The Classification of the Finite Simple Groups, Number 8

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MATHEMATICAL

Surveys and Monographs

Volume 40, Number 8

The Classification of the Finite Simple Groups, Number 8 Daniel Gorenstein Richard Lyons Ronald Solomon

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10.1090/surv/040.8

The Classification of the Finite Simple Groups, Number 8 Part III, Chapters 12 –17: The Generic Case, Completed

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MATHEMATICAL

Surveys and Monographs Volume 40, Number 8

The Classification of the Finite Simple Groups, Number 8 Part III, Chapters 12 –17: The Generic Case, Completed Daniel Gorenstein Richard Lyons Ronald Solomon

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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. 2010 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-40.8

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

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

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 2018 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 

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For Sofia and Larry

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Contents Preface

ix

Chapter 12. Introduction 1. Theorems C7 and C∗7

1 1

Chapter 13. Recognition Theory 1. Curtis-Tits Systems and Phan Systems 2. The Gilman-Griess Theorem for Groups in Chev(2) 3. The Wong-Finkelstein-Solomon Method

9 9 14 15

Chapter 14. Theorem C∗7 : Stage 4b+. A Large Lie-Type Subgroup G0 for p=2 1. Introduction 2. A 2-Local Characterization of L± 4 (q), q Odd (q) 3. The Case K ∼ = L± 3 4. The Case K ∼ = G2 (q) or 3D4 (q) 5. The Non-Level Case 6. The Other Exceptional Cases 7. The Case K ∼ = P Sp2n (q) = Cn (q)a 8. The Spinn (q) Cases, n ≥ 7 9. The Sp2n Cases 10. The Linear and Unitary Cases 11. The Orthogonal Case, Preliminaries 12. The Orthogonal Case, Completed 13. The Cases in Which G0 is Exceptional 14. Summary: p = 2

29 29 33 39 41 47 58 62 74 80 87 112 122 141 154

Chapter 15. Theorem C∗7 : Stage 4b+. A Large Lie-Type Subgroup G0 for p>2 1. Introduction 2. A Choice of p 3. The Weyl Group 4. The Field Automorphism Case 5. Some General Lemmas 6. The Case mp (B) = 4 7. The Case AutK (B) ∼ = W (BCn ) or W (F4 ) 8. The Case AutK (B) ∼ = W (Dn ), n ≥ 4 9. Some Exceptional Cases 10. The Case K/Z(K) ∼ = P SL± m (q) 11. The Final Case: K/Z(K) ∼ = E6 (q)

155 155 156 157 160 167 178 187 193 206 207 219

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viii

CONTENTS

12. 13. 14. 15. 16. 17. 18. 19.

Identification of G0 : Setup −q (q) G0 ∼ = Sp2n+2 (q) or An+1 ± ∼ G0 = Dn+1 (q)  G0 ∼ = Lkq (q) ∼ G0 = E8 (q)  G0 ∼ = E6q (q) and E7 (q) The Remaining Cases for G0 ΓD,1 (G) Normalizes G0

224 225 232 248 252 257 264 277

Chapter 16. Theorem C∗7 : Stage 5+. G = G0 1. Introduction and Generalities 2. The Alternating Case 3. The Lie Type Case, p = 2: Part 1 4. The Lie Type Case, p = 2: Part 2 5. The Lie Type Case, p > 2

291 291 293 294 298 303

Chapter 17. Preliminary Properties of K-Groups 1. Weyl Groups and Their Representations 2. Toral Subgroups 3. Neighborhoods 4. CTP-Systems 5. Representations 6. Computations in Groups of Lie Type 7. Outer Automorphisms, Covering Groups, and Envelopes 8. p-Structure of Quasisimple K-groups 9. Generation 10. Pumpups 11. Small Groups 12. Subcomponents 13. Acceptable Subterminal Pairs 14. Fusion 15. Balance and Signalizers 16. Miscellaneous

321 321 335 355 376 383 387 406 408 415 423 455 464 470 474 480 481

Bibliography

485

Index

487

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Preface Volumes 5, 7, and 8 of this series form a trilogy treating the Generic Case of the classification proof. An overview of the general strategy for this set of volumes, along with a brief history of the original treatment of these results, is provided in the preface and Chapter 1 of Volume 5. We shall not repeat that here; rather, we refer the reader to Volume 5. By the end of Volume 7, we arrived at the existence in our K-proper simple group G, for a suitably chosen prime p, of one of the following: (a) (Alternating case) a subgroup G0 ≤ G such that G0 ∼ = An for some n ≥ 13, p = 2, and ΓD,1 (G) normalizes G0 for any root 4-subgroup D of G0 . Thus for any involution d ∈ G0 which is the product of two disjoint transpositions, CG (d) normalizes G0 ; or (b) (Lie-type case) an element x of order p whose centralizer CG (x) has a p-generic quasisimple component K = d L(q) ∈ Chev(r), where p, r are a pair of distinct primes with q a power of r, such that either p = 2 and r is odd, or r = 2 and p divides q 2 − 1. In Chapter 16 of this volume, we shall prove that in case (a), G = G0 , so that G is a K-group, as desired. However, almost all our attention in this volume will be on case (b), where in order to catch up with case (a), we construct a subgroup G0 ∈ Chev such that ΓD,1 (G) := NG (Q) | 1 = Q ≤ D normalizes G0 for a suitable subgroup D ∼ = Zp × Zp of G. The results of Volume 7 provide a lot more information in case (b). Thus, we know that CG (K) has a cyclic or quaternion Sylow p-subgroup, and if p > 2, then K itself contains a copy of Zp × Zp × Zp . Significantly, also, a family of centralizers “neighboring” CG (x) also have semisimple p-layers. Considerable additional information is known about these centralizers. All this is spelled out precisely in Theorem 1.2 in Chapter 12 of this volume. [Note: The numbering of chapters in this volume continues that of Volumes 5 and 7. In particular, Chapter 12 is the first chapter of this volume.] Roughly speaking, using the terminology of the initial treatment of the Classification Theorem, we are faced with “standard form problems” for components in Chev whose centralizers have p-rank 1. In this volume we complete the proof of Theorem C∗7 . Let G be a K-proper simple group. Assume that γ(G) = ∅ and that G does not possess a p-Thin Configuration for any prime p ∈ γ(G). Then G ∼ = G0 for some G0 ∈ K(7)∗ . ∗ (See p. 1 and p. 4 for the definitions of the set K(7) of “generic” known simple groups and the set γ(G) of primes associated to G.) In particular, in conjunction with the 2-Uniqueness Theorems in Volume 4 and the main theorems of Volume 6, ix Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

x

PREFACE

we have completed our treatment of simple groups of odd type, in the sense that we have proved the following theorem. Theorem O. Let G be a K-proper simple group. Assume that either m2 (G) ≤ 2 or Lo2 (G) ∩ (T2 ∪ G2 ) = ∅. Then either G is an alternating group or G ∈ Chev(r) for some odd prime r or G ∼ = M11 , M12 , J1 , M c, Ly, or O  N . Put another way, we have the following theorem. Theorem. Let G be a minimal counterexample to the Classification Theorem. Then G is of even type, i.e., the following conditions hold: (a) m2 (G) ≥ 3; (b) O2 (CG (t)) = 1 for all t ∈ I2 (G); and (c) Lo2 (G) ⊆ C2 . Theorem C∗7 also includes the following result when G is of even type. Theorem GE. Let G be a K-proper simple group of even type. Assume that γ(G) contains at least one odd prime and that G does not possess a p-Thin Configuration for any prime p ∈ γ(G). Then G ∈ Chev(2). Historically, the portion of Theorem O covered in Volumes 5, 7, and 8 was principally the work of M. Aschbacher, J. H. Walter, D. Gorenstein, J. G. Thompson, and M. E. Harris. In particular, Chapter 14 in this volume relies at many points on arguments originally given by Aschbacher in his Classical Involution Paper [A9]. His work has been a continual source of inspiration for us. Theorem GE was treated originally by the first two authors [GL1] in conjunction with a paper of Gilman and Griess [GiGr1]. Our work here follows a similar outline to this earlier work and at numerous points benefits from arguments found therein. However, although the earlier work relied almost exclusively on the Gilman-Griess Theorem (see Section 2 of Chapter 13), we employ a variety of recognition techniques, as discussed also in Sections 1 and 3 of Chapter 13. In particular, the theory of Curtis-Tits and Phan systems, developed in the last several decades, plays a very important role. Also useful is an identification method pioneered by W. J. Wong [Wo1] in the early 1970s for the recognition of classical groups in odd characteristic, and later extended by Finkelstein and the third author to many cases of classical groups in characteristic 2, e.g. [FinS1]. In barest outline, the strategies for the non-alternating generic case of Theorem O and for Theorem GE are identical. The main theorems of Volume 7 have produced, for some p ∈ γ(G), an element x ∈ Ip (G) and a component K of CG (x) with K ∈ Chev(r) and with mp (CG (K)) = 1, having an acceptable subterminal (x, K)-pair (y, L) (defined on p. 7), where D := x, y ∼ = Zp × Zp and L is a large component of CK (y). Moreover, we have some u ∈ D − x such that the pumpup Lu of L in CG (u) is quasisimple with L < Lu , and in general, K and Lu are defined over the same field. The strategy is to use one of the recognition theorems mentioned in the previous paragraph and described carefully in Chapter 13 to identify the group G0 := N(x, K, y, L) as a quasisimple group in Chev(r). Here, by definition, G0 is the group generated by K and all of the neighboring components Lu , u ∈ D# lying over the subcomponent L inside K. Because of our hypothesis that G has no proper p-uniqueness subgroup, G0 will turn out to be G, and we will have completed the identification of G. In practice, we treat the cases p = 2 and p odd separately. For the construction of G0 , the p = 2 case is handled in Chapter 14, while the p odd case is treated in

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PREFACE

xi

Chapter 15. The principal difference is that, when p = 2, the existence of commuting involutions in the centers of suitable fundamental SL2 (q) subgroups of K and of Lu permits these subgroups to be lined up properly for the verification of suitable Curtis-Tits or Phan relations. An analogue is not available when p is odd, but— following Gilman and Griess—we use a toral subgroup B of exponent p visible both in CG (x) and all CG (u), u ∈ D# , permitting the construction of a Weyl group W for G as a quotient of NG (B). Then G0 may be identified with K, N for suitable N ≤ NG (B) covering W . A version of this strategy of recognizing G0 as K, N was first implemented by W. J. Wong for the characterization of symplectic and orthogonal groups in odd characteristic. Moreover, in the Classical Involution Paper, M. Aschbacher constructed a toral 2-subgroup of G and an associated Weyl-type group (which he calls the Thompson group), building on ideas of J. G. Thompson. By the end of Chapter 15, we have constructed in case (b) a subgroup G0 of G with G0 ∈ Chev(r) and ΓD,1 (G) normalizing G0 . Finally, in Chapter 16, we prove that G = G0 in both case (a) and case (b) by showing that, if not, then NG (G0 ) is a proper strong p-uniqueness subgroup of G of component type, contrary to hypothesis. (Note that for p = 2, the simple groups with a proper strong 2-uniqueness subgroup are classified in Volume 4, using the arguments of Bender and Aschbacher.) In this volume, we have found it of great convenience, and sometimes of necessity, to use some known theorems not previously included in our set of Background Results, viz., L. E. Dickson’s classification of smallest-dimensional faithful modules for the symmetric groups [Di2]; J. McLaughlin’s characterization of groups generated by transvections on a finite vector space of odd order [McL2]; R. Gramlich’s exposition of the state of Phan theory [Grm2]; and theorems of R. Blok, C. Hoffman, and S. Shpectorov on amalgams of Curtis-Tits and Phan types [BHS]. The full set of Background Results consists now of these four references and the Background Results listed in our first volume [I1 ]. At an appropriate future moment, we shall add to these the Aschbacher-Smith tomes on quasithin groups. We continue the notational conventions established in Volume 2 of this series [IG ]. We refer to the chapters of this book as [III12 ], [III13 ], [III14 ], [III15 ], [III16 ] and [III17 ]. As in previous volumes, the last chapter [III17 ] collects the necessary K-group lemmas for the main chapters [III14 ]–[III16 ], and thus logically precedes them. As early as 1999, the authors began a discussion with Curtis Bennett and Sergey Shpectorov concerning the possibility of a modern treatment of the theorems of Phan characterizing the finite unitary groups in odd characteristic. They took up this problem and expanded its scope, later recruiting Corneliu Hoffmann and Rieuwert Blok as collaborators. Also, in the early 2000s, Ralf Gramlich began his dissertation work under the supervision of Aryeh Cohen. Gramlich visited Rutgers during part of the 2001–02 academic year, which the third author spent at Rutgers. Gramlich became a leader in the development of Phan Theory and recruited many others to this project. We extend our warmest thanks to all of these mathematicians for their contributions to the development of this body of recognition theorems. We also thank Bob Griess, Jon Hall, Len Scott, Gary Seitz, and Gernot Stroth for numerous helpful conversations and valuable suggestions. Special thanks once

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xii

PREFACE

again go to our wives, Lisa and Rose, for their patience during our many hours of mathematical distraction. And, lastly, we recall with deep regret the passing of our colleague and friend, Kay Magaard, with whom we had many hours of fruitful discussion of some the topics destined for future volumes of this series, along with many lighter moments of laughter and good fellowship. We will miss him often as the years go by. This volume, in tandem with Volume 7, has spent many years in preparation. Both of us wish to thank the National Science Foundation and the National Security Agency for their years of grant support. The third author also extends thanks to the Ohio State University Emeritus Academy for its support and to the Simons Foundation for the current collaborative research grant (Award ID 425816) which has funded several weeks of face-to-face collaboration both in Ohio and in New Jersey. Richard Lyons and Ronald Solomon August, 2018

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10.1090/surv/040.8/01

CHAPTER 12

Introduction 1. Theorems C7 and C∗7 In this eighth monograph, we continue the analysis of the Generic Case from the previous monograph [GLS7], and bring that analysis to a complete proof of Theorem C∗7 . Thus, our goal is: Theorem 1.1 (Theorem C∗7 ). Let G be a K-proper simple group. Assume that γ(G) = ∅, and that G does not possess a p-Thin Configuration for any prime p ∈ γ(G). Then G ∼ = G0 for some G0 ∈ K(7)∗ . The set K(7)∗ of target simple groups for Theorem C∗7 consists of alternating groups and groups of Lie type, as follows [III7 ]: K(7)∗ = {An | n ≥ 13} ∪ {An (q) | n ≥ 4} ∪ {A3 (q) | q ≡  (mod 8)} ∪ {Bn (q) | n ≥ 3, q odd} ∪ {Cn (q) | n ≥ 4} ∪ {C3 (q) | q odd} (1A)

∪ {Dn (q) | n ≥ 5} ∪ {D4 (q) |  = 1 or q odd} ∪ {F4 (q)} ∪ {E6 (q), E7 (q), E8 (q)} −   − {A4 (2), A− 4 (4), A5 (2), A6 (2), C4 (2), D4 (2), D5 (2), F4 (2),

E6− (2), B3 (3), D4 (3)}. In Theorem C∗7 , the statement that G is K-proper means that all nonabelian simple sections of all proper subgroups of G are in the set K = Chev ∪ Alt ∪ Spor of known (quasi)-simple groups [I1 , Table I, pp. 8-9]. The statement that γ(G) = ∅ is equivalent to the statement that G is of generic type [I2 , Def. 16.1, p.106] (see the formal definition (1F) below). The hypothesis on p-Thin Configurations, while important in earlier stages of the proof, is not used in this monograph. As we have already discussed Theorems C∗7 and C7 , and our strategy, in some detail in [GLS5] and [GLS7], we will be briefer here. We repeat, however, that Theorem C∗7 assumes that p-Thin Configurations do not exist in G when our Kproper simple group G is of generic type relative to the prime p, while Theorem C7 [I2 , p. 106] does not make this assumption. The proof of the nonexistence of 2-Thin Configurations in the generic case has been given in [GLS6, Theorem C∗2 ], and the analogous theorem for p > 2 will be proved, along with Theorem C6 , in the next volume of this series. At that point the proof of Theorem C7 , i.e., the classification of G in the Generic Case, will be complete. We refer the reader to the introductory chapters of [GLS5] and [GLS7] for further elaboration. 1 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

2

12. INTRODUCTION

We have divided the proof of Theorem C∗7 into stages, of which Stages 1, 2, 3a, 3b, and 4a have been proved in [GLS5] and [GLS7]; Stages 4b and 5 remain to be proved in this monograph. Now let us summarize the results of [GLS5] and [GLS7]. Under the assumptions of Theorem C∗7 , those results yield one of two possible outcomes. In both, there is a pair (x, K) ∈ Jp (G) with K a component of CG (x), and an acceptable subterminal (x, K)-pair (y, L) (with L a component of CK (y)). The results concern the neighborhood N(x, K, y, L), defined as the set of all groups Lu , u ∈ x, y # , where Lu in turn is defined as the subnormal closure of L in CG (u). We put   # G0 = Lu | u ∈ x, y = N(x, K, y, L) . One possible outcome is: G0 ∼ = An , n ≥ 13, p = 2, x, y is a root four-subgroup of G0 , Γx,y,1 (G) ≤ (1B) NG (G0 ), and CG (G0 ) has odd order. Moreover, for any (x, K) ∈ J∗2 (G), K ∈ Alt. In the complementary case, we reach the following situation: (1) There is (x, K) ∈ J∗p (G) for some p ∈ γ(G); (2) There is (y, L), an acceptable subterminal (x, K)-pair; and (1C) (3) K ∈ Chev(r) for some prime r, exactly one of r and p equals 2, and p splits K. In (1C), the elements x, y and subgroups K, L satisfy important additional conditions. By combining parts (but not all) of Stages 2, 3a, 3b, and 4a from [GLS5] and [GLS7], we have established the following: Theorem 1.2 (Some of Theorem C∗7 , through Stage 4a). Under the assumptions of Theorem C∗7 , either (1B) holds, or else there exists a prime p ∈ γ(G), a pair (x, K) ∈ J∗p (G), and an acceptable subterminal (x, K)-pair (y, L) such that (1C) holds. For any such p, (x, K), and (y, L), the following additional conditions hold: (a) If r = 2, then mp (K) ≥ 3; (b) For every (x , K  ) ∈ Jp (G), K  ∈ Chev(r  ) for some r  = p, and mp (C(x , K  )) = 1; # (c) For each u ∈ x, y , let Lu be the subnormal closure of L in CG (u). Then for some u ∈ x, y − x , Lu > L; (d) For each u ∈ x, y # , (1) Lu ∈ Chev(r), where r is as in (a); and (2) Either q(Lu ) = q(K), or else p = 2 and one of the following holds: (i) K ∼ = 3D4 (q) or G2 (q) for some q; or (ii) K ∼ = P Sp4 (q), or a non-universal version of A± 3 (q) for some q; and # (e) If K ∼ = A (q),  = ±1, then for any u ∈ x, y such that the pumpup Lu of L in CG (u) is nontrivial, we have Lu /Z(Lu ) ∼ = K/Z(K), except possibly in any of the following situations: (1) L ∼ = A−2 (q); or (2) p = 2,  = 4, and Lu /O2 (Z(Lu )) ∼ = B3 (q)u ∼ = Spin7 (q) for some # u ∈ x, y ; # (3) p = 2,  = 2, and for any u ∈ x, y , Lu ∼ = L, A2 (q), or A− 2 (q).

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1. THEOREMS C7 AND C∗ 7

3

Theorem C∗7 : Stage 4b, and Stage 5, as formulated in earlier volumes [GLS5], [GLS7], assert: Theorem 1.3 (Theorem C∗7 : Stage 4b). Under the assumptions of Theorem let p ∈ γ(G). Suppose that there are x, K, y, L satisfying (1C) with K ∈ Chev. Then, if necessary after replacing (x, K, y, L) by another quadruple satisfying (1C), and setting D = x, y , the following conditions hold: (a) G0 ∈ Chev; (b) G0 /Z(G0 ) ∈ K(7)∗ ; (c) ΓD,1 (G) ≤ NG (G0 ); and (d) CG (G0 ) is a p -group.

C∗7 ,

(The additional conclusions in [III7 , Theorem 6.15] are trivial consequences of the above conclusions.) Theorem 1.4 (Theorem C∗7 : Stage 5). Let p, (x, K, y, L), and G0 be as in Theorem 1.3. Then G = G0 . As a result, G ∈ K(7)∗ . We shall actually prove a slightly different pair of theorems – a weaker 4b and stronger 5 – with the same overall effect as the above pair of theorems. We shall call them Stages 4b+ and 5+. The main weakening of Stage 4b is that we may have to change to another prime p ∈ γ(G) in order to obtain the conclusions. The other weakening is that two extra groups, and an extra family of groups, show up as possibilities for G0 . These changes ease the proof of Stage 4b without affecting the proof of Stage 5, except that the extra groups are easily ruled out after it is proved that G = G0 , because x and K arise from nonsimple configurations when G0 ∈ K(7)+ − K(7)∗ . Specifically, we define the new set of target simple groups as (1D)

− K(7)+ = K(7)∗ ∪ {A3 (q), q odd } ∪ {A− 4 (4), E6 (2)}.

The aim of Chapters 14 and 15 is to prove the following theorem. Theorem 1.5 (Theorem C∗7 : Stage 4b+). Under the assumptions of Theorem suppose that there are p, x, K, y, L satisfying (1C) with K ∈ Chev. Then, if necessary after replacing (p, x, K, y, L) by another quintuple satisfying (1C), and setting D = x, y , the following conditions hold: (a) G0 ∈ Chev; (b) G0 /Z(G0 ) ∈ K(7)+ ; (c) ΓD,1 (G) ≤ NG (G0 ); and (d) CG (G0 ) is a p -group. Moreover, if p = 2, an explicit list of possible neighborhoods N = {Lu | u ∈ D# } is given for each G0 as above. If p > 2, an explicit list of possible neighbors Lu , u ∈ D# is given for G0 . C∗7 ,

We wait until Chapters 14 and 15 to list the possible neighborhoods and neighbors. See Table 14.1 in Chapter 14 and Table 15.3 in Chapter 15. Chapter 16 in turn is devoted to the proof of: Theorem 1.6 (Theorem C∗7 : Stage 5+). Let p, (x, K, y, L), and G0 be as in the conclusion of Theorem 1.5. Then G = G0 ∈ K(7)∗ .

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4

12. INTRODUCTION

For the convenience of the reader, we recall the terminology used in the above theorems, insofar as it is relevant to what follows. Full definitions can be found in [GLS5] and [GLS7]. Definition 1.7. (a) Ipo (G) = {x ∈ G | xp = 1 = x, mp (CG (x)) ≥ 3, and mp (CG (x)) ≥ 4 if p is odd}. (b) ILop (G) = {(x, K) | x ∈ Ipo (G), K is a p-component of CG (x)}. (c) Lop (G) = {K | for some x, (x, K) ∈ ILop (G)}. The principal meaning of the Generic Case is that (1E)

Lop (G) ∩ Gp = ∅ for some prime p.

More precisely, the set γ(G) of those primes p through whose p-local structure we study G is defined as follows [III1 , 3.1]:

(1F)

(1) If Lo2 (G) ∩ G2 = ∅, then γ(G) = {2}. (2) If G is of even type, then γ(G) consists of those odd primes p such that (a) Lop (G) ∩ Gp = ∅, and (b) G does not have a strong p-uniqueness subgroup. (3) Otherwise, γ(G) = ∅.

As discussed in [III1 , Section 3], the Generic Case is defined by the condition γ(G) = ∅. Remark 1.8. There are two types of strong p-uniqueness subgroup [I2 , Def. 8.7, p. 94], but only those of component type are relevant in the Generic Case. The existence of a strong p-uniqueness subgroup was a key obstruction in Stages 1 and 2 in [GLS5], and it will be an obstruction again to the final stage of the proof of Theorem C∗7 in Chapter 16. Definition 1.9. A subgroup M of G is a strong p-uniqueness subgroup of component type if and only if (a) M is a p-component preuniqueness subgroup of G, and mp (M ) ≥ 4 if p > 2; (b) M is almost strongly p-embedded in G; and (c) One of the following holds: (1) Op (M ) = 1; or (2) M = NG (K) for some K ≤ G such that K ∈ Chev(2) and mp (CG (K)) ≤ 1. Remark 1.10. Likewise there are four alternative possible types of almost strongly p-embedded subgroups M of G [I2 , Definition 8.4, p. 94]; but in this monograph only two are relevant. They are the strongly p-embedded type, defined by the condition that ΓP,1 (G) ≤ M for some P ∈ Sylp (G), and the strongly closed type, whose definition we repeat [III1 , Definition 1.2, p. 2]. Recall that a proper subgroup of G is a K-preuniqueness subgroup [II3 , Definition 1.1] if and only M has a p-component K such that CG (y) ≤ M for every element y of order p in CM (K/Op (K)).

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1. THEOREMS C7 AND C∗ 7

5

Definition 1.11. Let M be a K-preuniqueness subgroup of G and let P ∈ Sylp (M ) and Q = CP (K/Op (K)). Then M is of strongly closed type if and only if (a) p is odd; (b) Either K ∈ Chev(2) with mp (Q) = 1, or K/Op p (K) ∼ = Lp (q),  = ±1, q ≡  mod p with Ω1 (Q) ≤ K; (c) Ω1 (Q) (if nontrivial) is strongly closed in P with respect to G; (d) NG (X) ≤ M for any noncyclic X ≤ P and for any 1 = X ≤ P for which mp (CP (X)) ≥ 3. For each prime p, the set Kp of all quasisimple K-groups K such that Op (K) = 1 has been partitioned Kp = Cp ∪ Tp ∪ Gp into the sets of groups that are then called “characteristic p-like”, “p-thin”, and “p-generic.” Furthermore, Gp is partitioned into 8 or 6 subsets Gip , i = 1, 2, . . . , depending on whether p = 2 or p > 2 [III3 , (5A), (5B)]. We define m  p (K), for any K ∈ Kp , by (1G)

 − mp (Z(K)),  m  p (K) := mp (K)

 is a universal covering group of K. where K For clarity and convenience we state the full definitions of Gi2 and Gip , p odd, as modified and discussed on page 5 of [III7 ]. In these definitions, we again use the terminology that  is the universal covering group of the quasisimple group K. K All groups L here are assumed to be in Kp , that is, they are quasisimple with Op (L) = 1. For p = 2: G12 = {L | L ∼ = 2An for some n ≥ 12}; G22 = {L | L ∼ = M c, Ly, or ON }; G32 = {L | L ∼ = An for some n ≥ 12, or L ∼ = F5 }; G42 = {L | L ∼ = J1 , He or Co3 }; (1H)

G52 = {L | L ∼ = A11 };  L3 (q) for any odd q and  = ±1}; G62 = {L | L ∈ Chev ∩ G2 but L ∼ = G72 = {L | L ∼ = L3 (q) for some odd q > 3 and some  = ±1}; G82 = {L | L ∼ = A9 or A10 };

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6

12. INTRODUCTION

and for p > 2: G1p = {L | L ∼ = An for some n ≥ 4p + 2};  − mp (Op (L))  ≥ 5}; G2p = {L | L ∈ Chev ∩ Gp and mp (L) G3 = {L | L ∼ = A13 } if p = 3, and ∅ otherwise; p

G4p = {L | L ∈ Chev ∩ Gp , mp (L) ≥ 3 and L ∈ G2p };

(1I)

G5p = {L | L ∈ Chev ∩ Gp − Alt and mp (L) = 2}; and G6 = {L | L ∈ Spor ∩ Gp or L ∼ = An for some n ≤ 4p + 1 p

(but n ≤ 12 if p = 3)}. Definition 1.12. Let p ∈ γ(G). (a) If K ∈ Gp − {G}, then dp (K) = the value of i such that K ∈ Gip . (b) d = dp (G) = min{dp (K) | (x, K) ∈ ILop (G), K ∈ Gp }. (c) [III3 , Definition 5.9] Jp (G) = {(x, K) ∈ ILop (G) | dp (K) = d and (x, K) is p-terminal in G}. (See [IG , Section 6] for a discussion of p-terminality.) Next, we use a ranking function F, defined on Chev, to order elements (x, K) ∈ Jp (G) with K ∈ Chev. The function F is a combination of an integer-valued function f on Chev and an ordering of the types A–G of simple Lie algebras. Definition 1.13. Let K ∈ Lie and write (1J)

K = d L(q).

Let  be the rank of the simple complex Lie algebra L, i.e., the dimension of a Cartan subalgebra of L. (1K)

2

f (K) := q  .

Furthermore if K ∈ Chev, so that K/Z(K) = [L, L] for some L ∈ Lie with Z(L) = 1, then f (K) := f (L). (If K is of ambiguous characteristic, the relevant characteristic will be clear from context.) The function F is a refinement of f . Definition 1.14. Let K ∈ Chev, as in (1J). Define τ (K) = A, BC, D, E, F , or G, according to the type of the simple Lie algebra L over C. Here D is reserved for those D such that  ≥ 4, and BC for those B and C such that  ≥ 2. Let p ∈ γ(G) and suppose that p divides |K|. Then we set F(K) = (f (K), τ (K)), and use the lexicographic ordering: F(K) < F(L) ⇐⇒ f (K) < f (L), or f (K) = f (L) with τ (K) < τ (L). Here we order the six types of Lie algebras independently of rank as follows: (1L)

A < D < E < BC < F < G.

For technical reasons irrelevant in the current volume, we formed a set J2p (G) ⊆ Jp (G) in [GLS7], and then defined the important set J∗p (G) in terms of J2p (G). Here p is any prime in γ(G). In this volume, however, while we are proving Theorem C∗7 :

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1. THEOREMS C7 AND C∗ 7

7

Stage 4b+ by contradiction, we are assuming that (1B) fails. As a result of this and the monotonicity result [III7 , Prop. 2.3], we have a much simpler characterization of the set J∗p (G), which should be kept in mind throughout. For all (x , K  ) ∈ Jp (G), we have K  ∈ Chev. Then J∗p (G) is the set of all (x , K  ) ∈ Jp (G) for which F(K  ) is maximized. (1M) Next, since acceptable subterminal (x, K)-pairs are the object of much analysis in this volume, we reproduce their elaborate definition, at least in the case K ∈ Chev. Definition 1.15. Let p ∈ γ(G) and (x, K) ∈ J∗p (G), and let (y, L) be a subterminal (x, K)-pair with y = x and K ∈ Chev. Then (y, L) is acceptable if and only if (y, L) is an (A, x, K)-admissible pair for some allowable (x, K) p-source A, and one of the following holds: (a) p > 2, with K ∈ Chev(2) a classical group and V a natural module, and one of the following holds: (1) K/Z(K) ∼ = SLn−2 (q) = Ln (q) with p dividing gcd(n, q − ), n ≥ 5, L ∼  (with y ∈ K) or SLn−1 (q) and (x, K, y, L) is not ignorable [III7 , 6.7]; (2) K/Z(K) ∼ = SLn−1 (q); = Ln (q) with p dividing q −  but not n, and L ∼ − ∼ (3) K/Z(K) = A3 (q) with p = 3 and q ≡  (mod 3), and L ∼ = SL2 (q 2 );  ∼ (4) K/Z(K) = Ln (q) and none of the situations (a1)–(a3) hold, y ∈ Ip (K) maximizes d := dim(CV (y)) and L ∼ = SLd (q); and ± ∼ (5) K = Sp2n (q), n ≥ 2, or Ω2n (q), n ≥ 4, y ∈ Ip (K) maximizes d := η dim(CV (y)) and L ∼ = Spd (q) (d ≥ 2) or Ωd (q) (with (d, η) = (4, +1)) for some sign η; (b) p > 2, with K = d L(q) ∈ Chev(2) of exceptional Lie type, m0 is the least positive integer such that p divides Φm0 (q), and one of the following holds: (1) p = 3 and K ∼ = SL3 (q), q ≡  (mod 3); = G2 (q) or 3D4 (q), and L ∼ n 2 n ∼ 2 (2) K = F4 (2 ), n ≥ 3, p divides 2 ± 1, and L ∼ = SU3 (2n ) or L2 (2n ) according as p = 3 or p > 3; (3) mp (K) = 2, situations (b1) and (b2) do not hold, y ∈ K and q(L) = q(K); (4) K ∼ = = E6 (q) or E7 (q) for some sign , Φm0 (q) = Φ3 (q), and L ∼ 3 D4 (q) or A5 (q), respectively; (5) K ∼ = E8 (q), m0 = 4, and L ∼ = D6− (q); (6) mp (K) ≥ 4 but case (b5) does not apply, y ∈ K, and L is an inductive associate of K, the triple (K, L, m0 ) is one of (F4 (q) (q > 2), C3 (q), 1 or 2), (E6 (q), A5 (q), 1 or 2), (E7 (q), D6 (q), 1 or 2), (E8 (q), D7 (q), 1 or 2), (E8 (q), E6 (q), 3), (E8 (q), E6− (q), 6);  → (c) p = 2, K ∈ Chev(r) ∩ G2 is a classical group of level q(K) = q, K  K/Z(K) is a covering such that K acts faithfully on a natural K-module  (only if y ∈ K) and one of the following V , y is the preimage of y in K holds: ± ∼ (1) K ∼ = L± 3 (q), Sp4 (q) or SL4 (q), and y ∈ L = SL2 (q); ± ∼ (q), y ∈ L SL  has 4 eigen(2) K ∼ = = Ω5 (q), Ω6 (q), or P Ω± 2 (q), and y 6 values −1 on V ; (3) K ∼ = SL3 (q); = P Ω6 (q), q ≡  (mod 8), y ∈ K, L ∼ (4) K ∼ = SL2 (q); = Spin7 (q) and y ∈ L ∼ (5) K ∼ (q), y ∈ K, and y ∈ L ∼ = Spin− = SL2 (q 2 ) with Z(K) = Z(L); 8

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8

12. INTRODUCTION   ∼ (6) K/Z(K) ∼ = L± 2n (q), n ≥ 3, L = SL2n−2 (q) or SL2n−1 (q), and (x, K, y, L) is not ignorable;  ∼ (7) K/Z(K) ∼ = L± 2n+1 (q), n ≥ 2, and y ∈ L = SL2n (q); (8) K/Z(K) ∼ = Sp2n−2 (q); = P Sp2n (q), n ≥ 3, and y ∈ L ∼ (9) K is an orthogonal group, spin group or half-spin group, of dimension  is an ∈ K at least 7, but not Spin− 8 (q) y ∈ I2 (K) is such that y involution maximizing the dimension d of the −1-eigenspace of y; ± ∼ L∼ = Ω± d (q) or Spind (q) (unless d = 4, in which case y ∈ L = SL2 (q)); and (x, K, y, L) is not ignorable; (10) K ∼ = Dn± (q), n ≥ 4, y acts on K as a graph automorphism, and ∼ L = Bn−1 (q); (d) p = 2, K ∈ Chev(r) ∩ G2 is of exceptional Lie type of level q(K) = q, and one of the following holds: n (1) K ∼ = L2 (q 2 ); = 2 G2 (3 2 ), n ≥ 3, y ∈ K, and L ∼ hs ∼ ∼ (2) K = E8 (q), y ∈ K, and L = D8 (q) ; or n (3) K ∼  2 G2 (3 2 ) or E8 (q) for any n, J is a long root A1 (q)-subgroup =  of K, y = Z(J), and L = O r (CK (J)). The pair (K, L) is one of the following: (G2 (q), SL2 (q)), (3D4 (q), SL2 (q 3 )), (F4 (q), C3 (q)), (E6 (q), A5 (q)), or (E7 (q), D6 (q)).

The main feature of this definition that is of interest in this volume is the isomorphism type of L and its embedding in K, given (x, K). A discussion of the technicalities in the above definition can be found in [GLS7]. For K ∈ Chev and p ∈ γ(G), we use the following notion of “splitting”. Definition 1.16. Let p be a prime and let K be a quasisimple group. Then p splits K if and only if the following conditions hold: (a) K ∈ Chev(r) for some prime r; and (b) p = 2 if r is odd; and p divides q(K)2 − 1 if r = 2. As stated above, this volume will complete the proof of Theorem C∗7 . We note the following important corollary of Theorems C2 , C3 , and C∗7 . Recall that a Kproper simple group G is of odd type if G is not of restricted even type. (“Restricted even type” is a minor technical strengthening of “even type” [I2 , p. 95].) Theorem 1.17. Let G be a K-proper simple group of odd type. Then G ∈ K, i.e., G is a known simple group. Thus, a minimal counterexample to the Classification Theorem is a K-proper simple group of restricted even type.

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10.1090/surv/040.8/02

CHAPTER 13

Recognition Theory 1. Curtis-Tits Systems and Phan Systems Recognition of Chevalley groups by Curtis-Tits systems and Phan systems is a fundamental tool in Chapters 14 and 15. We begin the discussion of these systems by recalling the Curtis-Tits Theorem [IA , 2.9.3]. For the terminology, see [IA , 2.3.1, 2.3.6], and [IG , Sec. 29]. Theorem 1.1. Let K be the universal version of a finite Chevalley group of  fundamental system Π,  and root twisted rank at least 3 with twisted root system Σ,   groups Xα , α  ∈ Σ. For each ∅ = J ⊆ Π, let KJ be the subgroup of K generated by  with at most all root subgroups Xα , ± α ∈ J. Let D be the set of all subsets of Π two elements, and partially order D by inclusion. Then AK := {KJ | J ∈ D} is a defining K-amalgam.  D, and the amalgam AK be as in Theorem 1.1. Definition 1.2. Let K, Π, Let L be any finite group with subgroups LJ , J ∈ D, satisfying the following properties: (a) For each J ∈ D there is a surjective homomorphism φJ : KJ → LJ with ker(φJ ) ≤ Z(KJ ); and  φ{ (b) For each α , β ∈ Π,  (K{ α} (K{ α} ) = φ{ α} ). α,β}  is a weak CT- (or weak  ∈ Π, Then we say that the family L of subgroups Lα , α Curtis-Tits) system in L of type K. Moreover, if |L| = 2, we call L a standard CT-pair of type K. This includes pairs of type A1 × A1 . Some remarks are in order. First, a weak CT-system comes not only with the subgroups L{  and φ{  and mappings φα  . For α} , but also with subgroups L{ α,β} α,β} simplicity, however, we usually abuse notation by referring just to the subgroups   Lα as constituting the system. Next, since K{  = K{  , the definition α} , K{β} α,β}    The “weak”-ness of the implies that L{ L{ for all α , β ∈ Π.  =  α} , L{β} α,β} definition is that it does not require the φ’s to form a morphism of amalgams AK → AL , where AK is formed by the KJ , J ∈ D, and AL is formed by the LJ , J ∈ , D. To have a morphism of amalgams would necessitate that φ{  |K{α} α} = φ{  α,β} whereas condition (b) in the definition only requires these two mappings to have the same image. Nevertheless, we shall see below (Theorem 1.3) that in our context, the existence of a weak CT-system generally does imply that there is a morphism AK → AL of amalgams. But first we illustrate the definition. A weak CT-system in L of type SLn (q), n ≥ 4, for example, consists of subgroups L1 , . . . , Ln−1 , all isomorphic to SL2 (q), and such that Li , Li+1 , for 1 ≤ i < n − 1, is isomorphic to a central quotient of 9 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

10

13. RECOGNITION THEORY

SL3 (q), by an isomorphism taking Li (resp. Li+1 ) to the upper-left (resp. lowerright) 2×2 block SL2 (q), and such that [Li , Lj ] = 1 whenever 2 ≤ i+1 < j ≤ n−1. Informally, a CT-system can be described as a set of groups, each pair of which is a standard CT-pair. The relationship between weak CT-systems and amalgams is clarified by the following theorem of Blok, Hoffman and Shpectorov [BHS, Theorem 4.19]. Theorem 1.3. Let K be the universal version of a finite group in Lie of twisted rank at least 3 and level q. Adopt the notation of Theorem 1.1. Suppose that L is a group with subgroups LJ , J ∈ D, forming a weak CT-system of type K. Let AL be the amalgam based on D formed by the LJ ’s and the inclusion mappings among them. Then there is a morphism of amalgams from AK to AL such that each mapping KJ → LJ , J ∈ D, is surjective. As a result, the Curtis-Tits Theorem   yields a homomorphism, necessarily sur jective, from K to L = L{ | α  ∈ Π . By the simplicity of K/Z(K), L is a α} central quotient of K. This immediate consequence of Theorems 1.1 and 1.3 is our principal application of the Curtis-Tits theorem. Theorem 1.4. Let K be the universal version of a finite group in Lie of twisted rank at least 3. Let L be a weak CT-system of type K in some group L. Then L is a central quotient of K. Equally useful is the Phan variation on this theory, in particular the analogues of Theorems 1.1, 1.3, and 1.4, as well as Definition 1.2, in Phan theory. The Phan variation makes sense only for certain groups K. The idea is again to generate a group of Lie type by SL2 (q) subgroups corresponding to the nodes of a (this time untwisted) Dynkin diagram, but the behavior of non-commuting pairs of SL2 (q)’s is a twisted version of the behavior in CT-systems. We begin by describing this non-commuting behavior, following [BHS]. In the next definition, 3-dimensional unitary groups and 4-dimensional symplectic groups are assumed to preserve the forms given by the matrices ⎡

⎡

⎤ 1 0 0 I = ⎣0 1 0⎦ 0 0 1

and

0 0 ⎢0 0 I0 = ⎢ ⎣ 0 −1 −1 0

0 1 0 0

⎤ 1 0⎥ ⎥ , respectively. 0⎦ 0

Definition 1.5. Let q be a prime power and let L1 and L2 be subgroups of some finite group L1,2 , with L1 and L2 centrally isomorphic to SL2 (q). Then L1 and L2 form a standard Phan pair (or standard P-pair) in L1,2 if and only if one of the following holds: (a) (Type A1 × A1 ) [L1 , L2 ] = 1 and L1,2 = L1 L2 ; (b) (Type A2 ) There is an isomorphism L1,2 → SU3 (q) or U3 (q) such that L1 and L2 map isomorphically to the 2 × 2 diagonal block SU2 (q) subgroups in the upper left and lower right, respectively; (c) (Type B2 ) L2 and L1 form a P-pair of type C2 ;

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1. CURTIS-TITS SYSTEMS AND PHAN SYSTEMS

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∼ Sp4 (q) or P Sp4 (q), and there is an injec(d) (Type C2 ) L1,2 = L1 , L2 = tion L1,2 → Sp4 (q 2 ) or P Sp4 (q 2 ) such that the images of L1 and L2 , respectively, consist of all elements of the form ⎤ ⎡ ⎤ ⎡ 1 0 0 0 a b 0 0 ⎢0 ⎢−bq aq 0 0⎥ a b 0⎥ ⎥ , respectively, ⎢ ⎥ ⎢ and q q ⎣0 −b a 0⎦ ⎣ 0 0 a −b⎦ 0 0 0 1 0 0 bq aq such that a, b ∈ Fq2 and a1+q + b1+q = 1. To be sure, the condition for Type B2 can be interpreted in terms of 5dimensional orthogonal groups. See [III17 , 4.3]. Note that L1 , L2 = L1,2 in all cases except for type A2 with q = 2. Underlying Phan twisting is the following elementary lemma. The group ΓK is introduced in [IA , 1.15.2b, 1.15.5]. Lemma 1.6. Let K be the universal version of a simple algebraic group over F r , where r is a prime. We use standard notation, so that (B, T ) is a Borel-torus pair invariant under σq , where q is a power of r. Then NKΓ (T )/T contains a K

unique element i inverting T elementwise. Specifically, either i ∈ W = NK (T )/T , or i = i0 γ where i0 ∈ W and γ ∈ I2 (ΓK ). The latter case occurs if and only if K∼ = A ,  > 1, D ,  odd,  ≥ 5, or E6 . In any case, i normalizes every subgroup of T , and in particular every subgroup of Z(K). Only certain groups of Lie type fit naturally with this twisting. Definition 1.7. Let K, T and i be as in Lemma 1.6. Take a preimage i of i in Aut0 (K) and set σ = (σq )i and K = CK (σ). (Since i is determined modulo the connected group T , Lang’s theorem implies that a different choice of i would lead to a conjugate automorphism σ and an isomorphic group K.) The groups K that arise from this process, and have untwisted rank  ≥ 3, we shall call Phan-able. They are the universal versions of (−1)

A−  (q), B (q), C (q), D

(q), F4 (q), E6− (q), E7 (q), and E8 (q).

Lemma 1.8. Let K, as just discussed, be Phan-able. Let Π be a fundamental system in the untwisted root system Σ of K, so that |Π| ≥ 3. Let σ = (σ q )i be as in Definition 1.7. For each ∅ = J ⊆ Π, set K J = X α , X −α | α ∈ J , and  KJ∗ = O r (CK J (σ)). Also let ΣJ be the root subsystem generated (over Z) by J. Then the following conditions hold: (a) For any ∅ = J ⊆ Π, K J is σ-invariant; ∗ ∼ (b) K{α} = SL2 (q) for every α ∈ Π; and ∗ ∗ (c) For any distinct α, β ∈ Π, K{α} and K{β} form a standard P-pair in   ∗ , whose type is the same as that of the root subsystem generated K{α,β} by α and β. i

Proof. Since i inverts T elementwise, X α = X −α for any α ∈ Σ. This implies (a). Let α ∈ Π and set T α = T ∩K {α} , a maximal torus of K {α} . Then i normalizes

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12

13. RECOGNITION THEORY

K {α} and inverts T α . Since T α is self-centralizing in Aut(K {α} ) and is inverted in K {α} , there exists iα ∈ NK {α} (T α ) such that i and iα have the same action on  K {α} . Therefore K ∗ = O r (C (σq iα )) ∼ = SL2 (q), the last since K is universal, {α}

K {α}

proving (b). We have not yet used the assumption that  ≥ 3. In (c), we let T {α,β} = T ∩ K {α,β} , a maximal torus of K {α,β} . We similarly see that there exists i{α,β} ∈ NAut0 (K {α,β} ) (T {α,β} ) with the same action as i on

K {α,β} . This observation, and the assumption  ≥ 3, reduce us to proving (c) when  = 2 but Σ is not G2 . Thus, Π = {α, β} has one of the four types appropriate for a standard P-pair. If Σ ∼ = A1 × A1 , both factors invariant under i, and (c) = A1 × A1 , then K ∼ follows from (b) in this case. If Σ ∼ = A2 , we identify K with SL3 (Fq ) in the usual way, with X α and X β identified with the root subgroups I + Fq e12 and I + Fq e23 , respectively, and T identified with the diagonal subgroup of K; here e12 and e23 are matrix units. Now i inverts T and so i = γt, where t acts as conjugation by a diagonal matrix and γ is the transpose-inverse automorphism. Thus σq i acts on K as σq γt, which is T -conjugate to σq γ by Lang’s theorem. Therefore (1A)

∗ K{α,β} = CK (σq γt) ∼ = CK (σq γ),

the isomorphism given by conjugation by some g ∈ T . One easily sees that the righthand group is SU3 (q), with respect to the identity matrix form (i.e., with respect to an orthonormal basis), and CK {α} (σq γ) is the upper-left block diagonal SU2 (q) ∼ = SL2 (q)-subgroup; similarly for β. Thus, the isomorphism in (1A), combined with the identification CK (σq γ) ∼ = SU3 (q), satisfies the conditions required in Definition 1.5. Similarly, if Σ = C2 or B2 , we may assume that Σ = C2 . We identify K with Sp4 (Fq ) as usual, so that X α and X β are identified with the root subgroups I + Fq e23 and I + Fq (e12 − e34 ) and again T is the diagonal subgroup of K; again the eij are matrix units. This time i ∈ W and we take i to be conjugation by the ∗ matrix I0 introduced before Definition 1.5. We have ((σq )i)2 = −σq2 , so Kα,β = 2 CK ((σq )i) ≤ CK (σq2 ) ∼ = Sp4 (q ), and a straightforward calculation shows that ∗ ∗ ∗ and K{β} of Kα,β are just the subgroups the fundamental SL2 (q) subgroups K{α} displayed in Definition 1.5d. The proof is complete.  Remark 1.9. There is an untwisted analogue of Lemma 1.8. If one changed σ ∗ ∗ from (σq )i to σq in Lemma 1.8, then the resulting subgroups K{α} and K{β} of the untwisted group CK (σq ) would form a standard CT-pair, instead of P-pair, for the group they generate. This follows easily from the definition of standard CT-pair. Several authors, starting with the pioneer Kok-Wee Phan [Ph1], [Ph2], and including R. Gramlich, M. Aschbacher, C. Bennett, R. Blok, A. Devillers, J. Dunlap, C. Hoffman, M. Horn, B. M¨ uhlherr, W. Nickel, S. Shpectorov, and S. Witzel, have established a twisted analogue of Curtis-Tits theory for most Phan-able groups. Generalizations of Theorem 1.1 are neatly summarized by Gramlich, with references, in [Grm2]. We shall need the following fairly general cases.

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1. CURTIS-TITS SYSTEMS AND PHAN SYSTEMS

13

Theorem 1.10. Let K be a Phan-able group in Lie. Let Π be a fundamental system in the untwisted root system of K, and let the groups KJ∗ , for J ⊆ Π, be as defined in Lemma 1.8. Let D be the set of all 1- and 2-element subsets of Π. Then A∗K = {KJ∗ | J ∈ D} is a defining K-amalgam in the following cases. (a) K ∼ = SUn (q), n ≥ 4, q > 3; (b) K ∼ = Bn (q), n ≥ 3, q odd, q > 3; (c) K ∼ = Cn (q), n ≥ 3, q > 2; (−1)n (q), n ≥ 4, q odd, q > 3; (d) K ∼ = Dn (−1)n ∼ (e) K = Dn (q), n = 4, 5, or 6, q even, q > 2; (f) K ∼ = E6− (q), q even, q > 2; or (g) K ∼ = F4 (q), q even, q > 8. For several cases of small fields, extra conditions on subgroups corresponding to certain 3-node and 4-node subsets of Π may be imposed to guarantee similar conclusions. First, let us define “weak Phan system.” Definition 1.11. Let K be a Phan-able group in Lie. Let Π, and the subgroups KJ∗ for J ⊆ Π and |J| = 1 or 2, be as in Lemma 1.8. Let D = {J ⊆ Π | 0 < |J| ≤ 2}. Let L be any finite group with subgroups LJ , J ∈ D, satisfying the following properties: (a) For each J ∈ D, there is a surjective homomorphism φJ : KJ∗ → LJ with ker(φJ ) ≤ Z(KJ∗ ); and (b) For each α, β ∈ Π, φα (Kα∗ ) = φ{α,β} (Kα∗ ). Then we say that the family L of subgroups Lα , α ∈ Π, is a weak P-system (or weak Phan system) in L of type K. The extra conditions to which we referred above in the case of small fields are captured – to the extent we shall need them – in the following definition. Definition 1.12. Let L = {Lα | α ∈ Π} be a weak P-system of type K ∈ Lie in the group L. We say that L is extendible if and only if the following extra conditions also hold. (a) If K = SUn (3), n ≥ 5, or K = Bn (3) = Spin2n+1 (3), n ≥ 4, or K = (−1)n (−1)n Dn (3) = Spin2n (3), n ≥ 4, then for any J ⊆ Π of type A3 or B3 there is a surjective homomorphism φJ : KJ∗ → Lα | α ∈ J such that ker(φJ ) ≤ Z(KJ∗ ); and (b) If K = SUn (2), n ≥ 7, then (1) For any J = {α, β, γ} ⊆ Π of type A3 , with (α, γ) = 0, there is a surjective homomorphism KJ∗ → Lα,β , Lβ,γ ; (2) For any J = {α, β, γ} ⊆ Π of type A1 × A2 , with (β, γ) = 0, [Lα , Lβ,γ ] = 1; and (3) For any J = {α, β, γ, δ} ⊆ Π of type A2 × A2 , with (α, β) = 0, [Lα,β , Lγ,δ ] = 1. (If K is not of one of these types then L is automatically considered to be extendible.) As with CT-systems, we need an analogue of Theorem 1.3 for Phan systems. This has been the subject of the work of many of the authors mentioned above. We use [BHS, Theorem 5.17].

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14

13. RECOGNITION THEORY

Theorem 1.13. Let K ∈ Lie be the universal version of a finite group of Lie type of untwisted rank at least 3. Suppose that K is Phan-able, and let D and A∗K be as in Theorem 1.10. Let L be a weak P-system of type K in a group L. Let AL be the amalgam based on D formed by the LJ ’s and the inclusion mappings among them. Then there is a morphism of amalgams from AK to AL such that each mapping KJ → LJ , J ∈ D, is surjective. It follows directly from Theorems 1.10 and 1.13 that if L is a weak P-system in L of type K ∈ Lie, and if K is one of the groups in Theorem 1.10, then there is a surjection K → L with kernel contained in Z(K). This fact can be generalized [Grm2] to include the groups K in Definition 1.12, if L is extendible. The result is our main tool from Phan theory. Theorem 1.14. Let K ∈ Lie be the universal version of a finite group of Lie type of untwisted rank at least 3. Suppose that K is Phan-able, and indeed that K is one of the groups listed in Theorem 1.10 or Definition 1.12. Let L be a weak Psystem of type K in some group L. If L is extendible, then L is a central quotient of K. We will use Theorems 1.4 and 1.14 extensively. Definition 1.15. By a weak CTP-system we mean either a weak CT-system or a weak P-system. By an extendible weak CTP-system we mean either a weak CT-system or an extendible weak P-system. 2. The Gilman-Griess Theorem for Groups in Chev(2) Again for terminology we refer to [IA , 2.3.1]. In particular, if K = d Σ(q) ∈ Lie but K is not a Suzuki group or a Ree group, and Π is a fundamental system in the root system Σ, then K has a σ-setup (K, γρ σq ); here ρ is an isometry of Π of order d, which extends to an isometry τ of Euclidean space V = RΣ = RΠ. The orthogonal

of Σ under this projection of V on V := CV (τ ) is written v → v , and the image Σ  of projection can be called the twisted root system of K. The further quotient Σ

defined as the set of rays through elements of Σ,

is sometimes called the twisted Σ, root system of K as well, as we have done in the previous section. The Weyl group  = CW (τ ), a W ≤ O(V ) of Σ is invariant under conjugation by τ , and we write W

 group that acts on both Σ and Σ. Definition 2.1. Let K = d Σ(q) and K1 = d1 Σ1 (q) be two groups of Lie type,  be the Weyl group of

and Σ

1 . Let W with corresponding twisted root systems Σ

and let Y be a subgroup of W  . Then K and K1 are compatible over Y if and Σ, only if the following conditions hold:

and Σ

1 are at least 3 and 2, respectively; (a) The ranks of Σ (b) K and K1 are either both twisted or both untwisted;

1 ) of Σ,

and

1 can be identified with a proper subsystem (also called Σ (c) Σ

both Σ and Σ1 are irreducible;

let α  = {

and (d) For any α

∈ Σ,  be the ray through α

. Let Σ α|α

∈ Σ} 

 Σ1 = { α|α

∈ Σ}. Then for any α  ∈ Σ1 , the root subgroups Xα of K and K1 are isomorphic; and

⊆Σ

there exists w ∈ Y such that {w(

1. (e) For any α

, β ∈ Σ, α), w(β)}

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3. THE WONG-FINKELSTEIN-SOLOMON METHOD

15

Here is the Gilman-Griess theorem [GiGr1], [IA , 2.9.9]. Theorem 2.2 ([IA ], 2.9.9). Let K and K1 be groups of Lie type of the same

as in Definition 2.1, even level which are compatible over the subgroup Y of W (Σ),   1 . with K1 = Xα | α ∈Σ Let K0 be a group containing K1 as well as a subgroup R for which there is a surjective homomorphism δ : R → Y . By this homomorphism we regard R as

Let W 1 be the (twisted ) Weyl group of Σ

1 , identified naturally with a acting on Σ. 

subgroup of the (twisted ) Weyl group W of Σ. Assume the following conditions: 1 , R ∩ K1 permutes the set {Xα | α  1 } the same (a) δ(R ∩ K1 ) = Y ∩ W ∈Σ

way it permutes Σ1 via δ, and R ∩ K1 acts transitively on the set of roots

1 of any given length; in Σ  in R normalizes the root subgroup Xα (b) For any α ∈ Σ1 , the stabilizer of α of K1 .  Set I = K1R . Then I ∈ Lie and I/Z(I) ∼ = K/Z(K). Assumption (b) allows one to make a consistent definition of root subgroups 

in the natural way. Namely, for any α

choose any

∈Σ

∈ Σ, Xα of K1R for all α g −1

g ∈ R such that g( α) =: β ∈ Σ1 , and set Xα = (Xβ) . Using the condition (e) in Definition 2.1, one verifies the Chevalley commutator relations by conjugation into K1 , and these relations imply the conclusion of the theorem.

3. The Wong-Finkelstein-Solomon Method In this section we extend results of W. J. Wong [Wo1] and Finkelstein-Solomon [FinS1] to give presentations of classical linear groups that will be important for us to use. Throughout we will be comparing a subgroup H = K, N of our K-proper simple group G with a classical group C = C(V ) of isometries of a vector space V of finite dimension > 4 over a finite field F. We shall assume that V is equipped with either the zero form (with C = GL(V )) or a nonsingular symplectic, unitary, or orthogonal (quadratic) form, according as C = Sp(V ), GU (V ), or O(V, Q) for some quadratic form Q. For consistency, in the case of GL(V ), we shall write ⊥ instead of ⊕ to indicate a direct sum; and references to “orthogonal” sums in that case should be interpreted to mean direct sums. We let κ denote the type of C with κ = L, U, S, or O, and correspondingly set |κ| = 1, 2, 1, or 1. We may use the notation C κ = C κ (V ) to emphasize the type of C. We also will denote L, U as L , with  = ±1, L+1 = L+ = L, and L−1 = L− = U . We set C o = [C, C], so that given that dim V > 4, C o ∈ Lie(r) where r is the characteristic of the field F. As such, C o has a level q. Then |F| = q |κ| in all four cases.

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16

13. RECOGNITION THEORY

We shall fix an orthogonal direct sum decomposition V = V0 ⊥ V1 ⊥ · · · ⊥ Vm

(3A)

of one of several particular types. In all cases, dim(Vi ) ≤ 2 for i = 0, 1, . . . , m, and Vi = 0 for i = 1, . . . , m, although V0 may be 0. Moreover, V1 , . . . , Vm will all be isometric. When char(F) is odd, i.e., q is odd, Wong treats all unitary and orthogonal spaces of suitably large dimension. We will only need the case in which V is an orthogonal space of dimension 8 and Witt index 3, so that C = O(V ) ∼ = O8− (q) in the latter case. Specifically, the decomposition we shall study with q odd is the following: κ = O, q > 3, q ≡ −1 (mod 4), dim V = 8, m = 7, Vi = Fvi , 0 ≤ i ≤ 7, (3B) Q(vi ) = 1, 1 ≤ i ≤ 7, and Q(v0 ) ∈ F× − (F× )2 , where Q is the quadratic form on V . When char(F) = 2, i.e., q is even, we will consider spaces V of any of the four types. In the case κ = O, we assume that dim(V ) is even and recall that given dim(V ), there are two possible quadratic forms Q, up to isometry, and our choice is specified by a sign , so that for dim(V ) = 2n ≥ 6, Ω (V ) ∼ = Dn (q). Here again + − 2 we write Dn (q) = Dn (q) and Dn (q) = Dn (q), and we use a similar convention for E6 (q). For q even, we let p be a prime divisor of q 2 − 1 and in this case let q = ±1 be such that p divides q − q . Thus p is odd. Let Vκ,p be a nonsingular F-vector space of type κ of minimum dimension such that |C κ (Vκ,p )| is divisible by p. Thus, dim(Vκ,p ) = 1 if κ = Lq , and otherwise, dim(Vκ,p ) = 2. (See [IA , 4.8.1].) In any case, Vκ,p is uniquely determined up to isometry by q, p, and κ. We shall study decompositions (3A) with Vi isometric to Vκ,p for all i = 1, . . . , m, when q is even, (3C) and the following specific cases will be our focus: κ = Lq , q even, q > 2, m ≥ 5, V0 = 0, dim Vi = 1, 1 ≤ i ≤ m; κ = U , q = 2, m ≥ 7, V0 = 0, dim Vi = 1, 1 ≤ i ≤ m; κ = L−q , q even, m ≥ 4, dim V0 = 0 or 1, dim Vi = 2, 1 ≤ i ≤ m; κ = S, q even, m ≥ 4 with strict inequality if q = 2, V0 = 0, dim Vi = 2, 1 ≤ i ≤ m; or (5) κ = O, q even, m ≥ 5, dim V0 = 0 or 2, V0 not isometric to V1 , dim Vi = 2, 1 ≤ i ≤ m, and dim V ≥ 12; (6) κ = O, q even, m = 5, dim V = 10, V0 = 0, V1 is of + type, and p divides q − 1 (in particular, q > 2). (1) (2) (3) (4)

(3D)

For any subset I ⊆ X := {0, 1, . . . , m}, we let VI = i∈I Vi , and we let I  = X −I. We set CI = CC (VI  )∩NC (VI ). Except for type L, CI = CC (VI  ) as then VI = VI⊥ . In any case, CI acts faithfully as a group of isometries of VI . Also, for each i = 0, . . . , m and each I ⊆ X, we write 

Ci = C{i} = CC (Vi ) ∩ NC (V{i} ), and CIo = C o ∩ CI . ∼ Σm , T perBy [III17 , 2.40], in the cases (3D), there is T ≤ C o such that T = mutes {V1 , . . . , Vm } faithfully, T centralizes V0 , and NT (Vi ) = CT (Vi ) for every (3E)

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3. THE WONG-FINKELSTEIN-SOLOMON METHOD

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i = 1, . . . , m. In the remaining case (3B), where m = 7, there is a subgroup T ≤ C o satisfying all these conditions, except that T ∼ = A7 , not Σ7 . We fix such a subgroup T , and define an action of T on X = {0, 1, . . . , m} by Vσ(i) = σ(Vi ) for all i ∈ X and σ ∈ T . Note that our conditions imply that for σ ∈ T and i ∈ X, (3F)

if σ(i) = i, then σ centralizes Vi .

Our main goal in this section is to prove the following theorem. Theorem 3.1. Suppose that H is a finite group generated by subgroups K and N . Let C = C(V ) be a classical group, and assume that V has a decomposition o (3A) satisfying either (3B) or one of the cases of (3D). Let Cm  be as in (3E), and o ∼ let T ≤ C with T = Σm or A7 be as fixed above. Suppose that there are surjective homomorphisms o f : Cm  → K and λ : N → T

and that the following conditions hold: (a) (b) (c) (d)

m ≥ 5; o ker f ≤ Z(Cm  ); o For all σ ∈ T ∩ Cm  , f (σ) ∈ N and λ(f (σ)) = σ; and For all I ⊆ X − {m} and all u ∈ N with σ = λ(u) and σ(I) ⊆ X − {m}, o ). we have f (CIo )u = f (Cσ(I)

Then there exists a homomorphism g : C o → H such that ker(g) ≤ Z(C o ). Moreover, the following conclusions hold: o o (a) If κ = O, then g|Cm  = f and [K, g(Cm )] = 1; o (b) If κ = O and q is even, then g(C{m−1,m} ) ≤ K; (c) If C o ∼ = Ω− 8 (q), q odd, then there is a 6-dimensional subspace U of V7 such that Ω(U ) ∼ )) = Ω− 6 (q) and g(Ω(U )) = f (Ω(U ≤ K;  (d) If V and K are of type Lq , then g(C o ) = K N ; and (e) If q is even, and C o = D5 (q) with the Vi all hyperbolic planes, or C o = o o D6 (q), q > 2, then g|Cm  = f and [K, g(Cm )] = 1.

We shall also prove a slight extension to the case m = 4. Theorem 3.2. Suppose that the hypotheses of Theorem 3.1 hold, except that the hypothesis m ≥ 5 is replaced by m = 4. In addition, assume the following: (a) λ is an isomorphism; and (b) [H{1,2} , H{3,4} ] = 1 (see Definition 3.3 below). Then f extends to a homomorphism g : C o → H such that ker(g) ≤ Z(C o ), g(C4o ) = K, and [K, g(C4o )] = 1. Note that because of our restricted set of examples (3B), (3D), the only cases in which Theorem 3.2 applies are (3D3, 4), i.e., for V a symplectic space of dimension 8 or a linear or unitary space of dimension 8 or 9 of type L−q . As we prove Theorem 3.1, we note which steps remain valid for m = 4. We begin the proof of Theorem 3.1, and explain the notation H{1,2} and H{3,4} in Theorem 3.2, by defining a family of subgroups of H.

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18

13. RECOGNITION THEORY

Definition 3.3. P is the set of all proper subsets D of {0, 1, . . . , m} such that dim(VD ) ≥ 2, but excluding {0} and {1, . . . , m}. (So if dim(Vκ,p ) = 2, then D is an arbitrary proper subset of {0, 1, . . . , m}, excluding {0} and {1, . . . , m}. However, if κ = Lq , q even, or C ∼ = O8− (q), q odd, then |D| ≥ 2 for all D ∈ P.) For each D ∈ P, we define a subgroup HD of H as follows: first choose F ⊆ {0, 1, . . . , m − 1} and δ ∈ T with δ(F ) = D; then choose nδ ∈ N with λ(nδ ) = δ; finally, set HD = f (CFo )nδ . We must establish that HD is well-defined, i.e., independent of the choices of F , δ, and nδ . Lemma 3.4. Let D ∈ P. Then the following conclusions hold: (a) (b) (c) (d) (e)

HD is independent of the choice of nδ ∈ λ−1 (δ); HD is independent of the choices of F and δ; o HD = f (CD ) if D ⊆ {0, 1, . . . , m − 1}; nσ HD = Hσ(D) for any σ ∈ T and nσ ∈ N with λ(nσ ) = σ; and o CD ∩ T ≤ λ(HD ∩ N ).

Proof. First, we argue that HD is independent of the choice of nδ ∈ λ−1 (δ). Thus we must argue that f (CFo )u = f (CFo ) for all u ∈ ker(λ) and all F ∈ P with F ⊆ {0, 1, . . . , m − 1}. But this is immediate by hypothesis (d) of Theorem 3.1. Next, assume that F  ⊆ {0, 1, . . . , m − 1} and δ  ∈ T satisfy δ  (F  ) = D. Then F = δ −1 (δ  (F  )). Choose n, n ∈ N with λ(n) = δ, λ(n ) = δ  . Since F and F  are subsets of {0, 1, . . . , m − 1} and λ is a homomorphism, it follows by hypothesis (d) again that 

−1

f (CFo ) = f (CFo  )n n , 

whence f (CFo )n = f (CFo  )n . Thus HD is independent of the choices of F and δ, proving (b). By (b), if D ⊆ {0, 1, . . . , m − 1}, then we may choose F = D and δ = 1, whence (c) holds. To establish (d), we observe that if λ(nδ ) = δ and λ(nσ ) = σ, then λ(nδ nσ ) = δσ, so Hσ(D) = f (CFo )nδ nσ = (HD )nσ . o o Finally, if m ∈ D, then T ∩ CD = λ(f (T ∩ CD )) ≤ λ(N ∩ HD ) by hypothesis (c) of the theorem and the definition of HD . If m ∈ D, on the other hand, then there is 0 < i < m such that i ∈ D, and we let σ = (i, m, j) ∈ T for some choice of j ∈ {0, i, m}. Then m ∈ σ(D). Choose any n ∈ N with λ(n) = σ. By the case just done, and conclusion (d), o o = (T ∩ Cσ(D) )σ T ∩ CD

−1

≤ (λ(N ∩ Hσ(D) ))σ −1

−1

= λ((N ∩ Hσ(D) )n ) = λ(N n

−1 −1

n ∩ Hσ(D) ) = λ(N ∩ HD ).

 We remark that Definition 3.3 and Lemma 3.4 remain valid if m = 4. The next two lemmas, which rely on basic generation results, are of great importance.

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3. THE WONG-FINKELSTEIN-SOLOMON METHOD

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Lemma 3.5. Suppose that m ≥ 5. Let D ⊆ X with |D| ≥ m/2. Let {i, j} ⊆ D with 0 < i < j ≤ m. If κ = O and |D| = 3, assume that D = {i, j, k} where Vk is 2-dimensional of + type. Then   o o o . CD = CD−{i} , CD−{j} Proof. Sufficient conditions for the desired conclusion are given in [III17 , 9.1] in the terms of κ, q, and lower bounds on dij = dim(VD−{i} ∩ VD−{j} ) = dim VD−{i,j} . Since m ≥ 5, |D| ≥ 3 and so dij > 0. This is sufficient for the types κ = L, U , and S, except for the case κ = U and q = 2, for which dij ≥ 2 is sufficient. But in that case m ≥ 7 by (3D), so |D| ≥ 4 and indeed dij ≥ 2. We may therefore assume that κ = O. Now, [III17 , 9.1] stipulates |D − {i, j}| ≥ 2, whether q > 3 is odd or q is even. If |D| ≥ 4 then this condition is met. Finally, if |D| = 3, then in view of our hypothesis on D = {i, j, k}, the desired generation result is [III17 , 9.2]. (Note that Ω(V{i,j,k} ) is involved in the K-proper group K in that case.) The proof is complete.  We remark that if κ = O and m = 5 then (3D6) holds. Thus all nonzero Vi ’s are of + type, so Lemma 3.5 applies. Lemma 3.6. Suppose that m ≥ 5. Let D, F ∈ P with D ∩ F = ∅. Then [HD , HF ] = 1 provided that one of the following conditions holds: (a) D ∪ F ∈ P; (b) κ = O; (c) |D| ≥ 4; or (d) κ = O and D = {i, j, k}, where 0 < i < j ≤ m and Vk is 2-dimensional of + type. Proof. First, suppose that D ∪ F ∈ P. Then there exist D∗ , F ∗ ⊆ {0, 1, . . . , m − 1} and δ ∈ T such that δ(D∗ ) = D and δ(F ∗ ) = F . Since [CD∗ , CF ∗ ] = 1 and f : C → H is a homomorphism, it follows that o o nδ = [HD , HF ], 1 = [f (CD ∗ ), f (CF ∗ )]

as desired. We may thus assume that D ∪ F ∈ P, and D ∪ F ⊇ {1, . . . , m}. Without loss, |D| ≥ |F |, so |D| ≥ m/2 and |D| ≥ 3. Choose i, j ∈ D with 0 < i < j and set X = HD−{i} , HD−{j} . By (a), [X, HF ] = 1, so it suffices to prove that  nδ o o X = HD . From the definition of HD it follows that X = f CD−{i} , CD−{j} −1 o where δ is chosen  with m ∈ δ (D). It therefore suffices to show that CD =  o o , CD−{j} . But since (b), (c), or (d) holds, Lemma 3.5 applies to complete CD−{i} the proof. 

The proofs of Lemmas 3.5 and 3.6 used the hypothesis m ≥ 5, so they will eventually have to be adapted to the situation of Theorem 3.2. From here through the proof of Lemma 3.10 we assume that (3D) holds. Thus T ∼ = Σm . We begin with the following consequence of Lemma 3.6. Lemma 3.7. Assume (3D). Then there is n(1,m) ∈ N with λ(n(1,m) ) = (1, m) and [n(1,m) , H{1,m} ] = 1.

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20

13. RECOGNITION THEORY

Proof. Let D = {1, m} and σ = (1, m). By Lemma 3.4e, there is n ∈ N ∩ HD such that λ(n) = σ. Provided that |D | ≥ 4 or κ = O, Lemma 3.6bc implies that [n, HD ] ≤ [HD , HD ] = 1, as desired. So we may assume that |D | = 3 and κ = O, whence m = 5 and V0 = 0. Thus (3D6) holds, so all Vk are of + type, and Lemma 3.6d yields the desired conclusion.  The proof of Lemma 3.7 depends on the hypothesis m ≥ 5 only to the extent that it quotes Lemma 3.6. Fix n(1,m) ∈ N as guaranteed by Lemma 3.7. For 1 < i < m, define n(i,m) = f ((1,i)) n(1,m) ∈ N . Notice that λ(f ((1, i))) = (1, i), so λ(n(i,m) ) = (1, m)(1,i) = (i, m). Also, as n(1,m) centralizes H{1,m} , (3G)

n(i,m) centralizes (H{1,m} )f ((1,i)) = H{i,m} ,

by Lemma 3.4d. We now define a function h extending our function f : Cm → K. Definition 3.8. Assume (3D). Let Γ = X − {0} = {1, . . . , m}. For i ∈ Γ, let i = {i} . Define h : ∪i∈Γ Cio → H by  o f (x) for x ∈ Cm ; h(x) = (i,m) n(i,m) o )) for x ∈ Ci , 0 < i < m. (f (x Lemma 3.9. Suppose that m ≥ 5. Then the function h is well-defined. In o particular, h is a well-defined homomorphism on CD for all D ∈ P. Proof. Let hi : Cio → H be given by the formula for h|Cio in Definition 3.8. If i and j are distinct elements of Γ, we must show that hi and hj agree on o o Cio ∩ Cjo = C{i,j}  . Let x ∈ C{i,j} . If j = m, then by definition, hm (x) = f (x). o On the other hand, if x ∈ C{i,m} , then hi (x) = (f (x(i,m) ))n(i,m) = f (x)n(i,m) = f (x), as desired, by (3G) and (3F). Suppose now that 0 < i < j < m. We apply Lemma 3.5 (using  the succeeding   o o remark if necessary) to D = {i, j} , and conclude that C{i,j} = C{i,j,m}  , (l, m) , o o o for any l ∈ {0, i, j, m} . Now, C{i,j,m}  = C{i,m} ∩ C{j,m} and so hi (x) = f (x) = o hj (x) for x ∈ C{i,j,m} . Since hi is a homomorphism for each i ∈ Γ, it remains only to show that hi and hj agree on (l, m). If i = 1, then by definition (3H)

hi (l, m) = f ((1, l))n(1,m) .

If 1 < i < m, then we argue that (3H) still holds. Namely, by definition, hi (l, m) = f ((l, m)(i,m) )n(i,m) = f ((i, l))f (i,1)n(1,m) f (i,1) = f ((1, l))n(1,m) f ((i,1)) . But f ((1, l))n(1,m) ∈ H{l,m} and i is distinct from 1, l, and m. As m ≥ 5, Lemma 3.6 applies with D = {l, m} and F = {1, i}, condition (a) holding. Thus [H{l,m} , H{1,i} ] = 1, and (3H) follows in this case as well. Hence, hi (l, m) is independent of i, and the lemma follows.  The proof of Lemma 3.9 will have to be adapted to the situation of Theorem 3.2.

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3. THE WONG-FINKELSTEIN-SOLOMON METHOD

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o Lemma 3.10. Assume (3D). Then h(CD ) = HD for all D ∈ P.

Proof. We may assume that D ⊆ {0, 1, . . . , m−1}. Then, for some i ∈ Γ−{m} and σ = (i, m), we have σ(D) ⊆ {0, 1, . . . , m−1}. By Lemma 3.4d and the definition of h, we have nσ o o σ nσ o h(CD ) = f ((CD ) ) = f (Cσ(D) )nσ = Hσ(D) = HD .

 The proof of Lemma 3.10 remains valid in the situation of Theorem 3.2. Now we define a suitable extension h of f in the case when (3B) holds, i.e., m = ∼ 7 and C o = Ω− 8 (q) with q odd, and T = A7 . Set σ = (2, 1, 6) ∈ T , γ = (5, 6, 7) ∈ T , and τ = σ γ = (2, 1, 7). For corresponding elements of N , set nσ = f (σ), choose n nγ ∈ N ∩ H{5,6,7} such that λ(nγ ) = γ (Lemma 3.4e), and set nτ = nσ γ . Thus by hypothesis (c) of the theorem, λ(nσ ) = σ, whence λ(nτ ) = τ . Note also that by Lemma 3.6, (3I)

[nγ , H{5,6,7} ] = 1.

o We have σ ∈ C{1,2,6} , so nσ ∈ H{1,2,6} , and conjugating by nγ , nτ ∈ H{1,2,7} by Lemma 3.4d. Thus by Lemma 3.6 again,

(3J)

[nτ , H{1,2,7} ] = 1.

Definition 3.11. Suppose case (3B) holds. For 1 ≤ i ≤ 7, let i = {i} . Define h : C7o ∪ C1o → H by  f (x) for x ∈ C7o ; h(x) = −1 for x ∈ C1o . (f (xτ ))nτ Lemma 3.12. In case (3B), the function h is well-defined. Proof. Let h0 : C1o → H be given by the formula in Definition 3.11 for o o x ∈ C1o . We must show that h0 (x) = f (x) for all x ∈ C{1,7}  . For x ∈ C{1,2,7} , we have −1 −1 h0 (x) = (f (xτ ))nτ = f (x)nτ = f (x), as desired, by (3F) and (3J). o o Now let x ∈ C{0,2} . Since στ −1 = (1, 6, 7) centralizes C{0,2} , it follows that σ τ nσ f (σ) σ τ = f (x ) = f (x ). Conjugating by nγ , which x = x . Then f (x) = f (x) o o σ ) = H0,2 and f ((C0,2 ) ) = H0,1 , by (3I), we see that centralizes both f (C0,2 f (x)nτ = f (xσ ) = f (xτ ), whence −1

h0 (x) = (f (xτ ))nτ = f (x)   o o o o by for all x ∈ C{0,2} . Now, as q > 3 by assumption, C{1,7} C{1,2,7}  =  , C{0,2}

o [III17 , 9.3]. Hence, h0 (x) = f (x) for all x ∈ C{1,7}  , as claimed, and the lemma follows. 

o Lemma 3.13. Assume (3B). Then h(CD ) = HD for all D ∈ P such that either 7 ∈ D or 1 ∈ D.

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Proof. If 7 ∈ D, this is immediate. So assume that 1 ∈ D. Then for τ = (2, 1, 7), 7 ∈ τ (D), so by Lemma 3.4d and the definition of h, we have −1

−1

−1

o o τ nτ h(CD ) = f ((CD ) ) = f (Cτo(D) )nτ = (Hτ (D) )nτ = HD .

 We now verify case-by-case the existence of a suitable function g. Theorem 3.14. Theorem 3.1 is valid in case (3D1), i.e., when V and K are q (q)). of type Lq , K has level q = 2a > 2, and m ≥ 5 (so that K/Z(K) ∼ = Lm−1 o Proof. Using Lemma 3.9, we shall verify that the map h : ∪m i=1 Ci → H can o be used to define a homomorphism g : C → H. By definition p divides q − q , so V = V1 ⊥ · · · ⊥ Vm with m ≥ 5, and dim(Vi ) = 1 for each i = 1, . . . , m. The Curtis-Tits (q = 1) and Phan (q = −1) presentations of C o ∼ = SLq (V ) in Theorems [III17 , 1.4 and 1.14] are in terms of generating subgroups Li ∼ = SL2 (q), o ∼ 1 ≤ i ≤ m − 1, of C o , which can be taken as Li = C{i,i+1} = SL(Vi ⊥ Vi+1 ), and relations asserting that [Li , Lj ] = 1 if |i − j| > 1, while Li,i+1 := Li , Li+1 ∼ = SL3 (q), with Li , Li+1 a standard pair in Li,i+1 . As m ≥ 5, for each 1 ≤ i < j < m, Li , Lj ≤ Cko for some k. Hence each of these relations holds not only within C o , but even within Cko for some k. As h|Cko is a nontrivial homomorphism for all k, this implies that there is a homomorphism g : C o → H such that g|Li = h|Li for all i = 1, . . . , m − 1, and ker g ≤ Z(C o ). o Since Cm  = Li |1 ≤ i < m − 1 and g and h are both homomorphisms when o o o o In particular, g(Cm Similarly, restricted to Cm  , g|Cm = h|Cm = f .  ) = K. as C1o = L2 , . . . , Lm−1 , we conclude that g(C1o ) = h(C1o ) = H1 . Let n ∈ N n n o with λ(n) = (1, m). Then H1 = Hm As C o = C1o , Cm  = K .  , im(g) = n N a a  let a ∈ N . Then K = Hm = Hi for some i. As K, K  ≤ K . Finally, Hi = H{1,i} , H{i,m} [III17 , 9.1], Hi ≤ im(g) for all i. Hence, K N ≤ im(g), and so equality holds, completing the proof. 

Theorem 3.15. Theorem 3.1 is valid in case (3D2), i.e., when C o = SUm (2) and m ≥ 7. Proof. Again, we have V = V1 ⊥ · · · ⊥ Vm , with dim(Vi ) = 1. As in the previous proof, we invoke the Phan presentation for SUn (2), as given in Theorem 1.14. In this case, the generating subgroups are Ui = SU (Vi ⊥ Vi+1 ), for 1 ≤ i ≤ m − 1, and Ui,i+1 = SU (Vi ⊥ Vi+1 ⊥ Vi+2 ), for i ≤ i ≤ m − 2. The relations to be verified are that [Ui , Uj ] = 1 for |i − j| > 1, as well as (a) Ui,i+1 , Ui+1,i+2 is isomorphic to a quotient of SU4 (2) for 1 ≤ i ≤ m − 2; (b) Ui and Uj,j+1 commute elementwise when i ∈ {j − 1, j, j + 1, j + 2}; and (c) Ui,i+1 and Uj,j+1 commute elementwise when i ∈ {j−2, j−1, j, j+1, j+2}. As m ≥ 7, each of these holds not only in C o but even in Cko for some k. o Again, as h|Cko is a homomorphism for all k, and as Cm  is generated by the Ui,i+1 that it contains, this implies that f extends to a homomorphism g : C o → H with we see that g agrees with h ker(g) ≤ Z(C o ). Exactly as in the previous theorem,  on Cio for all i, and we deduce that im(g) = K N , completing the proof. Theorem 3.16. Theorem 3.1 is valid in case (3D3), i.e., when q is even, V is − a linear or unitary space of dimension n ≥ 10 and C o ∼ = SLn q (q).

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Proof. Notice that if q = 2, then since p divides q − q , it must be that q = −1 so V is not a unitary space. Now we have V = V0 ⊥ V1 ⊥ · · · ⊥ Vm , with m ≥ 5, dim(V0 ) ≤ 1, and dim(Vi ) = 2 for i ≥ 1. We let V0 = Ff0 if V0 = 0 and we choose an (orthogonal) basis {ei , fi } for Vi , 1 ≤ i ≤ m. We define subgroups of C o as follows: Li = SL−q (Vi ), 1 ≤ i ≤ m (centralizing all other Vj ’s) and Li = SL−q (Ffi ⊥ Fei+1 ), 0 ≤ i ≤ m − 1 (centralizing all other ej ’s and fj ’s). The Curtis-Tits (q = −1) and Phan (q = 1) presentations of C o in Theorems 1.4 and 1.14 are in terms of these generating subgroups, and specify the relations that [Li , Lj ] = [Li , Lj ] = 1 for all i = j, [Li , Lj ] = 1 for j ∈ {i − 1, i},  − and Li−1 , Li ∼ = SL3 q (q) ∼ = Li , Li with Li−1 , Li (resp., Li , Li ) a standard pair in the group they generate. Now each Li and each Li lies in CIo for some I such that |I| ≤ 2. As m ≥ 5, each of the relations is satisfied within Cko for some k, o 1 ≤ k ≤ m. Again, as h|Cko is a homomorphism for all k, and Cm  is generated  by the Li and Li which it contains, f extends to a homomorphism g : C o → H o o o o with ker(g) ≤ Z(C o ). Hence, g(Cm  ) = K, and [K, g(Cm )] = g([Cm , Cm ]) = 1, completing the proof.  Theorem 3.17. Theorem 3.1 is valid in case (3D4), i.e., when q is even and Co ∼ = Sp(V ), where V is an Fq -vector space with dim(V ) ≥ 10. Moreover, in this case, there is an extension g of f such that for all i = 1, . . . , m, g|Cio = h|Cio , where h is as in Definition 3.8 and Lemma 3.9. Theorem 3.1 is also valid when q is even, C o ∼ = Ω+ (V ), V is an Fq -vector space with dim(V ) ≥ 10, and p is an odd prime divisor of q − 1. (This includes case (3D6).) In all these cases, m ≥ 5. Proof. As in the proof of Theorem 3.16, we can find Curtis-Tits generating subgroups Li ≤ C o with Li ∼ = SL2 (q) and Li ≤ CIoi for some Ii such that |Ii | = 2. o ∼ More precisely, in the symplectic case we choose L1 = C{1} = SL2 (q), a long o root subgroup, and then short root subgroups Li ≤ C{i−1,i} , 2 ≤ i ≤ m. In the o orthogonal case, we choose L1 and L2 as the (solvable) components of Ω(C{1,2} ), o and then further root SL2 (q)-subgroups Li ≤ C{i−1,i} , 3 ≤ i ≤ m. Then as usual as m ≥ 5, each presenting relation, as it involves only a pair of Li ’s, holds in Cko for some k, 1 ≤ k ≤ m. Thus, we conclude that, as h|Cko is a homomorphism for all k, there is a homomorphism g : C o → H such that g|Li = h|Li for all i, and ker(g) ≤ Z(C o ) = 1. o Moreover, L1 , L2 , . . . , Lm−1 = Cm  in both the symplectic and orthogonal o o o )] = cases, and we see that g(Cm ) = h(Cm ) = K and g extends f . Thus [K, g(Cm o o g([Cm , Cm ]) = 1. It remains to show that (3K)

g|Cio = h|Cio

for all i = 1, . . . , m in the symplectic case. We  have shown this to be the case for o o for any i = 1, . . . , m − 2, it suffices i = m, and since Cio = C{i,m−1} , C  {i,m} to show (3K) for i = m − 1. Now Lm−1 , Lm ∼ = SL3 (q) and there exists a ∈ Lm interchanging Vm−1 and Vm . Then L∗ := Lam−1 is supported on Vm−2 ⊥ Vm and (3L)

o a o = C{m−1} L1 , . . . , Lm−2 , L∗ = L1 , . . . , Lm−2 , Lm−1 a = (Cm ) .

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13. RECOGNITION THEORY

o Then g and h agree on Lm−1 , Lm ≤ C1o , hence on L∗ , and finally on C{m−1}  , in view of (3L). The proof is complete. 

Theorem 3.18. Theorem 3.1 is valid in case (3D5), i.e., when C o ∼ = Ω± (V ), with V a vector space of dimension n ≥ 12 over Fq , q even. (Thus, either V0 = 0 and m ≥ 5, or V0 = 0 and m ≥ 6.) Proof. First, suppose that p is a prime (odd) divisor of q − 1. By Theorem 3.17, we may assume that C o ∼ = Ω− (V ). Then V = V0 ⊕ V1 ⊕ · · · ⊕ Vm , with V0 an anisotropic plane and Vi a hyperbolic plane for all i ≥ 1. Now a Curtis-Tits presentation for Ω− (V ) = C o is of type Bm . We may take the generating subgroups o to be L1 = C{0,1} = Ω(V0 ⊥ V1 ) ∼ = L2 (q 2 ), a short root group, together with long o root groups Li ∼ = Ω(Vi−1 ⊥ Vi ) for 2 ≤ i ≤ m. = SL2 (q) such that Li ≤ C{i−1,i} o Again each Li is contained in CIi for some Ii with |Ii | = 2, and as m ≥ 5, every presenting relation can be verified in Cko for some k, 1 ≤ k ≤ m. As before, we conclude that, as h|Cko is a homomorphism for all k, there is a homomorphism g : C o → H such that g|Li = h|Li for all i and ker(g) ≤ Z(C o ) = 1. Moreover, as o o Cm  is of type Bm−1 , we have Cm = L1 , . . . , Lm−2 , Lm−1 , whence g extends f o ) = K, completing the proof in this case. and so g(Cm  Now suppose that p is a prime divisor of q + 1. Then V = V0 ⊥ V1 ⊥ · · · ⊥ Vm , with Vi an anisotropic plane for i ≥ 1, and either V0 = {0} or V0 is a hyperbolic plane. There are four cases: (1) V = (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−1 ⊥ Vm )  = (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−1 ⊥ Vm )

(2) V = (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−2 ⊥ Vm−1 ) ⊥ Vm (3M)

  = (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−2 ⊥ Vm−1 ) ⊥ Vm

(3) V = V0 ⊥ (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−1 ⊥ Vm )  = V0 ⊥ (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−1 ⊥ Vm )

(4) V = V0 ⊥ (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−2 ⊥ Vm−1 ) ⊥ Vm   = V0 ⊥ (V1 ⊥ V2 ) ⊥ · · · ⊥ (Vm−2 ⊥ Vm−1 ) ⊥ Vm   In each case, V2i−1 ⊥ V2i = V2i−1 ⊥ V2i with V2i−1 and V2i hyperbolic planes. In particular, in cases (1) and (2), we let {e, f } be a basis for V1 with e nonsingular, and let f  ∈ V2 with Q(f  ) = Q(f ). Then we may choose V1 := e, f + f  and V2 the orthogonal complement to V1 in V1 ⊥ V2 . As a result, V1 ∩ V1 contains the nonsingular vector e in cases (1) and (2). Moreover, in cases (3) and (4), we set V0 = V0 and assume that V0 = 0. Suppose that case (1) holds. Then C o = Ω+ (V ) ∼ = Dm (q), and we choose a weak CT-system {L1 , . . . , Lm } for C o with Li ∼ = L2 (q) for all i = 1, . . . , m, L1 L2 = L1 × L2 supported on V1 ⊥ V2 , and for 3 ≤ i ≤ m, Li supported on  ⊥ Vi . Vi−1 In case (1), we have a nonsingular vector e ∈ V1 ∩V1 , by our choice of V1 . There exists B = b1 , . . . , bm ∼ = Epm ≤ C o with [V, bi ] = Vi and Cio = E(CC o (bi )) for each i = 1, . . . , m, and we set CC o (e) =: Se ∼ = Sp(e⊥ /Fe) ∼ = Sp2m−2 (q) and CCio (e) =: ⊥ Si ∼ = Sp((e ∩ Vi )/Fe), i = 2, . . . , m. Moreover, if we put Bi := bj | 1 = j = i ,

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then Bi ≤ Si and NSi (Bi )/CSi (Bi ) ∼ = W (Cn−2 ). We also have λ(N ) ≥ Ne with Ne ∼ = Σm−1 fixing V1 pointwise and permuting {V2 , . . . , Vm } naturally. For each i, 2 ≤ i ≤ m, we have a homomorphism hi = h|Cio : Cio → H. The further restriction fi of hi to Si has image Se,i ∼ = Sp((e⊥ ∩ Vi )/Fe) and the Se,i ’s are permuted naturally by T . As dim(V ) ≥ 12, we have dim(e⊥ /Fe) ≥ 10, and so by Theorem 3.17, the fi extend to a homomorphism g1 : Se → H. As e ∈ V1 , we can verify all relations involving h(L3 ), . . . , h(Lm ) in g1 (Se ). Also, all relations o Now involving h(Li ) and h(Lj ) for 1 ≤ i < j < m − 1 are verified in h(Cm  ). [H{1,2} , H{m−3,m−2,m−1,m} ] = 1 in all cases by Lemma 3.6, as m ≥ 6. Thus [h(L1 L2 ), h(Lm−1 ), h(Lm ) ] = 1, completing the verification of the Curtis-Tits relations in case (1). Hence by Theorem 1.4, there exists a homomorphism g : C o → H such that ker g ≤ Z(C o ) o o and g|Li = h|Li for all i = 1, . . . , m. We have g(C{m−1,m}  ) = h(C{m−1,m} ) = o o (bm−1 ))) = L, as required for case (1). f (C{m−1,m} ) = f (E(CCm  Suppose that case (3) holds. The argument is similar. We have C o ∼ = Ω+ 2m+2 (q) and choose a weak CT-system {L0 , L1 , . . . , Lm } of SL2 (q) subgroups of C o such that L0 L1 = L0 × L1 is supported on V0 ⊥ V1 , and for i = 2, . . . , m, Li is sup ⊥ Vi . We choose a nonsingular vector e ∈ V0 = V0 and set ported on Vi−1 ∼ Se = CC o (e) = Sp(e⊥ /Fe) ∼ = Sp((e⊥ ∩ Vi )/Fe), = Sp2m (q) and CCio (e) =: Si ∼ o i = 1, . . . , m. Our homomorphisms hi : Ci → H restrict to homomorphisms fi : Cio ∩ Se → H, 1 ≤ i ≤ m, and as in case (1), the fi ’s in turn extend to a homomorphism g1 : Se → H, by Theorem 3.17. Then the Curtis-Tits relations on o h(L0 ), . . . , h(Lm−2 ) can be verified in Cm  ; those on h(L2 ), . . . , h(Lm ) in Se ; and finally [h(L0 )h(L1 ), h(Lm−1 ), h(Lm ) ] = 1 by Lemma 3.6. Hence by Theorem 1.4, there is a homomorphism g : C o → H such that g|Li = h|Li for all i = 0, . . . , m. o Since C{m−1,m}  = L0 , . . . , Lm−2 , g, h, and f agree on this subgroup, whose image is L, as asserted in our theorem. In cases (2) and (4), C o = Ω(V ) = Ω− (V ). We choose a weak CT-system of  ) for 1 ≤ i ≤ m − 2 (and also subgroups Li ∼ = SL2 (q) of C o with Li ≤ Ω(Vi ⊥ Vi+1  i = 0 in case (4)), and with Lm−1 = Ω(Vm−1 ⊥ Vm ) ∼ = Ω− 4 (q). In case (4), choose e ∈ V0 nonsingular. Let Se = CΩ(V ) (e) ∼ = Sp(e⊥ /Fe). As in previous cases, for each i, 1 ≤ i ≤ m, we have a homomorphism hi : Cio → H, the further restriction fi = hi |Se ∩ Cio , and with the help of Theorem 3.17, the fi extend to a homomorphism g1 : Se → H. Now, all Curtis-Tits relations among the o h(Li )’s, 0 ≤ i ≤ m−2 are verified in hi (Cm  ). Also, all the Li ’s for 1 ≤ i ≤ m−1 are contained in Se , and so the relations among the h(Li )’s for i > 0 may be verified in g1 (Se ). It remains to verify that [h(L0 ), h(Lm−1 )] = 1. Now, the support of L0 lies in V0 ⊥ V1 ⊥ V2 , and the support of Lm−1 lies in Vm−2 ⊥ Vm−1 ⊥ Vm As m ≥ 5, [H{0,1,2} , H{m−2,m−1,m} ] = 1 by Lemma 3.6. As usual Theorem 1.4 provides a homomorphism g : C o → H agreeing with h on each Li , i = 0, . . . , m. o To complete the proof in case (4) we show that g|Cm  = f . Call a subgroup o J ≤ C agreeable if J is in the domain of h and h|J = g|J. We need to prove o that Cm  is agreeable. Clearly, L1 , . . . , Lm are agreeable, subgroups of agreeable subgroups are agreeable, and for any i, any family of agreeable subgroups of Cio generates an agreeable subgroup.

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o  a Now L2 , L3 , L4 lies in Cm = V3 ,  and contains an element a such that (V1 )  a  o a o o a (V2 ) = V4 , and (C{3,4} ) = C{1,2} . As L1 ≤ Cm , L1 is agreeable. Set I = o ∼ − So is La1 , L4 , L5 , . . . , Lm . Then I = C{1,2}  = Ω2m−2 (q) and I is agreeable. o o o   I0 := I ∩ Cm = C{1,2,m} . Choose b ∈ L3 ≤ Cm  interchanging V2 and V3 . Then   o I1 := I0 , I0b ≤ I0 , b is agreeable. But by [III17 , 9.1], I1 = I0 , I0b = C{1,m} .   c Choosing c ∈ L2 interchanging V1 and V2 we similarly see that I1 , I1 is agreeable o and equals Cm  , as required. ∼ D− (q). Pick a weak CT-system {L1 , . . . , Lm−1 } for Last, in case (2), C o = m o ∼ SL2 (q) supported on V  ⊥ V  for i = 1, . . . , m − 2, and Lm = C with Li = i i+1 o   ∼ C{m−1,m} = Ω− 4 (q). We have a nonsingular vector e ∈ V1 ∩ V1 by our choice of V1 , ⊥ ∼ ∼ and set Se = CC o (e) = Sp(e /Fe) = Sp2m−2 (q) as usual. As in the previous cases,

the restrictions fi of hi to CCio (e) extend by Theorem 3.17 to a homomorphism g1 : Se → H. As e ∈ V1 , we can verify all relations involving h(L2 ), . . . , h(Lm−1 ) in g1 (Se ). Also, all relations involving h(L1 ) and h(Lk ) for 2 ≤ k ≤ m − 2 are o verified in h(Cm  ). As m ≥ 6, [H{1,2} , H{m−2,m−1,m} ] = 1 by Lemma 3.6. Thus [h(L1 ), h(Lm−1 )] = 1, completing the verification of the Curtis-Tits relations for case (2). As usual, there exists a homomorphism g : C o → H such that g|Li = h|Li o for all i = 1, . . . , m − 1. It remains to show again that Cm  is agreeable. Now m ≥ 6 and m is odd in case (2). We see that L2 , L3 ∼ = SL3 (q) is agreeable and contains an involution a interchanging V2 and V4 , centralizing V3 , and such that La2 , L5 ∼ = SL3 (q), centralizing V3 ⊥ V4 . Then L0 , L1 , La2 , L5 , L6 , . . . , Lm = o o o C{3,4} is agreeable. But C{1,2}  = L3 , . . . , Lm is agreeable. As m ≥ 7, Cm =   o o o by [III17 , 9.1], so Cm C{1,2,m}  is agreeable. This completes the proof  , C{3,4,m} of the lemma in all cases.  Theorem 3.19. Theorem 3.1 is valid in case (3B), i.e., when C o ∼ = Ω− 8 (q), q ≡ −1 (mod 4), and q > 3. Proof. We have V = V0 ⊥ V1 ⊥ · · · ⊥ V7 with dim(Vi ) = 1 for all i and Vi isometric to V1 for 1 ≤ i ≤ 7, but V0 not isometric to V1 . Choose a hyperbolic plane H2 ≤ V2 ⊥ V3 ⊥ V0 , and write U := V1 ⊥ V2 ⊥ V3 ⊥ V0 = W1 ⊥ H2 . o = Ω(U ) ∼ As V0 is not isometric to V1 , U has Witt index 1. We set L1 = C{0,1,2,3} = − Ω4 (q). Next, choose a hyperbolic plane H3 ≤ V4 ⊥ V5 ⊥ V6 , and write

U ⊥ = V 4 ⊥ V 5 ⊥ V 6 ⊥ V 7 = H3 ⊥ H4 , where H4 is a hyperbolic plane. Given the decomposition V = W 1 ⊥ H2 ⊥ H3 ⊥ H4 , ∼ SL2 (q) with L2 ≤ H2 ⊥ H3 and L3 ≤ H3 ⊥ H4 , so we may choose L2 ∼ = L3 = ∼ − that {L1 , L2 , L3 } is a weak CT-system for C o ∼ = Ω− 8 (q) with L1 , L2 = Ω6 (q), L2 , L3 ∼ = SL3 (q), and [L1 , L3 ] = 1. Taking D = {0, 1, 2, 3} and F = {4, 5, 6, 7}, we conclude from Lemmas 3.6 o ), h(CFo )] = 1, whence [h(L1 ), h(L3 )] = 1. As L1 , L2 ≤ C7o and 3.13 that [h(CD o and h|C7 = f is a homomorphism, it follows that {h(L1 ), h(L2 )} is a standard o o CT-pair in h(L1 ), h(L2 ) ∼ = A− 3 (q). Likewise, as L2 , L3 ≤ C1 and h|C1 is a

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3. THE WONG-FINKELSTEIN-SOLOMON METHOD

27

∼ A2 (q). homomorphism, {h(L2 ), h(L3 )} is a standard CT-pair in h(L2 ), h(L3 ) = Hence by Theorem 1.4, there is a homomorphism g : Spin(V ) → H extending the restriction of h to L1 , L2 ∪ L2 , L3 and with ker(g) ≤ Z(C o ). In particular, g extends f | L1 , L2 , so im(g) contains f (L1 , L2 ) = f (Ω(H4⊥ )) ∼ = Ω− 6 (q) with o  f (L1 , L2 ) ≤ f (C7 ) = K, completing the proof. Theorems 3.14, 3.15, 3.16, 3.17, 3.18, and 3.19 complete the proof of Theorem 3.1. Now we turn to the proof of Theorem 3.2. This requires us to provide proofs for the analogues of Lemmas 3.6 and 3.9, as well as Theorems 3.16 and 3.17 (the last only in the symplectic case). Assume then that the hypotheses of Theorem 3.2 hold. In particular, m = 4, and V is a linear or unitary space of dimension 8 or 9, or a symplectic space of dimension 8. In particular we do not have to deal with the troublesome orthogonal groups. The needed analogue of Lemma 3.6 asserts that if D, F ∈ P and D ∩ F = ∅, then [HD , HF ] = 1. The proof of Lemma 3.6, as it stands, covers all cases except |D| = |F | = 2. In that case σ(D) = {1, 2} and σ(F ) = {3, 4} for some σ ∈ T . By assumption [H{1,2} , H{3,4} ] = 1, so Lemma 3.4d implies that [HD , HF ] = 1, as required. In the proof of Lemma 3.9, the difficulty occurs in the case i = m = j, and we must prove that (3N)

f (x(i,m) )n(i,m) = f (x(j,m) )n(j,m)

o for all x ∈ C{i,j}  . Since λ is an isomorphism, we may assume that the nσ , σ ∈ Σ4 , are chosen so that nσ nτ = nστ for all σ, τ ∈ Σ4 . Write {1, 2, 3, 4} = {i, j, k, m}. o o , [x, C{i,j} ] = 1 and so Then for x ∈ C{k,m}

(3O)

f (x(i,m) )n(i,m) = f (x(i,j)(i,m) )n(i,j) n(j,m) n(i,j)

n  = f (x(i,j)(i,m)(i,j) )n(j,m) n(i,j) = f (x(j,m) )n(j,m) (i,j)

Now, f (x(j,m) )n(j,m) ∈ H{k,m} ≤ CH (H{i,j} ), by Lemma 3.6. Thus the last conjugation in (3O) has no effect, and (3N) is proved. To prove the analogue of Theorem 3.16 for dimensions n = 8 and 9, note again that Curtis-Tits and Phan generating subgroups Li , Li may be specified as in that proof, and each lies in CIo for some I with |I| = 2. Take any pair of these generating subgroups, lying, say, in CIo1 and CIo2 . If |I1 ∪ I2 − {0}| < 4 = m, then the presenting relations for this pair can be verified in CIo1 ∪I2 and hence in Cko for some k = 1, . . . , m. If I1 ∪ I2 ⊇ {1, 2, 3, 4}, on the other hand, then I1 ∩ I2 = ∅. As [HI1 , HI2 ] = 1 by the analogue of Lemma 3.6, the presenting relations in this case follow as well. Thus f extends to g : C o → H, as required in this case. The analogue of Theorem 3.17 for the symplectic case is proved similarly. This completes the proof of Theorem 3.2.

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10.1090/surv/040.8/03

CHAPTER 14

Theorem C∗7 : Stage 4b+. A Large Lie-Type Subgroup G0 for p = 2 1. Introduction This chapter and the next are analogous to the Chapter 10 of [GLS7], but here we aim for a target subgroup G0 ≤ G with G0 ∈ Chev. In the first-generation proof of the Classification, the identification of such a subgroup in the generic case was principally the work of Aschbacher in the case p = 2 [A9], and Gilman and Griess [GiGr1] in the case p > 2. We recall the basic notation and setup from [III10 ] based on the conclusions of Theorem C∗7 : Stages 3a and 3b (see [III12 , Theorem 1.2]). Thus we assume that (1A)

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

p ∈ γ(G); (x, K) ∈ J∗p (G), K ∈ Chev(r), and r = p; If p is odd, then mp (K) ≥ 3 and K ∈ Chev(2); and p splits K.

Given any such pair (x, K), by Theorem C∗7 : Stage 3b, after replacing it if necessary by a suitable pair (x , K  ) with K  /Z(K  ) ∼ = K/Z(K), there exists an acceptable subterminal (x, K)-pair. Thus we use the following notation:

(1B)

(1) (y, L) is an acceptable subterminal (x, K)-pair; (2) D = x, y ; (3) For each u ∈ D# , Lu is the subnormal closure of L in CG (u) (in particular, Lx = K); (4) De = {u ∈ E1 (D) | Lu > L}; (5) N = N(x, K,y, L) = {Lu | u ∈ D# }; (6) G0 = N = Lu | u ∈ D# ; (7) C = C(x, K); and (8) T ∈ Sylp (CG (x)), Q = C ∩ T ∈ Sylp (C), R = K ∩ T ∈ Sylp (K), z ∈ Ω1 (Z(R))# .

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

30

We have the following consequences of [III12 , Theorem 1.2]: (1) mp (C) = mp (Q) = 1; (2) For every u ∈ D − x , the pumpup Lu of L in CG (u) is either (1C) trivial or vertical, and Lu is quasisimple; and (3) For some u ∈ D − x , the pumpup Lu of L in CG (u) is vertical, i.e., u ∈ De . In addition we have the following property of the levels q(Lu ) of the pumpups Lu , u ∈ D − x : N is a level neighborhood, that is, for every u ∈ D − x , Lu ∈ Chev with q(Lu ) = q(K) = q(L), unless possibly p = 2 and one of the following holds: (1D) (a) K ∼ = G2 (q), q > 3, or 3D4 (q); and q is odd; (b) K ∼ = P Sp4 (q) or a nonuniversal version of L± 4 (q), for some odd q > 3. n Note that the case p = 2, K ∼ = 2 G2 (3 2 ), n odd, n > 1 has been ruled out by Theorem C∗7 : Stage 4a. In Section 5 of this chapter we shall clean up (1D) by proving (1E) For every u ∈ D − x , q(Lu ) = q(K) = q(L).

With this in hand, and assuming the basic setup (1A), (1B), (1C), (1D) throughout, we prove the following theorem in the remainder of this chapter and the next chapter. Theorem 1.1 (Theorem C∗7 : Stage 4b+). Under the assumptions of Theorem suppose that there are p, x, K, y, L satisfying (1C) with K ∈ Chev. Then, if necessary after replacing (p, x, K, y, L) by another quintuple satisfying (1C), and setting D = x, y , the following conditions hold: C∗7 ,

(a) (b) (c) (d)

G0 ∈ Chev; G0 /Z(G0 ) ∈ K(7)+ ; ΓD,1 (G) ≤ NG (G0 ); and CG (G0 ) is a p -group.

Moreover, the possible groups G0 /Z(G0 ), along with information about the corresponding neighborhoods N(x, K, y, L), are listed in Table 14.1 for p = 2 and in [III15 , Table 15.3] for p > 2. The following convenient notation when p = 2 is used in Table 14.1 and throughout this chapter. For any odd prime power q, q = ±1 is defined by q ≡ q

(mod 4).

We also use superscripts a (adjoint), hs (half-spin), and u (universal) to specify versions of groups of Lie type. Thus, our task in this chapter and the next is to show that for some p ∈ γ(G), some (x, K) ∈ J∗p (G) and some acceptable subterminal (x, K)-pair (y, L), if we put G0 = N(x, K, y, L) and D = x, y , then G0 ∈ Chev, G0 is ΓD,1 (G)-invariant, and if p = 2, one of the cases of Table 14.1 holds. [The condition that G0 /Z(G0 ) ∈ K(7)+ will be obvious, case by case.]

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1. INTRODUCTION

31

Table 14.1. The groups G0 = G0 /Z(G0 ) and neighborhoods N(x, K, y, L), p = 2 G0

K

L

{Ly , Lxy }

A3 (q)

A2 (q)

A1 (q)

{L, A2 (q)}

A (q),  ≥ 4

A−1 (q) A−2 (q)

A (q),  ≥ 5,  odd A−2 (q) A−3 (q) 

{L, A−1 (q)} {A−2 (q)} {L, B2 (q)}

B3 (q), q > 3

D3q (q)

B (q),  ≥ 4

Dq (q) B−1 (q)

{L, D± (q)}

C (q),  ≥ 3

C−1 (q)u C−2 (q)u

{L, C−1 (q)u }

D4 (q), q > 3

A1 (q)





D3q (q)

A1 (q)

−1

{L, D3± (q)}

D (q),  ≥ 5

q D−1 (q) B−2 (q)

± {D−1 (q)}

F4 (q)

B4 (q)u D4 (q)u

{B4 (q)u }

E6 (q)

D5 (q)u D4 (q)u

{D5 (q)}

E7 (q)

A7q (q)

E8 (q)







A5q (q) {E6q (q), D6 (q)hs }

D8 (q)hs D6 (q)u {D8 (q)hs , E7 (q)u }

Most of the time we work with an arbitrary p ∈ γ(G), and an arbitrary (x, K) ∈ J∗p (G) with an acceptable subterminal (x, K)-pair (y, L). However, occasionally we make more careful choices. switching from (x, K) to a more specifically defined pair in J∗p (G). Unless otherwise noted, however, (x, K) and (y, L) are arbitrary in what follows. Of course in this chapter, p = 2. We shall frequently use the maximality contained in the statement (x, K) ∈ J∗p (G). The following consequences of that maximality hold whether p = 2 or p > 2, and are its common applications. For compactness from now on, if X = Lp (X) is a p-component, we extend the definitions of the functions dp and F by setting (1F)

dp (X) = dp (X/Op (X)) and F(X) = F(X/Op (X))

whenever the right side of either equation is defined. Lemma 1.2. Assume (1A). Let (v, J) ∈ ILop (G). Suppose that K ∈ Chev(r) for some r = p and J/Op (J) ∈ Chev(r) ∩ Gp with dp (K) ≥ dp (J). Then the following conditions hold:

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

32

(a) F(J) ≤ F(K); (b) If F(J) = F(K), then mp (C(v, J)) = 1 and (v, J) ∈ J∗p (G); (c) If mp (C(v, J)) > 1, then there exists (w, I) ∈ ILop (G) such that I/Op (I) ∈ Chev(r) ∩ Gp , dp (I) = dp (K), I/Op (I) is a proper pumpup of some covering group of J/Op (J), and F(J) < F(I) ≤ F(K). Proof. We have J/Op (J) ∈ Gp and dp (J) ≤ dp (K). But dp (K) = dp (G) ≤ dp (J) since (x, K) ∈ Jp (G), so dp (K) = dp (J). Let (w, I) be any long pumpup  2Ar for any r = 9, 10, of (v, J) (in G). Note that if p = 2, then I/O2 (I) ∼ = or 11, by [III10 , Cor. 3.6]. Hence by [III11 , 1.2], I/Op (I) ∈ Gp with dp (I) ≤ dp (J) = dp (K) = dp (G). Therefore dp (I) = dp (G). Moreover, I/Op (I) ∈ Chev(r) by [III11 , 1.4, 1.5]. Let (w∗ , I ∗ ) be a p-terminal long pumpup of (w, I); such a pumpup exists by [III7 , Lemma 5.2]. As with (w, I) we have I ∗ /Op (I ∗ ) ∈ Gp ∩ Chev(r) and dp (I ∗ ) = dp (G), and so (w∗ , I ∗ ) ∈ Jp (G). Thus mp (C(w∗ , I ∗ )) = 1 by [III12 , Theorem 1.2]. Since (x, K) ∈ J∗p (G), it follows by [III7 , Def. 3.2] that F(I ∗ ) ≤ F(K). In view of [III11 , 12.4], we therefore have (1G)

F(J) ≤ F(I) ≤ F(I ∗ ) ≤ F(K).

This proves (a). Next assume that F(J) = F(I ∗ ). Then since J/Op (J) 2. Lemma 1.3. CG (G0 ) is a p -group. Proof. We have CG (G0 ) ≤ CG (K), with mp (C(x, K)) = 1 by (1C3), and x normalizes CG (G0 ). Thus, if p divides |CG (G0 )|, then x ∈ CG (G0 ). Hence [x, Lu ] = 1 for all u ∈ D# , contradicting the fact that N(x, K, y, L) is a nontrivial neighborhood (1C3).  Now we turn to the analysis of specific configurations in order to prove Theorem C∗7 : Stage 4b+. In this chapter, we complete the proof assuming that p = 2. That is, we identify G0 and show that ΓD,1 (G) ≤ NG (G0 ). In the next chapter, we assume that p > 2, identify G0 ∈ Chev(2), and prove that ΓD,1 (G) ≤ NG (G0 ).

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2. A 2-LOCAL CHARACTERIZATION OF L± 4 (q), q ODD

33

2. A 2-Local Characterization of L± 4 (q), q Odd As a preliminary to Sections 3 and 5, we give a characterization of L4 (q) in this section, for  = ±1 and q odd. We also consider a technical configuration that will arise in Section 5 as well as in the case K ∼ = P Sp4 (q). The setup for this characterization of L4 (q) already contains a great deal of 2-local information, some of which is redundant. It is as follows:

(2A)

(1) v, w is a four-subgroup of G; # (2) For each u ∈ v, w , Iu is a component of CG (u); (3) v ∈ Iv ∼ = Ivw /Z(Ivw ) ∼ = = SL2 (q), q > 3, q odd, while Iw /Z(Iw ) ∼ L3 (q),  = ±1; (4) Iv ≤ Iw ∩ Ivw , but v ∈ wG ∪ (vw)G ; (5) v, w ≤ P ∈ Syl2 (CG (w)) ∩ Syl2 (CG (vw)), v ∈ Z(P ), Qw = CP (Iw ) and Qvw = CP (Ivw ), and Rw = P ∩ Iw and Rvw = P ∩ Ivw ; and (6) Qw ∼ = Qwv is cyclic or quaternion of exponent dividing q − .

Proposition 2.1. Assume the conditions (2A). Define G1 = Iw , Ivw . Then Iv ≤ G 1 ∼ = L4 (q) and Γv,w,1 (G) ≤ NG (G1 ). To prove the proposition, we first show: Lemma 2.2. Choose v1 ∈ I2 (Rwv ) − {v} with v1 a square in Rwv if possible. Set E := w, v, v1 and V = E ∩ Rw . Then v1 ∈ V ∼ = E22 and wV = wNG (E) . Moreover, V ≤ E(CG (e)) for all e ∈ E − V . Proof. Since v ∈ Z(P ) ∩ Iv ≤ Z(P ) ∩ Iw ∩ Iwv by (2A3, 4, 5), v ∈ Z(Rw ) ∩ Z(Rwv ). In particular v, v1 is a four-group. By [III11 , 6.4f], there exists g ∈ NIwv (E) of order 3 with v g = vv1 , so that v1g = v. Then as g ∈ CG (wv), (2B)

wg = (wvv)g = wvv g = wvvv1 = wv1 .

If v1 ∈ Iw , then likewise there exists h ∈ NIw (E) with v1h = v, whence wgh = (wv1 )h = wv1h = wv. As NIw (E) is transitive on wV −{w} and on V −{1}, it follows from this and the fact (2A4) that v ∈ wG that wV = wNG (E) . Then V  NG (E) and as V ≤ E(CG (w)), it follows that V ≤ E(CG (e)) for all e ∈ E − V , proving the lemma. Thus it suffices to prove that v1 ∈ Iw . Suppose for a contradiction that v1 ∈ Iw . If v1 ∈ wIw , then wg = wv1 ∈ Iw , whence v ∈ wG , contrary to (2A4). Hence as w = Ω1 (Qw ), v1 ∈ Iw CG (Iw ). Thus v1 induces a non-inner automorphism on Iw . Suppose that q ≡  (mod 4), and write q −  = 2m > 2. Then Rwv has a subgroup of index 2 isomorphic to Z2m × Z2m by [III11 , 6.4a], so by the choice

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

34

of v1 , v1 ∈ m−1 (Rwv ). However by [III11 , 6.4a], we see in CG (w) that v1 ∈ m−1 (P ) ≤ Qw Rw , a contradiction. Hence q ≡  (mod 4), so by [IA , 2.5.12] all non-inner involutory automorphisms of Iw are graph automorphisms. In particular, CIw (v1 ) ∼ = P GL2 (q) (see [IA , 4.5.1]). Set w1 = wg , and J = E(CG (w1 , v)) = E(CG (w, v1 ))g ∼ = L2 (q). = E(CG (w, v1 )) ∼ Thus J ≤ E(CL2 (CG (v)) (w1 )) by L2 -balance and by (2B), w1 = wv1 ∈ CG (w, v ). Thus, w1 = wv1 normalizes Iv = E(CG (w, v )), and so J ≤ CL2 (CG (v)) (Iv ), i.e., JIv = J × Iv . Also w centralizes w1 , v . So w normalizes J × Iv , and w centralizes an involution of J. As w = Ω1 (Qw ) we have w, v = Ω1 (CP (w, v, Iv )) by [III11 , 6.4e]. Therefore either w or wv is in J. But then w centralizes a foursubgroup of J, a final contradiction showing that v1 ∈ Iw , and completing the proof of the lemma.  Now we can prove Lemma 2.3. G1 ∼ = L4 (q), NG (w, v ) ≤ NG (G1 ), and |CG (G1 )| is odd. Proof. We let E and V be as in Lemma 2.2, and count |S|, where S is defined by S = {(w1 , e, J1 ) | w1 ∈ wG ∩ E, e ∈ V # , e ∈ J1  E(CG (w1 , e )), J1 ∼ = SL2 (q)}. By Lemma 2.2 and (2A4), wG ∩ E = wV = wNG (E) and V ≤ E(CG (w1 )) ∼ = Iw for all w1 ∈ wG ∩ E. Thus |wG ∩E| = 4 and, for each w1 ∈ wG ∩E and each e ∈ V # , J1 = E(CG (w1 , e )) is uniquely determined and satisfies e ∈ J1 . Hence |S| = 12. Moreover, for any (w1 , e, J1 ) ∈ S, CE (J1 ) = w1 , e , so there is one other element (w1 e, e, J1 ) ∈ S with the same component J1 . Hence there are six distinct components occurring as third members in these triples. Since there are three conjugate choices for e, it follows that each e ∈ V # occurs as the center of exactly two such components. Let J0 := Iv and J2 be the two components occurring in this manner in E(CG (v)). Thus v ∈ J2 . Clearly E normalizes both J0 and J2 . Moreover, w and wv centralize J0 , while wv1 and wvv1 centralize J2 . Let I1 = E(CG (wv1 )). By Lemma 2.2, V ≤ I1 . Let J1 := E(CI1 (v1 )). Then [J1 , w] = 1 so J1 = E(CIw (v1 )). Hence (J0 , J1 ) is a canonically embedded pair of SL2 (q) subgroups of Iw ∼ = L3 (q), and (J1 , J2 ) is a canonically embedded pair of SL2 (q) subgroups of I1 , which is conjugate to Iw . Moreover, [J0 , J2 ] = 1. Thus J0 , J1 and J2 provide a weak CTP-system for L4 (q), so by [III13 , 1.4, 1.14] (note that q > 3 by (2A3)), J0 , J1 , J2 is a central quotient of SL4 (q). There is g ∈ Iw such that v1g = v. Now J0 , J1 = Iw and J1 , J2 = I1 = −1 −1 E(CG (wv1 )) = E(CG ((wv)g )) = (Iwv )g . Therefore G1 = Iw , Iwv = Iw , I1g = Iw , I1 g = J0 , J1 , J2 g . Clearly w, v normalizes both Iw and Iwv , so w, v normalizes G1 . Let U ∈ Syl2 (CG (G1 )). We may choose U to be w, v -invariant. Then U is a w, v -invariant

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2. A 2-LOCAL CHARACTERIZATION OF L± 4 (q), q ODD

35

subgroup of CG (Iw ). But since w = Ω1 (Qw ) and Qw ∈ Syl2 (C(w, Iw )), Qw ∈ Syl2 (CG (Iw )). As U is w, v -invariant, Ω1 (U ) ≤ Ω1 (Qw ) = w . Similarly Ω1 (U ) ≤ wv , so U = 1. Thus CG (G1 ) has odd order, as asserted. This also gives ∼ L (q), G1 = 4

as Z(SL4 (q)) is a 2-group. By (2A4), NG (w, v ) permutes {w, wv}. But G1 = Iw , Iwv = E(CG (w)), E(CG (wv)) , so NG (w, v ) normalizes G1 . The proof is complete.



Lemma 2.4. L2 (CG (v)) = E(CG1 (v)) ∼ = SL2 (q) ∗ SL2 (q). Proof. We have E(CG1 (v)) = Iv Ivt for some t such that t2 = 1 and Ivt = Iv . As Iv is a component of CG (v), Ivt is as well. Set M = L2 (CL2 (CG (v)) (Iv Ivt )); we prove that M = 1. Suppose false, and let M0 be a 2-component of M . By [III11 , 6.4g], CAut(Iw ) (Iv ) and CAut(Iwv ) (Iv ) are 2-nilpotent. Thus CCG (w,v) (Iv )/CG (w Iw ) and CCG (w,v) (Iv )/CG (wv Iwv ) are 2-nilpotent. We showed above that CG (Iw ) ∩ CG (Iwv ) has odd order, so it follows that (2C)

CCG (Iv ) (w) = CCG (w,v) (Iv ) is 2-nilpotent.

In particular w normalizes every 2-component of Ivt M . Suppose that w centralizes an involution v  ∈ M ∩ P − O2 2 (M ). Then as Ω1 (CP (Iv )) = w, v

(2D) 

by [III11 , 6.4e], we have v ∈ w, v . It follows that (2E)

w, v = v  , v ≤ CG (Iv Ivt ),

which is absurd as [w, Ivt ] = 1. Thus every involution of CM (w) lies in O2 2 (M ). This enables us to apply [IIK , 6.3a]. We conclude that M0 /O2 (M0 ) ∼ = SL2 (q0 ), q0 odd, or 2A7 , with w inducing an inner automorphism on M0 in the latter case; the only alternative in [IIK , 6.3a] would be for CM0 (w) to have a 2-component or solvable 2-component, which would violate (2C). If v ∈ M0 , then writing v  = P ∩ O2 2 (M0 ) we reach the contradiction (2E) as in the third paragraph of the proof. Thus v ∈ Syl2 (O2 2 (M0 )). Since [w, Ivt ] = 1 = [w, M0 ], w inverts commuting elements v1 , v2 of order 4 in Ivt and M0 , respectively, so w centralizes v  := v1 v2 . With (2D), we again get v  , v = w, v , so w ∈ Ivt M0 . But now there exist commuting elements w1 ∈ Ivt and w2 ∈ M0 of order 4 such that [wi , vi ] = v, i = 1, 2. Therefore w := w1 w2 is an involution centralizing w, v and Iv , so w ∈ w, v by (2D). Since w , v = v  , v , we must have w1 ∈ v1 v . But this is absurd as [w1 , v1 ] = 1. This final contradiction completes the proof of the lemma.  Lemma 2.5. Suppose that q ≡ − (mod 4). Then w induces a noninner automorphism on Ivt .

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

36

Proof. We have w, v acting on G1 ∼ = L4 (q) with Iw = E(CG1 (w)) and = E(CG1 (v)); moreover, Iv ≤ Iw by (2A4). Therefore Ivt w / v ∼ = P GL2 (q) by [III17 , 3.38], which proves the lemma.  Iv Ivt

Lemma 2.6. We have q ≡  (mod 4). Proof. Since Out(SL2 (q)) is abelian [III17 , 7.2a], we apply [III8 , 6.10] with Ivt , Iv and Inn(Ivt ) in the roles of X1 , X2 , and N1 there. Now w centralizes X2 = Iv , but if q ≡ − (mod 4), then by the preceding lemma, w does not map into N1 . Then [III8 , 6.10] yields w ∈ [CG (v), CG (v)]. On the other hand, by (2A4), v is weakly closed in Ω1 (Z(P )) = w, v with respect to G. Therefore by [III8 , 6.3], w ∈ [G, G], contradicting the simplicity of G. The lemma follows.  Now we can complete the proof of Proposition 2.1. First we use Lemma 2.6 to show that CG (v) ≤ NG (G1 ). Indeed by Lemma 2.4, E(CG (v)) = E(CG1 (v)); the same then holds for all G1 -conjugates of v, which, as q ≡  (mod 4), include all involutions of E(CG (v)), by [III17 , 11.21]. Therefore CG (v) normalizes Go1 := E(CG1 (v0 )) | v0 ∈ I2 (E(CG (v))) , and we argue that Go1 = G1 . Obviously Go1 ≤ G1 . For the reverse inclusion, choose any four-group  E ≤ E(CG (v)) and apply [IA , 7.3.3] to the action of E on G1 . It yields G1 = E(CG1 (v0 )) | v0 ∈ E # ≤ Go1 , proving our claim. More easily, as I2 (Iw ) = v Iw and Iw ≤ G1 by construction, with Iw  CG (w), we have CG (w) ≤ Iw CG (w, v ) ≤ NG (G1 ). A similar argument shows that CG (wv) ≤ NG (G1 ). Since NG (w, v ) ≤ NG (G1 ) by Lemma 2.3, we have proved that Γw,v,1 (G) ≤ NG (G1 ). Together with Lemma 2.3 this yields all the assertions of Proposition 2.1. To finish this section we consider the following configuration of involution centralizers relating to P Sp4 (q), q > 3, q odd:

(2F)

(1) (w, I) ∈ IL2 (G), I ∼ = P Sp4 (q), q odd, q > 3, and m2 (C(w, I)) = 1; (2) y0 ∈ I2 (I), y0 ∈ E(CI (y0 )) = L1 Lt1 ∼ = SL2 (q) ∗ SL2 (q), t ∈ I2 (CI (y0 )) ∩ y0I , Lt1 = L1 ; (3) If M1 is the subnormal closure of L1 in CG (y0 ), then either M1 = L1 or M1 ∼ = SL2 (q 2 ), with w inducing a field automorphism on M1 in the latter case; (4) R ∈ Syl2 (CG (w)) and R ≤ S ∈ Syl2 (CG (y0 )); and (5) One of the following holds: (a) R ∈ Syl2 (G); (b) M1 ∼ = SL2 (q 2 ); or (c) The subnormal closure of L1 in CG (wy0 ) is isomorphic to I.

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2. A 2-LOCAL CHARACTERIZATION OF L± 4 (q), q ODD

37

Proposition 2.7. Assume (2F). Then the following conditions hold: (a) If M1 = L1 , then w, y0 ∈ Syl2 (CG (L1 Lt1 )) and L1 Lt1 = E(CG (y0 )) = L2 (CG (y0 )); (b) If M1 > L1 , then y0 ∈ Syl2 (CG (M1 M1t )); (c) If f ∈ CG (w) induces a field automorphism of order 2 on I, with f normalizing L1 and [f, t] = 1, then the following hold: (1) f is not an involution; and (2) There is y1 ∈ y0G ∩ CG (w) such that I y1 ∼ = Inndiag(I) ∼ = SO5 (q); and (d) w ∈ Syl2 (C(w, I)), y0 is weakly closed in w, y0 with respect to G, and either y0 or w, y0 is a Sylow 2-center in G. To prove the proposition, we set E0 = w, y0 and first prove (d): Lemma 2.8. Assume (2F). Then w ∈ Syl2 (C(w, I)). Moreover, y0 is weakly closed in E0 with respect to G, and either y0 or E0 is a Sylow 2-center in G. Finally, (2F5c) holds. ∼ P Sp4 (q), [III17 , 8.8d] implies that Z(R) = y0 × CZ(R) (I), Proof. Since I = and as m2 (C(w, I)) = 1, CZ(R) (I) is cyclic with involution w. First assume (2F5a). As R ∈ Syl2 (G), w is not weakly closed in Ω1 (Z(R)) = E0 with respect to G. Now w and y0 are not G-conjugate; indeed by (2F1, 2, 3), E(CG (y0 )) has at least two components of 2-rank 1, while E(CG (w)) has at most one. Therefore there must exist g ∈ NG (R) ∩ CG (y0 ) such that wg = wy0 . Furthermore, g normalizes E(CG (E0 )) = L1 Lt1 H1 , where H1 ≤ C(w, I); thus y0 ∈ H1 , as w is the unique involution of C(w, I). As g centralizes y0 ∈ L1 ∩ Lt1 , L1 Lt1 = (L1 Lt1 )g ≤ I g . As a result, (2F5c) holds. Likewise if (2F5b) holds, then w is M1 -conjugate to wy0 and (2F5c) holds. So it suffices to prove the lemma assuming (2F5c). Let Iwy0 be the component of CG (wy0 ) isomorphic to I and containing L1 , and let Qw ∈ Syl2 (C(w, I)). Then w is the unique involution of Qw and [w, Iwy0 ] = [y0 , Iwy0 ] = 1, so Qw embeds in CAut(Iwy0 ) (L1 Lt1 ) ∼ = Z2 , whence Qw = w , as t asserted. Consequently E0 ∈ Syl2 (CG (w L1 L1 )) = Syl2 (CG (wy0 L1 Lt1 )). Since C(wy0 , Iwy0 ) ≤ CG (wy0 L1 Lt1 ), it follows that wy0 ∈ Syl2 (C(wy0 , Iwy0 )). Now we have I = E(CG (w)) and Iwy0 = E(CG (wy0 )), whereas M1  E(CG (y0 )), so y0 is weakly closed in E0 with respect to G. Since Z(R) = E0 , this immediately implies that y0 is 2-central in G, and either y0 or E0 is a Sylow 2-center in G. The proof is complete.  We remark that as a consequence of (2F5c), our setup (2F) also holds with wy0 in place of w, for a suitable choice of t1 ∈ I2 (I1 ) in place of t. Here I1 ∼ = I is the subnormal closure of L1 in CG (wy0 ). As a result of this symmetry between w and wy0 , wy0 ∈ Syl2 (C(wy0 , I1 )).

(2G)

Next let C1 = and let Q0 be a Sylow 2-subgroup of C1 w containing E0 . Let Q1 = Q0 ∩ C1 , so that y0 ∈ Q1 , Q0 = Q1 w , and |Q0 : Q1 | = 1 or 2 according as M1 = L1 or M1 > L1 . We next prove (a) and (b) of Proposition 2.7. Note that y0 ∈ CG (M1 M1t ), so if Q0 = E0 , then Q1 = y0 or E0 according as M1 > L1 or M1 = L1 . CG (M1 M1t )

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38

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Lemma 2.9. Assume (2F). Then Q0 = E0 , and M1 M1t = E(CG (y0 )) = L2 (CG (y0 )). Proof. It suffices to prove that Q0 = E0 . For then since y0 ∈ Q1 ∈ Syl2 (C1 ) and y0 ∈ Z(C1 ), C1 w is 2-nilpotent and L2 (C1 ) = 1, implying L2 (CG (y0 )) = M1 M1t . Now y0 is 2-central in G. By [III17 , 8.8d] and the fact that Qw = w , C1 w ∩ R = Ω1 (Z(R)) = E0 . Thus CQ0 (w) = E0 , so Q0 is dihedral or semidihedral. By Lemma 2.8, y0 is weakly closed in Ω1 (Z(R)). Thus by [III8 , 6.3], (2H)

w ∈ [CG (y0 ), CG (y0 )].

As y0 ∈ Z(C1 ), one of the following holds by [III17 , 11.3]: (1) C1 = O2 (C1 )Q1 ; or (2) There exists J0  C1 with J0 /O2 (J0 ) ∼ = SL2 (r) for some odd (2I) prime power r, or 2A7 . Moreover, w induces an outer automorphism on J0 /O2 (J0 ), which is a diagonal automorphism if J0 /O2 (J0 ) ∼ = SL2 (r). Consider first the case (2I1). Then M1 M1t = E(CG (y0 ))  CG (y0 ), so C1  CG (y0 ). We have Q1 ≤ Q0 ∼ = D2n or SD2n . If Q1 ∼ = Q8 or Q1 is cyclic, then w ∈ Q1 , so w induces an outer automorphism on Q1 . But Out(Q1 ) ∼ = Σ3 or Out(Q1 ) is abelian, so it follows that w ∈ [CG (y0 ), CG (y0 )], contradicting (2H). Similarly if Q1 is a fourgroup, then Q0 = Q1 w = E0 if w ∈ Q1 , and the conclusion of the lemma holds; if w ∈ Q1 we again get w ∈ [CG (y0 ), CG (y0 )]. In the remaining subcases of (2I1), Q1 has a unique cyclic maximal subgroup Y1 , and [w, Y1 ] = 1 as Q1 has maximal class. We let C2 = O2 (C1 )Y1  CG (y0 ). Then Hy := CCG (y0 ) (C2 /O2 (C2 )) contains [CG (y0 ), CG (y0 )]. Since w ∈ Hy , again w ∈ [CG (y0 ), CG (y0 )], a contradiction. Thus (2I2) holds. If J0  CG (y0 ), then as Out(J0 /O2 (J0 )) is abelian, J0 C(y0 , J0 ) contains [CG (y0 ), CG (y0 )] and does not contain w, again a contradiction. So J0  CG (y0 ). It follows that J0 ∼ = M1 and CG (y0 ) transitively permutes the three 2-components J0 , M1 , M1t of CG (y0 ). We apply [III8 , 6.10] to obtain a contradiction, with J0 here playing the role of X1 there, and Inn(J0 )Φ playing the role of N1 , where Φ ≤ Aut(J0 ) is generated by a field automorphism of maximal order. We know that w acts trivially or as a field automorphism on M1 and M1t , but the image of w in Out(J0 ) is the nontrivial element of Outdiag(J0 ). As the image of Φ in Out(J0 ) is a direct complement to Outdiag(J0 ), [III8 , 6.10] indeed yields w ∈ [CG (y0 ), CG (y0 )]. This again contradicts (2H), and the lemma is proved.  We now bring t into play. By (2F2), t ∈ y0I . Therefore wt ∈ (wy0 )G . Let Jwt be the component of CG (wt) isomorphic to I, so that wt ∈ Syl2 (C(wt, Jwt )) by (2G). Lemma 2.10. Either y0 induces an outer diagonal automorphism of Jwt , or q ≡ ±3 (mod 8). Proof. Set M1,2 = E(CM1 M1t (t)) ∼ = L2 (q) or L2 (q 2 ). With Lemma 2.9 and L2 -balance, M1,2 = E(CG (y0 , t )) ∼ = E(CG (y0 , wt )) = E(CJwt (y0 )).

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± 3. THE CASE K ∼ = L3 (q)

39

∼ P Sp4 (q). Thus by [IA , 4.9.1, 4.5.1], y0 induces a non-2-central innerBut Jwt ∼ =I = diagonal automorphism on Jwt , which we may assume to be inner (otherwise we are done). Again by [IA , 4.5.1], M1,2 ∼ = L2 (q), so M1 = L1 , and CJwt (y0 M1,2 ) has order q − q . As wt ∈ Jwt , |CG (y0 , wt, M1,2 )| is divisible by 2(q − q ). However, computing in CG (y0 ) and using Lemma 2.9, we find that CG (y0 , wt, M1,2 ) ≤ O2 (CG (y0 )) w, y0 , t with | w, y0 , t | = 8. Therefore q ≡ q (mod 8), whence q ≡ ±3 (mod 8). The proof is complete.  Now we can complete the proof of Proposition 2.7. Suppose that f ∈ CG (w) and f induces a field automorphism of order 2 on L1 with [f, y0 ] = [f, t] = 1. Then q is a square, so by Lemma 2.10, the image of y0 in Aut(Jwt ) lies in Inndiag(Jwt ) − Inn(Jwt ). Thus Jwt y0 ∼ = SO5 (q). Conjugating wt to wy0 and using the symmetry between w and wy0 , we obtain part (c2) of the proposition. Now suppose for a contradiction that f 2 = 1. Since f acts on M1 , we cannot have M1 > L1 , for then f 2 = 1. Thus, M1 = L1 . Let L = L12 = E(CL1 Lt1 (t)). Then [L12 , w] = 1, so L12 ≤ L2 (C(y0 , wt)) by L2 -balance. Therefore L12 = E(C(y0 , wt)) = E(CJwt (y0 )), on which f induces a nontrivial field automorphism. As f centralizes w, [f, wt] = 1, and then by [IA , 4.5.1, 4.2.3], f induces a field automorphism of order 2 on Jwt ∼ = P Sp4 (q). Now [f, y0 ] = 1 as f normalizes L1 . Thus the image of y0 in Aut(Jwt ) lies in CInndiag(Jwt ) (f ), which in turn lies in Inn(Jwt ) by [III17 , 12.8]. This contradicts the previous paragraph and completes the proof of Proposition 2.7. 3. The Case K ∼ = L± 3 (q) In this section we assume the setup (1A), (1B), (1C), and (1D), with p = 2. We begin a long case analysis of the different possibilities for K by proving: Proposition 3.1. If p = 2 and K/Z(K) ∼ = L3 (q), q > 3,  = ±1, then for a suitable choice of (y, L), G0 = N(x, K, y, L) is isomorphic to L4 (q), and ΓD,1 (G) ≤ NG (G0 ). Since q is odd, it follows from [IA , 6.1.4] that K∼ = L3 (q) or SL3 (q). By the definition of acceptable subterminal pair [III12 , Def. 1.15], (3A)

L∼ = SL2 (q).

Moreover, replacing y by xy if necessary, we may assume that (3B)

y ∈ L.

Recall that D = x, y , Q ∈ Syl2 (C(x, K)), x ∈ Q, and m2 (Q) = 1; in particular K  CG (x). Also Q ≤ T ∈ Syl2 (CG (x)), with R = T ∩ K. As K has just one conjugacy class of involutions and cyclic Sylow 2-centers [III11 , 6.4cd], we may assume that y = Ω1 (Z(R)). By (1D), for all u ∈ D# , Lu ∈ Chev with q(Lu ) = q. ∼ Lη (q) for some η = ±1. Moreover, Lemma 3.2. Ly = L and Lxy /Z(Lxy ) = 3 ∗ m2 (C(xy, Lxy )) = 1 with (xy, Lxy ) ∈ J2 (G).

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

40

Proof. By assumption (x, K) ∈ J∗2 (G) ⊆ J2 (G), so d2 (G) = d2 (K) = 7 by [III7 , (1E)]. As a result, for each u ∈ D# , Lu ∼ = L± = SL2 (q) or Lu /Z(Lu ) ∼ 3 (q), with Z(Lu ) of odd order in the latter case. As y ∈ L ≤ Ly , Ly ∼ = SL2 (q) so Ly = L. Then since Lu > L for some u ∈ x, y − x , Lxy has the asserted structure. The final sentence follows by Lemma 1.2b.  By definition G0 = K, Ly , Lxy , so Lemma 3.2 yields (3C)

G0 = K, Lxy .

We next prove: Lemma 3.3. T ∈ Syl2 (CG (xy)). Proof. By [III11 , 6.4c], and the fact that m2 (Q) = 1, we conclude that x, y = Ω1 (Z(T )). Therefore T ≤ CG (xy). But y ∈ L ≤ Lxy , so as m2 (C(xy, Lxy ) = 1 and Lxy has one class of involutions [III11 , 6.4d], y is 2-central in CG (xy). As T is a  Sylow subgroup of CG (x, y ) = CG (y, xy ), the lemma follows. Now we have symmetry between (xy, Lxy ) and (x, K), and we set Qxy = CT (Lxy ) and Rxy = T ∩ Lxy . Note that Lxy  CG (xy). Lemma 3.4. Q is cyclic of order dividing (q − η)2 . Moreover, Q ≤ Lxy CG (Lxy ) and Q acts faithfully on Lxy . Proof. We have Ω1 (Q) = x and Ω1 (Qxy ) = xy . Therefore Q ∩ Qxy = 1 and so Q acts faithfully on Lxy . Since CG (xy) and CG (x) share the Sylow 2-subgroup T by Lemma 3.3, Q  T and Rxy  T . But x, y ∩ Lxy = y , so Q ∩ Rxy = 1. Therefore [Q, Rxy ] = 1. But by [III11 , 6.4c], the image of Z(Rxy ) in Aut(Lxy ) is a Sylow 2-subgroup of  CAut(Lxy ) (Rxy ), and is cyclic of order (q − η)2 . The lemma follows. The same argument applies with x and xy interchanged to prove: Lemma 3.5. Qxy is cyclic of order dividing (q−)2 . Moreover, Qxy ≤ KCG (K) and Qxy acts faithfully on K. Lemma 3.6. y ∈ xG ∪ (xy)G . Proof. Suppose that y ∈ xG . Then as y ∈ L   CG (y), x ∈ J0  E(CG (x)) for some J0 ∼ = SL2 (q). As K  E(CG (x)), it follows that [J0 , K] = 1. But Q ∈ Syl2 (CG (K)) and Q is cyclic, a contradiction. Thus y ∈ xG , and the same argument with xy, Lxy , and Qxy in place of x, K, and Q shows that y ∈ (xy)G .  Lemma 3.7. K/Z(K) ∼ = Lxy /O2 (Lxy ), η = , and Q ∼ = Qxy . Proof. First, by Lemma 3.4, Q ≤ Rxy × Qxy . As Ω1 (Q) = x , Q ∩ Rxy = Q ∩ Qxy = 1. In particular Q embeds in Qxy . By symmetry, Qxy embeds in Q, so Q∼ = Qxy . Next suppose by way of contradiction that η = . Then by the symmetry between x and xy, we may assume that  = +1. By Lemma 3.3, Rxy  T . As Rxy ∩ Q = 1, Rxy acts faithfully on K, with Rxy ∼ = Rxy Q/Q  T /Q ≤ Aut(K).  This contradicts [III17 , 11.11], however, and the lemma is proved.

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4. THE CASE K ∼ = G2 (q) OR 3D4 (q)

41

We are now in a position where the conditions (2A) are satisfied, with the roles of v, w, wv, Iv , Iw , and Iwv there played by y, x, xy, L = Ly , K, and Lxy here. By Proposition 2.1, G0 = K, L, Lxy = K, Lxy ∼ = L4 (q) and ΓD,1 (G) normalizes G0 , completing the proof of Proposition 3.1. 4. The Case K ∼ = G2 (q) or 3D4 (q) We continue to assume the setup (1A), (1B), (1C), and (1D). Proposition 4.1. Suppose that p = 2. Then K ∼  G2 (q) or 3D4 (q). = We suppose false and proceed in a sequence of lemmas. Because K ∼ = G2 (q) or D4 (q) and K ∈ G2 , we have the following:

3

K∼  G2 (3); = K has a single conjugacy class of involutions; Z(R) = z ; CK (z) = I1 I2 R with I1 ∼ = SL2 (q), I2 ∼ = SL2 (q j ), j = 1 or 3, Ii  CK (z), i = 1, 2, |CK (z) : I1 I2 | = 2, CK (z)/I2 ∼ = P GL2 (q),  CK (z)/I1 ∼ = P GL2 (q j ), and F ∗ (CAut(K) (z)) = O 2 (I1 )I2 ; (5) T /QR embeds in Out(K), which is cyclic; any involution in T − QR induces a field or graph-field automorphism on K; (6) Ω1 (Z(T )) = x, z .

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

Indeed by [IA , 2.5.12], Out(K) is cyclic and any involution of Out(K) is the image of a field or graph-field automorphism of K, so (4A5) holds. Since K ∈ G2 , (4A1) holds [I2 , 12.1]. By [IA , 4.5.1], (4A2, 4) hold. By [III11 , 6.3e], CAut(O2 (I1 )) (R ∩ I1 ) and CAut(I2 ) (R ∩ I2 ) have odd order, so Z(R) ≤ CR (F ∗ (CAut(K) (z))) ≤ Z(F ∗ (CAut(K) (z))) = z , proving (4A3). Finally for any involution v ∈ T − QR, |CR (v)| ≤ |G2 (q 1/2 )|2 , |2 G2 (q 1/2 )|2 or |3D4 (q 1/2 )|2 by (4A5), according to the isomorphism type of K, which implies that |CR (v)| < R. Then Ω1 (Z(T )) ≤ QR = Q × R and (4A6) follows as m2 (Q) = 1 (see (1C1)). We set Z = Ω1 (Z(T )) = x, z , CZ = CG (Z), Cx = CG (x), and Cz = CG (z). We first prove Lemma 4.2. x is weakly closed in neither QR nor Z. Proof. Suppose false. Since Z ≤ QR and all involutions in R are K-conjugate, x is weakly closed in both QR and Z. As Z char T it follows that T ∈ Syl2 (G). By the Z ∗ -Theorem [IG , 15.3], there is g ∈ G such that x1 := xg ∈ T − QR, whence x1 induces a field or graph-field automorphism on K by (4A5). Replacing g by a suitable element of Cx gCx we may assume first that x1 is extremal in T with −1 respect to Cx , that is, CT (x1 ) ∈ Syl2 (CCx (x1 )), and then that CT (x1 )g ≤ T . Set 1 1 1 R1 = CR (x1 ). Then as CK (x1 ) ∼ = G2 (q 2 ), 2 G2 (q 2 ) or 3D4 (q 2 ) (see [IA , 4.9.1]), −1 −1 R1 contains an E23 -subgroup A (by [IA , 5.6.3]). Then Ag ≤ T1g ≤ T . As −1 m2 (Q) = 1 and m2 (T /QR) = 1 as well, Ag ∩ R = 1. Fix a ∈ A such that −1 ag ∈ R. By [IG , 4.9.1], all involutions in the coset x1 K are K-conjugate, and so

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42

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2 −1

−1

x1 = xg is K-conjugate to x1 a = (xag )g . Therefore x is G-conjugate to xag , contrary to our condition that x is weakly closed in QR. The lemma is proved.  Lemma 4.3. The following conditions hold: (a) Q = x ; (b) T ∈ Syl2 (G); (c) Ω1 (Z(S)) = z ; (d) x ∈ [Cz , Cz ]; (e) T /R is abelian, and x ∈ [Cx , Cx ]; and (f) xG ∩ Z = {x, xz}. Proof. Suppose that z = xg for some g ∈ G. Since T ∈ Syl2 (Cx ), it would follow that T ∈ Syl2 (Cz ) and so g could be chosen in NG (T ). Note however that in T , indeed in R ∩ I1 I2 , there is a subgroup H ∼ = Q8 ∗ Q8 such that Z(H) = z . But −1 −1 g −1 g −1 then H ≤ T with Z(H ) = x , whence H g ∩ R = 1 and so H g embeds in T /R. This is absurd as m2 (H) = 3 while m2 (T /R) ≤ m2 (Q) + m2 (T /QR) ≤ 2. We have shown that z ∈ xG . Therefore by Lemma 4.2, xG ∩Z = {x, xz}, which is (f). By [III8 , 6.3], (d) holds and |NG (T ) : NG (T ) ∩ Cx | = 2, so (b) holds as well. Moreover, Z(S) ≤ CS (x) = T so Ω1 (Z(S)) ≤ CZ (NS (T )) = z , proving (c). Furthermore, if (a) holds, then Cx /O2 (Cx )K is an extension of the (nontrivial) image of x by a subgroup of Out(K). As Out(K) is cyclic, Cx /O2 (Cx )K is abelian, and (e) is a consequence. It remains to prove (a). Fix g ∈ NS (T ) − T , so that xg = xz. Suppose by way of contradiction that Q > x . Then there is V ≤ Q with |V | = 4 and V  T . As m2 (Q) = 1, V ∼ = Z4 . Set U = V g . Thus U  T , U ∼ = Z4 , and Ω1 (U ) = xz . Consequently [U, R] ≤ U ∩ R = 1.   Thus U maps into O 2 (CAut(K) (R)), which in turn acts trivially on O 2 (I1 )I2 = F ∗ (CK (z)) by [III11 , 6.3e]. But F ∗ (CK (z)) = F ∗ (CAut(K) (z)) by [III17 , 11.13], and so U C/C ≤ Z(F ∗ (CK (z)))C/C = z C/C. (Recall that C = C(x, K).) Thus U ≤ z Q, so xz = Φ(U ) ≤ Φ(z Q) = x , a contradiction. The proof is complete.  The following fusion-theoretic fact will combine with Lemma 4.3d to restrict the structure of Cz and eventually lead to a contradiction. Lemma 4.4. I2 (xI1 I2 ) ⊆ xCz . Proof. By Lemma 4.3f and (4A2), all involutions of xI1 I2 lie in xG . Let u ∈ I2 (xI1 I2 ) and choose g ∈ G such that ug = x. Since [x, I1 I2 ] = 1, we have u = xy with y 2 = 1 and y ∈ I1 I2 . We wish to alter g in Cx to obtain z g = z. If y = 1 there is nothing to prove, so assume that y = 1. Since Sylow 2-subgroups of I1 and I2 are quaternion, with [I1 , I2 ] = 1 and I1 ∩ I2 = z , CI1 I2 (y) contains a nonabelian dihedral 2-subgroup P such that z ∈ [P, P ], by [III8 , 1.6]. In particular P g ≤ CG (ug ) = Cx . But Cx /K has abelian Sylow 2-subgroups by Lemma 4.3e, so [P g , P g ] ≤ K. Thus z g ∈ K. Again as K has only one class of involutions, there is h ∈ K such that z gh = z. Then ugh = xh = x and gh ∈ Cz , and the proof is complete.  

Now let L1 and L2 be the subnormal closures in Cz of O 2 (I1 ) and I2 , respectively. By L2 -balance, L2 is an x-invariant product of one or two 2-components of Cz interchanged in the latter case by x, and in either case I2 is a component

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4. THE CASE K ∼ = G2 (q) OR 3D4 (q)

43

∼ 3D4 (3), in which case by of CL2 (x). A similar statement holds for L1 unless K = solvable L2 -balance [IG , 13.8], either L1 ≤ O2 2 (Cz ) or L1 is a single x-invariant  2-component. In any case, as F ∗ (CAut(K) (z)) = O 2 (I1 )I2 by (4A4), we have   O 2 (CCz (x O 2 (I1 )I2 )) = x, z .  that Z = x, z and that x and xz are Cz -conjugate. We set V =  CRecall Q z = xCz . Then     (4B) O 2 (I1 )I2 ≤ xO 2 (I1 )I2 ≤ V, 

by the fact that O 2 (I1 )I2 is generated by its involutions, and by Lemma 4.4. Set C z = Cz /O2 (Cz ). Now we can prove Lemma 4.5. The following conditions hold: (a) The only nontrivial normal subgroup of C z centralized by x is z ; (b) O2 (C z ) is cyclic or of symplectic type; and (c) The only normal elementary abelian 2-subgroup of C z is z .   Proof. Suppose that N  C z and [N , x] = 1. Then N centralizes xC z = V , 

whence N ≤ CC z (x O 2 (I 1 )I 2 ) by (4B). By Lemma 4.3a, C = O2 (Cx ) x , and then by (4A4) and the F ∗ -Theorem [IG , 3.6], 

CCx (O 2 (I1 )I2 ) = O2 (Cx ) x, z = O2 (Cx )Z.  Hence N ≤ Z. But xCz is nonsolvable and N  C z , so x ∈ N . As x and xz are Cz -conjugate, N ≤ z , proving (a). If (b) fails, then by Philip Hall’s theorem [IG , 10.3], O2 (C z ) has a noncyclic characteristic elementary abelian subgroup, which is then normal in C z . So to complete the proof of (b) and (c) it suffices to assume that Y is elementary abelian, Y  C z , and prove that Y = z . If x ∈ Y , then (a) yields Y = z , a contradiction as x = z, so x ∈ Y . As Y  C z , we deduce that Y ∩ Z = z . Now CY (x) ≤ O2 (CC z (x)), whereas the only involutions of O2 2 (CCz (x)) = O2 2 (CCx (z)) lie in Z by (4A4). Thus CY (x) ≤ Z, whence CY (x) = z . If Y > z , then Y is therefore a four-group and the image of x in Aut(Y ) ∼ = Σ3 is nontrivial, so x ∈ [C z , C z ]. This contradicts Lemma 4.3d and completes the proof of the lemma.  We consider for a moment the exceptional situation I1 ∼ = SL2 (3), which only occurs if K ∼ = SL2 (33 ). = 3D4 (3) and I2 ∼  Lemma 4.6. If I1 ∼ = SL2 (3), then O 2 (I 1 ) ≤ E(C z ). 

Proof. Set Y = O2 (C z ) and Q1 = O 2 (I1 ) and assume, for a contradiction, that Q1 ≤ E(C z ). Then by solvable L2 -balance [IG , 13.8], Q1 ≤ Y . We set C1 = CCz (I2 ). Then Y ≤ C 1 , and so a Sylow 2-subgroup of CC 1 (x), being isomorphic to a 2-subgroup of CCx (I2 ) containing Q1 , equals Q1 or Q1 x . As Y is of symplectic type by Lemma 4.5, [III8 , 7.9] applies with C z , x, and an element of I 1 of order 3 in the roles of X, x and y there. The alternative conclusions (a) and (b) of that lemma are ruled out by Lemmas 4.3d and 4.5a, respectively. Therefore conclusion (c) of that lemma holds, so x ∈ Y , i.e., x ∈ O2 2 (Cz ), whence xCz ≤ O2 2 (Cz ). As

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44

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

xQ1 ≤ O2 2 (Cz ) and I2 (xQ1 I2 ) ≤ xCz by Lemma 4.4, it follows that I2 ≤ O2 2 (Cz ), a contradiction completing the proof.  Now we can prove Lemma 4.7. F ∗ (C z ) = E(C z ) = L1 L2 . Proof. We remark that no claim is being made that L1 = L2 ; these components might be equal. By Lemma 4.5c, Z(E(C z )) = z . In particular E(C z ) does not contain two disjoint components with nontrivial centers. This fact and L2 -balance, together with the preceding lemma and solvable L2 -balance to deal with the case I1 ∼ = SL2 (3), imply that L1 and L2 are (possibly equal) single x-invariant components of C z . Set L = L1 L2 and C 0 = CC z (L), and D = CC 0 x (x). Then x normalizes C 0 , and D is covered by CCx (I1 I2 ). By (4A4), Z ∈ Syl2 (D). Then [III17 , 11.4] implies either that C 0 x is a 2-group of maximal class or that C 0 has a normal subgroup M0 ∼ = = SL2 (r) for some odd r with CC 0 (M 0 ) = Z(M 0 ) = z and M 0 x / z ∼ P GL2 (r). Suppose that C 0 is 2-closed. Then L = E(C z ) so C 0  C z . If O2 (C 0 ) = z , then F ∗ (C z ) = L, as desired. So assume that O2 (C 0 ) > z . Then as D = CO2 (C 0 )x (x), O2 (C 0 ) x has maximal class, but is not a four-group by Lemma 4.5. If x ∈ O2 (C 0 ), then by [III8 , 1.7], the image of x in Aut(O2 (C 0 )) lies outside the commutator subgroup, and so x ∈ [C z , C z ], against Lemma 4.3d. Thus, x ∈ O2 (C 0 ). Then either O2 (C 0 )/Φ(O2 (C 0 )) is a four-group on which x acts nontrivially, or C 0 is a cyclic group on which x acts nontrivially. In either case, we again reach the contradiction x ∈ [C z , C z ]. It remains to assume that C 0 is not 2-closed and derive a contradiction. In this case let M 0 = E(C 0 ) ∼ = SL2 (r), r odd, as described above, so that M 0 x / z ∼ = P GL2 (r). As x ∈ [C z , C z ], M 0  C z . Now Out(M 0 ) = Bd × Bf where Bd ∼ = Z2 is the image of P GL2 (r) and Bf is a cyclic group of field automorphisms; moreover any involution in Aut(M 0 )−Inn(M 0 ) maps into Bd or Bf , by [III17 , 7.2b]. If L1 ∼ = M 0, 2 then we may similarly decompose Out(L1 ). Since CL1 (x) contains O (I 1 ) and x2 = 1 the (possibly trivial) image of x in Out(L1 ) lies in the field automorphism direct factor. A similar statement holds for L2 . Therefore by [III8 , 6.9], applied with A = Out(L), x ∈ [Cz , Cz ], again a contradiction. The proof is complete.  Lemma 4.8. We have [x, Li ] = 1 for each component Li of F ∗ (C z ). Proof. Suppose that x centralizes some Li , i = 1, 2. Then in view of Lemma 4.5a, x acts nontrivially on L3−i and Li  C z . Consequently L1 and L2 are interchanged in C z . In particular Li ∼ = L3−i . Thus I i = Li while I 3−i = CL3−i (x) < 1 ∼ SL2 (r) for some odd r, while I 3−i ∼ L3−i . It follows that Li = I i = = SL2 (r 2 ). However by (4A4), the field of definition of I 2 is a degree 1 or 3 extension of the field of definition of I 1 . This contradiction completes the proof.  



Now since O 2 (F ∗ (CG (x, z ))) = O 2 (I1 )I2 , we can apply [III17 , 10.33] to the action of x on Li , to obtain one of four possible outcomes. Conclusion (d) of that lemma, that Li ∼ = 2An for some n ≥ 7 or Li /Z(Li ) ∼ = L3 (4) with Z(Li ) of exponent 4, violates [III10 , Cor. 3.6]. We next rule out conclusion (c) of [III17 , 10.33], proving:

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4. THE CASE K ∼ = G2 (q) OR 3D4 (q)

45

Lemma 4.9. L1 = L2 , and for i = 1, 2, Li ∼ = A1 (qi )u or B2 (qi )u for some qi . Proof. Otherwise, by [III17 , 10.33], L1 = L2 satisfies x ∈ [Aut(L1 ), Aut(L1 )]. ∗

But F (C z ) = L1 in this case, so x ∈ [C z , C z ], contradicting Lemma 4.3d and completing the proof.  Lemma 4.10. z G ∩ Cx = z Cx . Proof. Suppose false and let u ∈ z G ∩ Cx − z Cx . As K has one class of involutions, u ∈ K; Lemma 4.3f implies that I2 (xK) ⊆ xG , so u ∈ xK. As m2 (Q) = 1, it follows that u induces a non-inner automorphism on K. Hence E(CK (u)) ∼ = 1 1 1 G2 (q 2 ), 3D4 (q 2 ), or 2 G2 (q 2 ), by [IA , 4.9.1]. As u ∈ z G , some composition factor of Cz involves E(CK (u)). However, since F ∗ (C z ) is the product of two components, we see using [IA , 7.1.1] that the only nonsolvable composition factors of Cz are those of F ∗ (C z ). These, however, are restricted in isomorphism type by [III17 , 10.33], and consequently by [III17 , 10.34] cannot involve E(CK (u)). This is a contradiction, and the proof is complete.  Lemma 4.11. xG ∩ Cz = xCz . Proof. This is immediate from Lemma 4.10 and [III8 , 6.7].



∗

Recall that L = L1 L2 and L = F (C z ). We set M = LCZ ≤ Cz and next prove Lemma 4.12. K ∼ = G2 (q), and the following conditions hold: (a) L1 and L2 are C z -conjugate; (b) xG ∩ Cz = I2 (Lx); (c) L1 ∼ = L2 ∼ = SL2 (r) for some r, and x induces a field automorphism on L1 and on L2 ; (d) x ∈ [M , M ]; and (e) |C z : M | = 2. Proof. First, x induces an outer automorphism on each Li , by Lemma 4.9 and [III17 , 10.33], which also imply that Out(L1 ) and Out(L2 ) are abelian. As x ∈ [C z , C z ], it follows that Out(L) cannot be abelian, so (a) holds and ∼ Out(L1 )  Z2 . Out(L) = The base group of this wreath product being abelian, it follows that every involution of [C z , C z ]L/L lies in Z(C z /L). Now by Lemma 4.11, xG ∩ Cz = xCz , so the left side of (b) is contained in the right side. Since the image of x is central in C z /L, C z permutes by conjugation the set X of all L-conjugacy classes of involutions y ∈ Lx such that CL (y) ∼ = CL (x), and xCz is the union of some of these classes. Let d be the number of such classes comprising X , and let E be the stabilizer in C z of the class xL . We conclude that E = LCC z (x) = LC Z = M , and so (4C)

|C z : M | ≤ d.

 Set H = L ∩ C Z . Of course CZ ∩ LO2 (Cz ) ≥ O 2 (I1 )I2 O2 (CZ ). Thus M /L ∼ =  C Z /H is a quotient of CZ /O 2 (I1 )I2 O2 (CZ ). But regarding CZ as CCx (z), we see

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46

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

that this quotient has abelian Sylow 2-subgroups and a normal 2-complement. As x ∈ L, therefore x ∈ [M , M ], proving (d). By Lemma 4.3d, x ∈ [C z , C z ]. It follows that M < C z . By (4C), d > 1, and then by [III17 , 10.36], (c) holds and d = 2, so equality holds in (4C). Hence (e) holds, and xCz = ∪X , proving the reverse inclusion in (b). Finally by (a), I 1 ∼  = I 2 so K ∼ = G2 (q). The proof is complete. Lemma 4.13. The following conditions hold: (a) There exists u ∈ Cx of order 4 and inducing a field or graph-field automorphism of order 2 on K; (b) Cx /K has cyclic Sylow 2-subgroups; and (c) R ≤ L. Proof. Recall that T ∈ Syl2 (Cx ) and R = T ∩ K. Note first that R = T ∩ I1 I2 b , where b is an involution such that Ii b /Z(Ii ) ∼ = Inndiag(Ii ), i = 1, 2. Note also that there is an involution b ∈ NL (I 1 I 2 ) ∩ CL (x) with the same property,  by Lemma 4.12c and [III17 , 11.10a]. Thus the images of b and b in Out(I 1 I 2 ) coincide. But CCx (I1 I2 ) = Z × O2 (Cx ). As [b, O2 (Cx )] = 1 it follows that b ∈ b Z. On the other hand, as b ∈ K, we have b ∈ z G . Thus b ∈ b Z ∩ z G ⊆ L x ∩ z G ⊆ L, the final inclusion by Lemma 4.12b. This proves (c). Since Q = x , T /R x embeds in Out(K), so T /R x is cyclic. Therefore T /R is abelian. On the other hand, as CL (x)/CLi (x) ∼ = P GL2 (q) for each i = 1, 2, we have CK (z) ≤ L and so M /L is a quotient of Cx /K. As |C z : M | = 2, this implies that |C z : L|2 ≤ 2|T : R| = 4|T : T ∩ KQ|. But x ∈ [C z , C z ] so Sylow 2-subgroups of C z /L are nonabelian. Thus, |C z : L|2 ≥ 8, whence T ≤ KQ. Let v ∈ T − KQ with v 2 ∈ KQ. By [III17 , 11.12], T /Q splits over RQ/Q, so we may choose v such that v 2 ∈ x , and then v acts on K as either a field automorphism or a graph-field automorphism according as q is not or is an odd power of 3. If v 2 = x, then (a) holds, and as Cx /K has abelian Sylow 2-subgroups, (b) holds as well. Thus we assume v 2 = 1 and argue to a contradiction to complete the proof of the lemma. Consider the action of the group U := v, x of order 4 on L. If v induces a field automorphism on K, then CL (U ) has 2-components or solvable 2-components 1 isomorphic to SL2 (q 2 ). As the components of L are isomorphic to SL2 (q 2 ), U is cyclic in this case, contradiction. Thus, v induces a graph-field automorphism on K, whence q is an odd power of 3, T = RU , and v 2 = 1. We have |C z /L|2 = 2|M /L|2 = 8. As x ∈ [C z , C z ] − L, we must have S/S ∩ L ∼ = D8 , with v interchanging L1 and L2 . Let S1 be the subgroup of index 2 in S such that S 1 ∼ = Z4 . Then by the Thompson transfer lemma [IG , 15.16], v is conjugate to some involution of Ω1 (S1 ) ≤ (S ∩ L) x . By Lemma 4.12b, Lx ∩ I2 (G) ⊆ xG . By [III17 , 11.10b], all involutions of L are conjugate to involutions of K, and hence all are G-conjugate to z. But Ω1 (S1 ) ⊆ L x . Thus, v is G-conjugate to either x or z. Set Kv = L2 (CK (v)), n so that Kv ∼ = 2 G2 (3 2 ) where q = 3n , n odd, n > 1. By L2 -balance, Kv has a pumpup Lv in L2 (CG (v)). Thus, Lv /O2 (Lv ) ∼ = SL2 (q 2 ) or G2 (q) according as

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5. THE NON-LEVEL CASE

47

v ∈ z G or v ∈ xG . But m2 (Kv ) > 1 = m2 (SL2 (q 2 )) and so the only possibility is that v ∈ xG , whence Kv := L2 (CG (v)) ∼ = K and v ∈ Syl2 (CCG (v) (Kv )). As all G elements of z ∩ Cx induce inner automorphisms on K (Lemma 4.10), z itself must induce such an automorphism on Lv . Thus, E(CC (v)) = E(CG (v, z )) = E(CC (v) (z)) ∼ = SL2 (q) ∗ SL2 (q) z

G

Given the structure of L = L (Cz ), the only possibility is that v induces a field automorphism on each Li , i = 1, 2, just as x does. Thus, vx ∈ LCC z (L) = L z = L, which is absurd as v and x are distinct modulo L. The lemma is proved.  2

Now choose a complement D to R in T containing x. Thus D is cyclic and ∗ DL/L ∈ Syl2 (M /L). Let L∗ be the preimage in M of O2 (M /L). Thus DL = M . Then ∗ (1) |C z : DL |2 = |C z : M |2 = 2 with D ∩ L∗ = 1; (4D) (2) xG ∩ Cz = xCz ; and (3) [M, M ] ≤ L∗ . Indeed Lemmas 4.12e and 4.11 yield (4D1) and (4D2), and (4D3) follows from (4D1) as D is cyclic. The hypotheses of [III8 , 6.6] therefore hold with G, Cz , L∗ [Cz , Cz ] and D in the roles of X, C, N , and y there. But the conclusion of that lemma, (4E)

VG→(S/S∩L∗ [Cz ,Cz ]) (y) = 1, where D = y ,

contradicts the simplicity of G. This completes the proof of Proposition 4.1. 5. The Non-Level Case In this section we rule out the final cases in which p = 2 and the neighborhood N(x, K, y, L) is non-level, thereby proving: Proposition 5.1. Let (x, K) and (y, L) satisfy the conditions (1A)–(1C). Then the neighborhood (x, K, y, L) is level, i.e., for every u ∈ x, y −x , Lu ∈ Chev with q(Lu ) = q(K) = q(L). ∼ 3D4 (q) or G2 (q) for some By (1D), if N(x, K, y, L) is not level, then p = 2, K = odd q, or ∼ P Sp4 (q) or a non-universal version of A± (q), (5A) K= 3

The first two possibilities have been ruled out in Proposition 4.1, so we need only consider the groups in (5A). Since K/O2 (K) ∈ G2 , in fact, q > 3. There are similarities between the two cases of (5A), and to take advantage of this we prove a preliminary result (Lemma 5.3 below) useful in both cases. The setup is as follows. (1) K is a non-universal version of C2 (q) or A3 (q), q odd, q > 3; (2) y0 ∈ I2 (K) and y0 ∈ E(CK (y0 )) = L0 Lt0 , t ∈ I2 (K), L0 a component of CK (y0 ) isomorphic to SL2 (q); (5B) (3) The pumpup L1 of L0 in CG (y0 ) satisfies L1 ∼ = SL2 (q 2 ). (4) T ∈ Syl2 (CG (x)) with y0 ∈ Z(T ), and T ≤ S ∈ Syl2 (CG (y0 )); moreover S0 = S ∩ C(y0 , L1 Lt1 ); (5C)

If K ∼ = P Sp4 (q). = A3 (q), then for no 2-element w ∈ CG (x) is E(CK (w)) ∼

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48

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Lemma 5.2. Assume (5B). Then x ∈ Syl2 (C(x, K)). Moreover, if (5C) also holds, then the following conditions hold: (a) S ∈ Syl2 (G) and x ∈ [CG (y0 ), CG (y0 )]; (b) CS0 (x) is cyclic of order 2 or of order dividing (q − )2 , according as K∼ = A3 (q); and = P Sp4 (q) or K ∼ t (c) L1 L1  CG (y0 ). Proof. We have m2 (C(x, K)) = 1 by (1C1), so if x ∈ Syl2 (C(x, K)), then there is v ∈ C(x, K) with v 2 = x. As [v, K] = 1, we have [v, L0 ] = 1 = [v, y0 ]. Thus v normalizes L1 . But since L1 ∼ = = SL2 (q 2 ) and L0 ∼ = SL2 (q), we have CAut(L1 ) (L0 ) ∼ 2 Z2 . As v maps into CAut(L1 ) (L0 ), x = v centralizes L1 . But L0 < L1 and L0 is a component of E(CK (y0 )), hence a component of E(CL1 (x)), a contradiction. Thus x ∈ Syl2 (C(x, K)). Assume that (5C) holds. By the previous paragraph, (5C), and [III17 , 8.8bc], Ω1 (Z(T )) = x, y0 . Since x induces a field automorphism on L1 , x is L1 -conjugate to xy0 ; but x ∈ y0G by the structures of the centralizers of these involutions. Hence y0 is 2-central in G and weakly closed in Z(T ) with respect to G. Then [III8 , 6.3] implies (a). Of course x ∈ S0 since x induces nontrivial field automorphisms on L1 and Lt1 . Then CS0 (x) embeds in CCG (x) (L0 Lt0 )/ x and hence, as x ∈ Syl2 (C(x, K)), in CAut(K) (L0 Lt0 ). But by [III17 , 11.14], a Sylow 2-subgroup of the latter group is of order 2 or dihedral of order 2(q − )2 , according to the isomorphism type of K, and when K ∼ = A± 3 (q), all the involutions w other than the image of y0 satisfy ∼ CK (w) = P Sp4 (q). By (5C), CS0 (x) contains no such involutions w, and so (b) holds. Suppose next that (c) is false, in which case there exists a component Lh1 of CG (y0 ), h ∈ CG (y0 ), which is distinct from L1 and Lt1 but contains y0 . If x (x) contains L2 (q 2 ), contradicting (b). Thus does not normalize Lh1 , then CLh1 Lhx 1 h x normalizes L1 . Again by (b), x does not centralize Lh1 , so by [III11 , 1.17], there is u ∈ Lh1 of order 4 such that ux = uy0 = u−1 . If there is a second component   Lh1 distinct from L1 and Lt1 , then there is an analogous element u ∈ Lh1 with  x  −1  (u ) = (u ) , and so x centralizes the four-subgroup y0 , uu , again contradicting C (y ) (b). Thus L1 G 0 = {L1 , Lt1 , Lh1 }. As CS0 (x) is cyclic of order dividing (q − )2 , a proper divisor of (q 2 − 1)2 , with Lh1 ∼ = SL2 (q 2 ), x cannot induce either an inner automorphism or a field automorphism on Lh1 . Hence the image of x in Out(Lh1 ) generates Outdiag(Lh1 ). Since x induces a nontrivial field automorphism on L1 and Lt1 , we may apply [III8 , 6.10], with N1 the subgroup P ΣL2 (q 2 ) of Aut(L1 ) generated by all inner automorphisms and a cyclic group of field automorphisms. We conclude that x ∈ [CG (y0 ), CG (y0 )], contradicting (a). This establishes (c), so the proof is complete.  Lemma 5.3. Assume (5B), (5C), and that for some w ∈ I2 (C(y0 , L1 )), the η pumpup Iw of L1 in CG (w) satisfies Iw /O2 (Iw ) ∼ = L3 (q 2 ), η = ±1. Then such a w exists satisfying the additional conditions [w, x] = 1 and wy0 ∈ wCG (y0 ) . Moreover, η = +1. Proof. We proceed by contradiction to prove the first statement. First we argue that (5D)

m2 (C(w, Iw )) = 1.

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5. THE NON-LEVEL CASE

49

By [IG , 6.10], there exists a chain (5E)

(w, Iw ) = (w1 , I1 ) < (w2 , I2 ) < · · · < (wn , In )

with (wn , In ) 2-terminal in G. Clearly each Ii ∈ Chev and indeed Ii ∈ G62 ∪ G72 . Accordingly by [III12 , Theorem 1.2] or [III4 , 20.3], m2 (C(wn , In )) = 1. Therefore if (5D) fails, some pumpup in the chain is nontrivial; say the first nontrivial pumpup is η (wi , Ii ) < (wi+1 , Ii+1 ). Thus Ii /O2 (Ii ) ∼ = Iw /O2 (Iw ) ∼ = L3 (q 2 ) and if (wi+1 , Ii+1 ) is a vertical pumpup of (wi , Ii ), then Ii+1 /O2 (Ii+1 ) lies in G72 and is isomorphic to L3 (q 4 ); otherwise by [III17 , 10.30d], Ii+1 /O2 (Ii+1 ) ∈ G62 with F(Ii+1 /O2 (Ii+1 )) > F(K), contrary to Lemma 1.2a. Now there exists yi = y0g ∈ y0G such that Lg1 ≤ Ii and [yi , wi+1 ] = 1. Thus according as Ii+1 is a diagonal or vertical pumpup, Lg1 lies g in a subgroup H ≤ CIi+1 (yi ) such that H/O2 (H) ∼ = SL2 (q 2 ) × SL2 (q 2 ) (with L1 diagonally embedded) or H/O2 (H) ∼ = SL2 (q 4 ). In either case, this contradicts the g fact that L1 is a component of CG (yi ). This establishes (5D). In particular Iw  CG (w). Obviously w = y0 . Thus CLt1 (w) maps into CAut(Iw /O2 (Iw )) (L1 ) with kernel of odd order, so by [III11 , 6.4g], CLt1 (w) is 2-nilpotent. In particular w acts nontrivially on Lt1 and does not induce a field automorphism on it, so w induces a nontrivial inner-diagonal automorphism on Lt1 . Therefore, wy0 ∈ wCG (y0 ) ⊆ wG . Notice that AutCG (w) (Iw /O2 (Iw )) has a cyclic Sylow 2-center whose involution is the image of y0 . It is visible that CG (w) ∼ = CG (y0 ), whence y0 ∈ wG and y0 is weakly closed in a Sylow 2-center of CG (w). Therefore by simplicity of G and [III8 , 6.3], w ∈ [CG (y0 ), CG (y0 )]. As w centralizes L1 and L1 Lt1  CG (y0 ), and as Out(L1 ) is abelian, it follows that w induces an inner automorphism on Lt1 . Thus, replacing w by a CG (y0 )-conjugate, we may assume that w = w1 v where w1 ∈ Lt1 has order 4 and v ∈ S0 with v 2 = y0 . Replacing w by an Lt1 -conjugate, we may arrange at our convenience that x centralizes w1 , or, if needed, that x inverts w1 . In particular if x normalizes v then we may arrange that [w1 , x] = [v, x] and so [w, x] = 1, as required. So x does not normalize v . Since we can replace w by any NCG (y0 ) (L1 )-conjugate, x does not normalize any NCG (y0 ) (L1 )-conjugate of v . Note that CS0 (w) = CS0 (v) does not contain w, so CS0 (v) embeds in CAut(Iw /O2 (Iw )) (L1 ), and hence is cyclic or quaternion, by [III11 , 6.4g]. Since v ∈ Z(CS0 (v)), CS0 (v) is cyclic. We argue next that we may assume that S0 has maximal class. If not, then S0 has a normal four-subgroup U ; but [v, U ] = 1 by the previous paragraph, and so v ∈ Φ(S0 ). As CS0 (v) is cyclic it follows that CS0 (v) = v , so S0 has maximal class in any event. We claim that C(y0 , L1 Lt1 ) has a normal 2-complement. Suppose not; then as S0 is of maximal class, it follows with the help of [III17 , 11.3] that C(y0 , L1 Lt1 ) has a characteristic subgroup H such that H/O2 (H) ∼ = SL2 (r) or 2A7 for some odd r, and |S0 : H ∩ S0 | ≤ 2. Since x ∈ [CG (y0 ), CG (y0 )], x then

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50

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

induces an inner automorphism on H/O2 (H), corresponding to an element u ∈ H of order 4. Conjugating w by some h ∈ H, and using the fact that H has one class of Z4 -subgroups and each is normal in some Sylow 2-subgroup of C(y0 , L1 Lt1 ), we may assume that u  u, v . This implies that u, and hence x, normalizes v , contrary to what we showed above. This proves our claim. Now x, S0 ≤ S is a 2-group. Since x ∈ [CG (y0 ), CG (y0 )], it follows that the xinvariant four-group S0 /Φ(S0 ) is centralized by x. Thus |CS0 (x)| ≥ |S0 /Φ(S0 )| = 4. In particular, K ∼ = L4 (q). = P Sp4 (q), so K/Z(K) ∼ Since CS0 (x) is cyclic, x centralizes an element v0 ∈ S0 of order 4. Let w0 ∈ CLt1 (x) be of order 4, and set w = w0 v0 , an involution centralizing x. Then in Aut(K), the image of w is the product of two commuting elements of order 4 whose squares are each equal to the image of y0 . One of these elements of order 4 (the image of w0 ) lies in Lt0 , while the other maps into CInndiag(K) (L0 Lt0 ), by [III17 , 11.19c], whence E(CK (w )) ∼ = L3 (q), again by [III17 , 11.19c]. Let Iw be the subnormal closure of L0 in CG (w ). Then Iw is equally well the subnormal closure in CG (w ) of L1 and that of E(CK (w )). As a pumpup of both of these, Iw /O2 (Iw ) ∼ = L3 (q 2 ), by [III17 , 10.49]. Thus w is the required element, contrary to our assumption that no such element exists. This completes the proof of the first statement. Suppose finally that η = −1. Then L0 , as a component of E(CK (y0 , w)), lies in a component Kw of E(CK (w)), and Iw is the pumpup of Kw in CG (w), i.e., Kw is a component of CIw (x). As η = −1, Iw admits no involutory field automorphisms, so Kw has level at least q 2 , by [IA , 4.2.2]. On the other hand, K is a quotient of Ω6 (q), so the only possibility, by [IA , 4.9.1, 4.5.1], is that Kw ∼ = L2 (q 2 ). But 2 L0 ∼ = SL2 (q) is not embeddable in L2 (q ), which has dihedral Sylow 2-subgroups. This contradiction shows that η = 1 and completes the proof of the lemma.  We first apply this to the case K ∼ = P Sp4 (q). According to [III12 , Def. 1.15], we may assume (replacing y by xy if need be) that y ∈ K, with E(CK (y)) ∼ = SL2 (q) ∗ SL2 (q). Recall the notation (1A)–(1C). We choose u ∈ x, y − x such that q = q(K) = q(L) = q(Lu ). Lemma 5.4. Suppose that K ∼ = P Sp4 (q), q > 3, q odd. Then (5B) and (5C) hold with y, L, and Ly in place of y0 , L0 , and L1 . Moreover, there exists an involution w ∈ C(y, Ly ) such that the pumpup Iw of Ly in CG (w) satisfies η Iw /O2 (Iw ) ∼ = L3 (q 2 ), η = ±1. Proof. Trivially (5C) holds. By [III12 , Def. 1.15], y ∈ L ∼ = SL2 (q). Thus in fact CK (y) = LLt t , where [L, Lt ] = 1 and t2 = 1. Therefore L is a component of CLu (x). If u = xy then L is a component of CLu (y) so by [III11 , 13.4], the only nonlevel possibility for the pumpup L < Lu is Lu ∼ Lemma = 3D4 (q 1/3 ). But then F(Lu ) = q 16/3 > q 4 = F(K), contradicting   t is the 1.2a. Therefore u = y ∈ Ly . We quote [III17 , 10.71] to conclude that Ly central product of two copies of SL2 (q 2 ) or of Sp4 (q 1/2 ). Indeed, the alternative conclusions of [III17 , 10.71] are impossible: Ly ∼ = 2An or [X]L3 (4), with center of exponent 4, would contradict [III10 , Cor. 3.6], and f (Ly ) > q 4 = f (K) with Ly ∈ G62 would contradict Lemma 1.2a.

