, F i24 ). Proof. Without loss Op (X) = 1. Let K0 = Lp (CK (x)). Since K0 = 1 by assumption and mp (CX (x)) ≤ 3, while K0 centralizes v, x ∼ = Ep2 , we have (8A)

mp (K0 ) − mp (Z(K0 )) = 1.

First suppose that mp (K) = 1, whence K ∼ = L2 (p) or Ap (p ≥ 5), or K ∼ = L2 (8) 5 or B2 (2 2 ) (p = 3 or 5 respectively). Since Lp (CK (x)) = 1, [x, K] = 1 and so mp (CX (K)) ≥ 2. Since ΓE,2 (X) ≥ K, mp (CE (K)) = 1 and so mp (Aut(K)) ≥ 5 2, whence K ∼ = L2 (8) or 2B2 (2 2 ) and some a ∈ E # induces a nontrivial field automorphism on K. Now the image of Ω1 (R) in Aut(K) is abelian, so R0 := [Ω1 (R), Ω1 (R)] ≤ CX (K). If R0 = 1 then as mp (CX (K)) and mp (K a) are both at least 2, mp (Ω1 (R)) ≥ 4, contradicting mp (CX (x)) = 3. Therefore R0 = 1. If R0 ∩ Ω1 (Z(R)) = v, then since Z(R) ∩ K = 1, conclusion (b) holds. So we may assume that R0 ∩ Ω1 (Z(R)) ≤ v. As v ∈ Z(R) and Z(R) meets K nontrivially, mp (Z(R)) ≥ 3. As a ∈ Z(R), this implies that mp (CX (x)) > 3, contrary to assumption. Hence the lemma holds if mp (K) = 1. 2

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Assume henceforth that mp (K) ≥ 2. Suppose next that K ∈ Chev(p) and Z(K) = 1. By the Borel-Tits Theorem and Lp -balance, Z(K0 ) is a p -group. If mp (K) = 2 then K ∼ = L2 (p2 ) or L± 3 (p), so since Lp (CK (x)) = 1 we must have [x, K] = 1. But then CX (x) ≥ K × v, x, contradicting mp (CX (x)) = 3. So mp (K) > 2, whence [x, K] = 1 as mp (CX (x)) = 3. Since K0 := Lp (CK (x)) is nontrivial, x is either a field, graph, or graph-field automorphism on K. As K0 ∈ Chev(p), we see from (8A) and [IA , 6.1.4, 3.3.3] that K0 ∼ = L2 (p), so by [IA , 4.9.1, 4.9.2], K ∼ = L2 (pp ). Thus conclusion (a) holds. Suppose that K ∈ Chev(p) but Z(K) = 1. By [IA , 6.1.4, 2.5.12], Out(K) is a p -group, so x induces an inner automorphism on K. But K0 = 1 so by the Borel-Tits Theorem, [x, K] = 1, whence K = K0 and mp (K) − mp (Z(K)) = 1. By Lemma 9.8, this forces p = 3 and K ∼ = 3A6 , which, however, is not a C3 -group. This completes the proof if K ∈ Chev(p). We may therefore assume henceforth that K ∈ Chev(p). If K ∈ Alt, then K ∼ = Akp , k = 2 or 3. But if k = 3 then as Z(X) = 1, mp (CX (x)) > 3, contradiction. Thus k = 2, whence mp (Z(R)) ≥ 3, so x ∈ Z(R) and Ω1 (Z(R)) = Ω1 (R), forcing x ∈ E, contrary to assumption. Thus, K ∈ Alt. If K ∈ Chev(2), then since ΓE/v,1 (K) ≤ ΓE,2 (K) < K, K and p must be as in [IA , 7.3.3]. Since K ∈ Chev(p) and mp (K) > 1, and K ∈ Cp , 1 we get K ∼ = Sp6 (2) with p = 3, or K = 2F4 (2 2 ) with p = 3 or 5. In the first case m3 (CK (x)) = 3 by [IA , 4.10.3], so m3 (CX (x)) > 3, contradiction. In the other two cases Lp (CK (x)) = 1, contradiction. Thus, K ∈ Chev(2). From the definition of Cp it remains to consider the cases in which K ∈ Spor. We use the tables [IA , 5.3] freely. For p = 3 or 11, the conditions that K0 exists  and mp (K0 ) = 1 + mp (Z(K0 )) reduce us to the case K ∼ = Co3 and O 3 (CK (x)) ∼ = Z3 × Aut(L2 (8)); but then m3 (CK (x)) = 3 so m3 (CX (x)) > 3, contradiction. The same considerations rule out all the groups K ∈ Cp for p = 5 and 7 for which mp (K) ≥ 3, except for K ∼ = Co1 with K0 ∼ = A5 ; but then again, mp (CK (x)) = 3, so mp (CX (x)) > 3, contradiction. We may therefore assume that p = 5 or 7, and mp (K) = 2. If R∩K ∼ = Ep2 , then mp (Z(P )) ≥ 3, leading to the contradiction x ∈ E as above. As K0 exists, we are reduced to the following cases in addition to the cases in conclusion (c): p = 5 and K ∼ = Co2 or Co3 . It remains therefore to rule these out. We have R∩K ∼ = 51+2 and there are precisely two K-conjugacy classes of subgroups of K of order 5: the 5-central class, represented by Z(R ∩ K), and a second class, represented by y, with E(CK (y)) ∼ = A5 . Moreover NK (Z(R ∩ K)) = NK (R ∩ K) acts transitively on E1 (R ∩ K/Z(R ∩ K)). Let E0 be the projection of E on K and set Γ = ΓE0 ,1 (K). Then Γ ≤ ΓE,2 (K) < K. Since Z(R ∩ K) is weakly closed in R ∩ K with respect to K, and thanks to the fusion in NK (Z(R ∩ K)), Γ contains CK (g) for all g ∈ R ∩ K # , so K must have a strongly 5-embedded subgroup. This,  however, contradicts [IA , 7.6.1]. The proof is complete. Lemma 8.18. Let Y ≤ X ≤ X ∗ = O8+ (2) with X ∈ K and Y ∼ = Z3 × U4 (2). Let D ∈ E32 (Y ) and suppose that ΓD,1 (X) normalizes Y . Then Y  X. Proof. Let U = O3 (Y ). Then E(Y ) = O 2 (CX ∗ (U )) = O 2 (CX (U )) and U  NX ∗ (E(Y )). By Lemma 7.1, O3 (X) = 1, as D4 (2) ∈ C3 and m3 (X) = 4 = m3 (D4 (2)). By this last condition and L3 -balance, E(Y ) ≤ L for some 3component L  X. If E(Y ) = L, then X normalizes LU , as desired. So assume

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that E(Y ) < L, whence U4 (2) ↑3 L. Thus L is as in Lemma 3.21a. As |L| divides |D4 (2)|, we must have L ∼ = D4 (2). But then by [IA , 7.3.3], L = ΓD,1 (L) so by hypothesis E(Y )  L, a contradiction. The proof is complete.  Lemma 8.19. Let K = F ∗ (X) = O 3 (X) = Un (2), n = 5 or 6, or Co2 . In the latter case, assume that K is a K-group. Let B0 be a hyperplane of an element of E3∗ (X), and suppose that [B0 , z] = 1 for some 2-central involution z ∈ K. Then K = E(CK (w)) | w ∈ B0 ∩ K, E(CK (w)) ∼ = U4 (2). Proof. Suppose first that K ∼ = Un (2). If B0 ≤ B ∈ E3∗ (X), then B is (the image of) a diagonal subgroup of X with respect to an orthonormal basis F of the underlying unitary space. The hypothesis on z implies that B0 is defined by the equality of, say, the first two eigenvalues. Then any four vectors in F that include the first two span an eigenspace of some element of B0 , and the desired assertion follows by repeated use of [IA , 7.3.2]. Now suppose that K ∼ = Co2 , so that X = K. We use [IA , 5.3k] freely. Let C = CK (z) and C = C/O2 (C) ∼ = Sp6 (2). Then B0 = w1  × w2  × w3  where wi ∈ I3 (C), CO2 (C) (wi ) = z, and CC (wi ) ∼ = Z3 × Sp4 (2), i = 1, 2, 3. Hence wi 2 is of class 3B, i.e., O (CK (wi )) = E(CK (wi )) ∼ = U4 (2). Also if w = w1 w2±1 , then ∼ O2 (CK (w, z)) ≥ CO2 (CK (z)) (w) = Q8 ∗Q8 , which implies that w is also of class 3B.  Let K0 be the right side of the desired equation. Then K0 ≥ O 2 (CC (w0 )) | w0 ∈  # w1 , w2   = C. But also K0 ≥ O 2 (CK (w1 )), so z ∈ Z ∗ (K0 ) and thus K0 > C. If O2 (K0 ) = 1 then O2 (C) acts faithfully on it, as O2 (C) = 1. But then K0 has sectional p-rank at least 16 for some p > 2, which is absurd. Hence, O2 (K0 ) = 1. As C is irreducible on O2 (C)/ z, O2 (K0 ) = 1 and every normal subgroup of C is indecomposable. Therefore E(K0 ) is simple and C ≤ E(K0 ). Since E(K0 ) ∈ K and CE(K0 ) (z) = C, it follows easily by [IA , 4.5.1, 5.2.8, 5.3] that K0 = K, as desired, unless possibly K0 ∈ Chev(2). But in the latter case, the structure of the parabolic subgroup C would force K0 = Sp2n (2), n ≥ 4, or F4 (2), and so |K0 |2 = 218 = |C|2 , a contradiction. This proves the lemma.  Lemma 8.20. Let E ≤ K = F2 with E ∼ = E32 and let Γ = ΓE,1 (K). Assume that K is a K-group. Then Γ = K. Proof. Suppose first that E contains a 3-central element e1 of K. The information in [IA , 5.3y] shows that [IA , 7.5.6] applies, and so L := F ∗ (Γ) is simple. Thus − CL (e1 ) has a subgroup L1 of index at most 2 with L1 of the shape 31+8 21+6 − Ω6 (2). 2 If L ∈ Chev(3), then since a Sylow 3-center of L1 has order 3, L1 = O (P ) where P is a parabolic subgroup of L. But this is impossible by the structure of L1 . If L ∈ Chev(r) for some r = 3, CL (e1 ) must be the permutable product of one or more Lie components with an abelian group, which is again not the case. By the structure of CL (e1 ), if L ∈ Alt∪Spor, then the only possibility is L = K, as desired. So we may assume that CK (x) = x × Kx with Kx = Aut(Lx ) and Lx ∼ = F i22 , # 2   . As O (C (x)) = 1, and |K| < |L | , L -balance implies that for each x ∈ E 3 x 3 3 3 K  Γ0 := Kx | x ∈ E # lies in a single component L of Γ. Since m3 (CK (Kx )) = 1 we cannot have L ∼ = K, so Γ = K, and the = Kx , so F i22 ↑3 L. By Lemma 3.9, L ∼ lemma is proved. 

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410

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Lemma 8.21. Suppose that K is a K-group isomorphic to Suz, Co1 , F i23 , F2 , or D4 (2). Then the following conditions hold: (a) K is outer well-generated for the prime p = 3; (b) If K ∼ = Co1 or F2 , E ∈ E32 (Aut(K)), and no element of E # is 3-central in Inn(K), then K = ΓE,1 (K); (c) If K ∼ = Suz and T ∈ Syl3 (K), then K = ΓZ(T ),1 (K); indeed, K = E(CK (x)) | x ∈ Z(T ), E(CK (x)) ∼ = 3U4 (3); and (d) In (c), J(T ) ∼ = E35 , T /J(T ) ∼ = E32 , and for all x ∈ T − J(T ), CJ(T ) (x) = Z(T ) ∼ = E32 . Proof. Part (a) holds for K = D4 (2) by [IA , 7.3.3], and is trivial for K ∈ Spor, since | Out(K)| = 2 is relatively prime to 3. In (b), we may assume by Lemma 8.20 that K = Co1 . The hypothesis implies, in view of [IA , 5.3l], that Ke := E(CK (e)) = L3 (CK (e)) = 1 for every e ∈ E # . Moreover, CO3 (Γ) (e) ≤ O3 (CK (e)) = 1, so O3 (Γ) = 1 as E is noncyclic. By L3 balance, therefore, Ke ≤ E(Γ). Considering the possibilities for Ke , from [IA , 5.3l], we see first that |Ke /Z(Ke )|33 > |K|3 , so the normal closure Le of Ke in E(Γ) is a single component. If Le = Ke for some e, so that Kf ≤ Ke for all f ∈ E # , we have Ke ∼ = [3 × 3]U4 (3) = 3Suz for a unique e ∈ E1 (E), with Kf = E(CKe (f )) ∼ and E = Z(Kf ) for all f ∈ E − e. But then by [IA , 5.3l], AutK (E) ∼ = D8 so e is not unique, contradiction. This shows that Le > Ke for all e ∈ E # , whence Ke ↑3 Le < K. As ↑3 (3Suz) = {Co1 , 3F i22 }, while 3F i22 does not embed in Co1 (because 6U6 (2) does not so embed) we may assume that Ke ∼ = [3×3]U4 (3) or A9 for all e ∈ E # , with Le ∼ = 3Suz or A9+3k , k ≥ 1. (Though A9 ↑3 F5 , |F5 | does not divide |Co1 |.) In either case, by the structure of CK (e), Le = Lf and hence Ke ∼ = Kf for all f ∈ E # . In the A9+3k case this implies that k ≥ 3, but then |Le | does not divide |K|, contradiction. In the 3Suz case, Z(Le ) ≤ Z(Ke ). Moreover, AutK (Z(Ke )) ∼ = D8 , so AutCK (e) (Z(Ke )) does not normalize any subgroup of Z(Ke ) of order 3 centralizing a 3Suz subgroup of K. As Z(Le )  Γ, this is a contradiction. This completes the proof of (b). Let K = Suz. From [IA , 5.3o], J(T ) ∼ = E35 and A := AutK (J(T )) ∼ = M11 . The action of a D10 -subgroup shows that |CJ(T ) (t)| = 33 for every t ∈ I2 (A). Then under a subgroup SL2 (3) ∼ = I ≤ A, J(T ) is the sum of a faithful 2-dimensional Imodule and a nonfaithful I-module. As A contains A4 and has one class of elements of order 3, such elements have a free summand on J(T ), and (d) follows. It remains to prove the second assertion of (c). Using Burnside’s lemma, we see that Z(T ) contains two conjugates each of subgroups of order 3 generated by elements of types 3A and 3B [IA , 5.3o]. Let Γ0 be the subgroup of K generated by the two 3U4 (3) subgroups centralizing subgroups of type 3A. Set Γ0 = Γ0 /O3 (Γ0 ). Using L3 -balance, we find that 3U4 (3) ↑3 L for some 3-component L of Γ0 . Hence L∼ = K or 3F i22 , by Lemma 3.9. By order considerations, L ∼ = K, so Γ0 = K. The proof is complete.  ∼ L3 (4), K  X, and A ∈ E3 (NX (K)) with CA (K) = Lemma 8.22. Suppose K = 3 1 and some element of A inducing a non-inner automorphism on K. Then K = ΓA,2 (K).

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Proof. This follows directly from the fact that CA (K) = 1 and from [IA , 7.3.5b2].  Lemma 8.23. Let K = Co1 or F2 . Then there exists D ≤ K, D ∼ = E52 , such that D contains no 5-central element of I5 (K). ∼ 2 2E6 (2) be an involution centralizer in K. Then ∼ F2 , let H = Proof. If K = 5 by [IA , 4.8.6] there is D ∈ E2 (H) such that CH (d) involves A8 for all d ∈ D# . As 5-central elements of order 5 do not centralize elements of order 7 in K [IA , 5.3y], D satisfies our requirements. If K ∼ = E53 . Then E = CK (E) and = Co1 , let R ∈ Syl5 (K) and E = J(R) ∼  ∼ O 5 (NK (E)/E) ∼ A (5) acts naturally on E. The 5-central elements in E = 5 = Ω3 are the isotropic vectors. We then take D ≤ E to be a 2-dimensional anisotropic subspace. The lemma is proved.  Lemma 8.24. Suppose (p, K) = (3, A9 ), (3, Sp6 (2)), (5, Co1 ), (5, F2 ), or (7, F1 ). Let E ∈ Ep2 (K) and suppose that E contains no p-central element of Ip (K). Then K = ΓE,1 (K). Proof. Set Γ = ΓE,1 (K). If for appropriate e1 , e2 ∈ E # , (8B)

E(CK (e1 )), E(CK (e2 )) = K,

then the lemma will follow. In the A9 case, no e ∈ E # is the product of three disjoint 3-cycles, by assumption. It follows quickly that E = e1 , e2  for some disjoint 3-cycles e1 , e2 , and (8B) holds by [IA , 7.5.2b]. Similarly if K = Sp6 (2) the hypothesis on 3-central elements yields E = e1 , e2  with e1 and e2 having disjoint supports of dimension 2 on the natural K-module. Again (8B) holds, this time by [IA , 7.3.2]. In the remaining cases there exist independent e1 , e2 ∈ E such that E(CK (e1 )) ∼ = E(CK (e2 )) ∼ = J2 , HS, or He, according to the respective isomorphism type of K. (In the Co1 case, since E has no 5-central elements, it lies in some A ∼ = E53 ,  and when A is considered as a natural O 5 (AutK (A)) ∼ = Ω3 (5)-module, E is an anisotropic plane, so it contains independent anisotropic 1-dimensional subspaces of both isometry types, and hence independent elements of class 5A.) Using [IA , 7.7.1b] we see that Op (CK (e)) = 1 for all e ∈ E # . Therefore Op (Γ) = 1. For i = 1, 2, E(CK (ei )) ≤ Lp (Γ) = E(Γ) by Lp -balance; indeed |E(CK (ei ))|pp > |K|p so E(CK (ei )) lies in a single component of E(Γ). Since ei  ∈ Sylp (CK (E(CK (ei )))), but mp (E(CK (ei ))) = 2, H := E(CK (e1 )), E(CK (e2 )) lies in a single component J of E(Γ). But then E(CK (e1 )) ↑p J. Given the isomorphism type of E(CK (e1 )) and the value of p, it follows in each case that J ∼ = K, whence Γ = K. The proof is complete.  Lemma 8.25. For p ≥ 3, L2 (pp ) is outer well-generated with respect to p. Proof. Suppose that K ∼ = L2 (pp ) and E ≤ Aut(K) with E ∼ = Ep2 but E ≤ # Inn(K). Then some e ∈ E induces a nontrivial field automorphism on K. Hence  K = ΓE,1 (K) by [IA , 7.3.8]. Lemma 8.26. Suppose that K = F4 (2), Sp8 (2), Sp6 (8), U7 (2), U6 (2), U5 (2), D4 (2), 2E6 (2)a , or 2 D5 (2). Suppose that B ∈ E3∗ (K) and b ∈ B # with I := E(CK (b)) ∼ = Sp6 (2), Sp6 (2), Sp4 (8), SU6 (2), U4 (2), U4 (2), U4 (2), U6 (2), or D4 (2), respectively. Assume that K is a K-group. Then K = I g | g ∈ K, bg ∈ B.

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Proof. In the cases that K is a classical group, the embedding of B is determined by [IA , 4.8.2a], and the conclusion follows from [IA , 7.3.2]. Suppose then that K ∼ = F4 (2) or 2E6 (2)a . Let B ∈ E3∗ (K), so that m3 (B) = 4 or 5, respectively. A routine argument descending from the algebraic group shows that AutK (B) ∼ = W (F4 ) or W (E6 ), respectively. In particular AutK (B) is irreducible on B. Let H = I g | g ∈ K, bg ∈ B. Then H is NK (B)-invariant. By L3 -balance, H = L3 (H). By the irreducibility noted above, B ≤ H. Set H = H/O3 (H). Then H is a nontrivial pumpup of I, and as m3 (I) = m3 (K) − 1 [IA , 4.10.3], H is quasisimple and I ↑3 H. By assumption, H ∈ K. Using [IA , 4.9.6, 5.2.9, 5.3, 2.2.10] we see that H ∈ Chev(2). Note that |K|3 is less than |Sp6 (8)|3 or |U6 (8)|3 , respectively, so with [IA , 4.9.1, 4.9.2], I ↑3 H via an inner-diagonal automorphism. In particular by [IA , 4.2.2], q(H) = 2. By [IA , 4.2.2], since I is a component of CH (b) and b ∈ H, the untwisted Dynkin diagram of H, or the extended diagram if a node is marked 3, contains the C3 -diagram or the A5 -diagram properly. In the F4 (2) case, if it is the F4 -diagram, we are done; otherwise it is the Cn -diagram for some n ≥ 4. But m3 (K) = 4 so the only other possibility is H ∼ = Sp8 (2), which is impossible because then AutAut(H) (B) ∼ = W (C4 ) does not contain W (F4 ). In the 2E6 (2) case, as U6 (2) ↑3 H and m3 (H) = 5, H is not a classical group, as that would entail m3 (H) ≥ 6, by [IA , 4.8.2]. Since q(H) = 2, the condition m3 (H) = 5 then determines H as isomorphic to K. The proof is complete.  Lemma 8.27. Let K = Suz, Co1 , F i22 , F i23 , F i24 , F2 , or F1 . Let B ∈ E3∗ (K) and b ∈ B # with I := E(CK (b)) ∼ = 3U4 (3), 3Suz, U4 (3), Ω7 (3), D4 (3), F i22 , or 3F i24 , respectively. Assume that K is a K-group. Then there is g ∈ K such that bg ∈ B and K = I, I g . ∼ Suz, this follows from Lemma 8.21c. Otherwise, let P ∈ Proof. If K = Syl3 (CK (b)) and T ∈ Syl3 (K) with P ≤ T . We check that P < T , choose g ∈ NT (P ) − P , and set X = I, I g  and X = X/O3 (X). In every case Z(P ) ≤ b I, g so [b, I g ] = 1. By L3 -balance, X = E(X) > I, the last as [b, I ] = 1. Since B ∈ E3∗ (K), B normalizes every component of X involving I, by [IG , 8.7]. Hence by L3 -balance, I ↑3 X. By Lemma 3.9 and the fact that |X| divides |K|, we conclude that X ∼  = K. The lemma follows. Lemma 8.28. Let K = F ∗ (X) = Sp2n (2), n ≥ 4, Sp6 (8), Dn± (2), n ≥ 5, D4 (2), Un (2), n ≥ 5, F4 (2), or 2E6 (2)a , and assume that m3 (X) ≥ 4. Then there is B0 ∈ E34 (X) and a hyperplane B1 of B0 such that IK (B1 ; 2) = {1}. Proof. Since Dn± (2) ≤ Sp2n (2), Sp8 (2) ≤ F4 (2), and U6 (2) ≤ 2E6 (2)a , the possible minimal counterexamples satisfy K ∼ = D4 (2), or U5 (2), or Sp6 (8), with X ≥ Z3 × Sp6 (2) in the last case as m3 (X) ≥ 4. In the respective cases X contains the direct product of four copies of Σ3 , or Σ3 × GU3 (2), or Σ3 × Sp4 (2) × Z3 . This implies the result.  Lemma 8.29. Let K = 3D4 (2) and let D ∈ E32 (K). If E(CK (x)) ∼ = L2 (8) for some x ∈ D# , then K = E(CK (d)) | d ∈ D# . Proof. For each d ∈ D# , let Kd = E(CK (d)). Then by [IA , 4.7.3A], CK (x) = x × Kx with Kx ∼ = L2 (8). Let Q ∈ Syl3 (CK (x)). Then D = Ω1 (Q) and as

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4 ∼ |K| to Ω1 (Q), so Kd = Kx for all d ∈ D − Kx . Let H =  3 = 3 , x shears E(CK (d)) | d ∈ D# and H = H/O3 (H), and suppose that H < K. Then using L3 -balance and the fact that m3 (K) = 2, we see that H is quasisimple and 3  L2 (83 ) or 2 G2 (3 2 ) or Co3 . Hence by L2 (8) ↑3 H. By order considerations H ∼ = [IA , 4.9.1, 4.9.2, 5.3], H ∈ Chev(2) with D inducing inner-diagonal automorphisms on H, and q(H) = 8 or 2 by [IA , 4.2.2]. Again by [IA , 4.2.2], if q(H) = 2, then the untwisted Dynkin diagram of H must contain an A1 × A1 × A1 subdiagram, so it must contain a D4 subdiagram. Hence in this case, |H| ≥ |D4 (2)|2 = 212 = |K|2 , so H lies in a parabolic subgroup of K by Tits’s lemma [IA , 2.6.7]. But then H∼ = L2 (8), contradiction. Therefore, q(H) = 8. As |K|2 = 212 , H/Z(H) ∼ = L± 3 (8) or Sp4 (8). But none of these is possible, by order considerations. The proof is complete. 

Lemma 8.30. Let K = F1 , x ∈ Ip (K), C = CK (x), and (p, E(C)) = (3, 3F i24 ) or (5, F5 ). Let w ∈ Ip (E(C)) with E(CC (w)) ∼ = D4 (3) or U3 (5), respectively. Set D = x, w and assume that K is a K-group. Then K = E(CK (u)) | u ∈ D# . Proof. Write E1 (D) = {d0  = x , d1  , . . . , dp }, and set Ki = E(CK (di )), i = 0, . . . , p. Let H = Ki | 0 ≤ i ≤ p, a D-invariant subgroup of K; we must prove that H = K. Now for any i, Lp (CK (di )) ≥ Lp (CK (D)) ≥ E(CC (w)) = 1 by Lp -balance, so F ∗ (CK (di )) = di  Ki = F ∗ (CDH (di )) from [IA , 5.3z]. In particular, Op (CH (di )) = 1. So Op (H) = 1. Then by Lp -balance, H = E(DH) = E(H). Moreover, since K does not contain the direct product of p copies of P ∈ Sylp (E(CK (D))), each Ki lies in the same single component of H, so E(H) is  quasisimple. As O p (CK (Ki )) = di  for all i, CD (H) = 1, and thus E(C) ↑p H. Using [III11 , 1.1ab] we conclude that H ∈ Spor, and then H = K by inspection of [IA , 5.3]. The proof is complete.  Lemma 8.31. Let K = O  N be a K-group and E ∈ E34 (K). Then K = ΓE,2 (K) = E(CK (e)) | e ∈ E # .  Proof. Let H = E(CK (e)) | e ∈ E # . Since  F ∗ (CK (e)) = O 3 (CK (e)) ∼ = E32 × A6

for all e ∈ E # , H ≤ ΓE,2 (K) and it suffices to show that H = K. For the same reason, ICK (e) (EE(CK (e)); 3 ) = {1} for each e ∈ E # , so CO3 (H) (e) = 1 and thus O3 (H) = 1. By L3 -balance, H = E(H), and as m3 (K) < 3m3 (A6 ), each E(CK (e)), e ∈ E # , lies in a single component of H. Since H is NK (E)invariant and NK (E) is transitive on E1 (E), H is quasisimple and E ≤ H. Thus F ∗ (CH (e)) = F ∗ (CK (e)) for all e ∈ E # . By Lemma 3.5, H ∼ = A6+9k , k ≥ 1, Suz,  3Suz, or K. As |H|3 = 34 , H = K and we are done. Lemma 8.32. Suppose that F ∗ (X) = K = L3 (q 7 ) with q a power of 2 and  = ±1. Let u ∈ X act on K as a field automorphism of order 7, and let D ∼ = E32 be a subgroup of X which is the image of a nonabelian 3-subgroup of GL3 (q 7 ). (Thus q ≡  (mod 3).) Then  [u, O3 (CK (d))] | d ∈ D# is not a 3 -group.

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Proof. Suppose the assertion is false. Since D is the image of a nonabelian subgroup of GL3 (q 7 ), D is self-centralizing in P GL3 (q 7 ). First assume that D induces inner automorphisms on K. Let r be a prime divisor of q 7 − 1 not dividing 2m − 1 for any lower power 2m < q 7 . Then for each d ∈ D# , [u, O3 (CK (d))] contains a Sylow r-subgroup Rd of K; Rd is homocyclic abelian of rank  2 and CD (Rd ) = d. Since we are assuming the lemma fails, the group W = Rd | d ∈ D# is a 3 -group. In particular by [IG , 11.21], the various Sylow r-subgroups Rd of W are CW (D)-conjugate. Hence CW (D) permutes the groups CD (Rd ) = d, d ∈ D# , transitively, which is absurd. Thus we may write D = d0 , d1  with d0  = D ∩ K and d1 ∈ P GL3 (q 7 ) − K. This time let s be a prime dividing q 21 −1 but not dividing 2m −1 for any 2m < q 21 . Then for d ∈ D − d0 , [u, O3 (CK (d))] contains a cyclic Sylow s-subgroup Sd of K, and CD (Sd ) = d. This time let W = Sd | d ∈ D − d0 . Again because W is a 3 -group and the groups Sd are Sylow in K, hence in W , CW (D) permutes the Sd transitively by conjugation and so permutes E1 (D) − {d0 } transitively, a contradiction.  Lemma 8.33. Let X be a K-group with F ∗ (X) = K ∼ = M12 or L3 (4). Suppose ∼ (2). Then there exists D ∈ E32 (H) such that that X contains a subgroup H P GU = 3  # K = O3 (CK (d)) | d ∈ D . ∼ 31+2 . Suppose first that K ∼ Proof. Let P ∈ Syl3 (H), so that P = = M12 . 2 From [IA , 5.3], H = O (H) ≤ K and there exists D ∈ E2 (P ) such that for all d ∈ D − Z(P ), CK (d) = d × Ad with Ad ∼ = A4 . Moreover, Z(P ) is weakly closed in D. Set K0 = O3 (CK (d)) | d ∈ D − Z(P ). Then K0 is NG (D)-invariant. But then K1 := NG (D)K0 ≤ K. By [IA , 5.3], K1 contains NK (d) for any d ∈ D# , as well as P . Since each NK (d) is maximal in K, K1 = K and then by simplicity of K, K = K0 , as required. Suppose then that K ∼ = L3 (4). Since H ≤ X we may assume that X = P GL3 (4). Choose d0 ∈ P − K such that CK (d0 ) is a Frobenius group of order 7 · 3, and set D = Z(P ) d0 . Then for all d ∈ D − Z(P ), CK (d) ∼ = CK (d0 ). Let K0 = O3 (CK (d)) | d ∈ D − Z(P ). Again K0 is NK (D)-invariant, and so K1 := K0 NK (D) has order divisible by 32 7. By [IA , 6.5.3], K1 = K so K = K0 . The proof is complete.  Lemma 8.34. Let p = 3 and K = 3D4 (2) or L3 (q 3 ),  = ±1, q = 2n ≡  (mod 3),  = ±1. Let a ∈ I3 (Aut(K)) be a graph automorphism or field automorphism, respectively. Then the following conditions hold:  (a) There exists E ∈ E32 (CK (a)) such that K = L3 (CK (e)) | e ∈ E # ; and (b) Let D ∈ E32 (CK (a)) with L3 (CK (d)) = 1 for all d ∈ D# . Suppose that t ∈ I2 (Aut(K)) and [t, D a] = 1. Then t centralizes an a-invariant Sylow 3-subgroup of K containing D and a subgroup E satisfying (a). Proof. Let P ∈ Syl3 (CK (a)). and R ∈ Syl3 (K) with P ≤ R. Suppose first that K ∼ = 31+2 . Hence |R : P | = 3. Choose = 3D4 (2). Then with [IA , 4.7.3A], P ∼ x ∈ R such that CK (x) ∼ = Z3 × L2 (8) and x is extremal in R. If x ∈ P , then CR (x) = x CP (x) = Ω1 (CR (x)), a contradiction since L2 (8) has elements of order 9. Hence x ∈ P . Take E ∈ E32 (P ) containing x; then the conclusion of (a) holds by Lemma 8.29. On the other hand, no parabolic subgroup of K has 3-rank > 1, and

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Out(K) has odd order, so by the Borel-Tits theorem, no D exists as in (b), i.e., (b) is vacuous in this case. Now assume that K ∼ = P GL3 (q) contains E ∼ = E32 = L3 (q 3 ). Then CK (a) ∼  which is the image of a diagonal subgroup of GL3 (q). Then E is also the image of a diagonal subgroup of GL3 (q 3 ), and (a) follows immediately (as q 3 > 2). In (b), if D is the image of a diagonal subgroup of SL3 (q 3 ), then L3 (CK (d)) = 1 for some d ∈ D# , contrary to hypothesis. Therefore, D is the image of a nonabelian 3-subgroup of SL3 (q 3 ). Also, K has one class of involutions z, and m3 (CK (z)) ≤ 1. The same inequality holds for graph automorphisms z (of order 2), by [IA , 4.9.2]. Hence, t induces a field or graph-field automorphism on K ∼ = L3 (q 3 ), and q is a 1 3 square. Let δ = ±1 be such that q 2 ≡ δ (mod 3). Then CK (t) ∼ = Lδ3 (q 2 ), so CK (t) shares a Sylow 3-subgroup with K, and a induces a field automorphism on CK (t). 1 1 Let E ≤ CK (t, a) ∼ = P GLδ3 (q 2 ) be the image of a diagonal subgroup of GLδ3 (q 2 ). We may choose E to normalize D. As in the previous paragraph, (a) holds for E. Let Q ∈ Syl3 (CKa (t)) with DE a ≤ Q; then Q ∩ K satisfies the requirements of (b). The lemma is proved.  8.4. Generation with respect to Non-elementary p-Groups, p > 2. Lemma 8.35. Suppose that K ∈ Chev(2), m3 (Aut(K)) ≥ 3, and K has level at most 2. Suppose that K is a component of a group X with m3 (X) ≥ 4. Let P ∈ Syl3 (X). Then there exists B ∈ E34 (P ) such that K ≤ ΓB,2 (X). Proof. Since m3 (Aut(2F4 (2 2 ) )) = m3 (2F4 (2 2 ) ) = 2 and 2B2 (2 2 ) is solvable, K must have level 2. The conclusion follows from a condition which holds in many cases: 1

(8C)

1

1

For some B ∈ E34 (P ) and some hyperplane A of B, IK (A; 2) = {1}.

Indeed if (8C) holds, then by [IA , 7.3.1], K ≤ ΓA,∗−1 (X) ≤ ΓB,2 (X). In addition, by [IA , 7.3.5], for any B ∈ E34 (P ), we have K ≤ ΓB,∗−1 (X) ≤ ΓB,2 (X) unless possibly K is one of the following groups: 2 An (2), n ≥ 3, Bn (2), n ≥ 3, D2n (2) and 2 D2n+1 (2), n ≥ 2, F4 (2), En (2), 2 E6 (2). If K = D4 (2) there is x of order 3 such that CK (x) ∼ = Z3 × U4 (2); but U4 (2)– for which 3 is a good prime–contains a parabolic subgroup H with Levi subgroup Z3 × L2 (2). A Sylow 3-subgroup of H embeds in a E33 -subgroup of U4 (2) by [IA , 4.10.3], so (8C) holds for Z3 × U4 (2) and hence for D4 (2). It then holds as well for any overgroup of Z3 × U4 (2) or D4 (2), which covers all the groups listed in the previous paragraph with the exception of K = U4 (2) and Sp6 (2). But if K has one of these forms, then m3 (Aut(K)) = 3 so X contains Z3 × K. As U4 (2) = Ω− 6 (2)  embeds in Sp6 (2), the proof is complete. Lemma 8.36. Let X be a K-group, p an odd prime, P ∈ Sylp (X), with Op (X) = 1, mp (P ) ≥ 4 and Z(X) = 1. Set Γ = Γoo P,2 (X) and let K be a p-component of X such that K ≤ Γ but K ∈ Cp . Then K ∼ = L2 (pn ), n ≥ 3, U3 (pn ), n ≥ 2, or (for n 2 2 p = 3) G2 (3 ), n ≥ 3. Moreover, K ∩ Γ is a Borel subgroup of K. Proof. We suppose false and choose a counterexample with K minimal and then P minimal.

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Let B ∈ E∗ (P ) and suppose first that B does not normalize K. Then by [IG , 8.7(iii)], K is not simple, so in particular mp (K) > mp (Z(K)). Since there  are at least p − 1 ≥ 2 conjugates of K centralizing K, we easily find E ∈ E4 (P ∩ K P ) such that mp (CE (K)) ≥ 3. Then K ≤ ΓE,2 (X) ≤ Γ, contradiction. Hence B normalizes K. If mp (Aut(K)) ≤ 2, then as mp (B) ≥ 4, mp (CB (K)) ≥ 2 so again K ≤ ΓB,2 (X) ≤ Γ, contradiction. Thus, mp (Aut(K)) ≥ 3. Suppose that K ∈ Chev(p) and K is simple. If K has twisted rank 1, then since mp (Aut(K)) ≥ 3, K is as in the conclusion of the lemma, mp (P ∩ KZ(X)) ≥ 4, so NK (P ∩ K) ≤ Γoo P,2 (K) < K. As NK (P ∩ K) is maximal in K, the lemma holds in this case. So assume that K has twisted rank at least 2. Let P0 be the subgroup of P inducing inner automorphisms on K. If some element of P induces a field or graph-field automorphism on K, then mp (P0 ) ≥ 2p ≥ 4; likewise if p = 3 and some element of P induces a graph automorphism on K, then m3 (P0 ) ≥ 5 by [IA , 3.3.3]. Hence by our minimal choice, K ≤ Γoo P0 ,2 (X) ≤ Γ, contradiction. So no such elements exist, whence P = P0 . Let Z be the high root subgroup of P ∩ K and set N = NK (Z), a parabolic subgroup of K corresponding to the set of nodes of the Dynkin diagram not connected to the lowest root. Then Z ≤ Z(P ), so NK (Z) = NK (ZZ(X)) ≤ Γ. Let the minimal parabolic subgroups containing P ∩ K and corresponding to the other nodes of the Dynkin diagram be H1 (and H2 , if K ∼ = An (q)). Then N, H1  (or N, H1 , H2 ) equals K, so without loss, H1 ≤ Γ. In particular mp (Op (H1 )Z(X)) ≤ 3 so mp (Op (H1 )) ≤ 2. Consequently K has level p. If Z(Op (H1 )) consists of a single root group, then that root group is normal, hence central, in P , so Z(Op (H1 )) = Z and H1 ≤ N , contradiction. So Z(Op (H1 )) contains at least two root groups. As mp (Op (H1 )) ≤ 2, Op (H1 ) ∼ = Ep2 , whence K ∼ = L3 (p) and m3 (Aut(K)) = 2, a contradiction. Thus if K ∈ Chev(p), then K is not simple, whence by [IA , 6.1.4], K∼ = 3U4 (3), 3Ω7 (3), 3G2 (3), or 32 U4 (3). In the last case we have m3 (K) = 6 by [IA , 6.4.4]. Hence m3 (O3 (H)) ≥ 4 for every H ≤ K such that Z(K) ≤ H and H/Z(K) is a proper parabolic subgroup containing P . Hence all such parabolics H lie in Γ, so K ≤ Γ, contradiction. We may therefore assume that Z(K) ∼ = 3Ω7 (3), then m3 (K) ≥ = Z3 . Similarly if K ∼ m3 (3U4 (3)) = 5, so Γ contains all minimal parabolics containing P ∩ K, hence K ≤ Γ, contradiction. In the other two cases, notice that for any U  P with U ∼ = E32 , we have m3 (CP (U )) = m3 (P ) ≥ 4, so NK (U ) ≤ Γ. Thus Γ covers NK/Z(K) (Z/Z(K)) for every subgroup Z/Z(K) of Z(P ∩K/Z(K)) of order 3. If K ∼ = 3G2 (3), then such normalizers include both maximal parabolic subgroups containing P ∩ K, and so Γ ≥ K, contradiction. And if K ∼ = 3U4 (3), then NK (Z), NK (J(P ∩ K)) = K ≤ Γ, contradiction. Therefore K ∈ Chev(p). ∼ A3p as mp (Aut(K)) ≥ 3. Then for B ∈ Ep∗ (X), B ∩ K = If K ∈ Alt, then K = g1 , g2 , g3  where the gi are disjoint p-cycles. Then CK (gi ) = CK (gi  Z(X)) ≤ Γ. But K = CK (gi ) | 1 ≤ i ≤ 3, so K ≤ Γ, contradiction.

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Note that if p = 3, K is simple, and m3 (K) ≥ 4, then using Konvisser’s theorem [IG , 10.17] we have m3 (CP ∩K (x)) ≥ 3 for all x ∈ I3 (P ∩ K), so NK (x) ≤ NK (x Z(X)) ≤ Γ. Hence in this situation Γ ∩ K is strongly 3-embedded in K, and [IA , 7.6.1] gives the contradiction m3 (Aut(K)) ≤ 2. Thus if p = 3 and K is simple, then m3 (K) ≤ 3. Suppose next that K ∈ Chev(2) − Chev(p). Thus by definition of Cp [V3 , 1.1], and in view of the preceding paragraph and the condition mp (Aut(K)) ≥ 3, K ∼ = SU6 (2), Sp6 (2), 3D4 (2), or Sp4 (8) with p = 3, or 5 K∼ = 2F4 (2 2 ) with p = 5. In the first two cases there is B ∈ E34 (X) such that for some parabolic subgroup H of K, B0 := B ∩ HZ(X) has 3-rank at least 3. (We can take, for example, H of type U4 (2) or Sp4 (2).) Then Γ ≥ ΓB,2 (K) ≥ ΓB0 ,2 (K) = K by [IA , 7.3.1], contradiction. In the last three cases mp (K) = 2 so there exists B ∈ Ep4 (X) containing an element b inducing a graph or field automorphism on K, 1 and such that CK (b) ∼ = G2 (2), Sp4 (2), or 2F4 (2 2 ), respectively. Then some element b1 ∈ B ∩ K centralizes an involution of CK (b). We take B0 = b, b1  Ω1 (Z(X)) and conclude that mp (B0 ) = 3, B0 ≤ B, and IK (B0 ; 2) = 1. Again [IA , 7.3.1] yields the contradiction Γ ≥ K. We are reduced to the case K ∈ Spor. The conditions mp (Aut(K)) ≥ 3, and m3 (K) = 3 if p = 3 and K is simple, restrict us to K ∼ = Co1 , Ly, F5 , F2 , or = J3 , 3J3 , 3M c, 3Suz, 3F i22 , or 3F i24 if p = 3; K ∼ F1 if p = 5; and K ∼ = F1 if p = 7 [IA , 5.6.1]. If K/Z(K) ∼ = J3 with p = 3, then by [IA , 5.3], m3 (C(P ∩K)Z(X) (x)) ≥ 4 for all x ∈ I3 (P ∩ K) − Z(K). Hence NK (x Z(K)) ≤ Γ. Moreover every element of K/Z(K) of order 3 splits over Z(K), so (Γ∩K)Z(K)/Z(K) is strongly 3-embedded in K/Z(K). But K/Z(K) has no such subgroup, by [IA , 7.6.1]. In the remaining cases we choose z ∈ P such that zZ(K)/Z(K) ∈ Z(P/Z(K)). (In the case K ∼ = 3Suz, we choose z so that CK (z) is solvable.) Then z Z(X) contains an Ep2 -subgroup U  P , whence Γ ≥ NK (U ). Thus Γ ≥ N , where N/Z(K) = NK/Z(K) (z Z(K)). We then choose an appropriate subgroup H ≤ K with P ∩H ∈ Sylp (H), and argue that H ≤ Γ and that K = N, H, so that K ≤ Γ, a contradiction. For p = 3, according as K/Z(K) ∼ = M c, Suz, F i22 , or F i24 , we choose ∼ H/Z(K) = U4 (3), 3U4 (3), Z3 × U4 (3), or Z3 × D4 (3). In every case m3 (H) ≥ 4 (see [IA , 6.4.4, 6.1.4, 3.3.3]), so by our minimal choice of counterexample, H ≤ ∼ Γoo P ∩H,2 (K) ≤ Γ. If K = 3M c, H is a maximal subgroup of K and H does not contain N by the structure of N [IA , 5.3], so K = H, N  as required. If K/Z(K) ∼ = Suz, then because of our choice of z, N does not normalize H. If K/Z(K) ∼ = F i22 or F i24 , then Z(H/Z(K)) is not 3-central in K/Z(K), so some element of P shears Z(H/Z(K)) to a p-central element of E(H/Z(K)). Hence if K/Z(K) ∼ = Suz, F i22 , or F i24 , then Lp (H, N  /Z(K)) contains a proper pumpup of 3U4 (3), U4 (3) or D4 (3), respectively. Since H, N  /Z(K) embeds in K/Z(K) and is a K-group by hypothesis, we get H, N  = K, as required, by Lemma 3.9. The remaining cases are similar; according as K ∼ = Co1 , Ly, F5 , F2 , or F1 with p = 5, we take H ∼ = F1 = Z5 × J2 , G2 (5), Z5 × U3 (5), Z5 × HS, or Z5 × F5 . If K ∼ with p = 7, we take H ∼  = Z7 × He. The proof is complete.

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Lemma 8.37. Suppose that K ∈ Spor ∩ C2 and mp (Aut(K)) ≥ 3, for some odd prime p. Suppose that K is a component of a K-group X with mp (X) ≥ 4. Let ∼ P ∈ Sylp (X). Then K ≤ Γoo P,2 (X), or else p = 3, K = Co2 , P normalizes K, CP (K) = 1, and for any B ∈ Ep∗ (P ), ΓB,2 (K) ∼ = U6 (2) · 2. Proof. As K ∈ C2 ⊆ K2 , O2 (K) = 1. Clearly we can then pass to K/O2 (K) and assume that K is simple. Since mp (Aut(K)) ≥ 3 and K ∈ Spor ∩ C2 , we have (p, K) = (3, J3 ), (3, Suz), (3, Co1 ), (3, Co2 ), (3, F in ), 22 ≤ n ≤ 24, (3, Fi ), 1 ≤ i ≤ 3, (5, Co1 ), (5, F2 ), (5, F1 ) or (7, F1 ) [IA , 5.6.1]. Let Γ = Γoo P,2 (X) and suppose that K ≤ Γ. If mp (Aut(K)) = 3, then CP (K) = 1. These are the cases (p, K) = (3, J3 ), (5, Co1 ), (5, F2 ), and (7, F1 ). Since CP (K) = 1, we may apply Lemma 8.36 in each case to KCP (K) and conclude that K ≤ Γ, contradiction. Likewise if (p, K) = (5, F1 ) we choose subgroups N = NK (Z(P ∩ K)) and H ∼ = Z5 × F5 of K with P ∩ H ∈ Sylp (H). By Lemma 8.36, H ≤ Γ. Also O5 (N ) ∼ = 51+6 so m5 (O5 (N )) = 4, whence N ≤ NG (O5 (N )) ≤ Γ. Then L5 (H, N ) is a pumpup of F5 (with respect to the prime 5), and a proper pumpup at that, since N does not normalize E(H) (as |N |5 > |H|5 ). As L5 (H, N ) is embeddable in K ∼ = F1 , it must equal K, so K = H, N  ≤ Γ, contradiction. In all the other cases p = 3. ∼ Suz, Co1 , F i22 , F i23 , F i , F3 , F2 , or F1 , choose H ≤ K such According as K = 24 ∼ 3U4 (3), 3Suz, 3 × U4 (3), 3 × Ω7 (3), 3 × D4 (3), that P ∩ H ∈ Syl3 (H) and H = 3 × G2 (3), 3 × F i22 , or Z3 × F3 , respectively. Then E(H) ∈ C3 and m3 (H) ≥ 4 and Z(H) = 1 [IA , 6.4.4, 3.3.3, 5.6.1]. By Lemma 8.36, H ≤ Γ. Now Z3 ∼ = Z(H) ∈ Syl3 (CK (H)) [IA , 5.3]. In all cases except K ∼ = Suz, Z(H) is not 3-central. Hence Z(H) is conjugate to an element of Z(P ∩H) under the action of NP (P ∩H). In the Suz case, in fact, P ≤ H, Z(P ) ∼ = E32 , and Z(H)  NK (P ). Hence in every case L := L3 (NK (P ), H) has a component which is a nontrivial pumpup of E(H). Since L embeds in K, it follows in every case that L = K. But NK (P ) ≤ Γ, so K = L ≤ Γ, contradiction. The case p = 3, K ∼ = Co2 remains. In this case we take H = NK (H0 ) with ∼ H0 = U6 (2) and |H : H0 | = 2, again such that P ∩ H ∈ Syl3 (H). There exists such a maximal subgroup H of K, see [CCPNW1]. As |Co2 |3 = 36 = |U6 (2)|3 , P is isomorphic to a Sylow 3-subgroup of U6 (2), and thus B := J(P ) ∼ = E34 is a diagonalizable subgroup of H0 . As H0 contains Σ3 × U4 (2), B has a hyperplane B0 centralizing an involution of H0 , and so Γ ≥ ΓB0 ,2 (H) = H by [IA , 7.3.1]. As Γ ∩ K < K, Γ ∩ K = H, completing the proof.  Lemma 8.38. Suppose that O3 (X) = 1, and K is a component of X isomorphic to D4 (4), F4 (4), E6 (2), or E6± (4). Assume that P ∈ Syl3 (X) and m3 (P ) ≥ 4. Then K ≤ Γoo P,2 (X). Proof. In each case a parabolic subgroup H of K exists such that m3 (H) ≥ 4 (a Borel subgroup, in fact, except for K ∼ = E6 (2), when H can be taken of type  D4 (2)). Let B ∈ E34 (H). Then by [IA , 7.3.1], K ≤ ΓB,∗−1 (X) ≤ Γoo P,2 (X). Lemma 8.39. Let K = Suz, Co2 , Co1 , F i22 , F i23 , F i24 , F3 , F2 , or F1 , and assume that K is a K-group. Let P ∈ Syl3 (K). Then Γoo P,2 (K) = K.

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Proof. Let Γ = Γoo P,2 (K). We consider the various K’s separately. Suppose that K = F i22 . Let L ≤ K with L = CK (Z(L)) ∼ = 2U6 (2). Without E loss, P ∩ L ∈ Syl3 (L). Then m3 (L) = 4 and a subgroup B ∼ = 34 of P is diagonalizable with respect to an orthonormal basis B of the natural module V . For any three-dimensional subspace W of V spanned by vectors in B, the SU3 (2)-subgroup LW of L supported on W satisfies m3 (CB (LW )) = 2, so LW ≤ Γ. Letting W vary and using [IA , 7.3.2], we conclude that L ≤ Γ. By [IA , 5.3t], the permutation representation of K on L is rank 3, with suborbit decomposition 3510 = 1 + 693 + 2816. As no nontrivial subsum divides 3510, L is maximal in K. But |P | = 39 > |P ∩ L|, so Γ ≥ L, P  = K, as claimed. Inductively, similar arguments handle the cases K∼ = 2F i22 , and then K ∼ = F i24 , using L ∼ = F i23 . = F i23 , this time using L ∼ Suppose that K = Suz. By Lemma 8.21c, it is enough to show that if x ∈ I3 (Z(P )) with Kx := E(CK (x)) ∼ = 3U4 (3), then Kx ≤ Γ. But the maximal parabolic subgroups K1 and K2 of Kx containing P are contained in NK (Z(P )) and NK (J(P )), respectively, with Z(P ) and J(P ) noncyclic and elementary abelian. (See [IA , 6.4.4].) Hence Γ ≥ K1 , K2  = Kx , as claimed. Suppose next that K = Co2 . Let B = J(P ). Since K ≥ M ∼ = M c and |K|3 = |M |3 , it follows by [IA , 5.3n] that B ∼ = E34 and AutK (B) ≥ M10 ≥ A6 . Let X be the set of all x ∈ B # such that Kx := E(CK (x)) ∼ = U4 (2). We argue that for any x ∈ X, NK (x) ≤ Γ. This is because Kx = ΓB∩Kx ,1 (Kx ). (The right-hand side contains NKx (B ∩ Kx ) and NKx (y) for 3-central 1 = y ∈ B ∩ Kx , and these are two maximal parabolic subgroups of Kx ∼ = P Sp4 (3).) Thus Γ contains Γ0 := Kx | x ∈ X as well as NK (B). Note also that Γ contains F ∗ (CK (y)) = O3 (CK (y)) for all 3-central y ∈ B # . It follows that O3 (Γ) ∩ CK (y) = 1 for all such y, as well as all y ∈ X, and so O3 (Γ) = 1. Thus by L3 -balance, Γ0 ≤ E(Γ). The action of NK (B) on B is irreducible and primitive so E := E(Γ) is a single simple component, with U4 (2) ↑3 E and [NK (B), NK (B)] inducing inner automorphisms on E. Next, let z ∈ I2 (K) be 2-central, set Cz = CK (z) and C z = Cz /O2 (Cz ), and let A ∈ E33 (Cz ). Let A = {a ∈ A# | CC z (a) ≥ Σ6 }. Let a ∈ A; then a must be in class 3B (like the elements of X). Expand A to a maximal elementary abelian subgroup B0 of CK (a); then B0 is conjugate to B, and replacing z by an appropriate conjugate, we may assume that A ≤ B. Then A ⊆ X and Cz = CCz (a0 ), ΓA,2 (Cz ) | a0 ∈ A ≤ Γ, indeed Cz ≤ E. As E is a K-group, E = K, so Γ = K, as claimed. If K = Co1 , the previous argument shows that Γ contains a subgroup L ∼ = Co2 . Let z ∈ I2 (L) be 2-central in L. It is then clear from [IA , 5.3l] that z is 2-central in K, O2 (CK (z)) ≤ L ≤ Γ, and CΓ (z)/O2 (CK (z)) contains J/O2 (CK (z)) ∼ = Sp6 (2) and lies in CK (z)/O2 (CK (z)) ∼ (2). As shown in the previous paragraph, an = Ω+ 8 element A ∈ E33 (J) lies in some A∗ ∈ E34 (L) ⊆ E34 (Γ). Hence ΓA,2 (K) ≤ ΓA∗ ,2 (K) ≤ Γ, so A ≤ A1 ≤ Γ for some A1 ∈ E34 (CK (z)). Then Γ contains ΓA1 ,2 (K), which in ∼ turn contains CCK (z) (x) for any x ∈ A# 1 such that CCK (z)/O2 (CK (z)) (x) = Z3 ×U4 (2), as argued in the previous paragraph. But such centralizers generate CK (z) modulo O2 (CK (z)), by [IA , 7.3.2]. We have proved that CK (z) ≤ Γ. As this holds for any 2-central z ∈ I2 (L), and Γ is a K-group, it follows directly that O2 (Γ) = O2 (Γ) = 1 and then that Γ ∼ = K, so Γ = K.

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∼ U4 (3) or If K = F2 or F1 , choose D ∈ E32 (P ) such that I := E(CK (D)) = D4 (3), respectively, and P ∩ I ∈ Syl3 (I). Let z be any 2-central involution of I. Then z is 2-central in K and D ≤ B ∈ E34 (CK (z)) for some B. Then B ≤ Γ. Now CK (z)/O2 (CK (z)) ∼ = Co2 or Co1 , so with the previous results, CK (z) ≤ Γ. As z was arbitrary in I it follows easily that E(Γ) is simple and CE(Γ) (z) = CK (z). But Γ is a K-group, and so this implies that Γ ∼ = K, as claimed. Finally, if K = F3 , choose a P -extremal x ∈ I3 (P ) with O 2 (CK (x)) = x × Ex , Ex ∼ = G2 (3). Set R = CP (x) ∈ Syl3 (CK (x)). For each of the two maximal parabolic subgroups M of Ex containing R, we see from the commutator formula that m3 (O3 (M )) ≥ 4. Hence M ≤ Γ and so Ex ≤ Γ, whence Ex ≤ H := L3 (Γ). But also |R| = 37 < |P | so there exists g ∈ NP (R) such that [Ex , xg ] = 1 and xg ∈ Z(R). Then by a similar argument, Exg ≤ H. We then quickly find that Ex , Exg  lies in a single 3-component H0 of H, and G2 (3) ↑3 (H0 /O3 (H0 )). This condition, together with Lemma 3.9 and the condition that |H0 | divide |K|, forces  H0 = K and hence Γ = K. The proof is complete. Lemma 8.40. Let K = L± 4 (3) or Co2 . Let P ∈ Syl3 (K). Then K = ΓP,2 (K). Proof. Let Γ = ΓP,2 (K). L± 4 (3).

Suppose that K = Then K is generated by the minimal (non-Borel) parabolic subgroups H of K containing P . For any such H, since F ∗ (H) = O3 (H) and H is not 3-closed, we have m3 (O3 (H)) ≥ 2, so H ≤ Γ. Thus, Γ = K in this case. Suppose that K = Co2 . We argue that if Γ < K, then K has a strongly 3-embedded subgroup, contradicting [IA , 7.6.1]. As Γ controls K-fusion in P , it suffices to show that if x ∈ I3 (P ) and x is extremal in P , i.e., R = CP (x) ∈ Syl3 (CG (x)), then CG (x) ≤ Γ. But from [IA , 5.3], either O3 (CG (x)) ∼ = 31+4 , in ∼ which case certainly CG (x) ≤ Γ, or CG (x) = Z3 × U4 (2), in which case again CG (x) ≤ Γ as U4 (2) ∈ Chev(3) is generated by parabolic subgroups containing a given Sylow 3-subgroup. The result follows.  Lemma 8.41. Suppose that X is a K-group with mp (X) ≥ 4 for some odd prime p and Op (X) = 1. Let P ∈ Sylp (X). Then any ΓP,2 (X)-invariant p -subgroup of X is trivial. Proof. Suppose false and let X be a minimal counterexample, with 1 = W ∈ IX (Γ; p ) and Γ = ΓP,2 (X). By [IG , 23.3(ii)], W ≤ Op (P W ), whence [W, Op (X)] = 1. Hence there exists a component K of X such that [W, K] = 1. By minimality we have X = K ∗ W P , where K ∗ = K W P . Thus K does not have the weak signalizer property with depth 3 (see [IA , p. 394]). Moreover, K ≤ Γ, for otherwise [W, K] is a p -group and hence [W, K] = 1, contradiction. This implies that mp (Aut(K)) ≥ 3, for otherwise for any B ∈ E4 (P ), K ≤ ΓB,∗−2 (X) ≤ Γ, yielding [W, Γ] = 1, contradiction. By [IA , 7.7.17], one of the following holds: p = 3 and K is one of the groups in Lemma 8.38; K ∼ = F i22 , with p = 5. Lemma 8.38 rules out = An , n > p2 ; or K ∼ the first possibility, and the last one is out since m5 (Aut(F i22 )) = 2. Suppose finally that K ∼ = An , n > p2 . Let B ∈ E∗ (P ), so that mp (B) ≥ 4. Then B normalizes K by [IG , 8.7(iii)]. It suffices to prove that K = ΓB,∗−2 (K), for then K ≤ Γ, a contradiction. Passing to KB/CB (K), we lose the condition mp (B) ≥ 4, but it suffices still to show K = ΓB,∗−2 (K). As B ∈ E∗ (P ), B ≤ K

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and B is the direct product of groups generated by disjoint p-cycles b1 , . . . , bk . Then for all 1 ≤ i < j ≤ k, ΓB,∗−2 (K) contains the alternating subgroup Kij supported on the union of the fixed point set of B with the support of bi , bj , since [Kij , b ] = 1 for all i =  = j. But then ΓB,∗−2 (K) contains Kij | 1 ≤ i < j ≤ k, which clearly equals K. Therefore K ≤ Γ, a final contradiction.  Lemma 8.42. Let X be a K-group with O3 (X) = 1 and P ∈ Syl3 (X). Assume that m3 (P ) = 3 and Z(X) = 1. Let K be a component of X and K0 a subgroup of K such that ΓP,2 (K) ≤ K0 < K. Then the following conditions hold: (a) K ∼ = A6 , M11 , L3 (4), SL3 (4), U3 (3), or L2 (8); (b) K0 is solvable; (c) If in addition Y is a P -invariant subgroup of X of order 2, then K/Z(K) ∼ = L3 (4) and a generator of Y induces a unitary automorphism on K. Proof. If Z(K) = 1, then as Z(X) = 1, ΓP ∩K,1 (K) ≤ ΓP,2 (K) ≤ K0 < K, so K has a strongly 3-embedded subgroup contained in K0 . If in addition m3 (K) > 1, then by [IA , 7.6.1, 7.6.2], conclusions (a) and (b) hold with K ∼ = M11 , A6 , L3 (4) or U3 (3). Suppose that m3 (K) = 1. Clearly m3 (CX (K)) = 1, for otherwise K = ΓP,2 (K), contradiction. Hence some x ∈ I3 (X) induces a non-inner automorphism on K. By [GL1, I.7-13(2)], x induces a field automorphism, and then by [IA , 7.3.8], K∼ = L2 (8) and K0 is a Sylow 3-normalizer, so (a) and (b) hold in this case as well. Suppose then that Z(K) = 1 and set K = K/Z(K). If K ∈ Alt ∪ Spor, then every element of I3 (K/Z(K)) is real and so splits over Z(K). Hence ΓP,2 (K) contains ΓP ∩K,1 (K), so K and K have strongly 3-embedded subgroups. Thus by [IA , 7.6.1, 7.6.2], K0 is solvable and K ∼ = SL3 (4) or 3A6 . In the last case as m3 (Aut(K)) = 2 < m3 (X) we must have m3 (CX (K)) = 2, a contradiction. Thus (a) and (b) hold in this case. Suppose, on the other hand, that K ∈ Chev(r) − Alt for some r. If r = 3 then by [IA , 6.1.4], K has twisted rank at least 2, so K is generated by its maximal parabolic subgroups containing P ∩ K, which lie in ΓP,2 (K). Thus K = K0 , contradiction, and so r = 3. Using [IA , 7.3.5, 6.1.4] and the fact that m3 (X) = 3, we find that r = 2 and K ∼ = SL3 (q), q = 2n ≡  (mod 3), 3  = ±1. Let E ∈ E3 (X) and write E = E0 × Z(K). Then ΓE0 ,1 (K) ≤ K0 , so by [IA , 7.3.4], K ∼ = SU3 (2). This completes the proof of (a) and = SL3 (4) with K0 ∼ (b). As for (c), it follows from Lemma 7.1 if K ∈ Chev(3) and from [IA , Table 5.6.2] if K∼ = = M11 (which is complete by [IA , 5.3]). Hence we may assume that K/Z(K) ∼ L3 (4). Then m2,3 (K/Z(K)) = 1 by inspection of the parabolic subgroups of K and the Borel-Tits theorem; so Y embeds in Out(K) and centralizes P ∩ K. By [IA , 4.9.1, 4.9.2] the only possibility is |Y | = 2, with CK/Z(K) (Y ) ∼ = U3 (2). The proof is complete.  Lemma 8.43. Let X = E(X) x, where x ∈ Ip (X) for some odd prime and Op (X) = 1, and suppose that x permutes the components of X transitively. Suppose that CX (x) has a component K ∼ = L2 (pp ). Let A ∈ Sylp (K). Then X = K, ΓA,2 (X). Proof. Let X0 = K, ΓA,2 (X). Note that the Schur multiplier of K is a p -group, by [IA , 6.1.4]. In view of Lemma 3.11d we can only have X ∼ = L2 (pp )  Zp

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∼ L2 (pp2 ) x with x inducing a nontrivial field automorphism on E(X) in or X = the latter case. Either way, CX (A), and in particular ΓA,2 (X), contains a Sylow 2 p-subgroup P of E(X). If E(X) ∼ = L2 (pp ), and if X0 < X, then by Tits’s Lemma [IA , 2.6.7], X0 ≤ NX (P ), which is absurd as K ≤ X0 and K is not p-closed. Thus the lemma holds in this case. If X ∼ = K  Zp , then K is a maximal subgroup of E(X) by [IG , 3.25], but P ≤ K, so X = K, P, x = X0 . The proof is complete.  Lemma 8.44. Let K be a K-group isomorphic to Co2 , Co1 , F i22 , or F2 . Let P be a 3-group acting on K with |P | ≥ 39 . Then K = ΓP,1 (K). Proof. Let Γ = ΓP,1 (K). If K = Co2 , Co1 , or F i22 , then |K|3 ≤ 39 , and so if Γ < K, then K has a strongly 3-embedded subgroup, contradicting [IA , 7.6.1].  If K = F2 , this follows from Lemma 8.20. 8.5. Other. Lemma 8.45. Let X be a K-group with a strongly 3-embedded subgroup M . Let P ∈ Syl3 (M ) and let P1 and P2 be subgroups of P which are strongly closed in P with respect to M (or equivalently X). Then P1 ∩ P2 = 1. Proof. Without loss O3 (X) = 1, and as M is strongly 3-embedded in X, O3 (X) = 1. If E(X) has distinct components K1 , K2 , then Ki ≤ NX (P ∩ K3−i ) ≤ M and so X = E(X)NX (P ∩E(X)) ≤ M , contradiction. Therefore F ∗ (X) = E(X)  is simple. Now [IA , 7.6.1, 7.6.2] apply, and the result is straightforward. 5

Lemma 8.46. Suppose that (p, K0 ) = (3, L2 (8)), (5, 2B2 (2 2 )), (5, A5 ), (7, L2 (7)), or (17, L2 (17)). Let X be a K-group with a normal component K and an element x of order p such that K0 is a component of CK (x). Suppose that Z ≤ CX (x) with Z = z0 , z1  ∼ = Ep2 , z1 ∈ K0 , [z0 , K0 ] = 1, and x ∈ Z. Let Γ = ΓZ,1 (X), and suppose that mp (Γ) ≤ 3, K ≤ Γ, but K0 covers a p-component of Γ/Op (Γ). Then (p, K0 ) = (5, A5 ), K ∼ = A5 and = A10 or Co1 , and ΓZ,1 (K) ≥ K0 × K1 with K1 ∼ [x, K1 ] = 1. Moreover, K0 and K1 are conjugate in NG (Z x ∩ K). Proof. Since K ≤ Γ, K = K0 . Suppose that K ∈ Chev(r) for some r. Then K0 ∈ Chev(r) by [IA , 4.9.6]. If r > 2, then the only choice is r = p. But then x induces a nontrivial field automorphism on K, so mp (K) ≥ p ≥ 3. In particular 3 p = 3 and K ∼ = 2 G2 (3 2 ). But then m3 (CK (z1 )) > 3 by [IA , 3.3.3] for the element z1 ∈ Z ∩ K0 , so m3 (Γ) > 3, contrary to assumption. Thus, r = 2. Now as none of the exceptional groups in [IA , 7.3.4] have a component isomorphic to K0 in the centralizer of an automorphism of order p, that result implies that K = ΓZ,1 (K), contradiction. Suppose next that K ∈ Spor. Since K0 is a component of CK (x), we see from [IA , 5.3] that (p, K) is one of the following: (3, Co3 ), (5, J2 ), (5, Co3 ), (5, Co2 ), (5, Co1 ), (5, HS), (5, Suz), (5, He), (5, Ru), (5, F i22 ), (7, Co1 ), (7, He), or (7, F3 ). Recall that Z = z0 , z1  with [z0 , K0 ] = 1 and z1 ∈ K0 . As K ≤ Γ, [z0 , K] = 1 so z0 projects onto an element y ∈ Ip (CK (K0 )). In many cases y is not p-central in K, but y, z1  is a Sylow p-center in CK (x), with z1 weakly closed in y, z1 . When that occurs, y g = yz1 for some g ∈ CK (z1 ). Thus g ∈ Γ and g normalizes K 0 = K0 Op (Γ)/Op (Γ). As [y, K0 ] = 1, also [yz1 , K 0 ] = 1. But this is absurd as z1 ∈ K0 − Z(K0 ). This argument rules out (p, K) = (3, Co3 ), (5, Co3 ), (5, Co2 ), (5, HS), (5, Ru), and (7, He).

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In other cases a Sylow p-subgroup R of K is isomorphic to Ep2 , and z1  has at least three conjugates under AutΓ (R). This again is a contradiction since K 0 is a component of Γ = Γ/Op (Γ), so it has at most two images under AutΓ (R). This argument rules out (p, K) = (5, J2 ), (5, Suz), (5, He), (5, F i22 ), (7, Co1 ) and (7, F3 ). The only remaining sporadic case is (p, K) = (5, Co1 ). But in that case CK (x) contains two A5 components interchanged in CK (x), and the lemma holds. Finally suppose that K ∈ Alt. Then K0 ∈ Alt so p = 5 and K0 ∼ = A5 . As mp (Γ) ≤ 3, K ∼ = An , n = 10 or 15. Now z1 ∈ K0 is a 5-cycle and Γ contains A, the alternating group on the fixed points of z1 . Also, there exists z2 ∈ Z # fixing pointwise the support of z1 on n letters. If z2 is not a 5-cycle, then n = 15 and CK (z1 z2 ) contains an element g moving the support of z1 to a set disjoint from it. As CK (z1 z2 ) ≤ Γ, it follows that Γ ≥ A, Ag  = K, contradiction. Thus, z2 acts on K as a 5-cycle. If n = 15, we easily see that Γ ≥ CK (z1 ), CK (z2 ) = K, contradiction. So n = 10, and the remaining statements follow easily.  Lemma 8.47. Let K = F ∗ (X) ∼ = L3 (64), U3 (83 ), 3D4 (4), or 3D4 (8). Let x ∈ I3 (X) induce a field automorphism of K or graph automorphism of K of order 3, such that I := E(CK (x)) ∼ = L3 (4), U3 (8), L3 (4), or U3 (8), respectively. Let V ≤ I with V ∼ = E32 be the image of a subgroup of SL3 (4) or SU3 (8) isomorphic to 31+2 . Then IK (V x ; 3 ) is not a 3 -group. Proof. Set W = IK (V x ; 3 ) and assume by way of contradiction that W is a 3 -group. Note that all elements of V # are I-conjugate, hence K-conjugate. Moreover ∼ (x) CK = P GL3 (4) or P GU3 (8), whence all elements of V # are 3-central in K. Let v ∈ V # . Then CK (v) is V x-invariant. If K ∼ = L3 (64), then Wv := O7 (CK (v)) ∼ = E72 is a Sylow 7-subgroup of K. In particular Wv ∈ Syl7 (W ). Moreover V , as the image of a nonabelian 3-subgroup of SL3 (64), is self-centralizing in K. Thus CV (Wv ) = v. Since W is a 3 -group, all V -invariant Sylow 7-subgroups of W are CW (V )-conjugate [IG , 11.21]. But then the groups CV (Wv ) = v, v ∈ V # , are all CW (V )-conjugate, which is absurd. Thus the lemma holds in this case. If K ∼ = U3 (83 ), a similar argument can be made, with 19 in place of 7. Suppose that K ∼ = 3D4 (q), q = 4 or 8, and accordingly set  = 1 or −1. Fix # v ∈ V and let P ∈ Syl3 (CK (v)) with V ≤ P . From [IA , 4.7.3A] we deduce that |P : P ∩Kv | = 3, where Kv = CK (v)(∞) ∼ = SL3 (q). Here P ∩Kv is of maximal class, and has a diagonalizable subgroup P1 of index 3 which is homocyclic abelian of rank 2 and exponent q −. On the other hand a maximal torus of K corresponding to the element ±1 of the Weyl group according as  = ±1 is isomorphic to Zq3 − × Zq− , and thus contains an abelian 3-subgroup P0 of type (3(q − ), q − ). Replacing P0 by a conjugate we may assume that P0 ≤ P . Then P0 ∩ Kv is an abelian maximal subgroup of P ∩ Kv . Replacing P1 by a NKv (P ∩ Kv )-conjugate we may assume  that P0 ∩ Kv = P1 . Now O 3 (CK (v))/ v ∼ = P GL3 (q) has a Sylow 3-subgroup of exponent q − . Hence v is the group of (q − )th powers in P0 . We argue that v is weakly closed in P0 with respect to K.

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Namely, v ∈ [P1 , P1 ] and P/ v ∈ Syl3 (P GL3 (q)) is of maximal class, so [P, P ] contains Ω1 (P0 ) ∼ = Ep2 . Hence P0 is the unique abelian maximal subgroup of P (otherwise [P, P ] would be cyclic). In particular by [IG , 16.9], NK (P0 ) controls K-fusion in P0 ; but it normalizes v by the previous paragraph. This proves our assertion. As any two subgroups of Kv of order 3 other than v are K-conjugate, v is weakly closed in Kv with respect to K. On the other hand V ∼ = E32 has been chosen in I so that its elements are all conjugate in I. Hence V = v, v1  where v1 ∈ P −Kv . In particular as Ω1 (P0 ) ≤ Kv , v1 ∈ P0 . Thus v1 does not centralize a maximal torus of Kv of type (q − , q − ). It follows that W1 := O3 (CKv (v1 )) has order r = (q 3 − )3 = 7 or 19, according to the value of q, and [g, W1 ] = W1 for any g ∈ Ω1 (P0 ) − v. By construction, [v1 , W1 ] = 1. Now v1 is conjugate in P to v1 g for some such g, and we set W2 := O3 (CKv (v1 g)) ∼ = W1 . As with W1 , g acts nontrivially on W2 , while [v1 g, W2 ] = 1, so [v1 , W2 ] = W2 . Note that by [IA , 4.7.3A], Y := Or (CK (v)) ∼ = W1 . We set Yi = Y Wi , i = 1, 2. Thus Y1 and Y2 are V -invariant Sylow r-subgroups of K, with CV (Y1 ) = V and CV (Y2 ) = v < V . If Y1 , Y2  were a 3 -group, then Y1 and Y2 would be CK (V )conjugate, hence CV (Y1 ) = CV (Y2 ), a contradiction. This completes the proof of the lemma.  Lemma 8.48. Let K = F i22 or F i24 , and let t ∈ Aut(K) be such that L := E(CK (t)) ∼ = D4 (2) or F i23 . Let Γ = L, CK (t0 ) | t0 ∈ I2 (L). Then Γ = K. Proof. By [IA , 5.3tv], L is standard in Aut(K) (and t ∈ Inn(K)) and CAut(K) (L) = t. Moreover, some 2-central involution of K lies in L. Thus, |Γ|2 = |K|2 > |CK (t)|2 . Therefore NΓ (L) does not cover Γ := Γ/O2 (Γ), so by L2 -balance, M > L, where M is the subnormal closure of L in Γ. As |L|5 = |K|5 , we must have L ↑2 M . If M ∼ = K, then Γ = K, as desired. Otherwise, M ∼ = D4 (4) ∼  and K = F i22 . But |D4 (4)| does not divide |K|, so the proof is complete. 9. p-Structure, p odd 9.1. Self-centralizing Ep2 -Subgroups, p ≥ 5. Lemma 9.1. Suppose X is a K-group and p ≥ 5 is a prime such that Op (X) = 1 and mp (X) ≥ 4. Suppose that there exists D ∼ = Ep2 , D ≤ X, such that D ∈ Sylp (CX (D)). Let D ≤ P ∈ Sylp (X). Then the following conditions hold: (a) mp (Lp (X)Op (X)) ≥ 3; and ∗ p ∼ (b) If Lp (X) ≤ Γoo P,2 (X), then Lp (X) = F (X) = L2 (p ), and some element d ∈ D# induces a nontrivial field automorphism on Lp (X). Proof. Let L = Lp (X) and Q = Op (X) and suppose first that mp (LQ) ≤ 2. ∗

If L = 1, then F (X) = Q has p-rank at most 2, whence mp (X) ≤ 3 by [V2 , 4.9], contradiction. So L = 1, whence L∩Z(P ) = 1 and in particular D∩L∩Z(P ) = 1. If Z(L) = 1, or if p divides | Outdiag(L)|, then by [IA , 6.1.4, 2.5.12], some component K of L satisfies K/Z(K) ∼ = Ln (q), q −  ≡ n ≡ 0 (mod p), and so mp (K) ≥ 3 by [IA , 4.10.3], a contradiction. So Z(L) = 1 = Op (Outdiag(L)). Of course if P does not normalize some component of L, then mp (L) ≥ p ≥ 3, contradiction. Now if L has more than one component, or if Q = 1, then Z(P ) meets that component

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or Q, respectively, and so Z(P ) is noncyclic. But this implies that D ≤ Z(P ), so mp (P ) = mp (CK (D)) = 2, again a contradiction. We conclude that L = F ∗ (X) is simple and mp (L) ≤ 2. Since p ≥ 5, L has no graph automorphism of order p, and as Op (Outdiag(L)) = 1, a Sylow p-subgroup of Out(L) can only consist of images of field automorphisms. Thus mp (X/L) = 1 and so mp (X) ≤ 3, a contradiction proving (a). To prove (b), set Γ = Γoo P,2 (X). Note first that if P does not normalize every component of L, then as p ≥ 5, mp (CP (K)) ≥ p − 1 ≥ 4 for every component K of L, so L ≤ Γ, contrary to assumption. Hence P normalizes every component of L. If K is a component of L with Z(K) = 1, then K/Z(K) ∼ = Ln (q), q −  ≡ n ≡ 0 (mod p), and mp (K) > n − 2 ≥ p − 2 ≥ 3. Taking A ∈ E∗ (P ∩ K) and applying [IA , 7.3.6], we have K ≤ ΓA,2 (X) ≤ Γ. Since L ≤ Γ by assumption, there exists a component K of L such that Z(K) = 1. Then D∩Z(P )∩K = 1. Since D ≤ Z(P ) as mp (X) ≥ 4, there cannot exist another component of L, nor can Q be nontrivial. Thus, K = F ∗ (X) is simple. Since p ≥ 5, [IA , 7.3.6] then yields that K ∈ Chev(r) for any r = p, for otherwise K ≤ ΓA,2 (X) ≤ Γ for A ∈ E4 (P ), contradiction. If K ∼ = An , then P ≤ K since Out(K) is a 2-group. As mp (P ) ≥ 4 but mp (CP (D)) = 2, D must have a regular orbit in its action on n letters. So n ≥ p2 . Then choose A ∈ E∗ (P ), so that A is the direct product of groups generated by p-cycles, and mp (A) ≥ p ≥ 5. As in the proof of Lemma 8.41 we see that K ≤ ΓA,2 (X) ≤ Γ, a contradiction. Suppose that K ∈ Spor. Then again P ≤ K and so by [IA , Table 5.6.1], p = 5 and K ∼ = 51+6 by a = F1 . But then for Z := Z(P ), CK (Z) is an extension of S ∼ covering group of J2 . Let d ∈ D − Z, so that D = d Z. Thus |CCK (Z) (d)|5 = 52 . If d ∈ S, however, then |CS (d)| = 56 , contradiction. So d ∈ S. Since |S/Z| = 56 , |CS/Z (d)| ≥ 52 , so |CCK (Z) (d)|5 ≥ |CS (d)|| d | ≥ 5|CS/Z (d)| ≥ 53 , again a contradiction. Thus, K ∈ Chev(p). If some d ∈ D induces a non-inner automorphism on K, then as p ≥ 5, d induces a field automorphism on K. But then |CK (d)|p = p, so CK (d) ∼ = L2 (p), K ∼ = L2 (pp ), and conclusion (b) holds. Thus we may assume that D ≤ K. Let R = P ∩ K. As mp (D) < mp (P ), we again have D = Z d, where Z = Ω1 (Z(P )) ≤ R. Hence |R/Φ(R)| = |CR/Φ(R) (d)| ≤ |CR (d)| = p2 . In particular |R/Φ(R)| < pp , so K admits no field automorphism of order p. This implies that R = P , so mp (R) ≥ 4. But again the fact that |R/Φ(R)| ≤ p2 implies that K ∼ = L2 (p2 ), L± 3 (p), B2 (p), or G2 (p). In every case, mp (K) < 4 by [IA , Table 3.3.1], a final contradiction. The lemma is proved.  Lemma 9.2. Suppose that p is a prime ≥ 5, X is a K-group with Op (X) = 1, and K = F ∗ (X) is simple with K ∈ Cp . Suppose that D ≤ X with D ∼ = Ep2 , D ∈ Sylp (CX (D)), and D ≤ K. Let d ∈ D − K. Then either mp (K) = 1 with d inducing a field automorphism on K, or K ∼ = Lp (q),  = ±1, q ≡  (mod p), and d induces a diagonal automorphism on K. In the latter case mp (X) ≥ 4. Proof. Since | Out(K)| is divisible by p ≥ 5, K ∈ Chev(r) for some r, and d induces an inner-diagonal or field automorphism on K. Obviously mp (CK (d)) = 1, and as K ∈ Cp , r = p. Let q be the level of K. Suppose that d is a field

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automorphism. Since q 1/p ≡ q (mod p), it follows from [IA , 4.10.3a] that mp (K) = mp (CK (d)), so mp (K) = 1, as required. On the other hand if d induces an innerdiagonal automorphism, we see from [IA , 4.8.2, 4.8.4] that as mp (CK (d)) < p − 2, we have K ∼ = Ln (q), q ≡  (mod p), n ≡ 0 (mod p),  = ±1, and if n > p, then CK (d) contains a split extension of Ln/p (q p ) by a field automorphism of order p. This last configuration is impossible since mp (CK (d)) = 1, so n = p. It follows that | Outdiag(K)| = p, so X ∼ = Inndiag(K), and hence mp (X) = p − 1 ≥ 4. The proof is complete.  9.2. Sylow p-Subgroups P , and Z(P ) and J(P ). Lemma 9.3. If q = pn , then a Borel subgroup (or Sylow p-normalizer) of L2 (q) is a Frobenius group of order pn · (pn − 1)/2. Proof. This is immediate from the Chevalley relations [IA , 2.4].



Lemma 9.4. Let p be an odd prime, K = L2 (pp ), and X = Aut(K). Let P ∈ Sylp (X). Then the following conditions hold: (a) P ∼ = Epp ; = Zp  Zp and P ∩ K ∼ (b) IX (P ∩ K; p ) = {1}; (c) CX (P ∩ K) = P ∩ K. Proof. This is well-known. We have X = P ΓL2 (pp ) (see [IA , 2.5.12]) so X = P GL2 (pp ) f  where f is a field automorphism of order p normalizing P ∩ K. Clearly P ∩ K ∼ = L2 (p), |CP ∩K (f )| = = Epp and P ∩ K is a TI-set in X. As CK (f ) ∼ p so P = (P ∩ K) f  ∼ = Zp  Zp . A straightforward computation shows that CX (P ∩ K) = P ∩ K. Then if 1 = W ∈ IX (P ∩ K; p ), the group W (P ∩ K) contains more than one Sylow p-subgroup of K; but any two Sylow p-subgroups of K generate K, so K = W (P ∩ K). As K is not p-solvable, this is a contradiction and the lemma follows.  Lemma 9.5. Suppose that Op (X) = 1, where p is an odd prime, and mp (X) ≤ 3. Let K ∈ K be a component of E(X) with Z(K) = 1. Let P ∈ Sylp (X). Then [Ω1 (Z(P )), K] = 1. Proof. Obviously Ω1 (Z(P )) normalizes K. If K ∈ Alt ∪ Spor, then by [IA , 6.1.4, 5.6.2], K ∼ = 3A6 , 3A7 , 3M22 , or 3O  N . In every case Out(K) is a 3 -group and P is extraspecial, so the result holds in these cases. If K ∈ Chev(r) − Alt, r = p, then by [IA , 6.1.4, 4.10.3] and the fact that mp (X) ≤ 3, the only possibility is K ∼ = SL3 (q),  = ±1, q ≡  (mod 3), with p = 3. # Let z ∈ Ω1 (Z(P )) . If z induces a field automorphism on K, then one calculates that |CK (z)|3 < |K|3 , contradiction. So z maps into Inndiag(K). As GL3 (q) maps onto Inndiag(K) and P ∩ K acts absolutely irreducibly on the natural K-module, z acts on K as conjugation by a scalar matrix, i.e., trivially, as desired. Finally suppose that K ∈ Chev(p) − Alt. By [IA , 6.1.4], p = 3 and K/Z(K) ∼ = U4 (3), Ω7 (3), or G2 (3). In the first two cases by [IA , 6.4.4], m3 (K) = m3 (K/Z(K))+ m3 (Z(K)) ≥ m3 (U4 (3)) = 4, contrary to assumption. So K ∼ = 3G2 (3). By Lemma 1.5, m3 (K) ≥ 4, contradiction. The proof is complete.  Lemma 9.6. Let K = SL3 (q), q ≡  (mod 3),  = ±1. Let P ∈ Syl3 (K). Then O (CAut(K) (P )) = 1. 3

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Proof. Since P is absolutely irreducible on the natural K-module, CGL3 (q) (P ) consists of scalars, and so CInndiag(K) (P ) = 1. Hence if the lemma fails, then some field automorphism φ of K of order 3 centralizes P , by [IA , 4.9.1]. But |CK (φ)|3 = |SL3 (q 1/3 )|3 < |K|3 , a contradiction.  Lemma 9.7. Let K = [3 × 3]U4 (3) and P ∈ Syl3 (K). Then Z(P )/Z(K) = Z(P/Z(K)) ∼ = Z3 . Proof. We have K ≤ L := 3Suz with 3 not dividing |L : K|, so P ∈ Syl3 (L). As Z(P/Z(K)) ∼ = Z3 (a long root subgroup, see [IA , 3.3.1b]), it suffices to show that |Z(P )| = 33 . But by [IA , 5.3o], |Z(P/Z(L))| > 3, and for any xZ(L) ∈ I3 (Z(L)), x does not shear to Z(L); thus, x ∈ Z(P ). Hence |Z(P )/Z(L)| > 3 and the result follows.  Lemma 9.8. Suppose that K ∈ Chev(p)∩Kp , p odd, and Z(K) = 1. If mp (K)− mp (Z(K)) = 1, then p = 3 and K ∼ = 3A6 . Proof. By [IA , 6.1.4], if the conclusion fails, then K ∼ = 3U4 (3), 32 U4 (3), 3Ω7 (3), or 3G2 (3). In the first two cases m3 (K) − m3 (Z(K)) = 4 by [IA , 6.4.4]. If ∼ K∼ = 3Ω7 (3), then since Ω7 (3) ≥ Ω− 6 (3) = 2U4 (3), again m3 (K) − m3 (Z(K)) ≥ 4. Then Lemma 1.5 completes the proof.  Lemma 9.9. Suppose that K ∈ C2 and K is unambiguously in Chev(r) for some odd prime r. Let p be an odd prime different from r. Then mp (K) ≤ 1. Proof. Our assumptions imply that K ∈ Chev(2). Hence by definition of C2 , K∼ = L2 (q), q ∈ FM, or r = 3 with K/Z(K) ∼ = L± 4 (3), L3 (3), or G2 (3). In all cases the assertion is easily checked.  Lemma 9.10. Let K = Suz and T ∈ Syl3 (K). Then E32 ∼ = Z(T ) ≤ [T, T ] and T is indecomposable. Proof. With Lemma 8.21d, it remains only to prove indecomposability. Suppose that T = T1 × T2 is a nontrivial decomposition. Since E32 ∼ = Z(T ) ≤ [T, T ], both T1 and T2 are nonabelian. In particular T1 ≤ J(T ) and we choose x ∈ T1 − J(T ). By Lemma 8.21d, Z(T ) ≤ [T, x] ≤ T1 , a contradiction. The lemma is proved.  Lemma 9.11. Let K = U3 (pn ) and P ∈ Sylp (K). Then P is indecomposable. Proof. By a matrix calculation or the Chevalley relations [IA , 2.4], AutK (P ) is transitive on the non-identity elements of Z(P ) = [P, P ]. Hence the result follows  from the Krull-Schmidt theorem [IG , 3.22]. Lemma 9.12. Let K = D4 (3)a and let W ∈ Syl5 (K). Then the following conditions hold: (a) AutK (W ) is a 2-group; and (b) K has three classes of elements of order 5, and they are fused by any outer automorphism of K of order 3. Proof. By [IA , 4.10.3a], W ∼ = E52 . Let V be a natural F3 K-module. Using  [IA , 4.8.2], we see that for any w ∈ W # , O 3 (CK (w)) ∼ = A6 , and there are at most 5 three K-conjugacy classes in E1 (K). Also, for a unique K-conjugacy class C of subgroups w ∈ E51 (K), CV (w) = 0; and W = W1 × W2 for the only W1 , W2 ∈ C

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lying in W . In particular, NK (W ) has a subgroup NK (W )o of index at most 2 normalizing W1 and W2 . As AutK (Wi ) is a 2-group, i = 1, 2, (a) follows. Indeed o ∼ Ω− 4 (3) = A6 , and NK (W ) has an element inducing maps of determinant −1 on both [V, W1 ] and [V, W2 ], so NK (W )o induces Σ6 on each of these commutator spaces, whence AutK (Wi ) ∼ = Z4 . Since K has a triality automorphism t, it suffices to show that t does not normalize C. Suppose for a contradiction that t normalizes C. Then as AutK (W1 ) =  Aut(W1 ), CKt (W1 ) covers K t /K. As O 3 (CK (W1 )) ∼ = A6 , and Out(A6 ) is a 3 -group, CKt (W1 ) must contain an element t ∈ I3 (K t − K). In particular 5 divides |CK (t )|. However, by [IA , 4.9.2beg], CK (t ) embeds in G2 (3), which is a  5 -group. This contradiction completes the proof. Lemma 9.13. Suppose K = D4 (2) and b ∈ I3 (Aut(K)) with E(CK (b)) ∼ = U4 (2). Then b is a 3-central element of Inn(K). Proof. We have |CK (b)|3 ≥ 3|E(CK (b))|3 = 35 = |K|3 .



Lemma 9.14. Let p be a prime, K ∈ Kp , and U ∈ Sylp (K). In the following cases, J(U ) is elementary abelian: (a) p = 3, K ∼ = U4 (2), U5 (2), Sp6 (2), F4 (2), or Co2 ; (b) p = 5 and K ∼ = Co1 . Proof. If K ∼ = Sp6 (2), then K ≥ L2 (2)  Z3 , so U ∼ = Z3  Z3 and the result holds. As 3 does not divide |Sp6 (2) : Ω− (2)|, it holds for K ∼ = U4 (2) ∼ = Ω− 6 6 (2).  Next, U5 (2) contains U4 (2) × Z3 with 3 -index, so the result holds in that case. If K ∼ = F4 (2), then from [IA , 4.7.3A] we see that m3 (U ) = 4, Z(U ) ∼ = Z3 , and U has a maximal subgroup U0 ∼ = E32 . For any = 31+4 with Y /Z(U ) := Z(U/Z(U )) ∼ E34 ∼ = E ≤ U , it follows that E ∩ U0 = Y and then E = CU (Y ), proving the result. If K ∼ = Co2 , a similar argument works; note that if C = CK (Z(U )), Q = O3 (C), and x ∈ I3 (C) − Q, then x lies in an SL2 (3) subgroup of C whose involution inverts Q/Z(Q), and hence |CQ/Z(Q) (x)| = 32 , as needed. Finally if K ∼ = Co1 and p = 5, it is evident from [IA , 5.3] that U has a maximal  subgroup isomorphic to E53 , and Z(U ) ∼ = Z5 . These facts imply the result. Lemma 9.15. Let K = 2E6 (2)a , P ∈ Syl3 (K), and A = J(P ). Then A ∼ = E35 and NK (A)/A contains W (E6 ) ∼ = SO5 (3). Proof. With standard notation [IA , 2.10], there is a σ-setup (K, σ = φ2 γ−1 ) for K such that K = E6a , φ2 is the Frobenius field automorphism, and γ−1 is a graph automorphism inverting the maximal torus T elementwise. It follows that T := CT (σ) ∼ = E36 is a maximal torus of Inndiag(K) and N := CN (σ) is an extension of T by W (E6 ). From [IA , 4.7.3A] we deduce that |T : T ∩ K| = 3. We have CK (T ∩ K) = T , so NK (T ∩ K) = N and then NK (T ∩ K) = N ∩ K. By [III17 , 1.14], the representation of [N, N ]/T on T is the natural representation of Ω5 (3). As |N ∩ K|3 = 39 = |K|3 , it remains to show that T ∩ K = J(P ) for  P ∈ Syl3 (N ∩ K). But this holds by [III17 , 2.3]. Lemma 9.16. Suppose that K = U4 (3) and let y ∈ I2 (K). (a) Let x ∈ I3 (CK (y)). Then m3 (CK (x)) = 4;  ×  ∈ K3 be a covering group of K. Then O 3 (C  (y)) ∼ (b) Let K = Z(K) K (SL2 (3) ∗ SL2 (3)); and

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(c) Let E ∈ E34 (K) and E ≤ P ∈ Syl3 (K). Then E = J(P ) = CAut(K) (E) and  (∞) ∼ NAut(K) (E)/E = A6 . 

Proof. O 3 (CK (y)) is the central product of two long root SL2 (3) subgroups corresponding to orthogonal positive roots α1 and α 2 . As the twisted root system  is of type B2 , the group F := Xα1 , Xα2 , X 21 (α1 +α2 ) is elementary abelian of order 34 , and contains a Sylow 3-subgroup of CK (y). This proves (a). Part (b) follows  is elementary abelian, by [IA , 6.4.4]. because the preimage of F in K Moreover, NK (F ) is a parabolic subgroup of type   X 21 (α1 −α2 ) , X 21 (−α1 +α2 ) ∼ = A6 , = L2 (9) ∼ so to prove (c) it suffices to show that F = J(P ), where P ∈ Syl3 (NK (F )). By [V2 , 6.1] it therefore suffices to show that no element of P/F of order 3 acts quadratically on F . But all subgroups of P/F of order 3 are conjugate in NAut(K) (F ), and NK (F )/F contains A4 . Therefore one – and hence all – elements of P/F fail to act quadratically on F . Hence, (c) holds as well.  Lemma 9.17. If K = F ∗ (X) ∼ = U6 (2) and A ∈ E34 (X), then A acts on K like a diagonalizable subgroup of GU6 (2).   and A  be the Proof. Without loss, X = O 3 (Aut(K)) ∼ = P GU6 (2). Let X ∼   inverse images of X and A in GU6 (2). Then A = A/Z, where Z = Z(X) ∼ = Z3 .  of order 9, then by [IA , 4.8.4], CX (a) ∼ If some a ∈ A has a preimage in A =  Z3 × P ΓL2 (8), which has 3-rank 3, contrary to m3 (A) = 4. Thus A has exponent 3.  A]  = Φ(A)  ≤ Z. If Φ(A)  = 1, then the conclusion of the lemma holds, so Clearly [A,   assume that Φ(A) = Z. Then A has an normal subgroup E ∼ = 31+2 with Z(E) = Z   = EC (E). It follows that the natural X-module is the direct sum of two and A A absolutely irreducible faithful E-modules. Hence CA(E) embeds in GL2 (F3 ), so is abelian of rank at most 2 (and contains Z(E)). Therefore m3 (A) ≤ 3, contradiction. The proof is complete. 

Lemma 9.18. Let K = F4 (2) or O  N , let P ∈ Syl3 (K), and let A = J(P ). Then the following conditions hold: (a) A ∼ = W (F4 ) or 21+4 = E34 and AutK (A) ∼ − D10 , respectively; (b) Every element of I3 (P ) is K-conjugate into A; and (c) If K = O  N , then no element of NK (A) acts as a reflection on A. Proof. First let K = O  N . Then (a) and (b) are immediate from [IA , 5.3s]. If t ∈ NK (A) acts as a reflection on A, choose a ∈ CA (t). Then t ∈ CK (a) ∼ = (Σ3 Y Σ3 ) × A6 . No element of A6 normalizes and induces a reflection on a Sylow 3-subgroup of A6 , so (c) follows. Now let K = F4 (2) and let Z = Z(P ). From [IA , 4.7.3A], C := CK (Z) = (L1 ∗ L2 )P with L1 ∼ = L2 ∼ = SU3 (2), |P : P ∩ L1 L2 | = 3, P ∩ L1 L2 ∼ = 31+4 , ∼ and C/Li = P GU3 (2), i = 1, 2. A routine argument shows that K contains an extension of E34 by W (F4 ) acting faithfully. (With standard notation, let (K, σ2 ) be a standard σ-setup for K, let w ∈ N be the involution inverting T , and observe  that by Lang’s theorem, K ∼ = K1 := O 2 (CK (σw)); the desired subgroup is easily seen in K1 .)

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Since m3 (L1 L2 ) = 3 < m3 (P ), there exists y ∈ I3 (P ) − L1 L2 such that A0 := CP ∩L1 L2 (y) contains, and then is isomorphic to, E33 . Passing to P/Z we see that every element of E4 (P ) must contain A0 , and then equals CP (A0 ). Thus A ∼ = E34 . From the construction above of a 34 W (F4 ) subgroup in K1 , we see that CK (A) is a maximal torus T of K, so NK (A) = NK (T ). Then (a) follows. Finally, (b) holds since A is elementary abelian and by inspection of [IA , 4.7.3A], m3 (CK (x)) = 4 for  all x ∈ I3 (K). The proof is complete. Lemma 9.19. Let P ∈ Syl5 (F2 ). Then P is connected. Proof. By [IA , 5.3], F2 has a subgroup X = P Q with Q ∼ = Q8 ∗ D8 , R := F (X) ∼ = 51+4 = [R, Z(Q)], Z := Z(R) = Z(X) and RQ  X. Moreover, P/R is inverted in CF2 (Z), an extension of RQ by A5 . Therefore NX (Q) = QNP (Q) with NR (Q) = Z and NP (Q) ∼ = E52 . Write NP (Q) = z, x with z = Z. Set P = P/Z. By Theorem B of Hall and Higman [G1], x acts indecomposably on R. Hence P has a unique normal E52 -subgroup U , and U ≤ R with U = CR (x). Notice that NQ (P ) = Z(Q) centralizes NP (Q) and normalizes U , inverting U . The action of NQ (P ) on U x forces U x to be abelian. We can now choose generators y1 , y2 , y3 , y4 of R such that yix = yi yi+1 for i = 1, 2, 3. Then y4 ∈ U , so y4x = y4 . Now x normalizes CR (y4 ), whence CR (y4 ) = y2 , y3 , y4 , the only x-invariant maximal subgroup of R. We next argue that ∗

[y1 , y4 ] = [y3 , y2 ] = 1.

(9A)

Namely, since y4 ∈ Z(R), [y1 , y4 ] = 1. Also, [y1 , y3 ] = [y1 , y3 ]x = [y1 y2 , y3 y4 ], and (9A) follows by bilinearity of commutation on R, together with the fact that [y2 , y4 ] = 1. The crucial point is that (y1 x−1 )5 = 1.

(9B) 2

Indeed, (y1 x−1 )5 = y11+x+x (y1 x

+x3 +x4

−1 5

. Writing i instead of yi , we get

2

) = 1 2122 3(122 3232 4)2 232 4342 .

Moving y1 terms to the left and y4 terms to the right, we get (y1 x−1 )5 = [y2 , y1 ]10 [y3 , y1 ]5 [y4 , y1 ]15 23 322 3232 22 3232 233 45 = [y4 , y1 ][y3 , y2 ]2 , whence (y1 x−1 )5 = [y1 , y4 ] = 1 by (9A), proving (9B). Since y1 centralizes R, CR (y1 x−1 ) ≤ U . But [U, x] = 1 = [U, y1 ], so CR (y1 x−1 ) = Z. Thus |CP (y1 x−1 )| = 52 , so the coset y1 x−1 [P, P ] is completely fused in P . Hence by (9B), it contains no element of order 5. Furthermore, from [IA , 5.3] we see that NF2 (P ) = NNF2 (Z) (P ) = P S with |S| = 16. No involution of F2 centralizes P , so S acts faithfully on P/[P, P ] ∼ = E52 , normalizing R/[P, P ] and CP (U )/[P, P ] since R = F ∗ (CF2 (Z)) and U char P . Thus S ∼ = Z4 × Z4 fuses all the cosets of [P, P ] in P outside R and CP (U ). It follows that I5 (P ) ⊆ R ∪ CP (U ). As R is obviously connected, every E52 -subgroup of P is connected to U . The proof is complete.  9.3. Other p-Subgroups. ∼ Lemma 9.20. Let N ∈ K3 and suppose that N contains V × Q1 , where V = E32 and Q1 is a 3-group with a normal but not central subgroup E ∼ = E32 . Then N ∼ = U5 (2), U6 (2), D4 (2), D5 (2), or Sp8 (2).

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Proof. Suppose false. Using [IA , 4.10.3] we find that for P ∈ Syl3 (N ), J(P ) ∼ = E34 , in every case. Moreover, using [IA , 4.8.2a], J(P ) = CN (J(P )). If N ∼ = U6 (2), then J(P ) is diagonalizable with respect to an orthogonal basis of the natural module, and A := NN (J(P ))/J(P ) ∼ = Σ6 . In particular, every x ∈ I3 (A) lies in an A4 -subgroup of A, so x has a free summand on J(P ). Thus |CJ(P ) (x)| ≤ 32 , which implies the desired result. This inequality is inherited by U5 (2), as well. Finally, D4 (2), Sp8 (2), and D5 (2) share the same P so we may assume K = D4 (2). Then NN (J(P )) permutes transitively the four irreducible J(P )-constituents of the  natural N -module, so P ∼ = Z3 × (Z3  Z3 ) and the lemma follows. Lemma 9.21. Let K ∈ C2 with m3 (K) ≤ 3. Then [Aut(K), Aut(K)] does not contain a 31+4 subgroup. Proof. Without loss K is simple, and we regard K ≤ Aut(K). Suppose that the lemma is false. If K ∈ Spor, then | Out(K)| ≤ 2, so K contains 31+4 and in particular m3 (K) = 3. By [IA , 5.6.1], K ∼ = J3 ; but then by [IA , 5.3h], Ω1 (P ) is abelian for P ∈ Syl3 (K), a contradiction. Thus, K ∈ Spor. Clearly, K ∼  L2 (q), q odd. If K ∈ Chev(3) ∩ C2 , then in every case | Out(K)| is a = power of 2 [IA , 2.2.10, 2.5.12], [V3 , 1.1], so again m3 (K) = 3; but then K ∼ = U4 (2) [IA , 3.3.3], and |K|3 = 34 , an impossibility. Therefore K ∈ Chev(2). Let Q ≤ [Aut(K), Aut(K)] with Q ∼ = 31+4 . If Q ≤ K, then m3 (K) = 3. Using [IA , 4.10.3] we conclude that K is a classical group. If K ∼ = L3n (q), q = 2m ≡  (mod 3), then m3 (K) = 2 for n = 1 and m3 (K) > 3 for n > 1, so K does not have this form. Hence some covering group of K by a 3 -group has a natural module of dimension k, say, over a field of characteristic 2. Because of Q, k ≥ 9. But then with [IA , 4.8.2a], m3 (K) ≥ 4, contradiction. Therefore Q ≤ K. Since Q ≤ [Aut(K), Aut(K)], and K ∼  D4 (q) (as = m3 (D4 (q)) > 3), we have Q ≤ Inndiag(K) [IA , 2.5.12]. Using [IA , 4.10.3], we see that m3 (K) ≤ 3 forces K ∼ = L3 (q), but then m3 (Inndiag(K)) = 2 as well. This  contradicts m3 (Q) = 3 and completes the proof. Lemma 9.22. Let K = U5 (2), U6 (2), Sp8 (2), D4 (2), or F4 (2). Then Aut(K) contains no 31+6 subgroup. Proof. Let P ∈ Syl3 (Aut(K)) and suppose that P ≥ Q ∼ = 31+6 . Then |P | ≥ 37 , which for the given K’s, implies K = U6 (2) and P = Q ∈ Syl3 (P GU6 (2)). But P GU6 (2) acts faithfully on V ∧ V ∧ V , of dimension 20, where V is the natural GU6 (2)-module (and this action is visible in a parabolic subgroup of Inndiag(2 E6 (2))). As 31+6 requires 27 dimensions for a faithful representation in characteristic 2, the lemma follows.  Lemma 9.23. Suppose that K ∼ = U6 (2), 3D4 (2), or Sp4 (8) with p = 3, or  5 2 ∼ K = F4 (2 2 ) with p = 5. Let H = O p (Aut(K)). Then H has no subgroup R∼ = Zp2 × Epk , where k = 4, 2, 2, or 2, respectively. ∼ U6 (2), so that H ∼ Proof. Suppose that R exists. Suppose first that K = = P GU6 (2). Then a Sylow 3-subgroup of H containing R lies in DW , where W ∼ = Σ6 , and D ∼ = E35 is diagonalizable and is the quotient of the natural permutation module for W by its 1-dimensional trivial submodule. Then R ≤ D. Choose x ∈ R − D. One calculates that |R ∩ D| ≤ |CD (x)| ≤ 33 , so R covers a Sylow 3-subgroup of DW/D. But then for some y ∈ R − D, |R ∩ D| ≤ |CD (y)| = 32 , contradicting |R| = 35 .

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In the other three cases, mp (K) = 2 and |H : K| = p so there exists x ∈ R − K 1 of order p. By [IA , 4.9.1, 4.9.2], CK (x) ∼ = G2 (2) or P GU3 (2), Sp4 (2), or 2F4 (2 2 ), and in every case a Sylow p-subgroup of CK (x) has exponent p, so R has exponent p, contradiction.  Lemma 9.24. Suppose that K is a 3-component of X isomorphic to D4 (4), 3 F4 (4), E6 (2), or E6± (4). Assume A ∈ E3 (X). Then for some A0 ∈ E2 (A),  that ∗ A m3 (CK ∗ (A0 )) > 3, where K = K . Proof. If some a ∈ A does not normalize K, let K1 be a component of CK ∗ (a). Then by induction on |X|, the result holds with CX (a) and K1 in place of X and K, which yields the desired assertion. Hence we may assume that A normalizes K and X = AK. Let P ∈ Syl3 (X) with A ≤ P . We may assume that A ∈ S3 (X), for otherwise m3 (CX (A)) ≥ 4 and the conclusion is obvious. In particular Ω1 (Z(P )) ≤ A. If K ∼ = D4 (4), then let A1 be the subgroup of A inducing inner-diagonal automorphisms on K, so that |A : A1 | ≤ 3. By [IA , 4.10.3], m3 (CK (A1 )) = m3 (K) = 4, so we can take A0 ≤ A1 . In all the remaining cases apart from K ∼ = E6 (4), Out(K) is a 3 -group and K is simple, and moreover K has a subgroup K0 ∼ = F4 (2) which shares a Sylow 3-subgroup with K. Hence it suffices to establish the desired conclusion under the assumptions K ∼ = F4 (2) and A ≤ K. Then |P | = 36 and by [IA , 4.7.3A], K has a subgroup H ∼ = Z3 × Sp6 (2). Without loss we may assume that P ∩ H ∈ Syl3 (H). Then |P : P ∩ H| = 3 and we set A1 = A ∩ H, so that A1 is noncyclic. Then by [IA , 4.10.3], m3 (CH (A1 )) = m3 (H) = 4 and we can take A0 ∈ E2 (A1 ). Now consider the remaining case, K ∼ = E6 (4). If K is simple, choose z ∈  Z(P ) ∩ K of order 3 and consider C := CK (z). We have C0 := F ∗ (C) = O 2 (C) = L1 ∗ L2 ∗ L3 with Li ∼ = SL3 (4), i = 1, 2, 3, and Z(L1 ) = Z(L2 ) = Z(L3 ) [IA , 4.7.3A]. As | Out(K)|3 = 3, there is a ∈ A − z inducing an inner automorphism on K, and we take A0 = z, a. Let a1 ∈ K induce the same automorphism on K as a. If a1 cycles L1 , L2 , L3 , then E(CC0 (a1 , z)) ∼ = L3 (4), so m3 (CK (A0 )) ≥ 4. Otherwise, for each i = 1, 2, 3, a1 normalizes a subgroup Ui ∼ = E32 of Li with z ∈ Ui . Set U = U1 U2 U3 , so that a1 centralizes U/ z and U ∼ = E34 . Then a1  CU (a1 ) centralizes A0 and has rank 4, as required. Finally assume that K ∼ = E6 (4)u . Then Z(K) ≤ A so it suffices to find a ∈ A − Z(K) such that m3 (CK (a)) ≥ 4. Again | Out(K)|3 = 3, and we again take a to induce an inner automorphism on K. From [IA , 4.7.3A] we see that m3 (E(CK (a))) ≥ 4, which completes the proof.  ∼ SU6 (2) and m3 (CX (K)) ≥ 2. Let Lemma 9.25. Let X = KCX (K) with K = E ≤ P ∈ Syl3 (X) with E ∼ = E32 . Then m3 (CP (E)) ≥ 4. ∼ E35 , Proof. Set R = P ∩ K. Then R is a split extension AW where A = ∼ E32 . Moreover, Z := Z(K) ∼ A  R, and W = = Z3 , Z ≤ A, and every w ∈ W # satisfies [A, w, w] ∩ Z = 1 = [A, w, w]. (This is because AutK (A/Z) ∼ = Σ6 , in which every element of order 3 lies in an A4 subgroup, hence has a free constituent on A/Z.) It follows that CA (E) ≤ Z. Hence if E ∩A = 1, then ECA (E) ∼ = E34 and the desired conclusion holds. So we may assume that E ∩A = 1, so that E = a, b with a ∈ A. Let U  CP (K) with U ∼ = E32 and Z(K) ≤ U . Then |CAU/Z(K) (b)| ≥ 33 .