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5. THE NON-LEVEL CASE

51

Now whatever the isomorphism type of Ly , (y, Ly ) cannot be 2-terminal. For ∼ SL2 (q 2 ), 2-terminality would yield a 2-Thin Configuration, contrary to if Ly = the assumption of Theorem C∗7 ; and if Ly ∼ = Sp4 (q 1/2 ), then 2-terminality would imply that (y, Ly ) ∈ J2 (G), whence m2 (C(y, Ly )) = 1 by [III12 , Theorem 1.2], contradicting the facts that Lty ≤ C(y, Ly ) and m2 (Ly ) > 1. We conclude with the help of [III8 , 2.7] that there exists a vertical pumpup (y, Ly ) < (w, Iw ), for some w ∈ I2 (C(y, Ly )). Notice that Ly is an intrinsic component of CIw (y). Now by [III11 , 13.4] and [III17 , 10.52], f (Iw /O2 (Iw )) ≥ (q 2 )4 or (q 1/2 )9 , according to the isomorphism type of Ly , so f (Iw /O2 (Iw )) > q 4 = f (K). Therefore Iw /O2 (Iw ) ∈ η G62 , by Lemma 1.2a. By the same lemmas, Ly ∼ = SL2 (q 2 ) and Iw /O2 (Iw ) ∼ = L3 (q 2 ), η = ±1. In particular, (5B) holds and the proof is complete.  Lemma 5.5. We have K ∼  P Sp4 (q), q > 3, q odd. = Proof. By Lemmas 5.3 and 5.4 there exists w ∈ I2 (CG (D)) such that [w, L1 ] = 1 and the pumpup Iw of L1 in CG (w) satisfies Iw /O2 (Iw ) ∼ = L3 (q 2 ). Then L1  CIw (y), and as [x, w] = 1 and x induces a nontrivial field automorphism on L1 , x induces a field or graph-field automorphism on Iw /O2 (Iw ). Let H = L2 (CIw (x)). Then H/O2 (H) ∼ = L± 3 (q). However, H is also a 2-component of CK (w), by L2 -balance. But K ∼ = P Sp4 (q) has no involutory automorphism whose centralizer has an L± (q) component, by 3  [IA , 4.5.1, 4.9.1]. This contradiction proves the lemma. For the remainder of this section we consider the one remaining case (5F) K∼ = Ω (q) or P Ω (q), q > 3, q odd. 6

6

By [III12 , Def. 1.15e2,e3], y ∈ K and one of the following holds: (1) y ∈ L ∼ = SL2 (q) and q ≡  (mod 8); (5G) (2) L = E(CK (y)) ∼ = SL3 (q) and q ≡  (mod 8). In either case we fix an involution y0 ∈ L and set L0 = E(CL (y0 )) ∼ = SL2 (q). Thus [y0 , D] = 1 and (5H)

E(CK (y0 )) = L0 Lt0 where t ∈ I2 (CK (y0 )) and Lt0 = L0 .

In case (5G1), y0 = y and L0 = L. Lemma 5.6. In the two cases of (5G), we have, respectively, (a) u = y = y0 , L ∼ = SL2 (q) and Ly ∼ = SL2 (q 2 ); or  ∼ ∼  (b) L = SL3 (q) and Lu /O2 (Lu ) = L3 (q 2 ). In both cases x induces a nontrivial field or graph-field automorphism on Lu . Proof. Since L is a component of CLu (x), the final statement follows immediately from the earlier statement. First suppose that L ∼ = SL3 (q). As q > 3, every pumpup of L lies unambiguously in Chev(r), where q is a power of the prime r. If Lu ∈ G72 , then as L < Lu is nonlevel, the only possibility is Lu /O2 (Lu ) ∼ = L3 (q 2 ). So assume that Lu ∈ G72 , 6 whence Lu ∈ G2 . Thus F(L) < F(Lu ) ≤ F(K) = (q 9 , A), by Lemma 1.2a. However, this is impossible, by [III17 , 10.31c]. Suppose then that y ∈ L ∼ = SL2 (q), so that D = x Z(L). Then Lxy possesses the involution y which has the intrinsic component L ∼ = SL2 (q) in CLxy (y). Hence

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52

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

∼ 3D4 (q 1/3 ). The latter case is imposby [III11 , 13.4], Lxy has level q, or else Lxy = sible, since if it occurred, CG (D) = CCG (xy) (y) would have a subnormal subgroup  H ∼ = SL2 (q 1/3 ) with y ∈ H; but y ∈ K − Z(K), so O 2 (H) ≤ K by (solvable) L2 -balance; however, CK (D) = CK (y) has no subnormal SL2 (q 1/3 ) subgroup or (for q = 33 ) subnormal Q8 -subgroup. Thus Lxy is a level pumpup of L, whence the nonlevel pumpup L < Lu occurs just for u = y. Now y ∈ L, the pumpup L < Ly is not level, AutCG (y) (L) ≥ AutCK (y) (L) ∼ = Inndiag(L), and if Ly /O2 (Ly ) ∈ G62 , then f (Ly ) ≤ f (K) = q 9 . Therefore by [III17 , 10.69a, 10.70] – and using [III10 , Cor. 3.6] – Ly ∼ = SL2 (q 2 ), as desired, or else η η (1) Ly ∼ = SL4 (q 1/2 ) ∼ = Spin6 (q 1/2 ), η = ±1, or (5I) η 1/2 ∼ (2) Ly = D4 (q ), η = ±1. Before we consider these cases, recall the involution t ∈ CK (y) from (5H), such that Lt = L. Note that as a result, if L is the only SL2 (q)-component of CLy (x) containing y, then Lty = Ly . η Suppose that Ly ∼ = SL4 (q 1/2 ). Then by [III17 , 10.70] and the previous remark, t Ly = Ly . In particular m2 (C(y, Ly )) > 1 so by Theorem C∗7 : Stage 3a, (y, Ly ) is not 2-terminal in G. As y ∈ Ly , it follows that there is (z, I) ∈ ILo2 (G) such that I := I/O2 (I) is a vertical pumpup of Ly and then (q 9/2 , A) = F(Ly ) < F(I) ≤ F(K) = (q 9 , A). Moreover, as Ly is a component of CI (y), y ∈ I − Z ∗ (I). By [III17 , 10.31a] the η only possibility is I ∼ = L5 (q 1/2 ). In particular Ly = L2 (CI (y)) and CAut(I) (Ly ) has cyclic Sylow 2-subgroups, by [III17 , 6.2]. On the other hand, X := CLt ,z (z) y

centralizes Ly and normalizes I, and by [III17 , 11.20] either m2 (X) ≥ 3 or y ∈ Lo2 (X). In the latter case as CAut(I) (Ly ) has cyclic Sylow 2-subgroups, we see that y must centralize I, which is absurd. Thus m2 (X) ≥ 3, whence m2 (C(z, I)) ≥ 2. Now Lemma 1.2c, with I in place of J there, produces a pumpup I1 of I such that  SLη4 (q 1/2 ). F(I) < F(I1 ) ≤ F(K), contradicting [III17 , 10.30e]. Thus, Ly ∼ = ± 1/2 t ∼ If Ly = D4 (q ) and Ly = Ly , then applying [IA , 4.5.1] to the action of x  on Ly Lty , we see that O r (CLy Lty (x LLt )) has sectional 2-rank at least 4. On the 

other hand, as y ∈ Ly , O r (CG (x LLt )) lies in C(x, K) and hence has 2-rank 1 and thus sectional 2-rank at most 2, a contradiction. Thus t normalizes Ly . Note that if Ly ∼ = D4− (q 1/2 ), then from [IA , 4.5.2], CLy (x) has a unique SL2 (q) component containing y, so Ly = Lty , contradicting what we just saw. Thus to complete the proof of the lemma it remains to consider the case (5J)

Ly = Lty ∼ = D4+ (q 1/2 ).

We rule this out in several paragraphs. As L and Lt are distinct components of CLy (x) containing y, it follows by [IA , 4.5.1] that x induces an inner-diagonal automorphism on Ly , whence [x, Z(Ly )] = 1. By [IA , 6.1.4], and as y ∈ Ly , Z(Ly ) ∼ = Z2 or E22 .

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5. THE NON-LEVEL CASE

53

∼ E22 . Then m2 (C(y, Ly )) ≥ 2 so Lemma 1.2c and Suppose that Z(Ly ) = [III17 , 10.31b] this time provide a pumpup (z, I) of (y, Ly ) with I := I/O2 (I) ∼ = Spin9 (q 1/2 ). Then we can take z to be the involution in Z(Ly ) ∩ Z ∗ (I), and in particular [x, z] = 1. Given the action of x on Ly we must have L2 (CI (x)) = J1 J2 with J1 /O2 (J1 ) ∼ = SL2 (q) and J2 /O2 (J2 ) ∼ = Sp4 (q 1/2 ). But by L2 -balance, J2 is a 2-component of CL2 (CG (x)) (z). By [IA , 4.5.1], J2 ≤ K, so [J2 , K] = 1. As m2 (J2 ) > 1 while m2 (C(x, K)) = 1, we have a contradiction. Therefore Z(Ly ) = y . We note for later use that by [III17 , 11.23], (5K)

x is Ly -conjugate to xy.

We next argue that (5L)

Ly  CG (y).

Suppose false and let J be the (nonempty) product of all components of E(C(y, Ly )) isomorphic to Ly . Then J is x-invariant and Lo2 (CJ (x)) lies in CCG (x) (LLt ), which is an extension of C(x, K) (of 2-rank 1) by a subgroup of CAut(K) (LLt ). The latter group in turn is dihedral by [III17 , 11.19b]. Consequently m2 (Lo2 (CJ (x))) ≤ 1. But examination of [IA , 4.5.1] shows that m2 (Lo2 (CJ (x))) > 1, a contradiction proving (5L). Then given the structure of E(CLy (x)) and the fact that |Z(Ly )| = 2, it follows from [III17 , 11.23] that the image of x in Aut(Ly ) lies outside [Aut(Ly ), Aut(Ly )]. Thus we have (5M)

x ∈ [CG (y), CG (y)].

Let T ∈ Syl2 (CG (D)). Since y is 2-central in CG (x), T ∈ Syl2 (CG (x)). If y were weakly closed in Z(T ) with respect to G, then (5M) and [III8 , 6.3] would imply x ∈ [G, G], contradicting the simplicity of G. Thus there exists y  = y g = y ∈ Z(T ) for some g ∈ G. Of course x is not conjugate to y as E(CG (x)) ∼ = E(CG (y)). In view of (5K), y  ∈ Z(T )−x, y . Clearly y  , which centralizes T ∩L ∈ Syl2 (L), must centralize LLt , so by [III17 , 11.19b] H := E(CK (y  )) ∼ = P Sp4 (q). On the other hand, Lgy ∼ = D4 (q 1/2 ) so by [IA , 4.5.1], E(CLgy (x)) has no component isomorphic to H. Thus by L2 -balance, [H, Lgy ] = 1. We have CLgy (x) ≤ CCG (x) (H), and CCG (x) (H) is an extension of the group C(x, K), in which x is the unique involution, by y  , as CAut(K) (H) ∼ = Z2 (see [IA , 4.5.1]). It follows that CLgy (x)/ y   has the image of x as its unique involution. Therefore O r (CLgy /yg  (x)) has a unique involution. But Lgy / y g is the simple group D4 (q 1/2 ), and by inspection of [IA , 4.5.1], no such involutory automorphism of D4 (q 1/2 ) exists. This contradiction completes the proof of the lemma.  In both cases of Lemma 5.6, since y0 is an involution of CLu (x), xy0 is an involution of Lu x, and as Lu is 2-saturated, it follows from [IA , 4.9.1] that (5N)

x is G-conjugate to xy0 .

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

54

We next prove Lemma 5.7. Let L1 be the subnormal closure of L0 in CG (y0 ). Then y0 ∈ L1 ∼ = SL2 (q 2 ) and x induces a nontrivial field automorphism on L1 . Moreover, (5B) holds, x ∈ Syl2 (C(x, K)), and L1 Lt1  CG (y0 ). Proof. The last sentence is an immediate consequence of Lemma 5.2 and the previous sentence, which we proceed to prove. If Lemma 5.6a holds, there is nothing to prove, so suppose that Lemma 5.6b holds. L0 = E(CL (y0 )) ≤ E(CG (D y0 )). Since L =  L Note that as L is quasisimple, # L0 , it follows that for any v ∈ D , the subnormal closure of L0 in CG (v) is the same as that of L, viz., Lv – which is quasisimple. It follows by [III8 , 2.13], with y0 , L0 , D, and y0 here in the roles of y, L, B1 , and u, that L1 ≤ E(CG (y0 )). Now y0 ∈ L1 , and L0 ∼ = SL2 (q) is a component of CL1 (x). On the other hand, as Lu /Z(Lu ) ∼ = SL2 (q 2 ) containing L0 , = L3 (q 2 ), CLu (y0 ) has a component M1 ∼ 2 ∼ and AutCLu (y0 ) (M1 ) = P GL2 (q ). In particular, M1 is a component of CL1 (u). In particular L1 cannot be the product of two x-conjugate copies of L0 , so L1 is quasisimple. Given the isomorphism types of L0 and M1 , and as y0 ∈ L0 ≤ M1 , it follows from [III17 , 10.43] that L1 ∈ G62 . But then F(L1 ) ≤ F(K) = (q 9 , A) by Lemma 1.2a, and [III17 , 10.43] implies that (5O)

L1 ∼ = SL2 (q 2 ) or SL± 4 (q).

It remains to rule out the case L1 ∼ = SL± 4 (q). Note that in that case F(L1 ) = F(K) and d2 (L1 ) = 6 = d2 (K) so by Lemma 1.2, (y0 , L1 ) is 2-terminal in G and m2 (C(y0 , L1 )) = 1. We have x mapping to an element of Out(L1 ) outside [Out(L1 ), Out(L1 )], again by [III17 , 10.43], so  K, so y0 is x ∈ [CG (y0 ), CG (y0 )]. By (5N), x is G-conjugate to xy0 , and L1 ∼ = weakly closed in x, y0 . By [III8 , 6.3] and the simplicity of G, y0 is not weakly closed in Z(T ). Therefore by [III17 , 8.8b], there exists y0 = y0g ∈ y0G ∩ Z(T ) with −1 E(CK (y0 )) ∼ = P Sp4 (q). Conjugating by g −1 , we see that E(CL1 (xg )) has a com± ponent isomorphic to P Sp4 (q). But as L1 ∼ = SL4 (q), there is no such component in the centralizer of any involution of Aut(L1 ) [IA , 4.9.1, 4.5.2]. This contradiction completes the proof of the lemma.  Observe that t ∈ CG (y0 ) with [Lt0 , L0 ] = 1, which easily implies that Lt0 centralizes L1 . Consequently Lt1 is a component of CG (y0 ) distinct from L1 , and as [t, x] = 1, (5P)

x induces a nontrivial field automorphism on Lt1 as well as on L1 .

Lemma 5.8. There exists no (w, I) ∈ ILo2 (G) such that I/O2 (I) ∼ = P Sp4 (q 2 ), x centralizes w, and L0 , t ≤ I. Proof. Suppose on the contrary that such a pair (w, I) exists. We shall reach a contradiction by quoting Proposition 2.7, with q 2 here in place of q there. Since x centralizes L0 , x normalizes I. Note that as y0 ∈ L0 , the involution ty0 satisfies the same requirements as t. Replacing t by ty0 if necessary, we may assume by [III17 , 11.15b] that (5Q)

t ∈ y0I .

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5. THE NON-LEVEL CASE

55

We argue first that (5R)

m2 (C(w, I)) = 1.

Clearly d2 (I/O2 (I)) = 6 = d2 (G) so if m2 (C(w, I)) > 1, Lemma 1.2c shows that there exists a pumpup I1 of some covering group of I/O2 (I) such that F(I/O2 (I))< F(I1 ) ≤ F(K); but no such I1 exists, by [III17 , 10.30f]. Thus (5R) holds. In particular I  CG (w); moreover, (w, I) is 2-terminal in G, so (w, I) ∈ J2 (G) and so by Theorem C∗7 : Stage 2, I is quasisimple. Thus (2F1) holds. Conditions (2F2, 3) hold by (5Q) and the paragraph before this lemma. Condition (2F4) is just notation, since y0 is 2-central in I. We complete the verification of (2F) by letting R ∈ Syl2 (CG (w)) with y0 ∈ Z(R), and showing that (5S)

R ∈ Syl2 (G).

By [III17 , 8.8d], Z(R) = y0 × CZ(R) (I), and as m2 (C(w, I)) = 1, CZ(R) (I) is cyclic. Now if R ∈ Syl2 (G), then R ∈ Syl2 (CG (y0 )). As CAut(I) (L1 Lt1 ) ∼ = Z2 by [III17 , 11.15b], CR (L1 Lt1 ) = y0 × CR (I) and in particular m2 (CR (L1 Lt1 )) = 2 and y0 ∈ Φ(CR (L1 Lt1 )). Therefore L1 and Lt1 are the only components of CG (y0 ) containing y0 , and so L1 Lt1  CG (y0 ). Since x induces a nontrivial field automorphism on I  CG (w), the image of x lies outside [AutR (I), AutR (I)]. But AutR (I) maps onto AutR (L1 Lt1 ) with kernel generated by the image of y0 ∈ [R, R]. Therefore [AutR (I), AutR (I)] maps onto [AutR (L1 Lt1 ), AutR (L1 Lt1 )] and so the image of x in AutR (L1 Lt1 ) lies outside the commutator subgroup. As Out(L1 Lt1 ) has a normal 2-complement, and L1 Lt1  CG (y0 ), it follows that x ∈ [CG (y0 ), CG (y0 )]. Consequently by [III8 , 6.3], y0 is not weakly closed in Z(T ) with respect to G, where T ∈ Syl2 (CG (x)) with y0 ∈ Z(T ). By (5N), xy0 ∈ xG , and by the structure of their centralizers, x and y0 are not G-conjugate. Hence by [III17 , 8.8b], y0 is G-conjugate to some y1 ∈ Z(T ) with E(CK (y1 )) ∼ = P Sp4 (q). Therefore CG (y0 ) contains a subgroup H ∼ = P Sp4 (q). But as L1 Lt1  CG (y0 ), and H is not involved in L1 , we have H = H (∞) ≤ C(y0 , L1 Lt1 ). Thus a Sylow 2-subgroup of H is embeddable in CR (L1 Lt1 ), which we have seen has 2-rank 2. As m2 (H) > 2 by [III11 , 3.9a], this is a contradiction, and (5S) is proved. Since (2F) holds, Proposition 2.7c1 is contradicted by the involution x: namely, [x, w] = [x, t] = 1 and x induces a nontrivial field automorphism on L1 . This completes the proof of the lemma.  Lemma 5.9. There exists no 2-element w ∈ CG (x) with E(CK (w)) ∼ = P Sp4 (q). Proof. Suppose first that such an involution w exists and set Kw = E(CK (w)). Replacing w by a conjugate we may assume that w centralizes L0 , t ; we may also suppose that t was chosen in Kw . In particular [w, y0 ] = 1 and we consider the subgroup w, x of CCG (y0 ) (L0 , t ). Clearly w, x then normalizes the subnormal closure L1 Lt1 of L0 Lt0 in CG (y0 ), and as [w, x , t] = 1 and |CAut(L1 ) (L0 )| = 2, we must have |Cw,x (L1 , t )| = 2. Since x acts nontrivially on L1 , it follows that we may assume, replacing w by wx if necessary, that [w, L1 ] = 1. Let I be the subnormal closure of Kw in CG (w). Then I is also the subnormal closure of L0 in CG (w), so L1 ≤ I. We have t ∈ Kw ≤ I. Consider the action of x, y0 on I. On the one hand, Kw is a component of CI (x); on the other, L1 and Lt1 are components of CI (y0 ). Since CI (y0 ) has at least two components, y0 normalizes each 2-component of I. If I were the product

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56

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

∼ Kw × Kw ; and as Kw of two 2-components interchanged by x, then I/O2 2 (I) = does not involve L1 , we would have y0 ∈ Z(L1 ) ≤ Z ∗ (I). But L1 is a component of CG (y0 ), so I = L1 Lt1 , which is absurd as I involves Kw . Thus I is a single 2-component. Then we quote [III17 , 3.32] and the structures of CI (x) and CI (y0 ) to obtain I ∼ = P Sp4 (q 2 ). But this contradicts Lemma 5.8. Thus no involution w satisfies the condition of the lemma. If w is of order 2n , n > 1, then by [III17 , 11.18], the image of w in Aut(K) is an involution, and so w2 ∈ C(x, K). By Lemma 5.7, w2 = x. Again replacing w by a K-conjugate we may assume that w centralizes L0 , t . Then w ∼ = Z4 embeds in the group CAut(L1 Lt1 ) (L0 , t ), which however has order 2. This contradiction completes the proof of the lemma.  Lemma 5.10. Let w ∈ I2 (C(y0 , L1 )). Let Iw be the subnormal closure of L1 in CG (w). Then either Iw is a trivial pumpup of L1 or Iw /O2 (Iw ) ∼ = L3 (q 2 ). Proof. Since y0 ∈ L1 ≤ Iw , Iw cannot be a diagonal pumpup of L1 . So we may assume that it is a vertical pumpup. Let W be the set of all involutions w ∈ I2 (C(y0 , L1 )) such that the pumpup Iw of L1 in CG (w) is nontrivial and not isomorphic to L3 (q 2 ). Thus, we must show that W = ∅. Assume false and let w ∈ W be arbitrary. By Lemmas 5.7 and 5.9, (5B) and (5C) hold, so Lemma 5.3 applies. By that lemma, Lemma 1.2a, and [III17 , 10.41], and the definition of W, I w := Iw /O2 (Iw ) is isomorphic to P Sp4 (q 2 ), Sp4 (q 2 ), or G2 (q 2 ). It follows from Lemma 1.2c and [III17 , 10.30f] that m2 (C(w, Iw )) = 1, (w, Iw ) is 2-terminal in G, and then as d2 (G) = 6 = d2 (Iw /O2 (Iw )), that (w, Iw ) ∈ J2 (G). Hence by Theorem C∗7 : Stage 2, Iw is quasisimple. We have L2 (CIw (y0 )) = L1 L2 where L1 ∼ = L2 , and |CAut(Iw ) (L1 L2 )|2 = 2. So y0 is complemented in any Sylow 2-subgroup of CCG (y0 ) (w L1 L2 ). We argue that (5T)

L1 Lt1 ≤ Iw ,

i.e., that L2 = Lt1 and w ∈ C(y0 , L1 Lt1 ). Suppose false, so that L2 ≤ C(y0 , L1 Lt1 ) and so CLt1 (w) ≤ CCG (y0 ) (w L1 L2 ). Thus y0 is complemented in a Sylow 2subgroup of CLt1 (w), which implies by [III17 , 11.8] that w induces an outer diagonal t automorphism on Lt1 . Moreover, wy0 ∈ wL1 by [III11 , 1.17]. As m2 (C(w, I)) = 1 it is obvious that CG (w) ∼ = CG (y0 ), so w ∈ y0G . Thus y0 is weakly closed in w, y0 with respect to G. If Iw ∼ = Sp4 (q 2 ), we would have w = Z(Iw ) and so y0 w ∈ y0Iw , a contradiction. Thus Iw ∼ = P Sp4 (q 2 ) or G2 (q 2 ), and consequently by [III17 , 8.8c], w, y0 is the exponent 2 subgroup of a Sylow 2-center in CG (w). Now, given the action of w on Lt1 (and its trivial action on L1 ), and since L1 Lt1  CG (y0 ), we have w ∈ [CG (y0 ), CG (y0 )]. But then w ∈ [G, G] by [III8 , 6.3], contradicting the simplicity of G. This establishes (5T). Notice that [w, x] = 1. For otherwise, as an element of CG (x), w maps into CAut(K) (L0 Lt0 ). Since E(CK (w)) ∼ = P Sp4 (q), by Lemma 5.8, the only possibility is that w is CG (x)-conjugate into x, y0 , by [III17 , 11.19b]. But x is G-conjugate to xy0 , and by centralizer structure, w is G-conjugate to neither x nor y0 . Thus, x centralizes no element of W.

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5. THE NON-LEVEL CASE

57

Recall that S ∈ Syl2 (CG (y0 )). Set S0 = CS (L1 Lt1 ) and observe that W ∩S ⊆ S0 by (5T). Since W = ∅ and L1 Lt1  CG (y0 ), W ∩ S0 = ∅. We use the idea of semirigidity [IG , Sec. 7] to argue next that S0 is of maximal class. Suppose not and choose w ∈ W ∩ S0 such that CS0 (w) has maximal order. As CS0 (w) centralizes t L1 Lt1 , CS0 (w) maps into CAut(I w ) (L1 L1 ), which in turn is equal to the image of y0 , by [III17 , 8.8c]. Thus CS0 (w) = y0 × S1 where [S1 , I w ] = 1. As neither P Sp4 (q 2 ) 2 nor G2 (q 2 ) is involved in L± 3 (q ), the only other possible nontrivial pumpup of L1 , it follows that I2 (S1 ) ⊆ W. Hence by our maximal choice, |CS0 (w)| ≥ |CS1 (w )| for all w ∈ I2 (S1 ). Also by the previous paragraph, CS1 (x) = 1, whence S1 ∩ S1x = 1. Since S0 is not of maximal class, |CS0 (w)| > 4. Thus |S1 | ≥ 4 and in particular as S1 ∩ S1x = 1, |S0 : S1 | ≥ 4 and CS0 (w) < S0 . Choose a ∈ NS0 (CS0 (w)) − CS0 (w) with a2 ∈ CS0 (w). Then as |S1 | ≥ 4, 1 = S1 ∩ S1a  CS0 (w) a . Hence S1 ∩ S1a ∩ Z(CS0 (w) a ) = 1, contradicting the maximality of |CS0 (w)|. Thus S0 is of maximal class, as asserted. As w, y0 ≤ S0 , S0 is dihedral or semidihedral. Suppose that S0 is dihedral. Since Z(S0 ) ≥ y0 ≤ Z(C(y0 , L1 Lt1 )), C(y0 , L1 Lt1 ) has a normal 2-complement, by [III8 , 6.11]. If |S0 | = 4 then since [x, w] = 1, x ∈ [CG (y0 ), CG (y0 )], contradicting Lemma 5.2a. If |S0 | > 4 then the unique cyclic maximal subgroup of S0 maps to a normal subgroup of CG (y0 )/O2 (CG (y0 )) not centralized by w, and so wy0 ∈ wG and w ∈ [CG (y0 ), CG (y0 )], whence w ∈ [G, G] by [III8 , 6.3], contradicting the simplicity of G. Thus, S0 is semidihedral. In particular S0 has a unique class of noncentral involutions. By a Frattini argument, CG (y0 ) = C(y0 , L1 Lt1 )NCG (y0 ) (S0 ) = C(y0 , L1 Lt1 )CCG (y0 ) (w), so the image A of CG (y0 , w) in Out(L1 Lt1 ) equals OutCG (y0 ) (L1 Lt1 ). We have x ∈ [CG (y0 ), CG (y0 )] by Lemma 5.2a, whence A is nonabelian. On the other hand, L1 Lt1 ≤ Iw  CG (w), so A is isomorphic to the image of CG (y0 , w) = CCG (w) (y0 ) t (L1 L ). This last group, as Iw ∼ in Out = P Sp4 (q 2 ) or G2 (q 2 ), is abelian by Aut(I w )

1

[III17 , 6.16]. Thus we have a contradiction and the proof is complete.



Now we can begin to see an A3 (q 2 ) subgroup of G emerging. Lemma 5.11. There exists w ∈ CG (x, y0 , L0 ) such that for each v ∈ w, y0 # , the subnormal closure Iv of L0 in CG (v) satisfies the following conditions: (a) Iy0 = L1 ∼ = SL2 (q 2 ); (b) Iw /Z(Iw ) ∼ = Iwy0 /Z(Iwy0 ) ∼ = L3 (q 2 ); CG (y0 ) ; and (c) wy0 ∈ w (d) C(w, Iw ) and C(wy0 , Iwy0 ) have isomorphic cyclic or quaternion Sylow 2-subgroups of exponent dividing q 2 − 1. Proof. If (y0 , L1 ) were 2-terminal in G, we would have a 2-Thin Configuration by definition, contrary to the hypothesis of Theorem C∗7 . Thus by [III8 , 2.7], there exists w ∈I2 (C(y  0 , L1 )) such that the pumpup Iw of L1 in CG (w) is nontrivial. L1 As L1 = L0 , Iw is also the subnormal closure of L0 in CG (w). As noted previously, (5B) and (5C) hold, so Lemma 5.3 applies. By that lemma and Lemma 5.10, w ∈ CG (x, y0 , L0 ) exists satisfying (b) and (c), with Iw /Z(Iw ) replaced by Iw /O2 (Iw ) (and similarly for Iwy0 ); and (a) follows from Lemma 5.7. It then suffices to show that Iw is quasisimple, and verify (d).

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

58

Now [w, x, y0 ] = 1 and for each v ∈ x, y0 # , L0 ≤ E(CG (v)).

(5U)

Indeed this is clear for both v = x and v = y0 . For v = xy0 , note that xy0 = xg for some g ∈ L1 ; g then normalizes L0 = CL1 (x) and (5U) follows by conjugation by g. Therefore Iw is quasisimple by [III8 , 3.2], applied with w, y0 , and L0 in the roles of y, z, and I there. It remains to prove (d). Since w and wy0 are conjugate it is enough to show that a Sylow 2-subgroup of C(w, Iw ) is cyclic or quaternion of exponent dividing q 2 − 1. We first show that m2 (C(w, Iw )) = 1. To prove this, note first that since L1   CG (y0 ), (w, Iw ) cannot have a diagonal pumpup; nor can it have a pumpup (v, J) with J ∼ = L3 (q 4 ). Such pumpups, in fact, would force L1 to be embedded in J   CG (v) with J = L2 (J), y0 ∈ O2 2 (J), and J/O2 (J) ∼ = L1 × L1 (with L1 embedded diagonally) or J/O2 (J) ∼ = SL2 (q 4 ). In either case the subnormality L1   CG (y0 ) would be violated. Thus any nontrivial pumpup (w, Iw ) < (v, J) would be vertical with J/O2 (J) ∈ G62 ; but then F(Iw ) < F(J/O2 (J)), and by [III17 , 10.30d], the maximality of F(K) (Lemma 1.2) would be violated. A similar argument shows that in any chain of pumpups starting with (w, Iw ), all pumpups are trivial. Taking such a chain terminating with a 2-terminal pair (w∗ , Iw∗ ), we conclude from [III4 , 20.3] that m2 (C(w∗ , Iw∗ )) = 1, and then as all the pumpups in the chain are trivial, m2 (C(w, Iw )) = 1, as desired. Let Qw ∈ Syl2 (C(w, Iw )), so that w is the unique involution of Qw . But Qw centralizes both L1 and wy0 , so Qw embeds in a Sylow 2-subgroup of CAut(Iwy0 ) (L1 ), which by [III11 , 6.4g] is quaternion of order 2(q 2 − 1)2 . Therefore Qw is cyclic or  quaternion of exponent dividing q 2 − 1. Now we can at last prove: Lemma 5.12. We have K ∼  A3 (q),  = ±1. = Proof. Assume false and continue the above argument. By Lemma 5.11, the conditions (2A) are satisfied, with the roles of v, w, vw, Iv , Iw , and Ivw there being played by y0 , w, wy0 , L1 , Iw , and Iwy0 . Proposition 2.1 furnishes a subgroup G1 ∼ = L4 (q 2 ) which is invariant under CG (y0 ) and in which E(CG1 (y0 )) is the product of L1 and some G1 -conjugate of L1 . As L1 Lt1  CG (y0 ), we must have E(CG1 (y0 )) = L1 Lt1 . Thus OutCG (y0 ) (L1 Lt1 ) ≤ OutAut(G1 ) (L1 Lt1 ), in which the image of x, which induces a nontrivial field automorphism on each of L1 and Lt1 , lies outside the commutator subgroup by [III17 , 11.19d]. Therefore x ∈ [CG (y0 ), CG (y0 )]. This contradicts Lemma 5.2a, so the lemma is proved.  In view of (5A), and Lemmas 5.5 and 5.12, Proposition 5.1, and (1E) are proved. 6. The Other Exceptional Cases In this section we continue the assumptions and notation of (1A), (1B) and (1C), as well as (1E), which has been proved in the previous section. We continue to assume that p = 2. In particular (x, K) ∈ J∗2 (G). We shall prove Proposition 6.1. If p = 2, then K is a classical group.

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6. THE OTHER EXCEPTIONAL CASES

59

This perhaps surprising result is an artifact of our ordering function F. While K = E6± (q) and E7 (q) are standard components in E7 (q) and E8 (q), respectively, they are not the standard components with the highest F-value; rather, they are exceeded by K = A± 7 (q) and D8 (q), respectively. So the target groups E7 (q) and E8 (q) will emerge from an analysis of the latter groups K. Although the analyses in all the cases of Proposition 6.1 are closely related, they differ in enough details for us to proceed case by case. We need only rule out the groups K of exceptional Lie type not covered in Proposition 4.1 and [III10 , Prop. 3.1], namely ∼ F4 (q), E  (q), E7 (q), E8 (q), q = r n , r an odd prime. (6A) K= 6

By [III12 , Theorem 1.2] and Proposition 5.1, (6B)

For any acceptable subterminal (x, K)-pair (y, L) and any u ∈ x, y −x , the pumpup Lu of L in CG (u) satisfies q(Lu ) = q(K).

Lemma 6.2. K ∼  F4 (q). = Proof. Suppose that K ∼ = F4 (q). By [III12 , Def. 1.15] and [IA , Table 4.5.1],  ∼ L = C3 (q)u and O r (CK (y)) = J ∗ L, with J ∼ = A1 (q) and Z(J) = Z(L). We may assume that Z(L) = y . Choose any u ∈ D − x with Lu > L. With (6B) and 2 [IA , 4.2.2], and since f (Lu ) ≤ f (K) = q 4 by Lemma 1.2b, we see that Lu ∼ = C4 (q) or K. Note that y ∈ L ≤ Lu with L ∩ Z(Lu ) = 1 in either case, so u = xy and  Ly = L. Moreover, from the structure of O r (CK (y)) and (solvable) L2 -balance  we have O 2 (JL) ≤ Lxy , which implies by [IA , Table 4.5.2] that Z(Lxy ) = 1. Let z be a short root involution of L. Thus by [III17 , 12.9b], CL (z) has a component I ∼ = C2 (q)u with z ∈ I. Set Lz = E(CK (z)), and note that CK (z) = u ∼ Lz = B4 (q) by [III17 , 12.9a]. Let Iz be the subnormal closure of I in CG (z). Then Lz = I Lz ≤ Iz so Iz is a pumpup of Lz with z ∈ I ≤ Iz . If Iz is a vertical pumpup of Lz , then we have F(B4 (q)) = F(Lz ) < F(Iz /O2 (Iz )) ≤ F(F4 (q)) by Lemma 1.2. But the only possibility then is that Iz /O2 (Iz ) ∼ = F4 (q), which is absurd as F4 (q) is simple and 2-saturated, while z ∈ Iz . Thus Iz is a trivial or diagonal pumpup of Lz . Let I0 be a 2-component of Iz , so that I0 /O2 (I0 ) ∼ = B4 (q)u . We claim that (6C) I0 is the unique 2-component of CG (z) with I0 /O2 (I0 ) ∼ = B4 (q)u . Suppose false. Then m2 (C(z, I0 )) > 1, so by [III12 , Theorem 1.2], (z, I0 ) ∈ J2 (G). As B4 (q) ∈ G62 and d2 (G) = 6, (z, I0 ) is not 2-terminal in G. Then [III8 , 2.7] implies that there is a vertical pumpup (z, I0 ) < (z1 , I1 ). We have F(B4 (q)) = F(I0 ) < F(I1 ) ≤ F(F4 (q)), and again this implies that I1 /O2 (I1 ) ∼ = F4 (q). Hence (z1 , I1 ) ∈ J∗2 (G), whence m2 (C(z1 , I1 )) = 1. On the other hand, if I0∗ is the product of all 2-components of CG (z) isomorphic modulo core to B4 (q)u and distinct from I0 , then by [IA , Table 4.5.2], CI0∗ (z1 ) contains a component I 0 with m2 (I 0 ) > 1. This implies that m2 (C(z1 , I1 )) > 1, a contradiction. Thus (6C) holds. By (6C), Iz is a trivial pumpup of Lz and Iz  CG (z). Thus x centralizes Iz /O2 (Iz ). If Lxy ∼ = K, then by the symmetry between x and xy, we conclude that xy also centralizes Iz /O2 (Iz ). Then D = x, xy

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

60

centralizes Iz /O2 (Iz ). However, Lz ≤ Iz and [D, Lz ] = 1 as Lz does not embed in E(CK (D)) ≤ JL. This contradiction shows that Lxy ∼ = K, whence Lxy ∼ = C4 (q). But then Lxy ∼ = P Sp8 (q), so there exists an involution t ∈ Lxy such that [I, I t ] = 1 and [t, z] = 1. Then like Iz , Izt is a 2-component of CG (z). By our claim above, Izt = Iz . Thus the central product II t embeds in Iz /O2 (Iz ). However,  since I ∼ = Spin5 (q), O r (CIz /O2 (Iz ) (I)) ∼ = Spin4 (q) does not involve I t . This contradiction completes the proof of the lemma.  Lemma 6.3. K ∼ = E6± (q). ∼ E  (q),  = ±1. By [IA , 6.1.4], |Z(K)| = 1 or 3. By Proof. Suppose that K = 6 [IA , 4.5.1] and [III12 , Def. 1.15], the only type of acceptable subterminal (x, K)-pair  (y, L) occurs for L ∼ = A5 (q), with y ∈ L, Z(K) ≤ Z(L), and O r (CK (y)) = J ∗ L, with y ∈ J ∼ = A1 (q). We fix such a y, J, and L. Lemma 1.2a implies that (6D)

For any u ∈ D# , Lu has untwisted rank at most 5 or is isomorphic to ± ± A± 6 (q), D6 (q), or E6 (q).

We claim that (1) Either Lxy = L or Lxy ∼ = E6 (q); and (6E) ∼ (2) Ly = D6 (q). 

For, y = J ∩ L, and so if Lxy > L, then LO 2 (J) ≤ Lxy , by (solvable) L2  balance, and LO 2 (J)   CLxy (x). This fact, together with (6D) and [IA , 4.5.1], implies that Lxy ∼ = E6 (q), so (6E1) holds. If (6E2) fails, then as y ∈ L ≤ Ly , we have Ly = L, by (6D) and [III17 , 10.55]. There is g ∈ K such that if we put y1 = y g , L1 = Lg and J1 = J g , then y1 ∈ L − Z(L) and J1 ≤ L, with J1 a root A1 (q) subgroup of both K and L. Then  O r (CL (y1 )) = HJ1 with H ∼ = A3 (q). Since Ly = L, L1   CG (y1 ) and thus by L2 -balance, L1 is the subnormal closure of H in CG (y1 ). It follows that the subnormal closure H1 of H in CCG (xy) (y1 ) centralizes x and hence centralizes y.   But then H1 = H H1 ≤ O r (CG (x, y )) and so H1 ≤ L, i.e., H1 = H. On the other hand, y1 ∈ J1 with J1 a root A1 (q) subgroup of L, hence a root A1 (q) subgroup of Lxy . Since Ly = L, we know that Lxy = L, and hence Lxy ∼ = E6 (q). Thus H1 ∼ = L ∼ = H, a contradiction. This completes the proof of (6E). By [IA , 4.5.1], L is a quotient of SL6 (q) by a central subgroup of odd order, and we have y ∈ L. Let y1 and H be as in the previous paragraph, and set v = yy1 . Then v = Ω1 (Z(H)) and we consider CG (v). Since y1 ∈ y K ∩ L, we have [y1 , x, y ] = 1 and hence [v, x, y ] = 1. Let Hx , Hy and Hxy be the pumpups of H in CG (x, v), CG (y, v), and CG (xy, v), respectively, and let H ∗ be the pumpup of H in CG (v). By [III17 , 12.15ab], we have Hx ∼ = D5 (q)u , Hy ∼ = D4± (q), and Hxy ∼ =   u  A3 (q) or D5 (q) according as Lxy = L or Lxy ∼ = Hxy , x acts = E6 (q). Since Hy ∼ nontrivially on H ∗ . But then by [III17 , 10.54] and Lemma 1.2 the only possibility is η H∗ ∼ = A3 (q) = D6 (q), η = ±1, with E(CH ∗ (xy)) containing distinct components E ∼ η ∗ ∼ ∗ and E = A3 (q). Then Lxy = L; as EE does not embed in L, L2 -balance implies that [E ∗ , L] = 1 in CG (xy). As y ∈ L, E ∗ ≤ CG (x, y L). On the other hand, CAut(K) (y L) = CInn(K) (y L) ∼ = SL2 (q), while m2 (C(x, K)) = 1. Hence CG (x, y L), and then E ∗ , is an extension of a group of 2-rank 1 by a subgroup of SL2 (q). This contradicts [III17 , 11.22] and completes the proof of the lemma. 

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6. THE OTHER EXCEPTIONAL CASES

61

Lemma 6.4. K ∼ = E7 (q). ∼ E7 (q). Then |Z(K)| ≤ 2. By [III12 , Def. 1.15] and Proof. Suppose K =  [IA , 4.5.1, 4.5.2], y ∈ I2 (K) and O r (CK (y)) = JL with J ∼ = A1 (q), L ∼ = D6 (q)hs u or D6 (q) according as |Z(K)| = 1 or 2, respectively, [J, L] = 1, Z(J) ≤ Z(L), and Z(J) ∩ Z(K) = 1. By (1C), choose u ∈ D − x such that Lu is a vertical pumpup of L, and subject to this condition, if possible, so that (6F)

+ L u / u ∼ = Ω+ 12 (q) or P Ω12 (q).

If in fact (6F) holds, then by Lemma 1.2a and [III17 , 10.56c], L u /u ∼ = Ω+ 12 (q) ± ∼ and Lu /Z(Lu ) = P Ωn (q), n > 12. In particular Z(K) = 1 and D = Z(L). Using [IA , 4.5.2], we see that JL/ u is the central product of the images of J and L, and the centers of these images both equal D/ u = x, u / u . As [x, Lu ] = 1, J embeds in CAut(Lu ) (L). But J ∼ = SL2 (q) and therefore by [III17 , 10.56b], n ≥ ± (q), m ≥ 8. In any case F(Lu ) > F(K), 15. Thus Lu ∼ = Bm (q), m ≥ 7, or Dm contradicting Lemma 1.2a. Thus (6F) must fail, whence (6G)

L u / u ∼ = D6 (q)hs .

As Lu has level q, it follows from [III17 , 10.56a] that Lu /Z(Lu ) ∼ = K/Z(K). As usual, with Lemma 1.2b, we have symmetry between x and u. Note that either L∼ = D6 (q)hs with u ∈ L, whence L u = L x . In = D6 (q)u with D = Z(L), or L ∼ + hs ∼ either case L x / x = D6 (q) and therefore L ux / ux ∼ = Ω+ 12 (q) or P Ω12 (q). Since the choice (6F) is impossible, we must have Lux = L. Choose t ∈ I2 (L) such that E(CL (t)) has a component I ∼ = D4 (q)u with t ∈ I. r Then t ∈ I2 (K) and by [III17 , 12.15c] and [IA , 4.5.1], O (CK (t)) = I0 ∗ Jx with I0 ∼ = A1 (q) and Jx /Z(K) ∼ = D6 (q)hs . Consequently Jx is the subnormal closure of I in CK (t). By symmetry the subnormal closure Ju of I in CLu (t) is isomorphic to Jx . Let J be the subnormal closure of I in CG (t). Note that t ∈ I ≤ L ≤ CG (D), so D normalizes J. Now Jx = I Jx ≤ J so Jx is a component of CJ (x), and similarly Ju is a component of CJ (u). On the other hand, since L = Lux , I is a component of CJ (ux). Since Jx ∼ = L, (t, Jx ) is an acceptable subterminal (x, K)-pair by definition [III12 , Def. 1.15]. Thus by [III12 , Theorem 1.2], J is not a diagonal pumpup of Jx . Furthermore as I < Ju , we have [x, Ju ] = 1, so J is a vertical pumpup of Jx , and again by [III12 , Theorem 1.2], J ∈ Chev has level q. Moreover, F(J) ≤ F(K) by Lemma 1.2a. Arguing as after (6F), but with t in place of u, we conclude that Jx t / t ∼ = D6 (q)hs , and then with [III17 , 10.56a] that J ∼ = E7 (q). But then [IA , 4.5.1] shows that CJ (ux) has no component isomorphic to D4 (q), contradicting the existence of I and completing the proof.  Lemma 6.5. K ∼  E8 (q). = Proof. Suppose K ∼ = E8 (q). We have L ∼ = D8 (q)hs by [III12 , Def. 1.15], and without loss we take y ∈ Z(L). Let u be any element of x, y − x such that Lu > L. By [III17 , 10.64], Lu ∼ = K, whence Z(Lu ) = 1, u = xy, and = E8 (q) ∼ Ly = L. By Lemma 1.2b we have symmetry between x and xy. Choose t ∈ I2 (L) such that E(CL (t)) = I1 I2 , with I1 ∼ = I2 ∼ = Spin+ 8 (q) and K r t ∈ I1 ∩I2 . Then t ∈ y . Indeed otherwise by [IA , Table 4.5.1], O (CK (t)) = J1 ∗J, J1 ∼ = E7 (q)u , and by L2 -balance, I1 ≤ J. This would imply that I1 = A1 (q), J ∼

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

62

is a component of CJ (y); but Aut(J) has no such involutory automorphism, by [IA , 4.5.1, 4.9.1]. Thus t ∈ y K , as asserted. Let It = E(CK (t)) ∼ = L. The first paragraph applies, with t and It in place of y and L. The analogue of the equation Ly = L is that It   CG (t). Similarly t ∈ L ≤ Lxy and we let It∗ = E(CLxy (t)), concluding by symmetry that It∗   CG (t). Therefore the subnormal closure of I1 in CG (t) equals both It and It∗ , so It = It∗ . In particular It is centralized by x, xy = x, y . But by [IA , 4.5.1], m2 (CAut(K) (It )) = 1, so the only involutions centralizing x It lie in x, t . Hence x, y = x, t so [t, L] = 1, contrary to our choice. This completes the proof. 

7. The Case K ∼ = P Sp2n (q) = Cn (q)a In this section we continue to assume the assertions and notation of (1A), (1B) and (1C), as well as (1E). We shall prove: Proposition 7.1. Suppose that p = 2 and K/O2 (K) ∼ = P Sp2n (q) for some n ≥ 2 and odd q. Then there exists (x , K  ) ∈ J∗2 (G) such that K  ∼ = Sp4 (q), Spin7 (q), or Sp8 (q 1/4 ). We suppose that the proposition fails. Thus P Sp2n (q) ∈ G2 , whence n ≥ 2, with strict inequality if q = 3.

(7A)

By [IA , 6.1.4] the Schur multiplier of P Sp2n (q) has order 2, so K∼ = P Sp2n (q). Since (x, K) ∈ J2 (G), (7B)

d2 (G) = 6.

By [III12 , Def. 1.15] and [IA , Table 4.5.1], our acceptable subterminal (x, K)-pair (y, L), which we may assume is chosen so that y ∈ K, satisfies O 2 (CK (y)) = K1 L, K1 ∼ = SL2 (q), L ∼ = Sp2n−2 (q), [K1 , L] = 1, and (7C) y = Z(K1 ) ∩ Z(L). Throughout this section we write Cx = CG (x), Cy = CG (y), and Cxy = CG (xy). Let u ∈ x, y − x with Lu > L; at least one such u exists, by (1C3). Thus Lu is a single component in Chev with q(Lu ) = q. As y ∈ L ≤ Lu , either |Z(Lu )| is  even or O 2 (K1 ) ≤ Lu , by (solvable) L2 -balance. In either case Lu /Z(Lu ) ∼  L± = 3 (q), so d2 (Lu ) = 6. Therefore by Lemma 1.2a, (7D)

F(Lu ) ≤ F(K) for all u ∈ {y, xy}.

We have T ∈ Syl2 (Cx ) and Q = CT (K) with m2 (Q) = 1; we may assume that T was chosen so that Ty := CT (y) ∈ Syl2 (CG (x, y )). We first reduce to the case n = 2, assuming through Lemma 7.4 that n ≥ 3.

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7. THE CASE K ∼ = P Sp2n (q) = Cn (q)a

63

Lemma 7.2. If n ≥ 3, then Ly = L. Proof. Suppose on the contrary that Ly > L. By (1C), Ly is a single component. We have y ∈ L ≤ Ly , so y ∈ Z(Ly ), and n ≥ 3. Together with (7D) and [III17 , 10.69c], this yields that n = 3 and one of the following holds:

(7E)

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

x x x x

induces induces induces induces

∼ Sp4 (q 2 ); a field automorphism on Ly = a graph automorphism on Ly ∼ = SL± 4 (q); 1 an outer diagonal automorphism on Ly ∼ = Sp8 (q 2 ); or an inner-diagonal automorphism on Ly ∼ = Spin7 (q).

We argue that (1) (7E4) does not hold; (2) xy ∈ xG ; and (7F) (3) x ∈ [Cy , Cy ]. Suppose in fact that (7E4) holds, so that Ly ∼ = B3 (q). Then f (Ly ) = q 9 = f (K), so F(Ly ) = F(K) as types B and C are considered equivalent by the function F. Therefore (y, Ly ) ∈ J∗2 (G) by Lemma 1.2b. Hence Proposition 7.1 holds with (x , K  ) = (y, Ly ), contrary to our indirect assumption. Thus one of (7E1, 2, 3) holds. It follows by [IA , 4.9.1], [III11 , 6.9], and [III17 , 11.24] that xy ∈ xLy ⊆ xG . On the other hand, as m2 (Q) = 1 and Z(K) = 1 = Z(Ly ), Cx has no component isomorphic to Ly , so y ∈ xG . Thus, y is weakly closed in x, y with respect to G. Since K ∼ = P Sp6 (q), y is 2-central in Cx . By our choice of T , y ∈ Z(T ). Then as m2 (Q) = 1, Ω1 (Z(T )) = x, y , by [III17 , 8.8e]. As y is weakly closed in x, y with respect to G, it follows that y is 2-central in G. As G is simple, [III8 , 6.3] then yields x ∈ [Cy , Cy ], completing the proof of (7F). But by (7E), x induces a non-inner automorphism on Ly . Moreover, if Ly is a symplectic group, then Out(Ly ) is abelian by [III17 , 7.2a], while if Ly ∼ = (q), then x induces a graph automorphism on L , and so the image of x in SL± y 4 Out(Ly )/ Outdiag(Ly )ΦLy ∼ = ΓLy ∼ = Z2 is nontrivial. Therefore (7F3) and [III8 , 6.10] imply that Ly  Cy , and more specifically that there exists a Cy -conjugate J of Ly such that J = Ly and one of the following holds: (1) J x = J; or (7G) (2) x induces an outer automorphism on J, and if Ly ∼ = SL± 4 (q), then x induces a graph or graph-field automorphism on J. We use this condition to contradict the fact that C := CJJ x (x) ≤ CG (x, y, L ) =: C0 = CCx (y L). Note that by [III17 , 13.5], CAut(K) (L) equals the image of K1 ∼ = SL2 (q) in Aut(K), with y ∈ K1 . On the other hand, x is the unique involution of C(x, K). Therefore C0 /O2 (C0 ) is the direct product of two groups of 2-rank 1, one of which is isomorphic to SL2 (q) and contains the image of y, and the other of which contains the image of x. Furthermore, x ∈ JJ x since when J = J x , x maps to a nontrivial element of Out(J). Therefore y is the unique involution of C. However, J ∼ = Ly . If J x = J then C contains a central quotient of J, which in view of the isomorphism type of Ly clearly has 2-rank larger than 1. This is a contradiction, so x normalizes J. But then by [III17 , 12.14], CJ (x) is not isomorphic  to SL2 (q), again a contradiction. This completes the proof of the lemma. Lemma 7.3. Suppose that n ≥ 3. E(CG (x, y )) of its isomorphism type.

Then L is the unique component of

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

64

Proof. As L ∼ = Sp2n−2 (q) with n > 2, m2 (L) > 1. Let L∗ be any component of E(CK (x, y )) isomorphic to L. By L2 -balance, E(CG (x, y )) = E(CCx (y)) = E(C(x, K))E(CK (y)) with [E(C(x, K)), E(CK (y))] = 1, and so L∗ ≤ E(C(x, K)) or L∗ ≤ E(CK (y)). The first case is impossible as m2 (C(x, K)) = 1, and the second implies that L∗ = L, since E(CK (y)) = J ∗ L with J ∼ = SL2 (q) (or if q = 3,  E(CK (y)) = L). The lemma follows. Lemma 7.4. We have n = 2. Proof. Suppose false. By Lemma 7.2, Ly = L, so Lxy > L and Lxy is quasisimple by (1C). Then x acts on Lxy nontrivially, as does y. As y ∈ L = Ly with y the unique minimal normal subgroup of CK (y) = K1 L, K1 L  CCxy (y) and CK1 L (Lxy ) = 1. As K1 ∼ = SL2 (q) and L ∼ = Sp2n−2 (q), it follows by [III17 , 10.57] ∼ (q) K. By Lemma 1.2b, (xy, Lxy ) ∈ J∗2 (G), and so P Sp and (7D) that Lxy ∼ = = 2n we have symmetry between (x, K) and (xy, Lxy ). Choose any y1 ∈ L such that y1 is the involution in a long root SL2 (q) subgroup of L. Then y1 has two eigenvalues −1 on the natural K-module, so y1 ∈ y K . Then by Lemma 7.3, CG (x, y1 ) has a unique component L1 ∼ = Sp2n−2 (q), and L1 is a r component of CG (y1 ). Note that O (L1 ∩ L) is the unique (solvable) component  of CG (x, y, y1 ) with yy1 ∈ O r (L1 ∩ L). Moreover, y ∈ K1 ≤ L1 . By symmetry we obtain a unique component L∗1 ∼ = Sp2n−2 (q) of CLxy (y1 ) with analogous properties; in particular L∗1 is a component of CG (y1 ). Moreover,   O r (L∗1 ∩ L) = O r (L1 ∩ L) by the uniqueness asserted at the end of the last  paragraph. (Note that xy, y, y1 = x, y, y1 .) Let H = O r (L∗1 ∩ L). Therefore L1 ∗ and L1 both equal the subnormal closure of H in CG (y1 ), so L1 = L∗1 . However, by construction [L1 , x] = 1 and [L∗1 , xy] = 1. Therefore L1 is centralized by x, xy and hence by y. Thus L1 is a component of CG (x, y, y1 ) with y1 ∈ L1 . Hence L1 ≤ K; but CK (x, y, y1 )’s nonsolvable composition factors are all isomorphic to L2 (q) or P Sp2n−4 (q), and none are isomorphic to  P Sp2n−2 (q) ∼ = L1 /Z(L1 ). This contradiction completes the proof. Thus for the remainder of this section, we have

(7H)

(1) K ∼ = Ω5 (q), q > 3 (as K ∈ G2 ); and = P Sp4 (q) ∼ (2) CK (y) = L, t = LLt t , t2 = 1, L ∼ = SL2 (q), [L, Lt ] = 1, and Z(L) = Z(Lt ) = y .

By [III17 , 11.15b], we may choose t not only to satisfy (7H) but also so that (7I)

t ∈ yK .

We proceed to a contradiction via a sequence of lemmas. Note that results proved for L will yield analogues for Lt , by conjugation. Lemma 7.5. If (w, I) is any long pumpup of (y, Ly ) or (xy, Lxy ), then I/O2 (I) is isomorphic to a member of the set S defined by S = {SL2 (q m ), L3 (q m ), U3 (q m ), m ≥ 1, P Sp4 (q)}. Proof. Let u ∈ {y, xy} and take a pumpup chain (u, Lu ) = (u0 , L0 ) < (u1 , L1 ) < · · · < (uN , LN )

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7. THE CASE K ∼ = P Sp2n (q) = Cn (q)a

65

from (u, Lu ) to a 2-terminal pair (uN , LN ), passing through (un , Ln ) = (w, I) for some 0 ≤ n ≤ N (see [IG , 6.10]). Suppose that the lemma fails and choose a counterexample with n minimal. Suppose that for all i = 0, . . . , N , (7J)

(ui , Li ) ∈ C2 ∪ T2 ∪ G72 ∪ G82 .

Since G has no 2-Thin Configuration, there is then a first r > 0 such that  SL2 (q m ) for any m. At that point, by [III11 , 13.5] and [III10 , Cor. Lr /O2 (Lr ) ∼ = m 7 3.6], Lr /O2 (Lr ) ∼ = L± 3 (q ) for some m. In particular Lr /O2 (Lr ) ∈ G2 , which ± mi ∼ implies together with (7J) and [III11 , 1.2] that Li /O2 (Li ) = L3 (q ), mi ≥ m, for all i ≥ r. Thus (w, I) has the required form in this case. Suppose then that for some i, Li /O2 (Li ) ∈ C2 ∪ T2 ∪ G72 ∪ G82 . By (7B), d2 (G) = 6. Thus by the definition of d2 (G), Li /O2 (Li ) ∈ G62 . Consequently LN /O2 (LN ) ∈ G62 by [III11 , 1.2]. Hence (uN , LN ) ∈ J2 (G). As (x, K) ∈ J∗2 (G) we conclude from [III11 , 12.4] and Lemma 1.2 that (7K)

F(Lk /O2 (Lk )) ≤ F(LN /O2 (LN )) ≤ F(K)

for all 1 ≤ k ≤ N . In particular f (Lk /O2 (Lk )) ≤ f (K) = q 4 , but  Sp4 (q m ), m ≥ 1, or G2 (q m ), m ≥ 1. (7L) Lk /O2 (Lk ) ∼ = Indeed in (7L), the second-order term in the definition of F [III7 , 2.2] rules out G2 (q), and our assumption that Proposition 7.1 fails rules out Sp4 (q) (with the help of Lemma 1.2b). Likewise if Lk /O2 2 (Lk ) ∼ = P Sp4 (q) for some k, then by [III11 , 12.4] and (7L), Lj /O2 2 (Lj ) ∼ = P Sp4 (q) for all k ≤ j ≤ N . Therefore we may assume that  P Sp4 (q). (7M) Ln−1 /O2 2 (Ln−1 ) ∼ = m By the minimality of n, Ln−1 /O2 (Ln−1 ) ∼ = SL2 (q m ) or L± 3 (q ) for some m ≥ 1. Accordingly, we apply [III17 , 10.51b] or [III17 , 10.51c] to the pumpup (un−1 , Ln−1 ) < (w, I). Given the restrictions (7K) and (7L), and our assumption 1 that I/O2 (I) ∈ S, we conclude that I/O2 (I) ∼ = Sp4 (q 2 ) with Ln−1 /O2 (Ln−1 ) ∼ = SL2 (q). Part d2 of the same lemma, together with (7L) and [III11 , 12.4], then 1 implies that for i > n, Li /O2 (Li ) ∼ = I/O2 (I) or Sp8 (q 4 ). Accordingly

LN /O2 (LN ) ∼ = Sp2k (q k ) with k = 2 or 4. 1

If k = 4, then F(LN /O2 (LN )) = F(K), so by Lemma 1.2, (uN , LN ) ∈ J∗2 (G) and Proposition 7.1 holds with (x , K  ) = (uN , LN ), contrary to our assumption that the proposition fails. Therefore k = 2. Since (uN , LN ) ∈ J2 (G) = J22 (G) (see [III7 , 3.3]), [III12 , Theorem 1.2] yields m2 (C(uN , LN )) = 1, whence uN ∈ LN and LN  CG (uN ). As Li /O2 (Li ) ∼ = LN /O2 (LN ) for all n ≤ i ≤ N , it follows that un = un+1 = · · · = uN . So we may assume that n = N and I = LN . By [III17 , 10.51b], LN −1 /O2 (LN −1 ) ∼ = SL2 (q) with uN ∈ LN −1 , and

(7N)

(7O)

uN −1 ∈ [CG (uN ), CG (uN )].

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66

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Let U ∈ Syl2 (CG (uN )) with uN −1 ∈ U . By [III17 , 11.17a], Z(U ) has trivial image in Aut(LN /O2 (LN )), whence Z(U ) ≤ C(uN , LN ). In view of (7N), Ω1 (Z(U )) = uN . Therefore uN is 2-central in G. Moreover, by [III17 , 11.24], uN−1 is CG (uN )-conjugate to uN−1 uN and setting CG (uN ) = CG (uN )/O2 (CG (uN )), CAut(LN ) (LN −1 ) is of order 2, generated by the image of uN −1 . Hence if we put

= H/O2 (H), then H

= 

0 , where H := C(uN −1 , LN −1 )∩CG (uN ) and H uN −1 × H

u

N is the unique involution of H0 and H0 = H ∩ C(uN , LN ). As uN ∈ LN −1 , it follows that uN ∈ Syl2 (Z ∗ (L2 (CG (uN −1 )))). Let V ∈ Syl2 (CG (uN −1 )) with uN ∈ V . Then uN ∈ Z(V ) and so we may assume that U was chosen so that V ≤ U . Then Ω1 (Z(V )) ≤ H by [III17 , 11.17a], so Ω1 (Z(V )) = uN , uN −1 . As uN ∈ L2 (CG (uN )), uN ∈ uG N −1 , and so uN is weakly closed in Ω1 (Z(V )). Hence by (7O) and [III8 , 6.3], uN −1 ∈ [G, G], contradicting the simplicity of G. The proof is complete.  Now that the isomorphism types of pumpups of Ly and Lxy have been restricted, our next goal is to prove that Ly = L, which will imply that Lt is also a component of Cy . It will also imply, as we shall see, that x, y ∈ Syl2 (CG (LLt )). Lemma 7.6. Either Ly = L or Ly ∼ = Lty ∼ = SL(2, q 2 ). Proof. As Ly is isomorphic to a member of S, by Lemma 7.5, and as y ∈ L ≤  Ly , the result is immediate. Lemma 7.7. We have Ly = L. Proof. Suppose false. Then by Lemma 7.6, Ly ∼ = Lty ∼ = SL(2, q 2 ). The setup (2F) holds, with x, y, K, L, and Ly playing the roles of w, y0 , I, L1 , and M1 there, and with (2F5b) holding. By Proposition 2.7b, y ∈ Syl2 (CG (Ly Lty )), so that F ∗ (CG (y)/O2 (CG (y))) ∼ = Ly Lty . We reach a contradiction by arguing that (y, Ly ) is 2-terminal in G, so that G has a 2-Thin Configuration, which is prohibited by the hypothesis of Theorem C∗7 . Since y ∈ Ly it suffices by [III8 , 2.7] to show that for any involution w ∈ CG (Ly ) with w = y, Ly has a trivial pumpup in CG (w). Suppose that w is such an involution; by [III17 , 7.7], w induces an involutory field automorphism on Lty . Let M0 = CLty (w), and let Mw be the pumpup of Ly in CG (w). Suppose that Mw is a nontrivial pumpup of Ly . Then y ∈ Z ∗ (Mw ). But Mw /O2 (Mw ) is isomorphic to a member of S by Lemma 7.5, and M0 Ly ∼ = SL2 (q) ∗ SL2 (q 2 ) embeds faithfully into  Mw /O2 (Mw ), a contradiction. The proof is complete. This has many consequences. Lemma 7.8. The following conclusions hold: (a) Lxy = E(CG (xy)) ∼ = K; (b) x ∈ Syl2 (CG (K)); (c) xy ∈ Syl2 (CG (Lxy )); (d) x, y ∈ Syl2 (CG (LLt )); (e) y is weakly closed in x, y with respect to G; and either y or x, y is a Sylow 2-center in G. Proof. By (1C) and Lemma 7.7, Lxy > L, whence [y, Lxy ] = 1. Therefore by L2 -balance, LLt ≤ Lxy with LLt ∩ Z(Lxy ) = 1. But Lxy is isomorphic to a member of S, and so Lxy ∼ = K.

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7. THE CASE K ∼ = P Sp2n (q) = Cn (q)a

67

Now with Lemma 7.7, we see that the setup (2F) holds, with x, y, K, L, and Ly playing the roles of w, y0 , I, L1 , and M1 there. (The previous paragraph implies that (2F5c) holds.) Therefore Proposition 2.7a gives conclusion (d), and as CG (Lxy ) ≤ CG (LLt ), this implies |C(xy, Lxy )|2 = 2, whence Lxy = L2 (CG (xy)) and (a) and (c) hold. Proposition 2.7d implies that (b) and (e) hold.  In the next series of lemmas we rule out the case that there exists f ∈ Cx such that f induces an involutory field automorphism (7P) on K. Lemma 7.9. Assume that (7P) holds. Then f may be chosen so that f 2 ∈ x , and then the following conditions hold: (a) f 2 = x; (b) y G ∩ LLt x = {y}; (c) |S : T | = 2 and y = Z(S), where T ∈ Syl2 (CG (x, y)) and T ≤ S ∈ Syl2 (CG (y)) ⊆ Syl2 (G); and (d) For some w ∈ y G ∩ Cx , K w ∼ = Inndiag(K) ∼ = SO5 (q). Proof. By Sylow’s theorem we may choose f ∈ T , so that f 2 ∈ Q = x . Furthermore, we may choose f so that y, t ≤ CK (f ) and y is 2-central in CK (f ). Now (2F) holds with x, y, K, and L in the roles of w, y0 , I, and L1 = M1 there, and we apply Proposition 2.7c to obtain (a) and (d). We may fix w ∈ I2 (T ) satisfying (d). Let H0 = O2 (Cx )K w , where w is as in (d), so that H0 x  Cx . Recall that R = T ∩ K, so that Rw = T ∩ H0 ∈ Syl2 (H0 ). As Out(K)/OutH0 (K) is a cyclic group (of images of field  automorphisms), O 2 (Cx )/H0 x ∼ = T /R w, x is cyclic; moreover, if H0  Cx ,  2 ∼ then O (Cx )/H0 = T /R w is cyclic as f 2 = x. Next, suppose that (b) fails, so that with Lemma 7.8e, y g ∈ (LLt − y ) x for some g ∈ G. As all involutions of LLt − y are K-conjugate, and f induces field automorphisms on L and Lt , we may assume that [f, y g ] = 1. Thus x is a square in CG (y g ). Now CG (y g ) ∼ = CG (y) has exactly two components, both isomorphic to SL2 (q), so x normalizes them both; and then every component of L2 (CCG (yg ) (x)) contains y g . Therefore by L2 -balance, y g lies in every component of L2 (CK (y g )). But this last group is isomorphic to L2 (q) by [III17 , 11.15cf] and thus is simple. This is a contradiction, so (b) holds. By (a), no involution of CG (x) induces an involutory field automorphism on K. In particular, the coset H0 x f contains no involution, so f is of minimal order in the coset R w, x f . Suppose next that T = S. Since T /R w, x is cyclic, the previous paragraph and the Thompson Transfer Lemma (see [IG , 15.15]) imply that there is g ∈ G such that f g ∈ R w, x , f g is extremal in T , and CT (f )g ≤ T . Let T0 = CT (f ). By [III17 , 11.17c], y ∈ [T0 , T0 , T0 ] and [T, T, T ] ≤ LLt . Thus y g ∈ y G ∩ LLt = {y} by (b), so g ∈ Cy . But CG (y) = O2 (CG (y))NCG (y) (x, y ) = O2 (CG (y))CCG (y) (x) by Lemma 7.8d and the fact that S = T . Therefore f g ∈ f Cx . But this is absurd as f g ∈ R w, x ≤ K w, x with f ∈ K w, x  Cx . Therefore T < S. By Lemma 7.8de, (c) holds, proving the lemma.

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

68

By [IA , 4.5.1], the involutions in K w ∼ = Inndiag(K) are partitioned into four K w -conjugacy classes, all of which are Aut(K)-conjugacy classes as well. Representatives of these classes are (7Q)

y, u ∈ I2 (LLt ) − {y}, w, and v,

where E(CK (w)) and E(CK (v)) are isomorphic, in some order, to L2 (q) and L2 (q 2 ). By L2 -balance, and since w ∈ y G , E(CK (w)) embeds in L2 (Cy ) = E(Cy ) = LLt , so E(CK (w)) cannot be isomorphic to L2 (q 2 ). Thus, (7R)

E(CK (w)) ∼ = L2 (q), E(CK (v)) ∼ = L2 (q 2 ), and v ∈ y G .

Lemma 7.10. Assume (7P). Then the following conditions hold: (a) y G ∩ Cx = y K ∪ wK ; and (b) xG ∩ Cy = {x, xy} ∪ (xv)Cy . Proof. We argue that it suffices to show that (7S)

xv ∈ xG and xw ∈ (xu)G .

Namely, by Lemma 7.9a, no involution of Cx induces a nontrivial field automorphism on K. Since Cx /O2 (Cx )K w, x maps injectively into the cyclic subgroup of Out(K) consisting of images of field automorphisms, it follows with the help of [IA , 4.9.1] that every involution of Cx lies in K w, x . We have seen in Lemma 7.9b and (7R) that y G ∩ K w is the union of the classes y K and wK . Moreover, by Lemma 7.9bc, y G ∩ K w x contains none of x, xy, or xu. Therefore (7S) will imply that xv ∈ y G and xw ∈ y G , and by (7Q) we will have y G ∩ K w x = ∅, whence (a). Now the number of Cx -conjugacy classes in (a) must equal the number of Cy -conjugacy classes in (b), since both numbers equal the number of G-orbits on commuting pairs in xG × y G . Note also that as O2 (Cy ) x, y  Cy , xv and x are not Cy -conjugate. Thus (7S) will imply (b) as well. By Lemma 7.9c, we may fix g ∈ S − T , and y = Z(S). As x, y = CS (LLg )  S we have xg = xy. Let C x = Cx /C(x, K), and identify C x with AutCx (K). Let A = O2 (Cx )K w, f , the full preimage in Cx of the four-subgroup of Out(K) generated by all involutions  in Out(K). Thus A = Inndiag(K) f . t ∼ K ∼ 4 3 By [III11 , 6.5ab] there  exists E ≤ R with E = E2 , E ∩ LL = E2 , E ∩ y = K ∼ ∼ AutK (E) = Σ5 , the last since q is a square. f = E25 , and ∅ = E ∩u , CA (E) = E 

Let E0 = E ×f , so that I2 CG (E0 ) = CA (E). Let E1 = E0g ≤ T g = T . Since ∼ E1 / x ∼ Φ(E0 ) = x and xg = xy, E 1 = = E23 ×Z4 . Moreover, Φ(E1 ) = y ∈ K so E 1 ≤ A. But CA (E) does not contain a copy of E 1 , so Ω1 (E 1 ) is not  K-conjugate, or even Aut(K)-conjugate, to E. (Note that A = Inndiag(K) f  Aut(K).) Consequently by [III11 , 6.5a], Ω1 (E 1 ) ≤ K. As xg = xy = y ∈ K, (7T)

E g ≤ K x .

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7. THE CASE K ∼ = P Sp2n (q) = Cn (q)a

69

Now g ∈ Cy ≤ NG (LLt ), so (E ∩LLt )g ≤ LLt . It follows that (E −LLt )g ⊆ K x w (note that all involutions of Cx lie in O2 (Cx )K w, x ). As AutK (E) is irreducible on E, E − LLt contains some y1 ∈ y K and u1 ∈ uK . Therefore we may replace w by y1g and v by ug1 and assume that w = y1g and v = ug1 . By [III11 , 6.5c], y K ∩ E − LLt consists of four elements of y K and four elements of uK ; moreover yu1 ∈ y K . Therefore yv ∈ wK , and yw ∈ v K . It follows that −1 xw ∈ (xyv)K and xv ∈ (xyw)K . But (xyv)g = xu1 ∈ y G , so xw ∈ y G . Likewise −1 (xyw)g = xy1 ∈ (xy)K ⊆ xG , so xv ∈ xG . This proves (7S) and the lemma.  Now let H = O2 (Cy )LLt t, w , so that R w ∈ Syl2 (H) and x ∈ H. Lemma 7.11. Assume (7P). Then the following conditions hold: (a) H  Cy ; (b) xG ∩ Cy ⊆ Hx; and (c) Cy /H x is not cyclic.   Proof. Lemma 7.10 implies that y G ∩ Cx = y K ∪ wK = K w . There G fore, y ∩ CG (x, y ) ≤ CKw (y) = LLt t, w . As both t and w lie in y G , with LLt = [w, LLt ], in fact  G y ∩ CG (x, y ) = LLt t, w . As x, y O2 (Cy )  Cy , Cy = O2 (Cy )NCy (CCy (x, y )) = O2 (Cy )NCy (CG (x, y )),  which normalizes O2 (Cy ) y G ∩ CG (x, y ) = H, proving (a). Now H x = O2 (Cy ) x, y H  Cy , so Hx ∈ Z(Cy /H). As xG ∩Cy = {x, xy}∪ (xv)Cy by Lemma 7.10, with v ∈ CKw (y) ≤ H, (b) follows. Finally suppose that Cy /H x is cyclic. Then Cy /H is abelian, so x ∈ [Cy , Cy ]. However, as y is weakly closed in Z(T ) = x, y , x ∈ [Cy , Cy ] by [III8 , 6.3]. This contradiction completes the proof of (c) and the lemma.  This allows us to prove Lemma 7.12. Assume (7P). Then Cy = CG (x, y )CCy (L). Proof. Suppose false and for this proof set C y = Cy /O2 (Cy ). As x, y  C y , it follows that CC y (L) ≤ CG (x, y ); moreover as K is simple, O2 (Cx ) ≤ O2 (Cy ). But CCG (x,y) (L) ≤ CCx (L), and CCx (L)/O2 (Cx ) x embeds in CAut(K) (L), which by [III17 , 13.5] equals the image of Lt . Thus CCx (L) ≤ O2 (Cx )Lt x ≤ H x , so CC y (L) ≤ H x , and CCy (L) ≤ H x .

(7U)

Now Cy stabilizes the set {L, L }, and as t ∈ H we have Cy = NCy (L)H. As H x  Cy , and L ≤ H, it follows with (7U) that Cy /H x ∼ = AutC (L)/AutH (L) ∼ = OutC (L)/OutH (L). t

y

y

But by [III17 , 7.2a], Out(L) = Outdiag(L) × F , where F is cyclic, and as w ∈ H, OutH (L) contains Outdiag(L). Thus Cy /H x is a subquotient of F and so must be cyclic. However, this contradicts Lemma 7.11c, so the proof is complete. 

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70

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

We next prove Lemma 7.13. Assume (7P). Suppose that U ≤ S, y = Ω1 (Z(U )), g ∈ G, and U g ≤ Cy . Assume also that U is not embeddable in D8 . Then g ∈ Cy . Proof. We show that there exists N  Cy such that Cy /N is embeddable in D8 and y G ∩ N = {y}. It will follow, as U g is not embeddable in D8 , that y g = Ω1 (Z(U g )) ≤ N , whence y g = y, as desired. Let N0 = O2 (Cy )LLt x . By Lemma 7.9b, y G ∩ N0 = {y}. As O2 (Cy ) x, y = CG (LLt ), Cy /N0 embeds in Out(LLt ) ∼ = Out(L)  Z2 . Let M be the subgroup of Out(LLt ) which is the direct product of the field automorphism subgroups of Out(L) and Out(Lt ), and let N/N0 be the preimage of M in Cy /N0 . Then Out(LLt )/M ∼ = D8 , so Cy /N embeds in D8 . Notice that it also follows that every noncyclic chief factor of Cy of even order lies in LLt . It remains to show that y G ∩N −N0 = ∅. Accordingly let s = y g ∈ y G ∩N −N0 . From the construction of N , s induces a trivial automorphism or an involutory field automorphism on L. Then CL (s) has a noncyclic chief factor of even order, which lies in (LLt )g by the preceding paragraph, whence y ∈ (LLt )g ≤ N0g . By the second paragraph, y = y g = s, completing the proof.  Lemma 7.14. Assume (7P). Suppose that s = y g ∈ y G ∩ S − {y} and s normalizes L. Then s induces outer diagonal automorphisms on both L and Lt . Proof. Suppose that s induces an inner automorphism or field automorphism on L. Let U ∈ Syl2 (CL (s)); accordingly, U contains Z8 or Q8 and in any case U is not embeddable in D8 . −1 But U g ≤ Cy , whence g ∈ Cy by Lemma 7.13, contradicting the assumption that s = y. Thus s induces an outer diagonal automorphism on L ∼ = SL2 (q) and  similarly on Lt . Lemma 7.15. Assume (7P). Then there is no b ∈ CCy (L) inducing an involutory field automorphism on Lt . Proof. Suppose to the contrary that such a b exists, and choose b of minimum order. Then b is a 2-element and without loss we may take b ∈ S. Then b2 ∈ CS (LLt ) = x, y . By [III17 , 13.5], no element of CT (L) = CS (L x ) induces an involutory field automorphism on Lt , so x, b ∼ = D8 and b is an involution.  1 Write L∗0 = CLt (b) ∼ = SL2 (q 2 ). By (solvable) L2 -balance LO 2 (L∗0 )   CG (y, b) normalizes every 2-component of CG (b); but by [III17 , 10.40] it cannot act faithfully on any such component. As L ≤ L2 (CG (b)) by L2 -balance, we must have y ∈ Z ∗ (L2 (CG (b))) and then y ∈ Syl2 (Z ∗ (L2 (CG (b)))), so y ∈ Z ∗ (CG (b)). We claim that y G ∩ CG (b) = y CG (b) . Suppose that s ∈ y G ∩ CG (b, y ). As b centralizes L but not Lt , s must normalize L and Lt . Thus, if s = y, then s induces an outer diagonal automorphism on Lt by Lemma 7.14. But then Lt s, b is an envelope of Lt ∼ = SL2 (q) mapping to the subgroup of Out(Lt ) generated by its involutions; such an envelope does not exist, by [III17 , 7.3]. This contradiction proves that y G ∩ CG (b, y ) = {y}. Since CG (b) = O2 (CG (b))CG (b, y ), it follows that y G ∩ CG (b) = y O2 (CG (b)) . But now since CG (b) is transitive on y G ∩ CG (b), Cy must be transitive on G b ∩ Cy . Let J = O2 (Cy )CG (x, y ), so that T ∈ Syl2 (J). As x, y O2 (Cy )  Cy and [b, x] = 1, bG ∩ S = bCy ∩ S ⊆ S − T . Thus by the Thompson Transfer Lemma, b ∈ [G, G], contrary to the simplicity of G. The lemma follows. 

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7. THE CASE K ∼ = P Sp2n (q) = Cn (q)a

71

At last we can prove Lemma 7.16. There is no element f ∈ Cx with f 2 ∈ x such that f induces a field automorphism on K. Proof. Suppose false and continue the argument of the last seven lemmas. Set

 C0 = CCy (L)CCy (Lt ) = CCy (L)Cy  Cy .

Since O2 (Cy )CG (x, y )  Cy there exists, by Lemma 7.12, v1 ∈ CCy (L) ∩ S − T . The subgroup of Cy inducing inner automorphisms on LLt is O2 (Cy )LLt x , so v1 induces an outer automorphism on LLt and hence on Lt . In view of Lemma 7.15 and [III11 , 6.2], v1 induces an outer diagonal automorphism on Lt . On the other hand, the image of CCG (x,y) (L) in Aut(Lt ) lies in the image of CAut(K) (L) in Aut(Lt ), and the latter image is Inn(Lt ), by [III17 , 13.5]. It follows that OutC0 (Lt ) = OutCCy (L) (Lt ) is generated by the image of v1 . Thus OutC0 (Lt ) = Outdiag(Lt ). In particular f ∈ C0 . Now C/C0 ∼ = CG (x, y )/CC0 (x, y ) by Lemma 7.12. As f ∈ C0 there exists N1  Cy such that C0 t, w ≤ N1 and Cy /N1 is a cyclic 2-group whose involution is f N1 . As w ∈ N1 , also v ∈ N1 and so by Lemma 7.10b, xG ∩ Cy ⊆ N1 . By [IG , 15.15] and the simplicity of G, it follows that there is g ∈ G such that f g is an extremal conjugate of f , f g ∈ N1 , and CS (f )g ≤ S. But CS (f ) contains the quaternion group U = CL∩S (f ), whose involution is y, so by Lemma 7.13, g ∈ Cy . As f ∈ N1  Cy but f g ∈ N1 , this is a contradiction. The lemma follows.  Corollary 7.17. Let Hx = O2 (Cx ) × x × K. Then Cx = Hx F w where F induces a cyclic group of odd order field automorphisms on K, w ∈ CT (t, y)∩NT (L), w2 ∈ K x , and one of the following holds: (a) w = 1; or (b) CKw,x (K) = x and the image ω of w in Out(K) is either a generator of Outdiag(K), or the product of such a generator with the image of a field automorphism of K of order 2. Moreover, x ∈ [Cx , Cx ]. Proof. This is clear from Lemma 7.8b, Lemma 7.16, and [III17 , 7.2] and  [III17 , 11.16]. Lemma 7.18. We have xy ∈ xCy and y is 2-central in G. Moreover, x ∈ [Cy , Cy ]. Proof. By Lemma 7.8e, y G ∩ x, y = {y}, and if xy ∈ xCy , then x, y is a Sylow 2-center in G. In particular x is 2-central in G. By Burnside’s Lemma [IG , 16.2] and the fact that y is weakly closed in x, y , each involution of x, y is weakly closed in x, y . By [III8 , 6.3], xy ∈ [Cx , Cx ], but Corollary 7.17 implies that xy ∈ [Cx , Cx ]. This contradiction proves that xy ∈ xCy . By [III8 , 6.3], the final statement of the lemma holds.  Set C y = Cy /O 2 (Cy ). Lemma 7.19. We have [C y , C y ] = x = 1.

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72

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Proof. By Lemma 7.18, x ∈ [C y , C y ]. By Lemma 7.8d, O2 (Cy )x, y  Cy . By Lemma 7.18, O2 (Cy )CG (x, y ) has index 2 in Cy . It follows from Corollary y = Cy /O2 (Cy )LLt , then C y has a normal cyclic subgroup F 7.17 that if we set C y /F is a 2-group of order at most 8|CK (y) : LLt |2 = 16. of odd order, and C y ∼ =C In particular x = 1. If |C y | = 8, or if |Z(C y )| > 2, then |[C y , C y ]| ≤ 2 and we are done. Suppose then that |C y | = 16, |Z(C y )| = 2 and |[C y , C y ]| = 4, so  that w = 1 in Corollary 7.17. Let D0 = x, w, t . Since [w, t] = 1 = [w, x] and x, t ∼ = x, t ∼ = E22 , D0 is noncyclic abelian. Moreover, D0 = T . Then for s ∈ C y − D0 , CD0 (s) = Z(C y ) ∼ = Z2 . But then |CC y (s)| = 4 so C y is of maximal class by [IG , 10.24], contrary to the structure of D0 . The lemma is proved.  Lemma 7.20. There is v ∈ S such that v normalizes L and Lt , Cy = O2 (Cy )CG (x, y ) v , v 2 ∈ LLt x, w , tv = ty, xv = xy, and [txy, v] = 1. Proof. Choose v ∈ S − T . Replacing v by vt if necessary we may assume that v normalizes L and Lt . As x, y  S, we have |S : T | = 2, so v 2 ∈ NT (L) ≤ LLt x, w . Now Lemma 7.19 implies that [S, S] ≤ LLt x , so v normalizes S ∩ LLt x and S ∩ LLt x, t . Since t interchanges the two direct factors S ∩ L/ y and S ∩ Lt / y of S ∩ LLt / y ∼ = S ∩ LLt x /x, y , we may in fact choose v ∈ NS (x, y, t ). If tv = txy, then as xv = xy, (tx)v = t. But t ∈ y K and xt ∈ (xy)K ⊆ xG , a contradiction. Hence tv ∈ {t, ty}. Suppose finally that v ∈ CG (t). As t ∈ y K , we know that O2 (CG (t))x, t is normal in CG (t). Hence xy = xv ∈ O2 (CG (t))x, t . But x, y, t ∼  = E23 , a contradiction. Thus tv = ty and the proof is complete. Set L12 = E(CLLt (t)) ∼ = L2 (q). Lemma 7.21. The following conclusions hold: (a) There exists w satisfying Corollary 7.17b such that the image of w in Out(L12 ) is either a generator of Outdiag(L12 ), or the product of such a generator with a field automorphism of L12 of order 2; (b) There exists v satisfying the conclusions of Lemma 7.20 such that v induces an involutory field automorphism of L12 and v 2 ∈ y ; and (c) q ≡ 1 (mod 8). Proof. Let v be as in Lemma 7.20. If w = 1 in Corollary 7.17, then Cy = O2 (Cy )LLt x, t, v , with [t, x] = 1 and [t, v] = y = [x, v]. Then C y is abelian, contrary to Lemma 7.19. Hence Corollary 7.17b holds. Given the action of w on K and the fact that w centralizes t, y, x , it follows from [III17 , 13.5] that the action of w on L12 is as claimed in (a). Now NG (L) = O2 (Cy )LLt x, w, v . As v ∈ S normalizes t, y , v 2 ∈ CLLt x,w (t) = L12 y, x, w . By [III17 , 7.2a] the image of w in Aut(L12 ) is not a square in Aut(L12 ), whence v 2 ∈ x, y × L12 . Since xv = xy, in fact v 2 ∈ y × L12 . By Lemma 7.8d,

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7. THE CASE K ∼ = P Sp2n (q) = Cn (q)a

73

x, y ∈ Syl2 (CG (LLt )), and so x, y, t = CS (L12 ). The image of v, w in Out(L12 ) is therefore a four-group, in which the image of w is not the image of a field automorphism. Hence, there is v0 ∈ v(S ∩ L12 w ) such that v0 induces an involutory field automorphism on L12 . As L12 w centralizes t, x, y , v0 and v have the same action on t, x, y , so v0 satisfies the conditions of Lemma 7.20. Thus (b) holds with v0 in place of v. Finally, as L12 admits an involutory field automorphism, q is a square so (c) holds, completing the proof.  Lemma 7.22. L12 is a component of CG (txy). Proof. By [III17 , 11.15], ty is not a 2-central involution of K, and we have that L12 = E(CCx (txy)) = E(CCy (txy)). Also, as t = y h for some h ∈ K, we have L2 (CG (t)) = E(CK (t)), and we see that E(CCxy (txy)) = E(CCG (t) (xy)) = E(CE(CK (t)) (y)) = L12 . Now let H be the subnormal closure of L12 in L2 ((CG (txy))). Then H is x, y -invariant and E(CH (z)) = L12 for all z ∈ x, y # . It follows first that H is quasisimple, and then, if H = L12 , [III11 , 1.16] and Lemma 7.21c yield that q = 9 and H ∼ = Σ6 . However, by [III17 , 11.15c], = A10+4k , k ≥ 0. Thus AutCG (txy) (L12 ) ∼ AutCCx (txy) (L12 ) contains the image of L12 w, v , which contains P GL2 (9). Hence it does not embed in Σ6 , a contradiction. Thus H = L12 , completing the proof.  Using [III17 , 11.15d] and Lemma 7.21c, we choose an involution u ∈ CK (ty) with tu = y, y u = t, and [L12 , u] = 1. We set ρ = uv. Then tρ = y v = y, xρ = xv = xy, and y ρ = tv = ty. Hence ρ ∈ CG (txy) and ρ transitively permutes t, y # . Also, we choose E r ≤ CL12 (v) with E r ∼ = E22 . We set = A4 and E ∼ A = x, y, t × E ∼ = E25 . Thus, [r, A] = E, [ρ, A] = t, y , and [r, ρ] = 1. Lemma 7.23. The following conclusions hold: (a) A ∈ Syl2 (CG (A)); and (b) NG (A)/CG (A) ∼ = Σ6 . Proof. We have A ∩ K = y, t, E ∼ = E24 and A = (A ∩ K) × x . Then [III11 , 6.5bc] implies that AutCx (A) ∼ Σ , = 5 and the NCx (A)-orbits on (A ∩ K)# are  of lengths 5 and 10, represented by y and yt respectively. Also since O 2 (OutCx (K)) contains no image of a field automorphism by Corollary 7.17, it follows by [III11 , 6.5b] that A ∈ Syl2 (CCx (A∩K)), whence (a) holds. By choice of E, u, v ≤ NG (A). As y ρ = ty, (A ∩ K)# ⊆ y NG (A) . As xv = xy, but txy ∈ xG by Lemma 7.22, it follows that |xNG (A) | = 6, and (b) follows.  We can now reach a final contradiction completing the proof of Proposition 7.1. We have [ρ, r] = 1, and we let ρ1 be a generator of a Sylow 3-subgroup of ρ , so that [ρ1 , r] = 1. We have [A, ρ1 ] = y, t , CA (ρ1 ) = txy E, E = [A, r], and CA (r) = x, y, t . Let ρ1 , r ≤ R0 ∈ Syl3 (NG (A)). Then CA (R0 ) = txy and, by a Frattini argument, there exists s ∈ NG (A) ∩ NG (R0 ) interchanging ρ1 and r modulo CR0 (A), and hence centralizing txy. But then as E = [A, r], E s = [A, ρ1 ] = y, t . So J := Ls12 is an L2 (q)-component of CG (txy) containing y, t . However,

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

74

CCy (L12 ) = O2 (Cy )y, x, t . Hence y, t ∈ Syl2 (J) and q ≡ ±3 (mod 8), contrary to Lemma 7.21a, a final contradiction. 8. The Spinn (q) Cases, n ≥ 7 In this section we continue to assume the assertions and notation of (1A), (1B) and (1C), as well as (1E). We assume that p = 2 and we shall prove the following “Spin Theorem”: Proposition 8.1. Suppose that p = 2 and K/O2 (K) ∼ = Spinn (q) or HSpinn (q) for some odd q, some n ≥ 7, and (when n is even) some sign . Then one of the following holds: u ∼ ∼ (a) n = 16, K ∼ = Spin+ = HSpin16 (q), L ∼ 12 (q), and Ly = Lxy = E7 (q) . (E8 (q) case.) (b) n = 10, K ∼ = Spin10 (q) for some  = ±1, L ∼ = Spin+ 8 (q), D = Z(L), D# ⊆ xG , and AutLu (D) ∼ = AutK (D) ∼ = Z2 for all u ∈ {y, xy}. (E6 (q) case.) # (c) n = 9, K ∼ ⊆ xG , and = Spin9 (q), L ∼ = Spin+ 8 (q), D = Z(L), D ∼ ∼ AutLu (D) = AutK (D) = Z2 for all u ∈ {y, xy}. (F4 (q) case.) (d) n = 8 and K ∼ = HSpin8 (q) ∼ = Ω+ 8 (q). As noted earlier, the E7 (q) case with n = 12 is missing since F(D6 (q)) < F(A7 (q)). We assume that the proposition fails. Note that by [IA , 6.1.4], (8A) O2 (K) = 1 unless K/O2 (K) ∼ = Spin7 (3) and O2 (K) ∼ = Z3 . As usual, m2 (C(x, K)) = 1 by (1C1), so Ω1 (O2 (Z(K))) = x . In particular, it is not possible that K ∼ = Spin+ 4k (q) for any integer k > 1. Notice that if n = 8, then because of (d) we may assume that  = −1. But in that case by [III12 , Def. 1.15], L ∼ = SL2 (q 2 ), a non-level case that has been ruled out by [III9 , Theorem 5, p. 46]. Thus we may assume that n = 8. We assume that n≥9 through the proof of Lemma 8.6. By [III12 , Def. 1.15] and [III17 , 13.4b], either ∼ + (1) x, y = Z(L), L ∼ = Spin+ m (q), and L/ x = Ωm (q), where m is the largest multiple of 4 less than n; or (8B) (2) n is even, n ≥ 10, and y acts on K/Z(K) ∼ = P Ωn (q) like a reflection. In either case, by [IA , 6.2.1] and [III17 , 7.5], y ∼NK (L) yx so Ly ∼ = Lxy . We fix any u ∈ x, y − x , so that Lu > L, i.e., (8C)

[x, Lu ] = 1.

We first prove: Lemma 8.2. Assume n ≥ 9. Then condition (8B1) holds.

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8. THE Spinn (q) CASES, n ≥ 7

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Proof. Suppose that (8B2) holds. Note that y induces a graph automorphism of order 2 on K. By [III17 , 7.4], K is not a half-spin group. Hence K ∼ = Spinn (q). Consequently L = E(CK (y)) ∼ = Spinn−1 (q), with x = Ω1 (Z(K)) ≤ Z(L). As L < Lu , we have x ∈ Z(Lu ), but x ∈ Lu . By [III17 , 10.31e], as n − 1 is odd, n = 10, L ∼ = F4 (q). As CAut(K) (L) is the image of y by = Spin9 (q) and Lu ∼ [III17 , 6.7], m2 (CG (x, y L)) = 2. Therefore m2 (C(u, Lu )) = 1. In particular Lu  CG (u). Using [IA , 4.5.1] we see that x is 2-central in CG (u). Indeed for an appropriate T ∈ Syl2 (CG (u)), the image of Ω1 (Z(T )) in Aut(Lu ) is the image of x , by [III17 , 8.9g], so Ω1 (Z(T )) = x, u . On the other hand, as remarked above, u ∈ {y, xy} is K-fused to ux. As x and u are clearly not conjugate in G, it follows that x is 2-central in G and x is weakly closed in Z(T ) with respect to G. Hence by [III8 , 6.3], u ∈ [CG (x), CG (x)]. But K  CG (x) and u induces a graph automorphism on K, so u ∈ [CG (x), CG (x)], contradiction. The proof is complete.  Thus, L and m are as in (8B1). In particular, as n ≥ 9, we have m ≥ 8. We apply [III17 , 10.64ab] to the pumpup L < Lu . Since x ∈ L  E(CLu (x)) and L/ x ∼ = Ω+ m (q), with m = 4k, one of the following holds: u ∼ (1) L ∼ = Spin+ 12 (q) and Lu = E7 (q) , (8D) + (2) L ∼ = Spin8 (q) and Lu ∼ = Spin± r (q) or HSpinr (q) for some r > 8. The following lemma imposes further restrictions. Lemma 8.3. Assume that n ≥ 9 and let (w, I) ∈ ILo2 (G). (a) If I/O2 2 (I) ∼ = P Ω± r (q) for some r, then r ≤ n; ∼ (b) If I/O2 (I) = E6± (q), then n ≥ 13; and (c) If I/O2 (I) ∼ = E7 (q), then n ≥ 15. Proof. If the lemma is false, then whichever conclusion fails, we have d2 (I) =  6 = d2 (K) and F(I) > F(K), and so Lemma 1.2a is contradicted. We next prove Lemma 8.4. Suppose that n = 11 or 12. Then x ∈ Syl2 (C(x, K)). Proof. Suppose false, so that as m2 (C(x, K)) = 1, x = t2 for some t ∈ C(x, K). By [III17 , 13.4d], there is w ∈ CK (L) such that w2 = x and such that  if we let v = tw and L∗ = E(CK (v)), then L ≤ L∗ ∼ = Spin10q (q). Now D = x, y ≤ L ≤ CG (v), so v normalizes Lu , while centralizing L. We consider first the action of x, v on Lu . Note that by (8D) and Lemma 8.3a, Lu /Z(Lu ) ∼ = P Ω± r (q), 8 < r ≤ n ≤ 12. There must exist v  ∈ {v, xv} such that L is not a component of CLu (v  ). For   # otherwise L  O r (CLu (s)) for all s ∈ v, x , whence L  Γrv,x,1 (Lu ) = Lu by [IA , 7.3.1], a contradiction. Replacing t by t−1 if necessary, we may assume that v  = v, i.e., L is not a component of CLu (v). Given the isomorphism type of Lu , it is evident (by orders, for example) that Lu does not involve L/Z(L) × L/Z(L), and hence by L2 -balance, L < J for some component J of CLu (v). As L is a component of CLu (x), we have [J, x] = 1. equals the subnormal Let L∗v be the subnormal closure  which  ∗ of L in CG (v), closure of L∗ in CG (v) as L∗ = LL . Then J = LJ ≤ L∗v , so [L∗v , x] = 1. By  L2 -balance, L∗ is a component of CL∗v (x). Moreover, L∗ ∼ = Spin10q (q) with x ∈ L∗

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76

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

and L∗ / x ∼ = Ω10q (q). If L∗v is a single 2-component, i.e., a vertical pumpup of L∗ ,  then by [III17 , 10.64ab], L∗v /O2 (L∗v ) ∼ = E6q (q). However, this violates Lemma 8.3b ∗ x as n ≤ 12. Therefore Lv = II where I/O2 2 (I) ∼ = L∗ /Z(L∗ ), I x = I, and L∗ is a x ∗ diagonal of II . In particular, since L ≤ L , L is a subgroup of a diagonal of II x . As [v, D] = 1, D normalizes L∗ , the unique component of CK (v) of its isomorphism type. Therefore D normalizes L∗v and so there is u ∈ {y, xy} such that u x   normalizes both I and I . But [L, u ] = 1, so there is H ≤ CII x (u ) such that H and L maps onto a diagonal of H/O2 2 (H) ∼ L≤H = L = L/Z(L) × L/Z(L). In particular the subnormal closure Lu of L in CG (u ) contains H. But Lu is a trivial or vertical pumpup of L. As we noted above for Lu , Lu /Z(Lu ) ∼ = P Ω± r (q), 8 < r ≤ n ≤ 12, so Lu does not involve L/Z(L) × L/Z(L). This contradiction completes the proof.  

Lemma 8.5. If 9 ≤ n ≤ 12, then Proposition 8.1 (b) or (c) holds. Proof. First suppose that n ≥ 11. By Lemma 8.4, x ∈ Syl2 (C(x, K)). On the other hand, if K ∼ = Spin− 12 (q) or Spin11 (q), then by [IA , 4.5.2], Lo2 (CK (y))O2 2 (CK (y)) = L ∗ E, 2 ∼ a central product, with x ∈ L ∩ E, L ∼ = Spin+ 8 (q), and E = SL2 (q ) or SL2 (q), according as n = 12 or 11, respectively. By (8D2) and Lemma 8.3, Lu /Z(Lu ) ∼ = 2 P Ω± r (q), 8 < r ≤ n ≤ 12. Then as x ∈ L < Lu , we must have O (E) ≤ Lu by (solvable) L2 -balance. In fact if q = 3, then |Out(Lu )| is a 3 -group so E induces inner automorphisms on Lu , and then CLu (x) has a solvable component  containing O 2 (E). In any case, CLu (x) has a second component or solvable component E  ∼ = E, by [III17 , 10.66]. By (solvable) L2 -balance, and the fact that  |C(x, K)|2 = 2, O 2 (E  ) ≤ L2 (CK (y))O2 2 (CK (y)) = LE. This is absurd, how  ever, since [O 2 (E  ), LO 2 (E)] = 1. We have ruled out the cases K = Spin− 12 (q) and Spin11 (q). Suppose next that K∼ = HSpin12 (q).

Again x ∈ Syl2 (C(x, K)) by Lemma 8.4, whence E := Lo2 (CG (x, y )) and ∼ E = E/O2 (E) satisfy E = E 1 ∗ E 2 , with E 1 ∼ = Spin+ 8 (q), E 2 = SL2 (q) × SL2 (q), and Z(E 1 ) = Z(E 2 ) = x, y = x, u . Moreover, x is the diagonal involution of Z(E 2 ) and so u lies in one of the SL2 (q) direct factors of E 2 . We put C0 =  O 2 (O 2 (CG (x, y ))) and conclude that C0 = L(M1 × M2 ) where [L, M1 M2 ] = 1,  M1 ∼ = M2 ∼ = O 2 (SL2 (q)), and u ∈ M1 . In view of (8D) and Lemma 8.3a, [III17 , 10.67] applies to give Lu ∼ = Spin+ 12 (q) or HSpin12 (q), with Z(Lu ) the diagonal subgroup of Z(M1 )Z(M2 ) in the latter case. In either case we have Lu /Z(Lu ) ∼ = K/Z(K) so by Lemma 1.2b, m2 (C(u, Lu )) = 1. Therefore u = Z(Lu ), so u is on  HSpin12 (q). the diagonal of M1 × M2 , contradiction. Therefore K ∼ = The remaining cases are K ∼ = Spin10 (q),  = ±1. We still have = Spin9 (q) or K ∼ NK (L) y∈L∼ by [IA , 6.2.1e]. Therefore u ∈ L ≤ Lu . = Spin+ 8 (q). Moreover, xy ∈ y Set Lu := Lu /u ; then E(CLu (x)) has the component L ∼ = Ω+ 8 (q) with x = Z(L). ± Now Lu ∼ = Spin9 (q) or Spin10 (q) by (8D) and Lemma 8.3a. Hence Z(Lu ) = u and xu ∈ xNLu (L) by [IA , 6.2.1e], i.e., xu ∈ uNLu (D) . But u = y or xy, and so x, y, and xy are all NG (D)-conjugate, and conclusion (b) or (c) of Proposition 8.1 holds, as claimed. 

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Lemma 8.6. We have n = 7. Proof. Since we are assuming that Proposition 8.1 fails, the only alternative is that n > 12, by Lemma 8.5. But n ≤ m + 4, so m > 8, and by (8D), m = 12 and Lu ∼ = E7 (q)u . In particular, n ≤ 16. By Lemma 8.3c, n ≥ 15. o If n = 15, or if K ∼ = Spin− 16 (q), we have L2 (CK (u)) = LE1 , where x ∈ 2 ∼ E1 = SL2 (q) or SL2 (q ), by [IA , 4.5.2], and L/ x ∼ = Ω+ 12 (q). Since [x, Lu ] = 1 and 2 x ∈ E1 , we have O (E1 ) ≤ Lu by (solvable) L2 -balance. But then as Lu ∼ = E7 (q)u , + it follows from [III17 , 10.64c] and the fact that L/ x ∼ = Ω12 (q) that x ∈ E1 , a contradiction. The only remaining case is K ∼ = HSpin16 (q). By [IA , 6.2.1], ux ∈ uK , so G xy ∈ y and conclusion (a) of Proposition 8.1 holds, contrary to our assumption that the proposition fails. The lemma is proved.  Now that n = 7, it follows from [IA , 4.5.2] that (8E)

O 2 (CK (y)) = L1 L2 L3 O2 (Z(K)), L1 ∼ = L2 ∼ = L3 ∼ = SL2 (q), Z(L1 ) = y , Z(L2 ) = xy , Z(L3 ) = x . 

Moreover, if we put K = K/Z(K) and regard K ≤ Aut(K), and set Mi = O 2 (Li ), i = 1, 2, 3, then we have (8F)

F ∗ (CAut(K) (y)) = (M 1 ∗ M 2 ) × M 3 .

This follows from [IA , 4.5.1], which gives the analogous result in Inndiag(K), and [IA , 4.2.3]. Recall that u ∈ {y, xy} and T ∈ Syl2 (CG (x)). By [III12 , Def. 1.15], x, y ≤ K, and so we still have y ∼K yx by [IA , 6.2.1]. We first prove: Lemma 8.7. The following conditions hold: (a) x ∈ O 2 (CG (x, v )) for all v ∈ I2 (CG (x)); (b) Lu /O2 (Z(Lu )) ∼ = Spin7 (q); (c) x = Ω1 (Z(T )) and T ∈ Syl2 (G); (d) I2 (K) ⊆ xG . Proof. Part (a) is immediate from [III17 , 11.26]. Let T ∈ Syl2 (CG (x)). By [III17 , 8.9e], Ω1 (Z(T )) ≤ CT (K) ≤ C(x, K) so Ω1 (Z(T )) = x . As x char T , T ∈ Syl2 (G), proving (c). Note that y ∈ K, so u ∈ K. Then all involutions of K − Z(K) are K-conjugate to u (see [IA , 4.5.2, 6.2.1e]). It now suffices to prove (b). For if (b) holds, then by Lemma 1.2b, (u, Lu ) ∈ J∗2 (G), so there is symmetry between x and u. In particular (c) applies to u , so u ∈ xG , and then I2 (K) = {x} ∪ uK ⊆ xG , which is (d). As T ∩ K is clearly not of maximal class, there is U  T with U ∼ = E22 , by [IG , 10.11]. In view of the conjugacy of all involutions of K − Z(K), we may assume that D = x, y  T . If y is 2-central in G, then y ∈ xG by (c), so (b) holds. Therefore we may assume that y is not 2-central in G. As |CT (D)| = |T |/2 we must have CT (D) ∈ Syl2 (CG (y)), so (8G)

x is 2-central in CG (y).

We shall derive a contradiction. Now as (y, L) is an acceptable subterminal (x, K) pair, y ∈ L = L1 by definition [III12 , Def. 1.15]. We have O 2 (L1 ) = M1 ≤ Ly ,

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

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x ∈ M3 , and [x, Ly ] = 1 = [xy, Ly ]. Hence by (solvable) L2 -balance, M3 ≤ Ly , and similarly M2 ≤ Ly . Thus M1 M2 M3 ≤ CLy (x). If q > 3, then [III11 , 13.4] applies to M3 = L3 ∼ = SL2 (q) and, as there are three Mi ’s, we get Ly ∈ Chev with q(Ly ) = q.

(8H)

∼ Q8 , and in On the other hand, if q = 3, then [III11 , 13.1] applies to each Mi = view of (8G) and the fact that there are three Mi ’s, Ly ∈ Chev with q(Ly ) = 3, so (8H) holds in general. Again as there are three Mi ’s, it follows easily that Ly ∈ G62 , so by Lemma 1.2a, f (Ly ) ≤ f (K) = q 9 . This fact, together with (8H), the existence of the three subnormal subgroups Mi of CLy (x) with x y / y 2-central in Ly / y , imply that Ly /O2 (Z(Ly )) ∼ = Spin7 (q), as can be seen from an examination of [IA , 4.5.1]. But then x is not 2-central in Ly , by [IA , 6.2.1e], a contradiction. The proof is complete.  We quickly deduce: Lemma 8.8. We have x ∈ Syl2 (C(x, K)). Proof. Suppose false. Since y ∈ xG by Lemma 8.7d, there exists w ∈ C(y, Ly ) such that w2 = y. Then as x, y ≤ M1 M2 M3 ≤ Ly , w ∈ CCG (x) (y M1 M2 M3 ). Passing to CG (x) = CG (x)/C(x, K) ≤ Aut(K), we get w ∈ CAut(K) (M 1 M 2 M 3 ) = Z(M 1 M 2 M 3 ) = y , the first equality by (8F) and the F ∗ -Theorem. Consequently y = (w)2 = 1, so y ∈ C(x, K), which is absurd. The lemma is proved.  Lemma 8.9. For any g ∈ NG (K) of prime order, if g induces a (possibly trivial) field automorphism on K, then g = x. Proof. Let g be as in the lemma, assume g = x, and let r be the order of the image of g in Aut(K). Of course g centralizes O2 (Z(K)) = x . Set Cg = 1 CG (g), Cx,g = Cg ∩ CG (x), and Kg = Cg ∩ K. Then E(Cx,g ) = Kg ∼ = Spin7 (q r ) (or 3Spin7 (3) with r = 1), by [IA , 4.9.1]. As Cg is a K-group, L2 -balance and [III17 , 10.68] imply that (8I)

x ∈ Z ∗ (Cg ) and Kg O2 (Cg )  Cg .

An application of [III17 , 8.9e] yields that for any S ∈ Syl2 (Cx,g ) we have Ω1 (Z(S)) ≤ CCx,g (Kg ). But g maps onto CAut(K) (Kg ) by [IA , 7.1.4c], so Ω1 (Z(S)) ≤ x, g in view of Lemma 8.8. If r is odd, then x = Ω1 (Z(S)). If r = 2, then g induces a nontrivial field automorphism on K by Lemma 8.8, so g ∈ [CG (x), CG (x)]. Regardless of the value of r, x ∈ [S ∩ Kg , S ∩ Kg ] by [IG , 15.12(i)]. So x = Ω1 (Z(S)) ∩ [S, S] char S in every case. Thus S ∈ Syl2 (Cg ). Since K has a unique conjugacy class of involutions outside Z(K), we may replace g by a K-conjugate and assume that g centralizes x, y . Then g induces a field automorphism of order r on each Li , 1 ≤ i ≤ 3, in (8E), by [IA , 4.2.3]. If r > 1, then L1 L2 L3 = E(CG (x, y )) = E(CLy (x)), and so [III17 , 10.1] shows that g induces a field automorphism of order r on Ly . The argument of the previous paragraph then applies with y in place of x to show that y = Z(S ∗ ) ∩ [S ∗ , S ∗ ] for some S ∗ ∈ Syl2 (Cg ). Hence y ∈ xCg by Sylow’s Theorem, contradicting (8I). Likewise if r = 1, then as CAut(Ly ) (L1 L2 L3 ) is generated by the image of x (cf. (8F)),

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g or gx centralizes Ly , and it is g unless possibly g is an involution. As g = x, by replacing g by gx if necessary, we may then assume that Ly ≤ Cg . But then again, x and y are Cg -conjugate as they are Sylow 2-centers in Cg , and (8I) is again contradicted. The proof is complete.  As a consequence of Lemma 8.9, O2 (Z(K)) = O2 (CG (x)) = 1. Now we can prove Lemma 8.10. K = F ∗ (CG (x)) = O 2 (CG (x)), I2 (CG (x)) = I2 (K), and G has one conjugacy class of involutions. Proof. By Lemma 8.7cd, the last assertion follows from the previous one. By Lemmas 8.9 and 8.8, CG (K) = x , and no element of CG (x) of odd prime order induces a field automorphism on K. In particular F ∗ (CG (x)) = K. Now Out(K) = Outdiag(K)×Φ, where Φ is the image of a cyclic group of nontrivial field automorphisms of K and Outdiag(K) ∼ = Z2 , by [IA , 2.5.12]. Thus K = O 2 (CG (x)). Indeed Lemma 8.9 implies (with r = 2) that no involution of CG (x) induces a nontrivial field automorphism on K. It follows with [IA , 4.9.1] that (8J)

All involutions of CG (x) map into Inndiag(K).

Assume now that I2 (CG (x)) ⊆ K, and let t ∈ I2 (CG (x)) − K. By Lemma 8.7ac, x ∈ O 2 (CG (x, t )), and CG (x) contains a Sylow 2-subgroup T of G. Let T0 be the preimage of Φ under the natural mapping T → Out(K). Then Ω1 (T0 ) ≤ K and |T : T0 | ≤ 2. By the Thompson Transfer Lemma, tG ∩ K = ∅, so t ∈ xG by Lemma 8.7d. Thus Lemma 8.7a applies with t and x in reversed roles. We conclude that t ∈ O 2 (CG (x, t )), which contradicts the fact that t ∈ O 2 (CG (x)).  Thus, I2 (CG (x)) ⊆ K, and the lemma is proved. Thus, 2-local analysis does not provide a contradiction; we are in the vicinity of an (exotic) Solomon 2-fusion system Sol(q). We write q = ra with r an odd prime and consider the r-local structure of G. Lemma 8.11. Let P be a parabolic subgroup of K with Levi decomposition P = QM , Q = Or (P ) ∼ = Eq5 and [M, M ] ∼ = B2 (q). Let U ∈ Sylr (P ) and ∗ P = NCG (x) (Q) = NCG (x) (P ) = QM ∗ with M ≤ M ∗ and M ∗ ∩ Q = 1. Then the following conditions hold: (a) AutG (Q) = AutCG (x) (Q) = AutP ∗ (Q); (b) There exists S ∈ Sylr (G) such that Q = Z(J(S)); and (c) P ∗ controls G-fusion in Q. Proof. Set N = NG (Q) and CQ = CG (Q). Then x ∈ CQ and CCQ (x) = CCG (x) (Q) = CK (Q) = Q × Z(K) = Q x . Therefore x ∈ Syl2 (CQ ). By a Frattini argument, N = CQ (N ∩ CG (x)), which implies (a). Let W = O2 (CQ ). We have Q x = CCQ (x), so CQ = W x with x inverting W/Q. In particular W is nilpotent. Let Wr = Or (W ) ≥ Q, so that N = O{2,r} (CQ )Wr P ∗ . Since K = O 2 (CG (x)), Sylr (P ) ⊆ Sylr (CG (x)). Let U ∈ Sylr (P ) and S = U Wr ; then S ∈ Sylr (N ).

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

If Wr = Q, then S = U , and Q = J(S) = Z(J(S)) by [III17 , 6.15a]. Hence in proving (b) we may assume that Wr > Q. Let V = x, t be a four-subgroup of [M, M ] ∼ = Spin5 (q). Then t and tx are M -conjugate, so CWr /Q (t) ∼ = CWr /Q (tx) = 1. Notice also that Y := CQ (t) = CQ (tx) is a (short) root subgroup of K. Let Nt = CN (t). Then Nt = O{2,r} (Nt )CWr (t)CM ∗ (t) with x inverting  CWr (t)/Y . Moreover, O r (CM (t)) = It × Itx , where t ∈ It ∼ = SL2 (q) and tx ∈ Itx ∼ = SL2 (q). Set Kt = O 2 (CG (t)) ∼ = K. By the Borel-Tits Theorem [IA , 3.1.3], there is a parabolic subgroup Pt ≤ Kt such that CWr (t) ≤ Or (Pt ) and CM (t) ≤ Pt . Since Z(Kt ) = t ≤ It , the only possibility by [III17 , 6.15bc] is that |Or (Pt )| = q 7 , COr (Pt ) (x) = Z(Or (Pt )) ∼ = Eq , and CM (t) acts irreducibly on Or (Pt )/Z(Or (Pt )). It follows that Z(Or (Pt )) = Y and CWr (t) = Or (Pt ). As t and tx are M conjugate, we see also that CWr (tx) = Or (Ptx ), where Ptx is N -conjugate to Pt . As CWr (x, t ) = CQ (t) = Y , this yields |Wr | = q 5+6+6 = q 17 . We know also that CM (t) is irreducible on CWr (t)/Y and CM (tx) on CWr (tx)/Y , so M is irreducible on Wr /Q, as well as on Q [III17 , 6.15a2,b2]. Therefore (8K)

Q is the only nontrivial abelian characteristic subgroup of Wr .

Hence to complete the proof of (b) it suffices to prove that (8L)

J(S) ≤ Wr .

For then Z(J(S)) = Z(J(Wr )) = Q by (8K). As Q char S and S ∈ Sylr (NG (Q)), it will follow that S ∈ Sylr (G) and (b) will hold. Again by (8K), Z(Wr ) = Q. Hence by [III8 , 7.7], if J(S) ≤ Wr , then in the semidirect product P = QM ∗ ∼ = QAutN (Q), we would have J(U ) ≤ Q. However, J(U ) = Q by [III17 , 6.15a1]. This contradiction establishes (8L) and completes the proof of (b). Finally by (b) and [IG , 16.9], G-fusion in Q is controlled in NG (J(S)), and hence in NG (Q). By (a), this fusion is controlled in P ∗ . Thus (c) holds as well.  Now we can complete the proof of Proposition 8.1. Since CG (x) has a single class of involutions outside x , and its Sylow 2-subgroups are not of maximal class (having 2-rank > 2), x, y  T for some Sylow 2-subgroup T of CG (x) by [IG , 10.11]. By Lemma 8.7c, T ∈ Syl2 (G). Then Lemma 8.7cd and [IG , 16.21] # imply that NG (x, y ) contains a 3-element g cycling x, y . In view of (8E), g cycles L1 , L2 , and L3 . But Sylow r-subgroups of Li are short root subgroups of K for i = 3 and long root subgroups of K for i = 1, 2. Hence long and short root subgroups of K are G-conjugate. Let P and Q be as in Lemma 8.11. By [III17 , 6.15a], Q contains both long and short root subgroups; moreover, they are not Aut(K)-conjugate. On the other hand, by Lemma 8.11c, they must be P ∗ conjugate and hence Aut(K)-conjugate. This contradiction completes the proof of Proposition 8.1. 9. The Sp2n Cases In this section we continue to assume the assertions and notation of (1A), (1B) and (1C), as well as (1E). We assume that p = 2 and we shall prove the following “Sp2n Theorem”: Proposition 9.1. Assume that p = 2 and K/O2 (K) ∼ = Sp2n (q) for some n ≥ 2 and some odd q. If n ≥ 3, assume that there is no (x , K  ) ∈ J∗2 (G) such that

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9. THE Sp2n CASES

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K ∼ = P Sp2n (q). Then we have G0 ∼ = P Sp2n+2 (q) and ΓD,1 (G) ≤ NG (G0 ), where D = x, y . It will be evident from the proof that row 6 of Table 14.1 applies. Note that by [IA , 6.1.4], O2 (K) = 1, so that K ∼ = Sp2n (q). Through Lemma 9.8 we assume that n=2 and aim to prove that (9A)

G0 ∼ = P Sp6 (q) and G0 is ΓD,1 (G)-invariant.

We assume that (9A) fails. Lemma 9.2. The following conditions hold: (a) x, y, and xy are conjugate in NG (D), where D = x, y ; (b) T ∈ Syl2 (G); and (c) There exists M   C(x, K) such that x ∈ K ∩ M , M ∼ = SL2 (q), and [K, M ] = 1. Moreover, KM = E(CG (x)) if q > 3. By [III17 , 11.17a], and since Proof. Recall that T ∈ Syl2 (CG (x)). m2 (C(x, K)) = 1 by (1C), x = Ω1 (Z(T )) so T ∈ Syl2 (G). We next argue that (9B)

y ∈ xG .

Suppose false. By the Z ∗ -Theorem [IG , 15.3], there exists v ∈ xG ∩ T − {x}. Since (9B) fails and m2 (C(x, K)) = 1, v ∈ y K and v acts nontrivially on K. Therefore by [IA , 4.5.1, 4.9.1], Lo2 (CK (v)) has a unique component or solvable component H, and x ∈ H. Choose g ∈ G such that v g = x. As H   CG (x, v), H g   CG (xg , x). Hence by (solvable) L2 -balance and the fact that x ∈ H, xg lies in a (solvable) component of CCG (x) (xg ), and in particular either in C(x, K) or in K. As x = v, we have xg = x, and so xg ∈ K. By [III17 , 15.2b] this forces xg ∈ y K , completing the proof of (9B). Clearly T ∩ K is not of maximal class, so by [IG , 10.11] and [III17 , 15.2b] we may assume that T was chosen so that D = x, y  T . Then by [IG , 16.21], AutG (D) ∼ = Σ3 , and (a) follows. Next, we see in CG (x) that the components and solvable components of CK (D) are My ∼ = SL2 (q), Mxy ∼ = SL2 (q), with y ∈ My and xy ∈ Mxy . As CAut(K) (My Mxy ) = AutD (K) by [IA , 4.5.1, 4.2.3], any further (solvable) component of CG (D) must lie in C(x, K). As m2 (C(x, K)) = 1, and in view of (a), it follows that C(x, K) has a subnormal SL2 (q)-subgroup Mx with x ∈ Mx . Hence (c) holds.  By Lemma 9.2, for each u ∈ D# , CG (u) has a unique component Ku such that ∼ Ku = K, and Kx = K. Furthermore E(CG (u)) ≤ Ku Mu with Mu ∼ = SL2 (q), and equality holds if q > 3. As a result, either Lu = L or Lu = Ku . In particular G0 = K, Ly , Lxy ≤ K, Ky , Kxy . Indeed without loss Ly = Ky , and since y ∼K xy, Kxy ≤ G0 and (9C)

G0 = K, Ky , Kxy .

We next prove: Lemma 9.3. O2 (CG (x)) = O2 (CG (y)) = O2 (CG (xy)) = 1.

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Proof. Suppose false and set W = O2 (CG (x)) = 1 (see Lemma 9.2a). Since D ≤ K, [D, W ] = 1. Thus W ≤ O2 (CG (D)); indeed by [III17 , 15.2a], we have O2 (CAutCG (x) (K) (D)) = 1, so W = O2 (CG (D)). Conjugating in NG (D) we see that W = O2 (CG (u)) for all u ∈ D# . Then G0 = K, Ky , Kxy ≤ CG (W ). Now if W = 1, then G0 is a K-group. By [III17 , 3.34], G0 ∼ = P Sp6 (q). As Ku is the unique component of CG (u) of its isomorphism type, for any u ∈ D# , NG (D) permutes the set {Ku | u ∈ D# } and normalizes G0 . Finally by [III17 , 15.2b], for each u ∈ D# , CG (u) = Ku NCG (u) (D) ≤ NG (G0 ), so ΓD,1 (G) ≤ NG (G0 ). Thus the conclusions of Proposition 9.1 hold, contrary to assumption. The lemma follows.  By Lemmas 9.2 and 9.3, all involutions of D are conjugate in NG (D); and for each u ∈ D# , Lo2 (CG (u)) = Ku Mu with Ku ∼ = Sp4 (q), [Ku , Mu ] = 1, Mu ∼ = SL2 (q), Z(Ku ) = Z(Mu ) = u , and D ≤ Ku . Moreover, Mu  CG (u), since Mu = E(C(u, Ku )) or O 2 (C(u, Ku )) according as q > 3 or q = 3. For any u ∈ D# , write Mu∗ = Lo2 (CKu (D)). Then by [IA , 4.5.2], Mu∗ = Mu1 ×Mu2 with E1 (D) = {u , u1 , u2 }. Thus Lo2 (CG (D)) is the central product Mx My Mxy with D = Z(Lo2 (CG (D))). Lemma 9.4. For each u ∈ D# , C(u, Lu ) = Mu ∼ = SL2 (q). Proof. Since NG (D) cycles D# we need only prove C(x, K) = Mx . As m2 (C(x, K)) = 1 and Mx  C(x, K), with O2 (C(x, K)) = 1 by Lemma 9.3, we have CC(x,K) (Mx ) = x . Thus if C(x, K) > Mx , there would exist h ∈ C(x, K) inducing an outer automorphism on Mx . Then h centralizes My and Mxy , and in particular centralizes D. So h ∈ CG (y), and h maps into CAut(Ly ) (Mxy ), which by [III17 , 13.5] equals the image of Mx in Aut(Ly ). Hence h induces an inner automorphism on Mx , a contradiction. This proves the lemma.  Now choose g ∈ NG (D) cycling the elements x, y, xy of D# , so that g cycles the three groups Mx , My , Mxy . Since Mx ∼ = SL2 (q) has one class of elements of order 4, we may choose tx ∈ Mx of order 4 and replace g by a suitable element 3 of gMx ⊆ NG (D) and assume that tgx = tx . We then set ty = tgx and txy = tgy . Replacing g by a suitable power we may also assume that g is a 3-element. For each u ∈ D# , t2u = u. Set V = tx , ty , txy , a g-invariant abelian 2-group, and v = tx ty txy ∈ CV (g). Thus v 2 = 1 and Ω1 (V ) = D × v . Moreover, looking in CG (u) for any u ∈ D# we see that vtu ∈ Ku has order 4 and (vtu )2 = u. Set   Ju = O r (CKu (v)) = O r (CKu (vtu )) for each u ∈ D# , where q is a power of the prime r. Lemma 9.5. The following conditions hold for any u ∈ D# : (a) Ju ∼ = SL2 (q) and according as q ≡ 1 or −1 (mod 4), Ju and Mw form a weak CT-system or weak P-system for Ku , for any w ∈ D − u ; in particular Ku = Ju , Mw ; (b) If q > 3, then E(CG (u, v )) = L2 (CG (u, v )) = Ju ; (c) If q = 3, then O 2 (CG (u, v )) = Ju ; and (d) If q = 3, then x is 2-central in CG (v). Proof. Assertions (a)–(c) are direct consequences of our construction together with [III17 , 4.2]. Suppose then that q = 3. Now V g = v × V0 g with V0 = [V, g] ∼ = Z4 × Z4 . If v ∈ xG , then CG (x) would contain an isomorphic copy of V0 g disjoint from x , which is easily seen to contradict the fact that m2 (K) = 2.

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∼ Q8 by Lemma 9.4. We also have Thus, v ∈ xG . Now C(x, K) = Mx and Q = |T : RQ| ≤ |Out(K)| = 2. Thus by the Thompson transfer lemma [IG , 15.15], v has an extremal G-conjugate v ∗ (with respect to T ) in RQ. Using [IA , 4.5.2] we see that v ∗ is K-conjugate either to v or to x, and we have seen above that the latter is impossible. It follows that v is extremal in some Sylow 2-subgroup of CG (x). Therefore x is 2-central in CG (v), completing the proof of (d) and the lemma.  

Let Iu be the subnormal closure of O 2 (Ju ) in CG (v), for each u ∈ D# . Then by Lemma 9.5cd, g cycles Ix , Iy , and Ixy .

(9D)

Lemma 9.6. The following conditions hold: (a) Ju and Iu are V -invariant for each u ∈ D# ; and (b) Ix = Iy = Ixy is a single g-invariant 2-component of CG (v). Proof. As V is abelian, [V, u, v ] = 1 and so Ju and then Iu are V -invariant, proving (a). We next claim that (9E)





[O 2 (Ju ), O 2 (Ju )] = 1 for all u = u ∈ D# .

Indeed if q > 3 and [Ju , Ju ] = 1, then with Lemma 9.5, Ku = Ju , Mu and Ku = Ju , Mu commute elementwise, which is absurd as Muu ≤ Ku ∩ Ku and Muu is nonabelian. If q = 3 and [O2 (Ju ), O2 (Ju )] = 1 for some u = u ∈ D# , then [O2 (Ju ), O2 (Ju )] = 1 for all u = u ∈ D# by conjugation by g , which centralizes v. Let R = O2 (Jx )O2 (Jy )O2 (Jxy ) v ; then D ≤ Z(R) and R/D ∼ = E27 . However, as |Out(K)| = 2, we see in CG (x) that |CT (D)| ≤ 29 ; as CT (D) ∈ Syl2 (CG (D)), this yields R ∼ = E27 . But from [IA , 4.5.1], CT (D) contains an = CT (D) and CT (D)/D ∼ element inducing outer automorphisms on My and Mxy , so CT (D)/D is nonabelian, a contradiction. This proves (9E). It follows immediately that (9F)

[Iu , Iu ] = 1 for each u = u ∈ D# .

Next suppose that q > 3. Then by L2 -balance, Iu is a product of 2-components of CG (v) permuted transitively by u , for any u ∈ D# . On the other hand, by Lemma 9.5b, L2 (CG (u, v )) is a single 2-component. Thus any u ∈ D# has at most one nontrivial orbit on the set of 2-components of CG (v). In particular D has no regular orbit. Likewise D can have no orbit of length 2, for if it did there would be three distinct g-conjugate orbits of length 2, and each Ju would be a diagonal in the product of components in one of these orbits, contradicting (9E). Thus D normalizes every 2-component of CG (v), and so (9F) and (9D) imply (b) in this case. Finally suppose that q = 3. Then by solvable L2 -balance and the action of g, either Iu is a single 2-component of CG (v) for each u ∈ D# , in which case (b) holds by (9F), or else Iu = O2 (Ju ) ≤ O2 (CG (v)) for all u ∈ D# . So assume that the latter holds. In particular D ≤ O2 (CG (v)). By Lemma 9.5e, x ∈ Z(O2 (CG (v))). Conjugating by g we see that D ≤ Z(O2 (CG (v))). As Ju  CG (u, v ) for any u ∈ D# by Lemma 9.5, we conclude that Iu = O2 (Ju )  O2 (CG (v)). But then choosing u ∈ D − u , we have [Iu , Iu ] ≤ Iu ∩ Iu = 1, contradicting (9F). This completes the proof. 

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

84

Set I = Ix = Iy = Ixy . We next prove:  Lemma 9.7. We have I/Z(I) ∼ = L3q (q). Moreover, Jx and Jy form a standard CT-pair or standard P-pair for I, according as q = 1 or −1.

Proof. We quote [III17 , 3.35], with CG (v)/O2 (I), I/O2 (I), and [V, g] in the roles of the groups X, K, and B there. We use the facts that for each u ∈ D# , Ju is the unique subnormal SL2 (q)-subgroup of CG (u, v), that [V, g] ∼ = Z4 × Z4 embeds in Aut(I/O2 (I)), and, if q = 3, that |CG (u, v)|2 = 3. We conclude with  [III17 , 3.35] that I/O2 (I) ∼ = L3q (q), and the second assertion of the lemma holds for I/O2 (I). It remains to show that H = Z(I), where H = O2 (I). As H is solvable of  odd order and I = O 2 (I), if I acts trivially on each I-chief factor of H, then H = Z(I) by [IG , 11.5]. So, by way of contradiction, we may assume that A is an I-chief factor of H with [A, I] = A. Since [D, v] = 1, D acts on H with Hu := CH (u) ≤ O2 (CCu (v)) for each u ∈ D# . As F ∗ (Cx ) ≤ Mx K, it follows from [III17 , 15.2] that Hx ≤ CG (K)Xx,v , where Xx,v is a cyclic subgroup of K of order dividing q − q . Now, Hx ∩ CG (K) ≤ Hu for all u ∈ D# . As CG (K) acts faithfully on Ky , it follows that Hx ∩ CG (K) ≤ Xy,v ∼ = Xx,v . We conclude that |A| = sa , where s is a prime divisor of q − q and a ≤ 3. In particular s = r and q ≥ 5. In particular, H is a q  -group. Let U ∼ = Eq be a preimage in I of a root subgroup of I/H. A Frattini argument shows that NI (U ) acts transitively on U # . But  we know that q − 1 ≥ 4; hence by Clifford’s Theorem, U ≤ CI (A). But then I = U I centralizes A, a final contradiction.  Lemma 9.8. n ≥ 3. Proof. Suppose that n = 2 and continue the above argument. By Lemmas 9.7 and 9.5a, Jy and Jx form a CT-pair or P-pair in I, according as q = 1 or −1, of I, and Jx and My do the same for K. Hence as [Jy , My ] ≤ [Ky , My ] = 1, Jy , Jx and My form a weak CT-system or weak P-system of type C3 (q), according to the value of q . By [III13 , 1.4, 1.14], G1 := Jy , Jx , My is a quotient of Sp6 (q). As m2 (C(x, K)) = 1, actually G1 ∼ = P Sp6 (q). For each u ∈ D# , the only possible noncyclic composition factor of CG (u)/Ku is L2 (q), so Ku is the unique subgroup of CG (u) of its isomorphism type. Now K = Jx , My ≤ G1 so K  CG1 (x). It follows quickly that D# is fused in G1 , so G0 = K, Ky , Kxy ≤ G1 . Thus by [III17 , 9.1], G0 = G1 , whence G0 is NG (D)-invariant, as NG (D) permutes {K, Ky , Kxy }. Finally, since D ≤ Ku for every u ∈ D# , and Ku has a unique class of four-subgroups [III17 , 15.2], CG (u) = Ku NCG (u) (D) ≤ NG (G0 ). Thus ΓD,1 (G) ≤ NG (G0 ), completing the proof of the proposition.  Now we encounter the general case. Lemma 9.9. If n = 3, then Lu ∼  Spin7 (q) for any u ∈ D − x . = Proof. If n = 3 and Lu ∼ = Spin7 (q), then F(Lu ) = F(K), and so (u, Lu ) ∈ by Lemma 1.2b. But then Proposition 8.1 gives a contradiction. 

J∗2 (G)

We have our acceptable subterminal (x, K)-pair (y, L); by [III12 , 1.15], xy is the involution in a long root SL2 (q)-subgroup Mxy ≤ K, (9G)

CK (y) = L × Mxy , y ∈ L ∼ = Sp2n−2 (q), and Z(L) × Z(Mxy ) = x, y .

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Of course L is a component of CLy (x) and CLxy (x). Note that since x ∈ LMxy ≤ Lo2 (CK (y)) = Lo2 (CK (xy)), it follows by (solvable) L2 -balance that (9H)

x induces an inner automorphism on Ly and Lxy .

Lemma 9.10. We have Ly = L and Lxy > L. Proof. If Ly > L, then in view of (9G) and (9H) and Lemma 1.2a, we may conclude from [III17 , 10.58a] that n = 3 and Ly ∼ = Spin7 (q), contradicting Lemma  9.9. Hence, Ly = L, and the lemma follows from (1C). We next prove Lemma 9.11. The following conditions hold: (a) xy ∈ xG and Mxy  CG (xy); (b) C(x, K) has a subgroup Mx  CG (x) with x ∈ Mx ∼ = SL2 (q); and (c) For any g ∈ G such that xg = xy, Mxg = Mxy . Proof. We have xy ∈ Mxy and by (9G), Z(L) = y . Then the vertical pumpup Lxy must satisfy F(Lxy ) ≤ F(K) by Lemma 1.2a; hence as n ≥ 3, Lxy ∼ = P Sp2n (q) or Sp2n (q), by [III17 , 10.58b], with x acting on Lxy as an involution of Lxy . Again by Lemma 1.2b, m2 (C(xy, Lxy )) = 1 and (xy, Lxy ) ∈ J∗2 (G). By hypothesis, Lxy ∼ = Sp2n (q). Then CLxy (x) = L × Mx with = P Sp2n (q), so Lxy ∼ (q), [L, M ] = 1, and xy ∈  M Mx ∼ SL = 2 x x . As K  CG (x) and Lxy  CG (xy), and n ≥ 3, both Mx and Mxy are normal subgroups of CG (x, y ). But then as xy ∈ Mx , Mx ∩ Mxy = 1 and [Mx , Mxy ] = 1. Then Mxy maps into CAut(Lxy ) (LMx ) ∼ = Z2 (see [IA , 4.5.1]) so [Lxy , Mxy ] = 1. Similarly [K, Mx ] = 1. In particular as m2 (C(x, K)) = 1, x ∈ Mx , and similarly xy ∈ Mxy . Now x, y ≤ K, and C(x, K) ≤ CG (K) as K is quasisimple. Therefore C(x, K) ≤ CG (x, y ), whence Mx  C(x, K). Indeed CAut(K) (y) covers Out(K), so CG (x) = KCG (x, y ) ≤ NG (Mx ), and (b) holds.  Since L ∼ = Sp2n−2 (q), we may choose an involution z ∈ L such that O r (CL (z)) =   J1 × J2 with J1 ∼ = O r (Sp2n−4 (q)), J2 ∼ = O r (SL2 (q)), and z ∈ J2 . Then J2 is Kr conjugate to O (Mxy ) and Lxy -conjugate to Mx . In particular Mxy = Mxg for some g ∈ G, whence xg = xy. Since L is a component of CG (y), Lg is a component of CG (y g ) and lies in L2 (CG (xy)) by L2 -balance. It follows immediately that Lgh = L for some h ∈ Lxy . Thus y gh = y and so gh interchanges x and xy. Hence gh interchanges CK (y) and CLxy (y), whence it interchanges Mx and Mxy . As Mx  CG (x), (c) follows, as does Mxy  CG (xy). Thus (a) holds and the proof is complete.  Next, taking all the long root SL2 (q)-subgroups relative to a fixed root system for L, we obtain involutions z1 , . . . , zn−1 ∈ L ∩ xG and conjugates Mzi ∼ = SL2 (q) of Mx , i = 1, . . . , n − 1, such that zi ∈ Mzi  CG (zi ) and [Mzi , Mzj ] = 1 for all 1 ≤ i < j ≤ n−1. Then also [Mzi , Mx ] ≤ [K, Mx ] = 1 and similarly [Mzi , Mxy ] = 1. Write x = zn and xy = zn . Thus with [III17 , 7.6], Z(L) = z1 · · · zn−1 . But Z(L) = y = zn zn , so z1 z2 · · · zn zn = 1. L As in the previous proof Mzi ∈ MyK ∩ Mx y , and this implies that for any g  g ∈ G and indices i and j = 1, . . . , n, n , zi = zj if and only if Mzgi = Mzj . Set M = Mz1 ∗ · · · ∗ Mzn ∗ Mzn and ∼ E2n . V0 = Z(M ) = z1 , . . . , zn−1 , zn = z1 , . . . , zn−1 , zn =

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

∼ Σn permuting z1 , z2 , . . . , zn natuThen AutK (V0 ) contains a subgroup S0 = rally and fixing zn = x, and similarly AutLy (V0 ) contains S1 ∼ = Σn permuting z1 , z2 , . . . , zn naturally and fixing zn . Hence the subgroup S := S0 , S1 ∼ = Σn+1 of AutG0 (V0 ) permutes z1 , z2 , . . . , zn , zn naturally. Also S permutes the Mzi along with the zi , so S normalizes M . This quickly yields Lemma 9.12. We have Lo2 (CG (y)) = E(CG (y)) = LI with I ∼ = Sp4 (q), Z(I) = Z(L) = y , and y ≤ Mx × Mxy ≤ I. Proof. Let E ∗ := zn , z1 z2 ≤ V0 . Then for some g ∈ NG (V0 ), (E ∗ )g = E := z1 , zn zn . Let H ∗ = Lo2 (CG (E ∗ )) and H = Lo2 (CG (E)), so that H = (H ∗ )g . Within CG (x) = CG (zn ) we see that E(H ∗ ) = E(CCG (zn ) (z1 z2 )) has a component M12 ∼ = Sp4 (q), such that z1 z2 ∈ Mz1 Mz2 ≤ M12 . We conjugate this cong figuration by g, letting Mnn = M12 . Thus Mnn ∼ = Sp4 (q) is a component of E(H) = E(CCG (y) (z1 )) with y = x(xy) ∈ Mx My ≤ Mnn . By L2 -balance and the fact that z1 ∈ L ≤ E(CG (y)), the subnormal closure I of Mnn in CG (y) is a single 2-component of CG (y). As [Mx , L] = 1 and Mx ≤ Mnn , I = L and so since L  E(CG (xy)), [I, L] = 1. Thus [z1 , I] = 1 and as Mnn is a component of CCG (xy) (z1 ), I = Mnn is a component of CG (y). Calculating in CG (x), H ∗ has a (solvable) component of type SL2 (q), a (solvable) component of type Sp2n−4 (q), and a component of type Sp4 (q). Thus H does as well. As LI ≤ E(CG (y)) with z1 ∈ L, it follows easily that CCG (y) (LI) has no components or solvable components. The lemma is proved.   Lemma 9.13. G0 = N = E(CG (u)) | u ∈ D# ∼ = P Sp2(n+1) (q). Proof. For each i = 1, 2, . . . , n, n , choose vi ∈ Mzi such that v 2 = zi , and set V = v1 , v2 , . . . , vn , vn , an abelian 2-group. We have v1 × · · · × vn ≤ Mz1 ×· · ·×Mzn−1 ×My , but v := v1 · · · vn vn is an involution as v 2 = z1 · · · zn zn = 1. Thus V1 := Ω1 (V ) = v, z1 , . . . , zn = v × V0 ∼ = E2n+1 . For each u ∈ D# set Ju := E(CE(CG (u)) (v)) (or, if n = q = 3, Ju := O (CE(CG (u)) (v))). Since vn ∈ Mx , v acts on K like vvn−1 = v1 · · · vn ∈ K, and  it acts on Mx like vn ∈ Mx . Thus with [IA , 4.5.2] we see that Jx = E(CK (v)) ∼ =  SLnq (q) and similarly Jxy = E(CLxy (v)) ∼ = Jx . On the other hand, v acts on I q ∼ like vn vn ∈ I and on L like vvn−1 vn−1  ∈ L, so Jy = J1 J2 , with J1 = SLn−1 (q) and J2 ∼ = SL2 (q). Moreover, as L ≤ K ∩Lxy , J1 ≤ Jx ∩Jxy and indeed J1 = E(Jx ∩Jxy ),  or O 3 (Jx ∩ Jxy ) if n = q = 3.   3

Now let J be the subnormal closure of J1 in CG (v), so that Jx = J1Jx

≤J

and similarly Jxy ≤ J. As J1 ≤ L ≤ E(CG (u)) for all u ∈ D , J ≤ E(CG (v)), by [III2 , 2.5] applied with J1 , Jx , v, and y here in the roles of I, L, y and z there. We   have E(CJ (x)) = Jx ∼ = SLnq (q) and E(CJ (xy)) = Jxy ∼ = SLnq (q). Moreover, J1 is a component or solvable component of CJ (y). Therefore by [III17 , 3.33], J/Z(J) ∼ = q (q). Given the structures of Jx , Jy and J1 , it follows that there are groups Ln+1 H12 , H23 , . . . , Hn−1,n , Hn,n , all isomorphic to SL2 (q) and with Z(Hij ) = zi zj , and  such that Hn,n = J2 , forming a weak CTP-system of type Anq (q) for J (of CTtype or P-type according as q = 1 or −1). In addition H12 , H23 . . . , Hn−2,n−1 ≤ L, Hn−1,n ≤ Lxy , and Hn,n = J2 ≤ I. But then Mz1 , H12 , H23 , . . . , Hn−2,n−1 form #

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a weak CTP-system for L, and these groups together with Hn−1,n yield a weak CTP-system for Lxy . The commutativity relation [Hn,n , Mz1 ] ≤ [I, L] = 1 is then the final relation showing that M1 , H12 , H23 , . . . , Hn−1,n , Hn,n form a weak CTP-system of type Cn+1 (q) for the group G1 := Mz1 , H12 , H23 , . . . , Hn−1,n , Hn,n . Thus G1 is a quotient of Sp2n+2 (q) by [III13 , 1.4, 1.14]. As m2 (C(x, K)) = 1, G1 ∼ = P Sp2n+2 (q). By [III17 , 9.1], G1 = K, Lxy . But G0 = K, Ly , Lxy = K, Lxy = G1 , since Ly = L ≤ K. The proof is complete.  Now we complete the proof of Proposition 9.1. By Lemma 9.13, it suffices to show that ΓD,1 (G) ≤ NG (G0 ). Note that L is the unique component of CG (D) of its isomorphism type. Hence NG (D) normalizes L, and so it permutes the pumpups Lx = K, Ly , and Lxy of L. Hence NG (D) normalizes G0 . Let w ∈ (xy)CG (x) . As xy ∈ K  CG (x), w ∈ K and CK (w) ∼ = CK (xy), by an isomorphism taking w to xy. From [IA , 4.5.1] we see that w ∈ (xy)K . Therefore CG (x) = KCG (x, xy ) ≤ G0 NG (D) ≤ NG (G0 ). Similarly CG (xy) ≤ NG (G0 ). Finally consider Cy = CG (y). Whether n = 3, in which case L and I are interchanged in Cy ∩ G0 , or n > 3, in which case L and I are non-isomorphic and hence normal in Cy , we have Cy = (Cy ∩ G0 )(NCy (L) ∩ NCy (I)), so it is enough to show that NCy (I) ≤ G0 . Note that D = x, y ≤ I ∼ = Sp4 (q). But it follows from [IA , 4.5.2] that I has a unique conjugacy class of four-subgroups. Thus NCy (I) ≤ INCy (D) ≤ G0 NG (D) ≤ NG (G0 ). The proof of Proposition 9.1 is complete. 10. The Linear and Unitary Cases We continue to assume the assertions and notation of (1A), (1B) and (1C), as well as (1E). We consider the case p = 2 and K/Z(K) ∼ = Ln (q), n ≥ 4,  = ±1, q a power of the odd prime r, omitting the subcase K/O2 (Z(K)) ∼ = Ω6 (q) or P Ω6 (q).  That is, when n = 4 we assume that K/O2 (Z(K)) ∼ = SL4 (q). Since (y, L) is an acceptable subterminal (x, K)-pair, we have L ∼ = SLn−1 (q)  or SLn−2 (q). The former may hold regardless of conditions on n and Z(K), but it necessarily holds if (q − )2 > n2 and |Z(K)| is odd. In particular, if n is odd, then L∼ = SLn−1 (q). Our goal is the following result: Proposition 10.1. Assume that p = 2 and K/Z(K) ∼ = Ln (q), n ≥ 4,  = ±1.  Moreover, if n = 4, assume that K ∼ = SL4 (q). Then one of the following holds: (a) K is 2-saturated and G0 /Z(G0 ) ∼ = Ln+1 (q) or Ln+2 (q). Moreover, ΓD,1 (G) ≤ NG (G0 ); or (b) K/Z(K) ∼ = L8 (q),  = q , K is not 2-saturated, L ∼ = SL6 (q), and with notation chosen so that y ∈ L, (Ly , Lxy ) = (HSpin12 (q), E6 (q)). Moreover, m2 (C(d, Ld )) = 1 for all d ∈ D# . Remark 10.2. We shall show in Proposition 13.1 below that if conclusion (b) holds, then G0 ∼ = E7 (q) with ΓD,1 (G) ≤ NG (G0 ). We begin with some preliminary lemmas.

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Lemma 10.3. If u ∈ {y, xy} and L < Lu , then F(Lu ) ≤ F(K). Proof. From the isomorphism type of K, we see that d2 (G) = 6. If Lu /O2 (Lu ) ∈ G62 , then the lemma holds by Lemma 1.2a. Therefore we may assume that Lu ∈ G62 . As N is a level neighborhood and Lu > L, Lu /Z(Lu ) ∼ = L± 3 (q) or ±  L (q), n ≥ 4, the L4 (3), with q = 3 in the latter case. But since K/Z(K) ∼ = n conclusion is then trivial.  Lemma 10.4. Suppose that L ∼ = SLn−1 (q) and u ∈ {y, xy} with L < Lu . Then  ∼ either Lu /Z(Lu ) = Ln (q) or n − 1 = 4 and Lu ∼ = Spin7 (q) with u ∈ Z(Lu ) and q ≡ − (mod 4). Proof. Except for the assertion u ∈ Z(Lu ), this follows directly from [III12 , Theorem 1.2], or [III9 , Theorem 6, p. 47]. Now m2 (C(x, K)) = 1 and m2 (CAut(K) (L)) = 1, so m2 (CG (DL)) = 2. As [DZ(Lu ), L] = 1, we have Z(Lu ) ≤ D, whence u ∈ Z(Lu ) as [x, Lu ] = 1.  ∼ SL (q) with n even and n ≥ 6. Let u ∈ Lemma 10.5. Suppose that L = n−2 {y, xy} with L < Lu . Then one of the following holds: (a) Lu /Z(Lu ) ∼ = Ln (q) and u ∈ L; or (b) Lu /Z(Lu ) ∼ = Ln−1 (q), u ∈ L, and L = E(CLu (x)) with CLu (L) a cyclic  r -group; (c) Lu ∼ = Sp2n−4 (q), Ω+ 2n−4 (q) or HSpin2n−4 (q), u ∈ L, L = E(CLu (x)), and CAut(Lu ) (L) is a cyclic r  -group; or (d) L ∼ = SL6 (q) and Lu /Z(Lu ) ∼ = E6 (q) with u ∈ L and with CLu (L) having ∼ a normal subgroup J = SL2 (q) with ux ∈ Z(J) ≤ Z(L); or (e) L ∼ = SL4 (q) ∼ = Spin7 (q), Spin− = Spin6 (q) and Lu ∼ 8 (q), or Spin9 (q), with   u ∈ L ∩ Z(Lu ), and either L = O r (CLu (x)) or m = 9 and O r (CLu (x)) = J ∗ L with J ∼ = SL2 (q) and u ∈ J.  r Finally, if O (CK (u)) = J × L with J ∼ = SL2 (q), then [O 2 (J), Lu ] = 1. Proof. As K/Z(K) ∼ = Ln (q) has untwisted Lie rank n−1, Lemma 10.3 implies ∼ that either Lu /Z(Lu ) = L± n (q) or Lu has untwisted Lie rank at most n − 2. By examination of [IA , Table 4.5.1], the only possibilities for L and Lu besides those listed are L ∼ = SL8 (q) with Lu ∼ = SL4 (q) with Lu ∼ = E7 (q), and L ∼ = Spin+ 8 (q). + ∼ [Note that Ω8 (q) = HSpin8 (q).] However, in the Lu ∼ = E7 (q) case, |Z(E(CLu (x)))| ≤ 12 (8, q − ), whereas  ∼ |Z(L)| = (8, q − ), a contradiction. In the Lu ∼ = Spin+ 8 (q) case, since L = SL4 (q) is a component of CLu (x), x induces a nontrivial inner-diagonal automorphism on Lu . Thus Z(Lu ) is a four-subgroup of CG (x) not containing x. As m2 (Q) = 1, Z(Lu ) maps injectively into CAut(K) (L). It follows that for some v ∈ Z(Lu )# , Kv := E(CK (v)) ∼ = SL5 (q). But then by definition [III7 , Def. 6.7], (x, K, y, L) is ignorable and so by [III12 , Def. 1.15], (y, L) is not an acceptable subterminal (x, K)-pair, a contradiction. Hence, L and Lu are as in (a)–(e). In cases (a) and (b), CZ(L) (Lu ) = 1 so u ∈ L. In case (c), the involution z ∈ Z(L) is either u or ux, and it lies in Z(Lu ) (see [IA , 4.5.1]). Hence z, u ≤ CG (Lu ) so z = u as [x, Lu ] = 1. Similarly, in case (e), u ∈ L∩Z(Lu ). In case (d), Z(Lu ) has odd order so u ∈ L. Clearly x ∈ L so ux = O2 (Z(L)). The remaining assertions in (a)–(e) follow from inspection of [IA , 4.5.1, 4.5.2]. Finally, suppose that J ×L  CK (x), and let J be the image of J in Aut(Lu ). By   L2 -balance, O 2 (J) ≤ CInn(Lu ) (L). Suppose that O 2 (J) = 1. Since J normalizes

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Lu , O 2 (J) is noncyclic. Therefore (b) and (c) do not occur. Neither does (d), for in that case Z(J) ≤ Z(L). Likewise (e) does not occur since u ∈ J ∩ L in that case. Thus (a) holds, so u ∈ L. Looking in K we see that x is on the diagonal of Z(J) × Z(L). Therefore u = Z(J). It follows that J × x ∼ = L2 (q) × Z2 lies in  CAut(Lu ) (L), which is impossible as m2 (CAut(Lu ) (L)) = 2. Therefore O 2 (J) = 1, completing the proof.  Lemma 10.6. Suppose that L ∼ = SL2 (q) and u ∈ {y, xy} with L < Lu . Then either Lu /Z(Lu ) ∼ (q) or L has untwisted Lie rank 2, i.e., Lu /Z(Lu ) ∼ = L± = L± u 4 3 (q), P Sp4 (q) or G2 (q). ∼ SL± (q) = A± (q). By Proof. The structure of L forces K/O2 (Z(K)) = 4 3 9 Lemma 10.3, F(Lu ) ≤ F(K) = (q , A). As q(Lu ) = q, the result follows directly.  Lemma 10.7. Let J ∼ = SLk (q) with k ≥ 3, q = r m , r an odd prime. Let J  N and let w be an involution in J. Then wN = wJ .  Proof. Let O r (CJ (w)) = Jw × Lw with Jw ∼ = SL (q), Lw ∼ = SLk− (q), g r w ∈ Jw . Let g ∈ N , v = w . Then O (CJ (v)) = Jv × Lv with v ∈ Jv ∼ = SL (q).  J  Necessarily,  =  . But then v ∈ w , as claimed. We now begin the proof of Proposition 10.1. First we analyze the case leading to G0 = E7 (q). Lemma 10.8. Suppose that L ∼ = SLn−2 (q) with n ≥ 5. Then (q − )2 ≤ n2 if |Z(K)| is odd. In particular, n is even. If K is not 2-saturated and notation is chosen, as it may, so that y ∈ L, then n = 8,  = q , and (Ly , Lxy ) ∼ = (HSpin12 (q), E6 (q)); moreover, m2 (C(u, Lu )) = 1 for all u ∈ x, y # . Proof. By assumption, (y, L) is an acceptable subterminal (x, K)-pair. Since L∼ = SLn−2 (q), (x, K, y, L) is not ignorable by [III12 , Def. 1.15], and in particular K contains no involution acting as a reflection on K. It follows that n is even, and (q − )2 ≤ n2 if |Z(K)| is odd, as claimed, by [IA , Table 4.5.1]. For the remainder of the proof, suppose that K is not 2-saturated. Then  we have O r (CK (y)) = J ∗ L with J ∼ = SL2 (q) and L ∼ = SLn−2 (q) and with y ∈ Z(J) ≤ Z(L). Suppose first that (10A)

L = Ly . 

Then L < Lxy . As y ∈ J, we also have O 2 (J) ≤ Lxy , by (solvable) L2 -balance. Also, xy ∈ L, and so by Lemma 10.5, either Lxy /Z(Lxy ) ∼ = Ln (q) or n = 8 and  ∼ Lxy = E6 (q). Now, we may choose g ∈ K so that y1 = y g ∈ L − Z(L), [y, y1 ] = 1, y1 is a classical involution in L (i.e., the involution in a root SL2 (q)-subgroup of L), and L ∩ L1 ∼ = SLn−4 (q), where L1 = Lg . Because of (10A), L1 is a component of E(CG (y1 )). Let L2 be the subnormal closure of L ∩ L1 in CLxy (y1 ). Then [L1 , L2 ] = 1, whence L2 ≤ L1 . As [L1 , x] = 1, we have CL1 (xy) = CL1 (y). Hence  O r (CL1 (xy)) = J ∗ (L ∩ L1 ). Thus, L2 = L ∩ L1 ∼ = SLn−4 (q) is a component  ∼ ∼ of CLxy (y1 ). If Lxy /Z(Lxy ) = E6 (q), then L2 = SL4 (q); but E6 (q) has no such component in the centralizer of an involution, by [IA , Table 4.5.1], contradiction.  Hence, Lxy /Z(Lxy ) ∼ = Ln (q) with J ∗ L = O r (CLxy (x)). As y1 is a classical involution in L, it is one in Lxy as well, so L2 ∼ = L1 , a contradiction. Hence, (10B)

L < Ly .

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90

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Then by Lemma 10.5, as y ∈ Z(Ly ), either we have Ly isomorphic to a member of the set {Sp2n−4 (q), Ω+ 2n−4 (q), HSpin2n−4 (q)} with [J, Ly ] = 1, or we have n = 6 ± ∼ and Ly /Z(Ly ) = P Ωm (q) for some m with 7 ≤ m ≤ 9. In particular, x ∈ y G . Suppose first that n = 6. As K is not 2-saturated, |Z(K)| is odd, and hence q ≡ − (mod 4) by the first paragraph. By [III17 , 8.8g], we may choose T ∈ Syl2 (CG (x)) with Ω1 (Z(T )) = x, y . We claim that y is weakly closed in x, y . Namely, we have already noted that x ∈ y G . If T ∈ Syl2 (G) then NG (T )/T controls fusion in x, y by Burnside’s Lemma, so y is weakly closed. And if T ∈ Syl2 (G) then NG (T ) fuses x to xy. Thus, our claim holds in any case. By [III8 , 6.3], (10C)

x ∈ [CG (y), CG (y)]

with L a component of CLy (x). − Since n = 6, we have Ly ∼ = Sp8 (q), Spin7 (q), HSpin8 (q) ∼ = Ω+ 8 (q), Spin8 (q), or Spin9 (q), and we argue that Ly  CG (y). Otherwise there exists a component L0 of C(y, Ly ) such that L0 ∼ = Ly . Then CL0 ,x (x) embeds in CG (x Ly ), which is an extension of C(x, K), whose unique involution is x, by a subgroup of CAut(K) (L) ∼ = GL2 (q). Thus CL0 ,x (x) is itself an extension of a group whose only involution is x by a subgroup of GL2 (q). In particular x must normalize L0 , and [III17 , 12.13] yields a contradiction. Thus Ly  CG (y), as claimed. Set C y = Cy /Ly C(y, Ly ), naturally identified with a subgroup of Out(Ly ). By (10C), (10D)

x ∈ [C y , C y ].

Suppose that Ly ∼ = Sp8 (q), Spin7 (q), or Spin9 (q). As q ≡  (mod 4), it follows from [IA , Table 4.5.1] that x induces an outer diagonal automorphism of Ly , so x = 1. But Out(Ly ) and hence C y are abelian in these cases, and so (10D) is contradicted. As Z(Ly ) = y , 3 does not divide Next, suppose that Ly ∼ = Ω+ 8 (q). |Out(Ly )|, and hence Out(Ly ) ∼ = D8 . It follows that |[C y , C y ]| ≤ 2 and [CG (y), CG (y)]C(y, Ly )/C(y, Ly ) is isomorphic to a subgroup of P SO8+ (q). Therefore x does not act as a similitude on Ly . But then, as q ≡  (mod 4), it follows from [IA , 4.5.1, 4.5.3] that E(CLy (x)) ∼  SL4 (q), a contradiction. = Ω6 (q) ∼ = ∼ The only remaining case for n = 6 is Ly = Spin− 8 (q). As Out(Ly ) is abelian it follows that x induces a non-trivial inner automorphism on Ly /Z(Ly ). Then by [IA , Table 4.5.2], x = wv with w ∈ CG (Ly ), v ∈ Ly and w2 = v 2 = y. Now,  O r (CLy (L)) = 1, whence [J, Ly ] = 1. As [J, v] = 1, [J, w] = [J, x] = 1. As w2 = y, w acts faithfully on K and centralizes J ∗ L. But CInndiag(K) (J ∗ L) is cyclic of order q −  and since (q − )2 = 2, it follows that the image of w in Aut(K) is in the coset of a graph automorphism of K. But then w induces a graph automorphism on L, a contradiction. Thus, the assumption n = 6 leads to a contradiction, and we have n ≥ 8. Hence, as noted after (10B), we have Ly ∼ = Sp2n−4 (q) or Ω+ 2n−4 (q) or HSpin2n−4 (q), with [J, Ly ] = 1. Again choose g ∈ K such that y1 := y g satisfies [y1 , y] = 1, y1 ∈ L,  and O r (CL (y1 )) = J1 × L1 with J1 ∼ = SL2 (q), L1 ∼ = SLn−4 (q), and y1 ∈ J1 . We may assume g was chosen in NK (L1 ). Since y1 = y g and L1 = Lg1 , the subnormal closure of L1 in CG (y1 ) is the component Ly1 := Lgy .

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We next show that Sylow 2-subgroups of C(y, Ly ) are quaternion or semidihedral. Since J ≤ C(y, Ly ) we may assume that m2 (C(y, Ly )) > 1 in doing this. Choose a pumpup chain from (y, Ly ) to a 2-terminal pair (y ∗ , L∗y ). Since Ly ∈ G62 , and d2 (G) = 6, (y ∗ , L∗y ) ∈ J2 (G). By [III12 , Theorem 1.2], m2 (C(y ∗ , L∗y )) = 1. Hence by [III8 , 2.4], there is a first vertical pumpup in the chain. Since y ∈ Ly , all previous pumpups in the chain are trivial, and then all previous 2-components M in the chain contain a conjugate of y in Z ∗ (M ). Hence there exists (not necessarily in the chain) a vertical pumpup (y, Ly ) < (z, I). Choose such a pumpup to maximize the order of Sy,z , a Sylow 2-subgroup of CC(y,Ly ) (z). As (x, K) ∈ J∗2 (G), F(I) ≤ F(L∗y ) ≤ F(K); but F(Ly ) < F(I). By [III17 , 10.31f], I ∼ = Bn−2 (q), and ∼ Ly =  HSpin2n−4 (q). Let I = I/O2 (I). Notice that CAut(I) (Ly ) has Sylow 2group generated by the image of y, by [III17 , 6.7]. Therefore Sy,z = S0 × y where S0 = CSy,z (I). It is immediate by L2 -balance that for every involution z  ∈ S0 , there is a vertical pumpup (y, Ly ) < (z  , I  ). Hence by our maximal choice, |CC(y,Ly ) (z  )|2 ≤ |Sy,z | for all z  ∈ S0 . But y lies in a quaternion subgroup of C(y, Ly ) and is outside the Frattini subgroup of Sy,z , so |Sy,z | ≤ |C(y, Ly )|2 /4. Thus |CC(y,Ly ) (z  )|2 ≤ |Sy,z | ≤ |C(y, Ly )|2 /4 for all z  ∈ S0 . If Sylow 2-subgroups of C(y, Ly ) have noncyclic center then such a center meets S0 nontrivially, contradicting this inequality. Otherwise y is the unique 2-central involution of C(y, Ly ), which forces |Sy,z | = 4. Hence Sylow 2-subgroups of C(y, Ly ) are semidihedral, as claimed. We next show that Ly ∼ = HSpin2n−4 (q). Suppose false. Then by [III17 , 12.11], y1 lies in a unique normal subgroup H1 ∼ = Sp4 (q) or Ω+ 4 (q) of CLy (y1 ), and [H1 , L1 ] = 1. As Ly  CG (y), H1  CG (y, y1 ). Similarly, reversing the roles of y and y1 , we see that y lies in a subgroup H0  CG (y, y1 ) with H0 ≤ Ly1 , H0 ∼ = H1 , and [H0 , L1 ] = 1. As y and y1 are the unique minimal normal subgroups of H0 and H1 , respectively, we have [H0 , H1 ] = H0 ∩ H1 = 1. Therefore H0 maps into  CAut(Ly ) (H1 L1 ), which is an r  -group, whence H0 = O r (H0 ) ≤ C(y, Ly ). As Sylow 2-subgroups of C(y, Ly ) are quaternion or semidihedral, this is a contradiction. Therefore Ly ∼ = HSpin2n−4 (q). We can now show that (10E)

m2 (C(y, Ly )) = 1.

If false, then as in the paragraph before last, there is a vertical pumpup (z, I) of (y, Ly ). Again F(I) ≤ F(K), but now because Ly is a half-spin group, we get a contradiction from [III17 , 10.31f]. This proves our claim. In particular, Ly  CG (y). Recall that [J, Ly ] = 1. Now, there is an involution w = w1 w2 ∈ CJLy (x), q with w1 ∈ J and w2 ∈ Ly of order 4, such that Lw := E(CLy (w)) ∼ (q), = Spin2n−6  E(CLw (x)) ∼ = An−4 (q), and y ∈ Z(Lw ). Note that by (10E) and L2 -balance, Lw = L2 (CG (y, w)). Let L∗w be the pumpup of Lw in CG (w). Suppose first that L∗w is a trivial pumpup of Lw . Then y ∈ Ω1 (O2 (Z(Lw ))) ≤ Ω1 (Z(Sw )), where Sw ∈  Syl2 (O2 2 (CG (w))). Recall that y1 ∈ y K ∩ L with O r (CL (y1 )) = J1 × L1 , L1 ∼ =  SLn−4 (q), and yy1 ∈ Z(L1 ). Then O r (CLy (L1 )) ∼ = Spin+ 4 (q), and we may assume that w2 was chosen in CLy (L1 ). Then L1 ≤ L∗w . But also, L1 ≤ Ly1 ∼ = Ly ; let  w = O r (CL (w)). Then L  w is the central product of Spin+ (q) and Spin+ (q). L y1 2n−8 4

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92

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

 w ] = 1. If q > 3 then L  w ≤ L2 (CG (y, w)) = Lw by L2 -balance; By [III8 , 3.1], [y, L  2  2 3  w ) ≤ CL (y1 ). if q = 3, then O (Lw ) ≤ O (O (CG (y, w))) = Lw . Thus O 2 (L w However, this is a contradiction, since it is obvious from the isomorphism types of  w , and from [IA , 4.5.1], that O 2 (L  w ) is not involved in Lw . Thus, L∗w is Lw and L a nontrivial pumpup of Lw in CG (w), and as y ∈ Lw , the pumpup is vertical. Then Lw Z ∗ (L∗w )/Z ∗ (L∗w ) is an intrinsic component of CL∗w /Z ∗ (L∗w ) (y) of type q (q); its spin involution, that is, is the image of y. As n ≥ 8, 2n − 6 > 8, Spin2n−6 and it follows easily from [IA , Table 4.5.1] that 2n − 6 ∈ {10, 12, 16}. But n is even,  so n = 8, and then from [IA , 4.5.1] again, L∗w /O2 (L∗w ) ∼ = E6q (q). To recap the   r whole picture, K/Z(K) ∼ = HSpin12 (q), and O (CG (y)) = J ∗ Ly with = L8 (q), Ly ∼ J∼ = SL2 (q). We next show that (10F)

 = q .

Suppose by way of contradiction that  = −q . First, let Sy ∈ Syl2 (CG (y)) and z ∈ Ω1 (Z(Sy ))# . As Ly ∼ = HSpin12 (q), it follows from [III17 , 8.9e] that [z, Ly ] = 1, so by (10E), z = y. Thus, y = Ω1 (Z(Sy )) char Sy and Sy ∈ Syl2 (G). On the other hand, let T ∈ Syl2 (CG (x)). Then as  = −q and K is not 2-saturated, Z(K) has odd order. By [III17 , 8.8f], and as m2 (C(x, K)) = 1, we have Ω1 (Z(T )) = x, t ∼ = E22 , where t ∈ K has four eigenvalues equal to −1. Expand T to S ∈ Syl2 (G). As  CG (y)), Ω1 (Z(S)) = t , S and Sy are conjugate but x ∈ y G (as clearly CG (x) ∼ =   G where t = t or xt and t ∈ y . Again since x is not 2-central in G, x is not weakly closed in x, t with respect to G, and so t is weakly closed. Let Lt be the (normal) component of CG (t ) isomorphic to Ly ∼ = HSpin12 (q). Now E(CLt (x)) contains E(CK (t )), which is the central product of two components isomorphic to SL4 (q) ∼ = Spin6 (q). From [IA , Table 4.5.1], since  = −q , x induces an outer (diagonal) automorphism of Lt . As Lt is a half-spin group, Out(Lt ) is abelian by [III17 , 7.4], and so x ∈ [CG (t ), CG (t )]. But then by [III8 , 6.3], x ∈ [G, G], a contradiction, and (10F) is proved. We show next that xy ∈ wG . Recall that [x, w] = 1 and E(CL∗w (x)) ≥   E(CLw (x)) ∼ = A4q (q). Using [IA , Table 4.5.1] applied to L∗w ∼ = E6q (q), and La  q q r grange’s theorem, we see that O (CL∗w (x)) ∼ = A1 (q) ∗ A5 (q) or D5 (q). Hence by L2 -balance and the fact that m2 (C(x, K)) = 1, CK (w) has a component isomor   phic to D5q (q) or A5q (q). As K ∼ = A7q (q), the latter must hold. Therefore w acts on K like an element y1 ∈ K of the form y g , g ∈ K. As y1 ∈ K and m2 (C(x, K)) = 1, w ∈ x, y1 = x, y g . But from the centralizer structures, w is clearly not conjugate to either x or y, so w ∈ (xy)G , as asserted. Finally, it remains to show that (10G)

m2 (C(xy, Lxy )) = 1.

But as m2 (C(x, K)) = 1 and (x, K, y, L) is not ignorable, m2 (CCG (D) (L)) = 2. As m2 (E6 (q)) > 2, this forces Lxy ∼ = E6 (q). Now as E(CLxy (x)) has L as a component, x induces an inner automorphism on Lxy . Moreover, Z(Lxy ) has odd  order. Therefore m2 (CCG (D) (L)) ≥ 1 + m2 (C(xy, Lxy )), and (10G) follows. The next five lemmas present detailed information in different cases about not only the neighborhood N = N(x, K, y, L), but also centralizers of certain other involutions in CG (x). We shall analyze the cases L ∼ = SLn−1 (q) and L ∼ = SLn−2 (q)

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separately. We shall also separate the cases n odd and n even, and the case n = 4 will require special attention as well. Lemma 10.9. Suppose that L ∼ = SLn−1 (q), and let u ∈ D# . Then the following conditions hold: (a) Either Lu /Z(Lu ) ∼ = Ln (q) or L = Lu ; (b) Lu  CG (u) and Lu is the only component of CG (u) of its isomorphism type; (c) If Lu /Z(Lu ) ∼ = Ln (q), then C(u, Lu )∩CG (D) has cyclic Sylow 2-subgroups and m2 (C(u, Lu )) = 1; (d) m2 ([CG (x L), CG (x L)]) ≤ 1 and m2 (CG (x L)) ≤ 2; (e) If n is odd, then CG (K) and CG (x L) are solvable with abelian Sylow 2-subgroups of ranks 1 and 2, respectively. Proof. First, as n ≥ 4, CAut(K) (L) is cyclic by [III17 , 6.2], and (d) is an immediate consequence as m2 (C(x, K)) = 1. By Lemma 10.4, (a) holds unless n = 5, L ∼ = SL4 (q), q ≡ − (mod 4), and ∼ Lu = Spin7 (q). We assume the latter, with y ∈ Z(L), whence u = y and Z(Lu ) = y . By [III17 , 8.8h], if T ∈ Syl2 (CG (x)) with y ∈ Z(T ), then Ω1 (Z(T )) = x, y . As L = E(CLy (x)), it follows from [IA , Tables 4.5.1, 4.5.2] that xy ∈ xCG (y) . But as m2 (C(x, K)) = 1, CG (x) has no component isomorphic to Ly , so y is weakly closed in x, y with respect to G. Hence, by [III8 , 6.3], x ∈ [CG (y), CG (y)]. On the other hand, we use (d) to show that (10H)

Ly  CG (y).

For otherwise, there would exist a component L1 ∼ = Ly of CE(CG (y)) (Ly ), and then C1 := CL1 Lx1 (x) ≤ CG (x L). However, m2 ([C1 , C1 ]) > 1; this is clear if L1 = Lx1 and follows by examination of [IA , 4.5.2] otherwise. Thus (d) is contradicted and (10H) is established. As Out(Spin7 (q)) is abelian, it follows that x ∈ Ly CG (Ly ), and then by [IA , 4.5.1], q ≡  (mod 4), a contradiction. This proves (a). Next we prove (c). It follows from Lemma 1.2ab that m2 (C(u, Lu )) = 1 and (u, Lu ) ∈ J∗2 (G). Moreover, there is v ∈ D−u such that Lv > L; this is part of our basic setup if u = x, while if u = x we can take v = x. We have Lv /Z(Lv ) ∼ = Ln (q) by (a). Let Qu be a D-invariant Sylow 2-subgroup of C(u, Lu ). Then CQu (D) ≤ CG (v), and as [u, Lv ] = 1 and u is the unique involution of Qu , CQu (D) embeds in the cyclic group CAut(Lv ) (L). Thus (c) holds. Now in (e), since n is odd, a Sylow 2-subgroup Q1 of the cyclic group CAut(K) (L), which induces inner-diagonal automorphisms on K, actually induces inner automorphisms on K. Also |Z(K)| is odd, so D = x × (D ∩ K). Thus [D, Q] = 1, so Q is cyclic by (c), whence C(x, K) is solvable. Moreover, a Sylow 2-subgroup of CG (x L) embeds in Q × Q1 . As CAut(K) (L) and C(x, K) are both solvable, CG (x L) is solvable, proving (e). It remains to prove (b), which is obvious from (c) if L < Lu . So assume that Lu = L. Let L1 be a component of C(u, L) isomorphic to L; we derive a contradiction. Let CL = CL1 ,x (x). Then m2 (CL ) > 1 and CL u ≤ CG (L x ). If n is even, then Z(L) has odd order, which quickly yields m2 (CL u ) > 2, contradicting (d). Thus, n is odd and in particular n − 1 ≥ 4. The argument used to prove (10H) will lead to a contradiction provided L has the following property: for any

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94

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

∼ SL (q) involutory automorphism ξ of L, m2 ([CL (ξ), CL (ξ)]) > 1. But as L = n−1 with n − 1 ≥ 4, the desired property is easily seen (with [IA , 4.5.2, 4.9.1]) to hold except in one situation: namely n − 1 = 4 and E(CL1 Lx1 (x)) ∼ = SL2 (q 2 ). In this exceptional case, CG (x L) contains a quaternion group; but n is odd, so (e) is contradicted. The proof is complete.  It is convenient to work with other four-groups than D in many cases. Initially our analyses will focus on a four-group E such that E = w, d ∼ = E22 for some d ∈ D and some classical involution w ∈ K (10I) such that [w, D] = 1. Table 14.2 lists our choice of E in the specific cases. As we analyze CG (e) for e ∈ E # ∪ D# , we will check the following important property, which will hold in many cases:

(10J)

For all e ∈ E # , there exists e ∈ E − e and a component Le of CG (e) such that (1) e ∈ Le ∼ = SLk (q)/Z for some Z of odd order; and (2) Le  CG (e).