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It follows by [IG , 9.16] that |CAUb (b)| ≥ 34 . But CAUb (b) = b × CAU (b) is elementary abelian and centralizes E = a, b, so the proof is complete.  Lemma 9.26. Suppose K = U4 (3) and y ∈ I2 (Aut(K)) with E(CK (y)) ∼ = A6 . Let x ∈ I3 (CK (y)) and assume that m3 (CK (x)) ≤ 3. Then CK (x) contains a subgroup P ∼ = 31+2 such that x ∈ P . Proof. Let I = E(CK (y)) and Q ∈ Syl3 (I). It follows from [IA , 4.5.1] that AutAut(K) (I) contains P GL2 (9), and so all elements of E1 (Q) are Aut(K)conjugate. It therefore suffices to find some ξ ∈ I3 (Q) satisfying the conclusion. Our hypothesis implies that all ξ ∈ I3 (Q) satisfy m3 (CK (ξ)) ≤ 3. Expand Q to R ∈ Syl3 (K) and let H be the parabolic subgroup of type SL2 (3) containing R. Set S = O3 (R). Then S is extraspecial of order 31+4 and exponent 3. In particular S ∩ Q contains some ξ of order 3. Since m3 (CK (ξ)) ≤ 3, ξ = Z(S).  Hence CS (ξ) = ξ × P with ξ ∈ P ∼ = 31+2 , as desired. Lemma 9.27. If K = Sp6 (2) or F4 (2), then Aut(K) does not contain Z9 × Z9 × Z9 . Proof. We have |Sp6 (2)|3 < 36 = |F4 (2)|3 , and m3 (F4 (2)) = 4 by [IA , 4.10.3a]. The result follows.  Lemma 9.28. Suppose K = F ∗ (X) ∼ = SU6 (2). Then X contains no subgroup ∼ Y = Z9 × Z9 × Z9 . Proof. Suppose false. Without loss X = K x with x3 ∈ Z(K), X/Z(K) ∼ = P GU6 (2), and x acts as diag(ω, 1, 1, 1, 1, 1). Let X1 = X ∗ y where y 3 = x−3 , and set X ∗ = K xy. Then X ∗ ∼ = GU6 (2). Now the group Y ∗ := (Y y) ∩ X ∗ is abelian of order |Y | and Y ∩ K = Y ∗ ∩ K ∼ = Z9 × Z9 × Z3 . But Sylow 3-subgroups of X ∗ are direct products of two copies of Z3  Z3 , and so do not contain Z9 × Z9 × Z3 , contradiction. The lemma is proved.  Lemma 9.29. Suppose that F ∗ (X) = IQ, where I is a component of X such that I/O2 (I) ∼ = L± 4 (3), and Q  X with m2 (Q) = 1 and O2 (Q) = 1. Let x ∈ I3 (X) and suppose that CX (x) has a subnormal subgroup C0 such that C0 /O3 (C0 ) ∼ = U3 (2). Suppose also that m3 (CX (x)) ≤ 3. Then a Sylow 3-subgroup of CI (x) contains Z3 × 31+2 . Proof. As Out(I) is a 2-group, x induces an inner automorphism on I; since I has a normal E34 -subgroup, m3 (CI (x)) = 3. Since m3 (CX (x)) = 3, CX (I) is a 3 group and x ∈ I. Set X = X/CX (I) and let P ∈ Syl3 (CX (x)) with R := CP (x) ∈ Syl3 (CX (x)). It is enough to show that R contains Z3 × 31+2 . By assumption CX (x) contains x × ES, where E ∼ = E32 and S is a 2-subgroup of NX (E) such that S/CS (E) ∼ = Q8 . Set A = x, E ∼ = E33 . By the Borel-Tits Theorem there exists a parabolic subgroup H ≤ I such that if we put H 0 = O3 (H), then A ≤ H 0 and S ≤ NH (H 0 ). Note that CS (E) ≤ CS (A) ≤ CS (H 0 ) since A ∈ S3 (H 0 ). But CS (H 0 ) = 1 by the Borel-Tits theorem. Thus S ∼ = Q8 acts faithfully on E. Suppose that I∼ = L4 (3). Then a Sylow 3-normalizer in Aut(I) has D8 Sylow 2-subgroups, so H 0 ∈ Syl3 (I). Now H is not a maximal parabolic subgroup of I, since that would entail either H0 ∼ = E34 or x being 3-central in I, contradicting m3 (R) = 3 in either case. Also if

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∼ Z3 ×31+2 , so we may assume that H 0 does not have this H0 ∼ = 31+4 , then CH 0 (x) = form. Up to an automorphism of I this leaves a unique possibility for H: namely, |H 0 | = 35 , Z(S) inverts J(H 0 ) ∼ = E34 and centralizes x. But this forces x to be a root element, whence x is 3-central in I, contradicting m3 (R) = 3. We may assume then that I∼ = U4 (3). In this case if H is a maximal parabolic subgroup of I, then there are two choices only: H 0 ∼ = E34 , contradicting m3 (R) = 3, or 31+4 , in which case the desired conclusion holds. The only other possibility is that H is a Borel subgroup. Now a 3-complement in NAut(I) (H 0 ) has a maximal subgroup Z0 × Z1 ≤ NInndiag(I) (H 0 ) ∼ Z2 and Z1 = ∼ Z8 . This subgroup meets S in a group of exponent 4, with Z0 = which implies that Z(S) is conjugate to Ω1 (Z1 ). But it is visible in the L2 (9)parabolic containing H 0 that CH 0 (Ω1 (Z1 )) ≤ J(H 0 ) ∼ = E34 . Hence x ∈ J(H 0 ), a final contradiction to m3 (R) = 3. The lemma is proved.  9.4. Other. Lemma 9.30. Let X be a K-group and p an odd prime. Suppose that Op (X)  = 1, P ∈ Sylp (X), and x is a weakly closed subgroup of P of order p. Let Y = xX . Then the following conditions hold: (a) mp (Y ) ≤ 2; and (b) If x ∈ [P, P ], then mp (Y ) = 1, and either Y is a p-component or Y = x. Proof. Clearly x ∈ Z(P ). If x ∈ Op (X), then x  X so Y = x. So we may assume that [F ∗ (X), x] = 1, whence [E(X), x] = 1. As x ∈ Z(P ), there is a component K = [K, x] of E(X). Applying [IA , 7.8.2] to K x /Z(K), we conclude that x ∈ P ∩ K (whence Y = K), and either P ∩ K is cyclic or K/Z(K) ∼ = U3 (p), G2 (q) (p = 3, q ≡ 0 (mod 3)), or one of six sporadic groups satisfying P ∩K ∼ = p1+2 .  Since SL3 (q) ≤ G2 (q), where  = ±1 and q ≡  (mod 3), Z(P ∩ K) ≤ [P, P ] in all cases unless P ∩ K is cyclic. Also by [IA , 4.10.3a], mp (K) ≤ 2. The lemma is proved.  Lemma 9.31. Let K ∈ C2 and suppose that mp (Aut(K)) ≥ 4 for some prime p ≥ 5. Then K ∈ Chev(2), or p = 5 and K ∼ = F1 . Proof. This can be checked using the definition of C2 [V3 , 1.1], together with [IA , 5.6.1] for sporadic groups and [IA , 4.10.3a, 2.5.12] for groups in Chev − Chev(2).  ∼ L± (3). Suppose that L ≤ K Lemma 9.32. Suppose K ∈ C2 with K/Z(K) = 4 with L ∼ = P Sp4 (3). If x ∈ I3 (L) is 3-central in L, then m3 (CK (x)) = 4. Proof. Since K ∈ C2 , Z(K) is a 2-group so we may pass to K/Z(K) and assume that K is simple. Let x ∈ Z(R) with R ∈ Syl3 (L), and let R ≤ Q ∈ Syl3 (K). Then R ∼ = Z3  Z3 , so x ∈ [R, R] ≤ [Q, Q]. But K has a parabolic subgroup P  containing Q of the form E34 A6 . Hence [Q, Q] ≤ O3 (P ), implying the result. Lemma 9.33. Suppose that p is an odd prime, Op (X) = 1, Op (Z(X)) = 1, and E(X) possesses p isomorphic p-components K1 , . . . , Kp ∈ Kp . Let E ∈ Ep3 (X) with E ∩ Z(X) = 1 and E normalizing each Ki , 1 ≤ i ≤ p. Then mp (ΓE,2 (X)) ≥ 4.

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Proof. Let x ∈ Ip (E ∩ Z(X)). Without loss, X = K1 · · · Kp E. Let P ∈ Sylp (X) with E ≤ P . If the Ki are all simple, then Z(P ) ∩ Ki contains an element xi of order p for each i, and ΓE,2 (X) contains x1 , . . . , xp , x, of rank p + 1 ≥ 4. Likewise if the Z(Ki ) are cyclic, then each P ∩ Ki contains Ui ∼ = Ep2 , with Ω1 (Z(Ki ) ∩ P ) ≤ Ui  P . Then E normalizes Ui so mp (CE (Ui )) ≥ 2, whence Ui ≤ ΓE,2 (X). As mp (U1 · · · Up ) ≥ p + 1 ≥ 4, we are done in this case. We may therefore assume, by [IA , 6.1.4], that p = 3 and K1 ∼ = 32 U4 (3). Then by [IA , 6.4.4], each P ∩ Ki has a characteristic subgroup Vi ∼ = E36 , containing Z(Ki ). Thus V := V1 V2 V3  P with m3 (V ) ≥ 14. Choose any e ∈ E − x. Then CV (e) = CV (x) ≤ ΓE,2 (X), and m3 (CV (e)) ≥ 5, so the lemma follows.  Lemma 9.34. Let M = 2 D5 (2), P GU6 (2), U7 (2), or 2E6 (2). Suppose that w ∈ I3 (M ) with m3 (CM (w)) < m3 (M ). Then M ∼ = P GU6 (2), w ∈ E(M ), and E(CM (w)) ∼ = L2 (8). Moreover, if M ∼ = P GU6 (2) and v ∈ I3 (M ) − wM , then |CM (v)| is not divisible by 7. Proof. If M ∼ = 2 D5 (2) or U7 (2), then w is the image of an element of Ω− 10 (2) or SU7 (2) of order 3, so by [IA , 4.8.2a], m3 (CM (w)) = m3 (M ), a contradiction. The same holds if M ∼ = P GU6 (2) unless w is the image of an element of GU6 (2) of order 9, in which case w ∈ E(M ), E(CM (w)) ∼ = L2 (8), and w is determined up to conjugacy, by [IA , 4.8.4]. That case apart, CM (v) is a homomorphic image of a subgroup of a direct product of unitary groups GUi (2), i < 6, so 7 does not divide |CM (v)|. Finally, suppose that M ∼ = 2E6 (2). Then m3 (M ) = 5 or 6 according as M is the adjoint or universal version. As w ∈ M , we see from [IA , 4.7.3A] that w is of class t3 , t4 , or t1,6 , and in every case m3 (CM (w)) ≥ m3 (W ). The lemma follows.  Lemma 9.35. Let K = G2 (4) and let P ∈ Syl3 (K). Then P ∼ = 31+2 . Let  x ∈ P # . If x ∈ Z(P ), then O 2 (CK (x)) ∼ = SL3 (4), while if x ∈ Z(P ), then  O 2 (CK (x)) ∼ = A5 . Proof. By [IA , 4.7.3A], K has two classes of elements of order 3, with centralizers as claimed. In SL3 (4), Sylow 3-subgroups are isomorphic to 31+2 , and all noncentral elements of order 3 are conjugate. This implies the lemma. 

10. 2-Structure 10.1. Orthogonal Groups over F3 . We consider the 2-structure of the groups (10A)

− − + K = Ω+ n (3), 5 ≤ n ≤ 7, K = Ω6 (3), K = P Ω6 (3) and P Ω8 (3).

− ∼ ∼ Of course Ω5 (3) ∼ = U4 (2), Ω+ 6 (3) = L4 (3), and Ω6 (3) = 2U4 (3). Our emphasis will be on the case n = 6. With n and  chosen to fit the particular group under consideration in (10A), we use the following notation:

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(1) V = Vn = F3 w1 ⊥ · · · ⊥ F3 wn is a natural module for Ωn (3), where (wi , wi ) = 1 for 2 ≤ i ≤ n, and (w1 , w1 ) = −1 if n = 6 and  = 1, and (w1 , w1 ) = 1 otherwise; (2) vi ∈ O(V ) is the reflection invering wi and v i its image (10B) in P O(V );  (3) For any I ⊆ {1, . . . , n}, vI = i∈I vi ∈ O(V ), and v I is its image in P O(V ). We have Aut(K) ∼ = P GO(V ) for the groups in (10A), = P GO(V ) and Aut(K) ∼ + ∼ except for K = P Ω8 (3), for which | Aut(K) : P GO(V )| = 3. Depending on whether the group K or K is being considered, we (10C)

(1) Set z = v{2,3,4,5} ∈ K, or z = v {2,3,4,5} ∈ K; (2) Fix R ∈ Syl2 (CK (z)), or R ∈ Syl2 (CK (z)); and (3) Set R0 = O2 (O 2 (CK (z))), or R0 = O2 (O 2 (CK (z))).

For the particular case n = 6, we define (10D)

V = {v ∈ I2 (CAut(K) (R0 )) | E(CK (v)) ∼ = Ω5 (3) ∼ = P Sp4 (3)}.

Note that Aut(K) ∼ = Aut(K) in this case. So if v ∈ V, then v acts on K, centralizing R0 (see (a) or (b) of the next lemma), and E(CK (v)) ∼ = Ω5 (3). Lemma 10.1. Consider K = 2U4 (3) and K = U4 (3), so that n = 6 and  = −1 in (10B). Adopt the notation (10C) and (10D). Let t = Z(K). Then the following conditions hold: (a) R ∈ Syl2 (K), z Z(K) = Z(R), z ∈ R0 ∼ = Q8 ∗ Q8 , R0 ∩ Z(K) = 1, R0 maps isomorphically on R0 , and CAut(K) (R) = CAut(K) (R) ∼ = Z2 ; (b) CAut(K) (R0 ) = CAut(K) (R0 ) ∼ = D8 , and all noncentral involutions in CAut(K) (R0 ) are NAut(K) (R0 )-conjugate and lie in V; (c) O 2 (CK (z)) = K1 ∗ K2 with K1 ∼ = K2 ∼ = SL2 (3), and O 2 (CK (z)) maps isomorphically on O 2 (CK (z)); (d) All involutions of K are conjugate, and for all w ∈ I2 (R0 − {z}) there exists an element of K interchanging z and w by conjugation; (e) If u ∈ I2 (Inndiag(K)) with Ku := E(CK (u)) ∼ = A6 , then I2 (Ku ) ⊆ (tz)K ; (f) If v ∈ V, then v is CK (z)-conjugate to vzt in K v but v is not Kconjugate to vz; and (g) All involutions of K/Z(K) split over Z(K). Proof. Now, z and zt = v{1,6} are not conjugate under the action of GL(V ) or GO(V ), and so (10E)

CK (z) = α(CK (z)) and CAut(K) (z) = CAut(K) (z),

where α : K → K/Z(K) is the canonical mapping. Now, CK (z), CK (z), and CAut(K) (z) are the stabilizers in K, K, and Aut(K) of the decomposition V = [V, z] ⊥ CV (z). Their subgroups centralizing CV (z) are normal and are K0 , α(K0 ), and α(K0 ), all isomorphic to Ω([V, z]) ∼ = SL2 (3) ∗ SL2 (3), and all containing z or z. As GO(CV (z)) ∼ = GO2− (3) is a 2-group, (c) holds and R0 ∼ = R0 ∼ = Q8 ∗ Q8 , Note that by Witt’s lemma or [IA , 4.5.1], all involutions of K are K-conjugate. By (10E), R ∈ Syl2 (K). Also, all involutions of R0 − {z} are K 0 -conjugate. Hence to prove (d) we may assume that z, z    R. Then (d) follows from [IG , 16.21].

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For (e) and (f), we work in K and O(V ). In (e), u is the image of an involution of O(V ) one of whose eigenspaces U on V is of dimension 4 and − type, since ⊥ A6 ∼ = Ω− 4 (3). Then Ku centralizes U , and any r ∈ I2 (Ku ) must have 2-dimensional K eigenspaces on U . Thus r ∈ (tz) , as asserted. In (f), since [R0 , v] = 1, v must have a 5-dimensional eigenspace U containing [V, z], say U = [V, z] ⊥ W, with eigenvalue η = ±1 and dim W = 1. Then dim CV (v) = dim CV (vz) so v ∼K vz. On the other hand, CV (z) = W ⊥ W  with W and W  isometric, and the η eigenspace of tvz is U  := [V, z] ⊥ W  . Using Witt’s lemma, we find g ∈ K interchanging W and W  , hence g ∈ CK (z) and v g = tvz. Thus (f) holds. Part (g) is a consequence of [IA , 4.5.2]. Finally consider (a) and (b). By (10E), α(R) = R and by (c), α(R0 ) = R0 . As R projects onto both a Sylow 2-subgroup of O([V, z]) and O(CV (z)), of which it is a crown product, both CAut(K) (R) and CAut(K) (R) induce scalars on [V, z] and CV (z). It follows that CAut(K) (R) = CAut(K) (R) is the image of z. Therefore, z Z(K) = Z(R). Since [CV (z), R0 ] = 1, R0 ∩ Z(K) = 1, proving (a). For (b), let C = CAut(K) (R0 ). By (10E) and (d), and as R0 = α(R0 ), C = CAut(K) (R0 ). Then as R0 is absolutely irreducible on [V, z], C is the image mod scalars of a subgroup C ≤ GO(V ) inducing scalars, indeed acting trivially, on [V, z]. Thus, C ≤ O(V ) and C ∼ = O(CV (z)) ∼ = O2− (3) ∼ = D8 . Moreover, NGO(V ) (C) has elements outside O(V ) (acting on [V, z] and CV (z)). Hence NGO(V ) (C) is transitive on the set of 1-subspaces of CV (z). Now noncentral involutions of C are reflections on CV (z) and hence reflections on V . Thus they lie in V and are permuted transitively by NGO(V ) (C), hence by NAut(K) (R0 ). This proves (b), so the proof is complete.  Lemma 10.2. Let K = L4 (3), and use the notation of (10B), (10C), and (10D), with n = 6 and  = +. Let S ∈ Syl2 (Aut(K)) with R ≤ S. Then the following conditions hold: (a) z = Z(R) = Z(S) with R ∈ Syl2 (K); (b) CS (R) = CS (R0 ) = CAut(K) (R0 ) = z, v for some v, and V = {v, vz}; (c) Let v ∈ V. Then CS (v) = R v, and v h = vz for all h ∈ S − R v; (d) O 2 (CK (z)) = K1 ∗ K2 with K1 ∼ = K2 ∼ = SL2 (3); (e) For any v ∈ V, there exists u ∈ I2 (K) − z G such that v ∼K vu; (f) If E22 ∼ = y, z  R with y ∈ z G , then yz ∈ E(CK (y)) ∼ = A6 ; (g) If y ∈ I2 (S) and E(CK (y)) = 1, then the image of y in Out(K) either is trivial or is the image of a generator of the subgroup ΓK of Aut(K); (h) R = R0 z K ∩ R ; and (i) K1 and K2 are interchanged in R, and R/ z is an extension of R0 / z ∼ = E24 by a four-group acting freely on R0 / z.  ∼ Ω+ (3), we see from [IA , 4.5.1] that z = Z(R) Proof. Since O 3 (CK (z)) = 4 and R ∈ Syl2 (K). Moreover, thinking of K in its natural 4-dimensional representation for a moment, we see that C := CInndiag(K) (Z(R)) is an extension of O 2 (C) = K1 ∗ K2 ∼ = D8 . Here D = u1 , u2  where u1 in= SL2 (3) ∗ SL2 (3) by D ∼ terchanges K1 and K2 , u2 induces an outer diagonal automorphism on K1 and centralizes K2 , and u21 = u22 = 1. Moreover R0 = R1 ∗ R2 where Ri = O2 (Ki ), i = 1, 2, and R is an extension of R0 by the four-subgroup u1  Z(D) of D, which acts freely on R0 /Z(R0 ), proving (i). It is also clear from [IA , 4.5.1] that V ∩ Inndiag(K) = ∅.

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Now think of K again as Ω(V ) as in (10B). Notice that −1 ∈ Ω(V ) so we can identify elements of Ω(V ) with their images modulo −1. Now v1 and v6 lie in V and v6 = v1 z. As R0 is absolutely irreducible on its support in V , and O(V1 ⊥ V6 ) ∼ = O2+ (2) ∼ = E22 , we have CAut(K) (R0 ) = v1 , z. In particular V = {v1 , v1 z}. Given the structure of R ∩ Inndiag(K), it follows that v1 , z = CS (R0 )  S. Also as w1 and w6 are not isometric, v1 and v1 z are not conjugate in Inn(K) v1 , but they are conjugate by a similarity in Aut(K) − Inn(K) v1 . It follows that v1 , z ≤ CS (R). Now (a), (b), (c), and (d) follow, with v = v1 in (b). Moreover, it suffices to prove (e) for v = v6 . We take u = v5 v6 ∈ K, so that vu = v5 ∼K v. In (f), since F ∗ (CK (z)/ z) = R0 / z, y, z ≤ R0 and hence y, z centralizes w1 and w6 . As K0 permutes I2 (R0 ) − {z} transitively we may take y = v{2,3} . Then E(CK (y)) = CK (w2 , w3 ) contains v{4,5} = yz, as asserted. In (g),  we may assume by [IA , 4.5.1] that O 3 (CK (y)) ∼ = A4 × A4 . But ΓK = γ where E(CK (γ)) = P Sp4 (3). By [IA , 4.5.1], y and γ lie in the same coset of Inn(K), as claimed. Finally, it is evident from viewing K as L4 (3) and z as a diagonal matrix that there is a conjugate z  of z, also diagonal, inducing outer diagonal automorphisms on K1 and K2 . Also by viewing K as Ω(V ) we see that v{1,2} ∈ z K interchanges K1 and K2 . The images of z  and v{1,2} generate R/R0 , so (h) holds. The proof is complete.  Lemma 10.3. Let F ∗ (X) = K = L4 (3), R ∈ Syl2 (K), and R ≤ T ∈ Syl2 (X). Let v ∈ I2 (T ) with E(CK (v)) ∼ = P Sp4 (3) and CT (v) ∈ Syl2 (CX (v)). Then the following conditions hold: (a) E∗ (R) = {E, E0 , E1 , E2 } with m2 (R) = 4; (b) E and E0 are normal in R and conjugate in T , while E1 and E2 are R-conjugate; (c) If Z(R) = z, then |z G ∩E| = 5 and |z G ∩E1 | = 9, and any four elements of z G ∩ E generate E; and (d) E = J(R ∩ E(CK (v))), and NK (E) centralizes v with NK (E)/E ∼ = Σ5 , with all involutions of [NK (E), NK (E)] acting freely on E; (e) For some y ∈ E # , E(CK (y)) = 1; and (f) m2 (X) = 4 or 5, and m2 (Aut(K)) = 5. Proof. We continue the notation of Lemma 10.2 and its proof. Let A ∈ E24 (K). Then −1 A ∈ E25 (SO(V )). Hence the irreducible A-submodules of V form the unique orthogonal frame FA stabilized by A; conversely, every orthogonal frame in V is stabilized by a unique element A of E24 (K), and A = CK (A). Moreover, as Aut(K) ∼ = P GO(V ), CAut(K) (A) must stabilize FA and so CAut(K) (A) ∼ = E25 . As m2 (Out(K)) = 2, it follows that Aut(K) does not contain a copy of E26 , and so m2 (Aut(K)) = 5. This implies (f). By [IA , 4.5.1], the Aut(K)-conjugacy class of v is uniquely determined by the structure of E(CK (v)). Hence by Lemma 10.2b, and as CT (v) ∈ Syl2 (CX (v)), we have [R, v] = 1. Without loss we may assume that v = v1 . Let  E = v{i,i+1} | 2 ≤ i ≤ 5 be the unique element of E24 (K) stabilizing the frame formed by w1 , . . . , w6 . For any i = 2, 3, 4, 5 let xi be the involution in O(V ) interchanging wi and wi+1 , but fixing all other wj . Let yi = v1 xi ∈ K. Then H := y2 , y3 , y4 , y5  ∼ = Σ5 and

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NK (E) = EH has odd index in K, so we may assume that E  R. Since the majority of 1-dimensional E-submodules of V are of + type, while T − R consists of non-isometry similarities, and as |T : R| = 2, E T = {E, E0 } for some E0 , proving (b) for E. Because z = v2 v3 v4 v5 , (c) holds for E. Also [NK (E), NK (E)] = E[H, H] and a four-subgroup of [H, H] acts freely on E, which implies (d). In (e) we can take y = v{1,2} . Let Rv = R ∩ E(CK (v)). Then Rv /E is a four-subgroup of R/E ∼ = D8 , and J(Rv ) = E. Hence J(R)/E ≤ R1 /E, where R1 is the other four-subgroup of R/E. We may assume that R1 = E y2 , y4 . Note that [E, y2 ] ∼ = [E, y4 ] ∼ = Z2 and [E, y2 y4 ] = [E, y2 ] × [E, y4 ]. One then checks that E24 (R) consists of E, CE (y2 , y4 ) y2 , y4 , and Ei := CE (yi ) yi  for i = 2 and 4, the last two of which are obviously R-conjugate. Hence (a) and (b) hold. It remains to check (c) for E1 . But from the first paragraph, E1 must correspond to an orthogonal frame in V consisting of three subspaces F3 w with (w, w) = 1 and three with (w, w) = −1. Then (c) follows easily. The lemma is proved.  Lemma 10.4. Let K = L4 (3) and adopt the notation of Lemma 10.3. Then CAut(K) (E) = E v. Moreover, for any h ∈ T such that E h = E0 , v h = vz. Proof. Continuing from Lemmas 10.2 and 10.3, we know that [R, v] = 1. Hence NT (E) = R v = CT (v). As v T = {v, vz} by Lemma 10.2c, the second assertion holds. Also, as CK (E) = E and R v ∈ Syl2 (NAut(K) (E)), CAut(K) (E) = CR (E) v = E v, as claimed.  Lemma 10.5. Suppose that K = F ∗ (X) = L4 (3) and T ∈ Syl2 (X). Then any nontrivial normal cyclic subgroup of T has order 2. Proof. We regard X ≤ Aut(K) = P GL4 (3) γ, where γ is a graph automorphism. Suppose that the lemma fails, and Y  T with Y ∼ = Z4 . Let z ∈ I2 (Z(T ∩K)) and R0 = O2 (CK (z)) ∼ = Q8 ∗ Q8 . Then [R0 , Y ] < Y so [R0 , Y ] ≤ Z(R0 ) = z. Hence Y ≤ R0 CT (R0 ). But CT (R0 ) = z or z, v as in Lemma 10.2b. Notice by Lemma 10.2i that no Z4 -subgroup of R0 is normal in R. Hence we may assume that CT (R0 ) = z, v. Therefore Y ≤ R0 ×v and as z, v = CT (R0 )  T , the projection Y0 of Y on R0 is isomorphic to Z4 and normal in T , hence in R, contradiction. The lemma is proved.  Lemma 10.6. Let F ∗ (X) = K = L4 (3) and t ∈ I2 (X) with E(CK (t)) ∼ = U4 (2). Then CK (t) ∼ = Aut(U4 (2)). Proof. Using Lemma 10.2b and the notation (10B), we may assume that X = Aut(K) and t = v6 . Then E(CK (t)) acts naturally on the span of w1 , . . . , w5 ,  and v{5,6} ∈ CK (t) is the needed outer automorphism of E(CK (t)). Lemma 10.7. Suppose that K = F ∗ (X) ∼ = U4 (3). Let z ∈ I2 (X) and suppose that O2 (CX (z)) contains Q8 ∗ Q8 ∗ D8 . Then X contains a copy of P O6− (3). Proof. Since Out(K) ∼ = D8 [IA , 2.5.12], z ∈ K and F ∗ (CK (z)) ∼ = Q8 ∗ Q8 . By [IA , 4.5.1], O2 (CInndiag(K) (z)) ∼ = Q8 ∗ Q8 ∗ Z4 , so as | Aut(K) : Inndiag(K)| = 2, O2 (CX (z)) ∼ = Q8 ∗ Q8 ∗ D8 and X covers Ω1 (Outdiag(K)). Moreover, there is  an involution t ∈ CX (O 3 (CK (z))) inducing a graph automorphism on K. From  [IA , 4.5.1], the only possibility is that CK (t), which contains O 3 (CK (z)) ∼ = SL2 (3)∗ − SL2 (3), is isomorphic to Ω5 (3). Hence t is a reflection in P O6 (3), and the lemma follows. 

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Lemma 10.8. Suppose that K ∼ = 2U4 (3), K  X, and t ∈ I2 (X) with L := E(CK (t)) ∼ = A6 and t inducing an inner-diagonal automorphism on K. Let y ∈ I2 (L) and z = Z(K). Then t ∼K tyz. Proof. We may identify K with Ω(V ) where V is a 6-dimensional orthogonal space of − type over F3 , and t with an involution of O(V ) such that [V, t] is 2dimensional of + type. Then L is identified with Ω(CV (t)), and [V, y] = [CV (t), y] is 2-dimensional of − type. Consequently [V, tyz] = CV (t, y) is isometric to [V, t] and the lemma follows.  Lemma 10.9. Let K = F ∗ (X) ∼ = 2U4 (3) and let v ∈ I2 (X) with L := E(CK (v)) ∼ = P Sp4 (3). Let T ∈ Syl2 (X) with CT ∩K (v) ∈ Syl2 (CK (v)). Let F = J(T ∩ L) and E = CX (F ). Then the following conditions hold: (a) F ∼ = E24 and E ∼ = E26 ; (b) E = F Z(K) v; (c) NKv (E) = EH with H ∼ = A6 , and E is a standard permutation module for H; moreover, NAut(K) (E) is a split extension of the image of E in Aut(K) by a copy of Σ6 ; (d) If v1 ∈ I2 (X) and CK (v1 ) contains an extension of E24 by A5 , then E(CK (v1 )) ∼ = P Sp4 (3); (e) AutL (F ) ∼ = A5 ; and (f) For any y ∈ F # , F interchanges the two SL2 (3) solvable components of O 2 (CK (y)). Proof. We use the notation (10B), and may assume that v = v1 . Let FF = {F3 wi | 2 ≤ i ≤ 6}, a frame in the space w1⊥ . Note that L = CK (w1 ) ∼ = Ω(w1⊥ ). ∼ The stabilizer M in L of FF has the form M = F H0 , where F = E24 normalizes all the lines F3 wi (i > 1) and H0 ∼ = A5 permutes them. In particular M contains a Sylow 2-subgroup RL of L, and a Sylow 2-subgroup of H0 acts freely on F , so F = J(RL ). As FF is the unique frame stabilized by F , CAut(K) (F ) ∼ = E25 is generated by the images of F and v. This implies (a), (b), and (e). If y is as in (f), then one of the two eigenspaces of y on V is W , generated by four of the wi ’s, and O 2 (CK (y)) induces Ω(W ) on W and centralizes W ⊥ . Since m2 (F Z(K)) = 5, some element of F Z(K) has determinant −1 on W , and hence so does some f ∈ F . Hence f interchanges the solvable components as claimed in (f). Similarly N := NKE (E) preserves the frame FE := {F3 wi | 1 ≤ i ≤ 6} and N = EH with H ∼ = A6 permuting FE . Note that |EH|2 = |K v |2 . Therefore E is a standard permutation module for H, so (c) holds. Moreover, NAut(K) (E) is the stabilizer in P GO(V ) of FE , so it is the split extension of the image of E by a copy of Σ6 . Finally, (d) is immediate from the choices for CK (v1 ) as given in [IA , 4.5.1].  Lemma 10.10. Suppose that K = F ∗ (X) ∼ = U4 (3). Let T ∈ Syl2 (X) and v ∈ I2 (T ) with L := E(CK(v)) ∼ = P Sp4 (3) and CT (v) ∈ Syl2 (CX (v)). Let F = J(T ∩ E(CK (v))) and E = F , v . Then the following conditions hold: ∼ E24 and E = ∼ E25 ; (a) F = (b) E = CAut(K) (F ); (c) AutK (F ) ∼ = AutK (E) ∼ = A6 , AutAut(K) (E) ∼ = Σ6 , and AutL (F ) ∼ = A5 . As A6 -modules, F and E are, respectively, the core of a standard permutation

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module, and the quotient of a standard permutation module by its unique 1-dimensional submodule; (d) The preimage of F in 2U4 (3) is isomorphic to the trace-0 submodule of a standard permutation module for A6 ; and # (e) For any y ∈ F , F interchanges the two Q8 -subgroups of O2 (CK (y)). Proof. Let K = 2U4 (3), so that Aut(K) is identified with Aut(K) by the natural correspondence. Recall that all involutions of K split over Z := Z(K). Hence the full preimage A of an elementary abelian 2-subgroup A ≤ K is elementary abelian. Moreover, by repeated use of (10E), CAut(K) (A) = CAut(K) (A). Another consequence is that the full preimage L of L in K is isomorphic to Z × L. Let E and F be the full preimages of E and F in K and E(L), respectively. Then K, F , and E are set up for us to apply Lemma 10.9. Our (a), (b), and (e) follow from Lemma 10.9abf. By (b) and the fact that F = E ∩K, NX (F ) = NX (E). Then the first sentence of (c) follows from Lemma 10.9ce. By Lemma 10.9c, E is a standard permutation module for AutK (E), and the rest of (c) and (d) follow directly.  In the next several lemmas we will focus on U4 (3), and for simplicity we change notation, assuming the following. ∼ U4 (3); (1) K = F ∗ (X) = (2) T ∈ Syl2 (X) and R = T ∩ K; (3) F ∈ E∗ (R) and E = CX (F ); (10F) (4) Z(R) = z and R0 = O2 (CK (z)) ∼ = Q8 ∗ Q8 ;  = Ω− (3) and for each x ∈ K, x  is a preimage of x in (5) K 6  K. Remark 10.11. Throughout this book we have used the term “frame in an ndimensional vector space V ” to mean a set {V1 , . . . , Vn } of 1-dimensional subspaces of V spanning V . The stabilizer of such a frame in GL(V ) is thus a full monomial subgroup of GL(V ). In the next lemma we speak of an “ordered frame,” which means an ordered n-tuple (V1 , . . . , Vn ) of such subspaces; its stabilizer in GL(V ) is a maximal diagonalizable subgroup of GL(V ). Lemma 10.12. Assume (10F). Then the following conditions hold:  ∼ (a) F ∼ = E24 and the preimage of F in K = Ω− 6 (3) preserves an ordered orthogonal frame F in the natural module V , consisting of isometric lines; (b) AutK (F ) ∼ = A6 and AutAut(K) (F ) ∼ = Σ6 , permuting F faithfully; (c) E∗ (R) = {F, F1 }, with F and F1 conjugate in Aut(K) ∼ = P GO(V ) but not conjugate in P O(V ); moreover, F and F1 act quadratically on each other; (d) CAut(K) (F ) ∼ = E25 is the stabilizer in Aut(K) of F; also, E = F or E = CAut(K) (F ); and (e) m2 (K) = 4, and m2 (X) = 4 or 5.  by Lemma 10.1g, the full Proof. As all involutions in K split over Z(K) ∼    of an  preimage F of F lies in E∗ (F ). Therefore F = E25 is the stabilizer in K  ordered orthogonal frame F in the natural K-module V . In particular, (a) holds. Again, repeated use of (10E) shows that CAut(K) (F ) centralizes F and so stabilizes F. It follows that CAut(K) (F ) ∼ = E25 , whence E = CAut(K) (F ) or F , proving

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(d). In any case every ordered frame in V stabilized by F or E is a permuted copy of F. As a result, AutAut(K) (F ) ∼ = Σ6 and AutK (F ) contains A6 . Since |K|2 = 27 , AutK (F ) ∼ = A6 , and (b) holds. Next, since m2 (Out(K)) = 2, if A ≤ X with A ∼ = E26 , then the above applies to F = A ∩ K and yields A ≤ E, a contradiction as |E| < |A|. Hence m2 (X) ≤ 5, and (e) holds. Fix any g ∈ Aut(K)−P O(V ). Then Fg is an ordered orthogonal frame consisting of lines not isometric to the lines constituting F. Hence the stabilizer F1 of Fg in K is isomorphic to F but not K-conjugate to F . Without loss we may assume that g was chosen so that F, F1  ≤ S for some S ∈ Syl2 (NK (F )) ⊆ Syl2 (K). As NK (F )/F ∼ = A6 has one class of involutions, and contains elements of order 3 fixed-point-free on F and inverted by involutions, every involution of S/F acts freely on F . As F1 = F , |F1 ∩ F | ≤ |CF (F1 )| = 4, so F1 F/F ∼ = E22 . Let U = CF (Z(S/F )) ∼ = E22 ; then F1 ∩ F = U . But by Lemma 10.1, |Z(S)| = 2. Hence one of the four-subgroups of S/F acts freely on F and the other is F1 F/F = CS (U )/F . Now for any y ∈ F1 − U and x ∈ F − U , CF (y) = U so xy has order 4. It follows that I2 (F1 F ) = F1# ∪ F # and so E∗ (S) = {F, F1 }, with both F and F1 weakly closed with respect to K, indeed with respect to P O(V ). In particular F1  S, so  [F, F1 , F1 ] = 1. Thus (c) holds and the proof is complete. Lemma 10.13. Assume (10F). Let t ∈ I2 (X) with J := E(CK (t)) ∼ = A6 . Then Sylow 2-subgroups of CX (J) are elementary abelian. Proof. Without loss, X = Aut(K). Note that Aut(K) = Aut0 (K) [IA , 2.5.12]. By [IA , 4.5.1], if t ∈ Inndiag(K), then CInndiag(K) (J t) = 1 so t ∈ Syl2 (CX (J)). Thus by [IA , 4.5.1], we may assume that t ∈ Inndiag(K), and CInndiag(K) (J t) = t, so t ∈ Syl2 (CInndiag(K) (J)). Hence |CX (J)|2 ≤ 4 and it suffices to find an involution in CX (J) outside Inndiag(K). Identify K with P Ω(V ) where V is an orthogonal space of − type and dimension 6 over F3 . Let t1 and t2 be images of reflections in O(V ) whose centers are orthogonal to each other but not isometric. Then t1 t2 is conjugate to t, and the corresponding conjugate of  t1 fills our need. This completes the proof. Lemma 10.14. Assume (10F). Let u and z be commuting involutions of X such that O 2 (CK (u)) ∼ = A4 × A4 , z ∈ K, and u interchanges the two SL2 (3) solvable components of CK (z). Then E(CK (uz)) ∼ = P Sp4 (3). ∼ Ω(V ), where V is a 6-dimensional orthogonal F3  = 2U4 (3) = Proof. Let K space which is the orthogonal sum V1 ⊥ · · · ⊥ V6 of isometric lines. Without loss, z  such that CV ( is the image of an involution z ∈ K z ) = V5 ⊥ V6 . In particular, z is not conjugate to − z , so u is the image of an involution u  ∈ O(V ) commuting with z. u, z-invariant. The fact We may therefore assume that the Vi ’s were chosen to be  that u interchanges the two SL2 (3) central factors of O 2 (CK (z)) ∼ = Ω((V5 + V6 )⊥ ) u) = −1. Hence u  inverts either 1 or 3 of V1 , . . . , V4 . On the means that det[V,z ] ( other hand, by the structure of CK (u), dim([V, u ]) = 3. It follows directly that the involution u z has an eigenspace on V of dimension 5, whence the lemma.  Lemma 10.15. Assume (10F). Suppose that AutX (F ) ∼ = Σ6 . Let N = NX (E)∩ CX (E/F ). Then there is x ∈ I2 (X) such that the following hold: (a) x ∈ CN (z) induces transvections on both E and F ; (b) [E, x] = z;

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(c) 3 divides CK∩N (x); and (d) E(CK (x)) = 1. Proof. By Lemma 10.10b, E = CX (F ). As |E/F | = 2 and F = K ∩ E, N = NX (E) = NX (F ). Let Aut(K) = Aut(K)/ Inn(K) = Out(K) ∼ = D8 . We have Outdiag(K) ∼ = Z4 . Now E = F v where v induces a graph automorphism on K, so E is a noncentral subgroup of Out(K) of order 2. Therefore N = E y where y ∈ Outdiag(K) and y 2 = 1. Since [F, v] = 1, AutKE (F ) ∼ = AutK (F ) ∼ = A6 by Lemma 10.10c. Hence by hypothesis, N > E so y = 1. We may assume that X = N . Thus we may identify X with P O(V ), where the orthogonal F3 -space V is the orthogonal sum of six isometric lines F3 wi , 1 ≤ i ≤ 6, and E is the joint stabilizer in X of these six lines. Let vi , 1 ≤ i ≤ 6, be the image in X of the reflection on V inverting wi . Then we may assume that E = vi | 1 ≤ i ≤ 6, F = vi vj | 1 ≤ i < j ≤ 6, v = v1 , and z = v2 v3 v4 v5 . Let x ∈ X be the image in X of the involution ξ ∈ GO(V ) interchanging w2 and w3 and centralizing the remaining wi ’s. Then CV (ξ) = w1 , w2 + w3 , w4 , w5 , w6  so E(CK (x)) ∼ = Ω5 (3) = 1 and E(CK (x)) contains an element of order 3 cycling w4 , w5 , and w6 , proving (c) and (d). Finally, [E, x] =  [F, x] = v2 v3 , yielding (a) and (b). Lemma 10.16. Assume (10F) with X = K. Then AutR0 (F ) is a four-group acting freely on F . Proof. Since R/F embeds in A6 , R/F ∼ = D8 . Thus m2 (R/F ) = 2 and so |R0 ∩ F | ≥ 23 . Indeed equality must hold as R0 ∼ and F is abelian. By = 21+4 + Lemma 10.12c, F1 F/F acts quadratically on F , so as |Z(R)| = 2, the other foursubgroup of R/F acts freely on F . Hence we are done unless R0 F = F1 F . But Z(R0 F ) = z and Z(F1 F ) = F1 ∩ F ∼ = E22 , so R0 F = F1 F . The lemma is proved.  Lemma 10.17. Assume (10F) with X = Aut(K). Identify P O6− (3) with a subgroup J ≤ X, so that J/K is a four-group. Let J0 = J ∩ Inndiag(K), and let J1 and J2 be such that Ji /K, i = 0, 1, 2, are the subgroups of J/K of order 2. Then there exist F1 and F2 such that the following conditions hold: (a) E∗ (R) = {F1 , F2 } and NX (Fi ) maps onto J/K; (b) For i = 1, 2, Fi centralizes an involution vi ∈ Ji − K, but no involution of J3−i − K. Moreover, vi may be chosen so that E(CK (vi )) ∼ = P Sp4 (3); (c) If s ∈ I2 (X) and O 2 (CX (s)) ∼ = A4 × A4 , then s ∈ Ji for some i ∈ {1, 2}, and some K-conjugate of s centralizes Fi vi . Proof. Part (a) is immediate from Lemma 10.12ac. It follows from Lemma 10.12d that CAut(K) (Fi ) is generated by F and a reflection vi . As det vi = −1, vi induces a graph automorphism on K and so maps to a noncentral involution in Out(K) ∼ = D8 . We may assume that v1 and v2 are conjugate by an element of Aut(K) − P O6− (3), so their images in Out(K) are distinct conjugates. This implies (b). Finally, an involutory preimage of s in O6− (3) has 3-dimensional eigenspaces on the natural module, so it stabilizes some ordered orthogonal frame F consisting of  with 2-dimensional eigenspace isometric lines. Therefore for some involution  h∈K for −1, [s, h] = 1 and sh is a reflection, K-conjugate to v1 or v2 . This implies (c). 

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Lemma 10.18. m2 (Aut(L± 4 (3))) = 5. Proof. For U4 (3), see Lemma 10.12de. For L4 (3), see Lemma 10.3f.



Lemma 10.19. Suppose K ∼ = P Sp4 (3), L± 4 (3), or 2U4 (3). Let z ∈ I2 (K) be 2-central. Then CAut(K) (z) covers Out(K). Proof. By [IA , 4.2.2j, 4.5.1], K/Z(K) has a unique class of involutions y such that CK/Z(K) (y) ∼ = CK/Z(K) (z). This implies the result if K is simple. For K = 2U4 (3), it implies the resuult with the help of (10E). The proof is complete.  Lemma 10.20. Let K = Ω− 6 (3) or Ω7 (3). Let V be the natural F3 K-module and let z ∈ I2 (K) have four eigenvalues −1 on V . Then z is 2-central in K, and z lies in the commutator subgroup of F ∗ (CK (z)) = O2 (CK (z)). Proof. For n = 6 this is contained in Lemma 10.1a, so assume n = 7 and use the notation (10B) with n = 7. Write V = V+ ⊥ V− as the sum of the eigenspaces of z, with z supported on V− . Then CK (z) is the joint kernel of the determinant and spinorial norm in O(V+ ) × O(V− ) ∼ = O3 (3) × O4+ (3). Thus + 1 9 |CK (z)|2 = 4 |O3 (3)|2 |O4 (3)|2 = 2 = |K|2 . The final assertion holds in K because it holds in Ω(V− ).  Lemma 10.21. Suppose X = Ω7 (3), t ∈ I2 (X), and K := E(CX (t)) ∼ = 2U4 (3). Let R ∈ Syl2 (K). Then there exist A, A∗  R such that the following conditions hold: (a) E25 (R) = {A, A∗ }; (b) There is D ∈ E26 (X) such that A ≤ D and F ∗ (AutX (D)) ∼ = A7 ; and (c) CX (A∗ ) = A∗ . Proof. Let V be a natural F3 X-module. Then V = [V, t] ⊥ V0 with [V, t] a natural K-module. It follows by Lemma 10.12ac that E2∗ (R/Z(K)) = E24 (R/Z(K)) = {A/Z(K), A∗ /Z(K)} where A, A∗  R are stabilizers of non-isometric ordered orthogonal frames F, F∗ , in [V, t], each frame consisting of six isometric lines. Then (a) follows. Say F = (V1 , . . . , V6 ) and F∗ = (V1∗ , . . . , V6∗ ). Since V1 and V1∗ represent the isometry classes of lines, we may assume that V0 and V1 are isometric. Let D be the stabilizer in X of (V0 , V1 , . . . , V6 ); then (b) is clear. On the other hand, A∗ is the full stabilizer of (V0 , V1∗ , . . . , V6∗ ) in X = Ω(V ); involutions in this stabilizer must have determinant 1 and then must centralize V0 as their spinorial norm is trivial. Hence, (c) holds and the proof is complete.  Lemma 10.22. Let X = SO7 (3) or P SO8+ (3), and let t ∈ I2 ([X, X]) with E(CX (t)) ∼ = Ω− 6 (3). Then there exists u ∈ I2 (X) such that [t, u] = 1, H := L E(CX (u)) ∼ = 4 (3), and E(CH (t)) ∼ = Ω5 (3). Proof. Since SO7 (3) embeds in P SO8+ (3), centralizing an anisotropic vector, it suffices to do the case X = SO7 (3). Let V be the natural F3 X-module, which we may take to have an orthonormal basis {v0 , . . . , v6 }. We may take t to be supported on v0⊥ . There exists a 6-dimensional subspace W of V containing v0 and of + type. Then W = F3 v0 ⊥ W1 , where W1 = W ∩ v0⊥ . Let u ∈ I2 (X) be supported on ∼ ∼ W . Then H = Ω(W ) ∼ = Ω+ 6 (3) = L4 (3) and E(CH (t)) = Ω(W1 ) = Ω5 (3), as claimed. 

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Lemma 10.23. Let K be the adjoint version of D4 (3). Then a Sylow 2-subgroup S of K has cyclic center and has a normal four-subgroup V whose involutions are cycled by a 3-element of NK (V ). Proof. There is a copy L of the adjoint version of D4 (2) embedded in K [IA , Table 6.2.2], and |K : L| is odd. So it suffices to prove the result for L instead of K. Use standard notation for L [IA , 2.10]. Let α∗ be the highest root and α the unique fundamental root such that α∗ − α is a root. Then we take S = Xβ | β > 0 and V = Xα∗ , Xα∗ −α , and have Z(S) = Xα∗ [IA , 3.2.2],  V = Z2 (S), and Xα , X−α  ∼ = Aut(V ) acting faithfully on V . Lemma 10.24. Let K = D4 (3)a , and let z ∈ I2 (K) be 2-central. Then 2 ∼ O3 (CK (z)) = O2 (CK (z)) ∼ = 21+8 + , and O (CK (z)) = L1 L2 L3 L4 , where z ∈ Li = SL2 (3) for all i = 1, 2, 3, 4, and [Li , Lj ] = 1 for all 1 ≤ i < j ≤ 4. Moreover, ∼ for all i = j and all tij ∈ I2 (Li Lj ) − {z}, E(CK (tij )) ∼ = Ω− 6 (3) = 2U4 (3) and 10 |CK (tij )|2 = 2 . Proof. This follows from [IA , 4.5.1]; z is in the conjugacy class of t2 in that ∼ table, and every inner involution t not conjugate to t2 satisfies E(CK (t)) ∼ = Ω− 6 (3) = − O (3). So it suffices to show that t is not conjugate to z. 2U4 (2) and CK (t) ∼ = 6 ij Now the Li are root SL2 (3)-subgroups corresponding to the four outer nodes of the extended D4 Dynkin diagram. Hence the preimage of Li Lj in the universal version  of K is isomorphic to Li × Lj × Z2 . Thus z pulls back to an involution of K,  K  while tij pulls back to an element of order 4. This implies the result. Lemma 10.25. Let K = D4 (3)a ∼ = P Ω+ 8 (3). Let z ∈ I2 (K) and let B0 ∈ E3∗ (CK (z)) act faithfully on a 2-subgroup R ≤ K of symplectic type with z ∈ Z(R). Then m3 (B0 ) = 4.  Proof. Since z ∈ K, B ≤ L := O 3 (CK (z)) ∼ = SL2 (3) ∗ SL2 (3) ∗ SL2 (3) ∗ SL2 (3) or Ω− (3), with z ∈ L in either case. In the first case the lemma holds; so 6 (3). Then z ∈ [R, B0 ] ≤ L and [R, B0 ]/ z assume for a contradiction that L ∼ = Ω− 6 is elementary abelian. But every involution in L splits over z by Lemma 10.1g, so R is elementary abelian, contradiction. The proof is complete. 

Lemma 10.26. Let K, z, and tij be as in Lemma 10.24. Suppose that K = F ∗ (X) and |X : K| is even. Then there exist 1 ≤ i < j ≤ 4 such that |CX (tij )|2 > 210 . Proof. We use the notation of Lemma 10.24. Without loss, |X : K| = 2. For any 1 ≤ i < j ≤ 4, Li Lj has 18 noncentral involutions, permuted transitively by Li Lj . It follows that if we set C = CX (z), Nij = NC (Li Lj ) and C = C/L1 L2 L3 L4 , then |CX (tij )|2 = 28 |N ij |2 . As |X : K| = 2, |C| = 24 , and C permutes {L1 , L2 , L3 , L4 }. Whatever the image of the permutation action, there is an unordered pair {Li , Lj } stabilized by a subgroup of some Sylow 2-subgroup of C of index 2, and hence |N ij |2 ≥ 23 , yielding the desired inequality.  Lemma 10.27. Let X = Aut(U4 (2)) and let H ≤ X with H ∼ = SL2 (3). Then the involution of H is 2-central in X.  Proof. Identify K := O 2 (X) with U4 (2) ∼ = P Sp4 (3). We have H = O 3 (H) ≤ 

O 3 (CK (t)), where t = Z(H). By [IA , 4.5.1], if t is not 2-central in K, then  O 3 (CK (t)) ∼  = A4 . The lemma follows.

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Lemma 10.28. Let K = Ω± 4 (3) have natural module V . Write V = V0 ⊥ V1 ⊥ V2 where V0 is 2-dimensional of − type, and dim V1 = dim V2 = 1. For i = 1, 2, let Hi = CK (Vi ) ∼ = A4 . Then O2 (H1 ) = O2 (H2 ). Proof. This is immediate as [V, O2 (Hi )] = V0 ⊥ V3−i , i = 1, 2.



10.2. Sylow 2-Subgroups and Their Overgroups in Quasisimple KGroups. Lemma 10.29. Let K ∈ K. Then K has dihedral Sylow 2-subgroups if and only if K ∼ = L2 (q) for some odd q ≥ 5 or K ∼ = A7 . Proof. See [IA , 5.6.3, 4.10.5bc, 5.3a, Table 2.4(IV)].



Lemma 10.30. Let K ∈ C2 with Z(K) = 1. Then Sylow 2-subgroups of X are neither quaternion nor semidihedral. Proof. This is immediate from Lemma 2.8.



Lemma 10.31. Suppose that K ∈ K is simple, S ∈ Syl2 (K), and Ω1 (S) is n abelian. Then K ∼ = L2 (q), q ≡ ±3 (mod 8), 2 G2 (3 2 ), n ≥ 3, or J1 , or K has a strongly embedded subgroup. Proof. If K ∼ = L2 (5). If K ∈ = An , then n = 5 as A6 contains D8 , so K ∼ Spor, it is evident from the tables [IA , 5.3] that unless K ∼ = J1 , K contains a n nonabelian dihedral 2-group. If K ∈ Chev(2) and K ∼  L2 (2n ), 2B2 (2 2 ), or U3 (2n ) = for some n > 1, then using the commutator formula we can find two noncommuting root involutions corresponding to positive roots, so Ω1 (S) is nonabelian. Finally n if K ∈ Chev(r), r odd, and K ∼  L2 (r n ) or (for r = 3) 2 G2 (3 2 ), then K contains = ± (S)L3 (q), Sp4 (q), P Sp4 (q), or G2 (q) for some odd q. Accordingly, K contains GL± 2 (q); SL2 (q)  Z2 ; SL2 (q) ∗ SL2 (q); and SL3 (q). Thus K contains D8 . So does L2 (q), q ≡ ±1 (mod 8), completing the proof.  Lemma 10.32. Suppose that K ∈ C2 with m2 (K) ≤ 3, and K has sectional 2-rank at most 4. Then Z(K) = 1. Proof. Suppose that Z(K) = 1 and K ∈ C2 . If K ∈ Chev(2), then by [IA , 6.1.4] and definition of C2 [V3 , 1.1], K/Z(K) ∼ = L3 (4) with Z(K) elementary 3 abelian, or K/Z(K) ∼ = 2B2 (2 2 ), G2 (4), Sp6 (2), D4 (2), U6 (2), F4 (2), or 2 E6 (2). For all but the first two, m2 (K/Z(K)) ≥ 6 by [IA , 3.3.3], contrary to assumption. In both the first two cases, m2 (K) ≥ 4, thanks to [IA , 6.4.1], a contradiction. If K/Z(K) ∼ = SL2 (q) ∈ C2 , contradiction. If K/Z(K) ∈ = L2 (q), q odd, then K ∼ Chev(3) − Chev(2), then as Z(K) = 1, K ∼ = 2U4 (3) and m2 (K) = 5, contradiction. Therefore K ∈ Spor. Using the 2-ranks given in [IA , Table 5.6.1], we see that the only possibility is K ∼ = 2J2 . Let K = K/Z(K), z ∈ K a 2-central involution, z a preimage of z, Q = O2 (CK (z)), and Q the full preimage of Q. Then Q∼ and CK (z) permutes transitively the elements of Q of order 4. As z is = 21+4 − a 2-central involution of K [IA , 5.5.8], Φ(Q) = x2 | x ∈ Q has order 2. Therefore  m2 (Q/Φ(Q)) = 5, contrary to hypothesis. The proof is complete. Lemma 10.33. Let K ∈ C2 with Z(K) = 1, and let S ∈ Syl2 (K). Then S/Z(K) does not embed in the direct product of an elementary abelian group and a dihedral group.