To unify the notation in the cases q > 3 and q = 3, we introduce the following functors, which depend on the value of q such that K/Z(K) ∼ = Ln (q).  L2 (X) if q > 3; Lq2 (X) = Lo2 (X) if q = 3.  (10K) E(X) if q > 3; E q (X) = Lo2 (X) if q = 3 and O2 (X) = 1. Thus when q = 3, E q (X) is the product of all components and subnormal SL2 (3)subgroups of X, but this notation applies only when it is known that O2 (X) = 1. Lemma 10.10. Suppose that n is odd. Choose notation so that y ∈ L (∼ =  SLn−1 (q)). Let Jw be any fundamental SL2 (q)-subgroup of L and write Z(Jw ) = w and E = y, w . Then the following conditions hold: (a) Ly = L  CG (y) and L is the only component of CG (y) of its isomorphism type; (b) Lxy ∼ = K and xy ∈ xCG (L) ;   (c) O 2 (Lq2 (CG (y))) = J 1 ∗ L with J 1 ∼ = O 2 (SL2 (q)) and y = Z(J 1 ) ≤ Z(L); moreover, J 1 ≤ Jy  C(y, L) for some Jy ∼ = SL2 (q), and there is a homomorphism φ : C(y, L) → GL2 (q) which is injective on Jy ; (d) w and y are interchanged by an element of G; (e) Lq2 (CG (wy)) = HLwy where H is a component such that w, y ≤ H ∼ = SL4 (q), and Lwy is a (solvable) component = H such that L covers Lwy /O2 (Lwy ) ∼ = SLn−3 (q) and wL ∩ Lwy = ∅; (f) C(x, K) is 2-nilpotent; and (g) (10J1) holds. So does (10J2), unless possibly n = 7 and e = wy. Proof. Since y ∈ L ≤ Ly , Ly /Z(Ly ) has a Schur multiplier of even order, ∼ Ln (q) as n is odd [IA , 6.1.4]. Therefore (a) follows from Lemma so Ly /Z(Ly ) = 10.9ab, and Lxy /Z(Lxy ) ∼ = K/Z(K). Moreover, Lemma 10.9e implies (f).

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Table 14.2. The subgroup E = d, w n

L

d

odd

SLn−1 (q); y ∈ L

y

even SLn−1 (q); L < Lxy even

SLn−2 (q)

x E=D

We next establish the following point, which will be used presently: For any 2-component H of CG (L) such that H/O2 (H) ∈ Chev(s) for some (10L) odd s, we have H/O2 (H) ∼ = A1 (sn ) for some n. Indeed, CH,x (x), which lies in CG (L x ), is solvable with abelian Sylow 2subgroups of rank at most 2, by Lemma 10.9e. This implies that x must normalize H and CH (x) x has abelian Sylow 2-subgroups of rank at most 2. The only possibility, by [III17 , 11.5], is H/O2 (H) ∼ = A1 (sn ) for some n, so (10L) holds. Next, let y0 ∈ y K and let L0 = E(CK (y0 ))  CG (y0 ). We choose y0 , as we may, so that [y, y0 ] = 1 and L ∩ L0 ∼ = SLn−2 (q). Then L ∩ L0 = E(CL0 (xy))  E(CLxy (y0 )). Given the isomorphism type of L ∩ L0 , it follows that (10M)

 O r (CLxy (y0 )) = J0 × (L ∩ L0 ) with J0 ∼ = SL2 (q).

We will eventually see that J0  CG (y0 ), but for now, by (solvable) L2 -balance, we  have O 2 (J0 ) ≤ L2 (CG (y0 ))O2 2 (CG (y0 )). The structure of CLxy (y0 ) implies that y0 induces an inner automorphism on Lxy . It follows that y0 ∈ xy × Z(J0 ), as m2 (C(xy, Lxy )) = 1 (Lemma 10.9c). Now J0 normalizes L0 and maps into the r  -group CAut(L0 ) (L ∩ L0 ); hence, as  J0 = O r (J0 ), J0 centralizes L0 . If xyy0 ∈ Z(J0 ), then y centralizes L0 , which is not the case. Therefore, y0 ∈ Z(J0 ). Let C0 = C(y0 , L0 ). We know that a Sylow 2-subgroup of CG (L x ) is abelian of rank 2; as L0 ∈ LK , the same can be said for a Sylow 2-subgroup U of CC0 (x). Let U ∗ ∈ Syl2 (C0 ) with U ≤ U ∗ , so that U = CU ∗ (x). Since J0 ≤ C0 , U ∗ is nonabelian and so U < U ∗ . As y0 ∈ Z(C0 ) and Ω1 (U ) = x, y0 , any element of NU ∗ (U ) − U fuses x to xy0 . Therefore xy0 ∈ xCG (L0 ) . Conjugating y0 back to y in K, we see that xy ∈ xCG (L) . Thus (b) holds. Before proceeding we return to the structure of CG (L). We argue that (10N) y ∈ H for any (solvable) 2-component H of Lo2 (CG (y)). Indeed we know that y ∈ L, so suppose that H ≤ C(y, L). If H is a 2-component, we have seen already that x normalizes H and CHx (x) has an abelian Sylow 2subgroup R0 of rank at most 2, with Ω1 (R0 ) ≤ x, y . So if y ∈ H then R0 is cyclic, containing either x or xy. However, this forces H to have a cyclic Sylow 2subgroup, which is absurd. So (10N) holds if H is a 2-component. If H is a solvable 2-component, it must again be x-invariant, for otherwise CHH x x (x) would have a Sylow 2-subgroup which would be of rank at least 3 or nonabelian. Again if y ∈ H, then Z ∗ (H) would have to contain x or xy and so CH (x) would contain a nonabelian 2-group, contradiction. This establishes (10N).

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96

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Next, as y0 ∈ Z(J0 ), y0 is a “classical involution” of Lxy by (10M), i.e., y0 has two eigenvalues equal to −1 on the natural Lxy -module. We fix L

w ∈ y0 xy ∩ (L ∩ L0 ). Thus w is K-conjugate to yy0 . As y0 ∈ y K and CG (y0 ) ≥ J0 L0 , CG (y) ≥ J1 L with y ∈ J1 ∼ = SL2 (q) and [J1 , L] = 1; moreover, w ∈ y G .  We may write O r (CK (w)) = Jw × L2 with w ∈ Jw ∼ = SL2 (q), Jw  CL (w), and wy ∈ L2 ∼ = SLn−2 (q). Let Lw be the pumpup of L2 in CG (w). By (10L), the pumpup is not diagonal. We claim that Lw is the unique component of E(CG (w)) isomorphic to L. ∼ G2 (4). Indeed, Lw /O2 (Lw ) ∈ Chev(r) unless r = 3, n = 5, and Lw /O2 2 (Lw ) = In the former case, since Lw /O2 (Lw ) is clearly not of the form A1 (r n ) for any n, our claim follows from (10L). In the latter case it follows from Lemma 10.9e by a similar argument using [III17 , 11.30]. In particular wy ∈ Lw  CG (w). This implies that C(w, Lw ) ≤ CG (w, y).  Moreover, Jw = O r (Jw ) maps into CAut(Lw ) (L2 ), an r  -group, so [Jw , Lw ] = 1. Notice that for each u ∈ D# , Lu  CG (u) by Lemma 10.9b, and as w is a classical involution in Lu , Jw is the unique subnormal SL2 (q)-subgroup of CLu (w) containing w. Therefore Jw  CG (u, w), and so Jw  ΓD,1 (CG (w)).   In particular [O 2 (Jw ), O2 (CG (w))] = 1. By (solvable) L2 -balance, O 2 (Jw ) ≤ E(CG (w))O2 (CG (w)). Furthermore, as Jw  CG (w, y), we have Jw  C(w, Lw ), and as Lw  CG (w), Jw   CG (w). As w ∈ y G , CG (y) has a subnormal SL2 (q)-subgroup Jy , normal in C(y, L), with y ∈ Jy . Computing in K we see that L2 ∩ L ∼ = SLn−3 (q). Since L is quasisimple with w ∈ L, C(y, L) ≤ CG (w). Therefore the action of C(y, L) on Lw gives a mapping φ : C(y, L) → CAut(Lw ) (L2 ∩ L) ∼ = GL2 (q). We have [y, L2 ] = 1, so [y, Lw ] = 1 and the image of y is nontrivial. By (10N), every component and solvable component of C(y, L) must be isomorphic to SL2 (q1 ) for some q1 ≤ q, and must contain y. Thus φ is injective on the product of all such components and solvable components, so there is at most one of them. As Jy is a component or solvable component of C(y, L), we conclude that 



O 2 (Lo2 (C(y, L))) = O 2 (Jy ). Thus (c) is proved. For (d), (e), and (g), if the conclusions are valid for one w, then they obviously are valid for any L-conjugate of w. So we keep the same w as above. Now (g) follows directly from (d) and (e) and what we have proved above. Namely, for e = y we can take e = w and Le = L to satisfy (10J); by (d), (10J) is then satisfied for e = w; and for e = wy, we can take e = w and Le = H, unless H  CG (wy), which can only happen if H ∼ = Lwy and n = 7. Moreover, (e) implies that y and w are interchanged by an element of H. It therefore remains to prove (e). We study CG (wy). As w is central in Jw , a fundamental SL2 (q) subgroup of L, we see that the −1-eigenspace for wy on a standard module for K has dimension n − 3. We may then write  O r (CK (wy)) = Jwy × Lwy ∼ Lwy ≤ L. ∼ SL (q), Jwy ∩ L = Jw , and SL (q) = with w ∈ Jwy = 3

n−3

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We claim that Lwy is a (solvable) component of CG (wy). If n = 5, then wy ∈ wK and the claim holds. So suppose that n > 5. We have Lo2 (CG (y, wy )) = Jy Jw Lwy (times O2 (CG (y, wy )) if q = 3), in which Lwy is unique of its isomorphism type. Likewise Lwy is the only component of its isomorphism type in Lo2 (CK (wy)), and hence in Lo2 (CG (x, wy)). As xy ∈ xCG (L) and Lwy ≤ L, a similar statement holds in CG (xy, wy), so Lwy  ΓD,1 (CG (wy)). Since n ≥ 7 and Lwy ∼ = SLn−3 (q), the subnormal closure I of Lwy in CG (wy) is (modulo O2 (I)) a product of groups in Chev(r), and by [IA , 7.3.3] and [IG , 3.28(ii)], I = ΓD,1 (I). Thus Lwy  I so Lwy = I is a component of CG (wy), as claimed. Set M = CG (Lwy ). Then, CM (y)   Jy Jw = Jy × Jw ∼ = SL2 (q) × SL2 (q). Moreover, since CG (K) and CG (Lxy ) are solvable by Lemma 10.9 and symmetry, E(CM (xy)) ∼ = SL3 (q) and Jw ≤ Jwy . As Jw Lwy ≤ L and = E(CM (x)) = Jwy ∼ CG (L) , we have Jw ≤ E(CM (xy)) as well. Let H be the subnormal closure xy ∈ x of Jw in M , which equals the subnormal closure of Jwy in M . By L2 -balance H is a product of 2-components of M in an x, y -orbit. Let W = O2 (M ) and Ww =    [O 2 (Jw ), CW (w)]. By [IG , 4.3(i)], Ww = [O 2 (Jw ), Ww ]. But O 2 (Jw )   CG (w),   so Ww ≤ O 2 (Jw ) ∩ O2 (M ) ≤ O2 (Jw ) = 1. Using O 2 (Jw ) ≤ E(CM (u)) for u = x   and xy, we similarly obtain [O 2 (Jw ), CW (u)] = 1. It follows that [O 2 (Jw ), W ] = 1, and so H is a product of components of M . If H is not a single component, then from the isomorphism types of Jw and E(CM (u)) for u ∈ {x, xy}, H = H1 H2 with y normalizing each Hi . But then Jw ≤ E(CM (x)) lies on the diagonal of H, whereas by (solvable) L2 -balance, Jw ≤ Hi for some i, contradiction. Therefore, H is quasisimple. Given the structures of CH (u) for u ∈ D# , y acts nontrivially on   H, and hence by (solvable) L2 -balance, O 2 (Jy ) ≤ H. Thus w, y ≤ O 2 (Jw Jy ) ≤ H. Moreover, by [III17 , 3.30], H ∼ = SL4 (q). To complete the proof of (e) we q need L2 (CG (wy)) = HLwy . As w, y ≤ H this follows from the structure of Lo2 (CG (w, y )) above. Finally, w is a classical involution in L ∼ = SLn−3 (q) is a = SLn−1 (q) and Lwy ∼ L component of CL (w). As n ≥ 5, there exists w1 ∈ w ∩ Lwy , completing the proof of (e) and the lemma.  Lemma 10.11. Suppose that n is even, n ≥ 6, and L ∼ = SLn−1 (q). We assume notation chosen so that Lxy is a vertical pumpup of L.  Then there exists w ∈ I2 (K) such that [w, D] = 1, O r (CK (w)) = J2 × Lw , with w ∈ J2 ∼ = SL2 (q) and SLn−2 (q) ∼ = Lw ≤ L. Let v be the central involution of Lw , and set E = x, w . Then the following conditions hold: (a) Lxy /Z(Lxy ) ∼ = K/Z(K), both K and Lxy are 2-saturated, and v = xw; (b) Lw is a component of CG (xw); and (c) One of the following conclusions holds: (1) Lq2 (CG (x)) = K, Ly = L, and Lq2 (CG (xw)) = I × Lw with I ∼ = SL3 (q), y ∈ I, and w ∈ y I ; or   (2) O 2 (Lq2 (CG (x))) = O 2 (J0 ) ∗ K with x ∈ J0 ∼ = SL2 (q) and [J0 , K] = 1; moreover, L2 (CG (xw)) = I ∗ Lw with E ≤ I, I/O2 (Z(I)) ∼ = SL4 (q), and w ∈ xI ; and (d) (10J1) holds. So does (10J2), unless possibly e = xw, n = 6 and (c2) holds. Proof. Since y acts on K as a reflection, it is clear that an involution w ∈ K exists with the properties asserted in the first sentence, except that at the moment

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98

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2 



we do not know that O r (CK (w)) = J2 × Lw , but rather just that O r (CK (w)) = J2 Lw with [J2 , Lw ] = 1. If K is 2-saturated, as we shall presently see is the case, then this product will be direct, as asserted, and moreover, x ∈ K and v = xw. Let u ∈ D# . By Lemma 10.9a, we have either Lu = L or Lu /Z(Lu ) ∼ = Ln (q); moreover, Lu  CG (u). As Lw ≤ L, we have [Lw , u] = 1, and as n ≥ 6, we see that Lw is the unique component of CLu (v) of its isomorphism type. Therefore Lw  CG (u, v ). Let L∗w be the subnormal closure of Lw in CG (v). In particular Lw  CL∗w (u). Since [Lw , D] = 1, L∗w is D-invariant and Lw  ΓD,1 (L∗w ). Now L∗w = K1 · · · Kr is a product of 2-components of CG (v), with each Ki /Z ∗ (Ki ) isomorphic to Lw /Z(Lw ) or to a proper pumpup of Lw . In particular each Ki /Z ∗ (Ki ) lies 1 in Chev(r) and clearly is not of the form A1 (r n ) or P Sp4 (3) or 2 G2 (3 2 ) , so by [IA , 7.3.3] and [IG , 3.28(ii)], L∗w = ΓD,1 (L∗w ). Thus, Lw  L∗w , whence Lw = L∗w   CG (v). Hence (b) will follow once K is proved to be 2-saturated. That is our next aim, i.e., to show that v ∈ Z(J2 ).

(10O) To this end, set

H = CG (Lw ), so that x ∈ H. Also note that by choice of notation and Lemma 10.9a, Lxy /Z(Lxy ) ∼ = r ∼ (C (v)) = J L with J (q), [J , L ] = 1, and x SL K/Z(K). Hence, O = L 1 w 1 2 1 w xy   1 0 , so that acting on J1 as 0 −1 (10P)



[x, O 2 (J1 )] has even order, J1 = [x, J1 ], and x ∈ O2 2 (CG (v)).

Suppose that (10O) fails. Then v = w is a classical involution of K and we may choose w1 ∈ wK ∩ Lw . Let Lw1 be the counterpart of Lw in CG (w1 ), that is, Lw1   CG (w1 ) with Lw1 ≤ K and Lw1 ∼ = Lw . As w1 ∈ Lw , H ≤ CG (w1 ), and moreover CLw1 (H) ≥ Lw1 ∩ Lw ≤ Z(Lw1 ). Hence H normalizes Lw1 , so CH (Lw1 )  H. But x ∈ CH (Lw1 ) ≤ CG (Lw , Lw1 ) = CG (K), of 2-rank 1. Therefore x ∈ Z ∗ (H). Set H ∗ = HLw ; since [x, Lw ] = 1, x ∈ Z ∗ (H ∗ ). On the other hand, 

H ∗ ≥ L2 (CG (w))O2 2 (CG (w)) ≥ O 2 (J1 ), 

the last by Lo2 -balance. Therefore [x, O 2 (J1 )] has odd order, contradicting (10P) and establishing (10O). From (10O) it follows that K is 2-saturated, as promised in the first paragraph. Moreover, (xy, Lxy ) ∈ J∗2 (G) and (x, L) is an acceptable subterminal (xy, Lxy )-pair, so by symmetry, Lxy is 2-saturated and (a) holds. As remarked above, (b) then  holds and v = xw. Now O r (CK (xw)) = J2 × Lw with w = Z(J2 ). Also, by  symmetry, O r (CLxy (xw)) = J1 × Lw . Moreover, we write Z(J1 ) = z1 , and then z1 = (xy)(xw) = yw. We continue to consider H = CG (Lw ). We still have (10P), and [J1 , x] = J1 ≤ H. Thus x ∈ O2 2 (H), and so w = (xw)x ∈ O2 2 (H). 

Hence, by (solvable) L2 -balance, O 2 (J2 ) lies in a 2-component I of CG (xw), and as [J2 , Lw ] = 1, [I, Lw ] = 1. Let L∗w be the subnormal closure of Lw in CG (w). As xw ∈ Lw  CL∗w (xw), it follows that L∗w is a single 2-component. By the

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Borel-Tits theorem, Lw∗ /O2 (L∗w ) ∈ Chev(2), and then, by inspection, we have Lw∗ /O2 (L∗w ) ∈ Chev(r). As before, choose w1 ∈ wK ∩ Lw . Then, L∗w1 ∼ = L∗w and  J2 < Lw1 ≤ L∗w1 . As O 2 (J2 ) < I ≤ L2 (CG (w1 )), I ≤ L∗w1 and, indeed, I is a 2-component of CL∗w (xw), whence I/O2 (I) ∈ Chev(r). 1 We next prove (c). Consider first the case (10Q)

q > 3.

If K = L2 (CG (x)), we show that (c1) holds. We have J2 = L2 (CH (x)) = L2 (CH (w)) with w ∈ Z(J2 ) but w ∈ Z ∗ (H). It follows from [III17 , 10.39] that 3 3 I/O2 (I) ∼ = L± 3 (q) or D4 (3) (with q = 3 in the second case), and x induces an inner automorphism on I with E(CI (x)) = J2 . As w, x ≤ I × wx , we have I = L2 (H). Moreover, J1 = E(CH (xy)) and so yw = z1 ∈ I. Thus, as w ∈ J2 ≤ I, y ∈ I and w ∈ y I , as I has a unique class of involutions. We may choose t ∈ Lw ∩ wK , and write Lt × Jt = E(CK (t)) with Lt ∼ = Lw and with t ∈ Jt ∼ = J2 . Now, J2 ≤ I ∩ Lt . As t ∈ y G , the pumpup L∗t of Lt in CG (t) must be isomorphic to the pumpup of L in CG (y), i.e., L∗t ∼ = L or L∗t /Z(L∗t ) ∼ = Ln (q).  ∼ As I is a 2-component of CL∗t (xw), it follows that I = SL3 (q). To complete the proof of (c1) in this case, it remains to show that Ly = L. Suppose that Ly < L, so that Ly /Z(Ly ) ∼ = K/Z(K) by Lemma 10.9a. Since w ∈ y G , CG (w) has a corresponding component which we call Kw , i.e., with Kw ∼ = Ly . But then E(CKw (xw))   E(CG (xw, w)) = E(CG (x, w)) = E(CK (w)) = J2 Lw , and the centers of these two components are xw and w. But visibly in Kw , w does not lie in any component of E(CKw (xw)). This contradiction completes the verification of (c1) if K = L2 (CG (x)). Thus, still assuming that q > 3, we now assume that L2 (CG (x)) = J0 ∗ K with J0 /O2 (J0 ) ∼ = 2A7 . Then L2 (CH (x)) = J0 J2 . = SL2 (q0 ), q0 odd, q0 > 3, or J0 ∼ Note that x ∈ J0 while w ∈ J2 . Thus, x, w ∈ Syl2 (Z ∗ (L2 (CH (x)))) and L2 (CH (x))/O2 (L2 (CH (x)))xw ∼ = SL2 (q0 ) ∗ SL2 (q) or 2A7 ∗ SL2 (q). As x ∈ Z ∗ (H), it follows by [III17 , 10.47] that J0 J2 pumps up to I in CG (xw) with ∗ ∗ I ∼ = Sp4 (q) or SL± 4 (q). Moreover, xw ∈ Z (I) and x ∈ I − Z (H). We conclude I  that w ∈ x . As in the case K = L2 (CH (x)), we choose t ∈ Lw ∩ wK , and set Lt × Jt = E(CK (t)) with Lt ∼ = Lw and t ∈ Jt ∼ = J2 . As t ∈ xG , the pumpup, L∗t , of ∗ ∼ Lt in CG (t) satisfies Lt = K. As I is a 2-component of CL∗t (xw), we conclude that I∼ = SL4 (q), and so (c2) holds. Suppose next that q = 3. 

As w ∈ O 2 (J2 )  CI (w) and I/O2 (I) ∈ Chev(3), we have q(I/O2 (I)) = 3 by [III11 , 13.1] and x acts as does w on I. Suppose that I/O2 (I) ∼ = L± 3 (3). We argue as in the paragraph following (10Q). Thus, w, x ≤ I xw and so I = L2 (H).  By the symmetry between x and xy, I is also the subnormal closure of O 2 (J1 ) in CG (xw), so yw ∈ I, and as I has one class of involutions, y ∈ I and w ∈ y I . Again choosing t ∈ Lw ∩ z2K , letting Lt be the SLn−2 (3) component of CK (t) and L∗t its pumpup in CG (t), we see that I is a 2-component of CL∗t (w) and hence I∼ = L3 (3). As in the paragraph following (10Q), we again argue that Ly = L, using 3 O (CKw (wx)) instead of E(CKw (wx)). Moreover, in this case x ∈ O2 2 (CG (xw)) and x ∈ I, so by (solvable) L2 -balance, K = Lq2 (CG (x)), so that (c1) holds.

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100

If I/O2 (I) ∼ = L± 3 (3), then I/O2 (I) ∈ Chev has level 3, and by [IA , 4.5.1] we see that CH (x), which contains CI (x), must contain a further (solvable) 2-component I0 (in addition to J2 ). But then as |CH (x) : J2 C(x, K)| ≤ 2 and CH (x)/C(x, K) embeds in GL2 (3), J2 I0 /O2 (J2 I0 ) is the direct product of the images of J2 and I0 ,  the involution in I0 is x, and xw ∈ O 2 (J2 I0 ) ≤ I. Again by [IA , 4.5.1], the only possibilities are I/O2 (I) ∼ = Sp4 (3) and SL± 4 (3). Reasoning with t as in the case q > 3, we find that I/O2 (Z(I)) ∼ = SL4 (3), so (c2) holds. This completes the proof of (c). Finally, whether q > 3 or q = 3, we have (10J) holding for e = x with e = w and Le = K; for e = xw with e = w and Le = I, unless possibly n = 6, Lw ∼ = I, and (c2) holds; and for e = w with e = xw ∈ Le , where Le is the normal closure of Lw in E(CG (w)). Note that if (c1) holds, then as w ∈ y I ⊆ y CG (Lw ) , Le is the unique component of Lw isomorphic to L; while if (c2) holds, then w ∈ xI so Le is  the unique component of Lw isomorphic to K. We pause to note that for the remainder of the proof of Proposition 10.1, we may assume that (10R)

K is 2-saturated.

Indeed this is trivial if n is odd. If n is even and n ≥ 6, it follows from Lemma 10.8 or 10.11, according as L ∼ = SLn−2 (q) or SLn−1 (q). If n = 4, then K/O2 (Z(K)) ∼ =  SL4 (q) by the hypothesis of Proposition 10.1. Lemma 10.12. Suppose L ∼ = SLn−2 (q) with n ≥ 6. Write O r (CK (y)) = J × L, (q) and xy ∈ L. Then Lo2 (CG (y)) = JLy or JLy O2 (CG (y)) where y ∈ J ∼ SL = 2 according as q > 3 or q = 3. Moreover, Lxy = L, and one of the following holds: (a) y ∈ xG , there is J0   C(x, K) with J0 ∼ = SL2 (q) and x ∈ Z(J0 ) ≤ Z(K),  and Lq2 (CG (xy)) = I ∗ L with O 2 (J0 × J) ≤ I ∼ = SL4 (q) and with xy ∈ Z(I) ∩ Z(L); or q (b) K = Lq2 (CG (x)), Ly /Z(Ly ) ∼ = Ln−1 (q), and L2 (CG (xy)) = I × L with  I∼ = SL3 (q). In all cases, D ≤ d E(CG (d)) for all d ∈ D. 



Proof. For any u ∈ {y, xy}, O r (CK (u)) = J × L with y ∈ Z(J). Hence  by Lemma 10.5, [O 2 (J), Lu ] = 1. Hence [x, y , Lxy ] = 1, so Lxy = L and thus  L < Ly with [O 2 (J), Ly ] = 1. As y ∈ L, Ly satisfies (a), (b), or (d) of Lemma 10.5. If it satisfies (d), i.e.,  ∼ Ly = E6 (q), then xy ∈ Z(I) with I ∼ = SL2 (q)  CLy (x). But then O 2 (I) ≤ CK (L), forcing Z(I) = y , a contradiction. Hence, by Lemma 10.5, Ly /Z(Ly ) ∼ = Ln−1 (q)  or Ln (q). Suppose that Ly /Z(Ly ) ∼ = Ln (q) with Ly not 2-saturated. Then (y, Ly ) ∈ ∗ J2 (G) and (xy, L) is an acceptable (y, Ly )-subterminal pair. Thus, permuting the roles of x, y and xy, we deduce from Lemma 10.8 that L < Lxy , a contradiction. Hence, if Ly /Z(Ly ) ∼ = Ln (q), then Ly is 2-saturated. # For each u ∈ D , write Cu for CG (u). As xy ∈ L < Ly , we have that C(y, Ly ) ≤ CCy (L) = Cy ∩ CCx (L). In particular, C(x, K)  CCy (L) and J  CCy (L). Moreover, (10S)

CCy (L)/(C(x, K) × J) maps injectively into CAut(K) (L)/J,

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which is cyclic of order dividing q −  by [III17 , 12.10]. Consider first the case in which (10T)

C(x, K) is 2-nilpotent,

whence CCy (L)/J is solvable. Suppose that Ly /Z(Ly ) ∼ = Ln (q). Then, as y ∈  r J  CCy (L), and H := O (CLy (L)) ∼ = SL2 (q) contains x, O r (CCy (L)) ≥ J × H. As CCy (L)/J is solvable, this implies that q = 3. But then |CCx (J × L)/C(x, K)| divides 4, whence CCx (J × L) is 2-nilpotent, contradicting H ≤ CCy (J × L). Hence, Ly /Z(Ly ) ∼ = Ln−1 (q). Let y1 = y g ∈ L with g ∈ K, and set Ly1 = Lgy . Let L1 be  the SLn−4 (q)-subgroup of L supported on the −1-eigenspace of xyy1 on the natural K-module. We may assume that g was chosen in CK (L1 ). Then xyy1 ∈ L1 = Lg1 ≤ Lg ≤ Lgy = Ly1 . Now, L1 is a (solvable) component of CLy1 (xyy1 ) with xyy1 ∈ Z(L1 ). It follows that 

(10U)

O r (CLy1 (xy))E(CLy1 (xy)) = 

O r (CLy1 (xyy1 ))E(CLy1 (xyy1 )) = I × L1

with I ∼ = SL3 (q) and with L1 ≤ L. Hence, I × L  E(CG (xy)); indeed as y1 ∈ L and E(CG (xy)) = E(C(xy, L))L, we have E(CG (xy)) = I × L. Now CCxy (y) = CCx (y) has just one (solvable) SL2 (q)-component, so y ∈ I. Hence x centralizes C(xy, IL)/O2 (C(xy, IL)), and it follows that IL = Lq2 (CG (xy)). Thus (b) holds in this case. Next, consider the case when (10T) fails, so that Lo2 (Cx ) = J0 K, with x ∈ J0 and J0 /O2 (J0 ) ∼ = SL2 (q0 ), q0 odd, or 2A7 . As y ∈ K, [J0 , D] = 1, and since [x, Ly ] = 1, X0 := CJ0 (Ly ) ≤ O2 (J0 ). Now, J0 /X0 embeds into CAut(Ly ) (L). If Ly /Z(Ly ) ∼ = Ln−1 (q), then CAut(Ly ) (L) is cyclic, yielding a contradiction. Hence  Ly /Z(Ly ) ∼ = Ln (q) and as shown above Ly is 2-saturated. Then Jy := O r (CLy (L)) ∼ = SL2 (q). Moreover, as xy ∈ Z(L) and y ∈ Z(Ly ), it follows that x ∈ Z(Jy ). Then   by Lo2 -balance, O 2 (Jy ) ≤ C(x, K). In particular J maps into CAut(Ly ) (O 2 (Jy )L), which is cyclic of order dividing q − , so [J, Ly ] = 1. It follows that CLy (L) = CLy (J × L)   CCy (J × L) = CCx (J × L), and as after (10S), CCx (J × L)/CCx (K) is cyclic of order dividing q − . Thus, if q > 3, Jy   L2 (CCx (J × L)) = L2 (C(x, K)) = J0 , 



while if q = 3, then Jy   O r (CCx (J × L)) = O r (C(x, K)). Hence, regardless of    L q, O 2 (J0 ) = O 2 (Jy ) ∼ = O 2 (SL2 (q)). Moreover, x ∈ y1 y for y1 ∈ y K ∩ L. Hence,  y ∈ xG . Again, xyy1 ∈ Z(L1 ) where I × L1  O r (CLy1 (xyy1 )), L1 ∼ = SLn−4 (q),   ∼ ∼ and I = SL4 (q). Thus (10U) holds, this time with I = SL4 (q), with L1 ≤ L, and  with O 2 (J0 × J) ≤ I. As y1 ∈ L and x ∈ I, we conclude that Lq2 (CG (xy)) = I ∗ L. As x ∈ J0 , y ∈ J, and xy ∈ Z(L), we have xy ∈ Z(I) ∩ Z(L). Thus (a) holds. Now, in all cases, D ≤ K and xy ∈ L ≤ E(CG (d)) for all d ∈ D# . Hence to prove the final assertion it remains to show that D ≤ xy E(Cxy ). But in both   case (a) and case (b), y ∈ O 2 (J) ≤ I ≤ E(Cxy ). The proof is complete. Lemma 10.13. Suppose that K/O2 (Z(K)) ∼ = SL4 (q). Then y ∈ K with O (CK (y)) = J2 × J4 , where J2 ∼ = J4 ∼ = SL2 (q), y ∈ J2 , and xy ∈ y K . Moreover, one of the following holds: r

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102

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2 

(a) K = Lq2 (CG (x)), C(x, K) is 2-nilpotent, J4  CG (xy) and O 2 (J2 ) ≤ q Lxy with Lxy /Z(Lxy ) ∼ = L3 (q). Furthermore, L2 (CG (xy)) ≤ (Lxy × J4 )O2 (CG (xy)); or (b) There is J0  C(x, K) with x ∈ J0 ∼ = Σ3 . = SL2 (q), and AutG (x, y ) ∼ In all cases, (10J) holds. 

Proof. By the structure of K, xy ∈ y K . Set Qi = O 2 (Ji ), i = 2, 4. By [III5 , Definition 1.2], for some d ∈ {y, xy}, Q2  CLd (x) with Ld a component ± of E(CG (d)). By Lemma 10.6, Ld /Z(Ld ) ∼ = L± 4 (q), P Sp4 (q), G2 (q), or L3 (q). Suppose that d = y. Then J4 acts faithfully on Ly , whence by L2 -balance or solvable L2 -balance, Q2 × Q4  CLy (x). It follows (see [IA , 4.5.1, 4.5.2]) that Ly ∼ = Sp4 (q) or SL± 4 (q) with x ∈ Ly . As y ∈ Q2 , this is impossible. Hence, d = xy and Q2 ∩ Z(Lxy ) = 1. Moreover, by symmetry, Q4 ≤ Lxy , whence [Q4 , Lxy ] = 1 by (solvable) L2 -balance. Suppose first that C(x, K) has a 2-component or solvable 2-component J. Thus  x ∈ J and m2 (J) = 1. Let Q = O 2 (J). By L2 -balance or solvable L2 -balance, either [Q, Lxy ] = 1 or Q ≤ Lxy . As [x, Lxy ] = 1, we must have Q × Q2  CLxy (x) with x ∈ Q and y ∈ Q2 . In particular, xy ∈ Z(Lxy ). Hence, Lxy ∼ = Sp4 (q) or (q). Thus, y and x are interchanged in L . Similarly y and xy are interchanged SL± xy 4 in K, so it follows that AutG (D) ∼ = Σ3 . Moreover, CG (D), viewed in CG (x), clearly has a unique normal SL2 (q) subgroup J2 such that y ∈ J2 . Conjugating in Aut(D), CG (D) has a unique normal SL2 (q) subgroup J0 such that x ∈ J0 . Clearly  [J0 , J2 J4 ] = 1 so J0 = O r (J0 ) ≤ C(x, K) ≤ CG (D), and conclusion (b) holds. Finally, suppose that Lo2 (C(x, K)) = 1, so that as m2 (C(x, K)) = 1, C(x, K) is 2-nilpotent. Then J2 × J4  CG (D), and J2 × J4 covers Lo2 (CG (D)) modulo O2 (CG (D)). As [Q4 , Lxy ] = 1, we conclude that Lo2 (CLxy (x)) has exactly one (solvable) 2-component. In view of the possibilities for Lxy , we conclude that Lxy /Z(Lxy ) ∼ = Lδ3 (q) for some sign δ. If δ =  then conclusion (a) holds. Thus to complete the proof in this case we assume that Lxy /Z(Lxy ) ∼ = L− 3 (q), and derive a contradiction. Suppose that q ≡  (mod 4). Then by [III11 , 6.4g], |CAut(Lxy ) (J2 )|2 = 4. However, CK (J2 ) contains a cyclic subgroup K0 of order q −  with x ∈ K0 . If (q − )2 = 4 we may assume K0 chosen so that Z := O2 (Z(K)) ≤ K0 . Now, a Sylow 2-subgroup T0 of K0 maps injectively into CAut(Lxy ) (J2 ). Thus |T0 | = (q − )2 = 4, so T0 = Z. Expand T0 to T1 ∈ Syl2 (CG (y, x)). Thus, T1 normalizes J2 , and as T1 ≤ CG (x), T0  T1 . But by [III11 , 6.4h] and the fact that |Z| = 4, y = Z(J2 ) ≤ [T0 , T1 ]. Therefore y ∈ T0 ; but since Ω1 (T0 ) = Ω1 (Z) = x , this is a contradiction. Suppose, on the other hand, that q ≡ − (mod 4). Then CLxy (J2 ) = CLxy (J2 J4 ) contains a cyclic subgroup Qxy of order 4 with y ∈ Qxy . Thus Qxy ≤ CG (x) and Qxy maps injectively into A := CAut(K) (J2 J4 ). But by [III17 , 11.19b], A is dihedral of order 2(q − ). Hence, Sylow 2-subgroups of A are four-groups, contradicting Qxy ∼ = Z4 . Thus, δ = , completing the proof in this case. The final assertion of the lemma is obvious for e = x. By NG (E)-conjugation it remains only to check e = xy when (a) holds. We take e = y and Le = Lxy . This completes the proof of the lemma. 

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We turn to the identification of G0 . The case q = 3 is delicate and requires special attention. Lemma 10.14. If q = 3, u0 ∈ E # and we set X = O2 (CG (u0 )), then G0 , D, ΓE,1 (G) ≤ NG (X). Proof. We consider the four cases described in Lemmas 10.10, 10.11, 10.12, and 10.13. The proof is essentially the same in all these cases. For each u ∈ E # , let E b (CG (u)) be the product of all components of E(CG (u)) that are locally 1-balanced with respect to the prime 2. We claim that (10V)

E ≤ u E b (CG (u)) for all u ∈ E # .

Indeed in the situations of Lemmas 10.10, 10.11, and 10.13, (10J1) holds, and Le is locally balanced for the prime 2 by [III17 , 12.1], implying (10V) in these cases. Similarly, in the situation of Lemma 10.12, E = D. We have y ∈ K ≤ E b (CG (x)), y ∈ I ≤ E b (CG (xy)), and xy ∈ L ≤ Ly ≤ E b (CG (y)), yielding (10V). It follows immediately from [IG , 20.6] that G is balanced with respect to E. Moreover, [E, O2 (CG (u))] ≤ [u E(CG (u)), O2 (CG (u))] = 1 for all u ∈ E # . Combined with balance, this implies that X = O2 (CG (u)) for all u ∈ E # . It then follows at once that ΓE,1 (G) ≤ NG (X). In particular, DE ≤ NG (X). Moreover, as G0 ≤ ΓD,1 (G), we are done if D = E. Assume then that D = E. Now N is a level vertical neighborhood, so each Ld , d ∈ D# , lies in Chev(3) and has level 3. Notice that if L ∼ = SL2 (3), then K/O2 (Z(K)) ∼ (3) and y = w, so E = x, w = D, contradiction. Hence = SL± 4 (3), k ≥ 3, and as a result [I , 7.3.3] implies that for each d ∈ D, L ∼ = SL± A k Ld ≤ ΓE,1 (G) ≤ NG (X). Thus G0 ≤ NG (X), and the lemma is proved.  Lemma 10.15. Suppose that q = 3, and that O2 (CG (u0 )) = 1 for some u0 ∈ E # . Then G0 is quasisimple, G0 /Z(G0 ) ∼ = Ln+1 (3) or Ln+2 (3), and ΓE,1 (G) ≤ NG (G0 ). Proof. Let X = O2 (CG (u0 )) = 1 and Y = D, ΓE,1 (G), G0 . By Lemma 10.14, Y ≤ NG (X) < G, so that Y is a K-group. For any d ∈ D# such that Ld is nonsolvable, Ld is a component of CG (d), hence of CY (d), and therefore Ld ≤ L2 (Y ), by L2 -balance. It follows that G0 ≤ L2 (Y ). (In the one case in ∼ which Ld is not a component, K/O2 (Z(K)) ∼ = SL± 4 (3) and Ld = L = SL2 (3); but  # Lemma 10.13 shows that there then is d ∈ D such that L < Ld and Ld is a component of CG (d ), so L ≤ L2 (Y ).) Now if n is even, then x ∈ K as K is 2-saturated and m2 (C(x, K)) = 1; while if n is odd, then y ∈ L ≤ Ly . Hence either K or Ly is an intrinsic 2-component of CL2 (Y ) (x) or CL2 (Y ) (y), respectively, and so K or Ly lies in a single 2-component H of L2 (Y ). In particular L ≤ H and it follows that G0 ≤ H.  Let d ∈ D# and set Md = O 2 (Ld ), so that either Md = Ld is a component of CG (d) and of CY (d), or Md ≤ O2 (CG (d)), as N is a semisimple neighborhood. Let W = O2 (Y ). For any d ∈ D# set Wd = [CW (d), Md ]. By [IG , 4.3(i)], Wd = [Wd , Md ]. Now Md   CG (d) so Wd ≤ Md . As Wd ≤ W = O2 (Y ), Wd ≤ O2 (Md ) ≤ Z(Md ), so Wd = 1. As L ≤ Ld , and d was arbitrary, we conclude  that [W, O 2 (L)] = 1. It follows, in view of the above remark about the case Ld ∼ = SL2 (3), that [W, Ld ] = 1 for all d ∈ D# , whence [W, G0 ] = 1. Therefore H is

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a component of Y . As (x, K) is 2-terminal in G, while K < G0 (as N is a vertical neighborhood), it follows that CY (H) has odd order. In particular, O2 (H) = 1 and H = E(Y )  Y . Now K is a component of CH (x) and is unambiguously in Chev(3) of level 3. Hence, H ∈ Chev(3) (see [III11 , 1.1, 5.11]). Moreover, as already mentioned, K or Ly is an intrinsic component of level 3, so by [III17 , 12.1], H has level 3. As |Z(H)| is odd, and K is a 2-saturated terminal component of CH (x) with K/Z(K) ∼ = Ln (3), we see from [IA , Table 4.5.1] that either (10W)

H/Z(H) ∼ = Ln+a (3), a = 1 or 2, with n even in the second case,

or we have H ∼ = Cn (3), with n odd, n ≥ 5, or H ∼ = E6 (3), with n = 6. ∼ If H = Cn (3), then by [III17 , 12.12], one of Ly and Lxy must be isomorphic to Cn−1 (3). This, however, contradicts Lemma 10.10ab. Similarly if H ∼ = E6 (3), then by Lemmas 10.11 and 10.12, some involution t of H has the property that E(CH (t)) is embeddable in a commuting product of two copies of SL4 (3). However, this is impossible by Lagrange’s theorem, according to the isomorphism type of E(CH (t)) given in [IA , Table 4.5.1]. Thus (10W) holds. It remains to show that G0 = H. Now Ld is a (solvable) component of CH (d) for each d ∈ D# . Let V be the natural (projective) H-module. Since K contains x and has level 3, x acts on V diagonalizably, with, say, a −1-eigenspace Vx− of dimension n. As n > (n + a)/2 = dim V /2, y stabilizes Vx− and the complementary eigenspace Vx+ of x. By construction y acts on Vx− diagonalizably, and so diagonalizably on V , respecting the eigenspaces of x, and with its own eigenspaces Vy± . We write V+,+ = Vx+ ∩ Vy+ , V−,+ = Vx− ∩ Vy+ , etc. We may assume that L is supported on V−,− , of dimension m, where L/Z(L) ∼ = Lm (3). Then Ly /Z(Ly ) ∼ = Lr (3) is supported ∼ on V+,− ⊕ V−,− , of dimension r, say, and Lxy /Z(Lxy ) = Ls (3) is supported on V−,− ⊕ V+,+ , of dimension s. Hence n + a = dim V = n + r + s − 2m. On the other hand, a similar calculation on the natural module for G0 , of dimension n + a , say, shows that n + a = n + r + s − 2m. Hence a = a and G0 = H. The proof is complete.  Lemma 10.16. Suppose that q = 3 and O2 (CG (u)) = 1 for all u ∈ E # . Then the following conditions hold: (a) O2 (CG (d)) = 1 for all d ∈ D# ; and (b) O2 (CG (x , v)) = 1 for all x ∈ xG and all v ∈ I2 (CG (x )). Proof. In (a), obviously we may assume that D = E. It suffices to find e ∈ E # ∩ L such that all components of CG (e) are in Chev(3) of level 3. For then all such components are balanced for the prime 2 [IA , 7.7.8]. Hence, for any d ∈ D# , O2 (CG (d)), which centralizes Ld and hence L, equals O2 (CG (d)) ∩ CG (e) ≤ O2 (CG (e)) = 1 by [IG , 20.6] and our hypothesis. We take e = y, wx, and w in the situations of Lemmas 10.10, 10.11, and 10.13 with D = E. Thus (a) is proved. For (b), we may assume that x = x. Note that since q = 3, it follows from Lemmas 10.10–10.13 that K is the only 2-component of CG (x). Then K is balanced for the prime 2 by [IA , 7.7.8] and O2 (CG (x)) = 1 by (a), so (b) follows from [IG , 20.6].  To identify G0 in general we use weak CT-systems if q = +1 and weak Psystems if q = −1.

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Lemma 10.17. Assume that if q = 3, then O2 (CG (u)) = 1 for all u ∈ E # ∪D# . Then the following conditions hold: and G0 /O 2 (Z(G0 )) ∼ (a) G0 is quasisimple = Ln+a (q), a = 1 or 2;  q # q and E (CG (u)) = E q (CG0 (u)) for all u ∈ (b) G0 = E (CG (u)) | u ∈ E # E ; (c) The assumptions of Lemma 10.12 do not hold; and (d) G0 and the neighborhood N(x, K, y, L) conform to one of the rows in Table 14.1. We remark that because of the special assumption when q = 3, we will be able to use the notation E q (X) for X = CG (u), u ∈ D ∪ E, as well as some other subgroups as in Lemma 10.16. Proof. If q = 3, our assumption implies that Lo2 (CG (u)) = E q (CG (u)) for any u ∈ E # ∪ D# ; and, by Lemma 10.16, Lo2 (CG (U )) = E q (CG (U )) for any four-group U containing an element of xG . We choose fundamental SL2 (q) subgroups J2 , . . . , Jn of K forming a weak CTP-system for K, and let Z(Ji ) = zi for i = 2, . . . , n. Thus, z2 , . . . , zn is elementary abelian. We may assume that J3 , . . . , Jn is a weak CTP-system for L if L ∼ = SLn−1 (q), while J4 , . . . , Jn is a weak CTP-system for L if L ∼ = SLn−2 (q). If  (q), then n is even and we may assume that y = z , which commutes L∼ SL = 2 n−2  SL (q), then [y, J ] = 1 for 3 ≤ k ≤ n. Let with zk for all k, 2 ≤ k ≤ n. If L ∼ = k n−1 V be the standard (projective) module for K. As [V, J4 , . . . , Jn ] = CV (J2 ) and [V, J2 ] = CV (J4 , . . . , Jn ), y fixes the eigenspaces for z2 on V , and so [y, z2 ] = 1. Hence, in all cases, [y, z2 , . . . , zn ] = 1. We subdivide the analysis into several subcases, guided by Lemmas 10.10–10.13. Case 1: n odd. In this case, y ∈ L ∼ = SLn−1 (q) and, by Lemma 10.10, E q (CG (y)) = J1 ∗ L with y =: z1 ∈ Z(J1 ) and J1 ∼ = SL2 (q); moreover, zn ∈ y G . Given the structure q of CG (y), we have E (CG (zn )) = Jn ∗ Ln with Ln ∼ = L and Jn ∼ = SL2 (q). Now q q (C (x, z )) = E (C (z )) = J × L , where L := J2 , . . . , Jn−2 ∼ E = SLn−2 (q). G n K n n 0 0 By L2 -balance, L0 ≤ L2 (CG (zn )) and then as n ≥ 5, L0 ≤ Ln . Then Jn acts on  Ln centralizing L0 , so [Jn , Ln ] = 1 as Jn = O r (Jn ). Hence Jn , Jn ≤ C(zn , Ln ). If q > 3, then by L2 -balance, (10X)

Jn = Jn .

We argue that (10X) holds if q = 3. Indeed in that case O2 (CG (x, zn )) = 1 by Lemma 10.16b. Since Jn  CG (zn ), it follows from [III8 , 7.1] that either Jn = Jn , as desired, or [Jn , Jn ] = 1. So assume the latter. By Lemma 10.10c, there is a homomorphism ψ : C(zn , Ln ) → GL2 (3) mapping zn → −1. This is impossible as zn ∈ Jn ∩ Jn , the unique minimal normal subgroup of Jn ∗ Jn , so (10X) is proved. Thus, E q (CG (zn )) = Jn ∗ Ln . Next, let L1 = E q (CL0 (y)) = J3 , . . . , Jn−2 ∼ = SLn−3 (q), so that E q (CL (zn )) = Jn × L1 . Using L2 -balance (and [III8 , 7.2] if q = 3) we get first that E q (CG (y, zn )) = J1 Jn L1 and then that E q (CLn (y)) = J1 × L1 with J1 ∼ = SL2 (q). (Note if q = 3 that as y ∈ J1 , y induces an inner automorphism on Ln .)

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

As [y, zk ] = 1 for all k, 2 ≤ k ≤ n, it follows that in the action of Ln on its natural (projective) module, {y, z2 , . . . , zn−2 } is simultaneously diagonalizable. It then follows that {J1 , J2 , . . . , Jn−2 } is a weak CTP-system for Ln . In particular, J1 , J2 , J3 ∼ = SL4 (q). So {J1 , J2 , . . . , Jn } is a weak CTP-system for G1 := J1 , J2 , . . . , Jn , and it is an extendible P-system if q = 3 and  = −1. Hence, G1 /Z(G1 ) ∼ = Ln+1 (q), by [III13 , 1.4, 1.14]. We will complete the proof of (a) below by showing that G0 = G1 . As for (b), note that y and w ∈ L are classical involutions in G1 , and by Lemma 10.10 and our construction above, G1 ≥ E q (CG (y)) ∼ = SL2 (q) ∗ SLn−1 (q). Hence (10Y)

E q (CG (u)) = E q (CG1 (u))

for u = y, and then for u = w as y ∈ wG1 , and even for u = wy with n = 5, since then uw ∈ y G1 . For u = wy and n > 5, we may take w = z3 , whence the right side of (10Y) equals J1 , J2 , J3 , J5 , . . . , Jn ; both sides are the product of an SL4 (q) component and an SLn−3 (q) component. By L2 -balance J3 , Jn ≤ E(CL (wy)) ≤ L2 (CG (wy)) ≤ E q (CG (wy)), and it follows that E q (CG1 (wy)) ≤ E q (CG (wy)), whence equality holds. Thus (10Y) holds for all u ∈ E # . To complete the proof inCase 1 it remains to check that G1 = G0 . Since D centralizes E, D normalizes E q (CG (u)) | u ∈ E # = G1 . By construction K ≤ G1 , so K is a component of CG1 (x). Likewise L is a component of CG1 (y). As L ≤ K and G1 /Z(G1 ) ∼ = Ln+1 (q), it follows that E(CG1 (xy)) ∼ = SLn (q), whence E(CG1 (xy)) = Lxy by the structure of CG (xy) (recall that xy ∈ xG ). It is then clear that G1 = K, Lxy = K, L = Ly , Lxy = G0 , completing the proof in Case 1. Note that Lemma 10.10 implies that the second row of Table 14.1 applies, with  = n odd. Case 2: n even, n ≥ 6, L ∼ = SLn−1 (q). In this case, Lemma 10.11 applies. We may choose w = z2 and set Lw = J4 , . . . , Jn ∼ = SLn−2 (q) and J0 = E q (C(x, K)). Then as w ∈ K, E q (CG (x, w )) = q E (C(x, K))E q (CK (w)) = J0 (J2 × Lw ), by L2 -balance (and Lemma 10.16 and [III8 , 7.2] if q = 3). On the other hand, by Lemma 10.11, Lw is a component of CG (xw), and E(CG (xw)) = I ∗ Lw with I ∼ = SL3 (q) or SL4 (q), according q q as K = E (CG (x)) or K < E (CG (x)), or equivalently according as J0 = 1 or J0 ∼ = SL2 (q). Moreover, w ∈ I, with E q (CI (w)) ∼ = J0 × J2 . Indeed by L2 -balance (and [III8 , 7.2] if q = 3) J2 is a (solvable) component of CI (w). In particular, J2 ≤ I. Subcase 2a: K = E(CG (x)), so that Lemma 10.11c1 applies.  In particular y ∈ I ∼ = SL3 (q) and [y, w] = 1. Let J1 = O r (CI (y)) ∼ = SL2 (q). Thus J1 and J2 form a weak CTP-system in I, and as w ∈ y I , J1 is a (solvable) component of CG (y). In particular as L = Ly , J1 normalizes L ∼ = SLn−1 (q). But  r J1 centralizes Lw , an SLn−2 (q)-subgroup of L, and as J1 = O (J1 ), we conclude that [J1 , L] = 1. Hence [J1 , Ji ] = 1 for all i = 3, . . . , n, so that {J1 , . . . , Jn } is a weak CTP-system of type SLn+1 (q). Moreover, when q = 3 and  = −1, this CTP-system is extendible, as we now show. Every subsystem of type A3 except {J1 , J2 , J3 } lies in L and so generates a group isomorphic to SU4 (3). And J1 , J2 , J3 lies in CG (Z(Jn )). There exists g ∈ K such that Jig = Jn+2−i for all i = 2, . . . , n. As Lw = J4 , . . . , Jn ≤ L ∼ = SUn−1 (3), J2 , J3 , . . . , Jn−2 ≤ Lg . Then g as J1 , J2 = I ∼ = U3 (3), J1 ≤ L . Hence in Lg we see that J1 , J2 , J3 ∼ = SU4 (3),

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proving extendibility. Thus, G1 := J1 , . . . , Jn is a quotient of SLn+1 (q), by [III13 , 1.4, 1.14]. As in Case 1, we next establish (10Y) for all u ∈ E # . Since K ≤ G1 , this is obvious for u = x, both sides equalling K. Next, w is a classical involution in K, so E q (CG1 (xw)) ∼ = SL3 (q) × SLn−2 (q) ∼ = E q (CG (xw)) by Lemma 10.11. q Moreover, E (CG1 (xw)) contains J2 and Jn , subgroups of K, so using L2 -balance as in Case 1 (but with x and K in place of y and L) we conclude that (10Y) holds for u = xw. For the final element of E = x, w , we have E q (CG1 (w)) = J2 × H with H ∼ = SLn−1 (q) and H ≥ Lw = J4 , . . . , Jn . Since Lw ≤ L and I CG (Lw ) w ∈ y ⊆ y , H must be the component of CG (w) isomorphic to L (see Lemma 10.9b). Then as wx ∈ Lw ≤ H, E q (CG (w)) = HE q (CG (w H)). We have CG (w H) = CCG (x) (w H) ≤ CCG (x) (w Lw ), which has J2 as its unique (solvable) component, as is seen in CG (x). Hence (10Y) holds for u = w as well. As in Case 1, completing the proof in this case now simply requires that we prove that G1 = G0 . As in that case, D normalizes G1 with K ∼ = SLn (q) and  ∼ L = SLn−1 (q) being components of CG1 (x) and CG1 (y), respectively, and L ≤ K, so E(CG1 (xy)) ∼ = K. But Lxy /Z(Lxy ) ∼ = Ln (q) and m2 (C(xy, Lxy )) = 1 by Lemma 10.9c, so Lxy is the only quasisimple subgroup of CG (xy) with central quotient Ln (q). Hence E(CG1 (xy)) = Lxy . Clearly then G1 = K, Lxy = K, Ly , Lxy = G0 , completing the proof in this subcase. Again we are on the second row of Table 14.1, this time with  even. ∼ SL2 (q), so that Lemma 10.11c2 Subcase 2b: E q (CG (x)) = J0 ∗K with x ∈ J0 = applies. Now, by Lemma 10.11c2, I ∼ = SL4 (q) with x, w ≤ I and w ∈ xI . Moreover, G w = z2 ∈ J2 , and so zk ∈ x for all k, 2 ≤ k ≤ n. As zn ∈ xG , E q (CG (zn )) = Jn ∗Kn  with zn ∈ Jn ∼ = SL2 (q) and Kn ∼ = K, and x ∈ O 2 (J0 ) ≤ Kn by (solvable) L2 balance. Then by L2 -balance and [III8 , 7.2], we see that Jn is a (solvable) 2component of CG (x, zn ) and hence of CK (zn ). Therefore Jn = Jn . Similarly J0 is a (solvable) component of E q (CKn (x)) = J0 J2 , . . . , Jn−2 . Thus, z0 := x and the zk ’s, 2 ≤ k ≤ n − 2, form a set of mutually commuting classical involutions in Kn , a covering group of Ln (q). We also let Ln = J4 , . . . , Jn−2 ∼ = SLn−4 (q), a (solvable) component of CLw (zn ). Considering CG (xw, zn ), we see that E q (CKn (xw)) = I × Ln . Computing on the natural (projective) module for Kn , we see that there is J1 ≤ I such that J1 ∼ = SL2 (q), [J1 , J3 ] = 1, and {J0 , J1 , J2 } is a weak CTPsystem for I. Therefore {J0 , J1 , J2 , . . . , Jn−2 } is a weak CTP-system for Kn . As [I, Lw ] = 1, we have [J1 , Jn−1 Jn ] = 1. Hence, {J0 , J1 , . . . , Jn } is a weak CTPsystem for G1 := J0 , J1 , . . . , Jn , and extendibility follows from our systems for Kn and K. Thus G1 /Z(G1 ) ∼ = Ln+2 (q), by [III13 , 1.4, 1.14]. As K is 2-saturated, Z(G1 ) has odd order. The remainder of the proof in this case is similar to those in the prior cases. Clearly (10Y) holds for u = x, and then for u = w since w ∈ xG1 . We have E q (CG (xw)) = ILw by Lemma 10.11c2, and this equals E q (CG1 (xw)) since I = J0 , J1 , J2 and Lw = J4 , . . . , Jn by construction. It remains to show that G1 = G0 . As usual, D normalizes G1 , and L2 (CG1 (x)) and L2 (CG1 (xy)) contain the respective components K and Lxy , isomorphic to K ∼ = SLn (q). Hence x and xy act on G1 as commuting classical involutions of SLn+2 (q); as E(CK (xy)) =

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

L ∼ = SLn−1 (q), the −1-eigenspaces of x and y on the natural SLn+2 (q)-module have a 1-dimensional intersection. Therefore y also acts as a classical involution on G1 , whence Ly ∼ = K and Ly is a component of CG1 (y). As a consequence, G1 = K, Ly , Lxy = G0 , completing consideration of Case 2. In this case row 3 of Table 14.1 applies, with  = n + 1 odd. Case 3: n even, n ≥ 6, L ∼ = SLn−2 (q), xy ∈ L, y ∈ J2 , so that Lemma 10.12 applies and we may take J2 as J in that lemma. By that lemma, we have E q (CG (xy)) = I ∗ L with I ∼ = SL3 (q) or SL4 (q). q Moreover, E (CG (y)) = J2 ∗ Ly , with Ly a 2-saturated central extension of Ln−1 (q) or Ln (q), respectively. We see from CG (x) that E q (CG (x, y )) = J0 ∗ J2 ∗ L, where J0 = 1 or J0 ∼ = SL2 (q) in the respective cases. Comparing this with the structure of CG (xy) we deduce that J2 is a (solvable) component of CI (y). In particular, J2 ≤ I, so y ∈ I. Subcase 3a: K = E q (CG (x)), so that Lemma 10.12b holds and I ∼ = SL3 (q). As usual, {J4 , . . . , Jn } is a weak CTP-system for L, and {J2 , . . . , Jn } for K.  The strategy is to choose a classical involution z1 of I and set J1 = O r (CI (z1 )) ∼ = SL2 (q), in such a way that {J1 , J2 } is a weak CTP-system for I and [J1 , J3 ] = 1. Since [I, L] = 1, this will yield that {J1 , . . . , Jn } is a weak CTP-system for G1 = J1 , . . . , Jn , which is therefore a covering group of Ln+1 (q). To define J1 we argue as follows. We have y = z2 and zi ∈ z2K for all i = 2, . . . , n. We let Ln be the component of CG (zn ) conjugate to Ly . Set M = J4 , . . . , Jn−2 ∼ = SLn−4 (q), a (solvable) component of CL (zn ). Also set Kn = Ln ∩ K, so that Kn ≥ J2 × M and Kn ∼ = SLn−2 (q). By L2 -balance and the ∼ structure of CG (zn ) = CG (y), the component I of CG (xy), which centralizes zn ∈ L, lies in Ln . Thus I × M ≤ Ln . Let z be the central involution of M . Then CLn (z) ≥ I × M . As Ln /Z(Ln ) ∼ =  Ln−1 (q), it follows from [IA , Table 4.5.1] that E q (CLn (z)) = I × M . We calculate on the standard (projective) Ln -module V to find z1 and J1 . We have V = VI ⊕ WI where VI = CV (z) = [V, I] and WI = [V, z] = CV (I). Set V2 = [V, z2 ]. As z2 ∈ J2 ≤ I, V2 ≤ VI and CV (z2 ) = CVI (z2 ) ⊕ WI . Let z0 = z2 z ∈ Z(Kn ). Then [V, z0 ] = [V, z2 ] ⊕ WI is the support of Kn on V . Let V3 = [V, z3 ]. As J3 is an Ln -conjugate of J2 , V3 = [V, J3 ] with dim(V3 ) = 2, and V2 ∩ V3 = VI ∩ V3 with dim(V2 ∩ V3 ) = 1. Let V1 = CVI (z3 ). Then dim(V1 ) = 2 and V1 is z2 -invariant.  Let z1 ∈ I2 (I) have support V1 on V and centralize [VI , z3 ]. Set J1 = O r (CI (z1 )). Then as V1 is z2 -invariant, [z1 , z2 ] = 1 so {J1 , J2 } is a weak CTP-system for I. Furthermore, as V1 = CVI (z3 ) = CVI (J3 ), it follows that [J1 , J3 ] = 1. Thus {J1 , . . . , Jn } is a weak CTP-system of type Ln+1 (q). If q = 3 and  = −1, then zn = y g for some g ∈ K interchanging zi with zn+2−i , 2 ≤ i ≤ n. Then J1 , . . . , Jn−2 ≤ Lgy , so J1 , . . . , Jn−2 form a weak CTP-system in Lgy . In particular J1 , J2 , J3 ∼ = SU4 (3). Hence {J1 , . . . , Jn } is extendible, whence G1 = J1 , . . . , Jn is a covering group of Ln+1 (q), by [III13 , 1.4, 1.14]. This case actually leads to a contradiction. Indeed K is a component of CG1 (x), by construction. There then exists a classical involution y  of G1 lying in CG1 (x) and such that (10Z)

L := E(CK (y  )) ∼ = SLn−1 (q).

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Therefore (x, K, y, L) was ignorable, so our original subterminal (x, K)-pair (y, L) was not acceptable, contrary to our choice of (y, L). Subcase 3b: E q (CG (x)) = J0 ∗ K with x ∈ J0 ∼ = SL2 (q). In this case we can argue by repeating the second paragraph of Subcase 2b, with Lxy and D substituting for Lxw and E there, and otherwise with y substituting for w there. We thereby construct a D-invariant group G1 = J0 , J1 , . . . , Jn which is a covering group of Ln+2 (q). Just as in Subcase 3a, however, we then can find an involution y  ∈ CG1 (x) satisfying (10Z), so again (y, L) is not acceptable, a contradiction. ∼ SL (q), so that Lemma 10.13 applies. Case 4: K = 4 Here {J2 , J3 , J4 } is a weak CTP-system for K with z2 , z3 , z4 being mutually commuting involutions. Subcase 4a: E q (CG (x)) = J0 ∗ K with x ∈ J0 ∼ = SL2 (q), i.e., Lemma 10.13b applies. As L ∼ = SL2 (q), we may take y = z2 and xy = z4 , both in xG . Let K4 be the pumpup of J2 in CG (z4 ). Thus, E q (CG (z4 )) = J4 ∗ K4 with ∼ K4 = K and with E q (CK4 (x)) = J0 × J2 . Let L3 = E q (CK4 (z3 )). Then considering the action of z3 on E q (CG (x)), we see that E q (CL3 (x)) = J0 . This condition, together with x ∈ J0 ≤ L3 , implies that 1 + ∼ 2  SL±  L2 (q 2 ). If L3 ∼ L3 ∼ = SL2 (q) × = 4 (q ), Sp4 (q), or Ω4 (q). Also, L3 /Z(L3 ) = SL2 (q), then x ∈ Z(L3 ), a contradiction. It follows from [IA , Table 4.5.2] that L3 ∼ = SL3 (q). As [z0 , z3 ] = 1 = [z3 , z2 ], K4 has a weak CTP-system {J0 , J1 , J2 }, with {J0 , J1 } a weak CTP-system for L3 . As L3 ≤ K3 , where E q (CG (z3 )) = J3 ∗ K3 , we conclude that [J1 , J3 ] = 1. Thus, {J0 , J1 , J2 , J3 , J4 } is a weak CTP-system of type A5 (q). Again if q = 3 and  = −1, we must establish that it is extendible. Since J2 , J3 , J4 = K ∼ = K4 = J0 , J1 , J2 , we need only show that J1 , J2 , J3 ∼ = K. ∼ Now J2 , Ji = U3 (3) for i = 1, 3, and so z2 centralizes elements t1 ∈ J1 and t3 ∈ J3 of order 4, so that t2i = zi , i = 1, 3. As [z1 , J2 ] = J2 = [z3 , J2 ], the images of t1 and t3 in Aut(J2 ) ∼ = Σ4 are commuting elements of order 4. Hence [z1 z3 , J2 ] = 1. Since Ji is the unique normal subgroup of CG (zi ) of its isomorphism type, and [J1 , J3 ] = 1, we have z1 = z3 . Hence z1 z3 is an involution and J1 , J2 , J3 ≤ CG (z1 z3 ). On the other hand, J1 and J3 are each conjugate to J2 , hence to each other. In  particular O 3 (CG (z1 )) = J1 ∗K1 with K1 ∼ = K. As [J1 , J3 , J4 ] = 1, J3 , J4 ≤ K1 . Thus there is a weak P-system {J3 , J4 , J5 } for K1 with [J3 , J5 ] = 1. Writing  Z(J5 ) = z5 we have z1 z3 = z5 ∼ z4 . So O 3 (CG (z5 )) = J5 ∗ K5 with K5 ∼ = K. Therefore J1 , J2 , J3 ≤ K5 , whence equality holds and J1 , J2 , J3 ∼ = K, proving extendibility. Putting G1 = J0 , J1 , J2 , J3 , J4 , we see that G1 /O2 (Z(G1 )) ∼ = L6 (q), by [III13 , 1.4, 1.14]. Note that E q (CG1 (x)) ∼ = SL2 (q) ∗ SL4 (q) ∼ = E q (CG (x)). But E q (CG (x)) is the only subgroup of CG (x) of its isomorphism type, as m2 (C(x, K)) = 1. Thus (10Y) holds for u = x. Also note that NG1 (E) is transitive on E # , so (10Y) holds for all u ∈ E # . As D = E, we have G1 = K, Ly , Lxy = G0 , completing the proof in Subcase 4a. This corresponds to row 3 of Table 14.1, with  = 5.

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Subcase 4b: E q (CG (x)) = K, so that Lemma 10.13a applies. We may take y = z2 and L = J2 , so that xy = z4 ∈ J4 . Then Ly = L and Lxy /Z(Lxy ) = L3 (q) with y ∈ E q (CLxy (y)) = J2 (see [III8 , 7.2]). We have E q (CG (xy)) = Lxy × J4 . We wish to show, as in the previous case, that (10AA)

there is v ∈ I2 (CG (x)) such that E(CK (v)) ∼ = SL3 (q).

Suppose that (10AA) fails. By the Z ∗ -theorem [IG , 15.3], there is v ∈ xG ∩ CG (x) with v = x, and we set E q (CK (v)) = Lv . If Lv ∼ = SL2 (q 2 ), then x ∈ Z(Lv ) and Lv is a component q of E (CKv (x)), where Kv = E q (CG (v)) ∼ = K. But then by [IA , Table 4.5.1], x ∈ Ω1 (Z(Kv )) = v , a contradiction. If v acts on K as an involution of K, then  E q (CG (x)). Since (10AA) fails, we v ∈ (xy)G , a contradiction as E q (CG (xy)) ∼ = conclude that v induces an outer involutory automorphism on K, and v centralizes an involution of K − Z(K), which we may take to be xy. If xy ∈ Lv , then xy ∈ Kv , and so xyv ∈ (xy)Kv . Then E q (CG (xyv)) = I0 × J0 with xyv ∈ J0 ∼ = SL2 (q) and I0 /Z(I0 ) ∼ = L3 (q). Now, E q (CI0 ×J0 (x)) = E q (CK (xyv)). But xyv ∈ K so xyv ∈ E q (CK (xyv)), whence E q (CJ0 (x)) = 1. Now, by [IA , Table 1 2 4.5.1], E q (CI0 /Z(I0 ) (x)) is isomorphic to one of L3 (q), L± 3 (q ), L2 (q), or SL2 (q). 1 ± 2 On the other hand, E q (CK (xyv)) is isomorphic to one of L± 4 (q ), Sp4 (q), Ω4 (q), 2  q SL2 (q ), SL2 (q) × SL2 (q), or SL3 (q). The only possibility is that E (CK (xyv)) ∼ = SL3 (q), a contradiction. We have proved (10AA). Replacing v by a K-conjugate, we may assume that v, D acts diagonalizably on K and DJ4 ≤ CK (v). Hence v induces a nontrivial inner-diagonal automorphism on J2 . Note that vxy does as well, and also satisfies J4 ≤ E(CK (vxy)) ∼ = SL3 (q), like v. Let g ∈ K exchange y and xy by conjugation, and consider E q (CG (y)) = J2 × Lgxy , which is v-invariant. We have J4 ≤ Lgxy and indeed J4 ≤ CLgxy (v). Hence v maps into CAut(Lgxy ) (J4 ), whose only involution is the image of xy ∈ Z(J4 ). In view of the previous paragraph we may then replace v by vxy if necessary and assume that [v, Lgxy ] = 1. Since [y, Lgxy ] = 1 but [D, Lgxy ] = 1 we have [x, Lgxy ] = 1. Let Lv be the subnormal closure of J4 in CG (v). Then I g ≤ Lv so [x, Lv ] = 1. Hence Lv is a nontrivial pumpup of E(CK (v)) ∼ = SL3 (q). Furthermore, Lv is D-invariant and J4   CLv (xy). This implies that Lv is a vertical pumpup of E(CK (v)). Moreover, since y ∈ xG , x does not shear to xy. By Lemma 1.2a and [III17 , 10.50], Lv /O2 (Lv ) ∼ = K/O2 (K). In particular by Lemma 1.2b, m2 (C(v, Lv )) = 1, so v ∈ Lv . Now xy ∈ J4 ≤ CLv (x) = CLv (xv). Hence xv acts nontrivially on the other SL2 (q) (solvable) component J of CLv (xy), so xv is J-conjugate, hence CG (J4 )conjugate, to xv(xyv) = y. In particular as y ∈ Lxy , y, xv is a four-subgroup of Lxy . The rest of the proof is mostly like that in Subcase 2a. We may assume that E(CK (xv)) = J3 , J4 . Now J2 = L ≤ Lxy , indeed J2 = E q (CLxy (y)). Therefore if we let J1 = E q (CLxy (xv)), {J1 , J2 } is a weak CTP-system for Lxy . Now J1  normalizes E(CK (xv))  CG (xv), [J1 , J4 ] = 1, and J1 = O r (J1 ). It follows that [J1 , J3 , J4 ] = 1. As {J2 , J3 , J4 } is a weak CTP-system for K, we have proved that

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{J1 , J2 , J3 , J4 } is a weak CTP-system for G1 := J1 , J2 , J3 , J4 of type A4 (q). If q = 3 and  = −1 we again check extendibility. We must show that H := J1 , J2 , J3 is a quotient of K. Arguing as in Subcase 4a we see that H centralizes [xy, xvz3 , where z3 = Z(J3 ). Moreover,  xvz3 ] = 1 and Lxy = J1 , J2 is a  J ,J 

component of CH (xy). As J3 ≤ J2 2 3 , H is the subnormal closure of Lxy in H. Let H ∗ be the pumpup of Lxy in CG (xvz3 ), so that H ≤ H ∗ . Since y ∈ J2 ≤ Lxy and xy is not conjugate to x = xyy, Lemma 1.2a and [III17 , 10.50] imply that H ∗ /O2 (H ∗ ) ∼ = K. Hence to prove extendibility it is enough to show that W := O2 (CG (xvz3 )) = 1. But all components of CG (u), u ∈ D# , are isomorphic to quotients of SU4 (3) and U3 (3), and hence are balanced (with respect to p = 2). Hence for all such u, CW (u) ≤ O2 (CG (u)) = 1 by assumption, and therefore W = 1, as desired. As a result, G0 is a quotient group of SL5 (q), by [III13 , 1.4, 1.14]. As J1 , J2 = Lxy and J2 , J3 , J4 = K, while Ly = J2 = L, we have G1 = Lxy , K = G0 . Moreover, E q (CG (x)) = K = E q (CG0 (x)) and E q (CG (xy)) = Lxy × J4 ∼ = E q (CG0 (xy)), and E q (CG (xy)) is the unique subgroup of CG (xy) of its isomorphism type and E q (CG (xy)) = E q (CG0 (xy)). As y and xy are K-conjugate, part (b) of the lemma follows in this case. Row 2 of Table 14.1, with  = 4, corresponds to this subcase. This finally completes the proof of Lemma 10.17.  Now, we quickly establish the following corollary, which together with Lemma 10.17 will complete the proof of Proposition 10.1. Corollary 10.18. Suppose that K is 2-saturated and K/Z(K) ∼ = Ln (q), n ≥ 4. Then ΓD,1 (G) ≤ NG (G0 ). We proceed in a short sequence of lemmas. Recall our choice (Table 14.2) of E in the various cases. Lemma 10.19. ΓE,1 (G) ≤ NG (G0 ). Proof. If q = 3 and O2 (CG (e)) = 1 for some e ∈ E # , then the result holds by Lemma 10.15. So assume that if q = 3, then O2 (CG (e)) = 1 for all e ∈ E # ∪D# (see Lemma 10.16). By [IA , 7.3.3] and Lemma 10.17b,   G0 = E q (CG0 (e)) | e ∈ E # = E q (CG (e)) | e ∈ E # . Hence, NG (E) ≤ NG (G0 ). Now let e ∈ E # , and let e ∈ E − e and Le be as in (10J1). Then (e )Aut(Le ) = (e )Le by Lemma 10.7. Hence if we put N = NCG (e) (Le ), then N = Le NN (E) ≤ NG (G0 ). In particular if (10J2) also holds, then CG (e) ≤ NG (G0 ). It remains to check that CG (e) ≤ NG (G0 ) in the two cases where (10J2) fails. In both, Le ∼ = SL4 (q) and G0 ∼ = L8 (q). But then |CG (e) : N | = 2 = |CG0 (e) : N ∩ G0 |. Thus, CG (e) ≤ G0 N ≤ NG (G0 ). The proof is complete. 

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Finally we prove: Lemma 10.20. ΓD,1 (G) ≤ NG (G0 ). Proof. By Lemma 10.19, we may assume that D = E, so L ∼ = SLn−1 (q) by # Table 14.2. Let M = G0 , ΓE,1 (G) ≤ NG (G0 ) and let d ∈ D . In every case, E ∩ L = 1. Choose e ∈ E ∩ L# . We have L ≤ Ld  CG (d) and Ld ≤ G0 . Now eG ∩ Ld = eLd by Lemma 10.7. Therefore CG (d) ≤ Ld CCG (d) (e) ≤ M . It remains to show that NG (D) ≤ M . If L  NG (D), then NG (D) = LCNG (D) (e) ≤ M and so we are done in that case. So assume that L  NG (D). Then L is not the unique (solvable) component of CG (D) = CCG (x) (y) of its isomorphism type, whence K ∼ = SL4 (q) and L ∼ = SL2 (q). But L ∼ = SLn−1 (q), a contradiction completing the proof of the lemma.  This completes the proof of Corollary 10.18 for the cases when K is 2-saturated. Likewise it completes the proof of Proposition 10.1.

11. The Orthogonal Case, Preliminaries In this section and the next we continue our standing assumptions and notation (1A), (1B) and (1C), as well as (1E). Our goal, to be reached in this section and the next, is: Proposition 11.1. Assume that K is a quotient of Ωn (q), q odd, n ≥ 6,  = ±1. If q = 3, assume that n ≥ 7. Assume finally that there is no (u, H) ∈ J∗2 (G) such that H/O2 (H) ∼ = SL± 4 (q). Then n is even, and G0 ∼ = Ωn+1 (q) or P Ω± n+2 (q). Moreover, ΓD,1 (G) ≤ NG (G0 ). Of course the value of  is immaterial if n is odd, and we omit it where convenient. The assumed nonexistence of SL± 4 (q) 2-components is justified by Proposition 10.1. We set k = [(n − 1)/2] and m = 2k, the largest even integer strictly less than n. Since (y, L) is an acceptable subterminal (x, K)-pair, it follows from [III12 , Def. 1.15] that one of the following holds: (1) n ≥ 7 and y ∈ L ∼ = ΩmL (q), where L = kq ; (2) n is even, n ≥ 8, y induces a graph automorphism on K, and L∼ = Ωn−1 (q); (11A) (3) n = 6, q > 3, and y ∈ L ∼ = SL2 (q), with E(CK (y)) = LLt for some t ∈ CK (y) − NG (L); or (4) n = 6, q ≡  (mod 8), y ∈ K ∼ = SL3 (q). = L4 (q), and L ∼ In the first two cases, L has a nonzero fixed point on the natural Ωn (q) module, so (11B)

x ∈ L.

Indeed (11B) clearly holds in the last two cases of (11A), so (11B) holds in all cases.

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We choose our quadruple (x, K, y, L) satisfying (1A), (1B), (1C), and (1E), so that (1) the hypotheses of Proposition 11.1 hold; (11C) (2) Subject to (1), if possible, (11A4) does not hold; and (3) Subject to (1) and (2), if possible, Z(K) has even order. We shall eventually see that under the hypotheses of Proposition 11.1, the conditions in (11C2, 3) are indeed possible. We choose an involution v ∈ CK (y) such that CK (v) has a component (or product of solvable components) J such that (11D) v∈J ∼ = ΩJ (q) (if n > 6) or SL2 (q) (if n = 6), m

kq .

Specifically, if y satisfies (11A1) or (11A3), we choose v = y and where J = J = L. In any case, we may choose (v, J) so that J ≤ L and [v, y] = 1. The following observations will be useful. Recall that C = C(x, K). Lemma 11.2. Suppose that n ≥ 7. Then (a) CCG (x) (J)/C is solvable; (b) If n is odd and Lxy ∼ = K, then C = O2 (C) x and CCG (x) (J) = O2 (CG (x)) x, y . Proof. In (a), as Out(K) is solvable, it suffices to consider CK (J). Let H = On (q) = O(V ), where V is a natural H-module. Without loss, we may consider J ≤ H and it suffices to show that CH (J) is solvable. Now CH (J) = Z(O([V, J])) × O(CV (J)), which is solvable as dim(CV (J)) ≤ 2. Thus (a) holds. If n is odd, then dim(CV (J)) = 1 and CH (J) is a four-group. By [III17 , 6.7], CAut(K) (J) ∼ = Z2 . Likewise in (b), CAut(Lxy ) (J) ∼ = Z2 . Since [x, Lxy ] = 1 and m2 (C) = 1, C = O2 (C) x . Then as CCG (x) (J)/C embeds in CAut(K) (J), CCG (x) (J) = O2 (CG (x)) x, y , completing the proof.  We first show that the case (11A4) can be avoided. Lemma 11.3. (x, K, y, L) satisfies (11A1), (11A2), or (11A3). Proof. By our hypothesis (11C2), it is enough to show that if K ∼ = L4 (q), q ≡  (mod 8), i.e., (11A4) occurs, then there exists another quadruple (x , K  , y  , L ) satisfying (1A), (1B), (1C), (1E), and K  /Z(K  ) ∼ = L4 (q) with Z(K  ) = 1. ∼ ∼ Write E(CK (v)) = J1 J2 , with J1 = J2 = SL2 (q) and [J1 , J2 ] = 1. We may assume that [v, y] = 1 and J1 ≤ L. Hence, [y, J2 ] = J2 . Let E ∼ = E23 be the elementary abelian image in K of a diagonalizable subgroup of SL4 (q), chosen so that y, v ≤ E. Thus y, v = CE (J1 ). Choose u ∈ {y, xy} such that L < Lu . Then since N is a level neighborhood, Lu has untwisted Lie rank at least 3, by [III17 , 10.48]; moreover, F(Lu ) ≤ F(K) = (q 9 , A) by Lemma 1.2, so Lu /Z(Lu ) ∼ = L4 (q). Then Lemma 1.2b implies that (u, Lu ) ∈ J∗2 (G). But then it follows easily from [III12 , Def. 1.15] that if Z(Lu ) = 1, there exists (y  , L ), a subterminal (u, Lu )-pair, such that (u, Lu , y  , L ) satisfies all the hypotheses of the proposition. Hence if Z(Lu ) = 1, we are done; so we may assume that Z(Lu ) = 1, whence Lu ∼ = K. Then since v ∈ J1 ≤ L ≤ Lu , we have E(CLu (v)) ∼ = J1 J2 with J1 a component of CLu (v). Thus E(CLu (v)) = J1 ∗ J3 with [x, J3 ] = J3 . Write x, y # = {x, u, u1 };

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we argue that J1 is a component of E(CLu1 (v)) as well. Indeed if L < Lu1 , this holds by the same argument that established it for Lu , while if Lu1 = L ∼ = SL3 (q), then it is obvious. Hence, J1 is a component of E(CG (v, d)) for all d ∈ x, y # . Let H be the subnormal closure of J1 in CG (v). Notice that since Ld is qua# sisimple for all d ∈ x, y , H too is quasisimple, by [III8 , 3.3] applied with J1 in the role of J there. Then by [III11 , 1.16], either J1   CG (v) or J1 ∼ = 2Ar for some r ≥ 9. But the latter is impossible by [III9 , 2.8]. We have proved that J1 is a component of CG (v). By symmetry, so is J2 ; and indeed since Lu satisfies all the conditions that K does, J3 is also a component of CG (v). Since [x, J2 ] = 1 = [x, J3 ], J2 = J3 , so J1 , J2 , and J3 are distinct. In fact the symmetry between J1 and J2 allows us to construct a fourth component J4 ∼ = SL2 (q) of CG (v). Namely, there exists y  ∈ E # ∩ y K such that [y  , J2 ] = 1. Write y  = y g , g ∈ K. Then L = E(CK (y  )) ∼ = SL3 (q), and as L has one class of involutions we may assume g was chosen so that J1g = J2 . Set J4 = J3g . Then J2 J4 = E(CLgu (v)) and J1 , J2 , and J4 are distinct, and [x, J4 ] = J4 . Also set u = ug . As g ∈ K and u ∈ {y, xy}, uu = yy  . Suppose that J3 = J4 . Since x acts nontrivially on both J3 and J4 , there exist by [III11 , 6.3a] elements t3 ∈ J3 and t4 ∈ J4 of order 4 inverted by x. Set t = t3 t4 . As v is the unique involution of both J3 and J4 , t is an involution in CCG (x) (J1 J2 x ). Note that by [III17 , 11.19b], CAut(K) (J1 J2 ) is a dihedral group. Also, as CAut(Lu ) (L) is cyclic of exponent q − , and x acts nontrivially on Lu , a Sylow 2-subgroup of C(x, K) embeds in Zq− . If t ∈ v, x then t = x or vx and replacing t4 by t−1 we may assume that 4 t = x. But then t4 is not inverted by x, contradiction. Therefore t ∈ v, x . As v ∈ K, t ∈ v C(x, K). Therefore by [III17 , 11.19b], CK (t) ∼ = P Sp4 (q). Let 2n be the 2-part of q − , so that n ≥ 3. Then a Sylow 2-subgroup of CJ3 J4 (t) contains an extension of an abelian group A of type (2n , 2n−1 ) by an involution t inverting A. Moreover, v = n−1 (A). Consider the subnormal closure I of I0 := CK (t) ∼ = P Sp4 (q) in CG (t). As A t centralizes J1 J2 , it normalizes I, and |A : CA (I0 )| ≤ 2. Since J1 is a component of CI0 (v) as well as of CG (v), J1 is a component of CI (v), so I is not a diagonal pumpup of I0 . Let I = I/O2 (I). If I is a trivial pumpup of I0 , then as |CAut(I) (J 1 J 2 )| = 2, |A : CA (I)| ≤ 2. Hence v ∈ Φ(A) ≤ CA (I), which is absurd. Therefore I is a vertical pumpup of I0 . But F(I/O2 (I)) ≤ F(K) = (q 9 , A), so by [III11 , 12.6b], |CAut(I) (I 0 )| ≤ 2. As noted above, |A : CA (I0 )| ≤ 2, and combining these inequalities gives |A : CA (I)| ≤ 4. Therefore again we reach the contradiction v ∈ C(t, I). Therefore J3 = J4 . We have [u, J3 ] = 1. Conjugating by g, [u , J4 ] = 1. In particular [uu , J3 ] = 1, and we have seen above that uu = yy  . Let v  = yy  . Then v  ∈ v K , [v, v  ] = 1, and J3 is a component of CG (v, v  ). Hence, writing v = (v  )h , h ∈ K, and setting w = v h , we see that [v, w] = 1 and J3h is a component of CG (v, w ) with w ∈ J3h . Moreover, w centralizes neither J1 nor J2 . Hence J3h acts faithfully on both J1 and J2 . By L2 -balance, J3h must be a diagonal of J1 and J2 , but then J3h ∼ = L2 (q), a contradiction as J3 ∼ = SL2 (q). The proof is complete. 

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Lemma 11.4. The following conditions hold: #

(a) The pumpup of J in CG (w), for any w ∈ v, x , is semisimple; and (b) J   CG (v). Proof. By Lemma 11.3, (11A4) does not hold. If (11A1) or (11A3) holds, then v = y and (a) holds by our basic setup (1C). If (11A2) holds, then (a) follows from the semisimplicity of the neighborhood N and [III8 , 3.3], with w in the role of z in that lemma. Now suppose that (b) is false and let Jv be a component of the subnormal closure of J in CG (v). Set Cv = CG (v) and C v = Cv /O2 (Cv ). Thus J v = J. If (v, Jv ) is a diagonal pumpup of J, then the subnormal closure Jv∗ of J in CG (v) is the product of two components interchanged by x. But N(x, K, y, L) is a vertical neighborhood, so v ∈ x, y , whence (11A2) holds and in particular n ≥ 8. At any rate y ∈ CG (v) and J ≤ L ≤ CG (y), so x, y normalizes Jv∗ . Hence some d ∈ {y, xy} centralizes Jv∗ , in which J is diagonally embedded. But by L2 -balance, L2 (CJv∗ (d)) ≤ Ld . In particular Ld is a nontrivial pumpup, hence a vertical pumpup, of L, as N is a vertical neighborhood. Now L ∈ G62 = Gd2 , so Ld ∈ Gd2 and hence F(Ld ) ≤ F(K) by Lemma 1.2. But then by [III17 , 10.62], Ld cannot contain a subgroup with two components factors centrally isomorphic to J/Z(J), a contradiction. Since we are assuming the lemma false, Jv is a vertical pumpup of the 2-component J. We remark for later use that this argument did not use the fact that v ∈ Jv , and so it is valid with vx replacing v: (11E)

the subnormal closure Jvx of J in CG (vx) is a single 2-component.

If (11A1) occurs, then v = y and J = L. Then by (1C2), Jv = Ly > L. But d(G) and F(Ly ) > y ∈ L ≤ Ly , and so [III17 , 10.61] yields Ly /O2 (Ly ) ∈ G62 = G2 F(K), as n > 6. This contradicts Lemma 1.2, however. Similarly, if (11A2) occurs, then still v ∈ J ≤ Jv and n > 6, and by [III17 , 10.61] F(Jv ) > F(K), once again contradicting Lemma 1.2. Finally suppose that (11A3) holds, so that v = y, whence Jv = Ly has level q > 3. Moreover, J = L < Ly = Jv , which implies that Jv ∈ G2 . As v ∈ Jv , d 9 Jv /Z(Jv ) ∼ = L± 3 (q), so Jv ∈ G2 . Consequently F(Ly ) ≤ F(K) = (q , A) by Lemma 1.2. Then one of the following holds by [III17 , 10.38]: (1) Ly ∼ = A± 3 (q), Ly is 2-saturated and x induces a graph automor(11F) phism on Ly ; or (2) Ly ∼ = Spin5 (q), and L = E(CLy (x)). Suppose that (11F2) holds. As Lt = L it follows that Ly = Lty . In particular m2 (C(y, Ly )) = 1. If (y, Ly ) is not 2-terminal in G, then using y ∈ Ly and [III8 , 2.7], there exists u ∈ I2 (C(y, Ly )) such that the pumpup (u, H) of Ly in CG (u) is nontrivial. Since y ∈ Ly we conclude first that H is a vertical pumpup of Ly and then, by [III17 , 10.52], that H/O2 (H) ∈ G62 and F(H) > F(K). But then for a 2-terminal long pumpup (u∗ , H ∗ ) of (u, H), H ∗ ∈ G62 and F(H ∗ ) ≥ F(H) > F(K) by [III11 , 12.4], contradicting (x, K) ∈ J∗2 . Hence (y, Ly ) is 2-terminal in G. But Ly ∈ G62 , so (y, Ly ) ∈ J2 (G) and then m2 (C(y, Ly )) = 1 by [III12 , Theorem 1.2], a contradiction. Therefore (11F1) holds. But F(Ly ) = F(K), so by Lemma 1.2b, (y, Ly ) ∈ J∗2 (G). As Ly /O2 (Ly ) ∼ = SL± 4 (q), this contradicts the hypothesis of Proposition 11.1 and the lemma is proved. 

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We choose notation so that in case (11A2), (11G)

L < Lxy .

Since y = v in the cases (11A1, 3), (11G) holds in all cases by Lemma 11.4. Lemma 11.5. Suppose that (11A2) holds. Then the following conditions hold: (a) Lxy is a quotient of Ωζn (q) for some sign ζ; and  (b) Either Ly = L or Ly is a quotient of Ωζn (q) for some sign ζ  . Proof. We know that Lxy is a level vertical pumpup of L ∼ = Bk (q) with 2 F(Lxy ) ≤ F(K) = (q (k+1) , D) by Lemma 1.2. As L is a quotient of Ωn−1 (q), Lxy is not of type F4 , and so by [IA , 4.5.1, 4.5.2], Lxy must have untwisted Lie rank k + 1 and must be of type D, proving (a). As Ly is a level nondiagonal pumpup of L, part (b) follows similarly.  Lemma 11.6. Let Jxv be the pumpup of J in CG (xv). Then the following conditions hold: (a) Jxv is a quotient of Ωηn (q) for some n and some sign η; and (b) One of the following holds: (1) n = n ; (2) n is even and n = n − 1. (Recall that in cases (11A1, 3), Jxv = Lxy .) Proof. First suppose that n ≥ 7. We have J ∼ = ΩmJ (q) with J = q k by (11A). It suffices to show that the pumpup Jxv of J in CG (xv) is vertical and level. For then by [III17 , 10.65], applied to this pumpup, one of the following holds: (1) Part (a) of the lemma holds with n > m; (2) Jxv is a 2-saturated quotient of SL± (11H) m (q); or (3) J xv := Jxv /O2 (Z(Jxv )) ∈ G62 and F(Jxv ) > F(K). However, (11H3) contradicts Lemma 1.2a, so (11H1) or (11H2) holds. Even then J xv ∈ G62 and so (11I)

F(Jxv ) ≤ F(K).

Suppose that (11H1) holds, so that m < n . According as n is even or odd we have m = n − 2 or n − 1. However, (11I) forces n ≤ n. For n odd, m < n ≤ n yields n = n and (b) holds; while for n even, n = n − 1 or n and again (b) holds. On the other hand, if (11H2) holds, then the untwisted ranks of J xv and K are m − 1 and at most 1 + m 2 , respectively, with m ≥ 6, so F(Jxv ) > F(K), a contradiction. When n ≥ 7 it remains to show that Jxv is a vertical and level pumpup of J∼ = ΩmJ (q). By Theorem C∗7 : Stage 3a and Proposition 5.1, this has been shown if v = y, i.e., in case (11A1). So we may assume that (11A2) holds and in particular n ≥ 8. By (11E), Jvx is a single 2-component. Suppose that Jxv is a vertical but non-level pumpup of J. As J ∈ Gd2 , also J xv ∈ Gd2 and so by Lemma 1.2, 2 2 2 F(Jxv ) ≤ F(K) = (q (k+1) , D), with k ≥ 3. Since (q 2 )k > q (k+1) , we see that x cannot induce a field or graph-field automorphism on J xv . The only other nonlevel possibilities, by [IA , Table 4.5.1], are Jxv ∼ = A7 (q 1/2 ) with k = 3, and Jxv ∼ = ± 1/2 D2k (q ), both of which give the contradiction F(Jxv ) > F(K). Therefore we only need rule out the possibility that Jxv is a trivial pumpup of J. For this, recall that

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[v, x, y ] = 1 and consider the action of x, v on Lxy . By Lemma 11.5, x acts like a reflection on Lxy /Z(Lxy ) ∼ = P Ωζn (q), centralizing L ∼ = Ωn−1 (q). Also v ∈ Z(J) ≤ L so the supports of v and x on the natural Lxy -module are orthogonal. We have J/Z(J) ∼ = P Ω± n−2 (q), so Lxv := E(CLxy (xv)) is a quotient of Ωn−1 (q) (and is not centralized by x). As J < Lxv , J has a nontrivial pumpup in CG (xv), as required. Finally suppose that n = 6. Then (11A3) holds, whence v = y and so xv = xy and Jxv = Lxy . Then L ∼ = SL2 (q), q > 3. Also, E(CK (y)) is the product of L and a conjugate component L1 = Lg for some involution g ∈ CK (y) ∩ y K , by [III17 , 11.19a]. Since [x, Lxy ] = 1, we have [y, Lxy ] = 1 and so LL1 ≤ Lxy by L2 -balance. Again as F(Lxy ) ≤ F(K) and K ∼ = A± 3 (q), Lxy as well has untwisted ± ∼ rank at most 3; indeed either Lxy = A3 (q) or Lxy has untwisted rank at most 2. Now LL1 is a product of components of CLxy (y) by L2 -balance. Thus [III17 , 10.42] implies that Lxy is a quotient of Ωηn (q) with n = 5 or 6, in which cases the lemma holds, or Lxy ∼ = G2 (q), q = 3n , with g inducing a graph-field automorphism of Lxy . Recall that q > 3, so n > 1. It remains to rule out this last configuration. But if it holds, then by L2 -balance and the fact that g a = y for some a ∈ G, CG (y) has a 2-component H which is a n n ∼2 ∼2 pumpup of E(CLxy (g))a ∼ = E(C Lxy (g)) = G2 (3 2 ). By [III17 , 10.32], H = G2 (3 2 )  or Lxy . Let H ∗ = H CG (y) . Then [H ∗ , LLg ] = 1. Consequently L2 (CH ∗ (x)) ≤ CCG (x) (LLg )(∞) . Since CAut(K) (LLg ) is solvable, L2 (CH ∗ (x)) ≤ C(x, K) and so x is the unique involution of x L2 (CH ∗ (x)). Given the isomorphism type of H, however, this is impossible by [IA , 4.5.1]. The proof is complete.  As a consequence of Lemma 11.6, we have: Lemma 11.7. The following conditions hold: (a) A Sylow 2-subgroup Q of C(x, K) is cyclic; (b) A Sylow 2-subgroup Qxv of C(xv, Jxv ) is cyclic or dihedral; and (c) K = L2 (CG (x)) = E(CG (x)) and Jxv = L2 (CG (xv)) = E(CG (xv)). Proof. Since [Q, J] ≤ [Q, K] = 1, [Q, vx] = 1. Also x = Ω1 (Q) does not centralize Jxv . Therefore Q embeds in CAut(Jxv ) (J), which has dihedral or cyclic Sylow 2-subgroups. As Q has a unique involution, (a) holds. Similarly [Qxv , x] = 1 as v ∈ J ≤ Jxv , so Qxv embeds in CAut(K) (J), whose Sylow 2-subgroups are cyclic or dihedral, proving (b). As xv ∈ Z(C(xv, Jxv )), C(xv, Jxv ) is 2-nilpotent and (c) follows easily.  For the rest of this section, we assume that n > 6. The next lemma gets us part of the way to the condition in (11C3), which is our next objective. Lemma 11.8. If n > 6, then K ∼ = Ωn (q) (or 3Ω7 (3), with n = 7). ∼ P Ω (q) = ∼ Ω (q), and Proof. Suppose false. Then n = 2(k + 1) is even, K = n n  q k+1  = q . Moreover, v ∈ E(CK (v)) = J ∼ = Ωn−2 (q). As Z(K) = 1, there is v1 ∈ v K ∩ J, and v1 has 2-dimensional support on the natural K-module. Let J1 = E(CK (v1 )) ∼ = J; then J1   CG (v1 ) by Lemma 11.4,   as v1 ∈ v K . Set J0 = O r (CJ (v1 )) = O r (CJ1 (v)) ∼ = Ωn−4 (q). Since [x, J1 ] = 1, r J0 = O (CJ1 (xv))   CG (xv, v1 ). As J0 ≤ J ≤ Jxv , J0   CJxv (v1 ). But v1 has

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2-dimensional support on the natural J-module, hence on the natural Jxv -module,   and so J0 ∼ = Ω± n −2 (q). Thus n −2 = n−4. However, n = n or n−1, a contradiction proving the lemma.  Lemma 11.9. Suppose that n is even and n ≥ 8, and K ∼ = Ωn (q) has trivial − ∼  P Ωn (q). center. Then Then Jxv /Z(Jxv ) = Proof. Suppose false and take a 2-terminal long pumpup (v ∗ , J ∗ ) of (xv, Jxv ). Since F(J ∗ /O2 (J ∗ )) ≥ F(Jxv ) = F(K), Lemma 1.2b implies that (v ∗ , J ∗ ) ∈ J∗2 (G), whence m2 (C(v ∗ , J ∗ )) = 1. Then by [III8 , 2.4], m2 (C(xv, Jxv )) = 1. So (xv, Jxv ) ∈ J∗2 (G). If there is an acceptable (xv, Jxv )-subterminal pair (y  , L ), then (xv, Jxv , y  , L ) satisfies (1A) and (1B1). The rest of (1B) is just notation, and (1C) is then satisfied by Theorem C∗7 : Stages 3a, 3b (with notation changed appropriately). Then (1E) is satisfied by Proposition 5.1. Hence (xv, Jxv , y  , L ) competed with (x, K, y, L) in our choice (11C). Indeed, in (11C3), Z(Jxv ) has odd order (since Z(K) did). The previous development thus applies to (xv, Jxv ) instead of (x, K). In particular by Lemma 11.8, Jxv ∼ = Ω− n (q), so Z(Jxv ) has even order, contradiction. Thus, there is no acceptable (xv, Jxv )-subterminal pair. In particular (v, J) is not acceptable. So, by definition [III12 , Def. 1.15], (xv, Jxv , v, J) must be ignorable. This means that y acts as a reflection on K, but no involution of CG (xv) acts as a reflection on Jxv . In particular (x, K, y, L) is in case (11A2). We complete the proof by showing that this leads to a contradiction. Namely, we may assume without loss that y centralizes J, and in particular [y, xv] = 1. Then E := x, y, v ∼ = E23 and [E, J] = 1. Now E maps into CAut(Jxv ) (J), in which any four-group contains a reflection. So the image of E has order at most 2. But m2 (CE (Jxv )) ≤ m2 (C(xv, Jxv )) = 1, so |E| ≤ 4, a final contradiction.  Let VK be a natural K-module giving rise by restriction to natural modules VL and VJ of L and J. (Note that there is no possible triality ambiguity here, since K = Ω(VK ), L = Ω([VK , L]) and J = Ω([VK , J]).) Next, we choose an involution z4 ∈ J such that the −1-eigenspace of z4 on each of VJ , VL and VK is 4-dimensional of + type. Then the −1-eigenspaces of z4 on the natural Jxv - and Lxy -modules are also 4-dimensional of + type. (Recall that Jxv is quasisimple, by Lemma 11.4a.) We have

(11J)

O 2 (CJ (z4 )) = R0 IJ , O 2 (CK (z4 )) = R0 Ix , and O 2 (CJxv (z4 )) = R0 Ixv ,  where [R0 , IJ ] = [R0 , Ix ] = [R0 , Ixv ] = 1, R0 = O r (R0 ) ∼ = SL2 (q) ∗ J 2 2  ∼ ∼ SL2 (q) ∼ O O (q), I (Ω (q)), I (Ω (q)), and Ixv ∼ = = = = Ω+ J x n−4 4 n−6 η 2 O (Ωn −4 (q)).

Here R0 ∼ = Ω+ 4 (q) acts naturally on its supports on VK , VL , and VJ . Next, set C4 = CG (z4 ). Lemma 11.10. Suppose that q > 3. Then R0  E(C4 ). #

Proof. If q > 3 then by Lemma 11.4a, J ≤ E(CG (w)) for all w ∈ x, v . It follows that R0 ≤ E(CG (w)) for all such w. By [III8 , 3.3], R0 ≤ E(C4 ). Then L2 -balance implies that the normal closure R0∗ of R0 in E(C4 ) is an x, v -invariant

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product of components of C4 such that R0   CR0∗ (w) for all w ∈ x, v # . Thus by [III11 , 1.16], R0 = R0∗ unless q = 5 or 9 and C4 has a component isomorphic to 2An , n ≥ 9. But such alternating components do not exist by [III9 , 2.8], so the lemma follows.  For the next several lemmas, we consider two situations, one quite general, the other quite special. In each case we define a subgroup I: Either (1) n ≥ 8, K ∼  Ω+ = 8 (q), and I is defined as the subnormal closure of I in C , or x 4 (11K) ∼ − (2) K ∼ = Ω+ 8 (q), Jxv = Ω8 (q), and I is defined as the subnormal closure of Ixv in C4 . We fix any u ∈ I2 (R0 ) − z4 ; thus, the commutator of u on the natural module case (11K2)) is 2-dimensional of type q and lies is 4-dimensional of + type. We also set  q E(CK (u)) ∼ (q) = Ωn−2 Ku = −q ∼ E(CJxv (u)) = Ω6 (q)

for K (in case (11K1)) or Jxv (in in the commutator of R0 , which

in case (11K1) in case (11K2)

Lu = the subnormal closure of Ku in CG (u). Lemma 11.11. Assume (11K). Then Lu is a component of CG (u) and ∼ P Ω± (q) where  = n if n is even, and  = n − 1 if n is odd. Lu /Z(Lu ) =  Proof. By L2 -balance and [III8 , 3.3], with x, v , J, and u in the roles of D, L, and z in that lemma, Lu is a product of components of E(CG (u)). Suppose first that Lu is a diagonal pumpup of Ku . Now, u ∈ R0 ≤ J with Ju,v := O 2 (CJ (u)) ∼ = + (q) (Ω (q) if n = 8). Also J   C (u, v), since u ∈ J   C (v). But Ω± u,v G G n−4 4 v ∈ u Ju,v ≤ u Ku and so v normalizes both components of Lu , with Ju,v embedded diagonally in O 2 (CLu (v)). This contradicts Ju,v   CG (u, v), however, so Lu is a component of CG (u). Suppose that (11K1) holds; the proof in the other case is similar. As u ∈ R0 we see that E(CJxv (u)) ∼ = Ω± n −2 (q) is not centralized by x, and so Lu is a vertical 6 pumpup of Ku . As Ku ∈ G2 , also Lu ∈ G62 , so F(Lu ) ≤ F(K) by Lemma 1.2. Then ± by [III17 , 10.63], Lu ∼ = Ω± n−δ (q) or P Ωn−δ (q), δ = 0 or 1. We let Vu be the natural module for Lu . Let du = dim(Vu ) = n − δ. Then v and x act on Lu like simultaneously diagonalizable involutions of O(Vu ). By abuse of language we consider v and x to act on Vu . As Ku = E(CLu (x)), x has an eigenspace W of dimension n−2 and type q , which is the support of Ku . As v and uv ∈ Lu have the same action on Vu , the support of v on Lu is a nondegenerate subspace W1 of W of dimension m − 2. Write W = W1 ⊥ W2 . Then W2 has dimension n − m and is an eigenspace of xv. Hence E(CLu (xv)) ∼ = Ω± du −(n−m) (q). ∼ Ω± (q). Thus du − (n − m) = n − 2. If But E(CG (u, xv)) = E(CJxv (u)) = n −2 n is even, then n − m = 2 and du = n . If n is odd, then n − m = 1 and  du = n − 1 = n − 1.

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Lemma 11.12. Assume (11K). Then I is a component of C4 . Proof. The proof is similar in the two cases of (11K). Assume for specificity that (11K1) holds. Then Ix is quasisimple. Choose a four-group U ≤ R0 with u ∈ U but z4 ∈ U . Since R0 ∼ = SL2 (q) ∗ i SL2 (q), some g ∈ R0 of order 3 acts transitively on U # , and we define Lugi = Lgu ,  0 ≤ i ≤ 2. Now O r (CLu (U )) contains Ix , and [g, Ix ] ≤ [R0 , Ix ] = 1, so for all u ∈ U # , Ix ≤ Lu ≤ E(CG (u )). In particular as Lu ∈ Chev(r), r odd, Ix ≤ CLu (U z4 )(∞) = L2 (CLu (U z4 )) = L2 (CG (U z4 )), by L2 -balance. We apply [III2 , 2.2] with the roles of d, e, and R there played by z4 , u , and U z4 . We conclude that [Ix , CO2 (C4 ) (u )] ≤ O2 (CG (u )). But then [Ix , CO2 (C4 ) (u )] = [Ix , CO2 (C4 ) (u ), Ix ] ≤ [O2 (CG (u )), E(CG (u ))] = 1. As u ∈ U # was arbitrary, [Ix , O2 (C4 )] = 1. This implies that I ≤ E(C4 ). To complete the proof it suffices by L2 -balance to rule out the possibility, which we assume, that I = I1 I2 with Ii /Z(Ii ) ∼ = Ix /Z(Ix ) for i = 1 and 2. Observe that Ix ≤ Ku ≤ Lu . Thus, Ix ≤ Lu,4 := O 2 (CLu (z4 )) ∼ Here by Lemma 11.11, = Ω± −2 (q).   L  − 2 ≥ n − 3 ≥ 5 so Lu,4 is quasisimple. But then Lu,4 = Ix u,4 ≤ I, so I1 /Z(I1 ) involves Lu,4 . As I1 /Z(I1 ) ∼ = P Ω± n−4 (q), we have a contradiction and the lemma is proved in this case.  Replacing Ix by Ixv , we obtain a proof in case (11K2). Lemma 11.13. If (11K) holds, then [R0 , I] = 1 and R0   C4 . Proof. We give the proof in the case (11K1); a proof in the case (11K2) can be obtained by interchanging x and xv everywhere. First suppose that [u, I] = 1. Now R0 ≤ C4 and [R0 , Ix ] = 1, so R0 normalizes I and then as R0 ∼ = Ω+ 4 (q) with z4 = Z(R0 ), R0 / z4 acts faithfully on I. If q > 3 then by L2 -balance, R0 ≤ I and so E(CI (x)) = R0 Ix with Z(R0 ) = z4 ≤ Z(I) ∼ − and Ix ∼ = Ω± n−4 (q) with n − 4 ≥ 5 or Ix = Ω4 (q). By [III17 , 10.60], no such group I exists, contradiction. Hence, q = 3. By solvable L2 -balance, O2 (R0 ) ≤ I and so again z4 ∈ Z(I). We apply [III11 , 13.1]. As u ∈ O2 (R0 ) and CI (u) has a component (containing Ix ), I ∈ Chev(2), by the Borel-Tits theorem. Also the component Ix ∼ = Ωn−4 (q) is not isomorphic to 2An for any n, so neither is I. As z4 ∈ Z(I), the only alternative is that I ∈ Chev(3) with level 3. Let H0 = R0 I and H 0 = H0 /O2 (H0 ). By [III11 , 10.2], H 0 = I. But by [III17 , 10.60] again, there is no such group I, a contradiction. We have shown that [u, I] = 1. Therefore I is a component of CLu (z4 ), whence I = E(CLu (z4 )) ∼ = Ω± −2 (q). ± r ∼ Here  − 2 = n − 3 or n − 2. But [R0 , Ix ] = 1 with Ix = Ωn−4 (q), and R0 = O (R0 ),  so R0 maps into O r (CAut(I) (Ix )) = 1. That is, [R0 , I] = 1. By Lemma 11.10, it remains to assume that q = 3 and prove that R0   I. Now z4 v ∈ Ix ≤ I. Set H = E(C4 )CC4 (E(C4 )). Then H ≤ I, CC4 (I) ≤ I, CC4 (z4 v) = I, CCG (v) (z4 ) . As z4 ∈ J − Z(J) with J   CG (v), CCG (v) (z4 ) normalizes J and then CJ (z4 ), and then R0 . (Note that J ∼ = Ω± m (q) admits no triality automorphism when m = 8.) As [R0 , I] = 1, H normalizes R0 . As q = 3, Aut(R0 ) is solvable, so [R0 , E(C4 )] = 1 and R0 ≤ H. Thus R0  H  C4 , and the proof is complete.  Lemma 11.14. Suppose that (11K) holds and n is even. Then I ∼ = Ω± n −2 (q). η q If n = n, then I ∼ = Ωn−2 (q), and u and xv are K-conjugate. 

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Proof. By Lemma 11.13, [u, I] = 1 for our u ∈ I2 (R0 ) − Z(R0 ). Hence the component I of CG (z4 ) is a component of CG (u, z4 ) containing Ix or Ivx in cases (11K1) and (11K2), respectively. Accordingly, Lu contains Ix or Ixv , so I is a component of CLu (z4 ) = CLu (z4 u). But z4 u ∈ Ku has two eigenvalues −1, hence z4 u has a 2-dimensional eigenspace of type q on the natural Lu -module. Note also that both xv and u lie in K and their −1-eigenspaces on the natural K-module η are both 2-dimensional, so they are K-conjugate. Thus Lu ∼ = Jxv ∼ = Ωn (q). This implies the result.  Now we can attain our objective. Lemma 11.15. If n is even and n = 6, then Z(K) ∼ = Z2 . Proof. By Lemma 11.8, K ∼ . As n = 2(k + 1), we may = Ωn (q). Let π = k+1 q ∼ (q), a simple group. In particular if n = 8, then assume that  = π, that is, K = Ω−π n K∼ (q). Hence (11K1) holds, and so Lemmas 11.11–11.14 apply. Moreover, by = Ω− 8  P Ωπn (q). Thus, Jxv ∼ Lemma 11.9, Jxv /Z(Jxv ) ∼ = Ωn−1 (q) or Jxv ∼ = K. = Since Ix ∼ (q), x acts on I as = Ω± = Ωn−4 (q) is a component of CI (x), with I ∼  n −2 an involution with one or two eigenvalues −1 on the natural I-module, according as n = n − 1 or n. In any case there is an involution z0 ∈ CI (z4 , x) and a subnormal subgroup I0 ∼ = Ωπn−4 (q) of CI (z0 ) such that z0 = Z(I0 ) = z4 , I0 ∩ Ix ∼ = Ωn−5 (q), and x acts on I0 as a reflection. Set x = z4 z0 , so that x is an involution centralizing  x. Let K  be the subnormal closure of O 2 (R0 I0 ) in CG (x ). Now x, z4 acts on  R0 I0 K with R0 I0   CR0 I0 K  (z4 ); and CR0 I0 K  (x, z4 ) = R0 (I0 ∩ Ix ). We argue that z4 ∈ O2 2 (CG (x )). Namely, consider any u ∈ I2 (R0 ) − {z4 }. We know that I ≤ Lu   CG (u) with Lu /Z(Lu ) ∼ = Ω± n (q). Then z0 ∈ Lu acts on the natural Lu -module with (n − 4)dimensional support. As I = E(CLu (z4 )), the support of z4 on the natural Lu module is disjoint from that of z0 , and is 2-dimensional. Therefore E(CLu (x )) ∼ =  Ω± (q). Moreover, E(C (x )) is not centralized by z . In particular z ∈ L 4 4 u n−2 O2 2 (CG (x )), as asserted. As x = z4 z0 , also z0 ∈ O2 2 (CG (x )). Since z4 = Z(R0 ), (solvable) L2 balance implies that K  is a product of 2-components of CG (x ). Label the pumpups   of O 2 (R0 ) and O 2 (I0 ) in K  as KR and KI , respectively. Since z4 ∈ O2 2 (CG (x )), but z4 ∈ L2 (CG (x )), KR and KI are single 2-components. An application of [III8 , 3.3], with a four-subgroup U ≤ R0 playing the role of D (with U ∩Z(R0 ) = 1), shows that KI ≤ E(CG (x )). If KR = KI , then z0 centralizes KR /O2 (KR ), as does x = z4 z0 , so z4 ∈ Z ∗ (KR ) ≤ O2 2 (CG (x )), contradiction. Therefore KR = K I = K  , a single component of CG (x ). As z4 ∈ KR and z0 ∈ KI , x ∈ K  , so x ∈ Z(K  ).  Note that K  is a pumpup of E(CLu (x )) ∼ = Ω± n−2 (q), with u ∈ O2 (R0 ) ≤ K . Then  π if q > 3, we have R0 × I0   CK  (z4 ), and by [III17 , 10.59], K ∼ = Ωn (q). On the other hand, if q = 3, then by [III11 , 13.1], K  ∈ Chev(3) has level 3. It follows again from [III11 , 10.2] and [III17 , 10.59] that K  ∼ = Ωπn (3). To complete the proof it remains to replace K by K  , and then deduce from (11C3) that Z(K) = 1. To be able to do this, we must verify that (x , K  ) ∈ J∗2 (G) and that there exists an acceptable (x , K  )-subterminal pair. Since F(K) = F(K  ), Lemma 1.2b yields (x , K  ) ∈ J∗2 (G). Then, as in Lemma 11.9, we must argue that if

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K satisfies (11A2), then some involution of CG (x ) acts on K  like a reflection. But we know that x acts on K  , acting on I0 like a reflection and centralizing R0 . Then either x or xz4 acts on K  like a reflection, as needed. The proof is complete.  12. The Orthogonal Case, Completed We continue the setup of the previous section, including the quadruple (x, K, y, L) satisfying K ∼ = Ωn (q) and (11C), the involution v and the component J of CK (v). We shall treat first the case n odd, n > 6, then the case n even, n > 6, and finally the case n = 6. Through Lemma 12.11 we consider the case (12A) n odd, n ≥ 7. ∼ K, and then (xy, Lxy ) ∈ J∗ (G) by Lemma 1.2b. Thus By Lemma 11.6, Lxy = 2 there is symmetry between x and xy. Let 2a be the largest power of 2 not exceeding n. Recall that T ∈ Syl2 (CG (x)) and Q = CT (K). By [III17 , 8.9c], Z(T ) ∩ K contains an involution z and K has a subgroup Kz such that z ∈ Kz ∼ = Ω+ 2a (q) and Kz   CK (z). In fact (12B)



O r (CK (z)) = Kz Mz ,

where Mz ∼ = Ωn−2a (q) (and here, by convention, Ω1 (q) = 1). Since n is odd, n − 2a is odd, y = v, and L = J. We assume, as we may, that the support of z on the natural K-module is contained in that of y. Thus Kz ≤ L. Lemma 12.1. Assume (12A). Then the following conditions hold: (a) x ∈ Syl2 (C(x, K)) and xy ∈ Syl2 (C(xy, Lxy )); (b) K = L2 (CG (x)) = CG (x)(∞) ; and (c) z, x ∈ Syl2 (CCG (x) (Kz Mz )). Proof. Part (a) follows from Lemma 11.2 and the symmetry between x and xy, and part (b) is an immediate consequence. By [III17 , 8.9c], z maps to a Sylow  2-subgroup of CAut(K) (Kz Mz ), so (c) follows from (a). Through Lemma 12.6 we assume that (12C)

n = 7.

Lemma 12.2. Assume (12A) with n = 7. Then the following conditions hold: (a) L = L2 (CG (y)) = CG (y)(∞) ; (b) LC(y, L) = LO2 (CG (y))Qy , where Qy ∈ Syl2 (C(y, L)) is dihedral or semidihedral; (c) O 2 (CG (y)) = O2 (CG (y))O 2 (CG (x, y )); and (d) If q = 3, then O 2 (CG (y)) = O2 (CG (y)) × L. Proof. By Lemma 11.4, L   CG (y). Expand D to Qy ∈ Syl2 (C(y, L)). By Lemma 11.2b, CQy (x) = D. Hence by [IG , 10.24], Qy is dihedral or semidihedral. Suppose for the moment that (12D)

C(y, L) = O2 (CG (y))Qy .

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Then (a) and (b) are immediate. Set Cy = CG (y) and C y = Cy /O2 (Cy ), and let g ∈ C y be of odd order. Then g normalizes Qy ∼ = Qy and centralizes y, so g 2 centralizes Qy . Thus, O (C y ) ≤ CG (D), which implies (c). And if q = 3, then Out(L) is a 2-group, which implies (d) as L  Cy is simple. It therefore remains to prove (12D). Since y ∈ Z(C(y, L)), (12D) holds if Qy is dihedral, by [III8 , 6.11]. So assume that Qy is semidihedral, but that (12D) fails. By [III17 , 11.3], C(y, L) = M x , where M/O2 (M ) ∼ = 2A7 or SL2 (s) for some odd s, x induces a non-inner automorphism on M/O2 (M ), and y ∈ Z(M ). Since m2 (L) > 2, L  Cy and then M  Cy . As Out(M/O2 (M )) is abelian, x ∈ [Cy , Cy ]. Therefore by [III8 , 6.3] and the simplicity of G, y is not weakly closed in Z(T ). Of course xy ∈ xQy . By [III17 , 8.9d], Z(T ) = z, y, x . By the structure of E(CG (x)) and E(Cy ), y ∈ xG . Therefore there is d ∈ D such that zd ∈ y G . Let S ∈ Syl2 (Cy ). We may choose S, by [III17 , 8.8ab], so that Ω1 (Z(S)) ∩ [S, S] = z, y . If y is G-conjugate to either z or zy, it follows that y is NG (z, y )conjugate to z or zy. But in Lo2 (CG (z, y )) there are two (solvable) components whose unique involution is z, one (solvable) component whose unique involution is y, and no (solvable) components whose unique involution is yz. Therefore NG (z, y ) = CG (z, y ), contradiction. Consequently, y is G-conjugate to either zx or zxy. Indeed as xy ∈ xM ∩ S ⊆ xCM (z) , y is G-conjugate to both. Let Cxz = CG (xz, z ) = CG (x, z ) = CCG (xz) (z). Within CG (x) we see that Cxz has a subnormal subgroup H which is the product of three subgroups: two (solvable) components isomorphic to SL2 (q) and a third subnormal subgroup isomorphic to L2 (q). Moreover, xz ∈ H. Since xz ∈ y G , there must exist an involution y  ∈ Cy such that CG (y, y  ) has a subnormal subgroup H  ∼ = H with y ∈ H  . Then CH  (L) ≤ C(y, L), and as C(y, L) has semidihedral Sylow 2-subgroups with center y , |CH  (L)|2 = 2 so CH  (L) is 2-nilpotent. It follows that if q > 3, then H  , which is perfect, lies in L; and if q = 3, then H  , which equals O 2 (H  ), lies  in LO2 (Cy ). In either case O r (CL/Z(L) (y  )) has a subnormal subgroup isomorphic to H  . But this contradicts the structure of centralizers of involutions in Aut(L)  given in [IA , 4.5.1, 4.9.1]. The proof is complete. Recall from the previous section that z4 ∈ I2 (L) has 4-dimensional support of + type on the natural L- and K-modules, and that R0 = O 2 (CL (z4 )). Let z2 = z4 y ∈ L. Assuming (12C), let u ∈ R0 ∩ z2K and set Ku = E(CK (u)) ∼ = Ω5 (q). Let Lu be the subnormal closure of Ku in CG (u). Lemma 12.3. Assume that n = 7. Then Lu is a quotient of Ω± 6 (q), and corre± spondingly O 2 (CLu (z4 )) ∼ = Ω4 q (q). Moreover, Lu = L2 (CG (u)). Proof. Let L2 be the pumpup in CG (z2 ) of K2 := E(CK (z2 )). Since z2 ∈ uK , L 2 ∼ = Lu . Moreover, R0 ≤ L2 . By [III8 , 3.3] with u, D, and a (solvable) component of CK2 (y) in the roles of z, D, and J, L2 ≤ E(CG (z2 )). Also L2  is the subnormal closure in CG (z2 ) of O r (CK2 (y)) ∼ = SL2 (q) ∗ SL2 (q). Hence  by L2 -balance, E(CL2 (x)) ∼ = Ω5 (q), while O 2 (CK2 (y)) = = E(CL2 (xy)) ∼ = K2 ∼  O 2 (CL (z2 ))  CL2 (y) as L  CG (y). If L2 is a diagonal pumpup of K2 , then this last condition is contradicted as K2 lies on the diagonal of L2 . Using Lemma 1.2 and [III17 , 3.29], we then conclude that L2 is a quotient of Ωη6 (q) for some sign η. As Lu ∼ = L2 the first assertion holds.

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Next, z4 acts on the natural Lu -module with 2-dimensional support of the same type as that of u and z2 on the natural K-module, viz., type q . The second statement of the lemma follows immediately. Finally, z4 ∈ L2 and so if we let E2 = L2 (CG (z2 )), then E2 = L2 CE2 (z4 ). As L2 (CG (y, z2 )) ≤ R0 ≤ L2 , it follows that L2 = L2 (CG (z2 )), whence Lu = L2 (CG (u)) = E(CG (u)). The proof is complete.



Lemma 12.4. If n = 7, then q = 3. Proof. Suppose that q > 3. Let E4 = L2 (CG (z4 )). By Lemma 11.10, R0  E4 . As u ∈ R0 , E4 = R0 CE4 (u). Let L4 = O 2 (CLu (z4 )). Then L4 = L2 (CLu (z4 )). As Lu = L2 (CG (u)), it follows that E4 = R0 L4 . Since L2 (CG (x, z4 )) = E(CK (z4 )) ∼ = R0 × Ω3 (q) and [R0 , x] = 1, x acts on L4 ∼ (q) as a reflection. Thus, x ∈ = Ω± 4 [CG (z4 ), CG (z4 )]. Let T ∈ Syl2 (CG (x)) with y, z2 ≤ Z(T ). Then Z(T ) = x, y, z2 . As E(CK (z2 )) ∼ = Ω5 (q), it follows that z4G ∩ Z(T ) ⊆ {z4 , xz4 }. Suppose by way of contradiction that z4g = xz4 for some g ∈ G. Then E(CG (xz4 )) = R0g Lg4 with xz4 ∈ R0g . But R0 ≤ E(CG (xz4 )), and it follows quickly from L2 -balance that E(CG (xz4 )) = R0g × R0 . But then x ∈ E(CG (z4 )), a contradiction. We have proved that z4 is weakly closed in Z(T ). By [III8 , 6.3], x ∈ [G, G], contradicting the simplicity of G and completing the proof.  Continuing to assume that n = 7, we have q = 3 by Lemma 12.4, and Lemma 12.3 applies. We have T ∈ Syl2 (CG (x)) with Z(T ) = x, y, z4 . Moreover, Ku = E(CK (u)) ∼ = Ω5 (3) and Ku ≤ Lu with Lu = L2 (CG (u)) = E(CK (u)) and Lu a 2 ∼ ∓ quotient of Ω± 6 (3). Also L4 = O (CLu (z4 )) = Ω4 (3). 2 Now, O (CK (z4 )) = R0 × Ax with Ax ∼ = A4 and Ax ≤ Ku ≤ Lu . = L2 (3) ∼ 2 Thus, Ax ≤ O (CLu (z4 )) = L4 . Similarly, as Lxy ∼ = K with R0  CLxy (z4 ), we have O 2 (CLxy (z4 )) = R0 × Axy with A4 ∼ = Axy ≤ L4 . By [III17 , 9.1], Lxy = L, Axy . As [x, Lxy ] = 1 = [x, L], we have [x, Axy ] = 1. Also, uy = uz4 z2 ∈ Ku . As u ∈ L, it follows that Qy acts faithfully on Lu , centralizing Lu ∩ L, and that 3 H = O 2 (Lu ∩ L) ∼ = Ω+ 4 (3). By [III17 , 6.9], |Qy | ≤ |CAut(Lu ) (H)|2 = 2 , whence 3+7+3 13 |CG (y)|2 ≤ |Qy ||Aut(L)|2 = 2 =2 . Let K4 be the subnormal closure of L4 in CG (z4 ). Note that CG (z4 , d ) is # solvable for all d ∈ D# = x, y , and so D normalizes every 2-component of CG (z4 ). Lemma 12.5. If n = 7, then L4 ∼ = Ω+ 4 (3). ∼ Ω− (3). If K4 = J4 J u is a diagonal Proof. Suppose false, in which case L4 = 4 4 2 u pumpup of L4 , then O (CK4 (x)) = H4 H4 with H4 ≤ J4 and H4 /Z(H4 ) ∼ = Ax ≤ H4 H4u . But this is impossible as Ax  CG (x, z4 ). Thus, K4 is a trivial or vertical pumpup of L4 with L2 (CK4 (u)) = L4 and with Ax  CK4 (x). Moreover, z2 ∈ Ax ≤ K4 . As z4 , z2 = z4 , y has solvable centralizer in G, L2 (CG (z4 )) = K4 . Suppose that K4 is a vertical pumpup of L4 . Choose a four-subgroup U ≤ R0 with u ∈ uR0 for all u ∈ U # . Just as we argued for u, we have that Lu := L2 (CG (u )) is a quotient of Ωη6 (q) for some sign η depending on u ∈ U # . Now

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u ∈ CG (u, z4 ) normalizes L4 , and [u , L4 ] ≤ [CO2 (R0 ) (u), E(CCG (z4 ) (u))] ≤ [O2 (CG (z4 )), L2 (CG (z4 ))] = 1. Hence, L4 ≤ O 2 (CLu (z4 )), and so L4 = O 2 (CLu (z4 )). Thus E(CK4 (u )) = L4 , for all u ∈ U # . Since O2 (R0 ) is connected, it follows by similar arguments that (12E)

E(CK4 (u )) = L4 for all u ∈ I2 (R0 ) − {z4 }.

In particular, NCG (z4 ) (R0 ) normalizes L4 . Now R0 centralizes Ax ≤ L4 and R0 =  O 3 (R0 ) so [R0 , L4 ] = 1. Also, the two solvable components of R0 ∼ = SL2 (3)∗SL2 (3) are interchanged in CK (z4 ). Since K4 is a vertical pumpup of L4 , it follows that CR0 (K4 /O2 (K4 )) = z4 . By (12E) and [III11 , 1.16], K4 /O2 (K4 ) ∼ = A6+4i for some i ≥ 1. Since CK4 (z4 , x ) is solvable it follows from [IA , 5.2.8] that i = 1. But this is impossible since R0 / z4 ∼ = A4 × A4 must embed in the subgroup of Aut(K4 /O2 (K4 )) centralizing the image of L4 , which is isomorphic to Σ4k = Σ4 . We conclude that K4 is a trivial pumpup of L4 . Applying [III8 , 3.3] with U in the role of D, we see that K4 = L4 is quasisimple. Now R0  CG (z4 , y ) and CG (z4 , y ) is solvable. On the other hand, L4 ∼ = L2 (9) has one class of involutions, so NCG (z4 ) (L4 ) ≤ L4 CG (z4 , z2 ) = L4 CG (z4 , y ). Hence L4 = NCG (z4 ) (L4 )(∞) = E(CG (z4 )) = CG (z4 )(∞) and CG (z4 ) = L4 CG (z4 , y ). As [R0 , L4 ] = 1 we have R0  CG (z4 ). Since L4  CG (z4 ) and x acts on L4 ∼ = Ω− 4 (3) as a reflection, x ∈ [CG (z4 ), CG (z4 )]. 2 In particular x ∈ O (CG (z4 )). Considering the layers of involution centralizers as in the previous lemma, we see that z4G ∩ Z(T ) ⊆ {z4 , xz4 }. Suppose that xz4 = z4g for some g ∈ G. As R0 ≤ CG (xz4 ), R0 normalizes R0g . Hence R0 = O 2 (R0 ) maps into O 2 (Aut(R0g )) = g Inn(R0g ) ∼ = A4 × A4 . It follows that z4 = Z(R0 ) centralizes R0 . But then x ∈ g 2 R0 R0 ≤ O (CG (z4 )), contrary to the previous paragraph. Thus, z4 is weakly closed in Z(T ), so by [III8 , 6.3], x ∈ [G, G], a contradiction completing the proof of the lemma.  Lemma 12.6. We have n = 7. ∼ Ω+ (3) by Lemma 12.5. By [III17 , 11.9], Proof. If n = 7, then q = 3 and L4 = 4 O2 (L4 ) ≤ Ax , Axy and we have seen that Ax , Axy ≤ L4 ∩ CG (R0 ). Thus CG (z4 ) ≥ R0 ∗ (O2 (L4 ) w ) with w ∈ L4 of order 3 and [w, O2 (L4 )] = O2 (L4 ). Again, y, z4 = z2 , z4 , O 2 (CG (y, z4 )) = O2 (CG (y))×R0 , and R0  CG (z4 , d ) for all d ∈ D# . Now, O 2 (CG (u)) = Lu O 2 (C(u, Lu )) and z2 ∈ Lu . Hence, C(u, Lu ) ∩ CG (z4 ) ≤ CG (u, z2 , z4 ) ≤ CG (y, z4 ). Thus C(u, Lu ) ∩ CG (z4 ) is solvable. As L4 = O 2 (CLu (z4 )) is solvable, it follows that CG (u, z4 ) is solvable. Similarly, CG (u , z4 ) is solvable for all involutions u ∈ R0 − z4 . We next show that (12F)

E4 := L2 (CG (z4 )) = 1.

Suppose false, set C4 = CG (z4 ) and C 4 = C4 /O2 (C4 ), and let J be a 2-component of E4 . Then x normalizes J, as we saw before Lemma 12.5. Write R0 = R1 ∗ R2 with R1 ∼ = R2 ∼ = SL2 (3). Since CG (u, z4 ) is solvable, [O2 (R0 ), J] = 1 and so [O2 (Ri ), J] = 1 for some i = 1, 2, say i = 1. By solvable L2 -balance, O2 (R1 ) ≤ J

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

and in particular z4 ∈ J. Thus Z(J) = 1. Furthermore, ΓD,1 (J) normalizes R0 and hence R0 ∩ J  ΓD,1 (J). In particular ΓD,1 (J) = J. It follows from [III11 , 13.1] and [IA , 7.3.4] that J ∼ = 2An for some n ≥ 6 or J ∈ Chev(2). In the first case n = 6 by [III9 , 2.8], and then as R1 ≤ CJ (D), every element of D induces a field automorphism on J ∼ = SL2 (9), which is impossible as D is noncyclic and CD (J) = 1. So J ∈ Chev(2). The same conclusions hold for any component of C 4 . Since R0 has just two solvable components and any component of C 4 must contain at least one of them, C 4 has at most two components. Hence O 2 (C4 ) normalizes J. As Ax  CCG (x) (z4 ), we have Ax  CC 4 (x), and Ax = O 2 (Ax ) normalizes J. Since J ∈ Chev(2), Out(J) does not involve A4 (see [IA , 2.5.12]). Therefore O2 (Ax ) ≤ JC(z4 , J). But z2 ∈ O2 (Ax ) and CG (z2 , z4 ) = CG (y, z4 ) is solvable, so O2 (Ax ) acts faithfully on J. As Ax  CG (z4 , x ) it follows that O2 (Ax ) ≤ J. Since the same conclusion holds for any component of C 4 , we must have J = E(C 4 ). Now R1 and R2 are conjugate in CG (z4 , x ), hence in C4 , so O2 (R0 ) ≤ J. Thus CR0 O2 (Ax ) (J) = z4 , and as a consequence m2 (J/Z(J)) ≥ 6. Just as O2 (Ax ) ≤ J, we similarly see that O2 (Axy ) ≤ J and so O2 (R0 ) ∗ O2 (L4 ) ≤ J. It follows that m2 (J/Z(J)) ≥ 8. Now with [IA , 6.1.3, 3.3.1], we get J/Z(J) ∼ = U6 (2), F4 (2), or 2 E6 (2). Expand 16 T to S ∈ Syl2 (C4 ). Then |S| ≥ |J|2 ≥ 2 . On the other hand, Ω1 (Z(S)) ≤ Ω1 (Z(T )) = D z4 . By [IA , 6.4.1, 6.4.2a], Z(S) is noncyclic. Therefore Z(S)∩D = 1, and so |S| ≤ maxd∈D# |CG (d)|2 . But |CG (x)|2 = |CG (xy)|2 ≤ 2 · |Aut(K)|2 = 29 and |CG (y)|2 ≤ 213 , so |S| ≤ 213 , a contradiction. This proves (12F). Again with T ≤ S ∈ Syl2 (C4 ), let P = O2 2 (C4 ) ∩ S ∈ Syl2 (O2 2 (C4 )), so that P = F ∗ (C 4 ). Let P0 be a critical subgroup of P with Φ(P0 ) = [P0 , P0 ] ≤ Z(P0 ).  For any v ∈ I2 (C4 ) and any solvable component V of CC4 (v), O 2 (V ) ≤ P , by  solvable L2 -balance, and then O 2 (V ) ≤ P 0 by the A × B-lemma. Taking v = x, we conclude that O2 (R0 ) ≤ P 0 . Taking v = u, we conclude that O2 (L4 ) ≤ P 0 . We saw above that O2 (L4 ) ≤ Ax , Axy ≤ CG (R0 ), so [O2 (L4 ), O2 (R0 )] = 1 and P 0 ≥ P 1 := O2 (L4 )O2 (R0 ). We next let P1 be the preimage of P 1 in P0 and prove that (12G) P0 = P1 ∼ = 21+8 and CP (P0 ) = Z(P0 ) = z4 . +

Recall that T ∈ Syl2 (CG (x)) with Z(T ) = x, z4 , z2 . Then P ≤ P T ≤ S ∈ Syl2 (C4 ) and Z(S) ≤ Z(T ). Now, x acts on L4 with CL4 (x) = Z(L4 ) × Ax and with z2 ∈ O2 (Ax ). Thus O2 (L4 ) x ≤ P0 T with Z(O2 (L4 ) x ) = Z(L4 ) and Z(L4 ) ∩ x, z2 = 1. Hence Z(S) ∩ x, z2 = 1, so Z(S) = z4 = Z(O2 (L4 )). Therefore P1 ∼ = 21+8 + . To prove that CP (P0 ) ≤ P0 = P1 , we set P0∗ = P0 CP (P0 ) and calculate |CP0∗ (y)|. Note that as z2 ∈ O2 (L4 ) − Z(O2 (L4 )), y = z4 z2 ∈ P0 − Z(P0 ). In particular y is not a square in P0∗ , so |CP0∗ (L)|2 ≤ 4 as Qy ∈ Syl2 (C(y, L)) has maximal class. Suppose first that I2 (Qy − {y}) ⊆ xC4 . As CL4 (x) = z4 × Ax , x does not centralize P1 / z4 , and so xC4 ∩ P0∗ = ∅, whence CP0∗ (L) = y . Then CP0∗ (y)/ y embeds in O2 (AutC4 (L)), so (12H)

|CP0∗ (y)| ≤ 28 = |CP1 (y)|.

On the other hand, if I2 (Qy − {y}) ⊆ xC4 , then xC4 ∩ Qy = xQy . It follows that CG (y) = O2 (CG (y))Qy CG (x, y ) and |CP0∗ (y)| ≤ 2|CP0∗ (x, y )|. As

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CG (x)/O2 (CG (x)) is isomorphic to a subgroup of O7 (3), it suffices to consider CO7 (3) (y  ) ∼ = O6− (3) × O1 (3) and CO6− (3) (z  ) = O4+ (3) × O2− (3). As |O2− (3)| = 8, it follows that |O2 (CG (x, y, z4 ))| ≤ 29 . But x ∈ P0∗ and z2 ∈ [P0 , P0 ], so |CP0∗ (x, y )| ≤ 27 and we again have (12H). Thus, CP0∗ (y) = CP1 (y), whence Z(P0∗ ) = Z(P1 ) = z4 and then P0∗ = P1 , proving (12G). We argue next that (12I)

z4 is weakly closed in Z(T ).

If not, then since CG (v) is not 2-constrained for v ∈ Z(T )# − {z4 , xz4 }, we have xz4 = z4g for some g ∈ G, and CG (xz4 ) = C4g is 2-constrained. Then P0g is a critical subgroup of P g ∈ Syl2 (O2 2 (CG (xz4 ))). Since R0  CG (x, xz4 ), solvable L2 -balance   again gives O 2 (R0 ) ≤ O2 2 (C4g ), and then by criticality, O 2 (R0 ) ≤ P0gh for some g gh gh h ∈ O2 (C4 ). But then z4 ∈ [P0 , P0 ] = xz4 , contradiction. Thus (12I) holds. By (12I), [III8 , 6.3], and the simplicity of G, (12J)

x ∈ [C4 , C4 ].

We shall contradict this conclusion to complete the proof of the lemma. As CC4 (P0 )= z4 , it follows that C4 /P0 embeds into Out(P0 ) ∼ = O8+ (2). As C4 ≥ R0 , L4 with [R0 , O2 (L4 )] = 1, it follows that R0 L4 is the central product of four copies of SL2 (3). In particular C 4 contains a subgroup E ∼ = E34 . However, m2,3 (O8+ (2)) < 4 by [III17 , 11.28], so O2 (C 4 /P 0 ) = 1, i.e., P 0 = O2 (C 4 ) = F ∗ (C 4 ). There are no nontrivial p-subgroups of O8+ (2) for p > 3 that are invariant under an E34 -subgroup, so F (C 4 /P 0 ) is a (possibly trivial) 3-group. Suppose that C4 is nonsolvable. We have R0  CC 4 (y) = CC 4 (z2 ). In particular 2 E3 ∼ = R0 P 0 /P 0  CC 4 (y)P 0 /P 0 . Hence, CC 4 (y)P 0 /P 0 is isomorphic to a subgroup of (Σ3 Z2 )×D8 . Moreover, |y C 4 | = 18k ≤ |I2 (P 0 )−{z 4 }| = 270, whence |C 4 /P 0 | = 2a 3b 5c 7d , with 4 ≤ b ≤ 5 and c + d = 1, which imply a ≤ 7. If O3 (C 4 /P 0 ) = 1, then the only possibility is F ∗ (C 4 /P 0 ) = X × I ∼ = GU4 (2), and [X, P 0 ] ∼ = Q8 or P 0 . In view of the structures of E, CK (z4 ), and CLu (z4 ), C 4 has no normal Q8 -subgroup, so V := P 0 / z 4 is a natural GU4 (2)-module. As y 2 = 1, the image of y in V is an isotropic vector and so CX×I (y) is a proper parabolic subgroup of X × I. This contradicts the fact that R0  CC 4 (y), however. Therefore F (C 4 /P 0 ) = 1, and considering the restrictions on |C 4 |, F := F ∗ (C 4 /P 0 ) must be simple with m3 (Aut(F )) ≥ 4. By [III17 , 9.9] and [IA , 4.10.3], there is no such K-group F , contradiction. Therefore C4 is solvable. By [III17 , 11.29], EP 0  C 4 , whence C 4 permutes the four solvable components of R0 L4 = EP 0 . As x permutes these as a transposition, x ∈ [C4 , C4 ]. This contradicts (12J) and finally completes the proof of the lemma.  We continue to assume that n is odd. Write n = 2a +  with 0 <  < 2a . By Lemma 12.6, a ≥ 3. Recall also that z ∈ I2 (K) has 2a -dimensional support on the natural K-module, and Kz ∼ = Ω+ 2a (q) is a component of CK (z) with Kz ≤ L. Lemma 12.7. Assume (12A). Then Kz   CG (z). Proof. Let Ch(z) be the product of all 2-components of CG (z) that lie in Chev(r) and involve Kz / z . By L2 -balance and [III17 , 10.46], Kz ≤ Ch(z). As Ch(z)  CG (z), it suffices to prove that Kz  Ch(z).

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On the other hand, [D, z] = 1, so D normalizes Ch(z) and so Ch(z) = ΓD,1 (Ch(z)) by [IA , 7.3.3]. Thus it is enough to show that Kz  CCG (d) (z) for each d ∈ D# . As Ld ∼ = K or L by Lemmas 11.4 and 11.6, Kz is the unique component of its isomorphism type in CLd (z), so it is enough to show that Ld  CG (d). Clearly this holds for d = x and xy as m2 (C(x, K)) = 1 = m2 (C(xy, Lxy )). We have Ly = L, and it suffices to show that L  CG (y). But by Lemma 12.2b, m2 (C(y, L)) = 2, while since a ≥ 3, m2 (L) > 2. Hence C(y, L) has no component isomorphic to L, and the lemma is proved.  In the same vein, we can prove: Lemma 12.8. Suppose that n = 2a+1 − 1, a ≥ 3. Then L2 (CG (z)) has at least two 2-components H such that H/O2 2 (H) ∼ = P Ω+ 2a (q). Proof. We have E(CK (z)) = Kz × Hx and E(CLxy (z)) = Kz × Hxy with Hx ∼ = Hxy ∼ = Ω2a −1 (q), and E(CL (z)) = Kz × Hy with Hy ∼ = Ω± 2a −2 (q). As a ≥ 3,  the result follows from [III17 , 3.28]. Lemma 12.9. Assume (12A). Then z is weakly closed in Z(T ) with respect to G. Proof. Suppose false and let w = z g ∈ Z(T ) with g ∈ G and w = z. Set Cw = CG (w) and C w = Cw /O2 (Cw ). Since z ∈ Z(T ), [w, z] = 1, so w acts on Kz . As [w, T ∩ Kz ] = 1 and Kz ∼ = Ω+ 2a (q), [w, Kz ] = 1 by [III17 , 8.9b]. Thus there is a component Hw of CK (w) containing Kz . Moreover, the component Kz of CG (z) has a pumpup H in CG (w), which contains Hw . Since z ∈ Kz , H is a vertical or trivial pumpup of Kz . Also we set Kw = Kzg , so that w = Z(Kw ). Since w = z while CH (z) contains K z ∼ = Kw , we have Kw = H and so [Kw , H] = 1. Consider the action of x on L2 (Cw ). We have Kz ≤ L2 (L2 (CH (z)) ∩ CH (x)) ≤ L2 (CH (x)). 

Set M = E(CKw (x)) or M = O 3 (CKw (x)) according as q > 3 or q = 3. As x ∈ Syl2 (C(x, K)), M is a product of components or Lie components of L2 (CK (w))  or O 3 (CK (w)), respectively. Moreover, [M, Kz ] = 1. Since [w, Kz ] = 1, when we consider w acting on the natural K-module, the support of Kz lies in an eigenspace of w. Let b be the dimension of that eigenspace, so that b ≥ 2a . It follows that M∼ = Ω± n−b (q) if n − b ≥ 3, while M = 1 if n − b ≤ 2.  On the other hand, since M = E(CKw (x)) or O 3 (CKw (x)) with Kw ∼ = Ω+ 2a (q) and a ≥ 3, we have that M = 1 [IA , 4.5.1], so n − b ≥ 3. Moreover, since M is a a a single component isomorphic to Ω± n−b (q), 2 − (n − b) ≤ 2. Thus n − b ≥ 2 − 2. a a a But n − b ≤ n − 2 ≤ 2 − 1, since 2 is the largest power of 2 not greater than n. As a result, n − b = 2a − 1 or 2a − 2. Since n is odd and b ≥ 2a , n = 2a+1 − 1. By Lemma 12.8, CG (z), and hence its conjugate subgroup CG (w), has at least two 2-components H  with H  /O2 2 (H  ) ∼ = P Ω+ 2a (q). In the case of CG (w), one of  these is Kw ; fix a second one H . Suppose that b > 2a , so that b = 2a + 1. Then CK (w) has a component isomorphic to Ω2a +1 (q), whose pumpup H  in CG (w) is distinct from Kw and H  . Then, if q > 3, L2 (CL2 (CG (w)) (x)) must have components in each of Kw , H  and H  , so it is the commuting product, modulo core, of at least three orthogonal groups Ω± k (q), k ≥ 3. By L2 -balance, the same holds for L2 (CL2 (CG (x)) (w)) = L2 (CK (w)); but clearly the latter group is the product of at most 2 commuting orthogonal groups,

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a contradiction. Thus q = 3; but we obtain a similar contradiction by considering  O 3 (CG (x, w)). Therefore b = 2a . As a result, and since w = z, w = xz. But z ∈ Kz ≤ H and xz = w ∈ Kw , whence x = (xz)z centralizes Kw . Therefore CG (x) involves the commuting product of two Ω+ 2a (q) subgroups, which is impossible by order considerations as its unique nonsolvable composition factor is Ω2a+1 −1 (q). The lemma is proved.  Next, choose involutions u, u ∈ Kz whose supports on the natural Kz -module and the natural K- and L-modules are 2-dimensional and mutually orthogonal.  Let Ku = E(CK (u)) = O r (CK (u)) ∼ = Ωn−2 (q), so that u ∈ Ku . Let Lu be the pumpup of Ku in CG (u), so that Lu is also the subnormal closure of E(CKu (y)) =  E(CL (u)) ∼ = Ω± n−3 (q) in CG (u). As Kz ≤ L, we have u ∈ E(CL (u)) and set η  ∼ z4 = uu . By Lemma 11.11, Lu /Z(Lu ) = P Ωn−1 (q) for some sign η. The next lemma, combined with the previous one, will eliminate the case in which n is odd. Lemma 12.10. Assume (12A). Then x ∈ [CG (z), CG (z)]. Proof. Assume the contrary. In particular, n ≥ 9. We have n = 2a + , with   odd and 1 ≤  ≤ 2a − 1. Recall (12B) that O r (CK (z)) = Kz Mz with [Kz , Mz ] = 1 and Mz ∼ = Ω (q). Assume first that  > 3. Then Mz is a component of CG (x, z ), and we let its subnormal closure in CG (z) be Hz . Since Kz   CG (z) with [Mz , Kz ] = 1, also [Hz , Kz ] = 1 and therefore [Hz , u] = 1. As Mz ≤ Ku ≤ Lu , with Lu a component of CG (u), and Mz ≤ Hz , it follows that the pumpup of Hz in CG (u) is Lu . Therefore Hz is a component η of CLu (z). But Lu /Z(Lu ) ∼ = P Ωn−1 (q). Also uz ∈ Ku is an involution whose −1eigenspace on the natural Ku -module is 2a − 2-dimensional. Therefore Hz ∼ = Ω± m (q) a where m = n − 1 − (2 − 2) =  + 1. Let H = L2 (C(z, Kz )), so that Hz is a component of H. By Lemma 12.1c, x, z ∈ Syl2 (CCG (x,z) (Kz Mz )). Thus if H  is any 2-component of H other than Hz , we have that x, z is self-centralizing in a Sylow 2-subgroup of x, z H  , which then must be dihedral or semidihedral. Since z ∈ Z(CG (z)) this is only possible, by [III17 , 11.3], if m2 (H  ) = 1. Hence either Hz  CG (z) or Hz and Kz are CG (z)-conjugate; in the latter case, Kz Hz  CG (z) and b = 2a − 1. In any case x acts as a reflection on Hz and centralizes Kz . Therefore in the respective cases x maps nontrivially to the commutator quotient of Out(Hz ) or Out(Kz Hz ). Hence x ∈ [CG (z), CG (z)], as desired. Therefore we may assume that  ≤ 3. The argument is similar. First, the supports of z4 = uu on the natural K, L, and Lxy -modules are all 4-dimensional of + type. Moreover, CL (z4 ) has a subnormal  subgroup R0 ∼ = Ω+ 4 (q) containing u, u . With the notation of (11J), let I be the subnormal closure of Ix in CG (z4 ). By Lemmas 11.12 and 11.13, I is a component of CG (z4 ), R0   CG (z4 ), and [u, I] ≤ [R0 , I] = 1. In particular as Ix ≤ Ku , the pumpup of I in CG (u) is Lu . Therefore I is a component of CLu (u ), and so ηq (q). I∼ = Ωn−3

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

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Now since x ∈ Syl2 (C(x, K)), and CAut(K) (R0 Ix ) is the image of z4 by [III17 , 6.6], we see that CR0 I := CCG (z4 ) (R0 I) x has Sylow 2-subgroups of maximal class. We claim that CR0 I is 2-nilpotent. Suppose false. As z4 ∈ Z(CR0 I ), it follows from [III17 , 11.3] that there is a 2-element t ∈ O 2 (CR0 I ) such that xt = xz4 . Now z ∈ K so z ∈ R0 I, whence [z, CR0 I ] = 1. As [x, Kz ] = 1 = [z4 , Kz ], it follows that t does not normalize Kz . But t ∈ O 2 (CR0 I ) ≤ O 2 (CG (z)). Considering the permutation action of CG (z) on the CG (z)-conjugates of Kz , we see that there are at least four such conjugates. On the other hand, as  ≤ 3, all the chief factors of CC(z,Kz ) (x) = CCG (x) (Kz ) of even order are of order 2, except possibly for a single chief factor isomorphic to L2 (q) if b = 3. As C(z, Kz ) has at least three components isomorphic to Kz ∼ = Ω+ 2a (q), with a ≥ 3, this is absurd, and our claim is established. We argue that Kz is the unique component of CG (z) of its isomorphism type. Indeed z4 ∈ Kz . If q > 3, then R0 I = CG (z4 )(∞) , which implies our assertion given the isomorphism types of I and Kz . If q = 3, then Out(R0 I) is a 2-group,   so O 2 (R0 )I = O 2 (O 2 (CG (z4 ))), and again our assertion holds unless possibly   O 2 (O 2 (CKz (z4 ))) ≤ O 2 (R0 ), which, however, is not the case as a ≥ 3. We conclude that Kz  CG (z). Aut(Kz ) By [III17 , 14.2], z4 = z4Kz . Thus, CG (z) = Kz C0 where C0 = CG (z, z4 ). We have x ∈ C0 , and to prove that x ∈ [CG (z), CG (z)] it suffices to show that x ∈ (Kz ∩ C0 )[C0 , C0 ].

(12K)

Thus as C0 ≤ CG (z4 ), it is enough to show that x ∈ (Kz ∩ CG (z4 ))[CG (z4 ), CG (z4 )]. We do this by considering the image x of x and the image X of X = Kz ∩ CG (z4 ) in the abelian quotient       (12L) Out(O 2 (R0 ))/Outs (O 2 (R0 )) × Out(I)/O 2 (Out(I)) Outdiag(I) 

of CG (z4 ). Here O 2 (R0 ) is the central product of two copies of SL2 (q) or Q8 , and   Outs (O 2 (R0 )) is the subgroup of Out(O 2 (R0 )) stabilizing both these copies. Since  x centralizes O 2 (R0 ) and induces a reflection on I, x is a nontrivial element of the second factor. On the other hand, X is of order 2, generated by the image ξ of an involution ξ ∈ CKz (z4 ) interchanging the two copies of SL2 (q) or Q8 generating  O 2 (R0 ). In particular X is disjoint from the second factor in (12L), so x ∈ X. This immediately implies (12K) and completes the proof of the lemma.  As an immediate consequence we have: Lemma 12.11. n is even. Proof. Otherwise by Lemma 12.6, we have n ≥ 9 and the preceding development applies. By Lemma 12.9 and [III8 , 6.3], x ∈ [CG (z), CG (z)], contradicting Lemma 12.10. The lemma is proved.  Next, through Lemma 12.15, we consider the case (12M) K/Z(K) ∼ = P Ωn (q), n even, n ≥ 8. By Lemma 11.15, ∼ Ω (q) and Z(K) ∼ (12N) K= = Z2 , so that  = n/2 n q .

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By Lemmas 11.4 and 11.6, (1) v ∈ J   CG (v); J (2) J ∼ (q) where J = q (n−2)/2 = q ; and = Ωn−2 (12O) (3) Jxv is a quotient of Ωn−1 (q) or Ωηn (q) for some sign η; accordingly, we set n = n − 1 or n. Recall that k = n2 − 1. As n ≥ 8, we have k ≥ 3. Our argument is essentially due to Aschbacher [A9]. We shall consider two related orthogonal spaces over Fq : V = V1 ⊥ V2 ⊥ · · · ⊥ Vk ⊥ Vk+1 , and

(12P)

W = W0 ⊥ V1 ⊥ V2 ⊥ · · · ⊥ Vk ,

of respective dimensions n and n , where V1 , V2 , . . . , Vk+1 are isometric nonsingular 2-dimensional subspaces, and W0 is nonsingular of dimension 1 or 2. In the general case, but not always, we will regard V as a natural K-module, identifying K with Ω(V ) and J with Ω(V1 ⊥ · · · ⊥ Vk ), and we will regard W as a natural Jxv -module, identifying Jxv with Ω(W ). Let 1 = ±1 be the signature of the quadratic form on V1 (and hence on all Vi , 1 ≤ i ≤ k + 1). If dim W0 = 2, let 0 be the signature of the quadratic form on W0 . Let g ∈ Ω(V ) be a 2-element interchanging Vi with Vk+1−i , 1 ≤ i ≤ k/2, and (if k is even) normalizing V(k+2)/2 . By [III17 , 4.7], g may be chosen, and subgroups H1 ∼ = H2 ∼ = ··· ∼ = Hk ∼ = SL2 (q) of Ω(V ) found, such that each Hi is supported on Vi ⊥ Vi+1 , 1 ≤ i ≤ k; {Hi−1 , Hi } is a standard CT-pair (if 1 = +1) or Ppair (if 1 = −1) for SL31 (q) for all 1 < i ≤ k; Hi−1 , Hi , Hi+1 is a quotient of  SL4q (q), 1 < i < k; and Hig = Hk+1−i for all 1 ≤ i ≤ k. Then [Hi , Hj ] = 1 for all 1 ≤ i < j − 1 ≤ k − 1, and Z(H1 ), . . . , Z(Hk ) is abelian. Also, H1 , . . . , Hk−1 ≤ Ω(V1 ⊥ · · · ⊥ Vk ). We consider the main case first, proving: Lemma 12.12. Suppose that n is even and n ≥ 8. Then one of the following holds: (a) K, Jxv ∼ = P Ωn+1 (q) or P Ω± n+2 (q); or  (b) n = n, q = −1, and η = −. Proof. Since Z(K) = 1, the natural K-module is the orthogonal sum of n/2 = k + 1 two-dimensional subspaces of signature q . Therefore we may take 1 = q and identify K with Ω(V ). Then without loss, we may assume that v and J are each supported on V1 ⊥  · · · ⊥ Vk . Thus J ∼ = Ω(V1 ⊥ · · · ⊥ Vk ) and we may also regard J as O r (CΩ(W ) (W0 )). Likewise we regard the subgroups H1 , . . . , Hk , defined above, as subgroups of K, whence H1 , . . . , Hk−1 ≤ J. As g ∈ K and v ∈ K, t := v g ∈ K, and the support of t on V is V1⊥ = V2 ⊥ · · · ⊥ Vk+1 . By (12O1), t ∈ J g   CG (t). Set   F = O r (CJ (t)) = O r (J ∩ J g ), so that F is identified with Ω(V2 ⊥ · · · ⊥ Vk−1 ), and F is a product of components (solvable components if n = 8 and q = 3) of CG (x, v, t ). In particular F   CJ g (v). We next argue that F   CJxv (t).

(12Q) ∗

Namely, let F be the subnormal closure of F in CJxv (t). Since F ≤ J g   CG (t), we have F ∗ ≤ J g and so F ∗ ≤ CJ g (xv) = CJ g (v), the last as J g ≤ K ≤ CG (x). By the previous paragraph, F   F ∗ , whence F = F ∗ and (12Q) holds.

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

132

Since F ∼ = Ωn−4 (q), (12Q) and [III17 , 6.5] imply that 

O r (CJxv (t)) = F M with [F, M ] = 1,

(12R)

 where M ∼ = Ω±  (q),  = n − (n − 4) = 3 or 4. As M centralizes F , M normalizes   the component J g of CG (t). Therefore, M = O r (M ) maps into O r (CAut(J g ) (F )), which is trivial by [III17 , 6.8]. Thus, [M, J g ] = 1 and in particular

(12S)

g

[M, H2 , . . . , Hk ] = [M, H1 , . . . , Hk−1 ] ≤ [M, J g ] = 1.

We set G1 := H1 , . . . , Hk , M ≤ K, Jxv .

(12T)

Note that H1 , . . . , Hk−1 , M ≤ J, M ≤ Jxv . Then there is (even if Jxv ∼ = D4 (q)) J a natural module W for Jxv such that the support of J on W is a natural J ∼ = Ωn−2 module. Hence we may decompose W as in (12P) and identify the subspaces Vi , 1 ≤ i ≤ k, of W with the corresponding subspaces of V . Next, since [F, M ] = 1, M acts as scalars on [W, F ] = V2 ⊥ · · · ⊥ Vk . Hence as  M = O r (M ), M has support W0 ⊥ V1 and may be identified with Ω(W0 ⊥ V1 ). We can then apply [III17 , 4.3] and [III17 , 4.6] with Ω(W0 ⊥ V1 ⊥ V2 ) and M in the roles of K and I there. If M ∼ = Ω+ 4 (q), we have 0 = 1 = q and deduce from [III17 , 4.6] that M = Hk+1 Hk+2 where [Hk+1 , Hk+2 ] = 1, Hk+1 ∼ = = Hk+2 ∼ = SL2 (q), and Hk+i , H1 ∼ q SL3 (q), i = 1, 2. Thus G1 = H1 , . . . , Hk , Hk+1 , Hk+2 , and with (12S) we have verified that the Hi , 1 ≤ i ≤ k + 2, form a weak CT-system or weak P-system, according as 1 = +1 or −1, corresponding to the first diagram below: αk+1

◦

(12U) αk

◦

αk−1

◦

···

◦

αk α1 αk+2

◦

◦

αk−1

◦

···

α1

◦

αk+1

◦

The system, moreover, is extendible if q = 3, as A3 -type subsystems lie either in K or in Jxv . Likewise if M ∼ = Ω3 (q), we set Hk+1 = M and similarly, by [III17 , 4.3], H1 , . . . , Hk+1 form a weak CT-system or weak P-system corresponding to the second diagram above, according as 1 = +1 or −1. Again when q = 3, the extendibility condition on H1 , H2 , Hk+1 is satisfied since it can be calculated in Jxv . Hence, by [III13 , 1.4, 1.14] we have proved in these two cases for M that G1 is a homomorphic image of Spinδn+2 (q) or Spinn+1 (q), with δ = kq . Similarly if M ∼ = Ω− 4 (q) with q = 1, we set Hk+1 = M and conclude by [III17 , 4.6] that H1 , . . . , Hk+1 form a weak CT-system corresponding to the second diagram above. Hence, by [III13 , 1.4], G1 is a homomorphic image of Spin− n+2 (q) in this case. We claim that conclusion (a) of the lemma holds in these cases. Note that a similar argument shows that H1 , . . . , Hk−1 , M , which lies in Jxv , is a quotient of Jxv and hence H1 , . . . , Hk−1 , M = Jxv . Thus, Jxv ≤ G1 . Moreover, there  exists h ∈ Hk interchanging Vk and Vk+1 , and then with [III17 , 9.1], K = J, J h ≤ J, Hk ≤ Jxv , Hk ≤ G1 , so G1 = K, Jxv . To prove (a), it remains to show that Z(G1 ) = 1. But x centralizes t, v, and F ≤ J ≤ K, by construction. In particular  x normalizes Jxv and then normalizes M = O r (CJxv (t F )). As Hi ≤ K ≤ CG (x) for 1 ≤ i ≤ k, x normalizes G1 . By [IA , 6.4.1], Z(G1 ) is a 2-group. If Z(G1 ) = 1,

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12. THE ORTHOGONAL CASE, COMPLETED

133

then since m2 (C(x, K)) = 1, x ∈ Z(G1 ). This is absurd as [x, Jxv ] = 1, so (a) holds.  In the remaining case, M ∼ = Ω− 4 (q) with q = −1. Thus n = n, 0 = −1 = −q = 1, and η = 0 q k = −q k+1 = −, so (b) holds. The proof of the lemma is complete.



Lemma 12.13. Suppose that n ≡ 2 (mod 4) and n ≥ 8. Then K, Jxv ∼ = P Ωn+1 (q) or P Ω± (q). n+2 Proof. By Lemma 12.12, we may assume that q = −1. Since Z(K) = 1, it ∼ + follows that K ∼ = Ω− n (q). Again by Lemma 12.12, we may assume that Jxv = Ωn (q). − ∼ We show that G1 := K, Jxv = P Ωn+2 (q). Still with k = n2 − 1, we again consider a pair of spaces V, W as in (12P) and we identify J with Ω(V1 ⊥ · · · ⊥ Vk ). However, we interchange the roles of K and Jxv , and we take each Vi to have signature +1. Thus we identify Jxv with Ω(V ) and K with Ω(W ). As in the general case, we identify each subspace Vi of W with the subspace Vi of V . We then have 1 = +1 and 0 = −1. We repeat paragraphs 2–6 and 10 of the proof of Lemma 12.12, interchanging x with vx, and K with Jxv , but making no other changes. This yields G1 ∼ = P Ω− n+2 (q), as desired.  Lemma 12.14. Suppose that n ≡ 0 (mod 4) and n > 8. Then K, Jxv ∼ = P Ωn+1 (q) or P Ω± (q). n+2 Proof. As Z(K) = 1, K ∼ = Ω+ n (q). By Lemma 12.12 again, we may assume − ∼ that Jxv = Ωn (q), and q = −1. Thus, J ∼ = Ω− n−2 (q). We shall show that G1 := − ∼ K, Jxv = P Ωn+2 (q). Once again with k = n2 − 1, we consider a pair of spaces V, W as in (12P) affording natural modules for K, Jxv respectively, and such that J = Ω(V1 ⊥ · · · ⊥ Vk ). Rename W0 = V0 and let U = W + V = V0 ⊥ V1 ⊥ · · · ⊥ Vm with m = k + 1. Set C o = Ω(U ). We have homomorphisms f : Ω(Vm⊥ ) → Jxv and f1 : Ω(V0⊥ ) → K which agree on Ω((V0 + Vm )⊥ ). By hypothesis, m is even. For each odd i, 1 ≤ i < m, we can write Vi ⊥  with Vj a hyperbolic plane for all j. We set V0 = V0 . Then Vi+1 = Vi ⊥ Vi+1 we can choose a weak CT-system {Li | 0 ≤ i ≤ m} for C o with Li ∼ = SL2 (q) for  all i, and Li ≤ Ω(Vi ⊥ Vi+1 ) for 1 ≤ i ≤ m − 1, while L0 ≤ Ω(V0 ⊥ V1 ⊥ V2 )  ⊥ Vm ). In particular, Li ≤ Ω(W ) for 0 ≤ i ≤ m − 3, while and Lm ≤ Ω(Vm−1 Lj ≤ Ω(V ) for 1 ≤ j ≤ m. Let F be the group freely generated by the Li ’s. We can define a function g ∗ : F → H := Jxv , K by (12V)

(1) g ∗ (xi ) = f (xi ) for xi ∈ Li , 0 ≤ i ≤ m − 3; and (2) g ∗ (xj ) = f1 (xj ) for xj ∈ Lj , 1 ≤ j ≤ m.

ˆ i = g ∗ (Li ) for all i. As f = f1 on Ω((V0 ⊥ Vm )⊥ ), g ∗ is well-defined. We set L It follows that all CT relations for Spin(U ) are in the kernel of g ∗ , with the possible exception of the relations: [L0 , Lj ] = 1 for m − 2 ≤ j ≤ m.

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134

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

We may choose an element z8 ∈ CK (xv) with [V, f1−1 (z8 )] = Vm−3 ⊥ Vm−2 ⊥ Vm−1 ⊥ Vm . Then z8 ∈ J8 , a component of CK (z8 ) with J8 ∼ = Ω+ 8 (q). Moreover,  + r ∼ O (CK (z8 )) = K8 × J8 with K8 = Ωn−8 (q) and x ∈ Z(K8 J8 ). It follows from (solvable) L2 -balance that the pumpup of J8 in CG (z8 ) is a single 2-component J8∗ with J8 an anchored component of CJ8∗ (x). It follows from [III17 , 10.69d] that J8∗ = O2 (J8∗ )J8 . Indeed, as v, x acts on J8∗ , it follows from [III8 , 3.3] that J8∗ = J8 . Now, xv ∈ J8 and K8 ≤ L ≤ Lxv . As [W, f −1 (z8 )] = Vm−3 ⊥ Vm−2 ⊥ Vm−1 , we ∗ have K8 ≤ K8∗ = E(CJxv (z8 )) ∼ = Ω+ n−2 (q). Now, K8 ≤ E(CG (z8 )) by L2 -balance ∗  0 ≤ K8∗ and L  j ≤ J8 and, since [K8 , L8 ] = 1, we deduce that [K8 , L8 ] = 1. But L   for m − 2 ≤ j ≤ m, whence [L0 , Lj ] = 1 for all j, m − 2 ≤ j ≤ m, as desired. g : Spin(U ) → Jxv , K whose image It follows that g ∗ lifts  to a homomorphism   j | 1 ≤ j ≤ m . As x is the unique involution in CG (x, K), contains K, since K = L it follows that ker(g) = Z(Spin(U )), and so G2 := g(Spin(U )) ∼ = P Ω(U ). Then K ≤ G2 and K = E(CG2 (x)). Also, xv ∈ K ≤ G2 , and, by the embedding of K in G2 , it follows that E(CG2 (xv)) ∼ = Ω− n (q). Hence, Jxv ≤ G2 , whence G2 = G1 , completing the proof.  Lemma 12.15. Suppose that n = 8. Then K, Jxv ∼ = P Ω9 (q) or P Ω± 10 (q). ∼ Ω+ (q) as Z(K) = 1. By Lemma 12.12, we may assume Proof. We have K = 8 − that Jxv ∼ = Ω8 (q) and q = −1. Thus J ∼ = Ω− 6 (q). Again, from (11J) and Lemmas 11.12, 11.13, and 11.14, z4 ∈ J is an involution such that z4 ∈ R0   CG (z4 ) with R0 ∼ = SL2 (q) ∗ SL2 (q), I is a component of ∼ CG (z4 ) with [R0 , I x, v ] = 1, I ∼ (q) = Ω+ = L4 (q). 6 We write R0 = R1 ∗ R1∗ where R1 ∼ = R1∗ ∼ = SL2 (q). We shall produce a weak CT-system {Ri | 1 ≤ i ≤ 5} of type D5 (q). The argument again comes from [A9].  We may write O r (CK (z4 )) = (R1 ∗ R1∗ ) × (R4 ∗ R5 ), where R1 ∼ = R1∗ ∼ = R4 ∼ = ∼ R5 = SL2 (q). As C(x, K) has cyclic Sylow 2-subgroups by Lemma 11.7a, this is the product of all components and solvable 2-components of CG (z4 , x). Note that  O r (CI (x)) = R4 ∗ R5 . Now there is an involution g ∈ K interchanging R4 ∗ R5 with R1 ∗ R1∗ . Thus R0 = R1 R1∗ ≤ I g ∼ = L4 (q) and [I g , R4 R5 ] = [I, R1 R1∗ ]g = 1. As R1 is a (solvable) component of CG (z4 ), there is R2 ≤ I g such that R1 and R2 are a standard CT-pair in R1 , R2 ∼ = SL3 (q). To complete our weak CT-system of type D5 (q), it remains to find R3 centralizing R1 and such that R3 and Ri form a standard CT-pair in R3 , Ri ∼ = SL3 (q) for i = 2, 4, and 5. We write Z(Ri ) = ri for each i = 1, . . . , 5. Thus, r1 = z4 . From the structure of R1 , R2 , [r1 , r2 ] = 1 and r2 normalizes R1 . In particular, r2 normalizes I. We argue that + (12W) r2 acts on I like an involution in SO6 (q) whose −1-eigenspace on the natural module is 2-dimensional of + type. ∼ L4 (q), we see from [III17 , 4.10] that R2 could have been choSince I g = sen so that [r2 , u] = 1, which we henceforth assume. Let Y be the pumpup of I   CG (u, z4 ) in CG (u). Since u ∼G xz by Lemma 11.14, and L2 (CG (x, u)) = g ∼ − ∼ L2 (CK (u)) ∼ = Ω− 6 (q), we must have Y = Jxz = Ω8 (q). Moreover, u ∈ R4 R5 ≤ + g g I ≤ Y , so CG (u, u ) has a component W ∼ = Ω6 (q), and W ≤ Y ∩Y by L2 -balance. Note that W, R4 R5 = Y by [III17 , 9.1]. Also r2 centralizes ug ∈ R4 R5 , so r2 normalizes Y , Y g , and W . As [r2 , R4 R5 ] = 1, the action of r2 on Y is determined by

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12. THE ORTHOGONAL CASE, COMPLETED

135

its action on W . But the supports of r1 , r2 , and u on the natural Y g -module are all contained in the support of I g , where their action as a triple is determined up to conjugacy. Therefore on W = Y ∩ Y g = E(CY g (u)), the actions of r1 and r2 are uniquely determined. Hence their actions on Y = W, R4 R5 and I = E(CY (r1 )) are uniquely determined. We can therefore compute the action of r2 on I in Ω+ 10 (q), and this yields (12W).  Continuing to view I as L4 (q), note that O r (CI (r2 )) contains and then equals g R4 R5 . So r2 acts the same on I as r1 = r4 = r5 . By [III17 , 4.10], there exists R3 ≤ I such that [ri , r3 ] = 1 for all i = 1, . . . , 5, such that R4 , R3 , and R5 form a weak CT-system for I, and such that r2 induces an outer diagonal automorphism on R3 . Then by [III17 , 4.13], R2 and R3 form a standard CT-pair of type SL3 (q), and so R1 , . . . , R5 form a weak CT-system of type D5 (q), as desired. By [III13 , 1.4], G1 := R1 , . . . , R5 ∼ = D5 (q). g Now, R ≤ I for i = 1, 2, and Ri ≤ I for i = 3, 4, 5. Moreover, I = i    2 O (R4 R5 ), Ixv ≤ K, Jxv , so as g ∈ K, G1 ≤ I, I g ≤ K, Jxv . As we argued before, r4 = r1g ∈ I has 4-dimensional +-type support on the natural G1 -module, so CG1 (z4 ) has a component isomorphic to I and containing R1 . As R1 ≤ I g , this component can only be I g , so G1 = I, I g . In particular, G1 is g-invariant. As I is x-invariant, so is G1 . But R1 R1∗ R4 R5 ≤ R4 R5 , R4g R5g ≤ CG1 (x), so E(CG1 (x)) ∼ is isomorphic to Ω+ 8 (q) = K and then equal to K as m2 (C(x, K)) = 1. Hence, K ≤ G1 . Moreover, u ∈ R1 R1∗ ≤ G1 , and u and xv are conjugate in K, hence in ∼ G1 , so xv ∈ G1 with E(CG1 (u)) ∼ = Ω− 8 (q) = Jxv . Hence Jxv = E(CG1 (xv)) ≤ G1 ,  so K, Jxv = G1 , completing the proof of the lemma. To complete the proof of Proposition 11.1 when n ≥ 7, it remains to show: Lemma 12.16. Suppose that n is even and n ≥ 8. ΓD,1 (G) ≤ NG (G0 ).

Then G1 = G0 and

We postpone the proof of this until we have constructed an analogous subgroup G1 in the case n = 6. See Lemma 12.20 below. Now we treat the cases when n = 6, i.e., K∼ = Ω6 (q) or P Ω6 (q), q > 3.

(12X)

By Lemma 11.3, (11A3) holds. Thus, v=y∈J =L∼ = SL2 (q). By Lemma 11.4, (12Y)

L  E(CG (y)),

and hence also L∗ := Lw  E(CG (y)), where w ∈ CK (y), and LL∗ = E(CK (y)). Moreover, L < Lxy , and by Lemma 11.6, (12Z)

η Lxy ∼ = Ω5 (q) or Lxy is a quotient of Ω6 (q), η = ±1.

As y ∈ L ≤ Lxy but y ∈ Z(Lxy ), and as q > 3, we also have L∗ ≤ Lxy , by L2 -balance.

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136

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

There are three cases for K:  (1) K ∼ = Ω6q (q) (so that x ∈ K);  (2) K ∼ (12AA) = P Ω6q (q); − (3) K ∼ = Ω6 q (q) (which is simple). The following situation is somewhat exceptional and will require special arguments: η (12BB) q = −1, Lxy /Z(Lxy ) ∼ = P Ω (q), and η = −1. 6

We aim to prove: Lemma 12.17. Assume (12X), (12Y), and (12Z). Then, after interchanging K and Lxy if necessary, there exist g ∈ K and H ≤ Lgxy such that (a) L and Lg form a standard CT-pair (if q = +1) or standard P-pair (if  q = −1) in L, Lg ∼ = SL3q (q); (b) [L, H] = 1; (c) Lg and the component(s) of H form exactly one of the following for the group I = Lg , H :  (1) A weak CT-system, if I/Z(I) ∼ = P Ω6q (q) and q = +1;  q (2) A weak P-system, if I/Z(I) ∼ = P Ω6 (q) and q = −1; − (3) A standard CT-pair, if I ∼ = Ω5 (q) or Ω6 q (q), and q = +1; ∼ (4) A standard P-pair, if I = Ω5 (q) and q = −1; or − (5) A PP-pair, if I ∼ = Ω6 q (q) and (12BB) occurs; and (d) H = E(CI (y)) and I = Lgxy . Notice that since q > 3, there is no concern about extendibility for any weak CTP-system {L, Lg , H} that may result. Proof. First assume that (12AA1) holds. The argument is again due to Aschbacher [A9]. As q ≡ q (mod 4), we have Z(K) ∼ = Z2 and thus Z(K) = x . Fix an involution u ∈ LL∗ − y and set t = ux. Then as x = Z(K) and u has 2-dimensional support on the natural K-module of type q (so that the spinorial norm of u is trivial), t has 4-dimensional support of type +. Therefore (12CC)

t ∈ y K , whence u = xt ∈ (xy)K .

By [III17 , 4.8], there is g ∈ K such that 2

Conclusion (a) holds, Lg = L, y g = t, and y = tg . Let I be the subnormal closure of Lg in CG (u). As xy = ug , I = Lgxy ∼ = Lxy . Now w := yu = (txy)g = (uy)g ∈ (LL∗ − y )g = Lg (L∗ )g − t , and the actions of w and y on I = Lgxy coincide – so that y acts like an involution whose support on the natural Lgxy -module is 2-dimensional of type q . Setting H = E(CI (y)) = + ∼ E(CI (w)), we then have that H ∼ = Ω3 (q) or Ω− 4 (q) or Ω4 (q), according as I = Ω5 (q) −q q or Ω6 (q) or Ω6 (q). In all cases, (c) holds, as an application of [III17 , 4.6, 4.3]. It remains in this case to verify that (12DD)

[L, H] = 1.

But [H, y] = 1 by construction, so by L2 -balance, H ≤ L2 (CG (y)). By (12Y), Ly = L, so H normalizes L. Now u ∈ LL∗ − L∗ , so u induces a nontrivial inner automorphism on L, whence CL (u) is solvable. Also [u, H] = 1 by construction. If

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12. THE ORTHOGONAL CASE, COMPLETED

137

we then let H u be the image of H u in Aut(L), we have H ≤ CAut(L) (u)(∞) = 1, proving (12DD). Next suppose that (12AA2) holds, so that x ∈ K. This time choose t ∈ I2 (LL∗ − y ) and set u = tx.   Since K ∼  Ω6q (q), t ∈ y K , and since x ∈ K, u ∈ y K . By [III17 , 4.8] = P Ω6q (q) ∼ = again there exists g ∈ K interchanging y and t, and so again u = tx = (xy)g . Again let I be the subnormal closure of Lg in CG (u). Then again I = (Lxy )g ∼ = Ω5 (q) or −  Ω6 q (q) or (P )Ω6q (q). Since Z(Lg ) = t = Z((L∗ )g ), we also have (L∗ )g ≤ I. As g interchanges y and t, y ∈ (LL∗ )g − t . Thus, setting H = E(CI (y)), we again get (c) holding, as a consequence of [III17 , 4.6, 4.3]. Again it remains to prove that [L, H] = 1. To do this, we simply repeat the paragraph following (12DD), the only change being that now u ∈ LL∗ x − L∗ x; but still CL (u) is solvable. This establishes (a) in case (12AA2) holds. −  In the final case (12AA3), K ∼ = Ω6 q (q) is simple. If Lxy /Z(Lxy ) ∼ = P Ω6q (q), ∗ then F(Lxy ) = F(K), so by Lemma 1.2b, (xy, Lxy ) ∈ J2 (G). Interchanging the roles of x and xy, we are back in either case (12AA1) or case (12AA2), and so conclusions (a), (b), and (c) hold. Since we are allowed to make this interchange, − we may assume that Lxy ∼ = Ω5 (q) or Ω6 q (q). This time we choose t ∈ I2 (CK (y)) ∩ y K normalizing both L and L∗ and inducing outer diagonal automorphisms on them both. By [III17 , 4.9], we may choose t, and an element g ∈ K, such that g interchanges y and t by conjugation − and such that L, Lg ∼ = SL3 q (q). We again set u = xt = (xy)g and let I be − the subnormal closure of Lg in CG (u), so that I = Lgxy ∼ = Ω5 (q) or Ω6 q (q) and g ∗ g ∗ [y, u] = 1. Indeed as t ∈ L   CI (t), (LL ) = F (CI (t)). Now g interchanges y and t, so y induces an outer diagonal automorphism on each of Lg and (L∗ )g . Let VI be the natural I-module, and W = [VI , t]. Then W is 4-dimensional of + type and [W, y] is 2-dimensional of −q type. Then [VI , y] = [W, y] ⊥ [W ⊥ , y]. Note that if I ∼ = K then the action of y on (LL∗ )g guarantees that y induces an inner automorphism on I, so y ∈ u × I and so dim[W, y] is even. Thus, in any event, we set H = E(CI (y)) and conclude that H ∼ = Ω3 (q) or Ω+ 4 (q), according to the isomorphism type of I. It follows from [III17 , 4.6, 4.3] again that Lg and the component(s) of H satisfy (c). Note that they do not form a PP-pair in this case. As usual, we prove (b) in this case by repeating the paragraph following (12DD), noting that although u now induces an inner-diagonal automorphism on L instead of an inner automorphism, still CAut(L) (u) is solvable. This completes the proof. 