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Proof. Suppose false. Set K = K/Z(K). Then S has an abelian maximal subgroup not containing Z4 × Z4 , and [S, S] is cyclic. In particular S involves no extraspecial group E of width 2 or more, and no special group T with noncyclic center. Since U4 (3) involves Q8 ∗ Q8 , K ∼ = 2U4 (3), and so, with the definition of C2 and [IA , 6.1.4], K ∈ Chev(2) ∪ Spor. If K ∈ Spor, then often K obviously contains an E [IA , 5.3]; since K is not 2-saturated and lies in C2 , we are reduced to K ∼ = M22 or Ru, which contain special groups T of order 22+4 (in L3 (4)) and 23+8 , respectively, 3 contradiction. So K ∈ Chev(2) ∩ C2 . Likewise 2B2 (2 2 ) and G2 (4) contain T ∼ = 23+3 2+8 1+4 ∼ ∼ and T = 2 , respectively, while Sp6 (2) contains E = 2 in the centralizer of a long root involution. As Sp6 (2) is involved in F4 (2), and also in D4 (2), U6 (2), and 2E6 (2) [IA , 4.9.2], we have covered all the cases in [IA , 6.1.4], and the proof is complete.  Lemma 10.34. Let S ∈ Syl2 (Sp4 (4)). Then Z(S) = [S, S] ∼ = E24 and S is indecomposable. Proof. The fact that Z(S) = [S, S] ∼ = E24 follows from the Chevalley commutator formula [IA , 2.4.5]. Suppose that S = S1 × S2 with both factors nontrivial. Then [Si , Si ] = Z(Si ) and hence |Si : Z(Si )| ≥ 4, i = 1, 2. As |S : Z(S)| = 42 , |Si : Z(Si )| = 4. But then |[Si , Si ]| = 2, contradicting Z(S) = [S, S]. The lemma is proved.  Lemma 10.35. Let K = Sp6 (2) and S ∈ Syl2 (K). Then J(S) ∼ = E26 and S is indecomposable. Proof. Let Σ = {±2ai , ±(ai ± aj ) | i, j = 1, 2, 3; i = j} be a root system of type C3 , and let S ∈ Syl2 (K) be generated by all positive root groups (where a1 > a2 > a3 > 0). Let A ≤ S be generated by the root groups for all 2ai and all ai + aj . Clearly A is elementary abelian and we claim that A = J(S). Note that NK (A) = AL is a parabolic subgroup with L ∼ = L3 (2). The subgroup A0 generated by all ai + aj root groups is a natural module for L, and A/A0 is the dual module. Hence no involution of L acts as a transvection on A, and no four-subgroup of L acts quadratically on A. By the Thompson Replacement theorem [III8 , 1.1], A = J(S). Suppose that S = S1 × S2 is a nontrivial decomposition. Then A = A1 × A2 , where Ai = A∩Si , i = 1, 2, and D8 ∼ = S/J(S) ∼ = (S1 /A1 )×(S2 /A2 ). Hence without loss of generality, S2 = A2 , and so Z(S) ≤ [S, S]. But Z(S) is the product of the high root subgroup and the high short root subgroup, both of which lie in [S, S] by the commutator formula. This contradiction completes the proof.  Lemma 10.36. Let X = 2D4 (2). Then a Sylow 2-subgroup of X has normal 2-rank at least 7. Proof. We know that X embeds with odd index in L := D4 (3) ∼ = Ω+ 8 (3). With respect to an orthonormal basis of the natural L-module, the monomial subgroup M of L is an extension of E27 by A8 . Hence |M |2 = 213 = |X|2 , and the result follows.  Lemma 10.37. Aut(HS) has no abelian subgroup of order 28 .

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Proof. We use [IA , 5.3cm]. We have | Aut(HS) : HS| = 2 and |HS : M22 | = 100, so if the lemma were false, M22 would have an abelian subgroup of order 25 . But M22 has a subgroup N of odd index with F ∗ (N ) ∼ = E24 being a natural module for N/F ∗ (N ) ∼ = ΓL2 (4). In particular every involution of N/F ∗ (N ) acts freely on F ∗ (N ), and N/F ∗ (N ) has nonabelian Sylow 2-subgroups of order 8. These properties imply that N , and hence M22 , has no abelian subgroup of order 25 .  Lemma 10.38. Let K ∈ K with K/Z(K) ∼ = U6 (2). Then m2 (K) = m2 (Z(K))+ 9. Proof. This follows from [IA , 3.3.3, 6.4.4].



Lemma 10.39. Suppose that K ∼ = U4 (2), J3 , F i24 , or F3 , and Q is a quaternion subgroup of Aut(K). Let T ∈ Syl2 (K). Then T does not normalize Q. Moreover, any involution in CAut(K) (T ) lies in Inn(K). Proof. For the second assertion, we use the information in [IA , 4.5.1] (for ∼ P Sp4 (3)) and [IA , 5.3hvx] to see that if u ∈ I2 (Aut(K) − Inn(K)), then U4 (2) = |CK (u)|2 < |K|2 . Suppose the first assertion is false. In every case | Out(K)| ≤ 2, so Q ∩ T contains Y  T with Y ∼ = Z4 . But also T is not of maximal class, so by [IG , 10.11], there is X  T with X ∼ = E22 . Let z ∈ I2 (Z(T )) and set C = CT (z). From [IA , 5.3hvx] and [IA , 4.5.1], R := F ∗ (C) is extraspecial. In particular z = Z(T ) and in C = C/ z, E22 ∼ = XY ≤ CR (T ). If K ∼ = Q8 ∗ Q8 and T /R ∼ = Z2 interchanges the two Q8 factors, = U4 (2), R ∼ whence the preimage of Z(T ) is elementary abelian, contradicting Y ≤ Z(T ). If K ∼ = Q8 ∗ D8 and T /R ∼ = A5 acts via the core of the natural = J3 , then R ∼ permutation module on R. Thus Z(T ) ∼ = Z2 , contradicting XY ≤ Z(T ). If K ∼ = E211 and = F i24 , consider the subgroup N of K with S = F ∗ (N ) ∼ ∼    N := N/S = M24 [IA , 5.3v]. Then Y ≤ Z(T ), whence equality holds by [IA , 5.3e].  of order As Y  T , Y induces a transvection on S. Then Y inverts a subgroup W  ) containing an element of order 7), so [S, W ] ∼ 3 (with CN (Y W = E22 . But this is  inconsistent with the fact that a Sylow 3-subgroup of N is isomorphic to 31+2 and acts faithfully on S.  = C/R ∼ Finally, suppose that K ∼ = F3 . Let C = A9 and let w ∈ I3 (C) map  on a 3-cycle w.  Then in [IA , 5.3x], w is in class 3C and CR (w) = z. Now w centralizes a Frobenius subgroup F ∼ = F9.4 , whose Sylow 2-subgroups S therefore  w  = XY admits an automorphism of order satisfy CR (S)  ∼ = A4 . Therefore CR (S) 3. But XY ∼ = D8 or Z4 × Z2 admits no automorphism of order 3, a contradiction. This completes the proof of the lemma.  Lemma 10.40. Let K = Co2 , S ∈ Syl2 (K), and E = J(S). Then E = CK (E) ∼ = E210 and AutK (E) contains Aut(M22 ). ∼ Proof. By [A2, 26.1(6)], K contains a split extension H = F M ∗ where E210 = F = CH (F )  F M ∗ , M ∗ ∼ = Aut(M22 ), and F is the unique faithful M ∗ -composition factor of the restriction of the Golay code F2 M24 -module to the stabilizer M ∗ in M24 of a two-point set. Let T ∈ Syl2 (F M ∗ ), so that |T | = 218 and so T ∈ Syl2 (K). Then CK (F ) = F × W with W of odd order and it is clear from [IA , 5.3k] that W = 1. It then remains only to show that F = J(T ).

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∼ F with A ≤ F M ∗ , Suppose F = J(T ). Then by [V2 , 6.1], there exists A = ∗ A = F , and [F, A, A] = 1. Since M contains D10 and F11.5 and has one class of involutions, we have |CF (y)| ≤ 26 for every y ∈ I2 (A − F ). Set M = [M ∗ , M ∗ ] ∼ = M22 . It follows that A0 := AF ∩ M has order 2n ≥ 8, and [F, A0 , A0 ] = 1. Hence if F ∗ is the dual F2 M -module to F , then [F ∗ , A0 , A0 ] = 1. But by [A2, 23.10(5)], F ∗ is the 10-dimensional Todd module, and hence M permutes a subset F0 of F ∗ of size 22 naturally. By Lemma 5.8d, any 4-subset of F0 is linearly independent. Hence by quadratic action, every A0 -orbit on F0 has size 1 or 2. In particular there are at least 11 A0 -orbits. But every involution of A0 fixes 6 points at most, so by the Cauchy-Frobenius orbit-counting lemma, the number of A0 -orbits is at most 2−n (22 + (2n − 1)6) = 6 + 24−n < 11, a contradiction. This completes the proof.  Lemma 10.41. Let K = Co2 and T ∈ Syl2 (K). Then J(T ) ∼ = E210 and the following conditions hold: (a) There are x, y, z ∈ J(T )# such that I2 (K) = xK ∪ y K ∪ z K , |xK ∩ J(T )| = 616, |y K ∩ J(T )| = 330, and |z K ∩ J(T )| = 77; (b) NK (J(T ))/J(T ) ∼ = Aut(M22 ); and (c) y ∈ [CT (y), CT (y)] and z = Z(T ) ≤ [T, T ]. ∼ E210 , E = Proof. We use [IA , 5.3k] freely. By Lemma 10.40, E := J(T ) = ∼ Aut(M22 ). By [IG , 16.9], H ∗ controls CK (E), and H ∗ := AutK (E) contains H = K-fusion in E. Let z, y, and x be representatives of conjugacy classes 2A, 2B, and 2C in E. (Obviously x exists, and y and z exist since y is half 2-central– see [IG , 10.20(i)]—and z is 2-central.) Let A, B, and C be the cardinalities of 2A ∩ E, 2B ∩ E, and 2C ∩ E, respectively. Thus A = |H ∗ : CH ∗ (z)| ≥ |H : CH (z)| and similarly for the other two classes. In particular as E = O2 (CK (x)), C ≥ |H|/| Aut(A6 )| = 616, so A + B ≤ 407. Also E ∩ CK (y) ≤ O2 (CK (y)) so by Tits’ lemma, |CH (y)| = |P | for some parabolic subgroup P of A8 . As |CK (y)|2 = |K|2 /2, |P |2 ≥ 2| Aut(M22 )|2 /407 > 17 so |P |2 = 21, whence B ≥ 2| Aut(M22 )|2 /21 = 330. Hence, A ≤ 77. But CM (z) contains no element of order 7 or 11 – the former because M22 has no maximal subgroup of odd index of order divisible by 7 [IA , 5.3c]. Therefore A = 77, forcing B = 330 and C = 616, proving (a). Moreover, |H ∗ : CH ∗ (x)| = |H : CH (x)|. But CK (x) ≤ H and so H ∗ = H, whence (b) holds. As for (c), obviously |Z(T )| = 2 so z ∈ [T, T ], and since F ∗ (CK (y)) = O2 (CK (y)) and m2 (CK (g)) = 2 for g ∈ I7 (CK (y)), y = [O2 (CK (y)), O2 (CK (y))]. The proof is complete.  Lemma 10.42. Let K = Co2 , z ∈ I2 (K) a 2-central involution, C = CK (z), and b ∈ I3 (C) such that CC (b) has an A6 composition factor. Then CO2 (C) (b) = z. Proof. We use [IA , 5.3k]. Since CC (b) involves A6 , CK (b) = b × L with L∼ = Z6 × Σ6 . So O2 (CK (b, z)) = z, which = Aut(U4 (2)), and then CK (b, z) ∼ implies the result.  Lemma 10.43. Let K = F i24 , let z ∈ I2 (K) be 2-central, and let v ∈ I3 (O2,3 (CK (z))). Then CO2 (CK (z)) (v) = z. Proof. We use [IA , 5.3v]. Since CCK (v) (z) = CK (vz) = CCK (z) (v) has a 3component of type 3U4 (3), v must be of class 3C, and then from the structure of CK (v) we see that |O3,2 (CCK (z) (v))|2 = |O3,2 (CK (vz))|2 = 2. This implies the lemma. 

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Lemma 10.44. Let K = F3 and let E = {E ∈ E25 (K) | AutK (E) = Aut(E)}. If 1 = z ∈ E ∈ E, then E ≤ O2 (CK (z)). Proof. We use [IA , 5.3x]. Let N = NK (E), C = CK (z), N = N/E, and C0 = C ∩ N . Then CK (E) = E and C 0 is a parabolic subgroup of N ∼ = L5 (2) of shape 24 L4 (2). Hence |O2 (C0 )| = 29 and C0 /O2 (C0 ) ∼ = A8 . In particular |C : C0 | is odd so O2 (C) ≤ O2 (C0 ), and then by orders, O2 (C) = O2 (C0 ). As E ≤ O2 (C0 ), the lemma follows.  Lemma 10.45. Let K = Sp6 (2), let P be a parabolic subgroup of K, and let W ∈ Syl3 (P ). Then the following conditions hold: (a) If P is of type L3 (2), then O2 (P ) ∼ = E26 and for any involution x ∈ P − O2 (P ), CO2 (P ) (x) ∼ = E24 ; (b) If P is of type Sp4 (2), then CO2 (P ) (W ) is generated by a long root involution; and (c) If P is of type L2 (2) × L2 (2), then CO2 (P ) (W ) is generated by a short root involution. Proof. Use standard notation [IA , 2.10], and take Π = {a1 − a2 , a2 − a3 , 2a3 } as a fundamental system of roots. Then in (a), O2 (P ) is generated by the root subgroups for all ai + aj , 1 ≤ i ≤ j ≤ 3, and it is immediate that it is elementary abelian. As P/O2 (P ) has one class of involutions it is enough to check that |CO2 (P ) (Xa1 −a2 )| = 24 , and this follows directly from the Chevalley commutator formula. In (b), P = CK (X2a1 ) with |W | = 32 and O2 (P ) ∼ = E25 , so CO2 (P ) (W ) has order 2 and thus equals X2a1 , as claimed.  In (c), P = CK (Xa1 +a2 ) with W ∈ Syl3 ( X±(a1 −a2 ) × X±2a3 ). Let W0 =  W ∩ X±(a1 −a2 ) . Now O2 (P ) = O2 (P )/Xa1 +a2 has a W0 -invariant filtration with quotients isomorphic to X2a1 , X2a2 , Xa1 +a3 , Xa2 +a3 , and Xa1 −a3 , Xa2 −a3 . On each quotient, W0 has trivial fixed points. Therefore CO2 (P ) (W ) = Xa1 +a2 , as asserted.  Lemma 10.46. Let K = F4 (2) and let P = RL be a Levi decomposition of a parabolic subgroup of K of type Sp6 (2). Then [R, L] = R and Φ(R) ∼ = Z2 . Proof. We take the root system of K to consist of all ±ai , ±ai ± aj , and 4 1 2 3 4 := 12 ( i=1 i ai ), where i, j = 1, 2, 3, 4; i = j; and each i = + or −. Then {a2 − a3 , a3 − a4 , a4 , + − −−} is a fundamental system and (using Aut(K)) we may assume that the first three form a fundamental system for L. Hence, R is generated by all root subgroups for which a1 has a positive coefficient. As the characteristic is 2, two orthogonal short root subgroups commute. It follows with the Chevalley commutator formula and the fact that root subgroups have order 2  that Φ(R) = [R, R] = Xa1 ∼ = Z2 , as required. Lemma 10.47. Let K = D4 (2) and T ∈ Syl2 (K). Then the following conditions hold: ∼ E22 . Moreover, all elements of U # (a) T has a unique normal subgroup U = are K-conjugate; (b) m3 (CK (Z(T ))) = 3 and m3 (K) = 4; (c) For every x ∈ I2 (T ), m2 (CT (x)) ≥ 5; and

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(d) Let P be a maximal parabolic subgroup of K with P/O2 (P ) ∼ = A8 . Then for every x ∈ O2 (P )# , 3 divides |CP (x)|. Proof. In part (a), the uniqueness follows from [IA , 3.3.1a] and the fact that there is a unique positive root of height 4, the maximum height being 5. Since U is generated by two root elements, and |Z(T )| = 2, all elements of U # are conjugate. Part (b) follows from [IA , 4.10.3] and the fact that CK (Z(T )) is a parabolic subgroup of type A31 . Next, let P be as in (d). Then O2 (P ) ∼ = E26 is # a natural module for P/O2 (P ) ∼ are the = Ω+ 6 (2). The two orbits of P on O2 (P ) singular vectors and the nonsingular ones, with stabilizers 24 Ω4 (2) and Σ6 , both of order divisible by 3. Finally, since the Dickson invariant of each x ∈ I2 (P/O2 (P )) is trivial, dim[O2 (P ), x] is even. Hence |CO2 (P ) (x)| = 24 , implying (c). The proof is complete.  Lemma 10.48. Let K = E(K) = O3 (K) be a K-group. Let T ∈ Syl2 (K) be invariant under B ∈ E3 (Aut(K)). Then Ω1 (T ) is elementary abelian and CT (B) = 1. Proof. By Lemma 1.8, Ω1 (T ) is elementary abelian. In a minimal counterexn ample, B normalizes K ∼ = 2B2 (2 2 ) for some n. In particular Outdiag(K) = 1 [IA , 2.5.12]. Then B/CB (K) is cyclic and induces field automorphisms on K by n/3 [IA , 4.9.1, 4.9.2], so CK (B) contains 2B2 (2 2 ) and thus has even order. The lemma follows.  Lemma 10.49. Let K = U6 (2), U5 (2), or D4 (2), and let z ∈ K be a 2-central involution. Set R = O2 (CK (z)). Then CAut(K) (R) = z. Proof. Let K ∗ = Inndiag(K). In each case z is a high root subgroup and C := CK ∗ (z) is a parabolic subgroup with R = F ∗ (C) extraspecial and C/R isomorphic to GU4 (2), GU3 (2), and Σ3 ×Σ3 ×Σ3 , respectively. Let C0 = CAut(K) (z). Then in the unitary cases C0 = C γ where γ is a graph automorphism induced by a nontrivial automorphism of F4 . In the orthogonal case C0 = CΓ where Γ ∼ = Σ3 is a group of graph automorphisms permuting transitively the three direct factors of C/R. Thus, in all cases C0 /C induces outer automorphisms on C/R, which implies the result.  Lemma 10.50. Let K = D4 (2) or U6 (2), and let z ∈ K be a 2-central involution. and NK (R) acts irreducibly Let R = O2 (CK (z)) and V = R/ z. Then R ∼ = 21+8 + on V . Proof. CK (z) contains a Borel subgroup so is a parabolic subgroup. Indeed CK (z) = RL where a Levi complement is generated by root subgroups and their negatives corresponding to nodes of the twisted Dynkin diagram not connected to the lowest root in the extended diagram. From this, using the Chevalley commutator formula, one easily gets R ∼ = Σ3 × Σ3 × Σ3 or = 21+8 , with L = L1 × L2 × L3 ∼ L∼ = U4 (2), respectively. In the unitary case, z is a transvection on the natural module U , with center u, say, and L acts faithfully on u⊥ /F4 u, while R stabilizes the chain U ⊇ u⊥ ⊇ F4 u ⊇ 0. It follows that V is the restriction to F2 of the natural F4 -module for U4 (2). By considering the action of an element of I5 (U4 (2)), we see that if W ⊆ V is a proper F2 U4 (2)-module, then dim W = 4. But U4 (2) does not embed in L4 (2), so W cannot exist, proving irreducibility.

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In the orthogonal case, one can check that the fundamental root subgroup Xα , where α is at the center of the Dynkin diagram, generates V as a free T -module of rank 1, where E23 ∼ = T ∈ Syl2 (L). It follows that as a module for any Li , V is the direct sum of four natural L2 (2) ∼ = Li -modules. Therefore V ∼ = V1 ⊗F2 V2 ⊗F2 V3 as L-module, where Vi is a natural L2 (2) ∼ = Li -module. Then on V := F4 ⊗ V , O3 (L) is diagonalizable and V is the direct sum of eight 1-spaces corresponding to distinct weights, and permuted transitively by T . Hence V , and then V , is irreducible.  Lemma 10.51. Let P = QL be a parabolic subgroup of K ∼ = Sp6 (2) with Q ∼ = ∼   of Q in K  is E26 and L = L3 (2). Let K = 2Sp6 (2). Then the preimage Q isomorphic to 21+6 + .  ≤ J with Z(K)  = Proof. Let J = Spin7 (3) and J = Ω7 (3), and regard K  Z(J) [IA , Table 6.2.2]. Then when Q is regarded in J, it preserves a decomposition of the natural J-module as the orthogonal sum of seven nonsingular 1-spaces, which are all isometric since they are permuted by an element of I7 (L). Hence, NJ (Q)  is extraspecial. contains A7 and therefore acts irreducibly on Q. It follows that Q 1+6 ∼   As 7 divides | Aut(Q)|, Q = 2+ , and the proof is complete.  Lemma 10.52. Let K = F ∗ (X) ∼ = 2Suz and let S ∈ Syl2 (X) and C = CX (Z(S)). Then Z(S) ∼ = E22 and [O2 (C), O2 (C)] ∼ = Z2 . Proof. Let CK = C ∩K. From [IA , 5.3o] we see that Q/Z(K) := O2 (C/Z(K)) = and Z/Z(K) := Z(Q/Z(K)) = Z(S/Z(K)) ∼ O2 (CK /Z(K)) ∼ = 21+6 = Z2 splits over − ∼ Z(K). That is, Z = E22 , and we must show that Z ≤ Z(S) and that [Q, Q] < Z. Indeed, the latter condition suffices, since it implies Z = Z(K) × [Q, Q], and [Q, Q] ≤ Z(S) since Q  S. Let C0 = CK (Z/Z(K)). Then C0 /Z(K) ∼ = (Q/Z(K))#Ω− 6 (2) is perfect, so Z ≤ Z(C0 ) and in particular Z = Z(Q). For any z ∈ Z # , Q/ z is extraspecial or abelian, because of the absolutely irreducible action of C0 /Q on Q/Z. If it is abelian for some such z, then [Q, Q] ≤ z and we are done, so assume for a contradiction that Q/ z is extraspecial for each z ∈ Z # . Choose yz ∈ Z − z. q (u) Then for u ∈ Q, the equivalence u2 ≡ yzz (mod z) defines, for each z ∈ Z # , a C0 -invariant nondegenerate quadratic form qz on Q/Z with values in {0, 1} = F2 . However, by absolute irreducibility, there is a unique such quadratic form q. For any u ∈ Q of order 4, let z = u2 . Then q(u) = qz (u) = 0, so qy (u) = 0 for all y ∈ Z  and thus u2 = 1, a contradiction. The proof is complete. Lemma 10.53. Let K = 22E6 (2)a and b ∈ I3 (K) with I := E(CK (b)) ∼ = 2U6 (2). Then there exist Sylow 2-subgroups S and T of I and K, respectively, such that S ≤ T and Z(S) = Z(T ) ∼ = E22 . Proof. Let K = K/O2 (K). By [IA , 4.7.3A], I ∼ = U6 (2), so Z(K) = Z(I). Now b is determined up to conjugacy by the isomorphism class of I, so we may take b in the root SL2 (2) subgroup corresponding to the lowest root, and then I is a subsystem subgroup. By [IA , 3.3.1a], Sylow 2-centers of K/Z(K) and I/Z(I) are long root subgroups. But any long root subgroup of I is also one of K, so T and S may be chosen so that Z(T ) = Z(S). By [IA , 6.4.1, 6.4.2], the preimage of this group in K is a four-group equal to both Z(T ) and Z(S), as required. 

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Lemma 10.54. Let K = 22E6 (2)a and z ∈ I2 (K) with z mapping to a 2-central involution of K/Z(K). Let M = CK (z). Then F ∗ (M ) = O2 (M ) ∼ = Z2 × 21+20 + and M/O2 (M ) ∼ = U6 (2). Moreover, if V is the natural F4 [M/O2 (M )]-module, then O2 (M )/Z(O2 (M )) is an F2 -form of V ∧ V ∧ V . Proof. Let K = K/Z(K). Then M = CK (z) by Lemma 10.53. Hence M = QL, where Q = O2 (M ) = F ∗ (M ) and L ∼ = U6 (2). Routine calculations with the commutator formula show that Q ∼ , and the final assertion follows from = 21+20 + Lemma 5.3. It remains to show that the preimage Q of Q is isomorphic to Z2 × Q. Now Z(Q) = Z(K) z by Lemma 10.53. Since Q/Z(Q) ∼ = Q/Z(Q) is absolutely irreducible as L-module, by Lemma 5.3, it supports a unique nonzero symplectic F2 -form, so [Q, Q] ∼ = Z2 . Then the irreducibility implies that Q/[Q, Q] must be elementary abelian, so Φ(Q) = [Q, Q]. Hence Z(K) is a direct factor of Q. The proof is complete.  Lemma 10.55. Let F ∗ (X) = K ∼ = 2F i22 and T ∈ Syl2 (X). Then Z(T ) is a four-group, and Z(T /Z(K)) = Z(T )/Z(K) ∼ = Z2 . Proof. We use [IA , 5.3t] without comment. Let X = X/Z(K), and let z ∈ I2 (K) be of class 2A. By the structure of CAut(K) (z), Z(T ) ≤ K. Let R = O2 (CK (z)) = F ∗ (CK (z)) and E = Z(R) ∼ = E22 . Then Z(T ) ≤ E, and R = O2 (CK (E)). If Z(T ) = E, then all elements of E would be of class 2A, whence by Burnside’s lemma they would all be conjugate in NK (T ) and hence in NK (R). But E ∩ [R, R] ∼ = Z2 , so this is impossible., Therefore Z(T ) < E so |Z(T )| = 2. Finally, Z(T ) covers Z(T ) and is elementary abelian, since F i23 has a noncyclic Sylow 2center, one of whose involutions has centralizer isomorphic to 2F i22 [IA , 5.3u]. The proof is complete.  We prove the next two lemmas together. Lemma 10.56. Let K = 2F i22 . Let b ∈ I3 (K) with I := E(CK (b)) ∼ = 2U4 (3). Then there exists T ∈ Syl2 (K) with T ∩ I ∈ Syl2 (I), Z := Z(T ∩ I) = Z(T ) ∼ = E22 , . and CO2 (CK (Z)) (b) ∼ = E22 × 21+4 + Lemma 10.57. Let K = 2F i22 , T ∈ Syl2 (K), Z = Z(T ), R = O2 (CK (Z)), and E = Z(R). Then the following conditions hold: ∼ E22 and E ∼ (a) Z = = E23 ; and and CK (Z)/R ∼ (b) R = F ∗ (CK (Z)) ∼ = E22 × 21+8 = Aut(U4 (2)), acting irre+ ducibly and naturally on R/E. Proofs of Lemmas 10.56 and 10.57. Let K = K/Z(K) ∼ = F i22 , z ∈ I2 (I), and C z = CK (z). We claim z is 2-central in K. Now,     O 3 (CC z (b)) ∼ = O 3 (CI b (z)) ∼ = SL2 (3) ∗ SL2 (3) × Z3 . This rules out C z ∼ = 2U6 (2) by [IA , 4.2.2]. Also C (b)/O2 (C (b)) contains Z3 × Cz

Cz

Σ3 × Σ3 , which rules out the other non-2-central class in I2 (K) for z. Thus our claim holds. ∼ − ∼ Let R = O2 (C z ) ∼ = Z2 × 21+8 + . Then C z /R = O6 (2) so O2 (CC z (b)) = CR (b) = 1+4 Z2 × 2+ . Involutions in Z(R) are of class 2A or 2B since their centralizers in K are nonsolvable. Hence by [IA , 5.3t], the preimages of these involutions in K

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are involutions that do not shear in K to Z(K). Let R be the preimage of R and E = Z(R); it follows that E ∼ = E23 . The involutions in E all centralize the extension of R by U4 (2), but they are obviously not all conjugate in the normalizer of that extension. Hence E contains two involutions y in class 2A. Thus, CK (E) is a maximal parabolic subgroup in CK (y), and in particular R/E is a natural module for C z /R ∼ = Aut(U4 (2)). Now let X = F i23 ; we may regard Z(K) ≤ X and K as CK (Z(K)). Then CK (z) = CX (Z(K), z) is the centralizer of a Sylow 2-center in 1+4 ∼ X. Using [IA , 5.3u] we conclude that R ∼ = E22 × 21+8 + , whence CR (b) = E22 × 2+ . This completes the proofs of the lemmas.  Lemma 10.58. Let X = JD where F ∗ (X) = J ∼ = D4 (3)a , D ∩ J = 1, and ∼ := D ∼ Z or Σ . Let t ∈ I (Aut(J)) with I E(C (t)) = 3 = Ω− 3 2 J 6 (3) or Ω7 (3). Let z ∈ I2 (I) be supported on a 4-dimensional subspace of the natural I-module. Then the following conditions hold: ∼ 21+8 ; and (a) z is 2-central in J and F ∗ (CX (z)) = O2 (CJ (z)) = + (b) CX (z) = AE1 where A is the central product of 4 copies of SL2 (3) permuted transitively by E1 , A  CX (z), A ∩ E1 = 1, E1 ∼ = Z2 × E0 , where E0 ∼ = D. = A4 or Σ4 and E0 /O2 (E0 ) ∼ Proof. By [IA , 4.5.1], t is unique up to Aut(J)-conjugacy, so z is as well. We + may choose a natural Ω+ 8 (3)-module V on which a preimage of t in O8 (3) acts with two or one eigenvalues equal to −1, according to the isomorphism type of I. Then  := Ω(V ) is supported on a 4-dimensional subspace V−1 of a preimage of z in K V = V−1 ⊥ V1 . Now (a) follows from the information in [IA , 4.5.1]. We set  A := Ω(V−1 ) ∗ Ω(V1 ) = O 3 (CJ (z)). Let u  ∈ Ω(V ) be an involution interchanging the (isometric) spaces V−1 and V1 . Let u  ∈ Syl3 (A). Let S−1 ∈ Syl2 (NO(V−1 ) (P−1 )) P−1 ∈ Syl3 (Ω(V−1 )) and P = P−1 P−1 u  . (We consider S±1 ≤ O(V ).) Note that |Si | = 8 and z ∈ Z(Si ), and S1 = S−1 u) = z × F0 with i = ±1. Then S−1 S1 is a central product, and F := CS−1 S1 ( ∼ 2 3 . We set F = F u E , where u is the image of u  in J. Then F0 ∼ E = 2 = 2 1 0 F2 := z×F1 ∈ Syl2 (NJ (P )). By Frattini arguments, CX (z) = ANCX (z) (P ) = AN with N = NCX (z) (F2 ) and N/F2 ∼ = D. Moreover, a Sylow 3-subgroup of N maps to a triality in Out(J) and so permutes nontrivially the four solvable components of A. It follows that N/ z ∼ = Z2 × A4 or Z2 × Σ4 according to the isomorphism type of D. It remains only to show that z ∈ [N, N ], for then N = z × E1 with E1 having the properties claimed in (b). Furthermore, it is clear that z ∈ [O2 (N )O 2 (N ), O2 (N )O 2 (N )], so we need only consider the case D ∼ = Σ3 and show that for some involution v ∈ N − J, CO2 (N ) (v) ∼ = E23 . Since D maps on a Sylow 3-normalizer in Out(J) ∼ = Σ4 , JD contains some conjugate of any given involution in Aut(J) − O 2 (Aut(J)), by the Baer-Suzuki theorem. In particular, JD contains a reflection r. As r centralizes a J-conjugate of z, we may replace r by a J-conjugate and assume that r ∈ CX (z), r centralizes Ω(V−1 ), and r interchanges the two SL2 (3) central factors of Ω(V1 ). Hence, again replacing r by a conjugate if necessary, we may assume that r ∈ NCX (z) (P ) and then that r ∈ N . Since r is a reflection, [r, g] = z for any g ∈ O2 (N ). Therefore CO2 (N ) (r) covers CO2 (N )/z (r), which is a four-group. As z is a direct factor of F2 , CO2 (N ) (r) ∼  = E23 . This completes the proof.

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Lemma 10.59. Let J, I, t and z be as in Lemma 10.58. Let X be a group with K = E(X), t ∈ Z(X), X/ t ∼ = Aut(F i22 ) or F i23 , and t ∈ K in the F i22 case. Let b ∈ I3 (K) with E(CK (b)) = I ∼ = Ω− 6 (3) or Ω7 (3) according to the isomorphism type of K. Then the following conditions hold: (a) Q := F ∗ (CK (z)) ∼ = E22 × 21+8 + ; (b) [Z(Q), b] = 1 and [Q, b] ∼ = 21+4 + ; and (c) CX (z) is an extension of U := O2 (CX (z)) by V , where (1) Q/Z(Q) ∼ = E28 is a natural module for V (∞) ∼ = U4 (2); ∼ and V Aut(U (2) If K ∼ = 21+10 = = 2F i22 , then U/ t ∼ 4 (2)); and + 1+8 ∼ ∼ 3 (3) If K = F i23 , then U = E2 × 2+ and V is a crown product of Σ3 and Aut(U4 (2)). Proof. First assume that K ∼ = 2F i22 . By Lemmas 10.56 and 10.55, the involution z ∈ I is 2-central, so from Lemma 10.57, (a) and (c1) hold. As O2 (CK (bz)) has exactly two Q8 central factors, and CZ(Q) (b) contains the four-group Z(K)Φ(Q), (b) follows. Part (c2) then follows from the centralizer information and Note 2 in [IA , 5.3t]. Assume then that K ∼ = = F i23 , so that X = K × t. Since O2 (CK (zb)) ∼ Q8 ∗ Q8 × E22 , we see that z is in class 2C [IA , 5.3u], whence (a) and (b) follow immediately. Since 2F i22 is an involution centralizer in F i23 , the truth of (c1) follows from its truth in the F i22 case. Finally, as X = K × t, (c3) follows from  the information in [IA , 5.3u]. Lemma 10.60. Let K = F ∗ (X) ∈ C3 , z ∈ X a 2-central involution. Then CX (z) has no normal subgroup L = L1 ∗ L2 ∗ L3 such that z ∈ Li ∼ = SL2 (3) for all i = 1, 2, 3. Proof. Suppose that such a z and L exist. Obviously K ∼  M11 . Hence by = [III11 , 13.1], K ∈ Chev(3) with q(K) = 3. Inspection of [IA , 4.5.1] then completes the proof.  Lemma 10.61. Let L = A6 , L± 3 (3), 3G2 (3), or L2 (8). Let z be a 2-central involution of L. Then CAut(L) (z) does not involve U3 (2). Proof. If L = L2 (8), then Aut(L) has E23 Sylow 2-subgroups, whereas U3 (2) contains Q8 . Likewise in the first two cases, Out(L) is a 2-group and |O 2 (CL (z))|3 ≤ 3, whereas |U3 (2)|3 = 32 . Finally, in the 3G2 (3) case, a Sylow 3-subgroup P of C := CAut(L) (z) has order 32 but |NC (P )/CC (P )| = 4, whereas U3 (2) ∼ = F9.8 .   ∈ Cp be simple. Let z ∈ Lemma 10.62. Let p be an odd prime. Let J = F ∗ (X)    ∈ Ep∗ (C  ( z )) is of symplectic type. Let B I2 (X) and assume that R = O2 (CX ( X z ))  = r − 1, where r ≥ 4. Assume that C  (R)  = 1 and m2 (C  (B))  = 1. with mp (B) B R Let X be a group such that Op (X) = 1 and F ∗ (X) = JA with A a nontrivial  with cyclic p-group. Assume that there is a surjective homomorphism φ : X → X ∼   kernel A such that φ(J) = J, φ(B) = B for some B ≤ X with B = Epr , and φ(z) = z for some z ∈ I2 (CX (B)). If m2,p (X) = r, then the following hold: (a) Either p = 3 and J ∼ = P Ω+ 8 (3), U5 (2), (S)U6 (2), D4 (2), (3)Suz, Co2 ,  Co1 , (3)F i24 , F3 , F2 , or F1 , or p = 5, r = 4, and J ∼ = F1 ;  and (b)  z  is a Sylow 2-center of J and of X; ∗   = F ∗ (C  (    (c) R  (B; 2), R ≤ J, and R is extraspecial. X z )) ∈ IX

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 ≤ Aut(J).  By [IA , 2.5.12], Out(J)/  Outdiag(J)  has Proof. We regard X  abelian Sylow 2-subgroups when J ∈ Chev. Since z ∈ [R, R], it follows that z ∈  in this case, and certainly z ∈ J if J ∈ Alt ∪ Spor. Note that since Inndiag(J)  B]  is extraspecial of width at least 3; and  = 1, and by [V2 , 4.12], R 0 := [R, CB (R)     RI := [R, B ∩ Inndiag(J)] has width at least 1.  = O p (C  Suppose first that J ∈ Chev(p) and set L  (z)). Then L is Inndiag(J)

I ≤ O2 (L).  I a central product of Lie components by [IA , 4.2.2], and R As R  = 3. In particular is nonabelian, the only possibility is that p = 3 and q(J) 3 ∼    B ≤ Inndiag(J) unless possibly J = D4 (3) or D4 (3). If J ∼ = 3D4 (3), then by   [IA , 4.5.1], J has sectional 2-rank 4, so R has width at most 2, contradiction. If J ∼ = D4 (3), then (a) holds, and (b) and (c) follow easily as well. So we may assume  ≤ Inndiag(J).  It follows that C (z) has at least r−1 ≥ 3 that J ∼ = D4 (3) and that B J Lie components isomorphic to SL2 (3). From [IA , 4.5.1] we see that J ∼ = D4 (3), a contradiction. (Note that if J ∼ = B3 (3), then one of the Lie components of CJ(z) is of adjoint type.) Hence, the lemma holds if J ∈ Chev(p).  ≥ 3, J ∼  If J ∈ Alt ∩ Cp , then since mp (B) = A3p . But then CAut(J)  (B) has odd order and cannot contain z, contradiction.  ≥ 3, If J ∈ (Chev(2) − Chev(p)) ∩ Cp , then p ≤ 5. For p = 5, as mp (Aut(J)) 5 2 # ∼   we have J = F4 (2 2 ) and some element of B is a field automorphism. But then  ∩ J,  of by the Borel-Tits theorem, some parabolic subgroup of J must contain B 5-rank 2. However, this is not the case. Therefore p = 3 if J ∈ Chev(2). Again  must normalize P. This rules out Sp6 (2), for some parabolic subgroup P of J, B 1 3 2  D4 (2), F4 (2 2 ) , and Sp4 (8). Hence, (a) holds unless J ∼ = F4 (2). But in that case  2) with Z(O2 (P)) ∼ O2 (P) ∈ I∗X (B; = E28 , contrary to assumption. Thus, (a) holds  2). in this case, and (b) and (c) follow easily as O2 (P) ∈ I∗X (B;  ≥ 3, By definition of Cp , it remains to consider the case J ∈ Spor. As m2,p (X)    z )), which is either p = 3 or (J, p) = (F1 , 5). Since B acts faithfully on R = O2 (CJ(  of symplectic type, we see from involution centralizer structure in J [IA , 5.3] that (a), (b), and (c) hold. Notice that the possibility J ∼ = F i22 is ruled out by the  = 1. The proof is complete.  condition that m2 (CR (B)) 10.3. Other Involution Centralizers. Lemma 10.63. Suppose K = D4 (2), t ∈ I2 (Aut(K)), and CK (t) contains a copy of U4 (2). Then t is a graph automorphism and CK (t) ∼ = Sp6 (2). Proof. If the lemma fails, then by [IA , 4.9.2f] and the Borel-Tits theorem, CK (t) ≤ P for some parabolic subgroup P of K. Therefore P is nonsolvable and m3 (P ) ≥ m3 (U4 (2)) = 3. But K has no such parabolic subgroups. The lemma is proved.  Lemma 10.64. Let t ∈ ΦK be a graph automorphism of K = U4 (2) of order 2. Then CK (t) ∼ = Σ6 and some involution of CK (t) − O 2 (CK (t)) is 2-central in K. ∼ Sp4 (2) = ∼ Σ6 , and a long root Proof. From [IA , 4.9.2abd], we have CK (t) = involution z ∈ K is a long root element in CK (t). As root involutions in Sp4 (2) lie outside the commutator subgroup, and as z is 2-central in K, the lemma follows. 

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Lemma 10.65. Let K = D4 (2) or U6 (2), and let t ∈ I2 (Aut(K)) be a graph automorphism such that K0 := CK (t) ∼ = Sp6 (2). Let z be a 2-central (high root) involution of K which is also a 2-central (high long root) involution of K0 ∼ = B3 (2) or C3 (2), according as K = D4 (2) or U6 (2). Put R = O2 (CK (z)) and R0 = O2 (CK0 (z)). Also let E ∈ Syl3 (CK0 (z)). Then the following conditions hold: (a) R0 = K0 ∩ R; and (b) [R, E] = R. Proof. We may assume t ∈ ΓK or t ∈ ΦK , in the respectiive cases, by [IA , 4.9.2]. Set C = CK (z) and C0 = CK0 (z). Thus C is t-invariant and C0 = CC (t). Moreover, there is a t-invariant Levi decomposition C = RL. We shall regard C as C3 (2) in both cases, so that according as K ∼ = D4 (2) or U6 (2), z is a high short or high long root involution of K0 . Clearly R0 ≥ K0 ∩R. To prove the reverse containment, it suffices to show that O2 (CL (t)) = 1. But in the D4 (2) case, L ∼ = Σ3 × Σ3 × Σ3 with t interchanging two factors and centralizing the third, so CL (t) ∼ = Σ3 ×Σ3 . In the U6 (2) case, L ∼ = U4 (2) ∼ and CL (t) = Sp4 (2). Hence, (a) holds. If K ∼ = D4 (2), then the 3-element of each direct factor of L is fixed-point-free on R, and one such element lies in E, so [R, E] = R. If K ∼ = U6 (2), then R/ z is a natural module for L ∼ = U4 (2). Now NCL (t) (E) contains a Frobenius group EH with complement H ∼ = Z4 . It follows that EH acts irreducibly on R/ z as F4 -module. Therefore by Clifford’s theorem, CR/z (E) = 1, or equivalently [R, E] = R. The proof is complete.  Lemma 10.66. Let K ∈ C2 , K  X, and t ∈ I2 (X). Suppose that L ∼ = SL2 (3) is a solvable 2-component of CX (t) and that O2 (L) ≤ K. If t centralizes a Sylow 2-subgroup of K, then Z(L) ≤ Z(K). Proof. Suppose false and let K = K/Z(K). If K ∈ Chev(3) − Chev(2), then K∼ = 2U4 (3). But then all involutions of L/Z(K) split over Z(K) by Lemma 10.1g, contradiction. If K ∈ Spor, then from [IA , 5.3] we see that as |Z(K)| is even, there is no 2-central automorphism of K whose centralizer contains a subnormal A4 -subgroup. Thus K ∈ Chev(2), whence by [IA , 6.1.4] and the definition of C2 3 [V3 , 1.1], K ∼ = L3 (4), 2B2 (2 2 ), G2 (4), Sp6 (2), D4 (2), U6 (2), F4 (2), or 2 E6 (2). As t centralizes a Sylow 2-subgroup of K, we obtain a contradiction from [III11 , 13.1].  Lemma 10.67. Suppose that K = Ω+ 8 (2) and t ∈ I2 (Aut(K)). Then the following conditions hold: (a) If CK (t) contains a subgroup L ∼ = L4 (2), then t ∈ Inn(K) and CK (t) ∼ = Sp6 (2); and (b) |CK (t)|2 divides |Sp6 (2)|2 . Proof. If t ∈ Inn(K) in (a), then by the Borel-Tits theorem, CK (t) ≤ P for some parabolic subgroup, and the only possibility is that P/O2 (P ) ∼ = L4 (2). But then O2 (P ) is a natural L4 (2) ∼ (2)-module and so t ∈ Z(P ) = 1, a = Ω+ 6 contradiction. Therefore we may assume that t ∈ Inn(K). Hence we may identify K t with O(V ) for an 8-dimensional orthogonal F2 -space V of + type. Let W be the space of singular vectors in W1 := Rad(CV (t)). Note that W1 = 0, for otherwise V = CV (t) ⊥ X for some X = 0 and so CX (t) = 0, a contradiction.

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Also [V, t] ⊆ W1 , because for any w ∈ CV (t) and v ∈ V , (v + v t , w) = (v, w) + (v t , w) = (v, w)+(v t , wt ) = 0. If W = 0, then W1 = F2 v where v is nonsingular, and t is a transvection, whence CK (t) ∼ = Sp6 (2). Both parts of the lemma hold in this case, so we may assume that W = 0, and hence CK (t) ≤ NK (W ) =: P , a t-invariant parabolic subgroup of K. In (a), the only choice is that P = O2 (P )L, whence P t /O2 (P ) ∼ = Σ8 , a contradiction as [L, t] = 1. In (b), we see by inspection that |P |2 divides |Sp6 (2)|, so |CK (t)|2 divides |Sp6 (2)|2 . The proof is complete.   = L/Z(L). Let t ∈ Lemma 10.68. Let L ∈ C2 with |Z(L)| = 2. Set L I2 (Aut(L)) and assume the following conditions:  ≤ 4; (a) m3 (L) (b) F ∗ (CL (t)) has an elementary abelian 2-subgroup of index at most 2; and (c) CL (t) has at most one composition factor which is not a 2-group or 3group, and if it exists, it is isomorphic to U4 (2). ∼ Then L = G2 (4), Sp6 (2), D4 (2), U6 (2), F4 (2), or U4 (3). Proof. Using [IA , 6.1.4] and the definition of C2 [V3 , 1.1], as well as condition (a) (with the help of [IA , 4.10.3, 5.6.1]), we find that we must rule out the following  possibilities for L: (10G)

2

3

B2 (2 2 ), L3 (4), M12 , M22 , J2 , HS, and Ru.

Condition (c) rules out the last three of these, and limits us to the case that  or L ∼ t is 2-central in Aut(L), = U3 (2). (Use [IA , 4.9.1, 4.9.2, = L3 (4) with CL (t) ∼ 5.3bcgmr].) The latter is out as F ∗ (CL (t)) is not a 2-group. Condition (b) then gives a contradiction by the comparison m2 (F ∗ (CL (t))) + 1 < log2 (|F ∗ (CL (t))|).  = 3, 4, 3, and 4, respectively, while |F ∗ (C  (t))| = 26 , 26 , 25 , 26 , Indeed, m2 (L) L respectively. This completes the proof.  Lemma 10.69. Let X = H t with H ∈ C2 , Z(H) = 1, and t2 = 1. Then |CX (t)|2 > 8. Proof. Set X = X/CX (H) ≤ Aut(H). By [IG , 9.16], it suffices to show that |CX (t)|2 ≥ 24 . This can be checked using [IA , 4.5.1] if H ∼ = U4 (3) and [IA , 5.3] if H ∈ Spor, unless H ∼ = 2M12 and t induces an outer automorphism on H. Now suppose that H ∈ Chev(2), and let T ∈ Syl2 (X) with t ∈ T . In the cases H ∼ = Sp6 (2), D4 (2), U6 (2), T has a normal elementary abelian subgroup of order at least 26 , visible in a suitable parabolic subgroup, and this yields the desired inequality. The same holds if H ∼ = F4 (2) and t induces an inner automorphism. 2 1+20 If H ∼ E (2), T has a normal 2 -subgroup, whose Frattini quotient yields the = 6 ∼ desired inequality. If H = G2 (4), the product of the three highest root groups 3 is isomorphic to E26 , while if H ∼ = 2B2 (2 2 ), t is 2-central. Finally suppose that H ∼ = L3 (4), so that every inner involution is 2-central. Using [IA , 4.9.1, 4.9.2] we obtain the desired inequality unless t is a graph automorphism and CH (t) ∼ = L2 (4). This case, together with the outer involution cases for H ∼ = M12 and F4 (2), still need to be treated. But as Z(H) = 1, |T | > |L2 (4)|2 | t | = 8 in the L3 (4) and M12 1 cases, while |T | > |2F4 (2 2 )|2 > 8 by [IA , 4.9.1] in the F4 (2) case. This completes the proof. 

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Lemma 10.70. Let K ∈ C2 and t ∈ I2 (Aut(K)). Suppose that

(a) CK (t) has a normal subgroup E1 H1 with E1  E1 H1 . Moreover, E1 ∼ = E28 and H1 ∼ = U4 (2) or Sp6 (2) acting via the natural module or a spin module, respectively, or E1 ∼ = E26 and H1 ∼ = SU3 (2) acting naturally; and (b) CK (t) contains a hyperplane of some element of E3∗ (K).

Then there is no covering group L of K such that L ∈ C2 , |Z(L)| = 2, and the preimage of E1 in L is extraspecial. Proof. Suppose that such a covering group L exists. Then K ∼  L and as = |Z(L)| = 2, K is simple. Since CE1 (H1 ) = 1 and [t, E1 ] = 1, our hypotheses imply that m2 (K) ≥ 8 or 6, as the case may be, with strict inequality if Out(K) has odd order. Moreover, m2,3 (Aut(K)) ≥ m3 (K) − 1. If K ∈ Spor, these conditions imply, with [IA , Tables 5.6.1, 5.6.2], that K ∼ = F i22 or F2 , and by (b) and [IA , 5.3tuvy], t is of class 2A or 2D; or 2A, respectively. In every case |O2 (CK (t))| < 26 , contradiction. Therefore, K ∈ Spor. If K ∈ Chev(3) − Chev(2), then K ∼ = U4 (3) as K ∈ C2 , and m2 (K) = 4 < 6, again a contradiction. As L ∈ C2 , it follows that K ∈ Chev(2). We claim that K ∼ = U6 (2), F4 (2), or 2E6 (2). If m2 (K) ≥ 8, then as |Z(L)| = 2, this follows directly from [IA , 6.1.4, 3.3.3]. Otherwise there is a parabolic subgroup P ≤ K with E1 ≤ O2 (P ) and NAut(K) (P )/O2 (P ) covering H1 ∼ = SU3 (2). As |Z(L)| = 2, we see from [IA , 6.1.4, 2.5.12] that | Out(K)|2 ≤ 4, so P covers 3 O3 (H1 ) ∼ = 31+2 . This rules out K ∼ = L3 (4), G2 (4), 2B2 (2 2 ), Sp6 (2), and D4 (2), none of which have a parabolic subgroup containing a nonabelian 3-group. This proves our claim. Next, we shall argue in each case that Either F ∗ (CK (t)) is simple, or t acts on K and L like a long root involution of K. Since F ∗ (CK (t)) is not simple by assumption, it will follow that CK (t) is a parabolic subgroup P of K of type U4 (2), Sp6 (2), or U6 (2), respectively, and in E1 ≤ O2 (P ) ∼ particular, by assumption, E1 H1  P . In the first case = 21+8 ,  clearly P ∼ contradicting E1 = E28 or E26 . In the second case, H1 = P , so E1 = O2 (P ), a contradiction as |E1 | ≤ 28 < |O2 (P )|. The third case is impossible as P has neither a composition factor isomorphic to U4 (2) or Sp6 (2), nor a chief factor of order 32 . Thus, the lemma will follow from (10H). To prove (10H), we first choose B ∈ E∗3 (K) such that A := CB (t) is a hyperplane of B. Expand B to P ∈ Syl3 (K). In every case (10H)

B = J(P ) = CAut(K) (B) and B is a maximal torus of K. By computing the centralizer of A in the overlying  algebraic group we see that O 2 (CK (A)), if nontrivial, is the commuting product of Lie components of level q(K) = 2. As B = J(P ) there can be at most one such Lie component and it must be L2 (2) or L3 (2). In any case, replacing t by a CK (A)-conjugate if necessary, we may assume that t normalizes B. Thus t induces a reflection on B. If K ∼ = = 2E6 (2), then by Lemma 9.15, we have AutAut(K) (B) ∼ Z2 ×W (E6 ) ∼ = O5 (3), with B as a natural module. As O5 (3) has exactly two classes of reflections, there are, up to conjugacy, just two possibilities for t. They both can occur: either t is conjugate into the group ΦK and is a graph automorphism, with CK (t) ∼ = F4 (2) of rank 4 [IA , 4.9.2], or t is a long root involution, with

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CK (t) = P a parabolic subgroup of type U6 (2). Hence, (10H) follows in this case. A similar argument applies in case K ∼ = U6 (2), where NKt (B) is an extension of − ∼ B by Z2 × Σ6 = O4 (3), and the two possible classes for t are represented by an element of ΦK and a long root involution. Finally, suppose that K ∼ = F4 (2). If 1 t ∈ Inn(K), then F ∗ (CK (t)) ∼ = 2F4 (2 2 ) by [IA , 4.9.1]. If t ∈ K, then a similar but somewhat different argument applies. Namely, NK (B) is an extension of B by W (F4 ) ∼ = O4+ (3), with two classes of reflections, and t is either a long root involution or its conjugate – a short root involution – under an outer automorphism of K. This completes the proof of the lemma.  Lemma 10.71. Let K ∈ C2 and t ∈ I2 (Aut(K)) with L := E(CK (t)) ∼ = A6 . Then (a) If K/Z(K) ∼ = L± 4 (3) or U5 (2), then K has no abelian subgroup of order 6 2 ; and (b) If K ∼ = L4 (2), L5 (2), or P Sp4 (3), then no involution of L is 2-central in K. Proof. Suppose that K := K/Z(K) ∼ = L± 4 (3). It suffices to show that K has 5 no abelian subgroup of order 2 . Suppose that such a subgroup A exists. But by Lemmas 10.3d and 10.12b, there is E24 ∼ = A6 = F = CK (F ) such that AutK (F ) ∼ or Σ5 . Thus N := NK (F ) satisfies |N |2 = 27 = |K|2 so we may assume that A ≤ N . As N/F has D8 Sylow 2-subgroups and F = CN (F ), |A ∩ F | = 23 and |A/A∩F | = 4. Therefore there is a ∈ A mapping onto an involution of [N/F, N/F ]. Such an involution inverts an element of [N/F, N/F ] of order 5 so a acts freely on F . But then A ∩ F ≤ CF (a) ∼ = E22 , a contradiction.  The argument if K ∼ = U5 (2) is similar, using N := O 2 (CK (z)) for a 2-central ∗ involution z. Then N = QL with Q = F (N ) ∼ = Q8 ∗ Q8 ∗ Q8 and L ∼ = SU3 (2), acting faithfully via its natural F4 -module on Q/ z. We assume that A is abelian of order 26 and A ≤ N and derive a contradiction. Clearly |A ∩ Q| ≤ 24 , and since L has Q8 Sylow 2-subgroups, |A ∩ Q| = 24 with A/A ∩ Q ∼ = Z4 . Let a ∈ A map on a generator of A/A ∩ Q. Then a2 ∈ Q, so dimF4 (CQ/z (a)) = 1. This implies that |(A ∩ Q)/ z | ≤ 4, a contradiction. Hence (a) is proved. If K = Ln (2), n = 4 or 5, in (b), then by [IA , 4.9.1, 4.9.2beg] and the Borel-Tits theorem, we may take t ∈ ΓK . Thus in matrix terms, g t = J(g T )−1 J −1 , where Jij = δi+j,n . There is an embedding L4 (2) → L5 (2) obtained by adding a new basis vector as the third vector in the new basis. This embedding intertwines the two graph automorphisms, so it suffices to show that when n = 4, [CK (t), CK (t)] contains no transvection. But if we interpret L4 (2) as A8 , what we must show is that if t is a transposition, then the root A6 centralizing t contains no involution that is fixed-point-free on 8 letters, This is trivially true.  = Sp4 (3). By [IA , 4.5.2], C  (t) ∼ Finally suppose that K ∼ = P Sp4 (3). Let K = K  SL2 (9) so involutions of L do not split over Z(K). But again using [IA , 4.5.2], we  The lemma follows. see that 2-central involutions of K do split over Z(K)  Lemma 10.72. Let K = P Sp4 (3) or L4 (4), K  X and t ∈ I2 (X). Then m2 (t CK (t)) ≥ 4.

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Proof. Without loss X = K t ≤ Aut(K). Suppose K = P Sp4 (3); then we identify Aut(K) with SO5 (3) and let V be its natural module. Then V = [V, t] ⊥ CV (t) and there exists an ordered orthogonal frame F in V preserved by t. The stabilizer of F in SO5 (3) is isomorphic to E24 and contains t, so m2 (t CK (t)) ≥ 4. Suppose that K ∼ = L4 (4) and let V be its natural module. If t ∈ K then we see from the Jordan canonical form of t that t stabilizes a chain V > W > 0 with dimF4 (W ) = 2. As the full stabilizer of this chain is isomorphic to E28 , we are done in this case. If t ∈ K then by [IA , 4.9.1, 4.9.2beg], CK (t) ∼ = L4 (2) or U4 (2), or a Sylow 2-subgroup of CK (t) is isomorphic to one of Sp4 (4). In all cases  m2 (CK (t)) ≥ 4 by [IA , 3.3.3]. Lemma 10.73. Let K = G2 (4) and let x ∈ I2 (Aut(K)). If CK (x) is solvable, then CK (x) is a 3 -group. ∼ G2 (2) Proof. If x ∈ Inn(K), then x is a field automorphism and CK (x) = is nonsolvable [IA , 4.9.1]. So x ∈ Inn(K), and we regard x ∈ K. Suppose for a contradiction that y ∈ I3 (CK (x)). Then y lies in a Cartan subgroup of K,  and x ∈ L := O 2 (CK (y)) ∼ = SL3 (4) or L2 (4) [IA , 4.7.3A]. In the first case all involutions of L are conjugate and they are long root elements of K. In the second case the involutions of L are short root elements. Since K contains H1 × H2 with H1 ∼ = H2 ∼ = L2 (4), one a long root L2 (4) and the other a short root L2 (4), CK (x) contains L2 (4) in either case, a contradiction. The lemma is proved.  10.4. Other. Lemma 10.74. Let K = Suz, Co1 , F i22 , F i23 , F i24 , or F2 , and let u ∈ I2 (Inn(K)) with H   CK (u) and H ∼ = L3 (4), G2 (4), 2U6 (2), 2F i22 , 2F i22 , or 22E6 (2), respectively. Then the following conditions hold: (a) Let T ∈ Syl2 (CAut(K) (H)). Then T is elementary abelian of order 4, 4, 2, 2, 4, or 2, respectively; (b) If K ∼ = Suz, Co1 , or F i24 , then there is an involution g ∈ CInn(K) (u) such that [T,  g] = u; (c) Let N = H N be a product of components of some K-group X, and suppose that t ∈ I2 (Aut(N )) with H   CN (t). Then CAut(N ) (H) is 2-closed; and (d) Let X be a group with a component H0 ∼ = H and such that Z(X) has even order. Then for any t ∈ I2 (X), |CX (t)|2 > 8. Proof. The tables [IA , 5.3oltuvy] establish (a) and (b) directly. In (c), if N is not a single component, it is the product of two t-conjugate components by L2 -balance, and |CAut(N ) (H)| = 2 by [IG , 6.19]. Without loss, we may then assume that N ∈ K2 and H ↑2 N . Suppose first that N/Z(N ) ∼ = K. If K is as in (b), then (c) follows by the information in [IA , 5.3]. Otherwise, Z(H) ∼ = Z2 and CAut(K) (H) = CC (H) where C = CAut(K) (Z(H)); and (c) again holds by [IA , 5.3]. Suppose next that N/Z(N ) ∼  K. Then K ∈ Spor. By Lemma 3.1, the only = possibilities are H ∼ = L3 (42 ) and G2 (42 ), respectively. = L3 (4) and G2 (4), with N ∼ Then |CAut(N ) (H)| = 2 by [IA , 7.1.4c]. This completes the proof of (c). For (d) we quote Lemma 10.69 if Z(H0 ) = 1. We may then assume that H0 ∼ = L3 (4) or G2 (4); but still Z(X) has even order, so it suffices to prove that |CH0 (t)|2 ≥ 4, with strict inequality if t induces an inner automorphism on H0 . But if this fails, then CH0 (t) ≤ 4 in the inner case, and CH0 t (t) ≤ 4 in the outer case.

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Either way, Sylow 2-subgroups of H0 are of maximal class by [IG , 10.24], which is absurd. The proof is complete.  Lemma 10.75. Let K = HS and let t ∈ I2 (Aut(K)). Suppose that U ∼ = Z4 × Z4 × Z4 is a t-invariant subgroup of K such that [U, t] = Ω1 (U ). Then t ∈ Inn(K). Proof. Suppose false; then we may take t ∈ I2 (K). By [IA , 5.3m] there is A ≤ K such that A = CK (Ω1 (A)) ∼ = Z4 × Z4 × Z4 , and N/A ∼ = L3 (2), where N = NK (A). As |K : N | is odd we may assume that U t ≤ N . If Ω1 (U ) ≤ A, then Ω1 (U ) = Ω1 (A). But t centralizes U/Ω1 (U ), so it centralizes 1 (U ) = Ω1 (U ) = Ω1 (A), whence t ∈ A. This implies that m2 (A) > 3, a contradiction. Therefore Ω1 (U ) ≤ A. As N := N/A has dihedral Sylow 2-subgroups it follows that U ∼ = Z4 and U ∩ A ∼ = Z4 × Z4 . Write U = u with u ∈ U . Then U ∩ A ≤ CA (u), so as m2 (U ∩ A) = 2, u2 centralizes Ω1 (A). Hence u2 ∈ A and u2 = 1, contradiction. The lemma is proved.  ∼ L± (3) or 2U4 (3). Let t1 ∈ I2 (X) with t1 Lemma 10.76. Let K  X with K = 4 inducing a graph automorphism on K and E(CK (t1 )) ∼ = A6 . Let z be a 2-central involution of K such that z ∈ O2 (O 2 (CK (z))). Then there exists z1 ∈ z K ∩ CK (t1 ) such that t1 ∼K t1 z1 . Proof. Let V0 be a nondegenerate 3-dimensional orthogonal space over F9 , 2 with bilinear form v, w, v, w ∈ V0 . Let α ∈ F× 9 with α = −1 and set X0 = y O(V0 ) where yv = αv for all v ∈ V0 . Define (v, w) = TrF9 /F3 v, w to obtain a nondegenerate bilinear F3 -form on V , where V = V0 as a set but V is a F3 vector space by restriction of scalars from F9 . Let X = GO(V ). Notice that (yv, yw) = Tr αv, αw = Tr(− v, w) = −(v, w), so y induces a similarity on V and y 2 = −1. Thus there is an injection X0 → X and we consider X0 ≤ X.  := Ω(V ). As Ω(V0 ) ∼ Clearly O(V0 ) ≤ CX (y) and Ω(V0 ) ≤ CK (y) where K = ∼ Ω3 (9) = A6 , and Ω(V0 ) has trivial fixed subspace on V0 , we see from [IA , 4.5.1]  is a uniquely determined that the involutory automorphism induced by y on K  and E(C  (y)) ∼ graph automorphism up to conjugacy by Inndiag(K), = A6 . We K    may therefore identify K with K or K/Z(K), as the case may be, and identify t1 with the automorphism of K induced by y. We may write V0 = V1 ⊥ V2 ⊥ V3 where the Vi ’s are nonsingular lines, i = 1, 2, 3, and for i = 1, 2, Vi = F9 wi with (wi , wi ) = 1. By choosing V3 appropriately  Let z1 ∈ X0 be the involution inverting we can cover the two possibilities for K. V1 ⊥ V2 and centralizing V3 . Then as a subspace of V , V1 ⊥ V2 is 4-dimensional of + type, so z1 ∈ z K . It remains to show that t1 z1 = tg1 for some g ∈ K. This reduces to showing that for i = 1, 2, t1 ↓ Vi and −t1 ↓ Vi are conjugate in O(Vi ) by an element gi , when Vi is considered as an F3 -space (note that O(V1 ⊥ V2 )/Ω(V1 ⊥ V2 ) has exponent 2, so g := g1 g2 ∈ K). By symmetry it suffices to show this for i = 1. By our choice of V1 , O(V1 ), considering V1 as F3 -space, is isomorphic to O2− (3) ∼ = D8 . Hence |GO(V1 )| = 16. As GO(V1 ) embeds in GL2 (3), it is semidihedral of order 16, and hence t1 ↓ V1 , like every element of order 4, is real. This completes the proof.  Lemma 10.77. Let K ∈ C2 with K   X. Let J m3 (B), contrary to assumption. The lemma follows.  Lemma 11.8. Let L ∈ K3 with U4 (2) ↑3 L. Suppose K ↑2 L/O3 (L) and K ∼ = U6 (2), U5 (2), or D4 (2). Then L ∈ C3 . Proof. Since K ↑2 L/O3 (L), we see as usual from Lemma 3.1 and [IA , 2.2.10, 5.3] that L ∈ Chev(2) ∪ {(3)F i22 }, and then from [IA , 4.9.1, 4.9.2] and the BorelTits theorem that if L ∈ Chev(2), then L/O3 (L) ∼ = L6 (4), L5 (4), or D4 (4), none of which lie in C3 . Hence L/O3 (L) ∼ = F i22 . But by [IA , 5.3], ↓3 (F i22 ) = {3U4 (3)}  does not contain U4 (2), contrary to assumption. The proof is complete. ∼ U7 (2), 2E6 (2)a , or 2 D5 (2). Let L ∈ K3 and t ∈ Aut(L) Lemma 11.9. Let K = 2 with t = 1 and K   CL (t). Then L ∈ C3 . Proof. By Lemma 3.1 and [IA , 2.2.10, 5.3], L ∈ Chev(2). Then by the BorelTits theorem and [IA , 4.9.1, 4.9.2], L/Z(L) ∼ = L7 (4), E6 (4), or D5 (4). In any case  L ∈ C3 , as required.

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Lemma 11.10. Suppose that L ∈ K3 and 2 D4 (2) ↑2 L/O3 (L). Then L ∈ C3 . Proof. As usual, by Lemma 3.1, [IA , 2.2.10], and the tables [IA , 5.3], L ∈ Chev(2). Then L/O3 (L) ∼ = D4 (4) ∈ C3 , as desired, by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2].  Lemma 11.11. Let K = Suz, Co1 , F i22 , F i23 , F i24 , F2 , or F1 . Let b ∈ I3 (K) with I := E(CK (b)) ∼ = 3U4 (3), 3Suz, U4 (3), Ω7 (3), D4 (3), F i22 , or F i24 , respectively. Suppose that L ∈ K, t ∈ Aut(L) with t2 = 1, and K is a component of CL (t). If I is a component of CL (b), then t = 1 and L = K. Proof. Suppose that t is an involution. By Lemma 3.1, L ∈ Spor. Surveying [IA , 5.3], we see that the only possibility is K = F i23 and L ∼ = (3)F i24 . Then since CL (b) contains I ∼ = Ω7 (3) we see from [IA , 5.3v] that CL (b)(∞) ∼ = D4 (3). In particular, I is not a component of CL (b), contrary to assumption. The lemma follows.  Lemma 11.12. Let F ∗ (X) = K ∈ Chev(2) ∩ C2 with m2 (Z(K)) ≤ 1. Suppose X = KB with B ∈ E3∗ (X) and m3 (B) ≥ 4. Let b ∈ I3 (B ∩ K) and I = E(CK (b)), and suppose that I is a terminal component in K (relative to p = 3). Suppose that there is J ∈ C3 and t ∈ I2 (Aut(J)) such that CJ (t) has a 3-component I0 with I ∼ = (U6 (2), U4 (2)) or = I0 /W for some W ≤ O3 (I0 ). If B ≤ K, then (K, I) ∼ 2 ( E6 (2)a , D4 (2)). In the latter case, J/Z(J) ∼ = F i22 . Proof. By assumption there is a ∈ B − K. If a induces a graph or graphfield automorphism on K, then m3 (CK (a)) = 2 and K ∼ = D4 (q) or 3D4 (q) (see [IA , 4.7.3A]). In the D4 (q) case, m3 (K) = 4 and m3 (CX (a)) < m3 (X) = m3 (B), a contradiction, while in the 3D4 (q) case, m3 (X) = 3, contrary to assumption. Our argument shows that the subgroup B0 of B inducing inner-diagonal automorphisms on K has index 1 or 3 in B. Suppose that B = B0 , so that a induces a field automorphism on K. Let  q = q(O 2 (CK (a))), so that q(K) = q 3 , and as b ∈ K, q(I) = q 3k for some k ≥ 1, by [IA , 4.2.2]. Note that K is not a 3 -group, since m3 (B) ≥ 4; so I is not a 3 -group. If J ∈ Chev(3), then I/W ∈ Chev(3) ∩ Chev(2); the only possibility is I0 ∼ = L2 (8) (3). But then every element of B − B induces a nontrivial and J ∼ G = 2 0 √ field automorphism on I, and so as m3 (B) ≥ 4, m3 (CB0 (I)) > 1. Also q = 2 or 2, but 3 the latter is impossible as m3 (2F4 (2 2 )) = 2 [IA , 4.10.3]. Now if m3 (CB0 ∩K (I)) >  1, then by terminality in K, I  Γ2B0 ∩K,1 (K) = K as q(K) = 8 [IA , 7.3.3], a contradiction. So m3 (CB0 ∩K (I)) = 1. It follows that m3 (X) = m3 (B) = 4, and m3 (Inndiag(K)) = 3. But B0 ≤ K, so then K ∼ = U3n (8) or 2E6 (8), whence m3 (Inndiag(K)) = 3n − 1 or 6, a contradiction. Therefore J ∈ Chev(3). If J ∈ Chev(2) − Chev(3), then J is one of a few groups given by the definition of C3 [V3 , 1.1]; by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2] the only possibility 3 is J ∼ = 2B2 (2 2 ), a 3 -group, which we saw is impossible. There= Sp4 (8), whence I ∼ fore J ∈ Spor, but inspecting involution centralizers in [IA , 5.3], we see that it is impossible that q(I) be a perfect cube. Therefore, B = B0 . Thus, a induces an inner-diagonal but non-inner automorphism on K, whence K/Z(K) ∼ = L3k (q), k ≥ 2, or E6 (q). Here  = ±1 and q ≡  (mod 3). Note that if Z(K) = 1, then by [IA , 6.1.4] and our assumption, K ∼ = 2U6 (2) or 22E6 (2); but then by [IA , 6.3.1], elements of Inndiag(K/Z(K)) − Inn(K/Z(K)) do not lift

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to automorphisms of K, so the existence of a provides a contradiction. Hence, Z(K) = 1. In the linear-unitary case, since b ∈ B ∈ E3∗ (X), b acts on K like an element of GL3k (q) of order 3, and so I/O3 (I) ∼ = L3k−2 (q) or L3k−1 (q), in view of the terminality of I (see [IA , 4.8.4, 4.8.2]). In the E6 case, the facts that b ∈ B ∈ E3∗ (X) and I is terminal imply that I ∼ = L6 (q), D4 (q), or D5 (q) (see [IA , 4.7.3A]). If J ∈ Chev(3), so that I/W ∈ Chev(3), the only possibility is K ∼ = U6 (2) and (2), a desired conclusion. If J ∈ Chev(2) − Chev(3), then J is as in [V3 , I ∼ U = 4 1.1] and I/W is determined by [IA , 4.9.1, 4.9.2]. But then I/O3 (I) is not of one of the above types. Finally if J ∈ Spor, we again survey the involution centralizers in [IA , 5.3] and find that only I ∼ = D4 (2) is possible (with J ∼ = F i22 ). The proof is complete.  Lemma 11.13. Suppose that K ∈ C2 ∩ C3 is simple and m2 (CK (v)) = 1 for some v ∈ I2 (Aut(K)). Then K ∼ = A6 and K v ∼ = P GL2 (9). Proof. If v ∈ Inn(K), then m2 (K) = 1, which is absurd by the Z ∗ -theorem [IG , 15.3]. So v ∈ Inn(K). If K ∈ Spor, then a look at the tables [IA , 5.3] shows that m2 (CK (v)) > 1 for all v ∈ I2 (Aut(K) − Inn(K)). If K ∈ Alt, then K ∼  P GL2 (9), v is a field automorphism = A6 as K ∈ C2 ∩ C3 . Then if K v ∼ = and CK (v) ∼ = P GL2 (3), of 2-rank 2, contradiction. If K ∈ Chev(3) − Alt, then as ± K ∈ C2 , K ∼ = L2 (8), P Sp4 (3), L± 3 (3), L4 (3), or G2 (3). In the first case Out(K) has odd order; in the last case v must be a graph-field automorphism, whence m2 (CK (v)) = 3; and in the other cases, Aut(K) = Aut0 (K) and CK (v) can be found in [IA , 4.5.1], showing that m2 (CK (v)) > 1, contradiction. Finally if K ∈ Chev(2) − Chev(3) − Alt, then as K ∈ C3 and | Out(K)| is even, K ∼ = U5 (2), U6 (2), D4 (2), F4 (2), or Sp4 (8), by [V3 , 1.1] and [IA , 2.5.12, 3.3.2d]. In the last two cases by 1 3 [IA , 4.9.1, 3.3.3], m2 (CK (v)) = m2 (2F4 (2 2 )) = 5 or m2 (CK (v)) = m2 (2B2 (2 2 )) = 3. In the other cases, a v-invariant parabolic subgroup P is easily found such that  m2 (Z(O2 (P ))) ≥ 3, forcing m2 (CP (v)) > 1. The proof is complete. Lemma 11.14. Let X = KA z with K = F ∗ (KA) ∈ Cp − Chev(p) for some odd prime p, A ∼ = Epn for some n ≥ 3, z 2 = 1 and [A, z] = 1. Let L be a simple component of CK (z) and suppose that CA (L) = 1 and L ∈ Chev(2). Then p = 3 and L ∼ = U5 (2), U6 (2), D4 (2), F4 (2), Sp6 (2), 3D4 (2), or Sp4 (8); or p = 5 and 5 L=K ∼ = 2F4 (2 2 ). Moreover, if n ≥ 4, then p = 3 and L is one of the first four groups listed. Proof. First suppose that L = K. Then L ∈ Chev(2) ∩ Cp − Chev(p) and the possibilities for L, given that mp (Aut(L)) ≥ 3, can be read directly from the definition of Cp [V3 , 1.1] and our knowledge of ranks [IA , 2.5.12, 4.10.3a]. On the other hand, if L  CK (z) with z ∈ I2 (Aut(K)), we see the possibilities for K in  [V3 , 1.1], and the resulting possibilities for L from [IA , 4.9.1, 4.9.2, 5.3]. Lemma 11.15. Let K ∈ C2 ∩ C5 . Suppose that either m2 (Aut(K)) ≥ 6 or that 5 m5 (Aut(K)) ≥ 3. Then K ∼ = 2F4 (2 2 ), Co2 , Co1 , Ru, F2 , or F1 . Moreover for any involution t ∈ Aut(K), m2 (CK (t)) ≥ 3. Proof. If K ∈ C2 ∩ Chev(5), then K ∼ = A5 , which does not satisfy the hy5 5 potheses. Neither does 2B2 (2 2 ). Suppose that K ∈ C2 − Chev(5) ∩ C5 − {2B2 (2 2 )}. 5 Unless K ∼ = 2F4 (2 2 ), Out(K) is a 5 -group and K is one of the listed groups or

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F4 (2 2 ) , J2 , HS, or F3 . The first assertion then follows as | Out(K)| ≤ 2, and from the rank information in the tables [IA , 3.3.1, 5.6.1]. The final statement is 5 immediate from the tables [IA , 5.3] if K ∈ Spor; if K ∼ = 2F4 (2 2 ), then Out(K) = 1 and the 2-rank of a Sylow 2-center is 5. The proof is complete. 

2

1

Lemma 11.16. Suppose that K ∈ C2 ∩Cp for some odd prime p, and there exists a four-group E ≤ Aut(K) such that E ∩ Inn(K) = 1. Then p = 3 and K ∼ = L± 4 (3). Proof. Since K ∈ C2 ∩ Cp , K is simple. We may clearly assume that 4 divides | Out(K)|, so in particular K ∈ Spor. If K ∈ Alt the only possibility is K ∼ = A6 , and then no E exists, by [IIIK , 2.2]. So assume that K ∈ Chev(r) − Alt for some ± r. If r > 2, then as K ∈ C2 , K ∼ = L2 (8), L± 3 (3), G2 (3), P Sp4 (3), or L4 (3) (see [IA , 2.10]); only the last two groups have 4 dividing | Out(K)|, so the lemma holds in this case. Thus r = 2 and K is unambiguously in Chev(2). As K ∈ Cp , we 1 must have p = 3 with K ∼ = U5 (2), U6 (2), D4 (2), 3D4 (2), 2F4 (2 2 ) , F4 (2), Sp6 (2), or 5 1 5 Sp4 (8), or p = 5 with K ∼ = 2B2 (2 2 ) or 2F4 (2 2 ) or 2F4 (2 2 ). In no case does 4 divide | Out(K)|. The proof is complete.  12. {2, p}-Structure 12.1. Wide 2-Components. Lemma 12.1. Let X   K t  K ∈ C2 with t ∈ I2 (Z(X)), B ∈ Ep∗ (X) for some odd prime p, mp (B) ≥ 4, and CB (K) = 1. Suppose that m2 (CX (B)) ≤ 1. Then there is B ∗ ∈ Ep∗ (X) such that (a) CB ∗ (K) = 1; (b) IX (B; 2) = IX (B ∗ ; 2); (c) If p divides | Out(K)|, then B ∗ = B; and (d) One of the following holds: (1) There is b ∈ B ∗ ∩ K # and a component I of CK (b) which is terminal in K, a subgroup F of B ∗ containing b, and a component H of CI (F ) which is B ∗ -wide; or (2) p = 3 and K ∼ = L± 4 (3), 2U4 (3), G2 (3), Co1 , 2Co1 = Co0 , Suz, 2Suz, F i22 , 2F i22 , F i23 , F i24 , or F3 . Proof. Since K ∈ C2 , we may pass to X/O2 (X) and assume that O2 (X) = 1. By [IG , 8.7(iii)], B normalizes K. As B ∈ Ep∗ (X) and CB (K) = 1, CX (K) is a p group. Hence KB/O2 (K) embeds in Aut(K). If K is as in (d2), then B ∗ = B satisfies all the requirements. Assume henceforth that K is not as in (d2). If K ∈ Spor ∩ C2 , then since mp (Aut(K)) ≥ 4, K ∼ = Co2 , F2 , 2F2 , or F1 with p = 3, or F1 with p = 5. In these cases, we take F = b, H = I, B ∗ ∈ Ep∗ (CX (b)), and (I, mp (B ∗ /CB ∗ (H))) = (U4 (2), 3), (F i22 , 5), (2F i22 , 5), (3F i24 , 7), and (F5 , 3), respectively, using [IA , 5.3, 5.6.1]. Then the 2-rank of H t /D, for any subgroup D of O2 (Z(H)) t with cyclic quotient and t ∈ D, is at least 5, 11, 11, 12, and 7, respectively, by [IA , 5.6.1], so H is B ∗ -wide, and (d1) holds. In all these cases mp (B ∗ ) = mp (B) > m2,p (Aut(K)), so IX (B ∗ ; 2) and IX (B; 2) both consist of all 2-subgroups of CX (K). Hence all the required conditions are satisfied. As mp (Aut(K)) ≥ 4 and K is not as in (d2), we now have K ∈ Chev(2), but n n K ∼ = 3D4 (q), 2B2 (2 2 ) or 2F4 (2 2 ) for any q or n, as mp (Aut(3D4 (q))) ≤ 3 [IA , 2.5.12, a 4.10.3a]. We write q(K) = 2 . Since B ∈ Ep∗ (KB) and CB (K) = 1, no element of

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B # induces a graph or graph-field automorphism on K. Let X 0 be the preimage of Inndiag(K) in KB, and B 0 = B ∩ X 0 . Thus B = B 0 × f , where f = 1 or f has order p and induces a field automorphism on K. Moreover, either mp (X 0 ) > 3 or f = 1, in which case a ≥ 3. We shall take B ∗ = B. Suppose that K is a classical group. Choose b ∈ (B 0 )# to have as large as possible an eigenspace in an n-dimensional natural (projective) representation of K. Let I be the component of CK (b) supported on this eigenspace. Let d be the dimension of an orthogonally indecomposable subrepresentation of a subgroup of order p. We use [IA , 4.8.1, 4.8.2] and first assume that d > 1 and Z(K) = 1. In that case, we take H = I. As d > 1, p does not divide | Outdiag(K)|. Now by [IA , 4.10.3c], there is a unique elementary abelian p-subgroup of maximal rank in each Sylow p-subgroup of K, so B is determined up to conjugacy. We have mp (B 0 ) = [n/d] and m2 (I) ≥   a according as K is or is not of type A [IA , 3.3.3]. Hence, [(n − d)/2]2 a or (n−d)/2 2 as Z(K) = 1, if m2 (I t) ≥ mp (B/ b) + 2 fails, we have in the type A case [(n − d)/2]2 a ≤ mp (B) = [n/d] + mp (f ) ≤ [n/d] + 1 with [n/d] ≥ 3, and either strict inequality holds on the right or f = 1, in which   case p divides a ≥ 3. In the non-A case, the left side is to be replaced by (n−d)/2 a, 2 (n−d+2)/2 except that if K is of type C, we can use a. 2 Suppose that [n/d] = 3, so that n = 3d + k, 0 ≤ k < d. Then the right side is 4 and the A-case left side least d2 a ≥ 3d2 ≥ 12, contradiction. The  is at (n−d)/2 d+(k/2) a= a ≥ d(d − 1)a/2 ≥ d(d − 1)p/2. This non-A-case left side is 2 2 forces d = 2, n = 6, k = 0, and p = 3. Hence K ∼ = Sp6 (8), and so the left side is (n−d+2)/2 d+1 a ≥ a ≥ 9, a contradiction. 2 2 Thus [n/d] ≥ 4, so n = cd + k, c ≥ 4, 0 ≤ k < d. The right side is c + c , where c = 0 or 1 and a ≥ p if c = 1. In the A case, the left side is at least [(c − 1)d/2]2 a ≥ (c − 1)2 a ≥ (c − 1)2 > c + 1, contradiction. In the non-A case   the left side is at least (c−1)d/2 a ≥ (c − 1)(c − 2)a/2. If a > 1, then this yields 2 (c − 1)(c − 2) ≤ c + 1, contradiction. So a = 1 and c = 0. In these cases d is even [IA , 4.8.1]. If d ≥ 4, we get (2c − 2)(2c − 3)/2 ≤ c, contradiction. Thus d = 2, whence p = 3 (as a = 1), and K = Spn (2) or Ω± n (2). In the Spn (2) case we have n ≥ 8 and n(n − 2)/8 ≤ n/2, contradiction. In the Dn case, we have n ≥ 8 and (n−2)(n−4)/8 ≤ n/2, so n = 8. But then m2 (I) = 4, and 4+1 ≥ 3+2, as required. This completes the consideration of classical groups if d > 1 and Z(K) = 1. Still assuming Z(K) = 1, the remaining classical group case is d = 1, whence K∼ = Un (2), = Ln (q),  = 1, and q ≡  (mod p). Suppose first that q = 2. Then K ∼ n ≥ 5. In this case we can choose F so that m3 (B/F ) = 3 and E(CI (F )) ∼ = U4 (2). Then as m2 (U4 (2)) = 4, we have our desired inequality. Hence we may assume that q = 2a , a ≥ 2. We have m2 (I) ≥ m2 (Ln−2 (q)) = [(n − 2)/2]2 a if p divides n, and m2 (I) ≥ m2 (Ln−1 (q)) ≥ [(n − 1)/2]2 a if p does not divide n. Also mp (B) ≤ n ≥ 4, and if n = 4 then p divides a. If n = 4 and the desired inequaliy fails, then we get p ≤ a ≤ m2 (I) ≤ mp (B) = 4, so p = 3. If  = 1, then since p divides q −  = 2a − 1, a is even, so a = 6, contradiction. Therefore K = U4 (2a ), a odd. We may take F so that mp (B/F ) = 2 and E(CI (F )) ∼ = L2 (2a ) and the desired inequality is a + 1 ≥ 4,

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which holds. Now suppose that n ≥ 5. We then choose F so that mp (B/F ) = 3 and E(CI (F )) ∼ = L4 (q), of 2-rank 4a, as desired. To complete the classical case we now assume Z(K) = 1. As mp (B) ≥ 4, we have only K/Z(K) = D4 (2) and U6 (2) to consider [IA , 6.1.4]. In the U6 -case,  2 (K) by [IA , 6.4.4], so the above arguments go through. Likewise for m2 (K) = m K = D4 (2), B induces inner automorphisms on K as B ∈ E3∗ (X) by assumption, so [Z(K), B] = 1 and hence Z(K) is cyclic by assumption. Hence, m2 (K) = m2 (K/Z(K)) + 1 and m2 (I) = m2 (IZ(K)/Z(K)) + 1 = 5 for a suitable b ∈ I3 (B), while m3 (B/CB (I)) = 3, as desired. It remains to consider those K of exceptional Lie type. We use [IA , 3.3.3] for 2-ranks, [III11 , Table 13.1] for I’s, and [IA , 4.10.3a] for p-ranks. We choose F = b and choose I = H depending on the order m0 = ordp (q) and the cyclotomic polynomial Φm0 (q). Practically any choice of I is successful. Specifically, we can use the following triples (K, Φm0 (q), I): (E8 (q), Φ1 (±q), E7 (q)), (E8 (q), Φ3 (±q), E6± (q)), (E8 (q), Φ4 (q), D6− (q)), (E7 (q), Φ1 (±q), D6 (q)), (E7 (q), Φ3 (±q), 3D4 (q)), (E6 (q), Φ1 (q), A5 (q)), (E6 (q), Φ2 (q), A5 (q)), (E6 (q), Φ3 (q), 3D4 (q)), (F4 (q), Φ1 (±q), C3 (q)). It must be remarked that if K ∼ = 2F4 (2) and p = 3, then K contains C4 (2) and hence contains Z3 × Sp6 (2). Also, as U6 (2) ≤ 2E6 (2), m2 ([2 × 2]2E6 (2)) ≥ 11 by  [IA , 6.4.4]. This completes the proof of the lemma. Lemma 12.2. Let X = KB t  K ∈ C2 with t ∈ I2 (Z(X)), CB (K) = 1 and B ∈ Ep∗ (X) for some odd prime p. Then one of the following holds: (a) There exists B ∗ ∈ Ep∗ (X), F ≤ B ∗ , and a component H of CK (F ) such that H is B ∗ -wide and CB ∗ (K) = CB (K); (b) p = 5 and (K, mp (AutB (K))) = (2J2 , 2); (c) p = 3 and (K, mp (AutB (K))) = (A6 , 2), (L± 3 (3), 2), (M11 , 2), (F3 , 5), (2Sp6 (2), 3), (2Suz, 5), or (2J2 , 2); or (d) p = 3 and K/Z(K) ∼ = L± 4 (3) or G2 (3). Proof. If K ∼ = L2 (r) for some odd prime r, we take B ∗ = B, F = CB (K), and H = K in the definition of “B-wide” [V4 , 12.2], Since O2 (H) = 1, we must take D = 1. Then, as m2 (H t) = 3 and F is a hyperplane of B, the desired condition holds: (12A)

m2 (H t /D) ≥ mp (B/CB (H)) + 1, with strict inequality if D = 1.

Suppose next that K ∈ Spor. We again take B ∗ = B, F = CB (K), and H = K. By [IA , 6.1.4] and as K ∈ K2 , Z(K) is a cyclic 2-group. If Ω1 (O2 (K)) ≤ t, then we can only take D = 1 as D cannot contain t. Since Z(K) and Out(K) are 2-groups, (12A) certainly holds then if m2 (K t) > mp (K) + 1. This sufficient condition holds, by [IA , 5.6.1], unless (K, p) = (M11 , 3), (2J2 , 3), (2J2 , 5), (2Suz, 3), (F3 , 3), with mp (K) given by [IA , 5.6.1]. Assume then that Ω1 (O2 (K)) ≤ t, so there is more than one “test” subgroup D ≤ O2 (K) t with t ∈ D. As O2 (K) t is noncyclic, we must choose D = 1 as O2 (H) t /D is stipulated by definition to be cyclic. We need m2 (K t /D) ≥ mp (K) + 1. This holds, with the only possible exception being K ∼ = 2Suz and p = 3. By definition of C2 it remains to consider the case K ∈ Chev(2) ∩ K2 , except for 2L4 (2), 2U4 (2), SL2 (q), q ∈ FM9, and covering groups of L3 (4) with centers

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of exponent 4. Suppose first that K is simple, and we try B ∗ = B, F = CB (K), and H = K. Then the only test subgroup is D = 1, and we need m2 (K t) > mp (AutB (K)) + 1, or equivalently m2 (K) ≥ mp (AutB (K)) + 1. We use [IA , 3.3.3, 4.10.3a] to compute m2 (K) and bound AutB (K) above. Two of the cases in which K ∈ Lie(2) (K = Sp4 (2) , G2 (2) ) give rise to exceptions, as noted, with p = 3, so let us ignore them, as well as the solvable groups in Lie(2). If AutB (K) ≤ Inndiag(K) then mp (AutB (K)) ≤ , the untwisted Lie rank of K. The only cases in [IA , 3.3.3] in which m2 (K) ≤  are K = L3 (2), U3 (4), and U5 (2). If K = L3 (2) then K is B-wide since mp (Aut(K)) ≤ 1 for all odd p. If K = U3 (4) or U5 (2), then the only problem is for p = 5 or 3, respectively. We choose a different F , with |F : CB (K)| = p and H = E(CK (F )) ∼ = L2 (4) or U4 (2), respectively. Then in both cases m2 (H t) − mp (B/F ) = 2, so the desired inequality holds. On the other hand, if mp (AutB (K)) > mp (Inndiag(K)), then some b ∈ B induces a nontrivial field automorphism on K and mp (AutB (K)) ≤  + 1. But in [IA , 3.3.3], p now divides a, and the desired inequality is easily checked, except in one case: p = 3, K ∼ = L2 (8). Again, = U3 (8). This time choose F such that |B/F | = 9 and H ∼ m2 (H t) − m3 (B/F ) = 2, as desired. Finally, consider the cases in which Z(K) = 1. If K/Z(K) ∼ = L3 (4), then Z(K) is elementary abelian as K ∈ C2 . Hence by [IA , 6.4.4], m2 (K) − m2 (Z(K)) = 4 and the desired inequality holds as mp (Aut(K)) ≤ 2 for all odd primes p. As G2 (4) ≥ SL3 (4) [IA , 4.7.3A], and its automorphism group has p-rank at most 2 3 for all odd primes p, it too satisfies the desired inequality. If K/Z(K) ∼ = 2B2 (2 2 ), then m  2 (K) = 3 by Lemma 1.8, whereas mp (K) ≤ 1 for all odd p. Note that  2 (U6 (2)) = 9 and m  2 (F4 (2)) ≥ m  2 (Sp8 (2)) = 10 by [V2 , 4.2] and m  2 (2 E6 (2)) ≥ m [IA , 6.4.4, 6.1.4, 3.3.3], far exceeding the p-ranks of their automorphism groups. It remains to consider K = D4 (2). Since O2 (K) t /D is cyclic, we need only check that m2 (2D4 (2))−1 ≥ mp (Aut(D4 (2)))+2 for all odd primes p. But 2D4 (2) embeds with odd index in Ω+ 8 (3) by [IA , 6.2.2], so m2 (2D4 (2)) = 7. As mp (Aut(D4 (2))) ≤ 4  for all odd p (see [IA , 4.10.3a, 4.7.3A]), the proof is complete. 12.2. Subgroups of Order 2p. Lemma 12.3. Let K ∈ Spor ∪ Alt be simple and let p be an odd prime such that mp (K) > 1. Then K has an element of order 2p unless p = 3 and K ∼ = A6 . In any case m2,p (K) > 0. Proof. For K ∈ Spor, this is easily checked from the tables [IA , 5.3]. For K = An , we may assume (since A4 ≤ A6 ) that n ≥ 7. Since p + 4 ≤ max(7, 2p), some root involution of K centralizes some p-cycle. The lemma follows.  Lemma 12.4. Let p be an odd prime and K ∈ Cp . Suppose that K ∼ = L± 3 (p) 1 or that (p, K) = (3, 3G2 (3)) or (5, 2F4 (2 2 ) ). Then there is a 2-central involution z ∈ K and an element b ∈ Ip (K) − Z(K) such that [b, z] = 1. Proof. In 2F4 (2 2 ), every involution lies in 2F4 (2 2 ) , which is of index 2, and a 1 1 2-central involution of 2F4 (2 2 ) centralizes a parabolic subgroup of type 2B2 (2 2 ) ∼ = ± F5.4 , as desired. If K ∼ = L3 (p) or 3G2 (3), then by [IA , 4.5.1], K has one class of involutions, and every involution centralizes a GL2 (p) or GU2 (p) subgroup. In particular G2 (3) contains a Z2 × Σ3 subgroup, whose preimage in 3G2 (3) contains the desired element of order 3.  1

1

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2 2 Lemma 12.5. Suppose (p, K) = (3,L2 (8)), (3,A6 ), (3,L± 3 (3)), (5,A5 ), (5, B2 (2 )), 1 2  2 or (5, F4 (2 ) ). Let x ∈ Ip (Aut(K)). Then there is c ∈ Ip (CK (x)) such that NK (c) has even order. 5

Proof. If K ∼ = L± 3 (3), we take c ∈ CK (x) to be 3-central in K, so c lies in a SL2 (3)-subgroup of K and CK (c) has even order. In all the other cases, every  element of Ip (K) is real (see [IA , 4.8.7]). This completes the proof. Lemma 12.6. Let K be a K-group and let p be an odd prime. Under any of the following conditions, K contains a subgroup A × V with |A| = p and V ∼ = E22 : 3 ∼ (a) p = 3, K = Un (2), n ≥ 4, Sp6 (2), D4 (2), D4 (2), Sp4 (8), F4 (2), or Co2 ; or 5 (b) p = 5, K ∼ = 2F4 (2 2 ), Co1 , or F2 . In (a), if K = U4 (2) (with respect to the form I), we can take A =  Proof. −1 diag(ω, ω, ω , ω −1 ) , where ω 3 = 1 = ω. By [IA , 4.8.2, 4.7.3A], K = Sp6 (2), 3 D4 (2), and Sp4 (8) contain Z3 ×H with H ∼ = Sp4 (2), L2 (8), and L2 (8), respectively. 5 2 2 All the other groups in (a) contain U4 (2) ∼ = Ω− 6 (2), so (a) follows. In (b), F4 (2 ) 5 2 2 contains Z5 × B2 (2 ) by [IA , 4.8.7], and Co1 and F2 contain Z5 × A5 , by [IA , 5.3].  Lemma 12.7. Suppose that K ∈ Chev(2) is simple, and p is an odd prime dividing | Outdiag(K)|. Let a ∈ Ip (Inndiag(K))−Inn(K) and assume that |CK (a)| is odd. Then there exists b ∈ Ip (K) such that b is p-central in Aut(K) and b is real in K (i.e., conjugate to its inverse). Proof. Since CK (a) has odd order, it follows from [IA , 2.5.12, 4.7.3A] that K∼ = Lpk (2n ),  = ±1, 2n ≡  (mod p). For the same reason, we see by [IA , 4.8.2, 4.8.4] that k = 1. Let b = diag(1, ω, ω 2 , . . . , ω p−1 ) ∈ K (we take the unitary form to be I if  = −1). Here ω is a p-th root of unity in F2 . Then b is p-central in Aut(K) and b is inverted by an appropriate permutation matrix of order 2 in K, as required.   Lemma 12.8. Let K ∈ C2 ∩ (Spor ∪ Alt ∪ r>2 Chev(r)) with cyclic center, and let p = 3 or 5. Then for some x ∈ Ip (K) that is p-central in Aut(K), there is an involution t ∈ NK (x) − Z(K). ∼ An , n = 5, 6, or 8 and Proof. Let P ∈ Sylp (K). If K ∈ Alt, then K = the result is easily checked. If K ∼ = L2 (q), q ∈ FM9 − {5, 9}, or L2 (8), then P is cyclic and the result follows by Burnside’s transfer theorem. If K ∼ = L4 (3), U4 (3), 2U4 (3), P Sp4 (3), L± (3), or G (3), then a long root element of K/Z(K) of order 3 2 3 lies in an SL2 (3)-subgroup, proving the desired result for p = 3. For these groups, Sylow 5-subgroups are cyclic and every involution of K/Z(K) splits over Z(K), so Burnside’s theorem yields the desired conclusion. Suppose that K ∈ Spor. We use the tables [IA , 5.3] to check that in every case, x ∈ Ip (K) may be chosen so that |CK (x)|p = |K|p , and either T is not cyclic or dihedral for T ∈ Syl2 (NK/Z(K) (x Z(K)/Z(K)), or some involution of T splits over Z(K). Thus the preimage of T in K contains an involution and the desired conclusion holds. The proof is complete. 

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Lemma 12.9. Let K ∈ C2 be unambiguously in Chev(2) and suppose that Z(K) is cyclic and nontrivial. Let p = 3 divide |K| and assume that either mp (Aut(K)) > 2 or K ∼ = 2G2 (4). Then for some x ∈ Ip (K) that is p-central in Aut(K), there is an involution t ∈ CK (x) − Z(K). Proof. The conditions on Z(K) and mp (Aut(K)), together with [IA , 6.1.4, 4.10.3a] yield that K/Z(K) ∼ = Sp6 (2), D4 (2), U6 (2), F4 (2), 2 E6 (2), or G2 (4). Let x ∈ I3 (K) be 3-central. In every case, by [IA , 4.8.2, 4.7.3A], CK/Z(K) (x) = CK (x)/Z(K) contains a subgroup H/Z(K) ∼ = SU3 (2), or SL3 (4), which in turn contains SU3 (2). The involution of H then splits over Z(K) by [V2 , 4.3], as required.  Lemma 12.10. Let K ∈ C2 be unambiguously in Chev(2) and suppose that Z(K) is cyclic and nontrivial. Let p = 3 or 5 divide |K| and assume in the latter case that m5 (Aut(K)) > 1. Then for some x ∈ Ip (K) that is p-central in Aut(K), there is an involution t ∈ NK (x) − Z(K). Proof. Suppose first that K ∼ = 2L3 (4). Then the result holds since every element of I3 (L3 (4)) is real and all involutions of K split over Z(K) [IA , 6.1.4]. In view of Lemma 12.9 we may then assume that p = 5. By our assumptions and [IA , 6.1.4, 4.10.3a], K/Z(K) ∼ = G2 (4), D4 (2), F4 (2), or 2 E6 (2). Now, A5 × A5 ∼ = − − Ω4 (2) × Ω4 (2) ≤ D4 (2) ≤ F4 (2) ≤ 2E6 (2), and A5 × A5 embeds in G2 (4) as well, as the product of long and short root SL2 (4)’s. Hence for suitable x ∈ I5 (K), NK/Z(K) (x) contains E23 . Since Sylow 5-subgroups of Aut(K) are isomorphic to E52 , the lemma follows.   Lemma 12.11. Let K ∈ C2 ∩ (Spor ∪ Alt ∪ r>2 Chev(r)) with cyclic center, and let p = 3. Let P ∈ Syl3 (Aut(K)). Assume that m3 (P ) > 2. Then for some x ∈ I3 (K) centralized by Ω1 (P ), I2 (CK (x)) ⊆ Z(K). Proof. If K ∈ Alt ∪ Chev(r), r > 2, then as m3 (P ) > 2 and K ∈ C2 , K ∼ = L± 4 (3), 2U4 (3), P Sp4 (3), or G2 (3). As in the proof of Lemma 12.8, a long root SL2 (3)-subgroup contains the desired x and involution. So we may assume that K ∈ Spor and use the tables [IA , 5.3]. Again as m3 (P ) > 2, K/O2 (K) ∼ = J3 , Co2 , Co1 , Suz, F i22 , F i23 , F i24 , F3 , F2 , or F1 . If K ∼ = J3 , then Ω1 (P ) is abelian and there exist elements of order 6, as desired. In the other cases if Z(K) = 1, any 3-central x ∈ I3 (K) has |CK (x)| even, as required. The same argument handles K∼ = 2F i22 since in that case, all involutions of K/Z(K) split over Z(K). Finally, if K/Z(K) ∼ = Co1 , Suz, or F2 , and Z(K) = 1, then for any 3-central x ∈ I3 (K), a Sylow 2-subgroup T /Z(K) of CK/Z(K) (x) is neither cyclic nor dihedral, so T has an involution outside Z(K), completing the proof.  Lemma 12.12. Let K be one of the groups in [V4 , Lemma 16.4]. Let B ∈ E3∗ (K). Then there is b ∈ B # such that NK (b) − Z(K) contains an involution. Proof. This is obvious for K ∼ = L2 (17), L2 (8), or A6 . In most cases K is simple and NK (b) hass even order for a 3-central element b ∈ I3 (K), by one of [IA , 4.8.2, 4.7.3A, 5.3]. This applies to all the simple cases except L± 3 (3), for which a Borel subgroup contains the desired involution. For the nonsimple cases, |Z(K)| = 2, and it suffices to show that NK/Z(K) (b) contains either E23 or Q8 for a suitable b ∈ I3 (K), so that a Sylow 2-subgroup of NK (b) is neither cyclic nor quaternion. For K ∼ = 2U6 (2), 2Sp6 (2), 2HS, 22E6 (2), and 2F2 this applies

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∼ 2F i22 , K/Z(K) contains Ω7 (3), in which with b being 3-central. Finally if K = the normalizer of a high root group is a parabolic subgroup of type A1 × A1 , thus containing E23 or Q8 . The proof is complete.  Lemma 12.13. Let K = 2U6 (2) and let b ∈ I3 (K) be 3-central in Aut(K). Then there exists u ∈ I2 (K) − Z(K) such that [u, b] = 1. 

2 Proof. Let K = K/Z(K).  Then O (CK (b)) is a central product L1 ∗ L2 with L1 ∼ = L2 ∼ = SU3 (2) and b = Z(L1 ) = Z(L2 ). An involution u in L1 is a transvection, hence 2-central in K. Thus, u splits over Z(K) by [IA , 6.4.1]. The lemma follows. 

Lemma 12.14. Let K = U6 (2) or D4 (2) and let P ≤ K with |P | = 2n , n ≥ 8. Then there exists x ∈ I2 (P ) and y ∈ I3 (K) such that [x, y] = 1.  ∼ U4 (2). As Proof. There exists y ∈ I3 (K) such that Ky = O 2 (CK (y)) = 6 n+6 > |K|2 . This is the case for K = |Ky |2 = 2 , the result follows provided 2 D4 (2), so we may assume that K ∼ = U6 (2). Then for a 2-central involution z ∈ K, Q := O2 (CK (z)) ∼ = 21+8 . As |K|2 = 215 < |Q||P |, it suffices to show that for any u ∈ I2 (Q), CK (u) contains an element of order 3. Now we may identify AutK (Q) with U4 (2), and then as AutK (Q)-module, Q := Q/ z is the restriction to F2 of the natural 4-dimensional F4 -module V . Clearly z centralizes an element of AutK (Q) of order 3, so it suffices to show that every nonzero v ∈ V is centralized by some x ∈ I3 (Isom(V )) = SU (V ). By Witt’s Lemma the SU (V )-orbits on V # are represented by v0 and v1 , where (vi , vi ) = i ∈ F2 . In any case V contains a 2-dimensional nondegenerate subspace W orthogonal to vi , and CSU(V ) (W ⊥ ) ∼ = SU2 (2) ∼  = Σ3 centralizes vi . This completes the proof.

Lemma 12.15. Let K ∈ C2 ∩ C3 − {A6 , M11 , L2 (8)}. Then there exist b ∈ I3 (K) and a 2-subgroup U ≤ K such that [b, U ] = 1 and one of the following holds: (a) b is 3-central in Aut(K), and |U | ≥ 4; or (b) | Out(K)| ≤ 2; for some P ∈ Syl3 (K), b ∈ Z(J(P )); and U contains the exponent 2 subgroup of some Sylow 2-center in both K and Aut(K). Proof. If K ∼ = J3 or L± 3 (3), then we show that (b) holds. For all these groups, | Out(K)| = 2, and if Q ∈ Syl2 (Aut(K)), then Ω1 (Z(Q)) = Ω1 (Z(Q∩K)) ∼ = Z2 (see [IA , 2.5.12, 5.3, 4.5.1]). Moreover, in these cases, K has one class of involutions, so it suffices to show that for P ∈ Syl3 (K), I3 (Z(J(P ))) contains some b such that CK (b) has even order. If K ∼ = J3 , then Z(J(P )) = Ω1 (P ) contains some b with CK (b) containing A6 , hence of even order, as desired. Likewise if K ∼ = L± 3 (3), then and Z(P ) lies in an SL (3) subgroup of K. Hence (b) holds Z(J(P )) = Z(P ) ∼ Z = 3 2 in these cases. In all other cases (see Lemma 11.1), we show that (a) holds. In the sporadic cases, this is easily checked from [IA , 5.3]. In the classical group cases except Sp4 (8), a 3-central element exists centralizing an SU3 (2)-subgroup of K and hence a Q8 -subgroup, by [IA , 4.8.2]. If K is not classical, the desired statement holds by [IA , 4.7.3A]. Finally, if K ∼ = Sp4 (8), every element of I3 (K) centralizes an L2 (8)-subgroup, by [IA , 4.8.2]. The proof is complete.  Lemma 12.16. Let X be a group with O2 (X) = 1 and every component of E(X) a C2 -group. Let T ∈ Syl2 (X) and suppose that t ∈ Z(X) with t2 = 1 and

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t ∩ [T, T ] = 1. Let B ∈ E3∗ (X) with m3 (B) ≥ 2. Then one of the following holds: (a) There exists b ∈ B # and u ∈ I2 (X) such that u = t and [b, u] = 1; or (b) E(X) ∼ = A6 or L3 (q), q = 2a ≡  (mod 3). Proof. Let X be a minimal counterexample. Our assumptions imply that t ∩ [X, X] = 1. If O2 (X) = 1, then b and u obviously exists as B is noncyclic. So X has a simple component K ∈ C2 . Then B normalizes K with CB (K) = 1, for otherwise b and u obviously exist. Since CB (K) = 1 and B ∈ E3∗ (X), O3 (KB) = 1. Hence by minimality, X = KB. Since B ∈ E3∗ (X), B contains a 3-central element b of K. If K ∼ = L2 (q), q ∈ FM9, then as B is noncyclic, q = 9 and we are done. If ∼ K = L± (3), L± 3 4 (3), or G2 (3), then we may take b to be a long root element, whose centralizer has even order. If K ∈ Spor, then CK (b) has even order unless K ∼ = J3 , in which case B = Ω1 (P ) for some P ∈ Syl3 (K), and some x ∈ B has centralizer of even order. So we may assume that K ∈ Chev(2). By [IA , 7.3.3, 4.7.3A], CK (x) has even order for some x ∈ B # unless possibly (b) holds or K ∼ = U4 (2) or L2 (8). However, in the U4 (2) case B ≤ K acts diagonalizably on the natural module and so in fact CK (x) has even order for all x ∈ B # ; and in the L2 (8) case some x ∈ B # induces a field automorphism on K, so CK (x) has even order. The proof is complete.  1 Lemma 12.17. Let F ∗ (X) = K ∼ = 2F4 (2 2 ) , 3D4 (2), Sp6 (2), or F4 (2). Suppose that X = KA with A ∈ E3∗ (X). Then there is A1 ∈ E3 (X) such that m3 (A1 ) = m3 (A) − 1, [A1 , u] = 1 for some u ∈ I2 (K), and u = Ω1 (Z(T )) for any T ∈ Syl2 (K).

Proof. Except for the case K ∼ = 3D4 (2), Out(K) is a 3 -group [IA , 2.5.12], so 3 ∼ X = K. In case K = D4 (2) we may assume without loss that X = Aut(K) ∼ = K γ, where γ 3 = 1 and CX (γ) ∼ = G2 (2). In that last case γ centralizes some b ∈ I3 (K) of class t2 in [IA , 4.7.3A], so that Kb := E(CK (b)) ∼ = L2 (8) and γ induces a field automorphism on Kb . Thus A1 := b, γ centralizes an involution in a Sylow 2-subgroup of Kb , which is a short root involution. As m3 (Aut(K)) = 3 and a Sylow 2-center of K is a long root subgroup, the lemma holds in this case. 1 In case K ∼ = 2F4 (2 2 ) , the centralizer of a long root subgroup is a parabolic 1 subgroup of type 2B2 (2 2 ), and in particular this centralizer is a 3 -group. As m3 (K) = 2, we therefore need only find an element of order 6 in K, and indeed by [IA , 4.7.3A] the centralizer of an element of I3 (K) has even order, as required. For K = Sp6 (2) and F4 (2) the Sylow 2-centers are noncyclic, so that requirement about Ω1 (Z(T )) holds vacuously. As the 3-rank of K is 3 or 4, respectively, and F4 (2) contains Z3 × Sp6 (2) [IA , 4.7.3A], it suffices to consider the case K = Sp6 (2). But then K contains Z3 × Σ6 , so K contains E32 × Σ3 . The proof is complete.  Lemma 12.18. Let K ∈ C2 ∩ C3 . Then one of the following holds: (a) There exist a 3-central b ∈ I3 (K) and an involution u ∈ K such that [b, u] = 1 and m2 (CK (u)) ≥ 3; (b) K ∼ = J3 and there are b ∈ I3 (Z(J(P ))) for some P ∈ Syl3 (K), and u ∈ I2 (K), such that [b, u] = 1 and m2 (CK (u)) ≥ 3; or (c) K ∼ = A6 , M11 , L2 (8), or L± 3 (3).