Lemma 12.18. Let L, Lg , and H be as in Lemma 12.17, and set G1 = L, Lg , H . If conclusion (c5) of that lemma does not hold, then G1 = K, Lxy ∼ = Ω7 (q), − P Ω+ 8 (q), or Ω8 (q). Proof. Since one of the conclusions (c1)–(c4) of the previous lemma holds by assumption, {L, Lg , H} is a weak CT-system of type B3 (q) or D4± (q) (if q = 1), or a weak P-system of type B3 (q) or D4 (q) (if q = −1 but Lxy /Z(Lxy ) ∼ = P Ω6 (q) or Ω5 (q)). Therefore by [III13 , 1.4, 1.14], G1 is a homomorphic image of Spin± a (q), a = 7 or 8. As g ∈ K, we have  I = Lg , H ≤ G1 ≤ L, I ≤ K, I = K, Lgxy = K, Lxy .

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138

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

To show G1 = K, Lxy , it remains to show that K ≤ G1 . But by construction, I is a component of CG1 (u), so u acts on G1 as an involution with 2-dimensional support of type q on the natural G1 -module. Since t ∈ Lg ≤ I, t has 4-dimensional support of + type on the natural I-module, so x = tu has 6-dimensional support of type q  on the natural G1 -module. Hence E(CG1 (x)) ∼ = Ω6q (q) ∼ = K. As m2 (C(x, K)) = 1, this forces K = E(CG1 (x)) ≤ G1 . Therefore G1 = K, Lxy . Finally, as x is the only involution of C(x, K), and [x, G1 ] = 1, Z(G1 ) has odd order. As q > 3, it follows from [IA , 6.1.4] that Z(G1 ) = 1, and (a) follows, proving the lemma.  Lemma 12.19. Suppose that conclusion (c5) of Lemma 12.17 holds. Let G1 = K, Lxy . Then G1 ∼ = Ω− 8 (q). ∼ Proof. Since Lemma 12.17c5 holds, we have K/Z(K) ∼ = P Ω− 6 (q) and Lxy = g L, g, H, and I be as in Lemma 12.17. Then {L , H} is a PP-pair for

Ω+ 6 (q). Let I∼ = Ω+ 6 (q).

Let VI be the natural I-module and VC = CVI (Lg ), so that VC is 2-dimensional of + type. Write VC = V1 ⊥ V2 as the sum of two lines, and set Hj = CH (Vj ), j = 1, 2. By [III17 , 4.5], {Lg , Hj } is a standard P-pair for Ij := Lg , Hj ∼ = Ω5 (q), for j = 1, 2. Moreover, Ij = CI (Vj ). Since [L, Hj ] ≤ [L, H] = 1 for j = 1, 2, it follows that {L, Lg , Hj } is a weak P-system of type B3 (q) for Mj := L, Lg , Hj , j = 1, 2. As q > 3, it follows from [III13 , 1.14] that Mj is an x, y -invariant homomorphic image of Spin7 (q) for j = 1, 2. Now, CMj (x) contains L, Lg ∼ =  SU3 (q), so O r (CMj (x)) does as well; it follows from [IA , 4.5.1] that E(CMj (x)) is a homomorphic image of SU4 (q) for j = 1, 2. But K is the unique subgroup of CG (x) centrally isomorphic to SU4 (q), and so K = E(CMj (x)) ∼ = Ω− 6 (q) for j = 1, 2. In − particular, as K ∼ = Ω7 (q) for j = 1, 2. = Spin6 (q), Mj ∼ Let W1 be a natural module for M1 . Then the support WK of K on W1 is 6-dimensional and u ∈ K with [WK , u] 2-dimensional, whence E(CM1 (u)) ∼ = Ω5 (q), As H1 = H2 , it follows from [III17 , 9.1] that H = H1 , H2 . If M1 = M2 , then I = Lg , H ≤ M1 , a contradiction. Thus M1 = M2 . We may decompose WK as a sum WK = V1 ⊥ · · · ⊥ V6 of isometric lines such that W2 := WK ⊥ V7 is a natural module for M2 with V7 isometric to V1 . Altering our choice of W1 , if necessary, we may assume that the orthogonal complement, V0 , to WK in W1 is not isometric to V1 . With this choice, set V := W1 + W2 = V0 ⊥ V1 ⊥ · · · ⊥ V6 ⊥ V7 . By [III17 , 2.40], we may choose a subgroup N < M2 with N ∼ = A7 permuting the set {V1 , . . . , V7 } and such that N ∩ K is the centralizer in N of the subspace V7 , acting as A6 on {V1 , . . . , V6 }. Suppose that N ≤ M1 . By the action of N on {V1 , . . . , V7 }, M1 ∩ M2 contains a conjugate K1 = Ω(V1⊥ ) of K. Then by [III17 , 9.1], M1 ∩ M2 ≥ K, K1 = M2 , contrary to the previous paragraph. Hence, N ∩ M1 = N ∩ K. Now we fix an isomorphism f : Ω(V7⊥ ) → G2 := M1 , N with image M1 , and an isomorphism f1 : Ω(V0⊥ ) → M2 . Finally, we define λ : N → T , where T = f1−1 (N ) < Ω(V0⊥ ). Then λ is an isomorphism, and all the compatibility conditions of [III13 , Theorem 3.1] hold (with M1 , G2 in the roles of K, H there), yielding that there is a homomorphism g : Spin(V ) → G2 and a 6-dimensional subspace U of V7⊥ such that Ω(U ) ∼ = Ω− 6 (q) and g(Ω(U )) = f (Ω(U )) ≤ M1 . Now,

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12. THE ORTHOGONAL CASE, COMPLETED

139

∼ Ω− (q). Let x1 = Z(K1 ). Then x1 ∈ xM1 , and we may change f (Ω(U )) =: K1 = 6 notation so that x1 = x and K1 = K. Let G3 := im(g). As x ∈ Syl2 (CG (x, K)), we see that G3 ∼ = Ω(V ) ∼ = Ω− 8 (q). Also, K is the centralizer in G3 of a plane in the ∼ + natural module for G3 . As E(CK (y)) ∼ = Ω+ 4 (q) and E(CG (xy)) = Lxy = Ω6 (q), it follows that E(CG3 (xy)) = Lxy . Hence, G1 := K, Lxy ≤ G3 . Using [III17 , 9.1]  we see that G3 = K, Lxy = G1 , completing the proof. Recall that G0 = N(x, K, y, L) = K, Ly , Lxy . We set M = NG (G0 ). From Lemmas 12.11–12.15, 12.18, and 12.19, n is even, x ∈ K, and the subgroup G1 = K, Jxv of G is isomorphic to Ωn+1 (q) or P Ωηn+2 (q) for some sign η. (When n = 6 we interpret v = y, Jxv = Lxy .) Accordingly we let r = 1 or 2. To complete the proof of Proposition 11.1, we need only prove the following result. Lemma 12.20. Let G1 = K, Jxv and M1 = NG (G1 ). Then the following conditions hold: (a) ΓD,1 (G) ≤ M1 , where D = x, y ; (b) G1 = G0 and M1 = M ; and (c) N(x, K, y, L) satisfies row 4, 5, 7, or 8 of Table 14.1. For notational convenience when n = 6, we change notation by writing J = LL∗ = E(CK (v)). Set E = x, v . Note that since x ∈ K and v ∈ J ≤ K, we have E ≤ K. The main step in the proof of Lemma 12.20 is to show that ΓE,1 (G) ≤ M1 . We let V1 be a natural (projective) orthogonal module for G1 of dimension n + r. Then we may assume that K and J are embedded naturally in G1 , so that the support of K and x has dimension n, and contains the support of J and v, which has dimension n − 2. Hence, xv acts on V1 with a 2-dimensional eigenspace for −1. As xv ∈ K ≤ G1 , [V1 , xv] is a plane of type q . Also, the pumpup of J in E(CG1 (xv)) ηq ηq (q), so we see that Jxv ∼ (q), with is isomorphic to Ω(CV1 (xv)) ∼ = Ωn+r−2 = Ωn+r−2 r ∈ {1, 2}. (The sign only applies if r = 2.) First we require the following lemma. For e ∈ E # , set S(e) = {K1 | K1  E(CG (e)), K1 ∼ = Ωπm (q), m ∈ {n − 1, n}, π = ±1} Lemma 12.21. S(x) = {K}, S(v) = ∅, and S(xv) = {Jxv }. Moreover, (xv)K ∩ J=  ∅. Proof. Since J = Ω([V1 , v]) with dim[V1 , v] = n − 2 ≥ 4, and [V1 , xv] is a plane of type q , there exists w ∈ (xv)K ∩ J, proving the final statement. We have m2 (CG (K)) = 1, whence K is the unique component of CG (x) in S(x), as n ≥ 6. Let K1 ∈ S(xv) − {Jxv }, so that Jxv ∗ K1  E(CG (xv)). Now, x normalizes Jxv , J ≤ Jxv and C(x, J)/CG (K) is solvable with dihedral Sylow 2-subgroups, while x is the unique involution of CG (K). Since [x, Jxv ] = 1, x ∈ K1 K1x , so CK1 K1x (K) is of odd order. Thus CK1 K1x (x) is solvable with cyclic or dihedral Sylow 2-subgroups, and in particular x normalizes K1 . As K1 ∼ = Ωπm (q) for some m ≥ 5, and CK1 (x) is solvable with cyclic or dihedral Sylow 2-subgroups, it follows from [IA , 4.5.1, 4.9.1]  that K1 ∼ = L2 (3). But then K ∼ = Ω− = Ω5 (3) and O 3 (CK1 (x)) ∼ 6 (3), contradicting the hypothesis of Proposition 11.1. Hence, S(xv) = {Jxv }. In fact, the same

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

140

argument shows that CG (xv) has no 2-component K2 with K2 /O2 (K2 ) a vertical pumpup of some Ω± m (q), m ∈ {n − 1, n}. Finally, let K1 ∈ S(v). Then J ∗ K1  E(CG (v)). Let w ∈ (xv)K ∩ J. Then by the last sentence of the preceding paragraph, the pumpup of K1 in CG (w) must be trivial and indeed K1 is the pumpup in CG (w) of E(CK (w)) ∼ = J. But then CG (x, J ) ≥ E(CK (w)), contrary to the fact that CG (x, J )(∞) = CG (K)(∞) has 2-rank at most 1. This completes the proof of the lemma.  (∞)

Lemma 12.22. We have NG (E) ≤ M1 . Moreover, G1 = E(M1 ) = M1

.

Proof. NG (E) permutes {S(e) | e ∈ E # } = {K, Jxv }, by the preceding lemma. Hence NG (E) normalizes K, Jxv = G1 , and so NG (E) ≤ M1 . Since m2 (C(x, K)) = 1 and [x, G1 ] = 1, CG (G1 ) = CM1 (G1 ) has odd order. As  G1  M1 , the second statement follows immediately. Lemma 12.23. Let e ∈ {x, xv}. Then v CG (e) ⊆ v CG1 (e) and CG (e) ≤ M1 . Proof. If e = xv, we have v ∈ J ≤ Jxv  CG (e), by Lemma 12.21. Hence ∼ ± v ⊆ v Aut(Jxv ) . As Jxv ∼ = Ω± m (q), m ∈ {n − 1, n}, n ≥ 6, with v ∈ J = Ωn−2 (q), it follows from [III17 , 14.2] that v Aut(Jxv ) = v Jxv ⊆ v CG1 (e) . Hence v CG (e) ⊆ v CG1 (e) . The same inclusion holds for e = x, by the same argument with K in place of Jxv . Now let g ∈ CG (e). Then v g = v h for some h ∈ CG1 (e). Hence, gh−1 ∈ CG (v, e ) = CG (E) ≤ M1 . So, g ∈ M1 h = M1 and CG (e) ≤ M1 , and the proof is complete.  CG (e)

Lemma 12.24. We have ΓE,1 (G) ≤ M1 . Proof. Since NG (E) ≤ M1 and CG (e) ≤ M1 for e ∈ {x, xv}, it remains to prove that CG (v) ≤ M1 . We know that J  E(CG (v)) and there exists w ∈ (xv)K ∩ J. Hence, L2 (CG (v)) ≤ JCG (w, v ) ≤ M1 , as CG (w) ≤ M1 . Therefore J ≤ E(CG (v)) = L2 (CG (v)) ≤ E(CG1 (v)) = JE(Lv ), with Lv ∼ = Ω± r+2 (q) and ± ∼ r ∈ {1, 2}. In particular, as v ∈ J = Ωn−2 (q), we see that J  CG (v) unless G1 ∼ = P Ω+ 8 (q). Returning to our w ∈ (xv)K ∩ J, let w1 ∈ wCG (v) . Then w1 ∈ wJ if G1 ∼ = P Ω+ 8 (q), using [III17 , 14.2] again. Hence in that case CG (v) ≤ JCG (w) ≤ M1 . Suppose finally that G1 ∼ = P Ω+ 8 (q). Then q > 3 and E(CG (v)) is the central product of four copies of SL2 (q), on which CAut(G1 ) (v) induces the permutation group Σ4 . Hence w1 is CAut(G1 ) (v)-conjugate to w. Hence, in any case, CG1 (w1 )(∞) ∼ = CG1 (xv)(∞) = CM1 (xv)(∞) = CG (xv)(∞) ∼ = CG (w1 )(∞) , so CG1 (w1 )(∞) = CG (w1 )(∞) ; the same holds for w1 v in place of w1 . Thus by [III17 , 9.1], G1 = CG (w1 )(∞) , CG (w1 v)(∞) for all w1 ∈ wCG (v) . As CG (v) permutes the set {(w1 , w1 v) | w1 ∈ wCG (v) }, it follows that CG (v) ≤ NG (G1 ) = M1 , as claimed.  Finally we prove Lemma 12.25. We have ΓD,1 (G) ≤ M1 = M and G1 = G0 .

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13. THE CASES IN WHICH G0 IS EXCEPTIONAL

141

Proof. Now, E = D and G1 = K, Jxv = K, J, Jxv = K, Ly , Lxy = G0 except in case (11A2). As we have ΓE,1 (G) ≤ M1 by Lemma 12.24, we may assume that case (11A2) holds. By our choice of v (before (11D)), [v, y] = 1 and so [E, D] = 1. Let d ∈ {y, xy} and as usual let Ld be the pumpup of L = E(CK (d)) ∼ = Ωn−1 (q) in CG (d). As n ≥ 6, with strict inequality if q = 3, Ld ∈ Chev(r). Then by [IA , 7.3.4], Ld = ΓE,1 (Ld ) ≤ M1 . It follows that Ld is a component of CG1 (d) isomorphic to Ω± k (q), where either k = n or k = n − 1, the latter being possible only if G1 ∼ = Ωn+1 (q). Since v ∈ J ≤ L ≤ Ld and CG (v) ≤ M1 , we have E(CG (d)) ≤ Ld CG (v) ≤ M1 , whence Ld = E(CG (d)). In particular, Ld  CG (d). Using [III17 , 14.2] as before, since k > 4, we see that v CG (d) = v Ld and so CG (d) ≤ Ld CG (v) ≤ M1 , as desired. Moreover, again by [IA , 7.3.4], we have G1 = K, Ly , Lxy = G0 . As NG (D) permutes the set {K, Ly , Lxy } = {E(CG (d)) | d ∈ D# }, we have NG (D) ≤ M1 = NG (G0 ) = M as desired.  This completes the proof of Lemma 12.20ab. It follows that for each d ∈ D# , Ld is the pumpup of L in CG0 (d). If n = 6, this directly implies that row 4 or 7 of Table 14.1 applies. Otherwise n ≥ 8 and there exists y  ∈ I2 (CG0 (x)) such that CK (y  ) has a component isomorphic to Ωn−1 (q). Since (x, K, y, L) is not ignorable, it follows that L ∼ = Ωn−1 (q). Then it is an easy consequence that row 5 or 8 of Table 14.1 applies. So Lemma 12.20 is completely proved. As noted before that lemma, the proof of Proposition 11.1 is now complete.

13. The Cases in Which G0 is Exceptional In this section we consider the four cases remaining for the proof of Theorem C∗7 : Stage 4b in the case p = 2, after the case analysis carried out in the previous sections of this chapter. (A summary of all the cases is given in the next section.) We continue our assumptions and notation (1A), (1B) and (1C), as well as (1E). We shall prove: Proposition 13.1. Assume that for some odd prime power q and sign , one of the following holds: # G ∼ (a) K ∼ = Spin+ = Spin9 (q), L ∼ 8 (q), D = Z(L), D ⊆ x , and AutLu (D) = Z2 # for all u ∈ D ; # G ∼ (b) K ∼ = Spin+ = Spin10 (q), L ∼ 8 (q), D = Z(L), D ⊆ x , and AutLu (D) = Z2 # for all u ∈ D ;   (c) K ∼ = A5q (q), and for one of u = y, xy, = A7q (q), K is not 2-saturated, L ∼ q q u ∼ ∼ we have xu ∈ L = A5 (q) , Lu = E6 (q) and Lxu ∼ = D6 (q)hs ; moreover, # m2 (C(d, Ld )) = 1 for all d ∈ D ; u ∼ (d) K ∼ = HSpin16 (q), L ∼ = Spin+ 12 (q), and Lu = E7 (q) for all u ∈ D − x . Set G0 = N . Then G0 ∼ = F4 (q), E6 (q), E7 (q), or E8 (q), in cases (a)–(d), respectively. Moreover, in all cases, ΓD,1 (G) ≤ NG (G0 ). It will be clear in each case of Proposition 13.1 that the appropriate row of Table 14.1 applies. Throughout this section we assume that the hypotheses of Proposition 13.1 hold, but the conclusion of the proposition fails.

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142

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

We consider the F4 (q) and E6 (q) cases first, assuming through Lemma 13.3 that hypothesis (a) or (b) of Proposition 13.1 holds. + ∼ We label the alternatives K ∼ = Spin9 (q), K ∼ = Spin− 10 (q), and K = Spin10 (q) as Cases A, B, and C, respectively. In each of these cases, by Proposition 8.1, L∼ = Spin+ 8 (q), so that Z(L) is a four-group. Lemma 13.2. Assume that hypothesis (a) or (b) of Proposition 13.1 holds. Then the following conditions hold: ∼ Σ3 ; (a) D = Z(L), D# ⊆ xG , and AutG (D) = (b) Lu = E(CG (u)) = CG (u)(∞) for all u ∈ D# ; (c) C(u, Lu ) has cyclic Sylow 2-subgroups for all u ∈ D# ; (d) AutNG (D) (L) ≥ Inn(L)Γ, where Γ is a group of graph automorphisms of L isomorphic to Σ3 and Inn(L)Γ is Aut0 (L)-conjugate to Inn(L)ΓL ; and (e) G0 is NG (D)-invariant.

 Proof. Part (a) is immediate from the hypotheses of the proposition. Indeed, AutK (L), AutLy (L) induces all of Aut(D) ∼ = Σ3 on D and lies in Aut0 (L). In particular x, y, and xy are conjugate in NK,Ly  (L). It suffices to prove that C(x, K) has cyclic Sylow 2-subgroups. For then (c) G follows as D# ⊆ with the Schreier property, in turn implies (b),  x . This, together whence G0 = E(CG (u)) | u ∈ D# is NG (D)-invariant, which is (e). Also since Aut0 (L)/ Inn(L) ∼ = Σ4 by [IA , 2.5.12], (d) follows immediately. Recall that Q ∈ Syl2 (C(x, K)). Choose any u ∈ D − x . As D ≤ L ≤ K, Q ≤ CCG (u) (L). But u is the unique involution in C(u, Lu ), while Ω1 (Q) = x . Therefore CQ (Lu ) = 1, so Q embeds in CAut(Lu ) (L), which is cyclic or dihedral by [III17 , 6.10] as Lu ∼ = K. As m2 (Q) = 1, Q is cyclic, and the proof is complete.  We shall choose root SL2 (q)- and SL2 (q 2 )-subgroups Li of K and of other components. Some of these Li will eventually yield a weak CT-system for G0 (see (13E) below). Here i ranges over the index set I consisting of 6 together with the subscripts in the diagrams for K and Iz in (13A) below. Thus the set I is the same in Cases A and B, but different in Case C. We also set Z(Li ) = zi for each i ∈ I. Each Li is associated with the node(s) labelled αi in (13A) below. In every case we choose SL2 (q)- and SL2 (q 2 )-subgroups of K corresponding to the fundamental roots and the lowest root. We choose SL2 (q 2 ) subgroups only in Case B and then only for the short roots. We number these Li ’s 0 ≤ i ≤ 3, as well as L4 in Cases A and B, and L4 and L4 in Case C, in accordance with the extended Dynkin diagrams at the top of (13A). Here the blackened node α0 is the lowest root. Without loss we may assume that L = L0 , L1 , L2 , L3 . The highest root SL2 (q) subgroup of L is not shown, but it will come into play and we name it L6 . By [III17 , 6.17], z0 , z1 , z3 , z6 ∼ = E23 , z0 z1 z3 z6 = 1, and D# = Z(L)# = {z1 z0 , z1 z3 , z1 z6 }.

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13. THE CASES IN WHICH G0 IS EXCEPTIONAL

Cases A: K ∼ = Spin9 (q) and B: K ∼ = Spin− 10 (q)

Case C: K ∼ = Spin+ 10 (q) • α0 ◦α4

• α0 α1

K: ◦

(13A)

◦

α2

α3

◦

α4

α1

◦

◦

◦

α2

α4 α1

α3

◦

Iz :

143

◦

α4

◦

α5

◦

α1

◦

α3

◦

α3 α4

◦ ◦

◦

α5

◦

◦

α4

◦ α5

By [III17 , 6.17], L6 is centralized by ΓL , which permutes {L0 , L1 , L3 } faithfully; and each of the four subgroups L0 , L1 , L3 , L6 is centralized by some Aut(L)conjugate of ΓL that permutes the other three of the four subgroups faithfully. We may thus assume that Γ = v, w ∼ = Σ3 is an Aut(L)-conjugate of ΓL such that 2 3 Lv0 = L3 , Lv6 = L6 , Lw 3 = L6 , [L1 , w] = [L1 , v] = 1, v = w = 1.

(13B)

For convenience, and in Case C only, when i = 4 or 5, we define Li = Li Li and zi = zi zi . Let z = z1 . As L is a subsystem subgroup of K, L1 is a root SL2 (q) subgroup  in both L and K. Then by [IA , 4.5.2] we have O r (CL (z)) = L0 L1 L3 L6 and  O r (CK (z)) = L1 L0 Jx with L6 ≤ Jx , [L1 L0 , Jx ] = 1, and x ∈ Jx = L3 , L4 ∼ = Spin5 (q) ∼ = Sp4 (q), or Spin6 (q) ∼ = SL4 (q), according as K ∼ = B4 (q) or D5 (q). Moreover, the spin involution x of K is the same as that of Jx and also that of L1 L0 = L 1 × L0 ∼ = Spin+ 4 (q), namely the diagonal involution of this direct product.  Conjugating by w and w2 we get O r (CLu (z)) = L1 Lk Ju for each u ∈ D − x , with u ∈ Ju ∼ = Jx , Lk Ju = Lk × Ju and k = 3 or 6, depending on u. Let Hz be the subnormal closure of L0 L3 L6 in CG (z). As L3 , L6 ≤ Jx and L0 ≤ Ju for u ∈ D − x , Hz is equivalently the subnormal closure of Jx , Jy , Jxy in CG (z), with Ju a component of CHz (u) for each u ∈ D# . We conclude from [III17 , 3.31] that Hz /[Hz , O2 (Hz )] ∼ = Sp6 (q) or a quotient of SL6 (q) by a subgroup of order dividing 3, again according as K ∼ = B4 (q) or D5 (q). Moreover, Hz is quasisimple, by [III8 , 3.3] applied with L0 in the role of J there. In fact, then Hz = Jx , Jy , Jxy by [III17 , 9.1], so [Hz , L1 ] = 1. A diagram for L1 Hz =: Iz is included in (13A) and may be helpful in following our argument. We have SL2 (q) × SL2 (q) × SL2 (q) ∼ = L0 , L3 , L6 ≤ Hz and z = z0 z3 z6 ∈ Hz . Note that since w centralizes z, w cycles Jx , Jy , Jxy , and so w normalizes Hz . Furthermore w cycles z0 , z3 , z6 according to the cycle (036). It follows that the group A := w, zz0 is isomorphic to the alternating group A4 , with O2 (A) = Z(L) = D ≤ Hz . Indeed z2 centralizes z0 , z3 , z6 and so A ≤ CHz w (z2 ).

(13C) v

Since w = w (13D)

−1

2

= w , while Out(Hz )/ Outdiag(Hz ) is abelian by [IA , 2.5.12],

w induces an inner-diagonal automorphism on Hz .

We define w w L5 = L w 4 , and (in Case C only) L5 = L4 and L5 = L4 .

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

144

Thus in Case B, L5 ∼ = SL2 (q 2 ), and in Case C, L5 = L5 L5 . As L4 ≤ Jx ≤ Hz and w normalizes Hz , L5 ≤ Hz . Conjugating the relations [L0 , L4 ] = [L2 , L4 ] = 1 by w yields [L3 , L5 ] = [L2 , L5 ] = 1. Among L2 , L3 , L4 , and L5 , these relations, together with the relations on the Li from K, leave only the relations between L4 and L5 to be established in order to reach our goal that (1) {L2 , L3 , L4 , L5 } is a weak CT-system of type F4 (q) or E6− (q), in Case A or B, respectively; and (13E) (2) {L2 , L3 , L4 , L4 , L5 , L5 } is a weak CT-system of type E6 (q), in Case C. Since [z2 , z] = [z2 , L4 ] = 1, z2 normalizes Hz . We set 

H2 = O r (CHz (z2 )). Our conditions imply that (13F)

L4 , L5 ≤ H2 ,

and we shall identify L4 , L5 by identifying H2 . Note that as [A, z2 ] = 1, H2 is A-invariant. Notice also that as α1 and α2 are connected in the K-diagram, z, z2 ∼ = E22 and AutG (z, z2 ) = AutK (z, z2 ) ∼ = Σ3 . It follows that (13G) If z ∈ E(CHz (z2 )), then z2 maps into Inn(Hz ). Because z ∈ L2 (CG (z, z2 )), and as z and z2 are NG (z, z2 )-conjugate, z2 ∈ L2 (CG (z, z2 )) ≤ L2 (CG (z)) by L2 -balance. This implies (13G). At this point we separate the cases A, B, and C, although the arguments in the three cases are very similar. Suppose that we are in case A. Let V be the natural Hz ∼ = Sp6 (q)-module, and set Vi = [zi , V ], the support of Li on V , for i = 0, 3, and 6. Moreover, as |Outdiag(Hz )| = 2, w induces an inner automorphism on Hz by (13D), so w induces an isometry on V . Since L4 is a short root subgroup of Hz , its support on V is 4-dimensional; but [z0 , L4 ] = 1 so the supports of both L4 and z4 on V are V0⊥ . Furthermore, from the action of z, z0 , z3 , z6 on V , V is clearly the direct sum of two faithful irreducible (3-dimensional) A-modules. To compute L4 , L5 , observe first that the support of z5 = z4w on V is (V0⊥ )w = ⊥ V3 , so [z4 , z5 ] = 1. Next consider the action of z2 on Hz . This action cannot be that of an involution of Hz . For if it were, H2 would leave invariant a unique 2dimensional subspace V  of V , and then V  would have to be A-invariant, whence 3 would divide dim V  , a contradiction. Indeed, as L2 , L3 ∼ = SL3 (q), with [z2 , z3 ] = 1, z2 induces an inner-diagonal automorphism ζ2 on L3 . Calculating in SL3 (q), moreover, we see that ζ2 is inner if and only if q ≡ 1 (mod 4). It follows from [IA , 4.2.3] that z2 induces an innerdiagonal automorphism on Hz . Moreover, as AutHz (L3 ) = Inn(L3 ) by [IA , 4.5.2], and |Outdiag(Hz )| = 2, we conclude that q ≡ 1 (mod 4) ⇐⇒ ζ2 is inner ⇐⇒ z2 maps into Inn(Hz ). Since z2 does not act on Hz as an involution of Hz , the only possibility, by [IA , 4.5.1],  is that O r (CHz (z2 )) = CHz (z2 )(∞) = E(CHz (z2 )) ∼ = A2 (q). As [z4 , z5 ] = 1, it

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13. THE CASES IN WHICH G0 IS EXCEPTIONAL

145

follows that L4 and L5 form a standard CT-pair in L4 , L5 ∼ = A2 (q). This is the desired Curtis-Tits relation and so (13E) holds in Case A. Next suppose that Case B holds, whence Hz is a central quotient of SU6 (q) by a subgroup of order dividing 3. Let V be the natural SU6 (q)-module, so that Hz acts projectively on V . The action on V of 2-subgroups of Hz , as well as of long root SL2 (q) subgroups of Hz , is thus well-defined. Again Out(Hz )/ Outdiag(Hz ) is abelian, so w acts on Hz like a 3-element of GU (V ), i.e., an isometry. Again the support of the short root SL2 (q 2 )-subgroup L4 on V is 4-dimensional, so that support equals V0⊥ , and V is the direct sum of two faithful irreducible A-modules of dimension 3. As in Case A, any proper AH2 -invariant subspace of V is 3-dimensional. This makes it impossible that H2 has SU4 (q) or SU5 (q) as a component. Also since H2 contains L4 ∼ = SL2 (q 2 ), and L2 (q 2 ) is not involved in SU3 (q), H2 is not the central product of two SU3 (q)’s. If z2 induces a graph automorphism on Hz , then  Sp6 (q) by (13G). Hence H2 is a 6-dimensional orthogonal group over Fq , H2 ∼ = and V is the direct sum V1 ⊕ V2 of two natural Fq H2 -modules, by [III17 , 6.14]. But then L4 ∼ = SL2 (q 2 ) has a natural module on the support of Z(L4 ) on V1 , so L4 ± 2 ∼ embeds in O4 (q). As Ω± 4 (q) = SL2 (q) ∗ SL2 (q) or L2 (q ), this is a contradiction. The only possibility remaining is that H2 is a central quotient of SL3 (q 2 ). Then as [z4 , z5 ] = 1, L4 , L5 = H2 and (13E) holds. Now assume that Case C holds. Then Hz , a quotient of SL6 (q) by a 3-group, acts projectively on the natural SL6 (q)-module V . As in Case A, let Vi = [zi , V ] be the support of Li on V . By (13D), w acts projectively on V as well, cycling V0 , V3 , V6 according to the permutation (036). Also L4 = L4 × L4 , with dim V4 = dim V4 = 2 and dim V4 = 4. Now [z0 , L4 ] = 1 so the eigenspaces V0 and V3 ⊕ V6 of z0 are L4 -invariant. If, say, V0 ⊆ CV (z4 ), then since L4 is irreducible on V4 , it would follow that V4 = V0 . But then V3 ⊆ CV (L4 ), and so V3 would be L3 , L4 -invariant. As L3 , L4 ∼ = SL3 (q) and L3 acts nontrivially on V3 , with dim V3 = 2, this is absurd. We conclude that V0 is centralized by z4 , and similarly by z4 . Thus V4 ∩ V0 = 0. As L4 normalizes V3 + V6 , it follows that V4 = V3 + V6 and CV (L4 ) = V0 . Note that L4 and L4 are normalized by z3 and centralized by z1 and z0 , so they are normalized also by z6 = z0 z1 z3 . Hence z4 and z4 are centralized by z0 , z3 , z6 . So z4 and z4 respect the decomposition V = V0 ⊕ V3 ⊕ V6 . z4w

V4w

Now z5 = inverts = V0 + V6 and centralizes V0w = V3 . It follows from the previous paragraph that [z5 , z4 ] = [z5 , z4 ] = [z5 , z4 ] = 1. Again (13F) holds, and we consider the possible actions of z2 on Hz . Again since L2 , L3 ∼ = SL3 (q) with [z2 , z3 ] = 1, z2 induces an inner-diagonal automorphism of L3 , and hence z2 maps into Aut0 (Hz ), by [IA , 4.2.3]. We consult [IA , 4.5.1] and again deduce the structure of H2 by a process of elimination. As in Case A, any proper AH2 -invariant subspace of V is 3-dimensional. This rules out H2 ∼ = A3 (q)A1 (q) and H2 ∼ = A4 (q). As the supports of z4 and L4 on V are 2-dimensional, H2 ∼ = A2 (q 2 ). Similarly, the action of L4 on V shows that H2 is not an orthogonal group, and (13G) rules out H2 ∼ = Sp6 (q). Having ruled out all these cases, we conclude from [IA , 4.5.1] that H2 = H2 H2 with [H2 , H2 ] = 1 and H2 and H2 both isomorphic to A2 (q).

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

146

By (solvable) L2 -balance, and since A2 (q) does not contain the central product of two SL2 (q)-subgroups (or even Z3 × SL2 (3) when q = 3), we may assume that notation has been chosen so that L4 ≤ H2 and L4 ≤ H2 . Clearly H2 and H2 w are w-invariant, so L5 = Lw 4 and L5 = L4 lie in H2 and H2 respectively. As Z(H2 ) has odd order and z4 and z4 centralize z5 = z5 z5 , [z4 , z5 ] = [z4 , z5 ] = 1. Moreover, [L4 , H2 ] = 1 = [L4 , H2 ]. Therefore (13E) holds in Case C as well. We have given the main part of the proof of the following lemma. Lemma 13.3. Assume that hypothesis (a) or (b) of Proposition 13.1 holds. Then G0 ∼ = Spin9 (q) or Spin10 (q). = F4 (q) or G0 ∼ = E6 (q),  = ±1, according as K ∼ Moreover, ΓD,1 (G) ≤ NG (G0 ). Proof. Continuing the above argument, let G1 = L2 , L3 , L4 , L5 in either case, whence G1 has isomorphism type asserted for G0 by (13E) and [III13 , 1.4]. w As L2 , L3 , L4 ≤ K ≤ G0 and L5 = Lw 4 ≤ G0 = G0 (note that w ∈ NG (D) ≤ (∞) ∼ NG (G0 )), we have G1 ≤ G0 . But then by [IA , 4.5.1] and = Lemma 13.2b, CG1 (u) # # ∼ = G0 . Thus, G1 = G0 , and K = Lu for all u ∈ D , so G1 contains Lu | u ∈ D the first assertion follows. By Lemma 13.2e it remains to prove that CG (u) ≤ NG (G0 ) for each u ∈ D# , and for that it suffices to show that CG (u) ≤ G0 NG (D). But D ≤ Lu and by [IA , 4.5.1], DLu is the unique Lu -conjugacy class of four-subgroups E of Lu such that Z(Lu ) ≤ E and E(CLu (E)) ∼ = Spin+ 8 (q). As Lu  CG (u), CG (u) =  Lu NCG (u) (D), and the lemma follows. We next prove Lemma 13.4. If Proposition 13.1c holds, then G0 ∼ = E7 (q) and ΓD,1 (G) ≤ NG (G0 ). We assume hypothesis (c) of Proposition 13.1 through Lemma 13.12. Thus notation may be chosen so that    K∼ = E6q (q), and Ly ∼ = D6 (q)hs . = A5q (q), Lxy ∼ = A7q (q), L ∼

(13H) We first note

Lemma 13.5. The following conditions hold: (a) CG (D) has a normal subgroup which is a central product L ∗ J, with L ∼ = SL6 (q) and J ∼ = SL2 (q); (b) J ≤ C(y, Ly ); and (c) Lu  CG (u) for all u ∈ D# . Proof. By the hypothesis of Proposition 13.1, m2 (C(u, Lu )) = 1 for all u ∈  D . This immediately implies (c). In particular K  CG (x), and O r (CK (D)) is the required subgroup in (a).  By [IA , 4.5.1], L = O r (CLy (x)) and x induces an inner automorphism on Ly . If q > 3 it follows from L2 -balance that J ≤ C(y, Ly ), as desired. If q = 3, then CAut(Ly ) (DL) is a 2-group and so as J = O 2 (J), (b) holds in this case as well, and the lemma is proved.  #

Next (see [IG , 10.11]) fix Ty ∈ Syl2 (CG (y)), and U  Ty with U ≤ Ly and U ∼ = E22 .

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13. THE CASES IN WHICH G0 IS EXCEPTIONAL

147

Lemma 13.6. The following conditions hold: (a) Ω1 (Z(Ty )) = y ; (b) Ty ∈ Syl2 (G);  (c) O r (CLy (U )) is a central product (J1 × J2 ) ∗ I where J1 ∼ = J2 ∼ = SL2 (q), + (q), Z(J × J ) = U = Z(I), and y ∈  J ∪ J ; I∼ Spin = 1 2 1 2 8 (d) There is an involution of Ty ∩ Ly − CTy (U ) interchanging J1 and J2 and inducing a graph automorphism on I; (e) AutG (U ) ∼ = Σ3 . Proof. Let Z = Ω1 (Z(Ty )). Since C(y, Ly ) has quaternion Sylow 2-subgroups, [Z, F ∗ (C(y, Ly )/O2 (CG (y)))] = 1. By [III17 , 8.9e], [Z, Ly ] = 1, and hence by the F ∗ -Theorem, Z maps into Z(Cy /O2 (Cy )), which is generated by y. Thus, (a) holds, and (b) is an immediate consequence. By [III17 , 11.25a] the image of U in K/ y ∼ = P Ω+ 12 (q) is generated by an involution v which is the image of an involution v ∈ Ω+ 12 (q) with −1-eigenspace of dimension 8 and + type. Then CSpin+ (q) (v) is the commuting product of Spin+ 4 (q) and 12



r Spin+ 8 (q), with the spin involutions identified. On the other hand, O (CK/y (v)) is + + ∼ the commuting product of Ω4 (q) = SL2 (q)∗SL2 (q) and Ω8 (q). These observations imply all assertions in (c). Furthermore, if w ∈ Ω+ 12 (q) is an involution commuting with v and acting on the two eigenspaces of v with −1-eigenspaces of dimension 1 and 3, respectively, then the image of w in Ly satisfies the assertions of (d). To prove (e) it will suffice to show that all involutions of U # are 2-central in G. For then, since Z(Ty ) is cyclic by (a), (e) will follow from [IG , 16.21]. Assume then that (e) fails, let u ∈ U − y , and set T0 = CTy (U ) = CTy (u). Then u is Ty -conjugate to uy, and neither u nor uy is 2-central in G. As |Ty : T0 | = 2, T0 is a Sylow subgroup of both CG (u) and CG (uy). Moreover, (a) implies that U = Ω1 (Z(T0 )). Let Iu be the pumpup of I in CG (u). As y ∈ U , Iu is not a diagonal pumpup; indeed since y is weakly closed in Z(I) = U with respect to G, it follows from [III17 , 14.11] that Iu is a trivial pumpup of I. This yields y ∈ Z ∗ (CG (u)), and similarly y ∈ Z ∗ (CG (uy)), so y ∈ Z ∗ (CG (v)) for all v ∈ U # . By the Thompson Transfer Lemma [IA , 15.15] there is a Ty -extremal G-conjugate ξ = xg of x such that [ξ, U ] = 1. Let Tξ = CTy (ξ) ∈ Syl2 (CG (ξ)), so that U  Tξ . By [IA , 7.3.3],  K g = ELie (CK g (v)) | v ∈ U # ,

where ELie (X) is the product of all Lie components of X. But for each v ∈ U # , y ∈ Z ∗ (CG (v)) so [y, ELie (CK g (v))] = 1. Therefore [y, K g ] = 1. By L2 balance, and the fact that Ly  CG (y), K g must lie in either Ly or C(y, Ly ). As m2 (C(y, Ly )) = 1, we reach the conclusion that |K/Z(K)| = |L8 (q)| divides |Ly /Z(Ly )| = |P Ω+ 12 (q)|. But this is seen to be absurd, by considering a primitive  prime divisor of q 7 − 1 or q 14 − 1. So the lemma is proved. (A primitive prime divisor of r n − 1, where r is prime, is a divisor of r n − 1 that does not divide r i − 1 for any positive integer i < n. See [IG , 1.1].) This quickly yields Lemma 13.7. There is J0 ∼ = SL2 (q) such that (a) J0  CG (y); (b) NG (U )/CG (U ) ∼ = Σ3 permutes {J0 , J1 , J2 } faithfully and maps isomorphically to Aut0 (I)/ Inndiag(I), the group of graph automorphisms on I. Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

148

Proof. Let S = S1 ∪ S2 , where S1 is the set of all components of CG (U ), and S2 is empty unless q = 3, in which case S2 is the set of all normal SL2 (3)subgroups of CG (U ). Then {I, J1 , J2 } ⊆ S. We may assume that u = Z(J1 ), and then yu = Z(J2 ). As we have seen, Z(I) = U . Note that by [IA , 4.5.2], when q = 3, CLy (U ) contains a 2-element t inverting J1 J2 /O2 (J1 J2 ) elementwise. As Ly is quasisimple, any element of S2 containing u must, being t-invariant, equal J1 . When q > 3, it is obvious that J1 is the only SL2 (q)-subgroup in S containing u. Similar statements hold for uy and J2 . Likewise, I is the only element of S containing U , so NG (U ) normalizes I. Now AutG (U ) ∼ = Σ3 . Let H ≤ [NG (U ), NG (U )] be a 3-group mapping onto O3 (AutG (U )). Then H maps into [Aut(I), Aut(I)] = Inndiag(I) h , where h is a graph automorphism of order 3. Indeed the image of CG (U ) ∩ [NG (U ), NG (U )] lies between Inn(I) and Inndiag(I), and it follows that h may be chosen to be in the image of H. Thus AutG (I) contains a graph automorphism of order 3. Since it also contains a graph automorphism of order 2, we see the isomorphism AutG (U ) ∼ = Aut0 (I)/ Inndiag(I) of (b). N (U) By the transitive action of NG (U ) on U # , there is J0 ∈ S ∩ J1 G isomorphic to SL2 (q) and containing y. Any J0 obtained in this way must map into  O r (CAut(Ly ) (IJ1 J2 )) = 1 (see [III17 , 11.25c]) so J0 ≤ C(y, Ly ). As m2 (C(y, Ly )) = 1, J0 is unique and hence NCG (y) (U )-invariant. By [III17 , 11.25a], CG (y) = Ly NCG (y) (U ), and (a) follows as [J0 , Ly ] = 1.  Now J0 , being conjugate to J1 , is conjugate to all the root SL2 (q)-subgroups of Ly . We begin our construction of E7 (q) by choosing a weak CT-system {J2 , . . . , J7 } for Ly in accordance with the following D6 diagram. Here J1 is the high root SL2 (q) with respect to this system. We include an isolated node for J0 , which centralizes Ly . ◦ α1 α2

(13I)

Ly : ◦

◦

α3

◦ α7 α4

◦

◦

α5

α6

◦

α0

◦

Note that I corresponds to the subdiagram of nodes 4, 5, 6, and 7. Now there is an Weyl element w ∈ Ly which permutes the nodes above according to the permutation (12)(67) of the subscripts. It follows that under the isomorphism of Lemma 13.7b, the transposition interchanging J1 and J2 corresponds to the graph automorphism of I interchanging nodes 6 and 7. Let γ ∈ ΓI be the graph automorphism of I corresponding to the permutation (46) of the nodes, and let g ∈ NG (U ) be an involution effecting this automorphism. It follows from the previous paragraph and Lemma 13.7b that J0g = J1 or J2 . We may assume that notation was chosen so that J0g = J2 . Consequently, J1g = J1 . Define J8 = J3g . Then we have the following diagram for Lgy : ◦ α1 α0

(13J)

Lgy : ◦

◦

α8

◦ α7 α6

◦

◦

α5

α4

◦

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α2

◦

13. THE CASES IN WHICH G0 IS EXCEPTIONAL

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Flipping this from left to right and superimposing with (13I), we obtain a diagram with the following two subdiagrams: ◦ α7 α2

◦

(13K)

α3

◦

α4

◦

◦

α5

α6

α8

◦

α0

◦

◦

◦ α7 (13L)

◦

α5

α2

◦

α3

◦

α1

◦

α8

◦

α0

◦

Lgy

only omitted the vertices 3 and 8, respectively, Because the diagrams for Ly and the diagrams (13K) and (13L) will represent genuine weak CT-systems formed by the corresponding groups Ji , once it is established that [J3 , J8 ] = 1. We will actually prove a weaker statement which is still sufficient. Lemma 13.8. For some s ∈ J1 , [J3 , J8s ] = 1. Proof. Set H = J2 , J3 , J1 , J8 , J0 (cf. (13L)) and C5 = CG (J5 , J7 ), so that H ≤ C5 . We prove that (13M)

H/Z(H) ∼ = L6 (q).

Also set zi = Z(Ji ) for each i, so that in particular z0 = y, z1 = u, and z2 = uy. Then CC5 (y) can be computed in CG (y) to have the commuting normal subgroups  J0 ∼ = SL2 (q) and Jy = O r (CLy (J5 , J7 )) ∼ = SL4 (q) with Jy = J2 , J3 , J1 and Ω1 (Z(Jy )) = z2 z1 = y. Moreover, y is a noncentral element of J1 , J8 , J0 ∼ = A3 (q). It follows by [III17 , 10.53] that H is a component of C5 and (13M) holds.  Then (13M) and [III17 , 4.12] imply our lemma. Now let s be as in Lemma 13.8. As s ∈ J1 , which commutes with Ji for i ∈ {0, 2, 4, 5, 6, 7}, we have J8s , Ji ∼ = J8 , Ji for these values of i. Therefore J0 , Ji , 2 ≤ i ≤ 7, and J8s form a weak CT-system based on the diagram (13K). We have proved that {Ji | 2 ≤ i ≤ 8} is a weak CT-system of type E7 (q), and therefore putting G1 = J2 , . . . , J8 , we have by [III13 , 1.4]: Lemma 13.9. G1 ∼ = E7 (q). Lemma 13.10. The following conditions hold:   N (U) ; (a) G1 = Ly G (b) ΓU,1 (G) ≤ NG (G1 ); and (c) G1 is the unique subgroup of NG (G1 ) isomorphic to E7 (q). Moreover, CG (G1 ) has odd order. Proof. It is clear from the diagram (13K) that if we choose h ∈ NG (U ) such that y h = uy = z2 , then E(CG1 (z2 )) has a component isomorphic to Ly , which  by the structure of CG (y) must then equal Lhy ; moreover, Ly , Lhy contains all Ji , i = 0, 2 ≤ i ≤ 8, and so  G1 = Ly , Lhy .  h But there exists k ∈ NLy (U ) such that (uy)k = u, and so Lhk y ≤ Ly , Ly = G 1 . Thus G1 is generated by Ly and its h- and hk-conjugates. As these are the unique subgroups of CG (y), CG (uy), and CG (u) of their isomorphism types, (a) follows.

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

150

By (a), NG (U ) ≤ NG (G1 ) and as NG (U ) is transitive on U # , to prove (b) it suffices to show that CG (y) ≤ NG (G1 ). But as already noted, by [III17 , 11.25a], CG (y) = Ly NCG (y) (U ), and since Ly ≤ G1 , (b) follows. Finally, y is the unique minimal normal subgroup of CG (y) of even order, so as [y, G1 ] = 1, CG (G1 ) has odd order. In particular since the Schur multiplier of E7 (q) is a 2-group [IA , 6.4.1], G1 is simple. In N = NG (G1 )/O2 (NG (G1 )), we then have F ∗ (N ) = G1 ∼ = G1 . Hence any subgroup of NG (G1 ) isomorphic to E7 (q) maps onto G1 , and so lies in G1 × O2 (NG (G1 )) and hence in G1 . The proof is complete.  Lemma 13.11. G0 = G1 . Proof. We first show that D centralizes some conjugate U g , g ∈ G. We know that U  Ty ∈ Syl2 (CG (y)) ⊆ Syl2 (G). By the Thompson Transfer Lemma [IG , 15.15], some Ty -extremal conjugate x0 = xh of x centralizes U . Let T0 = CTy (x0 ) ∈ Syl2 (CG (x0 )). Thus U ≤ T0 . Moreover, Dh = x0 , y0 where y0 = y h ∈ E(CG (x0 )) ∼ = A7 (q). In particular y0 ∈ [CG (x0 ), CG (x0 )]. Now another application of the Thompson Transfer Lemma, to CG (x0 ) instead of G, yields that y0 has a CG (x0 )-conjugate y1 = y0k ∈ CT0 (U ). As x0 ∈ Z(T0 ) and U ≤ T0 , the conjugate x0 , y1 = Dhk centralizes U , proving our claim. Since [D, U g ] = 1, each CG (d), d ∈ D# is U g -invariant, and by the preceding # , lemma ΓU g ,1 (CG (d)) ≤ NG (G1 )g . But given the isomorphism types of Ld , d ∈ D we have Ld ≤ ΓU g ,1 (CG (d)), by [IA , 7.3.4]. Therefore G0 = N = Ld | d ∈ D# ≤ NG (G1 )g < G. Hence G0 is a K-group. Then by [III17 , 3.36] and the structure of g the neighborhood N, G0 ∼ = E7 (q). By Lemma 13.10c, G0 = G1 . Now CG (y) ≤ ΓU,1 (G) ≤ NG (G1 ). Thus D6 (q)hs ∼ = Ly ≤ G0 ∩ G1 and so y is 2-central in both G0 and G1 . By [III17 , 8.9f], Sylow 2-centers of G0 and G1 are cyclic. As G0 and G1 are conjugate in G, they are therefore conjugate in CG (y), whence G0 = G1 , completing the proof.  Lemma 13.12. ΓD,1 (G) ≤ NG (G0 ). Proof. Recall that m2 (C(d, Ld )) = 1 for all d ∈ D# by the hypothesis of Proposition 13.1. Thus for each d ∈ D# , Ld is the unique component of CG (d) of its isomorphism type. As the three Ld ’s are pairwise nonisomorphic, the three involutions of D are pairwise nonconjugate. Thus NG (D) = CG (D) normalizes each Ld and hence it normalizes G0 . It remains to prove that (13N)

CG (d) ≤ NG (G0 ) for all d ∈ D# .

As CG (y) ≤ ΓU,1 (G) ≤ NG (G1 ) = NG (G0 ), we may assume henceforth that d = x or xy. By Lemma 1.3, CG (G0 ) has odd order. Using the structures of Ld , d ∈ D# given in Proposition 13.1c, and the fact that  = q , we can conclude with [IA , 4.5.1] that D induces inner automorphisms on G0 . As CG (G0 ) has odd order, D ≤ G0 . Now y ∈ L ≤ Ld . As m2 (C(d, Ld )) = 1, d is the only involution in C(d, Ld ), and so D = d, y ≤ Ld d . Letting CG (d) = CG (d)/C(d, Ld ) ∼ = AutCG (d) (Ld ), we see that D is the unique four-subgroup of CG (d) mapping onto y .

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13. THE CASES IN WHICH G0 IS EXCEPTIONAL

151

Moreover, with the isomorphism types of Ld and L given in Proposition 13.1c, the image y of D in Aut(Ld ) is determined up to Inn(Ld )-conjugacy by the struc ture of O r (CG (D)), by [III17 , 14.10]. Hence DCG (d) = DLd , so (13O)

CG (d) = Ld NCG (d) (D) ≤ NG (G0 ).

This completes the proof.



Lemma 13.12 proves Proposition 13.1 if hypothesis (c) holds. The next result will complete the proof of the proposition. Lemma 13.13. If Proposition 13.1d holds, then the conclusion of Proposition 13.1 holds with G0 ∼ = E8 (q). We assume through Lemma 13.20 below that Proposition 13.1d holds, but the conclusion of Proposition 13.1 fails. We first set up the situation. We have Ly ∼ = D8 (q)hs , and L ∼ = Spin+ = Lxy ∼ = E7u (q), K ∼ 12 (q) with D = Z(L). We first show Lemma 13.14. m2 (C(u, Lu )) = 1 for all u ∈ D# . Also, xy ∈ y K . Proof. Suppose that m2 (C(y, Ly )) > 1 and choose a 2-terminal long pumpup (v, I) of (y, Ly ). As Ly ∈ G62 and d2 (G) = 6, the long pumpup I also is in G62 , so (v, I) ∈ J2 (G) and then m2 (C(v, I)) = 1 by [III12 , [Theorem 1.2]. By [III8 , 2.4], there is a first pumpup on the way from (y, Ly ) to (v, I) which is vertical. But by [III11 , 13.23], the only possible pumpups H of E7 (q) are isomorphic to E7 (q 2 ) or E8 (q). With [III11 , 12.4], we conclude that F(I) ≥ F(H) > F(D8 (q)) = F(K). This, however, contradicts (x, K) ∈ J∗2 (G). Thus m2 (C(y, Ly )) = 1. We know that m2 (C(x, K)) = 1 and y ∈ D ≤ L ≤ K, so xy ∈ y CG (x) by [IA , 6.2.1e]. The lemma follows.  Lemma 13.15. ΓD,1 (G) ≤ NG (G0 ). Proof. Lemma 13.14 implies that for each u ∈ D# , Lu is the unique component of E(CG (u)) of 2-rank at least 2. Therefore NG (D) permutes the set {Lu | u ∈ D# }, and normalizes G0 . Let u ∈ D# . We have D ≤ L  CLu (D), which condition determines D/ u as a subgroup of Aut(Lu ) up to conjugation by an element of Inn(Lu ), by inspection of [IA , 4.5.1]. Therefore CG (u) ≤ NG (Lu ) ≤  Lu , NG (D) ≤ NG (G0 ), as required. Lemma 13.16. For any u ∈ D# , O2 (CG (u)) = O2 (CG (D)) = 1. Proof. Let W = O2 (CG (D)), suppose W = 1, and take any u ∈ D# . We have D ≤ L ≤ Lu , and Lu is balanced with respect to D/ u by [III17 , 15.4]. Therefore W = O2 (CCG (u) (D/ u )) ≤ CCG (u) (Lu ). As u was arbitrary, G0 ≤ NG (W ) < G. As Lu  CG0 (u) for any u ∈ D# , it follows by [III17 , 3.37], applied to NG (W ), that G0 = E(NG (W )) ∼ = E8 (q). Then with Lemma 13.15, the conclusion of Proposition 13.1 holds, contrary to assumption. Hence O2 (CG (D)) = W = 1. Finally, for any u ∈ D# , [O2 (CG (u)), D] = 1 as D ≤ Lu ≤ E(CG (u)). Therefore  O2 (CG (u)) ≤ O2 (CG (D)) = 1, and the proof is complete. By Lemma 13.16, the term E q (X) is defined for all X  CG (D) or X  CG (u), u ∈ D# (see (10K)).

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14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

152

Lemma 13.17. Let N = E q (CK (D)). Then the following conditions hold: (a) N is a central product N = LIJ with I ∼ =J ∼ = SL2 (q), IJ = I × J, and D = Z(L) = Z(IJ); (b) N = E q (CG (D)); and (c) K = E q (CG (x)), I = E q (C(y, Ly )) contains F ∗ (C(y, Ly )), and J = E q (C(xy, Lxy )) contains F ∗ (C(xy, Lxy )). Proof. Part (a) follows directly from [IA , 4.5.1, 4.5.2], with Z(I) = Z(K) = Z(J). Without loss, Z(I) = y , which implies that Z(J) = xy . Again by [IA , 4.5.1, 4.5.2], E q (CLy (D)) ∼ = A1 (q)D6 (q)u with Z(Ly ) = y not contained in the A1 (q) component. It follows that E q (CLy (D)) = JL and then as D ≤ L ≤ Ly and m2 (C(y, Ly )) = 1 we have E q (CG (y)) = Ly I with I ∼ = SL2 (q), [I, Ly ] = 1, and Ly ∩ I = y . In particular E q (CG (D)) = E q (CCG (y) (D)) = E q (CLy (D))I = LIJ = E q (CK (D)) = N , so (b) holds. By Lemma 13.14, xy = y g for some g ∈ K, and then J g = I and Lgy = Lxy . Since D ≤ L ≤ K and CG (D) = CCG (y) (D), E q (C(x, K)) = 1. As m2 (C(u, Lu )) = 1 for u = y and xy, (c) holds and the lemma is proved.  We choose a set of SL2 (q)-subgroups Li of Ly corresponding to the nodes of its extended Dynkin diagram, as shown in the top diagram of (13P) below. Thus {Li | 0 ≤ i ≤ 6} is a weak CT-system of type E7 (q), and L−1 is the lowest root SL2 (q). We again write zi = Z(Li ) for each i. It follows from [IA , 4.2.2, 4.5.1] that we may take z−1 ∈ D, D = y, z−1 , and L = Li | 1 ≤ i ≤ 6 . By Lemma 13.17cd, J is the unique normal SL2 (q)subgroup of CLy (D); we deduce from the Ly diagram that J = L−1 . Similarly, there are additional subgroups Li of K, i = 7, 8, 9, so that L1 , . . . , L9 satisfy the Curtis-Tits relations with respect to the extended Dynkin diagram of K, as shown in (13P). ◦ α3 (13P)

Ly :

K:

α−1

◦

α1

◦

α0

α1

◦ ◦ α3 ◦ ◦

α2

α4

◦

◦

α2

α5

◦

α4

◦

α5

α6

◦

α6

◦ ◦ α8 ◦ ◦

α7

α9

◦

Then LIJ = E(CK (D)) = LL8 L9 so {L8 , L9 } = {I, J}. Since there is symmetry between L8 and L9 we may assume without loss L8 = I and L9 = J. Thus, we have (13Q)

y ∈ L8 and xy ∈ L9 = L−1 .

We first prove Lemma 13.18. G0 = Li | 0 ≤ i ≤ 7 . Proof. Let G1 = Li | 0 ≤ i ≤ 7 . By definition G0 = N contains Ly , K and hence G0 contains all Li , −1 ≤ i ≤ 9. In particular G1 ≤ G0 . On the other hand, L9 = L−1 ≤ Ly = L i | 0 ≤ i ≤ 6 ≤ G1 and then K = Li | 1 ≤ i ≤ 9, i = 8 ≤ G1 . ≤ G1 , completing the proof. As xy ∈ y K , Lxy ≤ LK  y

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13. THE CASES IN WHICH G0 IS EXCEPTIONAL

153

We superimpose parts of the diagrams (13P) to obtain an E8 -diagram. ◦ α3 α0

◦

α1

◦

◦

α2

α4

◦

α5

◦

α6

α7

◦

◦

With one exception, all of the Curtis-Tits relations on the subgroups Li in this diagram follow from the relations implicit in the diagrams (13P) for the groups Ly , K. The missing relation, which may not hold, is [L0 , L7 ] = 1. Instead we shall show: Lemma 13.19. [L0 , Lh7 ] = 1 for some h ∈ L9 . Proof. Set H0 = L2 , L3 , L4 , L5 and H = L0 , L9 , L8 , L7 , and let H ∗ be the weak closure of L1 in CG (H0 ). From the diagrams (13P) we find that [H, H0 ] = 1, whence H ≤ H ∗ . Similarly, and with (13Q) and the fact that [Ly , L8 ] = [Ly , I] = 1, H contains subgroups isomorphic to A3 (q) and A1 (q)A2 (q) with the following diagrams: (13R)

α8

◦

α7

◦

α9

◦

and

α8

◦

α9

◦

α0

◦

Set Hy = E(CH ∗ (y)) if q > 3, and Hy = O 2 (CH ∗ (y)) if q = 3. We argue that (13S) Hy = L9 , L8 , L0 ∼ = A2 (q) × A1 (q). To see this, suppose first that q > 3. Then clearly L9 , L8 , L0 ≤ CH ∗ (y)(∞) . On the other hand, by Lemma 13.17c, F ∗ (CG (y)) = Ly I and so CG (y)(∞) = Ly I = Ly L8 . Then CH ∗ (y)(∞) ≤ (H ∗ ∩ Ly )(∞) L8 ≤ CLy (H0 )(∞) L8 ∼ A2 (q) × A1 (q). = L−1 , L0 L8 = L9 , L0 L8 = We have used [III17 , 6.3a] to identify CLy (H0 )(∞) . Thus CH ∗ (y)(∞) = L9 , L0 , L8 and (13S) follows in this case. In the remaining case q = 3, we clearly have L9 , L8 , L0 ≤ Hy and O 2 (CG (y)) = Ly I and similarly find Hy = O 2 (CH ∗ (y)) ≤ O 2 (H ∗ ∩ Ly )L8 ≤ O 2 (CLy (H0 ))L8 ∼ A2 (q) × A1 (q), = L−1 , L0 L8 = using [III17 , 6.3b]. This establishes (13S). Setting Hx = E(CH ∗ (x)) and using a similar argument, again with [III17 , 6.3], we find that Hx = L7 , L8 , L9 ∼ = A3 (q), whether q = 3 or not. Then by [III17 , 4.11], (q) and there exists h ∈ L9 such that [L0 , Lh7 ] = 1. The proof is complete. H∼ A = 4  Now we can complete the proof of Lemma 13.13. Lemma 13.20. G0 ∼ = E8 (q). Proof. Continuing the above argument, we have [L9 , Li ] = 1 for all i ≥ 1, except for i = 7 and 9. Thus h, chosen as in Lemma 13.19, centralizes L1 , . . . , L6 and L8 , and of course normalizes L9 . Thus, we replace our original choice of generators for K by their h-conjugates, and leave L0 unchanged. Then {L0 , . . . , L7 } is a weak CT-system of type E8 (q), as a consequence of our choices of generators for K and Ly , and of Lemma 13.19. By Lemma 13.18 and [III13 , 1.4], G0 = Li | 0 ≤ i ≤ 7 ∼ =  E8 (q).

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154

14. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p = 2

Lemmas 13.15 and 13.20 imply Lemma 13.13, which, together with Lemmas 13.3 and 13.4, yields Proposition 13.1. 14. Summary: p = 2 The results of the preceding sections combine to prove Theorem C∗7 : Stage 4b+ in the case p = 2, namely: Proposition 14.1. Suppose that 2 ∈ γ(G) and for some (x , K  ) ∈ J∗2 (G), K ∈ Chev. Then there exists (x, K) ∈ J∗2 (G) and an acceptable subterminal (x, K)pair (y, L) such that G0 = N(x, K, y, L) ∈ Chev, CG (G0 ) has odd order, and Γx,y,1 (G) ≤ NG (G0 ). Moreover, G0 /Z(G0 ) ∈ K(7)+ . 

Proof. That G0 /Z(G0 ) ∈ K(7)+ will be an immediate consequence of the definition of K(7)+ [III12 , (1D)]. By Lemma 1.3, CG (G0 ) has odd order. Suppose that one of the other assertions fails, and choose (x , K  ) ∈ J∗2 (G) with K  ∈ Chev and q(K  ) minimal. Our setup is that (1A) holds with p = 2. By Theorem C∗7 : Stages 3ab, upon replacing (x , K  ) with a suitable (x, K) ∈ J∗2 (G), we may assume that (1B), (1C), and (1D) hold, and moreover K  /O2 2 (K  ) ∼ = K/O2 2 (K). Proposition 5.1 implies that (1E) holds. Let K = K/O2 (K). Keep in mind that since (x, K) ∈ J∗2 (G), K ∈ G2 . Then  L3 (q) – otherwise, by Proposition 3.1, (x, K) satisfies for any odd q and sign , K ∼ = the desired conclusions. In particular d2 (G) = 7, so d2 (G) = 6. Next, K ∼  G2 (q) or = 3 D4 (q) by Proposition 4.1; and K is not of any other exceptional type by Proposition 6.1. Thus K is a classical group. If K ∼ = Sp4 (q), then Proposition 9.1 gives a contradiction. Likewise if K ∼ = (q) or HSpin (q), n ≥ 7, then Propositions 8.1 and 13.1 give a contradiction. Spin± n n Next, if K ∼ = P Sp2n (q), n ≥ 2, then Proposition 7.1 implies that there exists (x∗ , K ∗ ) ∈ J∗2 (G) with K ∗ /O2 (K ∗ ) ∼ = Sp4 (q), Spin7 (q), or Sp8 (q 1/4 ). In the first ∗ two cases, q(K ) = q = q(K), and the existence of such (x∗ , K ∗ ) has already been shown to lead to a contradiction. The third possibility contradicts our minimal  P Sp2n (q) for any n ≥ 2. In turn, Proposition 9.1 choice of q(K). Thus K ∼ = implies that K ∼ = Sp2n (q) for any n ≥ 3. ± ± ∼ ∼ Suppose next that K/Z(K) ∼ = L± n (q), n ≥ 5, or K = SL4 (q) (= Spin6 (q)). If K is not 2-saturated, then Proposition 10.1b shows that the assumptions of Proposition 13.1c hold, and that proposition yields that G0 ∼ = E7 (q) and Proposition 14.1 holds, contrary to assumption. Therefore K is 2-saturated, and Proposition 10.1 ± shows that the conclusions of our proposition hold with G0 ∼ = L± n+1 (q) or Ln+2 (q). ± We are now reduced to the case that K/O2 (K) is a quotient of Ωn (q) for some ± n ≥ 6, except that the case Ω± 6 (3) does not occur as Ω6 (3) ∈ G2 . By Proposition ∗ ∼ 10.1, there is no (u, H) ∈ J2 (G) such that H/O2 (H) = SL± 4 (q). Hence Proposition 11.1 shows that n is even and our desired conclusions hold with G0 ∼ = Ωn+1 (q) or P Ω± (q), a final contradiction.  n+2

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10.1090/surv/040.8/04

CHAPTER 15

Theorem C∗7 : Stage 4b+. A Large Lie-Type Subgroup G0 for p > 2 1. Introduction We turn to the case p > 2, and review the setup (see [III12 , Sec. 1]). The reader may wish to review the statement of Theorem C∗7 : Stage 4b+, in the introductions to the previous two chapters. Our goal, once again, is to prove that G0 ∈ Chev(2), G0 /Z(G0 ) ∈ K(7)+ , ΓD,1 (G) ≤ NG (G0 ). We have already proved in [III14 , Lemma 1.3] that CG (G0 ) is a p -group. In view of Stages 2 and 3ab of Theorem C∗7 , we are provided with a prime p, elements x, y, v ∈ G, and subgroups D, K, L ≤ G such that the following conditions hold. (1) p ∈ γ(G) is an odd prime and x ∈ G has order p; (2) (x, K) ∈ J∗p (G) is a terminal pair with K ∈ Chev(2), q = q(K), mp (K) ≥ 3, and mp (C(x, K)) = 1; (3) (y, L) is an acceptable subterminal (x, K)-pair and q(L) = q(K); (1A) (4) D = x, y , v ∈ D − x , and the pumpup Lv of L in CG (v) is vertical; (5) For every u ∈ D# , the pumpup Lu of L in CG (u) is either trivial or vertical, and satisfies Lu ∈ Chev(2) with q(Lu ) = q; and (6) p splits K, i.e., p divides q 2 − 1. As usual, (1B)

 G0 = N(x, K, y, L) = Lu | u ∈ D# = Lu | u ∈ De .

We set (1C)

De = {u ∈ E1 (D) | Lu > L}.

Thus, v ∈ De , and x ∈ De with Lx = K. Like the terminology q in the previous chapter in the case p = 2, the following terminology will be convenient for the rest of this chapter, with q = q(K).  1 if q ≡ 1 (mod p) (1D) q = −1 if q ≡ −1 (mod p) As p divides q 2 − 1 by assumption, there is no third alternative. See [III12 , (1D)] and [III12 , (1A)] for the definitions of the set K(7)+ of (known) simple groups. We make a choice of (p, x, K, y, L) in the next section. 155 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

156

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

2. A Choice of p In order to avoid a couple of difficult fringe cases, as mentioned in the introductory chapter, we have to make a careful choice of p, x, K, y, L. The critical item to choose is p ∈ γ(G); that being chosen, we use [III12 , Theorem 1.2] to guarantee the existence of appropriate pairs (x, K) and (y, L). So we concentrate here on the choice of p. For convenience we restate the conditions [III12 , (1C)] from Chapter 12: (2A)

(1) (x, K) ∈ J∗p (G) for some p ∈ γ(G); (2) (y, L) is an acceptable subterminal (x, K)-pair; and (3) K ∈ Chev(r) for some prime r, exactly one of r and p equals 2, and p splits K.

We define Q(G) to be the set of all quintuples (p, x, K, y, L) such that p ∈ γ(G) and (2A) is satisfied. For any such quintuple, [III12 , Theorem 1.2] applies. For a given prime p0 , let Qp0 (G) = {(p, x, K, y, L) ∈ Q(G) | p = p0 }. Lemma 2.1. Suppose that 3 ∈ γ(G). Then Q3 (G) = ∅. Proof. There exists (3, x, K, y, L) satisfying (2A1, 2), and such that m3 (K) ≥ 3, by [III7 , 3.3] and Theorem C∗7 : Stage 3b. Also by Theorem C∗7 : Stage 3a, n n  2B2 (2 2 ) or 2F4 (2 2 ) for any n. Hence it is K ∈ Chev(2). Since m3 (K) ≥ 3, K ∼ =  trivial from the definition that 3 splits K, so (3, x, K, y, L) ∈ Q3 (G). Lemma 2.2. If G is of even type, then there exists (p, x, K, y, L) ∈ Q(G) satisfying none of the following: (a) p > 3, K/Z(K) ∼ = L6 q (q), q = 2n ≡ −q (mod 3), and 3 ∈ γ(G); (b) p = 5, K ∼ = K for all u ∈ De , and 3 ∈ γ(G); or = Sp6 (4), Lu ∼ (c) p = 3, K ∼ = K for all u ∈ De , 7 ∈ γ(G), and there is = Sp6 (8), Lu ∼ (v, J) ∈ J∗7 (G) such that J ∼ = K. −

Proof. We know by [III12 , Theorem 1.2] that Q(G) = ∅. Suppose that every element of Q(G) satisfies (a), (b), or (c). We first claim that some element of Q(G) satisfies (c). For otherwise, 3 ∈ γ(G) and so there is (p, x, K, y, L) ∈ Q3 (G) by Lemma 2.1. As p = 3, this quintuple must satisfy (c), a contradiction proving the claim. Therefore 7 ∈ γ(G) and there is (v, J) ∈ J∗7 (G) such that J ∼ =K∼ = Sp6 (8). ∗ In particular, 7 splits J, and so by Theorem C7 : Stage 3b, there are (x , K  ) ∈ J∗7 (G) with K  /Z(K  ) ∼ = J/Z(J), and (y  , L ) such that (7, x , K  , y  , L ) lies in Q(G),   and hence satisfies (a). But then J/Z(J) ∼ = K  /Z(K  ) ∼ = L± 6 (q ) for some q , a contradiction. The lemma is proved.  Let Q∗ (G) be the set of all (p, x, K, y, L) ∈ Q(G) not satisfying (a), (b), or (c) in Lemma 2.2. Thus Q∗ (G) = ∅, and we choose (2B)

(p, x, K, y, L) ∈ Q∗ (G).

Then (1A) is satisfied. We also fix the notation in (1A), (1C), (1D), and (1B) for the rest of this chapter.

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3. THE WEYL GROUP

157

3. The Weyl Group Central to our constructions is an elementary abelian p-group B which we now choose and fix. By (1A2), K  CG (x). Let C 0 be the preimage in CG (x) of Inndiag(K). First, we choose B ∗ to satisfy the following conditions: (1) B ∗ ∈ Ep∗ (CG (x)); (2) D ≤ B ∗ ; (3A) (3) Subject to (1) and (2), B ∗ ≤ C 0 , if possible; and  (4) If p = 3, K ∼ = E6q (q), B ∗ ≤ C 0 and m3 (B ∗ ) = 6, then B ∗ ≤ DL. By [III17 , 13.1], such a choice exists. Thus by definition, (3B)

mp (B ∗ ) = mp (CG (x)).

Then we set B = B∗ ∩ C 0 (3C) = {b ∈ B ∗ | b induces an inner-diagonal automorphism on K}. In this regard, we observe: Lemma 3.1. The following conditions hold: (a) mp (L) ≥ 2; and (b) B ∗ normalizes L and B induces inner-diagonal automorphisms on L. Proof. Part (a) follows quickly from [III12 , Def. 1.15] and [IA , 4.10.3a], in view of the fact that mp (K) ≥ 3 by (1A2). If B ∗ does not normalize L, we apply [IG , 8.7(iii)] to the group CG (D) and conclude that p = 3 and L has three conjugates  under B ∗ , and L ∼ = SL3q (q). Then CK (L) contains an E32 -subgroup E such that E ∩ x = 1. Thus E acts faithfully on K, and by definition of subterminal pair [III3 , 1.2, 1.3], L is a component of CK (e) for all e ∈ E # . But K ∈ Chev(2) with m3 (K) ≥ 3, so this last condition is impossible by [III11 , 1.16]. Thus B ∗ normalizes L. Since B induces inner-diagonal automorphisms on K, and L  E(CK (y)),  of K satisfies Z = part (b) holds by [IA , 4.2.2] unless the universal version K  Ω1 (Op (Z(K))) = 1 and y is sheared in K to Z. Here K is the algebraic group  But generally Z = 1, and in the two cases in which Z = 1, namely overlying K.   ∼ K/Z(K) = Lnq (q) or E6q (q) with p dividing n or p = 3, respectively, we have q  ∼ L = SLn−δ (q), δ = 1 or 2, or L ∼ = A5q (q), and it is easily seen from [IA , 4.8.2, 4.7.3A] that y does not shear to Z in K. The lemma is proved.  Lemma 3.2. Let E ∈ Ep∗ (CG (x)) with D ≤ E. Then for any u ∈ D# and e ∈ E, E induces an inner-diagonal or field automorphism of Lu . Proof. Suppose false, so that p = 3 and some e ∈ E induces a nontrivial graph or graph-field automorphism on Lu ∼ = D4 (q) or 3D4 (q) (possibly u = x and Lu = K). If Lu ∼ = 3D4 (q) then m3 (Lu ) = 2 so Lu = K and u = x . Then as   q(L) = q, the component L of CLu (x) is isomorphic to SL3q (q), L3q (q), G2 (q), 3 or D4 (q), by [IA , 4.7.3A]. But (y, L) is an acceptable subterminal (x, K)-pair. Using [III12 , Def. 1.15] and the condition m3 (K) ≥ 3, we see that the only possibility   is that K ∼ = GL3q (q), and q > 2 since K/O3 (K) ∈ G3 . = L4q (q), with CK (y) ∼

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

158

∼ K, so a Sylow 3-subgroup of Now CAut(K) (L) is a cyclic subgroup of Inn(K) = CCG (x) (L) is the direct product of two cyclic groups. Since CCG (u) (L) centralizes u × Z(L) ∼ = E32 , it follows that x, and then any element of CCG (u) (L) of order 3, lies in u Z(L). On the other hand, as Out(Lu ) is cyclic it follows that e induces a graph automorphism on Lu , and then by [IA , 4.7.3A] there is e = eg, also of order 3, with g ∈ Lu and such that CLu (e ) contains L. This is a contradiction, and we have proved that Lu ∼ = 3D4 (q). The only other possibility is Lu ∼ = D4 (q). In this case, choose some element e ∈ E inducing a nontrivial field automorphism on Lu , if possible; otherwise set  e = 1. Then CLu (e) ∼ = 3D4 (q), G2 (q), or P GL3q (q), and in any case m3 (CLu (e)) = 2. We have m3 (E/CE (Lu )) ≤ m3 (e CLu (e)) + m3 (e ) = 3 + m3 (e ). Suppose that x induces an inner-diagonal automorphism on Lu . Then m3 (CLu (x)) = 4, and so m3 (CG (x)) ≥ 4 + m3 (CE (Lu )) ≥ m3 (E). Hence equality holds by (3B), and e = 1. Indeed x induces an inner-diagonal automorphism on CLu (e ), so m3 (CG (x)) ≥ 4 + m3 (e CE (Lu )) > m3 (E), a contradiction. Since q(Lu ) = q = q(L) = q(K), x does not induce a field or graph-field automorphism on Lu . Therefore x induces a (nontrivial) graph automorphism on Lu , whence   L = O 2 (CLu (x)) ∼ = G2 (q) or L3q (q). However, again from [III12 , Def. 1.15], since (y, L) is an acceptable subterminal (x, K)-pair, L is not of either of these isomor phism types. (L could be isomorphic to SL3q (q), which has a nontrivial center.) This contradiction completes the proof.  Lemma 3.3. The following conditions hold: (a) For every u ∈ D# , B is the set of all elements of B ∗ inducing innerdiagonal automorphisms on Lu ; (b) Either B = B ∗ or B ∗ = B × f , where f induces a field automorphism of order p on every Lu , u ∈ D# , as well as on L; and (c) B ∈ Ep∗ (KB). Proof. Suppose f ∈ B ∗ and f induces a nontrivial field automorphism on Lu for some u ∈ D# . Then by [IA , 4.2.3], f induces a nontrivial field automorphism on L. Conversely if f induces a nontrivial field automorphism on L, then by [III17 , 12.16], it induces a field automorphism on every Lu , u ∈ D# , including Lx = K. This implies (a). Then (b) is immediate from Lemma 3.2, applied to B ∗ . Finally, suppose that B ∈ Ep∗ (KB). Then mp (KB) > mp (B). By (b), mp (CG (x)) = mp (B ∗ ) = mp (B) + 1 and so mp (KB) = mp (CG (x)). Hence by our choice of B ∗ , and an application of [III17 , 13.1] to C 0 , we have B ∗ ≤ KB ≤ C 0 and so B ∗ = B, a contradiction. Hence, (c) holds and the proof is complete.  Now we define (3D)

(1) A = AutG (B) ∼ = NG (B)/CG (B); and (2) W = the subgroup of A generated by all reflections in A.

We will repeatedly abuse notation in the following way: Let H be a subgroup of G normalized by B. Then NH (B)/CH (B) maps isomorphically onto to the subgroup AutH (B) of A = AutG (B), and we shall refer to AutH (B) as A ∩ H. The subgroup of A ∩ H generated by reflections will be denoted WH . In particular, the subgroups WK , WLu (u ∈ D# ), and WL will play important roles in our analysis of W .

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3. THE WEYL GROUP

159

Our main goal will be to identify a subgroup N ∗ of NG (B) such that the subgroup (3E)

W ∗ := N ∗ CG (B)/CG (B)

∗ ∼ of W contains WK , W ∗ is generated by reflections, and W = AutG0 (B), where G0 # is the eventual target group G0 = Ld | d ∈ D . Somewhat more will be proved, in Proposition 3.4 below. Note that W may be identified with a subgroup of SL± (B), the group of invertible linear operators on B of determinant ±1. We will sometimes consider the subgroup

(3F)

W + = W ∩ SL(B)

of index 2 in W . Furthermore, for any subgroup Y ≤ W , we write Y + = Y ∩ W + = Y ∩ SL(B). Finally, for any simple complex Lie algebra L, we write W (L) for its Weyl group. We fix all this notation through Section 11. In Sections 4–11, we shall prove the following proposition. Proposition 3.4. According to the structure of K, and in one case for a suitable choice of subterminal (x, K)-pair (y, L), there exists W ∗ ≡B W (L) as in (3E) such that WK ≤ W ∗  W , and one of the possibilities (a) − (n) of Table 15.1 holds. The clause W ∗ ≡B W (L) means that W ∗ ∼ = W (L), and that involutions of W corresponding to reflections in W (L) act as reflections on B. These conditions imply in most but not all cases that the representation of W ∗ on B/CB (W ∗ ) is uniquely determined up to equivalence, as the dual module Sp∗ := (Fp ⊗Z Λ)∗ , where Λ is the root lattice of L. In any case there are very few possibilities for that representation. See [III17 , 1.5, (1A), (1B)] for more details. ∗

Remark 3.5. Each row of Table 15.1 asserts that K, L, and B have the asserted structure (subject to the given condition), and W ∗ exists with W ∗ and W/W ∗ having the asserted structure. No assertion about G0 is implied, except that the asserted structure for it is our target, to be established in Sections 12–18, after Proposition 3.4 has been proved. As a corollary, we will know that K and L do not have strange Schur multipliers: Corollary 3.6. K ∈ Lie(2) and L ∈ Lie(2). Proof. Suppose that K ∈ Lie(2). As mp (K) ≥ 3 and K/Op (K) ∈ Gp , we have K/Z ∼ = F4 (2) or E6− (2), by [III17 , 7.8], where we have set Z = O2 (Z(K)) = 1. By Proposition 3.4, the only possibility is K/Z ∼ = A− = E6− (2) and L ∼ 5 (2). By [III17 , 16.11], Z ≤ L, and by [IA , 4.10.3a], p = 3. For any u ∈ D − x , Lu /O2 (Lu ) ∈ Lie(2) and so LO2 (Lu )/O2 (Lu ), a component of CLu /O2 (Lu ) (x), lies in Lie(2). Therefore O2 (L) ≤ O2 (Lu ) ≤ Z(Lu ). Hence [Z, Lu ] = 1, whence [Z, G0 ] = 1 as u was arbitrary. It follows easily from L3 -balance that G0 lies in a 3-component H of L3 (CG (Z)). Of course Z ≤ Z(H) and therefore in H = H/[H, O3 (H)], CH (x) has a component K with 1 = Z = O2 (Z(K)) and K/Z ∼ = E6− (2). But by [III17 , 10.29], H has Schur multiplier of odd order.

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160

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Table 15.1. Neighborhoods, Weyl groups, and target groups K/Z(K)

W∗

L









q (a) Lnq (q) SLn−1 (q)

W (An )

mp (B)conditionG0 /Z(G0 )|W/W ∗ | n

q (b) Lnq (q) SLn−1 (q) W (An+1 ) n    − −q (c) Ln q (q) SLn−2 (q) W C[ n+2 ] n+2 2 2

−q

(d) L6

−q

(q) L4



1



1†

n≥4

q Ln+1 (q)

n≥4

q Ln+2 (q)

n≥6

q Ln+2 (q)

−

−q

(q)

W (F4 )

4



SL3q (q)



W (D4 )

4



L4q (q)

W (E6 )

5

p=3

E6q (q)

≤2



SL7q (q)



W (E8 )

8

p=3

E8 (q)

1

(e) L4q (q) (f) L6q (q) (g) L9q (q)



(h) Sp2n (q) Sp2n−2 (q) W (Cn+1 ) (i) Sp6 (q)

Sp4 (q)

n q

n−1 q

−n

−n−1

W (F4 )

n+1

(q)

D4 (q) 

n ≥ 3 Sp2n+2 (q)

4

(j) Ω2n (q) Ω2n−2 (q) W (Dn+1 ) n+1 q (k) Ω2n q (q) Ω2n−2 (q)

E6

1

F4 (q) n≥4

n+1 q

Ω2n+2 (q) −n+1

q n ≥ 4 Ω2n+2 (q)

1 ≤6

1 1 ≤2

W (Cn )

n

W (E6 )

6

E6q (q)

≤2



Ω+ 8 (q)

(m) E6q (q)



A5q (q)

W (E7 )

7

E7 (q)

1

(n) E7 (q)

Ω+ 12 (q)

W (E8 )

8

E8 (q)

1

(l) Ω10q (q)



†



1

Possibly 2 if G0 /Z(G0 ) ∼ = L6q (q). 

This is a contradiction, and so K ∈ Lie. Therefore L ∈ Lie, by [IA , 4.2.2], un1 less L = [L∗ , L∗ ] with L∗ ∼ = Sp4 (2), G2 (2), or 2F4 (2 2 ). From Proposition 3.4, ∼ K = Sp6 (2) is the only possibility, but it does not actually occur as Sp6 (2) ∈ G3 . The proof is complete.  4. The Field Automorphism Case Our first goal is to reduce the proof of Proposition 3.4 to the case that B ∗ = B. By Lemma 3.3, this means dealing with the configuration B ∗ = B × f with f inducing a nontrivial field automorphism on each Lu , u ∈ D# . Proposition 4.1. If B ∗ > B, then the conclusions of Proposition 3.4 hold. We set q0 = q 1/p , where q = q(K). Lemma 4.2. If g ∈ NG (B ∗ ) induces a reflection on B ∗ , then g induces a reflection on B and centralizes B ∗ /B.

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4. THE FIELD AUTOMORPHISM CASE

161

Proof. It suffices to show that [g, B ∗ ] ≤ B, and since [g, B ∗ ] is cyclic it will be enough if [g, B ∗ ] ∩ B = 1. Since D ∈ E2 (B ∗ ), there is 1 = u ∈ D ∩ CB ∗ (g). If g normalizes Lu , then g normalizes B, the subgroup of B ∗ inducing inner-diagonal automorphisms of Lu , and centralizes B ∗ /B as f induces a nontrivial field automorphism on Lu . Thus we may assume that Lgu = Lu . Now, L ≤ Lu and there is b ∈ B ∩ L − Z(L), by [III17 , 13.2]. Then bg ∈ Lgu so [bg , Lu ] = 1 and thus bg ∈ B. Hence 1 = [b, g] ∈ [B ∗ , g] ∩ B, proving the lemma.  Lemma 4.3. Suppose that B ∗ > B. Let f ∈ B ∗ − B, so that f induces a nontrivial field automorphism on K. Let K o = E(CK (f )) and let J be the subnormal closure of K o in CG (f ). Then the following conditions hold: (a) J  E(CG (f )) and J is a vertical pumpup of K o of level q0 = q(K o ); (b) B acts faithfully as a group of inner-diagonal automorphisms on J; and (c) Let W ∗ = AutJ (B). Then one of the cases of Table 15.1 holds, i.e., K, L, B, and W ∗ have the asserted structure. Proof. We have [f, D] = 1 by (3A) as f ∈ B ∗ , and f induces a nontrivial  field automorphism on each Lu , u ∈ D# , by Lemma 3.3. Set Lou = O 2 (CLu (f ))  and Lo = O 2 (CL (f )), so that q(Lou ) = q(Lo ) = q0 for all u ∈ D# . Obviously K o is B ∗ -invariant, so J is as well. We claim first that (4A)

J is a nontrivial pumpup of K o .

Suppose false, so that [x, J/Op (J)] = 1. Recall from (1A4) that v ∈ De . The subnormal closure J of Lo in CG (f ) then contains Lov . Hence [x, Lov ] ≤ Op (Lov ). As Lv is quasisimple and f induces a field automorphism of Lv , this implies that the image of x in Aut(Lv ) is a power of the image of f , by [IA , 7.1.4c]. But [x, Lv ] = 1 so x induces a nontrivial field automorphism on Lv , which is absurd as q(Lv ) = q(L) = q. This proves (4A). Suppose next that J is a diagonal pumpup of K o . As D acts on K o , there is some u ∈ D# normalizing each p-component of J. Now f induces a field automorphism on Lu , which is D-invariant, so Lou is a D-invariant group in Lie(2). We have Lou ≥ Lo . As in the previous paragraph the subnormal closure of Lo in CG (f ) contains Lou and equals J. Hence it contains no D-invariant p-component of CG (u, f ). But if Lou is nonsolvable, it is or contains such a D-invariant component, contradiction. Therefore Lou is solvable. The only possibility, since mp (Lou ) = mp (Lu ) ≥ mp (L) ≥ 2 (Lemma 3.1), is that Lu = L ∼ = SU3 (8) and K ∼ = U4 (8), with p = 3. In this case, however, solvable L3 -balance [IG , 13.8]  implies that the solvable component Lo ∼ = 31+2 and = SU3 (2) satisfies O 3 (Lo ) ∼  O 3 (Lo ) lies in a 3-component of CG (f ), not on the diagonal CJ (D). This contradiction shows that J is a single p-component. In fact, J is a component of CG (f ). This follows from [III2 , Lemma 2.5], with f, K o , y, D, Lo in the roles of y, L, z, B, I there. We next argue that J ∈ Chev(2). For otherwise, by [IA , 2.2.10] and [III17 , 10.9], p = 3, K ∼ = U4 (8), and J ∼ = Co2 or P Sp4 (33 ). Then Z(L) =: u ≤ D. If Lu > L, then u ∈ Z(Lu ) with Lu ∈ Chev(2) of level 8; using [IA , 6.1.4] we see that Lu has untwisted Lie rank at least 5, whence Lu ∈ G3 and the maximality of F(K)

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

162

is violated [III14 , Lemma 1.2]. Thus, Lu = L. Let D ≤ Q ∈ Syl3 (C(u, L)). As CAut(K) (L) is a cyclic subgroup of Inn(K), CQ (x) is abelian of rank 2. It follows from Konvisser’s theorem [IG , 10.17] that m3 (Q) ≤ 3. Therefore L has at most 3 conjugates in CG (u), so m3 (CG (u)) ≤ 1 + m3 (NCG (u) (L)) ≤ 1 + m3 (Aut(L)) + m3 (C(u, L)) = 7. On the other hand, u is 3-central in K o , so if J ∼ = P Sp4 (33 ), then u is 3-central in J and m3 (CG (u)) ≥ m3 (J) = 9, contradiction. Thus J ∼ = Co2 , and 1+4 again u is 3-central, so F := F ∗ (CJ (u)) ∼ and CJ (u)/F is an extension of 21+4 =3 − by A5 . The perfect group CJ (u) then normalizes L, and then F centralizes L. Now F is x-invariant and CF x (x) ≤ CG (DL), which is of rank 2. Thus CF x (x) ∼ = E32 so F x is of maximal class, contradicting the structure of F . We have proved that J ∈ Chev(2). Suppose next that x induces a nontrivial field or graph-field automorphism on J, which then has level equal to q(K) = q ≥ 2p . As J is D-invariant and D is noncyclic, there is some u ∈ D# inducing an algebraic automorphism on J. Thus, by [IA , 4.2.2], every Lie component of CJ (u) has level at least q(J) = q. But f induces a nontrivial field automorphism on Lu by Lemma 3.3, so Lou has level    q0 . It follows that [O 2 (CJ (u)), O 3 (Lou )] = 1. But O 3 (Lo ) ≤ Lo ≤ K o ≤ J so     O 3 (Lo ) ≤ O 2 (CJ (u)) ∩ O 3 (Lou ). As O 3 (Lo ) is nonabelian, this is a contradiction. Therefore x induces an algebraic automorphism on J, so by [IA , 4.2.2], q(J) ≤ q(K o ) = q0 In particular, J ∼  K, and as K = Lp (CG (x)) it follows that f ∈ xG . Now every =  ∗ element f ∈ B − B induces a nonalgebraic automorphism on K, so the above development holds for f  in place of f . Therefore (4B)

xG ∩ B ∗ ⊆ B, and in particular, xG ∩ x, f ⊆ x .

It follows that x is p-central in CG (K o ).

(4C)

For, if we let Qo ∈ Sylp (CG (x K o )) and Q1 ∈ Sylp (CG (K o )) with Qo ≤ Q1 , then mp (C(x, K)) = 1, so by [IA , 7.1.4c], Qo = S × f with x = Ω1 (S). Hence by (4B), x is weakly closed in Qo = CQ1 (x), so Q1 = Qo , proving (4C). Therefore, S maps onto a Sylow p-subgroup of CCG (f ) (K o )/C(f, J), and is cyclic. (4D) We next show that q(J) = q(K o ) = q0 .

(4E) Suppose false; in particular,

q0 = q(K o ) > 2. Since each Ld , d ∈ D , is a level pumpup of L, each Lod is a level pumpup of Lo , with Lod a component of CJ (d) and Lo a component of CJ (D) (not a solvable component, as q(Lo ) = q(K o ) > 2). Then we can apply [III17 , 3.27] with q0 , J, K o , Lo , Lod , x, y, and D here in the roles of q, J, Jx , L, Jd , x, y, and D there. That lemma implies (4E), completing the proof of (a). Obviously B ∗ acts on J, and f ≤ CB ∗ (J) ≤ CB ∗ (K o ) = f, x , by [IA , 7.1.4c]. As J is a vertical pumpup of K o , [x, J] = 1, so B ∗ = Bf × f for some Bf containing x and mapping injectively into Aut0 (J). To prove (b) it remains to show that no element g ∈ B ∗ induces a nontrivial graph automorphism on J. But if g exists, then p = 3 and J ∼ = D4 (q0 ) or 3D4 (q0 ). As m3 (K) ≥ 3, also #

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4. THE FIELD AUTOMORPHISM CASE

163

m3 (J) ≥ m3 (K o ) ≥ 3, whence J ∼ = D4 (q0 ) and x maps into Inndiag(J). Thus m3 (CJ (x)) = 4. As CB ∗ (J) = f we have m3 (B ∗ ) ≤ 1 + m3 (CAut0 (J) (g)) = 1 + 3 < 1 + m3 (CJ (x)) ≤ m3 (B ∗ ), a contradiction. So (b) holds. We next determine the possibilities for K and L, the pumpup J of K o , and mp (B). We wish to show that K and L are listed in Table 15.1, that mp (B) is at least as large as in Table 15.1, and that J/Z(J) is to have level q0 but otherwise to be the same as G0 /Z(G0 ) in the table. We keep (4D) in mind, as well as our conditions that mp (K) ≥ 3 and p splits K, hence p splits K o and J. Most importantly, J ∈ Chev(2), and given K, the isomorphism type of L is restricted by the condition that (y, L) be an acceptable subterminal (x, K)-pair. If either J or K is of exceptional Lie type, then J and K o must appear as a pair (K, L) in [III11 , Table 13.1]. If K ∼ = F4 (q) or E8 (q), then there is no such J by [III11 , Table 13.1], eliminating these cases. As mp (K) ≥ 3, the only other exceptional cases are K ∼ = E6 (q) and ∼ E7 (q), and [III11 , Table 13.1] gives J = E7 (q0 ) or E8 (q0 ), as in cases (m) and (n) of Table 15.1. Note that if p = 3, then K o must be the universal version, and K o B covers Inndiag(K o ), by [IA , 4.7.3A]. We may then assume that K is a classical group. We again use [III11 , Table 13.1] to determine the exceptional possibilities for J, and [IA , 4.8.2] to determine the classical possibilities. If K ∼ = Sp2n (q), then n ≥ 3; the only possible classical J is Sp2n+2 (q0 ), and the only possible exceptional J is F4 (q0 ), occurring for n = 3, as in cases (h) or (i) of Table 15.1 holds. Next, assume that K ∼ = Ω2n (q) for some n ≥ 4 and some  = ±1. If J is a q (q0 ), as in (j) classical group, then it must be an orthogonal group, indeed Ω2n+2 or (k) of Table 15.1. If, on the other hand, J is of exceptional type, then from [III11 , Table 13.1], one of the following possibilities must occur:  (1) K ∼ = D7q (q) and J ∼ = E8 (q0 ); ∼ (2) K = D6 (q) and J ∼ = E7 (q0 );  (3) K ∼ (4F) = E6q (q0 ); = D4 (q), p = 3, and J ∼ − q − (4) K ∼ = D4 (q) and J ∼ = E6 (q0 ).   (5) K ∼ = D5q (q) and J ∼ = E6q (q0 ); In case (4F5), we are in case (l) of Table 15.1. In the other cases of (4F) we reach a contradiction by finding u ∈ D − x such that F(Lu ) > F(K) and Lu has Gp -depth dp (G), like K. This contradicts the fact that (x, K) ∈ J∗p (G), according to [III14 , Lemma 1.2]. Specifically, in the four respective cases, we use [III17 , 3.3] to   − find u ∈ D − x such that Lou ∼ = E7 (q0 ), E6q (q0 ), A5q (q0 ), and A5 q (q0 ). Then Lu is isomorphic to the corresponding group over Fq , and F(Lu ) > F(K), as asserted. The last group of cases to consider is for K/Z(K) ∼ = L (q). n

Suppose first that L ∼ = SLn−1 (q), so that  = q and then n ≥ 4 as mp (K) ≥ 3. By [III12 , Theorem 1.2], for each u ∈ D# , either Lu = L or Lu /Z(Lu ) ∼ = K/Z(K). q (q0 ). Then by [III17 , 3.26], applied with Therefore either Lou = Lo or Lou ∼ = An−1 J/Z(J), AutJD (J) and AutD (J) here in the roles of J, H, and D there, either n = 4 with J ∼ = Ln+1 (q0 ) or Ln+2 (q0 ), as in cases (e), (a) or = D4 (q0 ), or J/Z(J) ∼ (b) of Table 15.1, respectively.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

164

Therefore we may assume that (4G)

L∼ = SLn−2 (q).

Since mp (L) ≥ 2, n > 4. Consider the case (4H)

J/Z(J) ∼ = Ln (q0 ),

whence n > n (as K o < J). If  = q , then CJ (x) has a subgroup isomorphic  q (q0 ). Then E(CK (b)) ∼ to GLnq (q0 ), so there is b ∈ B ∗ with E(CK o (b)) ∼ = SLn−1 = q SLn−1 (q), so (x, K, y, L) is ignorable, contrary to the definition of acceptable subterminal pair. Hence if (4H) holds then  = −q . Consequently, from [IA , 4.8.2], −q (q0 ) and AutJ (B ∗ ∩ J) ∼ we see that J/Z(J) ∼ = W (C[ n+2 ] ), as in case (c) of = Ln+2 2 Table 15.1. Still assuming (4G), suppose that J/Z(J) is a symplectic or orthogonal group over Fq0 . Now K o /Z(K o ) ∼ = Ln (q0 ), with K o  CJ (x) and mp (C(f K o )/C(f, J)) = 1. Using [IA , 4.8.2] we see that as n > 4, we must have  = q and CJ (x)/Op (CJ (x))   contains a quotient of O p (GLnq (q0 )) by a p -group. This leads to the existence of b and a contradiction as in the previous paragraph. Thus J/Z(J) is of exceptional Lie type and we can read the possibilities for J and K o off [III11 , Table 13.1]. In particular, if n ≥ 7, we find that J ∼ = Em (q0 ), q m = 7 or 8, with K o ∼ (q0 ), unless p = 3 and m = 8, in which case = Am−1  Ko ∼ = A8q (q0 ). With the exception of this last case, it is clear from the extended  Dynkin diagram that K o lies in an Amq (q0 ) subgroup of J and so CJ (x) contains q GLm (q). As in the last two paragraphs, this leads to the existence of b and a contradiction. If n = 5 or 6, we find that J ∼ = A5 (q0 ). We reach = E6 (q0 ) and K o ∼ ∗ the same contradiction if there is b ∈ B − x, f centralizing an SL5 (q) subgroup of K o . If  = q , this leaves only the case p = 3, q ≡  (mod 9). And if  = −q , then we have case (d) of Table 15.1. We are thus reduced to two cases in which    p = 3: J ∼ = A8q (q0 ) or A5q (q0 ), respectively, as in = E8 (q0 ) or E6q (q0 ) with K o ∼ cases (g) and (f) of Table 15.1. Thus K, L, mp (B) and J are as claimed. The final step is to check that W ∗ := WJ , the reflection subgroup of AutJ (B), satisfies W ∗ ≡B W (L), where L is as in Table 15.1. In fact, we will check that the reflection subgroup R of AutJ (B ∗ ) is as claimed. This will suffice because of Lemma 4.2, which gives a surjective restriction mapping ϕ : R → W ∗ as B ∗ = B f with [J, f ] = 1; and ker ϕ stabilizes the chain B ∗ > B > 1 so lies in Op (R). But if R is as claimed, then Op (R) = 1 and so ϕ will be an isomorphism, as desired. We know that B ∗ = B f induces inner-diagonal automorphisms on J, with CB ∗ (J) = f . Also B ∗ ∈ Ep∗ (CG (x)), by construction. In particular, if p does not divide |Outdiag(J)|, then B ∗ induces inner automorphisms on J, so B ∗ = (B ∗ ∩ J) × f with mp (B ∗ ∩ J) having the value of mp (B) from the table. It is then clear from [III17 , 2.6], as AutJ (B ∗ ) ∼ = AutJ (B ∗ ∩ J), that R is as desired. So we may assume that p divides |Outdiag(J)|. We use the fact that B ∗ normalizes K o . q q (q0 ), then K o ∼ (q0 ), δ = 1 or 2, while if p = 3 and Note that if J/Z(J) ∼ = SLkp−δ = Lkp    J∼ = D5q (q0 ) or A5q (q0 ). Hence [III17 , 2.8b] applies, with B ∗ , J, = E6q (q0 ), then K o ∼ and x here in the roles of B, K, and x there, yielding AutJ (B ∗ ) ≡B ∗ /f  W (Akp−1 ) or W (E6 ). The proof is at last complete. 

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4. THE FIELD AUTOMORPHISM CASE

165

Having constructed W ∗ = AutJ (B) in accordance with Table 15.1 when B < B , we next check that WK ≤ W ∗  W , thereby completing the proof of Proposition 3.4 in the case B < B ∗ . ∗

Lemma 4.4. If B < B ∗ , then WK ≤ W ∗ . Proof. By Lemma 3.3c, B ∈ Ep∗ (KB), and so B ∈ Ep∗ (K o B). Hence using [III17 , 2.8], WK = AutK (B) is a reflection group depending on K and p, as is WK o = AutK o (B), and WK = WK o . Obviously WK o ≤ WJ = W ∗ , so the lemma holds.  In fact, W ∗ is very close to W , as the next two lemmas show. Suppose that B < B ∗ .  Lemma 4.5. W ∗ (WK ) ≤W .

Then W = NW (W ∗ )NW (x ) and

Proof. It suffices to prove the first conclusion. For then, as NW (x ) ≤   W NW (W ∗ ) ∗ ∗ NG (WK ) and WK ≤ W , WK = WK ≤W . Let t ∈ NG (B) map to a product τ ∈ W of reflections. Set z = xt , I = K t = Lp (CG (z)), and t = t−1 . We argue first that (4I)

f induces a nontrivial field automorphism on I.

First, f centralizes z ∈ B so f normalizes I. Since B induces inner-diagonal  automorphisms on K, it does as well on I. Now (B ∗ )t ∈ Ep∗ (CG (x)) so by Lemma 3.2,  every element of (B ∗ )t induces an inner-diagonal or nontrivial field automorphism  on K. As B < B ∗ it follows from our choice of B ∗ that elements of (B ∗ − B)t  induce nontrivial field automorphisms on K (otherwise we would have chosen (B ∗ )t instead of B ∗ ). Since f ∈ B ∗ − B, (4I) follows. Now set I o = E(CI (f )), so that I o ∼ = K o has level q0 . Applying Lemma 4.3 to    ∗ t t ∗ (B ) and f instead of B and f , we see that (I o )t = E(CI t (f t )) = E(CK (f t ))  pumps up to a single component of CG (f t ). Therefore I o pumps up to a single ∗ component H of CG (f ). As I is B -invariant, so is H. Suppose that H = J. Now, CB ∗ (J) ≤ CB ∗ (K o ) = f, x , but [x, J] = 1, so CB ∗ (J) = f . As B ∗ contains every element of order p in CG (B ∗ ), it follows that f ∈ H, so that f ∈ Z(H). Then there is an x-invariant Ep2 -subgroup U ≤ H. If [x, U ] = 1, then U ≤ CCG (x) (K o ), so U ≤ x, f and so x ∈ U , a contradiction as [x, J] = 1. Thus [x, U ] = f so x ∼ xf , which contradicts (4B). Therefore H = J. Then K o and I o are isomorphic components of CJ (x) and CJ (xt ), respectively. If x and xt have images in Aut(J) that are WJ = W ∗ conjugate, then x and xt are actually W ∗ -conjugate since both lie in B, using Lemma 4.3b. In that case τ w ∈ NG (x ) for some w ∈ W ∗ . If this holds for every t ∈ NG (B) mapping on a product τ ∈ W of reflections, then W = W ∗ NW (x ), as desired. We may therefore assume that we chose t so that the images of x and xt in Aut(J) are not WJ = W ∗ -conjugate. We reduce to the case in which p does not divide |Outdiag(J)|. If J ∼ = E6± (q0 ) with p = 3, then by [IA , 4.7.3A, 4.2.2j], the images of x and xt are q Inndiag(J)-conjugate and hence Inn(J)-conjugate. Likewise if J/Z(J) ∼ (q0 ) = Lkp  q o o ∼ t and K /Z(K ) = Lkp−1 (q0 ), the images of x and x are Inn(J)-conjugate. We q (q0 ) by using the fact that every reach the same conclusion even if K o ∼ = SLkp−2

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166

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

element of order p in CG (K o ) lies in x, f . Indeed xt has a J-conjugate x centralizing K o , so x, x ≤ x, f and then x and x have the same image in Inndiag(J). Then, [III17 , 2.10] implies that the images of x and xt are conjugate in NJ (J(Q)), where Q is a B-invariant Sylow p-subgroup of J. So they are conjugate in WJ by [III17 , 2.8a], a contradiction. Thus, we may assume that p does not divide |Outdiag(J)|. By [III17 , Remark 1.20], W ∗ is irreducible on B. Let R = CG (B ∗ /B) ≤ NG (B ∗ ) and R0 = CR (B)  R. Then as W ∗ centralizes f and normalizes B, the image of R in Aut(B) contains W ∗ and so acts irreducibly on B. Hence [R0 , B ∗ ], an R-invariant subgroup of B, equals B or 1. But since f induces a nontrivial field automorphism on K, [R0 ∩K, B ∗ ] = 1, as follows from [IA , 4.10.2]. Hence [R0 , B ∗ ] = B, and therefore (4J)

Bf i ⊆ (f i )CG (B) , i = 1, . . . , p − 1.

Furthermore, as E(CG (x)) ∼ = E(CG (f )), xG ∩ B ∗ ⊆ B, and then the irreducible ∗ action of W shows that B is weakly closed in B ∗ with respect to G. Let P ∈ Sylp (CG (B)) with B ∗ ≤ P , and set P1 = [Ω1 (P ), Ω1 (P )]. Then NNG (B) (P ) covers A = AutG (B) and in particular acts irreducibly on B. Suppose that B ≤ P1 . Then by irreducibility, B ∩ P1 = 1. Now B ∗ contains every element of order p in CP (B ∗ ). If Ω1 (P ) is abelian, then B ∗ = Ω1 (P ), while if Ω1 (P ) is not abelian, then as Z(P ) ∩ P1 = 1, we have B ∗ = Ω1 (Z(P )). In either case NNG (B) (P ) ≤ NNG (B) (B ∗ ), so by (4J), NNG (B) (P ) ≤ CG (B)CNG (B) (f ) ≤ CG (B)NNG (B) (J). Passing modulo CG (B) we get that W normalizes AutJ (B) = W ∗ , which implies the lemma. We may therefore assume that B ≤ P1 . Let Cx = CG (x), C x = Cx /C(x, K), x = Cx /K. If p does not divide |Outdiag(K)|, then P is an extension of and C  = 1, so the cyclic group C P (K) by a cyclic group of field automorphisms, with x x ∈ P1 , contradiction. So p divides |Outdiag(K)|. Then by [III17 , 2.15], Ω1 (P ) is abelian. But clearly B = 1 so B ≤ P1 . This contradiction completes the proof.  Lemma 4.6. Suppose that B < B ∗ . Then W ∗  W . Moreover, one of the following holds: (a) W = ZW ∗ with Z acting as scalars (on B) and |W : W ∗ | = |Z : Z∩W ∗ | ≤ 2; moreover, if W ∗ ∼ = W (Am ) for m > 5 or W (E7 ), then W = W ∗ ; ∗ ∼ (b) W = W (Dn ), n ≥ 4, W ≤ W (Cn ), and |W : W ∗ | = 2; or (c) W ∗ ∼ = W (D4 ) and |W/W ∗ | divides 6.  W . Then W0 ≤ W ∗ by Lemma 4.5. In all cases of this Proof. Let W0 = WK ∗ lemma, any reflection in W has a W ∗ -conjugate in WK (for example, in case (d), WK ∼ = W (C3 ) and in case (k), WK ∼ = W (Cn−1 )). Hence W ∗ = W0  W . ∗ Suppose that W < W , so that there is a reflection t ∈ W − W ∗ . In cases (a), (b), (f), (l), and (m), conclusion (a) holds by [III17 , 2.17]. In the remaining cases W∗ ∼ = W (F4 ), W (Dn ), W (Cn ) or W (E8 ), and by [III17 , 1.21, 1.22], conclusion (b) or (c) holds.  Lemmas 4.3 and 4.6 complete the proof of Proposition 4.1.

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167

5. Some General Lemmas By Proposition 4.1, we may and shall assume in the remainder of the proof of Proposition 3.4 that B ∗ = B,

(5A)

so that B is a largest elementary abelian p-subgroup of CG (x), and hence maximal with respect to inclusion in G. In some cases we determine W , a stronger result than just proving the existence of some W ∗  W satisfying the proposition. Our next goal is to shed more light on the structures of BK, BL, and BLu , for u ∈ D# . We first prove Lemma 5.1. The following conditions hold: (a) (b) (c) (d)

Ip (CG (LD)) ⊆ D; CB (L) = D; mp (CB (Lu )) = 1 for all u ∈ De ; and If u ∈ De and p does not divide |Outdiag(Lu )|, then B = (B ∩ Lu ) × u and mp (C(u, Lu )) = 1.

In Lemma 5.3 below, we show that mp (C(u, Lu )) = 1 if u ∈ De , without the extra hypothesis of (d). Proof. By (1A2), x = CB (K). Suppose there is z ∈ Ip (CG (LD)) − D and set E = y, z , so that CE (K) = 1 and mp (E) = 2. If z induces a nonalgebraic automorphism on K, then by [IA , 4.2.3], [z, L] = 1, contradiction. If z induces a  graph automorphism on K, then p = 3, K ∼ = D4 (q), and L ∼ = A3q (q) by definition of acceptable subterminal pair [III12 , Def. 1.15]. But then m3 (CK (z)) = 2 [IA , Table 4.7.3A, 4.10.3a] while m3 (L) = 3, so [z, L] = 1, contradiction. Consequently z maps into Inndiag(K). Since y does as well, by definition of acceptable subterminal pair, E maps (injectively) into Inndiag(K). As mp (E) = 2, (5B)

mp (CInndiag(K) (L)) ≥ 2.

We claim that L is a component of CK (e) for every e ∈ E # . Indeed if E ≤ KC(x, K), the claim is immediate from the definition of subterminal pair [III3 , 1.2,  1.3]. Otherwise p divides |Outdiag(K)|. But (5B) rules out K ∼ = E6q (q) with  p = 3 (see [IA , Table 4.7.3A]) and in the only other case, K/Z(K) ∼ = Lnq (q), we  q q must have L ∼ (q) = SLn−2 (q), and there is e ∈ E # such that CK (e) has an SLn−1 component. Hence (x, K, y, L) is ignorable, contrary to the acceptability of the subterminal (x, K)-pair (y, L). This proves the claim, which then implies (e.g. by [III11 , 1.16]) that E is cyclic, contradiction. Thus (a) holds, and (b) is an immediate consequence. Suppose that u ∈ De , i.e., Lu > L. Then [x, Lu ] = 1 so u ≤ CB (Lu ) < D, and (c) holds as well. If p does not divide |Outdiag(Lu )|, then p does not divide |Z(Lu )| and so B = (B ∩ Lu ) × u . As B contains every element of order p in CG (B), mp (C(u, Lu )) = 1, and the proof is complete.  Lemma 5.2. The isomorphism types of K, L, and any vertical pumpup Lu of L (u ∈ D# ) must be one of the possibilities listed in Table 15.2.

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168

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Table 15.2. Possible components Lu > L, u ∈ D# , p > 2 K

L



Lu





q Anq (q), n ≥ 3 An−1 (q)



Anq (q), n ≥ 4, p|n + 1 −

Anq (q) 



q An−2 (q)

−

n−1



q q Anq (q), An−1 (q), Cn−1 (q), Dn−1 (q), E7 (q) (n = 8, p = 3), E8 (q) (n = 9, p = 5)

−

−q

q An q (q), n ≥ 5 An−2 (q) An q (q), D4− (q) (n = 5), E6

Cn (q), n ≥ 3 Cn−1 (q) 

q Dn (q), n ≥ 4 Dn−1 (q)

F4 (q) 

E6q (q) −q

E6

(q)

Cn (q) 

Dn (q), A4q (q) (n = 4,  = 1)

C3 (q) 

A5q (q) −q

A5

(q) (n = 7)

C4 (q), F4 (q) 



A6q (q), D6 (q), E6q (q) −q

(q)

E6

(q)



E7 (q)

D6 (q)

D7q (q), E7 (q)

E8 (q)

E7 (q)

E8 (q)

Proof. By [III12 , Theorem 1.2], mp (K) ≥ 3. This condition limits us to the K’s listed in the table (see [IA , 4.10.3a]). The corresponding isomorphism types for L are then consequences of [III12 , Def. 1.15]. We need therefore to determine the possible Lu ’s for each given K and L. For each u ∈ D# such that L < Lu , x acts nontrivially on Lu as an inner-diagonal automorphism by (5A) and Lemma 3.3a, with L a component of CLu (x) by Lp balance. Recall also that Lu ∈ Chev(2) with q(Lu ) = q(K) = q by (1A5), and (5C)

F(Lu ) ≤ F(K)

by [III14 , Lemma 1.2], as long as Lu /Op (Lu ) and K/Op (K) have the same Gp depth [III3 , pp. 63-64]. Note that in the first row of the table, the Lu entry is a consequence of [III12 , Theorem 1.2], so we can ignore that case. Now, suppose that p does not divide |Outdiag(Lu )|.

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5. SOME GENERAL LEMMAS

169

Then by Lemma 5.1, B = (B ∩ Lu ) × u . Let d = D ∩ Lu ; we claim that mp (CLu (L)) = 1. Indeed, let w ∈ Ip (CLu (L)) with [w, d] = 1. As D = d, u , w ∈ Ip (CG (LD)) ⊆ D, so w = d , proving the claim. If Lu is of exceptional type, then L must therefore appear in [III11 , Table 13.1], and this, together with (5C), yields only entries that are recorded in Table 15.2.  For example if Lu ∼ = A7q (q) (or, if p = 3, = E8 (q), then by [III11 , Table 13.1], L ∼ q q A8 (q)), D7 (q), or E7 (q). Since F(D8 (q)) < F(E8 (q)), while both D8 (q) and E8 (q) have Gp -depth 2 (see [IA , 4.10.3a]), this Lu does not belong in the fifth row of Table 15.2. In the second row of Table 15.2, it does belong if n − 2 = 7, as long as p divides n + 1 = 10; it does not belong if n − 2 = 8 and p = 3, since then p must divide n + 1 = 11. Assume then that Lu is a classical group; the component L of CLu (x) must arise q (q), for example, as in [IA , 4.8.2, 4.8.4]. If Lu ∼ = Cm−1 (q) or Am−1 = Cm (q), then L ∼  explaining the C’s in rows 2, 4, and 6 of Table 15.2. Since F(C6 (q)) > F(E6q (q)), q however, and both C6 (q) and E6 (q) have Gp -depth 2, C6 (q) does not appear in row 7. The other classical entries are similarly determined. Finally, assume that p divides |Outdiag(Lu )|,  q ∼ Am (q) with p dividing m + 1, or Lu = ∼ E q (q) with p = 3. In the so that Lu = 6  q  former case L ∼ = Am (q), m < m. This condition and (5C) leave only the entries  in rows 2, 5, and 7. In the E6q (q) case with p = 3, the possibilities for L are found in [IA , 4.7.3A], and again with the help of (5C), the entries (and non-entries) of Table 15.2 are determined. This completes the proof.  As a result we can sharpen Lemma 5.1d. Lemma 5.3. If u ∈ De , then mp (C(u, Lu )) = 1. Proof. By Lemma 5.1d, we may assume that p divides |Outdiag(Lu )|. Let (w, I) be a p-terminal long pumpup of (u, Lu ). The argument is easiest if Lu /Z(Lu ) ∼ = K/Z(K). In that case, the maximality of (x, K) ∈ J∗p (G) guarantees that I/Op p (I) ∼ = K/Z(K) ∼ = Lu /Z(Lu ), by monotonicity of F [III11 , 12.4], and then mp (C(w, I)) = 1 by [III12 , Theorem 1.2]. But then by [III8 , 2.4], mp (C(u, Lu )) = 1, as desired. On the other hand, if Lu /Z(Lu ) ∼  K/Z(K), we see from Table 15.2 that as p =  divides |Outdiag(Lu )|, the only possibilities are K ∼ = D4 (q) with Lu ∼ = A4q (q) and   p = 5, or K ∼ = E6q (q) with Lu ∼ = A6q (q) and p = 7. It is immediate that K/Op (K) and Lu /Op (Lu ) then have the same Gp -depth (4 or 2 respectively), which equals dp (G) as (x, K) ∈ Jp (G). Now, the same reasoning as in the previous paragraph will yield the desired result even if Lu /Z(Lu ) ∼  K/Z(K), as long as we can argue = that I/Op p (I) ∼ = Lu /Z(Lu ). We prove something stronger: any (t, J) ∈ ILop (G) for which J/Op p (J) is isomorphic to Lu /Z(Lu ) has no vertical pumpup (t∗ , J ∗ ) in G. The proof is simply that F(J ∗ ) > F(K) by [III17 , 10.30gh], contradicting  [III14 , Lemma 1.2]. The lemma is proved. Recall from Lemma 3.3 that B induces inner-diagonal automorphisms on Lu for any u ∈ D# . The next lemma locates B a bit more precisely. Recall also that B is a maximal elementary abelian subgroup of CG (x), and hence B is maximal with respect to inclusion among all elementary abelian subgroups of G.

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170

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Lemma 5.4. Let u ∈ De , and set CG (u) = CG (u)/C(u, Lu ). Then one of the following holds: (a) B ∈ Ep∗ (Lu u ); or (b) p divides |Outdiag(Lu )| and mp (B) = mp (Inndiag(Lu )).

Proof. If p does not divide |Outdiag(Lu )|, then by Lemma 5.1, B = (B∩Lu )× u , with B ∩ Lu containing every element of order p in CLu (B ∩ Lu ). Moreover,

(5D)

mp (B ∩ Lu ) = mp (B) − 1 = mp (CG (x)) − 1

as B ∈ Ep∗ (CG (x)). If p is a good prime for Lu , then mp (B ∩ Lu ) = mp (Lu ) by [IA , 4.10.3e], so (a) holds. If Lu /Op (Lu ) ∼ = K/Op (K), then as B maps into Inndiag(K), mp (CG (x)) = mp (K)+1, so by (5D), mp (B ∩Lu ) = mp (K) = mp (Lu ), − proving (a) in this case. A similar argument proves (a) if K ∼ = = A7 q (q) and Lu ∼ −q ∼ E6 (q), both groups of p-rank 4 (see [IA , 4.10.3a]). And if Lu = En−1 (q), n = 8  q or 9, p = 3 or 5, respectively, with K ∼ (q); in this case = Anq (q), we have L ∼ = An−2 B ≥ D × (B ∩ L) has rank at least n = mp (Lu ) + 1 so again (a) holds. These are all the cases of Table 15.2 in which p does not divide |Outdiag(Lu )|. So suppose that p divides |Outdiag(Lu )|. If (p, |Outdiag(L)|) = 1, then also (p, |Z(L)|) = 1. As B contains every element of order p in CG (B), and CB (L) = D by Lemma 5.1b, we have B = D × (B ∩ L), with B ∩ L ∈ Ep∗ (L). This applies  to the cases Lu /Z(Lu ) ∼ = Anq (q), p|n + 1 of Table 15.2 and yields mp (B) ≥ n + 1  q q (q) or L ∼ (q). Since B ∩ L ∈ Ep∗ (L), and L is or n, according as L ∼ = An−2 = An−1 supported on a subspace of codimension 1 or 2 in the natural Lu -module, B must act diagonalizably on the natural Lu -module, that is, B lies in a conjugate of the q image of a full diagonal subgroup of BLu ≤ P GLn+1 (q). Hence B centralizes, and then contains, a full exponent-p diagonalizable subgroup B1 of Lu . Thus B1 u ≤ B. But by [III17 , 2.13], B1 u ∈ Ep∗ (Lu u ). Whether or not u ∈ Lu , we have mp (B1 u ) = mp (Lu u ) ≥ n, so if mp (B) = n, then (a) holds. On the other q hand, if mp (B) > n, then B ≤ Inndiag(Lu ) with mp (B) = n = mp (P GLn+1 (q)) = mp (Lu ), so (b) holds.  To establish the lemma it remains to consider the case p = 3, K ∼ = E6q (q),   q q Lu ∼ = E6 (q), L ∼ = A5 (q). Then m3 (B) = m3 (CG (x)) ≥ 6 as x ∈ C(x, K). If m3 (B) > 6, then as m3 (CB (Lu )) = 1 by Lemma 5.1c, and from [IA , 4.10.3a], m3 (B) = 6 and (b) holds. So we assume that m3 (B) = 6. Then by (3A4), B ≤ DL. But from [IA , 4.7.3A] and Lemma 5.3, D = u, v for some v ∈ CLu (L) − u of  order 3. Hence B ≤ u Lu . If u ∈ Lu , then Lu /O3 (Lu ) ∼ = E6q (q)u so m3 (Lu u ) = m3 (Lu ) = 6 = m3 (B) by [IA , 4.10.3a]. If u ∈ Lu , then Lu is of adjoint type, and hence so is L, by [IA , 4.7.3A]. Hence as B ≤ DL = D × L, m3 (L) = 4, whence q ≡ q (mod 9). But in that case by [III17 , 8.1], m3 (Lu ) = 5. Hence again  m3 (Lu u ) = 6 = m3 (B) and the proof is complete.

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Now we can prove Lemma 5.5. For each u ∈ D# , A ∩ Lu is generated by reflections, as is A ∩ L. Indeed A ∩ Lu is a Weyl group of a type depending on Lu and p, as in [III17 , Table 17.1]. Moreover, if u ∈ De − {x }, then CB (A ∩ Lu ) = u and x is not A ∩ Lu -invariant. Thus, A ∩ Lu = WLu , A ∩ L = WL , and A ∩ K = WK . Proof. Let u ∈ De , and set CG (u) = CG (u)/C(u, Lu ). Suppose that Lemma 5.4a holds. Then B ∩ Lu ∈ Ep∗ (Lu ). Indeed, there are two cases: either u ∈ Z(Lu ) with B ≤ Lu , or Z(Lu ) is a p -group with B = (B ∩ Lu ) × u . In either case, the assertions of the lemma about A ∩ Lu follow from [III17 , 2.8], applied to X = Lu B with Lu here in the role of K there. Next, assume that Lemma 5.4b holds. Let Q ∈ Sylp (C(u, Lu )) with B normalizing Q. Thus Q is cyclic, by Lemma 5.3. Consequently B = Ω1 (BQ) so NLu (B) = NLu (BQ). Therefore by a Frattini argument, NLu (B) maps onto NLu (B). Moreover, the stabilizer in Lu of the chain B > u > 1 acts trivially on B; this is obvious if u ∈ Lu and holds by [III17 , 2.12a] otherwise. Therefore A∩Lu = AutLu (B) maps isomorphically to AutLu (B), and as mp (B) = mp (Inndiag(Lu )), the assertions of the lemma about A ∩ Lu follow from [III17 , 2.12b]. In all the above cases the structure of A∩Lu is given in terms of Lu in [III17 , Table 17.1]. This completes the proof of all the assertions about Lu . Finally, the structure of A ∩ L is easier to determine because it descends from that of A ∩ K. If p does not divide |Outdiag(L)|, then B = D × (B ∩ L) with B ∩ L ∈ Ep∗ (L). If mp (L) ≥ 3, we can apply [III17 , 2.8] to L × x to deduce that A ∩ L is generated by reflections (indeed reflections of A ∩ Lu ). If mp (L) = 2, then from Table 15.2, K and L are classical groups, from which the structure of A ∩ L is easily calculated in the few cases of such small rank.  Suppose then that p divides |Outdiag(L)|. If p = 3 and K ∼ = E6q (q), so that  L∼ = A5q (q), then from [IA , 4.7.3A], D induces inner automorphisms on K and in fact for any d ∈ D ∩ K − Z(K), d is a direct factor of BL, from [IA , 4.7.3A]. As m3 (L) ≥ 3, the desired result follows from [III17 , 2.8] applied to BL/ d . The last  q (q). case (see Table 15.2) is L ∼ = Lm+1 = SLmq (q) with p dividing m, and K/Z(K) ∼ Thus x is a direct factor of BK, hence of BL, so if mp (L) ≥ 3 we are again done by [III17 , 2.8], applied to BL/ d for any d ∈ D − x − Z(L). Otherwise   L∼ = L4q (q), B ∩ K is a maximal diagonalizable exponent-3 = SL3q (q), p = 3, K ∼  subgroup of CK (D) ∼ = GL3q (q), and clearly A ∩ L ∼ = Σ3 is generated by reflections. The proof of the lemma is complete.  We remark that the relationship between L and A ∩ L is also governed by [III17 , Table 17.1], even in the case mp (L) = 2. Through Lemma 5.11 we use the notation NG (B) = NG (B)/CG (B). Next we associate to each reflection in WL a certain subgroup of L. Lemma 5.6. Let r ∈ WL be a reflection (in its action on B). Then there exists a 2-element r ∈ NL (B) and a subgroup Jr = Jr ≤ L such that the following conditions hold: (a) r ∈ NL (B) maps to r ∈ WL ; (b) If we set Br = CB (r), then [Br , Jr ] = 1 and Jr   CG (Br );

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(c) Jr Z(K)/Z(K) ∈ Lie(2), and if q > 2, then Jr = O p (E(CG (Br ))); (d) B normalizes Jr ; and (e) [B, r] r ≤ Jr . Proof. Without loss we may factor out Op (L) and assume Op (L) = 1. The possibilities for K and L are in Table 15.2, and the corresponding structure of WL is in [III17 , Table 17.1]. In particular, either Z(L) = 1 and B = D × (B ∩ L), or  p L∼ = Amq (q) with p dividing m + 1. With the same exception, as B ∩ L ∈ E∗ (L), B is uniquely determined up to L-conjugacy, by [IA , 4.10.3c]. First suppose that L is a classical group and let V be a natural L-module. If L∼ = Cn (q), n ≥ 2, then V =⊥ni=1 Vi ,

(5E)

where the Vi are hyperbolic planes which are minimal B-invariant nondegenerate subspaces of V . The stabilizer of this decomposition in L contains NL (B) and has the form SL2 (q)Σn = (H1 ×· · ·×Hn )H with Hi ∼ = Σn permuting the = SL2 (q), H ∼ Hi ’s, and B ≤ H1 ×· · ·×Hn , and we may assume H chosen so that B is H-invariant. We take an involution r1 ∈ NH1 (B ∩ H1 ), and an involution r2 ∈ H corresponding to the transposition (12), and argue that the conditions of the lemma are satisfied for each of them. Since WL ≡B∩L W (Cn ) has just two classes of reflections, this will establish the lemma in this case. (Notice that [r1 , B] is supported on V1 , and [r2 , B] on V1 ⊕ V2 , so r 1 and r 2 are not WL -conjugate.) Set Jr1 = H1 . Then CB (r1 ) = D × (B ∩ (H2 · · · Hn )) and the assertions of the lemma are clear. As for Jr2 , let b1 ∈ B ∩ H1# and set b12 = b1 br12 ∈ H1 H2 ; by [IA , 4.8.2], CL (b12 ) has a subnormal SL2 (q) subgroup, which we call Jr2 , supported on V1 ⊕ V2 . Then CB (Jr2 ) = D × b12 × (B ∩ (H3 · · · Hn )) = CB (r2 ), 

and it is clear that Jr2 = O 2 (CG (CB (r2 ))). Therefore r2 ∈ Jr2 , and the remaining assertions of the lemma are again clear. So the lemma holds if L ∼ = Cn (q). n q ∼ Next suppose that L = Dn (q), n ≥ 3. Then WL ≡B∩L W (Dn ) has a unique class of reflections, so we need only find a single r1 and corresponding Jr1 satisfying the requirements of the lemma. Now (5E) holds with the Vi ’s of dimension 2, Binvariant, and all isometric, of type q . The stabilizer of this decomposition in the full orthogonal group O(V ) is again a wreath product (H1 × · · · × Hn )H ∼ with H = Σn , and with Hi = O(Vi ) ∼ = D2(q−q ) acting trivially on Vi⊥ . Fix an isometry φ : V1 → V2 , let b1 ∈ B ∩H1 , and set b12 = b1 bφ1 . Let Jr1 be the SL2 (q) Lie component of CL (b12 ) supported on V1 ⊕ V2 [IA , 4.8.2], and let r1 be an involution in NJr1 (B ∩ Jr1 ). Then CB (Jr1 ) ≤ CB (r1 ) and as CB (Jr1 ) is a hyperplane of B, r 1 is a reflection satisfying our requirements. n ∼ Dn−q (q), then WL ≡B∩L W (Cn−1 ); (5E) holds with V1 , . . . , Vn−1 being If L = B-invariant of type q . Then Vn is of type −q , so that by [IA , 4.8.2], mp (L) = n−1 and B is supported on V1 ⊥ · · · ⊥ Vn−1 . Since mp (L) ≥ 2, n ≥ 3. We construct φ, b12 , Jr1 , and r1 as in the preceding paragraph. As Jr1 is a Lie component of CL (b12 ), all our conditions hold. Note that [B, r1 ] is supported on V1 ⊥ V2 . A second class of reflections (on B) is represented in Jr2 := CL (V1 ⊥ · · · ⊥ Vn−2 ) ∼ = 2 ∼ Ω(Vn−1 ⊥ Vn ) ∼ (q) L (q ), with r inverting B ∩ J , which is supported on = Ω− = 2 2 r2 4

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Vn−1 . Hence r 1 is not conjugate in WL to r2 . The required properties follow easily, as usual. − Suppose that L ∼ = Am q (q). Since mp (L) ≥ 2, m ≥ 3. Indeed as A3 = D3 , − we have just discussed the case m = 3. Interpreted on the natural L = SL4 q (q)module, our b12 above is supported on a 2-dimensional subspace V1 ⊆ V , and Jr1 ∼ = SL2 (q) is supported on an (orthogonal) complement to V1 . On the other hand, Jr2 ∼ = SL2 (q 2 ) above is a component of CL (b2 ) where b2 ∈ B is supported − on all of V . For any m > 3 we choose L0 ≤ L with L0 ∼ = A3 q (q) centralizing a subspace of codimension 4 on the natural module, and take r1 , r2 , Jr1 , and Jr2 within L0 . If  L∼ = Amq (q) with p not dividing m + 1, then Z(L) = 1 and we may take B ∩ L to be a diagonal subgroup of order pm and exponent p. Then we take b12 ∈ L to be a scalar multiple of diag(ω, ω, 1, . . . , 1), where ω p = 1 = ω, Jr1 to be the upper-left Lie component in CL (b12 ), isomorphic to SL2 (q), and r1 ∈ Jr1 . Only one class of reflections needs to be found in this case as  WL ≡B∩L W (Am ). On the other hand, if L ∼ = Amq (q) with p dividing m+1, then by q (q) Table 15.2, and with a single exception to be discussed below, K/Z(K) ∼ = Lm+2   q q ∼ and so BL = Zp × GLm+1 (q). Working in GLm+1 (q) instead of in L, we take b12 = diag(ω, ω, 1, . . . , 1), and choose Jr1 ∼ = SL2 (q) and r1 ∈ Jr1 as before. Again Jr1 and r1 satisfy our requirements. If L ∼ = E7 (q), then WL ≡B∩L W (E7 ), with a single class of reflections, which is represented in a subgroup D × Z × L0 ≤ D × L containing B and with Z ∼ = Zp and L0 ∼ = D6 (q). We can then find the required Jr1 and r1 within L0 , as in the D6 (q) case above. This leaves only the exceptional Am case, namely,   p = 3, L ∼ = E q (q). = A q (q), and K ∼ 5

6

There is just one class of reflections in this case as WL ≡(B∩L)/(B∩Z(L)) W (A5 ).  If m3 (B) = 6, then by (3A4), B ≤ DL. As either DL ∼ = E32 × L6q (q) or DL ∼ = q q Z3 × SL6 (q), we can find r1 and Jr1 within a D × Z3 × L4 (q) subgroup of DL in either case. So we may assume that mp (B) > 6. We may apply [III17 , 2.8] to BL/ y , and conclude that B contains a full exponent-3 diagonal subgroup of L. We take B1 ≤ B ∩ K to consist of all diag(ω, ω, ω1 , . . . , ω4 ) of determinant 1 with  ω 3 = ωi3 = 1 for all i. Then we can take Jr1 = O 2 (CK (B1 )) and r1 ∈ Jr1 . The proof is complete.  Define (5F)

 R = {r ∈ W  r is a reflection in (WK )A }

For each r ∈ R, we write r for any preimage of r in NG (B). We also set (5G)

W1 = R  W.

By ournotation, WK is generated by the reflections that it contains, and so WK ≤ W1 = (WK )A . By (1A4) and Lemma 5.5, v ∈ De − {x } and WLv does not normalize x . Thus W does not normalize x . Since x = CB (WK ) by Lemma 5.5, WK  W and in particular, Lemma 5.7. WK < W1 , and W1 does not normalize x .

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Lemma 5.8. Let r∈ WK be a reflection. Then some WK -conjugate of r lies in WL . Moreover, W1 = (WL )A .  Proof. Since W1 = (WK )A , the second statement is immediate from the first. If WK has a unique class of reflections then there is nothing to prove, since WL obviously contains some reflections. Otherwise, by [III17 , Table 17.1], K ∼ = Cn (q) −

−n

−

(n ≥ 3), F4 (q), An q (q) (n ≥ 5), Dn q (q) (n ≥ 4), or E6 q (q). In these cases K has two root lengths and there are correspondingly two classes of reflections in WK . −

−n−1

q q (q), Dn−1 (q), or By Table 15.2, the corresponding L’s are Cn−1 (q), C3 (q), An−2 −q A5 (q). In every case WL is of type C for some  ≥ 2. Choose a fundamental set of reflections {r1 , r2 } for a W (C2 )-subgroup of WL . Since r1 , r2 ∼ = W (C2 ), r 1 and r2 are not WK -conjugate. Thus both classes of reflections in WK are represented in WL . The lemma follows. 

Lemma 5.9. Let r 1 , r2 ∈ R. Then w := r1 r2 is a {2, 3}-element of W1 , and if q > 2, then w has order at most 4. Moreover, regardless of q, any element w1 ∈ W of odd order and inverted by r1 is a 3-element. w 

Proof. The first sentence immediately implies the second as r1 w1 ∈ r1 1 ⊆ R. So it suffices to prove the first sentence. Set H = r 1 , r2 ≤ A and BH = CB (H). Then CH (B/BH ) stabilizes the chain B > BH > 1 so is a p-group by [IG , 11.5]. As the r i are reflections, |B/BH | ≤ p2 , so H is an extension of a p-group by the subgroup AutH (B/BH ) of GL2 (p). In particular, if q = 2, then as p = 3, H is a {2, 3}-group, and the desired assertion holds. Suppose then that q > 2. By definition of R and then Lemma 5.8, r a1 ∈ WL for some a ∈ A. We may then assume without loss that r1 ∈ WL ; set J1 = Jr1 . Write r 2 = rg for some reflection r ∈ WL and some g ∈ NG (B). Choose r ∈ NL (B) mapping on r, and set J2 := Jrg , with Jr as in Lemma 5.6. Then B2 := CB (r 2 ) = CB (J2 ). As B2 is a hyperplane of B, D2 := B2 ∩ D = 1. Note that D2 = CD (r 1 , r 2 ). Let L2 be the pumpup of L in CG (D2 ). Note that r1 ∈ J1 ≤ L2 , and L2 is quasisimple by (1A5).  Since q > 2, J2 = O p (E(CG (B2 ))) by Lemma 5.6. As D2 ≤ B2 , it follows from Lp -balance that J2 ≤ I := Lp (CG (D2 )). As L2  IB, it follows that either J2 ≤ L2 or [J2 , L2 ] = 1. In the latter case, r1 r2 = r 2 r 1 has order 2. Thus we may assume that r2 ∈ J2 ≤ L2 . Hence, r 1 , r 2 ≤ WL2 . By Lemma 5.5 and [III17 , Table 17.1], WL2 is a Weyl group, and not of type G2 . Thus r 1 r 2 has order at most 4. The lemma is proved.  This enables us to prove: Corollary 5.10. O2 (W1 )Z(W1 )/Z(W1 ) is a 3-group. Proof. Suppose not. Then, for some prime p1 > 3, there is a Sylow p1 subgroup of O2 (W1 ) which is normalized but not centralized by some reflection r1 ∈ W1 . Hence there exists w ∈ O2 (W1 ) of order p1 with w inverted by r1 . By  Lemma 5.9, p1 = 3, a contradiction.

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We next prove Lemma 5.11. Let r i ∈ R, i = 1, 2, 3, and set W12 = r 1 , r 2 and W123 = W12 , r3 . Assume that W12 ≤ WL and W12 ∼ = W (A2 ) or W (C2 ). If q > 2, then one of the following holds: (a) W123 ∼ = W (A1 ) × W12 or W12 ; or (b) W123 ∼ = W (A3 ) or W (C3 ). In particular, a Sylow 3-subgroup of W123 has order at most 3. Proof. We may argue as in Lemma 5.9, but now with r3 = r w , J3 = J w , B3 = CB (r3 ), and D3 := D ∩ B3 . As r 1 , r 2 ≤ WL , we have that D3 = CD (W123 ) = 1. Now let L3 be the pumpup of L in CG (D3 ). As in the proof of Lemma 5.9, we conclude that either [J3 , L3 ] = 1 or J3 ≤ L3 . In the former case, since r3 inverts B∩J3 ≤ Z(J3 ), we see that W123 ∼ = W (A1 )×W12 . So, we may assume that J3 ≤ L3 , whence W123 ≤ WL3 . Now the lemma follows from Lemma 5.5 and [III17 , 1.2].  Next, to help analyze the normal structure of W , we record an elementary lemma about the structure of B as Fp -vector space and W1 -module. Set B0 = [B ∩ K, WK ] = [B, WK ]. The equality holds since [B, WK ] = [B, WK , WK ] ≤ [B ∩ K, WK ] (see [IG , 4.3(i)]). By [III17 , 2.11ad], B0 has a unique nontrivial WK -composition factor, which is absolutely irreducible of codimension at most 1 in B0 . Moreover, either B0 is irreducible or B0 is absolutely indecomposable with 1-dimensional socle, as WK -module. Lemma 5.12. Let S be the socle of B as Fp [W1 ]-module. Then the following conditions hold: (a) Whether regarded as Fp [W1 ]-module or Fp [W1+ ]-module, B is absolutely indecomposable, B0 ≤ S, and S is absolutely irreducible; (b) dim(B/B0 ) ≤ 2;  (c) If dim(B/B0 ) = 2, then either p divides dim(B) or p = 3, K ∼ = E6q (q), and dim(B) = 7. Proof. By Lemma 5.7, WK < W1 . As both WK and W1 are generated by + | = |W1 : WK | > 1. Suppose first that reflections, W1 = W1+ WK and |W1+ : WK (5H)

+ B = x ⊕ B0 as WK -module.

+ ]-submodule of B. By Then by [III17 , 2.11d], B0 is an absolutely irreducible Fp [WK + Lemma 5.7, x is not W1 -invariant, or W1 -invariant as W1 = WK W1+ . Therefore either S = B or S = B0 , and in either case, B is absolutely indecomposable. Therefore we may assume that (5H) fails, whence by [III17 , 2.11e], p divides |Outdiag(K)|, so that one of the following cases holds:

(5I)

 (1) K/Z(K) ∼ = Lnq (q), and p divides n;  (2) K/Z(K) ∼ = E6q (q) and p = 3.

In particular, as p is odd and K/Op (K) ∈ Gp , we have n ≥ 5 if K/Z(K) ∼ = Ln (q). The WK -composition factors in B are B0 and one or two trivial modules, with dim(B0 ) = n − 2 or 5 in the respective cases of (5I). Thus, (c) holds.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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+ If x ∈ B0 , then B0 and B/ x are absolutely indecomposable WK -modules by [III17 , 2.11acd], with B0 of codimension at most 1, and with B0 / x absolutely + irreducible for both WK and WK . Consequently B is absolutely indecomposable + for W1 and W1 . Again, x is not W1+ -invariant, so B0 ≤ S and (a) and (b) hold. Finally, suppose that x ∈ B0 . Then as (5H) fails and B0 = [B, WK ], B = + x ⊕ B0∗ as WK -module, with B0∗ an absolutely indecomposable WK -module and with B0 of codimension 1 in B0∗ , by [III17 , 2.11c]. Suppose that some W1+ -invariant irreducible summand S1 of S has dimension + -module, whence S1 = x , a 1 or 2. Since n ≥ 5 in (5I1), S1 is a trivial WK contradiction as usual. Hence in all cases S is irreducible and contains B0 . Since + -module, the irredim(B0 ) > dim(S)/2 and B0 is absolutely irreducible as WK + ducibility of S implies absolute irreducibility, as W1 -module. It follows that B is  an absolutely indecomposable W1+ -module, completing the proof.

Next we establish that, in most cases, O2 (W ) acts as scalar matrices on B. Lemma 5.13. Suppose that the socle S of B as W1 -module has codimension at most 1 in B. Then the following conclusions hold: (a) CA (W1 ) = CA (W1+ ) = Z(A) acts as scalars on B; and (b) O2 (W ) ≤ Z(A). Proof. To prove (a) it is enough to show that CA (W1+ ) acts as scalars on B. But this follows directly from absolute indecomposability (Lemma 5.12a) and [III8 , 5.3], with the roles of X and E there played by A and W1+ here. By (a), in order to prove (b) it suffices to show that [W1 , O2 (W )] = 1, so we assume by way of contradiction that (5J)

Y := [W1 , O2 (W )] = 1.

For any W1 -invariant subgroup X ≤ O2 (W ), [X, W1 ] = [X, r] | r ∈ R is generated by elements of odd order inverted by elements of R, i.e., by 3-elements, thanks to Lemma 5.9. Thus O{2,3} (W ) ≤ CO2 (W ) (W1 ) ≤ Z(W ), by (a). Write F (O2 (W )/O2 (Z(W ))) = F/O2 (Z(W )); it follows that F = O3 (W )O2 (Z(W )), and F ∗ (O2 (W )/F ) ≤ O3 (O2 (W )/F ) is centralized by W1 . As W1 is generated by R, [IG , 3.17(ii)] implies that Y = [W1 , O2 (W )] ≤ O3 (W ) ∩ W1 ≤ O3 (W1 ) and then by [IA , 4.3(i)], (5K)

Y = [W1 , Y ] = [W1 , O3 (W1 )].

Choose r1 ∈ R inverting an element g ∈ O3 (W1 ) of order 3. Replacing r1 by a W1 -conjugate we may assume by Lemma 5.8 that r 1 ∈ WL . Then as WL is the Weyl group of an indecomposable root system, there is r2 ∈ WL with W12 := r 1 , r2 of type A2 or C2 . Set r3 = r1 g. Then W123 := r1 , r2 , r3 ≤ W1 . So g ∈ O3 (W1 ) ∩ W123 ≤ O3 (W123 ). If q > 2, then as O3 (W123 ) = 1, Lemma 5.11 gives W123 = W12 ∼ = W (A2 ) or W123 ∼ = W (A1 ) × W12 ∼ = W (A1 ) × W (A2 ). Hence g ∈ W12 ∩ O2 (W1 ) ≤ O2 (W1 ∩ K) = O2 (WK ). But by [III17 , 2.7], O2 (WK ) = 1, a contradiction, proving (b) when q > 2. Now suppose that q = 2, so that p = 3. We still have (5J) and (5K). Moreover, as p = 3, Y centralizes the socle S. As Y = 1, we must have S < B, so by hypothesis |B/S| = 3. In particular, CD (Y ) = 1. Choose d ∈ CD (Y )# , if possible d = x. Suppose that d ∈ De . Then Y normalizes Ld , and as A normalizes

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Y , we have [Y, WLd ] ≤ Y ∩ W Ld ≤ O2 (WLd ) = 1 by [III17 , 2.7]. In particular, [Y, WL ] = 1. As W1 = (WL )A and Y is A-invariant, [Y, W1 ] = 1, a contradiction as [Y, W1 ] = Y = 1. Therefore, d ∈ De , so Ld = L. In particular, d = x, and so [Y, x] = 1. Since |B/S| = 3 and [S, Y ] = 1, it follows that [Y, x] = [Y, B] is Ainvariant, so S = [Y, x]. Again as |B/S| = 3 and AutB (L) is noncyclic, [S, L] = 1. Let Y be the preimage of Y in NG (B). Then [Y , x] = S, and Y normalizes L = Ld since B ∩ L ≥ [NL (B), B] ≤ Z(L) (Lemma 5.6) and Y acts quadratically on B. But x centralizes L, so S = [Y , x] centralizes L, a contradiction completing the proof.  We end this section with an important lemma for the case O2 (W1 ) = 1. Note that conditions (a)–(d) are satisfied if W2 = W1 , or if W2 = W . Lemma 5.14. Let W2 ≤ W satisfy the following conditions: (a) W2 contains either W1 or a Sylow 2-subgroup of W ; (b) Any W (A2 )-subgroup of WL generated by reflections is W -conjugate to a subgroup of W2 ; (c) W2 is generated by reflections; (d) W2 acts indecomposably on B; and (e) W2 has a noncyclic normal elementary abelian 2-subgroup E. Then W2 ≡B W (Dn ) or W (Cn ), where n = mp (B). Proof. By (c), (d), (e), and [III17 , 1.4], W2 = HΣ, where Σ ∼ = Σn permutes a basis B = {b1 , . . . , bn } of B naturally, and H stabilizes each Bi := bi , i = 1, . . . , n. It is now clear that W2 is irreducible on B. To complete the proof the main step is to locate all reflections in W2 . Now, W2 is isomorphic to a subgroup of M , the group of monomial matrices in GL(B) with respect to the frame {B1 , . . . , Bn }, and Σ stabilizes {b1 , . . . , bn }. Moreover, M = H ∗ Σ∗ with H ∗ the group of diagonal matrices and Σ∗ ∼ = Σn stabilizing {b1 , . . . , bn }. Then under the isomorphism, the image of H lies in H ∗ and Σ is mapped onto Σ∗ . Let F = Ω1 (O2 (H ∗ )). Then F Σ∗ ∼ = W (Cn ). If n is odd, then F = Z ⊕ V with Z = Z(M ) ∩ F and V irreducible as F2 [Σ∗ ]-module. If n is even, then F is indecomposable as F2 [Σ∗ ]-module and the only non-central proper submodule has codimension 1 and contains Z(M ) ∩ F . In any case we may assume that E is the full preimage of F in W2 , and we have that EΣ ∼ = W (Dn ) or W (Cn ). It therefore remains to show that H = E. Notice that if n is even, then E contains the −1 element, which generates the socle of the natural F2 Σ-permutation module. Set 2 = W2 /H. W We first argue that E = H in a special case, namely when ∼ W (C3 ) and WL = ∼ W (C2 ), with n = 4. (5L) WK = Let wK ∈ WK be the −1-element, and w0 the −1-element of Aut(B), which, as noted above, lies in W2 . By (a), we may assume, replacing W2 by an appropriate conjugate if necessary, that wK ∈ W2 . Let r0 be the reflection r0 := w0 wK ∈ W2 , K contains Σ3 and so with center x . Then [WK , r0 ] = 1. As WK ∼ = W (C3 ), W +  and r0 ∈ CW 2 (WK ) = 1. That is, r0 ∈ H, whence r0 ∈ E. In particular, E ≤ W so E ∼ = F . Moreover, r0 normalizes each Bi and so we may assume that x = B4 .

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Let r ∈ W2 − H be a reflection. We claim that the only reflections in the coset 2 must be a transposition, so without loss we may assume rH lie in rE. First, r ∈ W that r = r1 . Let h ∈ H and suppose that r  := r1 h is also a reflection. Since r  interchanges B1 and B2 , it centralizes B3 and B4 , and in particular centralizes x. Likewise x is centralized by r2 and hence by W123 := r1 , r2 , r  = r1 , r2 , h , with h ∈ W123 ∩ H. Hence W123 embeds in the reflection subgroup of AutAut(K) (B ∩ K), which by [III17 , 2.21] is WK ∼ = W (C3 ). Hence W123 ∩ H is an elementary abelian F 2-group, so h2 = 1. As E ∼ = , h ∈ E, as claimed. 2 , we Noting again that every reflection in W2 −H maps on a transposition in W  set W2 = W/E and conclude by the previous paragraph that the nontrivial images 2 of reflections in W2 are simply r 1 and its Σ-conjugates.

in W As these images 

 generate W2 , W2 = Σ, and so W2 = EΣ. Thus, when (5L) holds, the conclusion of the lemma holds. Now suppose that (5L) fails. As a result there are reflections s1 , s2 ∈ WL such that W12 := s1 , s2 ∼ = W (A2 ), as an examination of Table 15.2 and [III17 , Table 17.1] shows. Replacing W2 by a suitable conjugate, we may assume that W12 ≤ W2 , by hypothesis (b). We claim that W12 ∩ O2 (W2 ) = 1. If W2 ≥ W1 , then by (a), [O2 (W2 ), F ∗ (W )] = [O2 (W2 ), O2 (W )] = 1, so O2 (W2 ) = 1. If W2 ≥ W1 , then W2 ≥ WK . By [III17 , 1.1], s1 and s2 are fundamental reflections in WK with respect to some positive system, so there is s ∈ WK ≤ W2 such that s1 , s2 , s ∼ = W (A3 ) or W (C3 ). In either case, O2 (s1 , s2 , s ) = 1, proving our claim. Note now that if p = 3, then M ∼ = W (Cn ), and so W2 ∼ = W (Cn ) or W (Dn ), as desired. So assume that p > 3, whence q > 2. Let s3 ∈ s1 H be any reflection. Writing s3 = s1 h we have h ∈ H so h ∈ W123 ∩ H where we now have put W123 := s1 , s2 , s3 . By Lemma 5.11, W123 ∼ = W (A2 ), W (A3 ), W (C3 ), or W (A2 ) × W (A1 ). In any event W123 ∩ H is an elementary abelian 2-group and so h ∈ E. Passing 2 = W2 /E, we see again that all reflections in W2 have their images in again to W

Σ. Therefore again E = H, and the proof is complete.  We get an immediate corollary: Corollary 5.15. Let X = W1 or W . If X has a noncyclic normal elementary abelian 2-subgroup, then X ≡B W (Cn ) or W (Dn ), where n = mp (B).

6. The Case mp (B) = 4 In this section we study W assuming that mp (B) = 4. Notice that this hypothesis implies that mp (AutB (K)) = 3, B ≤ K x , and one of the following holds (see [IA , 4.10.3a] and Lemma 5.4): −q −q (1) K ∼ = Sp6 (q), Ω− 8 (q), L6 (q), or L7 (q);  q (6A) (2) K/Z(K) ∼ = L4 (q); or  (3) p = 5 and K/Z(K) ∼ = L5q (q). Our goal in this section is the following result. Proposition 6.1. Assume that mp (B) = 4. Then one of the following conclusions holds: ∼ Lq (q), F ∗ (W ) = E(W ) ∼ (a) K/Z(K) = = A5 , and W ≡B W (A4 ); 4

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(b) p = 3, K/Z(K) ∼ = L4q (q), F ∗ (W ) = Z(W ) × E(W ) with |Z(W )| ≤ 2 and E(W ) ∼ = A6 , and B ∈ Syl3 (CG (B)); moreover, W = W ∗ or W ∗ × Z2 with ∗ W ≡B W (A5 ); (c) W ≡B W (D4 ) or W (C4 ); or (d) W ≡B W (F4 ). 

The following corollary is immediate. Corollary 6.2. If K/Z(K) ∼ = L5q (q), then mp (B) = 5. 

Since K/Z(K) is as in (6A), it follows from [III17 , Table 17.1, 2.11d] that

(6B)

(1) WK ∼ = W (A3 ) or W (C3 ), or p = 5 and WK ∼ = W (A4 ) ∼ = Σ5 ; and (2) Either B ∩ K is an absolutely irreducible module for WK of  dimension 3, or p = 5, x ∈ K ∼ = SL5q (q) and B is an absolutely indecomposable WK -module with B/ x absolutely irreducible of dimension 3.

We first prove Lemma 6.3. The following conditions hold: (a) For all z ∈ Z(W1 ) ∪ Z(W1+ ), z 4 = ±1; (b) Z(W ) acts as scalars on B; (c) For W + = W ∩ SL(B), Z(W + ) is cyclic of order dividing 4; (d) The socle S of B as W1 -module is an absolutely irreducible module for W + ; and. (e) O2 (W ) = 1. Proof. Recall ((5F), (5G)) that W1 = R  W and WK ≤ W1 . With (6B) and Lemma 5.12, the socle S of B as W1 -module is absolutely irreducible of codimension at most 1 in B; and it is absolutely irreducible as well for W1+ . This implies (d). By Lemma 5.13, (b) holds and O2 (W )(Z(W1 ) ∪ Z(W1+ )) ⊆ Z(W ). Note that W1 is generated by reflections and hence W1 ≤ SL± (B). In particular, if z ∈ Z(W1 ), then z 4 = ±1, so (a) holds, and (e) is an immediate consequence. Likewise Z(W + ) ≤ CW (W1+ ) = Z(W ) by Lemma 5.13 and so (c) follows from (b), completing the proof.  Lemma 6.4. Suppose that E is a nontrivial product of simple components of E(W ). Then CW (E) = Z(W ). Also, either E ∼ = A5 , or E ∼ = A6 with p = 3. Proof. Now, E ≤ W + , and E ∩ Z(W ) = 1 by assumption. But m2 (W + ) ≤ m2 (SL4 (p)) = 3, so it follows that E char W , with m2 (E) = 2. Hence by [IA , Prop. 5.6.3], either E ∼ = L2 (q1 ), L3 (q1 ), or U3 (q1 ) for some odd q1 , or E ∼ = A7 , M11 , or U3 (4). If E ∼ = A6 , A7 , M11 , or U3 (4), then by [III17 , 5.3], E cannot be a subgroup of GL4 (p), unless E acts absolutely irreducibly, p = 3 and E ∼ = A6 , i.e., the second conclusion of the lemma holds. Hence we may exclude these cases from further consideration, so assume that ∼ L2 (q1 ) or L (q1 ), q1 odd,  = ±1. (6C) E= 3

∼ A4 . Then as mp (B) = 4, B = [B, O2 (H)] × Choose H ≤ E with H = CB (O2 (H)), of dimensions 3 and 1, with both factors invariant under and absolutely irreducible for H. Consequently [III8 , 5.2] is applicable with H and E here playing

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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the roles of A and H there, and we conclude that either E is completely reducible on B, with irreducibles of dimensions 1 and 3, or (6D)

CW (E) = Z(W ) acts as scalars on B.

But E  W , so complete reducibility of E would yield the contradiction that W is decomposable on B. Hence (6D) holds. As L2 (q1 ) contains a Frobenius group with kernel of order q1 and complement of order q12−1 , either q1 is a power of p or q12−1 ≤ 4, i.e, q1 ≤ 9, by [IG , 9.12]. As L2 (9) has been excluded and L2 (5) is one of the desired conclusions, the case E∼ = L2 (q1 ) only needs to be considered for q1 = 7 and for q1 a power of p. We choose a reflection r ∈ R ⊆ W1 . If r ∈ E × CW1 (E), then det(r) = 1, a contradiction. Hence r induces a noninner automorphism on E. Suppose that E ∼ = L2 (q1 ). If r induces a field automorphism on E, then r normalizes a Sylow p-subgroup P of E and inverts some g ∈ P # . But then g = r(rg) has order p, whence p = 3 by Lemma 5.9. Thus E ∼ = L2 (32n ), n ≥ 1. As one of the ∼ desired conclusions is E = A6 and p = 3, we may assume that n ≥ 2. Now r also normalizes some complement H to P in NE (P ), inverting a cyclic subgroup of order 3n +1 2 , whose odd share, as n ≥ 2, is at least 5, contradicting Lemma 5.9. Hence with [IA , 4.9.1, 2.5.12], if E ∼ = L2 (q1 ), then E r ∼ = P GL2 (q1 ). If q1 = 7, then r inverts a Sylow 7-subgroup of E, again contradicting Lemma 5.9. So q1 = pn and r inverts a cyclic torus in E r of odd order q12±1 , according as q1 ≡ ±1 (mod 4), by [III17 , 11.7a]. If g is a cyclic generator of this torus, then r is conjugate to rg, whence by Lemma 5.9, q12±1 ≤ 3 or p = 3. In the first case, q1 ≤ 7, contrary to what we saw above. In the second case, |GL4 (3)| is divisible by |E| and hence by q1 (q1 ± 1)/2, so as q1 > 9, we must have q1 = 33 . But this yields that 7 divides |E|, again a contradiction as |E| divides |GL4 (3)|. Thus, the possibilities in (6C) are reduced to (6E)

E∼ = L3 (q1 ).

Now r induces a graph, field, or graph-field automorphism on K. By [III17 , 6.19], r inverts an element g ∈ K of odd prime order > 3, contradicting Lemma 5.9 and completing the proof.  Lemma 6.5. If E(W ) ∼ = A5 in Lemma 6.4, then Proposition 6.1a holds. Proof. By Lemma 6.4, CW (E) = Z(W ) in this case. As WK ∩ Z(W ) = 1, and WK  W , by Lemma 5.7, WK is isomorphic to a proper subgroup of Σ5 , whence WK ∼ = W (C3 ) or W (A4 ). So, K/Z(K) ∼ = L4 (q). It remains to prove that Z(W ) = 1. Otherwise, as W is generated by reflections, we would have Z(W ) = z inverting B elementwise, W = z × Σ with Σ ∼ = Σ5 containing a reflection. But then for any involution t ∈ Σ, |[B, t]| = p or p2 , so there are no reflections in zΣ. Hence W = Σ, a contradiction. The lemma follows.  We next argue that in the case when (6F)

p = 3 and E(W ) ∼ = A6 ,

∼ L (q), |Z(W )| ≤ 2, and B ∈ Syl (CG (B)). [Our ultimate we must have K/Z(K) = 3 4 goal in this case will be to prove that G0 ∼ = L6 (q) with 3 dividing q − , but 9 not dividing q − .] We proceed towards our intermediate goal in a sequence of lemmas.

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Lemma 6.6. Assume (6F). Then B ∈ Syl3 (CG (B)), |Z(W )| ≤ 2, and W ∼ = Σ6 × Z(W ). Proof. Let B1 ∈ Syl3 (CG (B)). Then B1 ≤ CG (x) and B = Ω1 (B1 ). No element g ∈ CG (x) of order 3 induces a nontrivial field automorphism on K, since that would imply that m3 (CG (x, g)) ≥ m3 (x, g ) + m3 (CK (g)) = 5, contradicting our choice that B ∈ E3∗ (CG (x)). If B1 /B1 ∩ K were noncyclic, however, then as m3 (C(x, K)) = 1, there would exist g ∈ B1 − KC(x, K) with g 3 ∈ K, implying the existence of a field automorphism of order 3 in gK, contradiction. Therefore B1 /B1 ∩K is cyclic and in particular, [B1 , B1 ] ≤ B1 ∩K. But NG (B) acts irreducibly on B, so NG (B1 ) does as well by a Frattini argument. It follows that [B1 , B1 ] = 1 and then that B1 is homocyclic abelian, with B1 /Φ(B1 ) ∼ = B as E(W )-module. Then by [III17 , 5.2], B1 = B, as claimed. As W is absolutely irreducible on B and p = 3, we have Z(W ) ≤ 2. Finally we argue that W ∼ = Σ6 × Z(W ). Let r ∈ W be a reflection. Then r ∈ E(W )Z(W ) since det r = −1, and CW (r) contains no element of order 5 as 5 does not divide |GL3 (3)|. The only possibility is E(W ) r ∼ = Σ6 , completing the proof.   Still assuming (6F), we assume that K/Z(K) ∼  L4q (q) and aim for a contra= diction. Thus (6A1) holds. Let N be the preimage of W in NG (B) and let P ∈ Syl3 (N ). As E(W ) = O 2 (W ), |P | = 36 and the action of E(W ) (and hence, of P ) on B is determined up to quasiequivalence, by [III17 , 5.1]. In particular, we may choose a basis {b0 , b1 , b2 , b3 } for B with x = b0 and bi ∈ K, i ≥ 1, such that Z(P ) =: β = b1 b2 b3 . Moreover, we may assume that {b1 , b2 , b3 } is permuted transitively by some ρ ∈ P ∩ K − B. We define    SL3q (q) if K ∼ = Sp6 (q) or Ω− 8 (q); I1 := O 2 (CK (β)) ∼ = −q − 2 ∼ SL3 (q ) if K = A5 (q) or A6 q (q).  With [IA , 4.8.2], we see that CK (β) ≥ I1 b1 with I1 b1 / β ∼ = P GL3q (q) or P GL3 (q 2 ), respectively. Now b0 , β = Z(CP (x)) with |P : CP (x)| = 3, so b0 , β  P . Hence I1 = Lo3 (CG (β, b0 )) is P -invariant. Set R1 = P ∩ I1 ∈ Syl3 (I1 ), R2 = CP (I1 ), and R = R1 ∗ R2 . Thus, R1 and R2 are extraspecial of order 33 , and R1 has exponent 3. Indeed we may choose notation so that R1 = b−1 1 b2 , ρ and R2 = b0 , σ , where the images of ρ and σ in W are commuting elements of order 3 in E(W ) ∼ = A6 .

Lemma 6.7. Assume (6F). Then the following conclusions hold: (a) R1 and R2 are normal in P ; (b) R is characteristic in P ; (c) R2 ∼ = R1 ; and (d) P ∈ Syl3 (G) and B = J(P ). Proof. Since I1 is P -invariant, (a) holds. Now, suppose that Q = α(R) for some α ∈ Aut(P ). As β = Z(R) = Z(P ), β = Z(Q) and Q ∼ = E34 in P = P/ β . If Q = R, then b1 ∈ P − R acts as a transvection on R. But b1 acts non-trivially on both R1 and R2 , a contradiction proving (b). Indeed, if b1 centralized R2 , then some element of b1 b0 would centralize R2 , whence m3 (CB (σ)) = 3; but every element of E(W ) of order 3 lies in an A4 -subgroup of W and so σ has a free summand on B, contradiction.

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For the same reason no nontrivial subgroup of P/B acts quadratically on B, so by [IG , 26.8], B = J(P ) char P . As E(W )  A with E(W ) absolutely irreducible on B, A/Z(A) embeds in Aut(E(W )) so P ∈ Syl3 (NG (B)). This implies (d). Finally, let C ≤ NG (B) ∩ NG (P ) be cyclic with image Z4 in E(W ). Then C acts irreducibly both on P/B ∼ = R1 . = R/R∩B and on R∩B/ β . Suppose that R2 ∼ Then Ω1 (R) = b0 × R1 , and C leaves Z(Ω1 (R)) = b0 , β invariant, contradicting the irreducible action on R ∩ B/ β . Hence, R2 ∼ = R1 , as claimed, completing the proof.  Let C be as in the preceding paragraph, and let c be a generator of C. Thus  R1c = R1 . Also let H1 be the subnormal closure in CG (β) of O 3 (I1 ). 

Lemma 6.8. Assume (6F). Then O 3 (I1 )O3 (CG (β))   CG (β). Either L3 (CG (β))/O3 (L3 (CG (β))) has just two 3-components, interchanged by c, or c K∼ = Ω− 8 (2) and R1 R1 O3 (CG (β))  CG (β). Proof. Note that by Lemma 6.7d, |CG (β)|3 = |G|3 = 36 and m3 (CG (β)) = m3 (P ) = 4. Assume first that K ∼  Ω− = 8 (2), so that I1 = E(I1 ). If H1 is a diagonal pumpup of I1 , then H1 involves I1 × I1 × I1 so |CG (β)|3 ≥ 39 , a contradiction. By [III11 , 13.29], if H1 is a vertical pumpup of I1 , then b0 induces a field automorphism on H1 /O3 (H1 ). Thus P ∩ H1 has order 35 and rank 2, forcing |P | > 36 , contradiction. Hence H1 is a trivial pumpup of I1 . As c ∈ NG (β ) but c ∈ NG (R1 ), it follows that I1c = I1 . Besides these two, no further 3-component of CG (β) exists, inasmuch as |P | = 36 and Z(P ) = β . Thus the lemma holds in this case. ∼ Now suppose that K ∼ = Ω− 8 (2), so that I1 = SU3 (2). By solvable L3 -balance, 3 either R1 = O (I1 ) ≤ H1 ≤ O3 3 (CG (β)), or H1 is a 3-component of CG (β). In the former case, the irreducible actions of C on R/R ∩ B and R ∩ B/ β force R = R1 R1c ≤ O3 3 (CG (β)) and then by Lemma 6.7b, RO3 (CG (β))  CG (β). In the latter case, H1 /O3 (H1 ) ∼ = SU3 (8) by [III17 , 10.23], and b0 induces a nontrivial field automorphism, leading to a contradiction as in the previous paragraph. The proof is complete.  Lemma 6.9. Assume in Lemma 6.4 that p = 3 and E(W ) ∼ = A6 . Then Proposition 6.1b holds. Proof. Suppose false and continue the above argument. It is clear from our choice of acceptable subterminal (x, K)-pair [III12 , Def. 1.15] that there is a fourgroup V  WK acting freely on B such that y is V -invariant and x = CB (V ). Now NG (B) controls G-fusion in J(P ) = B. Therefore by [III11 , 22.2], xy is 3central in G, and hence conjugate to β. By Lemma 6.8, Lxy ∼ = I1 . In particular, − ∼  Ω8 (2). However, given the isomorphism type of K, we I1 is nonsolvable so K = see from Table 15.2 that neither L nor Lu can be isomorphic to I1 for any u ∈ D# . This contradiction completes the proof of the lemma.  In view of Lemmas 6.4, 6.5, and 6.9, we may assume for the rest of the proof of Proposition 6.1 that (6G)

E(W ) has no simple component.

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Lemma 6.10. F ∗ (W ) = O2 (W ). Proof. Suppose not. Then E(W ) has a quasisimple component E with |Z(E)| even, by Lemma 6.3e and (6G). Also m2 (E) ≤ m2 (SL4 (p)) = 3.

(6H) Moreover, by [III8 , 5.4], (6I)

|AutE (H)| ≤ 4 for all subgroups H ≤ E of prime order s = p.

We first show that (6J)

m2 (E) = 1.

First, using [IA , Table 5.6.1] and (6H), we see that if E ∈ Alt∪Spor, then E/Z(E) ∼ = An , n ≤ 7, in which cases (6J) holds, or E ∼ = 2J2 . But in the last case, any element g ∈ E of order 7 is rational, so |AutE (g )| = 6. By (6I), p = 7. As J2 contains E52 but GL4 (7) does not [IA , 4.10.3a], we have a contradiction. Thus (6J) holds if E ∈ Chev. Assume then that E ∈ Chev(s) − Alt for some prime s, and choose s to be odd if possible. The case E ∈ Chev(2) is then ruled out by [III17 , 5.6], so s is odd. By 1  2 G2 (3 2 ) , so E ∈ Lie(s). [IA , 6.1.4], E ∼ = Suppose that m2 (E) > 1. Then for any involution z ∈ E − Z(E), CE (z) ≤ CAut(B) (z) ∼ = GL2 (p) × GL2 (p), so we see that together, the components and solvable components of CE (z) are at most two in number; if there are two, then they are disjoint; and their isomorphism types are among SL2 (p), SL2 (5) and SL2 (3). It follows from [IA , Tables 4.5.1, 4.5.2] that E ∼ = Sp4 (r) or SL± 4 (r) for some r ∈ {3, 5, p}. Then E contains an extraspecial r-group of order r 3 . As E ≤ GL4 (p), r = 3 or r = p. Suppose first that r = p. As |SU4 (p)| does not divide |SL4 (p)|, we have E ∼ = SL4 (p) or Sp4 (p), acting transitively on the 1-subspaces of B, with point stabilizer a parabolic subgroup of E. Therefore F ∗ (NW (x )) is a p-group. However, NW (x ) has a normal subgroup isomorphic to W (A3 ), W (C3 ) or W (A4 ), a contradiction. We conclude that E ∼ = Sp4 (3) or SL± 4 (3), and p = 3. Then [III17 , 5.5] yields a contradiction. We have proved (6J). Therefore E ∼ = 2A7 . With = SL2 (sa ) for some odd prime s and some a, or E ∼ [IA , 2.5.12], Out(E) is abelian. Suppose that E acts irreducibly on B. Then CW (E) is cyclic. In particular, E  W and by the previous paragraph, W/ECW (E) is abelian. As WK ≥ Σ0 ∼ = Σ4 , we conclude that A0 ≤ ECW (E), where A0 = [Σ0 , Σ0 ]. But then V = O2 (A0 ) ≤ E, a contradiction. Hence E acts reducibly on B. Choose an E-invariant subgroup BE of B of minimal order on which Z(E) acts nontrivially. Thus |BE | ≤ p3 . Moreover, E normalizes [BE , Z(E)] so BE = [BE , Z(E)]. But Z(E) ≤ [E, E] so |BE | = p2 . Consequently E, as a subgroup of GL2 (p), satisfies E ∼ = SL2 (p) or SL2 (5). In particular, p > 3. Since E exists, F ∗ (W ) > O2 (W ), and so by Lemma 5.14, every normal elementary abelian 2-subgroup of W is cyclic. Therefore Z(E) = Ω1 (O2 (E(W )))  W so Z(E) inverts B as B is indecomposable for W . Now let r1 be a reflection in WK . Suppose that E r1 = E. Then [E, E r1 ] = 1. There exists g ∈ E of odd prime order m > 3. Then r1 inverts g −1 g r1 of order

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m, contrary to Lemma 5.9. Hence r1 normalizes E. If r1 centralizes E, then E leaves [B, r1 ] invariant, and hence E centralizes [B, r1 ], contrary to the fact that Z(E) inverts B. Thus [E, r1 ] = E. If E ∼ = SL2 (sa ), then by [III11 , 1.17], r1 is E-conjugate to r1 z, where z = Z(E). But z inverts B so r1 z is not a reflection, contradiction. Likewise if E ∼ = 2A7 , then there exists an r1 -invariant subgroup E1 ≤ E such that E1 ∼ = SL2 (9) and E1 = [E1 , r1 ], and we reach the same contradiction. The lemma is proved.  Lemma 6.11. Suppose that F ∗ (W ) = O2 (W ). Then one of the following holds: (a) W ≡B W (D4 ) or W (C4 ); or (b) W ≡B W (F4 ). Proof. Suppose false. By Corollary 5.15, W contains no non-cyclic normal abelian 2-subgroup. By P. Hall’s theorem [IG , 10.3], R := O2 (W ) is of symplectic type. As WK ≤ W and WK contains a subgroup isomorphic to Σ4 , R cannot have width 1; but as mp (B) = 4, R cannot have width greater than 2, indeed R ∼ = Q8 ∗ Q8 ∗ Z(R) or Q8 ∗ D8 . As R is then absolutely irreducible on B, we have F ∗ (W ) = RZ(W ) = R0 ∗ Z(W ) with R0 ∼ = Q8 ∗ Q8 or Q8 ∗ D8 . Up to quasiequivalence, F ∗ (W ) has a unique faithful representation in GL4 (p), so F ∗ (W ) is uniquely determined in Aut(B) ∼ = GL4 (p) up to conjugacy. Suppose first that R0  W with R0 ∼ = Q8 ∗ D8 . Then Out(R0 ) ∼ = A5 . = Σ5 and O 2 (W ) ≤ O 2 (NGL4 (p) (R0 )) =: N0 with N0 /R0 ∼ Now N0 is conjugate to a subgroup of Sp4 (p), whence m2 (N0 ) = 2. However, WK contains Σ0 ∼ = Σ4 and so O 2 (Σ0 ) =: A0 ≤ N0 with A0 ∼ = A4 and A0 ∩ Z(N0 ) = 1, a contradiction. Suppose next that ∼ Q2 = ∼ Q8 . R0  W with R0 = Q1 ∗ Q2 , and Q1 = ∼ E32 . Again let A0 = O 2 (Σ0 ). Then Thus, O 2 (W/R0 ) embeds in O 2 (Aut(R0 )) = 2 A0 ≤ O (W ), whence O2 (A0 ) ≤ R0 . Let g ∈ A0 be of order 3. As [O2 (A0 ), g] = O2 (A0 ) ∼ = E22 , g acts on R0 with CR0 (g) = Z(R0 ). If g ∈ Syl3 (W ), then O2 (W ) g  W , so the 2-group W/O2 (W ) g permutes the three g-invariant E23 subgroups of R0 by conjugation. Hence W has a normal E23 -subgroup, contradiction. Therefore, |W |3 = 32 , and O 2 (W ) ∼ = SL2 (3) ∗ SL2 (3). Since R0 is absolutely irreducible on B, CA (R0 ) = Z(A) consists of scalar mappings. Thus O2 (A) = R0 O2 (Z(A)). Let t ∈ I2 (NA (Q1 ) − Z(R0 )) and write z = Z(R0 ). We claim that t ∼A tz. This is clear if t induces an inner automorphism on R0 . Otherwise, t ∈ O2 (A) so t inverts an element u of order 3, and u acts nontrivially on Qi for some i = 1, 2. Then Qi t, u ∼ = GL2 (3) so Qi t is semidihedral. Thus t ∼A tz, as claimed. In particular, t cannot be a reflection. Hence each reflection in A interchanges Q1 and Q2 , and as a result centralizes a subgroup of order 3 in O 2 (W )/R0 . But O 2 (W )/R0 ∼ = E32 , so as W is generated by reflections, |W : O2,3 (W )| > 2. As Out(R0 )/O3 (Out(R0 )) ∼ = D8 and the stabilizer of Q1 in Out(R0 )/O3 (Out(R0 )) is a four-group, the image of W in Out(R0 )/O3 (Out(R0 )) must be the other four-group. Let r ∈ W be a reflection. Then r ∈ R0 , so Qr1 = Q2 . Thus r acts freely on R0 /Z(R0 ) ∼ = O2 (W )/Z(W ) ∼ = E24 . Hence all involutions in rO2 (W ) are R0 -conjugate, modulo Z(W ). But clearly r

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is the only reflection in rZ(W ), so all reflections in rO2 (W ) are R0 -conjugate. Thus W is generated by all the reflections in NGL(B) (R0 ). As R0 is determined up to conjugacy in GL(B), so is W . But the action of W (F4 ) on the root lattice reduces modulo p to give an embedding of W (F4 ) in GL(B) as a group generated by reflections. Now (b) follows from [III17 , 1.5], contrary to our indirect assumption. With the knowledge that R0 does not exist as in the last two paragraphs, we finish the proof by arguing to a contradiction. If O2 (W/O2 (W )) = 1, then [O2,2 (W ), O2 (W )] is contained in a W -invariant Q8 ∗ D8 or Q8 ∗ Q8 subgroup of O2 (W ), contradiction. Thus, O2 (W/O2 (W )) = 1 = Sol(W/O2 (W )). Also, R2 := Ω2 (O2 (W )) ∼ = Q8 ∗ Q8 ∗ Z4 ; as Out(R2 ) ∼ = Σ6 , F ∗ (W/O2 (W )) = ∼ E(W/O2 (W )) = A5 or A6 . Any involution of A5 or A6 or P GL2 (9) inverts an element of order 5 so cannot be induced by the action of a reflection. (Three suitable conjugates of that reflection would generate a 21+4 .5 subgroup, which would then have a fixed point on B, despite acting irreducibly and nontrivially on B.) Thus, W/O2 (W ) ∼ = Σ5 or Σ6 , acting faithfully on R2 /Φ(R2 ) ∼ = E25 . Since O2 (W ) is absolutely irreducible on B, Z := Z(O2 (W )) acts as scalars on B, and by Lemma 6.3, |Z| = 4 or 8. We argue that (6K)

R2 = O2 (W ), i.e., |Z| = 4.

Otherwise, W/R2 , which is generated by involutions, is isomorphic to Σm × Z2 , m = 5 or 6. Indeed as |Z| = 8 in this case, det z = −1 for any generator z of Z. But there are no involutions in O2 (W ) − R2 in this case; every element of O2 (W ) − R2 has order 8. Now W/[W, W ] is a four-group and the distinct nontrivial subgroups Z[W, W ]/[W, W ] and W + /[W, W ] of it contain no images of reflections. Hence W is not generated by reflections, a contradiction. Thus R2 = O2 (W ).  = W/R2 ∼ Set W = Σm , m = 5 or 6. Let W12 = s1 , s2 be a W (A2 )-subgroup 12 lies in a subgroup W 2 ∼ of W , where s1 , s2 are reflections. Then W = Σ4 or Z2 × Σ4  generated by W -conjugates of s 1 , according as m = 5 or 6. We let W2 ≤ W be the 2 in W , and W0 the reflection subgroup of W2 . We shall full inverse image of W apply Lemma 5.14 to W2 . We first claim that CR2 /Z (O3 (W12 )) = 1 and W2 = W0 . By construction  2 . Let w ∈ O3 (W12 )# . If CR /Z (w) = 1, then [w, R2 ]W12 ∼ W0 = W = GL2 (3) is 2 generated by three reflections so has a nontrivial fixed point on B; but it contains Ω1 (Z), contradiction. Therefore CR2 /Z (w) = 1. As a result, since W0  W2 , W0 contains [w, R2 ] ∼ = Q8 ∗ Q8 . As no such subgroup is normal in W , W/Φ(Z) does not split over Z/Φ(Z). Hence neither does the odd-index subgroup W2 /Φ(Z). But W0 covers W2 /Z, so Z ≤ W0 and W2 = W0 , proving the claim. Let r ∈ W − R2 be any reflection. We claim that r is uniquely determined up  -conjugacy. As argued four paragraphs above, r cannot invert an element of to W  of order 5, so r is an odd involution in W ∼ W = Σm , m = 5 or 6. Hence (using the outer automorphism of Σ6 ) we may assume that r is a transposition. Our assertion therefore holds if m = 5, so assume that m = 6 and elements of both classes of  are images of reflections. This implies that all elements of outer involutions of W W of order 3 are inverted by a reflection, and hence by the previous paragraph, are fixed-point-free on R2 /Z. As W has noncyclic Sylow 3-subgroups, this is absurd, and our claim is established. Moreover, we observe that any two elements of order

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

3 inverted by r are CW r)-conjugate. It follows that any W (A2 )-subgroup of WL  ( generated by reflections is W -conjugate into W2 . As W2 ≥ R2 , W2 is irreducible on B, but [O2 (W2 ), R2 /Z(R2 )] is not cyclic, owing to the action of w. Therefore [O2 (W2 ), O2 (W2 )] is noncyclic and consequently O2 (W2 ) is not of symplectic type. By P. Hall’s theorem, W2 has a noncyclic normal elementary abelian subgroup. Now Lemma 5.14 applies to give W2 ∼ = W (D4 ) or W (C4 ). In particular, |W2 |2 ≤ 27 . However, by construction W2 /R2 contains Σ4 so |W2 |2 ≥ 29 . This contradiction completes the proof of the lemma.  Now Lemmas 6.4, 6.5, 6.9, 6.10, and 6.11 combine to complete the proof of Proposition 6.1. Corollary 6.12. If mp (B) = 4, then one of Proposition 3.4abcdehik holds. Proof. If Proposition 6.1a holds, then since W is generated by reflections, while inner involutions invert 5-elements so cannot be reflections, W ∼ = Σ5 , indeed W ≡B W (A4 ) by [III17 , 1.5], and Proposition 3.4a holds. Similarly, if Proposition 6.1b holds, then again since W is generated by reflections, W ∼ = Σ6 or Σ6 × Z2 , and W ∗ = WK E(W ) ≡B W (A5 ). Hence Proposition 3.4b holds. We thus may assume for the rest of the proof of this corollary that W ≡B W (D4 ), W (C4 ), or  W (F4 ). In any case, W  W (D4 ), so if K ∼ = L4q (q), then Proposition 3.4e holds. Otherwise, given the possible isomorphism types of K, WK ∼ = W (C3 ) (centralizing x), and so W ≡B W (C4 ) or W (F4 ). Hence one of Proposition 3.4cdhik holds with − W ∗ = W , as desired, or W ≡B W (F4 ) with K/Z(K) ∼ = L7 q (q) or Ω− 8 (q). It remains therefore to rule out these last two cases. In either case, by Lemma 5.2, Lu /Z(Lu ) ∼ = K/Z(K) for all u ∈ De . By [III17 , 3.2bc], |De | = 4 and NW (D) has two orbits on De of length 2. Suppose that − K/Z(K) ∼ = L7 q (q). Let B0 = {b1 , b2 , b3 } be the frame for B ∩ K preserved by WK , and set b4 = x. Then B = B0 ∪ {b4 } is a frame for B whose stabilizer WB in W is a copy of W (C4 ), and |W : WB | = 3. Choose w ∈ W − WB of order 3 and let B ∗ = B w and x∗ = xw . By [III17 , 3.2d] we may assume notation has been chosen so that x∗ = b1 b2 b3 b4 . Let K ∗ = Lp (CG (x∗ )) ∼ = K. For any u ∈ D# , let Ku∗ be the − product of all components of Lp (CK ∗ (u)) isomorphic to SLk q (q) for k ∈ {3, 5, 7}. (Note that given u, there can exist at most one such component by [IA , 4.8.2], so Ku∗ is quasisimple or trivial.) Then since B induces inner automorphisms on K ∗ (as it does on K), repeated use of [III17 , 9.1] yields  K ∗ = Ku∗ | u ∈ D# . Let u ∈ D# with Ku∗ = 1. If u ∈ De , then either Ku∗ ≤ L or [Ku∗ , L] = 1. In the first case, Ku∗ is a component of CL (x∗ ); since x∗ = b1 b2 b3 b4 , Ku∗ ∼ = SL2 (q 2 ), ∗ ∗ ∗ a contradiction. In the second case, [Ku , D ] = 1 where D = B ∩ L. But by the structure of W , D∗ = Dg for some g ∈ W with g stabilizing B. Hence again Ku∗ ∼ = SL2 (q 2 ), contradiction. Therefore u ∈ De , and so Ku∗ is a component of − CLu (x∗ ). This time if u = x, clearly Lp (CK (x∗ )) ∼ = SL3 q (q 2 ), while if u = y, − we may assume u = b3 , and again Lp (CLy (x∗ )) ∼ = SL3 q (q 2 ). Hence u ∈ De −  ∗ ∗ ∗ {x , y } = {v , v }, and thus K = Kv , Kv . In particular, interchanging − v and v  if necessary, we may assume that Kv∗ ∼ = SL5 q (q). But then as Kv∗

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∼ L−q (q) as an element of is a component of CLv (x∗ ), x∗ acts on Lv /Z(Lv ) = 7 order p with 2-dimensional support on the natural Lv -module. On the other hand, − L = E(CLv (D)) ∼ = SL5 q (q) is canonically embedded in Lv , and x∗ acts on L as an element of order p with 4-dimensional support on the natural L-module, which is a submodule of the natural Lv -module. This is a contradiction, and so we are reduced to ruling out the case K∼ = Ω− 8 (q). We again let B = {b1 , b2 , b3 = y , b4 = x } be a frame whose stabilizer WB in W has index 3 in W . Then WB ∼ = W (C4 ), and so b1 , b2 , b3 lie in K with 2-dimensional mutually orthogonal supports on the natural K ∼ = Ω− 6 (q)-module. Now W has two classes C1 and C2 of reflections, and the bi ’s are each inverted by a reflection from, say, class C1 . By [III17 , 3.2e], the bi ’s may be chosen so that the subgroups of B of order p inverted by a reflection from class C2 are the twelve b , 1 ≤ i < j ≤ 4. It follows that WLb3 b4 ∼ subgroups bi bj and b−1 = W (C3 )  ±1 i j so De = {x , y , x y }. The twelve subgroups of B of order p inverted by a  ±1 ±1 reflection from class C1 are the four bi and the eight subgroups b±1 1 b2 b3 b4 . Take a pair of subgroups of B of order p, one from each of these  sets of twelve. Up to conjugacy we may assume that we chose x and some v = b±1 i bj , 1 ≤ i < j ≤ 4. −q If j = 4 then Lp (CG (v, x)) = Lp (CK (v)) = Lp (CK (bi )) ∼ Ω = 6 (q), while if j < 4,  2 then Lp (CG (v, x)) = Lp (CK (v)) ≤ O (CK (v)) = H × J with H ∼ = L2 (q) and J ∼ = L2 (q 2 ). Moreover, H ∩ B is generated by a WK -conjugate of v. But we could have analyzed CG (v, x) by picking v first, and then choosing x from the other class such that Lp (CG (v, x)) ∼ = L2 (q) × L2 (q 2 ). The same reasoning then would show that B ∩ H is generated by a W -conjugate of x. It follows that x and v are W -conjugate, which is a contradiction. The proof is complete.  7. The Case AutK (B) ∼ = W (BCn ) or W (F4 ) In this section we consider the cases WK ∼ = W (Cn ) and W (F4 ), aiming to determine W . The arguments are effective, moreover, in a more general situation needed later when WK ∼ = W (An ). The essential generalization is that we work with subgroups Do , Lo , and pumpups Lod , similar to D, L, and Ld in (1A), but not necessarily arising from a pair (x, K) ∈ J∗p (G) and an acceptable subterminal (x, K) pair. Specifically, we keep the notation (1A), and work with the objects in the following setup. These include p, B ∗ = B, and q as in (1A), (3A1), and (5A). (1) n = mp (B) ≥ 4; (2) Do ≤ B with Do ∼ = Ep2 ; o p (3) L = O (E(CG (Do ))) ∈ Chev(2) and q(Lo ) = q; (4) For every d ∈ (Do )# , the pumpup Lod of Lo in CG (d) lies (7A) in E(CG (d)) and is either trivial or vertical; moreover, Lod ∈ Chev(2) with q(Lod ) = q; (5) B induces inner-diagonal automorphisms on Lod for every d ∈ (Do )#; and (6) Do = d ∈ (Do )# | Lo < Lod . If p does not divide |Outdiag(Lod )| for any d ∈ D# , then (7A5) is equivalent to: B = (B ∩ Lod ) × CB (Lod ) for every d ∈ (Do )# .

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

When (7A) holds, in addition to the subgroups A = AutG (B) and W , the reflection subgroup of A, we consider the subgroups WLo and WLod , d ∈ D# , defined as the reflection subgroups of A ∩ Lo and A ∩ Lod , respectively. We also continue to use the notation r ∈ W for a reflection which is the image of an element r ∈ NG (B). Of course in view of (1A) and Lemma 3.3a, (7A) is satisfied by Do = D, o L = L, and Lod = Ld for each d ∈ D# . Notice also that since the level q is the same as in (1A), p splits Lo and Lod for every d ∈ D# . Lemma 7.1. Assume (7A). Suppose that WLo ∼ = W (Cn−2 ), n ≥ 5, and r 1 is a long root reflection in WLo . Suppose also that for any d ∈ D# with Lo < Lod , we have WLod ∼ = W (Cn−1 ). Then r1 ∈ O2 (W ). Proof. Let r 2 = rw 1 , w ∈ W , be any W -conjugate of r 1 . By the Baer-Suzuki theorem it is enough to show that r 1 , r 2 is a 2-group. Now, CDo (r2 ) = 1 and we choose any d ∈ CD (r2 )# , and a preimage r2 of r 2 in NG (B). Then r2 ∈ NCG (d) (B). As B is a maximal elementary abelian p-group in G, B ∩Lo is noncyclic, so B ∩Lod is as well, whence r2 normalizes Lod . Thus r2 normalizes WLod . Now by [III17 , 1.32] and our hypothesis on WLod , r1 ∈ O2 (WLod ), whence r1 , r2 is a 2-group, as required.  Proposition 7.2. Assume (7A). Suppose that WLo ∼ = W (Cn−2 ) and for every d ∈ (Do )# with Lo < Lod , we have W (Lod ) ∼ = W (Cn−1 ). Then one of the following holds: (a) W ≡B W (Cn ); or (b) n = 4 and W ≡B W (F4 ). Proof. First suppose that n = 4, so that Proposition 6.1 applies. Choose d ∈ (Do )# such that Lod > Lo . Then WLod ∼ = W (C3 ), so mp (B ∩ Lod ) ≥ 3. Therefore mp (C(d, Lod )) = 1 by (7A5), as B contains every element of order p in CG (B). Given the p-rank of Lod , it follows from [IA , 4.10.3a] that Lod has untwisted Lie rank 3 or  more. Moreover, Lod is not centrally isomorphic to L4q (q), since W (C3 ) ∼  W (A3 ). =   q  L4q (q), by [III14 , Lemma 1.2]. Therefore F(Lod ) > F(L4 (q)). It follows that K ∼ =  (Note that if K ∼ = L4q (q), then q > 2 since K ∈ Gp and p divides q − q .) With this possibility for K ruled out, Proposition 6.1 implies that either (a) or (b) holds. Now suppose that n ≥ 5. There exists a long root reflection r1 ∈ WLo ∼ = W (Cn−2 ), and by (7A4) and Lemma 7.1, r 1 ∈ O2 (W ) =: R. If R has a noncyclic characteristic abelian subgroup, then (a) holds by Lemma 5.14 and the fact that O2 (W ) contains reflections. So we may assume that R is of symplectic type, by P. Hall’s theorem, and we must argue to a contradiction. Let m be the width of R. Fix d ∈ (Do )# such that WLod ∼ = W (Cn−1 ). Since n ≥ 5, certainly m > 1. As W1 is indecomposable on B and CR (B) = 1, we have CB (Ω1 (Z(R))) = 1. Therefore n = mp (B) ≡ 0 (mod 2m ), and as n ≥ 5, we have n ≥ 8. Choose any reflection r ∈ WLod − O2 (WLod ), so that r ∈ R. Then r inverts some element g ∈ W of odd order, and so |B : CB (g)| ≤ p2 . Since every R-composition factor on B has rank at least 2m , [R, g] = 1. As R is of symplectic type, [R, g] is nonabelian. Therefore there exist r1 , r2 ∈ R such that H := g, g r1 , g r2 = g, [r1 , g], [r2 , g] contains Ω1 (Z(R)). Hence six conjugates of r generate a group containing Ω1 (Z(R)). It follows that pn = |B| = |B : CB (Ω1 (Z(R)))| ≤ |B : CB (H)| ≤ p6 , a contradiction completing the proof.

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7. THE CASE AutK (B) ∼ = W (BCn ) OR W (F4 )

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We quickly deduce: Proposition 7.3. Suppose that WK ∼ = W (Cn−1 ), n ≥ 4. Then one of the following holds: (a) W ≡B W (Cn ); (b) n = 4 and W ≡B W (F4 ); or − − − (c) n = 5, K ∼ = A7 q (q), L ∼ = E6 q (q). = A5 q (q), and for some v ∈ D# , Lv ∼ − −q Proof. By Table 15.2, K ∼ (q), δ ∈ {2, 3}; WL ∼ = Cn−1 (q), Dn q (q), or A2n−δ = # W (Cn−2 ); and assuming that (c) fails, every Ld , d ∈ D , for which Ld > L satisfies WLd ∼ = WK . As remarked above, (7A) holds with D, L, and Ld in place of Do , Lo ,  and Lod . So Proposition 7.2 applies, and (a) or (b) holds, as asserted. n

Next, we consider the case WK ≡B W (F4 ). However, again we shall later need  a result in the context like (7A), for use in the case WK ∼ = Anq (q). We prove this result first. Proposition 7.4. Suppose that n = 5 and there exist d ∈ B # and a component I of E(CG (d)) such that B = d × (B ∩ I) and WI ≡B W (F4 ). Then there exist no e ∈ B # and component J of E(CG (e)) such that B = e × (B ∩ J) and WJ ≡B W (C4 ). Proof. Suppose on the contrary that such d and e exist. Note that as  mp (B) = n = 5, mp (C(d, I)) = mp (C(e, J)) = 1 so I = O p (E(CG (d))) and  J = O p (E(CG (e))). In particular, d and e are not G-conjugate. Both W (F4 ) and W (C4 ) have centers of order 2, acting as −1 on the corresponding root lattices. Let zd = Z(WI ) and ze = Z(WJ ); thus CB (zd ) = d and CB (ze ) = e . One consequence is that zd and ze are not W -conjugate. Another is that CW (zd ) ≤ NW (d ) and CW (ze ) ≤ NW (e ). Now, Qd := O2 (WI ) is absolutely irreducible on [B, WI ] = B ∩ I, and similarly for Qe := O2 (WJ ) on B ∩ J. Let Sd ∈ Syl2 (NW (d )). Then CSd (Qd ) acts as scalars on B ∩ I. It follows that Ω1 (CSd (Qd )) = z or z, i , where i inverts B. In particular, zd is the unique involution of Z(Sd )+ . It follows that Sd ∈ Syl2 (W ). The identical argument shows that CW (e) contains a Sylow 2-subgroup Se of W and ze is the unique involution of Z(Se )+ . Hence by Sylow’s theorem, zd and ze are W -conjugate, a contradiction completing the proof.  We next prove Proposition 7.5. WK ≡B W (F4 ). Moreover, Proposition 7.3c does not hold. Proof. If the proposition fails, then WLu ≡B W (F4 ) for some u ∈ D# . Fix such an element u. Then n = 5 and WLu is absolutely irreducible on the hyperplane B ∩ Lu of B. Let Z = Z(W ), W = W/Z, Q = O2 (WLu ), and z = Z(Q). By Lemma 5.13, O2 (W ) ≤ Z, and Z acts as scalars on B. As n = 5 is odd, Z + = O2 (W ) has order 1 or 5. Let C = CW (z) and S ∈ Syl2 (C) with S ∩WLu ∈ Syl2 (WLu ). Then [Z(S), Q] = 1 so Z(S) acts as scalars on both [B, Q] = B ∩ Lu and CB (Q) = CB (Lu ) = u . Hence z = Ω1 (Z(S))+  NW (S) and S ∈ Syl2 (W ). Moreover, OutW (Q) has 2-power index in Out(Q) ∼ = Σ3  Z2 , and Q = O2 (WLu )  C, so O2 (C + ) induces inner automorphisms on Q. Hence, O2 (C + ) = QY where

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

[Y, Q] = 1. As Q is absolutely irreducible on [B, Q], Y is cyclic. Now if O2 (W + ) = 1, then O2 (W + ) is a WLu -invariant subgroup of Q ∗ Y containing z. It follows that z = Ω1 (Z(O2 (W + ))), whence C = W , contradicting the fact that W is indecomposable on B. Thus O2 (W + ) = 1. Since O2 (W ) ≤ Z = Z(W ) we conclude that Sol(W ) = Z. Let E = E(W ) = E(W + ). Then Z(E) ≤ Sol(W + ) ≤ Z + so |Z(E)| divides mp (B) = 5. We easily deduce that E = F ∗ (W ) is simple, z is 2-central in E, and O 2 (CW (z)) ∼ = SL2 (3) ∗ SL2 (3) with F ∗ (CW (z)) = Q ∗ Y . By [III17 , 1.28], ∼ W = SO5 (3), W ∼ = Z × W , p = 3, and B is a natural W -module. We consider the associated geometry of B and its subspace D. Now, for every v ∈ D# , WLv ∼ = W (C3 ), W (C4 ) or W (F4 ), as we see from Table 15.2. If some v ∈ D# is an isotropic vector, then F ∗ (NW (v )) ∼ = Z ×O3 (NW (v )), contrary to the fact that WLv  NW (v ). Hence, D is an anisotropic plane. But ∼ then for some v ∈ D# , [CW (v), CW (v)] ∼ = Ω− 4 (3) = A6 . As WLv  CA (v) and ∼ WLv = W (C3 ), W (C4 ) or W (F4 ), we have a contradiction, and the lemma is proved.  As an immediate corollary of Propositions 7.3 and 7.5, we have: ∼ W (Cn−1 ), n ≥ 4, or W (F4 ). Then one Corollary 7.6. Suppose that WK = of Proposition 3.4cdhik holds, with W ∗ = W . Proof. By Proposition 7.5, WK ∼ = W (Cn−1 ). By Corollary 6.12, we may assume that mp (B) ≥ 5. Hence, W ≡B W (Cn ) by Propositions 7.3 and 7.5. It follows immediately that one of Proposition 3.4chk holds, completing the proof.  Although the next proposition does not concern W (Cn ) or W (F4 ), its proof has similarities to that of Proposition 7.5, so we include it here.  ∼ D4 (q) and Lu ∼ Proposition 7.7. Suppose that K = = A4q (q) for some u ∈ D# . Then W ≡B W (D5 ) or W (C5 ). Proof. Again n = 5; this time WK ∼ = W (D4 ). Again let Z = Z(W ), W = W/Z, Q = O2 (WK ), and z = Z(Q). Again Q ∼ = Q8 ∗ Q8 , but now |O2,3 (WK )|3 = 3, with a Sylow 3-subgroup T of WK acting without fixed points on Q/ z . By Lemma 5.13 again, O2 (W ) ≤ Z, and Z acts as scalars on B. As n = 5, again Z + has order 1 or 5. The second paragraph of the proof of Proposition 7.5 applies with x and K now in place of u and Lu , providing S ∈ Syl2 (W ) with z = Ω1 (Z(S + )) and S ∩ WK ∈ Syl2 (WK ). If O2 (W + ) = 1, then z ∈ O2 (W + ) and as W is indecomposable on B and n = 5 is odd, Ω1 (Z(O2 (W + ))) > z . Hence by Lemma 5.14, W ≡B W (D5 ) or W (C5 ) and the conclusion of the lemma holds. Thus we may assume that O2 (W + ) = 1, whence O2 (W ) = Z + = Sol(W + ) and so E := E(W ) = 1. As |Z + | = 1 or 5, the only possible non-simple component of W would be X = SL5 (sa ) for some prime s; but then CX (z) would be nonsolvable, a contradiction as CW (z) is solvable. So every component of E is simple. We argue that E is simple. Since z = Ω1 (Z(S + )), S + permutes the components of E transitively. But m2 (S + ) ≤ 4 since S + embeds in SL(B), so there are at most 2 components. If E = E1 × E2 , then m2 (E1 ) = 2 so E contains A4 × A4 , which does not embed in SL(B). Thus E is simple, and as z = Ω1 (Z(S + )), z ∈ E. Then CW (E) ≤ Sol(W ) = Z, so S + embeds in Aut(E) as does QT .

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∼ Am then m ≥ 8 as We consider the various possibilities for E ∈ K. If E = QT = O 2 (QT ) embeds in E; but then E contains a Frobenius group F8.7 so E does not embed in SL5 (p). Likewise if E ∈ Spor then QT  CE (z), and from the tables [IA , 5.3] we see that the only possibility is E ∼ = M12 . But M12 contains the Frobenius group U3 (2) of order 9.8, contradicting E ≤ SL5 (p) unless p = 3. And if p = 3, we get a contradiction from the existence of a D10 subgroup of Z2 × Σ5 containing a conjugate of z. So E ∈ Chev(s)−Alt for some prime s. If E ∈ Chev(2) but E ∈ Chev(t) for any odd t, then as m2 (E) ≤ 4 and Ω1 (Z(S + )) = z is cyclic, we have E ∼ = L2 (24 ), L3 (4), U3 (4), U3 (24 ), or U5 (2) (see [IA , Table 3.3.1]). But QT embeds in Aut(E) and QT  CW (z). As Q ∼ = Q8 ∗ Q8 , this gives a contradiction in each case. Now we may assume that s is odd. Using the structure of C and [IA , 4.5.1] we ∼ see that E ∼ = P Sp4 (3), L± 4 (3), or G2 (3). Then by [III17 , 16.10], E = P Sp4 (3). The proof of the proposition is then completed by the next lemma.  ∗ Lemma 7.8. Suppose that K ∼ = Ω+ 8 (q), mp (B) = 5, and F (W ) = Z(W ) × H, (3), |Z(W )| ≤ 2, and Z(W ) inverts B in case of equality. Then where H ∼ P Sp = 4 there is no u ∈ D# such that the pumpup Lu of L in CG (u) satisfies Lu /Z(Lu ) ∼ =  L5q (q).

Proof. Suppose false. As Lu  CG (u), it follows that WLu ∼ = Σ5 is normal in CA (u). Suppose first that p = 3. Then W/Z(W ) ≤ Aut(H) with B a natural orthogonal module for H, by [III17 , 1.15]. But then either CA (u) is solvable or ∼ ∼ CA (u) ∼ = Ω− 4 (3) = A6 . As WLu = Σ5 with WLu  CA (u), we have a contradiction. Hence we may assume that p > 3 and derive a contradiction. As H is isomorphic to a subgroup of SL(B) ∼ = SL5 (p) and |H|3 = 34 , it follows that p > 5. In 2 particular, as p divides q − 1, q ≥ 8. Let P ∈ Sylp (CG (x)) with B ≤ P . As B is a maximal elementary abelian p-subgroup of CG (x), it contains every element of order p in CG (B). Hence there is no f ∈ P of order p inducing a field automorphism on K. Thus, as p > 3, B = Ω1 (P ) and as CG (B) ≤ CG (x), P ∈ Sylp (CG (B)). If h ∈ NG (B) is any p-element, then [h, W ] = 1, whence h ∈ Z(A) and then h ∈ CG (B). Hence, A is a p -group and P ∈ Sylp (NG (B)), whence P ∈ Sylp (G). In particular, mp (G) = 5. Consequently, as p ≥ 7, for any b ∈ B # and any p-component I of CG (b) with I = I/Op (I) ∈ Chev(r), r = p, (7B)

B = (B ∩ I) × CB (I) with mp (B ∩ I) = mp (I).

Indeed as mp (G) = 5, both Outdiag(I) and the Schur multiplier of I/Z(I) are p groups. So if (7B) failed, then some b ∈ B # would induce a field automorphism on I, and then AutI (B) would contain a nontrivial p-element, contradicting the fact that A is a p -group. Thus (7B) holds. Let t be a 3-central element of H of order 3, so that CH (t) = T S with T = F (CH (t)) extraspecial of order 33 and with S ∼ = SL2 (3). As CH (t) acts faithfully on B, we have B = D1 × D2 with D1 = CB (T ) ∼ = Ep2 and with D2 = [B, T ] = [B, Z(T )]. Also, Z(S) either inverts or centralizes D1 . By [IA , 7.3.3],    (7C) Lu = O 2 (CLu (d)) : d ∈ D1# .

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2 

As mp (B) = 5 and B ≤ Lu C(u, Lu ), it follows that for all d ∈ D1 −u , O 2 (CLu (d))  is a product of quasisimple components isomorphic to SLmq (q) for some m, 2 ≤ m ≤ 4, each of which centralizes a WLu -conjugate of D. Hence, in particular, we can find d ∈ D1# with E(CLu (d)) = 1, and T Z(S) ≤ NA (d ). If d ∈ u , then as mp (B ∩ Lu ) = 4, T Z(S) acts faithfully on B ∩ Lu , whence T maps injectively into WLu ∼ = Σ5 , a contradiction. Thus d ∈ u . We  choose d ∈ D1# and a component J ∼ = SLmq (q) of CLu (d) with m as large as possible. Let Jd be the subnormal closure of J in Lp (CG (d)) and set CG (d) = CG (d)/Op (CG (d)). As p ≥ 7 and mp (G) = 5, Jd is a trivial or vertical pumpup of J. Note that J d ∈ Chev(2). Indeed since q = 2a ≥ 8 and p ≥ 7, the only 7 alternative, by [IA , 2.2.10, 5.3], is that p = 7, J ∼ = L2 (8), and Jd ∼ = 2 G2 (3 2 ). But then u induces a nontrivial field automorphism on Jd and (7B) is contradicted. We remark that if B1 ≤ B and I is a p-component of CG (B1 ) such that B ∩ I ≤ Op p (I), then CG (B) normalizes I. Hence for any subset X ⊆ A, it is meaningful to ask whether X normalizes I. Suppose that D2 = [B, Z(T )] centralizes J d . Then d × D2 ≤ CB (J d ) and so mp (Jd ) = mp (B ∩ Jd ) = 1. As T normalizes d × D2 , T normalizes both Jd and B ∩ Jd = B ∩ J. As mp (J) = 1, J ∼ = SL2 (q) and B ∩ J ≤ D1 . But then in Lu we see that E(CLu (B ∩ J)) has a component J1 with J1 ∼ = SL3 (q). This contradicts our choice of d and J. Thus, [D2 , J d ] = 1. We next argue that (7D)

T Z(S) normalizes Jd .

  T Z(S) = J11 · · · J1s as a product of p-components Suppose false. Write Jd∗ := Jd with Jd = J11 . By (7B), mp (Jd ) = mp (B ∩J1i ) for all i = 1, . . . , s, so smp (Jd )+1 ≤ mp (B) = 5. If s = 2, then mp (B ∩ Jd ) ≤ 2. In this case T normalizes J11 and J12 , so T normalizes B ∩ J11 and B ∩ J12 . Given the irreducible action of T on ∗ D2 , the only possibility is that mp (Jd ) = 1 and B ∩ Jd∗ = D1 . But then CB (J d ), being T -invariant, must equal D2 , so [D2 , J d ] = 1, a contradiction. Hence s = 3 ∗ and mp (B ∩ J1i ) = 1, 1 ≤ i ≤ 3. Now CB (J d ) = D1 , and in particular, [D1 , J] = 1.  By our maximal choice of d, J  O 2 (CLu (d1 )) for all d1 ∈ D1# , so J  Lu by (7C), which is absurd. This establishes (7D). As a consequence, CB (J d ) and B ∩ Jd are T Z(S)-invariant. As T is irreducible on D2 while [J d , D2 ] = 1, D2 ≤ B ∩ Jd , in view of (7B). Thus, mp (B ∩ Jd ) ≥ 3. Let Ad := AutAut(J d ) (B ∩ Jd ) and Id := AutInndiag(J d ) (B ∩ Jd ), so that Id ≤ Ad and T Z(S) embeds in Ad . Since Jd ∈ Chev(2), [Out(Jd ), Out(Jd )] is cyclic, by [III17 , 7.1]. As Z(S) inverts T /Z(T ), we conclude that |T ∩ Id | ≥ 9. Consequently, Jd ∼  D4 (q0 ) for any q0 , and so Out(Jd )/ Outdiag(Jd ) is abelian, by [III17 , 7.1]. = Thus T embeds in Id . As noted after (7B), the Schur multiplier of Jd /Op p (Jd ) is a p -group. As p ≥ 7, it follows from [IA , 4.10.3e] that B ∩ Jd lies in a maximal torus of the algebraic group overlying J d . Then by a Frattini argument, Id embeds in the Weyl group Wa of that algebraic group. As T embeds in Id , Wa has (Lie) rank at least 6, and mp (B ∩ Jd ) ≥ 3. By [III14 , Lemma 1.2], and since mp (K) = 4, we have

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

193

q(J d )36 ≤ f (J d ) ≤ f (K) = q 16 . Hence q(J d ) < q, and then by [III17 , 10.2], J∼ = A1 (q). As noted above, [J, Dg ] = 1 for some g ∈ WLu . By [III11 , 1.15], there is d1 = dg0 ∈ (Dg )# such that the subnormal closure Jd1 of J in CJd (d1 ) is a vertical  pumpup of J. But then Jd1 is a component of E(CLgd (d)), and as Ld0 ∼ = A3q (q), 0



A4q (q), or D4 (q), we have q(Jd1 ) = q and mp (Inndiag(Jd1 )) > 1. Let j1 be the untwisted Lie rank of Jd1 , so that j1 ≥ 2. However, [III17 , 10.2] applies to Jd1 and  Jd and yields that j1 = 1, a contradiction. The proof is complete.

8. The Case AutK (B) ∼ = W (Dn ), n ≥ 4 As in the previous section, our results in this section need to be more general than just to apply to the subgroups WK and WLu , u ∈ D# , of W . So n−1 again we consider the situation described in (7A), now with Lo ∼ = D q (q), n ≥ u

n−1

n−2

q (q). Again we assume that the pumpups Lod 5, for some u ∈ (Do )# , and Lo ∼ = Dn−2 of Lo in CG (d) for d ∈ D# lie in Chev(2) with level q, and we have the subgroups WLo and WLod of W . As in the previous section, (7A) is satisfied in particular by Do = D, Lo = L, and Lod = Ld for each d ∈ D# . The basic setup for this section is: (1) (7A) holds with n ≥ 5; n−2 q (q); and (2) Lo ∼ = Dn−2 (8A) n−1 q (q); and (3) x ∈ D# and K o := Lox ∼ = Dn−1 n−1 o # o o (4) For all d ∈ (D ) with L < L , Lo ∼ = D q (q).

d

d

n−1

We remark that such a setup would be satisfied in our simple group G, for any n odd prime divisor p of q 2 − 1, if G ∼ = Dnq (q), but also if G ∼ = E8 (q) with n = 8, or  G∼ = E6q (q) with n = 6 (possibly n = 5, if p = 3). Our goal will be to prove that W has a form appropriate to these examples (see Proposition 8.2 below). A consequence of (8A3) is that B = B1 × x where B1 = B ∩ K o , and there exist b1 , . . . , bn ∈ B and subgroups F1 and Σ1 of WK o such that the following conditions hold: (1) B = b1 , . . . , bn , with x = b1 , y = b2 , B2 := B ∩ Lo = b3 , . . . , bn , B1 = b2 × B2 , and Do = x, y ; (2) WK o = F1 Σ1 , F1 = t23 , . . . , tn−1,n with tij inverting both bi (8B) and bj , and centralizing bk for all k = i, j, and Σ1 ∼ = Σn−1 permuting the set {b1 , b2 , . . . , bn } naturally as the stabilizer of b1 . To justify this, we simply rename x as b1 ; then as WK o ≡B1 W (Dn−1 ) and n−2  ∼ Dq (q), we may choose a frame B = {b2 , . . . , bn } L = O p (E(CK o (Do ))) = n−2 in B1 such that Do = b1 , b2 and B ∩ Lo = b3 , . . . , bn . We then rename b2 as y. Note that when n > 5, the unordered frame B is uniquely determined, consisting of all cyclic subgroups of B2 whose centralizer in K o has a component isomorphic to Lo . When n = 5, there are three such frames for which subgroups F1 and Σ1 exist satisfying (8B2), but only one containing the projection of Do on K o . o

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

194

Before we prove Proposition 8.2, we need to check that in the main case, in n−1 n−2 q ∼ Dq (q) and L ∼ (q), the conditions (8A) hold with which K = = Dn−1 n−1 Do = D = x, y , K o = K, Lo = L, and Lod = Ld for all d ∈ D# . We see in fact that they do not necessarily hold if n = 5, since by Table 15.2, some  Ld might be isomorphic to A4q (q). However, in that case, by making a different choice of acceptable subterminal pair (y, L), we can arrange that (8A) is valid. The original pair (x, K) does not change, nor does the isomorphism type of L; but y and the neighborhood N(x, K, y, L) do change. Such a replacement can be made, moreover without affecting the validity of the conclusions of [III12 , Theorem 1.2], on which our initial setup (1A) is based. n−1 q (q), then there exists an acceptable subterminal Lemma 8.1. If K ∼ = Dn−1 (x, K)-pair (y, L) such that if we put D = x, y , L = E(CK (D)) and Ld = the pumpup of L in CG (d), for each d ∈ D# , then (8A) is satisfied for K o = K, Lo = L, Do = D, and Lod = Ld . Proof. Assume that condition (8A4), which is the only issue, fails, so that  by Table 15.2, n = 5 and Ld ∼ = A4q (q) for some d ∈ D# . By Proposition 7.7, W ≡B W (D5 ) or W (C5 ). Thus W acts monomially on B, preserving some unique frame B = {b1 , . . . , b5 }. Since WK ∼ = W (D4 ) the only possibility for x is one of the bi ; without loss, x = b1 . We then choose our new y to be y = b2 , our new D to be x, y , and L and Ld , d ∈ D# , are defined as usual. It suffices to  assume that Ld ∼ = A4q (q) for some d in our new D, and derive a contradiction. Obviously d = x , and d = y since x and y are W -conjugate and so have isomorphic centralizers. Thus with respect to B, d projects nontrivially on b1 and b2 , and trivially on b3 , b4 , b5 . But WLd contains an element w of order 5. Then w must preserve {b1 , b2 }. Hence it normalizes each bi , and hence by the structure of W , w2 = 1, a contradiction. The lemma follows.  Our goal is to complete the analysis of A = AutG (B) and W , the reflection subgroup of A, by proving the following result. We let Z = Z(A). Proposition 8.2. Assume (8A). Then Z is cyclic, acting as scalars on B, and W contains a copy of W (Dn ) unless n = 6 and W ∼ = W (E6 ). Moreover, one of the following holds: (a) (b) (c) (d)

W ≡B W (Dn ) or W (Cn ); A = W × Z with W ≡B W (E6 ) and n = 6; W = W ∗ or W ∗ × Z2 , with W ∗ ≡B W (E6 ), p = 3, and n = 5; or A = W ∗ Z with W ≡B W (E8 ) and n = 8.

Recall that x = b1 and y = b2 , so that Do = b1 , b2 . We set NDo = NG (Do ). 

Also we have Lo = O p (E(CG (Do )))  NDo . We begin with a lemma identifying b3 in Lo in a fairly general situation. ∼ K o with b1 ∈ Lo . Lemma 8.3. Assume (8A). Suppose that n > 5 and Lob2 = b2 o Then b3 has 2-dimensional support on the natural Lb2 -module.

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

195

Proof. Let V be the natural Lob2 -module (which is unique up to isomorphism as n > 5). Since Lob2 ∼ = K o and E(CLob (b1 )) = Lo , dim[V, b1 ] = 2 and 2

n−3

q (q). Hence, [V, Lo ] = [V, b1 ]⊥ . We also know from K o that E(CLo (b3 )) ∼ = Dn−3 using [III17 , 2.24], we see that if the lemma fails, then n = 6 and [V, b3 ] = [V, Lo ]. Now we can refine V = [V, b1 ] ⊥ [V, Lo ] into an orthogonal decomposition of V as a sum

V = W1 ⊥ W3 ⊥ W4 ⊥ W5 ⊥ W6 of B-invariant planes isometric to W1 := [V, b1 ], and correspondingly decompose B2 = B ∩ Lo as B2 = B23 ⊕ B24 ⊕ B25 ⊕ B26 where B2k = CB2 (Wk⊥ ), 3 ≤ k ≤ 6. Let B23 =: b . There exists an involution s in NLob (B) inverting b1 , b and centralizing 2 B24 ⊕ B25 ⊕ B26 . In particular, [B2 , s] = b . By [III17 , 2.24], if VK o is the natural orthogonal module for K o , then [VK o , b] = [VK o , Lo ]. Thus b = bi33 bi44 bi55 bi66 with 1 ≤ ij < p for all j, 3 ≤ j ≤ 6. On the other hand, as s inverts b1 , s ∈ NG (b1 ), and hence s ∈ NG (K o ) ∩ NG (B). As n > 5, it follows that bs3 ∈ bi for some i, −1 r s s 3 ≤ i ≤ 6. Set bs3 = bri . Then b−1 3 b3 = b3 bi ∈ [B2 , s] = b , whence b3 = b3 . But s inverts b, a final contradiction proving the lemma.  The next lemma gets us off the ground. Lemma 8.4. Assume (8A). Then b1  NDo ∩ NG (B). Proof. It suffices to show that b1  NDo . For, if b1 = bg1 for some g ∈ NDo , then g normalizes Lo and by [IA , 4.10.3c], B∩Lo and B g ∩Lo are Lo -conjugate. Hence g may be modified by an element of Lo to guarantee g ∈ NDo ∩ NG (B). Suppose on the contrary that b1  NDo . Then NDo normalizes both K o and o D ∩ K o = b2 . By (7A6) and (8A3), there is v ∈ Do − b1 with Lov ∼ = K o . Let v  = Do ∩ Lov .  Then we can find an involution t ∈ NLov (B) with (v  )t = (v  )−1 . Thus t acts on Do with determinant −1. Since t normalizes both b1 and b2 , and since v = b1 by assumption, we may assume that v = b2 and v  = b1 . Now rename Lov = K1o and let V1 be a natural orthogonal module for K1o , chosen if n = 5 so that [V1 , b1 ] is a nonsingular plane. We have that b3 , . . . , bn are conjugate subgroups of Lo . If n = 5, then b3 has 2-dimensional support on the natural Lo -module and hence on V1 . Hence in any case, [V1 , b3 ] is 2-dimensional by Lemma 8.3. Hence, NK1o (B) induces Σn−1 on the set {bi | i = 2, 1 ≤ i ≤ n}. In particular, there exists an involution τ (1, 3) ∈ A interchanging b1 and b3 , while normalizing b2 . Since K o contains τ (2, 3) interchanging b2 and b3 while normalizing b1 , it follows that τ (1, 3)τ (2, 3)τ (1, 3) normalizes Do and interchanges b1 and b2 , contrary to the assumption that b1  NDo . The proof is complete.  The case n = 6 is anomalous, as we have already seen in the proof of Lemma 8.3, because Lo ∼ = D4 (q) can admit a triality automorphism in G. More precisely, we have the following dichotomy. (8C)

(1) NDo stabilizes the conjugacy class of b3 in Lo ; or (2) n = 6 and (1) fails.

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196

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2 n−2

q Indeed if n = 6, then in Lo ∼ (q), b3 belongs to the unique conjugacy = Dn−2 n−3  ∼ Dq (q). This follows class of subgroups b of order p such that O 2 (CLo (b)) = n−3 easily from [IA , 4.8.2]. For the moment, we assume that (8C1) holds. The following lemma will show how we use this hypothesis.

Lemma 8.5. Assume (8A) and (8C1). Let V be a natural orthogonal module for K o . Let g ∈ NDo ∩ NG (B) with u := b1 g = b1 , so that Lou ∼ = K o , and o o let X be a natural orthogonal module for Lu . Let Bu := B ∩ Lu . Then, subject to choices of V and X when n = 5, the following conclusions hold: (a) There is a unique B1 -invariant orthogonal decomposition V2 ⊥ · · · ⊥ Vn of V as a sum of isometric non-singular planes so that Vi = [V, bi ] for all i = 2, . . . , n; (b) There is a unique Bu -invariant decomposition X2 ⊥ · · · ⊥ Xn of X as a sum of non-singular planes isometric to the Vi , and such that X2 = [X, D ∩ Lou ] = CX (Lo ) and Xi = [X, bi ] for all i = 3, . . . , n; and  (c) E(CK o (bi )) ∼  O 2 (CLo (bi )) for all 3 ≤ i ≤ n. = E(CLou (bi )) ∼ = Lo ∼ = Proof. Since p does not divide |Outdiag(K o )| it follows from Lemma 5.1d that n−2 ∼ Dq (q). B = (B ∩K o )×CB (K o ) = B1 ×b1 . By hypothesis, E(CK o (b2 )) = Lo = n−2 It follows from [III17 , 2.25] that, subject to a choice of module when n = 5, V = [V, b2 ] ⊥ CV (b2 ) with [V, b2 ] =: V2 a nonsingular plane and with CV (b2 ) = [V, Lo ]. Next, since Σ1 ≤ K o transitively permutes the set {b2 , . . . , bn }, it follows that Vi := [V, bi ] is isometric to V2 for all 2 ≤ i ≤ n. As B1 = b2 , . . . , bn is abelian, the orthogonal decomposition given in (a) follows readily. We have Lou = (K o )g and B g = B, so B = (B ∩ Lou ) × u . Also, as g ∈ NDo , E(CLou (b1 )) = E(CLou (Do ∩ Lou )) = Lo . Let X be the Lou -module obtained by g transporting V to an Lou -module by conjugation by g. Note that b2 = Do ∩ Lou . Lo Also {b3 , . . . , bn } is the set of all elements of b3 lying in B, so it is mapped to itself by g, by (8C1). Moreover, by [IA , 4.8.1], X1 and V1 are isometric. Thus (a) implies (b). Now (c) follows directly from (a) and (b).  Our next main result implies that Lo pumps up vertically in CG (b2 ), assuming (8C1). In fact it makes a stronger assertion: There exists g ∈ NDo ∩ NG (B) interchanging b1 and b2 , and stabilizing (8D) {b3 , . . . , bn }. Lemma 8.6. Assume (8A) and (8C1). Then (8D) holds. Proof. Choose g ∈ NDo ∩NG (B)−NG (b1 ) and set u = bg1 , so that Lou ∼ = K o. o Let X be an orthogonal module for Lu chosen as in Lemma 8.5, so that X2 := [X, Do ∩Lou ] is a non-singular plane with X2 = CX (Lo ). Let B3 := b3 , . . . , bn ≤ Lo ,   on := so that g permutes {b3 , . . . , bn } by (8C1). Set Lo2n := O 2 (CLo (bn )) and L o o o  n properly contains L2n . In particular, [L  n , Do ] = E(CLou (bn )). By Lemma 8.5b, L  no be the pumpup of L  on in CG (bn ). Next set Lon := E(CK o (bn )) ∼ 1. Let K = Lo , and o o o o o let Kn denote the pumpup of Ln in CG (bn ). Note that L2n ≤ Ln ≤ Kn . We shall  on ≤ K  no , argue that Lon < Kno . Suppose on the contrary that Lon = Kno . As Lo2n ≤ L o o o o o o o o n ≥ L  n . Thus L  n ≤ Ln ≤ K o , it follows that [Ln , Kn ] = 1, whence Ln = Kn = K

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

197

 on , x] = 1. But also L  on ≤ Lou , whence [L  on , u] = 1. Therefore L  on commutes and so [L o o o with D = b1 , u , a contradiction. Hence, Ln < Kn , as claimed. There exists h ∈ K o with bhn = b2 = y and (Lon )h = Lo . Hence Loy = (Kno )h > o L . We now consider N2 := NG (b2 ). Note that Loy  N2 and |Do ∩ Loy | = p. Let d = Do ∩ Loy . Notice that t23 ∈ WK o with t23 normalizing Do , centralizing b1 , and inverting b2 = y . Thus t23 normalizes d = Do ∩ Loy . Since b1 and b2 are the unique t23 -invariant subgroups of D of cardinality p, it follows that b1 = d ≤ Loy and Lo = E(CLoy (b1 )). Now by Lemma 8.3, b3 has 2-dimensional support on the natural Loy -module. By Lo -conjugation, so does bn . Thus bn = b1 g1 for some g1 ∈ NLoy (B) ≤ NG (b2 ), and g1 stabilizes {b1 , . . . , bn }. On the other hand, by (8A), bn = b2 g2 for some g2 ∈ NK o (B) ≤ NG (b1 ), and g2 stabilizes {b1 , . . . , bn }. The lemma follows, with g = g1g2 .  Now we drop the hypothesis (8C1) and assume instead its consequence (8D), through Lemma 8.15. Corollary 8.7. Assume (8A) and (8D). After replacing b1 by a suitable generator of b1 , we have that WLob contains F2 · Σ2 , where 2

F2 = t13 , t34 , . . . , tn−1,n ∼ = E2n−2 , with t13 inverting both b1 and b3 and centralizing bi for 4 ≤ i ≤ n (and i = 2). Moreover, Σ2 ∼ = Σn−1 permutes the set {b1 , b2 , . . . , bn }, fixing b2 . Proof. By Lemma 8.5, there is a natural K o -module V such that dim[V, bi ] = 2 for all i = 2, . . . , n, and by (8B2), WK o = F1 Σ1 with F1 = t23 , . . . , tn−1,n ∼ = E2n−1 and Σ1 ∼ = Σn−1 permuting {b2 , . . . , bn } naturally and fixing b1 . Since g preserves {b1 , . . . , bn }, interchanging b1 and b2 , there is a natural (K o )g = Lob2 -module X such that dim[X, bi ] = 2 for all i = 1, 3, 4, . . . , n. Moreover, Σ1 ∩Lo ∼ = Σn−2 permutes {b3 , . . . , bn } naturally, fixing b1 and b2 . Hence b3 , . . . , bn have the same eigenvalues on X. Replacing b1 by a power of itself we may arrange that b1 has the same eigenvalues on X as b3 . Then after a possible further replacement of b1 by b−1  1 , the existence of F2 · Σ2 follows directly. Recall that A = AutG (B) ≤ GL(B). We let M := H ∗ · Σ denote the subgroup of GL(B) consisting of all monomial matrices with respect to the ordered frame {b1 , . . . , bn }. Here H ∗ is the normal abelian subgroup of diagonal matrices, and Σ ∼ = Σn permutes the set {b1 , . . . , bn }. Our next result identifies a subgroup of A ∩ M isomorphic to W (Dn ), putting together pieces from (8B2) and Corollary 8.7. Lemma 8.8. Assume (8A) and (8D). Then W contains the subgroup F := t12 , t23 , . . . , tn−1,n ∼ = E2n−1 , with t12 inverting D elementwise and centralizing B2 = B∩Lo . Moreover, NA (F ) ≤ A ∩ M , with |NA (F ) : Z.F ∗ .Σ| ≤ 2. Here Σ ∼ = Σn is the stabilizer of {b1 , . . . , bn }; Z = Z(A) acts as scalar matrices on B; and F ∗ := NA (F ) ∩ Ω1 (O2 (M )), with either F = F ∗ or F ∗ = t1 , . . . , tn ∼ = E2n (where ti inverts bi and centralizes bj for all j = i). In any case NA (F ) ≤ Z(M ) t1 , . . . , tn Σ, and F Σ ∼ = W (Dn ).

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Proof. By the previous lemma, there exists t13 ∈ W inverting both b1 and b3 , and centralizing bi for 1 ≤ i ≤ n, i = 2. Set F = t13 , F1 . Then clearly F is as described in the lemma. Moreover, B is completely reducible as F -module and B := {b1 , . . . , bn } is a complete set of pairwise non-isomorphic simple F -submodules of B. Thus NA (F ) ≤ NGL(B) (F ) = M . It is clear that Σ = Σ1 , Σ2 ≤ NA (F ), so NA (F ) = (A ∩ H ∗ )Σ. On the other hand, F contains H ∗ ∩ AutK o (B) and H ∗ ∩ AutLoy (B). Let T = A ∩ H ∗ . Then T normalizes bi , i = 1, 2, K o , and Loy . Thus [T, Σ1 ] ≤ H ∗ ∩ AutK o (B) ≤ F ≤ Ω1 (O2 (T )), and similarly [T, Σ2 ] ≤ Ω1 (O2 (T )). As Σ = Σ1 , Σ2 , it follows that [T, Σ] ≤ Ω1 (O2 (T )). As T ≤ H ∗ , the action of Σ on T implies that T ≤ Z(A)Ω1 (O2 (H ∗ )) with Z(A) acting on B as scalars. Summarizing, we see that Z · F ∗ · Σ ≤ NA (F ) ≤ Z(M ) · Ω1 (O2 (H ∗ )) · Σ. As Ω1 (O2 (H ∗ )) = t1 , . . . , tn , as described in the statement of the lemma, and  |Ω1 (O2 (H ∗ )) : F ∗ | ≤ 2, the lemma holds. As a consequence we can locate the reflections in NA (F ). Lemma 8.9. Assume (8A) and (8D). Let t ∈ NA (F ) be an involution with just one or two eigenvalues on B equal to −1. Then t ∈ F ∗ Σ. Proof. Let F ∗∗ = Ω1 (O2 (M )), N0 = NA (F )F ∗∗ , and N 0 = N0 /F ∗∗ . Then N 0 = Z × Σ, so we may write t = zg with z ∈ Z and g ∈ Σ. If z = 1 then t ∈ NA (F ) ∩ ΣF ∗∗ = ΣF ∗ , as required, so assume that z = 1. Then z is an involution. But Z ∩ F ∗∗ = ±1 , so p ≡ 1 (mod 4) and z induces a scalar mapping of order 4 on B. Likewise if g = 1 then t ∈ zF ∗∗ , so all eigenvalues of t are primitive fourth roots of unity, which is absurd. Thus g = 1. Now t permutes the frame {b1 , . . . , bn } nontrivially. As t has order 2 and at most two eigenvalues different from 1 on B, g is either a transposition or the product of two disjoint transpositions. In either case as n ≥ 5, t must fix some bi which is also fixed by g, and hence fixed by tg −1 . But tg −1 ∈ zF ∗∗ , so again the eigenvalues of tg −1 are all primitive fourth roots of unity, a contradiction completing the proof.  Lemma 8.10. If F  A, then W ≡B W (Dn ) or W (Cn ). Proof. By Lemma 8.8, F ≤ W . Then the lemma follows immediately from Lemma 5.14, with W in the role of W2 there.  Our next goal is to prove that in general F  A, and to identify the possibilities when F  A. Finally, we shall return to the case in which (8C2) holds. First we consider CA (t12 ). Lemma 8.11. Assume (8A) and (8D). Then CA (t12 ) ≤ NA (F ).

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

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Proof. Let B be the unordered frame {b1 , . . . , bn }, and let N be its stabilizer in A. Now F is the group of all g ∈ A centralizing or inverting each bi , and inverting an even number of them. Hence N ≤ NA (F ). Obviously NA (F ) ≤ N , so N = NA (F ). We must show that CA (t12 ) ≤ N . Now CA (t12 ) normalizes [B, t12 ] = Do . The proof is easiest when (8C1) holds.  In that case, let D∗ = bn−1 , bn and L∗ = O p (E(CG (D∗ ))). Then D∗ is Σconjugate to D by Lemma 8.8, so NA (D∗ ) stabilizes B ∩ E1 (L∗ ). Let C = CA (t12 ) and let  t ∈ NG (B) map on t12 . Then C is the image of  normalizes Do , Lo ,  of NG (B) which normalizes  tCG (B). Then C the subset C o and B ∩ E1 (L ). Moreover, C contains a copy C0 of Σn−2 permuting B ∩ E1 (Lo ) naturally and centralizing Do . Notice that C0 ≤ N . Let g ∈ C. Then g permutes B ∩ E1 (Lo ) and hence there is c ∈ C0 such that g1 := gc fixes each point of B ∩ E1 (Lo ). In particular, g1 normalizes D∗ , and therefore stabilizes B ∩ E1 (L∗ ). Since g and c stabilize B ∩ E1 (Lo ), g1 ∈ N , so g ∈ N as well. Thus C ≤ N , our desired conclusion. Now suppose that (8C2) holds. In particular, n = 6. Let H be the set of all reflections and all involutions with exactly two −1 eigenvalues in NA (Do ). Set H = H  NA (Do ); we claim that H ≤ N . Let h ∈ H and let  h ∈ NG (B) map on h. Then  h normalizes o p o h normalizes b1 L = O (E(CG (D ))). Now if h inverts or centralizes Do , then  and K o , and as n − 1 = 4, h stabilizes B ∩ E1 (K o ) and B. Hence, we may assume that h acts on Do as a reflection and does not normalize b1 . Thus h|B ∩ Lo is a reflection or the identity. Let D∗ = b5 , b6 ≤ B ∩ Lo . Then h centralizes b for some b ∈ (D∗ )# . Choose an involution g ∈ N interchanging {b1 , b2 } with {b5 , b6 } and set L∗ = (Lo )g and b0 = bg ∈ Do . Then L∗ is a component of CG (D∗ ) and Lb := Lgb0 is its pumpup in CG (b). Now {b5 , b6 } is part of a WLo -invariant frame in B ∩ Lo . Hence {b1 , b2 } is part of a WL∗ -invariant frame in B ∩ L∗ . Therefore by [III17 , 14.12], the conjugacy class of b1 in Aut(Lb ) has just two members in b1 , b2 , namely, b1 and b2 . Hence h interchanges b1 and b2 . But by (8D), there is g ∈ N interchanging b1 and b2 . Thus hg normalizes b1 and K o , and as usual hg ∈ N . Therefore h ∈ N , proving our claim. By Lemma 4.5, H ≤ F ∗ Σ. Indeed as H ≤ NA (Do ), it is easily seen that H = F ∗ (z × Σ1,2 ), where z ∈ Σ interchanges b1 and b2 and fixes the remaining bi ’s, while Σ1,2 ∼ = Σ4 is the stabilizer in Σ of b1 and b2 . (The groups F ∗ , z and Σ1,2 are all generated by involutions with at most two −1 eigenvalues on B.) Suppose that F ∗ = F . Note that H + /F ∼ = Σ4 and hence O2 (H + )/F is a four-group for which F is the direct sum of a trivial module and a free module. It follows easily that F = J(O2 (H + ))  NA (Do ), so CA (t12 ) ≤ NA (Do ) ≤ N , as desired. On the other hand, if F ∗ > F , then H is isomorphic to a subgroup of Σ12 of odd index. Let S ∈ Syl2 (H); by [III17 , 8.10ab], Z(J(S)) = t12 , t34 , t56 . Hence NA (S) preserves {t12 , t34 , t56 }, the set of all members of Z(J(S))# with minimal support on B. Notice that CA (t12 ) ∩ N contains an element g ∈ A ∩ Lo interchanging t34 and t56 . It suffices to show that CA (Z(J(S))) ≤ N . For then a Frattini argument will give C ≤ HNC (S) ≤ HNC (Z(J(S))) ≤ H g, CA (Z(J(S))) ≤ N , as required.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Finally, O2 (H) = t1 , t2 , z ×t3 , t4 , t5 , t6 O2 (Σ1,2 ), whence J(O2 (H)) = F ∗ z and Z(J(O2 (H))) = t12 , t3 , t4 , t5 , t6  CA (t12 ). Similarly, arguing with t34 in place of t12 , we find that Z1 := t34 , t1 , t2 , t5 , t6  CA (t34 ). Hence CA (Z(J(S))) normalizes Z(J(O2 (H)))Z1 = F ∗ and F = (F ∗ )+ , completing the proof of the lemma.  Lemma 8.12. Assume (8A) and (8D). subgroup of A.

Then NA (F ) contains a Sylow 2-

Proof. Let S ∈ Syl2 (NA (F )) and set S1 = S ∩ ΣF ∗ . Let V be the weak closure of F ∗ in S with respect to A. Since F ∗ is generated either by the reflections t1 , . . . , tn or the elements tij = ti tj , it follows from Lemma 8.9 that V ≤ S1 . It suffices to find V0 ≤ V such that (8E)

V0 char V, V0 ≤ F ∗ , and tij ∈ V0 for some 1 ≤ i < j ≤ n.

Indeed the only elements of F with [B, F ] = p2 are the various tk ’s, which are permuted transitively by NA (F ). If (8E) holds, then it follows that NA (S) ≤ NA (V ) ≤ NA (V0 ) ≤ NA (F )CA (tij ) = NA (F ) by Lemma 8.11. To establish (8E), observe that if V = F ∗ , we can choose V0 = V . On the other hand, if V > F ∗ , we now argue that V0 = [V, V ] satisfies (8E). First suppose that m2 (F ∗ ) = n, i.e., F ∗ is the natural permutation module for Σ ∼ = Σn . Then S1 is isomorphic to a Sylow 2-subgroup of Σ2n by [III17 , 8.11a], so by [III17 , 8.10], m2 (S1 ) = n and J(S1 ) is the direct product of [n/2] copies T12 , T34 , . . . of D8 and a group Z ∼ = Z2 or 1, according as n is odd or even. Moreover, [Ti,i+1 , Ti,i+1 ] = ti,i+1 for each i = 1, 3, . . . . Since V is generated by conjugates of F ∗ , which are all of rank n, and V > F ∗ , V /Z is the direct product of some of the Ti,i+1 Z/Z’s and an elementary abelian group. Clearly [V, V ] then satisfies (8E). Next suppose that m2 (F ∗ ) = n − 1, so that F ∗ = F is the trace-0 submodule of the natural permutation module for Σ∼ = Σn . Then by [III17 , 8.11b], S1 is isomorphic to a Sylow 2-subgroup of A2n . Assume that n is odd. By [III17 , 8.10d], J(S1 ) is the direct product of [n/2] copies of D8 with centers t12 , t34 , . . . . Again as V > F ∗ it is clear that [V, V ] satisfies (8E). Still assuming that F = F ∗ is the trace-0 module, suppose now that n is even. Let S2 = S1+ ∈ Syl2 (F [Σ, Σ]). As F ≤ A+ , V ≤ S2 . Since V > F there is g ∈ A such that F = F g ≤ S. Hence by [III17 , 8.11c], J(S2 ) = F, F g = V and (8E)  holds for V0 = [V, V ]. This completes the proof. Recall that M is the monomial subgroup of B preserving the frame {b1 , . . . , bn }. Corollary 8.13. Assume (8A) and (8D). Then the following conclusions hold: (a) O2 (A) ≤ Z(A); (b) O2 (A) ≤ ZΩ1 (O2 (M )); and (c) If A = NA (F ), then O2 (A) ≤ Z(A).

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

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Proof. By Lemmas 8.8 and 8.12, NA (F ) = (ZΩ1 (O2 (M )) ∩ A) · Σ contains a Sylow 2-subgroup S of A. Clearly O2 (A) ≤ S ∩O2 (NA (F )) ≤ O2 (Z)Ω1 (Z(O2 (M ))) since Σ ∼ = Σn with n ≥ 5. This proves (b). Moreover, if O2 (A) ≤ Z(A), then, in view of the action of Σ, F ≤ Ω1 (O2 (A)) and then A ≤ M . Hence A = NA (F ), proving (c). Next, by Lemma 8.8, CA (t12 ) = (ZΩ1 (O2 (M )) ∩ A) · CΣ (t12 ), with CΣ (t12 ) ∼ = Z2 × Σn−2 acting faithfully on F ∗ . In particular, O2 (CA (t12 )) = O2 (Z). Hence t12 inverts O2 (A)/O2 (Z). As Σ transitively permutes the set {tij : 1 ≤ i = j ≤ n}, it follows that tij inverts O2 (A)/O2 (Z) for all i, j. But then t12 = t13 t23 centralizes  O2 (A)/O2 (Z), whence this quotient is trivial, proving (a). For the next two lemmas we make the additional assumption that (8F)

A1 := NA (F ) = A.

Lemma 8.14. Assume (8A), (8D), and (8F). Then E(A) is quasisimple and F ∗ (A) = E(A)Z(A). Proof. As A1 = A by hypothesis, we see that O2 (A)O2 (A) = Z(A) by Corollary 8.13. Hence F ∗ (A) = E(A)Z(A). Let E1 , . . . , Em be the components of E(A) and assume, for a contradiction, that m > 1. Let r ∈ Σ be a reflection. Then r normalizes each Ei ; otherwise it would invert an element of prime order s ≥ 5, contradicting Lemma 5.9. Consequently Σ normalizes each Ei , so by the Schreier property [Σ, Σ] ≤ E(A). As NA (F ) ∩ E(A)  NA (F ), it follows that F ≤ E(A). Now E(CA (t12 )) = 1, so t12 projects nontrivially onto each E i in E(A) = E(A)/Z(E(A)). As NA (F ) contains a Sylow 2-subgroup of A by Lemma 8.12, and NA (F ) acts irreducibly on F , it follows that F ≤ E1 . Hence t12 projects  trivially on E2 , a contradiction completing the proof. In the next lemma, we identify the two possibilities for E(A). Lemma 8.15. Assume (8A), (8D), and (8F). Then either (a) n = 5, p = 3, and W ∼ = SO5 (3) or SO5 (3) × Z2 ; or (b) n = 8 and W ∼ W (E = 8 ). Proof. Let J = E(A), C = CA (t12 ), and A1 = NA (F ). Thus by Lemma 8.8, C = CA1 (t12 ) = Z(A)(QCΣ (t12 )) where Q  A1 = Z(A)QΣ, Q is abelian, and Ω1 (Q) involves the core of the natural Σ ∼ = Σn -permutation module. Here CΣ (t12 ) ∼ = Z2 × Σn−2 . Also A1 contains a Sylow 2-subgroup S of A. In particular, F ∗ (C) = Z(A)O2 (C). Suppose that J ∈ Chev(s), s odd. Since E(C) = 1, it follows [IA , 4.2.2] that C is solvable. Hence by the structure of C, and as n ≥ 5, C/Z(C)O2 (C) ∼ = Σ3 . Thus |C/Z(C)|2 = 3, and it follows from [IA , Table 4.5.1] that J/Z(J) ∼ = L2 (q) for some q, L3 (3), U3 (3), or P Sp4 (3). In any case Out(J) is abelian, so [A1 , A1 ] ≤ F ∗ (A) = JZ(A). But as [A1 , A1 ] has 2-rank at least 4, it follows that J/Z(J) ∼ = P Sp4 (3) since = P Sp4 (3), indeed J ∼ m2 (Sp4 (3)) = 2. Then AutJ (F ) ∼ = A 5 < Σn ∼ = SO5 (3), = AutA1 (F ), so as Aut(J) ∼ we have n = 5 and A/Z(A) ∼ = Σ5 contains an involution u ∈ = SO5 (3). As Σ ∼ [Σ, Σ], u ∈ JZ(A). Then A = J u × Z(A). Indeed u permutes {b1 , . . . , b5 } as a transposition, so u is a reflection. By [III17 , 1.14], SO5 (3) embeds in GL5 (p) only if p = 3, whence conclusion (a) holds in this case.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Suppose next that J ∈ Spor. Now C/O2 (Z(C)) is a 2-constrained extension of an abelian 2-group by Z2 × Σn−2 , and A1 is a 2-local subgroup of A with A1 /Z(A)O2 (A1 ) ∼ = Σn . Notice also that t12 ∈ [Σ, F ] ≤ [A1 , A1 ] so t12 induces an inner automorphism on J. By inspection of [IA , Table 5.3], and using [IA , Table 5.6.1] together with the fact that m2 (A) = m2 (A1 ) ≥ n − 1 ≥ 4, with strict inequality if O2 (Z(J)) = 1 and n is odd, we see that the only J with these properties would be J ∼ = 2M22 (n = 6). Since H contains a Frobenius group of order 11 · 10, we must have p = 11. However, 2M22 contains 2L3 (4) and hence a central extension of U3 (2) by Z2 , which in turn requires at least 8 dimensions for a faithful representation over F11 . This contradicts n = 6, so J ∈ Spor. If J/Z(J) ∼ = Am is an alternating group not in Chev, then by the size and rank of a Sylow 2-subgroup of A1 and A, m ≥ 2n − 2 or 2n according as n is odd or even, i.e., m ≥ 4[n/2] ≥ 8. On the other hand, if Z(J) has odd order, then J contains the direct product of [m/4] copies of A4 , so from the action on B, n ≥ 3[m/4] ≥ 3[n/2] ≥ (3n − 3)/2, a contradiction as n ≥ 5. If Z(J) has even order, then J contains the direct product of [m/8] copies of F8.7 , so n ≥ 7[m/8] = 7[[n/2]/2] = 7[n/4] ≥ 7(n − 3)/4, giving n = 7 and [m/8] = 1. Then m2 (J) ≥ m2 (F ) = 6, but by [IA , 5.2.10d], m2 (J) = 1+3[m/8] = 4, a contradiction. Finally suppose that J is a group of Lie type in characteristic 2 but not in any odd characteristic. As A1 contains a Sylow 2-subgroup of H, it follows that A1 ∩ J lies in a proper parabolic subgroup N of J, by Tits’ lemma [IA , 2.6.7]. Also as Out(J) is solvable, [Σ, Σ] ≤ A1 ∩ J. Let R = O2 (N ). Then R ≤ O2 (A1 ∩ J) = O2 (Z(A))Q and R  A1 ∩ J. Since [Σ, Σ] ∼ = An acts faithfully on R, Ω1 (R) must contain the trace-0-submodule F in Ω1 (O2 (M )). As Ω1 (R) ≤ Ω1 (O2 (M )), either Ω1 (R) = F or Ω1 (R) = Ω1 (O2 (M )). In either case NA (Ω1 (R)) acts monomially on B, so NA (Ω1 (R)) ≤ A1 . Hence N ≤ A1 and then N = A1 ∩ J. In particular, [Σ, Σ] ∼ = An is the last term of the derived series of a Levi complement of N . Thus An ∈ Chev(2), so n ∈ {5, 6, 8}. ∼ Σ8 acts as O + (2) on its unique Suppose that n = 8. Then A1 /O2 (A1 )Z(A) = 6 ∼ E2a , a = 7 or 8. The structure nontrivial composition factor within Ω1 (R) = of N then determines that J/Z(J) ∼ = Ω+ 8 (2). Indeed, as the trace-0 submodule of Ω1 (R) contains a unique involution of Z(A), we must have J ∼ = 2Ω+ 8 (2). Next, since + ∼ Σ , we must have A/Z(A) O (2). As Σ stabilizes I := {b1 , . . . , b8 }, AutA1 (F ∗ ) ∼ = 8 = 8 Σ contains a reflection t on B (acting as a transposition on I). Since CA (A1 ) = Z(A1 ) ≤ Z(A), the action of t on J is uniquely determined by its action on A1 . Hence the isomorphism type of J t is uniquely determined, so J t ∼ = W (E8 ). Therefore by [III17 , 1.5], J t ≡B W (E8 ). Note that J t Z(A) = A, so J t  A. By [III17 , 1.21], W = J t and A = W ∗ Z for some Z ≤ Z(GL(B)), as desired. Next, suppose that n = 6. Then A1 /O2 (A1 )Z(A) ∼ = Sp4 (2). It follows = Σ6 ∼ that J/Z(J) ∼ = Sp6 (2) in this case, so J contains a subgroup isomorphic to L2 (8) and hence a Frobenius subgroup of order 8.7. However, since mp (B) = n = 6, this case is ruled out. Finally, suppose that n = 5. Then A1 /O2 (A1 ) ∼ = Σ5 acting as O4− (2) on [A1 , A1 ] ∩ O2 (A1 ). It follows that J ∼ = U4 (2) ∼ = P Sp4 (3), so J ∈ Chev(3), a final contradiction. 

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

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We now examine the case in which (8D) fails. By Lemma 8.6, this implies that (8C2) holds. Hence, n = 6, L ∼ = D4 (q), and there is g ∈ NDo ∩ NG (B) such that g b3 = bi for any i = 3, 4, 5, 6. We set t = t34 t56 , and regard t both as an element of NK o (B) and as an element of A. Lemma 8.16. Assume (8A) and (8C2). Then CA (t) = NA (Do ). Proof. As Do = CB (t), we see that CA (t) ≤ NA (Do ). On the other hand,  L = O p (E(CG (Do )))  NDo and NLo (B) = CLo (B)F0 Σ , where F0 := t34 , t45 , t56 ∼ = E23 and Σ ∼ = Σ4 permutes {b3 , b4 , b5 , b6 }. Identifying F0 and Σ with subgroups of NA (Do ), we conclude that Lo ∩ NA (Do ) = F0 Σ  NA (Do ). As t = Z(F0 Σ ), we  conclude that t  NA (Do ), whence CA (t) = NA (Do ), as claimed. o

Now set M = NA (Do ). Then M acts on Do and on B2 = b3 , b4 , b5 , b6 =  B ∩ Lo = B ∩ O p (E(CG (Do ))). For any element or subset X of M , we write X|Do and X|B2 for the images of X in Aut(Do ) and Aut(B2 ), respectively.  Let B = {b3 , b4 , b5 , b6 }, a frame in B2 such that E(CLo (X)) ∼ = D3q (q) for all X ∈ B. There exist exactly three such frames in B2 , which we call “special frames”; and the group NAut(Lo ) (B2 ) permutes transitively the set of special frames, by [III17 , 2.26]. Let V be a natural Lo -module such that dim[V, bi ] = 2 for all i = 3, 4, 5, 6. Then the stabilizer in NAut0 (Lo ) (B) of B equals NO(V ) (B). Moreover,  we let t23 be the involution in K o ∼ = D5q (q) inverting b2 and b3 but centralizing b4 , b5 , and b6 . Then t23 ∈ M , t23 normalizes Lo and Lo t23 ∼ = O(V ). Lemma 8.17. Assume that (8A) and (8C2) hold, but that (8D) fails. Let MB be the stabilizer in M of B. Then the following conditions hold: (a) |M : MB | = 3; (b) MB |Do normalizes b1 and b2 and is abelian; and (c) NM (b1 ) ≤ MB . Proof. The two special frames in B2 other than B are interchanged by t23 ∈ M . Moreover, by (8C2), M does not stabilize B. Hence (a) holds. Let g ∈ NM (b1 ). Then a preimage g of g in NG (Do ) ∩ NG (B) normalizes K o , so stabilizes the frame b2 ∪ B in B ∩ K o . This implies (c). For (b), let g ∈ MB . Thus b6 g = bi for some i ∈ {3, 4, 5, 6}. There exists h ∈ AK o such that bj h = bj for j = 1, 2, and b6 h = bi . Replacing g by gh−1 , g we may assume that b6 = b6 . Let L6 = E(CK o (b6 )). If L6 is a component of CG (b6 ), then b1 = CDo (L6 ) and b2 = Do ∩ L6 are normalized by g. Suppose then that L6 has a vertical pumpup K6 in CG (b6 ). As b6 ∼K o b2 , K6 ∼ = K o by (8A3). Let V6 be the natural orthogonal module for K6 . As L6 = E(CK6 (b1 )), we see that dim([V6 , b1 ]) = 2 and [V6 , b1 ] ∩ [V6 , L6 ] = 0. It follows that b1 NM (b6 ) ⊆ {b1 , b2 } with equality only if dim([V6 , b2 ]) = 2. But in the case of equality we reach the hypothesis (8D), contrary to assumption. Hence, we again conclude that g normalizes both b1 and  b2 , and the lemma follows. The mapping M → Aut(Do )×Aut(B2 ) sending m → (m|Do , m|B2 ) is injective. It follows that M1 := CM (Do )CM (B2 )  M , and ∼ M/M1 ∼ (M |Do ) / (CM (B2 )|Do ) = = (M |B ) / (CM (Do )|B ) . 2

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2

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Lemma 8.18. Assume that (8A) and (8C2) hold, but that (8D) fails. Then the following conditions hold: (a) M/M1 ∼ = Σ3 × Y , where Y ≤ AutF2 (Fq ). In particular, Y is cyclic; and (b) t23 does not map into O2 (M/M1 ). Proof. Since |M : MB | = 3, there is a 3-element g ∈ M cycling the three special frames. As t23 interchanges two special frames, we may choose g to be inverted by t23 . Now since CM (Do ) ≤ CM (b1 ), CM (Do ) ≤ MB and hence the images of g, t23 in M |B2 /CM (Do )|B2 and M/M1 have a Σ3 -quotient. This proves (b). Moreover, CM (Do )|B2 contains AutLo (B2 ), so M/M1 is involved in Out(Lo ) ∼ = Σ3 × AutF2 (Fq ). Hence (a) holds as well.  Lemma 8.19. Assume that (8A) and (8C2) hold, but that (8D) fails. Let M0 be the joint stabilizer in M of all the special frames. Then the following conditions hold: (a) M0 ≤ MB and M/M0 ∼ = Σ3 ; (b) M0 |Do consists of scalar mappings; and (c) CM (B2 ) = z , where either z = 1 or z = −1B · t, according as −1B ∈ M or −1B ∈ M . Proof. Note that −1B ·t centralizes B2 and inverts Do elementwise; as t ∈ M , −1B · t ∈ M if and only if −1B ∈ M . Part (a) is clear from the structure of g, t23 . By Lemma 8.17b, M0 normalizes b1 and b2 . If M0 does not act as scalars on Do , then b1 and b2 are normalized by NM (M0 ) = M , contradicting Lemma 8.17c. Hence (b) holds. Let g ∈ CM (B2 ). Then g ∈ M0 , so by Lemma 8.17, g induces a scalar mapping on Do . Then a preimage g of g in NG (Do )∩NG (B2 ) normalizes K o , and then either inverts or centralizes b2 , by [III17 , 2.27]. Accordingly g = 1 or −1B · t, proving (c).  Lemma 8.20. Assume that (8A) and (8C2) hold, but that (8D) fails. Then M = W2 × Z(A) with W2 ∼ = W (F4 ) acting faithfully on B2 and inducing a Σ3 action on Do . Proof. Let M2 be a subgroup of M containing M1 and such that M2 /M1 ∼ = Σ3 . Then M2  M (see Lemma 8.18a). Moreover, M2 |B2 ∼ = W (F4 ) by [III17 , 1.26a]. As |CM (B2 )| ≤ 2 by Lemma 8.19, M2 |Do has order 6 or 12. Let T ∈ Syl2 (M2 ). If z = 1, then z ∈ [T, T ]. As t ∈ [T, T ], it follows that if −1B ∈ M , then −1B = zt ∈ [T, T ]. Hence −1B ∈ [M, M ], by [IG , 15.12(i)]. As a result, M2 ∼ = W (F4 ) or W (F4 ) × −1 . In particular, M |Do contains a normal Σ3 -subgroup X. By Lemma 8.17b, M |Do has an abelian subgroup Y of index 3. Hence AutA (Do ) = XY = X × CY (X), and as X is absolutely irreducible on Do , CY (X) acts as scalars on Do . On the other hand, M |B2 has normal subgroups subgroups M2 ∼ = W (F4 ) and AutLo (B2 ) ∼ = W (D4 ). It follows by [III17 , 1.26b] that M |B2 = M2 C, where C is a cyclic group of scalar transformations. In particular, the preimage of C in M lies in MB and so projects into Y . Consequently M = M2 Z1 , where Z1 induces scalar transformations in both Do and B. In particular, Z1 normalizes b1 so embeds in AutNG (b1 ) (B). Thus by [III17 , 2.27], Z1 ≤ −1B · t Z(A). Hence M ≤ M2 −1B Z(A) = M2 Z(A), whence M = M2 × Z with Z ≤ Z(A). The proof is complete. 

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8. THE CASE AutK (B) ∼ = W (Dn ), n ≥ 4

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Lemma 8.21. Suppose that M = W2 × Z(A) as in Lemma 8.20. Then n = 6 and A = W × Z(A) with W ≡B W (E6 ). Proof. By Lemma 8.16, M = CA (t). Let S ∈ Syl2 (M ). Then t = Z(S) ∩ [S, S], whence S ∈ Syl2 (A). Let Z = Z(A) and X = O2 (A). Then t inverts XZ/Z. However, t is not normal in A ∩ K o ∼ = E24 Σ5 . Hence, X ≤ Z. Let R = O2 (A). Then R ≤ O2 (W2 ) × Z. Hence R ≤ (O2 (A ∩ K o ) ∩ O2 (W2 )) × Z. Now O2 (A ∩ K o ) ∩ O2 (W2 ) ∼ = E23 contains no nontrivial A ∩ K o -invariant subgroup. ∗ Thus R = O2 (Z) and F (A) = H ∗ Z, where H = E(A). Clearly, H is quasisimple. Moreover, A ∩ K o is generated by reflections, so H ∩ K o ∼ = A ∩ K o. 6 Let H = HZ/Z. Then |H|2 = 2 and t is a central involution in H with centralizer isomorphic to a subgroup of W containing [W, W ]. We see easily from the structure of CA (t) that H is neither a sporadic group nor an alternating group. If H ∈ Chev(r), r odd, then we see, comparing CA (t) with [IA , Table 4.5.1], that H∼ = G2 (3) or P Sp4 (3). If H ∈ Chev(2), then WH is a maximal parabolic subgroup of H, and it follows that H ∼ = U4 (2) ∼ = P Sp4 (3). As |G2 (3)| is not divisible by 5, we conclude that H ∼ = P Sp4 (3). Thus A = W × Z(A) with W ≡B W (E6 ), as claimed.  The results of this section prove Proposition 8.2. Specifically, if (8D) holds, then Lemma 8.10 shows that Proposition 8.2a holds if F  A, while Lemma 8.15 shows that Proposition 8.2c or 8.2d holds if F  A. Finally, Lemmas 8.16, 8.20, and 8.21 show that if (8C2) holds, then Proposition 8.2b holds, completing the proof of the proposition in all cases. ∼ W (Dm−1 ), m ≥ 5, and that (y, L) has Corollary 8.22. Suppose that WK = been chosen to satisfy (8A). (Such a choice is possible by Lemma 8.1.) Then one of the following holds: (a) Proposition 3.4j holds with W ∼ = = W (Dm ) or W (Cm ) containing W ∗ ∼  W (Dm ) with index 1 or 2; moreover, if m = 5, then Lu ∼  A4q (q) for any = u ∈ D# ; (b) Proposition 3.4l holds with W ∗ = W ∼ = W (E6 ) and m = 6. m−1

q (q). Thus we may Proof. As WK ∼ = W (Dm−1 ), we must have K ∼ = Dm−1 apply Proposition 8.2. If Proposition 8.2a or 8.2b holds, then Proposition 3.4j or 3.4l holds, and accordingly, (a) or (b) of this corollary follows. It remains to rule out conclusions (c) and (d) of Proposition 8.2. First, if Proposition 8.2c holds, then B ∼ = = E35 is a natural module for [W, W ] ∼ Ω5 (3), by [III17 , 1.14]. Then by (8A), for any u ∈ De , Lu ∼ = D4 (q), so WLu ∼ = W (D4 ) has order 26 .3. But if D (as a subspace of B, the natural [W, W ] ∼ = Ω5 (3)module) contains a nonsingular vector u, then WLu contains all reflections in CW (u) 2 and hence contains Ω± 4 (3), whose order is divisible by 3 , contradiction. Hence D is totally singular. But then N[W,W ] (D) is a parabolic subgroup of [W, W ] and in particular cannot normalize the complement B ∩L to D in B, again a contradiction. Thus (c) cannot hold.  Suppose finally that Proposition 8.2d holds, so that K ∼ = D7q (q). Then by # ∼ [III17 , 3.3], there is u ∈ D such that CW (u) = W (E7 ), contradicting Table 15.2. This completes the proof of the corollary. 

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

206

9. Some Exceptional Cases In this section, we consider certain cases when Ld is an exceptional group for some d ∈ D# and C(d, Ld ) has p-rank 1. This, of course, includes the case when d = x and Ld = K. However, it will also be applied in certain cases when K/Z(K) ∼ = Ln (q). As usual, we set A := AutG (B) and let W denote the subgroup of A generated by reflections. Our setup is as follows. (1) (7A) holds with n = 8 or 9; (2) For some d ∈ (Do )# with Lo < Lod , Lod ∼ = En−1 (q); and (9A) n−1 q o o o (3) Either D = D and L = L, or K := Lox ∼ (q). = Dn−1 Here is our main result: Proposition 9.1. Assume (9A). Then n = 8 and A = W ∗ Z, where W ≡B W (E8 ) and Z = Z(A) acts as scalars on B. As immediate corollaries, we have: Proposition 9.2. Suppose that d ∈ De with Ld /Z(Ld ) ∼ = En−1 (q), n ∈ {8, 9}. Then n = 8 and A = W ∗ Z, where W ≡B W (E8 ) and Z = Z(A) acts as scalars on B. Corollary 9.3. We have K ∼  E8 (q). If K ∼ = E7 (q), then Proposition 3.4n = ∗ holds with W = W . We begin the proof of Proposition 9.1. Fix d as in (9A2). By (7A) and since p does not divide |Outdiag(Lod )|, B = (B ∩ Lod ) × CB (Lod ). Since mp (Lod ) = n − 1 = mp (B) − 1 and d ∈ B, mp (C(d, Lod )) = 1. Similarly, by (7A), B induces inner-diagonal automorphisms on K o . Now by [III17 , 1.20], WLod is absolutely irreducible on B ∩ Lod . But by Lemma 5.12, B is absolutely indecomposable as Fp [W ]-module with absolutely irreducible socle V of codimension at most 2. In particular, V ∩ Lod = 1 so V ≥ B ∩ Lod . Thus by Lemma 5.13 we have (9B)

O2 (W ) ≤ Z(W ),

and by indecomposability, d is not W -invariant. We first eliminate the case n = 9.  E8 (q). Lemma 9.4. Lod ∼ = Proof. Suppose on the contrary that Lod ∼ = E8 (q). Then B = d × Bd , where Bd = B ∩ Lod . Let A1 = AutLod (B) ∼ = W (E8 ) and Z(A1 ) = z . Then z inverts Bd and so CB (z) = d , whence CA (z) ≤ NA (d ). However, as Lod  CG (d), it follows that J := [A1 , A1 ] is an intrinsic component of CA (z). It follows from [III17 , 10.28] and (9B) that O2 (A)J ≤ Z(A)J  A. Hence J  A. But then, as d = CB (J), it follows that d is W -invariant, yielding a contradiction.  We complete the proof of Proposition 9.1 by proving the following lemma.

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10. THE CASE K/Z(K) ∼ = P SL± m (q)

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∼ E7 (q). Then A = W ∗ Z, with W ≡B W (E8 ) Lemma 9.5. Suppose that Lod = and with Z = Z(A) inducing scalars on B. Proof. Let A1 = AutLo (B) ∼ = W (E7 ). Let Z(A1 ) = z . Then CB (z) = d d

and so CA (z) ≤ NA (d ). As Lod  CG (d), it follows that J1 := [A1 , A1 ] is a normal component of CA (z) with J1 ∼ = Sp6 (2). By (9B) and L2 -balance, J1 is a component of E(CE(A) (z)). Let J denote the pumpup of J1 in E(A). It follows from [III17 , 10.27] and L2 -balance, and [IA , 6.1.4], that one of the following conclusions holds: (a) J = J1 ; (b) J = L1 ∗ L2 with Li /Z(Li ) ∼ = J1 ; (c) J ∼ = Sp6 (4); (d) J ∼ = L6 (2) or L± 7 (2); (e) J/Z(J) ∼ = U6 (2); (f) J ∼ = Ω− 8 (2); (g) J/Z(J) ∼ = Ω+ 8 (2); or (h) J/Z(J) ∼ = F i22 . Now, mp (B) = 8 and by [III17 , 1.20], J1 × Z2 , and hence J1 , acts absolutely irreducibly on the hyperplane Bd = B ∩ Lod . If J = J1 , then we conclude first that E(A) = J1 and then that d is W -invariant, a contradiction. Hence, J = J1 . If J = L1 ∗ L2 with Li /Z(Li ) ∼ = J1 , and B = B1 ⊗ B2 with Bi a nontrivial irreducible Li -module, i = 1, 2, then mp (Bi ) ≤ 2 for some i, which is absurd. But also if B = CB (L1 )CB (L2 ), then mp ([B, Li ]) ≤ 4 for some i, a contradiction as m2 (Sp6 (2)) = 6. Hence, J is quasisimple. In particular, as n = 8, m2 (J) ≤ m2 (SL8 (p)) = 7. Moreover, if J contains a subgroup T ∼ = 21+6 , then Z(T ) inverts B, whence Z(T ) ≤ Z(J). But by [IA , 6.4.4, ± 2 Table 3.3.1], Sp6 (4), L± 6 (2), L7 (2), 2U6 (2) and (2 )U6 (2) have 2-ranks ≥ 9, so they cannot occur, nor can their overgroups F i22 and 2F i22 . In all the other cases for J, except J/O2 (J) ∼ = 2D4 (2), J contains T ∼ = 21+6 with Z(J) of odd order, ruling them out. Thus, the only possibility is J/O2 (J) ∼ = 2D4 (2). By [IA , 6.1.4], J ∼ = 2D4 (2). Letting y = Z(J) we have that t := yz is a reflection on B with CJt (t) ∼ = Z2 × W (E7 ). With [III17 , 1.5, 1.21], W = J t ≡B W (E8 ) and A = W ∗ Z, as claimed.  Lemmas 9.4 and 9.5 establish Propositions 9.1 and 9.2. 10. The Case K/Z(K) ∼ = P SL± m (q) In this section we keep the notation n = mp (B) and work under the following additional hypothesis: (10A)

K/Z(K) ∼ = Lmq (q) with m ≥ 5. 

Lemma 10.1. The following conditions hold: q (q); (a) L ∼ = SLn−1 (b) One of the following holds: (1) m = n; or (2) m = n + 1 and p divides n + 1; and  (c) B/ x embeds in Inndiag(K) ∼ = P GLmq (q) as the image of a diagonalizable q subgroup of GLm (q).

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

208

Proof. Recall that D = x, y ≤ B with B an elementary abelian p-subgroup of CG (x) of maximal rank. By (5A) and Lemma 3.3a, B induces inner-diagonal automorphisms on K and on every Lu , u ∈ D# , and hence on L as well. If p does not divide m, then B = (B ∩ K) × x with B ∩ K ∈ Ep∗ (K); clearly then,  B ∩ K is diagonalizable. If p divides m, then by [III12 , Def. 1.15], L ∼ = SLkq (q) with k ∈ {m − 1, m − 2}, and we similarly get B = (B ∩ L) × CB (L), with B ∩ L diagonalizable on the natural L-module VL = [B ∩ L, VL ] (note that m ≥ 5). The natural K-module VK contains VL and as dim(VK /VL ) ≤ 2 and p is odd, B must act diagonalizably on VK /VL , hence on VK . Thus (c) holds.  Now CK (y) contains a B-invariant subgroup H ∼ = GLkq (q) with [H, H] = L. As B contains every element of order p in CG (B) and induces inner-diagonal automorphisms on L, it follows that mp (B/CB (L)) = k − 1. Since mp (CB (L)) = 2,  n = mp (B) = k + 1, and all parts of the lemma follow. We have N = NG (B), A = AutG (B) K) ∈ {n − 2, n − 1, n}, W is the subgroup and WK = AutK (B) ∼ = Σn . In view of the throughout that

∼ = N/CG (B), mp (B) = n, mp (B ∩ of A generated by all its reflections, results of Section 6, we may assume

n ≥ 5.

(10B)

∼ SLq (q). Notice also that if q = 2, In particular, if m = 5, then n = 5, whence L = 4 then p = 3 and so K/O3 (K) ∈ G3 . Thus with Lemma 10.1b, (10C)

If q = 2, then m ≥ 7 and so n ≥ 7.

Our goal is the following proposition. Proposition 10.2. Assume (10A). Then A = Z(A)W with Z(A) acting as a cyclic group of scalar matrices on B. Moreover, W = W ∗ × Z with |Z| ≤ 2, and one of the following conclusions holds: (a) Σn+1 ∼ = W ≡B W (An ), m = n, Z = 1, and B is isomorphic as W -module to the quotient of a standard permutation module by the 1-dimensional trivial module; (b) Σn+2 ∼ = W ∗ ≡B W (An+1 ), m = n, p divides n + 2, and B is isomorphic ∗ as W -module to the core of a standard permutation module; if Z = 1, then n = 4; # ∗ (c) n = 5, m = 6, p = 3, Lu ∼ = Ω+ 8 (q) for some u ∈ D , and W ≡B W (E6 ); (d) n = 8, m = 9, p = 3, and W ≡B W (E8 ). As usual, to establish that W or W ∗ ≡B W (L), it suffices by [III17 , 1.5] to determine the isomorphism type of W or W ∗ , and identify which involutions are reflections on B. As a major first step, we shall prove the following result. Proposition 10.3. One of the following holds: (a) For any u ∈ De , Lu /Z(Lu ) ∼ = La (q) for some a ∈ {n, n + 1}; # ∗ ∼ ∼ (b) n = 5, m = 6, p = 3, Lu = Ω+ 8 (q) for some u ∈ D , and W = F (A) = Z2 × Ω5 (3); (c) Proposition 10.2c holds; or (d) Proposition 10.2d holds. We assume that Proposition 10.3 is false and proceed in a sequence of lemmas. As (a) does not hold, [III12 , Theorem 1.2] implies that L ∼ = SLm−2 (q). In

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209

particular, m − 2 = n − 1, so (10D)

m ≥ 6 and p divides m.

Now by (1A), there is d ∈ De − {x }, and for any such d, Ld is a level vertical pumpup of L. Moreover, as (x, K) ∈ J∗p (G), the Gp -depth of Ld /Op (Ld ) is at least that of K/Op (K), and in case of equality, F(Ld ) ≤ F(K) by [III14 , Lemma 1.2]. It follows by [III17 , 10.19] that since (a) fails, d may be chosen so that one of the following holds:

(10E)

(1) Ld ∼ = Sp2m−4 (q), m ≥ 6; m q (2) Ld ∼ (q), m ≥ 6; or = Ω2m−4 ± (3) Ld /Z(Ld ) ∼ (q), m = 8, 9, or 10. = Em−2 m

q (q), m ≥ 7. Lemma 10.4. Ld ∼ = Sp2m−4 (q), m ≥ 6, or Ω2m−4  Proof. Suppose that Ld /Z(Ld ) ∼ (q), m ∈ {8, 9, 10}. Since p divides = Em−2 m, m = 8. Then Proposition 9.2 implies that Proposition 10.3d holds, contrary to our assumption that the proposition fails. It remains to rule out Ld ∼ = Ω+ 8 (q), m = 6. In that case mp (B ∩ Ld ) = 4, n = 5, m = 6, and p = 3 by (10D). Now B = d × Bd with Bd = B ∩ Ld , and WLd ∼ = W (D4 ). Let z = Z(WLd ) ∼ = Z2 . Then CB (z) = d and so CA (z) = NA (d ) and WLd  CA (z). Moreover, if S ∈ Syl2 (CA (d)), then z is the unique involution in Z(S) with |CB (z)| = 3. Hence, S ∈ Syl2 (A). If W has a noncyclic normal abelian 2-subgroup, then by Lemma 5.14, W ∼ = W (C5 ) or W (D5 ); but W ≥ WK ∼ = Σ6 , a contradiction. Thus, if O2 (W + ) = 1, then W ≤ CA (z), contradicting the indecomposability of W on B. So O2 (W + ) = 1. We now repeat paragraphs 4 through 6 of the proof of Proposition 7.7, to conclude that E(A) ∼ = Ω5 (3), whence E(A) is absolutely irreducible on B. As noted above, p = 3. Thus |Z(A)| ≤ 2. If there exists a reflection t ∈ A − F ∗ (A) outside F ∗ (A), then conclusion (c) of Proposition 10.3 holds with W = E(A) t ; otherwise conclusion (b) holds. In either case we have a contradiction, and the lemma is proved.  q (q), D ∩ K = 1, and indeed by [III12 , Def. 1.15], y ∈ K. Choose As L ∼ = SLm−2 w ∈ y K ∩B with E(CL (w)) ∼ = SLm−3 (q). Then by [III17 , 3.11], for some i, v := wxi m−1 ∼ Sp2m−6 (q) or Ωq satisfies Jv := E(CLd (v)) = 2m−6 (q). Moreover, v is K-conjugate ∼ L, and I := E(Kv ∩ Jv ) = E(CL (v)) = to an element of D, Kv := E(CK (v)) = q (q) (see (10C)). Let Lv be the pumpup of Kv in CG (v). As E(CL (w)) ∼ = SLm−3 v is K-conjugate into D, Lv is either a trivial or a level vertical pumpup of Kv containing Jv . Since‘ Jv ≤ Kv , the pumpup is level vertical. Now set Do := d, v and take any e ∈ (Do )# with Jv < Le , where Le is the pumpup of Jv in CG (e). We claim that m−3 q (q) here in the roles of (7A) holds with Do and Jv ∼ = Sp2m−6 (q) or Ω2m−6 (10F) o o D and L there.

We know that Ld and Lv are level vertical pumpups of Jv and to prove the claim we must show that for every e ∈ (Do )# with Jv < Le ,

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210

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

(10G)

Le is a quasisimple level vertical pumpup of Jv on which B induces innerdiagonal automorphisms.

For e = d , this is obvious, so assume that e = d . Note that [III2 , 2.5] applies to I, with x, e, and d here in the roles of x, y, and z there, to yield that Le ≤ E(CG (e)). Suppose  that Le is a diagonal pumpup of Jv . If x normalizes  first d all the components of Le , then mp (B) = mp (CG (x)) ≥ pmp (Jv ) = p(m − 3) ≥ 3(m − 3) > m ≥ n,   d contradiction. If x cycles the components of Le , then CG (x, e ) has a pcomponent H such that H/Op (H) ∼ = Jv ; by Lp -balance H   CK (e), which forces q ∼ Jv = Ω6 (q) and m = 6; but in the orthogonal case, m ≥ 7 by Lemma 10.4. Thus Le is x-invariant, a vertical pumpup of Jv . Suppose that d induces a field 2 2 or graph-field automorphism on Le . Then f (Le ) = (q p )(m−3) > q (m−1) = f (K) since m ≥ 6, which is a contradiction by [III14 , Lemma 1.2]. (Note that if m = 6, then both Le and K lie in G4p [III12 , (1I)], while if m > 7, then both lie in G2p .  Now [III14 , Lemma 1.2] does not apply if m = 7, K/Z(K) ∼ = L7q (q) ∈ G2p , but + Le ∼ = Sp8 (q p ) or Ω8 (q p ) ∈ G4p . But in that case, E(CK (e)) has level q and is not isomorphic to Sp8 (q) or Ω+ 8 (q). As E(CK (e)) is a component of E(CLe (x)), all of these alternatives are impossible.) By [III17 , 10.3], q(Le ) = q(Jv ). If B does not induce inner-diagonal automorphisms on Le , then p = 3 and Le ∼ = 3D4 (q) or D4 (q); but as m ≥ 7 in the orthogonal case, this forces Le = Jv , contrary to assumption. Thus, (10G) and (10F) hold. m

q (q), m ≥ 7. Lemma 10.5. We have Ld ∼ = Ω2m−4

Proof. Otherwise, Ld ∼ = Sp2m−4 (q), m ≥ 6, by Lemma 10.4. As Jv ∼ = Sp2m−6 (q) and Jv < Lv , the untwisted Dynkin diagram of Lv must contain that of Jv by [IA , 4.2.2], so we have that either Lv ∼ = Sp2k (q) for some k ≥ m − 2, or n = 6 and Lv ∼ = F4 (q). The latter is impossible by (10F) and Proposition 7.4. Hence, Lv ∼ = Sp2k (q) for some k ≥ m − 2. By [III14 , Lemma 1.2], we have that k = m − 2. Likewise, for every e ∈ (Do )# , either Le = Jv or Le ∼ = Sp2m−4 (q). Hence the hypotheses of Proposition 7.2 hold and we conclude that W ∼ = E2m−1 Σm−1 . But W ≥ W0 ∼ = Σm , a contra= W (Cm−1 ) ∼ = W (Am−1 ) ∼ ± ∼ diction as m ≥ 6. Hence, Ld = Ω2m−4 (q), as claimed.  Lemma 10.6. For all e ∈ (Do )# , either Le = Jv or Le ∼ = Ld . Proof. Assume that Le > Jv . By (10G), Le is a level vertical pumpup of Jv . Moreover, unless possibly m = 7 and mp (Le ) = 4, F(Le ) ≤ F(K), by [III14 , Lemma m−1 ∼ Ωq 1.2]. As Jv = (q) and m ≥ 7, it follows by [III17 , 10.20] that mp (Le ) ≥ 5, 2m−6 m

∼ Ωq ∼ q ∼ and either Le = 2m−4 (q) = Ld , as desired, or m ∈ {7, 8, 9, 10} with Le = E6 (q), E7 (q), or E8 (q). Assume the latter. As F(Le ) ≤ F(K) and K/Z(K) ∼ = Lm (q), we have m ≥ 8. As p divides m, we have m ∈ {9, 10}. If m = 9, then Le ∼ = E7 (q) and the hypotheses of Proposition 9.1 hold. Also, p = 3, whence conclusion (d) of Proposition 10.3 holds, contrary to assumption. If m = 10, then p = 5, Le ∼ = E8 (q)  and L ∼ = Ld = SL8q (q). As mp (B) = 9, Proposition 9.1 is contradicted. Thus, Le ∼ and the lemma is proved. 

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We can now complete the proof of Proposition 10.3. By Lemma 10.6, the hypotheses of Proposition 8.2 hold and so since m ≥ 6, one of the following conclusions holds: (a) W ∼ = W (Dm−1 ) or W (Cm−1 ); (b) m ∈ {6, 7} and W contains W (E6 ) with index ≤ 2; or (c) m = 9 and W ∼ = W (E8 ). As W ≥ WK ∼ = W (Am−1 ), the first case leads to a contradiction. If the second case holds, then as |WK | divides |W (E6 )|, m = 6. As p divides m, this case and the third case satisfy conclusion (c) or (d) of Proposition 10.3, and we have completed the proof of Proposition 10.3. Lemma 10.7. Proposition 10.3b does not hold. Proof. Suppose that it holds. By [III17 , 1.15], we may identify B with the natural [W, W ] ∼ = Ω5 (3)-module. For any u ∈ De , WLu is a normal subgroup of NW (u ). As the stabilizer 33 Σ4 of a singular vector has no normal subgroup which is a Weyl group of a group of Lie type, D must be an anistropic plane. As such, it  contains two u such that WLu ∼ = Σ6 (and so Lu ∼ = A5q (q)) and two u such that WLu ∼ = W (D4 ) (and so Lu ∼ = D4 (q)). Next, by Lemma 5.3, for each u ∈ D# , m3 (C(u, Lu )) = 1 and so C(u, Lu ) is 3-solvable. Likewise, by Lemma 5.1a, every element of CG (LD) of order 3 is central, and so CG (LD) is 3-solvable. Fix a B-invariant root SL2 (q)-subgroup I ≤ L. Then I is a root SL2 (q)subgroup in Lu for each u ∈ D# . Let D1 = {u ∈ D# | Lu ∼ = D4 (q)} and D2 = D# − D1 . Set C = BCG (I), and Cu = E(CLu (I)) for each u ∈ D# . Then  Cu ∼ = L2 (q) × L2 (q) × L2 (q) or L4q (q) for u ∈ D1 or D2 , respectively. As C(u, Lu )  is 3-solvable, Cu = O 3 (E(CC (u))) = L3 (CC (u)) for each u ∈ D# . Likewise CD := E(CL (I)) = L3 (CC (D)) ∼ = SL2 (q). Note that q > 2 by (10C). Set C = C/O3 (C). By L3 -balance, C has a 3-component J such that C u =  L3 (CJ (u)) ∼ = L4q (q) for all u ∈ D2 , and L3 (CJ (u)) is the direct product of one, two, or three copies of L2 (q), for all u ∈ D1 . Moreover, L3 (CJ (D)) ∼ = L2 (q). By  [III17 , 3.22], J/Z(J) ∼ = L6q (q). As IJ ≤ G and m3 (B) = 5, Z(J) = 1. But now AutI (B) contains a reflection t inverting B ∩ I and centralizing B ∩ J. Thus CW (t) contains a W (A5 ) ∼ = Σ6 subgroup. But Z2 × Ω5 (3) contains no such reflection, a contradiction completing the proof.  In view of Proposition 10.3 and Lemma 10.7, we may assume for the rest of the proof of Proposition 10.2 that conclusion (a) of Proposition 10.3 holds, i.e., that For all u ∈ De , we have Lu /Z(Lu ) ∼ = La (q) for some a ∈ {n, m} ⊆ (10H) {n, n + 1}. Now let V be an m-dimensional vector space over F = Fq (Fq2 if q = −1) affording a natural representation for the universal version of K. By Lemma 10.1c, we may choose a basis {e1 , . . . , em } for V (orthonormal if q = −1) such that B ∩ K acts diagonally with respect to this basis. (Note: the action of elements of B may only be determined up to scalars.) We let A0 = A ∩ K, so that A0 permutes B := {Fei | 1 ≤ i ≤ m}. We may assume that L stabilizes em and the span of {e1 , . . . , em−1 }, or L stabilizes em−1 and em and the span of {e1 , . . . , em−2 }. Let H be the SL2 (q)-subgroup of L stabilizing the span of {e1 , e2 } as well as each ei , 3 ≤ i ≤ m. Let t ∈ NH (B) be the involution interchanging e1 and e2 , and let τ

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denote its image in A0 . Then τ acts as a reflection on B. Set b = [B, τ ]. Then dim([V, b]) = 2 and b ∈ H ≤ L. Let 

Jb := O 2 (CL (b)). q q Then Jb = E(CL (b)) ∼ (q) and Jb < Kb := E(CK (b)) ∼ (q). More= SLn−3 = SLm−2 over, we have b, t ≤ H ≤ L with

Kb = E(CK (H)). We fix this choice of t, τ , b, Jb , and Kb , and use this information to study CA (τ ). ∼ Σk , k ∈ {m − 1, m}, Lemma 10.8. CW (τ ) has a subnormal subgroup Στ = whose derived group is a component or solvable 2-component Lτ isomorphic to Ak . Moreover, the unique nontrivial Lτ -composition factor in B is isomorphic to the core of the standard Fp -permutation module for Lτ . Proof. By construction, [B, t] = [B, τ ] = b . Hence, NG (B) ∩ CG (t) ≤ NG (b ). Let Nb = NG (b ), so that CA (τ ) = CA∩Nb (τ ). We will show that Nb  has a component Lb such that Lb /Z(Lb ) ∼ = Lkq (q), k ∈ {m − 1, m} and [t, Lb ] = 1. Set Στ = AutLb (B) = A ∩ Lb . As Lb   Nb , it will follow with [III17 , 2.16] that Σk ∼ = Στ   CA∩Nb (τ ) = CA (τ ). As transpositions of Στ act as reflections on B, Στ   CW (τ ). Thus CW (τ ) will have an alternating Ak component or 2-component Lτ = [Στ , Στ ] as stipulated in the statement of the lemma. Moreover, we may identify Lτ with a subgroup of Lb . Then [B, Lτ ] ≤ B ∩ Lb , and as Lτ -module, [B, Lτ ] will clearly be a section of the standard permutation module, completing the proof of the lemma. q (q). Let Lb be the pumpup of Kb Suppose first that m = n, so that L ∼ = SLm−1  # in CG (b). Now, for any u ∈ D , either Lu /Z(Lu ) ∼ = Lmq (q) ∼ = K/Z(K) or Lu = L. As b ∈ H ≤ L ≤ Lu , [III17 , 12.17] yields that accordingly, Kb,u := E(CLu (b)) ∼ = Kb q or Kb,u = E(CL (b)) ∼ (q). Moreover, in either case, [t, Kb,u ] ≤ [H, Kb,u ] = = SLm−3 1. Since there are at least two subgroups u ∈ E1 (D) such that Lu > L, Lb is a q nontrivial pumpup of Kb ∼ (q). = SLm−2 We now examine Lb from a different point of view. First, there exist NK (B)conjugate elements b1 , . . . , bm ∈ B # with B = b1 , . . . , bm CB (K), [b1 · · · bm , K] = q q (q) for each i = 1, . . . , m, and E(CK (bi , bj )) ∼ (q) 1, E(CK (bi )) ∼ = SLm−1 = SLm−2  for each 1 ≤ i < j ≤ m. Indeed, each bi acts on K like an element of GLmq (q) with just two eigenspaces on V , namely the span of ei and the span of all the ej , j = i. Replacing b2 by τ (b 1 ) ∈ b2 CB (K), we may assume that τ interchanges b1 and b2 and so b = b−1 1 b2 . We also have D = bm , x . Set D1 = b1 , x , so that D1 = Dg for some g ∈ NK (B). Also, let I be the subgroup of K supported on the span of e3 , e4 , . . . , em−1 and centralizing all the other ek ’s. Thus I ≤ E(CL (b)). We use D1 to investigate CG (b). We have  SL q (q) ∼ = Lg  E(CG (D1 )). m−1

D1# ,

the pumpup Ld of Lg in CG (d) is either equal to Lg Moreover, for any d ∈ or centrally isomorphic to K, since the corresponding property holds for D and L. Note that D1 b = x, b1 , b = x, b1 , b2 = D1 b2 . Let JD1 = E(CLg (b)) = q (q). Let J be the pumpup of JD1 in CG (b), E(CLg (b2 )) = E(CK (b1 , b2 )) ∼ = SLm−2 # and for each d ∈ D1 , let Jd be the pumpup of JD1 in CG (d, b ). Then I ≤ JD1 ≤ J, so J = Lb . Given the isomorphism type of JD1 (and the fact that m ≥ 5, with m ≥ 7 if q = 2), we have Jd ∈ Chev(2) and J ∈ Chev(2), for each d ∈ D1# . (Semisimplicity

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follows as usual with [III8 , 2.5].) Now x ∈ D1 and Jx ≤ E(CK (b)) ∼ = JD1 , so Jx = JD1 . Similarly, if d ∈ D1 − x and Ld = Lg , then Jd = JD1 . Let d = b1 xk ∈ D1 − x with Ld > Lg . Then Ld /Z(Ld ) ∼ = K/Z(K) so Jd , a component of CLd (b), is q (q). (Note that [b, Lg ] = 1, so [b, Ld ] = 1.) If Jd = JD1 isomorphic to JD1 or SLm−1 for all d ∈ D1# , then by [III17 , 9.12], J = JD1 . But we saw above that J = Lb is q a nontrivial pumpup of SLm−2 (q), so this is a contradiction. Thus Jd > JD1 for # at least some d ∈ D1 , and so J = Lb is actually a pumpup (possibly trivial) of q SLm−1 (q). Now Lb ∈ Chev(2). We also know that if Lb /Op (Lb ) and K/Op (K) lie in Gp with the same Gp -depth, then F(Lb ) ≤ F(K) by [III14 , Lemma 1.2]. As m ≥ 5 it q Lb ∼ (q) or Lb /Z(Lb ) ∼ follows from [III = K/Z(K). Moreover, = SLm−1  17 , 10.18] that # clearly Lb = Kb,u | u ∈ D , and we saw above that [t, Kb,u ] = 1 for all u ∈ D# . Therefore [Lb , t] = 1, completing the proof in case m = n. q Hence, we may assume that m = n + 1, whence L ∼ (q), b ∈ uNK (B) for = SLm−2 some u ∈ (D ∩ K) − x , and p divides m. Thus m = n + 1 ≥ 6. Moreover, if q = 2, then as p = 3 divides m > 6, we have m ≥ 9.   q (q) and We have O 2 (CL (b)) = Jb < Kb = O 2 (CK (b)), with Jb ∼ = SLm−4   q 2 ∼ Kb = SLm−2 (q). As L = O (CK (D)), it follows that D does not centralize Kb , and so CD (Kb ) = x .  There is v ∈ De − {x }. Then Lv /Z(Lv ) ∼ = Lrq (q), with r ∈ {n, n + 1}. Moreover, L = E(CLv (D)) is canonically embedded in Lv , whence  q Jb = O 2 (CL (b)) < E(CLv (b)) = E(CLv (H)) ∼ (q). = SLr−2 

Let Lb be the subnormal closure of Kb in CG (b). As O 2 (CL (b)) ≤ Lb ∩ E(CLv (b)), it follows that E(CLv (b)) ≤ Lb . Now 



O 2 (CKb (v)) = O 2 (CKb (D)) = Jb < Kb . Hence 



Kb > O 2 (CKb (v)) = Jb = O 2 (Kb ∩ Lv ) < E(CLv (b)). Thus, Kb ≥ E(CLv (b)) and so Kb < Lb . As b ∈ uNK (B) for some u ∈ D# ,  q Lb is a vertical pumpup of Kb , and indeed Lb /Z(Lb ) ∼ (q). We = Lnq (q) or Ln+1 may then assume that we chose v = u, and so b and v are NK (B)-conjugate. Moreover, Kb = E(CLb (x)) and E(CLv (b)) = E(CLb (v)) with x, v ≤ B, and B inducing a diagonalizable group of inner-diagonal automorphisms on Lb . Hence by [III17 , 9.14], if we set L1 := Kb , E(CLv (b)) , then either Lb = L1 , or Lb /Z(Lb ) ∼ = q (q). It follows from [III17 , 2.16] that AutLb (B) ∼ K/Z(K) with L1 ∼ = Σm−1 = SLm−1 or Σm . Now t inverts b, and so t ∈ NG (b ). As t centralizes Kb , it follows that t normalizes Lb centralizing both Kb and E(CLv (b)), whence L1 ≤ CLb (t). Suppose  that L1 = Lb , whence Lb /Z(Lb ) ∼ = Lmq (q). By [III17 , 6.23], there is no involutory q (q). Hence, automorphism of Lb centralizing a subgroup of Lb isomorphic to SLm−1 in any case, we conclude that t centralizes Lb , and the proof of the lemma is complete.  We now refine Lemma 5.12.

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Lemma 10.9. Either B is an absolutely irreducible W1 -module or S = Soc(B) is an absolutely irreducible W1 -module of codimension 1 in B. In any case, O2 (W ) ≤ Z(W ) = CW (W1 ) acts as scalars on B. Proof. If S has codimension at most 1 in B, this follows immediately from Lemmas 5.12a and 5.13. Hence, we may assume that S = B0 has codimension 2 in B, and hence has dimension n − 2. (Here, as before, B0 = [B ∩ K, WK ].) Then necessarily p divides m. Also, b = [B, τ ] ≤ B0 = S. It follows that S = b ⊕ CS (τ ) as CW1 (τ )-module. In particular, Στ acts faithfully on CS (τ ), and dim(CS (τ )) = n − 3. However, Στ ∼ = Σm−1 or Σm , and p divides m. Thus by [III17 , 1.8], Στ has no faithful module of dimension as small as m − 3 in characteristic p. As n ≤ m, this is a contradiction, and the lemma follows.  Now we can use the structure of Lτ to study E(A) and complete the proof of Proposition 10.2 when Lτ is non-solvable. Lemma 10.10. Suppose that Lτ ∼ = Σm+1 = Ak with k ≥ 5. Then W = E(A) τ ∼ or Σm+2 , and m = n. Proof. Recall from Lemma 10.8 and the hypothesis preceding it that τ ∈ A is a reflection on B, and Lτ is a component of CW (τ ), hence also of CA (τ ), with k ∈ {m − 1, m}. As O2 (W ) ≤ Z(W ), it follows from L2 -balance that either E(A) ≥ EE τ with E/Z(E) ∼ = Ak or E(A) has a component E normalized by τ and such that Lτ   E(CE (τ )). Now τ ∈ WK ∼ = Σm and CWK (τ ) contains a subgroup Kτ with Kτ ∼ = Am−2 and with Kτ ≤ Lτ . It follows that [WK , WK ] ≤ E(A), and [WK , WK ] maps isomorphically into Inn(EE τ ). Hence, E/Z(E) is not isomorphic to Ar for any r < m. Suppose that E = E τ . Then E/Z(E) ∼ = Ak and so by what we just saw, k = m. Suppose in addition that Z(E) = 1. As Lτ is Στ -invariant, so is EE τ . Then Σ := (Στ × τ ) ∩ NG (E) is isomorphic to Στ and contains Lτ . We may write Σ = Lτ σ for some involution σ. Then E σ ∼ = Σ ∼ = E τ σ . It follows τ by [III17 , 1.8] that as E-module as well as E -module, any nontrivial composition factor in B has dimension at least m − 2. Hence dim(B) = n ≥ 2(m − 2) > m, a contradiction. Therefore, Z(E) = 1. Since Lτ = E(CEE τ (τ )) is simple, z 1+τ = 1 for all z ∈ Z(E). As O2 (E) ≤ Z(W ), this implies that Z(E) is a 2-group; but |Z(E)| ≤ 2, so Z(E) = Z(E τ ) ∼ = Z2 . Now EE τ contains an extraspecial 2-group Q of width 2[k/4], and so k = m ≥ n ≥ 22[k/4] . This implies that n ≤ k ≤ 7. In particular, E(A) can contain neither an extraspecial group of width 3 nor the direct product of two extraspecial groups of width 2, so EE τ  A. As W is indecomposable on B, Z(E) inverts B elementwise. But then because of Q, n ≡ 0 (mod 4). As n > 4, this is impossible. Hence E = E τ with Lτ a component of E(CE (τ )). If E = Lτ ∼ = Ak , then [WK , WK ] lies in E × CA (E), projecting onto E. But τ centralizes E and τ does

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not centralize [WK , WK ], a contradiction. Hence Lτ < E. Note also that (10I)

n ≤ m ≤ k + 1.

If E ∼ = At for some t, then t = k + 2,  ≥ 1. By [III17 , 1.8], t ≤ n + 2 ≤ m + 2. Hence as k ≥ m − 1, t = k + 2 ∈ {m + 1, m + 2}, as claimed. Indeed, if m = n + 1 then t < m + 2 so t = m + 1. Moreover, any nontrivial E-composition factor in B has dimension ≥ t − 2 ≥ m − 1 ≥ n − 1 = mp (B) − 1, and as W is absolutely indecomposable on B by Lemma 5.12, E is absolutely indecomposable as well. This implies that A = Z(A) × E τ = Z(A) × W ∼ = Z(A) × Σk+2 . Suppose that m = n + 1. By Lemma 10.1, p divides m, and so p does not divide k + 2. But k + 2 ≥ n + 2 = 2 + mp (B), which is impossible by [III17 , 1.8] as W acts faithfully on B. Hence, m = n, and the lemma holds. Thus we may assume for the remainder of the proof that E ∼  At for any t, whence k ∈ {5, 6, 8, 10} by = [III11 , 1.6]. If k = 10, then by [III11 , 1.6], E ∼ = F5 , and by (10I), n = mp (B) ≤ 11. However, by [IA , Table 5.3w], E contains an extraspecial 2-subgroup of width 4 and hence has no faithful representation over Fp of degree less than 24 , a contradiction. Hence k = 10. Suppose k = 8. Then n ≤ 9 and by [III11 , 1.6], either E ∼ = L4 (4) or E/Z(E) ∼ = (4), let P be an end-node maximal parabolic subgroup of E. Then HS. If E ∼ SL = 4 O2 (P ) is elementary of order 26 , and P transitively permutes the non-principal characters of O2 (P ). Hence by Clifford theory, E has no faithful representation over Fp of degree less than 63, a contradiction. We next assume that E/Z(E) ∼ = HS with k ∈ {6, 8}. Accordingly n ≤ 7 or n ≤ 9. We refer freely to information in [IA , Table 5.3m]. If Z(E) = 1, then E has subgroup V  V H with V ∼ = A6 acting transitively on the = E24 and H ∼ non-principal characters of V . Hence E has no faithful representation over Fp of degree less than 15, a contradiction. Therefore, |Z(E)| = 2. Let Z(E) = z . Then z inverts B elementwise, whence n is even. As E contains a subgroup H ∼ = A8 , E contains a Frobenius subgroup of order 8.7. As p is odd, it follows by [IG , 9.12(ii)] that n ≥ 7, whence k = n = 8. Consequently E is absolutely irreducible on B, whence E = E(A). Moreover, E τ /Z(E) ∼ = Aut(HS), and CE (τ ) = Lτ v with v 2 = z. It follows that Z(A) contains an element w with w2 = z, so that CA (τ ) contains Lτ vw ∼ = Σ8 . Hence 4 divides p − 1. In particular, p = 11. On the other hand, E contains a subgroup R with |R| = 11 and |NEτ  (R)/CEτ  (R)| = 10. Hence E τ has no faithful representation over Fp of degree less than 10, yielding a contradiction in this case. The next case is that k = 6, but E/Z(E) ∼  HS. Then n ≤ 7. It follows from = [III11 , 1.6] and [III17 , 10.37] that one of the following possibilities holds: ∼ Sp4 (4); (a) E = ∼ SL± (2); (b) E = 5 (c) E ∼ = L± 3 (9); or (d) E/Z(E) is a projective simple orthogonal group over F3 or F9 of dimension at least 4. Moreover, in all cases but the last, τ induces an outer automorphism on E. Suppose that E ∼ = Sp4 (4). Then E contains a subgroup R with |R| = 17 and with

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|NEτ  (R)/CEτ  (R)| = 8, whence p = 17. However, Sp4 (4) contains an elementary abelian subgroup of order 52 and GLn (17) contains no such subgroup when n < 8, a contradiction. If E ∼ = SL± 5 (2), then E contains an extraspecial subgroup of order 27 , whence E has no faithful representation over Fp of degree less than 8, a contradiction. Next suppose that E ∼ = E32 = L3 (9) or U3 (9). Let U ∈ Syl3 (E). Then Z(U ) ∼ and NE (U ) is transitive on Z(U )# . Hence by Clifford theory, E has no faithful representation over Fp of degree less than 8, unless possibly p = 3. Thus p = 3. But |U3 (9)| does not divide |GL7 (3)|, so E ∼ = L3 (9). As both Σ6 and A7 contain Frobenius groups of order 20, it follows that L3 (9) contains neither group. Hence n = 6 and, as CA (τ ) contains a subgroup isomorphic to Σ6 , we conclude with [IA , 4.9.1, 4.5.1] that 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. Since p = 3 and τ is a reflection (on B), g = τ τ g therefore embeds in GL2 (3), which is absurd. Thus, if k = 6, then E/Z(E) is a projective simple orthogonal group over F3 or F9 of dimension at least 4, and ∼ Lτ ∼ = A6 ∼ = Ω− 4 (3) = Ω3 (9). Moreover, E cannot be the full spin group, since Lτ is not the spin group. Suppose ∼ that E/Z(E) ∼ = Ω− = P Ωr (9) with r ≥ 4. Then E contains E0 ∼ 4 (9) = L2 (81). Now, the Borel subgroup of E0 has no faithful representation over Fp of degree less than 40 unless p = 3; hence, p = 3. However, 41 does not divide 3n − 1 for any n < 8, a contradiction. Thus, E/Z(E) ∼ = P Ωr (3) with r ≥ 5. If r ≥ 6, then E contains an extraspecial 3-group of width 2, whence n ≥ 9 unless p = 3. Therefore, either r = 5 or p = 3. By [III17 , 6.18], if r ≥ 6, then τ inverts an element of E of order 5, a contradiction as τ is a reflection on B. Therefore r = 5, so E ∼ = Σm acts faithfully on E, and there = Ω5 (3). As WK ∼ H ≤ E acting faithfully on B, n ≤ m ≤ 6 and n ∈ {5, 6}. Moreover, is 24 A5 ∼ = ∼ ∼ A , E τ Aut(E) W (E ) (see [IA , 4.5.1]). Now CGL(B) (H) is as Lτ ∼ = 6 = = 6 (∞) abelian. Hence E = E(A) = A > [WK , WK ]. It follows from [III8 , 5.3] that CA (E) = Z(A), and so A = Z(A) × E τ and W = Z(W ) × E τ . Suppose that n = 5. By [III17 , 1.14], p = 3, and then with [III17 , 3.12], there is u ∈ D# such that CW (u) has a normal subgroup isomorphic to W (D4 ). In particular, [CW (u), CW (u)] ∩ Z(CW (u)) = z where z is an involution with CB (z) = u . On the other hand, E(CW (u)) ∼ = Aa , a ∈ {5, 6}, giving a = WLu ∼ contradiction. Thus, n = 6.  We reach a similar contradiction in this case. Namely, K ∼ = A5q (q), and indeed  Lu ∼ = A5q (q) for all u ∈ De . However, by [III17 , 1.17, 3.10], there is u ∈ D# such that CW (u) ∼ = W (D5 ). As WLu  CW (u) and WLu ∼ = Σ5 or Σ4 , this is impossible. Thus the lemma holds if k ≥ 6. The final step in the proof of the lemma is to derive a contradiction from the assumption k = 5.

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∼ A5 , it follows from [III11 , 1.6] that one of the Then n ∈ {5, 6}. Moreover, as Lτ = following possibilities holds: (a) E/Z(E) ∼ = M12 ; (b) E ∼ = J1 ; (c) E/Z(E) ∼ = J2 ; (d) E ∼ = L2 (16); (e) E ∼ = U3 (4); (f) E/Z(E) ∼ = L3 (4); or (g) E ∈ Chev(5). Suppose that E/Z(E) ∼ = M12 . We use freely the information in [IA , Table 5.3b]. As E/O2 (Z(E)) contains a subgroup isomorphic to E9 : Q8 and n ≤ 6, we must have p = 3. Also E contains a Frobenius group F11·5 , so an element of E of order 11 has distinct eigenvalues on B. In particular, CA (E) is abelian and so E = E(A)  A. If AutA (E) = Aut(M12 ), then A contains a rational element of order 11, which is impossible as n ≤ 6. Hence A = ECA (E) = EZ(A). Since W is indecomposable on B, Z(A) has order dividing p − 1 = 2. Now if Z(E) = 1, then A ∼ = 2M12 ; but then τ , mapping on a noncentral involution of E/Z(E), has order 4, contradiction. Thus, A = E × Z(A) with E ∼ = M12 . As det τ = −1 on B, we conclude that n is odd, so n = 5. On the other hand, let v be a 2-central involution of E. Since O2 (CE (v)) ∼ = Q8 ∗ Q8 , v has four eigenvalues equal to −1. Hence v, v g is a {2, 3}-group for any g ∈ E. But v inverts an element of order 5 in E, a contradiction. Next suppose that E ∼ = J1 . We use [IA , Table 5.3f] freely. E contains Frobenius groups of orders 11 · 10 and 19 · 6, so p = 11 and an element of E of order 19 has distinct eigenvalues, as n ≤ 6. Therefore A = E × Z(A). In particular, A has abelian Sylow 2-subgroups. Hence A cannot contain Σ5 , a contradiction. Suppose that E/Z(E) ∼ = J2 . By [IA , Table 5.3g], E contains no subgroup isomorphic to A6 . Hence m = n = 5. As E contains a rational class of elements of order 7, we must have p = 7. However, GL5 (7) does not contain a subgroup of order 52 , a contradiction. If E ∼ = L2 (16), then E contains a Frobenius group of order 16 · 15, yielding a contradiction to n ≤ 6. If E ∼ = U3 (4), then a Sylow 2-subgroup T of E is special with |Z(T )| = 4 and T / z ∼ = Q8 ∗ D8 for each z ∈ Z(T )# . Hence T requires more than 6 dimensions for a faithful representation over a field of odd cardinality, a contradiction. Suppose next that E/Z(E) ∼ = L3 (4). By Lemma 10.8, CA (τ ) has a subnormal subgroup isomorphic to Σ5 , and it follows that |NA (E)/ECA (E)| is divisible by 4. Thus there is s ∈ NA (E) inducing an involutory automorphism of E, and such that CNA (E) (s) contains a rational element x of order 7. Then as x is rational on the eigenspaces of s and s ∈ Z(A), it follows that n ≥ 7, whereas n ≤ 6, a contradiction. Now suppose that E ∈ Chev(5). Let R ∈ Syl5 (CE (τ )), so that R contains a Sylow 5-subgroup of Lτ and hence is nontrivial. Set N = NE (R) and Q = O5 (N ). Since τ inverts no element of Q# , by Lemma 5.9, [τ, Q] = 1. But by the Borel-Tits theorem and [IG , 5.12(i)], Q = F ∗ (N ). Then by [IG , 3.17(iii)], [τ, N ] = 1. Hence R ∈ Syl5 (N ), so R ∈ Syl5 (E). Hence [τ, E] = 1 by [IA , 2.6.5e], a contradiction as Lτ < E. The lemma is at last completely proved. 

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Having determined the structure of A when Lτ ∼ = Ak , k ≥ 5, it remains for us (for the proof of Proposition 10.2) to consider the case k = 4, i.e., Lτ ∼ = A4 . As n ≤ m ≤ k + 1 but n > 4 by hypothesis, we have m = n = 5 and the following setup: (1) K/Z(K) ∼ = P SL5 (q); (10J) (2) mp (B) = 5; and (3) τ ∈ WK is a reflection and CA (τ ) has a subnormal A4 -subgroup. ∼ Σ6 with B the quotient Lemma 10.11. Under the hypotheses (10J), either W = of the standard permutation module by the trivial module, or p = 7, W ∼ = Σ7 , and B is the core of the standard permutation module. Proof. By Lemma 10.9, O2 (W ) acts as scalars on B. Suppose first that O2 (W ) does not act as scalars on B. Since B is indecomposable as W -module, B is a homogeneous O2 (W )-module; as n is an odd prime, O2 (W ) is abelian but noncyclic. Thus by Lemma 5.14, W ∼ = Σ5 permuting a = W (D5 ) or W (C5 ), with a subgroup Σ ∼ basis B = {b1 , . . . , b5 } of B with respect to which O2 (W ) acts diagonally. We claim that [Σ, Σ] is conjugate to [WK , WK ]. Both are A5 -subgroups of O 2 (W ), which is an extension of Q ∼ = E24 by A5 . Since a Sylow 2-subgroup of O 2 (W )/Q acts freely on Q, there is a unique class of complements to Q in O 2 (W ), proving our claim. Consequently by replacing Σ and B by a conjugate, we may assume that [WK , WK ] permutes B. We may then assume that x = b1 b2 · · · b5 , and since y is fixed by an A4 -subgroup of [WK , WK ], we may assume that y = b1 b2 b3 b4 ba5 , for some a = 1, and so NO2 (W ) (u ) is an extension of Q by 0 ≤ a ≤ p − 1. Then u := x−1 y = ba−1 5 A4 . But, u ∈ D and either Lu = L or Lu /Z(Lu ) ∼ = K/Z(K), whence CW (u) has a subnormal subgroup isomorphic to A4 or A5 , a contradiction. Hence, F (W ) acts as scalars on B. Therefore E := E(W ) = 1. Now E embeds in SL5 (p), whence m2 (E) ≤ 4. Moreover, as n = 5, |Z(E)| divides 5. Thus, if E/O2 (E) is not simple, it is the direct product of two simple groups of 2-rank 2 and so E contains a copy of A4 × A4 . But this forces n ≥ 6, contradiction. Therefore E is quasisimple. Hence, since F (W ) ≤ Z(W ), W (∞) = E by the Schreier property. Again as n = 5, E has no subgroup isomorphic to a 2-transitive Frobenius group with kernel of order 2a > 4. Also, CW (τ ) has a subnormal subgroup isomorphic to Σ4 . Thus, if E ∼ = Ar for some r, then r ∈ {6, 7}, as claimed. By inspection of the involution centralizers tabulated in [IA , Tables 5.3a-h,j,m,n,q,s] for sporadic groups of 2-rank at most 4, we see that E ∈ Spor. If E ∈ Chev(2) − Alt, then as E has no Frobenius subgroup of order 2n (2n − 1) for any n > 4, it follows that q(E) = 2 or 4. Moreover, by inspection of the 2-ranks tabulated in [IA , Table 3.3.1], we are reduced to the possibilities L3 (2), L3 (4), U3 (4), U4 (2), U5 (2), and G2 (2) . As U5 (2) contains an extraspecial subgroup of order 27 , any faithful representation (∞) over a field of odd characteristic has degree at least 8. As WK ≤ E, E ∼ = L3 (2)  or G2 (2) . Consideration of a maximal parabolic subgroup of L3 (4) shows that any faithful representation in odd characteristic has degree at least 15. We showed in the

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previous proof that a Sylow 2-subgroup of U3 (4) requires more than 6 dimensions for a faithful Fp -representation. Hence, E ∼ = U4 (2) ∼ = Ω5 (3). Thus we may assume that E ∈ Chev(r) for some odd prime r; say q(E) = r a . Now, τ acts on E with J   CE (τ ), J ∼ = Σ4 . If E ∼ = L2 (r a ) or if τ induces a field or graph-field automorphism on E, then E ∼ = A6 . Otherwise, we can use = L2 (9) ∼ [IA , 4.2.2, 4.5.1] and deduce that r = 3 (if not, then [J, J] = O 2 ([J, J]) centralizes all Lie components of CE (τ ) and so is cyclic, contradiction). Then [J, J] is a Lie ± component of CE (τ ), whence E ∼ = L± 3 (3) or P Ωs (3), s ≥ 5. Since [WK , WK ] ≤ E, ± E ∼ = P Ωs (3). If p > 3 and s > 5 then we get a contradiction since E contains an extraspecial 3-group of width 2, forcing n ≥ 9. Thus if s > 5, then p = 3, and ∼ from the order of Aut(B) ∼ = P Ω+ = GL5 (3) we see that E ∼ 6 (3) = L4 (3) is the only possibility. But then E ≥ Z4 × A6 , which does not embed in GL5 (3), contradiction. Therefore, s = 5. If τ induces an inner automorphism on E, then by the structure of J, τ acts as a non-2-central involution of E, whence τ inverts an element of E of order 5, contradicting Lemma 5.9. Thus τ induces an outer automorphism on E, and so E τ ∼ = Aut(E). Since n = 5, it follows from [III17 , 1.14] = SO5 (3) ∼ that p = 3 and B is a natural module for E. As τ acts as a reflection on B, CE (τ ) ∼ = SO4± (3) does not contain a normal subgroup isomorphic to J, contradiction. The upshot is that E∼ = Ar , r ∈ {6, 7}. Now F ∗ (W ) = E × Z(W ) so W/Z(W ) embeds in Aut(Ar ). In this case involutions of E have conjugates in WK and invert elements of E of order 5, so τ ∈ EZ(W ). Therefore |W : E τ × Z(W )| ≤ |Out(Σr )| with E τ embeddable in W/Z(W ) ≤ Aut(E). If r = 7, then E τ ∼ = Σ7 . As n = 5, we must have p = 7, and E τ ≡B W (A6 ), so we are done. If r = 6, on the other hand, then since WK ≤ W , WK ≤ E τ ∼ = Σ6 . Since WK fixes x, the E τ -conjugates of x generate a E τ -submodule M of B isomorphic to a quotient of the natural permutation module. But there are only two such quotients of dimension at most 5 – the trivial module and the module of the lemma. As x is not W1 -invariant, B is as claimed. Then as nonconjugate subgroups of E of order 3 have inequivalent representations on B but are Aut(E)-conjugate, we must have W = E τ × Z(W ), |Z(W )| ≤ 2, and E τ ≡B W (A5 ). If −1 ∈ Z(W ) then −τ0 ∈ E τ for some reflection τ0 ; but no involution of E τ has a −1-eigenspace of codimension 1, contradiction. Therefore W = E τ , completing the proof.  Proposition 10.3 and Lemmas 10.7, 10.10, and 10.11 complete the proof of Proposition 10.2. As an immediate corollary, we note:  Corollary 10.12. Suppose that K/Z(K) ∼ = Lmq (q), m ≥ 5. Then one of the cases of Proposition 3.4abf g holds.

11. The Final Case: K/Z(K) ∼ = E6 (q) In this section, we complete the analysis of the cases when K is an exceptional group of Lie type, and thereby complete the proof of Proposition 3.4. Our goal is the following proposition.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

 Proposition 11.1. Suppose that K ∼ = E6q (q). Then A = W Z(A) with W ≡B W (E7 ), Z(A) acting as scalars on B, and mp (B) = 7. In particular, Proposition 3.4m holds with W ∗ = W .

Recall that n = mp (B). Lemma 5.4, together with [IA , 4.10.3a], has the following consequences. If p = 3, then n = 7 and B = x × B0 with B0 = B ∩ K. Moreover, B0 is an irreducible module for WK ∼ = SO5 (3). If p = 3, = W (E6 ) ∼ on the other hand, then n = 6 or 7. In the former case, B ≤ x, K and B/ x is an irreducible module for WK (see (3A4)). In the latter case, B/ x is an indecomposable WK -module with absolutely irreducible 5-dimensional socle.  We have L = E(CK (D)) with L/Z(L) ∼ = L6q (q). Moreover, CK (L) contains a subgroup J ∼ = SL2 (q) generated by a (long) root subgroup and a conjugate, as is clear from the extended Dynkin diagram of E6 . We may assume that y ∈ D# ∩ J. We first prove: Lemma 11.2. For any u ∈ De , one of the following holds:  (a) Lu /Z(Lu ) ∼ = L7q (q); q ∼ (b) Lu = E6 (q); or (c) Lu ∼ = D6 (q). Moreover, y ∈ De . Proof. For any u ∈ De , Lu is a level pumpup of L, and if Lu has Gp -depth 2, like K, then F(Lu ) ≤ F(K) by [III14 , Lemma 1.2]. By [III17 , 10.14a], Lu satisfies the conclusion of the lemma. Hence it is enough to show that Ly > L. Fix v ∈ De − {x }. Since L contains a long root SL2 (q)-subgroup there is g ∈ K such that J g ≤ L. Set b = y g ∈ L and Lb = Lgy ∼ = Ly ; without loss we may assume that b ∈ B. How ever, b ∈ D. Let I = E(CK (b, v )) = E(CL (b)) ∼ = SL4q (q). Then I ≤ E(CK (b)) = g g L ≤ Ly = Lb . On the other hand, I ≤ L ≤ Lv , so I is a component of CLv (b, x ). But Lv is of one of the isomorphism types in the lemma, so [III17 , 10.14b] implies that I < Iv for some component Iv of CLv (b). As I ≤ Lb , also Iv ≤ Lb and [Iv , x] = 1. Hence [Lb , x] = 1, so Lb > L. As Ly ∼ = Lb , Ly > L, completing the proof.  Now y is inverted by an involution t ∈ NJ (B) inducing a reflection τ ∈ W1 on B. In particular, y = [B, t] = [B, τ ]. Lemma 11.3. The socle of B as W1 -module is absolutely irreducible of codimension at most 1 in B. Moreover, CA (W ) = Z(A) induces scalar mappings on B, and O2 (W ) ≤ Z(A). Proof. By [III8 , 5.3] and Lemma 5.12, the first statement implies the others. Let S be the socle of B as W1 -module. We have WK ∼ = W (E6 ) but WL ∼ = W (A5 ); hence y is not normalized by W . Similarly, using Lemma 11.2, we see that W does not normalize x . If p > 3, or if x ∈ K (with p = 3), then the structure of B as WK -module implies that |B : S| ≤ p. On the other hand, if p = 3 with x ∈ K, then B0 = [B ∩ K, WK ] lies in S and contains [y , t] = y . Looking at WLy we conclude again that |B : S| ≤ p. Finally, B/ x is either absolutely irreducible or indecomposable with absolutely irreducible socle of codimension 1. As x is not W -invariant, S is absolutely irreducible, completing the proof. 

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Since y = [B, t], CA (τ ) ≤ A ∩ NG (y ). Moreover, as [x L, t] = 1, it follows from Lemma 11.2 and [III17 , 10.14c] that (11A)

[Ly , t] = 1, whence WLy  CA (τ ).

Our next goal is to eliminate two of the cases in Lemma 11.2.  Lemma 11.4. Ly /Z(Ly ) ∼  L7q (q). =

Proof. Suppose this is false. and let A1 be the subnormal closure of Aτ := [WLy , WLy ] ∼ = A7 in A. Since O2 (W ) ≤ Z(A), A1 ≤ E(A) by L2 -balance, and Aτ is a component of CA1 (τ ). In turn, Aτ contains a subgroup A6 ∼ = AD ≤ CA1 (D), with AD ≤ [WL , WL ]. And AD lies in a component Ax ∼ = Ω5 (3) of CA1 (x). As Ω5 (3) does not embed in A7 , A1 has a component A0 which is a vertical pumpup of A7 and contains Ω5 (3). But then by [III11 , 1.6], A0 ∼ = Am for some m ≥ 9. As A11 contains the direct product of two Z5 -subgroups each generated by a rational element, and it also contains a Frobenius group isomorphic to U3 (2) (with complement Q8 ), however, it requires at least 8 dimensions for a faithful representation. Thus m = 9. However, A9 has no subgroup of index |A9 |/|Ω5 (3)| = 7, a contradiction. The lemma is proved.   Lemma 11.5. Suppose that Ly ∼ = E6q (q). Let A1 be the subnormal closure of E(WLy ) ∼ = Ω5 (3) in A. Then A1 ∼ = Ω+ = L4 (3) ∼ 6 (3) and p = 3.

Proof. By (11A), SO5 (3) ∼ = WLy  CA (τ ). By L2 -balance and as O2 (W ) ≤ Z(A), the subnormal closure A1 of E(WLy ) in A lies in E(A). If E(WLy ) = A1 , then since m2 (E(A)) ≤ m2 (SL7 (p)) = 6 and m2 (A1 ) = 4, A1  A, so y is Ainvariant, contradiction. If A1 is a diagonal pumpup, then 52 divides |GL(B)|; as n ≤ 7, p > 3. As Ω5 (3) and its covering group Sp4 (3) contain respective subgroups E24 A5 and 21+4 , we see that n > 7, a contradiction. Thus A1 is a vertical pumpup of E(WLy ). By [III17 , 10.24], one of the following possibilities holds: (1) A1 ∼ = L4 (4); (2) A1 ∼ = L5 (3); (11B) (3) A1 ∼ = Ω5 (9); or (4) A1 /Z(A1 ) ∼ = P Ω± k (3) with k ≥ 6, and in the first three cases, all involutions of A1 τ are Inndiag(A1 )-conjugate to τ . But CA1 (τ ) ≥ E(WLy ), which contains a copy of D10 . Hence some Inndiag(A1 )conjugate of τ inverts an element of A1 of order 5, so τ does as well, contradicting Lemma 5.9. Thus (11B4) holds, and τ acts as an involution t0 ∈ Ok± (3) with a 5dimensional eigenspace V0 for the eigenvalue −1 on the natural A1 -module V . If k ≥ g g 7, there exists g ∈ Ω± k (3) such that V0 is t0 -invariant and dim(V0 ∩ V0 ) = 3. Then g g τ τ acts on V like the involution t0 t0 , whose −1-eigenspace V2 is 4-dimensional of + type. In particular, O2 (Ω(V2 )) is extraspecial of width 2. It follows that m3 ([B, τ τ g ]) ≥ 4, which is absurd as τ and τ g are reflections on B. Therefore k < 7, so k = 6. As A1 contains an extraspecial 3-subgroup of width 2 and acts faithfully on B with n ≤ 7, we must have p = 3. Suppose finally that A1 /Z(A1 ) ∼ = = U4 (3). As U4 (3) contains a subgroup M0 ∼ E24 A6 , which has no faithful representation in characteristic 3 of dimension less than 15, it follows that |Z(A1 )| = 2 and A1 ∼ = Ω− 6 (3). As Z(A1 ) ≤ SL(B), n = 6. # Choose z ∈ B centralized by a Sylow 3-subgroup P of A1 . Then [A1 , z] = 1, so by the Borel-Tits Theorem, N1 := NA1 (z ) satisfies F ∗ (N1 ) = O3 (N1 ) and lies in

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a maximal parabolic subgroup H of A1 . In particular, z ∈ D, for otherwise N1 would have a normal subgroup isomorphic to W (A5 ), W (A6 ), W (D6 ), or W (E6 ), and so F ∗ (N1 ) = O3 (N1 ), a contradiction. Now D = CB (f ) for some f ∈ A1 of order 5, so z centralizes neither f nor any A1 -conjugate of it. Also |A1 : H| = 280 or 112. Consequently | z A1 |, which is divisible by 5, is a multiple of 280. Hence equality holds and |E1 (B) − z A1 | = 84, so that | u A1 | ≤ 84 for all u ∈ D# . But using Lemma 11.2 we compute that the only possibility is that | u A1 | = 63 for all u ∈ D# . Some of the 21 remaining elements of E1 (B) must be normalized by an element of A1 of order 5, so they lie in D and hence in an orbit of length 63, a contradiction. This completes the proof of the lemma.  Now we can prove Lemma 11.6. Ly ∼ = D6 (q).  Proof. Suppose false, so that Ly ∼ = Ω+ = E6q (q), p = 3, and A1 ∼ = L4 (3) ∼ 6 (3). 2 Since 5 divides |A1 | but 5 does not divide |GL(B)|, A1  A. As L4 (3) contains no copy of either W (A6 )(∞) ∼ = A7 (by orders) or W (D6 )(∞) (which has 2-rank 5),  we deduce that for each u ∈ D# , Lu ∼ = E6q (q) or Lu = L. Now x acts on Ly with L  CLy (x), so x induces an inner automorphism on Ly [IA , 4.7.3A]. As t inverts y and centralizes x, x ∈ Ly . By (11A), [t, Ly ] = 1. In particular, y ∈ Ly , so Ly is simple. Also, we now have symmetry between x and y. Thus, K is simple and there is an involution t1 ∈ Ly inverting x and centralizing K, and inducing a reflection τ1 on B. Moreover, both x and y are conjugate in A to b ≤ L with  E(CL (b)) ∼ = L4q (q), so x and y are A-conjugate. Let ζ = τ τ1 , so that ζ inverts D = [B, ζ] elementwise and ζ centralizes B ∩ K ∩ Ly . Since A1  A and |Out(A1 )| has exponent 2, while τ and τ1 are Aconjugate, ζ ∈ A1 CA (A1 ). Notice that CA (ζ) contains WL = WK ∩ WLy ∼ = Σ6 , which centralizes D. Since K is simple, WK is irreducible on B0 ∼ = E35 and acts trivially on B/B0 . As W does not normalize x , there is a reflection ρ ∈ W such that ρ does not ρ . Set B1 = B0 B0ρ . Then [B, A1 ] ≤ normalize WK . By [III17 , 6.22], A1 ≤ WK , WK B1 , and as ρ is a reflection, m3 (B1 ) = 6. Therefore B1 is the socle of B as an Amodule. We have D ≤ B1 . If Lu = L for some u ∈ D# , then A

| u | = |A : A ∩ NG (u )| ≥ |L4 (3) : NL4 (3) (A6 )| ≥ 34 · 13 > 364 = |E1 (B)|, ∼ E q (q), as noted an obvious contradiction. Thus for any u ∈ D# , L < Lu = A

6

above, and as a consequence | u | = |A : A ∩ NG (u )| = 117. We claim that A xy ∈ x 1 . Namely, since D centralizes a Sylow 5-subgroup R ≤ CA1 (ζ), N1 := NA1 (R) ∩ NA1 (D) controls A1 -fusion in D. Now N1 preserves the decomposition B1 = CB1 (R) × [B1 , R]. But as [R, ζ] = 1, CB1 (R) = [B1 , ζ] = D and [B1 , R] = CB1 (ζ), so N1 ≤ CA1 (ζ). Now E(CA1 (ζ)) ∼ = A6 clearly centralizes [B1 , ζ] = D, and we compute that CA1 (ζ)/E(CA1 (ζ)) ∼ = D8 . Hence the action of CA1 (ζ) on E1 (D) has two orbits of length 2, proving the claim. The number of elements of E1 (B1 ) that are A1 -conjugate into D is therefore 2|A1 : NA1 (x )| = 234. Let A / u 1 for any u ∈ D}. E = {E ∈ E1 (B1 ) | E ∈ Thus |E| = 364 − 234 = 130. Some element E ∈ E1 (B1 ) must be normalized by a Sylow 3-subgroup of A, and clearly E ∈ E. By the Borel-Tits theorem, NA1 (E)

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lies in a parabolic subgroup P of A1 , and P < A1 since A1 is clearly irreducible on B1 . Then |E A1 | = |A1 : NA1 (E)| is a multiple of 130 (if P is of type A1 × A1 ) or of 80 (if P is of type A2 ). A second such orbit within E is thus impossible, so any further orbit, if one exists, has cardinality divisible by 3 · 5. As 3 · 5 does not divide 130 − 80, P must be of type A1 × A1 , and under the action of A1 , E1 (B1 ) has orbit lengths 117 + 117 + 130, with corresponding stabilizers isomorphic to SO5 (3), SO5 (3), and P . The orbits of WK on E1 (B1 ∩ K) are of lengths 40, 45, 36, with corresponding  elements of order 3 being of 3-central, D4 (q), L6q (q) types, respectively [IA , 4.7.3A]. Then E must be the union of the 40-orbit of WK with the WK × t1 orbit of length 2 · 45 consisting of elements on the K × x -diagonal. Choosing such a diagonal element c ∈ B1 we see that A1 ∩ NG (c ) = P , while Kc := E(CK (c)) ∼ = D4 (q). Let H be the subnormal closure of Kc in CG (c), so that H has the usual form by L3  balance. If H is a trivial or vertical pumpup, then H/O3 (H) ∼ = E6q (q) or D4 (q 3 ) or n

Dnq (q), n ≥ 4, by [IA , 4.9.1, 4.8.2, 4.7.3A]. In any case, A1 ∩ NG (c ) is nonsolvable or O2 (A1 ∩ NG (c )) = 1, a contradiction in either case. If H is a diagonal pumpup, then again O2 (A1 ∩ NG (c )) = 1, a contradiction. This completes the proof of the lemma.  We can now complete the proof of Proposition 11.1. Lemma 11.7. A = Z(A)W with W ≡B W (E7 ). Proof. By (11A) and Lemma 11.6, WLy  CA (τ ) with WLy ∼ = W (D6 ). Moreover, as p does not divide |Outdiag(Ly )|, n = mp (B) = 7. Write ζ = Z(WLy ) and observe that −1B = ζτ ∈ W . Recall that W + = W ∩ SL(B). If O2 (W + ) = 1, then as O2 (W + ) × −1B  W , W has a normal abelian noncyclic 2-subgroup. By Lemma 5.14, with W here in the role of W2 there, W ∼ = W (D7 ) or W (C7 ). In any event, |WK | does not divide |W |, which is absurd. Therefore, O2 (W + ) = 1. Since O2 (W ) ≤ Z(A), it follows that E(W ) = E(W + ) = 1, and indeed E(W ) is the central product of quasisimple groups with centers of odd order. Let H := E(W ). As H ≤ SL(B), H has no more than three components, by consideration of 2-ranks. Clearly, then, E(WK ) = E(WK )(∞) ≤ H. Furthermore, either E(WK ) acts irreducibly on B0 = [B ∩ K, WK ] with B0 a hyperplane of B, or p = 3. Note that CA (E(WK )) normalizes x = CB (E(WK )), and so CA (E(WK )) lies in CA∩NG (x) (E(WK )), which is solvable. Hence E(WK ) projects nontrivially on every component of E(A)/Z(E(A)). If E(A) has more than one component, then as n = 7 < 32 , E(WK ) must have a projective representation in characteristic p of degree at most 2, i.e., E(WK ) embeds in L2 (p), which is clearly not the case. (∞) Therefore H is quasisimple, H = E(A), and Hτ := WLy  CH (τ ) with Hτ an extension of E := O2 (Hτ ) ∼ = E25 by A6 . By inspection of involution centralizers [IA , Tables 4.5.1 and 5.3], H ∈ Spor ∪ Chev(r) for any odd r, so H ∈ Alt ∪ Chev(2). If H/O2 (H) ∼ = An , then by the structure of Hτ , n ≥ 12. But then H contains A4 × A4 × A4 and so H has no faithful representation in odd characteristic p of degree less than 9, a contradiction. Thus H ∈ Chev(2).

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Let N0 = NA (E) and Q = O2 (N0 ) ∩ H. As B is multiplicity-free for E, it is multiplicity-free for Q. Let F be a Q-irreducible constituent in B. If dim F > 1, constituents in B of dimensions 6 then as Q is a 2-group and N0 has irreducible and 1, we have dim F = 2 and dim F N0 = 6. But then the stabilizer of F must cover [WLy , WLy ], which however has no nontrivial representation of degree 2, a contradiction. Thus Q acts diagonalizably on B, preserving the same ordered frame B as that preserved by E. In particular, Q is abelian. The same argument shows that for any abelian overgroup E ∗ ≤ H of E invariant under WLy , O2 (NA (E ∗ ))∩H preserves B and is abelian. Thus if we inductively define Ei+1 = O2 (Ni ) ∩ H and Ni+1 = NA (Ei+1 ) ≥ Ni , i = 0, 1, . . . , there exists  such that N+1 = N , and E preserves B. Recall that H = E(A) ∈ Chev(2). As [WLy , WLy ] ≤ H we have [WLy , WLy ] ≤ Ni ∩H for all i. Therefore each Ni / Sol(Ni ) is isomorphic to one of A6 , Σ6 , A7 , or Σ7 , with Sol(Ni ) abelian. Moreover, by the Borel-Tits theorem, N ∩ H is a parabolic subgroup of H. The only possibility is that N ∩ H/ Sol(N ∩ H) ∼ = Sp4 (2). Indeed, as Sol(N ) is abelian, H ∼ = Sp6 (2). Then Out(H) = 1, and so A = Z(A) × H. In fact W = −1B × H ∼ = W (E7 ) and A = Z(A)W . As τ = −ζ is a reflection on B, W ≡B W (E7 ). The proof of the lemma is complete.  Now the proof of Proposition 3.4 is complete. Indeed, if B < B ∗ , the proposition holds by Lemma 4.3. If B = B ∗ , then mp (Inndiag(K)) ≥ 3 as mp (C(x, K)) = 1 and B maps into Inndiag(K) (Lemma 3.3a) with mp (B) ≥ 4. By Corollary 6.12, if mp (B) = 4, the one of Proposition 3.4abcdehik holds, so we may in fact assume  that mp (B) ≥ 5. If K ∼ = Aq (q), then  ≥ 4, and one of Proposition 3.4abfg holds, − by Corollary 10.12. If K ∼ = A q (q), then  ≥ 7 and either (c) or (d) of Proposition 3.4 holds, by Corollary 7.6. The same corollary implies that if K ∼ = C (q), then (h) or (i) of Proposition 3.4 holds. Likewise (j) or (k) or (l) of Proposition 3.4 holds if K∼ = D± (q), for a suitable choice of L, by Corollaries 7.6 and 8.22. Next, the cases − K ∼ = F4 (q), E6 q (q), and E8 (q) are impossible by Corollary 7.6 and Proposition  9.1, and the cases K ∼ = E6q (q) and E7 (q) lead to cases (m) and (n) of Proposition 3.4, by Propositions 11.1 and 9.1, respectively. 12. Identification of G0 : Setup Having established the structure of W , we turn to the question of identifying the group G0 . Treating the various cases of Proposition 3.4 will occupy us through Section 18. In these sections we shall prove: Proposition 12.1. G0 ∈ Chev(2) ∩ K(7)+ . Throughout these sections, in addition to assuming [III14 , (1A), (1B)], (1A), (3A), (3C), and (3D), we assume

(12A)

(1) CG (B) ≤ N o ≤ NG (B) and W o = AutN o (B) ∼ = N o /CG (B); (2) W is generated by reflections and contains all reflections of AutK (B); o (3) W o ≤W and W is generated by reflections; and Wo . (4) G1 = K

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− q 13. G0 ∼ = Sp2n+2 (q) OR An+1 (q)

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We consider various possibilities for K, L, and W o . In all but two cases we take W o = W . For any u ∈ D# we write WLou for W o ∩ Lu and WLu for W ∩ Lu . As usual, we also set De = {u ∈ E1 (D) | Lu > L}

(12B)

o Our goals are to identify G1 , andin the case W = W , to prove that G1 = G0 . # (Recall that G0 = N(x, K, y, L) = Lu | u ∈ D .) In any case, we let N be the full preimage of W in NG (B), so that N/CG (B) ∼ = W. Recall that by Corollary 3.6, K and L lie in Lie(2). −q 13. G0 ∼ (q) = Sp2n+2 (q) or An+1

Cases (h) and (c) of Proposition 3.4 are similar enough that we can treat them together in this section. In case (h), the setup is the following: (1) K ∼ = Sp2n−2 (q); = Sp2n (q), n ≥ 3, and L ∼ (13A) (2) WK ≤ W o ≡B W (Cn+1 ); and (3) Either W = W o , or n = 3 with W ≡B W (F4 ). In case (c), (13B)

− −q (q); and (1) K/Z(K) ∼ = SLn−2 = Ln q (q), n ≥ 6, and L ∼ (2) W = W o ∼ = W (C[ n2 ]+1 ).

Thus in this section we assume either (13A) (the “symplectic case”) or (13B) (the “linear case”). Accordingly we set m = n + 1 or m = [ n2 ] + 1, so that in either case, m = mp (B) ≥ 4 and W o ≡B W (Cm ). o We put G1 = K W and G0 = N(x, K, y, L) . We wish to prove −q Proposition 13.1. In the linear case, G1 /Z(G1 ) ∼ (q) and G0 = G1 . = Ln+2 ∼ In the symplectic case, G1 = Sp2n+2 (q), and if W = W o , then G0 = G1 . 

Case (i) of Proposition 3.4, leading to G0 ∼ = F4 (q), will be treated separately. It will make use of the above proposition in the symplectic case, but with W o < W . We assume the proposition false. Notice that as K ∈ Gp by assumption, the case K ∼ = Sp6 (2), then p = 3 and K ∈ C3 , a = Sp6 (2) is excluded, for if K ∼ contradiction. Thus in the symplectic case, if m = 4, then q > 2. We identify W o with W (Cm ) = F S, a monomial group on B permuting a frame b1 , . . . , bm , with F ∼ = E2m normalizing each bi and S ∼ = Σm permuting {b1 , . . . , bm }. We may assume that notation has been chosen so that x = bm , since the bi are the unique cyclic subgroups of B whose centralizers in W o contain a copy of WK ∼ = W (Cm−1 ). In a similar way, we may assume that D = bm−1 , bm and then that y = bm−1 ∈ K. We first prove Lemma 13.2. The following conditions hold:

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226

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

(a) x and y are the only elements of E1 (D) inverted by a reflection in O2 (W o ); (b) Ly ∼ =K∼ = Lu for any u ∈ De ; Wo (c) x ∩ E1 (D) = {x , y }; and (d) If W = W o , then De = {x , y }. Proof. There are two classes of reflections in W o : those in F inverting a single bi , and those conjugate to a transposition in S. The latter are outside O2 (W o ) since m ≥ 3. Thus (a) holds, and (c) follows quickly. Suppose that u ∈ De , so that Lu is a level vertical pumpup of L. By [IA , 4.2.2], the untwisted Dynkin diagram of Lu has that of L as a subdiagram. In the symplectic case this implies that Lu ∼ = Cr (q) for some r ≥ n, or Lu ∼ = F4 (q), with n = 4. In the latter case WLu ∼ = W (F4 ), whose order does not, however, divide |W | = |W (C5 )|. Thus Lu ∼ = Cr (q). By [III14 , Lemma 1.2], F(Lu ) ≤ F(K), and so Lu ∼ = K (we use [IA , 6.1.4] here). In particular, since y = bm−1 is W o -conjugate to x = bm , Ly ∼ = K, and (b) holds in the symplectic case. In the linear case, on the other hand, if Lu is a classical group, then it follows from [IA , 4.8.2, 4.8.4, 4.9.1] and the fact that p does not divide q − (−q ) that Lu /Z(Lu ) ∼ = K/Z(K). If Lu is of exceptional Lie type, then using the same results and [III11 , Table − 13.1], we see that the only possibility is m = 4, n = 6, with Lu ∼ = E6 q (q). But then W = W o ∼ = W (C4 ), whereas AutLu (B) ≥ W (F4 ) by [III17 , 2.6]. As |W (F4 )| > |W (C4 )|, this is impossible. Hence (b) holds in either case. Finally, suppose that W = W o . By (b), any u ∈ De is centralized by m − 1 reflections in O2 (WLu ), all of which lie in O2 (W ). Hence part (d) holds since for u ∈ D − x − y , u is centralized by at most m − 2 of the reflections in O2 (W ). The proof is complete.  q Lemma 13.3. G1 ∼ (q) = Sp2n+2 (q) in the symplectic case, and G1 /Z(G1 ) ∼ = Ln+2 in the linear case.

−

o Proof.  W o Suppose false. Since the proposition fails, W = W and G1 = G0 . As = G1 it follows by [III17 , 9.1] that G1 = K, Ly . By Lemma 13.2d, Ly ≤ K  G0 = K, Ly . Hence, G1 = G0 , contradiction. The lemma follows.

Let V be an Fq -vector space which is a standard module for our target group −q (q)). We may write (whether Sp2n+2 (q) or Ln+2 V = V0 ⊥ V1 ⊥ V2 ⊥ · · · ⊥ Vm , a direct sum. Here, in the symplectic case, m = n, V0 = 0, and V1 , . . . , Vm are nondegenerate planes; in the linear case, m = [ n2 ] + 1, dim V0 ≤ 1, and V1 , . . . , Vm are again planes, which are nondegenerate if K isa unitary group. We adopt the notation  from [III13 , Section 3], so that VI = i∈I Vi for every I ⊆ {0, 1, . . . , m}, I = {0, 1, . . . , m} − I, C o = Sp(V ) or SL−q (V ) in the respective cases, and CIo = NC o (VI ) ∩ CC o (VI  ). Thus for i = 1, . . . , m, o ∼ Cio := C{i} = SL2 (q). We write i for {i} . Choose ei ∈ Ip (Cio ), i = 1, . . . , m, in such a way that the eigenvalues of ei in Vi are the same as those of ej in Vj , for all 1 ≤ i < j ≤ m. (Equivalently, e1 , . . . , em are all C o -conjugate.) There are then isometries V1 → Vi by which e1 |V1 is transported to ei |Vi , 1 < i ≤ m. Moreover, each ei is inverted by an involution of Cio . We set E = e1 , . . . , em and conclude

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− q 13. G0 ∼ = Sp2n+2 (q) OR An+1 (q)

that (13C)

227

∼ W (Cm ), WC acts there is WC ≤ NC o (E) such that CWC (E) = 1, WC = monomially on E with respect to {e1 , . . . , em }, and WC centralizes V0 .

Let zC ∈ Z(WC ) be the involution inverting E elementwise, and define (13D)

T = the stabilizer in WC of {e1 , . . . , em }.

∼ Σm and T permutes {e1 , . . . , em } faithfully. (When m = 4, we shall Thus, T = choose WC to satisfy additional properties, but we do not make that choice right now.) There is a homomorphism with central kernel o f : Cm  → K,

(13E)

which we fix, such that f (ei ) = bi for all i = 1, . . . , m − 1. Trivially f extends to a homomorphism, which we also call f : o o Cm  E = Cm × em → K × x = KB o taking em to x. Notice that since D = bm−1 , bm , L = f (C{m−1,m}  ) = H{0,1,...,m−2} , in the notation of [III13 , Section 3]. We can now handle the general case.

Lemma 13.4. We have m = 4. Proof. Assume false, so that m ≥ 5. We have W = F S, and we let N1 be the full preimage of S in NG (B). We verify the hypotheses of [III13 , Theorem 3.1] for K and N1 . Hypotheses (a) and (b) are trivial. We have a sequence of homomorphisms N1 → N1 /CG (B) = S ∼ = Σ{e1 ,...,em } ∼ =T = Σ{b1 ,...,bm } ∼ where the isomorphism between symmetric groups is based on the bijection bi → ei , 1 ≤ i ≤ m. Let λ : N1 → T be the composite of the above homomorphisms. Thus for i = 1, . . . , m, and w ∈ N1 , w(bi ) = bj if and only if λ(w)(ei ) = ej . On the other o o hand, for any σ ∈ T ∩ Cm  , when we apply the homomorphism f : Cm E → KB, f (σ) permutes {b1 , . . . , bm−1 , bm } the way σ permutes {e1 , . . . , em }. Since N1 is the full preimage of S in NG (B), f (σ) ∈ N1 . It is then trivial that λ(f (σ)) = σ. This is hypothesis (c) of [III13 , Theorem 3.1]. For hypothesis (d) of the theorem, let I ⊆ {0, 1, . . . , m − 1} and u ∈ N1 with λ(u)(I) ⊆ {0, 1, . . . , m − 1}. We need to show that o f (CIo )u = f (Cλ(u)(I) ).

Let EI = ei | i ∈ I and BI = bi | i ∈ I . Assume first that dim(V ) is even, so that V0 = 0 and dim(V ) = 2m. Then if q > 2, 

CIo = O 2 (CC o (EI  )) = E(CO2 (CC o (em )) (EI  −{m} )), so f (CIo )u = E(CK (BI  −{m} ))u = E(CG (BI  ))u = E(CG (Bλ(u)(I  ) )), o which by a similar calculation equals f (Cλ(u)(I) ), as required. If q = 2, we replace 

the layer functor E with O 2 and make the corresponding calculation.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

228

When dim(V ) is odd (whence we are in the linear case), we argue similarly, using the following characterization of CIo :  E(CC o (EI  )) if 0 ∈ I o (13F) CI = E(CC o (E(CC o (EI )))) if 0 ∈ I 

Again if q = 2 we need to replace the E-functor by the O 2 -functor. This establishes hypothesis (d). Thus [III13 , Theorem 3.1] applies with N1 in the role of N there, and we conclude that G contains a homomorphic image H of C o such that K ≤ H. Indeed, o o in the notation of that theorem, K is the image of Cm  , which centralizes Cm , which has order divisible by p; therefore, p divides |CH (K)|. Since mp (C(x, K)) = 1, we have mp (CG (K)) = 1 and so replacing H by a CG (K)-conjugate we may assume that x ∈ H. Therefore B = (B ∩K) x ≤ H. Now A∩H is generated by reflections  A∩H , and isomorphic  W to W , so A∩H = W . It follows from [III17 , 9.1] that H = K = G1 , contradicting Lemma 13.3 and completing the proof. so H = K  In this final case m = 4, we shall use [III13 , Theorem 3.2] in most cases. In order to do that, we need further preparation. Let (13G)

T0,K = f (T ∩ C4 )

and W0,K = f (WC ∩ C4 ). Then T0,K ≤ NK (B) is isomorphic to Σ3 , T0,K preserves {b1 , b2 , b3 }, and T0,K projects isomorphically modulo CG (B) onto the stabilizer S0 of b4 in S. As T ≤ WC , T0,K ≤ W0,K . In order to obtain hypothesis (b) of [III13 , Theorem 3.2], we need the following lemma. Lemma 13.5. Let w ∈ NG (B) interchange b1 with b3 , and interchange b2 with b4 . Let L0 be a subgroup of L containing Dw and such that L0 ∼ = Sp4 (q) or − ] = 1. L4 q (q), in the symplectic or linear cases, respectively. Then [L0 , Lw 0 Proof. Since m = 4, we have q > 2 in the symplectic case. Also x = b4 and w y = b3 , while b1 × b2 ≤ L. Then D = b1 , b2 ≤ Lw . As q > 2 in the symplec (u)) | u ∈ D# . tic case, it follows by [IA , 7.3.3] that in any case, Lw 0 = E(CLw 0 Therefore it suffices to take an arbitrary u ∈ D# , set J = E(CLw (u)), and 0 show that [J, L0 ] = 1. Note that J, if nontrivial, is a single component, and  J ≤ O 2 (CLw (u))(∞) = E(CLw (u)). Suppose first that u ∈ De , so that Lu ∼ =K and Lu  CG (u). By Lp -balance, either [J, L0 ] ≤ [J, Lu ] = 1 or J ≤ Lu . In the latw w 2 w ter case, since [Lw 0 , D ] ≤ [L, D] = 1, J ≤ O (CLu (D )) ≤ CLu (L0 ), as desired. On the other hand, if u ∈ De , then L = Lu  Lp (CG (u)), so J normalizes L. If J = 1, then J ∩ D = 1. But [D, L] = 1 and J is a single component, so [J, L] = 1,  completing the proof as L0 ≤ L. We now show that [III13 , Theorem 3.2] applies (and gives a contradiction) if the following condition is true: (13H) W o splits over CG (B), and a complement exists that contains T0,K .

Lemma 13.6. Condition (13H) does not hold.

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− q 13. G0 ∼ = Sp2n+2 (q) OR An+1 (q)

229

Proof. Suppose false, so that T0,K ≤ W0 for some complement W0 splitting W o over CG (B). Then the stabilizer N0 of {b1 , . . . , b4 } in W0 maps isomorphically onto S and contains T0,K . Now there is an isomorphism θ : T → S such that for all t ∈ T ∩ C4 , θ(t) = f (t)CG (B), and such that for all i = 1, . . . , 4 and all t ∈ T , θ(t) f (eti ) = f (ei )θ(t) = bi . Composing the isomorphisms N0 → S and θ −1 : S → T gives an isomorphism (13I)

λ : N0 → T,

with the following two properties: (1) for all σ ∈ T ∩ C4 , f (σ) ∈ T0,K ≤ N0 and λ(f (σ)) = θ −1 (f (σ)CG (B)) = σ; and (2) for all i = 1, . . . , 4 and all u ∈ N0 , λ(u) bui = f (ei ). It is now routine to check the hypotheses of [III13 , Theorem 3.2], with N0 in place of N there. Hypothesis (a) of [III13 , Theorem 3.2] holds by our construction. Finally let us show that Lemma 13.5 implies hypothesis (b) of [III13 , Theorem 3.2], that [H12 , H34 ] = 1. Indeed, let w ∈ W0 interchange b1 and b3 , and interchange b2 and b4 . Then w ∈ N0 by definition of N0 , and so λ(w) interchanges e1 and w = H{3,4} by e3 , as well as e2 and e4 . Hence λ(w)({1, 2}) = {3, 4}, so H{1,2} [III13 , Lemma 3.4d]. Letting L0 = H{1,2} and applying Lemma 13.5, we have [H{1,2} , H{3,4} ] = [L0 , Lw 0 ] = 1, as required. Thus by [III13 , Theorem 3.2], there is a homomorphism g : C o → G such that H := g(C o ) contains K = g(C4o ), in the notation of that theorem. Thus CH (K) contains g(C4o ), so p divides |CH (K)|. But mp (CG (K)) = mp (C(x, K)) = 1 so replacing h by a CG (K)-conjugate, we may assume that x ∈ H. If W o = W , we repeat the last paragraph of the proof of Lemma 13.4 to contradict Lemma 13.3. We may thus assume that W o < W ≡B W (F4 ). Note that J(O2 (WH )) = −1, O2 (WK ) = J(O2 (W o )). So WH ≤ NW (J(O2 (W o ))) = W o . As WH ∼ = W o . But it follows quickly from [III17 , 9.1] that H = K WH , W (C4 ) ∼ = W o ,WH = o whence K = K W = G1 , again contradicting Lemma 13.3. This completes the proof.  This allows us to rule out the following exceptional configuration: (1) g ∈ I2 (CG (B)); (2) g induces a nontrivial field automorphism on K in the symplectic case, and a nontrivial graph automorphism on K in the linear (13J) 1 ∼ 2 case, with K (g) = Sp6 (q1 ) or Sp6 (q), respectively; and  CC(g) ∼ G 2 (3) Gg := B = Sp8 (q ) or Sp8 (q), respectively. Lemma 13.7. The configuration (13J) does not occur. Proof. We have B ≤ Gg and Gg is an 8-dimensional symplectic group in either case. Hence by [III17 , 2.2], there is a complement U to CGg (B) in NGg (B), and U ∼ = Wo ∼ = W (C3 ) = W (C4 ). Thus U splits W o over CG (B). Let Ux = CU (x) = CU (b4 ) ∼  1 and Gg,x = O 2 (CGg (x)) ∼ = Sp6 (q 2 ) or Sp6 (q), respectively. Thus, Ux ≤ Gg,x . Let Y = E(f −1 (Gg,x )) and V = Y ∩ f −1 (U ), so that Y ∼ = Gg,x and V ∼ = U via f . Moreover, there exists a field or graph automorphism φ, respectively, of K of order 2 such that Y = f −1 (E(CK (φ))). It follows that there is a field or graph automorphism Φ of C o such that f intertwines Φ|C4o and φ. Also, [V, Φ] = 1. We apply [III17 , 2.32] with C o , E, x, and Φ in the roles of K, B, x, and Φ there. Now V is a complement in NC4o (E) to CC4o (E), centralized by Φ, and we deduce that

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

230

V is contained in a complement to CC o (E) in NC o (E). We call this complement WC o . Then T is the stabilizer of {e1 , e2 , e3 , e4 } in WC o and T0,K = f (T ∩ C4o ), by definition. But then T ∩ C4o ≤ V , whence T0,K = f (T ∩ C4o ) ≤ f (V ) = U . As U splits W o over CG (B), (13H) holds, contradicting Lemma 13.6. The lemma is proved.  o Let K0 = f (C{0,m}  ). Of course if dim V is even, then K0 = K, and otherwise − q ∼ K0 = SLn−1 (q). Notice also that AutK (B) = AutK0 (B).