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Proof. The set C2 ∩ C3 is enumerated in Lemma 11.1. Since K ∈ C2 ∩ C3 , K is simple. Suppose that K is not as in (c). Then by [IA , 3.3.3, 5.6.1], m2 (K) ≥ 3. Hence by [V2 , 6.4], m2 (CK (u)) ≥ 3 for all u ∈ I2 (K), and so it suffices to find a suitable b ∈ I3 (K) with |CK (b)| even. For K ∈ Spor, this is visible in [IA , 5.3]; note that for K = J3 and P ∈ Syl3 (K), Ω1 (P ) = J(P ) is abelian, so (b) holds. For K ∈ Chev(2), in fact, CK (b) has even order for all b ∈ I3 (K), as follows from [IA , 4.8.2, 4.8.4, 4.7.3A]. The lemma follows.  12.3. Components for Permutable Subgroups of Orders 2 and p. ∼ U6 (2), b ∈ I3 (K), and t∗ ∈ I2 (X) be such Lemma 12.19. Let K = F ∗ (X) = ∗ that [t , b] = 1, I := E(CK (b)) ∼ = U4 (2), and t∗ induces an outer automorphism on ∗ I. Then t induces a graph automorphism on K and CK (t∗ ) ∼ = Sp6 (2). Proof. Modifying t∗ by an element of I, if necessary, we may assume that CI (t∗ ) ∼ = Σ6 , by [IA , 4.9.2]. If t∗ induces an inner-diagonal automorphism on K,  then I t∗  = O 2 (CInndiag(K) (b)) is a product of Lie components. Hence it is a Lie component; but I t∗  ∼ = Aut(U4 (2)) ∈ Lie, a contradiction. Therefore, with  Sp6 (2), then [IA , 2.5.12], t∗ induces a graph automorphism on K. If CK (t∗ ) ∼ = CK (t∗ ) ≤ P for some parabolic subgroup P ≤ K, by [IA , 4.9.2f] and the Borel-Tits theorem. Hence Z3 × Σ6 ∼ = CbI (t∗ ) ≤ P , so m3 (P ) ≥ 3 and P contains Z3 × A6 . But K has no such parabolic subgroup, a contradiction. Thus, CK (t∗ ) ∼ = Sp6 (2), and the proof is complete.  Lemma 12.20. Let K = F i24 and a ∈ I3 (K) with H := E(CK (a)) ∼ = D4 (3). Let z be a 2-central involution of H. Then there is t ∈ I2 (CK (a, z)) such that I := E(CK (t)) ∼ = 2F i22 , t ∈ I0 := E(CI (a)) ∼ = 2U4 (3), and z is 2-central in I. Proof. Refer to [IA , 5.3v]. Choose t ∈ I2 (H) such that t ∈ I1 := E(CH (t)) ∼ = 2U4 (3) and z is 2-central in I1 . Then since CCG (t) (a) = CCG (a) (t) has the component I1 , t is of class 2A, i.e., I ∼ = 2F i22 . Then I0 = I1 . Finally, since z is 2-central in I1 and H, C := CH (t, z) satisfies z ∈ O 2 (C) ∼ = SL2 (3) ∗ SL2 (3). Hence in I = I/ t ∼ = F i22 , z is a Sylow 2-center. But since t is not 2-central in K, a Sylow 2-center of I covers one of I. Hence, z is 2-central in I, completing the proof.  Lemma 12.21. Let K = F i23 and b ∈ I3 (K) with I := E(CK (b)) ∼ = Ω7 (3). Let t1 ∈ I2 (I) be such that E(CI (t1 )) ∼ = 2U4 (3). Then CK (t1 ) ∼ = 2F i22 . Proof. We refer to [IA , 5.3u] without comment. Let H = E(CI (t1 )). Then CK (t1 ) ≥ NK (b) ∩ CK (t1 ) ∼ = Σ3 × H, so |CK (t1 )|3 ≥ 37 . This forces CK (t1 ) ∼ =  2F i22 , as claimed. Lemma 12.22. Let K ∈ K, t ∈ Aut(K) with t2 = 1, and Kt = E(CK (t)) ∼ = A5 or A7 . Let b ∈ I3 (Kt ) and assume that E(CK (b)) ∼ = A6 and b ∈ Syl3 (CK(E(CK (b)))). Then K ∼ = A7 , then K t ∼ = A9 , and if Kt ∼ = Σ9 . ∼ A9 . If, furtherProof. If K ∈ Alt, then the assumptions on CK (b) force K = more, Kt ∼ = Σ9 , as claimed. So we = A7 , then t acts as a transposition so K t ∼ may assume that K ∈ Alt, and aim for a contradiction. The structure of E(CK (b)) yields in any case that K ∈ Chev(2)∪Chev(3)∪Spor. We apply [III11 , 1.6ac], using the structure of Kt , and conclude that Kt ∼ = A5 with 4 K/Z(K) ∼ (4), L (2 ), M , J , or J . The sporadic cases are ruled out by = L± 2 12 1 2 3

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the structure of CK (b) [IA , 5.3bg], so K ∈ Chev(2). Since b ∈ Kt ≤ K, and E(CK (b)) ∼ = A6 ∼ = B2 (2) , it follows from [IA , 4.2.2] that the untwisted Dynkin diagram of K (which is A1 or A2 ) contains a B2 subdiagram. This contradiction finishes the proof.  ∼ Lemma 12.23. Let F ∗ (X) = K ∼ = P Ω+ 8 (3) and t ∈ I2 (X) with I := CK (t) = Ω7 (3). Let w ∈ I5 (I). Then the following conditions hold: (a) H := E(CK (w)) t ∼ = Σ6 ; and (b) There exists t0 ∈ tH − {t} ∩ CX (t). For any such t0 , tt0 ∈ I, and E(CI (tt0 )) ∼ = 2U4 (3). Proof. Let V be an 8-dimensional orthogonal space over F3 of + type and  = O(V ) and K  = Ω(V ). Then we may choose a covering K  → K and set L    a reflection t ∈ L whose action on K corresponds to the action of t on K. Let  of order 5. Since 5 does not I = E(CK t)). Let w  be a preimage of w in K  ( +  and V2 := CV (w)  are 4-dimensional of − type. Since divide |O4 (3)|, V1 := [V, w] w ∈ I, the support F3 vt of  t on V lies in V2 . Also, E(CK (w)) is the image of  ∼ E(CK = A6 , and H embeds in, and then is isomorphic to, O(V2 )/ ±1 ∼ = Σ6 .  (w)) u   such that vt ⊥ vt . Let u be the image in H of u  and There is u  ∈ E(CK  (w)) set t0 = tu . Then tt0 is the image in K of −tˆtˆuˆ , whose support on V is a hyperplane  The lemma follows. of − type in the support of I.  Lemma 12.24. Let K = [2 × 2]E6 (2), let w ∈ I5 (K), and set Cw = CK (w). Then the following conditions hold: (a) I5 (K) = wK ; (b) Cw = w × Z(K) × L, where L ∼ = A8 and a 2-central involution of L is 2-central in K; (c) CAut(K) (w) ∼ = Z5 × (Σ3 Y Σ8 ); and (d) For some 2-central involution z of L, z = [O2 (CK (z)), O2 (CK (z))]. Proof. Set K = K/Z(K). Part (a) holds by [IA , 4.8.6], which implies also that CK (w) ∼ = Z5 × A8 . By (a), CInndiag(K) (w) covers Outdiag(K), which acts fixed-point-freely on Z(K) by [IA , Table 6.3.1]. Hence as the Schur multiplier of A8 is only Z2 [IA , 6.1.4], E(CK (w)) ∼ = A8 . Next, let z ∈ K be 2-central and let C z = CK (z) = QM , where Q = F ∗ (C z ) ∼ and M ∼ = 21+20 = U6 (2). In particular 5 + divides |M L| so we may assume z chosen so that w ∈ L. By Lemma 10.54, Q/ z is an F2 -form of ∧3 V , where V is the natural U6 (2)-module. One calculates then  that CQ (w) ∼ = Q8 ∗ Q8 . It follows that z is 2-central in O 2 (CK (w)) ∼ = A8 . Now (b) follows. To show that a suitable preimage z of z satisfies (d) it is enough to show that Z(K) is a direct factor of Q, the full preimage of Q in K. Thus, we need to show that Z(K) ∩ Φ(Q) = 1. But Q is absolutely irreducible for M , so Q supports a uniqiue nonzero M -invariant bilinear form. This forces [Q, Q] ∼ = Z2 , and so by irreducibility againn, Φ(Q) ∼ = Z2 . As Φ(Q) covers Z(Q), Z(K) ∩ Φ(Q) = 1, and (d) is proved. Finally, Aut(K) = Inndiag(K) γ, where γ 2 = 1 and CK (γ) ∼ = F4 (2). By (a), replacing γ by an Inn(K)-conjugate, we may assume that w ∈ CK (γ). Then by  [IA , 4.8.6], O 2 (CK (γ, w)) ∼ = Σ6 . It follows that γ acts on E(CK (w)) ∼ = A8 as a transposition, and so (c) holds. The proof is complete. 

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Lemma 12.25. Let K = D4 (2) and let b ∈ I3 (K) be such that I := E(CK (b)) ∼ = U4 (2). Then there is an involution u ∈ NK (b) such that u inverts b and u maps onto Out(I). Proof. Applying a triality if necessary [IA , 4.7.3A, 4.8.2], we see that on some natural module V , dim[V, b] = 2 and I = CK ([V, b]) ∼ = Ω(CV (b)) ∼ = U4 (2). We can take u = u1 u2 , where u1 is a transvection on [V, b] centralizing CV (b), and u2 is a transvection on CV (b) centralizing [V, b].  Lemma 12.26. Let K ∈ C2 and u ∈ I2 (Aut(K)) with N := E(CK (u)) ∼ = Sp6 (2). Let b ∈ I3 (N ) with H := E(CN (b)) ∼ = A6 . Then H is not a component of CK (b). Proof. Using [IA , 4.9.1, 4.9.2, 5.3], we see that K/Z(K), as a pumpup of ± ∼ Sp6 (2) in C2 , is isomorphic to L± n (2), n = 6 or 7, D4 (2), or F i22 . If K = F i22 , there is no A6 -component in the centralizer of any automorphism of order 3 [IA , 5.3]. Moreover, as b ∈ N ≤ K, we see from [IA , 4.2.2] that if K ∈ Chev(2), then CK (b) has no component whose untwisted Dynkin diagram has a double bond. This proves the lemma.  Lemma 12.27. Let K ∈ K2 with K/Z(K) ∼ = U5 (2), U6 (2), Sp8 (2), or D4 (2). Let u ∈ I2 (Aut(K)) and b ∈ I3 (Aut(K)) with [u, b] = 1, and set I = E(CK (b)), N = E(CK (u)) and H = E(CI (u)). Suppose that b ∈ Inn(K) and I is terminal in K. If H ∼ = A6 , then K/Z(K) ∼ = U6 (2) or D4 (2) and N ∼ = Sp6 (2). Proof. Since b ∈ K, either I ∼ = Sp8 (2) only) I ∼ = Sp6 (2), = U4 (2) or (with K ∼  by [IA , 4.8.2], and by terminality, in any case, I = O 2 (CK (b)). If u ∈ Inn(K), then u induces an inner automorphism on I, and so by the Borel-Tits theorem, [I, u] = 1 and H = I, a contradiction as H ∼  I. Thus, with [IA , 2.5.12], u is = a graph automorphism of K. In particular K ∼  Sp8 (2). If K ∼ = U5 (2), then = CInn(K) (b) = b × I, and b is real in Aut(K) = Inn(K) u but not in Inn(K), so CAut(K) (b) = CInn(K) (b), a contradiction as [b, u] = 1. Thus K/Z(K) ∼ = U6 (2) or D4 (2), as claimed. Next, if N ∼  Sp6 (2), then by [IA , 4.9.2f], F ∗ (CK (u)) is a = 2-group, so CK (u) lies in a parabolic subgroup P of K by the Borel-Tits theorem. In particular P must contain a copy of Z3 ×A6 , so m3 (P ) ≥ 3 and P is nonsolvable. This forces K/Z(K) ∼ = U4 (2). But U4 (2) does not contain = U6 (2) and P/O2 (P ) ∼ Z3 × A6 , contradiction. The proof is complete.  Lemma 12.28. Let K = F1 , let z ∈ I2 (K) be 2-central, and let B ∈ E53 (CK (z)). Then there is a ∈ B # such that H = E(CK (a)) ∼ = F5 and H ≥ CO2 (CK (z)) (a) ∼ = 1+8 2+ . Proof. See [IA , 5.3wyz]. Let a ∈ I5 (K) with H := E(CK (a)) ∼ = F5 , and let z ∈ I2 (H) be 2-central. Then O2 (CCK (z) (a)) = O2 (CCK (a) (z)) ∼ = 21+8 + . If CK (z) ∼ = 2F2 , however, O2 (CCK (z) (a)) = 1, contradiction; so z is 2-central in K and O2 (CCK (z) (a)) ≤ O2 (CK (z)). The lemma follows.  Lemma 12.29. Let K ∈ Kp , x ∈ Ip (K), and Kx = E(CK (x)). Let z be a 2-central involution of Kx . Then z is 2-central in K, in the following cases: (K, p, Kx ) = (F2 , 3, F i22 ) or (F1 , 5, F5 ). Proof. In the second case the assertion follows from Lemma 12.28. In the first case, CCz (x) = CCx (z) ∼ O6− (2). From = x × CAut(F i22 ) (z) ∼ = Z3 × 21+10 +

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the centralizers of elements of order 3 in [IA , 5.3y], we find that there are three conjugacy classes in K of elements x of order 6 whose centralizers C satisfy C := − C/ x O2 (C) ∼ = Ω− 6 (2) or O6 (2); moreover, for only one of these classes is it true − that C ∼ = O6 (2) and O2 (C) is extraspecial. This must be the class of xz. Since the centralizer of an involution in class 2B in K contains such an element of order 6, z must therefore lie in class 2B, the 2-central class. The proof is complete.  Lemma 12.30. Let K = 2Sp6 (2), 22E6 (2), 2F i22 , or 2F2 . Then there exists B ∈ E3∗ (K) and b ∈ B # such that E = E(CK (b)) is quasisimple and Z(K) ≤ E. Proof. Let K = K/Z(K). Choose b ∈ I3 (K) with E := E(CK (b)) as follows: (K, E) = (Sp6 (2), A6 ), (2E6 (2), U6 (2)), (F i22 , U4 (3)), or (F2 , F i22 ). We see from [IA , 4.10.3, 5.6.1] that m3 (E) = m3 (K) − 1, and so since Z(E) = 1, b lies in some B ∈ E3∗ (K). It therefore suffices to show that the full preimage E of E is quasisimple. In the last two cases this fact is contained in [IA , 5.3ty]. In the first two cases it follows from the structure of the centralizer of an appropriate element  of order 6 in Co3 or F2 , respectively. Lemma 12.31. Let K = Suz and let x ∈ I3 (K) with I = E(CK (x)) ∼ = 3U4 (3). Let t ∈ I2 (NAut(K) (x)). Then CI (t) does not contain a copy of P Sp4 (3). Proof. If false, then by order considerations t would have to be in class 2A by P Sp4 (3), whereas [IA , 5.3o]. But then CK (t) would be a split extension of 21+6 −  by [IA , 5.3] it is a nonsplit extension. Lemma 12.32. Let K = Suz with x ∈ I3 (K) of class 3A (see [IA , 5.3]). Then involutions in E(CK (x)) are of class 2A. Moreover, there exists an involution u ∈ NK (x) of class 2B such that xu = x−1 and E(CE(CK (x)) (u)) ∼ = A6 . Proof. Let z ∈ I2 (E(CK (x))). Then CK (xz)/xz = CCK (x) (z)/xz is iso morphic to A4 × A4 . But if z were in class 2B, then O 3 (CK (z)) ∼ = L3 (4) and so CK (xz) would be 3-closed, a contradiction. Hence z is in class 2A. By [IA , 5.3], there is an involution u of class 2B inverting x, and lying outside P O6− (3), when E(CK (x))/ x is viewed as P Ω− 6 (3). Now the coset labelled g in [IA , 4.5.1] for the group A− (3) contains an involution γ1 which is a reflection, when interpreted as 3 (3), and hence lying in P O6− (3). Hence u must be in the an automorphism of P Ω− 6   coset labelled g , not g, and so CE(CK (x)) (u) ∼ = 2 D2 (3) ∼ = A6 , as claimed. Lemma 12.33. Let K  X = KCX (K) with K ∼ = Co1 , Suz, F i22 , 2F i22 , or F i23 , and CX (K) ∼ = Z2 . Then there exist b ∈ I3 (K) and u ∈ I2 (X − CX (K)) such that the following conditions hold: (a) m3 (CK (b)) = m3 (K); (b) I := E(CK (b)) is quasisimple; and (c) u normalizes b; (d) I0 := E(CI (u)) = 1 and I0 ≤ H = E(CK (u)) with H quasisimple. The quadruples (K, I, I0 , H) are (Co1 , 3Suz, J2 , G2 (4)), (Suz, 3U4 (3), A6 , L3 (4)), (F i22 , U4 (3), U4 (3), 2U6 (2)), (2F i22 , 2U4 (3), 2U4 (3), [2 × 2]U6 (2)), and (F i23 , Ω7 (3), Ω7 (3), 2F i22 ). Finally, in case K/Z(K) ∼ = F i22 , u ∈ H.

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Proof. Refer to [IA , 5.3]. For each K, choose b of class 3A. Then I has the asserted isomorphism type, and (a) follows in each case from [IA , 5.6.1, 3.3.3]. (If K = 2F i22 , then we know that I/Z(I) ∼ = U4 (3). But I = E(CF i23 (y)) for some ∼ y ∈ F i23 of order 6, and this leads to I ∼ = Ω− 6 (3) = 2U4 (3).) The information in [IA , 5.3] provides an involution u inverting b. If K ∼ = Aut(Suz) so u may be chosen so that I0 ∼ = = Co1 , then I u /Z(I) ∼ J2 (note that J2 is 3-saturated). The only involutions in K in this case whose centralizers contain a copy of J2 are those of type 2B. (J2 contains 3A6 , but no centralizer of an element of order 3 in Ω+ 8 (2) contains 3A6 , see [IA , 4.8.2].) Hence (d) holds in this case. If K ∼ = Suz, then by Lemma 12.32, u can be chosen to satisfy (d). If K/Z(K) ∼ = F i22 then by Note 3 of [IA , 5.3t], u can be chosen to satisfy (d), and then lies in I0 . Finally if K ∼ = F i23 , then I0 = I is embeddable only in the involution centralizer CK (u) ∼ = 2F i22 by consideration of orders, so again (d) holds. The proof is complete.  Lemma 12.34. Let K, X, b, u, I, I0 , and H be as in Lemma 12.33. Let L ∈ C2 and t ∈ I2 (Aut(L)) with E(CL (t)) having a component H ∗ isomorphic to H. Assume moreover that if u ∈ H, then u corresponds to an element of Z(L). Let I0∗ be the subgroup of H ∗ corresponding to I0 . Then (a) The pairs (K, L) are (Co1 , G2 (16) or Co1 ), (Suz, L3 (16) or Suz), (F i22 , F i22 ), (2F i22 , 2F i22 ), (F i23 , F i23 or F i24 ); (b) If K ∼ = Suz and L ∼ = F i24 , then I0∗ normalizes no 3-nilpotent subgroup R of L of odd order in such a way that [R/O3 (R), I0∗ ] is nontrivial; and (c) If K ∼ = Suz, then there is no 3-nilpotent subgroup R of L of odd order such that I0∗ t normalizes R and acts faithfully on R/O3 (R). Proof. We identify H ∗ with H and I0∗ with I0 . Our hypotheses imply that H ↑2 L and if H/Z(H) ∼ = U6 (2), then Z(H)∩Z(L) = 1. If L ∈ Spor, then consulting [IA , 5.3], we see that L is as claimed in (a); the condition on U6 (2) rules out L ∼ = F i23 or F i24 . If L ∈ Spor, then K ∼ = G2 (4) or L3 (4), so L ∈ Chev(2) by [III11 , 1.1ab]; then (a) holds by [IA , 4.9.1, 4.9.2] and the Borel-Tits theorem. Suppose that I0 (or I0 t, if K ∼ = Suz) normalizes R as in (b) or (c). If ∼ K = Co1 , then I0 ∼ = J2 . Then |L|3 /|J2 |3 = 1 or 36 according as L ∼ = G2 (16) or Co1 , so only the latter is possible, with |R|3 ≤ 36 . As J2 contains E52 , this is impossible. Similarly if K ∼ = F i22 , 2F i22 , or F i23 , then by (a) and our assumption, L ∼ = K so |R|3 ≤ |L|3 /|I0 |3 = 33 , 33 , or 34 , which is impossible as I0 contains elements of order 5, 5, or 7, respectively. And if K ∼ = Suz and L ∼ = L3 (16), then |R|3 ≤ |L|3 /|I0 |3 = 1, an impossibility. Thus (b) holds, and we are reduced to proving (c) when K ∼ =L∼ = Suz. ∼ Now E(CL (t)) = L3 (4) so |CL (t)|3 = 32 = |I0 |3 . Therefore CR (t) = 1, where we have put R = R/O3 (R). So R is abelian, indeed elementary abelian since 5 divides |I0 |. If |R| ≤ 34 , then equality holds and by [III17 , 5.1], R is the core of a natural permutation module for I0 . It follows that |CR (Q)| = 32 or 3 for Q ≤ I0 with |Q| = 3 or 32 , respectively. Hence in this case R is the unique subgroup of the semidirect product RI0 isomorphic to E34 . Whether or not |R| ≤ 34 , on the other hand, |L|3 = 37 and for S ∈ Syl3 (L), J(S) ∼ = E35 , by [IA , 5.3]. Let T ∈ Syl3 (RI0 ) and assume, as we may, that T ≤ S. It follows that T ∩ R ≤ J(S), so J(S) ≤ CL (T ∩ R), and then by a Frattini argument, AutL (T ∩ R) embeds in AutL (J(S)), which is isomorphic to M11 . Since M11 does not contain Z2 × A6

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(its involution centralizer is a {2, 3}-group) we have a contradiction. The lemma is proved.  Lemma 12.35. Let L ∈ {Sp4 (4), L4 (2), L5 (2), HS, 2HS}. Suppose L  X and t ∈ I2 (X) with E(CL (t)) ∼ = A6 . Then there exists x ∈ I3 (CL (t)) such that NLt (x) ≥ x, s × I t, where x, s ∼ = Σ3 , I ∼ = A5 , and I t ∼ = Σ5 . ∼ Proof. If L = Sp4 (4), then t induces a field automorphism on L, with CL (t) ≥ H1 × H2 , H1 ∼ = H2 ∼ = Sp2 (2) ∼ = Σ3 , and H1 × CL (H1 ) ∼ = Σ3 × L2 (4) and t inducing a field automorphism on CL (H1 ), so the result holds in this case. If L ∼ = L4 (2) ∼ = A8 , ∼ then t is a graph automorphism, L t = Σ8 , and the result holds with x a 3-cycle. If L ∼ = L5 (2), then again t is a graph automorphism and there exists L0 ≤ L with L0 ∼ = L4 (2) and t inducing a graph automorphism on L0 . The result then follows from the L4 (2) case. If L ∼ = HS, then t induces an inner automorphism of class 2B [IA , 5.3m], a class distinguished from class 2A by centralizers containing an E32 -subgroup. Hence taking any x ∈ I3 (L), we have NL (x) = x, s × I t  with x, s ∼ = Σ3 , I t  ∼ = Σ5 , and t of class 2B as CI (t ) ∼ = Σ3 . Replacing x, s, and I by g-conjugates for appropriate g ∈ L, we can achieve t = t, so the result follows in this case. Essentially the same argument works if L ∼ = 2HS; note that L t ∼ = 2HS ∗ Z4 in this case since involutions of class 2B do not split over Z(L). t and t have the form t0 z, where [L, z] = 1, t0 ∈ L, and Thus the  2involutions 2 t0 = z = Z(L).  12.4. 2-Local p-Rank. Lemma 12.36. Let K ∈ Cp for some odd prime p, and suppose that m2,p (K) = 5 mp (Z(K)). Then (K, p) = (L2 (pn ), p > 3), (Ap , p > 5), (L2 (8), 3), or (2B2 (2 2 ), 5). 5 Proof. If mp (K) = 1, then K ∼ = Ap , L2 (p) (p > 3), L2 (8) (p = 3), or 2B2 (2 2 ) (p = 5), as K ∈ Cp . So the lemma holds in this case. Assume then that mp (K) > 1. Suppose that K ∈ Chev(p). If K/Z(K) ∼ = L2 (pn ), n > 1, then we may assume n that p = 3, but then K contains Z(K)×A4 . If K ∼ = 2 G2 (3 2 ) , then p = 3 and n ≥ 3 as mp (K) > 1, so K contains L2 (3n ). Otherwise K contains some L ∼ = SL2 (p), so m2,p (K) > mp (Z(K)) unless possibly Z(K) = 1; thus by [IA , 6.1.4], and the fact that Ω7 (3) contains U4 (3), we may assume that p = 3 and K ∼ = 32 U4 (3) or 3G2 (3). But then K/Z(K) contains GU2 (3) or GL2 (3), so m2,3 (K) > m3 (Z(K)). Suppose that K ∈ Alt ∪ Spor. Then every element of Ip (K/Z(K)) splits over Z(K), so m2,p (K/Z(K)) = 0. Hence by Lemma 12.3, mp (K) = 1, contradiction. Finally suppose that K ∈ Chev − Chev(p) − Alt. Refer to the definition of Cp [V3 , 1.1]. Thus, K ∈ Chev(2); and either K is simple or p = 3 and K ∼ = SU6 (2). It is easy to find a parabolic subgroup witnessing m2,p (K) > mp (Z(K)) in each case. 

Lemma 12.37. Let J ∈ Cp , p odd. Suppose that K ∈ C2 (in particular O2 (K) = 1), and either J/Z ∼ = K or J/Z ↑p K/O2 (K) for some Z ≤ Z(J). Then the following hold: (a) If m2,p (J/Z(J)) > 0, then m2,p (J) > mp (Z(J)); (b) If m2,p (J/Z(J)) = 0, then mp (J) = 1; and (c) m  p (J) ≥ mp (J/Z(J)) − 1, with strict inequality unless J/Z(J) ∼ = G2 (3). 

Proof. As J ∈ Cp , Z(J) is a p-group and J = O p (J). If Z(J) = 1, then (a) is trivially true, and (b) follows from Lemma 12.36, in view of the fact that

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if p > 3 and n > 1, then L2 (pn ) neither lies in C2 nor has a p-pumpup in C2 , by [IA , 4.9.6, 5.3] and the fact that mp (L2 (q)) ≤ 1 if q ∈ FM9. In (c), both sides of the inequality are unchanged if we replace J by its largest covering group with a p-group center, so in (c) we shall assume that J is p-saturated. Then, if Z(J) = 1, (c) is trivially true as well. Suppose then that Z(J) = 1. From [V3 , 1.1] and [IA , 6.1.4] we get p = 3 and J/Z(J) ∼ = U4 (3), Ω7 (3), G2 (3), U6 (2), J3 , M c, Suz, O  N , F i22 , or F i24 . Since either J/Z ∈ C2 or J/Z(J) ↑3 K/O2 (K) with K ∈ C2 , we have J/Z(J) ∼ = U4 (3), G2 (3), U6 (2), J3 , Suz, F i22 , or F i24 , using [V3 , 1.1] again for C2 , the BorelTits theorem, and [IA , 5.3]. From the 2-local structure of J/Z(J) it is clear that m2,3 (J/Z(J)) > 0, so (b) is vacuous in this case. Moreover, as m3 (G2 (3)) = 4 by [IA , 3.3.3], (c) follows from [IA , 6.4.4], Lemma 1.5, the existence of SU6 (2), and [IA , 5.6.1]. Moreover, (a) is a direct consequence of [IA , Table 5.6.2]. The lemma is proved.  Lemma 12.38. Let J = D4 (2), U5 (2), or U6 (2) and H = J or J < H ∼ = P GU6 (2). Then either m2,3 (H) = 3, or H ∼ = P GU6 (2) and m2,3 (H) = 4. In all cases, m3 (H) = 1 + m2,3 (H). Proof. The assertion concerning m2,3 (H) follows by easy inspection of the maximal parabolic subgroups of H, in view of the Borel-Tits theorem. Likewise,  m3 (H) is easily computed (see [IA , 4.8.2a, 4.10.3a]). Lemma 12.39. Suppose K ∈ C3 , and Aut(K) contains E ∼ = 31+6 as well as ∗ 1+2 E1 W with E1 = F (E1 W ) ∼ = E26 and W ∼ = 3 . Then m2,3 (K) ≥ 4, with strict inequality if Z(K) = 1. ∼  A9 , 3D4 (2), Proof. Suppose false. We have m3 (K) ≥ 4. In particular, K =

or 2F4 (2 2 ) . By Lemma 9.22 and [IA , 4.10.3a], K ∈ Chev(2) − Chev(3). Suppose that K ∈ Chev(3). If K is a classical group, let its natural module be V , and let L be a covering group of K/Z(K) acting on V . We claim that dim V ≥ 7. Otherwise, m2 (Z(L)) ≤ 1. Hence some (not necessarily full) preimage of E1 in L must be elementary abelian or extraspecial of order 21+6 . Accordingly, as K is perfect, n dim V ≥ 7 or 8, proving our claim. Likewise since m2 (K) ≥ 6, K/Z(K) ∼ = 2 G2 (3 2 ) or G2 (3n ), by the structure of a (2-central) involution. Now we can check from [IA , 4.5.1, 3.3.3] that if K is simple, K has an involution z such that m3 (CK (z)) ≥ 4. For example if K is an orthogonal group, z may be chosen so that CK (z) contains n ∼ Ω± 6 (3 ) for some sign. And in the one case K = 3Ω7 (3) in which Z(K) = 1, we use [IA , 6.4.4] to get m2,3 (K) > 4. It remains to consider the case K ∈ Spor. Then |Z(K)| ≤ 3. If the lemma fails, we have |K|3 ≥ 37 and m3 (K) ≥ 4; moreover, if |K|3 = 37 then m3 (K) = 4. Also, m2,3 (K) < 4 or 5 according as Z(K) = 1 or not. Using [IA , Tables 5.6.1, 5.6.2] we conclude that K ∼ = F3 and m2 (K) = 5. But this contradicts the existence of E1 . The proof is complete.  1

Lemma 12.40. Suppose that O3 (X) = 1, Z(X) = 1, and K := E(X) ∼ = F i24 , 3F i24 , or [3 × 3]U4 (3). Let m = 5, 6, 6, or 4, respectively. Then KZ(X) contains a subgroup E ∼ = E3m normalizing an extraspecial 2-subgroup T of K. Proof. Let z ∈ K be a 2-central involution. If K ∼ = P Ω± 8 (3), take ET ≤ + − 3 ∼ Z(X) × Ω (3)Ω (3) [IA , 4.5.1]. In the other cases we use O (CKZ(X) (z)) =

P Ω± 8 (3),

4

4

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the fact that m3 ([3 × 3]U4 (3)) = 6 [IA , 6.4.4], which implies that m3 (CK (z)) = m3 (CK/Z(K) (z)) + m3 (Z(X)). Since m3 (CK/Z(K) (z)) = m3 (3U4 (3)) = 5 or m3 (SL2 (3) ∗ SL2 (3)) = 2, and O2 (CK (z)) is extraspecial, the lemma follows.  Lemma 12.41. Let K = U4 (3), Ω7 (3), F i22 , or P Ω+ 8 (3), and let m = 2, 3, 3, or 4, respectively. Then K contains a subgroup E ∼ = E3m normalizing an extraspecial 2-subgroup T of K. Proof. Let z be a 2-central involution of K. We take E ≤ CK (z) and T to be [O2 (CK (z)), E] in the first, third, and fourth cases. In the Ω7 (3) case, K ≥ + Ω+  4 (3) × Ω3 (3) ≥ Ω4 (3) × Z3 , which completes the proof. Lemma 12.42. Let K = F ∗ (X) ∼ = Un (2), n = 5 or 6. Let B ∈ E3∗ (X). Then some hyperplane of B centralizes some 2-central involution of K. Proof. We may assume that X = K or X = P GU6 (2). In either case, B is diagonalizable with respect to an orthogonal basis of the underlying unitary module. The hyperplane defined by the equality of (say) the first two eigenvalues centralizes a transvection, which is a 2-central involution of K, as claimed.  Lemma 12.43. Let X be a K-group with O2 (X) = 1 and Z(X) = 1 and every component of E(X) a C2 -group. Let T ∈ Syl2 (X) and suppose that T has a quaternion or cyclic subgroup T0 with |T : T0 | = 2. Then m3 (X) ≤ 1. Proof. If E(X) = 1, then by Lemma 2.8, m2 (X) ≥ 2 + m2 (Z(X)) ≥ 3, so m2 (T0 ) ≥ 2, contradiction. Therefore E(X) = 1 and as T has sectional rank at most 3, the result follows.  Lemma 12.44. Let K and I be as in [V4 , Lemma 15.2]. Then the following conditions hold: (a) If O3 (I) = 1, then m3 (Aut(K)) ≤ m2,3 (I) + m3 (I), unless K ∼ = Sp6 (8) and I ∼ = Sp4 (8); and (b) If O3 (I) = 1, then m3 (Aut(K)) < m2,3 (I) + m3 (I) − m3 (Z(I)). Proof. This is a straightforward observation about 3-ranks and 2-local 3-ranks of groups in Chev(2) ∪ Chev(3) ∪ Spor. See [IA , 5.6.1, 5.6.2] for the sporadic groups; [IA , 3.3.3, 4.5.1] for 3-ranks, and sufficient information about 2-local 3-ranks, of groups in Chev(3); and [IA , 4.10.3] for 3-ranks of groups in Chev(2). Note that by the Borel-Tits theorem, if I ∈ Chev(2), then m2,3 (I) is the maximum of m3 (P ) as P ranges over all (proper) parabolic subgroups of I.  12.5. Mates. Lemma 12.45. Let F ∗ (X) = K ∈ C2 and let p be an odd prime such that mp (X) ≥ 4. If p = 3 assume that K/Z(K) ∼  L± = 4 (3) or G2 (3). p ∗ Let B ∈ E∗ (X). Then there exists B ≤ X and b ∈ (B ∗ ∩ K)# such that B ∗ ∼ = B, CK (b) has a component I which is terminal in K, and IX (B; 2) = IX (B ∗ ; 2). Moreover, if p divides | Out(K)|, then we may take B ∗ = B. Proof. We immediately reduce to the case O2 (K) = 1 and thus assume that K is simple. We identify X with a subgroup of Aut(K). First suppose that K ∈ Spor. We use [IA , 5.6.1, 5.6.2]. For one thing, by Lemma 9.31, either p = 3, or K ∼ = F1 with p = 5. For another, mp (K) > m2,p (K). Hence if there exists b ∈ Ip (K) such that mp (CK (b)) > m2,p (K), and CK (b) has

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a component I which is terminal in K, then the desired conclusion holds with B ∗ ∈ Ep∗ (CK (b)) and IK (B; 2) = {1} = IK (B ∗ ; 2). (Note that | Out(K)| ≤ 2 in this case.) Such an element b exists in every case; the pairs (K, I) are as follows: (Co2 , U4 (2)); (Co1 , 3Suz); (Suz, 3U4 (3)); (F i22 , U4 (3)); (F i23 , Ω7 (3)); (F i24 , D4 (3)); (F3 , G2 (3)); (F2 ,F i22 ); (F1 ,3F i24 ); and for p = 5, (F1 , F5 ). Since K ∼ = L± 4 (3) or G2 (3), by assumption, and mp (K) ≥ 4, K ∈ Chev(2). Also as mp (K) ≥ 4, Ep∗ (K) is a single conjugacy class, by [IA , 4.10.3]. We will achieve B ∗ = B in this, the only remaining case. If some element g ∈ B induces a graph or graph-field automorphism on K, then p = 3 and as m3 (K) ≥ 4, K ∼ = D4 (q). But then m3 (B ∩ K) ≤ m3 (CK (g)) ≤ 2 while m3 (K) = 4, and this makes it impossible that B ∈ E3∗ (X) whether or not B contains a field automorphism. Thus setting B0 = B ∩ Inndiag(K), we have B = B0 × f , where f = 1 or f is a field automorphism of order p. If X contains a field automorphism f1 of order p, then f1 centralizes an elementary abelian psubgroup E of Inndiag(K) of maximal rank. All such E are conjugate in Inndiag(K) by [IA , 4.10.3d]. It follows, whether or not E ≤ X, that f1 centralizes an elementary abelian p-subgroup E1 of Inndiag(K) ∩ X of maximal rank. Hence f = 1 if and only if f1 exists. And importantly, B ∩ Inndiag(K) ∈ Ep∗ (X ∩ Inndiag(K)). We next consider the case that p divides | Outdiag(K)|, so that K ∼ = Lkp (q) or (with p = 3) E6 (q), with q ≡  = ±1 (mod p). In the latter case there is b ∈ B ∩K # such that I := E(CK (b)) ∼ = L6 (q) and I is terminal in K (with respect to p = 3). In the former case, we choose b ∈ B ∩ K # such that I := E(CK (b)) is as large as possible. Then I ∼ = SLr (q) where r = kp − 1 or kp − 2, but in either case, I is terminal in K. Thus we are done if p divides | Out(K)|, with B ∗ = B. We now may assume that p does not divide | Outdiag(K)|. Suppose that there  exists b ∈ B ∩K # with a perfect Lie component I = O p (I), and choose such b and I such that |I| is maximized. Since p does not divide | Outdiag(K)|, B ∩K normalizes I by [IA , 4.2.2]. Indeed if f = 1, then q(I), being a power of q(K) by [IA , 4.2.2], is a pth power. Using [IA , 8.7(iii)], we deduce that in any case, B normalizes I. We use the maximal choice of I and [IA , 4.10.3d] to prove that mp (CK (I)) = 1, which immediately implies the terminality of I. Namely, if mp (CK (I)) > 1, choose a B-invariant Ep2 -subgroup U ≤ CK (I). Then B1 := U CB (U ) ∼ = B, so B1 ∩ K is conjugate to B ∩ K by [IA , 4.10.3d]. By Lp -balance and [IA , 4.2.2] again, for each b1 ∈ B1# , I ≤ I1 for some p-component I1 of CK (b1 ). By maximality of I (and the   conjugacy of B and B1 ), I = I1  O 2 (CK (b1 )). Thus I  Γ2B1 ,1 (K) =: Γ, and so Γ < K. But [I, B1 ] = 1 so Γ = K by [IA , 7.3.1], contradiction. Finally, assume that for all b ∈ B ∩ K # and all Lie components I of√CK (b), I is not perfect. Then I ∈ Lieexc so q(I) = 2. By [IA , 4.2.2], q(K) = 2 or 2, and the 1 latter is impossible as mp (Aut(2F4 (2 2 ) )) ≤ 2 for all odd primes p. Using [IA , 4.8.2] for classical groups, and [III11 , Table 13.1] for exceptional type groups, and the fact that mp (K) ≥ 4, we easily find b ∈ B ∩ K # and a perfect Lie component   I = O p (I) of CK (b), completing the proof of the lemma.

12.6. p-Components in C2 -Groups. The next two lemmas are very closely related, and we prove them together. The term Lcs (H) is defined in [V4 , 9.8].

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Lemma 12.46. Suppose that K  X = KA where K = F ∗ (X) ∈ C2 , A is an  elementary abelian p-group for p ∈ {3, 5}, K = O p (K), and mp (KA) = mp (A) ≥   3. For each a ∈ A# set E p (a) = O p (E(CK (a))). Then the following conditions hold: (a) One of the following holds: (1) K ∈ Chev(2), mp (Aut(K)) ≥ 3, and mp (A ∩ K) ≥ 2; (2) p = 3, A ≤ K, and K/Z(K) ∼ = L± 4 (3), G2 (3), J3 , Co2 , Co1 , Suz, F i22 , F i23 , F i24 , F3 , F2 , or F1 ; or ∼ (3) p = 5,  A ≤ K, and K= Co1 , F2 , or F1;  p # (b) Set Y = E (a) | a ∈ A and YK = E p (a) | a ∈ A ∩ K # . Also let  # . Suppose that K ∈ Chev(2). Then the Y1 = Y Lcs 3 (CK (a))] | a ∈ A following conditions hold: (1) Y1 = K; (2) Y = K unless possibly p = 3, m3 (A) = 3, Y = 1, and K ∼ = U4 (2), 3 D4 (2), or U3 (8); or (3) YK = K unless possibly p = 3, m3 (A) = 3, YK = 1, and K ∼ = U4 (2), 3 D4 (2), or L3 (q),  = ±1, q ≡  (mod 3); and (c) If K ∈ Chev(2), then CX (A)/Z(K) is a p-group. Lemma 12.47. Suppose that X = KB where K = F ∗ (X) ∈ C2 , B is an elementary abelian p-group for some odd prime, and mp (KB) = mp (B) ≥ 4. Then the following conditions hold: (a) One of the following holds: (1) K ∈ Chev(2) and mp (B ∩ K) ≥ 3; (2) p = 3, B ≤ K, and K/Z(K) ∼ = L± 4 (3), G2 (3), Co2 , Co1 , Suz, F i22 ,  F i23 , F i24 , F3 , F2 , or F1 ; or ∼ (3) p = 5,   B ≤ K, and K = F1 ; p (b) Let Y = O (E(CK (b))) | b ∈ B ∩ K # . If K ∈ Chev(2), then Y = K; and (c) If K ∈ Chev(2), then CX (B)/Z(K) is a p-group. Proofs of Lemmas 12.46 and 12.47. We refer to these lemmas as Lemma A and Lemma B, respectively. As K ∈ C2 , Z(K) is a 2-group. First suppose that K ∈ Spor. Then | Out(K)| ≤ 2 so A ≤ K (resp. B ≤ K). Thus A (resp. B) lies in Ep∗ (K). The restrictions in (a) are immediate from the conditions K ∈ C2 and mp (K) ≥ 3, resp. 4, and the assumption in Lemma A that p ≤ 5 [IA , 5.6.1]. Next, given the possibilities for K, one sees from [IA , 5.3] and [IA , 4.10.3, 5.2.10, 5.6.1] that mp (CG (x)) < mp (B) for all x ∈ Aut(K) of any prime order r = p. This implies (c), and (b) is vacuous in this case. Next, assume that K ∈ C2 − Chev(2) − Spor. As mp (Aut(K)) ≥ 3, K/Z(K) ∼ = ± L4 (3) or G2 (3), whence m3 (K) = 4 [IA , 3.3.3]. Again Out(K) is a 2-group so A (resp. B) lies in E3∗ (K). Only part (c) needs to be checked. Let C = CAut(K) (B). By the Borel-Tits theorem, C ≤ N where N = NAut(K) (P ) for some parabolic subgroup P of K, and F ∗ (N ) = O3 (N ). As m3 (B) = m3 (Aut(K)), O 3 (C) ≤ O3 (N ) = 1 by the Bender-Thompson lemma [V2 , 2.2], proving (c). We may now assume that K ∈ Chev(2). Suppose that (a) fails. Then A (resp. B) has a noncyclic subgroup D whose image in Aut(K) is disjoint from Inn(K).

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Then Out(K) has noncyclic Sylow p-subgroups. If some element d ∈ D# induces a graph or graph-field automorphism on K, then p = 3 and K ∼ = D4 (q) (Out(3D4 (q)) is cyclic). As m3 (K) = 4 but m3 (CK (d)) = 2, we can construct a bigger element of E3 (X) by dropping d and including an entire element of E3∗ (K). This contradicts the assumption that A (resp. B) is of maximal rank in KA (resp. KB). It follows, with [IA , 2.5.12, 4.9.1], that D = d, f  with d mapping nontrivially into Outdiag(K) and f mapping to a nontrivial field automorphism. In particular, p divides |Outdiag(K)| and mp (KA) and mp (KB) are at least r := mp (K d) = mp (Inndiag(K)). Thus K/Z(K) ∼ = Lpk (q) with r = kp − 1, or E6 (q) 5 with r = 6 [IA , 4.10.3a]. Here q ≡  (mod p),  = ±1, and p = 3 in the E6 case. As (a) fails, r ≤ 3 (resp. r ≤ 4), ruling out the E6 case. Hence either kp = 3, or we are in Lemma B with kp = 5. But then in either case by Lemma 3.77, CInndiag(K) (f ) ≤ Inn(K), contradicting the existence of d. This proves (a). It remains to prove (b), so assume that K ∈ Chev(2), and, without loss, that K is simple. Since mp (Aut(K)) ≥ 3, K ∈ Lie(2). As in the previous paragraph, elements of A (resp. B) do not induce graph or graph-field automorphisms on K if K ∼ = D4 (q). However, they can induce graph (“quasifield”) automorphisms if K∼ = 3D4 (q). In any case, some subgroup of A (resp. B) of index at most p maps into Inndiag(K). Let L0 be the set of all Lie components of CK (b) as b ranges over A ∩ K # (resp. B ∩ K # ), and let L∗0 be the set of all maximal elements of L0 under inclusion which have orders divisible by p. We claim that (12B)

L0  = L∗0  .

To see this, let L ∈ L0 . It suffices to show L ≤ L∗0 , which is trivially true if p divides |L|. Suppose then that L is a p -group. If r := mp (CA∩K (L)) > 1 (resp. r := mp (CB∩K (L)) > 1), then L ≤ L∗0  by [IA , 4.9.7]. So suppose that r = 1. As L is a p -group, mp (Aut(L)) ≤ 1 and thus A (resp. B) does not normalize L. Hence by [IA , 4.2.2, 4.9.1, 4.9.2], K is not p-saturated. If K ∼ = E6± (q), then p = 3, and by [IA , 4.7.3A], every L ∈ L0 has order divisible by 3. Likewise if K ∼ = Lkp (q),  = ±1, q ≡  (mod p), every element of L0 contains L2 (q) and so has order divisible by p. This proves (12B). Suppose that q := q(K) > 2. Then for any L ∈ L∗0 , L ∈ Lieexc and so L = p O (E(L)) ≤ YK . As mp (A ∩ K) ≥ 2 and mp (A) = mp (AK) (resp. mp (B ∩ K) ≥ 3 and mp (B) = mp (BK)), it follows from [IA , 7.3.3, 4.9.5, 4.9.7] and Lemma 8.12, applied to A ∩ K (resp. B ∩ K), that either K = L0  = L∗0  = YK = Y = Y1 , or we are in Lemma A, K ∼ = L3 (q), q ≡  = ±1 (mod 3), with A ∩ K the image of a nonabelian subgroup of GL3 (q), and YK = 1. In this exceptional case, m3 (Inndiag(K)) = 2, so some element of A# induces a field automorphism on K. If q > 8 then it follows from [IA , 7.3.8] that Y = K, so Y1 = K as well. If q = 8, then by [IA , 7.3.8] again, K is generated by normal complete (S)U3 (2)-components of CG (a), a ∈ A# , so Y1 = K, as desired. Thus (b) holds if q > 2. Suppose then that q = 2. If A (resp. B) does not map into Inndiag(K), then K ∼ = 3D4 (2), p = 3, m3 (Aut(K)) = 3, and we are in Lemma A. In this case, for  every a ∈ A# , it follows from [IA , 4.7.3A] that O 2 (CK (a)) ∼ = L2 (8) or CK (a) is a 2 complete (S)U3 (2)-component (of itself). Hence Y1 = ΓA,1 (K) = K by [IA , 7.3.3].  Moreover, by Lemma 8.29, if YK = 1, so that O 2 (CK (a)) ∼ = L2 (8) for some a ∈ A# ,

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∼ 3D4 (2), so we may assume that A then YK = K. Thus the lemma holds if K = (resp. B) maps into Inndiag(K). We next assume that K is p-saturated, and show that (12C)

mp (CA (L)) = 1 (resp. mp (CB (L)) = 1) for all L ∈ L∗0 .

Indeed, suppose that A0 := CA (L) (resp. B0 := CB (L)) is noncyclic. As K is # p-saturated, it follows that for any a ∈ A# 0 (resp. B0 ), A (or B) normalizes L and L lies in a Lie component of CK (a) by Lemma 4.2, so L is such a Lie component by   maximality. Therefore L  O 2 (CK (a)), whence NK (L) ≥ Γ2A0 ,1 (K) = K (or the corresponding statement for B0 ) by [IA , 7.3.1]. But then L  K, which is absurd. This proves (12C). In turn, as every L ∈ L∗0 is A-invariant (resp. B-invariant) by Lemma 4.2, we have mp (AutA (L)) ≥ 2 (resp. mp (AutB (L)) ≥ 3). Notice that since K is p-saturated, A (resp. B) induces inner automorphisms on K and so lies in K. Unless L ∈ Lieexc for some L∗ ∈ L∗0 , we conclude that YK = K, whence Y = Y1 = K. The only possible L ∈ Lieexc occur if mp (Aut(L)) = 2, whence we 1 are in Lemma A, p = 3, and L/Z(L) ∼ = U3 (2) or Sp4 (2). (G2 (2) and 2F4 (2 2 ) cannot occur as Lie components in L0 because of their diagrams, see [IA , 4.2.2].) If one of these occurs, then in view of (12C), K ∼ = U4 (2), Sp6 (2), or D4− (2). ∼ If K = U4 (2), then A is a full diagonal subgroup of K with respect to an orthonormal basis {v1 , . . . , v4 } of the natural F4 K-module V . For each i, 1 ≤ i ≤ 4, let ai ∈ A# fix vi and act as a scalar mapping of order 3 on vi⊥ . Then CK (ai ) ∼ = GU3 (2) is a complete (S)U3 (2)-component of itself, so Y1 ≥ CK (ai ) | 1 ≤ i ≤ 4 = K by [IA , 7.3.2], as desired. If K ∼ = D4− (2), then by [IA , 7.3.2], K is generated ∗ ∼ by two elements of L0 isomorphic to Ω+ 6 (2), so YK = K. And if K = Sp6 (2),  then by a Frattini argument, for each a ∈ E # for which O 2 (CK (a)) ∼ = Sp4 (2), 2 3 3 ∗ O (CK (a)) ≤ O (CK (a))NK (E), with O (CK (a)) ∈ L0 being NK (E)-invariant;  by [IA , 7.3.2], K is generated by the set of all such O 2 (CK (a))’s, a ∈ E # , and hence K = YK NK (E)  YK and YK = K. This completes the p-saturated case. Finally, if K is not p-saturated, then as q = 2, K ∼ = U3k (2), k ≥ 2, or E6− (2), with p = 3 in either case. We write K = K1 , K2  where K1 and K2 are A-invariant (resp. B-invariant) with either K1 ∼ = K2 ∼ = U3k−1 (2) or (K1 , K2 ) = (A5 (q), D4 (q)). Then inductively,    Ki ≤ O p (CKi (b))(∞) | b ∈ B # , i = 1, 2, which implies that YK = K, as 







O p (CKi (b))(∞) ≤ O p (CK (b))(∞) = O p (E(CK (b)) = E p (b). This completes the proofs of the lemmas.