Lemma 13.8. CCG (x) (K0 ) has odd order and acts faithfully on Ly . Proof. Let Y = CCG (x) (K0 ). We have Ly  CG (y) and y = b3 ∈ K0 . Hence Y acts on Ly . Suppose that 1 = t ∈ CY (Ly). Then G 0 :=  K0 , Ly , lying in CG (t), is a K-group. By [III17 , 1.31], W o = WK , WLy = AutK0 (B), WLy ≤ o AutG0 (B), so as Ly ∈ K W , K ≤ G0 . Thus, G0 = G1 . Now by [III17 , 3.16, −q 10.17], G1 is a homomorphic image of Sp2n+2 (q) or SLn+2 (q) (contradicting Lemma 13.3) or F4 (q), the last only in the symplectic case. But in the symplectic case, o since Ly ∈ K W , it follows from [III17 , 3.16] that G1 ∼ = Sp8 (q), a contradiction. Hence Y acts faithfully on Ly . Therefore Y embeds in CAut(Ly ) (K0 x ), which − embeds in the centralizer of an element of order p in SLk q (q), k ∈ {2, 3}, and thus has odd order. The lemma is proved.  We next prove Lemma 13.9. CG (B) has odd order. Proof. Suppose on the contrary that there is an involution g ∈ CG (B). Set    K o = O 2 (CK (g)), Loy = O 2 (CLy (g)), Lo = O 2 (CL (g)), and Go = K o , Loy . Thus Go ≤ CG (g), so Go is a K-group. By Lemma 13.8, [K, g] = 1. Thus by [III17 , 2.30], g has two possible actions 1 − on K: (a) K ∼ = Sp6 (q 2 ); (b) K/Z(K) ∼ = Ln q (q), n ∈ {6, 7}, and = Sp6 (q) and K o ∼ Ko ∼ = Sp6 (q). Similar arguments show that [Ly , g] = 1 and (a) or (b) holds for Ly 1 and Loy (recall that K ∼ = Ly ). Then by [III17 , 3.16], Go ∼ = Sp8 (q 2 ) or Sp8 (q). In either case, (13J) holds, contradicting Lemma 13.7, and completing the proof.  Using Lemma 13.6, we can make the following reduction. Lemma 13.10. The following conditions hold: − (a) K/Z(K) ∼ = L6 q (q), q > 2; and (b) p > 3 and q ≡ −q (mod 3). Proof. Suppose false. We have exact sequences 1 → CG (B) → N → W → 1 (13K)

1 → CG (B) → NK → WK → 1 1 → CK (B) → NK ∩ K → WK → 1,

where NK is the full preimage of WK in N . The last of these sequences splits, by [III17 , 2.6], and this implies that the second sequence splits as well. Since |W : WK | = |W (C4 ) : W (C3 )| = 23 and CG (B) is (solvable!) of odd order, the first sequence splits, by [III8 , 5.5]. Let W1 be a complement in the first sequence, and set T1 = CG (B)T0,K ∩ W1 . Then T1 is a complement to CG (B) in CG (B)T0,K . If CG (B) is filtered by abelian T0,K -invariant sections U of (odd) prime exponent

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− q 13. G0 ∼ = Sp2n+2 (q) OR An+1 (q)

231

such that H 1 (T0,K , U ) = 0, then T0,K = T1g for some g ∈ CG (B). Hence W0 := W1g will split W o over CG (B) and will contain T0,K , contradicting Lemma 13.6. Since we are assuming that (a) or (b) fails, [III17 , 2.36] implies that CG (B) is filtered by T0,K ∼ = Σ3 -modules M of one of three types: modules of order relatively prime to 3; trivial modules; and modules in which an element of t ∈ T0,K of order 3 acts freely. As T0,K is generated by involutions and |CG (B)| is odd, H 1 (T0,K , M ) = 0 for any M of the first type, by [IG , 9.24(i)]; of the second type, by [III8 , 5.7a]; and of the third type, by restriction to a Sylow 3-subgroup [IG , 9.24(i), 9.25(iii)]. This completes the proof of the lemma.  Now L ∼ = L4 q (q) and we fix t ∈ I3 (L) with minimal (i.e., 2-dimensional) support on the natural L-module, so that H := E(CL (t)) ∼ = L2 (q). Set I = − E(CK (t)) = Lp (CK (t)) = L3 (CK (t)) ∼ = L4 q (q). Also set Ct = CG (t) and J = I E(Ct ) . −

Lemma 13.11. The following conditions hold: (a) I ≤ E(Ct ), and J is the product of all components in a single x -orbit on the set of components of E(Ct ); (b) Every component of J lies in G3 ; and (c) m3 (Ct ) ≥ 4. Proof. This is mostly a repeat of part of the proofs of [III9 , 10.1, 10.3a]. First, I is a component of CCG (x) (t) = CCt (x), so if I ≤ E(Ct ), then (a) holds by Lp balance. Moreover, H is a component of CCG (D) (t) = CCt (D), so by Lp -balance, H ≤ Lp (C  t ). Thus, to prove (a) it is enough to  show that [H, Op (Ct )] = 1, for then I = H I ≤ E(Ct ). Set X = Op (Ct ) = CX (d) | d ∈ D# ; it is enough to  show that Xd := [H, CX (d)] = 1 for any d ∈ D# . As H = O p (H), Xd = [H, Xd ] by [IG , 4.3(i)]. As H ≤ L ≤ Ld   CG (d), this implies Xd ≤ Ld , so Xd ≤ Op (CLd (t)). We have Ld ∈ Chev(2), and q(Ld ) = q > 2 by Lemma 13.10. Therefore H =    O 2 (H) ≤ O 2 (CLd (t)) = E(CLd (t)) and Xd ≤ [Op (CLd (t)), O p (E(CLd (t)))] = 1. − So (a) holds. As I ∼ = L4 q (q) with q ≡ −q (mod 3), m3 (t I) = 1 + 3 = 4, and (c) follows. Finally I 2

232

By Lemmas 13.10 and 13.12, the conditions of Lemma 2.2a are satisfied, so (p, x, K, y, L) ∈ Q∗ (G). This, however, contradicts our choice (2B). Hence, the proof of Proposition 13.1 is complete. ± 14. G0 ∼ (q) = Dn+1

We next treat cases (j) and (k) of Proposition 3.4. Thus we assume q (1) K ∼ (q); = Dn (q), n ≥ 4,  = ±1, and L ∼ = Dn−1 (2) If  = q n , then mp (K) = n and W = W o ≡B W (Dn+1 ) or (14A) W (Bn+1 ); and (3) If  = −q n , then mp (K) = n − 1 and W = W o ≡B W (Bn ). Our goal in this section is the following proposition. q (q). Proposition 14.1. Assume (14A). Then G0 ∼ = Dn+1 We assume that Proposition 14.1 fails and aim for a contradiction in a sequence of lemmas.  D4 (2) ∈ C3 . Note that since K ∈ Gp and p divides q 2 − 1, K ∼ = We set m = n + 1 if  = q and m = n if  = −q . Thus in all cases, mp (B) = m and W is a split extension of an elementary abelian 2-group F by a group S ∼ = Σm . We make the following normalization. Let {b1 , . . . , bm } be a frame of B invariant under W . We may assume that F normalizes each bi , and S is the stabilizer in W of {b1 , . . . , bm }. Moreover, x = bi for some i, say i = m, and (bm−1 , E(CK (bm−1 ))) is an acceptable subterminal (x, K)-pair. So we may assume that y = bm−1 and L = E(CK (y)). The effect of this normalization is that y is F -invariant, and is W -conjugate to x . [We remark that if K ∼ = D4 (q), then there are other conjugacy classes of elements  y  ∈ Ip (K) such that E(CK (y  )) ∼ = A3q (q). Thus our normalization may entail a change in the subterminal pair (y, L). Such a change, however, does not change the essential property that (p, x, K, y, L) ∈ Q∗ (G).] (14B)

Lemma 14.2. We have De = {x , y }, and Ly is W -conjugate to K. Proof. Let u ∈ D − x . If u = y , then since x and y are W -conjugate, Lu is conjugate to K. It remains to show that if u projects nontrivially on both factors in D = x × y , then Lu = L. Note that as W preserves {b1 , . . . , bm }, we have |F : CF (u)| = 4. Also the image of CW (u) in W/O2 (W ) ∼ = S ∼ = Σm is isomorphic to either Σm−2 or Σm−2 × Z2 . Thus |CW (u)| divides (|F |/2)(m − 2)!, which divides 2m−1 (m − 2)!. On the other hand, WLu must embed in CW (u). Suppose that Lu ∼ = K. Then WLu ∼ = W (Bm−1 ) or W (Dm−1 ) and in any case |WLu | is a multiple of 2m−2 (m−1)!, which must then divide 2m−1 (m − 2)! by the previous paragraph. As m ≥ 4, this  K. is impossible. Thus, Lu ∼ = On the other hand, if u ∈ De , then we know that Lu is a level vertical pumpup 2 q of L ∼ (q). Moreover, by [III14 , Lemma 1.2], either F(Lu ) ≤ F(K) = q n , in = Dn−1 which case Lu has untwisted rank at most n, or dp (Lu ) > dp (K). Since Lu ∼ = K, we η conclude from [III17 , 10.21] that n = 4 and Lu ∼ = A4 (q), η = ±1. As mp (Lu ) ≥ 3, η = q . But then WLu ∼ = Σ5 . On the other hand, m = n or n + 1 so m − 2 ≤ 3 and CW (u) is solvable by the first paragraph. This contradiction finishes the proof of the lemma. 

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± 14. G0 ∼ = Dn+1 (q)

233

Since x and y are W -conjugate, we have symmetry between (x, K) and (y, Ly ). Furthermore, Lemma 14.2 implies that  (14C) G0 = K, Ly ≤ K W = G1 . q Lemma 14.3. G has no subgroup J ∼ (q) containing K and with p dividing = Dn+1 |CJ (K)|.

Proof. Since mp (C(x, K)) = 1, mp (CG (K)) = 1. Hence if J exists, then replacing J by a CG (K)-conjugate we may assume that KB ≤ J. Now y = D ∩K and n + 1 ≥ 4, with L = E(CK (y)) ≤ E(CJ (y)) ∼ = Ly . It follows that E(CJ (y)) =  E(C J (y)) L ≤ Ly . But E(CJ (y)) ∼ = Dn (q) ∼ = Ly , so Ly = E(CJ (y)) ≤ J. Using [III17 , 9.6] we get J = K, E(CJ (y)) = K, Ly = G0 , contrary to our assumption that Proposition 14.1 fails.





q Let C o := Dn+1 (q) and let V be the its natural module. Thus

V = V0 ⊥ V1 ⊥ · · · ⊥ Vm where V1 , . . . , Vm are all isometric 2-dimensional Fq -spaces of type q , and V0 is either 0 (if  = q n ) or 2-dimensional of type −q (otherwise). For any I ⊆ {0, 1, . . . , m} let VI , as usual be the sum of all Vj , j ∈ I, and VI  the sum of all other Vj . We write i for {i} . We also define CIo = CC o (VI  ). We choose ei ∈ Cio of order p, i = 1, . . . , m, in such a way that the ei are all conjugate in C o = Ω(V ), and set E = e1 , . . . , em . For every I ⊆ {0, 1, . . . , m}, set EI = ei | i ∈ I and BI = bi |, i ∈ I . The general case n ≥ 5 is easily handled. Lemma 14.4. One of the following holds: (a) K ∼ = D4− (q) and m = 4; or ∼ (b) K = D4 (q) and m = 5, with p dividing q + 1 (i.e., q = −1). Proof. Suppose false. Because (a) does not occur, m ≥ 5. If q = −1 and V0 = 0, then since (b) does not occur, n + 1 = m ≥ 6. If q = 1, V0 = 0, but n + 1 ≤ 5, then m = n + 1 = 5 so K ∼ = D4 (q), and so C o = Ω+ 10 (q). Thus, the dimension restrictions in [III13 , Theorem 3.1] are satisfied. Let N0 ≤ NG (B) be such that N0 /CG (B) ∼ = Σm , permuting {b1 , . . . , bm }. o There is an isomorphism Cm  → K taking e1 , . . . , em−1 to b1 , . . . , bm−1 , respectively. By [III17 , 2.40], NC o (E) contains a subgroup T ∼ = Σm faithfully permuting o ∼ {e1 , . . . , em } and such that T ∩ Cm  = Σm−1 is the stabilizer of em , with o f (T ∩ Cm  ) ≤ N0 . Thus there exists an isomorphism α : S → T such that any σ ∈ S permutes {b1 , . . . , bm } the way α(σ) permutes {e1 , . . . , em }. We let λ : N0 → T be the composite of α with the quotient map: λ : N0 → N0 /CG (B) → T. It is then immediate that for any σ ∈ T ∩ Cm , λ(f (σ)) = σ, which is hypothesis (c) of [III13 , Theorem 3.1]. To check hypothesis (d) of [III13 , Theorem 3.1], we let X = {0, 1, . . . , m}, as usual. Let I ⊆ X − {m} and u ∈ N0 with λ(u)(I) ⊆ X − {m}. We need to show

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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o that f (CIo )u = f (Cλ(u)(I) ). According as V0 = 0 or V0 = 0 we repeat the argument of paragraph two, or paragraphs three and four, of the proof of Lemma 13.4. Theorem 3.1, whose hypotheses (a) and (b) hold trivially, then applies and q yields that G has a subgroup J ∼ (q) containing L and with mp (CJ (L)) ≥ 2 = Dn+1 o )). (in the notation of Theorem 3.1, CJ (L) contains g(C{m−1,m} p As mp (J) = mp (L) + 2, there is D1 ∈ E2 (CJ (L)) with L = E(CJ (D1 )). Let D2 ∈ Ep2 (L) have minimal (4-dimensional) support on a natural L-module. Assuming for the moment that L ∼  D4 (q), D2 is then J-conjugate to D1 and W -conjugate = to D. Therefore D1 ∈ DG . Replacing J by a CG (L)-conjugate we may assume that #  D = D1 ≤ J.  For any u ∈ D , let Lu be the subnormal closure of L in CJ (u). Then  Lu = LLu ≤ Lu . But there exist exactly two u ∈ E1 (D) such that Lu > L, ∼ Lu . As |De | = 2, J = Lu | u ∈ De = G0 , as desired. We and for those, Lu = ∼ D4 (q). may therefore assume that L =  Thus K ∼ = D5q (q). If q > 2, then by Theorem 3.1, we have K ≤ J with p dividing |CJ (K)|, contradicting Lemma 14.3. Thus q = 2, in which case p = 3. Now there is w ∈ W such that xw ∈ B ∩ L. Hence CG (xw ) contains B as well as a subgroup B0 ≤ J with B0 ∼ = B. Since CG (x) has a unique conjugacy class of E36 subgroups, by [IA , 4.10.3], we may replace J by a conjugate and assume that B ≤ J. Now there exist exactly six u ∈ E1 (B) such that WCG (u) contains W (D5 ), and these include x and y. Likewise there are exactly six u ∈ E1 (B) such that AutJ (B) contains W (D5 ), so these must be the same six. Therefore K and Ly lie in J, and G0 = K, Ly = J, using [III17 , 9.1] for the last equality. The proof is complete. 

Lemma 14.5. There exist subgroups J− ∼ = J+ ∼ = L2 (q) such that the following conditions hold: (a) (b) (c) (d)

D ≤ J− × J+  CG (L); y ±1 x ∈ J± and O 2 (J± )  CG (y ∓1 x);  For any u ∈ E1 (D) − {x , y , yx , y −1 x }, Lp (CG (u)) = L; and L  CG (u) for all u ∈ E1 (D) − De .

Recall that x = bm and y = bm−1 , and that m = 4 or 5 according as (a) or (b) of Lemma 14.4 holds. Proof. Let K1 = Lp (CG (b1 )). Since b1 and bm = x are conjugate by the transposition (1m) ∈ S, we have K1 ∼ = K and D ≤ b2 , . . . , bm ≤ K1 , with E(CK1 (D)) embedded naturally as L ∩ K1 in K1 . Here (L ∩ K1 , K1 ) = − + + (Ω− 4 (q), Ω8 (q)) or (Ω4 (q), Ω8 (q)). We note that in the latter case, q > 2 since, as noted above, K ∼ = D4 (2). In any case, ∼ ∼ ∼ Ω+ 4 (q) = CK1 (L ∩ K1 ) = J− × J+ ≥ D with J− = J+ = L2 (q),  −1 and {B ∩ J− , B ∩ J+ } = {yx , y x }. In particular, as CG (L) ≤ CG (b1 ) and C(b1 , K1 ) is p-solvable, any non-p-solvable simple section of CG (L) must be involved in CK1 (L ∩ K1 ) and hence in L2 (q). So CG (L) contains no copy of L, whence L  CG (u) for all u ∈ E1 (D)−De , proving (d). In particular, L  CG (y ±1 x), so J±  normalizes and then centralizes L, as it maps into O 2 (CAut(L) (b1 (L ∩ K1 ))) = 1. (14D)

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loss we may assume that we have labelled J− and J+ so that B ∩J± =  ±1Without y x . As CG (L) ≤ CG (b1 ), CG (L) normalizes J− × J+ , and (a) follows. Also, part (c) follows from the structure of CG (L). Moreover, unless possibly q = 2, C(y ±1 x, L) has J± as its unique component so (b) holds. If q = 2, then L ∼ = L4 (2) so a Sylow 3-subgroup P of NG (L) lies in CG (b1 ). This implies that P embeds in Z3 × (Z3  Z3 ). But m3 (Z(P )) ≥ 3 so P ∼ = E34 , and this implies that even in this case, (b) holds. The proof is complete.  Lemma 14.6. K ∼ = D4− (q). Proof. Otherwise K ∼ = D4 (q), with p dividing q +1 and q > 2, as noted above. Also  = +1 = q 4 so m = 5 and W ∼ = W (D5 ) or W (B5 ). We construct a weak P-system of type D5− (q), leading to a contradiction with Lemma 14.3. For each u ∈ D# , write Lp (CG (u)) = Lu Ju with Ju a product of p-components of CG (u) and [Lu , Ju ] = 1. By Lemma 14.5,  ∼ Ju = 1 if and only if u = b±1 4 b5 , in which cases Ju = L2 (q) and  ∓1 (14E) B ∩ Ju = b4 b5 ; furthermore, E(CG (L)) = Lp (CG (L)) = Jb−1 b5 × Jb4 b5 . 4

Next, for 1 ≤ i ≤ 4, let ti ∈ S ∼ = Σ5 correspond to (i, i + 1), that is, ti interchanges bi and bi+1 . Set Li = Lp (CG (CB (ti ))). We claim that for i = 1, 2, 3, 4, (1) Li ∼ = L2 (q) is a component of Lp (CG (B{i,i+1} )) ∼ = Ω+ 4 (q); and (14F) (2) [B, ti ] = B ∩ Li . We have CB (ti ) = bi bi+1 B{i,i+1} . Thus, CB (t4 ) = B{1,2,3} b4 b5 , and using (14E) and Lp -balance, we find (B{1,2,3} )) = Jb b ∼ L4 = Lp (CL  (C (b b )) (B{1,2,3} )) = Lp (CLJ = L2 (q) p

G

4 5

b4 b5

4 5

∈ L4 . Moreover, the image of AutL4 (B) ∼ with = Z2 in AutG (B) fixes b1 ,b2 ,b3 and b4 b5 and inverts b−1 b , so equals t . Conjugating by powers of (12345) in S, 5 i 4 we obtain (14F). Similarly, for 2 ≤ i ≤ 4, Li−1 , Li ≤ Lp (CG (CB (ti−1 , ti ))). For example, L1 , L2 ≤ Lp (CG b1 b2 b3 D) = Lp (CL (b1 b2 b3 )) ∼ = SU3 (q), with L1 and L2 embedded in the standard way; so L1 , L2 ∼ = SU3 (q) as q > 2. Conjugating, Li−1 , Li ∼ = SU3 (q) for 2 ≤ i ≤ 4. Similarly, [L1 , L3 ] = 1, as is visible in K, and, conjugating appropriately we find [L1 , L4 ] = [L2 , L4 ] = 1. In particular, L1 , L2 , L3 form a weak P-system of type U4 (q) for the subgroup of K that they generate. Let s1 ∈ F ≤ W be the element inverting b1 and b2 but centralizing B{3,4,5} . Set r1 = t1 s1 and L∗1 = E(CG (CB (r1 ))). Then b1 b2 ∈ L∗1 ∼ = L2 (q) and L∗1 is ∗ ∗ conjugate to L1 . Let L = Lp (CG (b1 , b2 )). Then L is W -conjugate to L, and we see, using (14E), that Lp (CG (L∗ )) = L1 × L∗1 . Likewise, as L1 , L2 ≤ L, we have L3 , L4 ≤ L∗ , whence [L3 , L4 , L1 , L∗1 ] = 1. Also L∗1 , L2 ≤ CL (b−1 1 b2 b3 ) and we find as for L1 , L2 above that L∗1 , L2 ∼ = SU3 (q). It follows that L1 , L∗1 , L2 , L3 , L4 is a weak Phan system of type D5− (q) for J := L1 , L∗1 , L2 , L3 , L4 = K, L4 . Hence by [III13 , Theorem 1.14], J ∼ = D5− (q). By ∗ construction [b5 , J] ≥ [b5 , L4 ] = 1, but [b5 , L1 , L1 , L2 , L3 ] = 1. By [III13 , Theorem 1.14], L1 , L∗1 , L2 , L3 ∼ = K. It follows that p divides |CJ (K)|, contradicting Lemma 14.3. The proof is complete.  b−1 4 b5

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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In a similar way we can prove: Lemma 14.7. p divides q + 1, i.e., q = −1. Proof. Suppose false, so that p divides q − 1. By Lemma 14.6, K ∼ = D4− (q), 4 ∼ so  = −1 = q . Thus W = W (B4 ) and m = mp (B) = 4. We may assume that x = b4 and y = b3 .  Choose reflections t1 , t2 , t3 ∈ S and r1 ∈ F with [B, ti ] = b−1 i bi+1 for 1 ≤ i ≤ 3, and [B, r1 ] = b1 . Then WK = t1 , t2 , r1 . Set L∗1 = E(CG (CB (r1 ))), and Li = E(CG (CB (ti ))) for 1 ≤ i ≤ 3. 2 ∼ Since x = b4 centralizes WK , it is visible in K that L∗1 ∼ = Ω− 4 (q) = L2 (q ), ∼ ∼ while L1 = L2 = L2 (q). Note that as p divides q − 1, we have q > 2, so the Li ’s and L∗1 are all quasisimple. Moreover, by [III17 , 4.14], L∗1 , L1 , and L2 form a weak ∗ CT-system for K: L1 , L2 ∼ = SL3 (q), L∗1 , L1 = L ∼ = Ω− 6 (q), and [L1 , L2 ] = 1. Next, by Lemma 14.5, there are subgroups J− and J+ satisfying (1) and (2) of the following: (1) J− ∼ = J+ ∼ = L2 (q) and J− × J+ × L  NG (L); ±1 (14G) (2) b3 b4 ∈ J± ; and (3) L3 = J− ≤ CG (L1 , L∗1 ). In particular, J−  CG (b1 , b2 , b3 b4 ) and so J− = L3 . As J− × J+  CG (L), while L contains L1 and L∗1 , (14G3) also holds. Now we can prove that (14H)

L∗1 , L1 , L2 , L3 form a weak CT-system of type D5− (q).

To show this, in view of the relations already established, it remains only to prove that L2 and L3 are a standard CT-pair in L2 , L3 ∼ = SL3 (q). Let w = (14)(23) ∈ w w ∼ = t and t = t . Hence L , L S ∼ = Σ4 . Then tw 1 2 2 3 = L2 , L1 = SL3 (q) and 3 2 (14H) follows. Therefore by [III13 , Theorem 1.4], J := L∗1 , L1 , L2 , L3 = K, L3 ∼ = D5− (q). − ∗ By construction H := L1 , L1 , L2 ≤ CJ (b4 ), and H ∼ = D4 (q) ∼ = K by [III13 , Theorem 1.4]. As x = b4 , we have K ≤ J with p dividing |CJ (K)|, so Lemma 14.3 is contradicted. The proof is complete.  The analysis of the remaining case (14I)

K∼ = D4− (q), p | q + 1

is more difficult and will involve 2-local analysis. We will, in particular, have to deal with the tricky case q = 2. We will show that (14J)

G0 ∼ = D5 (q).

We assume that (14I) holds but (14J) fails, and proceed in a sequence of lemmas, first showing: Lemma 14.8. C(x, K) ∩ C(y, Ly ) = 1, and C(x, K) and C(y, Ly ) each embed in Zq+1 . Proof. The proof is similar for C(x, K) and C(y, Ly ) so we do it only for C(x, K). Suppose first that there is 1 = g ∈ C(x, K) ∩ C(y, Ly ). Set Cg = CG (g), so that Cg ≥ K, Ly ≥ B. By Lp -balance, K, Ly ≤ Lp (Cg ). By [III17 , 3.17], K, Ly lies in I, a p-component of Cg with K, Ly covering I/Op (I) ∼ = D5 (q).

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Given the structure of K and Ly and the fact that L  E(Cu ) for all u ∈ E1 (D) − {x , y }, we conclude with [III17 , 15.3a] that [L, COp (Cg ) (u)] = 1 for all u ∈ D# , whence I = K, Ly is quasisimple, i.e., K, Ly ∼ = D5 (q) by [IA , 6.1.4]. But again since De = {x , y }, G0 = K, Ly , so (14J) holds, contrary to assumption. Thus C(x, K) ∩ C(y, Ly ) = 1. Now C(y, Ly ) centralizes B = y × (B ∩ Ly ), so C(y, Ly ) centralizes x and then acts on K = Lp (CG (x)). Hence C(y, Ly ) embeds faithfully in CAut(K) (K ∩ Ly ) ≤ CAut(K) (L) ∼ = O2− (q) ∼ = D2(q+1) , by [III17 , 15.3b]. But y is central in C(y, Ly ), so C(y, Ly ) embeds in O 2 (D2(q+1) ) ∼ = Zq+1 . The proof is complete.  J±

Recall the normal subgroup J− × J+ of CG (L) from Lemma 14.5, with y ±1 x ∈ and O 2 (J± )   CG (J∓ ). Lemma 14.9. The following conditions hold: (a) F ∗ (CG (L)) = O 2 (J− × J+ ); and (b) F ∗ (CG (y ±1 x)) = H± × O 2 (J∓ ) × L, with y ±1 x ∈ H± ∼ = Zq+1 and H± ≤ J± .

Proof. By Lemma 14.5d, L  CG (u) for all u ∈ D − De . Write C± = CG (y ±1 x). By Lemma 14.5b, O 2 (J∓ )  C± , and by Lemma 14.5a, ∗ 2 J∓  CC± (L). It follows that we may write ± ) = H± ×  ±1F (C∓1 O (J∓ ) × L for some ±1 subgroup H± with y x ∈ H± . Clearly y x, y x, B ∩ L = B is centralized by H± , so H± ≤ CG (x) and H± maps into CAut(K) (L y ), which is in turn isomorphic to Zq+1 , as noted in the previous proof. Using the subgroup J− × J+ × L  LCG (L) and Lemma 14.8, we see that (b) holds, as does (a).  Lemma 14.10. F ∗ (CG (D)) contains a subgroup A×L  CG (D), with CG (AL) = ∼ A, where L ∼ = Ω+ 6 (q) and A = C(x, K) × C(y, Ly ) = Zq+1 × Zq+1 . Here D = Ω1 (Op (A)). Proof. Let A = CO2 (J+ J− ) (D) ∼ = Zq+1 × Zq+1 . Since L  CG (D) and J+ J−  CG (L), A  CG (D). Also CCG (L) (A) maps into CAut(J+ J− ) (A), which is the image of A. Hence CCG (L) (A) ≤ ACCG (L) (J+ J− ) = A, as asserted. Finally A ≤ CG (LD) ≤ CCG (x) (y L), which in turn maps into CAut(K) (y L) ∼ = Zq+1 with kernel C(x, K), which also embeds in Zq+1 . As |A| = (q+1)2 , we have A = CG (LD) and |C(x, K)| = q + 1. By symmetry, |C(y, Ly )| = q + 1. As C(x, K) × C(y, Ly ) ≤ CG (LD) = A, equality holds, completing the proof of the lemma.  Recall that B ∩ L = b1 , b2 and we have g ∈ NG (B) with bgi = b5−i , 1 ≤ i ≤ 4. Lemma 14.11. F (CG (bi )) ≤ L for i = 1, 2. Proof. As bi ∈ xG , F (CG (bi )) =: Xi ∼ = C(x, K) ∼ = Zq+1 for i = 1, 2. If q = 2, then Xi = bi and the lemma clearly holds. So, assume that q > 2. As g interchanges D and Dg = b1 , b2 , and as L = E(CG (D)) = CG (D)(∞) , we have (J− × J+ )g = E(CG (Lg )) by Lemma 14.9a. Now, D ≤ Lg and g g × J+ ≤ E(CG (D)) = L. D g ≤ A g ≤ J−

But also Ag × Lg ≤ CG (b1 ). So Ag ≤ CG (b1 Lg ), and CG (b1 Lg )/C(b1 , K g ) ≤ CAut(K g ) (Lg ) ∼ = O2− (q). It follows that Ag = F (CG (b1 Lg )) contains F (CG (b1 )). Therefore, F (CG (b1 )) ≤ L.  As b2 ∈ bL 1 , the same holds for b2 , completing the proof.

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∼ Since L ∼ = Ω+ 6 (q), there exists U ≤ L such that B normalizes U = Eq 4 , indeed 2 ∼ NL (U ) is a parabolic subgroup with O (NL (U )/O2 (NL (U ))) = L2 (q) × L2 (q). Then by the Borel-Tits theorem, U ≤ O2 (P4 ) for some parabolic subgroup P4 of K  with NL (U )(B ∩ K) ≤ P4 . As mp (P4 ) ≥ 3, we must have O 2 (P4 )/O2 (P4 ) ∼ = Ω− 6 (q) ∼ acting naturally on O2 (P4 ) = Eq6 . We may write O2 (P4 ) = U × X4 with X4 = [X4 , b3 ] ∼ = Eq2 and both factors invariant under a Levi factor of NL (U ). We set C0 = CG (U ). Note that x = b4 and y = b3 are interchanged in J− × J+ ≤ CG (L). Hence, the same reasoning yields a B-invariant parabolic subgroup P3 of Ly with O2 (P3 ) = U × X3 ∼ = Eq6 and with X3 = [X3 , b4 ]. Let Ai = C(bi , E(CG (bi ))), 1 ≤ i ≤ 4. Thus by Lemma 14.10, A = A3 × A4 with A3 = C(y, Ly ) and A4 = C(x, K). Lemma 14.12. For i = 3, 4, we have CC0 (bi ) = U × Xi A ti , with | ti | ≤ 2 and CA (Xi ) = Ai . In any case, CO2 (C0 ) (bi ) ≤ U × Xi for i = 3, 4. Proof. As x = b4 and y = b3 are conjugate in NJ− ×J+ (A), it suffices to prove the assertions for i = 4. We have CC0 (b4 ) = CCG (x) (U ). Clearly U contains a long root subgroup of K, so CAut(K) (U ) ≤ Aut0 (K) ∼ = O8− (q) by [III17 , 6.1]. Let ∗ ∗ P4 = NAut0 (K) (P4 ). Then P4 is a split extension of O2 (P4∗ ) = O2 (P4 ) = U × X4 by L1 × Z1 , where L1 ∼ = O6− (q) acts naturally on U × X4 , and where Z1 ∼ = Zq−1 acts as scalars on U . Considering U × X4 as a 6-dimensional orthogonal space, we see that X4 = [X4 , A] and U are nonsingular subspaces, and so CAut0 (K) (U ) embeds in U × X4 O(X4 ) ∼ = U × X4 O2− (q) and contains the Zq+1 -image of A. As C(x, K) = A4 , the lemma follows.  Lemma 14.13. Let u ∈ D − (x ∪ y ). Then CC0 (u) = U × C(u, L), and one of the following conclusions holds: (a) u ∈ J± and C(u, L) = H± × J∓ ; or (b) F ∗ (C(u, L)) = A and |C(u, L) : A| ≤ 2. In all cases, CO2 (C0 ) (u) = U . Moreover, A contains a Sylow p-subgroup of C0 . Proof. As L  CG (u) and CAut(L) (U ) = U , it follows that CC0 (u) = U × C(u, L). If q = 2, then F ∗ (CG (L)) = A ∼ = E32 by Lemma 14.9a, so A = CC(u,L) (A)  C(u, L). As A contains exactly two conjugates of x , it follows that F ∗ (C(u, L)) = A in this case, whence (b) holds. Suppose then that q > 2, so that F ∗ (CG (L)) = J− × J+ ∼ = Ω+ 4 (q). Now, b1 ∈ L and F (CG (b1 )) ≤ L by Lemma 14.11. Setting K1 = E(CG (b1 )), we have that C(b1 , K1 ) ∩ CG (L) = F (CG (b1 )) ∩ CG (L) ≤ Z(L) = 1, so CG (L) is isomorphic to a subgroup of CAut(K1 ) (L1 ), where L1 = E(CL (b1 )) ∼ = Ω− 4 (q). It follows by + [III17 , 6.25] that CG (L) is isomorphic to a subgroup of O4 (q) containing J− ×J+ ∼ = + ∼ Ω+ (q). Since O (q) L (q)Z , it follows that if u ∈  J or J , then C (u) = A = 2 2 − + J J − + 4 4 and |C(u, L) : A| ≤ 2 with A = F ∗ (C(u, L)), as claimed. Finally, if u ∈ J± , then clearly C(u, L) = CJ− J+ (u) = H± × J∓ . It follows in all cases that CO2 (C0 ) (u) = U . By Lemma 14.12 and the above, A contains a Sylow p-subgroup of CC0 (u) for all u ∈ D# , whence A contains a Sylow p-subgroup of C0 , completing the proof.  Corollary 14.14. We have O2 (C0 ) = 1.

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± 14. G0 ∼ = Dn+1 (q)

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Proof. Let X0 = O2 (C0 ). Suppose that A ∩ X0 =: A0 = 1. Then A0 ≤ CA (Xi ) for i = 3, 4. But CA (X3 ) ∩ CA (X4 ) = A3 ∩ A4 = 1. Hence, A0 = 1. In particular, X0 is a p -group by Lemma 14.13, and CX0 (u) ≤ A0 = 1 for all u ∈ D# .  Hence, X0 = 1, as claimed. Lemma 14.15. We have F ∗ (C0 ) = O2 (C0 ). Proof. Suppose not. Then E = E(C0 ) = 1. Suppose first that D ∩ E = 1. Then J− J+ ∩ E = 1 and so E(CE (u)) = 1 for all u ∈ D# . Hence, D normalizes each component E1 of E, and CD (E1 ) = 1 for all such E1 . But, as D ∩ E = 1, E1 is a p -group and so mp (Aut(E1 )) ≤ 1, a contradiction. Thus D ∩ E = 1. If x ∈ D ∩ E or y ∈ D ∩ E, then D ≤ E, since y ∈ xC0 . Hence, in any case, as CD (Xi ) = bi for i = 3, 4, it follows that X4 = [X4 , D ∩ E]  CE (x). Now, let D ≤ A0 ∈ Sylp (C0 ). Let J0 be a component of E with J0 ∩ D = 1. If J1 is another component of E, then J1 = J± , and then J0 = J∓ . But then X4 ≤ CJ− J+ (x), a contradiction. Hence E is quasisimple. If E ∈ Chev(2), then  O 2 (CE (x)) is a product of Lie components in characteristic 2. But then X4 ≤ O2 (CE (x)) = 1, a contradiction. As G is a group of even type, E ∈ C2 − Chev(2) and Aut(E) has abelian Sylow p-subgroups of rank 2. It follows immediately that E ∈ Spor with D ≤ E and [O2 (CE (x)), D] = 1. By inspection of the tables in [IA , 5.3], E ∼ = M22 with p = 3. But then D# ⊆ xE , a final contradiction completing the proof.  Now let R = F ∗ (C0 ) = O2 (C0 ) and Ri = U × Xi . Lemma 14.16. We have R = U × X3 × X4 = R3 R4 ∼ = Eq 8 . Proof. By Lemma 14.13, CR (u) = U for all u ∈ D − (x ∪ y ). Moreover, CR (x) ≤ U × X4 and CR (y) ≤ U × X3 . As A acts irreducibly on Xi for i = 3, 4, and y ∈ xC0 , it follows that either R = U or R = U X3 X4 . As U = CG (U ), the latter case holds. Let R = R/U . By [G1, Lemma 3.16], R = X 3 X 4 . As A acts irreducibly on X 4 , it follows that NR (X 3 ) = R, and likewise for X 4 . Hence, [X 3 , X 4 ] = 1. Now X4 U is normalized by X3 and centralized by A4 . As [X3 , A4 ] = X3 , it follows  that [X4 U, X3 ] = 1 and R = U × X3 × X4 ∼ = Eq8 , as claimed. Since R is abelian by Lemma 14.16, and CG (R) ≤ CC0 (R) ≤ R by Lemma 14.15, R = CG (R). We now set P = NG (R) and P = P/R. Lemma 14.17. We have R = F ∗ (P )  NG (R3 ), NG (R4 ) , and E(P ) = O 2 (F ∗ (P )) ∼ = Ω+ 8 (q). Moreover, R is a natural module for E(P ), and yx and −1 y x both map into root SL2 (q)-subgroups of E(P ). 

Proof. We set Ci = CG (Ri ) = CC0 (Xi ), i = 3, 4, and claim that R = O2 (Ci ) for i = 3, 4. As R is abelian, R ≤ O2 (Ci ). On the other hand, as in Lemma 14.16, we see that O2 (Ci ) ≤ R, proving our claim. Therefore NG (Ri ) ≤ P , i = 3, 4. As Ri = CR (bi ), NLbi (Ri )  CP (bi ). This implies that CP (bi ) has a component Hi ∼ = Ω− 6 (q) acting naturally on CR (bi ) = ∼ Ri = Eq6 and centralizing X7−i = [R, bi ]. We have (14K)

R = X3 ⊕ (R3 ∩ R4 ) ⊕ X4 , and Ri = Xi ⊕ (R3 ∩ R4 ), i = 3, 4.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

As Ri is a natural Hi -module, i = 3, 4, it supports an Fq -vector space structure preserved by the action of Hi . Also Ri is irreducible as F2 Hi -module by [III17 , 5.10], so the centralizer in End(Ri ) of Hi is a field, and contains a unique copy of Fq . Thus the Hi -invariant vector space structure on the abelian group Ri is unique up to Aut(Fq ). In the same way, again using [III17 , 5.10], R3 ∩ R4 is a natural H3 ∩ H4 ∼ = Ω+ 4 (q)-module, and supports a Fq -vector space structure, unique up to Aut(Fq ). It follows that Fq -vector space structures for R3 and R4 may be chosen so that they coincide on R3 ∩ R4 and commute with the actions of H3 and H4 on R3 and R4 , respectively. Extending linearly to R, we may consider R as an Fq H3 , H4 -module. Similarly, as R3 ∩ R4 is an absolutely irreducible Fq (H3 ∩ H4 )-module, it supports a H3 ∩ H4 -invariant quadratic form over Fq , unique up to scalar multiples. The same holds for Ri as Fq Hi -module, i = 3, 4. Hence there exist (nondegenerate) quadratic forms Qi on Ri , i = 3, 4, such that Q3 |R3 ∩R4 = Q4 |R3 ∩R4 . Using the decomposition (14K), we define Q(x3 + r + x4 ) = Q3 (x3 + r) + Q4 (x4 ) = Q3 (x3 ) + Q3 (r) + Q4 (x4 ) = Q3 (x3 ) + Q4 (r) + Q4 (x4 ) = Q3 (x3 ) + Q4 (r + x4 ) for all xi ∈ Xi and r ∈ R3 ∩ R4 . Then Q is a nondegenerate quadratic form on R preserved by H3 , H4 . As both Q3 and Q4 are of the same type, and Q3 |R3 ∩R4 is of + type, Q is also of + type. Let X = H3 , H4 . We have proved that X embeds ∼ + in Ω(R, Q) ∼ = Ω+ 8 (q). Indeed, by [III17 , 9.1], X = H3 , H4 = Ω(R, Q) = Ω8 (q). Furthermore, since De = {x , y }, NP (D) permutes {H3 , H4 } and so (14L)

NP (D) ≤ NP (X).

We next show that (14M)

ΓD,1 (P ) ≤ NP (X).

Notice that with [IA , 4.8.2], there are two elements u ∈ E1 (D) − {x , y } such that CX (u) is the direct product of three copies of L2 (q). Since L  CG (u), it follows therefore from Lemma 14.9ab that if we put F = F ∗ (CG (L)), then F ≤ P and F ≤ X, and that these two u ’s are y ±1 x . As D ≤ F and D = Ω1 (S) for some S ∈ Sylp (F ), a Frattini argument gives NG (L) ≤ F NG (D), so NP (L) ≤ F NP (D) ≤ NP (X).

 Now for any d ∈ D − x − y , L  NG (d ) so it is immediate that NP ( d ) ≤  NP (X). Note also that by [IA , 4.8.2], the isomorphism type of O 2 (CH4 (y)) ∼ = L2 (q) × L2 (q) distinguishes the H4 -conjugates of y from all other subgroups of H4 N (x) H of order p. Therefore y P = y 4 , and so NP (x ) ≤ H4 NP (D) ≤ NP (X). Similarly NP (y ) normalizes X, and (14M) follows. We conclude that [X, Op (P )] = 1. So, by Lp -balance, X = H3 , H4 ≤ E(P ); indeed as p divides |H3 ∩ H4 |, X lies in a single component E of P . As H3  CE (y) with H3 ∼ = Ω− 6 (q), it follows from [III11 , 1.1ab] and [IA , 2.2.10] that E ∈ Chev(2), or q = 2 with E ∼ = P Sp4 (33 ) or Co2 . But m3 (P Sp4 (3)) > 4, and Co2 contains 1+4 3 , so neither of these groups can act nontrivially on R ∼ = E28 , contradiction. Thus E ∈ Chev(2). Then by [IA , 7.3.4] and (14M), E = ΓD,1 (E) normalizes X, so X = E is a component of P . Since CG (R) = R and X acts absolutely irreducibly on R, CP (X) ∼ = Zs acts as  scalars on R, with s dividing q − 1. In particular, R = F ∗ (P ) and X = O 2 (F ∗ (P )),

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± 14. G0 ∼ = Dn+1 (q)

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as claimed. The final assertion of the lemma follows from the fact, noted above,  that CP (y ±1 x) contains the direct product of three copies of L2 (q). The arguments to complete the identification G0 ∼ = D5 (q) are quite different for q > 2 and q = 2. Through Lemma 14.23, we assume that q > 2. Lemma 14.18. Assume that q > 2. Then there exists Y4 ∼ = Zq−1 and I4 ∼ = D4 (q) such that Y4 ≤ L, Y4 × I4 ≤ P , CR (Y4 ) = 1, I 4 = E(P ), B ≤ I4 , and E(CI4 (D)) ∼ = Ω+ 4 (q). ∼ D4 (q) is absolutely irreducible on R = F ∗ (P ), Proof. Since E(P ) = CP (E(P )) is cyclic of order dividing q − 1 and acts Frobeniusly on R. For i = 3 and 4, let Pi = NLbi (Ri ), a parabolic subgroup of Lbi with Levi factor isomorphic to Zq−1 × Ω− 6 (q). Then for any X such that Lbi ≤ X ≤ Aut(Lbi ), O2,2 (NX (Pi )) = O2,2 (Pi ) = Ri Yi for some Yi ∼ = Zq−1 . Thus CPi (Yi ) covers E(CE(P ) (bi )). Replacing Y4 by an R4 -conjugate if necessary, we may assume that Y4 centralizes b3 . Then Y4 normalizes R3 = CR (b3 ) and so maps into NAut(Lb3 ) (P3 ), while centralizing E(CE(P ) (b3 , b4 )) ∼ (Ω+ = Ω+ 4 (q). But CAut(Ω− 4 (q)) has order 6 (q)) relatively prime to q − 1. Moreover, so does C(bi , Lbi ) ∼ = Zq+1 . This implies that Y3 may be taken to be equal to Y4 , and thus Y4 centralizes   E(CE(P ) (b3 )), E(CE(P ) (b4 )) = E(P ).  ∼ + Since O 2 (CP3 (b4 )) is an extension of a 2-group by Ω+ 4 (q), E(CP (D)) = Ω4 (q). By the first paragraph, CR (Y4 ) = 1, so CP (Y4 ) contains a subgroup I4 such that I4 ∼ = I 4 = E(P ). Replacing Y4 and I4 by suitable conjugates if necessary, we may arrange that B ≤ I4 . Finally Y4 ≤ CK (y) and |CK (y) : L| is relatively prime to  q − 1, so Y4 ≤ L. The proof is complete.

Fix Y4 ∼ = Zq−1 and I4 ∼ = D4 (q) as in Lemma 14.18, and set C4 = CG (Y4 ), so that B ≤ I4 ≤ C4 . Let ID = E(CI4 (D)) ∼ = Ω+ 4 (q). Then ID ≤ L. Lemma 14.19. Assume that q > 2. Let 1 = g ∈ Y4 . Then I4  CG (g). Proof. Set Cg = CG (g) and let u ∈ D# . If u = x or y , then ∼ Lp (CCG (u) (g)) = Lp (CLu (g)) ∼ = Ω− 6 (q) = Lp (CI4 (u)), so Lp (CI4 (u)) = Lp (CCG (u) (Y4 )) = Lp (CCG (u) (g)). Then by [III17 , 9.1] and Lp -balance,  I4 = Lp (CI4 (x)), Lp (CI4 (y)) = Lp (CCG (x) (g)), Lp (CCG (y) (g)) ≤ Lp (Cg ). Let X = Op (Cg ). We show that [I4 , X] = 1. It is enough to show that [ID , X] = 1, and so enough to show that [ID , CX (u)] = 1 for every u ∈ D# . Now g ∈ Y4 ≤ L. Then CL (g) = Y4 × ID . Hence if u ∈ E1 (D) − De , then as L  CG (u) we have [ID , CX (u)] = 1. If u = x, on the other hand, then CK (g) ∼ = Ω− = Y4 ×Id with Id ∼ 6 (q) and Id ≤ I4 . Hence [ID , CX (x)] ≤ [Id , CX (x)] = 1. Similarly [ID , CX (y)] = 1, completing the proof that [I4 , X] = 1. Thus, I4 ≤ E(Cg ).

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Now mp (CG (B)) ≤ mp (CG (x)) ≤ mp (Aut(K)) + 1 ≤ 6. Say that I4 projects nontrivially on k components of E(Cg ); then mp (CG (B)) ≥ 4k, so k = 1 and I4 lies in a component I0 of Cg . For a similar reason, I0  Cg . Since Y4 ≤ L, (Cg ∩ NG (L))(∞) ≤ J− × J+ × (C4 ∩ L)(∞) , by Lemma 14.9a. Using this and the structures of CG (x) and CG (y), it is straightforward to check that for any u ∈ D# , Lp (CCg (u)) ≤ I4 . In particular, Lp (CI0 (u)) = Lp (CI4 (u)), and Lp (CI0 (x)) ∼ = U4 (q). As q > 2, I0 ∈ Chev(2). Hence by [III17 , 9.5], I4 = I0 , completing the proof.  Lemma 14.20. Assume that q > 2. Let 1 = g ∈ Y4 . Then F ∗ (CG (g)) = Y4 I4 and I4 = CG (g)(∞) . Proof. By the previous lemma and the facts that C(x, K) ∼ = Zq+1 and x ∈ I4 , F (CG (g)) ≤ Y4 I4 . Hence [Y4 , F ∗ (CG (g))] = 1 so the lemma follows by the F ∗ theorem and the Schreier property.   There exists h ∈ L such that Y4 , Y4h = Y4 × Y4h ≤ CG (D) and Y4h ≤ E(CL (Y4 )) ≤ E(CK (Y4 )). Then Y4h ≤ CG (Y4 )(∞) = I4 . Let ∗

J4 = E(CI4 (Y4h )). h ∼ We have y ∈ CK (Y4 Y4h )(∞) ∼ = Ω− 4 (q) = CLy (Y4 Y4 )  x, these groups being components of CJ4 (x) and CJ4 (y), respectively. It follows that D ≤ J4 ∼ = Ω+ 6 (q), by ±1 ∼ [III17 , 6.36]. In particular, E(CJ4 (y x)) = L2 (q). Lemma 14.21. Assume that q > 2. Then for all g ∈ Y4 Y4h −Y4 −Y4h , J4  CG (g). ∼ L2 (q) × Proof. Let g ∈ Y4 Y4h − Y4 − Y4h , and set Jg = E(CL (g)). Then Jg = L2 (q) or 1, since g ∈ L ∼ = L4 (q) has no fixed points on the natural L-module. We have Lp (CG (g, x )) = E(CK (g)) = CJ4 (x) × Jg , and Lp (CG (g, y )) has similar structure. For u = y ±1 x, we have L  CG (u) and so Lp (CG (g, u )) = Jg J∓ = Jg E(CJ4 (u)), a commuting product. Finally for u ∈ D − (x ∪ y ∪ yx ∪ y −1 x ), Lp (CG (g, u )) = Jg . It follows by [III17 , 10.12] that Lp (CG (g)) = Jg J4∗ with J4∗ a normal p-component of CG (g) and J4∗ = J4 Op (J4∗ ). Using Lemma 14.13, if we put X = [Op (J4∗ ), D], then X = [CX (x), D][CX (y), D]. But y ∈ E(CCG (x) (g)), so [CX (x), y] = 1; also [CX (y), x] = 1, in a similar way, so X = 1. As D ≤ J4 , J4∗ is quasisimple so J4∗ = J4 . The proof is complete.  Within L we see that Y4 Y4h ≤ Jg1 , Jg2 for the two subgroups g1 and g2 of Y4 Y4h of order q − 1 such that Jgi = 1. But Jg1 ∼ = L2 (q) × L2 (q) contains an Ep2 subgroup D∗ . As Sylow p-subgroups of L are abelian of rank 2, D∗ is conjugate to b1 , b2 and hence to D, in G. As each Jgi centralizes J4 , [D∗ , J4 ] = 1. But CG (D) has a unique subgroup isomorphic to Ω+ 6 (q), namely L. Hence J4 is conjugate to L and Y4 Y4h to a subgroup a+ , a− of J+ × J− with a± ∈ J± . Let r be a prime divisor of q−1 and let a5 ∈ a+ , a− be of order r and conjugate to an element of Y4 . Let I = E(CG (a5 )) ∼ = D4 (q). Then E(CI (a+ , a− )) = L. Since J4  CG (g) for all g ∈ Y4 Y4h − Y4 − Y4h , there exist exactly two subgroups of a+ , a− of order r centralizing a copy of D4 (q) in G. (Otherwise Lemma 14.20 would be contradicted.) One of these we have named a5 ; let the other one be a4 . Now a5 does not lie in either J+ or J− , for if it did, say a5 ∈ J+ , then J− would be a component of CG (a5 ), contradicting Lemma 14.20. Therefore an involution i in a Sylow r-normalizer of J+ does not normalize a5 , so i must interchange a5

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with a4 . Furthermore, there is an involution of I inverting a+ , a− ∩ I, and it follows that a4 = Ω1 (Or (a+ , a− )) ∩ I. As E(CI (a+ , a− )) ∼ = Ω+ 6 (q), it follows from [III17 , 6.37] that there exist a1 , a2 , a3 ∈ E(CI (a+ , a− )) of order r such that if we put A = a1 , a2 , a3 , a4 , a5 , then NI (A) contains an element j cycling a1 , a2 , a3 , a4 and centralizing a5 . Then NI := i, j normalizes A and stabilizes {a1 , a2 , a3 , a4 , a5 }, inducing Σ5 on it. Lemma 14.22. Assume that q > 2. Then I, NI has a subgroup J ∼ = D5 (q) such that I ≤ J. Proof. Paragraphs three, four, and five of the proof of Lemma 14.4 may be repeated here with A in place of B, m = 5, V0 = 0, r in place of p, ai in place of  bi , I in place of K, and NI CG (A) in place of N0 . Lemma 14.23. q = 2, p = 3, C(x, K) = x , and C(y, Ly ) = y . Proof. Assume that q > 2. By construction, I = E(CG (a5 )) is conjugate to I4 = E(CG (Y4 )), which contains B. Hence a conjugate J ∗ of J contains B. Because W ≡B W (C4 ), there exist exactly four subgroups b ∈ E1 (B), including x and y , such that WCG (b) involves W (C3 ). On the other hand, there are four b ∈ E1 (B) such that E(CJ ∗ (b)) ∼ = D4− (q). Therefore these are the same four, whence E(CJ ∗ (x)) = K and E(CJ ∗ (y)) = Ly . Thus, G0 = K, Ly = J ∗ , the last equality by [III17 , 9.1]. This contradicts our assumption that the proposition fails, however. Hence q = 2. The other statements follow as p divides q + 1 and from Lemma 14.10.  Since E(P ) ∼ = D4 (2) is absolutely irreducible on R, F ∗ (P ) = E(P ). As P acts + ∼ on R, P = Ω8 (2) or O8+ (2). Set m = |P : E(P )| ≤ 2. Let S ∈ Syl2 (E(P )) and let S be the full inverse image of S in P . Set Z = z = Z(S) ∼ = Z2 , Cz = CG (Z), and Q = O2 (Cz ). We may assume that S was chosen so that z is a root element of L within U . Then + Cz ∩ P is a split extension of E26 by Ω+ 6 (2) or O6 (2), according as m = 1 or 2. Now CCz (x) contains C(x, K) × CK (Z) = x × CK (Z) with index m, and ∼ CK (Z) = Q4 H4 with Q4 ∼ = 21+8 + , Z(Q4 ) = Z, H4 = H × K4 , H = L2 (2), and − ∼ y ∈ K4 = Ω4 (2), with Q4 /Z the tensor product of natural modules for H and K4 . Moreover, CH4 (D) = H × K4,D with K4,D ∼ = Z3 . Similarly, CCz (y) ∼ = CCz (x) contains y ×CLy (Z) with index m, and CLy (Z) = Q3 H3 with Q4 ∼ = Q3 , Z(Q3 ) = Z, H3 = H ∗ × K3 , H ∗ ∼ = L2 (2), and K3 ∼ = K4 . In particular, O3 (CCz (x)) and O3 (CCz (y)) are 2-groups. Lemma 14.24. The following conditions hold: ∼ 21+12 ; and (a) F ∗ (Cz ) = Q = + (b) For any u ∈ E1 (D) − De , D centralizes every D-invariant 3 -subgroup of CG (u).

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Proof. By Lemma 14.9, F ∗ (CG (L)) = D. For any u ∈ E1 (D) − De , L  CG (u), so D = O3 (CG (L))  CG (u), which implies (b). In particular, O3 (Cz ) ≤ O3 (CCz (x))O3 (CCz (y)) so O3 (Cz ) is a 2-group. Thus, O3 (Cz ) = Q = CQ (x)CQ (y) ≤ Q3 Q4 . Now, Q4 K4 is a 3-component of CCz (x), so by L3 -balance and the (B3 )-property, Q4 ≤ O3 (Cz ) = Q. Similarly Q3 ≤ Q, so Q = Q3 Q4 . In particular, CQ (D) = Q3 ∩ Q4 . It follows that Q = [Q3 , x](Q3 ∩ Q4 )[Q4 , y]. We claim that this factorizais a central product [Q4 , y]CQ4 (y) = tion is a central product. First, Q4 ∼ = 21+8 + [Q4 , y](Q3 ∩ Q4 ), and as y ∈ K4 , both factors are isomorphic to 21+4 + . A similar factorization holds for Q3 . Also, [Q3 , x] = [Q, x]  Q and similarly [Q4 , y]  Q. Hence [[Q3 , x], [Q4 , y]] ≤ [Q4 , y] ∩ (Q3 ∩ Q4 ) = z , and the action of x, y forces [[Q3 , x], [Q4 , y]] = 1, proving our claim. It follows immediately that Q ∼ . = 21+12 + ∗  Finally, we know that O3 (Cz ) = Q, so to show that F (Cz ) = Q, it suffices to derive a contradiction from assuming that CCz (Q) has a nontrivial Sylow 3subgroup. But in this case [D, g] = 1 for some 3-element g ∈ CCz (Q). As CD (Q) = 1, g ∈ D. Then g normalizes L and centralizes Q3 ∩ Q4 = O2 (CL (z)). By the Borel-Tits theorem, [L, g] = 1. However, by Lemma 14.10, A = D = CG (DL), so g ∈ D, a contradiction. This completes the proof.  z = Cz /Q and prove We next set C  = I1 × Iz , with Lemma 14.25. There exists M  Cz such that Q ≤ M and M + ∼ ∼     I1 = L2 (2), Iz = Ω6 (2), and R ∈ Syl2 (I1 ). Moreover, S ∈ Syl2 (M ), and there are two Iz -composition factors in Q/Z, both natural modules. Proof. We have Cz ∩ P = CP (Z), an extension of R by a group with a subgroup of index m of the form 26 Ω+ 6 (2) and containing D. Set H = CP (Z)(∞) and S1 = O2 (CP (Z)), so that |S1 | = 214 . Under the action of H/O2 (H), S1 has just two non-trivial chief factors, both natural modules (of order 26 ). We argue that these nontrivial factors are both contained in Q. Suppose that k > 0 of them are outside Q. If k = 1, then |R ∩ Q| ≤ 27 , the maximum order of an elementary abelian subgroup of Q, and so a trivial module of order 2 is also involved in S1 /S1 ∩ Q. Thus in any case if k > 0, then |S1 /S1 ∩ Q| ≥ 27 and |S1 Q| ≥ 214+6k . But with Lemma 14.24b, S1 Q = CS1 Q (x)CS1 Q (y) with both factors of order at most 212 m. The intersection of the two factors is CS1 Q (D) = CQ (D)CS1 (D), with |CQ (D)| = |Q3 ∩ Q4 | = 25 and |CS1 /S1 ∩Q (D)| ≥ 22−k m. Therefore |S1 Q| ≤ 224 m2 /27−k m = 217+k m, so 217+k m ≥ 214+6k . As m ≤ 2 and k = 1 or 2, this is a contradiction. It follows that Q ≤ S1 . Hence by L2 -balance,  ≤ L2 (C z ). H Note that m3 (CCz (D)) ≤ m3 (CCz (x)) = 1 + m3 (CK (Z)) = 3. As D ≤ H it z .  lies in a single 2-component Iz of C follows that H We have CL (Z)  CCz (D), with |CCz (D) : CL (Z)| = 9m, O2 (CL (Z)) = ∼ Q3 ∩ Q4 ∼ = 21+4 + , and CL (Z)/O2 (CL (Z)) = Σ3 . For any u ∈ E1 (D) − De , ∗ CCz (u) = CCG (u) (Z) normalizes F (CG (L)) = D and since E1 (D) contains exactly 2 conjugates of x , O 2 (CCz (u)) ≤ CCz (D), so CCz (u) has abelian Sylow 3-subgroups of rank 3. Similarly CCz (x) = CCG (x) (z) and CCz (y) have abelian  ≤ H,  it Sylow 3-subgroups of rank 3, by inspection of O 2 (Aut(K)) ∼ = K. As D

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± 14. G0 ∼ = Dn+1 (q)

245

 centralizes every D-invariant  z . follows that D elementary abelian 3-subgroup of C z ) has a cyclic Sylow 3-subgroup and hence has a normal 3-complement. Thus O2 (C  z ) lies in the {2, 3}-group CC (D) Moreover, every D-invariant subgroup of O{2,3} (C z z ) is cyclic, whence Iz is quasisimple. (Lemma 14.24) and so is trivial. Thus O2 (C Set I1 = CL (Z); then Σ3 ∼ = I1  CCz (D). For u = x or y, CCz (u), the image of CCG (u) (Z) modulo Q, is equal to I1 × Iu , ∼ Ω− (2). ∼ Ω− (2) or O − (2) according as m = 1 or 2. Thus E(C  (u)) = with Iu = 4

4

Cz

4

On the other hand, for u ∈ E1 (D) − De , as L  CG (u) we have I1  CCz (u) but   C  (u). E(CCz (u)) = 1, and D Cz +  with |R|  = 2. It  ∼ But also H = Ω (2) is a D-invariant component of C  (R), 6

Iz

follows by [III17 , 10.26], the structure of CCz (u) given in the previous paragraph  for u ∈ D# , and the fact that m3 (CCz (D)) ≤ 3, that Iz = H.  z = Γ (C z ). As I1  C  (u) for all u ∈ D# , we conclude that Therefore C  D,1

Cz

z . I1  C  By our construction S = Finally, let M ≤ Cz be the full preimage of I1 H. O2 (SH)(S∩H) = S1 (S∩H), and we showed above that S1 ≥ Q with S1 centralizing Iz . Therefore S1 ∈ Syl2 (I1 ), so S ∈ Syl2 (M ). The proof is complete.  Since I1 × Iz acts faithfully on Q/Z ∼ = E212 by Lemma 14.24, and since the  centralizer ring of the Iz -natural module is F2 , it follows immediately from Lemma  and is 14.25 that Q/Z is the tensor product of natural modules for I1 and I, ∗  absolutely irreducible. Hence F (Cz ) = O3 (I1 )Iz , and z ∼ C = Σ3 × A8 or Σ3 × Σ8 . Moreover, if we let S ≤ S ∗ ∈ Syl2 (Cz ), then |S ∗ : S| ≤ 2 and Z(S ∗ ) = Z, so S ∗ ∈ Syl2 (G). Let P0 be the full preimage of E(P ) in P , so that P0 is an extension of R by E(P ) ∼ = D4 (2) and S ∈ Syl2 (P0 ). Lemma 14.26. P0 splits over R, and M splits over Q. Proof. O3 (I1 ) is fixed-point free on Q/Z, so M/Z splits over Q/Z. Let g ∈ I3 (M ) with  g = O3 (I1 ). If M does not split over Q, it follows that CM (g) has a component I ∗ ∼ = 2A8 , by [IA , 6.1.4]. But by definition, I1 ≤ L, so g ∈ L and hence g has a conjugate in B ∩ L = b1 , b2 . Under the action of W , therefore, g is conjugate into D, and indeed either to x or yx. Hence 2A8 is embeddable in CG (x)(∞) or CG (yx)(∞) . As the latter lies in L, 2A8 is in any case embeddable in K∼ = Ω− 8 (2). This contradicts the Borel-Tits theorem as no parabolic subgroup of K has a Levi factor containing a copy of A8 . We have proved that M splits over Q. It then follows from [III8 , 5.6] that the isomorphism type of M is uniquely  ∈ Syl2 (I1 ), we may determined. In particular, S is of type D5 (2). Moreover, as R  choose an involution h ∈ R such that  h is an involution of I1 . Then hQ = R, and this calculation may be mimicked in D5 (2). Therefore when we regard S ≤ D5 (2), R = O2 (PR ) for some parabolic subgroup PR , and hence S splits over R. Therefore  P0 splits over R, and the proof is complete.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Now P0 and M are isomorphic to maximal parabolic subgroups of our target group D5 (2), and have S as a Sylow 2-subgroup. We need to construct a further parabolic subgroup. Let LP be a complement to R in P0 , and LM a complement to Q in M . By [III17 , 6.40], S has exactly three elementary abelian subgroups of maximal order 210 ; call them Ai , 1 ≤ i ≤ 3. They are normal in S, and S splits over each of them. Two of them are unipotent radicals of maximal parabolic subgroups in D5 (2), and are interchanged in NAut(D5 (q)) (S). Set Mi = NG (Ai ). Since Z = Z(S), and Cz = CG (Z) satisfies F ∗ (Cz ) = Q = O2 (Cz ), F ∗ (Mi ) = O2 (Mi ), 1 ≤ i ≤ 3. We also have Ai = CS (Ai ) for each i by [III17 , 6.40], with S/Ai isomorphic to a Sylow 2-subgroup of SL5 (2). Two of the Ai ’s, say A1 and A2 , are found in P0 in the form Ai = Ui Qi , where Ui ≤ R is totally singular and |Ui | = 24 , while Qi ≤ LP and NLP (Qi ) =: Ni is a parabolic subgroup of type Ω+ 6 (2). Computing in D5 (2) we see that Ai = J(Ai R), i = 1, 2. Hence NP0 (Ai ) = NP0 (Ai R) = RNi for i = 1, 2, a split extension of Ai , as S splits over Ai . Similarly, two of the Ai ’s, say A1 and Ak (where k = 2 or k = 3) are found in M in the form Ai = Vi Ti , with Vi ∈ E7 (Q) and Ti ≤ E(LM ) ∼ = Ω+ 6 (2) and NE(LM ) (Ti ) =: Yi is a parabolic subgroup of type L3 (2). Again computing in D5 (2) we see that Ai = J(Ai Q), i = 1, k, so NM (Ai ) = NM (Ai Q) = Q(Y0 × Yi ) with Y0 ∼ = L2 (2), again a split extension of Ai , i = 1, k. Furthermore, Cz ∩ Mi = NCz (Ai ) = NCz (Ai Q) ≤ M since elements of Cz − M induce graph automorphisms on E(LM ) and hence cannot normalize Yi . Since z = Z(S), it follows that S ∈ Syl2 (Mi ), i = 1, k.

∼ 1 = M1 /A1 . Then it follows that (M1 ∩ P0 ) is an extension of R Set M = E24 + ∼

by Ω6 (2) = L4 (2), acting naturally. Likewise (M1 ∩M ) is an extension of Q ∼ = E26 by L2 (2) × L3 (2), acting via the tensor product of natural representations. Both of 1 ) ∩ S = 1. As S ∈ Syl2 (M1 ), these actions are irreducible, and it follows that O2 (M we have F ∗ (M1 ) = A1 . Indeed the only subgroups of S that are simultaneously Sylow subgroups of normal subgroups of (M ∩ M1 ) and (P0 ∩ M1 ) are S and 1, so for each odd prime 1 )] = 1 or CS (Or (M 1 )) = 1. In the latter case, as m2 (S)

= 6, also r, either [S, Or (M 1 )) ≥ 6 by Thompson’s dihedral lemma. But this is absurd as M 1 embeds mr (Or (M ∼   in Aut(A1 ) = GL10 (2). Therefore [S, F (O2 (M1 ))] = 1, so [S, O2 (M1 )] = 1. It 1 ) is quasisimple and contains (M ∩M1 ) and (P0 ∩Mi ) as 2follows easily that E(M 1 ) ∼ local subgroups of odd index. By [III17 , 16.8], E(M = L5 (2). Since S ∈ Syl2 (M1 ), |Out(L5 (2))| = 2, and S splits over A1 , M1 is a split extension of A1 by L5 (2). Lemma 14.27. There exist i ∈ {1, 2} and subgroups X ≤ P0 , Ji ≤ Mi , and H ≤ P0 ∩ Mi such that the following conditions hold: (a) P0 = RX, R ∩ X = 1, and B g ≤ X for some g ∈ G; (b) Mi = Ai Ji and Ai ∩ Ji = 1; and (c) H(S ∩ X) and H(S ∩ Ji ) are parabolic subgroups of X and Ji , respectively, in both of which H ∼ = L4 (2) is a Levi factor. Proof. By Lemma 14.26, there is a complement X to R in P0 , and then S ∩ X ∈ Syl2 (X). By construction, B normalizes U , then R = O2 (CG (U )), so B ≤ O 2 (NG (R)) = P0 . By Sylow’s theorem, B g ≤ X for some g ∈ R. Let

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± 14. G0 ∼ = Dn+1 (q)

247

P1 = NX (A1 R) = NX (A1 ), a parabolic subgroup of X containing S ∩ X and normalizing the totally singular 4-dimensional subspace E1 = A1 ∩ R of R. Then P1 has a Levi decomposition P1 = T1 H1 with T1 ∼ = E26 and H1 ∼ = L4 (2) acting irreducibly on T1 , E1 , and R/E1 ; indeed H1 normalizes a complement E2 to E1 in R, by [III17 , 6.38]. Clearly, A1 = T1 E1 and RP1 ≤ M1 . Since M1 splits over A1 we may choose a complement J1 , and then S ∩ J1 ∈ Syl2 (J1 ). Then RP1 ∩ J1 is a parabolic subgroup of J1 containing S ∩ J1 , with Levi decomposition T2 H2 , say, where T2 ∼ = E24 , H2 ∼ = L4 (2), and S ∩ H2 ∈ Syl2 (H2 ). Regarding T2 H2 ≤ P1 ≤ P , we must have T 2 = 1, i.e., T2 ≤ R. Since H2 normalizes T2 , it follows from [III17 , 6.38] that H 2 and H 1 are T 1 -conjugate. Replacing H1 by a conjugate we may thus assume that E2 = T2 and H 1 = H 2 . Then as H 1 (H 1 , Ei ) = 0 for i = 1, 2 by [III17 , 16.9], H1 and H2 are R-conjugate. Conjugating X by an element of R, we may finally assume that H1 = H2 . Then H = H1 satisfies the conditions of the lemma.  Now as H is a Levi factor in X, a weak CT-system {L2 , L3 , L4 } of H extends to a weak CT-system {L1 , L2 , L3 , L4 } in X of type D4 (2), with L3 the central node. Likewise as H is a Levi factor in J1 , {L2 , L3 , L4 } extends to a weak CT-system {L2 , L3 , L4 , L5 } in J1 of type L5 (2), with end nodes corresponding to L2 and L5 . Lemma 14.28. {L1 , L2 , L3 , L4 , L5 } is a weak CT-system of type D5 (2). Proof. It remains only to check that [L1 , L5 ] = 1. From the known connections among the Li ’s, all five of them are G-conjugate. By Lemma 14.17, yx maps into a root SL2 (2)-subgroup of P/R, and hence O3 (Li ) is G-conjugate to yx , i = 1, . . . , 5. We argue that Li  NG (O3 (Li )) for each such i. It suffices to prove this for i = 2. Set N2 = NG (O3 (L2 )). We (∞) have L = NG (yx )(∞) , so E2 := N2 is conjugate to L ∼ = L4 (2). But L3 (2) ∼ = L4 , L5 ≤ CJ1 (L2 ) ≤ CG (L2 ), so CE2 (L2 ) contains L3 (2). Hence L2 acts trivially on E2 , so L2 ≤ CN2 (E2 ). Moreover, in P we see that CP (L2 ) contains the direct product of three copies of L2 (2). Thus a Sylow 2-subgroup of L2 centralizes an E33 -subgroup of N2 . With Lemma 14.5ab, this implies that L2 is G-conjugate to J+ and so L2  N2 , as claimed. Now consider the embedding of L1 and L5 in N2 . We saw above that L5 ≤ E2 , and complete the proof by showing that [L1 , E2 ] = 1. Now in P0 = RX, L1 , L2 is a Levi factor for the centralizer in X of a totally singular plane in R. Hence m2 (CP0 (O3 (L1 , L2 ))) ≥ 4. This implies, given the structure of N2 , that O3 (L1 ) centralizes E2 . Indeed E2 is the unique subgroup of N2 of its isomorphism type, and N2 is conjugate in G to NG (O3 (L1 )), so as [O3 (L1 ), E2 ] = 1, we have E2 = (NG (O3 (L1 )))(∞) . As L1  NG (O3 (L1 )) and Sol(E2 ) = 1, [L1 , E2 ] ≤ L1 ∩ E2 = 1. The proof is complete.  At last we can show: Lemma 14.29. G0 ∼ = D5 (2). Proof. Suppose this is false and continue the above argument. Set G0 = L1 , L2 , L3 , L4 , L5 , so that G0 ∼ = D5 (2) by [III13 , Theorem 1.4]. Note that G0 ≥ g L1 , L2 , L3 , L4 = X ≥ B for some g ∈ G, by Lemma 14.27. As B is, up to conjugacy, the unique E34 -subgroup of CG (x), NG (B) controls the G-fusion of x in W B. But from the structure of W ∼ = W (C4 ), x is the unique orbit of W on E1 (B)

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248

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

with point stabilizers containing W (C3 ). It follows that E(CG0 (xg )) ∼ = D4− (2), i.e., −1 K g ≤ G0 . Thus, J := (G0 )g contradicts Lemma 14.3. The lemma follows.  This completes the proof of Proposition 14.1.  15. G0 ∼ = Lkq (q)

We next treat the cases (a) and (b) of Proposition 3.4. These lead to the target  group G0 /Z(G0 ) ∼ = Lkq (q) for some k. According to Propositions 10.2 and 6.1 there  are two cases, in both of which K/Z(K) ∼ = Lmq (q): q (1) m ≥ 4, mp (B) = m, W ∼ (q), and B is iso= Σm+1 , L ∼ = SLm−1 morphic as W -module to the quotient of a standard permutation module by the 1-dimensional trivial module; (15A) (2) m ≥ 4, mp (B) = m, p divides m + 2, W ∼ = Σm+2 (or possibly q (q), and B is isomorphic as Z2 × Σ6 if m = 4), L ∼ = SLm−1 W -module to the core of a standard permutation module. Note that when m = 4, Proposition 6.1 gives F ∗ (W ) = Z(W ) × Ak , k = m + 1 or m + 2. Thus F ∗ (W ) contains no reflections. But W is generated by reflections. If σ ∈ W is a reflection, then we must have E(W ) σ ∼ = Σk (P GL2 (9) is impossible since σ cannot invert an element of order 5 when p = 3). It follows that W ∼ = Σk (or Z2 × Σ6 ), as asserted in (15A). Furthermore, in (15A2) with m = 4, the only permutation nontrivial F3 [A6 ]-module of dimension 4 is the core of a standard module; and in (15A1) with m = 4, since WK ∼ = Σ4 , xW is a quotient of a standard permutation module, so it must have dimension at least 4, hence equal B. Proposition 15.1. If one of the cases (15A) holds, then G0 is a nontrivial q q (q) or SLm+2 (q), with p dividing m + 2 in the latter homomorphic image of SLm+1 case. Taking case (15A1) first, and using Propositions 10.2 and 6.1 and Lemma 5.2, we assume that  (1) K/Z(K) ∼ = Lmq (q), m ≥ 4, and if q = 2, then m ≥ 7; q ∼ (2) L = SLm−1 (q), and for all u ∈ De , Lu /Z(Lu ) ∼ = K/Z(K); and (15B) (3) W ∼ = Σm+1 , mp (B) = m, and B is W -isomorphic to the quotient of the natural permutation module for W by the 1-dimensional trivial submodule. We first prove: Lemma 15.2. If (15A1) holds, then G0 is a nontrivial homomorphic image of q (q). SLm+1 We assume that Lemma 15.2 is false. Since WK ∼ = Σm−1 and WK is the stabilizer of x in W , we may label xW as {b1 , . . . , bm+1 = x}. Lemma 15.3. We have b1 b2 · · · bm+1 = 1. Proof. W permutes {b1 , . . . , bm+1 }, while CB (W ) = 1 by (15B2).



Now let V = V1 ⊥ V2 ⊥ · · · ⊥ Vm be the natural Fqτ -module for C(V ) ∼ = q GLm (q), with dim(Vi ) = 1 for all i. Here τ = 1 or 2 according as q = 1 or −1, and if  q = 1, then “⊥” just indicates direct sum. Set C(V  )o = [C(V  ), C(V  )] ∼ = SLmq (q). 

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q 15. G0 ∼ = Lk (q)

249

∼ GLq (q), and Let V := V  ⊥ Vm+1 be the natural Fqτ -module for C(V ) = m+1  q likewise set C(V )o = [C(V ), C(V )] ∼ = SLm+1 (q). Fix ζ, a primitive p-th root of 1 in Fqτ . Also choose 0 = vj ∈ Vj , j = 1, . . . , m + 1. We specify an elementary abelian p-subgroup E = e1 , . . . , em+1 of C(V ). If p does not divide m + 1, then for each i = 1, . . . , m + 1, let ei ∈ C(V )o be defined by ei (vj ) = ζvj for all j = i, and ei (vi ) = ζ −m vi . Note that in this case, ζ −m = ζ. Then e1 e2 · · · em+1 = 1, and this generates all dependence relations among e1 , . . . , em+1 . On the other hand, if p does divide m + 1, then we cannot necessarily find ei ’s in C(V )o , and instead we simply define ei ∈ C(V ) by ei (vj ) = vj for all j = i, and ei (vi ) = ζvi . In this case, the ei are independent. In any case, we set E = e1 , . . . , em+1 , and there is a subgroup T ≤ NC(V )o (E), which we fix, such that T ∼ = Σm+1 acts faithfully on the set E := {e1 , . . . , em+1 }. Because of Lemma 15.3, there is a surjective homomorphism η : E → B such that η(ei ) = bi for all i = 1, . . . , m + 1. Then η extends to a surjection f : C(V  )o E → KB. (Note that em+1 |V  is a scalar mapping, so it lies in Z(C(V  )).) Let N be the preimage of W in NG (B). Thus, N/CN (B) = W ∼ = Σm+1 , and N faithfully permutes the set B := xN = {b1 , . . . , bm+1 }. As T permutes E, f (T ∩ C(V  )o ) permutes B and so lies in NG (B ) = N . Moreover, there is an isomorphism W → T which, together with η −1 , transports the W -action on B to the T -action on E. Thus there is a homomorphism λ : N → T such that λ(f (σ)) = σ for all σ ∈ T ∩ C(V  )o , which is hypothesis (c) of [III13 , Theorem 3.1]. We verify hypothesis (d) of [III13 , Theorem 3.1]. Let I ⊆ {1, . . . , m} and u ∈ N with λ(u)(I) ⊆ {1, . . . , m}. We need to show that o f (CIo )u = f (Cλ(u)(I) ).

(15C)

o   Here, as in [III13 , Section 3], we write  CI = CC(V  )o (VI  ) ∩ NC(V  )o (VI ) for any  subset I ⊆ {1, . . . , m}, where VI  = j∈I, j≤m Vj and VI = j∈I Vj . Thus if we put

EI  = ei | 1 ≤ i ≤ m, i ∈ I and BI  = bi | 1 ≤ i ≤ m, i ∈ I , 

we have CIo = O p (E(CC(V  )o (EI  ))), as long as when q = 2 (so that p = 3 and q = −1), we assume that |I| ≥ 4. Subject only to this restriction on I ⊆ {1, . . . , m}, we conclude that 



f (CIo ) = O p (E(CK (BI  ))) = O p (E(CG (BI  bm+1 ))).  The condition λ(u)(I) ⊆ {1, . . . , m} yields BIu bum+1 = B(λ(u)(I)) bm+1 , so    f (CIo )u = O p (E(CG (BI  bm+1 )))u = O p (E(CG (BIu bum+1 )))

(15D)



o = O p (E(CG (B(λ(u)(I)) bm+1 ))) = f (Cλ(u)(I) ),

as desired. It remains to prove (15C) when q = 2 and |I| ≤ 3. Since m ≥ 7 in −1 this case, we can find subsets of I1 , I2 of {1, . . . , m} ∩ {1, . . . , m}λ(u) such that |I1 | = |I2 | = |I| + 1 and I1 ∩ I2 = I. Note that it follows directly from the definition of CIo that (15E)

CIo1 ∩I2 = CIo1 ∩ CIo2 .

With this fact, (15C) follows quickly for the remaining values of |I| by descending induction on |I|.

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The other hypotheses of [III13 , Theorem 3.1] hold trivially, and so  q H := K N is a homomorphic image of SLm+1 (q). As a result, we easily deduce Lemma 15.2. Obviously H is N -invariant, and thus q B-invariant. The only subgroups F ∈ E2 (B) such that E(CK (F )) ∼ (q) are = SLm−1 the subgroups bi , bm+1 , 1 ≤ i ≤ m, so without loss y = bm and D = bm , bm+1 . Then CW (y) ∼ = CW (x) ∼ = Σm while CW (u) ∼ = Σm−1 or Σm−1 × Z2 for all u ∈ D − x − y . But if u ∈ De , then CW (u) ∼ = Σm by (15B2). Thus De = {x , y } so G0 = K, Ly = H0 , the last by [III17 , 9.1]. This completes the proof of Lemma 15.2. As we are assuming that Proposition 15.1 fails, (15A1) does not hold. Thus (15A2) must hold. Again with Propositions 10.2 and 6.1 and Lemma 5.2, we thereby assume (1) K/Z(K) ∼ = Lmq (q), m ≥ 4, and if q = 2, then m ≥ 7; moreover, p divides m + 2; q (q), and for all u ∈ De , Lu /Z(Lu ) ∼ (2) L ∼ = K/Z(K); and = SLm−1 ∗ ∼ (3) W = Σm+2 , mp (B) = m, and B is W -isomorphic to the core of a natural permutation module for W . 

(15F)

Note also that since p divides m+2, p divides neither m nor m+1. In particular, B = x × (B ∩ K). We consider W ∼ = Σm+2 to be permuting {1, . . . , m + 2}. Let B ∗ = (b1 × · · · × bm+2 )/ b1 · · · bm+2 be the quotient of the natural permutation module for W over Fp modulo its 1-dimensional trivial submodule Z. Here W permutes {b1 , . . . , bm+2 } naturally. Caution: B ∗ is not a subgroup of G. For each i = j set bij = b−1 i bj (mod Z). Then bij ∈ B ∗ and we identify B with the subgroup of B ∗ generated by all bij , 1 ≤ i < j ≤ m + 2. As m ≥ 4, it follows easily that the stabilizer Wij of the ordered pair (i, j) is the subgroup of W fixing bij , and CB ∗ (Wij ) is in turn generated by the images of bi and bj . In particular, CB (Wij ) = bij . As B = x × (B ∩ K) and WK ∼ = Σm , with p dividing m + 2, it follows from [III17 , 1.29] that x = bij and WK = Wij for some i = j. We assume without loss that x = bm+1,m+2 and WK = Wm+1,m+2 . Likewise [III17 , 1.29] implies that for any u ∈ De − {x }, u = bij for some 1 ≤ i < j ≤ m + 2. q (q) centralizes D = x, u , it follows that we may assume that As L ∼ = SLm−1 De ⊆ {x , bm,m+2 , bm,m+1 }. As these three subgroups of B are W -conjugate, equality must hold. Writing y = bm,m+2 , therefore, we have proved:  Lemma 15.4. In case (15A2), De = {x , y , xy −1 }. Let Wm+2 ∼ = Σm+1 be the stabilizer of m + 2 in W , and let Nm+2 be the full preimage of Wm+2 in NG (B). Lemma 15.5. G contains a B-invariant subgroup H1 such that K ≤ H1 and q (q). H1 is a homomorphic image of SLm+1 Proof. For each i = 1, . . . , m + 1, define fi = bi,m+2 . Then f1 · · · fm+1 = 1, and Wm+2 permutes {f1 , . . . , fm+1 } faithfully. We have fm+1 = x, and for 1 ≤ i ≤ m, CWm+2 (fi , x) ∼ = Σm−1 . It follows that the (nontrivial) action of fi on K is that of a diagonalizable gi ∈ K of order p with m−1 equal eigenvalues, say equal to ζi . Since CWm+2 (x) permutes {f1 , . . . , fm } transitively, ζ2 = · · · = ζm = ζ, say. The one

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remaining eigenvalue of gi is then necessarily ζ 1−m . Note also that the projections of f1 , . . . , fm on x are all equal, so fi−1 fj = gi−1 gj for all 1 ≤ i < j ≤ m. Moreover, gi−1 gj has two eigenvalues different from 1, namely ζ ±m . Next, let V be an m + 1-dimensional vector space of type Lq over the field Fqτ . Write (15G)

V = V1 ⊥ V2 ⊥ · · · ⊥ Vm+1

where dim Vi = 1 for all i, 1 ≤ i ≤ m + 1. Let C = C(V ) be the isometry group of o ∼ V , C o = [C, C], V(m+1) = m i=1 Vi , and C(m+1) = CC o (Vm+1 ) ∩ NC o (V(m+1) ) = q SLm (q). For each i = 1, . . . , m + 1, let ei ∈ C o preserve the decomposition (15G), with eigenvalue ζ −m on Vi and ζ on every other Vj . Set E = e1 , . . . , em+1 , an elementary abelian p-subgroup of C o in which all dependence relations on e1 , . . . , em+1 are consequences of the obvious relation e1 · · · em+1 = 1. There is a group T ≤ C o permuting faithfully the summands in (15G), with T ∼ = Σm+1 , and represented by permutation matrices with respect to a basis consisting of one vector from each Vi ,  −1 o ei ej | 1 ≤ i < j ≤ m , with e−1 i = 1, . . . , m + 1. Furthermore, C(m+1)  ∩ E = i ej having eigenvalues ζ m+1 = ζ −1 on Vi , ζ −m−1 = ζ on Vj , and 1 on all other Vk in (15G). q o o ∼ As C(m+1)  = SLm (q), there is a surjective homomorphism f : C(m+1) → K, which we may assume takes the stabilizer of (15G) into NK (B). Then for 0 ≤ i < −1 j < m, f (e−1 i ej ) has the same eigenvalues as ei ej . Fix an integer r such that −1 −1 r r mr ≡ m + 1 (mod p); we may in fact assume that f (e−1 i ej ) = (gi gj ) = (fi fj ) . r o Now define η : E → B by η(ei ) = fi . Then η and f agree on C(m+1) ∩E. Moreover, o ei acts on C(m+1)  by conjugation as a quasi-reflection (i.e., its 1-eigenspace has codimension 1) with eigenvalue ζ −m−1 , while fir , like gir , acts on K as a quasireflection with eigenvalue ζ −mr . It follows that there is a surjective homomorphism o C(m+1)  E → KB extending both η and f . Just as in the case (15A1), there is a homomorphism λ : Nm+2 → T with kernel CG (B) such that hypothesis (c) of [III13 , Theorem 3.1] is satisfied, now with Nm+2 in the role of N ; and λ and η transport the action of Nm+2 on {f1 , . . . , fm+1 } to the action of T on {e1 , . . . , em+1 }. A calculation just like that in the case (15A1) shows that hypothesis (d) 13 , Theorem 3.1] is also satisfied. That theorem  of [III then yields that H1 := K Nm+2 satisfies the conclusions of the lemma (note that B ≤ Nm+2 so H1 is B-invariant). Note also that the surjection f ∗ : C o → H1 provided by the theorem extends both f and η.  Now set V∗ = V ⊥ Vm+2 , where Vm+2 is isometric to V1 . Let C∗ = C(V∗ ) and C∗o = [C∗ , C∗ ]. We regard C o as CC∗o (Vm+2 ) ∩ NC∗o (V ). Thus E ≤ C o ≤ C∗o . We have just seen that there is a surjection f ∗ : C o → H1 extending f and η. As f ∗ (ei ) = fir = bri,m+2 for each i, ∗ r 1 ≤ i ≤ m + 1, it follows that if we define eij = e−1 i ej for i = j, then f (eij ) = bij . o Here again, rm ≡ m + 1 (mod p). (As E ≤ C , there is no need to define an analogue η ∗ of η.)  Lemma 15.6. Let N be the full preimage of W in NG (B) and set H2 := K N . q Then H2 is a homomorphic image of SLm+2 (q).

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

252

  Proof. Since H1 = K Nm+2 with Nm+2 ≤ N , H2 = H1N . Now, there is a subgroup T ∗ ∼ = Σm+2 of C∗o permuting {V1 , . . . , Vm+2 } faithfully and consisting of permutation matrices with respect to some basis {v1 , . . . , vm+2 } with vi ∈ Vi for each i = 1, . . . , m + 2. From the definition of the eij and the bij , it follows that there is a homomorphism λ∗ : N → T ∗ with kernel CG (B) such that f ∗ and λ∗ transport the action of W on the bij ’s to the action of T ∗ on the eij ’s. Thus again, hypothesis (c) of [III13 , Theorem 3.1] is satisfied. The verification of hypothesis (d) is a bit different from the usual. Let X = {1, . . . , m + 1}, and suppose that I ⊆ X and u ∈ N with J := λ(u)(I) ⊆ X. As usual, we need (15H)

f ∗ (CIo )u = f ∗ (CJo ).

This time we set BI  = bij | i, j ∈ (X − I) ∪ {m + 2} and we have f ∗ (CIo ) =  O p (E(CG (BI  ))), as long as when q = 2 we assume |I| ≥ 4. Again, subject only to this last condition, we get BIu = BJ  and (15H) follows. And again, if q = 2 and |I| ≤ 3, then since m ≥ 7, we can write I = I1 ∩ I2 where |I1 | = |I2 | > |I| and Ik ∪ λ(u)(Ik ) ⊆ X, k = 1, 2, so that (15H) follows by descending induction on |I|, using the analogue of (15E).  Now [III13 , Theorem 3.1] implies the lemma. 

q (q). Lemma 15.7. In case (15A2), G0 is a homomorphic image of SLm+2

Proof. Let H2 be as in Lemma 15.6. Then K = Lp (CH2 (x)) and L = Lp (CH2 (D)). By [III17 , 3.4], there are exactly three elements u ∈ E1 (D) such that L is not a component of Lp (CH2 (u)). As L   CG (u) for u ∈ E1 (D) − De , these three elements of E1 (D) must comprise De . By Lemma 15.4 and [III17 , 9.1],   G0 = K, Ly , Lxy−1 = H2 , and the proof is complete. Lemmas 15.2 and 15.7 now complete the proof of Proposition 15.1. 16. G0 ∼ = E8 (q) The group E8 (q) is the target group in cases (g) and (n) of Proposition 3.4. We treat these cases together in this section, assuming: (1) W ≡B W (E8 ); and   (16A) (2) Either (g) p = 3, K/Z(K) ∼ = SL7q (q), or (n) = L9q (q), and L ∼ K∼ = D6 (q). = E7 (q) and L ∼ We shall use the Gilman-Griess theorem [III13 , 2.2], and prove: Proposition 16.1. Assume (16A). Then G0 ∼ = E8 (q). For uniformity, we shall show that even in case (g), there is a semisimple neighborhood of the same type as in case (n) (although not containing an element of J∗p (G)). In the two cases, 

mp (B) = 1 + mp (E7 (q)) = 8 and m3 (B) = 1 + m3 (GL7q (q)) = 8, respectively. ∗ Let G∗ = E8 (q), with root system Σ∗ . Let (G , σ) be a σ-setup for G∗ . Several ∗ calculations in G may be made in G instead. Specifically, since W ≡B W (E8 ) and Aut(W ) = Inn(W ), the action of W on B is equivalent to the action in G∗ = E8 (q) on B ∗ ∈ Ep8 (G∗ ) by AutG∗ (B ∗ ) ∼ = W . We may therefore compute CW (E) for

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various subgroups E ≤ B by computing the counterpart in G∗ – we call this the “G∗ -trick.” As an application we show: Lemma 16.2. Suppose we are in case (g). Then |E1 (D)| = 4, and the elements  of E1 (D) may be labelled ui , 1 ≤ i ≤ 4, in such a way that Lu1 = K ∼ = A8q (q), q ∼ ∼ ∼ Lu2 = E7 (q), and Lu3 = Lu4 = D7 (q). Proof. By [IA , 7.4.1a4], for any E ∈ E2 (B ∗ ),  (16B) q 120 = |G∗ |2 = |CG∗ (E)|−p 2

|CG∗ (E1 )|2 .

E1 ∈E1 (E)  In case (g), p = 3 and L ∼ = A6q (q). Let E ∈ E2 (B ∗ ) correspond to D ∈ E2 (B). Then by the G∗ -trick, CW (E) ∼ = Σ7 , which implies that CG∗ (E) ∼ = WL ∼ = A6 and thus q 2 ∼ O (CG∗ (E)) = A6 (q). Similarly, for any E1 ∈ E1 (E), we find – using [IA , 4.7.3A]    – that O 2 (CG∗ (E1 )) ∼ = A8q (q), D7q (q), or E7 (q). Let n1 , n2 , n3 be the number of E1 ∈ E1 (E) with centralizers of these three respective types. Then n1 + n2 + n3 = 4 and by (16B), 120 = −3 · 21 + 36n1 + 42n2 + 63n3 . The only solution is n1 = 1, n2 = 2, n3 = 1. Hence by the G∗ -trick, we may label elements of D so that x = u1 , CW (u2 ) ∼ = W (E7 ), and CW (u3 ) ∼ = CW (u4 ) ∼ = W (D7 ). Since the Lui are level pumpups of L, the lemma follows. 

Lemma 16.3. Suppose that case (g) holds. Then there exists E ∈ E2 (B) whose subgroups of order 3 may be labelled ei , 1 ≤ i ≤ 4, in such a way that  L3 (CG (e1 )) ∼ = L3 (CG (e2 )) ∼ = E7 (q), L3 (CG (e3 )) ∼ = L3 (CG (e4 )) ∼ = D7q (q), and L3 (CG (E)) ∼ = D6 (q). ∼ D6 (q), as can Proof. In G∗ , there is E ∗ ∈ E2 (B ∗ ) such that L3 (CG∗ (E ∗ )) = be seen from [IA , 4.7.3A]. For e∗ ∈ (E ∗ )# , L3 (CG∗ (e∗ )) is a pumpup of D6 (q)  and again by [IA , 4.7.3A], L3 (CG∗ (e∗ )) ∼ = E7 (q) or D7q (q). An application of ∗ (16B) with E in the role of E shows that there are two subgroups e∗ of each type. Hence labelling is possible such that CW (e1 ) ∼ = CW (e2 ) ∼ = W (E7 ), CW (e3 ) ∼ = CW (e4 ) ∼ = W (D7 ), and CW (E) ∼ = W (D6 ), by the G∗ -trick. Now B ∗ = J(P ) for any Sylow 3-subgroup P of G∗ containing B ∗ , by [IA , 4.10.3d]. Thus G∗ -fusion in B ∗ is controlled in NG∗ (B ∗ ). By [IA , 4.7.3A], there is a unique orbit, in the action of W (E8 ) on E1 (B ∗ ), on subgroups with a given isomorphism type of component (modulo core) in their centralizers. As different components in this case have different Weyl groups, we conclude that W (E8 ) is transitive on cyclic subgroups 1 = b ≤ B ∗ for which CW (E8 ) (b) has a given isomorphism type. This translates directly to the action of W on B. Therefore, if e, e ∈ E and CW (e) ∼ = CW (e ), it follows that L3 (CG (e)) ∼ = L3 (CG (e )). In particular, since the Lui ’s in Lemma 16.2 are quasisimple, so are the Lei ’s. The lemma follows.  Now we can prove Lemma 16.4. Assume (16A). Then there exists D∗ ≤ B such that the following conditions hold: (a) D∗ ∼ = D6 (q); = Ep2 and L∗ := Lp (CG (D∗ )) ∼ (b) For every d ∈ (D∗ )# such that the pumpup L∗d of L∗ in CG (d) is nontrivial,  L∗d is quasisimple and isomorphic to D7q (q) or E7 (q);

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254

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

∼ E7 (q) and L∗ = ∼ Dq (q); and (c) There exist u, v ∈ (D∗ )# such that L∗u = v 7 (d) With u, v as in (c), there are reflections w1 , . . . , w8 ∈ W such that {w1 , . . . , w7 } and {w2 , . . . , w8 } are fundamental sets of reflections in AutL∗u (B) ∼ = W (E7 ) and AutL∗v (B) ∼ = W (D7 ), respectively. ∗ Moreover, in case (n), D = D satisfies the above conditions. Proof. If case (g) holds, let D∗ be the subgroup E of Lemma 16.3. That lemma implies (a), (b), and (c), and (d) follows directly. On the other hand, if case (n) holds, then as L = E(CG (D)), (a) holds with D∗ = D, and CW (D) ∼ = W (D6 ). We take u = x in this case, and compute in G∗ ∼ that CW (v) = W (D7 ) for some v ∈ D# , with fundamental reflections generating WK and CW (v) as stated in (d). Now for any d ∈ De , Ld is a vertical level pumpup of L with mp (Aut0 (Ld )) > mp (L) = 6, and F(Ld ) ≤ F(K) by [III14 , Lemma 1.2]. As the Dynkin diagram of Ld thus contains the D6 diagram, the only choices are  Ld ∼ = D7q (q) or K, so (b) holds. As CW (v) ∼ = W (D7 ), (c) holds as well. The proof is complete.  Now fix D∗ as in Lemma 16.4, and define De∗ = De in case (n), and De∗ = E1 (D∗ ) in case (g). Set GD∗ = L∗d | d ∈ De∗ . Thus in case (n), GD∗ = G0 . Note that as D∗ ≤ B, B ≤ CG (d) for all d ∈ (D∗ )# . In case (g), B = (B ∩ L∗d ) × d for all d ∈ (D∗ )# . In any case, GD∗ is B-invariant. We aim to prove that (16C) GD ∗ ∼ = E8 (q). Through Lemma 16.9, assume that (16C) is false. Lemma 16.5. C(u, L∗u ) and C(v, L∗v ) have odd order. Proof. Let T ∈ Syl2 (C(u, L∗u )) and let d ∈ De∗ . Then T centralizes B = (B ∩ L∗u ) u so T maps into CAut(L∗d ) (L∗ (B ∩ L∗d )), which has odd order by Lemma 16.4ab and [III17 , 6.24]. Therefore [T, L∗d ] = 1, and as d ∈ De∗ was arbitrary, GD∗ = L∗d | d ∈ De∗ ≤ CG (T ). If T = 1, then GD∗ < G and so by [III17 , 3.19], GD ∗ ∼ = E8 (q), contrary to our assumption. Thus, C(u, L∗u ) has odd order. The  proof that C(v, L∗v ) has odd order is similar. Let Σv be a root system of type D7 . We quote [III17 , 2.38] to obtain the following, after replacing B by a conjugate, if necessary. NL∗v (B) contains a subgroup Wv ∼ = W (Σv ) with CWv (B) = 1 and CB (Wv ) = v . Moreover, Wv may be chosen so that there is a B-invariant subgroup L∗1 ≤ L∗v such that L∗1 ∼ = D4 (q), ), and W permutes naturally a W (D B = (B ∩ L∗1 ) × CB (L∗1 ), W1 := Wv ∩ L∗1 ∼ = 4 1 set {Xα | α ∈ Σ1 } of root groups of L∗1 ; here Σ1 is the root system of L∗1 . Indeed, W1 = nα (1) | α ∈ Σ1 , using standard notation [IA , 2.10] for L∗1 . Furthermore, Wv may be chosen so that Wv ∩ L∗ ∼ = W (D6 ). The next lemma is the key result. Lemma 16.6. Assume (16A). Then there exists W0 ≤ NG (B) ∩ GD∗ such that (with standard notation for L∗1 ) the following conditions hold: (a) W0 CG (B)/CG (B) = W and W0 ∩ CG (B) = 1; (b) W0 ∩ L∗1 permutes naturally the system {Xα | α ∈ Σ1 } of root groups of L∗1 ; and (c) W0 ∩ L∗1 = nα (1) | α ∈ Σ1 .

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Proof. By the paragraph before the lemma, (b) and (c) hold with W1 in place of W0 ∩ L∗1 . It therefore suffices to produce W0 satisfying (a) and such that W1 = W0 ∩ L∗1 . By [III17 , 2.2ab], there exists a complement Wu to CG (B) in NL∗u (B), and Wu ∼ = W (D6 ). Then = W (E7 ). Let Wu,v = CWu (D∗ ) ∼ 



Wu,v = O 2 (Wu,v ) ≤ O 2 (CL∗u (D∗ )) = L∗ . In particular, Wu,v ≤ L∗v , so Wu,v ≤ NL∗v (B) = CL∗v (B)Wv . Similarly, let Wv,u = CWv (u), so that Wu,v ≤ CL∗v (B)Wv,u . So both Wu,v and Wv,u are complements to CL∗v (B) in CL∗v (B)Wv,u , and their images modulo CG (B) are both W ∩ L∗ . By [III17 , 2.2], CL∗v (B) has odd order, and by [III17 , 2.20], CL∗v (B) has a W ∩L∗ -invariant filtration all of whose quotients M are abelian and satisfy H 1 (W ∩ L∗ , M ) = 0. It follows that Wu,v and Wv,u are CL∗v (B)-conjugate. Replacing Wv by a CL∗v (B)-conjugate, therefore, we may assume that Wu ∩ Wv = Wu,v = Wv,u ∼ = W (D6 ). We may then choose involutions t1 , . . . , t8 , mapping on w1 , . . . , w8 ∈ W , respectively, such that t1 , . . . , t7 form a fundamental set of generators for Wu ; t2 , . . . , t8 do the same for Wv ; and t2 , . . . , t7 = Wu,v . We set W0 = t1 , . . . , t8 , so that W0 maps onto W . Then Wv ≤ W0 by construction, so W1 ≤ W0 ∩ L1 . It therefore remains to verify that W0 ∩ CG (B) = 1, which we do, following [GiGr1], by checking that the ti ’s satisfy the usual defining relations for a finite reflection group (of the form (ti tj )mij = 1). All but one of those relations hold in either Wu or Wv ; the single one that may not hold is [t1 , t8 ] = 1. From the action on B, we see that [w1 , w8 ] = 1. Let g = [t1 , t8 ] = (t1 t8 )2 ∈ CG (B). Then t1 and t8 invert g ∈ CG (B). We argue that g = 1 to complete the proof. Suppose first that g has even order and let z be the involution in g . Then t1 ∼ t1 z and t8 ∼ t8 z in t1 , t8 . As g ∈ CG (B), z ∈ CG (B) ≤ CCG (u) (B). Now L∗u ∼ = E7 (q), so CAut0 (L∗u ) (B ∩ L∗u ) has odd order. Moreover, C(u, L∗u ) has odd order by Lemma 16.5. Hence the image of z in Aut(L∗u ) lies outside Aut0 (L∗u ), so z induces a nontrivial field or graph-field automorphism on L∗u . A similar argument proves that z induces a nontrivial field or graph-field automorphism on L∗v . Consider C := CG (z) ≥ B. Now, Lp (CC (u)) ∼ = E7 (q 1/2 ), Lp (CC (v)) ∼ = D7± (q 1/2 ), ∼ D6 (q 1/2 ). Clearly the only possibility, using Lp -balance, is and Lp (CC (D∗ )) = that C has a p-component I with I/Op (I) ∼ = E8 (q 1/2 ) and B ≤ I. Set C = C/Op (C). As B is a maximal elementary abelian p-subgroup of G, I = F ∗ (C).  Hence, O p (C (∞) ) ≤ I. As [z, t1 ] = 1, z is Inndiag(L∗u )-conjugate to zt1 by [IA , 4.9.1]. Hence they are ∗ Lu -conjugate and so t1 ∼G z, whence CG (t1 ) ∼ = C. Now, t1 lies in a root SL2 (q) subgroup of L∗u , so CL∗u (t1 ) contains a p-component H such that H/Op (H) ∼ = D6 (q) (see [III11 , 6.19]). Thus, H/Op (H) is involved in I ∼ = E8 (q 1/2 ). But this contradicts Lagrange’s theorem as |H/Op (H)| is divisible by q 8 − 1, but |I| is not, by Zsigmondy’s theorem. We conclude that g has odd order. Suppose that q = 2. Then p = 3 and by [III17 , 2.33b], CL∗u (B) ≤ B, so g ∈ [t1 , CG (B)] ≤ CL∗u (B) ≤ B. Hence if g = 1, then [t1 , B] = g = [t8 , B]. Since [w1 , w8 ] = 1 this implies that CB (t1 ) = CB (t8 ) and so w1 = w8 , which is not

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the case as the wi ’s form a fundamental set of generating reflections for W , and CW (B) = 1. Thus, g = 1 in this case. Now assume that q > 2. Choose b4 ∈ B such that b4 = [B, t4 ]. As the action of W on B is the same in G as in E8 (q), we calculate that b4 is W -conjugate to  u . Let K4 = E(CG (b4 )) ∼ = L∗u . We see in L∗u that t1 ∈ O p (E(CL∗u (b4 ))), and in  L∗v that t8 ∈ O p (E(CL∗v (b4 ))). By Lp -balance, t1 , t8 ≤ K4 . Thus g = [g , t1 ] = [g , t8 ] ∈ [CK4 (B), t1 ] ∩ [CK4 (B), t8 ]. But the intersection on the right is trivial by [III17 , 2.33c], so g = 1. The proof is complete.  For any α ∈ Σ1 , let Wα be the preimage in W0 of the stabilizer of α in W . By Lemma 16.6c and a basic property of reflection groups [St1, p. 273, (36)],   (16D) Wα ∩ L∗1 = nβ (1)  β ∈ Σ1 ∩ α⊥ .  Lemma 16.7. Assume (16A). Let α ∈ Σ1 . Then Wα = (Wα ∩ L∗1 )Wα . Proof. We have Wα ∼ = W (E7 ). As E7 is simply laced, Wα has a unique conjugacy class C of reflections. But Wα ∩ L∗1 contains some reflections by (16D),   so (Wα ∩ L∗1 )Wα ≥ C = Wα . Lemma 16.8. Assume (16A). For any α ∈ Σ1 , let Lα = Xα , X−α . Then [Wα , Xα ] = [Wα , Lα ] = 1. Proof. Let α ∈ Σ1 . We first claim that Wα normalizes Lα . Indeed, Wα normalizes Bα := CB (wα ), a hyperplane of B containing u. Hence CG (Bα ) can be calculated in CG (u). Now B normalizes L∗1 and induces inner automorphisms on it, so B = (B ∩ L∗1 ) × CB (L∗1 ). As wα ∈ L∗1 , CB (L∗1 ) ≤ Bα . It follows that    L∗1 = O 2 (CL∗u (CB (L∗1 ))). Thus O 2 (CL∗u (Bα )) = O 2 (CL∗1 (Bα ∩ L∗1 )) = Lα , the last equation with the help of [IA , 4.8.2]. As C(u, L∗u ) has odd order by Lemma  16.5, and Out(L∗u ) = 1 when q = 2, we see that Lα = E(CG (Bα )) or O 2 (CG (Bα )), according as q > 2 or q = 2. In any case, Lα is Wα -invariant, as claimed. By the claim and Lemma 16.7, it suffices to show that (16E)

[Wα ∩ L∗1 , Lα ] = 1.

But (16E) holds by (16D) and the Chevalley relations in L∗u . The proof is complete.  Now we attain our objective. Lemma 16.9. Assume (16A). Then GD∗ ∼ = E8 (q).  ∗ W Proof. Let G2 = (L1 ) 0 . By [III17 , 1.3], any pair of roots in Σ∗ is W conjugate to a pair in Σ1 , so E8 (q) and L∗1 are compatible over W . By Lemmas 16.6b and 16.8, hypotheses (a) and (b) of the Gilman-Griess theorem [III13 , 2.2] hold. That theorem therefore yields G2 ∼ = E8 (q). We shall show that G2 = GD∗ . Now W0 ≤ L∗u , L∗v by construction,  ∗ W so W0 ≤ GD∗ and hence G2 ≤ GD∗ . On ∗ ≤ (L1 ) 0 as every root in Σ is W0 -conjugate to one the other hand, clearly L  ∗ W u 0 ≤ G2 . Likewise B ≤ (L∗u ∩ B)W0 ≤ G2 . Let d be any in Σ1 . Thus (Lu ) ∗ element of De ; to complete the proof we need only show that L∗d ≤ G2 . By Lemma  16.4b, L∗d ∼ = D7q (q) or E7 (q). Accordingly WL∗d ∼ = W (D7 ) or W (E7 ), and CW0 (d) has a subgroup isomorphic to WL∗d . As CW0 (d) acts irreducibly on B/ d , it does

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not normalize D∗ ; therefore, since D∗ = CB (L∗ ), CW0 (d) does not normalize L∗ . as G2 ∼ Hence L∗  CG2 (d). = E8 (q), we see from [III11 , Table 13.1] that Now q ∗ ∗ I := (L∗ )E(CG2 (d)) ∼ D (q) or E = 7 7 (q). As I ≤ Ld , it follows easily that I = Ld . ∗  Therefore Ld ≤ G2 and the proof is complete. Lemma 16.10. Assume (16A). Then G0 ∼ = E8 (q). Proof. In case (n) there is nothing to prove as GD∗ = G0 in that case. So assume that case (g) holds. Let d ∈ D# . The component Ld of CG (d) is given in Lemma 16.2, and as GD∗ ∼ = E8 (q), there exists b ∈ B # and a component J of ∼ CGD∗ (b) such that WJ = WLd (see [IA , 4.7.3A]). Now WGD∗ = W , so as in Lemma 16.3, this implies that d and b are GD∗ -conjugate. Hence Ld and J are GD∗ conjugate, so Ld ≤ GD∗ . We have proved that G0 ≤ G  D∗ . Finally, with Lemma  16.2 again, and with [IA , 7.3.3], GD∗ = Γ2D,1 (GD∗ ) ≤ Ld | d ∈ D# ≤ G0 . Thus  G0 = GD ∗ . This completes the proof of Proposition 16.1.  17. G0 ∼ = E6q (q) and E7 (q)

Next, consider case (m) of Proposition 3.4:   (17A) K ∼ = A5q (q), and W ≡B W (E7 ). = E6q (q), L ∼ Proposition 17.1. Assume (17A). Then G0 ∼ = E7 (q). We assume that Proposition 17.1 fails, and proceed almost exactly as in the proof of Proposition 16.1. In particular, we use the Gilman-Griess theorem and follow [GiGr1]. Let Σ be the root system of K, and let Σ∗ be the root system of type E7 . We regard Σ as a subsystem of Σ∗ . There exists a B-invariant root SL2 (q)-subgroup J ≤ CK (L), and B ∩ J = D ∩ J ∼ = Zp . We define v  ∈ D by (17B)

v  = D ∩ J.

One difference from the E8 -case is that the roles of K and x there will be taken by Lv  ∼ = D6 (q) and v  as in the following lemma. Lemma 17.2. Assume (17A) and (17B). Then the following conditions hold:   (a) For any u ∈ De , Lu ∼ = A6q (q), D6 (q), or E6q (q);  (b) There is v ∈ D# such that Lv ∼ = A6q (q); moreover, there is g ∈ CG (B)  ∼ such that Lv g /CLv g (Lv ) = P GL7q (q); (c) Lv ∼ = D6 (q); (d) There are reflections w1 , . . . , w7 ∈ W such that w1 , . . . , w6 form a fundamental set of reflections in WLv and w2 , . . . , w7 form a fundamental set of reflections in WLv . Proof. Let G∗ = E7 (q). As in the E8 case, since W ≡B W (E7 ) and p is odd, the G∗ -trick applies, i.e., we may compute CW (E) for various subgroups E ≤ B by computing the counterpart in G∗ . As L = E(CG (D)), CW (D) ∼ = W (A5 ). We compute in G∗ that CW (v) ∼ = W (A6 ) for some v ∈ D# and CW (v  ) ∼ = W (D6 ), with a reflection in W ∩ J inverting v  . Moreover, there are fundamental reflections generating CW (v  ) and CW (v) as stated in (d), with WL = w2 , . . . , w6 . Now for any u ∈ De , Lu is a vertical level pumpup of L with F(Lu ) ≤ F(K) by

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258

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

[III14 , Lemma 1.2]. As the untwisted Dynkin diagram of Lu thus contains the  A5 diagram by [IA , 4.2.2], the only choices are Lu ∼ = A6q (q), D6 (q), or K, so (a)  holds. As CW (v) ∼ = W (A6 ) and CW (v ) ∼ = W (D6 ), (c) and the first statement of (b) hold as well. (Note that it is not possible that Lv ∼ = D6− (q), as this would entail mp (B ∩ Lv ) = 5, and so B ∩ Lv could not admit W (D6 ) faithfully.) It remains to prove the second part of (b). If 7 does not divide q − q , then P GL7 (q) ∼ = L7 (q) and we can take g = 1. So suppose that 7 divides q − q . We have AutLv (B) ∼ = Σ7 , and we choose a 7-element t ∈ NLv (B) − CLv (B). We can choose t so that CAut(Lv ) (t) covers a Sylow 7-subgroup of Aut(Lv )/Aut0 (Lv ), i.e., t is centralized by a Sylow 7-subgroup of the group ΦLv of field automorphisms of Lv . Then if g does not exist as asserted, a t-invariant Sylow 7-subgroup P of CG (B) has the property that Ja (P ) = J0 (Ja (P ) ∩ C(v, Lv )), where J0 = Ja (P ∩ Lv ). Here J0 /J0 ∩ C(v, Lv ) is abelian of rank 6, exponent 7b , say, and order 76b−1 . Hence t acts on P := P/Φ(P )Ωb−1 (Ja (P )) with minimal polynomial a divisor of (t − 1)6 . But the image t of t in AutG (B) lies in a Frobenius group of order 8.7. Hence t has a free summand on P, a contradiction. The proof is complete.  Lemma 17.3. C(x, K), C(v  , Lv ), and C(v, Lv ) have odd order. Proof. The proof is like that of Lemma 16.5, now using [III17 , 3.20] and [III17 , 6.24b].  Let Σv be the root system of Lv . By [III17 , 2.2, 2.39], there exists Wv ≤ NLv (B) with Wv ∩ CG (B) = 1 and Wv = nα (1) | α ∈ Σv ∼ = W (D6 ), in standard notation for Lv . Similarly, there is Wv ≤ NLv (B) with Wv ∩ CG (B) = 1, Wv ∼ = W (A6 ), and Wv ∩ L ∼ = W (A5 ). Lemma 17.4. Assume (17A), and let notation be as in Lemma 17.2. Use standard notation for Lv . Then there exists W0 ≤ NG (B) ∩ Lv , Lv with the following properties: (a) W0 CG (B)/CG (B) = W and W0 ∩ CG (B) = 1; (b) W0 ∩ Lv permutes naturally a system {Xα | α ∈ Σv } of root groups of Lv  ; (c) W0 ∩ Lv = nα (1) | α ∈ Σv . Proof. The proof is similar to that of Lemma 16.6, but with Lv now in the roles of both K and L1 there. By the paragraph before the lemma, (b) and (c) hold with Wv in place of W0 ∩ Lv . It suffices to produce W0 satisfying (a) and such that Wv = W0 ∩ Lv , or equivalently Wv ≤ W0 . Let Wv ,v = CWv (D) ∼ = W (A5 ). Then 



Wv ,v = O 2 (Wv ,v ) ≤ O 2 (CLv (D)) = L. In particular, Wv ,v ≤ Lv , so Wv ,v ≤ NLv (B) = CLv (B)Wv . Similarly, let Wv,v = CWv (v  ), so that Wv ,v ≤ CLv (B)Wv,v . So both Wv ,v and Wv,v are complements to CLv (B) in CLv (B)Wv,v , and their images in W , i.e., modulo CG (B), are both WL . By [III17 , 2.20], CLv (B) has a WL -invariant filtration for which every quotient M is abelian and satisfies H 1 (WL , M ) = 0. Hence, Wv ,v and Wv,v are CLv (B)conjugate. Replacing Wv by a CLv (B)-conjugate, therefore, we may assume that ∼ W (A5 ). Wv ∩ Wv = Wv ,v = Wv,v =

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We may then choose involutions t1 , . . . , t7 , mapping on w1 , . . . , w7 ∈ W , respectively, such that t1 , . . . , t6 form a fundamental set of generators for Wv ; t2 , . . . , t7 for Wv ; and t2 , . . . , t6 for Wv ,v . We set W0 = t1 , . . . , t7 , so that W0 maps onto W , and just as in the proof of Lemma 16.6, complete the proof by verifying that [t1 , t7 ] = 1. From the action on B, we see that [w1 , w7 ] = 1. Let g = [t1 , t7 ] = (t1 t7 )2 ∈ CG (B). Then t1 and t7 invert g ∈ CG (B). Suppose first that g has even order and let z be the involution in g . Then t1 ∼ t1 z and t7 ∼ t7 z in t1 , t7 . As g ∈ CG (B), z ∈ CG (B) ≤ CCG (v ) (B). Now Lv  ∼ = D6 (q), so CAut0 (Lv ) (B ∩ Lv ) has odd order. Moreover, C(v  , Lv ) has odd order by Lemma 17.3. Hence the image of z in Aut(Lv ) lies outside Aut0 (Lv ), so z induces a nontrivial field or graph-field automorphism on Lv . A similar argument proves that z induces a nontrivial field or graph-field automorphism on Lv and K. Consider C := CG (z) ≥ B. Now, Lp (CC (x)) ∼ = E6± (q 1/2 ), Lp (CC (v)) ∼ = ± 1/2 ± ± 1/2  1/2 ∼  D (q ), and L (C (D)) A (q ). Clearly the only A6 (q ), Lp (CC (v )) ∼ = 6 = 5 p C possibility, using Lp -balance, is that C has a p-component I with I/Op (I) ∼ = E7 (q 1/2 ) and B ≤ I. Set C = C/Op (C). As B is a maximal elementary abelian  p-subgroup of G, I = F ∗ (C). Hence, O p (C (∞) ) ≤ I. As [z, t1 ] = 1, z is Inndiag(Lv )-conjugate to zt1 , by [IA , 4.9.1]. Hence they are Lv -conjugate and so t1 ∼G z, whence CG (t1 ) ∼ = C. Now, t1 lies in a root SL2 (q) subgroup of Lv , so CLv (t1 ) contains a p-component H such that H/Op (H) ∼ = D4 (q) (see [III11 , 6.19]). Thus, H/Op (H) is involved in I. As I ∼ = E7 (q 1/2 ), this contradicts Lagrange’s theorem as |H/Op (H)| is divisible by (q 4 − 1)2 , but |I| is not, by Zsigmondy’s theorem. We conclude that g has odd order. Suppose that q = 2. Then p = 3 and by [III17 , 2.33b], CLv (B) ≤ B, so g ∈ [t1 , CG (B)] ≤ CLv (B) ≤ B. Hence if g = 1, then [t1 , B] = g = [t7 , B]. As in the E8 (q) case this implies w1 = w7 , a contradiction. Thus, g = 1 in this case, as desired. Now assume that q > 2. Choose b4 ∈ B such that b4 = [B, t4 ]. As the action of W on B is the same in G as in E7 (q), we calculate that b4 is W -conjugate to  v  . Let L4 = E(CG (b4 )) ∼ = K. We see in Lv that t1 ∈ O p (E(CLv (b4 ))), and in  Lv that t7 ∈ O p (E(CLv (b4 ))). By Lp -balance, t1 , t7 ≤ L4 . Thus g = [g , t1 ] = [g , t7 ] ∈ [CL4 (B), t1 ] ∩ [CL4 (B), t7 ]. But the intersection on the right is trivial by [III17 , 2.33c], so g = 1. The proof is complete.  Again, for any α ∈ Σv , let Wα be the preimage in W0 of the W -stabilizer of α. As with (16D), we have   (17C) Wα ∩ Lv = nβ (1)  β ∈ Σv ∩ α⊥ .  Lemma 17.5. Assume (17A). If α ∈ Σv , then Wα = (Wα ∩ Lv )Wα . Proof. The proof of Lemma 16.7 may be repeated, with W (D6 ) and Lv in  place of W (E7 ) and L1 , respectively. Lemma 17.6. Assume (17A). For any α ∈ Σv , let Lα = Xα , X−α . Then [Wα , Xα ] = [Wα , Lα ] = 1. Proof. As in the proof of Lemma 16.8, Lemma 17.5 and (17C) reduce the proof to showing that Wα normalizes Lα .

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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Now, Bα := CB (wα ) is a Wα -invariant hyperplane of B containing v  . Hence   CG (Bα ) can be calculated in CG (v  ). We have O 2 (CLv (Bα )) = O 2 (CLv (Bα ∩ Lv )) = Lα , the last equation with the help of [IA , 4.8.2]. As C(v  , Lv ) has odd order by Lemma 17.3, we see that if q > 2, then Lα = E(CG (Bα )), so Lα is Wα  invariant, as desired. So suppose that q = 2. If Lα = O 2 (CG (Bα )), then again Lα is  Wα -invariant, as desired. So assume that Lα < O 2 (CG (Bα )). We have Lα ∼ = L2 (2)  and Lα = O 2 (CLv (Bα ))  CG (Bα ). Therefore CG (Bα ) = Lα × Y , where |Y |2 = 2 and so Y contains an involution g inducing a nontrivial graph automorphism on Lv . But then as B = O3 (Lα ) × Bα , g centralizes B. This contradicts [III17 , 2.31] and completes the proof of the lemma.  ∼ E7 (q). Lemma 17.7. Assume (17A). Then G0 =   0 Proof. Let G2 = LW . By [III17 , 1.3], any pair of roots in Σ∗ is W v conjugate to a pair in Σv , so E7 (q) and Lv are compatible over W . By Lemmas 17.4b and 17.6, hypotheses (a) and (b) of the Gilman-Griess theorem [III13 , 2.2] hold, and so G2 ∼ = E7 (q). Now W0 ≤ Lv , Lv by construction, so W0 ≤ G0 and hence G2 ≤ G0 . Let u be any element of De ; to complete the proof we need only show that Lu ≤ G2 . Of   course L ≤ Lv ≤ G2 . By Lemma 17.2a, Lu ∼ = A6q (q), D6 (q), or E6q (q). Accordingly WLu ∼ = W (A6 ), W (D6 ) or W (E6 ), and CW0 (u) has a subgroup isomorphic to WLu . As CW0 (u) has nontrivial irreducible socle on B/ u , it does not normalize D; C G2 (u). Now therefore, since D = CB (L), CW0 (u) does not normalize L. Hence  E(CL (u))  ∼ G2 as G2 ∼ (q), we see from [III , Table 13.1] that I := L E = A6q (q), = 7 11 q D6 (q), or E6 (q). As I ≤ Lu , it follows that I = Lu , and the lemma follows.  This completes the proof of Proposition 17.1. Next, consider (17D)

 K∼ = D5q (q), L ∼ = D4 (q), and W ∼ = W (E6 ).

 Proposition 17.8. Assume (17D). Then G0 ∼ = E6q (q).

We assume false and proceed as usual in a sequence of lemmas. Lemma 17.9. For all u ∈ De , Lu ∼ = K. Moreover, |De | = 3. Proof. For any u ∈ De , Lu is a level vertical pumpup of L ∼ = D4 (q), and by  p (Lu /Op (L)) ≥ 5, so that Lu /Op (Lu ) ∈ G2p . Therefore F(Lu ) ≤ [III17 , 10.15], m  F(K) by [III14 , Lemma 1.2]. It follows directly that Lu ∼ = D5q (q), the sign forced by our rank condition. Since CB (WL ) = D and CB (WLu ) = u , |De | equals the number of subgroups W5 ≤ W isomorphic to W (D5 ) containing a given subgroup W4 ∼ = W (D4 ). There is just one conjugacy class of subgroups isomorphic to W5 , and W5 contains 5 conjugate subgroups isomorphic to W4 and generated by reflections, so |De | = 5|W : NW (W5 )|/|W : NW (W4 )| = 5 · 33 /5 · 32 = 3, as asserted.  Lemma 17.10. There exists a 3-element t normalizing D, inducing a triality automorphism (of order 3) on L, and cycling De .

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q 17. G0 ∼ = E6 (q) AND E7 (q)

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∼ W (E6 ). Let Z = Z(WL ) ∼ ∼ W (D4 ), with WL ≤ W = Proof. We have WL = = Z2 and C = CW (Z) ∼ = W (F4 ). Then C acts faithfully on [B, Z] = B ∩ L and acts on CB (Z) = D. It follows that some 3-element t ∈ NG (B) ∩ NG (D) maps to a generator of O3 (Aut0 (L)/ Inn(L)), and the image of t in W has order 3. Since AutAut(K) (L) contains no triality, however, t does not normalize x and so the orbit of x under t has order 3. This proves the lemma.  We remark that any element of order 3 of the form ht with h ∈ LCG (LD) satisfies the conclusions of Lemma 17.10 just as well as t. We shall make a specific choice of such an element ht later. For now, however, let t be any element satisfying Lemma 17.10. Set v = xt . Thus v ∈ De . Lemma 17.11. C(x, K) and C(v, Lv ) have odd order. Proof. It suffices to consider C(x, K), since v = xt and thus Lv = K t . Let T ∈ Syl2 (C(x, K)). Then [T, B] = 1 so for all u ∈ D − x , T ≤ CCG (u) (LD). 17 , 6.24c] and Lemma 17.9, T ≤ C(u, Lu ). Therefore T centralizes Hence by [III Lu | u ∈ D# = G0 , so if T = 1, then G0 is a K-group. By [III17 , 3.21], G0 ∼ =  E6q (q), contrary to our assumption that the proposition fails. Thus, T = 1, proving the lemma.  The case q = 1 proceeds smoothly via the Gilman-Griess theorem [III13 , 2.2], much like the cases leading to G0 ∼ = E8 (q) or E7 (q). Thus we next prove: Lemma 17.12. We have q = −1, i.e., p divides q + 1. Proof. Suppose that q = 1; we show that G0 ∼ = E6 (q), a contradiction. First, we construct a complement W0 = t1 , . . . , t6 that splits W over CG (B), where the ti are involutions inducing reflections on B, and Wx := t1 , . . . , t5 and Wv := t2 , . . . , t6 are complements splitting WK and WLv over CG (B), respectively. As q = 1, the complements Wx and Wv exist by [III17 , 2.2] and we may take Wx = nα (1) | α ∈ Σ , where Σ is the root system of K ∼ = D5 (q), for which we use standard notation. The construction of W0 then proceeds as in Lemma 16.6, with the following modifications. To ensure that these two complements overlap in a copy of WL (after a suitable conjugation of one of them), we need to know that CLv (B) has odd order and has a WL -invariant filtration whose quotients M all are abelian and satisfy (17E)

H 1 (WL , M ) = 0.

Here WL ∼ = W (D4 ). These facts are provided by [III17 , 2.2, 2.20] and Lemma 17.11. We have involutory generators t1 , . . . , t5 for Wx ; t2 , . . . , t6 for Wv ; and t2 , . . . , t5 for Wx,v = CWx (D). It remains to check that [t1 , t6 ] = 1, which equation clearly holds modulo CG (B). Set g = [t1 , t6 ]. If g contains an involution z, then setting C = CG (z), we have 1 B ≤ C and Lp (CC (u)) ∼ = D5± (q 2 ) for each u ∈ De . Then C has a p-component 1 I whose image in C = C/Op (C) is F ∗ (C) ∼ = E6± (q 2 )a , by [III17 , 3.21]. Moreover, from the structure of CK (t1 ) we see that CI (x) has a p-component H ∼ = A3 (q). This contradicts the structure of I, so g has odd order. As in the proof of Lemma 16.6, we obtain the contradiction w1 = w6 if q = 2, while if q > 2 we write [B, t4 ] = b4 and K4 = E(CG (b4 )) ∼ = K and find that g ∈ [CK4 (B), t1 ]∩[CK4 (B), t6 ] = 1, the last

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by [III17 , 2.33c]. This completes the construction of the complement W0 splitting W over CG (B) and containing Wx . Since K is untwisted, the analogues of parts (b) and (c) of Lemma 16.6 follow from the Chevalley relations, with K now playing the role of L∗1 there. The proofs of Lemmas 16.7 and 16.8 easily adapt to the present situation, and we conclude as  in Lemma 16.9, with the help of [III17 , 1.3], that G2 := K W0 ∼ = E6 (q). Clearly, W0 ≤ K, Lv ≤ G0 so G2 ≤ G0 . Conversely, every Lu , u ∈ D# , with Lu > L is W0 -conjugate to K, as mentioned in the first paragraph, so G0 ≤ G2 . Hence G0 = G2 ∼ = E6 (q), contradicting our assumption that Proposition 17.8 fails. The lemma is proved.  Lemma 17.13. CG (B) has odd order. Proof. By Lemma 17.11, C(x, K) has odd order. Now K ∼ = D5− (q) and p divides q + 1, so mp (K) = 5, and by [III17 , 2.2], CK (B ∩ K) has odd order. Thus any involution t ∈ CAut(K) (B∩K) must induce a graph automorphism on K. Hence by [IA , 4.9.2f], if t exists, then either mp (B4 (q)) = 5 or m2,p (K) = 5. But both of these equations are false, by [IA , 4.10.3a] and the Borel-Tits theorem. Hence t cannot exist, and the lemma follows.  For the case q = −1 we use the following decorated F4 diagram as a guide. ◦α0 α0

◦

◦

α1

α2

◦

α3

◦

α4

◦

Here α0 , α1 , . . . , α4 give an extended F4 diagram. We regard α0 , α1 , α2 , and α3 as giving the twisted diagram for K ∼ = 2 D5 (q), and adding α0 gives the extended diagram for K. Then in K, the root groups corresponding to ±α0 , ±α1 , ±α2 , and ±α0 generate a D4 (q) subgroup, which without loss we can take to be L. With standard notation for K, we set Li = Xαi , X−αi ∼ = L2 (q) for i ∈ {0, 1, 2, 0 }. Then for a certain 4-dimensional +-type subspace V+ of the natural K-module V , the stabilizer M of V+ in K has a normal subgroup M0 = L0 × L0 × Ix , with Ix ∼ = Ω(V+⊥ ) ∼ = U4 (q) and D ∩ Ix = 1. We assume, as we may, that V+ has been chosen to be B-stable. Note also that M0 ∩L = L0 ×L0 ×L2 ×L5 with L5 ∼ = L2 (q).  Let Bi = B ∩ Li . Then B0 is supported on V+ and so O 2 (CK (B0 )) = L0 × Ix =  O 2 (CK (L0 )). Notice that Ix is CG (B)-invariant. Moreover, we can now specify the desired action of t on L. It should cycle L0 to L2 to L5 to L0 , with L0 , L1 ≤ CL (t) ∼ = G2 (q). This can be achieved by modifying t by a suitable element of L. See Lemma 17.10 and the remark following it. Lemma 17.14. Let J be the pumpup of Ix in CG (B0 ). Then J ∼ = U6 (q) or SU6 (q), and [J, L0 ] = 1.  i Proof. For any u ∈ De , u = xt for some i, by Lemma 17.9. Then Lu ∼ = K and since [t, L0 ] = 1 and t cycles L0 , L2 , and L5 , we similarly calculate that   O 2 (CLu (B0 )) = O 2 (CLu (L0 )) = Ju × Iu with Iu ∼ = Ix ∼ = U4 (q) and Ju ∼ = L2 (q). Here Ju = Li for some i ∈ {0 , 2, 5}, and L0 L2 L5 ≤ Ju Iu . Thus Iu ∩ Ix contains Li for either i = 2 or i = 5, so J is also the pumpup of Iu in CG (B0 ), with Lp (CJ (u)) having the component Iu .

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q 17. G0 ∼ = E6 (q) AND E7 (q)

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For any u ∈ E1 (D) − De , on the other hand, L0 L2 L5  CJ (u). If J ∼ = U6 (q) or SU6 (q), then as L0 centralizes Iu | u ∈ De , [L0 , J] = 1 by [III17 , 9.1]. Suppose next that J/Z(J) ∼  U6 (q). By [III17 , 3.24], q = 2, p = 3, and J ∼ = Co2 = or P Sp4 (33 ); and accordingly, CAut(J) (Ix ) ∼ = Σ3 or Z3 . Now L0 × x ∼ = Σ3 × Z3 maps to CAut(J) (Ix ), with the image of x nontrivial. It follows that [L0 , J] = 1.

(17F)

Now B0 B0 contains an element of order 3 which has minimal support on V and which therefore is K-conjugate to some v ∈ D∩K. Hence by L3 -balance and (17F), Cv := CG (v) has a 3-component H such that in C v = Cv /O3 (Cv ), H is isomorphic to J or a pumpup of J. In particular, H ∼  Lv so H ≤ C(v, Lv ). We obtain a = contradiction from the 3-rank m := m3 (C(v, Lv ) ∩ CG (x)). On the one hand, m ≥ m3 (CH,x (x)) ≥ 3 by [III17 , 16.7]. On the other hand, m ≤ m3 (CG (L, x )) = 2, contradiction, the last because m3 (C(x, K)) = 1. Finally, by [IA , 6.1.4], we must rule out the possibility that J/Z(J) ∼ = U6 (2) 2 Z or E . Then p = 3. Now B ≤ C (B ) so B normalizes with Z := O2 (Z(J)) ∼ = 2 G 0 2 J. But CG (B) has odd order by Lemma 17.13, so Z = [B, Z] ∼ = E22 . Indeed B = B0 × B0 × (B ∩ Ix ) × x , with B0 = [B0 , L0 ] and L0 normalizing J. As Out(J) is abelian, B ∩ K = B0 B0 (B ∩ Ix ) maps into Inn(J) and so centralizes Z. Therefore Z = [Z, x]. Next, as |De | = 3, D∩K ∈ De , whence L  CG (D∩K). Hence Z normalizes L and maps into CAut(L) (B ∩ L), which has odd order by [III17 , 2.2]. h h Consequently [Z, L] = 1. Similarly, for any h ∈ NK (B ∩ K), L  CG ((D ∩ K) ) h NK (B∩K) = K. Similarly, CG (Z) and we get [Z, L ] = 1. Hence C  L G (Z) contains 2

2

contains K t and K t , so G0 = K, K t , K t ≤ CG (Z) < G. Using [III17 , 3.21], we get G0 ∼  = E6− (q), a contradiction, and the lemma is proved.

In terms of the diagram above, Ix is generated by the subgroups L2 = X±α2 and L3 = X±α3 of K. Here Xα3 ∼ = Eq2 is a short root group and L3 ∼ = L2 (q 2 ). We next define L4 = Lt3 and prove Lemma 17.15. The following conditions hold: (a) [L0 , L3 ] = [L0 , L3 ] = [L1 , L3 ] = 1; (b) [L1 , L4 ] = [L2 , L4 ] = 1; and (c) L3 , L4 is a D t -invariant homomorphic image of SL3 (q 2 ) in which {L3 , L4 } is a standard CT-pair. Proof. Part (a) holds in K, as is immediate from the diagram. Conjugating by t yields (b). Let L01 = L0 , L1 ∼ = SL3 (q), so that L01 ≤ L ≤ K. By [III17 , 6.26],  L3 = O 2 (CK (L01 )) = E(CK (L01 )) = Lp (CK (L01 )) ∼ = L2 (q 2 ),

while CL (L01 ) is solvable. By Lp -balance, Lp (CM01 (x)) ∼ = L2 (q 2 ), where we have put M01 = Lp (CG (L01 )). Similarly, CM01 (D) is p-solvable. For each u ∈ D# , set M01u = Lp (CM01 (u)). Now M01 , like L01 , is t-invariant. Hence M01u ∼ = L2 (q 2 ) for 2 each u ∈ De , and in particular, L4 = M01v ∼ = L2 (q ).

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Let u ∈ De . As B0 ≤ L0 ≤ L01 , M01u ≤ CCG (u) (B0 )(∞) ≤ Ju Iu as in the proof of Lemma 17.14. Since Ju ∼ = L2 (q) but M01u ∼ = L2 (q 2 ), M01u ≤ Iu ≤ J. Let J01 = Lp (M01 ∩ J) ≥ M01u | u ∈ De , a t-invariant group. Then J01 /Op (J01 ) is a pumpup of M01x ∼ = L2 (q 2 ), J01 ≤ J ∼ = (S)U6 (q), D acts faithfully on J01 , and NG (D)∩NG (J01 ) contains a 3-element cycling De . By [III17 , 3.25], J01 /Y ∼ = L3 (q 2 ) or SL3 (q 2 ), where Y = [J01 , Op (J01 )] is a 2-group; also, if Y = 1, then L3 and L4 form a standard CT-pair in J01 = L3 , L4 . Thus, if Y = 1, then (c) holds. On the other hand, if Y = 1, we argue to a contradiction. In that case, J01 ≤ J2 for some  parabolic subgroup J2 of J of type A2 (q 2 ). By [III17 , 6.39a], O 2 (J2 ) ≤ J01 . Now if  we put X = [M01x , CY (x)], then X = [X, M01x ] is a 2-subgroup of O 2 (CK (L01 )) ∼ = M01x . Thus X ≤ O2 (M01x ) = 1. On the other hand, as O2 (J2 ) ≤ J01 , this implies that [M01x , CO2 (J2 ) (x)] = 1, which contradicts [III17 , 6.39b]. Thus Y = 1 and we have proved: J01 ∼ = L3 (q 2 ) or SL3 (q 2 ). As J01 is D t -invariant, the lemma follows.



At last we reach our goal. Lemma 17.16. G0 ∼ = 2 E6 (q). Proof. In G2 , {L1 , L2 , L3 , L4 } is a weak CT-system of type 2 E6 (q), so by [III13 , 1.4] G2 := L1 , L2 , L3 , L4 ∼ = 2 E6 (q),  and we show that G0 = G2 . Since t permutes De transitively, G0 = K t . As Li ≤ K for i = 1, 2, 3, and L4 = Lt3 , G2 ≤ G0 . On the other hand, Lp (CG2 (L2 )) ∼ = U6 (q) or SU6 (q). Since L2 and L0 are K-conjugate, Lp (CG (L2 )) ∼ = Lp (CG (L0 )) ∼ = U6 (q) or SU6 (q). Therefore Lp (CG (L2 )) ≤ G2 , and in particular, L0 ≤ G2 . Thus K = Li | 0 ≤ i ≤ 3 ≤ G2 . Now G2 = L0 , L1 , L2 , L3 , L4 ≤ K, L3 , L4 = L, L3, L4 ≤ G2 , which shows that G2 is t-invariant, by Lemma 17.15c. Hence G0 = K t ≤ G2 , completing the proof.  This completes the proof of Proposition 17.8. 18. The Remaining Cases for G0 In this section we consider the remaining cases of Proposition 3.4, viz., cases (d), (e), (f), and (i). We first consider case (e):   (18A) K ∼ = SL3q (q), and WK ≤ W ∗  W with W ∗ ≡B W (D4 ). = L4q (q), L ∼ Proposition 18.1. If (18A) holds, then G0 ∼ = D4 (q) and AutG0 (B) = W ∗ . Since K ∈ Gp , p divides q 2 − 1, and U4 (2) ∈ G3 , we have q > 2. As usual we set De = {u ∈ E1 (D) | Lu > L}. Lemma 18.2. For any u ∈ De , Lu ∼ = K. Moreover, |De | = 3. Proof. Since u ∈ De , we know that Lu ∈ Chev(2) is a vertical level  pumpup of L ∼ = SL3q (q), and either mp (Lu ) = 2 or F(Lu ) ≤ F(K) = (q 9 , A) by [III14 , Lemma 1.2]. As q > 2, the only possibilities other than Lu ∼ = K are Lu ∼ = G2 (q) or 3D4 (q) with p = 3.

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18. THE REMAINING CASES FOR G0

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Suppose that one of these exceptional cases holds. Note that D = Z(L) x , so m3 (CG (LD)) = m3 (CCG (x) (L)) = 2 as m3 (C(x, K)) = 1. Similarly as O3 (Z(Lu )) = 1, u ∈ Z(L), so D = Z(L) u and 2 = m3 (CG (LD)) = m3 (CCG (u) (L)), whence m3 (C(u, Lu )) = 1. But m3 (Inn(Lu )) = 2. Hence, m3 (B) = 4 and some element f ∈ B induces a non-inner automorphism on Lu . If Lu ∼ = G2 (q), then f induces a field automorphism on Lu and hence on L. But B = (B ∩ K) × x maps into Inndiag(L), contradiction. Hence, Lu ∼ = 3D4 (q) and f induces a graph automorphism on Lu , contradicting Lemma 3.2. Thus, Lu ∼ = K. Note also that Lu is generated by the conjugates of L under WLu (and L itself is CG (B)-invariant). So if u, v ∈ D# and WLu = WLv , then Lu = Lv . It follows that |De | is the number of subgroups W0 ≤ W ∗ such that WL ≤  W0 ∼ = Σ4 . This number is easily calculated to be 3, and the lemma follows. By Lemma 18.2, we have symmetry between x and u, for any u ∈ De . We now complete the proof of Proposition 18.1. Let M be a B-invariant SL2 (q)subgroup of L, and let z ∈ WL ≤ W ∗ be the reflection generating AutM (B). Thus [B, z] = M ∩ B. For each u ∈ De , Mu := E(CLu (M )) ∼ = M and we let zu ∈ AutG (B) be the reflection generating AutMu (B). Then zu is NLu (B)conjugate to z, so zu ∈ W ∗ . Moreover, L, Lzu = Lu , so the three involutions zu (as u varies over De ) are distinct, and distinct from z. Now, [z, zu ] = 1 for all u ∈ De . But CW ∗ (z) ∼ = E24 and so CW ∗ (z) is the direct product of z and the three zu ’s. Hence, z and the three zu ’s are the only reflections in W ∗ commuting with z. Notice also that if we set Bu = CB (zu ), then u ∈ Bu and consequently  Mu = O p (E(CG (Bu ))). By construction, [M, Mu ] = 1 for all u ∈ De . A similar argument applies with zx in place of z and Mx in place of M . We conclude that [Mx , Mu ] = 1 for all u ∈ De − {x }. We fix such a u , and let M0 be a Binvariant SL2 (q) subgroup of L such that {M, M0 } is a weak CTP-system for L. Then we have that {M, M0 , Mx , Mu } is a weak CTP-system of type D4 (q). Hence by [III13 , 1.4, 1.14], as q > 2, G2 := M, M0 , Mx , Mu ∼ = D4 (q). Clearly K = L, Mx and Lu = L, Mu , so G2 = K, Lu ≤ G0 . By [III17 , 3.5], there are exactly three subgroups u ∈ E1 (D) such that L is not a component of CG2 (u ). These must then be the three elements of De , and so G2 = G0 . Finally, let z0 be the reflection generating WM0 . Then z and z0 are conjugate in z, z0 , and as the Dynkin diagram of D4 is connected, we see that W (D4 ) ∼ = WG0 = z, z0 , zx , zu ≤ W ∗ . As W ∗ ∼ = W (D4 ) by hypothesis, the proof of Proposition 18.1 is complete. The next remaining case of Proposition 3.4 is (f):   p = 3, K/Z(K) ∼ = L4q (q), and W = W ∗ × Z with W ∗ ≡B = L6q (q), L ∼ (18B) W (E6 ), |Z| ≤ 2.  Proposition 18.3. Assume (18B). Then G0 ∼ = E6q (q).

Notice that q > 2; otherwise p = 3 and K/Z(K) ∼ = U6 (2) ∈ C3 , whereas by our main setup, K/Z(K) ∈ Gp .

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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Lemma 18.4. The following conditions hold: ∼ E35 and B is a natural [W, W ] ∼ (a) B = (B ∩ K) x = = Ω5 (3)-module; (b) There exists a reflection z ∈ W ∗ such that [B, z] = x ; (c) In the action of W ∗ on B, W ∗ has a unique conjugacy class of reflections; and  (d) K ∼ = L6q (q). Proof. Since L ∼ = L4q (q) and (x, K, y, L) is not ignorable (by definition of  acceptable subterminal pair), no b ∈ B # satisfies E(CK (b)) ∼ = SL5q (q). It follows that m3 (AutB (K)) = 4, whence B = (B ∩ K) x has 3-rank 5. Then (a) holds by [III17 , 1.14]. We regard B as a F3 -vector space with a [W, W ]-invariant symmetric bilinear form. Now W ∗ = [W, W ] τ for some reflection τ . Thus [W, W ] −τ ≤ SO(B), so equality holds by orders. If ρ ∈ W ∗ is any reflection, then as −1 ∈ W ∗ , −ρ ∈ SO(B) − [W, W ] = SO(B) − Ω(B). Computing the spinorial norm of −ρ we see that on the axis Bρ of ρ, the determinant of the bilinear form is −1, so Bρ is a nonsingular hyperplane of − type. Hence by Witt’s lemma, Bρ is unique up to ∼ ∼ [W, W ]-conjugacy, which proves (c). Moreover, Ω(Bρ ) ∼ = Ω− 4 (3) = A6 = [WK , WK ]. ∗ Therefore W contains a reflection z whose axis is [B, [WK , WK ]]. The center of z must then be CB ([WK , WK ]) = x , proving (b).  Finally, if (d) is false, then by [IA , 6.1.4], and as q > 2, K ∼ = SL6q (q). Then B is isomorphic to the trace-0 submodule of the natural permutation module for WK . As such, it is an indecomposable [WK , WK ]-module. But this is impossible  as [WK , WK ] centralizes z. The proof is complete. 

Since (y, L) is an (x, K)-subterminal pair with L ∼ = L4 (q), we must have (18C)

q ≡ q

(mod 9).

Lemma 18.5. The following conditions hold: (a) (b) (c)

There exists a unique u ∈ E1 (D) − {x } such that Lu /Z(Lu ) ∼ = L6q (q); ∗ u ∈ x W ; and  −1 W∗ ∈ ux . ux 

Proof. As shown above, W ∗ = [W, W ] z where SO(B) = [W, W ] −z . By Lemma 18.4b, x is nonsingular (as an element of B) and CB (z) = x⊥ . We have NW ∗ (x ) = z × WK . Let u = D ∩ K. Then u is inverted in K by a reflection z  ∈ WK which interchanges the two nontrivial eigenspaces of u (as an element of  W∗ L6q (q)). As W ∗ has one class of reflections it follows that u ∈ x . Moreover,  −1 ∗ W since u is inverted in CW ∗ (x), ux ∈ ux . It remains to show that it  q is not the case that Lux /Z(Lux ) ∼ = L6 (q). But if this isomorphism held, then (ux, Lux ) ∈ J∗3 (G), so there would be symmetry between x and ux. In particular, ∗ by Lemma 18.4bc, ux ∈ x W . Thus D = x ⊥ u would be a nondegenerate plane in the space B, and the quadratic form q(v) = (v, v) would be constant on D# . There is, however, no such 2-dimensional orthogonal space. This contradiction completes the proof.  ∗ Let Σ = {H ≤ W ∗ | H ∼ = Σ3 and H ∩ z W = ∅}. Make Σ an undirected graph by joining H1 , H2 ∈ Σ if and only if [H1 , H2 ] = 1.

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Lemma 18.6. The graph Σ is a disjoint union of triangles. Moreover, W ∗ is transitive on the vertices of Σ as well as on the edges. Proof. For this proof it is easiest to use the isomorphism W ∗ ∼ = O6− (2), under W∗ which z maps to the unique class of transvections on the natural O6− (2)-module V . Hence a Σ3 -subgroup H ≤ W ∗ lies in Σ if and only if H centralizes a 4dimensional subspace VH ⊆ V of + type (and acts naturally on VH⊥ , of − type). As VH has exactly two 2-dimensional subspaces of − type, the first statement of the lemma follows. The second statement then is implied by Witt’s lemma.  Fix u as in Lemma 18.5. Choose a B-invariant subgroup M ≤ L such that  M ∼ = SL3q (q) and M is L-conjugate to a block-diagonal subgroup of L. Then WM ∈ Σ and we put Mx = E(CK (M )). As a result, M Mx is B-invariant and is Kconjugate to a block-diagonal subgroup of K. Moreover, [WM , WMx ] = 1. Working analogously in Lu we set Mu = E(CLu (M )) and deduce that [WM , WMu ] = 1. By Lemma 18.6, [WMu , WMx ] = 1. Making the same construction but starting with Mx instead of M , we see that M now plays the role of Mx , and some group Mu plays the role of Mu . We deduce that [WMu , WM ] = 1 and so by Lemma 18.6, WMu = WMu . 

Lemma 18.7. M = O 3 (E(CG (CB (WM )))). The same statement holds with Mu , Mx , or Mu in place of M . 

Proof. Let BM = CB (WM ) and M ∗ = O 3 (E(CG (BM ))). Then x ∈ BM   so M ∗ = O 3 (E(CCG (x) (BM ))) = E(CK (BM )). Now M ∼ = SL3q (q) and an E33 subgroup of Inndiag(M ) is a natural permutation module for WM ∼ = Σ3 . A direct  calculation in K then verifies that M ∗ = M , as required. As a result of Lemma 18.7, Mu = Mu . Therefore [Mx , Mu ] = 1. Now choose weak CTP-systems {L1 , L2 , L3 , L4 , L5 } and {L1 , L2 , L3 , L6 , L7 } for K and Lu , respectively, such that L1 , L2 , L3 = L, L1 , L2 = M , L4 , L5 = Mx , and L6 , L7 = Mu . It is then immediate that {Li |1 ≤ i ≤ 6} is a weak CTP-system of type E6 , and so by [III13 , 1.4, 1.14], G2 := Li | 1 ≤ i ≤ 6 ∼ = E6q (q), 

in fact, the adjoint version, in view of Lemma 18.4d. Moreover, since u ∈ x W and AutG2 (B) ≥ W ∗ , we have L7 ≤ G2 , and so

∗

G2 = K, Lu ≤ G0 . We can now quickly complete the proof of Proposition 18.3. Let v = ux or ux−1 . Then by Proposition 10.2, Lv ∼ = = D4 (q). By (18C) and [III17 , 3.9], E(CG2 (v)) ∼ D4 (q). Hence Lv = E(CG2 (v)) ≤ G2 . It follows directly that G0 ≤ G2 , so G0 = G2 . This completes the proof of Proposition 18.3. In the two remaining cases of Proposition 3.4, mp (B) = 4, W ≡B W (F4 ), B = (B ∩ L) × D, and WL ∼ = W (B2 ) ∼ = D8 . (18D) ∼ Moreover, for every u ∈ De , WLu = W (B3 ).

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

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These statements follow from Tables 15.1 and 15.2. The next lemma therefore applies in both cases. Lemma 18.8. Assume (18D). Then the following conditions hold: (a) |De | = 4; (b) For each u ∈ De , there is u = u such that De ∩ u W = {u , u }; (c) For each u ∈ De , WLu lies in a unique subgroup Wu∗ ∼ = W (C4 ) of W generated by reflections; moreover, the four commuting independent reflections in O2 (Wu∗ ), on which Wu∗ induces the permutation group Σ4 , consist of three such generators of O2 (WLu ) and a fourth reflection inverting u and centralizing WLu ; (d) For each u = v ∈ De , Wu∗ = Wv∗ ⇐⇒ v = u . Proof. We have Q := F ∗ (W ) ∼ = Q8 ∗ Q8 acting irreducibly on B, and W/Q ∼ = Σ3 × Σ3 . Write Z(Q) = z and Z(WL ) = zL . Since W/Q has abelian Sylow g = zL z. 2-subgroups, zL ∈ Q, but zL = z. Hence there is g ∈ I2 (Q) such that zL Then g interchanges D = CB (zL ) with its complement D∗ := [B, zL ] = B ∩ L. Let WD be the group generated by all reflections in W stabilizing D. Then WD normalizes L and D∗ , and it follows that WD = WL × WLg , the two factors having complementary supports on B. Note also that T := WD g ∈ Syl2 (W ), and since B = D × Dg , all reflections in T lie in WD , hence in WL ∪ WLg . Every reflection t ∈ W with v := [B, t] ≤ D then lies in WD and hence in WLg . Moreover, CW (t) ∼ = t × W (B3 ), acting naturally on B/ v , so v ∈ De . Conversely for any u ∈ De , W (Lu ) ∼ = W (B3 ) by (18D), so Z(W (Lu ))z contains a unique reflection tu , and tu has center u . Hence |De | is the number of reflections in WLg , or equivalently in WL ∼ = W (B2 ); this proves (a). Moreover, the four reflections in WL fall into two WL -conjugacy classes. Hence if (b) were false, then all reflections in T would be W -conjugate, by the previous paragraph. As T ∈ Syl2 (W ), all reflections in W would be conjugate. But W has two classes of reflections, contradiction. This proves (b). Since W has an automorphism interchanging the class of long root reflections with the class of short root reflections, it suffices to prove (c) for some u ∈ De . It is clear from the extended Dynkin diagram of F4 that W contains a subgroup W∗ ∼ = W (C4 ) containing four long root reflections t1 , . . . , t4 and all twelve short root reflections. Indeed |W : W ∗ | = 3 and W ∗ is the stabilizer in W of {t1 , . . . , t4 }. As |W ∗ |2 = |W |2 we may take W ∗ to contain T . Then since t1 , . . . , t4  W ∗ , t1 , . . . , t4 ≤ T , indeed by what we saw above, {t1 , . . . , t4 } = tW u ∩ T for some u ∈ De . Since CW (tu ) = tu × WLu ∼ = CW ∗ (tu ), WLu ≤ CW (tu ) ≤ W ∗ . Hence the three long root reflections in WLu are t1 , . . . , t4 , excepting tu . Now if WLu ≤ W1∗ ≤ W with W1∗ ∼ = W (C4 ), then W1∗ also contains Z(WLu )z and ∗ hence contains tu ; thus W1 = NW ({t1 , . . . , t4 }) = W ∗ , proving (c). Finally, as  tu W = tu W , (d) follows as well. We now consider case (d) of 3.4: (18E)

K/Z(K) ∼ = L6

−q

(q), L ∼ = L4

−q

(q), W ≡B W (F4 ).

We shall prove: − Proposition 18.9. Assume (18E). Then G0 ∼ = E6 q (q).

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Note that if q = 2, then p = 3, q = −1, and our target group is E6 (2). As remarked above, (18D) holds, and by Table 15.2, we have: Lemma 18.10. Assume (18E). Let u ∈ De . Then Lu /Z(Lu ) ∼ = D4− (q) or K/Z(K). In either case of Lemma 18.10, it is clear that WLu ∼ = W (C3 ). So Lemma 18.8 applies, and we name the four elements of De as follows: u1 = x , u2 = u1 ∈ W W W x , u3 ∈ x , and u4 = u3 ∈ u3 . Thus, Lu2 ∼ = K. Lemma 18.11. Assume (18E). Then D ∩ K = u2 and D ∩ Lu2 = u1 . Proof. We prove the first assertion; the second follows by a similar argument. Since L = Lp (CK (D)) = Lp (CK (D ∩ K)), D ∩ K lies in a (long) root SL2 (q)subgroup J of K such that J is B-invariant. There are NK (B)-conjugates J1 , J2 of J such that J, J1 , J2 = J ×J1 ×J2 and WNK (JJ1 J2 ) = WK ∼ = W (C3 ). But as −1 ∈ W ∼ W (F ), there is a reflection z ∈ W inverting x. Then by Lemma 18.8c, z × = 4 WNK (JJ1 J2 ) lies in a W (C4 )-subgroup of W inducing Σ4 on {z , WJ , WJ1 , WJ2 }. In particular, z and WJ are interchanged in W , as are x = [B, z] and D ∩ K. Hence, E(CG (D ∩ K)) contains a copy of K, so D ∩ K ∈ De . By Lemma 18.8b,  moreover, D ∩ K = x = u1 = u2 , completing the proof. − Let M ≤ L with M ∼ = SL3 q (q), M centralizing a (nondegenerate) 3-dimensional subspace U of the natural K-module, and normalizing a complement to U . Set H = CG (M ). As in the case of Proposition 18.3 above, set Mui = Lp (H ∩Lui ), i = 1, 2. As noted above, Lu2 ∼ = K. Thus Mu1 ∼ = Mu2 ∼ = M . Also Mui ∼Lui M , i = 1, 2. We shall show that

[Mu1 , Mu2 ] = 1,

(18F)

which will quickly lead to a proof of the proposition. Lemma 18.12. Assume (18E). Then the following conditions hold: (a) Mui = Lp (CH (ui )), i = 1, 2; (b) u3−i ∈ Mui , i = 1, 2; (c) Lp (CH (u)) has no component isomorphic to L2 (q 2 ) for any u ∈ D# ; (d) mp (CH (ui Lp (CH (ui )))) ≤ 1, i = 1, 2; and − − (e) H is embeddable in L6 q (q) or SL6 q (q). Proof. Since Lui = Lp (CG (ui )), i = 1, 2, (a) is immediate. Lemma 18.11 implies (b). Suppose by way of contradiction that u ∈ D# and Lp (CH (u)) has a component I∼ = L2 (q 2 ). Then u = ui for any 1 ≤ i ≤ 4, so u ∈ De , i.e., L is a component of CG (u). Since mp (CG (x)) = mp (B) it follows that I is B-invariant and then that WI is generated by a reflection zI . Again as −1 ∈ W ∼ = W (F4 ) we conclude that u is inverted by a reflection zu ∈ W , and zu is conjugate to a reflection in WL . But any reflection in WL is W -conjugate to a reflection inverting some element of De , and so u is W -conjugate to ui for some 1 ≤ i ≤ 4, which is absurd as CG (ui ) contains no component isomorphic to L. Thus (c) holds. For (d), we may assume i = 1. Then CH (x Mu1 ) ≤ CCG (x) (Mu1 M ) =: CM  C(x, K), and CM /C(x, K) is involved in the p -group −

CGL− q (q) (SL3 q (q) × SL−epsq (q)). 3 6

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Finally by Lemma 18.11, M contains a W -conjugate v of x. Hence H ≤   O p (CH (v)(∞) ) ∼ = O p (CH (x)(∞) ) = K. Thus (e) holds and the proof is complete.  Let H0 = Lp (H). By Lp -balance, the assertions of Lemma 18.12 hold with H0 in place of H. Now (18F) is an immediate consequence of these assertions and [III17 , 3.23]. Now, just as in the proof of Proposition 18.3, we choose weak CTPsystems {L1 , L2 } for M , {L1 , L2 , L3 } for L, {L1 , L2 , L3 , L4 , L5 } for K, and {L1 , L2 , L3 , L6 , L7 } for Lu2 , in such a way that Mu1 = L4 , L5 and Mu2 = L6 , L7 . Set G2 = Li | 1 ≤ i ≤ 6 . With (18F) we conclude from [III13 , 1.4, 1.14] that − G2 ∼ = E6 q (q) (either universal or adjoint). Moreover, G2 ≤ K, Lu2 ≤ G0 . By [III17 , 3.8], Lp (CG2 (u3 )) ∼ = D4− (q), which is not embeddable in L± 6 (q) by Lagrange’s theorem and Zsigmondy’s theorem, so Lu3 = Lp (CG2 (u3 )) ≤ G2 , and similarly Lu4 ≤ G2 . Hence G0 ≤ G2 , so equality holds. This completes the proof of Proposition 18.9. We finally consider case (i) of Proposition 3.4, assuming (18G)

K∼ = Sp6 (q), q > 2, L ∼ = Sp4 (q), and W ≡B W (F4 ).

Our goal, the final recognition result of this long chapter, is: Proposition 18.13. If (18G) holds, then G0 ∼ = F4 (q). We assume the proposition fails. Again Lemma 18.8 applies, and Table 15.2 implies: Lemma 18.14. Assume (18G). Let u ∈ De . Then Lu ∼ = Sp6 (q). For each u ∈ De , let Wu∗ be as in Lemma 18.8cd, and set   W∗ Iu = Lu u . Lemma 18.15. Assume (18G). For each u = u ∈ De such that Wu∗ = Iu = Lu , Lu ∼ = C4 (q). There are two such unordered pairs (u , u ), say  (u1 , u1 ) and (u2 , u2 ), and G0 = Iu1 , Iu2 . Moreover, Iu1 = Iu2 .

Wu∗ ,

Proof. Since WLu ≤ Wu∗ ∼ = W (C4 ), the setup (13A) holds, with the roles of K and W o there played by Lu and Wu∗ here. Then Proposition  13.1 implies that Wu∗ Wu∗  ∼ Iu = C4 (q). By Lemma 18.8d, u ∈ u so Lu ≤ Lu = Iu . Then by [III17 , 9.1], Iu = Lu , Lu . The final assertions are then immediate from Lemma  18.8 and the definition of G0 . Now set I = Ix ∼ = C4 (q). Let {J1 , J2 , J3 , J4 } be a weak CTP-system for I, reading the C4 Dynkin diagram from (short root) left to (long root) right, such that for each i = 1, . . . , 4, Ji is B-invariant and Bi := CB (Ji ) is a hyperplane of B (see [III17 , 4.15]). For any subset A ⊆ {1, 2, 3, 4}, let  JA = Ji | i ∈ A and BA = i∈A Bi ≤ CB (JA ). We write J234 , B234 , etc., instead of J{2,3,4} , B{2,3,4} , etc. We may assume that J234 = K, so that B234 = x .

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Lemma 18.16. There exists a B-invariant subgroup J5 ≤ G such that {J2 , J3 , J4 , J5 } is a weak CT-system or weak P-system of type F4 (q), according as q = +1 or −1. Proof. As x ∈ B23 and C(x, K) is p-solvable,    SL3q (q) ∼ = J23 = O p (E(CK (B23 ))) = O p (E(CG (B23 )))  NG (B23 ).

Conjugating in NI (B) we see that J12  NG (B12 ). Let B 12 = B ∩ J12 and W12 = W ∩ J12 ∼ = Σ3 , so that W12 acts indecomposably on B 12 ∼ = Ep2 . Then by [III17 , 1.33], the subgroup W 12 of W generated by all reflections in CW (W12 ) satisfies W 12 ∼ = Σ3 and CB (O3 (W 12 )) = B 12 . 12 Now, [B, W12 ] = B and CB (W12 ) = B12 . Hence B12 is W 12 -invariant. Indeed 12 W acts indecomposably on B12 ; otherwise, B12 ≤ CB (O3 (W 12 )) = B 12 , so B12 = B 12 ≤ Z(J12 ), which is absurd as |B12 | = p2 . Let N 12 be the preimage of W 12 in NG (B); then N 12 normalizes J12 . Set J0 = NG (J12 ) ∩ CG (B 12 ). Then BJ4 ≤ J0 and N 12 ≤ J0 . As J4 is a component of CG (B4 ), J4 ≤ Lp (J0 ) by Lp -balance. Let J ∗ be the normal closure of J4 in Lp (J0 ); thus, J4 is a component of CJ ∗ (B4 ). As Zp ∼ = B ∩ J4 ≤ B12 and N 12 acts ∗ ∼ 2 indecomposably on B12 = Ep , J is a nontrivial pumpup of J4 . Since CAut(J12 ) (B 12 ) is solvable, [J ∗ , J12 ] = 1. This implies that WJ ∗ ≤ W 12 by definition of W 12 . But as W 12 ≤ J0 , WJ ∗   W 12 ∼ = Σ3 . Since Z2 ∼ = WJ4 ≤ WJ ∗ , this forces WJ ∗ = W 12 . In particular, ∗ 2 |B ∩ J | ≥ p . Set B 1 = B ∩ J1 ; then J ∗ ≤ CG (J1 ) ≤ CG (B 1 ). Notice that in I, it is clear that 1 B is conjugate to a subgroup of D. Hence Lp (CG (B 1 )) has a unique p-component I1 of p-rank larger than 1, and I1 ∼ = Sp4 (q) or Sp6 (q). As WJ ∗ = W 12 , the only  ∗ ∼ possibility then is J ≤ I1 = Sp6 (q) with J ∗ /Z(J ∗ ) ∼ = L3q (q). Note also that J3 ≤ CG (J1 ) ≤ CG (B 1 ), and J3 ≤ Lp (CG (B 1 )) by Lp -balance. We next argue that there is J5 ≤ J ∗ such that the following hold: (1) J5 ∼ = L2 (q) and J5 is a B-invariant component of CJ ∗ (b) for some (18H) b ∈ B # ; and (2) [J5 , J3 ] = 1. This is obvious if [J ∗ , J3 ] = 1, so assume that [J ∗ , J3 ] = 1. Then by Lp -balance, J3 ≤ I1 . Now J ∗ is B-invariant and so by [III17 , 6.30], J5 := E(CJ ∗ (B ∩ J3 )) satisfies (18H). By (18H2), J5 = J4 . But J4 and J5 are L2 (q)-components of centralizers of subgroups of B in J ∗ , so they form a standard CT-pair or standard  P-pair in J ∗ , according as q = +1 or −1. The lemma follows. Unfortunately, at the moment, there is no Phan-type recognition theorem for F4 (4) and F4 (8). So we have to settle at this point for the following lemma, and we have further work to do to identify G0 . Lemma 18.17. We have q = 4 or q = 8, with p = 5 or 3, respectively. Proof. Suppose false. Since q > 2 by assumption, it follows that either q = 1 or q > 8. Set G∗0 = J2 , J3 , J4 , J5 ; thus by [III13 , 1.4, 1.14], G∗0 ∼ = F4 (q). We show that G∗0 = G0 , contrary to our assumption that Proposition 18.13 fails.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

272

As each Ji is B-invariant, G∗0 is B-invariant. We have K = J234 ≤ G∗0 . By [III17 , 3.7], there exist exactly four subgroups u ∈ E1 (D) such that L ≤ E(CG∗0 (u)) ∼ = Sp6 (q). For any such u, E(CG∗0 (u)) therefore lies in the subnormal closure Lu of L in CG (u). Therefore E(CG∗0 (u)) = Lu and u ∈ De . Since |De | = 4, G0 = Lu | u ∈ De ≤ G∗0 . In particular, G0 is a K-group. As K is a component of CG0 (x) and |De | = 4 it follows from [III17 , 3.15] that G0 = G∗0 . This completes the proof.  Now set r = q − 1, so that (18I)

r = 3 or 7, respectively. Recall that for any u ∈ De , Iu = Lu , Lu = Iu ∼ = Sp8 (q) and WIu = Wu∗ (see Lemma 18.15). Lemma 18.18. Let u ∈ De and v ∈ Ir (L) with E(CIu (v)) ∼ = Sp6 (q). Then E(CIu (v))   CG (v). Proof. Let Ev = E(CIu (v)) and C = CG (v). First, the proof of [III9 , 10.3a] shows that (18J)

Ev ≤ E(C).

For clarity we repeat the details. Set H = E(CIu (v D)) = E(CL (v)) ∼ = L2 (q). Thus H is a component of Lp (CC (D)), so H ≤ Lp (C) by Lp -balance. Set X = to show that [H, X] = 1, for then H ≤ E(C) Op (C). To prove (18J)it suffices and consequently Ev = H Ev ≤ E(C). It then suffices to fix d ∈ D# , set Xd =  [H, CX (d)], and show that Xd = 1. Since H = O p (H), Xd = [H, Xd ] by [IG , 4.3(i)]. Then as H ≤ L ≤ Ld   CG (d), Xd ≤ Ld . Also Xd ≤ X so Xd ≤ Op (C ∩ Ld ) = Op (CLd (v)). As H = [H, H] and Lu ∈ Chev(2) we use [III11 , 10.3] to get 

Xd = [H, Xd ] ≤ [O p (CLd (v)(∞) ), Op (CLd (v))] = 1. Thus (18J) holds.  Let d ∈ D# . Since Ld   CG (d), E(CLd (v))   CC (d). Let Hd = H E(CLd (v)) , a product of components of CC (d). If d ∈ De , this implies that Hd = H   CC (d). If d ∈ De , then L < Ld ∼ = Sp6 (q), with v ∈ L ∼ = = Sp4 (q), and so Hd = H or Hd ∼ Sp4 (q). Moreover, if d = u or u , then Hd ∼ = Sp6 (q). = Sp4 (q) as E(CIu (v)) ∼ Finally if Hd ∼ = Sp4 (q), then as d ∈ De , mp (C(d, Ld )) = 1 (otherwise, CG (x) would contain an Ep5 -subgroup inducing inner automorphisms on L, contradicting mp (C(x, K) = 1). Hence mp (CC (H d )) = 1 for all d ∈ De suchthat H d ∼ = Sp4 (q). E(C) Therefore by [III17 , 3.13], H E(C) ∼ Sp and these (q). As E(C (v)) ≤ H = 6 Iu two groups are isomorphic, they are equal, which is the assertion of the lemma.  Lemma 18.19. There exists E ∼ = Er4 such that E ≤ Iu for all u ∈ De . Moreover, if we set L∗ = Lp (CG (B ∩ L)), then L∗ ∼ = L, LL∗ = L × L∗ , and ∗ ∗ ∼ ∼ E = (E ∩ L) × (E ∩ L ) with E ∩ L = E ∩ L = Er2 . Proof. Let I ∗ = Iu1 ∩Iu2 = ∩u∈De Iu . Then I ∗ ≥ ∩u∈De Lu ≥ L. For every u ∈ De , L = E(CIu (D)) = Lp (CG (D)) and L ∼ = E(CIu (L)) = E(CIu (B ∩ L)) = Lp (CG (B ∩ L)) as B ∩ L is conjugate (even in Iu ) to D. Set L∗ = Lp (CG (B ∩ L)).  We conclude that L × L∗ ≤ I ∗ . As mr (L) = mr (L∗ ) = 2, the lemma follows. Lemma 18.20. G0 = G, CG (LL∗ ) = 1, and CG (Iu ) = 1 for every u ∈ De .

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Proof. If G0 < G, then G0 is a K-group, and by the structure of the neighborhood N(x, K, y, L) and [III17 , 3.15], G0 ∼ = F4 (q), contrary to our assumption that the proposition fails. Thus, G0 = G. Finally, since LL∗ ≤ Iu for all u ∈ De , it suffices to show that CG (LL∗ ) = 1. Let g ∈ CG (LL∗ ). Then [g, D] = 1 and so g normalizes Iui , i = 1, 2. But by [III17 , 6.29], (18K)

CAut(Iui ) (LL∗ ) = 1, i = 1, 2.

Therefore g centralizes Iu1 , Iu2 = G0 , so g = 1, completing the proof.



Lemma 18.21. We have q = 8, p = 3, r = 7 ∈ γ(G), and De = E1 (D). Proof. Recall that q > 2. As |De | = 4, it suffices to prove that p = 3 and r = 7 ∈ γ(G). Now, it follows from Lemma 18.18 that Lor (G) ∩ Gr = ∅, whence one of the following holds: (1) r ∈ γ(G); or (18L) (2) G possesses a strong r-uniqueness subgroup M . Suppose that (18L2) holds. Replacing M by a conjugate, we may assume that LL∗ ≤ M . In particular, B = (B ∩ L)(B ∩ L∗ ) ≤ M . Let u ∈ De . Then L < Lu ∼ = K. Now E ∩ L ∼ = Er2 lies in some F ∈ Er3 (Lu ). As M is a strong r-uniqueness subgroup of G, and L ≤ M , M contains NG (E ∩ L), then F , then ΓF,2 (Lu ), which equals Lu by [IA , 7.3.3]. As u was arbitrary, G0 ≤ M . But M < G so this contradicts Lemma 18.20. Thus, r ∈ γ(G). It remains to rule out the case p = 5, r = 3. In that case, (p, x, K, y, L) satisfies Lemma 2.2b. This contradicts our choice in (2B), and the lemma follows.  Lemma 18.22. Let u, v be as in Lemma 18.18. Then E(CG (v)) ∼ = Sp6 (8) and CG (v E(CG (v))) = v . Moreover, (v, E(CG (v))) ∈ J∗7 (G). Proof. As in Lemma 18.18, u ∈ De and CG (v) has a component Jv = E(CIu (v)) ∼ = Sp6 (8). Let g ∈ CG (v Jv ); we must prove that g ∈ v . By Lemma 18.19 we may assume that v ∈ E = (E ∩ L) × (E ∩ L∗ ). Given the structure of E(CIu (v)), it follows that v ∈ L ∪ L∗ . For specificity assume that v ∈ L. Then L∗ ≤ Jv so [g, D] ≤ [g, L∗ ] = 1. Hence g normalizes Lu for all u ∈ De , whence g normalizes Iu . Thus g maps into CAut(Iu ) (v Jv ), which is the image of v . As CG (Iu ) = 1 by Lemma 18.20, g ∈ v and the first sentence is proved. Finally, if (v, E(CG (v))) ∈ J∗7 (G), then (7, x, K, y, L) satisfies Lemma 2.2c, contrary to our choice in (2B). This completes the proof.  Lemma 18.23. CG (E) = E. Proof. Let u and v be as in Lemma 18.22, and let Ev = E(CG (v)). Then CG (E) normalizes Ev and maps into CAut(Ev ) (E ∩Ev ), which is the image of E ∩Ev . Hence, CG (E) ≤ CG (v Ev )(E ∩ Ev ) = v (E ∩ Ev ) = E.  Lemma 18.24. Let v ∈ E # ∩ L∗ with the subnormal closure Lv of L in CG (v) a nontrivial pumpup of L. Then Lv ∼ = Sp6 (8) and Lv ≤ Iu for some u ∈ De . Proof. By Lemma 18.18, there is v  ∈ E # ∩L∗ and u ∈ De such that Sp6 (8) ∼ = E(CG (v  )) ≤ Iu . If we can take v  = v, then we are done.

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274

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Otherwise v, v  = E ∩ L∗ and L is a component of CCG (v) (v  ), so L has a pumpup Lv in CG (v) by Lr -balance, and this pumpup is vertical or diagonal by hypothesis. Suppose it is a vertical pumpup. As L ∼ = Sp4 (8) and CG (E) = E, E ∩ L∗ contains its centralizer in CCG (v) (L). So the only possibilities are Lv := Lv /O7 (Lv ) ∼ = 7  Sp6 (8) and Sp4 (8 ), with v inducing a field automorphism on Lv in the latter case. Since there exists an involution in AutG (E) inverting E elementwise, the field automorphism case is not possible. Thus Lv ∼ = Sp6 (8). Since L  CLv (v  ), L is canonically embedded in Lv . There are subgroups Li ∼ = L2 (8) of L, i = 1, 2, such that [L1 , L2 ] = 1 and for any y ∈ Ip (Li ), L3−i = E(CL (y)) ≤ E(CLv (y)) ∼ = Sp4 (8). Let bi ∈ B # ∩ L centralize L3−i . Then CLv (bi ) has an Sp4 (8) 7-component Ni meeting L∗ and L nontrivially. So CG (bi ) has an Sp6 (8) component Mi containing Ni , b3−i , and L∗ . It follows that M1 , M2 ≤ Iu for some u ∈ De . Moreover, N1 , N2 ≤ M1 , M2 ≤ Iu and [N1 , N2 , v] = 1 and N1 , N2 covers Lv . So CIu (v) involves Sp6 (8), as desired. It remains to rule out the case in which Lv is a diagonal pumpup of L. In that case, let D ≤ D∗, a Sylow 3-subgroup of Lv . Then D∗ is abelian of 3-rank 14. But m3 (CG (x)) ≤ 5, a contradiction.  Now let N = NG (E) and N = N/E, and for each u ∈ De , set N u = N ∩ Iu ∼ = W (C4 ). We regard N as a subgroup of Aut(E). Also, let N 0 be the subgroup of N generated by all its reflections. Thus N u ≤ N 0 ≤ SL± (E). We aim to prove that N0 ∼ = W (F4 ). We first prove that N is a 7 -group: Lemma 18.25. E ∈ Syl7 (G). Proof. We assume that 7 divides |N |, and derive a contradiction. Now, N u is irreducible on E, and N u ≤ N 0 ≤ N , so N 0 and N are irreducible on E as well. In particular, O7 (N ) = 1. Suppose first that N contains a transvection t. Let E = {e ∈ E # | E(CG (e)) ∼ = K}. Then CE (t) = ∅. For if e ∈ CE (t), then t maps to a transvection in AutAut(E(CG (e))) (E ∩ E(CG (e))), which, however, is a {2, 3}-group. Choose any 1 = e ∈ CE∩L∗ (t); then by Lemma 18.24, the pumpup Le of L in CG (e) is trivial, so t normalizes E ∩ L∗ = CE (L) and E ∩ L. But E ∩ L = ∅ = E ∩ L∗ . Hence CE (t) = ∅, a contradiction. We have proved that N contains no transvections.  If J is a component of O 7 (E(N )) with |Z(J)| odd, then by [III17 , 16.5], J ∼ = L2 (7) or L2 (72 ). In either case as J does not embed in SL2 (7), the J-composition factors in E are of dimensions 3 and 1, or 4. Let C = CN (J); it follows that the only involution in C ∩ SL(E) is the involution z inverting all of E. On the other hand, N u ≤ N and N u ∼ = O 2 (W (D4 )) with = W (C4 ) has a subgroup N D = QH ∼ ∼ ∼ Q = [Q, H] = Q8 ∗ Q8 and H = Z3 . As Out(J) is a 2-group, N D ≤ J × C. Clearly Z(Q) = z ∈ C, so N D projects isomorphically onto a subgroup N 1 ≤ C. Then Q1 = O2 (N 1 ) = [Q1 , N 1 ] ≤ C ∩ SL(E). Hence by what we saw above, I2 (Q1 ) = {z}. But this is absurd since Q1 ∼ = Q8 ∗ Q8 . =Q∼

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Thus if O 7 (E(N)) = 1, then for any component J of O 7 (E(N )), Z(J) has even order. If m2 (J) > 1, then it follows from [III17 , 16.6] that J contains transvections, contradiction. Hence, m2 (J) = 1. If there is z ∈ I2 (J) with z not inverting every element of E, then J ∼ = SL2 (7) contains a transvection, contradiction. Hence, Z(J) inverts E elementwise. If J acts irreducibly on E, then CN (J) is cyclic and N /JCN (J) is abelian. Hence [[N u , N u ], [N u , N u ]] ≤ J, whence J has sectional 2-rank at least 4, a contradiction. So J acts reducibly on E. So J acts faithfully on a 2-dimensional subspace E1 ≤ E, whence J ∼ = SL2 (7). Let r 1 be a reflection not centralizing E1 . As CN (r1 ) ∼ = Z2 × W (C3 ) is a 7 -group, r 1 normalizes but does not centralize J. Therefore by [III11 , 6.3a], r 1 shears to Z(J), which is impossible as r1 Z(J) contains a non-reflection. Thus, E(N ) ≤ O7 (N ). As O7 (N ) = 1, F ∗ (N ) is a 7 -group. For some prime R ≤ r = 7, an element h ∈ N of order 7 acts nontrivially on some r-subgroup F ∗ (N ). We choose R minimal subject to this condition. Since R h ≤ SL4 (7), R is nonabelian. Hence Sylow r-subgroups of SL4 (7) are noncyclic, and by minimality, r > 2. Thus r = 3. But then |R| ≥ 36 > |SL4 (7)|3 , a final contradiction.  Lemma 18.26. F ∗ (N 0 ) = O2 (N 0 ) ∼ = Q8 ∗ Q8 . Proof. As in the previous proof, N u acts irreducibly on E, so N does as well. Let J be a component of N . We argue that J contains a Q8 ∗Q8 subgroup. Namely, Out(J) is 2-nilpotent. (Otherwise by [III11 , 2.1a], J ∼ = D4 (q) for some odd q, so q 12 divides |GL4 (7)|, which is absurd.) In N u ∼ = W (C4 ) is a subgroup QH with H∼ = Z3 and Q = [Q, H] ∼ = Q8 ∗ Q8 . It follows that Q ≤ JCN (J). On the other hand, by the pa q b -theorem, Lemma 18.25, and [IA , 4.10.3a], J has a nontrivial cyclic Sylow s-subgroup S for some s ∈ {5, 19}. We see that J  N , CAut(E) (S) is abelian, and no two involutions of CAut(E) (S) are conjugate in Aut(E). The same holds for CN (J), and it follows that Q = [Q, H] ≤ J, as claimed. In particular, as CQ (E) = 1, Q is absolutely irreducible on E, so Z(Q) ≤ Z(J) ≤ Z(Aut(E)). Of course m2 (J) ≤ m2 (SL(E)) = 3 so m2 (J) = 3 by the previous paragraph. Also by [III8 , 5.4], |AutJ (H)| ≤ 4 for all subgroups H of J of order 19, and |J| divides |SL(E)|. If J ∈ Alt ∪ Spor, then using these facts and [IA , Table 5.6.1], we see that J∼ = 2J2 . But this is impossible since m5 (N ) ≤ 1 while m5 (J2 ) > 1. Thus J ∈ Chev. 1  2 G2 (3 2 ) as Likewise, by [III17 , 5.6], J ∈ Chev(p1 ) for some odd p1 . Indeed J ∼ = Z(J) = 1, so J ∈ Lie(p1 ). Let z ∈ I2 (J)−Z(J). Then CJ (z) embeds in CAut(E) (z) ∼ = GL2 (7)×GL2 (7), so is a {2, 3}-group by Lemma 18.25. Hence p1 = 3, indeed from [IA , 4.5.1, 4.5.2] and Lemma 18.25, J ∼ = Sp4 (3) or SL4 (3). As m2 (J) = 3, only SL4 (3) is possible. But |SL4 (3)|3 = 36 > 34 = |SL4 (7)|3 , a contradiction. We have proved that E(N ) = 1. In particular, E(N 0 ) = 1. Next, let X = O2 (N 0 ) and suppose that X = 1. As m7 (E) is a power of 2 and N 0 ≤ SL± (E), no scalar mappings lie in X. Therefore CX (Q) = 1. Now AutQH (X) contains A4 , so as |SL(E)|3 = 34 , X ∼ = E33 is a Sylow 3-subgroup of a Cartan subgroup of SL(E). Therefore there is g ∈ X such that g acts as a scalar on

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[E, g] ∼ = E73 . Let g ∈ I3 (N ) map on g and choose 1 = v ∈ [E, g]∩L∗ . Let Lv be the subnormal closure of L in CG (v). If Lv ∼ = Sp6 (8), then g normalizes Lv and E ∩ Lv , with eigenvalues ω, ω, 1 on E ∩ Lv (here ω 3 = 1 = ω). But AutAut(Lv ) (E ∩ Lv ) contains no element of order 3 with exactly two distinct eigenvalues on E ∩ Lv , contradiction. It follows that g normalizes E ∩ L. Similarly, we can argue that g normalizes E ∩ L∗ . But then g induces a scalar mapping on either E ∩ L or E ∩ L∗ . In the first case we obtain a contradiction from Lemma 18.18; in the second case, by interchanging L and L∗ , we obtain a similar contradiction. We have proved that X = 1, so F ∗ (N 0 ) = O2 (N 0 ). Suppose that N 0 has a noncyclic abelian normal 2-subgroup F . Then by [III17 , 1.4], N 0 preserves a frame in E. But for each i = 1, 2, N ui ∼ = W (C4 ) preserves a unique frame Ei in E. As N ui ≤ N 0 , E1 = E2 . But Iui = E(CG (v)) | v ∈ Ei , so Iu1 = Iu2 , a contradiction. Thus by Philip Hall’s theorem, R := F ∗ (N 0 ) is of symplectic type. As N 0 ≤ GL(E) with 7 − 1 not divisible by 4, R is extraspecial of width at most 2. If the lemma fails, then as QH ≤ N 0 , R ∼ = Q8 ∗ D8 and then N 0 /R embeds in Σ5 ∼ = Out(R), with QH mapping onto a Sylow 2-normalizer in [Out(R), Out(R)] ∼ = A5 . Hence Q/Q ∩ R acts freely on R/Z(R); also Z(Q) inverts E, so Z(Q) = Z(R). But now Q centralizes Q ∩ R/Z(Q), contradicting the free action of Q/Q ∩ R. This completes the proof.  Lemma 18.27. N 0 ∼ = W (F4 ). Proof. We simply repeat the third paragraph of the proof of Lemma 6.11, making the following substitutions: the roles of p, B, W , A, R0 , and Σ0 there are now to be taken, respectively, by 7, E, N 0 , AutG (E), O2 (N 0 ), and a Σ4 -subgroup of CN 0 (e) for any e ∈ E # for which E(CG (e)) ∼  = Sp6 (8). Lemma 18.28. Let (e, H) ∈ IL7 (G) with e ∈ CE (L). Then H/O7 (H) ∼ = Sp6 (8), Sp4 (8), or L2 (8). ∼ Proof. If L pumps up nontrivially in CG (e), then by Lemma 18.24, H = Sp6 (8). So assume that L pumps up trivially in CG (e), say to H0 . If H = H0 , then the result holds as H0 /O7 (H0 ) ∼ =L∼ = Sp4 (8). Otherwise, [H, H0 e /O7 (H0 )] = 1, and in particular, m7 (H) = 1. We choose f ∈ E ∩ H0# such that F ∗ (CG (f )) ∼ = Z7 × Sp6 (8). Then H/O7 (H) is isomorphic to a component H1 of CF ∗ (CG (f )) (e), and it follows easily from [IA , 4.8.2] that H1 ∼  = L2 (8). The lemma is proved. Now by Lemma 18.21, 7 ∈ γ(G). Hence by [III12 , Theorem 1.2], there is (t, I) ∈ J∗7 (G), and I ∈ Chev(2) with m7 (I) > 2. Moreover, if we let v and J be as in Lemma 18.18, then by Lemmas 18.25 and 18.22, (v, J) ∈ J7 (G) − J∗7 (G). By definition of J∗7 (G) this implies that (18M)

2

F(I) > F(J) = (83 , BC).