Lemma 12.48. Suppose that K  X = KA where K = F ∗ (X) ∈ C2 , K is  simple, A is an elementary abelian p-group for some odd prime p, K = O p (K), and mp (KA) = mp (A) ≥ 3. If mp (A) = 3, assume that p ∈ {3, 5}. Suppose that J ∈ C2 and for some involution t ∈ Aut(J), K is a component of CJ (t). Then one of the following holds: (a) There exists a ∈ A ∩ K # and a p-component L of CK (a) which is not a component of CJ (a); (b) p = 3, K ∼ = A∗ ≤ K, 1 = a∗ ∈ A∗ , and a = F i23 and there exists A ∩ K ∼ ∗ 3-component L of CK (a ) which is not a component of CJ (a);

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12. {2, p}-STRUCTURE

489

∼ U4 (2), and one of the following holds: (c) p = 3, K = (1) there is a ∈ A ∩ K # such that O3 (CK (a)) ≤ O3 (CJ (a)) and a is 3-central in K; (2) J/Z(J) ∼ = L± 4 (3); or 3 (d) p = 3, K ∼ = D4 (2) or L3 (q), q ≡  = ±1 (mod 3), and L3 (CK (a)) = 1 for all a ∈ A ∩ K # . Proof. If K ∈ Spor, then J ∈ Spor by Lemma 3.1. The only possibility, by [IA , 5.3] and the simplicity of K, is (K, J) = (F i23 , F i24 ). In particular p = 3, and we take a∗ such that CK (a∗ ) = a∗  × L with L ∼ = Ω7 (3), and A∗ ∈ E3∗ (CK (a∗ )). ∗ By [IA , 3.3.3, 5.6.1], m3 (A ) = 6 = m3 (K), as claimed. Next, assume that K ∈ C2 − Chev(2) − Spor. As mp (Aut(K)) ≥ 3, K/Z(K) ∼ = ± L4 (3) or G2 (3). On the other hand, these groups are not components of involution centralizers in any group in C2 , by Lemma 3.10, so J cannot exist, contradiction. Thus we may assume that K ∈ Chev(2). Suppose that Lp (CK (a)) = 1 for some a ∈ A ∩ K # , i.e., in the language of Lemma 12.46b, YK = 1. There are few exceptions to this and we deal with them at the end of the proof. If J ∈ Spor, then as K ∈ Chev(2) is simple with mp (Aut(K)) ≥ 3, we have p = 3 and J/Z(J) ∼ = F i22 or F2 . In both cases J has no automorphism b of order 3 such that CJ (b) has a component in Chev(2), so (a) holds. If J ∼ = L± 4 (3), G2 (3), or L3 (3), then again, J has no automorphism b of order 3 such that CJ (b) has a component, by the Borel-Tits theorem. Thus we assume that J ∈ Chev(2). If t ∈ Aut0 (J), then there exist σ-setups of K and J/Z(J) of the form (K, σ) and (K, σ 2 ), respectively, with t acting on J/Z(J) as σ. Choose any b ∈ B # ∩ K such that CK (b) has a nonsolvable Lie component L of order divisible by p. Then there is a product L = L1 · · · Ln of (algebraically) simple components of CK (b) such that σ cycles L1 , . . . , Ln and L = CL (σ) maps onto CL1 (σ n ). Set L∗ = CL (σ 2 ), a product of Lie components of CJ (b). If n is even, then L∗ has two Lie components in whose product L is a diagonal. If n is odd, then L∗ is a single Lie component mapping on CL1 ((σ n )2 ). In either case, L(∞) is not subnormal in (L∗ )(∞) , yielding the desired assertion. So when J ∈ Chev(2), we may assume that t induces a graph automorphism on J. Thus J and K are given in [IA , 4.9.2]. In particular the untwisted Dynkin diagram of J has only single bonds, while that of K has a double bond. In each case there is b ∈ B # ∩ K such that CK (b) has a component L whose untwisted diagram has a double bond, whereas by [IA , 4.2.2] the untwisted diagram of CJ (b) has only single bonds. For example if K ∼ = F4 (q), then since mp (K) ≥ 3, q ≡ ±1 (mod p), and b can be chosen so that K∼ = Sp6 (q) [III11 , Table 13.1]. Thus, (a) holds. Now return to the exceptional case YK = 1. By Lemma 12.46b, either (d) holds or p = 3 and K ∼ = U4 (2), which we assume. Thus U4 (2) ↑2 J, so by Lemma 3.10, J ∼ (3), 2U (3), or L4 (4). By (c2) we may assume that J ∼ = L± = L4 (4). Then 4 4 3 3 for b ∈ I3 (K) a 3-central element, |O3 (CK (b))| = 3 but O (CJ (b)) ∼ = GL3 (4), so  |O3 (CJ (b))| = 3 and (c1) holds. The proof is complete. Lemma 12.49. Let L = 2 D5 (2), P GU6 (2), U7 (2), or 2E6 (2). Let D ≤ L with ∼ D = E32 . Let LD = E(CL (D)) and suppose that LD ∼ = U5 (2) if L = U7 (2), and LD ∼ = U4 (2) otherwise. Then D = O 2 (CL (LD )). ∼ Un (2) with n ∈ {6, 7}, then C ∼ Proof. Set C = CL (LD ). If [L, L] = = ∼ Z3 × SL2 (2), and the claim holds. Likewise if L ∼ GU2 (2) = = 2 D5 (2), then either

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8. PRELIMINARY PROPERTIES OF K-GROUPS

490

∼ ∼ ∼ C∼ = Ω+ 4 (2) = Σ3 ×Σ3 or C = Z(GU4 (2))×Z3 = E32 . In either case, the claim holds. Finally, suppose that L = 2E6 (2). From the information in [IA , 4.7.3A], for every d ∈ D# , E(CL (d)) ∼ = U6 (2) or D4 (2), and D = F ∗ (CC (d)). On the other hand, since LD contains an element of order 5, O 2 (C) embeds in A8 , by Lemma 12.24, so D ∈ Syl3 (C). This in turn implies that O3 (C) = 1. Hence if the lemma fails, then E(C) = L3 (C) = 1. By Lemma 12.24 again, since m3 (LD ) = 3, C must be a 5 group. The only possibility is E(C) ∼ = L3 (2). But then there is 1 = d ∈ CD (E(C)),  so CC (d) is not 3-closed, a contradiction. The proof is complete. 12.7. Other. Lemma 12.50. Let J ∈ C3 be as in [V5 , Table 7.1]. Suppose that J  X = JZ(X)B with O3 (X) = 1, Z(X) = CX (J) ∼ = Z3 , and B ∈ E3∗ (CX (z)) for some 2-central involution z of J. Then the following conditions hold: (a) m = m3 (B) is as given in [V5 , Table 7.1]; (b) For some a ∈ B − Z(X), there is a component JD of CJ (a) containing z as a 2-central involution and having the isomorphism type given in [V5 , Table 7.1]; (c) B/ a, Z(X) acts faithfully on CO2 (CJ (z)) (a); and (d) F ∗ (CJa (a)) = JD a Z(J). Proof. Our hypotheses imply that Z(X) ≤ B. First suppose that J is a classical group as in the first four rows of [V5 , Table 7.1]. If B maps into Inn(J), then since CJ (z) =: P is a parabolic subgroup of type GU3 (2), Z(J) × U4 (2), U4 (2), or L2 (2) × L2 (2) × L2 (2), respectively, and since Z(X) ≤ CX (z), m = 4, as in the table. The existence of a satisfying (b) follows easily from [IA , 4.8.1, 4.8.2]. In the unitary cases, with respect to a suitable ordered orthogonal frame preserved by B, and with ω 3 = 1 = ω, a = diag(ω, ω, ω, ω, ω −1 ) if K = U5 (2), and a = diag(1, 1, 1, 1, ω, ω −1 ) if K/Z(K) ∼ = U6 (2), with z a transvection in the first two dimensions and trivial in the remaining dimensions. Then (d) follows directly. In (c), let S = CO2 (CJ (z)) (a) ∼ = 21+4 + , in each unitary case; then CB (S) = a Z(X), so (c) follows too. On the other hand if B does not map into Inn(J), then Inndiag(J) ∼ = P GU6 (2) and we can take a = diag(1, 1, 1, 1, 1, ω), with z again in the first two dimensions, and the desired statements hold. In the case J ∼ = D4 (2), z is a root involution and we take a so that on a natural module V , dim[V, a] = 2 and [V, a, z] = 1. Then again S = CO2 (CJ (z)) (a) ∼ = 21+4 + , and (c) and (d) follow directly. Now consider the nine possible J ∈ Spor. If Z(J) ∼ = Z3 then every element of I3 (J/Z(J)), being real, splits over Z(J) by [IA , 5.3]. With the help of [IA , 6.4.4] in the case J ∼ = 3F i24 , we see in every case that m = 1 + m3 (CJ/Z(J) (zZ(J)/Z(J))), and the values in [V5 , Table 7.1] then are correct, using [IA , 5.3]. In K = Co2 , let x ∈ I3 (K) be of type 3B and let y ∈ I2 (CK (x)) be 2central. Then m3 (CK (xy)) = 3 so y is 2-central in K. As CK (y) has one class of E33 -subgroups by [IA , 4.10.3], we may take B ≤ CKZ(X) (xy), z = y and a = x. Then CB (O2 (CK (az))) = a Z(X), and all parts of the lemma follow. A similar argument handles other cases as well, namely, (K, x, m3 (CK (xy))) = (Co1 , 3A, 4), (Suz or 3Suz, 3A, 3), (F3 , 3A, 3), and (F1 , 3A, 6), in the last case using the fact [IA , 5.3l] that for P ∈ Syl3 (Co1 ), J(P ) ∼ = E36 .

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12. {2, p}-STRUCTURE

491

∼ F i or 3F i . We use [IA , 5.3v]. Suppose that Suppose next that K = 24 24  K∼ = F i24 . Again choose x to be in class 3A and y to be 2-central in CK (x), so that  O 3 (CK (xy)) = x × C, where C has a normal subgroup of index 3 which is the central product of four copies of SL2 (3) containing y. We argue that y is of class 2B. Otherwise, CK (y)/ y ∼ = Aut(F i22 ), in which the centralizer of x would have to have a normal E28 -subgroup. But no such element of I3 (F i22 ) exists [IA , 5.3t], so y is of class 2B. As O 2 (CK (y)/O2 (CK (y))) ∼ = 3U4 (3) has a unique class of elementary abelian 3-subgroups of maximal rank (see [IA , 6.4.4]) we can finish the  proof as in the previous cases. The proof for 3F i24 is similar; O 3 (CK (xy)) has an extra Z3 direct factor. Notice also that by [IA , 6.4.4], m3 ([3 × 3]U4 (3)) = 6. Finally, suppose that K ∼ = F2 [IA , 5.3y]. This time, begin with a 2-central involution y of K (class 2B), set Cy = CK (y) and C y = Cy /O2 (Cy ) ∼ = Co2 , and choose x ∈ I3 (CK (y)) such that NC y (x) ∼ = Σ3 × Aut(U4 (2)) [IA , 5.3k]. Let v be in class 3B. Then NK (v) has no Σ3 × Aut(U4 (2)) section containing an image of v [IA , 5.3y]. Hence x must be in class 3A. Now U4 (3) ≤ M c ≤ Co2 all share a Sylow 3-subgroup, so we can finish the proof as in the other cases.  Lemma 12.51. Let X be a group with K y ≤ X, y 2 = 1, and L := E(CK (y)) ∼ = A6 . Assume that K ∼ = A6 , or K ∼ = A6 × A6 , or K is simple with K ∈ ↑2 (A6 ) ∩ C2 − {L4 (3)}. Let x ∈ I3 (L). Then O3 (CK (x)) embeds in Z3 × 31+2 or E34 . Proof. Let P ∈ Syl3 (K) with CP (x) ∈ Syl3 (CK (x)) and CP (x) containing a Sylow 3-subgroup Q of CL (x). Using Lemma 3.10 we see that either P is elementary abelian of order at most 34 , in which case the result is trivial, or K ∼ = U4 (3), U4 (2), or U5 (2). In the last two cases P ∼ = (Z3  Z3 ) × P0 with |P0 | = 1 or 3; hence if x ∈ Z(P ), then CP (x) embeds in E34 . If x ∈ Z(P ), then x has (at least) three equal eigenvalues on the natural K-module, and O3 (CK (x)) embeds in Z3 ×31+2 , as claimed. Hence the lemma holds in these cases, and we may assume that K ∼ = U4 (3). Since x lies in an A4 subgroup of CK (y), it is not the image of a transvection in SU4 (3), i.e., it is not 3-central in K. Let E = J(P ) ∼ = E34 , with AutK (E) ∼ = A6 (Lemma 9.16). If V is a natural SL2 (9)-module, then E ∼ = V ⊗ V (3) as SL2 (9)module, where V (3) is a Galois twist of V . We compute that AutK (E) has orbits on E of length 10, 15, and 15. Hence if x ∈ E, then CNK (E) (x)/E ∼ = Z2 × Σ4 , whence O3 (CK (x)) = E ∼ = E34 , as desired. It remains to consider the case x ∈ P −E. From [IA , 4.5.1] we see that all elements of I3 (L) are Aut(K)-conjugate, hence none of them are conjugate into E. Therefore Q ∩ E = 1. Since AutK (E) ∼ = A6 contains A4 , |CE (Q)| ≤ 32 . Hence |CP (x)| ≤ 34 and CP (x) = Ω1 (CP (x)), which proves the lemma.  Lemma 12.52. Let K ∈ C2 with Z(K) = 1 and |K|5 = 5. Suppose K  X and Sylow 2-centers in X have order 2. Then | Aut(K)|2 ≤ 212 . Proof. Suppose that | Aut(K)|2 > 212 . This condition, together with the conditions |K|5 = 5 [IA , 4.10.3a], Z(K) = 1 [IA , 6.1.4], and K ∈ C2 [V3 , 1.1], leaves only the group K/Z(K) ∼ = U6 (2). Let T ∈ Syl2 (X); we complete the proof by  of K satisfies Z(K)  ∼ showing that Z(T ) ≤ Z(K). A full 2-covering group K = E22 , ∼   and Out(K) = Z2 acts faithfully on Z(K) by [IA , 6.3.1]. Moreover, by [IA , 6.4.1,  then Ω1 (Z(S)) ≤ Z(K).  Therefore a 2-element of Aut(K) − 6.4.2], if S ∈ Syl2 (K), Inn(K) centralizes an involution of Z(T ∩K)−Z(K), and the proof is complete. 

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492

8. PRELIMINARY PROPERTIES OF K-GROUPS

13. {2, 3}-Neighborhoods Lemma 13.1. Let K ∈ C3 . Suppose that b, t ≤ Aut(K) with [b, t] = 1, b3 = 1, and t an involution. Let Kb and Kt be components of CK (b) and CK (t), respectively. If Kt ≤ Kb , then one of the following holds: (a) b = 1; (b) K ∼ = D4 (q) or 3D4 (q), q = 3m > 3; Kb ∼ = G2 (q); and Kt ∼ = SL2 (q); 1 3m 3 ∼ ∼ (c) K = D4 (q), q = 3 > 3; Kb = D4 (q 3 ); and Kt ∼ = SL2 (q). Proof. Suppose that (a) is false, so that b has order 3. Let q = q(K). Suppose first that K ∈ Chev(3). Then b is a field, graph-field, or graph automorphism. In the graph case, K ∼ = = D4 (q) or 3D4 (q), q = 3m , and Kb = CK (b) ∼ G2 (q). As [b, t] = 1, either t ∈ Kb or t is a field automorphism. In the former case, (b) holds, while in the latter case, |Kt |3 = q 6 = |Kb |3 , but m3 (Kt ) < m3 (Kb ) by [IA , 3.3.3], so Kt ≤ Kb , contrary to assumption. If b is, on the other hand, a field or graph-field automorphism, then q(Kb ) = q 1/3 and Kt is a component of CKb (t) as well as of CK (t). If t induces an element of Aut0 (K) on K, then q(Kt ) = q k for some integer k ≤ 4 [IA , 4.5.1]; also, t must induce an element of Aut0 (Kb ) on Kb ,  1 1 so q(Kt ) = (q 3 )k with k ≤ 4. Therefore k = 3, and by [IA , 4.5.1], Kb ∼ = 3D4 (q 3 ), so K = D4 (q) and Kt ∼ = SL2 (q), so (c) holds. If t induces a field or graph-field automorphism on K, then |Kt |3 > |Kb |3 , contradicting Kt ≤ Kb . Next, suppose that K ∈ Spor. Just the conditions [b, t Kt ] = 1 are already impossible, except for K ∼ = He; but He ∈ C3 . Finally, assume that K ∈ Chev(2) − Chev(3). By the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], the only involutions t ∈ Aut(K) with E(CK (t)) = 1 occur for K∼ = U5 (2), U6 (2), D4 (2), F4 (2), and Sp4 (8); t is a graph automorphism in the first three cases and a graph-field automorphism in the last two. In the first three cases, CK (t) = Kt by [IA , 4.9.2], and Out(K) ∼ = Z2 or Σ3 by [IA , 2.5.12], so the condition [b, t Kt ] = 1 is again impossible. In the last two cases, CAut(K) (Kt ) is a 3 -group by [IA , 7.1.4c], a final contradiction.  Lemma 13.2. Let K ∈ C2 ∪ C3 , a ∈ I3 (Aut(K)), and t ∈ I2 (Aut(K)), with [a, t] = 1. Let L be a component of CK (a). Then L is not a component of CK (t). Proof. Without loss K is simple. By Lemma 13.1, we are done if K ∈ C3 , so assume that K ∈ C2 − C3 . If K ∼ = An , then n ≤ 9 and the result is easily checked. Likewise if K ∈ Spor, then K ∼  He and the result is easily checked from the tables = [IA , 5.3]. If K ∼ = L2 (q), q ∈ FM9, then CK (a) has no components. Hence we may assume that K ∈ Chev(2). If t is a field or graph-field automorphism, then by [IA , 7.1.4c], CAut(K) (L) is a 2-group, so [a, L] = 1, contradiction. By the Borel-Tits theorem, t ∈ Inndiag(K), so t is a graph automorphism. By [IA , 4.9.2], CK (t) is quasisimple, q(CK (t)) = q(K), and Z(CK (t)) is a 3 -group. Since [t, a] = 1 it follows that a ∈ Inn(K). If a is a field automorphism, then q(L) < q(K) so L cannot equal CK (t), contradiction. Thus a maps nontrivially to Outdiag(K), or K is untwisted and a maps nontrivially to ΓK . But t inverts O3 (Outdiag(K)) and  O3 (ΓK ), so [a, t] = 1, contrary to hypothesis. The proof is complete. Lemma 13.3. Let K ∈ C2 . Let B ∈ E3∗ (K) and b ∈ B # , and let I be a component of CK (b). Assume that I is terminal in K with respect to p = 3, and m := m3 (B) ≥ 4. Assume also that m2 (CK (B)) ≤ 1. Let X = J t be a group

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13. {2, 3}-NEIGHBORHOODS

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with t2 = 1 and J ∈ C3 , and suppose that CJ (t) has a component I1 = J which is a quotient of I by a {2, 3} -group. Then one of the following holds: m2,3 (X) − m3 (Z(X)) ≥ m; K/Z(K) ∼ = L± 4 (3) or G2 (3); (K, I, J) = (Co2 , U4 (2), L± 4 (3) or XU4 (3)), X = 3 or 3 × 3, and m = 4; (K, I, J) = (2Suz, 6U4 (3), 3Ω7 (3)), and m = 5; (K, I, J) = (F i22 , U4 (3), L4 (9)), and m = 5; (K, I, J) = (2F i22 , 2U4 (3), Ω7 (3) or P Ω± 8 (3)), and m = 5; (3)), with m = 6; (K, I, J) = (F i23 , Ω7 (3), P Ω± 8 (K, I, J) = (F3 , G2 (3), G2 (9)), with m = 5; (K, I, J) = (2F2 , 2F i22 , F i23 or (3)F i24 ), with m = 6; m = 4, I ∼ = (2)U6 (2), U5 (2), Sp8 (2), (2)D4 (2), or D5 (2), and = U4 (2), K ∼ J/O3 (Z(J)) ∼ (3); = L± 4 (k) (K, I, J) = (Sp8 (2) or (2)F4 (2), Sp6 (2), U6 (2) or D4 (2)), with m = 4; (l) (K, I, J) = ((2) 2 E6 (2), 2U6 (2) or D4 (2), (3)F i22 ), with m = 5; or (m) (K, I, J) = (2 D5 (2), D4 (2), (3)F i22 ), with m = 5. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

Proof. Because of conclusion (b) and the definition of C2 [V3 , 1.1], and as m = m3 (K) ≥ 4, we may assume that K ∈ Chev(2) or K/O2 (Z(K)) ∈ {Co2 , Co1 , Suz, F i22 , F i23 , F i24 , F3 , F2 , F1 }. Consider these sporadic cases for K first. We have m = 4, 6, 5, 5, 6, 7, 5, 6, or 8, respectively, and I/I ∩ Z(K) ∼ = U4 (2), 3Suz, 3U4 (3), U4 (3), Ω7 (3), P Ω+ 8 (3),  G2 (3), F i22 , or 3F i24 , respectively. Here I ∩ Z(K) is a 2-group as K ∈ C2 . In particular O{2,3} (I) = 1 and so I ∼ = I1 . Hence I/O3 (I) ↑2 J/O3 (J). Suppose that K ∼ = Co2 . Then Z(K) = 1 so U4 (2) ↑2 J/O3 (J) via t. Hence by ± ∼ ∼ Lemma 3.80, J ∼ = P Sp4 (9), L± 5 (3), or J/O3 (J) = P Ωn (3), n ≥ 6. If J/O3 (J) = ± P Ω6 (3), then conclusion (c) holds; otherwise, it follows easily from [IA , 4.5.1, 3.3.3, 6.4.4] that for some involution u ∈ J, m3 (CJ (u)) − m3 (Z(J)) ≥ 4 = m, so conclusion (a) holds. If K/Z(K) ∼ = Suz and so by Lemma 3.80, the condition = Co1 , then I/Z(I) ∼ I/O3 (I) ↑2 J/O3 (J) is not satisfied for any J, contradicting our hypothesis. A similar argument, again using Lemma 3.80, rules out K ∼ = F2 or F1 . (3), n = 6, 7, 8, so that m = n − 1. Then by Suppose that I/Z(I) ∼ = P Ω± n Lemma 3.80, J ∈ Chev(3). In particular if O3 (Z(I)) = 1, then O3 (Z(I)) ≤ O3 (J); but then by [IA , 6.1.4], J ∼ = 3Ω7 (3), whence I ∼ = 6U4 (3) and conclusion (d) holds. ± Hence we may assume that O3 (Z(I)) = 1, and it follows that I ∼ = Ω− 6 (3) or P Ωn (3)  ∼ according as K = 2F i22 or F i16+n . Now J ∈ Chev(3). If t induces a field or graph+ ∼ field automorphism on J, then I ∼ = P Ω± n (3) and J = P Ωn (9) as Z(J) has odd order. + But then if n = 6, J contains Ω6 (9) so m2,3 (J) ≥ 8 > n − 1, and (a) holds. Thus we may assume that I ∼ = U4 (3), whence conclusion (e) holds. Next, we may assume that t induces an algebraic automorphism on J. If t induces a graph automorphism on J ∼ = L± n (3), then J contains an involution with a four-dimensional eigenspace on the natural module, whose centralizer demonstrates that m2,3 (J) ≥ n − 1 = m, and (a) holds. If t induces a graph automorphism on J ∼ = P Ω± 2k (3), then n = 7; if + k = 4, then (g) holds, while if k ≥ 5, then Ω8 (3) ≤ J and (a) holds. So we may assume that t induces an inner-diagonal automorphism on J. If K ∼ = 2F i22 , so that ± ∼ (3), then J P Ω (3), k > 6, and the condition m (J) < m forces k = 7 I∼ = = Ω− 2,3 6 k or 8, so conclusion (a) or (f) holds. Otherwise Z(I) = 1, so the only possibility is

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∼ P Ω± (3), now with k > 8, and the we get m2,3 (J) ≥ m, so I ∼ = Ω7 (3); again J = k (a) holds. Next, if I ∼ = G2 (3), the only possibility for J, by [IA , 4.5.1, 4.9.1, 6.1.4, 5.3], is G2 (9), and conclusion (h) holds. Similarly, the final sporadic case K ∼ = 2F2 leads with [IA , 5.3] to conclusion (i). Now we may assume that K ∈ Chev(2), whence I ∈ Chev(2) by Lemma 3.1. Moreover, as I is terminal in K, m3 (CK (I)) = 1 by Lemma 3.4, so m3 (Aut(I)) ≥ m − 1 ≥ 3. In particular J ∼  A9 or A6 . = If J ∈ Chev(3), then again by Lemma 3.1, I ∈ Chev(3). With [IA , 2.2.10] and the previous paragraph, we see that I/Z(I) ∼ = U4 (2) and m = 4, and we may assume, in view of conclusion (a), that m2,3 (X) − m3 (Z(X)) ≤ 3. Note that I ∼ = Sp4 (3), since that would force J ∼ = Sp4 (9), (P )Sp2n (3), or SU4± (3) = 2U4 (2) ∼ by [IA , 4.5.1, 4.9.1]; and then m2,3 (X) − m3 (Z(X)) ≥ 4, contradiction. Hence, I∼ = Co2 above that J/O3 (J) ∼ = L± = U4 (2). It follows exactly as in the case K ∼ 4 (3). Moreover, I ↑3 K/Z(K) with Z(K) a 2-group. Thus as b ∈ K, it follows from [IA , 4.7.3A, 4.2.2] that K/Z(K) is a classical group of level q(K) = 2; and then by [IA , 4.8.2] and the fact that m = 4, K/Z(K) ∼ = U6 (2), U5 (2), Sp8 (2), D4 (2), or D5 (2). Using [IA , 6.1.4] and the hypothesis that m2 (CK (B)) ≤ 1, we see that conclusion (j) holds. Next, suppose that J ∈ Chev(2) − Chev(3), so that J/Z(J) ∼ = U6 (2), U5 (2), 1 3 2 2 Sp6 (2), D4 (2), D4 (2), F4 (2), F4 (2 ), or Sp4 (8). As J is a proper pumpup of I with respect to the involution t, and m3 (Aut(I)) ≥ m − 1 ≥ 3, t induces a graph automorphism on J and we have J/Z(J) ∼ = U6 (2) or D4 (2) with I ∼ = Sp6 (2) and ∼ m = 4. This yields K/Z(K) = Sp8 (2) or F4 (2), as b ∈ K, and by [IA , 4.8.2, 4.7.3A]. Hence, conclusion (k) holds. Finally suppose that J ∈ Spor. Since I/O2 (I) ∈ C2 ∩ Chev(2), J ∈ C3 , m3 (Aut(I)) ≥ m − 1 ≥ 3, and I/O2 (I) ↑2 J/O2 (J), the only possibilities for J, by [V3 , 1.1] and [IA , 5.3], are J/Z(J) ∼ = Co3 , F i23 , F i24 , F2 , and F i22 . Accordingly I/O2 (I) ∼ = 2Sp6 (2), [2×2]U6 (2), [2×2]U6 (2), 2 2 E6 (2), and one of 2U6 (2), D4 (2), and Sp6 (2). Since K ∈ Chev(2) and I is a terminal component of CK (b) with CB (I) = b and m3 (B) = m3 (K), the 2Sp6 (2) and 2 2 E6 (2) cases are impossible by Lemma 3.37. So are the [2×2]U6 (2) cases, since then K ∼ = [2 × 2]2 E6 (2) and so m2 (CK (B)) > 1, contrary to hypothesis. Therefore J/Z(J) ∼ = F i22 , and so m2,3 (X) − m3 (Z(X)) ≥ 4. Hence either conclusion (a) holds or m ≥ 5. In the latter case as m3 (Aut(I)) ≥ m − 1 ≥ 4, we cannot have I∼ = Sp6 (2). It then follows from [IA , 4.8.2, 4.7.3A] that conclusion (l) or (m) holds, completing the proof at last.  Lemma 13.4. Suppose that K ∈ C3 with O3 (K) = 1 and A5 ↑2 K. Then K ∼ = A9 . ∼ A6 or A9 , and only A9 has Proof. If K ∈ Alt, then by definition of C3 , K = the requisite subcomponent. There are no such groups K ∈ Spor, by [IA , 5.3]; note that J1 , J2 , and M12 are T3 -groups. If K ∈ Chev(r), then r = 2 or 5 by [IA , 4.9.6]. By [IA , 2.2.10] and the definition of C3 , K ∈ Chev(2). By the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], K ∼ = L2 (16) or L± 3 (4); but then K ∈ C3 , contradiction. This completes the proof. 

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14. Semirigidity 14.1. Conditions for Semirigidity. Lemma 14.1. Let K ∈ Chev(2), t ∈ I2 (Aut(K)), and J a component of CK (t). Then t ∈ Syl2 (CAut(K) (J)). Proof. If t is a field or graph-field automorphism, the result holds by [IA , 7.1.4c]. By L2 -balance and the Borel-Tits theorem, the only other case to consider is that t is a graph automorphism. By [IA , 4.9.2af], either K is untwisted with t conjugate into ΓK , or K is a Steinberg group and t is conjugate into ΦK . Without loss, t ∈ ΓK or ΦK , respectively. If the lemma fails, then there exists D such that t < D ≤ CAut(K) (t J) and |D| = 4. By the Borel-Tits theorem, D ∩ Inn(K) = 1. By the structure of Out(K), we may assume that D = t, u with u ∈ I2 (ΦK ) (in the untwisted case), or D = u ∼ = Z4 with u ∈ Φ(K) (the Steinberg group case). In the untwisted case, by [IA , 4.9.2, 4.2.3], u induces an automorphism of order 2 on J, contradiction. In the Steinberg group case, the same conclusion actually holds, but it is perhaps easier just to note that with standard notation [IA , 2.10], the high root subgroup Xα of K lies in J, and is isomorphic to the additive group ∼ F+ q where q = q(K); then ΦK / t = Gal(Fq /F2 ) acts faithfully on Xα . Hence u = Cu (J) ≤ Cu (Xα ) = t, contradiction.  Lemma 14.2. Let K ∈ C2 , t ∈ I2 (Aut(K)), and J a component of CK (t). In the following cases t ∈ Syl2 (CAut(K) (J)): (a) J ∼ = L3 (3); (b) J ∼ = P Sp4 (3); (c) J ∼ = 2U6 (2) or U6 (2); (d) J ∼ = U5 (2); or (e) J ∼ = D4 (2). Proof. Using [IA , 5.3, 5.2.8, 4.5.1], and the definition of C2 [V3 , 1.1], we see ∼ that if K ∈ Chev(2), then K ∼ = L4 (3) (case (a)), L± 4 (3) (case (b)), or K = F i22 (cases (c) and (e)), and in all cases the desired conclusion holds. If K ∈ Chev(2), we quote Lemma 14.1.  Lemma 14.3. Let K ∈ C2 , t ∈ I2 (Aut(K)), and J a component of CK (t) with  L4 (3). Then J is semirigid in K. J∼ = = A6 and K ∼ Proof. Let t ≤ D ∈ Syl2 (CAut(K) (J)). We may assume that t < D, for otherwise, semirigidity holds trivially [IG , Remark, p. 45]. The possibilities for K are given in Lemma 3.39. Since K ∈ C2 , options (a) and (d) of that lemma do not occur, and if (e) occurs, then K ∼ = U4 (3) or 2U4 (3). Consider these cases viewing K as a (projective) orthogonal group with natural module V . If t is a graph automorphism, then t = D by [IA , 4.5.1], contradiction. So t ∈ Inndiag(K), whence t is the image of an involution τ ∈ O(V ) such that dim(CV (τ )) = 4. By Lemma 10.13, D ∼ = E22 . Moreover D contains the image r of a reflection ρ ∈ O(V ) such that CV (τ ) ⊆ CV (ρ), so J is not a component of CK (r), proving semirigidity. If Lemma 3.39b holds, then likewise CAut(K) (J) = D ∼ = E22 , and for all d ∈ D −t, E(CK (d)) ∼ A , see [I , 5.3m]. Thus the lemma holds in this case. Finally, = 8 A if Lemma 3.39c holds, we quote Lemma 14.1. The proof is complete. 

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Lemma 14.4. Let J ∈ C2 and t ∈ I2 (Aut(J)), and suppose that CJ (t) has a component K. Let Q ∈ Syl2 (CAut(J) (K) ∩ CAut(J) (t)). Then one of the following holds: (a) There exists R ≤ Q such that Q = t R, t ∈ R, and CJ (R) has a tinvariant component I such that K < I; 3 (b) (J/Z(J), K/Z(K)) = (M12 , A5 ), (J2 , A5 ), (Suz, L3 (4)), (Ru, 2B2 (2 2 )), or (Co1 , G2 (4)), and m2 (Q) = 2; or (c) K = U3 (3) or A6 , J/Z(J) ∼ = U4 (3) or L4 (3), respectively, and Q ∼ = Z4 or Q has a cyclic subgroup of index 2, respectively. Proof. Without loss, J is simple. If J ∈ Chev(2), then (a) holds by Lemma 14.1. If J ∼ = L3 (3) or L2 (q), q ∈ FM9, then ↓2 (J) = ∅, contrary to the existence of K. Next, suppose that J ∈ Spor. The condition J ∈ C2 narrows down the possibilities for J. We use the tables [IA , 5.3] freely. If t induces a non-inner automorphism on J, then since | Out(J)| ≤ 2 we can set R = Q ∩ Inn(J), and (a) will hold provided that if R = 1, then K is not a component of CJ (R). From the tables we see that in every case, R is elementary abelian, and if R = 1, then J ∼ = A5 , so = M12 , F i22 , or F i24 . In the first case K ∼ (b) holds. If J ∼ = Sp6 (2) with |R| = 2, and the pumpup = F i22 , then as R = 1, K ∼ of K in CJ (R) must be nontrivial (namely, 2U6 (2)), so (a) holds. If J ∼ = F i24 , 2 then K ∼ = 2 U6 (2). In this case we set Z = Z(K) and argue that tZ contains an involution u of type 2C, i.e., such that L := CJ (u) ∼ = F i23 . For if not, then all involutions of tZ are F -conjugate to t, and hence NJ (K)-conjugate to t. Therefore NJ (K) ≥ NJ (K t) with |NJ (K t) : CJ (t)| = 4. But this is impossible since it implies that |CJ (K) : Z(K)| = 4, whereas a Sylow 2-subgroup of the centralizer of a subgroup of J of order 11 has order only 8. Thus u exists. Then for some v ∈ Z # , CL (v) ∼ = 2F i22 , by Lemma 3.63. We put R = u, v and have E(CJ (R)) ∼ = F i22 , so Z t = R × t, as desired. Now we may assume that t ∈ J. In particular, J is not of characteristic 2-type. If J ∼ = J2 , then Q ∼ = E22 and = M12 , again (b) holds. Likewise if J ∼ (b) holds. If J ∼ = Suz, Co1 , or Ru, then t is determined up to conjugacy by the condition E(CJ (t)) = 1, and (b) holds. Keeping in mind that J ∈ C2 we are reduced (still in the sporadic case) to (J, K) = (HS, A6 ), (F i22 , 2U6 (2)), (F i23 , 22 U6 (2)), (F i24 , 2F i22 ), (F2 , F4 (2)), or (F1 , 2F2 ). In the last case (a) holds with R = 1 as CJ (t) ∼ = 2F2 . In the F i22 case, CAut(K) (K) ∼ = 2U6 (2)2 so we can take R = 1. In the HS, F i23 , F i24 , and F2 cases, Q ∼ = E22 and there is R ≤ Q with R ∼ = Z2 and E(CJ (R)) ∼ = A8 , 2F i22 , F i23 , and 2 2 E6 (2), respectively. Thus the lemma holds if J ∈ Spor. ∼ The only remaining cases are J ∼ = G2 (3) or L± 4 (3). If J = G2 (3), then t must be a graph-field automorphism, whence CAut(J) (K) = t by [IA , 7.1.4c]. Thus ∼ (a) holds with R = 1. Finally assume that J ∼ = L± 4 (3). If K = P Sp4 (3), then t is a graph automorphism and by [IA , 4.5.1], Q = t, so (a) holds. If K ∼  A6 = or U3 (3), the only other possibility, by [IA , 4.5.1], is K ∼ = L3 (3) and J ∼ = L4 (3), and again Q = t and (a) holds. Notice that if K ∼ = U4 (3), then = U3 (3) and J ∼ Q∼ = A6 . We refer to [IA , 4.5.1], where = Z4 , so (c) holds. Finally suppose that K ∼ we may assume that t is of class t2 (if it is of class γ2 , then |Q| = 2 and we are done). If Q ∩ J = 1, then J ∼ = L4 (3), t ∈ Q ∩ J is a projective involution, and Q ∩ Inndiag(J) is cyclic. As | Aut(J) : Inndiag(J)| = 2, (c) follows in this case. If

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∼ U4 (3) and Q∩Inndiag(J) ∼ Q∩J = 1, then as t ∈ Inn(J), J = = Z2 by [IA , 4.5.1]. As Outdiag(J) ∼ = D8 , Q is elementary abelian. Hence by [IA , 4.5.1], = Z4 and Out(J) ∼ if |Q| > 2, then Q = t × R where I = E(CJ (R)) ∼ = P Sp4 (3), and (a) holds. The proof is complete.  Lemma 14.5. Let J ∈ C2 and t ∈ I2 (Aut(J)), and suppose that CJ (t) has a component K ∈ C2 with K/O2 (K) ∈ Cp for some odd prime p. Let Q ∈ Syl2 (CAut(J) (K) ∩ CAut(J) (t)) Then one of the following holds: (a) There exists R ≤ Q such that Q = t R, t ∈ R, and CJ (R) has an x-invariant component I such that K < I; (b) (K, p) = (A5 , 5), J/Z(J) ∼ = M12 or J2 , and m2 (Q) = 2; or (c) p = 3, K = U3 (3), J/Z(J) ∼ = U4 (3), and m2 (Q) = 1; or (d) p = 3, K ∼ = L4 (3), and m2 (Q) = 2. = A6 , J/Z(J) ∼ Proof. If Lemma 14.4a holds, then conclusion (a) of this lemma holds. Otherwise, one of Lemma 14.4bc holds, and we get our stronger conclusion from the assumption that K/O2 (K) ∈ Cp .  14.2. Z4 -Semirigidity. The notion of Z4 -semirigidity is defined in [V2 , 8.2, 8.3]. Lemma 14.6. Let (t, K) ∈ ILo2 (G), where G is a group of even type, and K ∈ C2 ∩ C3 . Let Q ∈ Syl2 (C(t, K)) and Q ≤ R ∈ Syl2 (CG (K)). Let Q1 = Q unless the pumpup of K in CG (y) is trivial for all y ∈ I2 (Q), in which case set Q1 = R. Accordingly set Y = {x} or Y = I2 (Z(R)). Then K is Z4 -semirigid in G unless possibly K ∼ = A6 and for some z ∈ Y , (z, K) < (u, J) ∈ ILo2 (G) for ∼ some u and J = L4 (3) with CQ1 (z)/CQ1 (J z) embeddable in D8 . Moreover, if m3 (Aut(K)) ≥ 3, then K is semirigid in G. Proof. Suppose that the lemma fails. According to the definitions of semirigidity in G [IG , 7.2] and Z4 -semirigidity in G [V2 , 8.3], there then is z ∈ Y and a vertical pumpup (z, K) < (u, J) ∈ ILo2 (G) such that K is not semirigid in J, and if m3 (Aut(K)) < 3, then Sylow 2-subgroups of CAut(J) (K) are not isomorphic to Z4 . We apply Lemma 14.4. If conclusion (a) of that lemma holds, then K is semirigid in J, contradiction. If Lemma 14.4b holds, then K ∈ C3 , contradiction. So Lemma 14.4c holds. If K ∼ = U3 (3), then m3 (Aut(K)) = 2 and Sylow 2-subgroups of CAut(J) (K) are isomorphic to Z4 , contradiction. Therefore K ∼ = A6 and J ∼ = L4 (3). If t induces a graph automorphism on J, then CAut(J) (t K) ∼ = Z2 by [IA , 4.5.1], contradiction. So t induces an inner-diagonal automorphism on J. Thus regarding + J as P Ω+ 6 (3), t acts on J like an involution of Ω6 (3) with a 2- and 4-dimensional − ∼ eigenspaces of − type. Hence CAut(J) (K) = O2 (3) ∼ = D8 , which is the exception allowed in the lemma. The proof is complete.  15. Miscellaneous Lemma 15.1. Let K be a simple K-group and D ∈ Syl3 (K). Assume that D = d1  × d2  ∼ = A5 or Σ5 , i = 1, 2, and = E32 , CK (di ) = di  × Xi with Xi ∼ D  CK (d1 d2 ) with index 2 or 4, for  = ±1. Assume also that |K|{2,3} divides 35. Then K ∼ = A8 .

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∼ An , then n = 8 by the structure of CK (di ). Likewise this Proof. If K = structure implies that K ∈ Spor [IA , 5.3]. Assume that K ∈ Chev(r). By [IA , 4.2.2] and the structure of CK (di ), r = 2 or 5. If r = 5, then as |K|5 = 5, K ∼ = A5 , which is absurd. Thus, r = 2 and by [IA , 4.2.2], q(K) = 2 or 4. Since |K|3 = 32 , K is not of exceptional type, so K is a classical group, indeed (using also |K|5 = 5 and |K|{2,3} divides 35) K ∼ = L3 (4) or L4 (2). By the structure of CK (di ), K ∼ = L4 (2) ∼ = A8 , as asserted.  Lemma 15.2. Let L ≤ H with L ∼ = L3 (4) and H ∼ = U6 (2). Then CAut(H) (L) = 1. Proof. Let x ∈ I7 (L). Then it is easily computed that CInndiag(H) (x) = x, y, where y ∈ I3 (Inndiag(H)−Inn(H)) and CH (y) ∼ = P ΣL2 (8). Then using [IA , 4.9.2], it follows that CAut(H) (x) = x, y, γ, where γ 2 = 1 and CH (γ) ∼ = Sp6 (2). Neither Sp6 (2) not P ΣL2 (8) contains L, as L contains Frobenius groups of order 9 · 8 and hence has no faithful representation over F2 of dimension less than 8. Hence,  CAut(H) (L) = 1, as claimed. + Lemma 15.3. Let H ∈ K be a component of a group X ≤ O12 (2) with |O2 (X)| ≤ 3 and O2 (X) = 1. Let b ∈ I3 (H) and w ∈ I5 (H). Suppose that CX (b) is an extension of E35 by Z2 × A, and CX (w) ∼ = Z5 × Σ3 × A∗ , where either A ∼ = A4 and ∗ ∗ ∼ ∼ A = 1, or A = Σ4 and A = Z2 . Then the following conditions hold: (a) H = F ∗ (X) ∼ = 3U4 (3); (b) Z(H) = O2 (H) is inverted in X; (c) X/H is elementary abelian of order 2|A∗ |; and (d) On the natural Ω+ 12 (2)-module, H is irreducible and Z(H) is fixed-pointfree.

Proof. We have CX (H) ≤ O 5 (CX (w)) so CX (H) embeds in Σ3 × A∗ . In particular m3 (Aut(H)) ≥ 4. Also w ∈ Syl5 (X). If H ∈ Spor, then by the previous paragraph and [IA , 5.6.1], H/Z(H) ∼ = O N ; but then CX (b) is nonsolvable by [IA , 5.3s], contradiction. Likewise if H/Z(H) ∼ = An , then n ≥ 12 so m5 (H) > 1, contradiction. Hence, H ∈ Chev(r) for some r. 2 ∼ If r > 3 then either r = 5 or mr (H) ≤ mr (Ω+ 12 (2)) ≤ 2, so H = L2 (r) or L2 (r ) and m3 (Aut(H)) = 1, contradiction. Suppose that r = 2. Since m5 (X) = 1 and m3 (Aut(H)) ≥ 4, it follows easily from [IA , 4.10.3a] that H is a classical group of untwisted Lie rank n ≥ m3 (H) ≥ 3 + and level q, say. Since X embeds in O12 (2), |H|2 ≤ 230 (by Tits’s Lemma [IA , 2.6.7]) and m2 (X) ≤ 15. Suppose that m3 (H) = 3. Then H admits a field automorphism of order 3 so q ≥ 8. The rank and 2-order conditions imply that the only possibility 2 is H ≥ L± 4 (8) (see [IA , 3.3.3, 4.10.3a]). But then H ≥ L2 (8 ) ≥ Z13.5 , contradiction. Therefore, n ≥ m3 (H) ≥ 4. Likewise if q ≥ 8, then we reach a similar contradiction. Thus q = 2 or 4. If q = 4, then as n ≥ 4, we deduce with [IA , 4.10.3a] that m5 (H) > 1, contradiction. Therefore q = 2. If H is a linear, symplectic, or orthogonal group, then as m3 (H) ≥ 4, H contains L8 (2) or D4 (2) so m5 (H) > 1, contradiction. Hence, H/Z(H) ∼ = Un (2) for some n ≥ 5. If n = 5, then m3 (H) = 4 so m3 (CX (H)) = 1. But 3 divides |CH (w)| in this case, so m3 (CX (w)) > 1, contradiction. If n > 6, then m3 (CH (w)) ≥ m3 (SUn−4 (2)) > 1, contradiction. Therefore, n = 6. Likewise, if m3 (CX (H)) ≥ 1, then m3 (CX (w)) > 1, contradiction. So F ∗ (X) ∼ = U6 (2).

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∼ P GU6 (2) of index at most 2, and b ∈ X0 is As m3 (X) ≥ 5, X contains X0 = diagonalizable. Let B be a full diagonal subgroup of X0 containing b. Then by our assumptions, B  CX0 (b) and |CX0 (b)/B|3 = 3. But then b must have an eigenvalue of multiplicity at least 3, so CX (b) involves SU3 (2). However, by our assumptions, CX (b) does not involve Q8 . This contradiction completes the proof that r = 3. We have + 5 ≤ m3 (CX (b)) ≤ m3 (X) ≤ m3 (O12 (2)) = 6 + 8 and |X|3 ≤ |O12 (2)| = 3 . Again, m3 (H) ≥ 4. If H ∼ = L2 (3n ), then n = 4 or + 2 ∼  B2 (32 ). As 13 does 5, so 41 or 11 divides |O12 (2)|, contradiction. Likewise H = + n not divide |O12 (2)|, H/Z(H) ∼ = Lm (3 ), m ≥ 3, or G2 (3). If H ∼ = U3 (3n ), then n = 2 and again Lagrange’s theorem is contradicted. Now our bounds on m3 (H) and |H|3 force H/Z(H) ∼ = U4 (3). If Z(H) = 1, then Clifford’s theorem applied to a parabolic subgroup P ∼ = + E34 A6 shows that H does not embed in O12 (2), contradiction. Since m3 (CX (w)) = 1, H = F ∗ (X) ∼ = E35 A6 , and = 3U4 (3), which is (a). By [IA , 6.4.4a] H contains P0 ∼ + Clifford’s theorem implies that P0 is an irreducible subgroup of O12 (2), implying (d). Now w Z(H) = CH (w) and w is rational in H so (c) follows by a Frattini argument, and as Z(H) ≤ Z(CX (w)), (b) follows as well. The proof is complete.  + Lemma 15.4. Suppose Y ≤ X ∼ (2) and Y has odd order. Let A = = O12 AutX (Y ). Then A does not involve U4 (2). 

Proof. Suppose false. Then A0 := O 2 (A) involves U4 (2). By [IG , 3.17(ii)], CA0 (F (Y )) has odd order. So for some odd prime r, AutX (Or (Y )) involves U4 (2), i.e., we may assume that Y is an r-group. But Y ≤ X, so mr (Y ) ≤ 2 unless r = 3. As GL2 (r) does not involve U4 (2), r = 3. Note that |X|3 = 38 and |U4 (2)|3 = 34 , so |Y | ≤ 34 . Since Aut(Y ) involves U4 (2), order considerations imply that Y ∼ = E34 . As Y ≤ X, Y has at most 6 irreducible nontrivial submodules V1 , V2 , . . . on the natural X-module V , all of dimension 2. The Vi are permuted by A, and as U4 (2) does not embed in Σ6 , the common stabilizer in A of all the Vi still involves U4 (2). But this common stabilizer embeds in the direct product O(V1 ) × O(V2 ) × . . . , a solvable group. This contradiction completes the proof.  Lemma 15.5. U4 (2) is not involved in A10 . Proof. |A9 |/|U4 (2)| = 7, but A9 has no subgroup of index 7. Hence if the lemma fails, then U4 (2) has a transitive action on 10 letters. By Tits’s lemma, a point stabilizer then lies in a (proper) parabolic subgroup (of P Sp4 (3) ∼ = U4 (2)). But no such parabolic subgroup exists of index dividing 10, contradiction.  Lemma 15.6. Let X be a K-group and suppose that every component of L := E(X) is a C2 -group. Suppose that |L|2 = 2n with 14 ≤ n ≤ 17. Suppose also that t ∈ I2 (X) and CX (t) has a subgroup H of index 2 isomorphic to a faithful extension of E29 by U4 (2) or Aut(U4 (2)), with a chief factor of order 28 . Then Z(L) = 1. Proof. Suppose that the lemma fails and let K be a component of L with Z(K) = 1. The structure of CX (t) implies that t normalizes K. By Lemma 10.69, |CKt (t)|2 > 8. Hence |H ∩ K t |2 > 4. It follows that the chief factor of H of order 28 lies in K ∩ H. Hence Aut(K) contains a faithful extension F of E2n , n ∈ {8, 9}, by U4 (2). Therefore, and by our hypothesis, |K|2 = 2n , 14 ≤ n ≤ 17. Also |K|3 ≥ |U4 (2)|3 = 34 . As Z(K) = 1 and K ∈ C2 ,

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we conclude that K ∼ = 2Suz, [2 × 2]D4 (2), 2U6 (2) or [2 × 2]U6 (2). In the last three cases, by the Borel-Tits theorem, some parabolic subgroup of K contains a copy of F . But it is easily seen that no such parabolic subgroup exists. Likewise if K ∼ = 2Suz, then m2 (K) < 8 by [IA , 5.6.1], contradiction. This completes the proof.  Lemma 15.7. Suppose that the K-group X has a subgroup H of odd index with + F ∗ (H) ∼ (2) with F ∗ (H) having = U4 (2) and m3 (X) = 3. If X embeds in Y := O10 8 a composition factor of order 2 on the natural F2 Y -module, then X = H. Proof. By Lemma 15.4, [F ∗ (H), F (X)] ≤ [F ∗ (H), O2 (X)] = 1. As H contains a Sylow 2-subgroup of E(X), F ∗ (H) normalizes every component of E(X), and then F ∗ (H) ≤ E(X) by the Schreier property. Since H embeds in Aut(F ∗ (H)) and has odd index in X, E(X) is a single component. Suppose that O2 (X) = 1 and let g ∈ O2 (X) have prime order r. As m3 (X) = 3, r > 3. But no element of order r in Y centralizes a U4 (2) subgroup of Y with an 8-dimensional composition factor on the natural F2 Y -module (see [IA , 4.8.2]). This contradiction to our hypothesis implies that O2 (X) = 1. Also as F ∗ (H) is simple and H has odd index, O2 (X) = 1. Let X0 = E(X). Then X0 is a simple K-group, U4 (2) ∼ = F ∗ (H) ≤ H ∩ X0 ≤ X0 ≤ Y , and H ∩ X0 is embeddable in Aut(U4 (2)) with |X0 : X0 ∩ H| odd. The conditions that |X0 | divides |Y | and m3 (X0 ) ≥ m3 (H) = 3 imply that m3 (X0 ) = 3 and X0 ∈ Spor, and if X0 ∈ Alt then X0 ∼ = A9 or A10 , which contradicts Lemma 15.5. Thus, X0 ∈ Chev(r) for some prime r. If r > 3, then as |X0 | divides |Y |, X0 ∼ = L2 (r n ) for some n, contradicting m3 (X0 ) ≥ 3. If r = 3 but X0 ∈ Chev(2), then as m3 (X0 ) = 3, X0 ∼ = L2 (33 ) by [IA , 3.3.3], a contradiction as 3 |H| does not divide |L2 (3 )|. Thus we may assume that r = 2, whence by Tits’s lemma, either X0 ≤ H or O2 (P ) ≤ H ∩ X0 ≤ P for some proper parabolic subgroup P of X0 . As X0 ∩ H ∼ = U4 (2) or Aut(U4 (2)), it must be that X0 ≤ H, which implies that X = H, as claimed. The proof is complete.  Lemma 15.8. Suppose that F ∗ (X) is a simple K-group. Let Y ≤ X with F (Y ) ∼ = L3 (4), |Y : F ∗ (Y )| ≤ 2, and |X : Y | = 22 or 792. Then |X : Y | = 22 and ∗ F (X) ∼ = M22 . ∗

Proof. Our hypotheses yield |X| = c|M22 |, where c = 1, 2, 36, or 72. In particular |X|3 = 32 or 34 ; in the first case |X|2 = 27 or 28 , while in the second case, |X|2 = 29 or 210 . Let F = F ∗ (X). If F ∈ Spor, these conditions imply F ∼ = M22 , so |X : Y | = 22. If F ∈ Alt, the conditions |X|5 = 5 and |X|11 = 11 are inconsistent. Thus, F ∈ Chev(r) for some r. If r > 3, then |X|r ≤ r 2 , so F ∼ = L2 (q) for some odd q, whence m2 (X) ≤ 3, a contradiction as X contains L3 (4). If r = 3, then since F ∼  L2 (q), the only possibility is F ∼ = P Sp4 (3), a contradiction as 11 = divides |X|. Therefore, F ∈ Chev(2) − Chev(3) − Alt. n Since Y ≤ X, clearly F ∼  L2 (2n ), 2B2 (2 2 ), or U3 (2n ). If q(F ) > 2, then = as |X|2 ≤ 210 , F ∼ = L3 (8) or B2 (4), again a contradiction as 11 divides |X|. So q(F ) = 2. The only possibilities are L5 (2) or Sp6 (2) (impossible because of 11), and U5 (2) (with 3-share 35 , contradiction). The proof is complete.  Lemma 15.9. Let K ∈ C2 and suppose that K  X and Z(X) has even order. Let u ∈ I2 (K ∪ O2 (X)). Then m2 (CX (u)) ≥ 3.

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Proof. If K is simple, then as Z(X) has even order, the result is trivial. So we may assume that Z(K) = 1. By Lemma 2.8, m2 (K) ≥ 3. Hence we may assume that u ∈ K. Let T ∈ Syl2 (K). As m2 (T ) ≥ 3, T has a four-subgroup U  T . As K is perfect, u has a K-conjugate v ∈ CT (U ), by the Thompson transfer lemma. If v ∈ U , then CT (v) ≥ v U ∼ = E23 . If v ∈ U , then for any E ∈ E3 (T ),  v ∈ CE (U )U ∼ = E2m , m ≥ 3. The proof is complete. Lemma 15.10. Let K ∈ C2 with m3 (K) = 1. Then K contains no element of order 6. If moreover t ∈ I2 (Aut(K)) and m2 (CK (t)Z(K)/Z(K)) = 1, then K∼ = L2 (q), q ∈ FM. Proof. As m3 (K) = 1 and K ∈ C2 , K/Z(K) ∼ = L2 (2n ), L3 (22n+1 ), or U3 (22n ), or L2 (q), q ∈ FM. The first conclusion then follows easily. If Z(K) = 1 then by [IA , 6.1.4], K ∼ = SL2 (q), q ∈ FM, a contradiction as K ∈ C2 . Therefore Z(K) = 1 and m2 (CK (t)) = 1. Hence t induces an outer automorphism on K: a field automorphism on K ∼ = L3 (22n+1 ) or = L2 (2n ), a graph automorphism on K ∼ 2n U3 (2 ), or a diagonal automorphism on K ∼ = L2 (q), q ∈ FM. Then by [IA , 4.9.1, 4.9.2], if K ∈ Chev(2), we have K ∼ = L2 (4) ∼ = L2 (5) or K ∼ = L3 (2) ∼ = L2 (7). The lemma follows.  Lemma 15.11. Let X be a 3 -group which is a K-group, and B a 3-group acting on X with Z2 ∼ = I2 (CX (B)) ≤ Z(X). If m2 (X) ≤ 2, then X is solvable. Proof. Without loss O2 (X) = 1. Since m2 (L) ≥ 3 whenever L ∈ K2 with n L/Z(L) ∼ = 2B2 (2 2 ), F ∗ (X) = O2 (X) =: Q. If A is any characteristic abelian subgroup of Q, then as Z(X) = 1 and m2 (X) ≤ 2, [A, B] = 1 so A is cyclic. Thus Q is of symplectic type by P. Hall’s theorem [IG , 10.13]. For any subgroup Y ≤ X of odd order, [Q, Y ] is extraspecial by [V2 , 4.12], and as m2 (Q) ≤ 2, |Y | = 5. Hence X is a {2, 5}-group, so X is solvable.  Lemma 15.12. There is no simple K-group L containing a subgroup M ∼ = M22 and such that |L : M | divides 210 and 210 | Aut(M22 )| divides | Aut(L)|. Proof. Such a group L, if it existed, would have |L| dividing n := 217 32 5.7.11 and | Aut(L)| a multiple of 2n. Then L ∈ Alt ∪ Spor, since (apart from A6 )  L2 (q) | Out(L)| ≤ 2, forcing |L| = n, which is impossible [IA , 5.3]. Obviously L ∼ = for any q, so L ∈ Chev(2). Now 11 divides |L|, but |L| has no divisor 25a − 1. As |L|2 ≤ 217 , this reduces the possibilities to L = Un (2), n ≥ 5; but then |L|3 > 32 , a final contradiction.  Lemma 15.13. Let t ∈ I2 (F3 ) and C = CF3 (t). Let X be a group with |X : C| = 2. Set Q = O2 (C). Then X = CCX (Q). Proof. Set X = X/Z(Q). We have C/Q ∼ = A9 , and if x ∈ I3 (C) maps on a 3-cycle in C/Q, then CQ (x) = 1. Suppose that X/Q ∼ = Σ9 . Then a transposition u ∈ C/Q inverts a conjugate of x so u acts freely on Q. Hence, CX/Q (u) = u × Y (Y ∼ = Σ7 ) acts on Q/[Q, u] ∼ = E24 and [Q, u] ∼ = E24 . But Y does not embed in ∼ L4 (2) = A8 , so [Y, Y ] stabilizes the chain Q > [Q, u] > 1. As [Y, Y ] is perfect and acts faithfully on Q, this is a contradiction. Thus X/Q ∼ = Z2 × A9 . As C/Q acts absolutely irreducibly on Q, CX (Q)/Q = O2 (X/Q) ∼ = Z2 . As Q is extraspecial, CX (Q) = QCX (Q) and the proof is complete. 