By Lemma 18.25, we may replace t and I by conjugates and assume that t ∈ E. Then E = t × (E ∩ I).

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19. ΓD,1 (G) NORMALIZES G0

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Lemma 18.29. q(I) = 8. 

Proof. As E ≤ CG (t), we have I = Γ2E∩L∗ ,1 (I) by [IA , 7.3.3]. In particular, CI (e) has a Lie component H for some e ∈ E ∩ L∗ , and using [IA , 4.9.7], we may assume that 7 divides |H|. By Lemma 18.28, the pumpup H ∗ of H in CG (e) is Sp6 (8), Sp4 (8), or L2 (8). But H is a component of CH ∗ (t), and so q(H) = 8, as follows from [IA , 4.8.2]. On the other hand, as E ∈ Syl7 (G) is abelian, E induces inner automorphisms on I and so q(H) is a power of q(I) by [IA , 4.2.2]. Thus q(I) = 8 or 2. Suppose that q(I) = 2. The previous paragraph shows that for any e ∈ E ∩ L∗ , and any component H of CI (e) of order divisible by 7, H ∼ = Sp6 (8), Sp4 (8), L2 (8), or a subcomponent of one of these, which adds only the possibility H ∼ = L3 (8). If I is a symplectic group, however, we see using [IA , 4.8.2] that there exists e ∈ E ∩ L∗ and a component H of CI (e) with H ∼ = [Sp2n (2), Sp2n (2)], n ≥ 2, contradiction.  F4 (2), so the Dynkin diagram of I, and hence of any H, Since m7 (I) = 3, I ∼ = is simply laced. Thus only H ∼ = L2 (8) and L3 (8) are actually possible. If I is an orthogonal group, then as with the symplectic groups, some e and H exist with H ∼ = Ω± 2n (2), n ≥ 3, contradiction. Similarly if I is a linear or unitary group we find e and H with H/Z(H) ∼ = Ln (2), n ≥ 3, or Un (2), n ≥ 6, contradiction. Hence I is not a classical group. The conditions m7 (I) = 3 and q(I) = 2 yield only I ∼ = E6 (2) and E7 (2). These are out because they contain (A2 (2))3 and so yield m3 (AutI (E)) ≥ 3, whereas m3 (AutG (E)) = m3 (W (F4 )) = 2. The proof is complete.  Now we can reach a final contradiction completing the proof of Proposition 18.13. By Lemma 18.29 and (18M),  ≥ 4, where  is the untwisted Lie rank of I. Using Lemma 18.29, the fact that m7 (I) = 3, and [IA , 4.10.3], we see that the only possibilities are I/Z(I) ∼ = D4− (8) or Un (8), n = 6 or 7. Therefore the reflection subgroup of N ∩ I is isomorphic to W (C3 ), and so, as N 0 ∼ = W (F4 ), it is N 0 -conjugate to N ∩ CG (e) for some e ∈ E with E(CG (e)) ∼ = Sp6 (8). Consequently t and e are conjugate, so I ∼ = Sp6 (8), a contradiction. The proof of Proposition 18.13 is complete. Taken together, Propositions 15.1, 15.1, 13.1, 18.9, 18.1, 18.3, 16.1, 13.1, 18.13, 14.1, 14.1, 17.8, 17.1, and 16.1 cover all the cases (a)–(n) of Proposition 3.4, in alphabetical order. Hence Proposition 12.1 is proved. In every case, G0 has turned out to have the predicted structure of the “target group” in Table 15.1. 19. ΓD,1 (G) Normalizes G0  Having identified the isomorphism type of G0 = Ld : d ∈ D# , we set M = NG (G0 ). Our goal in this section is to complete the proof of Theorem C∗7 : Stage 4b+ by proving: Proposition 19.1. ΓD,1 (G) ≤ M . We continue to assume the setup (1A) and other notation introduced in Section 3. Recall that by Proposition 3.4, G0 , K, and L are as described in one of the cases (a)–(n) of Table 15.1, as is G0 , as just noted. We give more details, viz., isomorphism types of all possible groups Lu , u ∈ De , in Table 15.3 below. Note

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

278

that Lu is a component of CG0 (u). In the cases that G0 is a classical group, these groups Lu can be found by using [IA , 4.8.2] and the values of K/Z(K) and L to locate D in G0 , and then using [IA , 4.8.2] and the fact that (x, K) ∈ J∗p (G) to determine the possible Lu ’s. In the cases that G0 is of exceptional type, we have actually listed all Lu with the following properties: L/Op (L) ↑p Lu /Op (Lu ) L. The proof is complete.  Now we can prove Lemma 19.4. NG (D) ≤ M . Proof. Consider the set L of all pairs (u, H) such that u ∈ D# , H is a component of E(CG (u)), and for some t ∈ D# − CG (H), CH (t) has a component isomorphic to L. Clearly NG (D) permutes L, so NG (D) normalizes L . But by Lemma 19.3, L = {Lu | u ∈ D# , Lu > L}, so L = G0 by definition of G0 . The proof is complete.  What remains is more complex: to show that NG (u ) ≤ M for all u ∈ D# . For a given u, we focus on Lu , which we are able to do because almost always, Lu  NG (u ). Lemma 19.5. Let u ∈ D# and suppose that Lu  NG (u ). Then p = 3,   Lu = L ∼ = SL3q (q), and G0 ∼ = L6q (q) with q > 2 (case (b) of Table 15.3). Moreover, t x ∈ L for some t ∈ NG (u ).

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19. ΓD,1 (G) NORMALIZES G0

279

Table 15.3. The possible pumpups Lu > L, u ∈ De G0 /Z(G0 )K/Z(K)

L

condition

q (a) Ln+1 (q)



q Lnq (q) SLn−1 (q) n ≥ 4



q Lnq (q) SLn−1 (q) n ≥ 4

q (b) Ln+2 (q)

−

Lu /Z(Lu )





Lnq (q)





Lnq (q)

−

 

−

−

q q (c) Ln+2 (q) Ln q (q) SLn−2 (q) n ≥ 6

−q

(d) E6

−q

(q) L6

−q

(q) L4

−q

(q)

L6



(q), D4− (q)



SL3q (q)



L4q (q)

p = 3 L6q (q), D4 (q), A4q (q)



SL7q (q)



p = 3 L9q (q), D7q (q), E7 (q)

L4q (q)

(e)

D4 (q)

(f)

E6q (q)

L6q (q)

(g)

E8 (q)

L9q (q)



Ln q (q)





L4q (q) 







(h) Sp2n+2 (q) Sp2n (q) Sp2n−2 (q) n ≥ 3

Sp2n (q)

(i)

Sp6 (q)

F4 (q)

Sp6 (q)

n+1 q

n q

−n+1

−n

(j) Ω2n+2 (q) Ω2n (q)

†

Sp4 (q) n−1 q

Ω2n−2 (q)

n

n≥4

q Ω2n (q)

−n−1

−n

q q (k) Ω2n+2 (q) Ω2n q (q) Ω2n−2 (q) n ≥ 4





Ω+ 8 (q)

E6q (q)



A5q (q)

E6q (q), D6 (q), A6q (q)

E7 (q)

Ω+ 12 (q)

E7 (q), D7q (q)

(l)

E6q (q)

Ω10q (q)

(m)

E7 (q)

(n)

E8 (q)

†

Ω2n q (q)





Ω10q (q) 





If n = 4, then D ∩ K ∈ x G0 (see the normalization (14B)).

Proof. Notice that K = Lp (CG (x)) as mp (C(x, K)) = 1, so u = x . Let g ∈ NG (u ) with Lgu = Lu . Let L∗ be the product of all components of E(CG (u)) isomorphic to, but distinct from, Lu . Then Lgu ≤ L∗ . Since [D, L] = 1, D normalizes Lu and L∗ . Set C = CL∗ u (D) = CL∗ (x) u . Then C ≤ CM (Lu ), by Lemma 19.4. Since CG (G0 ) is a p -group (19A), a Sylow p-subgroup of C embeds in CAut(G0 ) (Lu ). In particular, (19B)

mp (C) ≤ m := mp (CAut(G0 ) (Lu )).

Before considering several cases, we remark that (19C)

Lu /Z(Lu ) ∼  K/Z(K). =

For if Lu /Z(Lu ) ∼ = K/Z(K), then by [III14 , 1.2b], mp (C(u, Lu )) = 1. But this is absurd as L∗ ≤ C(u, Lu ), and L∗ contains a copy of L with mp (L) ≥ 2. Hence (19C) holds. We next prove (19D)

Lu = L.

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280

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Suppose by way of contradiction that Lu > L. Consider the subcase in which p does not divide |Outdiag(Lu )|. Then p does not divide |Z(Lu )|, so L∗ u = L∗ × u and so mp (C) ≥ 2. Thus m ≥ 2. As Lu > L, it follows with [III17 , 10.4] that m = 2, and q    (q), SLnq (q)) or (3, E6q (q), SL5q (q) or D4 (q)). (p, G0 /Z(G0 ), Lu ) = ((n + 2)/k, Ln+2 q (In the first case, p divides n + 2 because K is a terminal SLn (q)-component in G0 /Z(G0 ).) Since m = 2, mp (C) ≤ 2 and so mp (CL∗ (x)) = 1. But then as mp (L∗ ) ≥ mp (Lu ) > 1, x cannot induce an inner automorphism on L∗ . As p does not divide |Outdiag(K)|, x induces a graph, field, or graph-field automorphism on each component of L∗ , so mp (CL∗ (x)) > 1 by [III17 , 8.6], contradiction. q  (q) or E6q (q), with Thus, p divides |Outdiag(Lu )|, whence Lu /Z(Lu ) ∼ = Lpk p = 3 in the latter case. In the E6 case, Table 15.3 shows that K/Z(K) ∼ = Lu /Z(Lu ), contradicting (19C). Hence  (19E) Lu /Z(Lu ) ∼  K/Z(K), = L q (q) ∼ = pk

 and we are in case (m) of Table 15.3. Thus, Lu ∼ = A6q (q), p = 7, and G0 ∼ = E7 (q). In particular, dp (Lu ) = 2 = dp (K). Now choose a p-terminal long pumpup (v, I) of (u, Lu ) [IG , 6.22]. Then (v, I) ∈ Jp (G) so mp (C(v, I)) = 1 by [III12 , Theorem 1.2].  But mp (C(u, Lu )) ≥ mp (L∗ ) > 1, so by [III8 , 2.4], I/Z(I) ∼  Lu /Z(Lu ) ∼ = = A6q (q). q But then by [III17 , 10.30h], F(I) > F(E6 (q)) = F(K), contradicting [III14 , Lemma 1.2]. This completes the proof of (19D). Next, we show that

(19F)

p divides |Outdiag(L)|.

Suppose that (19F) fails. Since B induces inner-diagonal automorphisms on L, B = (B∩L)×CB (L), with u, x = D = CB (L) since (y, L) is an (x, K)-subterminal pair. We have DL∗ ≤ C(u, L). Let Cx = CDL∗ (x) = CL∗ (x)D. Then Cx ≤ CCG (x) (LD) so by [III17 , 6.28], every element of Ip (Cx ) induces an inner-diagonal automorphism on K. By [III17 , 8.6], mp (CL∗ (x)) ≥ 2, whether x induces an inner or outer automorphism on L∗ . Hence as u ∈ L∗ , mp (CCG (x) (LD)) ≥ 3. Let F ∈ Ep3 (CG (LD)) and B∗ = (B ∩ L)F . Then B∗ is elementary abelian and induces inner-diagonal automorphisms on K, with mp (B∗ ) > mp (B). Hence B∗ should have been chosen rather than B ∗ (see (3A)), a contradiction proving (19F). Next, we claim that ∼ SLq (q), and K = ∼ Lq (q), q > 2. (19G) p = 3, L = 3

4

Indeed, as p divides |Outdiag(L)|, we see from Table 15.3 and (19B) that p = 3,  and one of the cases (a), (b), or (e) holds with L ∼ = SL3q (q), or case (m) holds  q with L/Z(L) ∼ = SL6 (q) (see [IA , 4.7.3A] for the versions of K and L in the last case). In the last case, m3 (B ∗ ) ≤ 8 by Table 15.1, and as m3 (L) − m3 (Z(L)) = 4, we have m3 (CDL∗ (x)) ≤ 4. In particular, this implies first that x normalizes every component of L∗ , and then, using [IA , 4.8.2, 4.8.4], that E(CL∗ (x)) ∼ = SL2 (q 3 ). ∗ But E(CL (x)) ≤ CG (LD) = CCG (x) (LD), which is 3-solvable by [III17 , 6.28], a  contradiction. Thus, L ∼ = SL3q (q), which together with Table 15.3 implies (19G). In K we see that D = x Z(L). Hence CCG (x) (L) = CG (LD) is 3-solvable. If u ∈ Z(L), then D = u Z(L) centralizes L∗ , so L∗ ≤ CCG (x) (L), which is 3-solvable, a contradiction. Therefore u = Z(L) = Z(Lt ) for some t ∈ NG (u ) with Lt ≤ L∗ . In CG (x) we see also that a Sylow 3-subgroup of CG (L x ) is abelian

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19. ΓD,1 (G) NORMALIZES G0

281

of rank two, containing D. In particular, the only elements of CG (L x ) of order 3 lie in D. Let P ∈ Syl3 (CG (u)) with B ∗ ≤ P . Let L 1 = L t , L 2 , . . . , Lr be the components of L∗ . We argue that r = 1. Now, x normalizes each Li , as CG (x L) is 3-solvable; hence Z(Li ) ≤ CG (LD) for each i = 1, . . . , r so Z(Li ) ≤ D. Again as CG (x L) is 3-solvable, Z(Li ) = u for all i = 1, . . . , r. Let Ui be an x-invariant E32 -subgroup of Li containing u. Since D contains I3 (CG (L x )), [Ui , x] ≤ u . If r > 1 then m3 (CU1 U2 (x)) > 1 so x ∈ U1 U2 , and then U1 U2 ≤ D, contradiction. Thus, r = 1, as claimed. A similar argument shows that F ∗ (NG (u )/O3 (NG (u ))) is the central product of the images of L and L1 and a cyclic 3-group.  If case (b) of Table 15.3 holds, then as G0 ∼ = L6q (q), NG0 (u )(∞) = LLt1 for  some t1 ∈ NG0 (u ); and x ∈ Lt1 as E(CG0 (x)) = K ∼ = L4q (q). Given the structure of NG (u ) in the previous paragraph, it follows that Lt1 = L1 , so the conclusion of the lemma holds. Thus, we may assume that case (a) or case (e) of Table 15.3 holds. As r = 1, P normalizes L1 = L∗ , so U1 could have been chosen to be P invariant. Then U1 normalizes B (Lemma 3.3a). If U1 does not centralize B, then any g ∈ U1 − u induces a transvection on B with center u . Also WK ∼ = Σ4 with elements of I3 (WK ) having a free summand on B. Thus, |A|3 ≥ 32 . If case (a) of Table 15.3 holds, then as Σ5 is complete, A contains W × Z3 , which is impossible as A acts faithfully on B ∼ = E34 . Thus case (e) of Table 15.3 holds. But then by [III17 , 1.23], A contains no transvections, a contradiction. Thus, U1 centralizes B. Hence U1 ≤ CG (LD) so D = U1  P . Therefore P ≤ NG (D) ≤ M , so P embeds in Aut(G0 ). Since x (P ∩ L) ≤ G0 but G0 does not contain a 31+4 -subgroup, there is g ∈ I3 (P ∩ L∗ ) such that g ∈ G0 . Thus g induces an outer automorphism on G0 , centralizing L. By [IA , 4.2.3], g induces a graph automorphism on G0 , and by [IA , 4.7.3A] the only possibility is that CG0 (g) ∼ = G2 (q). On the other hand, all elements of I3 (L∗ − Z(L∗ )) are   ∗ L -conjugate, so O 3 (CG (g)(∞) ) ∼ = K. As K ∼ = L4q (q) does not contain G2 (q), by order considerations, we have a final contradiction completing the proof of the lemma.  Lemma 19.6. Suppose that u ∈ D# and Lu = L  NG (u ). Assume that CAut(G0 ) (L) has abelian Sylow p-subgroups of rank 2. Then the following conditions hold: (a) NG (u ) = C(u, Lu )NNG (u) (D) = C(u, Lu )NM (u ); (b) C(u, Lu ) has an abelian Sylow p-subgroup Q with D = Ω1 (Q); and (c) If L contains a G0 -conjugate or M -conjugate of D, then NG (u ) ≤ M . Proof. Let Q ∈ Sylp (C(u, Lu )) with D ≤ Q. Then NQ (D) ≤ M by Lemma 19.4. As CM (G0) is a p -group, NQ (D) embeds in a Sylow p-subgroup of CAut(G0 ) (L), so NQ (D) is abelian of rank 2. Hence D = Ω1 (NQ (D)) and then NQ (D) = Q, proving (b). As C(u, Lu )  NG (u ), (a) then follows by a Frattini argument. Finally if Dg ≤ L for some g ∈ M , then C(u, Lu ) ≤ CG (Dg ) = CG (D)g ≤ M by Lemma 19.4, and (b) implies (c). This completes the proof. 

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

282

We subdivide the proof of Proposition 19.1 into several cases. To begin, we assume that (19H) one of the conclusions (c), (d), (f), (h), (i), (j), (k) or (n) of Table 15.3 holds. By Lemma 19.5, Lu  NG (u ) for all u ∈ D# . Also, in these cases, Op (K) = 1, by [IA , 6.1.4] and Lemma 19.2; and y induces an inner automorphism on K, so we may assume that y = D ∩ K. Then these cases are unified by the following properties, according to [III17 , 14.6] (applied with A = AutNG (x) (K) and then with A = AutNG (u) (Lu )). Lemma 19.7. Assume (19H) with y = D ∩ K. Then y ∈ x G0 and = y NG0 (x) . Moreover, there exists y1 ∈ x G0 ∩ L, and for all y N (u) N (u) = y1 M . u ∈ D# , y1 G NG (x)

Proof. This follows directly from [III17 , 14.6] except possibly for those u ∈ D# such that Lu = L, CAut(G0 ) (L) has abelian Sylow p-subgroups of rank 2, and L contains a G0 -conjugate of D. But for those u’s, NG (u ) ≤ M by Lemma 19.6.  Hence NG (u ) = NM (u ) and the lemma follows. Note that in case (j) with Lu ∼ = D4 (q) for some u ∈ D# , we have W ≡B W (D5 ) or W (C5 ). Hence AutW (B ∩ Lu ) ∼ = W (C4 ). As AutW (B ∩ Lu )  AutA (B ∩ Lu ), no r ∈ NG (u ) can induce a triality automorphism on Lu ; this yields the “triality-free” hypothesis of [III17 , 14.6]. Lemma 19.8. Assume (19H). Then NG (x ) ≤ M . g

h

Proof. Let g ∈ NG (x ). Then y = y for some h ∈ NG0 (x ) by Lemma gh−1 gh−1 = y and x = x . So gh−1 ∈ NG (D) ≤ M , and then 19.7. Hence y g ∈ Mh = M.  Lemma 19.9. Assume (19H). Then ΓD,1 (G) ≤ M . g

h

Proof. Let u ∈ D# and g ∈ NG (u ). Then, y1 = y1 for some h ∈ G NM (u ) by Lemma 19.7. Thus gh−1 ∈ NG (y1 ). As y1 ∈ x 0 , again by  Lemma 19.7, we have NG (y1 ) ≤ M by Lemma 19.8. Hence, g ∈ M h = M . We next assume that (19I) Case (a) or (b) of Table 15.3 holds. ∼ L (q) in case (a), while in case (b), Thus, K ∼ = SLn (q); and G0 /Z(G0 ) = n+1  G0 /Z(G0 ) ∼ = Ln+2 (q). We set u = y in case (a), and u = x in case (b). Thus, in either case, we may assume that u = diag(α, α−1 , 1, . . . , 1) ∈ G0 , where α is a primitive p-th root of 1. (Of course, u is really only defined up to a scalar multiple.) Then, there exists a permutation matrix t ∈ G0 such that w := ut ∈ L. We fix t and set Lw := Ltu . By Lemma 19.5, Lv  NG (v ) for all v ∈ D# , unless n = 4, G0 /Z(G0 ) ∼ = L6 (q) and p = 3. In the latter case, u = x and we choose t and w more precisely. Namely, we may assume that y = diag(α, α, α, 1, 1, 1),

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19. ΓD,1 (G) NORMALIZES G0

283

and by Lemma 19.5, we may choose a permutation matrix t ∈ NG0 (y ) with xt = w ∈ L. Notice that y is then the unique subgroup in E1 (D) such that L is a component of CG (y). Lemma 19.10. Assume (19I). Then the following conclusions hold: (a) K, Lw = Lu , Lw = G0 ; (b) DNG (v) ⊆ DCG (Lv )NG0 (v) for v = x, and in case (a) for v = y; (c) w NG (v) = w NG0 (v) for all v ∈ D# . Proof. Part (a) follows from the definitions of u and w and [III17 , 9.1]. For (b), suppose first that mp (C(v, Lv )) = 1. Then C(v, Lv ) has a normal p-complement and a cyclic Sylow p-subgroup. Also Lv  NG (v ) by Lemma 19.5. Therefore if we set N = NG (v ) and N = N/C(v, Lv ), it is enough to show N

Lv

that D = D . But as D is generated by a diagonalizable element of order p, this follows from [III17 , 14.1]. Thus in proving (b) we may assume that mp (C(v, Lv )) > q 1. In particular, v = y and by hypothesis, G0 /Z(G0 ) ∼ (q). Thus Ly ∼ = = Ln+1 q SLn−1 (q) and D ≤ C(y, Ly ). By Lemma 19.6, NG (y ) ≤ C(y, Ly )NG (D) so DNG (y) = DCG (yLy ) , completing the proof of (b). For (c), let v ∈ D# . Now w ∈ L ≤ Lv . By [III17 , 14.1], w Aut(Lv ) ⊆ N (v) . Hence if g ∈ NNG (v) (Lv ), then there is h ∈ NG0 (v ) such that w G0 g h w = w . In particular, if Lv  NG (v ), then the conclusion of (c) holds. By Lemma 19.5 and the remarks before this lemma we may therefore assume that  p = 3, n = 4, G0 ∼ = L6q (q), v = y , Ly = L has exactly two conjugates in NG (y ), and g ∈ NG (y ) − NG (L). But then we chose w = xt where t ∈ NG0 (y ) gt h does not normalize L, so gt ∈ NG (L). Thus, w = w for some h ∈ NG0 (y ). g ht−1 , with ht−1 ∈ NG0 (y ) as required. Thus (c) holds as well Hence w = w and the lemma is proved.  Lemma 19.11. Assume (19I). If NG (w ) ≤ M , then ΓD,1 (G) ≤ M . g

h

Proof. Let v ∈ D# and g ∈ NG (v ). Then, by Lemma 19.10c, w = w for some h ∈ G0 . Hence, gh−1 ∈ NG (w ) ≤ M , and then g ∈ M , proving the lemma.  Lemma 19.12. Assume (19I). Then CG (K) ≤ M and CG (Lu ) ≤ M . Proof. Let v ∈ {x, u}. As w ∈ L ≤ Lv , we have CG (Lv ) ≤ CG (w) ≤ NG (Lw ) for all v ∈ D# . So, CG (Lv ) normalizes Lv and Lw , and hence normalizes G0 , by Lemma 19.10a.  Lemma 19.13. Assume (19I). Then NG (x ) ≤ M . Proof. Let g ∈ NG (x ). By Lemma 19.10b, since CG (K) ≤ M by Lemma 19.12, there exists g0 ∈ M with Dg = Dg0 . Then gg0−1 ∈ NG (D) ≤ M , and so g ∈ M.  Lemma 19.14. Assume case (b) of Table 15.3 holds. Then ΓD,1 (G) ≤ M . Proof. As w ∈ xG0 ∩ L, NG (w ) ≤ M by Lemma 19.13. Then Lemma 19.11 completes the proof.  Henceforth, we may assume that case (a) holds and u = y.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

284

Lemma 19.15. Assume case (a) of Table 15.3 holds. Then ΓD,1 (G) ≤ M . Proof. First, let g ∈ NG (y ). By Lemma 19.12, we have CG (Ly ) ≤ M . Hence, by Lemma 19.10b, Dg = Dm for some m ∈ M , whence gm−1 ∈ NG (D) ≤ M . So, g ∈ M , whence NG (y ) ≤ M . As w ∈ y G0 in our case (a), we have  NG (w ) ≤ M . Again, Lemma 19.11 completes the proof. By Lemmas 19.14, 19.15, and 19.9, it remains to consider cases (e), (g), (l) and (m), to prove Proposition 19.1. Assume next that (19J)

Either case (g) or case (m) of Table 15.3 holds.

By definition of acceptable subterminal pair [III12 , 1.15], we have y ∈ K; indeed in these cases we may assume, by [III17 , 6.35], that y was chosen so that yZ(K) lies in a root SL2 (q) subgroup Xy Z(K)/Z(K) of K/Z(K). Here Xy is a root SL2 (q) subgroup of K, and we may assume that y ∈ Xy (since if Z(K) = 1, then Z(K) = x ). Then clearly there is t ∈ K such that Xw := Xyt is a root SL2 (q)-subgroup of L, and we set w := y t ∈ L and Lw = Lty . Lemma 19.16. If (19J) holds, then w NG (u) = w NG0 (u) for all u ∈ D# . Proof. Since Xw is a root SL2 (q)-subgroup of L  CLu (D), and given the possibilities for Lu in Table 15.3, Xw is a root SL2 (q)-subgroup of Lu . Hence by [III17 , 14.4], NAut(Lu ) (w ) covers Out(Lu ). By Lemma 19.5, Lu  NG (u ), and CG (Lu ) ≤ CG (w) as w ∈ Lu , so NG (u ) ≤ Lu NG (w ) ≤ NG (w )NG0 (u ), as required.  Lemma 19.17. Assume (19J). Then G0 = Ly , Lw . Proof. Let G2 = Ly , Lw . By Lemma 19.5, G2 is y, w -invariant. By Lp balance, G2 = Lp (G2 ). Set G2 = G2 /Op (G2 ). Since w ∈ L ≤ Ly and Lw ∼ = Ly , Lw ≤ Ly . Also, mp (G2 ) ≤ mp (G0 ) < 2mp (Ly ). Thus Ly ↑p G2 . In the two cases, Ly ∼ = E7 (q), resp. D6 (q), by [III17 , 6.35], and G2 embeds in G0 ∼ = E8 (q), resp. E7 (q), so by [III17 , 10.22], G2 ∼  = G0 , whence G2 = G0 . Lemma 19.18. Assume (19J). Then ΓD,1 (G) ≤ M . Proof. By Lemma 19.16, NG (u ) ≤ G0 NNG (u) (w ) for any u ∈ D# , and Lu ≤ G0 ≤ M by definition. For u = y we have NNG (y) (w ) ≤ NG (Ly , Lw ) = NG (G0 ) = M by Lemma 19.17, so NG (y ) ≤ M . Now let u be arbitrary. We have w ∈ y K ⊆ y G0 , also NG (w ) ≤ M . Therefore NG (u ) ≤ G0 NG (w ) ⊆ M . The proof is complete.  Next, we assume that case (l) holds: (19K)

  K∼ = E6q (q). = D4 (q), G0 ∼ = D5q (q), L ∼

We choose v ∈ B∩L lying in a long root SL2 (q)-subgroup of K, so that E(CK (v)) =  J × Kv with J ∼ = SL2 (q) and Kv ∼ = Ω6q (q). Then v lies in a long root SL2 (q)subgroup of G0 , so by [III17 , 6.35], the pumpup Lv of Kv in CG0 (v) satisfies  Lv /Z(Lv ) ∼ = L6q (q). Also, by [III17 , 9.1], L, Kv = K.  We let L∗ be the pumpup of Kv in CG (v). Thus Lv = KvLv ≤ L∗ . We prove: Lemma 19.19. L∗ = Lv is a normal component of CG (v) and NG (v ).

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19. ΓD,1 (G) NORMALIZES G0

285

Proof. Note that there exists g ∈ WK such that Kvg ≤ L. An application of [III8 , 2.13], with v g , Kvg , D, and v g in the roles of y, L, B1 , and u, shows that Kvg ≤ E(CG (v g )). Hence Kv ≤ E(CG (v)), so L∗ = E(L∗ ). Since Lv ≤ L∗ , Kv ↑p L∗ .  Suppose that Lv < L∗ . Since Kv ↑p L∗ with Kv ∼ = L4q (q), either L∗ ∈ Chev(2) or q = 2, p = 3, and L∗ ∼ = Co2 or B2 (33 ), by [III17 , 10.9]. But Lv ≤ L∗ with q ∼ Lv = A5 (q), so it is impossible that L∗ ∼ = B2 (33 ). We rule out L∗ ∼ = Co2 as follows. We have m3 (B) = 6 and B normalizes G0 , with CG (G0 ) a 3 -group. Therefore Lv lies in a P GU6 (2) subgroup of CG0 (v) normalizing L∗ . However, Aut(Co2 ) does not involve P GU6 (2) by [III17 , 16.3], contradiction. We have proved that L∗ ∈ Chev(2). Next, if m  p (L∗ ) ≥ 5, then [III11 , 21.1] and [III14 , Lemma 1.2] give (q 25 , A) = F(Lv ) < F(L∗ ) ≤ F(K) = (q 25 , D), forcing F(L∗ ) = (q 25 , D). In particular, L∗ is an orthogonal group, indeed L∗ ∼ = 2 ± (q 1/k ) for some integer k. Since Lv ≤ L∗ , it is impossible by [III17 , 16.1] D5k that k = 1, so k > 1 and q > 2. Then the minimum dimension of a nontrivial  Fq1/k2 -representation of Lv ∼ = A5q (q) is at least 6k2 , since q 6 − 1/q − 1 divides |Lv |. Therefore 10k ≥ 6k2 so k < 2, a contradiction. Thus, we may assume that   m  p (L∗ ) ≤ 4. Hence p = 3 and by [III17 , 10.13], L∗ ∼ = A5q (q 3 ). As Kv ∼ = A3q (q) is a component of CL∗ (x), this is a contradiction. This proves that Lv is a component of CG (v). As mp (CCG (v) (x)) ≤ mp (CG (x)) ≤ 7 and E(CCG (v) (x)) = E(CK (v)) = J ×Kv , it follows easily that Lv is the unique component of CG (v) of its isomorphism  type, and so Lv  CG (v). The proof is complete. Lemma 19.20. If (19K) holds and u ∈ D# , then G0 = Lu , Lv . Proof. Let G2 = L, Lv . Then G2 ≤ Lu , Lv ≤ G0 , so it suffices to prove that G2 = G0 . Since L, Kv = K, G2 = K, Lv . As v ∈ B ∩ L, we have [x, v] = 1, and x centralizes Kv and normalizes Lv , by Lemma 19.19. Thus G2 is x, v -invariant. By Lp -balance, G2 = K, Lv ≤ Lp (G2 ). As Lv is quasisimple with Kv ≤ K ∩ Lv but Kv ≤ Z(Lv ), it follows that G2 = K G2 . As K does not involve Lv , G2 is a single p-component. Set G2 = G2 /Op (G2 ). Then as [x, Lv ] = 1,   G2 ↓p K ∼ = D5q (q). Since G2 is involved in G0 ∼ = E6q (q), it follows from [III17 , 10.16] q that G2 is a version of E6 (q). Hence G2 = G0 . The lemma is proved.  By Table 15.3, for any u ∈ D# , Lu = L or Lu ∼ = D5q (q). In either case v lies in a long root SL2 (q)-subgroup of Lu , so by [III17 , 14.4], 

(19L)

v

Aut(Lu )

= v

Lu

for all u ∈ D# .

Lemma 19.21. If (19K) holds, then ΓD,1 (G) ≤ M . Proof. Let u ∈ D# . We have NG (u ) = NNG (u) (Lu ) ≤ Lu NNG (u) (v ) ≤ Lu (NG (Lu ) ∩ NG (Lv )) ≤ Lu NG (Lu , Lv ) = M, where we use, in order, (19L), Lemma 19.19, and Lemma 19.20 (and the fact that  Lu ≤ G0 ). It remains to treat case (e):   (19M) K∼ = SL3q (q), G0 ∼ = L4q (q), L ∼ = D4 (q), q > 2. Note that q > 2 since otherwise p = 3 and K ∈ G3 .

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

286

In this case, we may assume that y ∈ K. Then L = Ly , as can be seen in G0 . Moreover, by Lemma 19.6, there is an abelian Sylow p-subgroup Q of C(y, Ly ) such that D = Ω1 (Q). By a Frattini argument, CG (y) ≤ C(y, Ly )NG (D), so CG (y) = C(y, Ly )CM (y). This immediately implies part (b) of the following lemma. Lemma 19.22. Assume (19M). Then the following conclusions hold: (a) If u ∈ De , then DNG (u) = DNG0 (u) ; and (b) bCG (y) = bCM (y) for all b ∈ B ∩ L. Proof. It remains to prove (a). By Table 15.3, Lu ∼ = K. Then D = u, v Aut(Lu ) Lu with v = D ∩ Lu . Hence, v = v , by [III17 , 14.1], which implies the result as Lu ≤ G0 by definition.  We may choose a basis {b1 , b2 , b3 , b4 } for B and a standard module V for G0 such that Vi := [V, bi ] is 2-dimensional, V = V1 ⊥ V2 ⊥ V3 ⊥ V4 , x = b1 , and  −1 b , b y = b2 b3 b4 ∈ K. Then, B ∩ L = b−1 3 3 b4 . 2 Lemma 19.23. If (19M) holds, then NG (u ) ≤ M for any u ∈ De , and  indeed for any u ∈ Ip (G0 ) such that E(CG0 (u)) ∼ = L4q (q).  Proof. By [III17 , 14.13], any u ≤ B with E(CG0 (u)) ∼ = L4q (q) is G0 conjugate to an element of De , so we may assume that u ∈ De . Let g ∈ NG (u ). By Lemma 19.22a, there exists h ∈ NG0 (u ) with Dg = Dh . Then  gh−1 ∈ NG (D) ≤ M , and so g ∈ M .

As an immediate consequence, we have  Lemma 19.24. If u ∈ Ip (G0 ) and E(CG0 (u)) ∼ = L4q (q), then E(CG0 (u)) = O p (E(CG (u))). 

We now subdivide case (e) into two subcases. For the next two lemmas, we assume that p > 3. The significance of this for us is that, in this subcase,  −1 b2 ∈ b2 b3 b4 , b−1 2 b3 , b3 b4 = y, B ∩ L . Note that b2 ∈ x G0 and so NG (b2 ) ≤ M by Lemma 19.23. Lemma 19.25. Assume (19M) with p > 3, and let β = b−1 3 b4 . Then CG (β) ≤ M . Proof. We first prove that NG (B) ∩ NG (β ) ≤ M. We have CG (B) ≤ CG (D) ≤ M and W ∼ = W (D4 ), W (B4 ), or W (F4 ), with W (D4 ) ≡B AutG0 (B)  W by Propositions 18.1 and 3.4. Let  B = {b ∈ B # | E(CG (b)) ∼ = L q (q)}. (19N)

0

4



For any b ∈ B, let Eb = E(CG0 (b)) = O p (E(CG (b))) (see Lemma 19.24). Now let g ∈ NG (B) ∩ NG (β ), and let g be its image in AutG (B). Since g normalizes β , g normalizes AutG0 (B) and permutes B, by [III17 , 1.30]. So g normalizes Eb | b ∈ B , which contains and then equals Lu | u ∈ De = G0 . This establishes (19N).

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19. ΓD,1 (G) NORMALIZES G0

287

∼ Ω+ (q) and We may write E(CG0 (β)) = J0 × J1 , where J0 = Ω(V1 ⊥ V2 ) = 4 J1 ∼ = SL2 (q). Observe that b3 b4 ∈ J1 . Note that J0 = J0+ × J0− with J0+ ∼ = J0− ∼ = SL2 (q) and b± b ∈ J . We shall argue that 0± 1 2 (19O)

E(CG (β)) = J0 × J1 ,

and so B ≤ β × E(CG (β)). Hence, CG (β) ≤ E(CG (β))(NG (B) ∩ CG (β)) ≤ M by a Frattini argument and (19N), so the lemma will be proved. Let L1 and L0± be the subnormal closures of J1 and J0± in CG (β), and set L0 = L0+ L0− . Now, b4 ∈ β, J1 ≤ β, L1 , and b4 ∈ xG0 , whence CG (b4 ) ≤ M . Hence, if J0 ≤ CG (β, L1 ), then L0+ L0− ≤ CG (β, L1 ) ≤ CG (b4 ) ∩ CG (β, L1 ) ≤ CM (β, L1 ) ≤ CM (β, J1 ) whence L0 = J0  E(CG (β)). Then, as b1 ∈ J0 and CG (b1 ) ≤ M , we have L1 ≤ CG (J0 ) = CM (J0 ), whence L1 = J1 and E(CG (β)) = J0 × J1 , as desired. Another possibility is that J0 ≤ L1 . Then J0 ×J1 ≤ L1 with J0 ×J1 ∼ = SL2 (q)× SL2 (q) × SL2 (q). Since [J0 , J1 ] = 1, L1 is a single p-component, by Lp -balance. Now J0 is a product of components of CM (b3 , b4 ) = CG (b3 , b4 ) = CCG (β) (b4 ), indeed  ±1 J0 = O p (E(CCG (β) (b4 ))). Notice also that b±1 1 b2 β ∈ B, so CCG (β) (b1 b2 ) ≤ M . Let L1 = L1 /Op (L1 ). By [III17 , 10.6], it follows that L1 ∈ Chev(2) has level q(L1 ) = q, or f (L1 ) > q 9 with mp (L1 ) ≥ 3. The latter case implies that F(L1 ) > F(K) and violates [III14 , Lemma 1.2]. In the former case, again in view of [III14 , Lemma  1.2], and as K ∼ = A3q (q), either L1 ∼ = Aδ3 (q) or L1 has Lie rank 2. In either case, L1 cannot contain a subgroup isomorphic to A1 (q) × A1 (q) × A1 (q) (see [III17 , 11.27]). Hence, this case cannot occur. Finally, the only other alternative is that one component of J0 , say J0+ , lies in −1 −1 L1 , while J0− does not. Let h = b−1 1 b2 b3 b4 = b1 b2 β ∈ β J0− . Then h ∈ B, so q ∼  as shown above, Eh = Lp (CG (h))) = L4 (q) with mp (C(h, Eh )) = 1. Thus by Lp balance, L1 ≤ Lp (CG (β, h)) ≤ Eh ≤ M . As J1 and J0+ are distinct components  of CM (β), this is a contradiction, and the lemma is proved. Lemma 19.26. If (19M) holds with p > 3, then ΓD,1 (G) ≤ M . Proof. By Lemmas 19.5 and 19.23, it suffices to show that NG (u ) ≤ M for all u ∈ D# such that Lu = L  NG (u ). But CG (L) ≤ CG (β) ≤ M by Lemma 19.25. As p > 3, CM (L) has abelian Sylow p-subgroups of rank 2, so Lemma 19.6 completes the proof.  It remains to consider case (e) with p = 3. In this case we have  −1 y = Z(L) ≤ B ∩ L = b−1 2 b3 , b3 b4 . Lemma 19.27. Suppose that (19M) holds with p = 3, g ∈ G, and B g ≤ M . Then B g ∈ B G0 . Proof. Since G0 has a unique class of E34 -subgroups by [IA , 4.10.3], it is g on B, we have enough to show that B ≤ G0 . Since NG (B) acts irreducibly  G x ∩ B = B, so it is enough to show that if h ∈ G and xh ∈ M , then xh ∈ G0 . But if xh ∈ M − G0 then x induces an outer automorphism on G0 as CG (G0 )  is a p -group. Hence E(CG0 (xh )) ∼ = G2 (q), L3q (q), D4 (q 1/3 ) or 3D4 (q 1/3 ), by [IA , 4.7.3A,4.9.1]. None of these groups, however, embeds in CG (xh ) ∼ = CG (x), as   O p (CG (x)(∞) ) = K ∼  = L4q (q) (see [III17 , 11.27]). This proves the lemma.

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15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

288

Lemma 19.28. If (19M) holds with p = 3, then NG (B) ≤ M . Proof. Let g ∈ NG (B), so that B ≤ Gg0 . We argue that (19P)

E(CGg0 (u)) = E(CG0 (u)) = Lu for all u ∈ D − y .

 Since u ∈ B ≤ Gg0 , and p = 3, either E(CGg0 (u)) ∼ = L4q (q) or L2 (q) × L2 (q) × L2 (q),    or u ∈ E(CGg0 (u)) ∼ = SL3q (q). However, O 3 (CG (u)(∞) ) = Lu ∼ = L4q (q), which is centerless and does not contain L2 (q) × L2 (q) × L2 (q) by [III17 , 11.27]. Hence E(CGg0 (u)) ∼ = E(CG0 (u)) and then (19P) holds. Now y ∈ De , so by definition and (19P), G0 ≤ Lu | u ∈ D − y ≤ Gg0 .  Hence Gg0 = G0 and g ∈ M , proving the lemma.

Lemma 19.29. Assume (19M) with p = 3. If g ∈ G and B g ≤ M , then g ∈ M . Proof. By Lemma 19.27, B g = B h for some h ∈ G0 . Then g = (gh−1 )h ∈ NG (B)G0 ≤ M , by Lemma 19.28.  Lemma 19.30. Assume (19M) with p = 3. Then no element of I3 (M ) induces a field or graph-field automorphism on G0 . Proof. Suppose that t were such an element. Then t induces a nontrivial automorphism on G0 as CG (G0 ) is a 3 -group. Hence t ∈ [M, M ]. However, replacing t by an M - conjugate, we may assume that [t, B] = 1. By Lemma 19.29 and  [III8 , 6.1h], t ∈ [M, M ], a contradiction. Let S ∈ Syl3 (NG (B)). In the following lemma, Ja (S) is the subgroup of S generated by all abelian subgroups of maximal order. Lemma 19.31. If (19M) holds with p = 3, then the following are true: (a) A = AutG (B) ≤ GO4+ (3); (b) CG (B) = O3 (CG (B))T where T is a homocyclic abelian 3-group of rank 4 with B = Ω1 (T ); (c) S ∈ Syl3 (G) and B = Ω1 (Ja (S)); and (d) S/T is elementary abelian of order 3 or 9 and does not act quadratically on B; and for all s ∈ S − T , |[B, s]| = 32 ; and (e) W ≡B W (D4 ), W (C4 ), or W (F4 ). Proof. We have Q8 ∗ Q8 ∼ = O2 (W )  A. Thus [III17 , 1.27] applies and yields (a), (d), and (e). As NG (B) ≤ M and M/O3 (M ) embeds in Aut(G0 ) ∼ = Aut(D4 (q)), but contains no field automorphisms of order 3, (b) follows by [III17 , 2.35]. We argue that T = Ja (S), which immediately implies (c). Suppose that T = Ja (S) and let T0 ≤ S be abelian with |T0 | ≥ |T |. By [III8 , 1.1] we may choose T0 to act quadratically on T . Hence by (d), in the quotient S = S/T , |T 0 | = 3 and |[T, T 0 ]| > 3. As T0 is abelian, this implies that |T0 | < |T |, again a contradiction.  Thus, T = Ja (S) and the lemma is proved. Lemma 19.32. If (19M) holds with p = 3, then L = L3 (CG (y)). Proof. Suppose J is a 3-component of CG (L). As CB (L) = D, we have B ∩ J ≤ D. Suppose first that y ∈ Z(J).

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19. ΓD,1 (G) NORMALIZES G0

289

Then S ∩ J is non-abelian. Suppose that O 2 (W ) ≤ W (D4 ). Then S = T w with w ∈ L. Let w0 ∈ S ∩ J with w0 ∈ T . We may assume that w0 ∈ T w. So, w0 w−1 ∈ T . But [T ∩ L, w−1 ] = [T ∩ L, w0 w−1 ] = 1, a contradiction. Hence, O 2 (W ) ∼ = O 2 (W (F4 )). As NG (B) ≤ M , there exists v ∈ M of order 3 with CG0 (v) ∼ = G2 (q). We may assume that v ∈ S and CS (v) ∈ Syl3 (CG0 v (v)). In particular, CS (v) ∩ G0 ∈ Syl3 (CG0 (v)). As y ∈ Z(J) ∩ Z(L) and S ≤ (S ∩ JL)T , we have y ∈ Z(S) and so y ∈ Z(CS (v) ∩ G0 ). So E(CG0 (v, y )) =: J0 ∼ = SL3 (q). As δ E(CG0 (y)) = L, J0 = L and v ∈ CS (L). Suppose that J ∼ = SL3 (q0 ) for some q0 and δ with q0 ≡ δ (mod 3). Let T1 ≤ CT (L) of exponent exp(T ) and with x = Ω1 (T1 ). Then T1 acts faithfully on J and so (q0 − δ)3 ≥ (q − )3 . Hence, CT (L) ≤ J. Then CS (L) = CT (L) v ≤ J, since S ∩ J is non-abelian. But then, as x ∈ J − Z(J) and v ∈ J − Z(J), v ∈ xJ and L3 (CG (v)) ∼ =K∼ = L4 (q). But CG0 (v) ≤ L3 (CG (v)) with CG0 (v) ∼ = G2 (q), a contradiction by [III17 , 11.27]. Now, as m3 (J) = 2 and y ∈ Z(J), we see by [III17 , 11.2] that J ∼ = 3A6 , 3A7 or 3M22 . As T1 acts faithfully on J, T1 = x and T = B. Now, L b2 / y ∼ = P GL3 (q) and b2 induces an 2 inner automorphism on J centralizing x. As O (CJ (x)) = x, y , we conclude that b := b2 d−1 ∈ CG (J) for some d ∈ D. Then L b × v ≤ CG0 v (v, y ) = L × v , a contradiction proving that y ∈ Z(J). Therefore B ∩ J = d for some d ∈ D − y . If w ∈ W is of order 3, then [B, w] is 2-dimensional by Lemma 19.31d. So, S ∩ J = T ∩ J is cyclic. Also, d  S and so d, y = D = Z(S), whence AutG (B) ∼ = W (D4 ) or W (C4 ). Also, there exists r ∈ J inverting T ∩ J. Suppose that b2 induces an inner automorphism on J. Then we may assume b2 ∈ CB (J) or b2 = dc with c ∈ CS (JB) of order 3, whence c ∈ CB (J). In either case, B = d × CB (J) and r induces a reflection on B inverting d ∈ D − y . It follows that W ∼ = W (C4 ). Suppose that d = x . Then r centralizes b2 and so b2 ∈ CB (J). But then J ≤ CG (b2 ) ≤ M , a contradiction. If d = x , then using Lemma 19.23 and replacing b2 by any W -conjugate of d, we reach a similar contradiction. Hence, b2 induces an outer automorphism on J. As m3 (J) = 1, b2 acts as a field automorphism on J, by [III17 , 11.1]. Then [S ∩ J, b2 ] contains d , so an element of S ∩ J induces a transvection on B, contradicting Lemma 19.31d and completing the proof of the lemma.  Lemma 19.33. If (19M) holds with p = 3, then ΓD,1 (G) ≤ M . Proof. By Lemmas 19.4 and 19.23, it suffices to prove that NG (y ) ≤ M , or equivalently, by Lemma 19.32, NG (L) ≤ M . Since B induces inner-diagonal automorphisms on L, we have B ≤ CG (L)LB  NG (L), and hence by Lemma 19.29, it suffices to show that CG (L)LB ≤ M , or equivalently CG (L) ≤ M . Note that any hyperplane F of B containing B ∩ L contains one of the elements −1 −1 b1 , b2 , b−1 1 b2 b3 b4 , or b1 b2 b3 b4 . All these elements are G0 -conjugate to x, xy, or −1 x y, so their centralizers lie in M by Lemma 19.23. Therefore, CG (F ) ≤ M , and so O3 (CG (L)) ≤ M . Let R = S ∩ O3 3 (CG (L)), so that CG (L) = O3 (CG (L))NCG (L) (R) by Lemma 19.32. Let T1 = CT (L), a homocyclic abelian 3-group of rank 2 containing D. Then CS (L) = T1 v where either v = 1 or v has order 3, inducing a graph automorphism on G0 with CG0 (v) ∼ = G2 (q), and normalizing T1 with CT1 (v) = y . If possible we choose v ∈ R.

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290

15. THEOREM C∗ 7 : STAGE 4B+. A LARGE LIE-TYPE SUBGROUP, p > 2

Since m3 (CG (L)) ≥ m3 (T1 ) = 2, also m3 (R) ≥ 2. Let U ≤ R with U  S and U ∼ = E32 . If we can choose U = D, then an element of U − D induces a transvection on B, contradicting Lemma 19.31d. Thus U = D. If Ω1 (R) ≤ T1 , then D = Ω1 (R) and NG (R) ≤ NG (D) ≤ M by Lemma 19.4, as desired. So assume that Ω1 (R) ≤ T1 . Then R = (R ∩ T1 ) v with v of order 3. Every h ∈ I3 (R) − D induces a graph automorphism of order 3 on G0 with [h, L] = 1, hence with E(CG0 (h)) ∼ = G2 (q). Consequently such an h is not G-conjugate to any   u ∈ D − y ; if it were, G2 (q) would embed in O 3 (CG (u)(∞) ) ∼ = L4q (q), which is impossible by [III17 , 11.27]. Therefore the set D − y is strongly closed in R with respect to G. Hence, NG (R) ≤ NG (D − y ) = NG (D), and the proof is complete.  Lemmas 19.9, 19.14, 19.15, 19.18, 19.21, 19.26, and 19.33 complete the proof of Proposition 19.1. In view of [III14 , Proposition 14.1] and Proposition 12.1, and (19A), the proof of Theorem C∗7 : Stage 4b+ is now complete.

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10.1090/surv/040.8/05

CHAPTER 16

Theorem C∗7 : Stage 5+. G = G0 1. Introduction and Generalities The proof of Theorem C∗7 is completed in this chapter. Suppose that G satisfies the assumptions of Theorem C∗7 . By Stages 3b, 4a, and 4b+, we may fix a prime p ∈ γ(G) and a pair (x, K) ∈ J∗p (G) such that one of the following holds:

(1A)

 (1) p = 2, K ∼ = = An for some n ≥ 9 and D = O 2 (CG (K)) = x, y ∼ E22 for some involution y. Furthermore, Dg is a root foursubgroup of K for some g ∈ G, G0 = K, K g ∼ = An+4 ∈ K(7)∗ , CG (G0 ) has odd order, and ΓD,1 (G) ≤ NG (G0 ); or (2) K ∈ Chev(r) for some prime r = p such that 2 ∈ {p, r}, mp (CG (K)) = 1, and there is an acceptable subterminal (x, K)pair (y, L) such that G0 := N(x, K, y, L) ∈ Chev(r), CG (G0 ) is a p -group, G0 /Z(G0 ) ∈ K(7)+ , and ΓD,1 (G) ≤ NG (G0 ).

For uniformity of notation in this chapter, we have changed the name of the subgroup E of Theorem C∗7 : Stage 4a to D, in (1A1). In (1A2), recall that by definition of neighborhood,  G 0 = Lu | u ∈ D # , 

where for each u ∈ D# , Lu is the subnormal closure of O p (L) in CG (u). With this notation, we shall prove: Theorem 1.1 (Theorem C∗7 : Stage 5+). G = G0 ∈ K(7)∗ . It suffices to prove that G = G0 . For then G ∈ K(7)+ by (1A). If G ∈ K(7)∗ , then either p = 2 with G ∼ = L4 (q), q ≡  (mod 8), or p = 5 with G ∼ = U5 (4), − or p = 3 with G ∼ = E6 (2). In the first case, Lo2 (G) consists of SL2 (q) and L2 (q 2 ) [IA , 4.5.1], so Lop (G)∩Gp = ∅, contrary to the assumptions of Theorem C∗7 . Likewise in the third case, Lo3 (G) = {U6 (2), D4 (2)} [IA , 4.7.3A] and so again Lop (G)∩Gp = ∅, contradiction. Finally in the second case, m5 (G) = 3 so ILop (G) = ∅, contrary to the assumptions of Theorem C∗7 . Accordingly, for the remainder of this chapter, we assume that (1B)

G0 < G,

an assumption that will eventually lead to a contradiction. We set (1C)

M = NG (G0 ),

fix this notation, and first prove: 291 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

16. THEOREM C∗ 7 : STAGE 5+. G = G0

292

Lemma 1.2. The following conditions hold: (a) M is a maximal subgroup of G;  (b) G0 = O p (E(M )); and (c) CM (G0 ) = CG (G0 ) = Op (M ). Proof. We have M < G, for otherwise G = G0 by simplicity, contrary to (1B). Let M ∗ be a maximal subgroup of G containing M . We have DOp (M ∗ ) ≤ ΓD,1 (G) ≤ NG (G0 ). Hence [G0 , Op (M ∗ )] ≤ Op (G0 ) ≤ Z(G0 ), so [G0 , Op (M ∗ ), G0 ] = 1. As G0 is perfect by (1A), the Three Subgroups Lemma gives [G0 , Op (M ∗ )] = 1.

(1D)

Since K is a component of CG (x), it is a component of CM ∗ (x), and Lp -balance yields that K ≤ Lp (M ∗ ). By (1A), G0 is quasisimple and hence G0 = K G0 ≤ Lp (M ∗ ). Then (1D) implies that G0 ≤ E(M ∗ ). Since K is a component of CG (x) and K/Z(K) is a proper section of G0 /Z(G0 ), the normal closure of K in E(M ∗ ) is not the product of p covering groups of K cycled by x. Thus by Lp -balance, K lies  in a single NG (K)-invariant component G1 of E(M ∗ ), whence G0 = K G0 ≤ G1 . But now as ΓD,1 (G) ≤ NG (G0 ), and mp (Aut(K)) ≥ 3 if p > 2 (Theorem C∗7 : Stage 3a), it is immediate from [III17 , 9.17, 9.16] that G0 = G1 . As CG (G0 ) is a p -group

(1E) 

by (1A), G0 = O p (E(M ∗ ))  M ∗ , so M ∗ = M , proving (a) and (b). Finally (1D) and (1E) imply (c), and the lemma is proved.  We shall make use of the set

(1F)

  U(G, M, p) := A ≤ M  A is a p-group and

 for all g ∈ G, Ag ≤ M ⇐⇒ g ∈ M .

When p = 2, our strategy will be to apply [III8 , Corollary 6.2], a corollary of Theorem ZD of Aschbacher [II2 , p. 21]. Thus we shall aim to find an involution z ∈ M such that (1G)

z ∈ U(G, M, 2) and zz1 ∈ U(G, M, 2) for all z1 ∈ z M ∩ CG (z) − {z}.

By [III8 , Corollary 6.2], the existence of such an involution z will imply that M is solvable or M = G. As G0 ≤ M , M = G, contrary to our assumption that M < G. When p is odd, we have to work harder, aiming to show that M is almost strongly p-embedded in G, which will contradict the assumption of Theorem C∗7 that p ∈ γ(G). Our condition ΓD,1 (G) ≤ M is quickly strengthened as follows. Lemma 1.3. Suppose that G0 ∈ Chev, i.e., (1A2) holds. Then D ∈ U(G, M, p), and any E ∈ Ep2 (G) such that ΓE,1 (G) ≤ M lies in U(G, M, p). Proof. Suppose that E ≤ M g . By [III17 , 9.15], since K   CG0 (x) with K ∈ Gp , we have Gg0 = ΓE,1 (Gg0 ) ≤ M . But by Lemma 1.2, M/G0 is an extension of the p -group CM (G0 )/Z(G0 ) by a subgroup of Out(G0 ), which is solvable. Thus  Gg0 ≤ O p (M (∞) ) = G0 , whence g ∈ NG (G0 ) = M . The lemma follows. 

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2. THE ALTERNATING CASE

293

Numerous elementary properties of groups in U(G, M, p) (see [III8 , 6.1]) will be useful in obtaining fusion-theoretic and transfer-theoretic information about M . 2. The Alternating Case In this section we consider case (1A1), proving: Proposition 2.1. If G0 ∼ = Am for some m ≥ 13, then G = G0 . We suppose false and consider M = NG (G0 ), and first prove Lemma 2.2. The following conditions hold: (a) M = G0 × O2 (M ); (b) x ∈ U(G, M, 2); (c) M contains a Sylow 2-subgroup of G; and (d) M controls the G-fusion of its involutions. Proof. By (1A1), D ∈ Syl2 (CG (K)) so D ∈ Syl2 (CM (K)). By Lemma 1.2c, M/O2 (M ) embeds in Aut(G0 ) ∼ = Σm . But some transposition in Aut(G0 ) centralizes K, so |CM (K)|2 = |D||M/G0 O2 (M )|2 and then M = G0 O2 (M ). As O2 (G0 ) = 1, (a) holds. Since CG (x) ≤ ΓD,1 (G) ≤ M , to prove (b) it is enough to show that xG ∩ M = M x . Suppose false and let g ∈ G with z := xg ∈ M − xM . Then z ∈ G0 , so z is the product of 2k disjoint transpositions, k > 1. Then CG0 (z) contains a subgroup X = AΣ where A ∼ = E22k−1 is the trace-0 submodule of the natural permutation module for Σ ∼ = Σ2k . Hence z is the unique minimal normal subgroup of X. Let −1   Y = X g . Then Y = O 2 (Y ) ≤ O 2 (CG (x)) ≤ CG0 (x) and x is the unique minimal normal subgroup of Y . As x ∈ K, Y ∩ K = 1 and so Y embeds in CG0 (x)/K ∼ = D8 , contradicting the fact that k > 1. Thus (b) holds. By (a), (b) and [III8 , Lemma 6.1c], G0 contains a Sylow 2-subgroup S of G, so (c) holds. Finally it is clear that every involution of M centralizes an M -conjugate  of x, so (d) follows from (b) and [III8 , Lemma 6.1d]. The proof is complete. Lemma 2.3. Let x1 and x2 be unequal commuting elements of xM . Set t = x1 x2 . Then t ∈ U(G, M, 2). Proof. It suffices in view of Lemma 2.2d to show that CG (t) ≤ M . Clearly x1 and x2 are root involutions in M . Therefore either t is a root involution in M , in which case CG (t) ≤ M by Lemma 2.2b, or t has support of size 8 in the natural representation of G0 ∼ = Am on m letters. We may thus assume that the latter case holds. Set C = CG (t). Then CM (t) = O2 (M ) × (R × J) u , where J ∼ = Am−8 , Σ is isomorphic to the centralizer of a 2-central involution J u ∼ = 21+4 = Σm−8 , R ∼ 3 + in the A8 subgroup of G0 with the same support as t, and R u is isomorphic to the corresponding centralizer in Σ8 . Moreover, by [III17 , 16.4], O2 (R) contains a four-subgroup V all of whose involutions are root involutions and hence conjugate to x. Then J is a component (recall that m ≥ 13) of CM (v, t ) for each v ∈ V # ; moreover, as CG (v) ≤ M , J is a component of CC (v) for each such v. It follows by [III8 , 2.8] and [III11 , 1.16ab] that the subnormal closure Jt of J in C is a component with Jt /Z(Jt ) ∼ = Ar for some r ≡ m (mod 4), or J2 or M12 , with m = 13 in the last two cases. In all cases, if we take any root involution x ∈ J, we

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

294

have Jt = J, CG (x ) ≤ M , by [III11 , 17.8] and Lemma 2.2b. But x ∈ Jt   C,  so C ≤ M by [III8 , Lemma 6.1a]. The proof is complete. Now Lemmas 2.2b and 2.3 show that (1G) holds with x playing the role of z there. As discussed following (1G), this contradicts our assumption that M < G, thereby completing the proof of Proposition 2.1. 3. The Lie Type Case, p = 2: Part 1 In this section we begin the treatment of case (1A2) with the eventual goal of proving: Proposition 3.1. If G0 ∈ Chev(r) and p = 2, then G0 = G. As in the preceding two sections, we suppose that the proposition fails and consider M = NG (G0 ). Throughout this section, we shall assume (3A) G0 ∈ Chev(r) ∩ K(7)+ but G0 /Z(G0 ) ∼  P Ωn (q) for any n ≥ 6 and odd q. = Since K(7)+ , in odd characteristic, already excludes all groups of twisted rank 3 ∼ 1, as well as A± 2 (q), B2 (q) = Ω5 (q), D4 (q), and G2 (q), (3A) implies ∼ A± (q) (n ≥ 4), Cn (q) (q ≥ 3), F4 (q), E ± (q), G0 = n 6 (3B) E7 (q), E8 (q), q = r a , all with |Z(G0 )| odd. By (1A2) and Lemma 1.3, for D = x, y , we have (3C)

ΓD,1 (G) ≤ M and D ∈ U(G, M, 2).

By [III8 , Lemma 6.1h], it follows that D ≤ [M, M ].

(3D)

We focus on involutions c0 ∈ G0 that are classical involutions in the sense of Aschbacher [A9]. For the groups G0 in (3B), these are described as follows. Lemma 3.2. If G0 is as in (3B) with level q, then the involutions lying in some long root SL2 (q)-subgroup of G0 are all G0 -conjugate, and they are the only involutions c ∈ G0 for which there is a subgroup J0 (c) ∼ = SL2 (q) such that c ∈ J0 (c)  CG0 (c). Moreover, for any such c, J0 (c) is the unique subgroup of CG0 (c) with these properties. Proof. This is [III17 , 12.3].



We fix an involution c0 in a long root SL2 (q) subgroup of G0 , and define

(3E)

0 C = cG 0 ,     B = cc  c, c ∈ C, cc = c c ∈ cG 0 ∪ {1} , and 

J(c) = O 2 (J0 (c)) for all c ∈ C. We tabulate the following facts: Proposition 3.3. One of the following conclusions holds, possibly after interchanging y and yx: (a) G0 /Z(G0 ) ∼ = Ln (q), n ≥ 5, x acts on G0 like a reflection, and xy ∈ C; (b) G0 /Z(G0 ) ∼ = Ln (q), n ≥ 6, and D# ⊆ C;

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3. THE LIE TYPE CASE, p = 2: PART 1

295

∼ P Sp2n (q), n ≥ 3, x, y ∈ C, and xy ∈ B if B = ∅ while otherwise (c) G0 = xy ∈ C; (d) G0 ∼ = F4 (q) or E6 (q), all involutions of D are 2-central and lie in B, and NG0 (D) contains a Sylow 2-subgroup of G0 ; (e) G0 ∼ = E7 (q), D ∩ C = ∅ and B = ∅; or (f) G0 ∼ = E8 (q), D ∩ C = ∅ = D ∩ B, and D# ⊆ C ∪ B. In all cases, C ∩ CG (D) = ∅, and if B = ∅, then B ∩ CG (D) = ∅. Proof. If G0 /Z(G0 ) ∼ = Ln (q), then n ≥ 5 by (3B). Moreover, any involution c ∈ M such that E(CG0 (c)) has an SLn−2 (q) component belongs to C. This, together with the information in [III14 , Table 14.1], yields (a) or (b). A maximal diagonalizable elementary abelian 2-subgroup of G0 D contains elements of C and B as well as a conjugate of D, proving the final assertion of the lemma in this case. If G0 ∼ = P Sp2n (q), then for commuting c = c ∈ C, CG0 (c) = J0 (c) ∗ I with  c ∈ I centralizing the commutator of c on the natural module, so CG0 (cc ) has an Sp4 (q) component. Thus c, c is conjugate to D. If n = 3 it follows that cc ∈ C so B = ∅. If n > 3 then it follows that cc ∈ B, and (c) holds. The final assertions of the lemma follow easily in this case as well. If G0 ∼ = F4 (q) or E6 (q), then from the information in [III14 , Table 14.1], D# consists of 2-central involutions of G0 (none of which lies in C), and x is the unique involution centralizing K ∼ = Spinm (q), m = 9 or 10. As x is 2-central and C ⊆ G0 , there exists c ∈ C ∩ CG0 (x). By [IA , 6.2.1d], cx ∈ cK so x = c · cx ∈ B. Now L ≤ CG (D) with L ∼ = Spin8 (q) by [III14 , Table 14.1], so y maps to a 2-central involution of CG0 (x)/ x by [III17 , 8.9c]. Hence NG0 (D) contains a Sylow 2-subgroup of G0 . By the Thompson Transfer Lemma, every class of involutions of G0 meets CG0 (D). Hence the lemma holds in these cases. If G0 ∼ = E7 (q), then from [III14 , Table 14.1] and [IA , Tables 4.5.1, 4.5.2], D contains representatives of all the conjugacy classes of involutions of G0 . Moreover,  replacing y by xy if necessary, y ∈ C and then O r (CG0 (y)) = J(y)Ly with Ly ∼ = hs D6 (q) and Ly = CCG0 (y) (J(y)). Let c ∈ C with [c, y] = 1 and c = y. Then either [c, J(y)] = 1 or c ∈ Ly , and in either case cy ∈ cCG0 (y) . So c · cy = y ∈ C, proving that B = ∅. Finally suppose that G0 ∼ = D8 (q)hs and x is 2-central in G0 . = E8 (q). Then K ∼ As in the F4 (q)-case this quickly yields that for c ∈ C ∩ K, cx ∈ cG0 , so x ∈ B. As G0 has only two classes of involutions, the remaining statements of the lemma follow easily.  Using this, we can establish the following lemma. Lemma 3.4. For all c ∈ C, J(c)  CG (c) ≤ M . Proof. Let c ∈ C and C = CG (c). Note that E(M ) = G0 and CM (E(M )) = O2 (M ), by Lemma 1.2. Now either J(c) = O2 (Lo2 (CG0 (c))) or J(c) is the unique component of E(CG0 (c)) of 2-rank 1. In either case, as G0  M , J(c)  CM (c). 0 Thus it will suffice to prove that C ≤ M . Since C = cG 0 , it is enough to show that CG (c) ≤ M for some c ∈ C. If C ∩ D = ∅, then we are done by (3C). Thus, by Proposition 3.3, we may assume that G0 ∼ = F4 (q) or E6 (q), in which case D  T for some T ∈ Syl2 (G0 ). Replacing c by a G0 -conjugate and using Thompson Transfer, we may assume that c ∈ CT (D) and CT (c) ∈ Syl2 (CG0 (c)).

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

296

Let X = O2 (C). Then X ≤ ΓD,1 (G) ≤ M , whence [X, E(CM (c))] = 1. Let I = J(c) be a component of CM (c). Thus by [IA , 4.5.1], I ∼ = Sp6 (q) or an  2 even covering group of L6 (q), and I = O (CG0 (J0 (c))). Now since D  CT (c) and c ∈ D, we have [D, T ∩ J0 (c)] = 1, whence [D, J0 (c)] = 1 by [III11 , 1.17]. Therefore D ≤ I. Thus J0 (c)  CG0 (D c ) = CCG0 (D) (c). But L = CG0 (D)(∞) if q > 3 and O 2 (CG0 (D)) if q = 3, so J0 (c) ≤ L. Thus c is a classical involution  in L ∼ = D4 (q), whence H := O r (CL (c)) is the central product of four copies of SL2 (q), and Z(H) = D c . This determines the embedding of D in I; indeed by [III17 , 9.4],  I = E0 (CI (d)) | d ∈ D# and II (D; r) = 1, where E0 (CI (d)) is the Sp4 (q) or SL4 (q) component of CI (d). It follows by L2 balance that I ≤ L2 (C) and, since [X, I] = 1, the normal closure I ∗ of I in L2 (C) is a product of components. Since E0 (CI (d)) ∩ E0 (CI (d )) contains SL2 (q) for all d = d ∈ D# , L2 -balance actually implies that I ∗ is a single component. Then CI (x)   CCM (c) (x) = CCG (c) (x) so CI ∗ (x), like CI (x), has an Sp4 (q) or SL4 (q) component. Thus I ∗ is unambiguously in Chev(r) by [III17 , 10.45]. Since  II (D; r) = {1} it follows from [IA , 7.3.1] that I ∗ = CI ∗ (d) | d ∈ D# ≤ M , whence I ∗ = I   C. But D ≤ I ≤ M with D ∈ U(G, M, 2), by Lemma 1.3. Thus C ≤ M  by [III8 , Lemma 6.1a], completing the proof. Now we can prove Lemma 3.5. Let b ∈ C and g ∈ G, and suppose that J(b) ≤ M g . Then g ∈ M . −1

−1

−1

Proof. Let b1 = bg . Then b1 ∈ J(b)g  CG (b1 ) and J(b)g ≤ M . Since E(M ) = G0 and CM (E(M )) = O2 (M ) it follows by [III17 , 12.4] that b1 ∈ C. −1 Therefore bg = b1 = bm for some m ∈ G0 ≤ M . It follows that mg ∈ CG (b), so mg ∈ M by Lemma 3.4. Hence g ∈ M as required.  Since J(c)  CG (c) for all c ∈ C we can extend the definition to all b ∈ cG 0 by defining J(cg0 ) = J(c0 )g for all g ∈ G.  Lemma 3.6. Suppose that c and c are distinct elements of cG 0 with [c, c ] = 1.  G   / c0 or there exist R ∈ Syl2 (J(c)) and R ∈ Syl2 (J(c )) such Suppose that either cc ∈ that [R, R ] = 1. Then [J(c), J(c )] = 1.

Proof. Since J(c )  CG (c ), c normalizes J(c ). Let R be a c-invariant Sylow 2-subgroup of J(c ). If [c, R ] = 1 then by [III11 , 1.17], there is g ∈ R such that cc = cg ∈ cG 0 . Hence our hypothesis is contradicted in this case. Therefore [c, R ] = 1. Again by [III11 , 1.17], [c, J(c )] = 1. Thus J(c ) normalizes J(c). By a symmetric argument we may assume that J(c) normalizes J(c ). As c and c are disjoint and are the unique minimal normal subgroups of J(c) and J(c ) respectively, we have [J(c), J(c )] ≤ J(c) ∩ J(c ) = 1. The proof is complete.  Now we can prove Lemma 3.7. Let a ∈ B. Then CG (a) ≤ M . Proof. We may write a = cc with c, c ∈ C and a ∈ cG . By Lemma 3.6, [J(c), J(c )] = 1. By [III17 , 12.6] there is a single G0 -conjugacy class of such pairs (c, c ), so B is a single conjugacy class. Thus if B ∩ D = ∅, then a is G0 -conjugate

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to an element of D, and as ΓD,1 (G) ≤ M , we are done. We may therefore assume that B ∩ D = ∅, but of course that B = ∅. By Proposition 3.3, we may assume that G0 ∼ = Ln (q) for some n ≥ 5, with c ∈ D, but a ∈ D. Then a has four eigenvalues equal to −1 on the natural SLn (q)-module. Thus if n = 5, a acts like a reflection on G0 , so a is again G0 conjugate to an element of D and we are done. So we may assume that n ≥ 6. Let H be the component of CG0 (a) with H ∼ = SL4 (q) and c ∈ H. Note that there 2 # G0 exists B ∈ E2 (H) with B ⊆ c . Then by Lemma 3.4, ΓB,1 (G) ≤ M , and so B ∈ U(G, M, 2) by Lemma 1.3. Furthermore, as n ≥ 6 there exists c1 ∈ cM such that [c1 , H] = 1. Then H is a component of CG (c1 , a), so its subnormal closure H ∗ in CG (a) lies in L2 (CG (a)) and is a product of 2-components J such that J/O2 (J) is unambiguously in Chev(odd), by [III17 , 10.45]. Hence by [IA , 7.3.4], H ∗ ≤ ΓB,1 (G) ≤ M . Thus H   CG (a). As B ≤ H, CG (a) ≤ M by [III8 , Lemma 6.1a].  We next prove Lemma 3.8. If c ∈ C and b = cg ∈ M , then g ∈ M . Proof. Suppose false and choose c ∈ C and g ∈ G − M with b = cg ∈ M . Let Q ∈ Syl2 (CM (b)) and Q ≤ S ∈ Syl2 (M ). Observe that for any m ∈ M , we may replace c by cm and g by m−1 g to get another counterexample to the lemma, with b unchanged. Making such a change if necessary, we may assume that R := J(c) ∩ S ∈ Syl2 (J(c)). Let R = R1 , R2 , . . . , Rn be the distinct G-conjugates of R lying in S, so that by [IA , Theorem 4.10.6b], [Ri , Rj ] = 1 for all 1 ≤ i < j ≤ n. Given the possible isomorphism types of G0 it is clear that n > 1. Notice that Q permutes {R1 , . . . , Rn }. Write Z(Ri ) = ci , so that c = c1 . Let U = c1 , c2 ∼ = E22 . Then CG (u) ≤ M for all u ∈ U # by Lemmas 3.4 and 3.7. Suppose first that b normalizes Ri for each i = 1, . . . , n, so b centralizes U . Thus U normalizes J(b). As ΓU,1 (G) ≤ M , U ∈ U(G, M, 2) by Lemma 1.3. Applying [III8 , Lemma 6.1f] with X = b and A1 = A = U we conclude that g ∈ M , as desired. Now consider the case that for some i = 1, . . . , n, b does not normalize Ri . We may then assume notation to have been chosen so that R1b = R2 . Consequently J(c1 )b = J(c2 ). Since R1 R2 = R1 × R2 , J(c1 )J(c2 ) = J(c1 ) × J(c2 ) and X := CJ(c1 )J(c2 ) (b) ∼ = J(c) with Z(X) = c1 c2 . But X normalizes J(b) and so by [III17 , 11.6], [c1 c2 , J(b)] = 1. Thus J(b) ≤ CG (c1 c2 ) ≤ M by Lemma 3.7. Then g ∈ M by Lemma 3.5, completing the proof of the lemma.  By Lemmas 3.4 and 3.8, c ∈ U(G, M, 2) for all c ∈ C. Finally we prove Lemma 3.9. Let a ∈ B. Then a ∈ U(G, M, 2). Proof. It suffices by Lemma 3.7 to show that aG ∩ M = aM . Let I be the set of all involutions of M centralizing some element of C. By [III8 , Lemma 6.1f], aG ∩ M ∩ I ⊆ aM so it suffices to show that aG ∩ M − I ⊆ aM . We will in fact show that aG ∩ M − I is empty. Let a0 ∈ aG ∩ M − I. Then by [III17 , 12.7] and (3B), a0 induces an outer automorphism on G0 ∼ = P Sp2n (q), with

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298

16. THEOREM C∗ 7 : STAGE 5+. G = G0

 ∼ P Spn (q 2 ). Moreover, there exists M0 ≤ M with |M : M0 | = 2 and O r (CG0 (a0 )) = M = M0 a0 , and there exists a1 ∈ I with M = M0 a1 . In particular I2 (M0 ) ⊆ I. G M By [III8 , Lemma 6.1f] again, it follows that aG 1 ∩ M0 = a1 ∩ M0 ∩ I ⊆ a1 ∩ M0 = ∅. Thus by the Thompson Transfer Lemma, a1 ∈ [G, G], contradicting the simplicity of G and completing the proof. 

Continuing the argument, we have seen that c ∈ U(G, M, 2). Then Lemma 3.9 implies that whenever c1 = c2 ∈ cM with [c1 , c2 ] = 1 and a := c1 c2 , then a ∈ U(G, M, 2). Hence, as M < G, the hypotheses of [III8 , Corollary 6.2] are satisfied, so G has a solvable strongly embedded subgroup N , which is absurd as then G0 = ΓD,1 (G0 ) ≤ N . This contradiction completes the proof of Proposition 3.1 in the case that (3A) holds. 4. The Lie Type Case, p = 2: Part 2 In this section we complete the proof of Proposition 3.1. Thus we assume that (3A) fails, so that (4A) p = 2, G0 /Z(G0 ) ∼ = P Ω± n (q), q odd, n ≥ 6, and we argue that G0 = G. As in the preceding section, we suppose that G0 < G and consider M = NG (G0 ). We first observe: Lemma 4.1. The following conditions hold: (a) Z(G0 ) = 1; and (b) If q = 3, then n ≥ 8. Proof. Suppose that q = 3 with n = 6 or 7. Then using [IA , 4.5.1] we see that E(CG0 (x)) has no component in G2 . However, CG (x) ≤ M and the component K of E(CG (x)) = E(CG0 (x)) does lie in G2 , contradiction. This proves (b). Finally by Lemma 1.2c, Z(G0 ) has odd order. Hence by [IA , 6.1.4], if Z(G0 ) = 1, then by G0 /Z(G0 ) ∼ = P Ω− 6 (3) or Ω7 (3) (and Z(G0 ) is a 3-group). This contradicts (b), so (a) holds, completing the proof.  Now we treat the special case when G0 ∼ = P Ω6 (q). Lemma 4.2. G0 is not isomorphic to P Ω6 (q), q odd; that is, n ≥ 7. Proof. Suppose on the contrary that G0 ∼ = P Ω6 (q) ∼ = L4 (q). Then by  ∼ [III14 , Table 14.1], D = x, y with K = SL3 (q) and y ∈ K with E(CG (y)) ∼ = SL2 (q) ∗ SL2 (q). Note that q > 3 by Lemma 4.1b. We shall apply [III8 , Corollary 6.2] to the class xG . Since ΓD,1 (G) ≤ M , we have CG (x) ≤ M and CG (y) ≤ M . Moreover, if x ∈ xM ∩ CG (x) − {x}, then x and x are commuting reflections (considering G0 as L4 (q)), whence xx ∈ y M and so CG (xx ) ≤ M . We shall argue that M controls G-fusion of its 2-elements. For that, since y is 2-central in G0 , it suffices to show that y G ∩ M = y M , as then y ∈ U(G, M, 2) and the desired control of fusion follows from [III8 , Lemma 6.1g]. Suppose then that w ∈ y G ∩ M . Then CG0 (w)(∞) = E(CG0 (w)) is isomorphic to a subgroup of CG (y)(∞) = E(CG (y)) ∼ = SL2 (q) ∗ SL2 (q). But if w ∈ y M , then by [IA , Table 4.5.1,  1/2 4.9.1], E(CG0 (w)) ∼ ), none = L3 (q), L2 (q 2 ), L2 (q) × L2 (q), or P Sp4 (q), or L± 4 (q of which is embeddable in E(CG (y)), by order and 2-rank considerations. Hence y G ∩ M = y M , as desired, and M controls 2-fusion in G.

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In particular x and y lie in U(G, M, 2). Now if x1 , x2 ∈ xM with [x1 , x2 ] = 1, then CG (x1 x2 ) ≤ M , as argued above. Thus [III8 , Corollary 6.2] implies that M is solvable, which is absurd. The proof is complete.  For the remainder of this section, we shall therefore assume (4B) G0 ∼ = P Ω (q), n ≥ 7,  = ±1, with q > 3 if n = 7. n

The case n = 8,  = + requires special care because of the existence of the triality automorphism of G0 . Hence in much of the succeeding argument, we assume (4C)

G0 ∼  P Ω+ = 8 (q),

noting explicitly when we do so. When (4C) holds, there is a unique natural module V for G0 , and the image P On (q) of On (q) = O(V ) in Aut(G0 ) is characteristic in Aut(G0 ). For any involution t ∈ M , we shall say that t is of type −1a 1b if and only if a + b = n, the image t of t in Aut(G0 ) lies in P On (q), and t is the image of an involution  t ∈ On (q) = O(V ) whose −1 eigenspace on V has dimension a. Using − t in place of  t, we see that t is of type −1a 1b if and only if it is of type −1b 1a . By a classical involution in G0 (as in (4A)), we mean – following Aschbacher – an involution t of type −14 1n−4 whose −1-eigenspace on V is of + type. Thus there is Jt such that t ∈ Jt   CG0 (t) and Jt ∼ = Ω+ 4 (q). We tabulate the following facts, which follow immediately from the description of N(x, K, y, L) in [III14 , Table 14.1]. Proposition 4.3. Assume (4C). Then D lifts to a four-subgroup of On (q). Moreover, after interchanging y and xy if necessary, one of the following conclusions holds: (a) n = 7, x is of type −11 16 , y is of type −13 14 , and xy is of type −12 15 ; (b) n = 8, x and y are of type −12 16 , and xy is of type −14 14 ; furthermore, some eigenspace of xy has + type, and if  = −1, then x and y are not Aut(K)-conjugate; (c) n > 8, n is odd, x and y are of type −11 1n−1 , and xy is of type −12 1n−2 ; (d) n > 8, n is even, and x, y, and xy are all of type −12 1n−2 . Using Lemma 1.3 and transfer considerations, we easily prove Lemma 4.4. If n is odd, then I2 (M ) ⊆ G0 . Proof. By Lemma 1.3 and [III8 , 6.1h], D ≤ [M, M ]. By [III17 , 7.2a], since n  is odd, Out(G0 ) is abelian. Therefore D ≤ O 2 ([M, M ]) ≤ G0 . As D has support of dimension at most 3 on the natural module by Proposition 4.3, any involution u ∈ I2 (M ) has an M -conjugate centralizing D, by [III17 , 6.12]. Thus we may  assume that u ∈ CM (D). But then u ∈ O 2 ([M, M ]) ≤ G0 by [III8 , Lemma 6.1h], completing the proof.  Our goal will be to verify the hypotheses of [II2 , Corollary ZD] with Z the G-conjugacy class of a suitable choice of involution z of type −11 1n−1 or −12 1n−2 . First consider the case z ∈ D# , so that CG (z) = CM (z), and consider a Gconjugate w = z g ∈ M . Our first aim is to show that g ∈ M . Lemma 4.5. Assume (4C). Let z ∈ D# be of type −11 1n−1 or −12 1n−2 . Let g ∈ G be such that w := z g ∈ M . Then wM ∩ CG (D) = ∅.

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Proof. First suppose that w lifts to an involution w  of On (q) = O(V ). Let  be a four-subgroup of O(V ) lifting D. Except in case (b) of Proposition 4.3, D  V ] ≤ 3. But w dim[D,  has an eigenspace on V of dimension at least 4, which  V ]a for some a ∈ M . Then [wa−1 , D] = 1 so the conclusion of therefore contains [D, the lemma holds in this case. A similar argument works if Proposition 4.3b holds, unless possibly the eigenspaces of w  are both of dimension 4. But in that case, by (q) and xy ∈ K ≤ G0 , and so w is conjugate to xy ∈ D, again (4C), G0 ∼ = Ω− 8 yielding the desired conclusion. Now suppose that w does not lift to an element of On (q). By [IA , 4.5.1, 4.9.1], one of the following holds: (a) w induces an involutory field or graph-field automorphism on G0 ; or (b) G0 ∼ = P Ω2m (q) and E(CG0 (w)) ∼ = Ωm (q 2 ); or  ∼ (c) G0 = P Ω2m (q) and w is a projective involution with E(CG0 (w)) ∼ = J/Ω1 (O2 (Z(J))), where J = SLηm (q), η = ±1. Moreover, if m is even, then  = +, while if m is odd, then η = . As w ∈ z G and CG (z) ≤ M , we see that CG (w) ∼ = CM (z). Let Kz = E(CG (z)) = E(CM (z)) = CM (z)(∞) , Kw = E(CG (w)) ∼ = Kz , and Iw = E(CM (w)) = E(CG0 (w)) = CM (w)(∞) . Then Iw is isomorphic to a subgroup of Kz . We shall argue that such a containment is impossible. As Kz ∼ = Ωn−1 (q) or Ωn−2 (q), Iw must have a faithful orthogonal Fq -representation of dimension n − 1. In case (a),  1 Iw ∼ = P Ωn (q 2 ). Then by [III17 , 5.7], Iw has no faithful representation of dimension less than n, a contradiction. In case (b), Iw ∼ = Ωn2 (q 2 ). In this case, [III17 , 5.8] rules out a representation of dimension less than n, since n2 ≥ 4. Hence, case (c) holds, and Iw is isomorphic to SLn (q), n ≥ 10, or (for n ≡ 0 (mod 4)), its quotient by 2 ±1 . Then by Lemma [III17 , 5.9] and (4C), Lw has no faithful representation in Ωn−1 (q), yielding a contradiction in this case as well. This completes the proof.  Lemma 4.6. Assume (4C). Then there exists z ∈ D# , with z of type −11 12k if n = 2k + 1 and with z of type −12 1n−2 if n is even. For any such z, z ∈ U(G, M, 2). Proof. Whether n is even or odd, the existence of z ∈ D# of the appropriate type follows from Proposition 4.3. Then CG (z) ≤ ΓD,1 (G) ≤ M so it remains to show that z G ∩M = z M . Let g ∈ G with w := z g ∈ M ; we prove g ∈ M . By Lemma 4.5 we may replace g by gm for suitable m ∈ M , and assume that [w, D] = 1. Then D ≤ CG (w) = CG (z)g ≤ M g . But D ∈ U(G, M, 2) by Lemma 1.3. Thus g ∈ M and the lemma follows.  Lemma 4.7. Suppose that n is odd. Let z be as in Lemma 4.6, let z1 ∈ z M ∩ CG (z) − {z}, and let u = zz1 . Then u ∈ U(G, M, 2). Proof. Since z is of type −11 1n−1 , u is of type −12 1n−2 . As I2 (M ) ⊆ G0 by Lemma 4.4, the conjugacy class of u in G0 is uniquely determined by its type. But D contains an element of the same type as u, and likewise D# ⊆ I2 (M ) ⊆ G0 , so D contains a conjugate of u. Since ΓD,1 (G) ≤ M it follows that CG (u) ≤ M . It remains to show that uG ∩ M = uM . Let g ∈ G with ug ∈ M . Then ug ∈ G0 by Lemma 4.4, and so ug centralizes an element of z M , by [III17 , 6.11], with u or −u in the role of D there. Therefore by [III8 , 6.1f], g ∈ M . The proof is complete. 

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Lemma 4.8. n is even. Proof. Suppose that n is odd. We let z be as in Lemma 4.6 and apply [III8 , Corollary 6.2] to the conjugacy class z G . Since z ∈ U(G, M, 2), Lemma 4.7 yields the hypothesis of that corollary. We conclude that M is solvable, which is absurd, and the proof is complete.  Lemma 4.9. G0 ∼ = P Ω± 8 (3). Proof. If G0 ∼ = P Ω± 8 (3), then since K is a component of CG0 (x) and K/O2 (K) ∈ G2 , the only possibility is K ∼ = Ω7 (3) (see [IA , 4.5.1]). But then  x acts on G0 like a reflection, which contradicts Proposition 4.3 as n = 8. The next lemma is valid without (4C), i.e., it applies even if G0 ∼ = D4 (q). Lemma 4.10. Suppose that t ∈ G0 is a classical involution. If CG (t) ≤ M , then M controls G-fusion of its involutions, and t ∈ U(G, M, 2). Proof. We choose any natural module V for G0 . By [III17 , 6.13], there exists a four-subgroup E ≤ G0 such that E # ⊆ tG0 , some four-group preimage of E in Ωn (q) has 6-dimensional support on the natural module, and every involution in η u ∈ M centralizes a G0 -conjugate of E, unless possibly E(CG0 (u)) ∼ = SLn/2 (q), in which case we call u exceptional. We argue that no exceptional involution u ∈ M lies in tG . For if it did, then η E(CG0 (u)) ∼ = SLn/2 (q) would embed in CG (t)(∞) = CM (t)(∞) = E(CM (t)) ∼ = SL2 (q) ∗ SL2 (q) ∗ Ωn−4 (q). But n ≥ 8 and so such an embedding would contradict Lagrange’s Theorem, using Zsigmondy’s Theorem [IG , 1.1] and a suitable prime divisor of q k ± 1, where k = n/2 or (n/2) − 1. Thus if g ∈ G and tg ∈ M , then tg centralizes a G0 -conjugate of E. Since CG (t) ≤ M , we have ΓE,1 (G) ≤ M and so E ∈ U(G, M, 2) by Lemma 1.3. Then tG ∩ M = tM by [III8 , Lemma 6.1d]. Consequently t ∈ U(G, M, 2). But by [III17 , 6.13d], every element of I2 (M ) centralizes an element of tM , so M controls G-fusion of its involutions by [III8 , 6.1d]. This completes the proof.  Lemma 4.11. G0 ∼ = Ω− 8 (q). Proof. Suppose G0 ∼ = Ω− 8 (q). Then by Proposition 4.3, D contains representatives of both conjugacy classes of involutions of Aut(K) of type −12 16 , as well as a classical involution. By Lemmas 4.6 and 4.10, d ∈ U(G, M, 2) for every d ∈ D# . Choose d ∈ D# of type −12 16 and with support of + type on the natural module. If u ∈ M , g ∈ G and u = ddg with [d, dg ] = 1, then g ∈ M and so u is either M conjugate to d or a classical involution. In either case u has an M -conjugate in D, so u ∈ U(G, M, 2). Hence by [III8 , Corollary 6.2], M is solvable, a contradiction completing the proof.  Lemma 4.12. Assume (4C). Let t ∈ I2 (G0 ) be of type −12 1n−2 . Then CG (t) ≤ M. Proof. By (4C) and Lemmas 4.8 and 4.11, n = 2k ≥ 10, and so by Proposition 4.3, every element of D# is of type −12 1n−2 . We may assume all d ∈ D# have support of the same type – namely q – as one another, for otherwise D contains representatives of all conjugacy classes of involutions of type −12 1n−2 , and the

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lemma holds as ΓD,1 (G) ≤ M . Thus we may assume that the support of t on the natural G0 -module is of −q type. −q  (q). As the support of D is 3-dimensional and Let I = E(CM (t)) ∼ = Ωn−2 the support of each d ∈ D# is of type q , I contains an M -conjugate of D, so we assume without loss that D ≤ I. Fix z ∈ D# and let I0 = E(CI (z)) ∼ = Ω− n−4 (q). As  CG (z) ≤ ΓD,1 (G) ≤ M , I0 ≤ L2 (L2 (CG (z))∩CG (t)) ≤ L2 (CG (t)) by L2 -balance. Let J be the subnormal closure of I0 in CG (t), which is also the subnormal closure of I in CG (t). Now as n ≥ 10, every component of J := J/O2 (J) is unambiguously in Chev(r) (see [IA , 2.2.10] and [III11 , 1.1b]), so by [IA , 7.3.4], J ≤ ΓD,1 (G) ≤ M . Thus J = I. As D ≤ J   CG (t), with D ∈ U(G, M, 2), CG (t) ≤ M by [III8 , 6.1a]. The lemma is proved.  Similarly we show Lemma 4.13. Assume (4C). Let t ∈ I2 (G0 ) be a classical involution. Then CG (t) ≤ M . Proof. As in the proof of Lemma 4.12, n ≥ 10, and CM (t) has a component I ∼ = Ωn−4 (q) containing an E22 -subgroup E all of whose involutions are of type −12 . Thus by Lemma 4.12, ΓE,1 (G) ≤ M , so E ∈ U(G, M, 2). There also exists z ∈ I2 (G0 ) of type −12 1n−2 with support contained in the support of t (on the natural module). Hence CG (z) ≤ M with I ≤ L2 (CG (z)), so I ≤ L2 (CG (t)) by L2 -balance. Let J be the subnormal closure of I in CG (t), so that I is a component of CJ (z). Then every component of J/O2 (J) is unambiguously in Chev(r), so J = ΓE,1 (J) ≤ M by [IA , 7.3.4], and then I = J. Then as E ≤ J   CG (t), CG (t) ≤ M by [III8 , 6.1a]. The proof is complete.  Now we are reduced to a single configuration. Lemma 4.14. We have G0 ∼ = D4+ (q), q > 3. Proof. Assume false and continue the above argument. By Lemmas 4.9, 4.13, and 4.10, M controls G-fusion of its involutions. Then Lemmas 4.12 and 4.13 yield that every involution u ∈ G0 that is either of type −12 1n−2 or a classical involution satisfies u ∈ U(G, M, 2). Choose u ∈ G0 of type −12 1n−2 and let u ∈ uM with [u, u ] = 1. Then uu ∈ G0 is either of type −12 1n−2 or a classical involution, so uu ∈ U(G, M, 2). Then by [III8 , Corollary 6.2], M is solvable, a contradiction. The lemma follows.  We proceed to a final contradiction in a short sequence of lemmas. Lemma 4.15. Let t ∈ G0 be a classical involution. Then t ∈ U(G, M, 2). Moreover, M controls G-fusion of its 2-elements, and t is 2-central in M . Proof. As D contains a classical involution, by Proposition 4.3, and as ΓD,1 (G) ≤ M , we have CG (t) ≤ M . Lemma 4.10 immediately implies the first statement. By [III17 , 8.9a], t is 2central in M . Let T ∈ Syl2 (M ) with t ∈ Z(T ). Then T ∈ Syl2 (G) by [III8 , Lemma  6.1c] and M controls G-fusion in T by [III8 , Lemma 6.1g]. Lemma 4.16. Let v ∈ I2 (G0 ). Then v ∈ U(G, M, 2).

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Proof. By Lemma 4.15, we may assume that v is not a classical involution.  Let I = E(CG0 (v)). Then by [IA , 4.5.1], v ∈ I ∼ = Ω6q (q). As classical involutions are 2-central in M , it follows from Lemma 4.15 and [III8 , 6.1b] that a Sylow 2subgroup T of CM (v) is a Sylow 2-subgroup of CG (v) as well. Thus I, which is T -invariant, lies in a 2-component J of CG (v), by [IG , 5.30]. Now there is a natural G0 -module V such that under the projection π : Ω(V ) → P Ω(V ) ∼ = G0 , v lifts to an involution v of type −16 12 . There exists a four-subgroup  of Ω(V ) with the same support on V as v, and such that if we let E = π(E),  E # # then E consists of classical involutions of G0 . Then for all e ∈ E , E(CI (e)) ∼ = SL2 (q) ∗ SL2 (q). By Lemma 4.15, CG (e) ≤ M for all such e, and so E(CI (e)) = E(CG (e, v)) = E(CJ (e)). As J is a single component, it follows from [III17 , 10.44] that J/O2 (J) ∈ Chev(r) unambiguously for some odd r, and hence by [IA , 7.3.4], J = ΓE,1 (J) ≤ M . Thus I = J   CG (v). But I contains a classical involution. So by Lemma 4.15 and [III8 , 6.1a], CG (v) ≤ M , and as M controls G-fusion of its involutions, v ∈ U(G, M, 2), as required.  Now we can apply [III8 , Corollary 6.2] to any involution z ∈ I2 (G0 ). By Lemma 4.16, z ∈ U(G, M, 2). Let g ∈ M be such that z g ∈ CG (z) − {z}. As G0  M , z g and zz g lie in G0 . Hence zz g ∈ U(G, M, 2) by Lemma 4.16. By [III8 , Corollary 6.2], M is solvable, which is absurd. This final contradiction completes the proof of Proposition 3.1, and with it the proof of Theorem C∗7 : Stage 5 in the case p = 2. We can afford a breather here to note that we have reached a milestone. Our results to this point prove, among other things, that a minimal counterexample G to the Classification Theorem is of restricted even type [III12 , Theorem 1.17].

5. The Lie Type Case, p > 2 We now treat the final case of (1A2). Our goal is the following proposition. Proposition 5.1. If p is odd and G0 ∈ Chev(2), then G0 = G. As in the preceding sections, we suppose that the proposition fails and consider M = NG (G0 ) < G. We continue the notation q = ±1 with q ≡ q (mod p). Lemma 5.2. G0 is not isomorphic to any of the following groups: L± n (2), n ≤ 7, n n Bn (2), n ≤ 4, 2B2 (2 2 ), D4± (2), D5− (2), G2 (q), F4 (2), or 2F4 (2 2 ) . Proof. We have K/Op (K) ∈ Gp and K is a component of CG0 (x). The  lemma follows by [III17 , 10.5]. By Theorem C∗7 : Stage 4b+, we have (5A)

ΓD,1 (G) ≤ M, where D = x, y .

We denote by B an elementary abelian p-subgroup of M of maximum rank containing D. Let P ∈ Sylp (M ) with B ≤ P . Let G1 be the subgroup of M inducing inner-diagonal automorphisms on G0 . For any u ∈ G1 of order p, we let Lu denote the central product of all Lie components of CG1 (u).

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Lemma 5.3. The following conclusions hold: (a) CG (G0 ) = Op (M ); (b) If B0 ≤ B, mp (B0 ) ≥ 2, and ΓB0 ,1 (G) ≤ M , then B0 ∈ U(G, M, p); in particular, D ∈ U(G, M, p); (c) NG (B) ≤ M and NG (P ) ≤ M ; (d) P ∈ Sylp (G); (e) NG (B) controls G-fusion in B, and B = J(P ); (f) P ≤ [M, M ];  (g) O p (AutM (G0 )) ≤ Aut0 (G0 ); (h) B induces inner-diagonal automorphisms on G0 ; (i) M contains a Sylow p-subgroup of CG (b) for all b ∈ B; (j) For any b ∈ B # , Lp (CM (b)) = Lp (CG0 (b)); and (k) For any B0 ≤ B, CG (B0 ) ≤ M implies NG (B0 ) ≤ M . Proof. First, (a) is given in Lemma 1.2c. In particular as M/G0 is an extension of the p -group CG (G0 ) by a subgroup of the solvable group Out(G0 ), M/G0 is p-solvable. Let B0 be as in (b), and suppose that (B0 )g ≤ M . By [IA , 7.3.4], G0 = Γ(B0 )g ,1 (G0 ) ≤ M g . As G0 = E(G0 ) and M g /Gg0 is p-solvable, it follows that G0 ≤ Gg0 . Therefore g ∈ NG (G0 ) = M , as claimed in (b). Now (c), (d) and (i) follow directly from (a) and (c) of [III8 , Lemma 6.1]. The p-solvability of M/G0 also implies (j) immediately. Let M0 be the preimage of Inndiag(G0 ) in M , and P0 = P ∩ M . We claim that (5B)

mp (CM0 (D)) = mp (M0 ) = mp (P0 ).

We follow the case division in [III15 , Table 15.3]. In cases (a) and (b) of the table, ei ther p does not divide n or p does not divide n−1. Accordingly, O p (Inndiag(K)) ≤  Inn(K) or O p (Inndiag(L)) ≤ Inn(L). In the first case, y ∈ K is diagonalizable and so mp (CDG0 (D)) ≥ 1 + mp (CK (y)) = 1 + (n − 1) = n; in the second case mp (CDG0 (D)) ≥ 2 + mp (L) = 2 + (n − 2) = n. It is then enough to show mp (DG0 ) ≤ n. This is obvious in case (a), and can only fail in case (b) q if there is x0 ∈ Ip (DG0 ) such that E(CG0 (x0 )) ∼ (q). We may take x0 = Ln+1 to be diagonal, indeed to centralize x and K. But then as mp (C(x, K)) = 1, x0 = x , an impossibility as E(CG0 (x0 )) does not embed in CK (x)(∞) . Thus conclusion (a) holds in these cases. In cases (c), (d), (g)–(l), and (n), p divides neither |Outdiag(G0 )| nor |Outdiag(L)|, and it follows easily from [IA , 4.10.3a] that mp (DL) = 2 + mp (L) ≥ mp (G0 ), implying the desired conclusion. The same argument applies to cases (e) and (m) if p > 3. We are therefore reduced to cases (e), (f), and (m), all with p = 3. In case  (e), since (y, L) is an acceptable subterminal (x, K)-pair, CK (y) ∼ = GL3q (q), so q m3 (CG0 (D)) ≥ 1 + m3 (GL3 (q)) = 4 = m3 (G0 ) = m3 (Inndiag(G0 )), as asserted.  In case (f), now since (y, L) is a subterminal (x, K)-pair but L ∼  SL5q (q), we must = have m3 (K/Z(K)) = 4, whence q ≡ q (mod 9). Furthermore, (x, K, y, L) is not ignorable, by definition of acceptable subterminal pair [III12 , Def. 1.15], and so  there exists no u ∈ I3 (CM0 (x)) such that E(CK (u)) ∼ = SL5q (q). Now if M0 > G0 , then such u exists by [IA , 4.7.3A], contradiction. So M0 = G0 and therefore  m3 (M0 ) = 5 = m3 (D) + m3 (L4q (q)) ≤ m3 (CM0 (D)), as required. Finally consider case (m). Let (K, σ) be a σ-setup for G0 ; in standard notation [IA , 2.10], we may take σ = σq or σ = σq w, where w ∈ N inverts T , according as q = 1 or −1. Then

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σ induces the power mapping t → tq q on T . There is a 2-dimensional subtorus T 0 ≤ T such that CK (T 0 ) = LT 0 with L ∼ = A5 , and for some 1-dimensional torus u ∼ T 1 ≤ T 0 , CK (T 1 ) = M T 1 with M = E6 . Then E7 (q) contains CT (σ) ≥ E37 as  well as elements x ∈ I3 (T 1 ) and y ∈ T 0 − T 1 such that E(CG0 (x)) ∼ = E6q (q) and  E(CG0 (x, y )) ∼ = A5q (q), as required. Thus (5B) holds in all cases. We next prove that B = J(P ). Let B0 = B ∩ M0 . Note that if G0 ∼ = D4 (q), then some element u ∈ D# satq isfies E(CG0 (u)) ∼ L (q), so by [I , = 4 A 4.7.3A], u is not centralized by any triality automorphism of G0 . This implies that any element of B − B0 (should it be nonempty) induces a field automorphism on G0 , and in particular |B : B0 | ≤ p. By [IA , 4.10.3c], P0 has a unique elementary abelian subgroup A0 of maximal rank. By (5B), mp (CM0 (D)) = mp (M0 ) = mp (P0 ), so replacing P by a conjugate, we may assume that D ≤ A0 . If B0 < B, then likewise by [III17 , 8.5], mp (CM0 (b)) = mp (M0 ) for any b ∈ B − B0 , and so b centralizes an M -conjugate of A0 and hence an M conjugate of D. Hence by [III8 , 6.1h], b ∈ [M, M ]. But b does not map into an element of [Aut(G0 ), Aut(G0 )], contradiction. Thus B = B0 . As B was an arbitrary element of Ep (M ) containing D and of maximal rank subject to that, we must have B = A0 . Again by [III17 , 8.5], any g ∈ M with g p = 1 and inducing a field automorphism on G0 centralizes a conjugate of B, so g = 1 by our choice of B. Thus Ω1 (P ) ≤ P0 , whence J(P ) = A0 = B. Then (h) follows as well. In particular, B is weakly closed in P with respect to G, and [P, B, B] = 1. Therefore (e) holds by [IG , 16.9]. Moreover, Yoshida’s Theorem [IG , 15.27] implies that NG (B) controls p-transfer in G, and (f) follows from the simplicity of G. Since Aut(G0 )/Aut0 (G0 ) is abelian by [IA , 2.5.12], (f) implies (g). Finally (k) holds by a Frattini argument and (e): for B ≤ Q ∈ Sylp (CG (B0 )), we have NG (B0 ) ≤  CG (B0 )NG (Q) ≤ CG (B0 )NG (B) ≤ M . The proof is complete. Lemma 5.4. Let b ∈ M be of order p with either b ∈ B or Pb := CP (b) ∈ Sylp (CG (b)). Then every component of CM (b) lies in a p-component of CG (b). Proof. If b ∈ B, let B ≤ Pb ∈ Sylp (CG (b)). Then Pb ≤ M . Thus, in any case, we may Pb ∈ Sylp (CG (b)) with Pb ≤ M . Let J be a component of CM (b).  choose Then J Pb is a Pb -invariant product of components, so by [IG , 5.30], J lies in a  p-component of CG (b), as required. Lemma 5.5. Suppose that p ≥ 5 and b ∈ B # . Let L0 be a p-component of CG (b). If L0 D has abelian Sylow p-subgroups, then L0 ≤ M . Proof. Note that B normalizes L0 by [IG , 8.7(iii)], so L0 D is a group. Suppose that the lemma fails, and choose a counterexample b for which Lp (CM (b)) is as large as possible. Let B ≤ Pb ∈ Sylp (CG (b)), so that Pb ≤ M . By [III17 , 9.11], 1 L0 /Op (L0 ) ∼ = L2 (pm ), m ≥ 2, or p = 5 and L0 /Op (L0 ) ∼ = 2F4 (2 2 ) . By [III8 , 8.7(iii)], B normalizes L0 and then as B  Pb , B induces inner automorphisms on L0 . Thus, B = (B ∩ L0 )CB (L0 /Op (L0 )). Hence there is Y ≤ NL0 (B) such that (B ∩ L0 )Y covers NL0 (B ∩ L0 ) modulo Op (L0 ). In particular [B, Y ] ≤ B ∩ L0 . Let Y0 = CY (B ∩ L0 ) = Y ∩ Op (CG (b)). Note also that by [IA , 7.6.2], M ∩ L0 = Op (L0 )NL0 (B ∩ L0 ) is p-solvable.

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

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Let B0 = CB (Lp (CM (b))). Then B ∩ L0 ≤ B0 , so Y normalizes B0 . On the other hand, by [IA , 4.2.2], the image Y of Y in M = M/Op (M ) satisfies [Y ∩ M 0 , B 0 ] = 1, where M0 is the full preimage in M of Inndiag(G0 ). Thus Y ∩ M 0 ≤ Y 0 . Suppose that Lp (CM (b)) = 1. Every component J of Lp (CM (b)) lies in a p-component J ∗ of CG (b) by Lemma 5.4. As M ∩ L0 is p-solvable, L0 = J ∗ . Thus [J, Y ] = 1, and so [Lp (CM (b)), Y ] = 1. This implies by [IA , 4.2.3] that no element of Y maps to the image of a nontrivial field automorphism in Aut(G0 )/ Inndiag(G0 ). Hence Y /Y0 embeds in the group of graph (outer) automorphisms of G0 , which embeds in turn in Σ3 . But Y /Y0 is cyclic of order 12 (pm − 1) or contains SL2 (3), with m ≥ 2, so this is a contradiction. Therefore Lp (CM (b)) = 1.  Suppose that mp (CB (L0 )) > 1. Then by [IA , 7.3.3, 4.9.7], O 2 (CG0 (g)) has a Lie component of order divisible by p for some g ∈ CB (L0 )# . In particular since 1 5  2F4 (2 2 ) or 2B2 (2 2 ), Lp (CM (g)) = 1. But also L0 ≤ CG (g) ≤ M . Let Lg G0 ∼ = be the pumpup of L0 in CG (g). By [III17 , 10.7], Lg /Op (Lg ) has abelian Sylow psubgroups. Hence by our original choice of L0 , Lg ≤ M . This is absurd as L0 ≤ Lg but L0 ≤ M . Therefore mp (CB (L0 )) = 1. As mp (B) ≥ 4, we have L0 ∼ = L2 (pm ) with m + 1 = mp (B) ≥ 4. Let n = mp (B). Let N0 = NCG (b) (B). From the action of N0 ∩L0 we have orbits O1 = {b }, O2 and O3 of size 12 (pm −1), exhausting E1 (B)−E1 (B ∩L0 ), as well as a single orbit O4 or two equal-size orbits O4 , O5 exhausting E1 (B ∩ L0 ). Consider how these orbits are fused under NG (B) ≤ M . Note that either NG (B) acts irreducibly on B or the NG (B)-socle S of B has codimension 1 and is irreducible, the latter occurring only q (q). In particular O1 fuses with some other Oi . if G0 ∼ = Lpk q (q) case, some of the Oi must fuse to form E1 (S). But In the exceptional Lpk 1 n−1 − 1) > |E1 (S)|, so E1 (S) = O4 ∪ O5 , and there exist at most |O2 | = |O3 | = 2 (p two G0 -orbits on E1 (B) − E1 (S). But one easily finds b1 , b2 ∈ B − S such that no two of b , b1 , and b2 are M -conjugate. Thus this exceptional case does not occur, so N is irreducible on B. As D ∩ L0 = 1, we may assume that for all g ∈ O4 , CG (g) ≤ M . Hence O1 and O4 do not fuse. Since O4 ∪ O5 = B, O4 must fuse to O2 (or O3 , a similar case). If O1 does not fuse to O3 , then it fuses to O5 only. But then O1 ∪ O5 is NG (B)-invariant; every basis of B contained in O1 ∪ O5 contains b , and b is unique in that regard, so b is N -invariant, contradiction. Therefore, O1 fuses to O3 . Finally O5 is not an NG (B)-orbit, by irreducibility. Hence NG (B) = NM (B) has exactly two orbits on E1 (B): One is the orbit of b , and Lp (CM (b)) = 1; the other consists of all g ∈ E1 (B) such that Lp (CM (b)) = 1. This fusion pattern in  M is impossible by [III17 , 2.41], so the proof is complete. Our goal is to establish that M is a strong p-uniqueness subgroup in G, contrary to hypothesis. By [IG , 17.11(iii)], [III12 , Definitions 1.9, 1.11 and Remarks 1.8, 1.10], and the preceding lemma, it will suffice to prove that one of the following holds: (1) CG (u) ≤ M for all u ∈ Ip (M ); or (2) G0 ∼ (5C) = Lp (q), p | q − , and NG (X) ≤ M for all X ≤ P ∈ Sylp (M ) such that either mp (X) ≥ 2 or mp (CP (X)) ≥ 3.

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First, we consider elements b ∈ B such that Lp (CG0 (b)) = 1. Lemma 5.6. Suppose that b ∈ B and Lp (CG0 (b)) = 1. If CG (b) ≤ M , then p = 3, G0 ∼ = U8 (2), U9 (2), or 2E6 (2), and b is 3-central in G0 . Proof. Assume that the lemma is false. Let C := CG (b) ≤ M , and set C0 = C ∩ M < C. By Lemma 5.3j, Lp (C0 ) = 1. By [III17 , 6.31], and as this lemma fails, p ≥ 5. Let B ≤ Pb ∈ Sylp (C0 ) ⊆ Sylp (C); see Lemma 5.3i. First suppose that Pb =: A is abelian, so that B = Ω1 (A). If L0 is a pcomponent of Lp (C), then L0 ≤ M by Lemma 5.5, so L0 ≤ Lp (CG0 (b)) = 1, contradiction. Hence Lp (C) = 1, so C is p-constrained, Op p (C) = Op (C)A  C, and C = Op (C)NC (B) ≤ M , contrary to assumption. Thus, by [III17 , 6.31] and Lemma 5.3, we may assume that Pb = P = A w contains an abelian subgroup A with index p, CA (w) = b , and G0 ∼ = Lp (q) with p > 3 and p dividing q − . Suppose that L0 is a p-component of C. Since Z(P ) = b , b ∈ Z(L0 ). Since p > 3, this implies that L0 /Z(L0 ) ∼ = Lpm (q1 ) for  some m ≥ 1 and q1 ≡  (mod p). But then L0 = ΓD,1 (L0 ) ≤ C0 by [IA , 7.3.3], a contradiction as C0 is solvable. Hence, Lp (C) = 1 and C is p-constrained. Let R = P ∩ Op p (C). We claim that A ≤ R. If R ≤ A, then A ≤ CP (R) ≤ R and we are done. So, we may assume that w ∈ R. Let A1 = A ∩ R. Then as R  P , A1 ≥ [A, w], with |[A, w]| = |A : CA (w)| = |A|/p. In particular |A1 | > p2 , and as |CP (w)| = p2 , it follows that A1 = J(R)  NC (R). Then |R/A1 | = p, so A stabilizes the chain R > A1 > 1, whence A ≤ Op (NC (R)/CC (R)) = RCC (R)/CC (R), and so A ≤ R, as claimed. But now we have that C = Op (C)NC (B) ≤ M , since Op (C) = ΓD,1 (Op (C)) ≤ M and NG (B) ≤ M . This completes the proof of the lemma.  Next, we establish control over the subnormal closure in CG (b) of p-components of CG0 (b). Lemma 5.7. Let b ∈ M be of order p. Suppose that B0 ≤ CB (b) with mp (B0 ) ≥ 2 and with ΓB0 ,1 (G) ≤ M . Suppose that J is a p-component of Lp (CG0 (b)) and Pb := CP (b) ∈ Sylp (CG (b)). Moreover, assume that either b ∈ B or J is not isomorphic to any member of the set {L2 (4), Sp4 (2) , U4 (2), Ω+ 8 (2)}. Then one of the following holds: (a) J  E(CG (b)); or  M for some component J ∗ of CG (b) with J ∗ ∈ (b) b ∈  B, and J ≤ J ∗ ≤ Spor − {F1 , F2 , 2F2 } and mp (J ∗ ) > 2. Proof. By Lemma 5.4, J ≤ J ∗ for some p-component J ∗ of CG (b). Moreover, Op (CG (b)) ≤ ΓB0 ,1 (G) ≤ M , and as J is a component of CM (b) we have [J, Op (CG (b))] = 1, whence J ∗ is quasisimple. We assume that (a) and (b) both fail, and we let J0 = J ∗ ∩ M , so that J0 < J ∗ . We assume that b and J were chosen subject to these conditions with |J| maximal. As ΓB0 ,1 (J ∗ ) ≤ J0 it follows by [IG , 3.28(ii)] that J ∗ is B0 -invariant. Since J ∗ is subnormal in Cb , we have P0 := Pb ∩ J ∗ ∈ Sylp (J ∗ ). Moreover, as Pb ≤ M , we have P0 ≤ J0 . As Op (J ∗ ) ≤ ΓB0 ,1 (G) ≤ M and J is a component of CM (b), it follows that [Op (J ∗ ), J] = 1 and so Op (J ∗ ) ≤ Z(J ∗ ).

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

Suppose first that J ∗ ∈ Chev(p). Then, as P0 ∈ Sylp (J ∗ ), J0 must lie in a proper parabolic subgroup X of J ∗ , by [IA , 2.6.7]. But F ∗ (X) = Op (X) ≤ J0 while J is subnormal in J0 , so [J, F ∗ (X)] = 1, a contradiction. Next suppose that J ∗ ∈ Spor. If mp (J ∗ ) ≤ 2, then by [III17 , 9.8], p = 5,  J∗ ∼ = F i22 , and J ∼ = D4 (2) has level q = 2. As b ∈ O p (M ), b induces an element of Aut0 (G0 ) on G0 by Lemma 5.3g, so by [IA , 4.2.2] applied to this action, G0 has level q = 2. As p divides q 2 − 1, we have a contradiction. Thus if J ∗ ∈ Spor, then mp (J ∗ ) ≥ 3. We argue that 

(5D)

if d ∈ B0# , H  CJ ∗ (d) and H = [H, H] = O p (H), then H is a commuting product of groups in Chev(2) (but not 3A6 ). 

Indeed H   CG (b, d ) = CM (b, d ) with H ≤ O p ([M, M ]) = G0 , so H   CG0 (b, d ), As G0 ∈ Lie(2), (5D) follows. ∗ ∗ We set J = J ∗ /Op (J ∗ ). If (J ; p) = (F1 ; 3, 5, 7), (F2 ; 3, 5), (F5 ; 3), or (Co1 ; 3), then (5D) is violated for every d ∈ D# , regardless of its action on J ∗ . In particular, J ∗ ∈ Spor − {F1 , F2 , 2F2 }. As conclusion (b) fails, we must have b ∈ B. Hence without loss, B0 = D. Note that for any d ∈ D# , we have CJ ∗ (d) = CJ0 (d) as ΓD,1 (G) ≤ M . We now argue that (5E)

if Op (J ∗ ) = 1, then for any R ∈ Sylp (J ∗ ), J(R) is elementary abelian and NJ ∗ (R) ≤ J0 .

Indeed B ≤ CG (b) as b ∈ B # , so B normalizes J ∗ by [III8 , 8.7(iii)]. Then since Op (J ∗ ) = 1 and |Out(J ∗ )| ≤ 2, B = (B ∩ J ∗ ) × CB (J ∗ ). Since B = J(P ), also B = J(Pb ) and it follows immediately that B ∩ J ∗ = J(P0 ). Similarly, P0 × CB (J ∗ ) is normalized by NJ ∗ (P0 ) and contains B, so NJ ∗ (P0 ) ≤ M by Lemma 5.3a, proving (5E). ∗

∗

Now we consider the remaining possibilities for J . If (J ; p) = (F3 ; 3), (F i23 ; 3), (F i24 ; 3), (F5 ; 5) or (Ly; 5), then mp (J ∗ ) = 5, 6, 7+m3 (Op (J ∗ )), 3, or 3, respectively ∗ [IA , Table 5.6.1], and J contains Z3 × G2 (3) ≥ E33 × 31+2 , D4 (3), O3 (J ∗ ) × Z3 × D4 (3), 51+4 , or 51+4 , respectively, so (5E) is contradicted. (Note that D4 (3) has three maximal parabolic subgroups Q containing P0 with O3 (Q) ∼ = E36 .) Suppose ∗ that (J ; p) = (Co1 ; 5). As p divides q 2 − 1, q (= q(G0 )) ≥ 4, and then q(J1 ) ≥ 4 for every component J1 of J, by [IA , 4.2.2]. Since J(P0 ) ∼ = E53 is self-centralizing in P0 and |J ∗ |5 > 5, the Weyl group of some component J1 of J has order divisible by 5. As q ≥ 4 and |J1 |2 ≤ |J ∗ |2 = 221 , the only possibility is J ∼ = L± 5 (4). But 5 |Co1 | is not divisible by 4 ± 1, contradiction. ∗ ∗ In all the remaining cases, p = 3. In three of the cases: J /O3 (J ) ∼ = Suz, ∗ ∗ Co3 , and Ly, AutJ ∗ (B ∩ J ∗ ) contains M11 , and in a fourth case, J /O3 (J ) ∼ = M c, AutJ ∗ (B ∩ J ∗ ) contains M10 . Accordingly the Weyl group of some component J1 of J has order divisible by 11 or 5. Hence |J1 |2 ≥ |L11 (2)|2 = 255 or |L5 (2)|2 = 210 ∗ respectively. However, |J1 |2 ≤ |J |2 = 214 , 210 , 28 , or 27 respectively, a contradiction. ∗ ∗ There remain the cases J /O3 (J ) ∼ = J3 , Co2 , F i22 , and O  N . In the F i22 ∗ ∗ case, O3 (J ) ≤ J(P0 ) with J(P0 )/O3 (J ) = J(P0 /O3 (J ∗ )) ∼ = E35 having automizer ∗ ∗ containing P Sp4 (3) (because Ω7 (3) embeds in J /O3 (J ); see [III17 , 6.15]). Hence as usual, some component of J1 has untwisted Weyl group containing P Sp4 (3). This forces |J1 |2 ≥ 236 > |J ∗ |2 , contradiction. In the J3 case, a similar argument

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∼ E33 and some component of J1 has untwisted Weyl group gives J((P0 )/O3 (J ∗ )) = 15 containing E32 , forcing |J1 |2 ≥ |L± > |J ∗ |2 , contradiction. 6 (2)| = 2 ∗ Suppose that J ∼ = Co2 . If some d ∈ D# is of type 3A in [IA , Table 5.3k],  (∞) 3 then CJ ∗ (d) = O (CJ ∗ (d)) is 3-constrained, contradicting (5D). Thus for all d ∈ D# , CJ ∗ (d)(∞) =: Jd ∼ = U4 (2). As m3 (U4 (2)) = 3 and m3 (Co2 ) = 4, J must be a single component. Now J ∗ ≥ M c ≥ 34 M10 , so by the usual argument the untwisted Weyl group of J contains A6 . As |J ∗ |2 = 218 , this forces J ∼ = L± 6 (2), in ∗  ∗ fact U6 (2) as 31 does not divide |J |. Since ΓD,1 (J ) ≤ J1 , we must have Jd  CJ (d) for all d ∈ D# . This however violates [III17 , 3.6]. ∗ ∗ ∗ Therefore J /O3 (J ) ∼ = O  N . If O3 (J ) = 1, then E(CJ ∗ (d)) ∼ = 3A6 for ∗ any d ∈ D# , violating (5D). So J ∼ = O  N . Pick any d ∈ D# and let I = NJ ∗ (P0 ), CJ ∗ (d) ≤ M ∩ J ∗ = J0 . Since NJ ∗ (P0 ) is transitive on P0# , J ∗ has a strongly 3-embedded subgroup, which is not the case by [IA , 7.6.1]. This is a final contradiction for the case J ∗ ∈ Spor. Suppose next that J ∗ ∈ Chev(r) with r = p. We have J   J0 with J ∈ Chev(2). In particular J0 is not p-constrained, whence [III17 , 9.10] implies that p = 3 and one of the following holds: (1) J ∗ ∼ = L3 (4), J0 ∼ = A6 , and AutB0 (J ∗ ) ∈ Syl3 (Inn(J ∗ )) ; ∗ ∼ (2) J (2) and J0 ∼ Sp (5F) = 6 = O6− (2); or (3) J ∗ ∼ = Sp8 (2) and J0 ∼ = O8+ (2). By hypothesis b ∈ B, so we may assume that B0 = D. If (5F1) holds, then D faithfully induces inner automorphisms on J ∗ , so B = (B ∩ J ∗ ) × CB (J ∗ ). Then NJ ∗ (B ∩J ∗ ) ≤ NJ ∗ (B) ≤ J0 . However, AutJ0 (B ∩J ∗ ) ∼ = Z4 while AutJ ∗ (B ∩J ∗ ) ∼ = Q8 by [III17 , 8.4], a contradiction. Hence one of (5F2, 3) holds, so J ∼ = Ω2m (2) and J ∗ ∼ = Sp2m (2) for some m ∈ {3, 4}. If G0 is of exceptional type then by [IA , 4.7.3A], G0 ∼ = 2E6 (2), m = 4, and some a ∈ I3 (CG0 (b)) induces a graph automorphism of order 3 on J. But then a normalizes J ∗ , which is impossible as |Aut(J ∗ )|3 = 35 = |J|3 . Thus, G0 is a classical group. We argue that (5G)

CB (J) = b .

Otherwise by [III17 , 9.13], there is b1 ∈ CB (J)# such that the subnormal closure J1 of J in CM (b1 ) is quasisimple and J1 > J. By our maximal choice of J, we have J1   CG (b1 ). However, since b1 ∈ CG (b J), b1 normalizes J ∗ and hence maps into  O 3 (CAut(J ∗ ) (J)), which is trivial by the facts that Aut(J ∗ ) = Inn(J ∗ ), Z(J) = 1  ∗ and m3 (J) = m3 (J ∗ ). Thus, [b1 , J ∗ ] = 1. But then J ∗ = J J ≤ J1 , which is a contradiction as J1 ≤ M . This proves (5G). Then by [III17 , 10.25], G0 ∼ = Un (2), n ≤ 6, Bn (2), n ≤ 4 or Dn± (2), n ≤ 5. However, all these are forbidden isomorphism types for G0 , by Lemma 5.2. This contradiction finishes the proof that J ∗ ∈ Chev. Finally, suppose that J ∗ ∼ = AΩ for some |Ω| = n ≥ 5. As Akp2 ∈ Chev for any k > 0, [III17 , 9.7] gives p ∈ {3, 5}, D acts on Ω = {1, . . . , n} with all orbits of size p, and Soc(J0 ) = J1 × · · · × Jr with Ji ∼ = Ap and n = rp with 2 ≤ r ≤ p + 1. As J is subnormal in J0 , it follows that p = 5 and J ∼ = L2 (4), = A5 ∼ whence b ∈ B. Moreover, as B = J(Pb ), it follows that B normalizes J ∗ , and then B = (B ∩ J0 ) × CB (J ∗ ), with B ∩ J0 ∈ Syl5 (J ∗ ). Then NJ ∗ (B ∩ J0 ) ≤ NG (B) ≤ M . We may assume notation chosen so that J = J1 , and we set B1 = CB (J1 J2 ). Then for all b1 ∈ B1# , the subnormal closure I of J in CG (b1 ) contains an A10 subgroup whose support on Ω is the same as that of J1 J2 . In particular, I ≤ M .

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

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Therefore by our maximal choice of J, all components of the pumpup of J in CM (b1 ) are isomorphic to J. But by [III17 , 9.13] this implies that B1 is cyclic. Hence m5 (B) = m5 (J1 J2 ) + 1 = 3, a contradiction as mp (B) ≥ 4. The proof of the lemma is complete.  This gives a simple sufficient condition for CG (b) ≤ M for an element b ∈ B # . Lemma 5.8. Let b ∈ B # . Suppose that b Lp (CG0 (b)) contains an element a of order p such that CG (a) ≤ M . Then CG (b) ≤ M . Proof. Let C = CG (b) and J = Lp (CG0 (b)). By Lemma5.7, with D in the role of B0 , J is a product of p-components of E(C). Set J ∗ = J C = J1 J2 · · · Jn , with each Ji a p-component of E(C) and J = J1 · · · Jk . As a ∈ b J, J ∗ CC (J ∗ ) ≤ JCC (a) ≤ M . In particular Lp (C) = Lp (C ∩ M ) so each Ji is a p-component of  = C/J ∗ CC (J ∗ ) ≤ Out(J ∗ ). Then by [III17 , 6.32bd], applied Lp (CG0 (b)). Let C  of with M/Op (M ) and G0 /Op (G0 ) in the roles of M and G0 there, the image D  D in C lies in Outdiag(J1 ) × · · · × Outdiag(Jn ). This direct product, however, is  ≤ Op (C).  Let Q be a Sylow a normal abelian subgroup of Out(J ∗ ). Therefore D p-subgroup of Op (C mod J ∗ CC (J ∗ )) containing D. Since D ∈ U(G; M ; p) and D   Q, we have NC (Q) ≤ M by [III8 , 6.1a]. But then by a Frattini argument,  C = J ∗ CC (J ∗ )NC (Q) ≤ M as claimed. We can now prove that ΓB,1 (G) ≤ M in all the cases in which G0 is a classical group, except for G0 ∼ = U8 (2) or U9 (2). Lemma 5.9. Suppose that G0 is a classical group. Let b ∈ B # . If G0 ∼ = U8 (2) or U9 (2), assume that b is not 3-central in G with L3 (CG0 (b)) = 1. Then CG (b) ≤ M . Proof. By Lemma 5.6 we may assume that Lp (CG0 (b)) = 1. Let J be a p-component of Lp (CG0 (b)). As ΓD,1 (G) ≤ M , Lemma 5.8 implies that we are done as long as (5H) J can be chosen so that Dg ∩ J = 1 for some g ∈ M . First suppose that G0 ∼ = Anq (q).

(5I)

∼ A q (q) unless there is no element z ∈ M of order p Thus by Table 15.3, K = n−1 q q with Lp (CG0 (z)) ∼ (q). We write K ∼ (q), = An−1 (q), in which case K ∼ = An−2 = An−a  q ∼ a = 1 or 2. Again by Table 15.3, L = An−a−1 (q). As p divides q − q , B acts on G0 as the image of some diagonalizable p-subgroup of GLn+1 (q), by [IA , 4.10.3c] and [III17 , 2.8]. Then D# contains an element d (d = x if a = 2, d = x if a = 1) such that d has a preimage d ∈ GLn+1 (q) of order p whose eigenvalues are 1 (multiplicity n − 1), ω and ω −1 , for some primitive pth root of unity ω. But now, taking any component J of Lp (CG0 (b)) = 1, we see that J is supported on an eigenspace of b, necessarily at least 2-dimensional. Hence dG0 ∩ J = ∅, so (5H) holds. Thus if (5I) holds, then the conclusion of the lemma holds. Suppose then that (5I) fails. By [IA , 4.8.2] and Table 15.3, if we consider the action of G0 on its natural module V (in case G0 = D4 (q) we have to make the right choice of the three natural modules), then x acts on V with 2-dimensional 

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support, which is the minimum possible for an element of order p. Moreover, as mp (M ) ≥ 4, we have dim(V ) ≥ 8. We consider various cases; first, suppose that q > 2. If b acts on V with a 1eigenspace of dimension at least 2 (if G0 is of type A or C) or 4 (if G0 is of type D), then Lp (CG0 (b)) has a p-component supported on that eigenspace and containing a G0 -conjugate of x, so CG (b) ≤ M . This includes all b ∈ B # whose support on V is at most 4-dimensional, since dim V ≥ 8. If, on the other hand, b is not of such a form, then by [IA , 4.8.2, 4.8.4], any p-component J of Lp (CG0 (b)) must be c

of the form Akq (q c ) for some k ≥ 2 and some c ≤ 2 (the parameters depending on J). Moreover, the support of J on V is a natural J-module or the sum of two dual natural modules. In any case J contains some a ∈ B # with 4-dimensional support on V . As we have just seen, CG (a) ≤ M , and so by Lemma 5.8, CG (b) ≤ M . Thus the lemma holds if q > 2. Suppose that q = 2, so that p = 3 with m3 (M ) ≥ 4, whence G0 ∼ = An (2), n ≥ 7, Cn (2), n ≥ 5, or Dn± (2), k ≥ 5, but G0 ∼  D5− (2) (see Lemma 5.2). (G0 is =  not a unitary group, since (5I) fails.) Accordingly O 3 (M ) = G0 and by Table 15.3, x has 2-dimensional support on V , and D has 4-dimensional support. Moreover, every element a of B # whose support has dimension at most 4 lies in a conjugate of D, whence CG (a) ≤ M . Hence we are done by Lemma 5.8 if some element of B ∩ J has support of dimension at most 4 on V . Now if dim CV (b) ≥ 4 (or 6 in the case G0 has type D), then J can be taken to be supported on CV (b), so that J is of type Ak (2), k ≥ 3, Ck (2) , k ≥ 2, or Dk± (2), k ≥ 3, and hence J contains some a ∈ B # with support of dimension at most 4, as desired. Suppose then that dim CV (b) < 4 (or 6 in the type D case). If J can be chosen so that its support is contained in the support of b, then J ∼ = Ak (4), k ≥ 3, or 2 Ak (2), k ≥ 3, so J contains an element a ∈ B # with 4-dimensional support on V , as desired. Finally, if the only choice for J has its support in CV (b), then J ∼ = A2 (2) or 2 D2 (2), and again J contains an element a ∈ B # with support of dimension 2 or 4 on V . The proof is complete.  In the next two lemmas we prove that CG (b) ≤ M with the following assumptions and notation: (1) G0 is of exceptional type; (2) M = M/Op (M );  (3) b ∈ B # and Lb = O 2 (CG0 (b)); and (4) One of the following holds: (5J)  (a) O p (Lb ) = Lp (CG0 (b)); or    (b) p = 3 and O 3 (CM (b)) ≤ b Lb = b × J × L3 (CG0 (b)) with J ∼ = L2 (2). Lemma 5.10. Assume (5J). If Lp (CG (b)) = Lp (CG0 (b)), then CG (b) ≤ M . Proof. Let C = CG (b) and I = Lp (CG (b)) = Lp (CG0 (b)). Let C1 = CC (I), so that Lp (C1 ) = 1, i.e., C1 is p-constrained. By Lemma 5.3i, we have B ≤ Pb ≤ M for some Sylow p-subgroup Pb of C. In particular B = J(Pb ). Let C2 be the subgroup of C normalizing each component of I and inducing inner-diagonal automorphisms on each component of I. Then (5K)

B ≤ C2 ∩ M = Lb T b , with T b abelian.

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

From [IA , Table 4.7.3A] we see that |C ∩ M : C2 ∩ M | = 1 or 3, the latter occurring if and only if G0 ∼ = SL3 (q)3 or D4 (q). = E6 (q), with p = 3 | q −  and I ∼ Let P2 = Pb ∩ C2 ∈ Sylp (C2 ), so that B ≤ P2 and hence B = J(P2 ) by Lemma 5.3e. As C2  C, C ≤ C2 NG (P2 ) ≤ C2 NG (B) ≤ C2 M ; see Lemma 5.3c. It therefore suffices to show that C2 ≤ M . Since the image of C2 in Aut(I) lies in Inndiag(I) by definition, C2 /C1 I is abelian. Therefore C2 /I is p-constrained. Moreover, C2 /I has abelian Sylow p-subgroups; if (5J4a) holds this follows since P 2 ≤ C2 ∩ M = Lb T b with Tb abelian, while if (5J4b) holds it follows directly from the given structure of  O 3 (C ∩ M ). Therefore, C2 /I has p-length 1. Let W = Op (C2 mod I) and W0 = [P2 , W ]. Then W0 I/I is a p -group and C2 = IW0 NC2 (P2 ); again as NG (P2 ) ≤ M , it suffices to show that W0 ≤ M . Since P2 maps into Op (Out(I)), W0 induces inner automorphisms on I. Then the projection W1 of W0 on CC (I) satisfies W0 ≤ IW1 , and W1 is a P2 -invariant p -group, so W1 ≤ ΓD,1 (W1 ) ≤ M . As I ≤ M , we have W ≤ M , and hence C ≤ M , completing the proof.  Lemma 5.11. If (5J) holds, then Lp (CG0 (b)) = Lp (CG (b)) and CG (b) ≤ M . 

Proof. Let C = CG (b) and this time set I = O p (Lb ). Choose B ≤ Pb ∈ Sylp (C), so that Pb ≤ M by Lemma 5.3i. Whichever part of (5J4) holds, CM (b)/Lp (I) and then CM (b Lp (I)) are p-solvable, in view of [IA , 4.2.2]. By Lemma 5.10, it suffices to prove that Lp (CG0 (b)) = Lp (CG (b)). Thus if the lemma fails, there is a p-component L0 of C with L0 ≤ Lb . Now the p-components of I are p-components also of C, by Lemma 5.7. Hence [L0 , Lp (I)] = 1. Then L0 ∩ M ≤ CM (b Lp (I)) is p-solvable, so ΓD,1 (L0 ) is p-solvable. Hence by [IG , 3.28(ii)], D normalizes L0 and acts faithfully on L0 /Op (L0 ). If (5J4b) holds, however, then p = 3 and |Pb : b (Pb ∩ Lp (I))| = 3 with [b (Pb ∩ Lp (I)), L0 ] = 1, whence CD (L0 ) = 1, a contradiction. Hence, we may assume that (5J4a) holds, so that I = Lp (I). Suppose first that p = 3. From [IA , 4.7.3A] we see that CPb (I) is cyclic or abelian of rank 2 (the latter only for G0 ∼ = D4 (q)). But b (Pb ∩ L0 ) ≤ CPb (I). = E6 (q) and I ∼ In particular L0 has abelian Sylow p-subgroups. As [b, L0 ] = 1 it follows from [IG , 15.12(iv)] that b ∈ L0 . Hence CPb (I) is abelian of rank 2 and Pb ∩ L0 is cyclic, and L0 is the unique p-component of C outside M . Thus Pb ∩ L0  Pb . It follows that Ω1 (CPb (I)) = b Ω1 (Pb ∩ L0 ) ≤ Z(Pb I). However from [IG , Table 4.7.3A] we see that Z(Pb I) is cyclic, a contradiction. Thus we may assume that p ≥ 5. In particular the Schur multiplier of G0 is  a p -group by [IA , 6.1.4]. As O p (AutM (G0 )) ≤ Aut0 (G0 ) by Lemma 5.3g and G0  has no graph automorphisms of order p, O p (AutM (G0 )) ≤ Inndiag(G0 ). Then [IG , 4.2.2] implies that a Sylow p-subgroup of CM (b)/I is abelian. In particular the Sylow subgroup Pb ∩ L0 of L0 is abelian. As B = J(Pb ) normalizes L0 by [IG , 8.7(iii)] and Pb /Pb ∩ I is abelian, we also have [B, Pb ∩ L0 ] ≤ L0 ∩ I ≤ Z(L0 ). Thus either B(Pb ∩ L0 ) is abelian or Pb ∩ Z(L0 ) = 1. Now as L0 ∩ M is p-solvable, we may apply [III17 , 9.11] to deduce that L0 /Op (L0 ) is simple, so Pb ∩ Z(L0 ) = 1. Thus B(Pb ∩ L0 ) is abelian, so L0 D has abelian Sylow p-subgroups, and by Lemma  5.5, L0 ≤ M , a contradiction. The proof is complete.

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313

Note: The case (5J4b), in which Lb > E(Lb ), arises for one class of elements of order 3 in each of G0 = 2E6 (2) and E7 (2). Corollary 5.12. Let 1 = B0 ≤ B. Then NG (B0 ) ≤ M , unless perhaps p = 3,  CG (B0 ) ≤ M , and one of the following holds, where we write Lb = O 2 (CG0 (b)) for any b ∈ B0# : ∼ U8 (2), U9 (2), or 2E6 (2), and for all b ∈ B # , b is 3-central in G0 (a) G0 = 0 with L3 (Lb ) = 1; # (b) G0 ∼ = En (2), n ∈ {6, 7, 8}, and, for all b ∈ B0 , we have |CG0 (b) : Lb | = 3 and Lb = J ∗ L, with J ∼ = SU3 (2) and L ∼ = SL3 (4), SU6 (2), or 2E6 (2)u , respectively. Proof. By Lemma 5.3k, it suffices to argue that CG (b) ≤ M for some b ∈ B0# ,

(5L)

since then CG (B0 ) ≤ M . By Lemma 5.9, (5L) holds if G0 is a classical group, unless (a) holds. So we may assume that G0 is of exceptional Lie type. # Suppose that G0 ∼ = 2E6 (2). Then by [IA , 4.7.3A], an element b ∈ B0 is 3central if and only if L3 (CG0 (b)) = 1. We may assume that conclusion (a) fails, whence B0# contains a non-3-central element b. Then by [IA , 4.7.3A] we see that (5J4) holds. Thus by Lemma 5.11, (5L) holds. Therefore we may assume that G0 ∼ = 2E6 (2). Note that G0 ∼ = F4 (2), since Sp6 (2) ∈ G3 . We then see from [IA , 4.7.3A] that for any b ∈ B0# , either hypothesis (5J4) holds or b satisfies the conditions in conclusion (b). Hence either conclusion (b) holds or by Lemma 5.11, (5L) holds. The proof is complete.  We shall presently establish control over G-fusion of elements of B in almost  all cases. For the case G0 ∼ = Lpq (q), we shall need the following lemma.  Lemma 5.13. Suppose that G0 ∼ = Lpq (q). If g ∈ G with CG (G0 ) ∩ CG (G0 )g = 1, then g ∈ M .

Proof. Since 4 ≤ mp (M ) ≤ mp (Aut0 (G)) by Lemma 5.3g, we have p ≥ 5. Then by Corollary 5.12, ΓB,1 (G) ≤ M . Let B0 = B ∩G0 . Thus mp (B0 ) ≥ p−2 ≥ 3 and in particular B0 ∈ U(G, M, p) by Lemma 5.3b. Let 1 = Y ≤ CG (G0 ) and CY = CG (Y ). Clearly B0 ≤ G0 ≤ CY . As B0 ∈ U(G, M, p), M contains some P0 ∈ Sylp (CY ). Then by [IG , 5.30], G0 lies in a single p-component I of CY . We have Op (CY ) ≤ ΓB0 ,1 (G) ≤ M , whence [G0 , Op (CY )] ≤ Op (CY ) ∩ G0 ≤ Op (G0 ). Hence [G0 , Op (CY )] = 1 and then I is quasisimple. Now there exists b ∈ B0# such that J := E(CG0 (b)) ∼ = SLp−2 (q), which, as p ≥ 5 and q ≡ q (mod p), is unambiguously in Chev(2). But CG (b) ≤ M , so J is a component of CG (b) and then also of CI (b). Hence by [III11 , 1.1ab], I ∈ Chev(2). By [IA , Theorem 7.3.6], I = ΓB0 ,1 (I) ≤ M , whence I = G0 . As CG (G0 ) is a  p -group, we have G0 = O p (E(CY )). Finally, if X := CG (G0 ) ∩ CG (G0 )g = 1, we apply the previous paragraph to −1 Y = X and Y = X g and conclude that 



O p (E(CG (X))) = G0 = O p (E(CG (X g Consequently g

−1

−1

))).

∈ NG (G0 ) = M by (1C), completing the proof.

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

314

We now establish control over G-fusion of elements of B in almost all cases. Lemma 5.14. Let b ∈ B with CG (b) ≤ M . Then b ∈ U(G, M, p), i.e., if g ∈ G g with b ≤ M , then g ∈ M . In particular, if d ∈ D# and g ∈ G with dg ∈ M , then g ∈ M. Proof. Let b ∈ B with CG (b) ≤ M , and let w = bg ∈ M . It will suffice to show that w ∈ bM . For then, bg = w = bm for some m ∈ M , whence gm−1 ∈ CG (b) ≤ M , and so g ∈ M . As M controls G-fusion in B, we may assume that wM ∩ B = ∅. By Lemma  5.3g, O p (AutM (G0 )) ≤ Aut0 (G0 ), so it follows from [III17 , 14.7] that one of the  following holds (we write Gw for O 2 (CG (w))): (1) G0 ∼ = Lpq (q), and CG0 (w) is a Frobenius group with kernel of order q p − q /p(q − q ) and complement of order p; q  (q), m > 1, and Gw /Op (Z(Gw )) ∼ (2) G0 /Op (G0 ) ∼ = Lpm = Lmq (q p );   q q (3) G0 ∼ = E6 (q), p = 3, and Gw ∼ = 3D4 (q) or L3 (q 3 ); or (4) G0 ∼ D (q), q > 2, p = 3, w induces a graph automorphism on = 4 q (q) or L (q). G G0 , and Gw ∼ = 2 3 

(5M)

In cases (5M2, 3) and the first case of (5M4), it follows from Zsigmondy’s Theorem (see [IG , 1.1]) that some prime divisor of q pm − q , q 8 + q 4 + 1 or q 9 − q , or q 6 − 1, respectively, does not divide |CG (b)(∞) | but does divide |E(CG0 (w))|, yielding a contradiction.  Suppose that G0 ∼ = L3q (q) with p = 3. = D4 (q), q > 2, and E(CG0 (w)) ∼  In this case, a component of CG (b) must be isomorphic to SL2 (q), SL3q (q) or q   q q L4 (q) ∼ = Ω6 (q); but by [III17 , 5.4], L3 (q) has no faithful Fq -representation of dimension less than 8, again a contradiction. Thus (5M1) holds. As mp (B) ≥ 4, p ≥ 5. Then P = A w with A abelian and B = Ω1 (A). Let z = Z(P ) = CA (w) = CB (w). Let r be a prime divisor of q p − q not dividing 2m − 1 for any 2m < q p (if q = 1) or 2m < q 2p (if q = −1); see [IG , 1.1]. In particular r = p. Moreover, as 2r−1 ≡ 1 (mod r), 2r > q. It follows that r does not divide |Out(G0 )|; nor does it divide |CG0 (b)|, by an elementary computation. Now by [IA , 4.8.4], we have CG0 (w) = C0 z , where C0 is cyclic, and R1 := Or (C0 ) satisfies R1 = [R1 , z]. Now by Alperin’s Theorem, we may assume that there exists Q ≤ P with CP (Q) ≤ Q and with w ∈ Q and wh ∈ B for some h ∈ NG (Q). Without loss, we may assume that wh = b. As z = Z(P ) ≤ Q, we have that z, w = CQ (w) and so z, b = CQ (b). Hence, A ∩ Q = z, b and Q = z, b, w . Suppose first that b ∈ z . Then Q is extraspecial, so z  NG (Q). But NG (z ) ≤ M by Corollary 5.12, so w and b are M -conjugate, a contradiction. Therefore b = z and Q = z, w . The facts that |NB (Q)| = p2 and h ∈ NG (Q) with wh = z imply that NG (Q)/CG (Q) contains a subgroup S with S ∼ = SL2 (p). Let τ ∈ NG (Q) be a 2-element mapping to the central involution of S. Then τ inverts z, and so τ ∈ M by Corollary 5.12. Also, τ inverts w, and so w, τ acts on R := CG (G0 ). Let R0 be a w -invariant Sylow r-subgroup of R. Assume first that R0 is cyclic. In that case, as NM (R0 ) covers M/R, w maps into [AutM (R0 ), AutM (R0 )] = 1. But then CM (w) contains R0 × R1 , whereas a Sylow r-subgroup of CG (b) is contained in R, and hence is cyclic, a contradiction.

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5. THE LIE TYPE CASE, p > 2

315

Thus, R0 is noncyclic. We argue that (5N)

M controls G-fusion of its elements of order r.

First, R∗ := R0 × R1 ∈ Sylr (M ), since as noted above, r does not divide |AutM (G0 )|, and R1 ∈ Sylr (G0 ). Since mr (R0 ) > 1 = mr (R1 ), it follows that for any g ∈ NG (R∗ ), R0 ∩ R0g = 1 and thus g ∈ M by Lemma 5.13. Thus, NG (R∗ ) ≤ M and R∗ ∈ Sylr (G). In particular by Burnside’s Lemma [IG , 16.2], M controls fusion in Ω1 (Z(R∗ )). Note that as R∗ = R0 × R1 , Z(R∗ ) is noncyclic. To establish (5N), it suffices to show that, for each T ≤ R∗ satisfying Z(R∗ ) ≤ T , M controls NG (T )-fusion in Ω1 (T ). We have T = R1 × T0 , where T0 ≤ R0 . If mr (T0 ) = 1, then Ω1 (T ) = Ω1 (Z(R∗ )) and we have already seen that M controls fusion in Ω1 (Z(R∗ )). If, on the other hand, mr (T0 ) > 1, then for any g ∈ NG (T ), we have T0g ∩T0 = 1, so g ∈ M by Lemma 5.13. This establishes (5N). In particular, both R0 = R∗ ∩ R and R1 = R∗ ∩ G0 are weakly closed in R∗ with respect to G. Finally, we have R0 ∈ Sylr (CM (z)) and CM (z) = CG (z). However, R1 ≤ CM (w). As z ∈ w G , therefore, R1 is G-conjugate to a subgroup of R0 . But R0 ∩ R1 = 1, so the weak closure of R1 in R∗ with respect to G is contradicted. The proof is complete.  In the next two lemmas, we complete the proof that B0 ∈ U(G, M, p) for all non-identity subgroups of B, except perhaps when B0 is a 3-central subgroup and G0 ∼ = U8 (2) or U9 (2). Lemma 5.15. Let B0 be a non-identity subgroup of B. Then B0 ∈ U(G, M, p), except perhaps when p = 3, G0 ∼ = U8 (2), U9 (2), or 2E6 (2), and every element # b ∈ B0 is 3-central and satisfies L3 (CG0 (b)) = 1. Proof. By Corollary 5.12, it remains to treat the cases when G0 ∼ = En (2), B0 = b , and CG0 (b) has a normal subgroup J ∗L0 of index 3, with b ∈ J ∼ = SU3 (2) and L0 ∼ = SL3 (4), SU6 (2), or 2E6 (2)u , according as n = 6, 7, or 8. In particular the conjugacy class of b is uniquely determined; if B0# contains any element not conjugate to b, then B0 ∈ U(G, M, 3). Note also that L0 is a component of CG (b) by Lemma 5.7. To prove that CG (b) ≤ M , it is enough to show that CG (b)∩NG (L0 ) ≤ M ; for then E(CG (b)) ≤ M , whence E(CG (b)) = E(CM (b)) = L0  CG (b). We may choose b1 ∈ B ∩ L0 such that CL0 (b1 ) = b1 × b × K1 , with K1 ∼ = 1, U4 (2), or D4 (2), according as n = 6, 7, or 8. Suppose that b1 ∈ bG0 and let  C = O 3 (CG0 (b1 )) = J1 ∗ L1 , with J1 ∼ = J and L1 ∼ = L0 . By solvable L3 -balance [IG , 13.8], O3 (J) ≤ L1 . (The alternative, that O3 (J) ≤ J1 , would force b = Z(O3 (J)) = Z(O3 (J1 )) = b1 , contradiction.) In particular, b ∈ L1 . As K1 ≤ L1 by L3 -balance, it follows that F ∗ (CL1 (b)) contains b1 × O3 (J) × K1 . But given the isomorphism types of K1 and L1 , this is impossible (see [IA , 4.7.3A, 4.8.2]). Hence b1 ∈ bG0 . But then by the first paragraph of the proof, b1 ∈ U(M, G, 3). Let g ∈ CG (b) ∩ NG (L0 ). Then bg1 ∈ Lg0 = L0 ≤ M , whence g ∈ M . Hence CG (b) ≤ M , as claimed.  We now eliminate the 2E6 (2) exceptional case. Lemma 5.16. Let 1 = B0 ≤ B. Then B0 ∈ U(G, M, p), except perhaps when # p = 3, G0 ∼ = U8 (2) or U9 (2) and every b ∈ B0 is 3-central in G0 and satisfies L3 (CG0 (b)) = 1.

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316

16. THEOREM C∗ 7 : STAGE 5+. G = G0

∼ 2E6 (2) and we may assume Proof. Suppose not. Then by Lemma 5.15, G0 = that B0 = Z(P ) = b , with P ∈ Syl3 (G0 ). Again the conjugacy class of b in M is uniquely determined, so b1 ∈ U(G, M, 3) for all b1 ∈ B # − bM . In fact, by [IA , 4.7.3A], every b1 ∈ I3 (G0 ) satisfies m3 (CG0 (b1 )) = 5 = m3 (G0 ), so b1 lies in an M -conjugate of B. Thus b1 ∈ U(G, M, 3) for all b1 ∈ I3 (G0 ) − bM . Let C = CM (b). Then by [IA , 4.7.3A], and as |Out(G0 ) : Outdiag(G0 )| = 2, C has a normal subgroup L0 = L1 ∗ L2 ∗ L3 with b ∈ Li ∼ = SU3 (2) for each i = 1, 2, 3 and with |C : L0 |2 ≤ 2. Moreover there is u ∈ C of order 3 cycling L1 , L2 , L3 by conjugation. Set Pi = O3 (Li ) ∼ = 31+2 , i = 1, 2, 3, and let Q1 ∈ Syl2 (L1 ), so that Q1 ∼ = 31+6 . = Q8 . Then O3 (C) = O3 (L0 ) ∼ M We claim first that b1 ∈ b for any b1 ∈ CO3 (C) (L1 ) − b . If on the contrary b1 ∈ bM , then C1 := CM (b1 ) has a normal subgroup J = J1 ∗ J2 ∗ J3 , with Ji ∼ = SU3 (2). But L1 ≤ CM (b1 ). As |C1 : J|2 ≤ 2, it follows that |Q1 ∩ J| ≥ 4. Hence, P1 = [P1 , Q1 ∩ J] ≤ O3 (C1 ). But then, b ∈ [P1 , P1 ] ≤ [O3 (C1 ), O3 (C1 )] = b1 , a contradiction proving the claim. Consequently there exists B1 ≤ O3 (L2 L3 ) with b ∈ B1 ∼ = E32 and with E1 (B1 ) ⊆ U(G, M, 3). Now, we claim that b ∈ Syl3 (O3 3 (C ∗ )), where C ∗ := CG (b). For, P ∈ Syl3 (C ∗ ), and so a Sylow 3-subgroup of O3 3 (C ∗ ) lies in one of O3 3 (C), namely O3 (L0 ). As C acts irreducibly on O3 (L0 )/ b , it follows that either our claim holds or O3 (L0 ) ≤ O3 3 (C ∗ ). In the latter case, as B1 ≤ O3 (L0 ) and B1 ∈ U(G, M, 3), C ∗ ≤ O3 (C ∗ )NG (O3 (L0 )) ≤ M , contrary to assumption. Choose any b1 ∈ B1# . As CG (b1 ) ≤ M , L1 is a solvable 3-component of CG (b1 ) ∩ C ∗ . By solvable L3 -balance [IG , 13.8], P1 lies in a 3-component K1 of C ∗ . If K1 is not u-invariant then we get L3 (C ∗ ) ≥ K1 ∗ K2 ∗ K3 with P2 P3 ≤ K2 K3 . But as CG (B1 ) ≤ M , this implies that M covers K1 /O3 (K1 ), a contradiction. Therefore K1 is u-invariant and so O3 (L0 ) ≤ K1 , with b ∈ K1 . Now K1 ∩ M < K1 contains both ΓB1 ,1 (K1 ) and O3 (L0 ); and P ∈ Syl3 (C ∗ ) normalizes K1 with |P | = |M |3 = 39 or 310 . But there are no such groups K1 and P , by [III17 , 9.18], a final contradiction.  Having handled all of the centralizers of elements of B # (except for the U8 (2) and U9 (2) 3-central cases), it remains to control the p-fusion and the centralizers of elements of order p not M -conjugate to B. Lemma 5.17. Let u ∈ M of order p. Assume that G0 ∈ {U8 (2), U9 (2)}.  Then either CG (u) ≤ M , or G0 ∼ = Lpq (q) and uG ∩ B = ∅.

Proof. If uM ∩B = ∅, then by Corollary 5.16, CG (u) ≤ M , as claimed. Hence, using Lemma 5.14, we may assume that uG ∩ B = ∅. Without loss, u normalizes B.   Lpq (q). As in Lemma 5.14, one of the following cases must Suppose that G0 ∼ = hold by [III17 , 14.7], where we write Gu = E(CG0 (u)): q  (1) G0 /Z(G0 ) ∼ (q), Gu /Op (Z(Gu )) ∼ = Lpm = Lmq (q p ), and m ≥ 2; or   (2) G0 ∼ = E6q (q), p = 3, and Gu ∼ = 3D4 (q) or L3q (q 3 ); or (5O) (3) G0 ∼ = D4 (q), p = 3, q > 2, and u induces a graph automorphism  on G0 , whence Gu ∼ = G2 (q) or L3q (q).

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5. THE LIE TYPE CASE, p > 2

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As m > 1, we see by [III17 , 14.9] that in all cases, mp (CB (u)) ≥ 2 and B ∩ Gu = 1. By Lemma 5.16, B ∩ Gu ∈ U(G, M, p). It suffices to show that Gu  E(CG (u)). For then, as Gu ≤ M and Gu   CG (u), we conclude that CG (u) ≤ M by [III8 , 6.1a], as desired. Now Lemma 5.7 implies that Gu  E(CG (u)) unless possibly Gu ≤ J ∗ for some component J ∗ of CG (u) with J ∗ ∈ Spor − {F1 , F2 , 2F2 } and mp (J ∗ ) > 2. So we must derive a contradiction from the latter conclusion. Set J0 = J ∗ ∩ M < J ∗ and let Pu ∈ Sylp (CM (u)) with B ∩ Gu ≤ Pu . Then Pu ∈ Sylp (CG (u)) as B ∩ Gu ∈ U(G, M, p).  Let Mu = O p (CM (u)) and M u = Mu /Op (Mu ). In case (5O2), we have p = 3  and M u ∼ = Z3 × 3D4 (q) f or Z3 × L3q (q 3 ) f , where f 3 = 1 and f induces a graph or field automorphism on Gu = E(M u ), respectively. Consequently m3 (J ∗ u ) = m3 (Pu ) = m3 (CM (u)) = m3 (M u ) = 4. From [IA , 5.6.1] we then see that the only possibility for J ∗ is J ∗ ∼ = J3 or 3J3 . But in those cases Ω1 (Pu ∩ J ∗ ) is abelian [IA , 5.3h] while Ω1 (P u ∩ Gu ) is nonabelian, a contradiction. If (5O3) holds, then m3 (CM (u)) = 3 and [IA , 5.6.1] yields J ∗ ∼ = 3O  N . Hence ∗ ∗ u = Z(J ) and Pu ∩ J / u is elementary abelian. As Gu contains 31+2 , this is a contradiction. Therefore (5O1) holds. Note that mp (Inndiag(Gu )) = m − 1, and u ∈ Gu , so (5P)

mp (J ∗ u ) = mp (CM (u)) ≤ m + 1

by [IA , 4.8.4]. On the other hand, (5Q)

m

|J ∗ |2 ≥ |M u |2 = q p( 2 ) ; in particular pn

m 2

≤ 21,

by [IA , 3.3.1, 5.3], where q = 2n . By (5Q), m ≤ 4, and if m = 4, then p = 3 and q = 2. Suppose that m = 4. Then |J ∗ |2 ≥ 218 , so J ∗ /O3 (J ∗ ) ∼ = Co1 , Co2 , F i23 , 3F i24 , or F i24 . Accordingly ∗ m3 (J u ) = 7, 5, 7, 8, or 8, by [IA , 5.6.1]. Hence J ∗ ∼ = Co2 and Gu ∼ = L± 4 (8). But then for some g ∈ I3 (Gu ), CJ ∗ (g) contains SL± 3 (8). This is inconsistent with the 3-centralizer structure of Co2 [IA , 5.3k], however. We conclude that m ≤ 3, so mp (J ∗ u ) ≤ 4. If p > 3, then from [IA , 5.6.1], p = 5, and so n is even as q ≡ ±1 (mod 5). Then since |Gu | and |J ∗ | are divisible by 42p − 1, they are divisible by 41, whence J ∗ /Z(J ∗ ) ∼ = F1 or F2 , contradiction. Thus, p = 3. Finally we have m3 (J ∗ u ) ≤ 4 and m3 (J ∗ ) ≥ 3, so J ∗ ∼ = J3 , 3J3 or 3O  N [IA , 5.6.1]. q 3 ∼ Now Gu /O3 (Z(Gu )) = L3 (q ) contains elements of order 9, so J ∗ does as well, ruling out J ∗ ∼ = 3O  N . Then we again reach the contradiction that Ω1 (Pu ∩ J ∗ ) is abelian [IA , 5.3h] while Ω1 (P u ∩ Gu ) is nonabelian. This completes the proof that  Gu  E(CG (u)) and thus, as argued above, the proof of the lemma. We now complete the proof of Proposition 5.1 in all but two cases. Lemma 5.18. Assume that G0 ∈ {U8 (2), U9 (2)}. Then M is a strong p-uniqueness subgroup of G. Proof. By the preceding lemma, either M is strongly p-embedded in G, or  G0 ∼ = Lpq (q). In the first case, (5C1) holds, and in the second, we verify (5C2). We have P = A w with A abelian, Ω1 (A) = B, and CP (w) = w, z , where z = Z(P ) ≤ B. Thus if 1 = X ≤ P with either mp (X) ≥ 2 or mp (CP (X)) ≥ 3,

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16. THEOREM C∗ 7 : STAGE 5+. G = G0

318

then X ∩ B = 1. By Lemma 5.16, X ∩ B ∈ U(G, M, p). Hence NG (X) ≤ M by  [III8 , 6.1a], completing the proof of the lemma. As G does not contain a proper strong p-uniqueness subgroup by our hypothesis, we have reached a final contradiction, completing the proof of Proposition 5.1, except in the two excluded cases. We shall now treat these cases by using our special case of Holt’s Theorem [II2 , Theorem SF, p. 24]. Note that by definition of γ(G) [III12 , (1F)] and the fact that p ∈ γ(G), (5R)

G is of even type.

Also it is obvious from the structure of M that G is not isomorphic to A9 or a simple Bender group, the groups in the conclusion of Holt’s Theorem. Therefore by Holt’s Theorem, (5S)

No 2-central involution of G lies in U(G, M, 2).

Under the assumption (5T)

G0 ∼ = Un (2), n = 8 or 9,

we shall, however, find a 2-central involution in U(G, M, 2), contradicting (5S) and completing the proof of Proposition 5.1. Lemma 5.19. Assume (5T). Then every involution of M centralizes an element b ∈ (B # )M such that E(CG0 (b)) = 1, whence CG (b) ≤ M . Proof. Let t ∈ G0 ∼ = Un (2) be an involution. Let r be the rank of [V, t], where V is the natural module for SUn (2), which covers G0 . Let  t be the involution of t) has a subgroup L0 ∼ SUn (2) lifting t. By [III17 , 6.34], CSUn (2) ( = SUn−2r (2) acting on [V, L0 ] as a natural module. If n − 2r > 1, then L0 contains an element b of order 3 with dim([V, b]) = 2, whence E(CG0 (b)) = 1, as desired. If n − 2r ≤ 1, then [III17 , 6.34] provides a subgroup L∗0 ∼ = SUr (2) acting on [V, L∗0 ] as the direct sum of two natural modules. We can then find an element b ∈ J of order 3 with dim([V, b]) = 4, whence CG0 (b) has a component Lb ∼ = SU (n − 4, 2), as desired. Now suppose that t ∈ M − G0 is an involution. Then by [IA , 4.9.2], CG (t) ∼ = Sp8 (2) or the centralizer of a transvection in Sp8 (2). In any case, CG0 (t) contains b of order 3 with dim([V, b]) = 2, and we are done as before.  Lemma 5.20. Assume (5T). Let S ∈ Syl2 (M ) and z ∈ Ω1 (Z(S))# . Then CG (z) ≤ M . Proof. Suppose on the contrary that CG (z) ≤ M . Let g ∈ G and suppose that z g ∈ M . By Lemma 5.19, z g centralizes b ∈ (B # )M with E(CG0 (b)) = 1. Now by Lemma 5.16, b ∈ U(G, M, 3). Hence by [III8 , 6.1f], it follows that g ∈ M , whence z ∈ U(G, M, 2). Consequently S ∈ Syl2 (G), so z is 2-central in G. But this contradicts (5S), and the lemma follows.  Lemma 5.21. Assume (5T). Then |CG (G0 )| is odd. Proof. Suppose not. Set R = CG (G0 ), let S ∈ Syl2 (M ), and let z ∈ Ω1 (Z(S))∩ R# . Then G0 ≤ Cz := CG (z). In particular, as O3 (Cz ) = ΓD,1 (O3 (Cz )), we have O3 (Cz ) ≤ M and then [O3 (Cz ), G0 ] = 1, whence O3 (Cz ) ≤ R. Let Pz ∈ Syl3 (CM (z)). Then Pz ∩ G0 ∈ Syl3 (G0 ); in particular we choose b0 ∈ Pz ∩ G0 such that E(CG (b0 )) ∼ = SUn−2 (2). By Lemma 5.16, b0 ∈ U(G, M, 3) and so Pz ∈ Syl3 (Cz ). Now by [IG , 5.30], G0 ≤ E for some component E of E(Cz ). But as CG (b0 ) ≤ M , E(CE (b0 )) = CM (b0 )(∞) = E(CG (b0 )) is unambiguously in Chev(2),

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5. THE LIE TYPE CASE, p > 2

319

so E ∈ Chev(2) by [III11 , 1.1ab]. Also there is a toral subgroup A ≤ G0 with A∼ = U4 (2) = 1, so ΓA,1 (G) ≤ M by Lemma 5.16. Hence = E34 and E(CG0 (A)) ∼ by [IA , 7.3.8], E = ΓA,1 (E) ≤ M , whence E = G0 . As b0 ∈ U(G, M, 3) and G0   Cz , we conclude that Cz ≤ M . This contradicts Lemma 5.20 and completes the proof.  Lemma 5.22. Assume (5T). Let z be a 2-central involution of G0 . Then CG (z) ≤ M . Proof. Let Cz = CG (z) and let Uz , Lz , and B0 be as in [III17 , 6.33]. By [III17 , 6.33d] and Lemma 5.15, ΓB0 ,1 (G) ≤ M. Set L0 = E(Lz ). We have O3 (Cz ) ≤ ΓB0 ,1 (G) ≤ M and then, O3 (Cz ) ≤ O3 (CM (z)) = RUz . But then by Lemma 5.21 and the fact that G is of even type (5R), R ∩ O3 (Cz ) ≤ O2 (Cz ) = 1, whence O3 (Cz ) ≤ Uz as |R| and |Uz | are coprime. Since Lz acts irreducibly on Uz / z , either O2 (Cz ) = z or O2 (Cz ) = Uz . We set E = E(Cz ). Also let Pz ∈ Syl3 (CM (z)) with B0 ≤ Pz . Then as B0 ∈ U(G, M, 3), Pz ∈ Syl3 (Cz ). Suppose first that O2 (Cz ) = z . Then F ∗ (Cz ) = z E. As O3 (Cz ) = z and Pz ∈ Syl3 (Cz ), it follows that 1 = Pz ∩ E, so as M ∩ E  CM (z), [III17 , 6.33c] yields Uz = [Uz , M ∩ E] ≤ E. Choose a component J of E such that [Uz , J] = 1. Then in particular J ≤ M . Since B0 is noncyclic and ΓB0 ,1 (G) ≤ M , we may apply [IA , 7.7.17] and conclude that J/Z(J) ∼ = L3 (4) or An , n ≥ 11. As G is of even type, each component J of E is a C2 -group, so the alternating groups are impossible and thus J/Z(J) ∼ = L3 (4). But then as m3 (B0 ) ≥ 3 and m3 (Aut(L3 (4))) = 2, CB0 (J) = 1, so J ≤ ΓB0 ,1 (G) ≤ M . This is a contradiction. Thus, O2 (Cz ) = Uz . Now, [Uz , Pz ∩ E] = 1, whereas CM (Uz ) ≤ R z , a 3 group. Therefore Pz ∩ E = 1, whence E ≤ O3 (Cz ) = 1. Hence, F ∗ (Cz ) = Uz and R = 1. Set C = Cz /Uz . As O3 (Cz ) = Uz , we have F (C) = O3 (C)  ≤ Pz . If F (C) = 1, it follows from [III17 , 6.33a] that F (C) = O3 (CM (z)) = b , with E(CG0 (b)) ∼ = SUn−2 (2). Therefore with Lemma 5.16, Cz = Uz NCz (b ) ≤ M , and we are done. Hence, we may assume that F (C) = 1. Since L0 is P z -invariant, (5U)

L0 ≤ J, a 3-component of C,

by [IG , 5.30]. Note that Pz ∩ L0 contains an element of B0# ⊆ U(G, M, 3) since |CM (z) : L0 |3 is at most 3. Thus, if equality holds in (5U), then Uz L0   Cz and so Cz ≤ M , and we are done. We may therefore assume that L0 < J. Let b0 , b1 ∈ B0# be as in [III17 , 6.33e] and set B1 = b0 , b1 . As L0 ≤ J and CG (b0 ) ≤ M , CJ (b0 ) has a U5 (2) component. This being unambiguously in Chev(2), it follows that J ∈ Chev(2). But again by [III17 , 6.33e], B 1 centralizes a nontrivial 2-subgroup of L0 and hence of J. Thus by [IG , 7.3.1], J ≤ ΓB ,1 (J). Since B1 ≤ B0 and O2 (Cz ) ≤ 1

M , this implies that J ≤ CM (z). But then J = L0 , a final contradiction.



Now Lemmas 5.18, 5.20 and 5.22 show that G has a strong p-uniqueness subgroup, which, as p ∈ γ(G), contradicts the definition of γ(G). Thus the proof of Proposition 5.1 is complete. This in turn completes the proof of Theorem C∗7 : Stage 5+, and with it the proof of Theorem C∗7 .