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Lemma 15.14. Suppose K, L ∈ C3 , L/Z(L) ∼ = U4 (3) or M c, and L 1; P Ωn (3 ), n ≥ 7 (since − m m3 (Ω6 (3)) = 4; see [IA , 6.4.4]); P Spn (3 ), n ≥ 6, or n ≥ 4 and m > 1; and all groups of exceptional type except G2 (3), whose order, however, is not divisible m ∼ by |A6 |. As m2 (E) ≥ 4, K/Z(K) ∼ = L± n (3 ) for any n ≤ 3. Hence K/Z(K) = ± + L4 (3). If K ∼ (3) = L4 (3) ∼ = Ω6 (3), then by Clifford’s theorem, the natural Ω+ 6 module V decomposes as the orthogonal sum of six lines permuted transitively by E. Hence the six lines are all isometric, so V is of − type, a contradiction. Therefore  K/Z(K) ∼ = U4 (3), as claimed.

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Lemma 15.17. Suppose F (X) = 1 and |X| = 2a 3b 5. Then X embeds in Aut(K) for K = A5 , A6 , or P Sp4 (3). Proof. This is immediate from the F ∗ theorem, the pa q b theorem, and [III17 , 9.9].  Lemma 15.18. Suppose that K ∈ C2 , T ∈ Syl2 (K), I ≤ K with I ∼ = 2U4 (3), T normalizes I, Z(I) ≤ Z(K), and |T | ≤ 215 . Let y ∈ I2 (T ∩ I) and assume that CK (y) is solvable. Then K = I. Proof. If K ∈ Chev(2), then by Tits’s lemma [IA , 2.6.7], T I ≤ P for some parabolic subgroup P of K. Then [I, O2 (P )] ≤ I ∩O2 (P ) ≤ Z(K) so [I, O2 (P )] = 1, contradicting [IA , 2.6.5e]. If K ∈ Chev(3) ∩ C2 , then |K| is not divisible by |I| unless I = K. By Sylow 2-structure, K ∼  L2 (q), q ∈ FM9. Hence K ∈ Spor. = As |T | ≤ 215 , m3 (K) ≥ m3 (I) = 4, and Z(K) has even order, the only possibility is K ∼ = 2Suz [IA , 5.3, 5.6.1]. But in that case for any y ∈ I2 (K), CK (y) is not solvable, contrary to hypothesis. The lemma is proved.  Lemma 15.19. Suppose that F ∗ (X) = O3 (X) and X = O3 (X)L with L ∼ = SL3 (4) and Z(L) ≤ Z(X). Then m2,3 (X) > 4. Proof. Suppose false and take a counterexample with Y := O3 (X) minimal. Then Y has exponent 3 and class at most 2, and if Y has class exactly 2, then L centralizes Z(Y ). Let P = NL (Q) be a parabolic subgroup of L of type L2 (4), so that Q ∼ = E24 . Now P permutes the hyperplanes of Q transitively, so for any involution t ∈ Q, m3 (CY /Φ(Y ) (t)) ≥ 7. In particular, we are done if Y is abelian, or even if |Φ(Y )| = 3, since then m3 (CY (t)) ≥ 5. On the other hand, if |Φ(Y )| ≥ 9, then by the Thompson dihedral lemma, X contains the direct product of Z(Y ) with four copies of Σ3 , and again m3 (CY (t)) > 4 for some involution t ∈ L. The proof is complete.  Lemma 15.20. Let K be a K-group with K ≤ L6 (2). Suppose that J ≤ K with |K : J| odd, and J ∼ = A5 or A7 . = A5 or Σ5 . Then E(K) ∼ ∼ Proof. Let J0 = [J, J] = A5 . As K ≤ L6 (2), K has sectional r-rank ≤ 3 for all odd primes r, with strict inequality if r > 3. On the other hand, J0 contains subgroups isomorphic to Z5 and A4 . It follows that [F (K), J0 ] = 1. Therefore [E(K), J0 ] = 1 and as J contains a Sylow 2-subgroup of K, J0 ≤ E(K). Let  A7 , so T ∈ Syl2 (E(K)), so that T ∼ = E22 or D8 . We may assume that E(K) ∼ = ∼ K = L2 (q) for some odd q by Lemma 10.29. But by [IA , 6.5.1], Σ5 does not embed in L2 (q), so |T | = 4 and q ≡ ±3 (mod 8). As |K|2 divides |L6 (2)|2 = 34 .5.72 .31, Lagrange’s theorem forces q = 5, as desired.  Lemma 15.21. Let P = P1 × P2 where P1 ∼ = E3n for some n, and P2 ∼ = 31+2r ∼ or 1 for some r. Suppose that u, t × L acts on P with L = L2 (8) acting faithfully and u, t ∼ = E22 . If |CP (t)| ≤ 3, then for some v ∈ I2 (t, u − t), m3 (CP (v)) ≥ 7. Proof. Let N be a Sylow 2-normalizer in L. Then N ∼ = F8.7 acts faithfully on P , hence faithfully on CP (v) for some v ∈ t, u − t, as |CP (t)| ≤ 3. Let P = P/Φ(P ) and let R be a faithful irreducible N -submodule of CP (v). Thus |R| = 37 . Let R be the full preimage of R. If R is abelian, then we are done, so assume that Φ(R) = 1. Then Φ(R) = Φ(P ) ∼ = Z3 and as |R| = 38 , R is not extraspecial. Therefore Z(R) > Φ(R). By irreducibility, Z(R) = R, contradiction, and the lemma follows. 

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Lemma 15.22. Let K ∈ K be a simple subgroup of GL12 (2), with a subgroup L∼ = M24 such that |K : L| is a subsum of 1 + 759 + 759 + 2576 containing the 1 term. Then K = L. Proof. The only subsums s > 1 which divide |GL12 (2)|/|L| are s = 1519 = 72 31 and s = 212 − 1 = 32 5.7.13. So |K| = s|M24 | = 210 33 5.73 11.23.31 or 210 35 52 72 11.13.23. Neither of these is the order of any simple group in Alt ∪ Spor or of any group of the form L2 (r a ) or L± 3 (r) for any odd prime, so K ∈ Chev(2). Because 23 divides |K|, |K| must have a factor of 211a − 1 for some a ≥ 1, but then  |K|2 > 210 , contradiction. The proof is complete. Lemma 15.23. Suppose K ∈ C2 is a p -group for some prime p, and x is a p-element of Aut(K) centralizing a Sylow 2-subgroup S of K. Then x = 1. Proof. Using the definition of C2 -group [V3 , 1.1], as well as [IA , 2.5.12, 5.3], we see that Out(K) is a 2-group if K ∈ C2 −Chev(2). So if x = 1, then K ∈ Chev(2).  But then CAut(K) (S) is a 2-group by [GL1, I.13–3], a contradiction. Lemma 15.24. Suppose that p is an odd prime, X is a K-group with Op (X) = 1, and K = Lp (X) = 1. Suppose that x ∈ Ip (Z(X)) and let L = K x. Let R ∈ Sylp (X) and S = R ∩ K x. Assume that mp (S) ≤ 3 and every component of K lies in Cp and has p-rank at least 2. Then for some noncyclic S1 char S, R = S1 CR (S1 ). Proof. Since x ∈ Z(S), the condition mp (S) ≤ 3 implies that K has at most 2 components, which then are R-invariant. If p does not divide | Out(K)|, then R = SCR (K) and we can take S1 = S. So we may suppose that p divides | Out(J)| for some component J of K. In particular J ∈ Chev(r) for some r. If r = p, then J admits a graph or field automorphism of order p. By [IA , 6.1.4, 2.5.12], Z(J) = 1. Moreover, by [IA , 4.9.1, 4.9.2, 3.3.3], mp (J) > 2 so mp (S) ≥ mp (J x) > 3, contradiction. Therefore, r = p. Then from the definition of Cp and the fact that mp (J) > 1 by assumption, r = 2 and (p, J) = (3, (S)U6 (2)), (3, D4 (2)), (3, 3D4 (2)), 5 or (5, 2F4 (2 2 )). The first two cases are out since mp (J x) ≤ 3 by assumption. In the last two cases, we can take S1 = Ω1 (Z(S)), which lies in Z(R). This completes the proof.  Lemma 15.25. Let P be the central product of n copies of D8 . Then Sp2n (q) contains no copy of P , for any odd q. Proof. For n = 1, Sp2 (q) = SL2 (q) has a unique involution. Inductively, for n > 1, suppose that P ≤ Sp(V ), where dimFq (V ) = 2n . Let t ∈ I2 (P ) − Z(P ). Then CP (t) = t × P0 where P0 is the central product of n − 1 copies of D8 . Also CV (t) is nondegenerate, and as t shears to Z(P ), dim(CV (t)) = 2n−1 and P0 embeds in Sp(CV (t)), a contradiction by induction. The lemma follows.  Lemma 15.26. Let S ∈ Syl2 (U3 (8)) and suppose that S acts faithfully on a 3-group P . Then for some involution t ∈ S, m3 (CP (t)) ≥ 5. Proof. Let Z = Z(S). Then CAut(S) (Z) (which contains an element of order 9) acts irreducibly on S/Z. Hence for any hyperplane Y ≤ Z, S/Y is extraspecial of order 26 , and so requires 8 dimensions for a faithful F3 -representation. We may assume that P has class at most 2 and exponent 3, and P = [P, Z]. Let P = P/Φ(P ) = P 1 × · · · × P 7 , where each P i = CP (Yi ) and Y1 , . . . , Y7 are the

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hyperplanes of Z. There exist two values of i, say i = 1, 2 such that P 1 = 1 = P 2 . Then |P i | ≥ 38 , i = 1, 2, so Q := CP (Y1 ∩ Y2 ) satisfies |Q/Φ(Q)| ≥ 316 . If the lemma fails, then m3 (Q) ≤ 4. In particular m3 ([Q, Q]) ≤ 4. If m3 ([Q, Q]) = 4 then obviously m3 (Q) > 4, contradiction. So m3 ([Q, Q]) ≤ 3, whence |Q : CQ (x)| ≤ 33 for every x ∈ Q. It follows that an abelian subgroup R ≤ Q exists with m3 (R) ≥ 4 and R∩[Q, Q] = 1. If [Q, Q] = 1 then m3 (R[Q, Q]) > 4, contradiction, so [Q, Q] = 1 and m3 (Q) ≥ 16, contradiction. The lemma is proved.  16. Preliminary Lemmas for Theorem C∗6 : Stage 1 16.1. p-Ranks, Tpi -Groups, and Flat TGp -Groups. Lemma 16.1. If K ∈ Tp , p an odd prime, then mp (K) ≤ 2, mp (Inn(K)) ≤ 2, and, if K ∈ Chev, then mp (Inndiag(K)) ≤ 2. Proof. This is immediate from the definition of Tp [V7 , (1D)], and from [IA , 4.10.3, 5.6.1].  Lemma 16.2. Let p be an odd prime and K ∈ Tp . Suppose that E ∈ Ep3 (Aut(K)). Then p = 3, K/Z(K) ∼ = L3 (q),  = ±1, q ≡  (mod 3), or G2 (8), and some e ∈ E # is a field automorphism. Proof. By Lemma 16.1, some e ∈ E induces an outer automorphism on K, whence K ∈ Chev and e induces a field, graph-field, or graph automorphism on K. From the definition of Tp [V7 , (1D)], p = 3 and K admits no graph or graph-field  automorphism, and K/Z(K) ∼ = L3 (q) or G2 (8), as claimed. Lemma 16.3. Let p be an odd prime and let the sets Tpi , 1 ≤ i ≤ 5, be defined by [V7 , Definition 1.1]. Suppose that (x, K), (y, L) ∈ ILp (G) and (x, K) < (y, L). If L/Op (L) ∈ Tpi for some i, then K/Op (K) ∈ Cp or K/Op (K) ∈ Tpj for some j ≥ i. Proof. Set K = K/Op (K). We may assume that Op (L) = 1. We freely use the definitions [V3 , 1.1] and [V7 , 1.1]. Notice that if i ≤ 4, then mp (L) > 1. If i = 5, then mp (L) = 1, whence mp (K) = 1. Hence K ∈ Cp ∪ Tp5 , as desired. If i = 4, then L ∼ = L3 (4) or F i22 ; thus mp (K) = 1 and again K ∈ Cp ∪ Tp5 . We may now assume that p = 3. If i = 3, then by [IA , 4.8.2, 4.8.4], K ∼ = L2 (q) for some q ≡ 0 (mod 3), so mp (K) = 1, yielding the desired result. If i = 2, the only additional possibility for K is L3 (q) for some q ≡  (mod 3),  = ±1, in which case K ∈ T32 ∪ T33 . Finally, similar reasoning applies if i = 1 unless (x, K) < (y, L) is a diagonal pumpup, in which case the lemma holds with j = 1, 2, 3, or 4.  Lemma 16.4. Let K ∈ TG3 be flat, i.e., such that whenever J ∈ K3 and J 2, ↑3 (SL3 (q)) contains G2 (q), 3D4 (q), F4 (q), and E6 (q 1/2 ), if q = q02 with q0 ≡ − (mod 3); also n SU3 (22n+1 ) ↑3 2F4 (2 2 ) for n ≥ 1, all by [IA , 4.7.3A]. As G2 (2) , 3D4 (2), F4 (2), and 1 2 F4 (2 2 ) are not G3 -groups, all remaining groups G3 -groups K of exceptional type satisfy m  3 (K) > 4. Thus, assume that K is a classical group. Again, if q > 2 and q ≡  (mod 3), then ↑3 (SL3 (q)) contains Sp2n (q) for n ≥ 3 and Dn± (q) for n ≥ 4, as well as Ln (q) for n ≥ 4. Also for q ≡ − (mod 3), ↑3 (SL3 (q 2 )) contains SL− n (q), n ≥ 6. The only alternative for K, since the lemma fails, is that q(K) = 2, and specifically, if  3 (K) ≤ 4 then rule out all K∼ = Ln (2), then n < 6. The conditions K ∈ G3 and m possibilities except those listed in the lemma, contradiction.  The final assertion follows from [IA , 4.10.3a, 5.2.10, 5.6.1]. 16.2. Pumpups and Subcomponents. Lemma 16.5. Suppose that p is odd and L ∈ Tp with Op (L) = 1 and mp (L) > 1.  L. Assume that every intermediate term Suppose that L 4 [IA , 4.10.3a] and = U3n (q). If K ∼ = E6− (q) or n ≥ 3, then m  SU6 (2), since it is not the so K ∈ G3 − TG3 , contrary to assumption. Also K ∼ = case that SU6 (2) ↓3 SU3 (8) [IA , 4.8.2, 4.8.4]. So K ∼ = SU6 (8) and by inspection, Z(J) ≤ Z(K), a contradiction. The proof is complete.  Lemma 16.9. Let K ∈ K3 be a C3 -group, a T3 -group, or a flat TG3 -group. Let b ∈ I3 (Aut(K)) and L = CK (b)/O3 (CK (b)). Then the following conditions hold: (a) E(L) has no component isomorphic to SU6 (2); and (b) If K/Z(K) ∼ = U6 (2), then L has no subgroup isomorphic to SU3 (2)×U4 (2). Proof. If (b) fails, then K/Z(K) contains U4 (2) × M with M ∼ = U3 (2) or SU3 (2). So m3 (K/Z(K)) ≥ 5. But m3 (U6 (2)) = 4, a contradiction. If (a) fails, then m3 (K) ≥ m3 (L) ≥ 5. But if K ∈ T3 or K is a flat TG3 -group, then m3 (L) < 5 (see [IA , 4.10.3a, 5.3]). Hence, K ∈ C3 . As U6 (2) ∈ Chev(r) ∪ Alt for any r > 2, K ∈ Chev(2) ∪ Spor. If K ∈ Chev(2), then as K ∈ C3 and m3 (K) ≥ 5, K ∼ = SU6 (2) ([V3 , 1.1], [IA , 4.10.3a]), which is absurd. Therefore K ∈ Spor, and by inspection  of [IA , 5.3], it is impossible that SU6 (2) ↑3 K. The lemma follows. Lemma 16.10. Let L = F ∗ (X) = L3 (q),  = ±1, q ≡  (mod 3), q > 2. Suppose t ∈ I3 (X) and t induces a nontrivial field automorphism on L. Then there exists b ∈ I3 (CL (t)) such that E(CL (b)) ∼ = SL2 (q). Proof. Since all field automorphisms of order 3 are conjugate to t or t−1 under Inndiag(L) [IA , 4.9.1], we may assume that L = L3 (r 3 ) (with respect to the form I if  = −1) and t raises all matrix entries to the rth or r 2 th power, according as  = 1 or −1. Since r ≡ r 3 ≡  (mod 3), q = r 3 ≡  (mod 9). Let ω be a primitive 9th root of unity; then the image b of diag(ω, ω, ω −2 ) modulo scalars lies in K and has the desired properties.  Lemma 16.11. Suppose that L ∼ = L2 (q), q > 8, and L ↑3 J via x ∈ I3 (Aut(J)), where J ∈ K3 is a C3 -group, a simple T3 -group, or a flat TG3 -group. Suppose that  = ±1 and q ≡  (mod 3). Then J is isomorphic to one of the following groups: −η 1/2 L2 (q 3 ), L3 (q), L− ), n = 4, 5. The last case occurs only 4 (q), P Sp4 (q), or Ln (q 1/2 ≡ η (mod 3), where η = ±1. In any case, every if q is a perfect square and q element of I3 (CInn(J) (x)) induces an inner automorphism on L. Moreover, unless J∼ = L2 (q 3 ), x induces an inner-diagonal automorphism on J, and indeed an inner automorphism if q is a perfect cube. Proof. If J ∈ Chev and x ∈ Inndiag(K), then by [IA , 4.9.1, 4.9.2], J ∼ = L2 (q 3 ) and L ∼ = CJ (x), so the result holds in this case. Hence we assume that if J ∈ Chev,

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then x ∈ Inndiag(J). In particular, J ∈ Chev(3), by the Borel-Tits theorem. Since x × L lies in Inndiag(J) (if J ∈ Chev) or Inn(J) (otherwise), m3 (J) > 1. Also, J ∈ C3 , by Lemma 3.2. If K is a simple T3 -group, then K ∼ = L3 (q), since A7 , M12 , M22 , J2 , and G2 (8) are not pumpups of L2 (q) (recall that q > 8), and neither is L± 3 (q0 ) for any q0 = q. Moreover, if q is a perfect cube, then Fq2 contains a 9th root of unity ω, and the action of x is by a conjugate of diag(ω, ω, ω −2 ), an inner automorphism. Thus the lemma holds in this case. Likewise as q > 8 and q is not a power of 3 by assumption, we see easily from Lemma 16.4, [IA , 4.8.2], and [IA , 5.3] that if K is a flat TG3 -group, then K is a classical group; that if we let V be its natural module and d = dim[V, x], 1 −η 2 then (K, d) = (L− 4 (q), 2), (P Sp4 (q), 2 or 4), or (Ln (q ) (n = 4, 5), 4), the last as stipulated in the lemma; and that I3 (CInn(J) (x)) induces inner automorphisms on L, as asserted.  Lemma 16.12. Assume that K ∈ C3 ∪ (T3 − T31 ) or K is a flat TG3 -group. Let K  X and suppose that x ∈ I3 (X) and E(CK (x)) has a component H ∼ = L2 (8). Then H = E(CK (x)) and K is isomorphic to one of the following groups: (a) L2 (83 ), U3 (8), Sp4 (8) or SL4 (8); (b) 3D4 (2) or U6 (2) or SU6 (2); or 3 (c) 2 G2 (3 2 ) or Co3 . Moreover, a Sylow 3-subgroup V of CAut(K) (H) is cyclic of order at most 9; in the cases (b) and (c), |V | = 3. Proof. If K ∈ Spor, then by inspection of [IA , Tables 5.3], K ∼ = Co3 and Sylow 3-subgroups of CAutK (H) have order 3, as asserted. Since H ∈ Alt and H ∈ Chev(r) for any r > 3, by Lemma 3.1, K ∈ Chev(2) ∪ Chev(3). If K ∈ 1 3 Chev(3), then since H ∼ = 2 G2 (3 2 ) = 2 G2 (3 2 ) is the only Chev(3)-guise of H, K ∼ and CAut(K) (H) ∼ = Z3 by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2, 7.1.4c]. Hence, we may assume that K ∈ Chev(2). If x induces a field automorphism on K, then K ∼ = L2 (83 ) and the lemma holds by [IA , 7.1.4c]. By [IA , 4.7.3A, 4.9.1], x does not induce a graph or graph-field automorphism on K. Hence we may assume that x induces an inner-diagonal automorphism on K, whence m3 (Inndiag(K)) > 1. Therefore m3 (K) > 1, and q(K) = 2 or 8 by [IA , 4.2.2], since H is a component of CK (x). The last reference also shows that if q(K) = 2, then the untwisted Dynkin diagram of K has at least three nodes disconnected from one another. If K ∈ T3 − T31 , then by definition of these sets and the previous paragraph, ∼ K = U3 (8), while if K is a flat TG3 -group, then K ∼ = L4 (8), L5 (8), Otherwise K ∈ C3 , so K ∼ = Sp4 (8), U6 (2), SU6 (2), D4 (2), or 3D4 (2). We use [IA , 4.8.2, 4.8.4, 4.7.3A] to identify centralizer components. Using these, we see that the existence of H forces K to have one of the desired isomorphism types and moreover, CAut(K) (H) has the desired structure. The lemma follows.  Lemma 16.13. Let K = U9 (2) and E32 ∼ = D ≤ K. Then for some d ∈ D# , ∼  U3 (8). E(CK (d)) =  = SU9 (2). By [IA , 4.8.2], any preimage d of Proof. Suppose false and let K  has order 9. But then the preimage of D in K  has a unique subgroup of d in K order 3, so is cyclic. Hence D is cyclic, contradiction. 

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Lemma 16.14. Let X = Inndiag(K) where K = 2E6 (2). Let y ∈ I3 (X) be such that CK (y) ∼ = 2 D5 (2). (y exists by [IA , 4.7.3A]). There exists b ∈ CK (y) such that CK (y, b) ∼ = GU5 (2). For any such b, the conjugacy class of b in K is uniquely determined, and E(CK (b)) ∼ = U6 (2). Proof. The element b exists by [IA , 4.8.2]. Then   O 2 (CK (b)) ≥ O 2 (CK (y, b)) ∼ = U5 (2). 

If the result were false, then by [IA , 4.7.3A], as b ∈ K, we would have O 2 (CK (b)) solvable or isomorphic to D4 (2), and in neither case would it contain a copy of  U5 (2), contradiction. The lemma follows. Lemma 16.15. Suppose that I ∈ C3 ∪ T3 ∪ TG3 and U5 (2) ↑3 I. Then I does not contain a subgroup H with H/O3 (H) ∼ = 2 D5 (2). Proof. By [IA , 5.3] and Lemma 3.1, since U5 (2) ↑3 I, we must have I ∈ Chev(2). If H exists then m  3 (I) ≥ 5, so I ∈ TG3 . But all Chev(2)-groups in C3 ∪ T3 have 3-rank at most 4 [V3 , 1.1], [V7 , (1D)], [IA , 4.10.3a], contradiction. This completes the proof.  Lemma 16.16. Let K = L3 (2n ),  = ±1, 2n ≡  (mod 3). Let x ∈ I3 (Aut(K)) and suppose that CK (x) is a solvable {2, 3}-group. Then n = 2 or 3. Proof. If x is a field automorphism, then n/3 = 1 by solvability. So assume  x ∈ Inndiag(K). By solvability, since n > 1, we have O 2 (CK (x)) = 1. By [IA , 4.8.2, 4.8.4], |CK (x)| is divisible by either 2n −  or (23n − )/(2n − ). Since CK (x) is a {2, 3}-group, and using Zsigmondy’s theorem [IG , 1.1], we see that the second case is impossible, and in the first case, n = 2 or 3, as claimed.  Lemma 16.17. Let J be one of the following groups: U3 (8), 3 D4 (2), 2 F4 (2) , C3 (2), C4 (2), D4 (2), 2 D4 (2), D5 (2), F4 (2), U5 (2), m m U6 (2), SU6 (2), Sp4 (2m ), L− n (2 ), 4 ≤ n ≤ 5, q = 2 ≡  (mod 3),  = ±1. Let x ∈ I3 (Aut(J)) and I  CJ (x) with I ∼ = P SU3 (2). Then J ∼ = U3 (8), 3 D4 (2), ∼ or D4 (2). Moreover, in no case is x ∈ F ≤ Aut(J) with F = E32 such that CJ (x1 ) has a normal subgroup isomorphic to I for all x1 ∈ F # .

Proof. Suppose x ∈ Aut(J) of order 3 with I  CJ (x), I ∼ = P SU3 (2). Suppose 3 ∼ (8), D (2), or D (2). By [I , 4.7.3A], J  F that J ∼ U = 4 (2). If J ∼ = U6 (2) or = 3 4 4 A SU6 (2), then by [IA , 4.8.2, 4.8.4], every Lie component of the centralizer of an automorphism of J of order 3 is either L2 (8) or SUn (2) for some n, ruling out these cases. In particular, x induces an inner automorphism on J, whence m3 (J) ≥ 3. This rules out 2 F4 (2), Ln (2m ), n ∈ {4, 5}, and Sp4 (2m ). In all of the remaining cases, J is a classical group with natural module of dimension at most 10. As I has no faithful representation in characteristic 2 of dimension less than 8, we must have J∼ = 2 D4 (2), Sp8 (2) or D5 (2). By [IA , 4.8.2], CJ (x) has no normal U3 (2)-subgroup, contradiction. Hence J ∼ = U3 (8), 3 D4 (2), or D4 (2). Suppose that F is as given. As Aut(J) contains no subgroup F ∼ = E32 disjoint from J, we may take x1 ∈ (F ∩ J)# . But then CJ (x1 ) has no normal subgroup isomorphic to I, again by [IA , 4.8.2, 4.7.3A], completing the proof. 

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∼ 3 D4 (2) and let x ∈ Aut(J) of order 3. If x ∈ J, then Lemma 16.18. Let J = CJ (x) ∼ = G2 (2) or P GU3 (2). If x ∈ J, then CJ (x) = x × H with H ∼ = L2 (8) or CJ (x)/ x ∼ = P GU3 (2), with F ∗ (CJ (x)) ∼ = 31+2 in the latter case. Moreover, if CJ (x) = x × H with H ∼ = L2 (8), then there exists an involution v ∈ NJ (x) inverting x and centralizing H. Proof. The assertions come from [IA , 4.7.3A].



16.3. 3-Structure. Lemma 16.19. Let X be a group such that K = F ∗ (X) ∼ = SL3 (q) or L3 (q), q ≡  (mod 3),  = ±1. Let P ∈ Syl3 (K) and let R be any 3-subgroup of X containing P . Then the following conditions hold: (a) CX (P ) = Z(P ) × Y , where Y is a 3 -group and Y ∩ K = 1. Moreover, P has maximal class or is elementary of order 9; and (b) Z(R) ∼ = Z3 , unless Z(K) = 1, q ≡  (mod 9), and R = P , in which case P ∼ = E32 . Proof. Notice first that if  = −1, then K embeds in K ∗ = SL3 (q 2 ) or L3 (q 2 ) with q 2 ≡ 1 (mod 3), and |K ∗ : K| is relatively prime to 3 since q ≡  (mod 3). Likewise P GU3 (q) embeds in P GL3 (q 2 ), as the fixed point subgroup of the product of the transpose-inverse map τ and the generator of Gal(Fq2 /Fq ). We have Aut(K ∗ ) = P GL3 (q 2 )Φ∗ τ  where Φ∗ acts as the group Aut(Fq2 ), commuting with τ . As Aut(K) = P GU3 (q)Φ, with Φ acting as Aut(Fq2 ), Aut(K) embeds in Aut(K ∗ ) with index relatively prime to 3. If the lemma holds for K ∗ , it follows that it holds for K. Hence, we may assume that  = 1. Next, it suffices to prove the lemma in the case Z(K) = 1. For by that case, if Z(K) = 1, then CX/Z(K) (RZ(K)/Z(K)) = Z(P/Z(K)) × Y0 /Z(K) where  Y0 /Z(K) is a 3 -group and Y0 ∩ K = Z(K). Hence O 3 (CX (R)) ≤ P , so as  P ≤ R, O 3 (CX (R)) ≤ Z(P ). But P is nonabelian, hence absolutely irreducible on the natural K-module, so Z(P ) consists of scalars, i.e., Z(P ) = Z(K). Then CX (R) = Z(K) × O3 (Y0 ), and (a) holds – the only part needing to be checked. Assuming now that  = 1 and Z(K) = 1, we regard X as a subgroup of Aut(K) containing Inn(K), which we identify with K. Without loss, X = P ΓL3 (q), a subgroup of Aut(K) of index 2. Let S ∈ Syl3 (X) with S ≥ R, and set P ∗ = S ∩ Inndiag(K). Then P  S so S contains a Sylow 3-subgroup of CX (P ). We may assume that P = P0 w and P ∗ = P0∗ w, where P0 and P0∗ are images of diagonal subgroups of SL3 (q) and GL3 (q), respectively, P0 ≤ P0∗ , and w is a permutation matrix of order 3. Let 1 = x ∈ CP0∗ (w). A simple calculation shows that x is the image modulo scalars of diag(1, ω, ω −1 ), where ω ∈ F× q has order 3. Hence, CP0∗ w (w) ∼ = E32 , so P ∗ and P are of maximal class (or E32 , in the case of P ) by [IG , 10.24]. Moreover, |P ∗ | ≥ 33 so P ∗ is nonabelian. Now if q ≡ 1 (mod 9), then P ∼ = E32 and K admits no field automorphism of order 3. It follows that CX (P ) = P × Y for some 3 -group Y , and CS (P ∗ ) ∼ = Z3 . We will verify below that Y ∩ K = 1, finishing the proof in this case. So assume that q ≡ 1 (mod 9), whence P is nonabelian and so Z(P ) = Z(P ∗ ) ∼ = Z3 . Also P0 is abelian of type (3n , 3n−1 ), where 3n = (q − 1)3 ≥ 9. Then S = P ∗ Φ where we may assume that Φ consists of field automorphisms. Suppose that z ∈ CS (P ) and write z = hwi φ with h ∈ P0∗ , i = 0 or ±1, and φ ∈ Φ. Then φ induces a power mapping on P0 , and in particular m3 (CP0 (φ)) = 2. But wi φ centralizes P0 and

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CP0 (w) ∼ = Z3 , so wi = 1. Then [P0 , φ] = 1, so φ = 1. But then z ∈ CP0 (w) = Z(P ). Therefore, Z(R) ≤ Z(P ) and (b) holds. Moreover, CX (P ) = Z(P ) × Y where Y = O3 (CX (P )). It remains to show that Y ∩ K = 1. But considering the 3 -group Y as acting on SL3 (q), we see that Y ∩ K centralizes the Sylow 3-subgroup P of SL3 (q), since it centralizes P/Z and Z, where Z = Z(SL3 (q)). As P is absolutely irreducible, Y ∩ K acts like scalars, i.e., trivially, on K. The proof is complete.  Lemma 16.20. Suppose that O3 (X) = 1 and K  X with m3 (CX (K)) = 1. Suppose that K/Z(K) ∼ = L3 (q), q ≡  (mod 3),  = ±1, or G2 (8). Let t ∈ I3 (X) and suppose that t induces a nontrivial field automorphism on K. Let T0 ∈ Syl3 (CX (t)) and U = Ω1 (Z(T0 )). Then (a) If Z(K) = 1, then U = Z(K) t; and (b) If Z(K) = 1, then U = x, z, t ∼ = E33 , where x ∈ I3 (CX (K)) and z is a Sylow 3-center in K. Proof. We treat the G2 (8) case at the end. We may assume that X = KT0 . Expand T0 to T ∈ Syl3 (X) and let T1 = CT (K) = CX (K)  X and x = Ω1 (T1 ). Also let Ct = CX (t) and T2 = CT1 (t), so that x = Ω1 (T2 ). Let Kt = CK (t). Then by [IA , 7.1.4c], CCt (Kt ) = t × T2 . Let C t = Ct / t × T2 , so that K t  C t . Write q0 = q 1/3 ≡  (mod 3). Then q ≡ 1 (mod 9), so by Lemma 16.19, z := Z(T ∩ K) ∼ = Z3 . Suppose that Z(K) = 1. We have K t  C t with K t ∼ = P GL3 (q0 ) and F ∗ (C t ) =  O 2 (K t ). By Lemma 16.19, U = z ∼ = Z3 . Therefore U ≤ z Ω1 (t × T2 ) = z, x, t. As z, x ≤ Z(T ), equality must hold, as desired. Suppose then that Z(K) = 1. Then Z(K) = x and as T1 is cyclic, [T1 , t] ≤ Z(K). Also Kt ∼ = SL3 (q0 ). This time, Lemma 16.19 gives U ≤ T 1 . Hence U ≤ T1 t so U ≤ Ω1 (T1 t) = Z(K) t. The reverse inclusion is obvious. Finally, suppose that K ∼ = G2 (8). Then Ct = Kt × t × CCX (K) (t), with ∼ Kt = G2 (2) ∼ = Aut(U3 (3)). Each direct factor has a Sylow 3-center of rank 1, containing z, t, and x, respectively. The lemma follows.   Lemma 16.21. Suppose that K ∼ = L ≤ K. Then = SL3 (q), and O 3 (SL3 (q 1/3 )) ∼ Z(L) ≤ Z(K).

Proof. Let P ∈ Syl3 (L). Then P is nonabelian, so P is absolutely irreducible on the natural K-module. Hence Z(P ) ≤ Z(K). As Z(L) ≤ Z(P ), the lemma holds.  Lemma 16.22. Let K = L3 (q),  = ±1, q ≡  (mod 3). Let R ∈ Syl3 (K). If 1 = Z  R and Z is cyclic, then |Z| = 3. Proof. By Lemma 16.19, R has maximal class or is isomorphic to E32 . Hence R has a unique normal subgroup of order 9, and it is noncyclic since R is not cyclic. The lemma follows.  Lemma 16.23. Suppose that K = F ∗ (X) = L3 (q), q ≡  (mod 3), and T ∈ Syl3 (X). Let E ∈ E2 (T ). If Ω1 (CT (E)) is nonabelian, then E = e Z(T ) for some e ∈ E # inducing a field automorphism on K.

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Proof. We regard X ≤ Aut(K). Suppose first that E # contains no e inducing a field automorphism on K. Then E ≤ Inndiag(K). If the preimage of E in  := GL3 (q) is nonabelian, then E = CInndiag(K) (E) and so as T /T ∩ Inndiag(K) K  is is cyclic, Ω1 (CT (E)) is abelian, contradiction. Thus, the preimage of E in K abelian, whence a Sylow 3-subgroup R of CAut(K) (E) is the semidirect product of Y × Y by a cyclic group F of field automorphisms, where Y ∈ Syl3 (Fq×2 ). But then F induces power mappings on Y × Y , so Ω1 (R) is abelian, again a contradiction. Therefore, there is e ∈ E # inducing a nontrivial field automorphism on K, and E = e y where y ∈ I3 (T ∩ Inndiag(K)). Notice that Z(T ) ∼ = Z3 . If y = Z(T ), then the previous paragraph applies to y Z(T ) in place of E, and we conclude that Ω1 (CT (E)) ≤ Ω1 (CT (y)) = Ω1 (CT (y Z(T ))) is abelian, contrary to assumption. Thus y = Z(T ) and the proof is complete.  Lemma 16.24. Let K = U3 (8) and P ∈ Syl3 (K). Then CAut(K) (P ) = Z(P ) is of order 3. Proof. By Lemma 16.19, CAut(K) (P ) = Z(P ) × Y with Z(P ) ∼ = Z3 and Y a 3 -group with Y ∩ Inn(K) = 1. Hence Y = 1 or Y = g where g is an involutory graph automorphism. But if Y = g, then CK (g) ∼ = L2 (8) by [IA , 4.9.2], whereas |P | = 34 . Thus |P | does not divide |CK (g)|, contradiction. Hence Y = 1 and the lemma is proved.  

Lemma 16.25. Let K ∼ = U4 (3), Ω7 (3), G2 (3), or J3 . Let x ∈ I3 (Aut(K)). Then |CK (x)| ≥ 33 . Proof. By [IA , 2.5.12, 5.3], Out(K) is a 3 -group, so X is conjugation by some y ∈ I3 (K). In every case m3 (K) ≥ 4, so by [IG , 10.17] y normalizes an E34 subgroup of K, whence |CK (y)| ≥ 33 , as claimed.  3

Lemma 16.26. Let K = 2 G2 (3 2 ) and X = K b = Aut(K), where b ∈ I3 (X) is a field automorphism. Let K0 = CK (b) ∼ = P ΣL2 (8) and let t0 ∈ I3 (K0 ) induce a field automorphism of order 3 on E(K0 ) ∼ = L2 (8). Then CX (t0 ) contains (A1 × ∼ 3 and A b Z  Z E A2 ) b, where Ai ∼ = 3 = 3 i 3 for i = 1, 2; moreover, t0 ∈ A1 and A2 is a Sylow 3-center in K. Proof. Let P be a b-invariant Sylow 3-subgroup of K. By [IA , Table 2.4], P has a central series P > Ω1 (P ) > Z(P ) > 1 with factors isomorphic to E33 , all acted on freely by b; moreover, Ω1 (P ) ∼ = E36 . As t30 = 1, CP (t0 ) contains Ω1 (P ). We can take A2 = Z(P ) and A1 a b-invariant complement to A2 in Ω1 (P ). The lemma follows.  16.4. Other. Lemma 16.27. Let K = Co3 and let x ∈ I3 (K) with L := E(CK (x)) ∼ = L2 (8). Let E ∈ E32 (x L). Then K = ΓE,1 (K). Proof. By [IA , 5.3j], NK (x) ∼ = Σ3 × P ΣL2 (8) is maximal in K. Let E ≤ P ∈ Syl3 (CK (x)). Then Z(P ) = E = x, z with z ∈ L∩[P, P ]. Since P ∈ Syl3 (K), xz = xg for some g ∈ K. Then ΓE,1 (K) contains NK (x) and Lg , so equals K.  Lemma 16.28. Let K = Co3 , P ∈ Syl3 (K), E = J(P ), and N = NK (E)/E. Then E = CK (E) ∼ = E35 and N ∼ = Z2 × M11 has just two orbits on E1 (E), of lengths 55 and 66.

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16. PRELIMINARY LEMMAS FOR THEOREM C∗ 6 : STAGE 1

513

Proof. We use [IA , 5.3j] freely. In particular the structures of E = CK (E) and N are as stated. By [IG , 16.9], N controls K-fusion in E. For x ∈ I3 (K), we have m3 (CK (x)) = 5 if x lies in class 3A or 3B, while m3 (CK (x)) < 5 if x lies in class 3C. Therefore E ∩ 3A and E ∩ 3B are the two orbits of N on E # . For x ∈ E ∩ 3B, E  CK (x) and so |xN | = |NK (E)|/|CK (x)| = 132. As |E1 (E)| = 121, the lemma follows. 

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Bibliography Michael Aschbacher and Stephen D. Smith, The classification of quasithin groups. I, Mathematical Surveys and Monographs, vol. 111, American Mathematical Society, Providence, RI, 2004. Structure of strongly quasithin K-groups. MR2097623 [ASm2] Michael Aschbacher and Stephen D. Smith, The classification of quasithin groups. II, Mathematical Surveys and Monographs, vol. 112, American Mathematical Society, Providence, RI, 2004. Main theorems: the classification of simple QTKE-groups. MR2097624 [A9] Michael Aschbacher, A characterization of Chevalley groups over fields of odd order, Ann. of Math. (2) 106 (1977), no. 2, 353–398, DOI 10.2307/1971100. MR498828 [A23] Michael Aschbacher, A characterization of some finite groups of characteristic 3, J. Algebra 76 (1982), no. 2, 400–441, DOI 10.1016/0021-8693(82)90222-8. MR661863 [A24] Michael Aschbacher, Finite groups of rank 3. II, Invent. Math. 71 (1983), no. 1, 51–163, DOI 10.1007/BF01393339. MR688262 [A2] Michael Aschbacher, Sporadic groups, Cambridge Tracts in Mathematics, vol. 104, Cambridge University Press, Cambridge, 1994. MR1269103 [A19] Michael Aschbacher, 3-transposition groups, Cambridge Tracts in Mathematics, vol. 124, Cambridge University Press, Cambridge, 1997. MR1423599 [CCPNW1] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, Atlas of finite groups, Oxford University Press, Eynsham, 1985. Maximal subgroups and ordinary characters for simple groups; With computational assistance from J. G. Thackray. MR827219 [F1] Walter Feit, The representation theory of finite groups, North-Holland Mathematical Library, vol. 25, North-Holland Publishing Co., Amsterdam-New York, 1982. MR661045 [Fin3] Larry Finkelstein, Finite groups in which the centralizer of an involution is ·0, J. Algebra 32 (1974), 173–177, DOI 10.1016/0021-8693(74)90179-3. MR439935 [Fin2] Larry Finkelstein, Finite groups with a standard component isomorphic to M23 , J. Algebra 40 (1976), no. 2, 541–555, DOI 10.1016/0021-8693(76)90210-6. MR414700 [Fi2] Bernd Fischer, Finite groups generated by 3-transpositions. I, Invent. Math. 13 (1971), 232–246, DOI 10.1007/BF01404633. MR294487 [FinS1] Larry Finkelstein and Ronald Solomon, A presentation of the symplectic and orthogonal groups, J. Algebra 60 (1979), no. 2, 423–438, DOI 10.1016/0021-8693(79)90091-7. MR549938 [GHN] R. Gramlich, M. Horn, and W. Nickel, Odd-dimensional orthogonal groups as amalgams of unitary groups. II. Machine computations, J. Algebra 316 (2007), no. 2, 591–607, DOI 10.1016/j.jalgebra.2007.03.035. MR2356846 [GL1] Daniel Gorenstein and Richard Lyons, The local structure of finite groups of characteristic 2 type, Mem. Amer. Math. Soc. 42 (1983), no. 276, vii+731, DOI 10.1090/memo/0276. MR690900 [GLS1] Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1994. MR1303592 , Outline of Proof, ch. 2 in [GLS1], American Mathematical Society, 1994. [I2 ] , Overview, ch. 1 in [GLS1], American Mathematical Society, 1994. [I1 ] [ASm1]

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516

[IG ]

[IA ]

[GLS4]

[II3 ]

[IIK ] [II2 ] [GLS5]

[III2 ] [IIIK ] [III1 ] [III3 ] [GLS6]

[IV1 ] [GLS7]

[III8 ] [III11 ] [III7 ] [GLS8]

[III17 ] [III13 ] [III14 ] [III15 ] [III16 ]

BIBLIOGRAPHY

Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 2. Part I. Chapter G, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1996. General group theory. MR1358135 Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 3. Part I. Chapter A, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1998. Almost simple K-groups. MR1490581 Stephen D. Smith, Applying the classification of finite simple groups, Mathematical Surveys and Monographs, vol. 230, American Mathematical Society, Providence, RI, 2018. A user’s guide. MR3753581 Anna De Simone, Decomposition theorems in orthomodular posets: the problem of uniqueness, Proc. Amer. Math. Soc. 126 (1998), no. 10, 2919–2926, DOI 10.1090/S0002-9939-98-04380-9. MR1452800 , Properties of K-Groups, ch. 4 in [GLS4], American Mathematical Society, 1999. , Strongly Embedded Subgroups and Related Conditions on Involutions, ch. 2 in [GLS4], American Mathematical Society, 1999. E. H. Rothe, Introduction to various aspects of degree theory in Banach spaces, Mathematical Surveys and Monographs, vol. 23, American Mathematical Society, Providence, RI, 1986. MR852987 , General Group-Theoretic Lemmas, ch. 2 in [GLS5], American Mathematical Society, 2002. , Properties of K-Groups, ch. 6 in [GLS5], American Mathematical Society, 2002. , Theorem C7 : General Introduction, ch. 1 in [GLS5], American Mathematical Society, 2002. , Theorem C∗7 : Stage 1, ch. 3 in [GLS5], American Mathematical Society, 2002. Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 6. Part IV. The special odd case, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2005. MR2104668 , General Introduction to the Special Odd Case, ch. 1 in [GLS6], American Mathematical Society, 2005. Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 7. Part III. Chapters 7–11. The generic case, stages 3b and 4a, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2018. MR3752626 , General Group-Theoretic Lemmas, ch. 8 in [GLS7], American Mathematical Society, 2017. , Properties of K-Groups, ch. 11 in [GLS7], American Mathematical Society, 2017. , The Stages of Theorem C∗7 , ch. 7 in [GLS7], American Mathematical Society, 2017. Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 8. Part III. Chapters 12–17. The generic case, completed, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2018. MR3887657 , Preliminary Properties of K-groups, ch. 17 in [GLS8], American Mathematical Society, 2018. , Recognition Theory, ch. 13 in [GLS8], American Mathematical Society, 2018. , Theorem C∗7 , Stage 4b+: A Large Lie-type Subgroup G0 for p = 2, ch. 14 in [GLS8], American Mathematical Society, 2018. , Theorem C∗7 , Stage 4b+: A Large Lie-type Subgroup G0 for p > 2, ch. 15 in [GLS8], American Mathematical Society, 2018. , Theorem C∗7 , Stage 5+: g = g0 , ch. 16 in [GLS8], American Mathematical Society, 2018.

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, Theorems C7 and C∗7 , ch. 12 in [GLS8], American Mathematical Society, 2018. [GrMeSeg] Robert L. Griess Jr., Ulrich Meierfrankenfeld, and Yoav Segev, A uniqueness proof for the Monster, Ann. of Math. (2) 130 (1989), no. 3, 567–602, DOI 10.2307/1971455. MR1025167 [G1] Daniel Gorenstein, Finite groups, 2nd ed., Chelsea Publishing Co., New York, 1980. MR569209 [Is1] I. Martin Isaacs, Character theory of finite groups, Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1976. Pure and Applied Mathematics, No. 69. MR0460423 [KMa1] Kenneth Klinger and Geoffrey Mason, Centralizers of p-groups in groups of characteristic 2, p-type, J. Algebra 37 (1975), no. 2, 362–375, DOI 10.1016/00218693(75)90084-8. MR390047 [McL3] Jack McLaughlin, Some subgroups of SLn (F2 ), Illinois J. Math. 13 (1969), 108–115. MR237660 [PoTh1] M. B. Powell and G. N. Thwaites, On the non-existence of certain types of subgroups in simple groups, Quart. J. Math. Oxford Ser. (2) 26 (1975), no. 102, 243–256, DOI 10.1093/qmath/26.1.243. MR442078 [Seg1] Yoav Segev, On the uniqueness of Fischer’s Baby Monster, Proc. London Math. Soc. (3) 62 (1991), no. 3, 509–536, DOI 10.1112/plms/s3-62.3.509. MR1095231 [T2] John G. Thompson, Nonsolvable finite groups all of whose local subgroups are solvable, Bull. Amer. Math. Soc. 74 (1968), 383–437, DOI 10.1090/S0002-9904-196811953-6. MR230809 [Tho70] John G. Thompson, Nonsolvable finite groups all of whose local subgroups are solvable. II, Pacific J. Math. 33 (1970), 451–536. MR276325 [Wo1] W. J. Wong, Generators and relations for classical groups, J. Algebra 32 (1974), no. 3, 529–553, DOI 10.1016/0021-8693(74)90157-4. MR379679 [III12 ]

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Index

Corollary C6 , 310 Corollary C∗6 , 312 C3 , C3 -group, 33 Cp , Cp -group, p odd, 33 Cop , 145 C∗p , 145 Curtis-Tits theorem, 9 C(x, K), 19

2-local p-rank, 2 ↓p , 24 ↑p , 24 almost strongly p-embedded subgroup, 35 Alt, 2 Aschbacher, M., 9, 10, 31, 95, 237, 245, 250, 253, 266, 272, 274 A-terminal p-component, 94

d, 310 degenerate neighborhood, 90 degenerate nonconstrained neighborhood, 6 dT,p , 310

B13 , 312 B3 -group, 311 Background Results, 9 balanced signalizer functor, 11 barely in Cp , 145 Bender, H., 3 Bender-Thompson lemma, 13 bicharacteristic group, 6 B-narrow component, 146 Bp (G), 34 Bp∗,c (G), 89, 104 Bp∗ (G), 34, 104 Bp∗,o (G), 89, 104 broad, 90 broad triple, 6, 146 B-terminal p-component, 94 BtK3exc (G), 5, 90 BtKp (G), 5, 90, 144 B-wide component, 146

E1 (P ), 11 Ep (X), Epm (X), 11 even type, 2 exceptional situation, 313 faithful symplectic pair, 89, 143 field triple, 44 Finkelstein, L., 9, 10, 253 Fischer, B., 31 flat TG3 -group, 312 frame, 441 Gilman ordering, refinement of, 313 Gilman, R., 95, 313 Gilman-maximal, 25 ΓK , 250 Γo1 P,2 (G), 93 Gp -depth, 312 ΓP,k (X), 11 ΓP,k (X), 11 Gramlich, R., 9 Griess, R. L., 10, 31

C2 , C2 -group, 33 Co2 , 2 C(b, J), 8 Chev, Chev(r), 2 complete (S)U3 (2)-component, 126 component preuniqueness hypothesis, 95 constrained {z, b}-neighborhood, 8, 209 constrained neighborhood of U4 (2)-type, 209 constrained neighborhood of sporadic type, 209 constrained neighborhood, 210 core of a constrained neighborhood, 209 core of a neighborhood, 8

Hb∗ , 327 Horn, M., 9 I3∗ (K), 121 Ipo -rigid pair, 311 Ipo -terminal Tp -pair, Tpi -pair, 311 Ipo -rigid, 25 Ipo -terminal, 25 519

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520

Ipo (X), 2 Ip (X), 2 J-p central element, 196 K, 2 KbIJp (G), 90 KbI p (G), 90, 146 Klinger, K., 3 Koehl, R., 9 Kp , 2 L2 (pp ) field triple, 37, 44 (H), 126 Lcs 3 LCp -uniqueness subgroup, 35 locally unbalancing triple, 13, 36 Lop (G), 2 m2,p (G), 2, 34 Mason, G., 3 mate, 5, 90, 148 maximal symplectic triple, 219 Meierfrankenfeld, U., 10, 31 m  p (K), 16, 312

I∗G (D; 2), 77 neighborhood of U4 (2)-type, 8 neighbors in a neighborhood, 8 I∗G (B; 2), 5 Nickel, W., 9 nonconstrained {t, b} neighborhood, 6, 90 nonconstrained neighborhood, 90, 210 nondegenerate constrained neighborhood, 8, 209 nondegenerate neighborhood, 90, 160 nondegenerate nonconstrained neighborhood, 6 IX (B; 2), 34 N (x, K), 311 one-step rigid, 25 ordered frame, 441 outer well-generated, 100 p-component preuniqueness subgroup, 35 Phan, K.-W., 9 Powell, M., 95 preuniqueness subgroup, 35 p-saturated, 374 p-Thin Configuration, 314 p-uniqueness subgroup, 35 q(K), for K ∈ Chev, 165 regular, 90 regular triple, 5, 146 regularly embedded, 90 regularly embedded triple, 146 restricted even type, 2

INDEX

saturated, 374 Segev, Y., 10, 31 σ(G), 2, 309 Smith, S. D., 10 Solomon, R., 9, 10 Spor, 2 sporadic type neighborhood, 8 Sp (X), 34, 89 strong p-uniqueness subgroup, 3, 34, 35 strongly p-embedded subgroup, 34, 35 strongly balanced, 39 strongly locally balanced component, 14, 15 symplectic pair, 89, 143 symplectic pair (trivial, faithful), 4 {t, b}-neighborhood, 160 TG3 -group, 311 TGp (G), 313 Theorem C5 , 1 Theorem C5 : Stage 1, 4, 34 Theorem C5 : Stage 2, 6 Theorem C5 : Stage 3, 9 Theorem C5 : Stage 4, 9 Theorem C6 , 310 Theorem C∗6 , 312 Theorem C7 , proof of, 314 Theorems C7 , C∗7 , 312 Θ1 (a), 11 Θ1 (G; A), 11 Thompson, J. G., 3 Thwaites, G., 95 TJp (G), 313 TJ∗p (G), 313 TJp (G), 313 Tp , 309 Tpi , 1 ≤ i ≤ 5, 310 trivial symplectic pair, 89, 143 unambiguously, 347 unbalancing triple, 13, 36 V1 , V2 , V3 , V4 , V5 , V6 , VK , 1 Walter, J. H., 3 weak Curtis-Tits system, 273 weak Phan system, 275 width, 220 width of an extraspecial or symplectic type group, 20 Wong, W., 9, 10 w(S), 20 Z4 -semirigid, 24 Z4 -semirigid in J, 24 {z, b}-neighborhood, 209

σ0 (G), 3, 309

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This book is the ninth volume in a series whose goal is to furnish a careful and largely self-contained proof of the classification theorem for the finite simple groups. Having completed the classification of the simple groups of odd type as well as the classification of the simple groups of generic even type (modulo uniqueness theorems to appear later), the current volume begins the classification of the finite simple groups of special even type. The principal result of this volume is a classification of the groups of bicharacteristic type, i.e., of both even type and of p-type for a suitable odd prime p. It is here that the largest sporadic groups emerge, namely the Monster, the Baby Monster, the largest Conway group, and the three Fischer groups, along with six finite groups of Lie type over small fields, several of which play a major role as subgroups or sections of these sporadic groups.

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