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10.1090/surv/040.8/06

CHAPTER 17

Preliminary Properties of K-Groups 1. Weyl Groups and Their Representations Lemma 1.1. Let Σ be the root system of a simple complex Lie algebra. Let Σ0 be a subsystem, and Π0 a fundamental system in Σ0 . If RΣ0 ∩ Σ = Σ0 , then there is a fundamental system in Σ containing Π0 . Proof. Expand Π0 to an ordered basis B of RΣ, putting the elements of Π0 last. Order RΣ by setting v > w if and only if the first nonzero coefficient of v − w, when expressed as a linear combination of the ordered basis B, is positive. Call an element v ∈ RΣ with v > 0 positive, and call a positive root α ∈ Σ indecomposable if and only if it cannot be expressed as the sum of two positive roots (in Σ). Then, as is well-known, the indecomposable positive roots form a fundamental system Σ1 for Σ. However, if α = β + γ with α, β, γ > 0, α ∈ Π0 , and β, γ ∈ Σ, we see that the first nonzero coefficients of β and γ must occur at elements of Π0 . Thus β ∈ RΠ0 ∩ Σ = RΣ0 ∩ Σ = Σ0 and similarly for γ, contradicting the indecomposability of fundamental roots. This proves that α ∈ Σ1 , so Π0 ⊆ Σ1 and the lemma is proved.  Lemma 1.2. Let W = W (L) be the Weyl group of a simple complex Lie algebra L, with L = G2 . Let r1 , r2 , r3 ∈ W be reflections such that W12 := r1 , r2 ∼ = W (A2 ) or W (C2 ). Let W123 = r1 , r2 , r3 . Then W123 ∼ = W12 , W (A1 ) × W12 , W (A3 ), or W (C3 ). Proof. As a finite group generated by reflections, W123 has rank 2 or 3. In the former case it equals W12 , since G2 has been excluded. In the latter case, if it is not W (A1 ) × W12 , then it is indecomposable, so it is isomorphic to W (A3 ) or  W (C3 ). The next lemma will be used in a Gilman-Griess characterization of groups in Lie of type E. Lemma 1.3. Let Σ be a root system of type En , 6 ≤ n ≤ 8. Let Σ1 be an indecomposable subsystem of Σ of rank at least 3. Then for any α, β ∈ Σ there exists w ∈ W (Σ) such that w({α, β}) ⊆ Σ1 . Proof. Replacing β by −β if necessary, we may assume that (α, β) ≤ 0. Since Σ1 has rank at least 3 there are α , β  ∈ Σ1 such that (α , β  ) = 0. But α⊥ ∩ Σ is indecomposable (of type A5 , D6 , or E7 ) and it follows that W (Σ) is transitive on orthogonal pairs of roots. Hence we are done if (α, β) = 0. Finally, if (α, β) < 0 then since Σ has rank at least 3, (Rα + Rβ) ∩ Σ is of type A2 . Hence by Lemma 1.1, the pair (α, β) is W (Σ)-conjugate to some pair of non-orthogonal fundamental roots, and then to any such pair. The same is true for any pair of non-orthogonal  fundamental roots of Σ1 , and the result follows. 321 Licensed to AMS. License or copyright restrictions may apply to redistribution; see https://www.ams.org/publications/ebooks/terms

322

17. PRELIMINARY PROPERTIES OF K-GROUPS

Lemma 1.4. Let p be an odd prime and V a finite-dimensional F-vector space, F a finite field of characteristic p. Let W be a subgroup of GL(V ) generated by reflections and acting indecomposably on V . If W has a noncyclic normal abelian 2-subgroup A, then V has a basis B such that W = F S  F , where S ∼ = ΣB permutes B naturally, and F contains A and is represented as diagonal matrices with respect to B. Proof. Let R be the set of reflections in W . By assumption A  W for some noncyclic 2-subgroup A. Then W preserves the decomposition V = ⊕m i=1 Vi of V into A-homogeneous constituents. As A is noncyclic, m > 1. Since W acts indecomposably on V , W permutes the Vi transitively. If dim(V1 ) > 1 then any r ∈ R fixes a vector in V1 and hence normalizes V1 , contradicting the transitivity of W = R . So dim Vi = 1 for all i = 1, . . . , m. Let F = ∩m i=1 NW (Vi ), the kernel of the natural homomorphism φ : W → Σ{V1 ,...,Vm } . Thus A ≤ F . Any reflection r ∈ R at most interchanges two Vi ’s, so φ(r) is either 1 or a transposition. Hence φ(W ) is a transitive group generated by transpositions, so it is all of Σ{V1 ,...,Vm } . In particular there exist reflections rij , 1 ≤ i < j ≤ m, mapping to the transpositions interchanging Vi and Vj , and then also centralizing all other Vk ’s. Choosing an arbitrary nonzero v1 ∈ V1 , we r define vi+1 = vi i,i+1 for i = 1, . . . , m − 1 and have that vi ∈ Vi , ri,i+1 interchanges vi and vi+1 , and ri,i+1 fixes all other vk ’s. Putting S = r12 , . . . , rm−1,m we see that S ∼ = Σm permutes the basis B = {v1 , . . . , vm } naturally. This completes the proof.  We come to a main result of this section, characterizing modules V (written additively) satisfying the following conditions: (1) W ∼ = W (L) is a Weyl group of a finite-dimensional simple complex Lie algebra L of rank n ≥ 3; and (2) p is an odd prime, F is a finite field of characteristic p, and V is (1A) a faithful FW -module; (3) The involutions of W corresponding to reflections in W (L) (on the root lattice Λ := ZΣ) also act as reflections on V . Finite groups generated by reflections in positive characteristic have been treated before, for example by Wagner [Wag1], [Wag2] and by Zalesski˘ı and Sereˇzkin [ZS]. However, we have chosen to derive what we need. In (1A), we define FW -modules Sp and Sp∗ by Sp = F ⊗Z Λ and Sp∗ = the dual module to Sp , (1B) and shall eventually prove: Proposition 1.5. Assume (1A) and (1B). Let V o = CV (W ) and V = V /V o . Then one of the following holds: (a) dim V = n. As FW -module, V is quasi-equivalent to Sp∗ , and indeed isomorphic to it unless possibly L = D4 ; or (b) dim V = n − 1. Moreover, L = An with p dividing n + 1, or L = E6 with p = 3 and n = 6. As W -module, V is isomorphic (quasi-equivalent if n = 5) to the unique nontrivial composition factor of Sp and that of Sp∗ . Definition 1.6. Whenever (1A) holds, we write W ≡V W (L)

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to indicate that one of the conclusions of Proposition 1.5 holds. If the module V is clear from context, we just write W ≡ W (L). The proof of Proposition 1.5 proceeds in a sequence of lemmas. Let R be the set of all reflections in W (on Λ). Since p is odd, we may pass to V and assume that V o = 0. Lemma 1.7. The FW -module V is indecomposable and dim V ≤ n. Moreover, with the possible exceptions given in (b) of the proposition, Sp ∼ = Sp∗ . Proof. If there is a decomposition V = V1 ⊕ V2 , then any element of R must centralize either V1 or V2 , and so R is not irreducible, contradicting the simplicity of L. Any k reflections jointly centralize a subspace of V of codimension at most k. Thus, since W is generated by n reflections and V o = 0, dim V ≤ n. Finally, with the exceptions given in (b) of the proposition, as p is odd, the mod-p reduction of  the Euclidean form on Λ is still nondegenerate and W -invariant, so Sp ∼ = Sp∗ . Now we start the case analysis. The following century-old result for Weyl groups of type A is due to L. E. Dickson. (For a modern reference interpreting Dickson, see [Wag3, 2.1].) Lemma 1.8. Let X = Σn , n ≥ 4, let p be an odd prime, and suppose that V is a faithful FX-module. If p divides n, then dimF (V ) ≥ n − 2, and if p does not divide n, then dimF (V ) ≥ n − 1. If dimF (V ) = n − 2 or n − 1, respectively, and transpositions in X are represented by reflections on V , then V is isomorphic to the core of a standard permutation module for X. Proof. This is proved in [Di2]. Observe that since |Aut(A6 ) : Σ6 | = 2, Σ6 has two nonisomorphic but quasi-equivalent standard permutation modules over F.  By the “core” of a standard permutation module we mean the unique nontrivial composition factor of that module. We recall the following structure theorem for a standard permutation module for Σn and An . Lemma 1.9. Let K = Σn or An , n ≥ 4. Let F be a field of characteristic p > 0, and let V be a standard permutation module 1 , . . . , vn } n for K over F, with n basis {v n = F ( v ) and V = α v | permuted naturally by K. Let V 1 i 2 i i i=1 i=1 i=1 αi =  0 . Then V1 and V2 are the only proper submodules of V . Both V1 and V /V2 are trivial 1-dimensional modules. If p divides n, then V1 ⊆ V2 and the core of V is V2 /V1 . If p does not divide n, then V = V1 ⊕ V2 and V2 is the core of V . In any case, the core of V is absolutely irreducible for K.  Proof. Obviously V1 ⊆ V2 if and only if p divides n. Let v = ni=1 αi vi ∈ V − V1 . Without loss, α1 = α2 . If K = Σn , let g = (12) ∈ K. Then [v, g] = (α2 − α1 )(v1 − v2 ) so the submodule U generated by v contains all vi − vj , whence U = V2 or V . Hence V1 and V2 are the only submodules. If K = A4 , then K is a Frobenius group with complement of order 3, so V2 is absolutely irreducible.  αi vi ∈ V − V1 again, and set h = (123). Then If K = An , n ≥ 5, let v = 3 u := [v, h] = i=1 βi vi ∈ V2 with β2 = α2 − α1 = 0. As [u, (245)] = β2 (v4 − v2 ), the submodule generated by v is V2 or V . Since these calculations are over an arbitrary

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324

17. PRELIMINARY PROPERTIES OF K-GROUPS

field of characteristic p, the core of V is absolutely irreducible, completing the proof.  Lemma 1.10. Proposition 1.5 holds if L = An , n ≥ 3. Proof. In this case W = W (An ) ∼ = Σn+1 . Suppose first that dim V = n. There exists a subgroup W1 ≤ W such that W1 ∼ = Σn and W1 is generated by only n − 1 elements of R. Hence W fixes a nonzero element v ∈ V , and so the 1  W -submodule v W is a nonzero quotient of a natural permutation FW -module U (when n = 5, the one in which elements of R act as reflections). Now if p does not divide n + 1, then it follows from Lemma 1.8 that V is the unique nontrivial composition factor of U , so we are done in that case. Assume then that p divides n + 1. If V were irreducible, it would be a quotient of U ; but U has composition factors of dimensions n − 1, 1, and 1, contradiction. Thus V is reducible. Now Lemma 1.9 implies that the composition factors of V are the core U0 of U , and a single 1-dimensional module U1 . Since elements of R act nontrivially on U0 they must, as reflections, act trivially on U1 . It follows that V is indecomposable extension of a submodule isomorphic to U0 by a 1-dimensional trivial module. Now U has a unique composition series by Lemma 1.9, so U0 is not a quotient of U , and hence V is the unique quotient of U of dimension n. Note that since Sp consists of all ai − aj with respect to a basis {ai } of U permuted by W , Sp has a trivial submodule when p divides n + 1, so V ∼ = Sp∗ . We have proved that if dim V = n, then conclusion (a) of the proposition holds. On the other hand, if dim V < n, then by Lemma 1.8, conclusion (b) of the proposition holds. (Note that U is self-dual; an invariant bilinear form is obtained by taking the ai above to be orthonormal.) In the exceptional case n = 5, there are two quasi-equivalent permutation modules of dimension 6, giving rise to quasiequivalent cores.  Lemma 1.11. Proposition 1.5 holds if L = Cn or Dn , n ≥ 3. Proof. Let E be a normal elementary abelian 2-subgroup of V of largest order. Then W/E ∼ = Σn , and except for L = D4 , E is uniquely determined by W ; for D4 , there are three choices. As n ≥ 3, |E| > 2. By Lemma 1.4, and as W/E ∼ = Σn in W that permutes naturally a basis = Σn , E has a complement P ∼ B of V . Thus W = EP , and with respect to B, P is represented as the group of all permutation matrices, while E is either the full exponent 2 subgroup E ∗ of the group of all diagonal matrices, or a maximal P -invariant subgroup of E ∗ . But E ∗ is the natural F2 P -module so it has a unique submodule of codimension 1. There are thus at most two choices for V , and these must be Sp∗ for W (Cn ) and W (Dn ), as desired.  Since the root system of type Bn is dual to that of Cn , Lemma 1.11 handles the case L = Bn as well. For the next case, L = F4 , we need a lemma. Lemma 1.12. Let φ : X → GL(V ) be a representation of the group X over a field F of characteristic p > 0, and suppose that φ|F ∗ (X) is absolutely irreducible. Then φ|[X,X]Op (X) is uniquely determined by φ|F ∗ (X) . (The uniqueness is as a mapping, and not just up to equivalence.) Likewise the composite mapping X → GL(V ) → P GL(V ) is uniquely determined.

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Proof. Let Q = F ∗ (X). Suppose ψ is another F-representation of X over F, with φ|Q = ψ|Q . Then for any x ∈ X, φ(x) and ψ(x) induce the same mapping on φ(Q) by conjugation, so by absolute irreducibility, λ(x) := φ(x)ψ(x)−1 is a scalar, proving the second statement. For any x, y ∈ X, φ([x, y]) = [φ(x), φ(y)] = [λ(x)ψ(x), λ(y)ψ(y)] = [ψ(x), ψ(y)] = ψ([x, y]). Finally, φ(xm ) = λ(x)m ψ(xm ) for any m. Thus if x ∈ X is a p-element, then so is λ(x), whence λ(x) = 1, as desired.  Lemma 1.13. Proposition 1.5 holds if L = F4 . Proof. We regard W ≤ GL(V ). Let Q = O2 (W ) = F ∗ (W ) ∼ = Q8 ∗ Q8 . Then up to equivalence, Q has a unique faithful F-representation of degree at most 4, and it is absolutely irreducible. By Lemma 1.12, the action of each element of W is determined up to a scalar mapping. In particular as no nontrivial scalar multiple of a reflection is a reflection (since n > 2), the action of each element of R is unique, proving the lemma.  For the case L = E6 we need more preparation, especially if p = 3. We have W (E6 ) ∼ = SO5 (3) ∼ = Aut(P Sp4 (3)) ∼ = Aut(U4 (2)). Lemma 1.14. Let X = SO5 (3) and let φ : X → GL5 (Fpa ) be a faithful representation for some odd prime p. Then p = 3, φ is writable over F3 , and the corresponding F3 Y -module is a natural module, where Y = [X, X] = Ω5 (3). Moreover, if φ(X) is generated by reflections, then φ is determined up to equivalence. Proof. First suppose that p > 3. Let z be a 3-central element of X of order 3. Then z = Z(P ) for some P ≤ X with P ∼ = 31+2 . Moreover, z is conjugated by some involution t ∈ X to z −1 . (This follows from the fact that z is a fundamental root subgroup with respect to a suitable torus, and X ∼ = Inndiag(Y ).) Then t normalizes P = O3 (NX (z)). As p > 3, z must have each primitive cube root of unity at least three times as an eigenvalue; but the degree of φ is only 5, contradiction. Thus, p = 3. We next show that V is a natural Y -module. Since X ∼ = Aut(Y ) [IA , 2.5.12], it suffices to show that φ(Y ) is uniquely determined up to conjugacy in GL(V ). Let y ∈ Y be a 2-central involution, so that CY (y) = QT u , where Q ∼ = Q8 ∗ Q8 , T ∼ = E32 , QT = H1 ∗H2 with H1 ∼ = SL2 (3), and u2 = 1 with u interchanging H1 and H2 . Since Q has a unique complex faithful representation of degree 4, φ|Q is determined up to conjugacy in GL(V ), and we show that φ|Q determines φ(Y ). Now, φ(Q) is absolutely irreducible on [V, Q], so given φ|Q , φ(u)|[V,Q] is determined up to a scalar, i.e., up to sign. Since Z(Q) inverts [V, Q], φQu is determined on [V, Q], and then as φ(Q u ) lies in SL(V ), φQu is determined. Similarly, using Lemma 1.12, φ|[T,u] is determined on [V, Q] and then on V . Now let S = Q u ∈ Syl2 (CY (y)). Then J(S) ∼ = E24 and the minimal J(S)-invariant subspaces of V form a frame F in V . Let N be the stabilizer of F in Y and N ∗ its stabilizer in GL(V ). Then φ(N ) ≤ N ∗ . But [N, N ] ≥ [NX (J(S)), NX (J(S))] ∼ = 24 A5 ∼ = [N ∗ , N ∗ ]. Hence, given φ|S , which has already been determined, F and then φ([N, N ]) = [N ∗ , N ∗ ] are uniquely determined. Since [T, u] does not normalize Q u = QJ(S), [T, u] ≤ N . But Y ∼ = U4 (2) ∈ Chev(2), so by [IA , 2.6.7], Y = Q[T, u], [N, N ] . Thus φ(Y ) is uniquely determined by φ(Q). Finally φ|Y determines φ|X−Y up to sign, by Lemma 1.12. If in addition φ(X) is generated by reflections, then det(x) = −1 for

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all x ∈ X − Y , which determines φ|X−Y as φ has odd degree. This completes the proof.  The last paragraph of the proof proves the following lemma: Lemma 1.15. Let K = Ω5 (3) ∼ = P Sp4 (3) and suppose that K is faithfully represented on a 5-dimensional F3 -vector space V . Then K preserves a nondegenerate bilinear symmetric form on V . The group SO5 (3) has a familiar action on E35 not as transformations of determinant 1, but as the reflection group W (E6 ). We will call this action or module the reflection module for SO5 (3). It is self-dual. Lemma 1.16. Let K = SO5 (3) and let M be a faithful F3a K-module of dimension 6, for some a, such that some involutions of K act as reflections on M . Suppose also that for a Σ6 -subgroup J of K centralizing an involution of K, the composition factors of M are the same as the composition factors of a natural F3a Jpermutation module. Then as F3a K-module, M has one trivial composition factor and one SO5 (3) composition factor which is the reflection module. Proof. Let tK ⊆ K be a class of involutions acting as reflections on M . Then t ∈ [K, K] as detM t = −1. Moreover, t cannot invert an element g ∈ K of order 5. (This is because two conjugates of t jointly centralize a subspace of M of codimension 2; on the other hand, g ∈ K is rational, so dim[M, g] ≥ 4.) These facts uniquely determine the conjugacy class of t in K, and so t ∈ SO5 (3) is a reflection on the reflection module and CK (t) = t × J with J ∼ = Σ6 and J generated by reflections. We may assume the isomorphism J ∼ = Σ6 to be such that transpositions in J act as reflections on M . As J-module, M has the trivial submodule M0 = [M, t], one other trivial composition factor, and a 4-dimensional composition factor which is the core of a natural permutation module. Since J preserves the decomposition M = M0 ⊕ CM (t), we see that by replacing M by its dual if necessary, we may assume that there is a 4-dimensional J-submodule M1 ≤ CM (t). Set M o = M0 ⊕ M1 . We claim that M o is K-invariant. Choose an involution u ∈ J which is the product of three disjoint transpositions; then dim([M, u]) = 3 and [M, u] ≤ M1 . Set z = tu, so that z ∈ [K, K], dim([M, z]) = 4, and [M, z] ≤ M o . It follows that z is a 2-central involution of [K, K]. Thus Q := O2 (C[K,K] (z)) ∼ = Q8 ∗ Q8 , and for P ∈ Syl3 (CK (z)), QP/ z ∼ = A4 × A4 . As dim(CM (z)) = 2, Q centralizes CM (z). Since [M, z] ≤ M o this implies that Q normalizes M o . Thus NK (M o ) contains H := Q, t, J . We show that H = K. Let P ∈ Syl5 (J). Then |NK (P )| = 5.23 and NH (P ) contains NJt (P ), of order 5.22 . Hence it follows from Sylow’s theorem that |H|/5 ≡ 3 or 4 (mod 5). However, because of Q, t, and J, |H|/5 = m.26 32 ≡ m (mod 5) for some m dividing 18. Therefore m = 1 or 2, so |K : H| = 18/m ≤ 6. As K does not embed in Σ6 , H = K. This proves our claim. Since t centralizes M/M o , so does K. Hence, the lemma follows from Lemma 1.14.  Lemma 1.17. Let φ : W = W (E6 ) → GL6 (pa ) be a faithful representation for some odd prime p, such that reflections map to reflections. If p = 3, assume that

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there is no trivial subrepresentation. Then the representation is determined up to equivalence. Proof. First suppose that p > 3. Let P ∈ Syl3 (W ). There is a 3-local subgroup L ∼ = Σ3  Z3 of L = Σ3  Σ3 in W and containing P . The subgroup P1 ∼ contains P and has a unique faithful complex representation of degree at most 6, and in particular φ|P1 and φ|P are determined up to conjugacy in GL6 (p). Let Q be the unique subgroup of P isomorphic to 31+2 . Then according as pa ≡ 1 or −1 (mod 3), φ(Q) either has two nonisomorphic (faithful) constituents or is irreducible over Fp . Correspondingly, C := CGL6 (p) (φ(Q)) is isomorphic to Zpa −1 × Zpa −1 or Zp2a −1 . Let N = CW (Z(Q)) = QS with Q  N , CS (Q) = 1, and S ∼ = SL2 (3). Notice that P1 covers W/[W, W ], and [N, N ](N ∩ P ) is a maximal (parabolic) subgroup of [W, W ] but does not contain P1 ∩ [W, W ], so P1 , [N, N ] = W . It therefore suffices to show that given φ|P1 , it follows that φ([N, N ]) is determined uniquely. For then φ will be determined up to quasi-equivalence, and in fact up to equivalence since W is complete. Now for any x ∈ S, φ(x) is uniquely determined modulo C. Hence M := φ(N )C is uniquely determined, and of course C  M . If C ∼ = Zp2 −1 , then [M, M ] centralizes C and φ([N, N ])/φ(Q) = [φ(P ), O2 ([M, M ]/φ(Q))] is uniquely determined. On the other hand, if C ∼ = Zp−1 × Zp−1 , C normalizes the two eigenspaces of Z(Q), as does φ(N ) since Z(Q) ≤ Z(N ). Moreover, C induces scalars on each such eigenspace, so [C, φ(N )] = 1. Then φ([N, N ]) = [M, M ] is again uniquely determined, completing the proof if p > 3. Now suppose that p = 3. From the extended Dynkin diagram there is a subgroup S ∼ = Σ6 of CW (w) for some reflection w ∈ W , such that S is generated by five reflections. Thus S has a trivial 1-dimensional subrepresentation modulo which, as argued in the proof of Lemma 1.10 for the case W (A5 ), there is another trivial composition factor and a 4-dimensional composition factor, the core of the natural permutation module. Hence by Lemma 1.16 and our assumption of no trivial subrepresentations, the representation φ is indecomposable with a 5-dimensional socle. The restriction φ|H of φ to a subgroup H ∼ = 24 Σ5 of W is uniquely determined up to a conjugacy in GL6 (3 ) by Lemma 1.11, since it contains reflections. Let z ∈ I2 (H) be 2-central in H and let C = CW (z) and C0 = O 2 (C) ∼ = SL2 (3) ∗ SL2 (3). Then |C0 : H ∩ C0 | = 3. Write H ∩ C0 = O2 (C0 ) t with t of order 3, and let u ∈ CC0 (t) − H be of order 3. Then W = H, u so it is enough to show that φ(u) is uniquely determined. Now, C has a 4-dimensional constituent ψ, on which O2 (C) = F ∗ (C) is absolutely irreducible, and a 2-dimensional constituent θ on which O2 (C), and z in particular, is trivial. These are uniquely determined up to H-conjugacy since z is. We have φ(u) = ψ(u) ⊕ θ(u). First, ψ(u) is uniquely determined by Lemma 1.12. Also, θ(u) is nontrivial, for otherwise φ(W ) = φ(H, u ) has a trivial constituent, contrary to hypothesis. Now θ(u) has a 1-dimensional trivial constituent, within the socle of φ. So θ(u) is determined, if one chooses the proper basis for the 1-dimensional trivial H-constituent. The proof is complete.  Now we can prove Lemma 1.18. Proposition 1.5 holds if L = E6 . Proof. Suppose that W = W (E6 ). If dim V < 6, then (b) of the proposition holds by Lemma 1.14. So assume that dim V = 6. Now W has no faithful

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4-dimensional representation, and V is indecomposable. But V has no trivial submodule by our original setup. Hence Lemma 1.17 implies that V is unique up to equivalence. An easy calculation shows that the module S3 has a trivial submodule, generated by α1 − α2 + α4 − α5 , where αi , i = 1, 2, 3, 4, 5 are the fundamental roots along the A5 subdiagram of the E6 Dynkin diagram (see (1B)). Hence, V ∼ = S3∗ , and (a) of the proposition holds.  The final step in the proof of the proposition is: Lemma 1.19. Proposition 1.5 holds if L = E7 or E8 . ∼ Z2 ×Sp6 (2). Let φ : W → Proof. Suppose that W = W (E7 ) = Z(W )×W0 = GL7 (pa ) be the corresponding representation. From the extended Dynkin diagram and Lemma 1.10, there exists WA ≤ W with WA ≡V W (A7 ). As WA ∼ = Σ8 , there is a subgroup H ≤ WA which is the holomorph of Q := O2 (H) ∼ = E23 . Let N = NW (Q). By the Borel-Tits theorem, N lies in a parabolic subgroup P of W , with Q ≤ O2 (P ). Since P contains an element of order 7, ∼ L ≤ P ∩ H. By Clifford’s we must have P = T L where T ∼ = E26 and L3 (2) 7 = theorem, Q is multiplicity-free on V , and V = i=1 Vi is the direct sum of the Q-homogeneous subspaces. Then as [T, Q] = 1, T must preserve every Vi , so T is uniquely determined as the largest exponent 2 subgroup of SL(V ) stabilizing all the Vi , 1 ≤ i ≤ 7. Then φ|WA ,T  is uniquely determined by φ|WA . But WA , T = WA , H, T = WA , P = W because P is a maximal subgroup of W and WA ≤ P . Finally, suppose that W ∼ = W (E8 ). Let φ : W → GL8 (pa ) be the corresponding representation. From the extended Dynkin diagram and Lemma 1.11, W has a subgroup W0 ≡V W (D8 ) of the form W0 = AS  A, where A ∼ = E27 acts diagonally and Σ8 ∼ = S, with transpositions in Σ8 corresponding to reflections in S. Then we may assume that φ(A) is the full determinant 1 and exponent 2 diagonal group, and φ(S) is the group of permutation matrices. Let T ∈ Syl2 (W0 ). As |W : W0 | = 135, T ∈ Syl2 (W ), and U := W0 ∩ [W, W ]/Z(W ) is a T -invariant maximal parabolic subgroup of [W , W ] := [W, W ]/Z(W ) ∼ = D4 (2) of shape E26 A3 (2). As elements of T −[W, W ] induce graph automorphisms on [W , W ], they interchange the other two maximal parabolic subgroups containing T ∩ [W, W ] and isomorphic to E26 A3 (2). Let P be one of these parabolic subgroups and P its preimage in [W, W ] ∼ = 2D4 (2). Also let U be the preimage of U in [W, W ]. Since 2D4 (2) ≤ 2D4 (3) ∼ = Ω+ 8 (3) + [IA , Table 6.2.2], we can regard P and U as subgroups of Ω8 (3). Then O2 (U ) ∼ = E27 contains some involutions with exactly 2 eigenvalues equal to −1 on the natural + Ω+ 8 (3)-module; these pull back to elements of Spin8 (3) of order 4, by [IA , 6.2.1]. + As there exists a triality automorphism of Spin8 (3) carrying P to U and acting nontrivially on Z(Spin+ 8 (3)) by [IA , 2.5.12j], it follows that O2 (P ) is not elementary abelian. Since P is irreducible on O2 (P ), O2 (P ) ∼ = 21+6 . Of course O2 (P ) ≤ T , and φ(O2 (P )) is absolutely irreducible. By Lemma 1.12, and as P = [P, P ], φ|P is uniquely determined by φ|W0 . As W = W0 , P and φ|W0 is determined up to equivalence, the lemma follows.  Proposition 1.5 follows from Lemmas 1.10, 1.11, 1.13, 1.18, and 1.19. Remark 1.20. An easy consequence of Proposition 1.5 is that Sp and Sp∗ are absolutely irreducible, unless L = An with p dividing n + 1, or L = E6 with p = 3. Moreover, again excluding these cases, W preserves the mod-p reduction of

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the Euclidean form (which is integral on Σ), which is nondegenerate, so Sp ∼ = Sp∗ . Explicit representations of the root systems Σ are given in [IA , 1.8.8]. In the next several lemmas we investigate the following situation.

(1C)

(1) W ≡V W (L) for some Fp W -module V , where p is an odd prime and L is a finite-dimensional simple complex Lie algebra of rank ≥ 3; we regard W ≤ GL(V ); and (2) W   Y ≤ GL(V ).

Lemma 1.21. Assume (1C), with W ≡V W (E8 ). Then the subgroup of Y generated by its reflections is W , and Y = W Z with Z ≤ Z(GL(V )). Proof. We have W/Z(W ) ∼ = O8+(2), a self-normalizing subgroup of Aut(D4 (2)). As W   Y , [W, W ] is a component of Y . Since W is absolutely irreducible on V and [W, W ] contains a Frobenius group of order 8.7, [W, W ] is also absolutely irreducible on V , whence CY ([W, W ]) consists of scalars. Thus [W, W ] = E(Y ) and Y = W ∗ Z where Z ≤ Z(GL(V )). In particular O2 (Y ) ≤ Z. Let r ∈ Y be a reflection and write r = wz with w ∈ W and z ∈ Z. As r 2 = 1, 2 z = ±1. Now r ∈ O2(Y ), so r inverts an element t ∈ [Y, Y ] = [W, W ] of odd prime order s. Also t = t−2 = [t, r] = r t r ≤ r t , r . As det t = 1, V0 := CV (t) satisfies dim(V /V0 ) = 2. On V /V0 , r t and r are distinct reflections and so t acts nontrivially on V /V0 . If p = s then V = V0 ⊕ V1 with V1 t-invariant and dim V1 = 2. If p = s then since t acts nontrivially on V /V0 , t has a single nontrivial Jordan block on V and it is of size at most 3. Now the reduction of the root lattice modulo 2 gives a natural module for W/Z(W ) ∼ = Ω+ 8 (2), and so the only possibility is that t has order s = 3, with CW/Z(W ) (t) = T × J, T ∼ = t , and J ∼ = O6− (2). As w inverts t and is a scalar on V0 , w centralizes J. Hence the image of w modulo t Z(W ) is uniquely determined. It follows that there is a unique conjugacy class of reflections in Y = W Z. Since W contains one reflection, it contains them all.  Lemma 1.22. Assume (1C), with W ≡V W (F4 ), W (Cn ) or W (Dn ), n ≥ 3. If W  Y and Y is generated by reflections, then one of the following holds: (a) Y = W ; (b) W ≡V W (Dn ) and Y ≡V W (Cn ); or (c) W ≡V W (D4 ) and |Y /W | divides 6. ∼ O2 (W )  Y and O2 (W ) is absolutely Proof. If W ≡ W (F4 ), then Q8 ∗ Q8 = irreducible on V . Thus F ∗ (Y ) = O2 (W ) ∗ Z with Z cyclic and acting on V as scalars. Let t ∈ Y be a reflection. Then obviously t does not shear to Z(O2 (W )). However, if t normalized a subgroup Q ∼ = Q8 ∗ Z4 or SD16 , = Q8 of W , then Q t ∼ and in either case t would shear to Z(O2 (W )), contradiction. We conclude that t interchanges the two Q8 subgroups of O2 (W ) and acts freely on O2 (W )/Z(W ). It follows that the image of t in Out(W ) ∼ = O4+ (2) lies in Ω+ 4 (2), which is the image of W . Therefore Y = W Z. Thus, ζt ∈ W for some scalar ζ, whence ζ 4 = 1. Suppose that ζ 2 = 1. By the correspondence between representations of 2-groups over Q and over Fp , p odd, ζt has eigenvalues i, i, i, −i in the natural representation of W , which contradicts the rationality of the representation. Hence ζ = ±1 so t ∈ W , as asserted.

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Suppose next that W ≡ W (Cn ). Then W has a unique normal E2n -subgroup E, so E  Y and E preserves an ordered frame F = (V1 , . . . , Vn ) in V . Consequently Y lies in the monomial group M preserving the unordered frame F0 = {V1 , . . . , Vn }. We have W = ES where S ∼ = Σn permutes a basis {v1 , . . . , vn } of V with vi ∈ Vi for all i. Also, M = SE ∗ where E ∗ is the stabilizer in M of F . Thus E = Ω1 (O2 (E ∗ )). Suppose by way of contradiction that t ∈ Y − W is a reflection and set W ∗ = W t . Then |W ∗ ∩E ∗ : E| = 2 and W ∗ ∩E ∗ is S-invariant. It follows from Lemma 1.9 that W ∗ ∩ E ∗ = E z , where z acts on v as a scalar mapping of order 4. Then zt ∈ W and the eigenvalues of zt are ζ with multiplicity n − 1 > 1 and −ζ with multiplicity 1, where ζ ∈ Fp and ζ 2 = −1. But zt is a 2-element of W , so again by the equivalence of representation theories of zt over Q and over Fp , the eigenvalues of zt in its natural representation are i and −i, with multiplicities n − 1 and 1, respectively. Again this contradicts the rationality of the natural representation of W. Finally suppose that W ≡ W (Dn ). With the notation of the previous paragraph, W = E0 S where E0 = E ∩ SL(V ). Let t ∈ Y −W be a reflection, chosen if possible not to preserve F0 . If t does preserve F0 , then by our choice, Y preserves F0 and hence normalizes EW = ES ≡ W (Cn ), whence Y ≡ W (Cn ) by the previous paragraph. Hence, we may assume that t does not preserve F0 , whence O2 (W ) ≤ E. Therefore O2 (S) = 1, so n = 4 and O2 (W ) ∼ = Q8 ∗ Q8 . Thus O2 (W ) is absolutely irreducible on V . But O2 (W )  Y , so CY (O2 (W )) =: Z consists of scalars, and Y /Z embeds in Out(O2 (W )) ∼ = Σ3  Z2 . Let u ∈ I3 (W ). Then [O2 (Y ), u] = O2 (W ) contains exactly three u-invariant E23 -subgroups Ui , 1 ≤ i ≤ 3, all normal in O2 (W ). If (1D)

u ∈ Syl3 (Y ),

then Y = O2 (W ) u T Z where T is a 2-group normalizing u . In this case T must normalize some Ui , and then Ui  Y . But then [Ui , u] ∼ = E22 preserves a unique ordered frame F = (V1 , . . . , V4 ) in V , so Ui = Z(O2 (W ))[Ui , u] does as well. Therefore Y preserves the corresponding unordered frame F0 , and Y acts monomially on V . Moreover, we let Y0 = NY (F ), so that Y /Y0 ∼ = Σ4 permutes F0 faithfully. We continue to assume (1D). Since Y is generated by reflections, there are reflections t ∈ Y − Y0 . As t centralizes a hyperplane of V and permutes F0 , it must induce a transposition on F0 . We can then choose reflections ti ∈ Y − Y0 , 1 ≤ i ≤ 3, interchanging Vi and Vi+1 but centralizing the other two Vj ’s. Then choosing 0 = v1 ∈ V1 , set v2 = v1t1 , v3 = v2t2 and v4 = v3t3 . It follows that {v1 , . . . , v4 } is a basis of V permuted faithfully by S := t1 , t2 , t3 ∼ = Σ4 . Still assuming (1D), if Ω1 (O2 (Y0 )) = E ∼ = E24 , then W (C4 ) ≡ W E  Y and Y ≡ W (C4 ) by the previous case. So assume that Ω1 (O2 (Y0 )) = E0 ∼ = E23 . Then E0 ≤ W  Y . As Y is generated by involutions, and S ≤ Y , Y0 /E0 has exponent 2 and is centralized by S. Therefore Y0 ≤ EZ0 , where Z0 is generated by a scalar mapping of order 4. In particular as E ≤ Y0 , |Y0 /E0 | ≤ 2. But this implies that |Y : W | ≤ 2, as desired. Hence we may assume that (1D) fails, so there is U ∈ Syl3 (Y ) such that U ∼ = E32 . As in the W (F4 ) section of the proof, we find that Y , being generated by ∼ reflections, maps into and then onto Ω(W/Z(W )) ∼ = Ω+ 4 (2) = Σ3 × Σ3 , with each

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∼ Σ3 × Σ3 × Z. We reflection mapping into a direct factor. Set Y = Y /O2 (W ) = need to show that Z = 1, or equivalently, there do not exist reflections t1 , t2 ∈ Y such that t1 t2 is an involution in Z. Suppose false, so that t1 t2 = zx for some z ∈ Z of order 4 and some x ∈ O2 (W ). Then ti inverts zx, i = 1, 2. Also,  × Z,  [ in Y := Y /Z(O2 (W )) = W ti , x ] = 1, i = 1, 2. We may assume that t1 normalizes U . Then [U, t1 ] = u ≤ U with [O2 (W ), u] = O2 (W ). This has two consequences: (a) the full preimage of CO2 (W  ) (t1 ) in O2 (W ) is elementary abelian, and (b) no commutator [x, t1 ] with x ∈ O2 (W ) generates Z(O2 (W )). By (a), x2 = 1. Consequently zx has order 4 and [zx, ti ] = (zx)2 = z 2 . But z ∈ Z(Y ) so [x, ti ] = z 2 , contradicting (b). This completes the proof.  Lemma 1.23. Assume (1C) with W ≡V W (D4 ) and p = 3. If W  Y , then Y contains no transvection. Proof. Suppose that t ∈ Y is a transvection. Let Q = O2 (W ) ∼ = Q8 ∗ Q8 . Then Q is absolutely irreducible on V , so [t, Q] = 1. If [t, Q] contains a four-group, then t ∈ A ∼ = A4 for some A, and t has a free summand on V , contradiction. Otherwise, [t, Q] ∼ = Q8 . As Z(CQ (t)) inverts V , CQ (t) has only = Q8 so CQ (t) ∼ 2-dimensional composition factors on V . But CQ (t) normalizes the cyclic group [V, t], again a contradiction. The lemma is proved.  Lemma 1.24. Let B ∼ = Epn for some odd prime p and some n ≥ 5. Let A ≤ Aut(B) and suppose that the reflection subgroup W of A satisfies W ≡B W (Cn ) or W (Dn ). Then the following conditions hold: (a) There is a unique frame B in B fixed pointwise by O2 (W ); (b) There is W1  A such that W1 is generated by reflections and W1 ∼ = W (Dn ); and (c) A = W1 NA (B0 ) for any B0 ∈ B. Proof. Since n ≥ 5, O2 (W ) is elementary abelian and for some frame in B, contains all possible sign changes of determinant 1 on that frame. This quickly implies (a). Since W (Cn ) contains W (Dn ), (b) is immediate. Finally, W is transitive on B, and B is A-invariant by (a) as W  A, so A = W NA (B0 ). Clearly W (Cn ) = W (Dn )O2 (W (Cn )), so W = W1 NW (B0 ), which completes the proof.  Lemma 1.25. Let p be a prime, X a group, and M an Fp X-module such that M = [Op (X), M ]. Then H 1 (X, M ) = 0. In particular, if p is odd and M = [M, O2 (X)], then H 1 (X, M ) = 0. Proof. There is no indecomposable Fp X-module M ∗ such that M := [X,M ∗ ]  M , for M ∗ = CM ∗ (Op (X)) ⊕ M . Hence, H 1 (X, M ) = 0.  ∗

Lemma 1.26. Suppose N  X, N ∼ = W (D4 ), X/N ∼ = Σ3 , and F ∗ (X) = O2 (N ). Suppose also that there is an involution s ∈ X − N such that N s ∼ = W (C4 ). Then the following conditions hold: (a) X ∼ = W (F4 ); and (b) If X  Y ≤ GL4 (p) for some odd p, and N  Y , then Y = XCY (X). Proof. Let R = F ∗ (X) ∼ = Q8 ∗Q8 and X = X/R, and consider X ≤ Out(R) ∼ = ∼ ∼ = O3 (N )  X with CR/Z(R) (O3 (N )) = 1. = Σ3  Z2 . As N = W (D4 ), Z3 ∼ Hence X = NOut(R) (O3 (N )) is uniquely determined up to conjugacy in Out(R). Let P ∈ Syl3 (N ) and T ∈ Syl2 (NN s (P )). Then T ∼ = E23 . Expand P to

O4+ (2)

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Q ∈ Syl3 (X). Then |NX (Q)|2 = |NX (P )|2 = 23 and P = Q ∩ N  NX (Q), so T ∈ Syl2 (NX (Q)). As T is elementary abelian and X = RNX (Q), X splits over R. As H 1 (X/R, R/Z(R)) = 0 by Lemma 1.25, it follows that the isomorphism type of X is uniquely determined. Since W (F4 ) satisfies the hypotheses, (a) is proved. In (b), X  Y /RCY (R) ≤ Out(R) with X = S 1 × S 2 , S 1 ∼ = S2 ∼ = Σ3 , and P ≤ S 1 . Since P  Y /RCY (R), the structure of Out(R) forces X ∼ Y = /RCY (R). Hence Y = XCY (R). As R is absolutely irreducible in GL4 (p), CY (R) consists of scalar matrices and (b) follows.  Lemma 1.27. Let B = E34 , let A ≤ Aut(B), and let W be the subgroup of A generated by all its reflections. Assume that W = 1, A has a normal subgroup isomorphic to Q8 ∗ Q8 , and 3 divides |A|. Then the following conditions hold: (a) A preserves, up to scale, a nonsingular quadratic form on B of + type; (b) Let 1 = S ∈ Syl3 (A). Then S ∼ = Z3 or E32 , S does not act quadratically on B, and |[B, s]| = 32 for all s ∈ S # ; and (c) W ≡B W (D4 ), W (C4 ), or W (F4 ). Proof. Let Q  A with Q ∼ = Q8 ∗ Q8 . As there is a unique faithful irreducible F3 Q-representation, which is of degree 4, Q is uniquely determined in GL(B) up to ∼ conjugacy. As Ω+ 4 (3) = SL2 (3) ∗ SL2 (3) contains Q8 ∗ Q8 , Q preserves a quadratic form q as in (a). The absolute irreducibility of Q implies that q is unique up to scale, which implies that N := NGL(B) (Q) preserves q up to scale. Thus (a) holds. Now + + + |N : Ω+ 4 (3)| = |N : O4 (3)| · |O4 (3) : Ω4 (3)| = 8, so O 2 (N ) = Ω+ 4 (3). Let r ∈ A be a reflection. Then r interchanges the two SL2 (3) central factors of O 2 (N ). Hence S, which is nontrivial by assumption, contains a diagonal element t of order 3 in this central product. Thus t lies in an A4 -subgroup of A, so t has a free summand on B. Hence S does not act quadratically on B. Since orthogonal groups in odd characteristic do not contain transvections, (b) follows. Note also that O4+ (3)/Q ∼ = Σ3 × Σ3 , and the image of any reflection in this quotient must centralize an element of order 3 and so must lie in one of the Σ3 direct factors. Now, a reflection which is a similarity must be an isometry, so W ≤ O4+ (3). Since S = 1, and by the Baer-Suzuki theorem [IG , 15.5], W contains a reflection r inverting some x ∈ S # . If |S|3 = 3, then either W = Q x, r or W = Q x T for some T ∈ Syl2 (NO+ (3) (x )). In either case W is determined up to N -conjugacy, 4 and we get W ≡ W (D4 ) or W (C4 ). Finally, if |S|3 = 32 , then since W is generated by reflections, W = O4+ (3) in view of the structure of O4+ (3)/Q. Hence W is unique,  and W ≡ W (F4 ). This completes the proof. Lemma 1.28. Suppose that E(X) = K ∈ K. Let z be a 2-central involution of X and suppose that C := CX (z) and R := F ∗ (C) = O2 (C) satisfy the following conditions: (a) R = Q ∗ Z where Q ∼ = Q8 ∗ Q8 and Z is cyclic; and (b) |OutX (Q)| = 36 or 72. ∼ If φ : X → P GL5 (p) is an embedding for some odd prime p, then p = 3, X = SO5 (3), X factors through a representation ψ : X → GL5 (p), and ψ(X) fixes a nondegenerate symmetric bilinear form.

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1. WEYL GROUPS AND THEIR REPRESENTATIONS

333

Proof. Our hypotheses imply that Z(Q) = z . Since X embeds in P GL5 (p) for some odd p, there is no Frobenius subgroup of X of order 8.7. Therefore K ∼ = Am for any m ≥ 8. As |Aut(Am )|2 < |CX (z)|2 for m < 8, K ∈ Alt. If K ∈ Spor, then as |Out(K)| ≤ 2, F ∗ (CK (z)) = Q ∗ Z0 for some cyclic Z0 , and O2,3 (CK (z)) ∼ = SL2 (3) ∗ SL2 (3) ∗ Z0 . The tables [IA , 5.3] show that these conditions do not hold for any K ∈ Spor. Thus, K ∈ Chev(s) − Alt for some s. Assumption (b) implies that [OutX (Q), OutX (Q)] contains a Sylow 3-subgroup of Out(Q). Hence either K ∼ = D4 (sa ) for some a, with O 2 (CK (z)) < SL2 (3) ∗ SL2 (3), 2 or else O (CK (z)) ∼ = SL2 (3) ∗ SL2 (3). The D4 case is impossible by the structure of C for s odd [IA , 4.5.1], and by the 2-rank for s even (m2 (X) = m2 (C) ≤ m2 (R) + m2 (C/R) ≤ 3 + 2 = 5, but m2 (D4 (2a )) = 6a [IA , Table 3.3.1]). Hence, O 2 (CK (z)) ∼ = SL2 (3) ∗ SL2 (3). Suppose that s = 2 but K ∈√Chev(s ) for any odd s . Then as a Sylow 2-center in K is cyclic, K has level 2 or 2. As noted above, m2 (K) ≤ 5, so by [IA , Table 1 3.3.1], the only possibilities are K ∼ = U5 (2), 3D4 (2), and 2F4 (2 2 ) . But in these groups, CK (z) involves SU3 (2), L2 (8), and D10 , respectively, a contradiction. Now we may assume that s is odd. Using the structure of C and [IA , 4.5.1] we see that K ∼ = P Sp4 (3), L± 4 (3), or G2 (3). As G2 (3) contains a Frobenius group of order 8.7, it does not occur. Similarly U4 (3) has an E24 -subgroup whose normalizer is transitive on its set of hyperplanes, so by Clifford’s theorem, it cannot occur. In the remaining two cases the Schur multiplier of X is a 2-group [IA , 6.1.4], so X embeds in SL5 (p). Now, suppose K ∼ = E33 : SL3 (3), and hence by = L4 (3). Then K ≥ P1 ∼ Clifford’s theorem, K has no faithful module in characteristic p > 3 of dimension less than 26. Hence, p = 3. But there is an involution t ∈ K such that CK (t) ≥ Z ×J ∼ = Z4 × A6 . As Z × J cannot embed in SL5 (3), we have a contradiction. Hence, K ∼ = Ω5 (3), whence X = K or X ∼ = SO5 (3). By (b), = P Sp4 (3) ∼ |X|2 ≥ 27 , so X ∼ SO (3). As noted above, X embeds via ψ in GL5 (p). Now = 5 Lemma 1.14 yields that ψ|K is equivalent to the inclusion K = Ω5 (3) ≤ GL5 (3). As K is absolutely irreducible, ψ(x) = ±x for x ∈ X − K, and either way, ψ(X)  preserves the form preserved by SO5 (3). Lemma 1.29. Let K = Σn , n ≥ 6, let p be an odd prime divisor of n, and let V be the natural permutation Fp K-module, with a basis {v1 , . . . , vn } permuted    naturally by K. Let V = V /  ni=1 vi , and let M = { ni=1 ai vi ∈ V | m i=1 ai = 0}, so that M is the unique nontrivial composition factor of M and has dimension n − 2. Write xi = v i and xij = xi − xj for all i = j between 1 and n. Let Kij be the stabilizer in K of the points i and j. If x ∈ M and CK (x) contains a subgroup H ∼ = Σn−2 that is generated by reflections, then H = Kij and Fp x = Fp xij for some 1 ≤ i < j ≤ n. Proof. Let g ∈ K act on M as a reflection. Then g has order 2. If g = (ij)(k) · · · then g inverts the 2-space spanned by xij and xk , contradiction. Thus g is a transposition in K. Since CK (x) is generated by reflections, it contains a L, with support S ⊆ {1, . . . , n}, say S = {3, . . . , n}. Then we root Σn−2 -subgroup  xi = 0, we may see that if x = ai xi ∈ CM (L), then ai = aj for all i, j ∈ S. As write x = a1 x1 + a2 x2 . Then since x ∈ M , x is a scalar multiple of x12 , completing the proof. 

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334

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Lemma 1.30. Let p be an odd prime, let B = Ep4 , let A ≤ Aut(B), and let W be the subgroup of A generated by all its reflections. Suppose that W0  W with W0 ∼ = W (D4 ) generated by reflections. Let g ∈ A and suppose that g normalizes b , the center of some reflection rb ∈ W0 . Then the following conditions hold: (a) g normalizes W0 ; and (b) g permutes {u ∈ B # | CW0 (u) ∼ = Σ4 }. Proof. We prove (a), from which (b) is immediate. Since W  A, we may assume that W0 < W . Let Q0 = O2 (W0 ) ∼ = Q8 ∗ Q8 . Then Q0  W and CA (Q0 ) ≤ Z := Z(GL(B)), whence W/W ∩ Z embeds in Aut(Q0 ). As W0 < W , either O2 (W ) contains E ∼ = E24 and W ≤ NGL(B) (E) =: M , a monomial group, or O 2 (W ) ∼ = SL2 (3) ∗ SL2 (3) with Q0 = O2 (O 2 (W ))  A. In the former case, as W0   A, it follows that in fact A ≤ W ∗ Z with W ∼ = W (C4 ), whence W0  A. Hence we may assume that W ∼ = W (F4 ) and W ∗ Z is a subgroup of A ∗ Z of index 2, since otherwise W0  A. Now, W contains exactly two classes of reflections. If g ∈ A normalizes b , the center of the reflection rb , then rbg is a reflection  inverting bg = b , whence rbg = rb . Hence, g fixes the class rbW and W1 := rbW ≤ W0 . As W0 = E0 S with E0 ∼ = E23 and S ∼ = Σ4 , and rb may be taken to be a transposition in S, we see that S ≤ W1 and then, since W1  W0 , W1 = W0 , completing the proof.  Lemma 1.31. Let W = W (Cn ) act faithfully on E ∼ = Epn , where p is an odd prime and n ≥ 4, with W stabilizing an unordered frame {e1 , . . . ,en }, and with {e1 , . . . , en } stabilized by a subgroup S of W isomorphic to Σn . Then W = CW (e1 ), CW (en ) . Proof. By our assumptions, W = AS with A = t1 , . . . , tn ∼ = E2n . Here ti is an involution inverting ei and centralizing ej for all j = i. Then, CW (e1 ) = t2 , . . . , tn Se1 and CW (en ) = t1 , . . . , tn−1 Sen . By [IA , 7.5.2a], S = Se1 , Sen . The lemma then follows immediately.  Lemma 1.32. Let W0 ≤ W1 with W0 ∼ = W (Cn−1 ) and W1 ∼ = W (Cn ), n ≥ 4. If r is a long root reflection in W0 , then r ∈ O2 (W1 ). Proof. Write W0 = R0 S0 with R0 ∼ = E2n−1 and S0 ∼ = Σn−1 permuting a basis of R0 , namely the long root reflections of W0 . In particular r ∈ R0 . Similarly write W1 = R1 S1 and let W 1 = W1 /R1 . If n = 5, then O2 (S0 ) = 1 and so S 0 ∼ = S0 is a maximal subgroup of W 1 . As S0 normalizes R0 , R0 ≤ O2 (W 1 ) and so R0 ≤ O2 (W1 ) in this case. Suppose finally that n = 5. First let Q = O2 (W0 ), an extension of R0 by a fourgroup acting freely on R0 . It follows easily that R0 = J(Q). On the other hand, |W1 : W0 | = 10 and W 0 contains a copy of Σ3 , so W 0 ∼ = Σ4 . Let Q0 = W0 ∩ R1 . Then Q0  W0 , Q0 ∼ = E24 , and W0 /Q0 ∼ = Σ4 . In particular Q0 ≤ J(Q) = R0 so  R0 = Q0 ≤ R1 . The proof is complete. Lemma 1.33. Let p be an odd prime, and let B be an Fp W -module such that W ≡B W (F4 ). Let Σ0 ⊆ Σ be root systems of types A2 and F4 , respectively, and regard W0 := W (Σ0 ) ∼ = W (A2 ) as a subgroup of W . Let W1 be the subgroup of CW (W0 ) generated by all the reflections that it contains. Then W1 ∼ = W (A2 ) and CB (O3 (W1 )) = [B, W0 ].

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2. TORAL SUBGROUPS

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Proof. By the equivalence of the representation theory of Z2 over Q and over Fp , an element t ∈ W is a reflection on QΣ if and only if it is a reflection on B. Let Π = {α1 , α2 , α3 , α4 } be a fundamental system in Σ, with each αi connected to its immediate predecessor and successor (if any). We may assume that α∗ = −2α1 − 3α2 − 4α3 − 2α4 is the lowest root, and α∗ is connected only to α1 in the extended Dynkin diagram. Let Γ0 = {α∗ , α1 } and Γ4 = {α3 , α4 }, both A2 fundamental systems. Let Π0 be a fundamental system in Σ0 . As Σ has no G2 -subsystem, Lemma 1.1 implies that replacing Π0 by a suitable W -conjugate, we can achieve Π0 ⊆ Π, in which case Π0 = {α1 , α2 } or Γ4 . Note further that {α1 , α2 } is W -conjugate to Γ0 . Hence we may assume that Π0 = Γ0 or Γ4 . Let Π1 be the other Γi , and Σ1 = ZΠ1 ∩ Σ. Then Σ0 ⊥ Σ1 so W 0 := W (Σ1 ) ≤ CW (W0 ). As W 0 is generated by two reflections, and every reflection in W has a root for a center, ⊥ W 0 ≤ W1 = W (Σ ∩ Σ⊥ 0 ). But Σ ∩ Σ0 is a subsystem of Σ of rank 2 containing Σ1 , so it must equal Σ1 , whence W1 = W 0 ∼ = W (A2 ), the first assertion of the lemma. Clearly [B, W0 ] = [B, t1 ], [B, t2 ] where t1 , t2 are reflections generating W0 . Now each [B, ti ] is cyclic and W1 -invariant, hence centralized by O3 (W1 ) ≤ [W1 ,W1 ]. Thus, [B, W0 ] ≤ CB (O3 (W1 )). Moreover, |[B, W0 ]| = p2 . On the other hand, W1 lies in a Σ4 -subgroup W4 of W . This is clear from the extended Dynkin diagram of W if Π1 = Γ0 , and otherwise follows by applying an automorphism of W interchanging long and short root reflections (e.g. arising from a graph-field automorphism of F4 (2n )). From the Frobenius action of O3 (W1 ) on O2 (W4 ) we see that O3 (W1 ) has a free summand on B. Thus |CB (O3 (W1 ))| ≤ p2 , whence [B, W0 ] = CB (O3 (W1 )), completing the proof.  2. Toral Subgroups Lemma 2.1. Let K ∈ Chev(2), let p be an odd prime, and let P ∈ Sylp (K) and B ∈ E∗ (P ). Assume that mp (B/B ∩ Z(K)) ≥ 3 and that p divides |Z(Ku )|, where Ku is the universal version of K. Then there is a σ-setup (K, σ) of K and a σ-invariant maximal torus T of K such that P ≤ NK (T ), B = Ω1 (Op (CT (σ))), and NK (B) = NK (T ). We prove this together with the following more general result. Lemma 2.2. Let K ∈ Chev(2) be of level q with overlying algebraic group K. Let p be a prime divisor of q 2 − 1. Let P ∈ Sylp (K) and B ∈ E∗ (P ) with mp (B/B ∩ Z(K)) ≥ 3, and let W be the Weyl group of K. Under any one of the additional conditions (a), (b), or (c) below, B = J(P ) and B lies in a maximal torus T = CK (B) of K. Moreover, CK (B) has odd order and NK (B) is a split extension of CK (B) by a group W1 ∼ = W such that CB (W1 ) ≤ Z(K). (a) p divides q − 1 and K is untwisted; (b) p divides q + 1, K is untwisted, and −1 ∈ W ; or − − (c) p divides q + 1 and K ∼ = A− n (q), n ≥ 3, D2k+1 (q), k ≥ 2, or E6 (q). Proofs of Lemmas 2.1 and 2.2. Let Ku be the universal version of K. We first prove the lemmas assuming that K = Ku . In that case there is no loss in assuming that K also is universal. Let (B, T ) be a σq -invariant Borel-torus pair and set N = NK (T ). We have CK (T ) = T . Let ΓK be the set of graph or graph-field automorphisms, as defined

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336

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in [IA , 1.15.5], normalizing B and T . In each case (a)–(c) of the lemma, we choose g ∈ N ΓK ∩CK (σq ) such that (K, σ) is a σ-setup for K, where σ = σq g; furthermore, we verify that T p := Ω1 (Op (CT (σ))) is conjugate to B. In case (a) we may take g = 1. In case (b) we choose g ∈ N mapping modulo T to −1 ∈ W . In case (c), ΓK N /T ∼ = Z2 × Σn+1 , W (C2k ), or Z2 × W (E6 ), respectively, and we choose g to map to the unique central involution of this quotient. Since σq , by definition, induces the qth power mapping on T , it follows in every case that T p , as defined above, is the largest exponent-p subgroup of T , and mp (T p ) = dim T , the rank of K. It is then straightforward to check in every case that mp (T p ) = mp (K), using [IA , 4.10.3] and the universality of K. (Note that in (b), for example, −1 ∈ W if and only if all the polynomial invariants of W are even.) But T p ≤ K by definition, so T p ∈ Ep∗ (K). Again since K is universal, it follows from [IA , 4.10.3d] that all elements of Ep∗ (K) are K-conjugate. Hence without loss we may take B = T p . Notice also that N contains a Sylow p-subgroup of K [IA , 4.10.2].  We claim that CK (B) = T . For any T -root α, let K α = X α , X −α ∼ = SL2 (F2 ). Then T contains a maximal torus T α of K α , and [B ∩ T α , X α ] = 1. Therefore X α ≤ CK (B). As α was arbitrary, CK (B) = CN (B). It therefore suffices to prove that CW (B) = 1. Now as Fp W -module, B ∼ = Sp , the mod-p reduction of the root lattice Λ, and CW (Λ) = 1. It follows that CW (B) ≤ Op (W ), since for any p -group the representation theories in characteristics 0 and p are equivalent. As mp (B/B ∩ Z(K)) ≥ 3, Op (W ) = 1, completing the verification of our claim. Hence T  NK (B) = N , as B is characteristic in T . In particular CK (B) = CT (σ) has odd exponent, hence odd order. Moreover, in cases (b) and (c), g inverts T elementwise, so W1 := CN (σ, g ) is a complement to CK (B) in CN (σ) = NK (B). In case (a), using standard notation [IA , 2.10], we may take W1 = nα (1) | α ∈ Σ as the desired complement, where Σ is the T -root system of K. This group clearly covers both N /T and NK (B)/CK (B), and as q is even, it is isomorphic to N /T [IA , 1.12.11]. Finally, if −1 ∈ W , it is obvious that CB (W1 ) = 1. Otherwise, let b ∈ CB (W1 )# . Then by Steinberg’s connectedness theorem [IA , 4.1.3], X := CK (b) is a connected reductive group with T as a maximal torus. As N ≤ X and X is connected, X ≤ N , and so X contains some ! T -root group U . Since −1 ∈ W , all roots are W -conjugate and so K =

U

N

= CK (b). Thus b ∈ Z(K), and the

lemma is proved, in case K = Ku . For the general case, assume now that K = Ku /Z for some Z = 1; write K u for the universal version of the algebraic group, and K = K u /Z. There is no loss q (q) or (with p = 3) in assuming that Z is a p-group. Since Z = 1, Ku ∼ = SLkp q E6 (q), where q = ±1 and q ≡ q (mod p). As mp (K/Z) ≥ 3, kp ≥ 5 in the SL case. Let Bu ∈ Ep∗ (Ku ), so that Bu = Ω1 (Op (CT u (σ))) for some maximal torus T u of K u , N u := NK u (Bu ) = NK u (T u ), and CK u (Bu ) = T u . Then T := T u /Z is a maximal torus of K with normalizer N := N u /Z. Let C u = CK u (Bu Z/Z). Then C u normalizes Bu and stabilizes the chain Bu > Bu ∩ Z > 1, so C u /T u = C u /CK u (Bu ) ≤ Op (N u /T u ) ∼ = Op (W ). But Op (W ) = 1, as observed above. Therefore C u = T u . It follows that if we put C = CK (Bu Z/Z), then C = T . In

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2. TORAL SUBGROUPS

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particular, NK (Bu Z/Z) ≤ NK (T ) = N . Since N u contains a Sylow p-subgroup of Ku , N contains a Sylow p-subgroup of K. We may assume then that B ≤ N . Note that Bu ∩Z = 1, since Z = 1. As a module for N /T ∼ =W ∼ = Σkp in the SL case, Bu Z/Z is the core of a standard permutation module for W (Akp−1 ) ∼ = Σkp , generated by vectors vi − vj , 1 ≤ i < j ≤ kp, permuted by Σkp according to  their indices, and subject to the relation kp i=1 vi = 0. Suppose that B ≤ T . Then [Bu Z/Z, B] = 1 so by [IG , 26.8], there exists g ∈ Ip (N /T ) such that [Bu Z/Z, g, g] = 1 = [Bu Z/Z, g]. Say g = (12 · · · p) · · · . Then [v1 − vp , g, g] is nontrivial if p ≥ 5. Hence p = 3; but then every element of I3 (Σ3k ), k ≥ 2, lies in an A4 -subgroup of Σ3k and so has a free constituent on Bu Z/Z, contradicting quadratic action. This proves that B ≤ T , whence B = Ω1 (Op (T )) by maximality. The same conclusion holds in the E6 case, for which Bu Z/Z is the reflection module for W (E6 ) ∼ = SO5 (3), by Lemma 2.3 below (with q = r = 3). Now B ≥ Bu Z/Z by maximality, so CK (B) = T . The remaining assertions of the lemmas follow easily. Since all elements of T have odd order, CK (B) has odd order. The surjection Ku → Ku /Z ∼ = K carries W1 to a complement to T in N , and carries a Sylow p-subgroup of Ku to a Sylow p-subgroup P of K. As B = Ω1 (Op (T )), B = J(P ). Also NK (B) ≤ NK (T ) as T = CK (B). The proof is complete.  Lemma 2.3. Let X = V W , where W = Ω5 (q), q is a power of the odd prime r, and V is the natural Fr W -module. Let Q ∈ Sylr (X). Then J(Q) = V . Proof. We use the method of Mal’cev [IA , 3.3.4]. In Ω7 (q), the stabilizer of a singular 1-space in the natural module is a parabolic subgroup P such that " r ∼ X = O (P ). So we can regard Q ∈ Syl3 (Ω7 (q)). Thus Q = α∈Σ+ Xα , where Σ is a root system of type B3 , which we explicitly realize as the set of all ±ai ± aj , 1 ≤ i < j ≤ 3, and all ±ai , 1 ≤ i ≤ 3; Σ+ consists of those roots for which the first coefficient is positive. We order Σ by a linear functional λ : Σ → Z, with λ(ai ) = N 3−i , N >> 1. Let S = {a1 , a1 ± a2 , a1 ± a3 }, so that S has the following two properties: (i) S is a final segment of Σ, and (ii) S is the unique abelian subset of the positive roots of (maximal) cardinality 5. Using the correspondence between abelian subgroups of U and tame abelian subalgebras of V (U ) as in [IA , 3.3.4], we note that the tame abelian subalgebra of V (U ) corresponding to Q is the unique such subalgebra of maximal order, by properties (i) and (ii). Hence by [IA , 3.3.4], Q = J(U ).  Lemma 2.4. Let K ∈ Chev(2) and let p be an odd prime such that mp (K/Z(K)) ≥ 3. Let P ∈ Sylp (K). Then J(P ) is elementary abelian. The same conclusion holds for any Sylow p-subgroup of any overgroup of Inn(K) in Inndiag(K). Proof. This follows immediately from [IA , 6.1.4] and [IA , 4.10.3c], with the  help of [IA , 4.10.3a]. Lemma 2.5. Let K ∈ Chev(2) be of level q with overlying algebraic group K. Let p be an odd prime divisor of |K| but not of |Outdiag(K)|. Also exclude K = 3D4 (q) if p = 3. Let B ∈ Ep∗ (K). Then B lies in a maximal torus of K.

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17. PRELIMINARY PROPERTIES OF K-GROUPS

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Table 17.1. AutK (B) ≡ W (L), K ∈ Chev(2), B ∈ Ep∗ (K), p > 2, q ≡ q (mod p), q = ±1, mp (B/CB (K)) ≥ 3 

−

n

−n q

Dn

Cn−1

K Anq (q) An q (q) Cn (q) Dnq (q) Dn L

An

C[(n+1)/2]

Cn



−q

(q) F4 (q) E6q (q) E6 F4

E6

(q) E7 (q) E8 (q)

F4

E7

E8

Proof. Let (K, σ) be a σ-setup for K. By [IA , 4.10.3e], the result holds if p is a good prime for K, and in particular if K is a classical group. In most cases where p is bad and p divides q 2 − 1, one finds a σ-invariant subgroup Z × I ≤ K such that Z is a 1-torus, I is a classical group, p divides the order of Z := CZ (σ),  and I := O 2 (CI (σ)) has p-rank mp (K) − 1 [IA , 4.10.3a]. Then mp (ZI) = mp (K) p and as E∗ (K) is a single conjugacy class, by [IA , 4.10.3d] and our hypotheses, ZI contains a conjugate of B, which we may assume is B itself. Since p is good for I, we obtain the desired embedding of B in a maximal torus of K. One can find Z × I and Z × I in [IA , Tables 4.7.1, 4.7.3AB], for K ∼ = F4 (q), E6± (q), E7 (q), and E8 (q). For example when p = 3 or p = 5 and K = E8 (q), Z × I ∼ = Zp × D7 (q) (q ≡  (mod p)) satisfies our conditions. In the remaining cases, similar arguments n are effective. If K = G2 (q) or 2F4 (2 2 ) with p = 3, we have m3 (K) = 2, and K   2 contains I ∼ = A2 (F2 ), with I = O 2 (CI (σ)) ∼ = A2q (q), q ≡ q (mod 3), or A− 2 (q ). And if K = E8 (q) with p = 5 and q ≡ ±2 (mod 5), then using a 2-dimensional torus Z and I = D6 (F2 ), we can find a subgroup ZI ∼ = Zq2 +1 × D6− (q) of K. The proof is complete.  Lemma 2.6. Let K ∈ Chev(2) have level q, and let p be an odd prime such that mp (K/Z(K)) ≥ 3. Assume that q = ±1 is such that q ≡ q (mod p). Let B ∈ Ep∗ (K). Then the following conditions hold: (a) AutK (B) is generated by reflections. Moreover, AutK (B) ≡B/B∩Z(K) W (L)

(2A)

for some simple complex Lie algebra L, which is given in Table 17.1 and depends on p and the isomorphism type of K; (b) CK (B) has odd order; and (c) NK (B) splits over CK (B). Proof. Let (K, σ) be a σ-setup for K. By Lemmas 2.2 and 2.5, B ≤ T for some maximal torus T of K. As T ≤ CK (B)o , which is σ-invariant, we may choose T to be σ-invariant. Then B = Ω1 (Op (CT (σ))). By a Frattini argument, NK (B) ≤ CK (B)N , where we put N = NK (T ). Then setting W = N /T , we have AutK (B) = AutCW (σ) (B)

(≤ Aut(B)).

This is straightforward to calculate, and the results are in Table 17.1. Moreover, reflections in the natural action of CW (σ) in characteristic 0 act as reflections on B = Ω1 (Op (CT (σ))), so AutK (B) is generated by reflections, as asserted; and AutK (B) ≡B/B∩Z(K) W (L) by Proposition 1.5. Thus (a) holds. Parts (b) and (c) follow from Lemma 2.2 except in the following cases: K ∼ = −

−n

−

An q (q), Dn q (q), and E6 q (q). In each case there is a graph automorphism γ ∈ Aut(K) of order 2 such that Kγ := CK (γ) ∼ = C[ n+1 ] (q), Cn−1 (q), and F4 (q), 2

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respectively, and such that mp (Kγ ) = mp (K) (by [IA , 4.10.3a]). By Lemma 2.4, we may replace B by a conjugate and assume that B ≤ Kγ . Then Lemma 2.2 provides an isomorphic copy of W (L) complementing CKγ (B) in NKγ (B), so (c) holds. As  for (b), there is in each case an element b ∈ B # such that Kb := O 2 (CK (b)) ∼ =  q SL[ n+1 ] (q 2 ), SLn−1 (q), and C3 (q), respectively, and with B ∩ Kb ∈ Ep∗ (Kb ), by 2





[IA , 4.8.2] and [III11 , Table 13.1]. Then O 2 (CK (B)) = O 2 (CKb (B)) = 1 by Lemma 2.2, so (b) holds and the lemma is proved.  Lemma 2.7. Let K and L be as in Table 17.1. If the rank of L is at least 3, then O2 (W (L)) = 1. A2 .

Proof. Indeed, F ∗ (W (L)) = O2 (W (L))E(W (L)) in all cases except L = 

The next few lemmas concern the following configuration. (1) K  X = KB with K ∈ Chev(2), p is an odd prime, p divides q(K)2 − 1, and X = X/CX (K) = AutX (K); (2) B ∈ Ep∗ (CX (x)) for some x ∈ B # , and mp (CB (K)) = 1; (2B) (3) B ≤ Inndiag(K) and mp (B) ≥ 3; (4) mp (B) ≥ mp (K), with strict inequality if Z(K) ≤ Op (K); and (5) P is a B-invariant Sylow p-subgroup of K. With this notation, we say that B is torally embedded in X relative to K if and only if the following conditions hold: (1) J(P ) ≤ B; (2) AutK (B) is generated by reflections, and AutK (B) ≡B/CB (K) (2C) W (L), where L is as specified in Table 17.1 and (2A); and (3) CB (K) is the unique AutK (B)-invariant cyclic subgroup of B. We need a bit of help in showing that (2B) implies (2C) when p = 3 and 3 divides |Outdiag(K)|, in the form of the assumption that one of the following holds: (1) B ∈ E3∗ (BK); or (2) According as K/Z(K) ∼ = L3k (q) or E6 (q),  = ±1, q ≡  (2D) (mod 3), CK (x) has a component L such that L/Z(L) ∼ = L3k−δ (q), δ = 1 or 2, or L6 (q) or D5 (q), respectively. Lemma 2.8. Assume (2B). Then the following conditions hold: (a) Assume p divides |Outdiag(K)|. Then J(P ) is elementary abelian, and in the case K ∼ = An (q) with q ≡  (mod p), the preimage of J(P ) in  SLn+1 (q) is diagonalizable and multiplicity-free on the natural module. Moreover, AutK (J(P )) is isomorphic to the Weyl group of the algebraic group K overlying K and is generated by reflections; (b) Assume that if p = 3 and 3 divides |Outdiag(K)|, then either (2D1) or (2D2) holds. Then B is torally embedded relative to K. Note that if p > 3, or if p = 3 and 3 does not divide |Outdiag(K)|, then Lemma 2.8b applies.

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Proof. We may assume that Op (X) = 1. If CB (K) ≤ K, then as mp (CB (K)) = 1, Z(K) = 1 and X = CB (K) × X0 with F ∗ (X0 ) = K. First we prove (b) in the cases in which p does not divide |Outdiag(K)|. Since X ≤ Inndiag(K), B = CB (K) × (B ∩ K) in this case. We argue that mp (B ∩ K) = mp (K). If p is a good prime for K, this follows from our assumption that B ∈ Ep∗ (CX (x)) and [IA , 4.10.3e]. Otherwise, we use [IA , Tables 4.7.3AB] and [IA , 4.10.3a] to check that whatever inner automorphism x induces on K, we have mp (CK (x)) = mp (K). By [IA , 4.10.3c], J(P ) = B ∩ K, and by Lemma 2.6, AutK (B) ∼ = AutK (B ∩ K) is generated by reflections. By Remark 1.20, AutK (B) is irreducible on B ∩ K, so all parts of the definition (2C) are satisfied. We may now assume that p divides |Outdiag(K)|. q  ∼ so that, as usual, K/Z(K) = Lkp (q) or E6q (q) (with p = 3), q ≡ q (mod p). Since mp (B) ≥ 3, kp ≥ 5 in the first case. We have P = PT PW ≤ PT W ≤ K, where PT lies in a maximal torus of the overlying algebraic group, and PW ∈ Sylp (W ), where W is a complement to CK (PT ) in NK (PT ). Moreover, J(P ) = Ω1 (PT ) and CP (J(P )) = PT . (See Lemmas 2.1 and 2.2 to justify these assertions and complete the proof of (a). Note, however, that the groups B and K here are not the B and K there.) To complete the proof of (b), observe that B normalizes J(P ) = Ω1 (PT ). But then as B ≤ Inndiag(K), BP = CPW where C = CBP (Ω1 (PT )) = CBP (PT CB (K)). Now C ∩ PW = 1 and Outdiag(K) is cyclic, so |C : PT CB (K)| = |CPW : PT PW CB (K)| = |BP : P CB (K)| ≤ p. Therefore, C is abelian. Suppose that B ≤ C. As B contains every element of order p in CX (B), J(P ) = Ω1 (PT ) ≤ B. Moreover, as [B, PT ] = 1, B = Ω1 (BPT ) and BPT ∈ Sylp (CX (J(P ))). Hence NX (J(P )) = CX (J(P ))NX (B) by a Frattini argument. Therefore AutK (B) ∼ = AutK (J(P )), whence AutK (B) ≡B/CB (K) W (L) as in (2C2). In the case K ∼ = E6± (q) and p = 3, we deduce (2C3) from [IA , 4.7.3A], by examining the centralizer of a putative AutK (B)-invariant subgroup of B. In the linear group case, the socle of B under the action of AutInndiag(K) (B) is the core of the natural W -permutation module so (2C3) holds. We have proved that if B ≤ C, then the lemma holds. Thus we may assume that B ≤ C, which we show leads to a contradiction. We have (2E)

1 = AutB (Ω1 (PT )) ≤ AutBP (Ω1 (PT )) = AutPW (Ω1 (PT )).

Consider the group BJ(P ). If B ∩C ≤ P CB (K), then B ∩C ≤ Ω1 (CP CB (K) (J(P ))) = J(P )CB (K). Thus CBJ(P ) (J(P )) = J(P )CB (K). By (2E), there exists a subgroup 1 = R ≤ PW such that AutB (J(P )) = AutR (J(P )). Hence BJ(P )CB (K) ∼ = RJ(P )CB (K). As RJ(P ) ≤ P , J(RJ(P )CB (K)) = J(P )CB (K) so J(BJ(P )) = J(P )CB (K). However, mp (B) ≥ mp (K) = mp (P ) by assumption (2B4), with strict inequality if CB (K) ≤ K, so J(BJ(P )) = BJ(P ). Hence B ≤ J(P )CB (K) ≤ C, a contradiction. We conclude that B ∩ C ≤ P CB (K).

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This implies that mp (C) > mp (P CB (K)), so q  (q) or E6q (q). mp (C) = mp (X) = kp or 7, according as K ∼ = Lkp

As |BP : P CB (K)| ≤ p, equality holds and B = (B ∩ C)(B ∩ P ), with mp (B ∩ P ) = mp (B) − 1 or mp (B) − 2 according as p divides |Z(K)| or not. Therefore mp (B ∩ P ) ≥ mp (K) − 1, by (2B4). Since B ≤ C, B ∩ P ≤ C. The last two statements are in conflict if p ≥ 5. Indeed in that case, let BW be the image of B ∩ P in W ∼ = Σkp . Suppose that BW has nontrivial orbits  O1 , . . . , O on kp letters, with |Oi | = pai , i = 1, . . . , . Then mp (BW ) ≤ i=1 ai  and mp (J(P )/CJ(P ) (BW )) ≥ −1 − c + i=1 (pai − 1). Here c = 0 unless Z(K) ≤ [J(P ), BW ], in which case c = 1. As mp (J(P )) = mp (K) ≤ mp (B ∩ P ) + 1 ≤ mp (BW ) + mp (CJ(P ) (BW )) + 1, we get  #

(pai − (ai + 1)) ≤ 2 + c

i=1

As p ≥ 5, the only solution is  = a1 = 1, c = 1, and p = 5. Moreover, since c = 1,  it follows that k = 1. But now K/Z(K) ∼ = L5q (q) and CC (BW ) ≤ KCB (K), so B ∩ C ≤ P CB (K), contrary to what we showed above. Thus, p = 3, and so (2D) applies.  Suppose that (2D2) holds. In the L3kq (q) case, since CX (x) has a component L with L/Z(L) ∼ = L3k−δ (q), δ = 1 or 2, x acts on K like a diagonalizable element  of GL3k (q). Hence x lies in conjugate C g of C. The same conclusion holds in the  E6q (q) case, by the structure of CX (x). As C is abelian, m3 (B) = m3 (CX (x)) ≥ m3 (C g ) = m3 (C) = m3 (X). Thus in any case, (2D1) holds, i.e., B ∈ E3∗ (X). Using [IG , 26.8] we see that some element of I3 (PW ) acts quadratically on J(P ), which gives a contradiction as in the last paragraph of the proof of Lemma 2.2. Thus (b) holds, and the proof is complete.  Similarly, using [IG , 26.8], one can prove (a) of the following lemma. Lemma 2.9. Assume (2B). Suppose that p divides |Outdiag(K)|. Let Q ∈ Sylp (Inndiag(K)). Then the following conditions hold: (a) J(Q) is elementary abelian and AutK (J(Q)), which is isomorphic to the Weyl group of the algebraic group overlying K, acts on J(Q) as a group generated by reflections; (b) AutK (J(Q)) leaves invariant no subgroup of J(Q) of order p; (c) If P ∈ Sylp (K) and the image of P mod Z(K) lies in Q, then  O p (CK (J(P ))) is centralized by J(Q). Proof. Part (b) follows from the same arguments that proved (2C3) in the  paragraph before (2E). Part (c) holds because O p (CK (J(P ))) and J(Q) arise from the toral parts of P and Q respectively, and a maximal torus of the adjoint version of a simple algebraic group is the image of a maximal torus of any other version.  Lemma 2.10. Assume (2B), and if p = 3, suppose that (2D2) holds. Then NK (J(P )) controls K-fusion in B. Proof. Expand B to R ∈ Sylp (X) and let P = R ∩ K. By Lemma 2.8b, J(P ) ≤ B. Let b ∈ B and g ∈ K with bg ∈ B. Then J(P ), J(P )g ≤ CK (bg ) so  J(P )gh = J(P ) for some h ∈ CK (bg ). So bg = bgh with gh ∈ NK (J(P )).

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Lemma 2.11. Assume (2B), and that B is torally embedded in X relative to + K, i.e., (2C) holds. Let WK = AutK (B) and WK = WK ∩ SL(B). Also set B0 = [B ∩ K, WK ]. Then B0 = [B, WK ] = [B0 , WK ] and the following conditions hold: + (a) B has a unique nontrivial WK - and WK -composition factor, whose codimension in B0 is at most 1, and which is absolutely irreducible; + - or WK -composition factors; (b) B has at most 2 trivial WK + (c) B/CB (K) is absolutely indecomposable for WK and WK , with absolutely irreducible socle not a trivial module, and at most one trivial composition factor; (d) Either the socle of B0 is a nontrivial absolutely irreducible module for WK + and WK or else the socle of B0 equals B ∩ Z(K), a 1-dimensional trivial + , and module, with B0 /B ∩ Z(K) absolutely irreducible for WK and WK + B0 absolutely indecomposable for WK and WK ; and (e) If p does not divide |Outdiag(K)|, then B0 = [B, WK ] is absolutely ir+ reducible for WK and WK , and B = B0 × CB (K) = B0 × CB (WK ) = + ). B0 × CB (WK

Proof. Since WK is generated by reflections, WK = O p (WK ) and so [B, WK ] = [B, WK , WK ] ≤ [B ∩ K, WK ] by [IG , 4.3(i)]. Hence, B0 = [B, WK ] = [B0 , WK ]. First suppose that p does not divide |Outdiag(K)|. Then p does not divide |Z(K)| and as B contains every element of Ip (CX (B)), B = (B ∩ K) × x . By Remark 1.20, B ∩ K is absolutely irreducible for WK . We + argue that B ∩K is absolutely irreducible for the subgroup WK of index 2. Suppose + false. Then by Clifford’s theorem, WK has a faithful summand V ⊆ Fp ⊗ (B ∩ K) with dimFp (V ) ≤ mp (B ∩ K)/2. We have WK ≡B∩K W (L) with L as in Table + contains a subgroup 2n−1 An , 17.1. If L = Cn , Dn , F4 , E6 , E7 , or E8 , then WK n−1 1+4 1+4 1+6 2 An , 2 , 2 , F8.7 , or 2 , respectively, which has no faithful characteristic + is the p representation of degree ≤ n/2. If L = An , then WK ∼ = Σn+1 and WK alternating group An+1 , acting on the trace-0 module V = B ∩ K, with n = dim V ≥ 3 by assumption. By Lemma 1.9, V is absolutely irreducible both for WK + and for WK . This implies (e), and the other parts follow easily if p does not divide |Outdiag(K)|. Now assume that p divides |Outdiag(K)|, q ∼ so that as usual K/Z(K) = Lkp (q) with WK = W (Akp−1 ) ∼ = Σkp and kp ≥ 5, or  q ∼ (q) and W = W (E ) SO (3); in both cases, q ≡ q = ±1 p = 3 and K ∼ E = 6 K 6 = 5 Σ , the exponent p-subgroup of a full diagonal sub(mod p). In the case WK ∼ = kp q group of GLkp (q) has a WK -invariant conjugate E which is isomorphic to a standard permutation module V for WK . The image E of E in Inndiag(K) contains the image B of B in Inndiag(K), and by Lemma 2.9, E is an indecomposable extension of the core V2 /V1 of V (see Lemma 1.9) by a 1-dimensional trivial quotient. Hence either B = E or B ∼ = V2 /V1 . Moreover, B0 = [B0 , WK ] ∼ = V2 or V2 /V1 , and the socle of V2 is V1 (see Lemma 1.9). These remarks imply the assertions in (a)–(d). Note that if Fp ⊗ E were decomposable, it would be isomorphic to Fp ⊗ [(V2 /V1 ) ⊕ Fp ],

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∼ (V2 /V1 ) ⊕ Fp by a standard result. This is a contradiction, and thus and hence E = in this context indecomposability automatically implies absolute indecomposability.  It remains to consider the case K ∼ = E6q (q), p = 3. Let B be the image of B  in Inndiag(K). Then Inndiag(K) contains a subgroup L2 (q) × P GL6q (q) which in turn contains B. Therefore AutK (B) contains Z2 × I, where I ∼ = Σ6 has two trivial composition factors and one nontrivial composition factor – the core of a natural permutation module – on B. Hence by Lemmas 1.16, 2.9b, and 1.14, the socle of the WK -module B is the reflection module for WK and the natural module for + . As it is of prime dimension and irreducible, the socle is absolutely irreducible. WK This implies (a), (b), and (c). To prove (d) it only remains to show that if K is + the universal version (i.e., |Z(K)| = 3), then B0 is indecomposable for WK . But 2 in that case O (NK (B0 )) contains a Sylow 3-subgroup P of K. From the structure of CK (t3 ) (see [IA , 4.7.3A]) we see that Z(P ) is cyclic. On the other hand, if B0 were decomposable, each summand would meet Z(P ) nontrivially, a contradiction. This completes the proof.  Lemma 2.12. Assume (2B). Suppose that B ∈ Ep∗ (Inndiag(K)). Then the following conditions hold: (a) The stabilizer in K of the chain B > CB (K) > 1 centralizes B; and (b) AutK (B) is generated by reflections, and leaves invariant no nontrivial cyclic subgroup of B. Proof. We may assume that Op (KB) = 1. If Outdiag(K) is a p -group, then Z(K) ≤ Op (K) = 1 by [IA , 6.1.4], so KB = K ×CB (K) and B = (B∩K)×CB (K). Then (a) is trivial and (b) follows from Lemma 2.8b.  So assume that p divides |Outdiag(K)|. As usual, K ∼ = Anq (q) with p dividing q ∼ n + 1, or K = E6 (q) with p = 3. Here q = ±1 is such that q ≡ q (mod p). Let Q ∈ Sylp (X). Since mp (B) = mp (Inndiag(K)), Lemma 2.9ab gives B = J(Q) and (b) holds. By Lemmas 2.9c and 2.4, B centralizes and hence contains J(Q∩K). As mp (K) ≥ 3 it follows from Lemma 2.8a that Op (AutK (J(Q∩K))) = 1, so the stabilizer of the chain J(Q ∩ K) > CB (K) ≥ 1 acts trivially on J(Q ∩ K). Hence if (a) failed there would exist a p-element of CK (J(Q ∩ K)) not centralizing B. But as B = J(Q), this would contradict Lemma 2.9c. This completes the proof.  q Lemma 2.13. Let p be an odd prime and K ∈ Gp with K/Z(K) ∼ (q), = Lkp q = ±1, q ≡ q (mod p). Let P ∈ Sylp (K). Then J(P ) is the exponent-p subgroup q (q). of the image of a maximal diagonalizable p-subgroup of SLkp

Proof. This follows from Lemma 2.8a. Specifically, notice that kp ≥ 5. For otherwise p = 3 and k = 1, whence K ∈ T3 by definition of T3 [I2 , 13.1], contrary to assumption. Set X = K or K × Zp , according as p divides |Z(K)| or not. Let B ∈ Ep∗ (X) and let P be a B-invariant Sylow p-subgroup of K. Then mp (AutB (K)) ≥ kp − 2 ≥ 3, and all the conditions of (2B) are satisfied, with x ∈ Ip (Z(X)). The lemma then follows from Lemma 2.8a.  Lemma 2.14. Let H = A x where A  H, A is an abelian p-group, p is an odd prime, xp = 1, and [x, Ω1 (A)] = 1. Then Ω1 (H) = Ω1 (A) x . Proof. Let H be a minimal counterexample. Then H = Ω1 (H) has class at least 3. Since the mapping a → ap injects Ω2 (A)/Ω1 (A) into Ω1 (A), x centralizes

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Ω2 (A)/Ω1 (A), so Ω2 (A) x has class at most 2. But by minimality, Ω1 (H/Ω1 (A)) ≤ Ω2 (A) x /Ω1 (A). Therefore H = Ω1 (H) ≤ Ω2 (A) x , of class at most 2, a contradiction.  Lemma 2.15. Let p be an odd prime and let K ∈ Gp ∩ Chev(2) with p dividing |Outdiag(K)|. Suppose that K  X1 and mp (CX1 (K)) = 1. Let X1 = X1 /CX1 (K). Let B ∗ ∈ Ep∗ (X1 ) and let B be the preimage in B ∗ of Inndiag(K). Assume that mp (B) ≥ 3 and that mp (B) = mp (BK). Let Q ∈ Sylp (CX1 (B)) with B ∗ ≤ Q. Then Ω1 (Q) is abelian. Proof. Without loss, Op (X1 ) = 1. We check that the conditions (2B) hold for X = KB. Indeed they are all trivial except for the condition that mp (B) ≥ mp (K), with strict inequality if Z(K) = 1. Now B ∗ , in place of B, satisfies this condition trivially, so we may assume that B ∗ > B. But then B ∗ = B × f where f induces a field automorphism on K. Then mp (B ∗ ) = mp (X1 ) ≥ mp (f CK (f )CX1 (K)) ≥ 1 + mp (CK (f )), with strict inequality if Z(K) = 1. But mp (CK (f )) = mp (K) so (2B) holds, as claimed. Then by Lemma 2.8b, J(P ) ≤ B where P is a B-invariant Sylow p-subgroup of K. We may assume P chosen to contain PK ∈ Sylp (CK (B)). But a Sylow p-subgroup R of CAut(K) (J(P )) is an extension of an abelian group T (Sylow in a maximal torus in Inndiag(K)) by a cyclic group centralizing Ω1 (T ). Hence by Lemma 2.14, Ω1 (R) is abelian. As CX1 (B) embeds in CAut(K) (J(P )), Q embeds in R, and the result follows.  Lemma 2.16. Assume that p is an odd prime, X = KB with Op (X) = 1, K = E(X), and B is maximal in Ep (X) with respect to inclusion. Suppose that  K ∼ = Lnq (q), n ≥ 4, where q = ±1 and q ≡ q (mod p). Let X = X/CX (K) ∼ = AutX (K). Assume that B is the image in Aut(K) of a diagonalizable p-subgroup  of GLnq (q). Then AutK (B) ∼ = NK (B)/CK (B) ∼ = Σn . 

Proof. Since B is the image of a diagonalizable p-subgroup of GLnq (q), CK (B) contains a full diagonal subgroup D of K. Hence by maximality, Ω1 (Op (D)) ≤ B. Since n > 3, this in turn implies that the image of CX (B) in Aut(K) is contained  in the image of a diagonal subgroup of GLnq (q), and in particular CK (B) = D. Let DK and DC be the full preimages in K and CX (K), respectively, of the projections of B ∩ KCX (K) on K/Z(K) and CX (K)/Z(K). Then as CK (B) = D, we have DK ≤ D, and so DK is abelian. Since B is abelian it follows that DC is abelian. Since X = KB, CX (K) = CX (K) ∩ KB ≤ KCX (K) ∩ KB = K(KCX (K) ∩ B). Thus, the projection of B∩KCX (K) on CX (K)/Z(K) is onto, and so CX (K) = DC is abelian. Furthermore, B centralizes D, which contains DK , so [B, DK ] = 1 and then [B, CX (K)] = 1. As X = KB, CX (K) = Z(X). Now B = Ω1 (Op (BZ(X))) is invariant under NK (BZ(X)), which covers NK (B). But, again as n > 3, AutK (B) ∼ = Σn . Hence AutK (B) contains Σn . But NK (B) ≤  NK (B ∩ K) with NK (B ∩ K)/D ∼ = Σn . As [B, D] = 1, the lemma follows. Lemma 2.17. Suppose that p is an odd prime, q is a power of 2, J is simple,   and J ≤ X ≤ Inndiag(J), with J ∼ = Amq (q), m ≥ 4, E7 (q), or E6q (q). Suppose p that q = ±1 and q ≡ q (mod p). Let B ∈ E∗ (X) and W = AutX (B), so that

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W ≡B W (Am ), W (E7 ), or W (E6 ), respectively. Let W  Y ≤ Aut(B) and suppose that Y is generated by reflections (on B). Then the following conditions hold: (a) Y = W × Z where |Z| ≤ 2 and Z acts as scalars on B; and  (b) If J ∼ = Amq (q), m > 5, or E7 (q), then Z = 1. Proof. Since Y is generated by reflections, it suffices to prove the lemma for Y = W y , with y ∈ Y − W a reflection. In particular |Y : W | = 2. In the Am case, by Lemma 2.11, W ∼ = Σm+1 is centerless and B is absolutely indecomposable as W -module, with socle the core V of the natural permutation module over Fp . In the E7 case, W ∼ = Z2 × Sp6 (2), [W, W ] is centerless, and as [W, W ] contains a Frobenius group of order 8.7, B is an absolutely irreducible [W, W ]-module. In the E6 case, W ∼ = SO5 (3) is centerless, and B is absolutely irreducible unless p = 3, in which case it is at any rate absolutely indecomposable. Thus, in all cases, CY (W ) consists of scalars. In particular Z(Y ) is cyclic. Hence, if Y = W CY (W ), then (a) holds, and if J ∼ = E7 (q), then (b) is clear. Note also that if Y = W (Am ) × −1 , then since Y = W y , −y ∈ I2 (W ) and 1 is an eigenvalue of −y with multiplicity 1. Say that −y is the product of k disjoint transpositions. Then the number of eigenvalues of −y equal to 1 on V is at least (m + 1 − k) − 2, so m − 1 − k ≤ 1. As k ≤ (m + 1)/2 this yields m ≤ 5, contradicting the hypothesis of (b). It remains to consider the case Y > W CY (W ). Then Aut(W ) ≥ AutY (W ) > Inn(W ). The only possibility, by [IA , 2.5.12, 5.2.1], is that W ∼ = Σ6 and Y ∼ = Aut(W ). As Y = W y with y 2 = 1, CW (y) contains an element v of order 5, and wy is a product of two disjoint 3-cycles for any 3-cycle w ∈ W . If p = 3, then v acts irreducibly on V , but respects the eigenspaces of the reflection y, a contradiction. Therefore p > 3, so |V | = p5 . Then |CV (w)| = p3 and as y is a reflection, |CV (w, y )| ≥ p2 . However, wy ∈ w, y and as p > 3, |CV (wy )| = p, a contradiction. The proof is complete.  For the next result (Lemma 2.20) we need two preliminaries. Lemma 2.18. Let Mn be the natural permutation module for Σn , n ≥ 2, over Fp , p an odd prime. Then Mn is self-dual and H 1 (Σn , Mn ) = 0. Proof. By making the natural permutation basis of Mn an orthonormal basis, we see that Mn supports a Σn -invariant nondegenerate bilinear form. Hence, Mn is self-dual. For n ≤ 3, H 1 (Σn , Mn ) = 0 by Maschke’s theorem if p > 3 or n = 2, and because O3 (Σn ) acts freely on Mn if p = n = 3. We proceed by induction on n. It suffices to show that any exact sequence 0 → X → M0 → Mn → 0 ∼ Σn−1 be of Σn -modules, with X a trivial 1-dimensional module, splits. Let Y = the stabilizer of a point in Σn . Then as Fp Y -modules, Mn ∼ = X1 ⊕ Mn−1 , with X1 a trivial module. The preimage of X1 in M0 is then a 2-dimensional Σn−1 module with two trivial composition factors, so it is a trivial Σn−1 -module as Σn−1 is generated by involutions and p is odd. Hence, there is m ∈ M0 − X fixed by Σn−1 . Then mΣn is n-dimensional and is the required complement to X in M0 . The lemma is proved. 

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Lemma 2.19. Let W be the Weyl group of Dn , n ≥ 3, p an odd prime, and M the Fp W -module ZΣ/pZΣ, where Σ is the root system of Dn . Then H 1 (W, M ) = 0. Proof. We have W = F S, where S ∼ = Σn permutes a basis {bi }ni=1 of M and ∼ each element of F = E2n−1 inverts an even number of the bi , centralizing the rest of the bi ’s. It follows that [O2 (W ), M ] = M . Hence H 1 (W, M ) = 0 by Lemma 1.25.  Lemma 2.20. Let K ∈ Chev(2) be of level q and let p be an odd prime dividing q − q , where q = ±1. Let K  X and let B ∈ Ep∗ (X) with mp (CB (K)) = 1, X = KCX (B) and CX (B) of odd order. Assume that B maps into Inndiag(K). Let u ∈ B and let L be a component of E(CK (u)) such that one of the following holds: q (a) (K, L) = (D2m+1 (q), D2m (q)), m ≥ 2;  (b) (K, L) = (A6q (q), A5 (q)), and there is g ∈ CX (B) such that if we put  K ∗ = K g , then AutK ∗ (K) ∼ = P GL7q (q). Let WL be a complement to CL (B) in NL (B). (WL exists by Lemma 2.2.) Then CX (B) has a WL -invariant filtration with abelian quotients M all satisfying H 1 (WL , M ) = 0. Moreover, each such quotient M is a trivial or faithful WL module. Proof. Note that the conditions (2B) are satisfied by X0 := KB. Moreover, 3 does not divide |Outdiag(K)|, so Lemma 2.8b applies. Let P be a B-invariant Sylow p-subgroup of K. We conclude in case (a) that B = CB (K) × (B ∩ K) with B ∩ K = J(P ), and in case (b) that J(P ) ≤ B ∩ K, with J(P ) a maximal  elementary abelian image of diagonalizable p-subgroups of SL7q (q). We now drop the assumption that mp (CB (K)) = 1. Since WL is generated by involutions, H 1 (WL , M ) = 0 for any abelian group M of odd order with trivial WL action. We can therefore pass to X/CX (K) and assume that CX (K) = 1, so that we regard X ≤ Aut(K). Likewise X/X ∩ Inndiag(K) has odd order by assumption and is filtered by trivial WL -modules, so we may assume that X ≤ Inndiag(K). Thus, q  (q) or P GL7q (q). X∼ = D2m+1 In case (a), using [IA , 4.8.1] we see that the natural X-module is the orthogonal sum V = V1 ⊥ · · · ⊥ V2m+1 of isometric nonsingular 2-spaces Vi , on each of which B is orthogonally indecomposable. Then CX (B) = T1 × · · · × T2m+1 , with each Ti ∼ = Zq−q , acting nontrivially on Vi only. As L  E(CX (u)), we may assume that ⊥ u is supported on V2m+1 , and then L = Ω(V2m+1 ). Then as WL -module, CX (B) is filtered by elementary abelian subgroups of rank 2m + 1 and exponent r dividing q − q . Each of these has a natural WL -composition factor Mr and a trivial WL composition factor. By Lemma 2.19, H 1 (WL , Mr ) = 0, and the lemma follows in this case. In case (b), CX (B) is again homocyclic abelian, and a similar filtration leads to odd-order quotients of rank 6 which are natural permutation modules for WL ∼ = Σ6 . This time, Lemma 2.18 completes the proof.  −

−

Lemma 2.21. Let K = F ∗ (X) = C3 (q), D4− (q), A5 q (q) or A6 q (q) for some q = 2m , and let p be a prime dividing q − q , for some q = ±1. Let B ∈ Ep∗ (K). Then the subgroup of AutX (B) generated by reflections is isomorphic to W (C3 ).

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Proof. Let AX be the reflection subgroup of AutX (B), and let AK = AutK (B). By Lemma 2.6, AK ∼ = W (C3 ), and as AK  AX , Lemma 1.22 implies that AX = AK , as claimed.  The next several lemmas concern certain classical groups, in the following setup. q = 2a and p is a prime dividing q − q , where q = ±1; − K = A q (q),  ≥ 5, or C (q) or D± (q),  ≥ 3; p B ∈ E∗ (K); V is a natural K=module, and V = CV (B) ⊥ V1 ⊥ · · · ⊥ Vn , n ≥ 3, where for each i = 1, . . . , n, Vi is a B-invariant subspace of dimension 2; (5) B = B1 × · · · × Bn , where |Bi | = p and [Vj , Bi ] = 0 for all i, j = 1, . . . , n with i = j; (6) Bi = bi , i = 1, . . . , n. (1) (2) (3) (4)

(2F)

Lemma 2.22. Conditions (2F1 − 3) imply that there are a natural K-module V and subgroups B1 , . . . , Bn of K satisfying the remaining conditions of (2F). 

Proof. This follows easily from [IA , 4.10.3, 4.8.1]. From [IA , 4.8.1, 4.8.2] and Lemma 2.6, we deduce:

Lemma 2.23. In (2F), the following conditions hold: (a) The Vi ’s, i = 1, . . . , n, are all isometric of dimension 2, the isometry type depending only on p, q, and whether K is a linear, unitary, symplectic, or orthogonal group; −q (b) One of the following holds: V0 = 0 with K = A2n−1 (q), Cn (q), Cn (q), n

−

or Dnq (q); dim V0 = 1 with K = A2n q (q); or dim V0 = 2 with K = −n+1 q

Dn+1 (q); and (c) AutK (B) ≡B W (Cn ), W (Cn ), W (Cn ), or W (Dn ), according as K is of −n+1

n

q (q), or Dnq (q), respectively. type A, C, Dn+1

n

Lemma 2.24. Assume (2F), with K = Dnq (q). Let C be the set of all subgroups  n−1 q (q). Then either C is a single b of K of order p such that O 2 (CK (b)) ∼ = Dn−1 conjugacy class in K with dim[V, b ] = 2 for all b ∈ C, or there is a K-conjugacy class C0 of subgroups of order p such that the following conditions hold: (a) n = 4; −1 (b) C = C0 ∪ C0ρ ∪ C0ρ , where ρ is a triality automorphism (so that Aut0 (K) = O8+ (q) ρ );   (c) For all b ∈ C0 , dim[V, b ] = 2, while [V, bρ ] = [V, bρ

−1

]=V.

Proof. By [IA , 4.8.2], for any b ∈ K of order p and component L of CK (b),  there is an integer m such that either L ∼ = Amq (q) is supported on a 2(m + 1)m q ∼ Dm dimensional subspace of [V, b], or L = (q) is supported on all of CV (b). In the latter case if m = n − 1, then dim[V, b] = 2 and O([V, b]) has cyclic Sylow psubgroups, so b is determined up to O(V )-conjugacy, and indeed as n ≥ 3, up to Ω(V ) = K-conjugacy. Hence if C is not a single conjugacy class, it must be because

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348

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Am = Dn−1 for some m, implying n = 4 and m = 3. The remaining statements  follow from the information in [IA , 4.7.3A]. n

 Lemma 2.25. Assume (2F) with K ∼ = Dnq (q). Let b ∈ B # be such that  n−1 q (q). Then, provided V was suitably chosen when n = 4 from O 2 (CK (b)) ∼ = Dn−1 among the three isomorphism types of natural K-modules, the following conditions hold: (a) b = Bj for some j = 1, . . . , n; and  (b) [V, O 2 (CK (Bi ))] = Vi⊥ for each i = 1, . . . , n.

Proof. By Lemma 2.24, and with a suitable choice of V if n = 4, we may  assume that dim[V, b] = 2, proving (a). Part (b) follows from [IA , 4.8.2]. Lemma 2.26. Let K = D4 (q), q even, let p be a prime divisor of q 2 − 1, let B ∈ Ep∗ (K) and set X = {b ∈ E1 (B) | E(CK (b)) ∼ = D3± (q)}. Then |X| = 12 and for every b ∈ X, there is a unique frame B for B such that b ∈ B ⊆ X. Moreover, NAut0 (K) (B) permutes X transitively. Proof. This follows easily from Lemma 2.24. If one such frame is {bi }4i=1 , then the others are {b11 b22 b33 b4 | 1 2 3 = 1} and {b11 b22 b33 b4 | 1 2 3 = −1}. Here the i range over {±1}.



Lemma 2.27. Assume (2F) with K = D± (2m ),  ≥ 5. Let k ∈ Z and x ∈ NAut(K) (B) with bxi = bki for all i = 1, . . . , n − 1. Then bxn = b±k n . Proof. We use the results [IA , 4.8.1, 4.8.2] freely. We may choose injections ψi : Zp → Bi , 1 ≤ i ≤ m, such that the resulting representations Ψi : Zp → Isom(V ) are all orthogonally equivalent. Let bi = ψi (1) and b = b1 · · · bm . Then  O 2 (CK (b)) ∼ = SLn (q).  Hence O 2 (CK (bx )) ∼ = SLn (q). By assumption x normalizes Bi for all 1 ≤ i < n. Also as  ≥ 5, Aut(K) acts as semilinear transformations on V . Therefore x normalizes Bn = CB ([V, B1 · · · Bn−1 ]). Hence bx = (b1 · · · bm−1 bcn )k for some x c ∈ F× p , and we must prove c = ±1. From the centralizer of b we conclude that the representations bm → Ψm (1) and bm → Ψm (c) are orthogonally equivalent, and this yields c = ±1, completing the proof.  Lemma 2.28. Assume (2F) with K ∼ = Sp2n (q), n ≥ 3. Then the following conditions hold: (a) CK (B) ∼ = (Zq−q )n ; (b) AutK (B) ≡B W (Cn ), and the involution of Z(AutK (B)) inverts CK (B) elementwise; ∼ (c) If n ≥ 4 and x ∈ B with Kx := E(CK (x))  = Sp2n−2 (q), then for any subgroup M ≤ NK (B) covering AutK (B), KxM = K; and (d) Finally, if g ∈ Aut(K) is a field automorphism of odd prime order s centralizing B, then g induces a power mapping on CK (B) with fixed points 1 isomorphic to (Zq0 −q )n , where q0 = q s . Proof. Since Vi = [V, Bi ], CK (B) = CSp(V1 ) (B1 ) × · · · × CSp(Vn ) (Bn ) = C1 × · · · × Cn , where Ci is cyclic of exponent q − q , proving (a). By Lemma

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2.23, AutK (B) ≡B W (Cn ), and its central involution is clearly t1 . . . tn with ti ∈ g have x = Bi and x = Bj AutSp(Vi ) (B) inverting Ci , proving In (c), we  M(b).  g ⊥ ⊥ for some i = j and g ∈ M , so Kx ≥ Kx , Kx = Sp(Vi ), Sp(Vj ) = K by [IA , 7.3.2] (see Lemma 9.1 below). Writing CK (B) in matrix form it is evident that some field automorphism f of K of order s induces a power mapping on CK (B), centralizing B and a homocyclic subgroup of CK (B) of exponent q 1/s − q . Then every field automorphism of K of order s centralizing B lies in CKf  (B) = CK (B) f , so has the same action on CK (B) as a nontrivial power of f . This implies (d) and finishes the proof.  −

q (q), δ = 0, 1, or Sp2n (q). Then Lemma 2.29. Assume (2F), with K = L2n+δ the following conditions hold: (a) mp (B) = n; (b) AutK (B) ≡B W (Cn ); and −q (c) If x ∈ B with Kx := E(CK (x)) ∼ (q) or Sp2n−2 (q) respectively, = SL2n+δ−2  then for any subgroup M ≤ NK (B) covering AutK (B), we have KxM = K.

Proof. For (a) and (b), see [IA , 4.10.3a] and Lemma 2.23. Part (c) holds in the symplectic case by Lemma 2.28, and as 2n ≥ 6 in the linear/unitary case, part (c) follows in that case by the same argument, again using [IA , 7.3.2] or Lemma 9.1.  −

q (q), δ = 0, 1, or Sp2n (q). Lemma 2.30. Assume (2F), with K = L2n+δ Let t ∈ I2 (Aut(K)) be such that mp (CK (t)) = mp (K). Then either t is a 1 field automorphism of K = Sp2n (q) (so that CK (t) ∼ = Sp2n (q 2 )), or t is a graph −q automorphism of K = L2n+δ (q), with CK (t) ∼ = Sp2n (q).

Proof. By our assumptions dim(V0 ) ≤ 1 in (2F). By Lemma 2.6, CK (B) has odd order. Since q = 2a , |Outdiag(K)| is odd, so t is a field, graph-field, or graph automorphism. In particular, if K = Sp2n (q), t must be a field automorphism and we are done. If t is a field or graph-field automorphism of K = L2n+δ (q), then CK (t) ∼ = 1 2 ), and q ≡ −1 (mod p) as − (q = 1. Then p does not divide q − 1 = SL± q 2n+δ 1 1 (q 2 − 1)(q 2 + 1), whence B ≤ CK (t), contrary to assumption. Hence, t is a graph −q automorphism of K = L2n+δ (q) and, with [IA , 4.9.2], CK (t) ∼ = Sp2n (q), as claimed, completing the proof.  Lemma 2.31. Let k ≥ 2 and let g be an involutory graph automorphism of K = D2k (2). Then m3 (CK (g)) < m3 (K). Proof. We have K g ∼ = O4k (2). Let V be the natural F2 K g -module, and let B ∈ E3∗ (K). Since V is of + type with dim(V ) ≡ 0 (mod 4), it follows from [IA , 4.8.1] that (2F) applies with n = 2k. As CO(Vi ) (Bi ) = Bi for each i, CO(V ) (B) = B. In particular B centralizes no involution of K g , completing the proof.  −

Lemma 2.32. Assume (2F) with K = Sp8 (q) or Ln q (q), n ∈ {8, 9}. Let Φ ∈ I2 (Aut(K)) be a field automorphism if K = Sp8 (q), or a graph automorphism − if K = Ln q (q), with CK (Φ) ∼ = Sp8 (q) in the latter case. Set K0 = CK (Φ) and

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17. PRELIMINARY PROPERTIES OF K-GROUPS

q suppose that B ∈ Ep4 (K0 ). Let K1 = E(CK (b1 )) ∼ (q), respectively. = Sp6 (q) or SLn−2 Let B234 = b2 , b3 , b4 . Then any complement to CK1 ∩K0 (B234 ) in NK1 ∩K0 (B234 ) is contained in a complement to CK (B) in NK (B).

−

Proof. We have K0 ∼ = Sp8 (q0 ), where q0 = q 2 or q according to the isomorphism type of K. Since B ≤ K0 , AutK0 (B) ≡ W (C4 ), so AutK0 (B) = AutK (B). Let U be a complement to CK1 ∩K0 (B234 ) in NK1 ∩K0 (B234 ). It therefore suffices to show that U lies in a complement U ∗ to CK0 (B) in NK0 (B). Now CK0 (E(CK0 (b1 ))) is isomorphic to SL2 (q0 ) and in particular contains a dihedral group b1 , t of order 2p. Let z be the central involution of U . Then z inverts B234 and centralizes b1 , so zt inverts B. Indeed CK0 (B) is homocyclic abelian of rank 4 and exponent q0 ± 1 (whichever is divisible by p), see [IA , 4.8.2]. Thus CK0 (B) lies in a direct product of four copies of SL2 (q0 ) in K0 , with zt projecting nontrivially on each factor. Hence, zt inverts CK0 (B) elementwise. Let U ∗ = CNK0 (B) (zt). Then U ≤ U ∗ and CU ∗ (B) = 1. Moreover, as zt maps on a central involution of AutK0 (B), U ∗ is the desired complement.  1

Lemma 2.33. Let q be a power of 2, p a prime divisor of q − q , q = ±1. Let n

K ∈ Chev(2) with either K = E7 (q) or K = Dnq (q), n ≥ 5. Use standard notation [IA , 2.10] for K. Let B ∈ Ep∗ (K). Then the following conditions hold: (a) CK (B) is abelian of odd order; (b) If q = 2, then B = CK (B); (c) Let t, u ∈ NK (B) be involutions whose images in AutK (B) ≡ W (E7 ) or W (Dn ) are distinct commuting reflections. Then [t, CK (B)]∩[u, CK (B)] = 1. Proof. Choose, by Lemma 2.2, a σ-setup (K, σ), a σ-invariant maximal torus T of K such that B = Ω1 (Op (T )), and a complement W ≤ K to CK (B) = T in NK (B) = NK (T ), such that σ induces the q qth power mapping on T . Then CK (B) ≤ CT (σ), which is homocyclic abelian of exponent q − q , proving (a) and (b). Suppose that (c) fails for some pair wα , wβ of distinct commuting reflections. As CK (B) = T , there is a prime divisor s of q−q such that the exponent s-subgroup Ts of T satisfies [Ts , wα ] ∩ [Ts , wβ ] = 1. But wα and wβ act as reflections on Ts , and as they commute, it follows that [Ts , wα ] = [Ts , wβ ] and then [Ts , wα wβ ] = 1. Since wα = wβ this contradicts the faithful action of NK (T )/T on Ts , the mod-s reduction of the T -root lattice for K. The proof is complete.  Lemma 2.34. Let K = E6 (2m ), E7 (2m ), or E8 (2m ). Let p > 3 be a prime dividing 2m − , where  = ±1. Let B be an elementary abelian p-subgroup of K of maximal rank. Then there exists a subgroup X ≤ K such that B normalizes X and X∼ = A (2m ), with  = 5, 7, or 8, respectively. Proof. By [IA , 4.10.3abd], as p > 3, any two elementary abelian p-subgroups of K of maximal rank r (r = 6, 7, 8, respectively) are conjugate, so it suffices to find X of the required isomorphism type and normalized by some Epr -subgroup B0 of K. We use subsystem subgroups X0 of K such that mp (X0 ) = r and X  X0 . The subgroups X0 are visible from the extended Dynkin diagram: X0 = A1 (2m )A5 (2m ),

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A7 (2m ), A8 (2m ), respectively. Note that in the last two cases the Weyl group m contains −1 so the subgroups A−   (2 ) are available. The proof is complete. Lemma 2.35. Let K = D4 (q), q = 2n , and suppose that K = F ∗ (X), but X contains no element of order 3 inducing a field automorphism on K. Let B ≤ K with B ∼ = E34 . Then CX (B) = O3 (CX (B))PT where PT is a homocyclic abelian 3-group of rank 4 containing B. Proof. As the Schur multiplier of K is a 3 -group [IA , 6.1.4], we may assume that K is simple. By Lemma 2.2 and [IA , 4.10.2], there is P ∈ Syl3 (K) such that P = PT PW , PT is homocyclic abelian of rank 4, |PW | = 3, [PW , B] = 1, and B = J(P ) = Ω1 (PT ). By Burnside’s normal p-complement theorem it is enough to show that PT ∈ Syl3 (CX (B)), since B = Ω1 (PT ) and PT is abelian. Let PT ≤ Q ∈ Syl3 (CX (B)) and suppose for a contradiction that PT < Q. Since X contains no field automorphism of order 3, Q = PT g where g 3 ∈ PT and g maps to the image of a graph or graph-field automorphism in Out(K). Then Z(P Q) ∩ P contains Z(P ) ∩ B, which is noncyclic. But P Q contains a graph or graph-field automorphism g0 of order 3, 3 so Z(P Q) ∩ P lies in a Sylow 3-center of CK (g0 ) ∼ = G2 (q), P GL± 3 (q), or D4 (q). As these groups have cyclic Sylow 3-centers, we have our contradiction. The lemma follows.  ∼ Sp6 (q), L−q (q), or L−q (q). Let Lemma 2.36. Assume (2F) with K/Z(K) = 6 7 C = CK (B). Let SK ≤ NK (B) with SK ∼ = Σ3 permuting {b1 , b2 , b3 } naturally and centralizing V0 . Then one of the following holds: (a) CK (B) has odd order and is filtered by SK -modules M each of which satisfies one of the following: (1) M is a 3 -group; (2) O3 (SK ) acts freely on M ; or (3) [M, SK ] = 1; or − (b) K ∼ = L6 q (q), q > 2, p > 3, and q ≡ −q (mod 3). Proof. Assume that (b) does not hold. Note that it is then not the case, by − our hypotheses, that K/Z(K) ∼ = L6 q (q) with q ≡ −q (mod 3). Hence the center of the universal version of K is a 3 -group, so we may assume that K is the universal version. By our assumptions, SK permutes the Vi = [V, bi ], 1 ≤ i ≤ 3. By Lemma 2.6, CK (B) has odd order. − Now suppose that K ∼ = Sp6 (q) or SL6 q (q). Then Ci := CCK (Vi⊥ ) (B) is cyclic of order q − q . Also C1 × C2 × C3 is SK -invariant; and in the SL6 case, we have − − q ≡ q (mod 3). Hence, GL2 q (q)/SL2 q (q) is a 3 -group. Hence in both cases, we see that if the SK -module M is outside C := C1 ×C2 ×C3 , then M is a 3 -group. On the other hand, O3 (SK ) cycles C1 , C2 , and C3 , so C can be filtered by SK -modules − on which O3 (SK ) acts freely. Thus the lemma holds in the Sp6 and SL6 q cases. − Finally, suppose that K ∼ = SL7 q (q). The same argument as above applies to  filter C. If q ≡ q (mod 3), then O 3 (CK (B)/C) is the image of a cyclic group acting faithfully on V0 and trivially on V1 ⊥ V2 ⊥ V3 . By the faithful action,  [O 3 (CK (B)/C), SK ] = 1, so we are done in this case. Thus we may assume that q ≡ −q (mod 3). But then a Sylow 3-subgroup P of CK (B) is a homocyclic abelian 3-group of exponent 3m = (q+q )3 and generated by elements xi , i = 1, 2, 3,

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centralizing Vj for j ∈ {1, 2, 3} − {i}, acting as diag(η, 1) on Vi for some η ∈ F× q2 of order 3m , and acting as η −1 on V0 . Thus, O3 (SK ) cyclically permutes x1 , x2 , x3 , and so acts freely on P , completing the proof in all cases.  root system Π. Use Lemma 2.37. Let K = D (F2 ),   even, with fundamental  × standard notation for K. Let T = hα (t) | α ∈ Π, t ∈ F2 and W = nα (1) | α ∈ Π , so that T is a maximal torus of K and NK (T ) = T W is a split extension. Then CK (W ) is connected. Proof. We may take K as the connected component of the identity in the  stabilizer in SL2 of the quadratic form i=1 ξi ξ2−i . Then we may take T as the subgroup of K of all its diagonal matrices, and W = W0 , W0t , where W0 stabilizes the subspaces defined by ξ1 = · · · = ξ = 0 and ξ+1 = · · · = ξ2 = 0 and acts on each as a permutation matrix, and t is a transvection interchanging ξ and ξ+1 but fixing every other ξi . Then g ∈ CK (W ) ⇐⇒ g ∈ CK (W0 ) ∩ CK (W0 )t . If we let J  be the permutation matrix corresponding to the involution (1, 2)(2, 2 − 1) · · · (,  + 1) ∈ Σ2 , then we compute that CK (W ) consists of all (2G)

αI + βJ + γJ  ∈ K.

 As  is even, the condition that αI + βJ + γJ  preserve i=1 ξi ξ2−i reduces by a straightforward calculation to the equations γ = α + 1 and β = α(α + 1), defining an irreducible variety.  

Lemma 2.38. Assume (2F), with K = D7q (q). Then there exist subgroups W0 , L, and L1 of K with the following properties: (a) W0 ∼ = W (D7 ) is a complement to CK (B) in NK (B), and CB (W0 ) = 1; (b) L ∼ = D6 (q) and B = (B ∩ L) × CB (L); (c) L1 ∼ = D4 (q), Z(L1 ) = 1, and B = (B ∩ L1 ) × CB (L1 ); (d) W1 := W0 ∩ L1 ≡B∩L1 W (D4 ) and W0 ∩ L ≡B∩L W (D6 ); and (e) With standard notation for L1 , W1 is L1 -conjugate to nα (1) | α ∈ Σ , where Σ is the root system of L1 . Proof. In (2F), we have n = 7, V0 = 0, and O(Vi ) ∼ = D2(q−q ) for each i = 1, . . . , 7. Let W0 = F0 S be a complement to CK (B) in NK (B) as in Lemma 2.40. Thus S ∼ = Σ7 permutes {V1 , . . . , V7 } and {b1 , . . . , b7 } naturally and (2H)

NS (Vi ) = CS (Vi ) for each i = 1, . . . , 7.

Also, there exist ti ∈ I2 (CO(V ) (Vi⊥ )), 1 ≤ i ≤ 7, permuted naturally by S, such ∼ that btii = b−1 i , and such that F0 is the subgroup of t1 , . . . , t7 = E27 consisting of all even products of t1 , . . . , t7 . In particular W0 satisfies (a). Set V16 = V7⊥ , V14 = (V5 + V6 + V7 )⊥ , L = CK (V7 ) ∼ = Ω(V16 ) ∼ = D6 (q) and ∼ ∼ L1 = CK (V5 +V6 +V7 ) = Ω(V14 ) = D4 (q). Then as B = b1 , . . . , b7 with bi ∈ Ω(Vi ) for each i, (b) and (c) hold. Using (2H), we see that L1 ∩ W0 ≡b1 ,b2 ,b3 ,b4  W (D4 ) and similarly for L ∩ W0 , so (d) holds. Using standard notation for L1 , let W2 = nα (1) | α ∈ Σ ∼ = W (D4 ). We argue that (2I)

W2 normalizes B g for some g ∈ L1 .

If q = 1, then W2 normalizes a Cartan subgroup H2 of L1 by the Chevalley relations, and Ω1 (Op (H2 )) ∼ = B ∩ L1 , whence Ω1 (Op (H2 )) is L1 -conjugate to B ∩ L1

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by [IA , 4.10.3f]. As B = (B ∩ L1 ) × CB (L1 ), our assertion follows in this case with g = 1. So assume that q = −1. Let (L1 , σq ) be a σ-setup for L1 , and (B, T ) a σq -invariant Borel-torus pair. We may assume that W2 normalizes T . As W2 is σq invariant, so is CL1 (W2 ). Let w2 be the central involution of W2 ; thus w2 inverts T elementwise. By Lemma 2.37, CL1 (W2 ) is connected, so by Lang’s theorem, w2 = σq−1 hσq h−1 for some h ∈ CL1 (W2 ). Thus σqh

−1

= σq w2 . Now, W2 = W2h

h

normalizes T as well as CT h (σq ), a maximal torus of Inndiag(K) isomorphic to −1 CT (σqh ) = CT (σq w2 ), which is homocyclic abelian of rank 4 and exponent q + 1. By [IA , 4.10.3] again, Ω1 (Op (CT h (σq ))) is L1 -conjugate to B∩L1 , and (2I) is proved. Finally, CL1 (B g ) ∼ = CL1 (B) is homocyclic abelian of rank 4 and (odd) exponent q − q , and both W1g and W2 are complements to CL1 (B g ) in NL1 (B g ). Moreover, whether q = 1 or −1 in the previous paragraph, Z(W2 ) = w2 inverts CL1 (B g ) elementwise, as a consequence of the Chevalley relations. By Sylow’s theorem Z(W2 ) and Z(W1g ) are conjugate in NL1 (B g ). Since CNL1 (B g ) (Z(W2 )) = W2 , it follows that W2 and W1g are NL1 (B g )-conjugate. Hence W1 and W2 are L1 conjugate, completing the proof of (e) and the lemma.  Lemma 2.39. Assume (2F), with K = D6 (q). Then there exists W0 ≤ K with the following properties: (a) W0 ∼ = W (D6 ) is a complement to CK (B) in NK (B); (b) With standard notation for K, W0 is K-conjugate to nα (1) | α ∈ Σ , where Σ is the root system of K. Proof. Set W2 = nα (1) | α ∈ Σ ∼ = W (D6 ). It suffices to argue that W2 normalizes B g for some g ∈ K,

(2J) −1

for then W0 = W2g satisfies the conclusion of the lemma. Note that CG (B) has odd order by Lemma 2.2, while O2 (W (D6 )) = 1, so that W0 will indeed be a complement to CG (B) in NG (B). If q = 1, then W2 normalizes a Cartan subgroup H2 of K by the Chevalley relations, and Ω1 (Op (H2 )) ∼ = B, whence Ω1 (Op (H2 )) is K-conjugate to B by [IA , 4.10.3f], and our assertion follows in this case with g = 1. So assume that q = −1. Let (K, σq ) be a σ-setup for K, and (B, T ) a σq -invariant Borel-torus pair. We may assume that W2 normalizes T . As W2 is σq -invariant, so is CK (W2 ). Let w2 be the central involution of W2 ; thus w2 inverts T elementwise. By Lemma 2.37, CK (W2 ) is connected, so by Lang’s theorem, w2 = σq−1 hσq h−1 for some h ∈ CK (W2 ). Thus σqh

−1

h

= σq w2 . Now, W2 = W2h normalizes T as well as CT h (σq ), a maximal torus −1

of Inndiag(K) isomorphic to CT (σqh ) = CT (σq w2 ), which is homocyclic abelian of rank 6 and exponent q + 1. By [IA , 4.10.3] again, Ω1 (Op (CT h (σq ))) is K-conjugate to B, and (2J) and the lemma are proved.  In the next lemma, q can be an odd prime power. Lemma 2.40. Let V = V0 ⊕ V1 ⊕ V2 ⊕ · · · ⊕ Vm , m ≥ 2, be an orthogonal decomposition of a a natural module for a classical group K of level q. Assume that either q is even or V is an orthogonal space. Suppose that V1 , . . . , Vm are isometric subspaces of dimension 1 or 2. Let C = ± C(V ) be the isometry group of V and set C o = [C, C] = SL± n (q), Sp2n (q), or Ωn (q), o o so that K/Z(K) ∼ = C /Z(C ).

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Then there is a subgroup S ≤ C such that the following conditions hold: (a) [V0 , S] = 0; (b) S ∼ = Σm permutes {V1 , . . . , Vm } naturally, NS (Vi ) = CS (Vi ) for each i = 1, . . . , m, and Am ∼ = [S, S] ≤ C o ; (c) If q is even, then S may be taken to be a subgroup of C o ; (d) Suppose that q is even, p is a prime dividing q − q , q = ±1, and C o ∼ = SL−q (V ), Sp(V ), or Ω(V ). Then S normalizes some elementary abelian p-subgroup B ≤ C o that is maximal with respect to normalizing each Vi and centralizing V0 ; (e) Let V , C o , and B be as in (d). Then there is a basis {b1 , . . . , bm } of B which is permuted naturally by S and such that the support of bi on V is Vi , i = 1, . . . , m; and (f) In the situation of (d) and (e), B lies in a complement H to CC o (B) in NC o (B). Proof. For each i = 1, . . . , m, fix an isometry τi : V1 → Vi , with τ1 = 1V1 . Then define σ1i ∈ C, for 1 < i ≤ m, by σ1i |V1 = τi , σ1i |Vi = τi−1 , σ1i |Vj = 1Vj for all j = 0, . . . , m except j = 1, i. Set S = σ1i | 1 < i ≤ m . Then S permutes {V1 , . . . , Vm } with kernel S0 , say, and S/S0 ∼ = Σm . If σ ∈ S and σ(V1 ) = V1 , then it is trivial to verify that σ|V1 = 1V1 . As the Vi are all S-conjugate, it follows that NS (Vi ) = CS (Vi ) for any 1 ≤ i ≤ m. In particular S0 = 1. Thus, S ∼ = Σm and S ≤ C. As [C, C] ≤ C o , [S, S] ≤ C o with [S, S] ∼ A . This proves (a) and (b). = m If the characteristic is 2, det σ1i = 1 for all i. Moreover, in the orthogonal case with q even, each σ1i is the product of an even number of transvections. Thus (c) holds. By our hypotheses in (d) (see [IA , 4.8.1]), the isometry group of each Vi has cyclic Sylow p-subgroups, lying in the commutator subgroup. Choose B1 = b1 ≤ τ −1

CC o (V1⊥ ) of order p and set bi = b1i and Bi = bi for each i = 1, . . . , m. Then {b1 , . . . , bm } and B = B1 × · · · × Bm are S-invariant, and S permutes the bi ’s and Bi ’s as it permutes the Vi ’s. This yields (d) and (e). Under the hypotheses in (d), Isom(V1 ) contains an involution t1 inverting b1 . Extend t1 to s1 ∈ Isom(V ) by having s1 act trivially on Vi for all i = 1. Then det s1 = 1 so unless V is an orthogonal space, in fact, s1 ∈ C o . In any case, for any i = 2, . . . , m, choose any gi ∈ S mapping 1 to i and define si = sg1i . Since NS (Vi ) = CS (Vi ) for any i, si is well-defined and S permutes {s1 , . . . , sm } naturally. Let F = s1 , . . . , sn ∼ = E2m , so that F is a natural permutation module for S. Also let F0 = s1 s2 , . . . , s1 sn , the trace-0 F2 S-submodule of F . Then F S is the complement to CK (B) in NK (B) required by (f), except in the orthogonal  group case, for which F0 S serves the purpose. The lemma is proved. Lemma 2.41. Let G0 ∈ Chev(2) be as in [III15 , Table 15.3], and let p > 3 be a prime divisor of q − q for some q = ±1. Assume that p does not divide |Outdiag(G0 )|. Let B ∈ Ep∗ (G0 ). Then there are at least two orbits in the action of NAut(G0 ) (B) on E1 (B). If there are exactly two, then the centralizer in G0 of any non-identity element of B is non-p-solvable. Proof. Since p > 3, q > 2. Suppose first that G0 is a classical group. Since p does not divide |Outdiag(G0 )|, we may replace G0 by a covering group that acts

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faithfully on a natural module V . If G0 = Lmq (q), then m ≥ 5 and the cyclic subgroups generated by diag(α, α−1 , 1, . . . ) and diag(α, α, α−1 , α−1 , 1, . . . ), where αp = 1 = α, are in distinct orbits and centralize a nonsolvable SL2 (q)-subgroup. Otherwise 2 is the dimension of minimal possible support on V for an element of G0 of order p; there are then elements of E1 (B) with supports on V of dimensions 2, 4, and 6, and these are in distinct orbits. If G0 is of exceptional type, note that by [IA , 4.10.3c], B is unique up to conjugacy and so it suffices to find enough nonconjugate elements of order p with centralizers of rank mp (B). If G0 ∼ = E7 (q) or E8 (q), then as p divides q 2 −1 there are clearly three end-node parabolic subgroups giving rise to nonconjugate subgroups of order p of parabolic type. Similar arguments show that for G0 ∼ = E6± (q) or F4 (q), there are two maximal parabolic subgroups giving rise to nonconjugate subgroups of order p of parabolic type with non-p-solvable centralizers. This completes the proof.  3. Neighborhoods In this section we examine toral subgroups D ≤ Inndiag(J), where J ∈ Chev(r) has level q, and D ∼ = Ep2 , for primes p and r, one of which is 2, such that p divides q 2 − 1. In this context a (Lie) component JD of CK (D) gives rise to a “neighborhood” N consisting of the pumpups Jd of JD in CJ (d) as d ranges over D# . The main question is how N and J determine each other. Most, but not all, of our results are about the case r = 2, and the following notation will be common, with q the level of some group in Chev(2): (3A)

(1) p is an odd prime and q is a power of 2; and (2) q = ±1 and q ≡ q (mod p).

For classical groups, the results [IA , 4.8.1, 4.8.2] clarify most neighborhood questions, when p is odd. So a good portion of our results in this section concern groups of exceptional type. If we expand D to B ∈ Ep∗ (Inndiag(K)), the structure of a neighborhood is naturally intimately connected with the reflection group AutK (B). Through Lemma 3.12 we consider how the group J determines the neighborhood N. Lemma 3.1. Let K ∈ Lie(2) be the adjoint version, let p be an odd prime, and 3 let E ≤ Inndiag(K) with E ∼ = A± = Ep2 . If p = 3, exclude K ∼ 2 (q) and D4 (q). Let (K, σ) be a σ-setup of K with K adjoint, so that CK (σ) ∼ = Inndiag(K). If E lies in an element of Ep∗ (Inndiag(K)), then E lies in a maximal torus of K. Proof. By Lemma 2.5, this is true if p does not divide |Outdiag(K)|. So assume that p divides |Outdiag(K)|. Thus by our assumptions we may assume q  (q), kp > 3, or K ∼ that K ∼ = E6q (q), with p = 3, using the notation (3A). By = Lkp p [IA , 4.10.3d], E∗ (Inndiag(K)) is a single Inndiag(K)-conjugacy class, so it suffices to show that for some torus T of K, Tp := Ω1 (Op (CT (σ))) has rank mp (Inndiag(K)). In the Lkp (q) case we may take σ = σq γ, where γ = 1 or γ is the transposeinverse graph automorphism (according as  = 1 or −1), and then the torus T of diagonal matrices in K satisfies the requirement. Similarly in the E6 (q) case with p = 3 dividing q − , there is a semisimple subgroup L ≤ K with L = A1 × A5 ,

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17. PRELIMINARY PROPERTIES OF K-GROUPS

invariant under the group ΓK of graph automorphisms of K. Again taking T to be a maximal torus of L arising from diagonal subgroups of A1 and A5 , and σ = σq γ for an appropriate graph automorphism γ ∈ ΓK K normalizing T , we see that  mp (Tp ) = 6, as required. The proof is complete. Lemma 3.2. Let p be an odd prime and let W = W (F4 ) act naturally on B, the mod-p reduction of the F4 -root lattice. Let X = {x ∈ E1 (B) | CW (x) ∼ = W (C3 )}. Let x1 , x2 ∈ X be such that D = x1 , x2 ∼ = Ep2 , and CW (D) ∼ = W (C2 ) is generated by reflections. Then the following conditions hold: (a) For any x ∈ X, NW (x ) = CW (x) × rx , where rx is a reflection inverting x; (b) |X ∩ E1 (D)| = 4; (c) AutW (D) ∼ = W (C2 ) has two orbits on X ∩ E1 (D) of length 2; (d) Let x ∈ X and let F = {B1 , B2 , B3 } be the CW (x)-invariant frame in [B, CW (x)]. Then the stabilizer in W of the frame F ∪{x } is isomorphic to W (C4 ). Moreover, generators bi for Bi may be chosen, i = 1, 2, 3, such that the centers of twelve W -conjugates of rx are x , B1 , B2 , B3 , and  the ±1 ±1 b b x ; and the eight groups b±1 1 2 3 (e) If the bi are chosen as in (d), and we set b4 = x, then the centers of the twelve reflections in W − rxW are b±1 b j , 1 ≤ i < j ≤ 4. i Proof. As the unique minimal normal subgroup of W inverts B, W acts faithfully on B. As noted in Remark 1.20, B carries a nonsingular W -invariant quadratic form. Let x ∈ X. If x ⊆ x⊥ , then CW (x) preserves a chain in B with factors of dimension 1, 2, and 1. But CW (x) contains W (C3 ) ∼ = Z2 × Σ4 , which requires at least 3 dimensions for a faithful representation in odd characteristic. Hence x × x⊥ = B, a decomposition preserved by NW (x ). As CW (x) acts faithfully and absolutely irreducibly on x⊥ , Z(CW (x)) ∼ = Z2 inverts x⊥ and so Z(CW (x))Z(W ) contains a reflection rx inverting x. Then clearly NW (x ) = CW (rx ) and (a) follows easily. As CW (r) ∼ = Z2 × W (C3 ) for any reflection r ∈ W , the correspondence r → [B, r] is a bijection between the set of reflections in W and X. Then assertions (d) and (e) can be deduced from an explicit description of the F4 root system [IA , 1.8.8]. Moreover, in (d), the frame {x , B1 , B2 , B3 } is orthogonal and is permuted naturally by some S ≤ W , S ∼ = Σ4 . Suppose D = x1 , x2 is as in the lemma. Since x is anisotropic for any x ∈ X, D = D⊥ . But then since CW (D) ∼ = D8 is nonabelian and completely reducible, we must have D ∩ D⊥ = 1; otherwise CW (D) would preserve an ordered frame in B, which is impossible as it is nonabelian. Then CW (D) contains 4 reflections, in two classes of size 2; therefore, X0 := X ∩ E1 (D⊥ ) has 4 elements, in two CW (D)-orbits of length 2. Let x ∈ X0 . From the descriptions in (d) and (e), we see that we may assume that  X0 = {x , B1 , b±1 1 x }. Thus D = B2 , B3 and so D and D⊥ are W -conjugate. Now (b) and (c) follow, completing the proof. 

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Lemma 3.3. Assume (3A). Let J ∈ Chev(2) and D ∈ Ep2 (Inn(J)), and put E(CJ (D)) = JD and E(CJ (d)) = Jd for all d ∈ D# . For the following quadruples (J, Jd , JD , Jd ), given d and D, there exists d ∈ # D with the given isomorphism type for Jd : (a) (b) (c) (d) (e)



(E8 (q),D7q (q),D6 (q),E7 (q));   (E7 (q), D6 (q), D5q (q), E6q (q)); (F4 (q), C3 (q), C2 (q), C3 (q)); and − − − (E6 q (q), D4− (q), D3 q (q), A5 q (q)); q q q (E6 (q), D4 (q), D3 (q), A5 (q)), with p = 3 and q ≡ q (mod 9).

Proof. We first show that in each case there is some D = d, d ∈ ∗ ∼ a component JD = JD , lying in comrespectively, with Jd∗ and Jd∗ centrally isomorphic to Jd and Jd . Let (J, σ) be a σ-setup for J. We use standard notation [IA , 2.10]. We take σ = σq in the first three cases if q = +1, and σ = wσq if q = −1, where w ∈ N (the monomial group) maps on the Weyl group element −1. In these cases σ-invariant subsystem subgroups J d , J d , and J D are visible from the Dynkin diagram, and ∗ . (In (c), since q is even, taking fixed points of σ yields the desired Jd∗ , Jd∗ , and JD C3 (q) ∼ = B3 (q).) Similar arguments handle the other two cases, with σ = γσq if q = 1 in (d) or q = −1 in (e) (where γ is the involution in ΓJ ), and with σ = γwσq in the remaining cases (where w ∈ γN maps on −1 ∈ W , i.e., −1 ∈ γ W ). To complete the proof, we show that there is a unique Aut(J)-class of subgroups d ≤ Inndiag(J) of order p whose centralizers in J have components isomorphic to Jd , and there is a unique Inn(Jd )-class of subgroups of Aut(Jd ) of order p whose centralizers in Jd have a component isomorphic to JD . Hence D will be determined up to Aut(J)-conjugacy, and the result will follow in these cases. In cases (b), (c), and (d), J is simple and d Jd contains some B ∈ Ep∗ (K), which is determined up to conjugacy by Lemma 2.4. In these cases B = d × (B ∩ Jd ), and NJd (B ∩ Jd ) contains an element inverting B∩Jd , while NJ (B) also contains an element inverting B. Therefore there is a reflection rd ∈ AutJ (B) such that [B, rd ] = d . As there is a unique class of reflections in AutJ (B) in cases (b) and (d), and in AutAut(J) (B) in case (c), d is unique up to Aut(J)-conjugacy, and a similar argument proves the uniqueness of D ≤ Jd d up to Jd -conjugacy with the appropriate subcomponent JD . In case (e), we use the table [IA , 4.7.3A]. The hypotheses on p = 3 imply that m3 (CInn(J) (Jd )) = 1 and also that some element g ∈ CInn(J) (d) induces a triality on Jd . Then Jd g permutes transitively the subgroups d ≤ Jd of order 3 with E(CJd (d )) having a D3 (q) component. The proof may now be finished as in the other cases.  Finally, in case (a), let d ∈ D# with CJ (d) having a component Jd ∼ = D7q (q). p Let B = d × (B ∩ Jd ) ∈ E∗ (J) and σ be as before. By Lemma 2.5, d ∈ T for some maximal torus T of J. Let J d be the semisimple part of CJ (d). Then Jd embeds in    a component of O 2 (CJ d (σ)), for which the only possibilities then are A7q (q), D7q (q)   and E7 (q). But D7q (q) does not embed in A7q (q) or E7 (q), by order considerations 4 3 (for E7 (q) consider the divisor (q − 1) of its order). Thus J d ∼ = D7 (q). But any fundamental D7 -subsystem of the E8 root system lies in a fundamental E8 -system by Lemma 1.1, so NJ (T ) is transitive on pairs (d , J d ) such that d ∈ T and J d Ep2 (Inndiag(J)) such that E(CJ (D)) has ponents Jd∗ and Jd∗ of CJ (d) and CJ (d )

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is a D7 (q) component of CJ (d). In particular, d is determined up to conjugacy. Then it is easily seen that d, d / d is also unique up to conjugacy, and the proof is complete.  

Lemma 3.4. Assume (3A). Let J = Lnq (q), with p dividing n and n ≥ 6 (strict q inequality if q = 2). Let x ∈ Ip (J) be such that Jx := E(CJ (x)) ∼ (q) and = SLn−2 q mp (CJ (Jx )) = 1. Let y ∈ Ip (CJ (x)) be such that L := E(CJx (y)) ∼ (q), and = SLn−3 let E = {u ∈ E1 (x, y ) | L is not a component of CJ (u)}. Then |E| = 3. Proof. From the structure of Jx , x must be the image mod scalars of diag(ω, ω −1 , 1, 1, . . . , 1) for some primitive pth root of unity ω, with respect to some (orthonormal) basis. Since q ≡ q (mod p), x and y are simultaneously diagonalizable with respect to some basis B. As mp (CJ (Jx )) = 1, y is uniquely determined modulo x by its action on Jx , and so without loss we may take y to be the image mod scalars of diag(1, ω, ω −1 , 1, 1, . . . , 1). Then E = {x , y , xy }, proving the lemma.  Lemma 3.5. Assume (3A). Let J = D4 (q), q = 2n > 2. Suppose that D =   x, y ∈ Ep2 (J), Jx := E(CJ (x)) ∼ = L4q (q), and JD := E(CJx (y)) ∼ = SL3q (q). Then there are exactly three elements u ∈ E1 (D) such that JD  CJ (u). Proof. Let De be the set of all u ∈ E1 (D) such that JD  CJ (u). It follows  easily from [IA , 4.8.2] that for all u ∈ D# , O 2 (CJ (u)) ∼ = JD or Jx , according as u ∈ De or not. As p > 2, p is a good prime for J, so D lies in a torus of the overlying algebraic group by [IA , 4.10.3f]. Hence by [IA , 7.4.1a4], 

|De |

q 12 = |L4q (q)|2



|De |−1

/|L3q (q)|2

= q 6|De |−3|De |+3 , whence |De | = 3. 

Lemma 3.6. Let K = U6 (2) and D ∈ E32 (K). Then for some d ∈ D# ,  U4 (2). E(CK (d)) ∼ =  = SU6 (2) have natural module V . For any Proof. Suppose false. Let K #  with respect to d ∈ D , therefore, d is the image of d := diag(ω, ω −1 , 1, 1, 1, 1) ∈ K   some orthonormal basis of V . Moreover, it is then clear that d is not K-conjugate to   which  any other element of dZ(K). Hence any d1 ∈ D − d has a preimage d1 ∈ K  d1 ] = 1. So d and also is diagonalizable with four eigenvalues equal to 1, and [d,  d1 are simultaneously diagonalizable, and we may assume that d1 is also diagonal. Then dd1 also must have an eigenspace of dimension 4. The only possibility is that this eigenspace corresponds to the eigenvalue 1. Up to replacing ω by ω −1 and permuting coordinates, we must then have d1 = diag(1, ω, ω −1 , 1, 1, 1). But then dd−1 1 has no eigenspace of dimension 4, a contradiction. The lemma is proved. 

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Lemma 3.7. Assume (3A). Let J = F4 (q), q > 2, and let B ∈ Ep∗ (J). Let D ∈ E2 (B) with JD := E(CJ (D)) ∼ = Sp4 (q). Then there exist exactly four u ∈ E1 (D) such that E(CJ (u)) ∼ = Sp6 (q). 



Proof. Since q > 2, JD = O 2 (CJ (D)). For each d ∈ D# , let Jd = O 2 (CJ (d)). If d ∈ D# is such that JD < Jd , then by [IA , 4.2.2] (and as B ≤ J by assumption), the only possibility, given the location of the C2 -subdiagram of the F4 diagram, is that Jd ∼ = Sp6 (q). Let m be the number of u ∈ E1 (D) such that JD < Jd . By Lemma 3.1, D lies in a maximal torus of the overlying algebraic group, so [IA , 7.4.1a4] applies. We get m−1 q 24 = |Sp6 (q)|m = q 9m−4(m−1) , so m = 4. 2 /|Sp4 (q)|2  −

Lemma 3.8. Assume (3A). Let J = E6 q (q), q > 2. Suppose that D = x, y ∈ − − p E2 (J), Jx := E(CJ (x)) ∼ = L6 q (q), and JD := E(CJx (y)) ∼ = L4 q (q). Then there are exactly two elements u ∈ E1 (D) such that the pumpup Ju of JD in CJ (u) is isomorphic to Jx , exactly two for which Ju ∼ = D4− (q), and no other u ∈ E1 (D) for which Ju > JD . ∼ Jx and Proof. By Lemma 3.3, there exist u1 , u2 ∈ D# such that Ju1 = ∼ D− (q). Expand D to B ∈ Ep (K) = Ep∗ (K). Then AutJ (B) = ∼ W (C3 ) Ju2 = ui 4 4 contains an involution inverting B ∩ Jui , and AutK (B) contains −1, so there exist reflections ri ∈ AutK (B) such that [B, ri ] = ui , i = 1, 2. By Lemma 3.2, u1 and u2 each have two AutK (B)-conjugates in D. It remains to show that Ju = JD for any u ∈ E1 (D) not conjugate to u1 or u2 . By Lemma 3.1, D lies in a torus of the algebraic group overlying K. Thus Wielandt’s formula [IA , 7.4.1a4] applies  to the action of D on K. Writing O 2 (CK (u)) = Ju Mu for u ∈ D, where Mu is a product of Lie components, we get |JD |2 |Ju1 |22 |Mu1 |22 |Ju2 |22 |Mu2 |22  |Ju |2 |Mu |2 1 = 36 q |JD |22 |JD |22 |JD |2  |Ju |2 |Mu |2 = |Mu1 |22 |Mu2 |22 , |JD |2 where the products are over all u ∈ E1 (D) except the conjugates of u1 and u2 . It follows that Mu = 1 for all u, and Ju = JD except for conjugates of u1 and  u2 , as asserted.  Lemma 3.9. Assume (3A), with p = 3 and q ≡ q (mod 9). Let J ∼ = E6q (q), 3 # and let D ∈ E2 (J). For any d ∈ D , write Jd = E(CJ (d)). Suppose that Jd ∼ =   L6q (q) for exactly two d ∈ E1 (D), and JD := E(CJ (D)) ∼ = = L4q (q). Then Jd ∼ D4 (q) for the other two d ∈ E1 (D).

Proof. By [IA , 4.7.3A], there is a unique Inn(J)-conjugacy class of subgroups  d of order 3 such that CJ (d) has a component Jd ∼ = A5q (q). Since q ≡ q (mod 9),  q AutJ (Jd ) ∼ = L6 (q). There then exists a unique NJ (d )-conjugacy class of sub groups D ≤ d Jd such that D ∼ = E32 and E(CJd (D)) ∼ = A3q (q). Since a quadruple # exists as in Lemma 3.3e, there exists d ∈ D with Jd ∼ = D4 (q). By Lemma 3.1, Wielandt’s formula applies [IA , 7.4.1a4], and it implies that for the fourth d ∈ E1 (D), |Jd |2 = q 12 . The only choice, by [IA , 4.7.3A], is Jd ∼ = D4 (q), completing the proof. 

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Lemma 3.10. Assume (3A). Let J = Inndiag(E6q (q)), and let T0 be a maximal torus of J containing a subgroup Ω1 (Op (T0 )) =: T ∼ = Ep6 . Set W = AutJ (T ) ∼ = W (E6 ). Let S0 ≤ S1 be subgroups of W generated by reflections on T , such that S0 ∼ = Σ5 and S1 ∼ = Σ6 . Let D = CT (S0 ) and De = {u ∈ E1 (D) | CW (u) >   S0 }. Then |De | = 4, and for u ∈ De , O 2 (CJ (u)) ∼ = D5q (q) (for two u ’s), q q A4 (q) × A1 (q), and A5 (q). Proof. Suppose first that p > 3. Then as W -module, T is isomorphic to the mod-p reduction of the E6 -root lattice. We regard the positive roots as ai ± aj , 8 "8 4 ≤ i < j ≤ 8, and 12 (a1 − a2 − a3 + i=4 i ai ) where each i = ±1 and i=4 i = 1. Here the ai ’s are an orthonormal basis of Euclidean 8-space, and T is the span of the fundamental system α1 = { 12 (a1 − a2 − a3 − a4 − a5 − a6 − a7 + a8 ), αi−2 = ai − ai+1 , (4 ≤ i ≤ 7), α6 = a7 + a8 }. Then we may take S0 to be generated by the reflections in α2 , . . . , α5 , and S1 to be generated by S0 and the reflection in α1 . It follows that D is spanned by β1 = a1 − a2 − a3 and β2 = a4 + a5 + a6 + a7 + a8 . Then letting Wm,n be the subgroup of W generated by reflections in all roots (mod p) orthogonal to mβ1 + nβ2 , we compute that W1,0 ∼ = W (D5 ), W1,1 = S1 ∼ = Σ6 , W1,3 ∼ = W (D5 ), ∼ and W5,3 = W (A4 ) × W (A1 ). Now the Wielandt formula [IA , 7.4.1a4] implies that Wm,n = S0 unless mβ1 + nβ2 ≡ β1 , β1 + β2 , β1 + 3β2 , or 5β1 + 3β2 , modulo p, as required. We may now suppose that p = 3. Consulting [IA , 4.7.3A] we see that the only   possible subgroups O 2 (CJ (d)), d ∈ D# , containing A4q (q), are 





A5q (q), A4q (q) × A1 (q), and D5q (q). 

Moreover, the only one of these with d in the hyperplane [AutJ (T ), T ] is A5q (q), so  A5q (q) occurs just once. There are just 3 other elements of E1 (D), and Wielandt’s  formula [IA , 7.4.1a4] implies that they are as asserted. n

q (q), n ≥ 5. Lemma 3.11. Assume (3A). Let K = Sp2n (q), n ≥ 4, or Ω2n p #  Let B ∈ E∗ (K), and let x, w ∈ B with L := E(CK (x)) ∼ = SLn (q),  = ±1, and E(CL (w)) ∼ = SLn−1 (q). Then there is v ∈ x, w − x such that E(CK (v)) ∼ =

n−1

q (q), according to the isomorphism type of K. Sp2n−2 (q) or Ω2n−2

Proof. Let V be the natural K-module. We then have the setup (2F). Then  (see [IA , 4.8.2], [GL1, §8]) there is an isomorphism φ : CK (x) ∼ = GLnq (q) such that φ(x) = ζI for some pth root ζ of unity in Fq2 , and φ(B) is a full exponent-p  q (q), φ(w) clearly has diagonal subgroup of GLnq (q). Then as E(CL (w)) ∼ = SLm−1 a b eigenvalues ζ = ζ of multiplicities 1 and p − 1, respectively. Then v = wx−b is the required element.  Lemma 3.12. Let J, p, T0 and T be as in Lemma 3.10, with p = 3. Let T0 be as there, and let T1 = [T, AutJ (T )] ∼ = E35 , so that AutJ (T1 ) ∼ = W (E6 ). Let ∼ ∼ Σ4 = S0 < S1 = Σ6 be subgroups of W such that dim CT1 (Si ) = 2 − i, i = 0, 1.  Then there exists d ∈ CT1 (S0 )# such that O 2 (CJ (d)) ∼ = D4 (q). Proof. By Lemma 1.14, we can identify T1 with the reflection module V for W ∼ = SO5 (3). Now, W has exactly 3 orbits on V # , represented by v0 , v1 , and v−1 , where q(vi ) = i for each i ∈ {0, 1, −1}. Corresponding elements t of T1# + ∗ ∼ ∼ have F ∗ (CW (t)) a 3-group, Ω− 4 (3) = A6 , and F (Ω4 (3)) = Q8 ∗ Q8 . Then from

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[IA , Table 4.7.3A] we see that v0 corresponds to t3 in the table, and v±1 correspond to t4 and t1,6 . Since S1 is contained in no 3-local subgroup of W , by the Borel-Tits theorem, CT1 (S1 ) corresponds to a (1-dimensional) nonsingular subspace V1 of V . Furthermore, S0 is contained in no 3-local subgroup of S1 , so CT1 (S0 ) corresponds to V1 ⊥ V2 , where V2 is a 1-dimensional nonsingular subspace of V1⊥ . As V1 ⊥ V2 is 2-dimensional and nondegenerate, it contains vectors of every q-value except possibly 0. In particular it contains a vector corresponding to a W -conjugate t ∈ T1 of  t1,6 , whence O 2 (CJ (t)) ∼  = D4 (q), as required. The next set of lemmas seeks to identify J from a knowledge of a neighborhood N. Typically, for J ∈ Chev(2), we will be in the following situation, with notation as in (3A).

(3B)

(1) J ∈ Kp and D ∈ Ep2 (Aut(J)); (2) For each d ∈ D# , Jd is a subnormal subgroup of CJ (d), Jd ∈ Chev(2) ∪ {SU3 (2)}, and Jd has level q; (3) There is a common subnormal subgroup I to all the groups CJd (D), I ∈ Chev(2) ∪ {SU3 (2)}, and I has level q; and (4) For some d0 ∈ D# , mp (CInn(J)D (Jd0 )) = 1.

Lemma 3.13. Assume (3A) and (3B), with p dividing q + 1 and q > 2. Suppose that I ∼ = L2 (q), and that for each d ∈ D# , either Jd = I, or Jd ∼ = Sp4 (q) with mp (CJ (Jd )) = 1. Suppose finally that Jd ∼ = Sp4 (q) for at least two d ∈ E1 (D). Then J ∼ = Sp6 (q). Proof. If J = Jd ∼ = Sp4 (q) for some d ∈ D# , then J = Jd for at least one d ∈ D − d . Hence [J, D] = 1, which is absurd. So Sp4 (q) ↑p J. By [III11 , 1.1ab] and [IA , 2.2.10], J ∈ Chev(2). Fix d ∈ D# with Jd ∼ = Sp4 (q). If there is d1 ∈ D# − Inndiag(J), then as ∼ L Jd1 ∼ (q) or Sp (q), J = 2 = L2 (q p ) or Sp4 (q p ). But then mp (Out(J)) = 1 so there 4 # is some d2 ∈ D ∩ Inn(J), whence Jd2 ∼ = L2 (q p ) or 1, contrary to assumption. Thus, D ≤ Inndiag(J). Then [IA , 4.2.2] yields several conclusions. For one, the untwisted Dynkin diagram for J must contain the C2 diagram, so it must be F4 or Cn for some n. Moreover, since it contains a unique double bond, q(J) = q. And as Z(J) = 1, every element of Ip (CJ (d)) induces an inner-diagonal automorphism on Jd . Finally, since mp (CJ (Jd )) = 1 whenever JD < Jd , we have mp (CJ (d)) ≤ mp (Inndiag(Jd ))+1 = 3 for such elements d ∈ D# . Moreover, mp (CJ (d)) = mp (J), by [IA , 4.10.3f] or – in case p = 3 and J ∼ = F4 (q) – by [IA , 4.7.3A]. It follows that mp (J) = 3, whence by [IA , 4.10.3b], J ∼  = Sp6 (q). The proof is complete. The next lemma develops the setup (3B). Lemma 3.14. Assume (3A) and (3B). Assume that mp (I) ≥ 2, and that   D4 (q) and I/Z(I) ∼  L3q (q), G2 (q), or 3D4 (q). Then the if p = 3, then Jd0 ∼ = = following conditions hold: (a) J ∈ Chev(2) and D ≤ Inndiag(J); (b) If J is a classical group, then q(J) = q; and (c) If J is of exceptional type, then q(J) = q unless possibly for all d ∈ D# , the untwisted Lie ranks rJ and rJd of J and Jd , respectively, satisfy rJ ≥ 2rJd .

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Proof. If J ∈ Chev(r) ∪ Alt ∪ Spor for some odd r, then   Jd0 ∈ Chev(r) ∪ Alt ∪ Chev(p) ∩ Chev(2), with the help of [III11 , 1.1ab]. As mp (I) ≥ 2 and D/ d0 acts faithfully on Jd0 , with I   CJd0 (D), the only possibility by [IA , 2.2.10] is Jd0 ∼ = C2 (3), = U4 (2) ∼ q = 2, p = 3 = r. But then I/Z(I) ∼ U (2), contrary to assumption. Thus, = 3 J ∈ Chev(2). As [I, D] = 1, D normalizes Jd for all d ∈ D# . Choose d1 ∈ D# ∩ Aut0 (J), which exists since mp (Aut(J)/Aut0 (J)) ≤ 1. Suppose first that D ≤ Aut0 (J) and let d ∈ D − Aut0 (J). Then by [IA , 4.2.3], d induces a nontrivial field or graphfield automorphism on Jd1 and so q(I) = q(Jd1 ), contrary to assumption. Hence, D ≤ Aut0 (J). If some d ∈ D# induces a graph automorphism on J, then p = 3 and by  [IA , 4.7.3A], Jd ∼ = G2 (q) or L3q (q). In particular m3 (I) ≤ m3 (Jd ) = 2 [IA , 4.10.3a]. Thus m3 (I) = 2. As q(I) = q and I   CJd (D), we must have Jd ∼ = G2 (q) and q I∼ SL (q), contrary to assumption. Thus, D ≤ Inndiag(J), proving (a). = 3 Next, set qJ = q(J) and suppose that q = qJ . Then by [IA , 4.2.2], qJ = q 1/a for some a > 1, with arJd ≤ rJ , so (c) holds. Finally, assume that J is a classical group. If [IA , 4.8.4] applies to some q  d ∈ D# , so that qJp = q and J/Z(J) ∼ (qJ ), then Jd /Z(Jd ) ∼ = Lkp = Lkq (q). Then mp (CAut(J) (Jd )) = 1 so I < Jd . As mp (I) ≥ 2, it follows that k ≥ 4. But the q (qJ ) is clearly noncyclic, while the kernel of the natural surpreimage of D in GLkp q jection GLkp (qJ ) → Inndiag(J) is cyclic, so as p is odd, some d1 ∈ D# is induced q (qJ )). It follows from [IA , 4.8.2] that q(Jd1 ) = qJ , a by an element of Ip (GLkp contradiction as q(Jd1 ) = q by (3B). Thus, [IA , 4.8.2] applies to each d ∈ D# in  its action on J. In particular if J/Z(J) ∼ = Lnq (qJ ) with qJ ≡ q (mod p), or if J 2 is a symplectic or orthogonal group with qJ ≡ 1 (mod p), then q(Jd ) = qJ = q for every d ∈ D# , a contradiction. Thus J does not have one of these forms. Now let V be the natural J-module and J = [Isom(V ), Isom(V )]. Then there is a covering J → J whose kernel is a p -group. Moreover, the minimal dimension m of [V, g], as  is at least 2 in the linear and unitary cases, and bigger than g ranges over Ip (J), 2 in the symplectic and orthogonal cases. Now as mp (CInn(J)D (Jd0 )) = 1 by (3B),   As and by [IA , 4.8.2], Jd0 ∼ = SLnq (q) where nm = dim[V, P ] for any P ∈ Sylp (J). # mp (I) ≥ 2, n ≥ 4. Moreover, as [D, I] = 1, some d1 ∈ D must centralize the support VI of I on V , and dim VI ≥ 3m. Let V1 = CV (d1 ). Thus, dim V1 ≥ dim VI ≥ 6. Hence Jd1 ∼ = [Isom(V1 ), Isom(V1 )] so q(Jd1 ) = qJ < q, a contradiction. The proof is complete.  Lemma 3.15. Assume (3A) and (3B), with q > 2. Suppose that I ∼ = Sp4 (q), Jd = I for all but four d ∈ E1 (D), and Jd ∼ = Sp6 (q) for those four d . Then J∼ = F4 (q). Proof. By Lemma 3.14a, J ∈ Chev(2). Moreover, by [IA , 4.2.2], J, like ∼ Sp6 (q), must have a double bond in its Dynkin diagram. Hence J is of type Jd0 = F4 or Cn , n ≥ 4. We argue that q(J) = q. Otherwise, by Lemma 3.14bc, J is of exceptional type and rJ ≥ 2r0 , these being the untwisted ranks of J and Jd0 , respectively. But in that case, J must be of type F4 . Therefore r0 ≤ 2, a contradiction as r0 = 3. Hence, q(J) = q.

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∼ Cn (q) for some n ≥ 4. But in that Thus the lemma holds unless possibly J = case, using mp (CInn(J)D (Jd0 )) = 1 and [IA , 4.8.2], we see that n = 4 and then Jd ∼ = Sp6 (q) for only two d ∈ E1 (D), contrary to hypothesis. This proves the lemma.  Lemma 3.16. Assume (3A) and (3B). Suppose that for some d1 = d2 ∈ E1 (D), Jd1 ∼ = Jd2 ∼ = Sp2n (q), n ≥ 3, and I ∼ = Sp2n−2 (q). Suppose also that mp (CInn(J)D (Jd1 )) = 1. Then J ∼ = Sp2n+2 (q), or n = 3 and J ∼ = F4 (q). In J the latter case, if J = Jd1 , Jd2 , then d1 ∈ d2 . Furthermore, in any case, CAut(Jd1 ) (I d ) has odd order, where d ∈ Jd1 has the same action on Jd1 as d2 . Proof. By Lemma 3.14a, J ∈ Chev(2). By [IA , 4.2.2], the untwisted Dynkin diagram or extended diagram of J contains a Cn subdiagram. We can repeat the second paragraph of the proof of Lemma 3.15 to show that q(J) = q. Hence either J∼ = F4 (q). The condition mp (CInn(J)D (Jd1 )) = 1 forces = Sp2m (q) for some m or J ∼ m = n + 1 in the first case, and n = 3 in the second case. It follows easily from [IA , 4.8.2] and [IA , 4.2.3] (or see Lemma 6.1 below) that C := CAut(Jd1 ) (I) ∼ = SL2 (q). Then d ∈ Ip (C) so |CC (d)| = q − q , proving the final statement. It remains to assume that J J = Jd1 , Jd2 with d2 ∈ d1 , n = 3, and J ∼ = F4 (q),

and derive a contradiction. By Lemma 3.2bc, there are exactly four d ∈ E1 (D) such that E(CJ (d)) ∼ = Sp6 (q), and they fall into two orbits O1 , O2 of length 2 under the action of NJ (B) ∩ NJ (D), with, say, d1 ∈ O1 . We must show that d2 ∈ O2 and that O1 and O2 are not fused in NJ (B). (Note that by [IA , 4.10.3c], B is weakly closed in a Sylow p-subgroup of J, so NJ (B) controls J-fusion in B by [IG , 16.9].) We claim that D ∈ E2 (B) is uniquely determined up to NJ (B)-conjugacy by  the condition that O 2 (CJ (D)) contains a subgroup JD ∼ = Sp4 (q). That condition implies that the group WD := AutJD (B) satisfies WD ∼ = D8 and D = CB (Z(WD )). But if W0 is any D8 -subgroup of WJ := AutJ (B) ∼ = W (F4 ) with Z(W0 ) = Z(WJ ), then Z(W0 ) ≤ O2 (WJ ) as WJ /O2 (WJ ) has abelian Sylow 2-subgroups, and all noncentral involutions of O2 (WJ ) are WJ -conjugate. This proves the claim. From the extended Dynkin diagram, and from the existence of the nontrivial graph-field automorphism of J, there exist subgroups J ∼ = Js ∼ = Sp8 (q) of J with long root subgroups of J lying in J (resp. Js ) as long (resp. short) root groups. By the previous paragraph, we may replace J and Js by conjugates to obtain D × I ≤ J ∩ Js . Then by [IA , 7.3.2] (see Lemma 9.1 below), J = J1 , J2 and Js = J3 , J4 , where the Ji are Sp6 (q)-subgroups containing I and centralizing elements ei ∈ D# . Clearly J1 and J2 are J -conjugate, but are not J-conjugate to J3 and J4 , which contain different numbers of long J-root groups. It follows that (without loss) O1 = {e1 , e2 } and O2 = {e3 , e4 }, and e1 ∈ e3 J . This completes the proof.



Lemma 3.17. Assume (3A) and (3B), with q = −1. Suppose that I ∼ = A3 (q) ∼ Jd = ∼ D− (q) for some d1 = d2 in E1 (D), but that Jd = I for all and Jd1 = 2 4 d ∈ E1 (D) − {d1 , d2 }. Then J ∼ = D5 (q).

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Proof. By Lemma 3.14a, J ∈ Chev(2) and D embeds in Inndiag(J). Suppose that q(J) = q. Then by Lemma 3.14bc, J is of exceptional type and untwisted rank rJ ≥ 2r(Jd0 ) = 8. We refer to [IA , 4.2.2]. Thus J ∼ = E8 (q 1/a ), and the extended Dynkin diagram of E8 contains a > 1 disjoint copies of the D4 diagram. This is obviously a contradiction; hence, q(J) = q. If J is a classical group, then as its Dynkin diagram contains the D4 diagram, J∼ = Dn (q) for some n ≥ 5 and sign . As p divides q + 1 and mp (CInn(J)D (Jd0 )) = 1 it follows with the help of [IA , 4.8.2] that J ∼ = D5 (q), as desired. Finally suppose that J is of exceptional type. Then (d0 , Jd0 ) is terminal in J [III11 , p. 291] and so it can be found in [III11 , Table 13.1], with p dividing q + 1. The only possibility in the table is J ∼ = A5 (q) = E6 (q). But then by Lemma 3.3, Jd ∼  for some d ∈ D# , contradiction. The proof is complete. Lemma 3.18. Assume (3A) and (3B). Suppose that there are exactly two sub− groups d = d1 , d2 ∈ E1 (D) such that Jd > I. Assume that q > 2, I ∼ = L4 q (q), − − q q and Jd1 ∼ = Jd2 ∼ = L6 (q). Then J ∼ = L8 (q). Proof. By Lemma 3.14a, J ∈ Chev(2) and D embeds in Inndiag(J). We have d0 = di for either i = 1 or i = 2. As Jd0 has untwisted rank 5, it follows from Lemma 3.14bc that q(J) = q. Moreover, mp (Jd0 ) = 3 so mp (CJ (d0 )) = 4. If J is a classical group, then it follows quickly from [IA , 4.8.2] that J ∼ = −q L8 (q). So assume that J is of exceptional type. Then (d0 , Jd0 ) is terminal in J [III11 , p. 291] and so it can be found in [III11 , Table 13.1], with p dividing q − q . − p The only choice is J ∼ = W (F4 ) = E6 q (q). Expand D to B ∈ E∗ (J). Then AutJ (B) ∼ and Lemma 3.2b is contradicted. This completes the proof.  Lemma 3.19. Assume (3A) and (3B) with I ∼ = D6 (q). Suppose that for each  d ∈ D # , Jd ∼ = I, D7q (q), or E7 (q), and that each of the last two actually occurs for some d ∈ D# . Then J ∼ = E8 (q). Proof. By Lemma 3.14a, J ∈ Chev(2). Since Jd ∼ = E7 (q) for some d ∈ D# ⊆ # Inndiag(J) , the Dynkin diagram of J contains the E7 diagram and so must be  the E8 diagram. Then Lemma 3.14c yields q(J) = q, whence the lemma.  Lemma 3.20. Assume (3A) and (3B) with I ∼ = A5q (q). Suppose that for each   q q d ∈ D # , Jd ∼ = I, D6 (q), A6 (q), or E6 (q), and that each of the last three actually occurs for appropriate d ∈ D# . Then J ∼ = E7 (q).

Proof. As in Lemma 3.19, J must be of type E7 or E8 . Again Lemma 3.14c yields q(J) = q. Since mp (CInn(J)D (Jd0 )) = 1, (d0 , Jd0 ) is terminal in J in the sense of [III11 , p. 291], whence Jd0 must appear as a Lie component of J in [III11 , Table 13.1], with p dividing q 2 − 1. For the given possible isomorphism types of Jd0 , only  J∼ = E7 (q) is possible. This completes the proof. Lemma 3.21. Assume (3A) and (3B) with I ∼ = D4 (q). Suppose that for each   d ∈ D # , Jd ∼ = E6q (q) or D6 (q). If the number of d ∈ = I or D5q (q). Then J ∼  E1 (D) such that Jd > I is at least 3, then J ∼ = E6q (q). ∼ Dq (q). By Lemma Proof. Since mp (CInn(J)D (Jd0 )) = 1, we must have Jd0 = 5 3.14a, J ∈ Chev(2), and it follows easily from the structure of Jd0 and Lemma 3.14bc that q(J) = q (cf. the first paragraph of the proof of Lemma 3.17). By Lemma 3.14a and [IA , 4.2.2], D ≤ Inndiag(J) and then the D5 -diagram of Jd0

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embeds in the Dynkin diagram of J. Hence J is of type Dn , n ≥ 5, or En , n = 6, 7, 8. In the Dn case the condition mp (CInn(J)D (Jd0 )) = 1, together with [IA , 4.8.2], yields J ∼ = D6 (q) and the number of d ∈ E1 (D) with Jd > I is 2. In the En case, (d0 , J0 ) is terminal in J (see [III11 , p. 291]), so appears in [III11 , Table   13.1] for J. Therefore J ∼ = E6q (q). The proof is complete. Lemma 3.22. Assume (3A) with p = 3 and q > 2. Let J ∈ K3 ∩ Chev(2) and D ∈ E32 (Aut(J)). Set Jd = L3 (CJ (u)) for all d ∈ D# . Suppose that for two  d ∈ E1 (D), Jd ∼ = L4q (q), and for two d ∈ E1 (D), Jd is the direct product of one, two, or three copies of L2 (q). Suppose also that L3 (CJ (D)) ∼ = L2 (q). Then  J/Z(J) ∼ = L6q (q). Proof. Let q0 = q(J). Suppose that some d ∈ D# induces a field or graphfield automorphism on J. Then q0 = q 3 . But Aut(J)/Aut0 (J) is cyclic and so there is 1 = v ∈ D ∩ Aut0 (J). Then by [IA , 4.2.2], for any component I of Jv , q(I) ≥ q0 > q. As q(I) = q by assumption, this is a contradiction, and so D ≤ Aut0 (J). Likewise if D ≤ Inndiag(J), or if J is of exceptional type, then  Jd ∼ = L4q (q) for any q, by [IA , 4.7.3A], a contradiction. Therefore J is a classical group and D ≤ Inndiag(J). Again by [IA , 4.2.2], q = q0a for some integer a > 0.  Choose d ∈ D# with Jd ∼ = L4q (q). One possibility is that d ∈ Inn(J), a = 3,  q and J ∼ = L12 (q0 ) [IA , 4.8.4]. But then q0 ≡ q (mod 3) and D ∩ Inn(J) = 1, whence by [IA , 4.8.2], for any v ∈ D# ∩ Inn(J), q(Jv ) = q0 , a contradiction. This possibility having been ruled out, the structure of Jd is determined by [IA , 4.8.2] for each d ∈ D# . Let V be the natural J-module. From [IA , 4.8.2] we see that as p = 3, one of the following holds: (1) J is a linear or unitary group, and V is the orthogonal sum of dinvariant 1-dimensional subspaces (nondegenerate in the unitary case); or (3C) (2) Elements of order 3 require 2 dimensions for a nontrivial classical representation on a subspace of V that is required to be nondegenerate in the unitary, symplectic, and orthogonal cases. If (3C1) holds, then q = q0 , Jd is supported on a 4-dimensional D-invariant eigenspace of d, and D lies in a maximal torus of the overlying algebraic group. Hence using [IA , 7.4.1a4], we find that |J|2 = q 11+b with 0 ≤ b ≤ 4. However, as q > 2, all other eigenspaces of d are at most 1-dimensional, so dim V ≤ 6. Thus  J/Z(J) ∼ = L6q (q), as desired, with b = 4. Suppose then that (3C2) holds. Suppose further that for some d ∈ D# , dim[V, d] ≥ 8. Then equality holds and Jd is supported on [V, d]. Furthermore, a = 2 in the linear and unitary cases, while a = 1 in the symplectic and orthogonal cases. Some element v ∈ D# then must satisfy dim(C[V,d] (v)) ≥ 4. Consequently, in the linear and unitary cases, Jv has a component centrally isomorphic to SL± k (q0 ) with k ≥ 4, a contradiction as q = q02 . And in the symplectic and orthogonal cases, Jv has a subnormal subgroup isomorphic to Spk (q) or Ω± k (q) with k ≥ 4, according to the type of J. One possibility is Jv ∼ (q). But again D is toral and so |J|2 = = Ω+ 4 q 11+b with 0 ≤ b ≤ 4. Therefore J ∼ = D4+ (q). But then L3 (CJ (D)) ∼ = SL2 (q),    a contradiction. And if Jv ∼ = L4q (q), then L3 (CJ (D)) has an L3q (q) = Ω6q (q) ∼ component, again a contradiction.

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The final possibility is that dim[V, d] ≤ 6 for all d ∈ D# , whence V is an   orthogonal space and dim CV (u) = 6 for those d ∈ D# with Jd ∼ = L4q (q). = Ω6q (q) ∼  There exist two such elements d, d that are independent, by hypothesis. But by assumption L3 (CJ (D)) does not contain SL2 (q) × SL2 (q). Whether dim CV (D) = 0, 2, or 4, this condition gives a final contradiction.  Lemma 3.23. Assume (3A). Let H = Lp (H) = E(H) be a K-group admitting − ∼ D = Ep2 , with |H| dividing |SL6 q (q)|. Assume that there are distinct subgroups − ui ∈ E1 (D), i = 1, 2, such that for i = 1 and 2, u3−i ∈ Lp (CH (ui )) ∼ = SL3 q (q) and mp (CH (Lp (CH (ui )))) = 1. Then one of the following holds: (a) Lp (CH (ui )) is a component of H for i = 1 and 2; or (b) For some u ∈ E1 (D) with u1 = u = u2 , Lp (CH (u)) has a component isomorphic to L2 (q 2 ). Proof. Set Hi = Lp (CH (ui )), i = 1, 2. Since u3−i ∈ Hi , i = 1, 2, we have D = u1 , u2 ≤ H. Thus D normalizes all components of H, so by Lp -balance, each Hi lies in a single component of H. If u1 and u2 lie in different components of H, then clearly H1 and H2 are distinct components of H, and (a) holds. So assume that H is quasisimple and that D ≤ H. Since H1 ∈ Alt, H ∈ Alt. If H ∈ Spor, then we see from [IA , 5.3] that the only possibilities are H ∼ = L3 (2). But then = M24 and He, with p = 3 and H1 ∼ by assumption |H| divides |L6 (2)|, a contradiction in either case. Therefore, H ∈ Chev(r) for some r. If r = 2 then H1 ∈ Chev(r) ∩ Chev(2), and the only possibility is r = 7, H1 ∼ = L3 (2) ∼ = L2 (7), and p = 3. Then |H|7 divides |L6 (2)|7 = 72 , so ∼ H/Z(H) = L2 (7), contradicting u1 H1 ≤ H. We have proved that H ∈ Chev(2). It follows now from [IA , 4.8.2, 4.8.4] that if H is a classical group, then q(H) = q − and H/Z(H) ∼ = Lm q (q). Here u2 ∈ H1 has 2-dimensional support on a natural H-module, and so m = 5. In any case the H-centralizer of some element of D − u1 − u2 has an L2 (q 2 )-component, so (b) holds. We may therefore assume that H is of exceptional type. As mp (CH (Hi )) = 1 by assumption, the subcomponent Hi must appear in [III11 , Table 13.1]. Though components A2 (q a ) appear in that table, the corresponding primes p never satisfy the condition that q a ≡ − (mod p). This is a contradiction, and the lemma is proved.  Lemma 3.24. Assume (3A) with q = −1. Let J be a K-group p-component with J  JB, where B ∼ = Ep2 . Assume that = Ep5 . Suppose that D ≤ B with D ∼ J1 × J2 × J3 ≤ CJ (D) with J1 ∼ = J2 ∼ = J3 ∼ = L2 (q), and that there are subgroups Di ∈ E1 (D), 1 ≤ i ≤ 3, such that for each E ∈ E1 (D), either (a) E = Di for some i, and writing {1, 2, 3} = {i, j, k}, we have Ji × Li  CJ (Di ), where Jj Jk ≤ Li ∼ = U4 (q). Moreover, B induces inner-diagonal automorphisms on Ji and Li , and is of maximal rank in CJB (Di ) subject to this condition; or (b) E = Di for any i, and J1 J2 J3  CJ (E). ∼ U6 (q), or q = 2 and J = ∼ P Sp4 (33 ) or Co2 . In the last two Then J/Z(J) = cases, CAut(J) (L1 ) ∼ = Z3 or Σ3 , respectively. Proof. It is a consequence of assumptions (a) and (b) that J1 centralizes COp (J) (E) for every E ∈ E1 (D), so [J1 , Op (J)] = 1 and hence J is quasisimple. It is also clear from conditions (a) and (b) that CD (J) = 1.

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Since L1 ∈ Alt, J ∈ Alt. If J ∈ Spor or Chev(r), r odd, then by [IA , 2.2.10] and [III11 , 1.1b], q = 2, so p = 3. Then with [IA , 5.3] and the Borel-Tits theorem, J∼ = Co2 or P Sp4 (33 ), with CJ (L1 ) as asserted. Thus we may assume that J ∈ Chev(2). Since B induces inner-diagonal automorphisms on J1 and L1 it maps to Aut0 (J), by [IA , 4.2.3]. Now mp (Aut0 (J)) ≥ mp (D1 J1 L1 ) = 5, so J ∼ = D4 (2n ) or 3D4 (2n ) for any n. Thus B maps to Inndiag(J) and mp (Inndiag(J)) ≥ 5. If p is a bad prime for J, then the structure of CJ (D1 ) must appear in some row of the tables [IA , 4.7.3AB]; but no such structure appears there, so p is a good prime for J. Suppose that J is of exceptional type. Then as Outdiag(J) is a p -group, B maps to Inn(J) and by [IA , 4.10.3e], its image lies in an element of Ep∗ (Inn(J)). Hence by our assumption, mp (Inn(J)) ≤ mp (B) = 5. Hence equality holds, and the only possibility, by [IA , 4.10.3a], is J ∼ = E6± (2n ) with p = 3; but then p is a bad prime, contradiction. Hence, J is a classical group. If J is a linear group, then by [IA , 4.8.2, 4.8.4], each Li is a linear group, contradiction. If J/Z(J) ∼ = Um (q0 ) for some even q0 ≡ −1 (mod p), then as CJ (d) has more than one Lie component of level q for all d ∈ D# , it follows by [IA , 4.8.2, 4.8.4] that q(J) = q, whence J/Z(J) ∼ = Un (q), n ≥ 6, and D is diagonalizable with respect to an orthonormal basis of the natural GUn (q)module V . Then the Lie components Ji and Li are supported on homogeneous constituents Wi and Xi , respectively, of Di on V , i = 1, 2, 3; moreover each such Wi is a homogeneous constituent of any E ∈ E1 (D) − {D1 , D2 , D3 }. It follows that Xi = Wj ⊥ Wk whenever {i, j, k} = {1, 2, 3}. If V = W1 ⊥ W2 ⊥ W3 , choose a D-invariant 1-dimensional subspace V0 ⊆ (W1 ⊥ W2 ⊥ W3 )⊥ . Let V1 ⊆ W1 be 1-dimensional. Then some d ∈ E1 (D) acts homogeneously on V0 ⊥ V1 , so SU (V0 + V1 ) ≤ CGUn (q) (d ). As either W1 , W1 + W2 , or W1 + W3 is invariant under CGUn (q) (d) by assumption, this is a contradiction. Therefore V = W1 ⊥ W2 ⊥ W3 and J/Z(J) ∼ = U6 (q), as desired. Hence, we may assume that J does not have the form in the previous paragraph, so that J is a symplectic or orthogonal group, or J/Z(J) ∼ = Un (q0 ) for some even q0 with q0 ≡ −1 (mod p). Let d ∈ D# . Again since CJ (d) has more than one Lie component by assumption, CJ (d) must be as stipulated in [IA , 4.8.2]. As mp (B) = 5, we can therefore write V = V0 ⊥ V1 ⊥ · · · ⊥ V5 with V0 = CV (B) and each Vj , 1 ≤ j ≤ 5, being orthogonally indecomposable under the action of B. By [IA , 4.8.2], the support of each Li on V must be the sum of four Vj ’s, 1 ≤ j ≤ 5; without loss we may assume that L1 is supported on ⊥4j=1 Vj , on which D1 is orthogonally homogeneous, and L2 is supported on ⊥5j=2 Vj , on which D2 is orthogonally homogeneous. Therefore L1 ∩ L2 contains an SU3 (q0 ) subgroup  supported on ⊥4j=2 Vj . However, L1 ∩ L2 ≤ O 2 (CL1 (D2 )) = J2 × J3 ∼ = SL2 (q) × SL2 (q) by assumption. As U3 (q0 ) is not embeddable in SL2 (q) × SL2 (q) (for instance by the structure of Sylow 2-subgroups), we have a contradiction. The proof is complete.  Lemma 3.25. Assume (3A), with q = −1. Let X  K = U6 (q), where q = 2n . Suppose that D ≤ X, with CD (K) = 1 and D ∼ = Ep2 . Let De ⊆ E1 (D) with |De | = 3 and De cycled by an element t ∈ I3 (NX (D)). Let M = Lp (M ) be a D t invariant subgroup of K such that Lp (CM (u)) has a component Mu ∼ = L2 (q 2 ) for

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∼ L3 (q 2 ), for which the images each u ∈ De , and [Mu , D] = 1. Then M/Op p (M ) = of any two Mu , u ∈ De , form a standard CT-pair. Moreover, either M ∼ = SL3 (q 2 ) 2 2 or L3 (q ), or M lies in a parabolic subgroup of K of type L3 (q ). Proof. For the definition of standard CT-pair, see [III13 , 1.2]. Without loss we may pass to X/CX (K) and assume that F ∗ (X) = K ∼ = U6 (q). 1 (∞) ∼ If some u ∈ De induces field automorphisms on K, then CK (u) = U6 (q p ) involves Mu ∼ = L2 (q 2 ). However, as p ≥ 3, this is impossible by order considerations, using Zsigmondy’s theorem [IG , 1.1]. Thus, D ≤ O 2 (Aut0 (K)) = Inndiag(K). Let V be the natural SU6 (q)-module and let u ∈ De . Then u acts on K like an isometry of V of order p. Indeed, otherwise, by [IA , 4.8.4], CK (u)(∞) ∼ = L2 (q p ), 2 which does not involve L2 (q ), contradiction. Thus by [IA , 4.8.2], the possibilities for components of CK (u)(∞) are SUn (q), 2 ≤ n ≤ 5, and for such a group to involve L2 (q 2 ) we must have n = 4 or 5 by order considerations. Now consider the possibilities for M := M/Op p (M ). Since [Mu , D] = 1 for each u ∈ De , CD (M ) = 1, and this immediately implies that CDt (M ) = 1. In particular, M is a nontrivial pumpup of M u . By Zsigmondy’s theorem [IG , 1.1], |K| is not divisible by |L2 (q 2 )|2 , so by Lp -balance, we must have L2 (q 2 ) ↑p M . Suppose first that M ∈ Chev(2) ∪ Spor. Then L2 (q 2 ) ∈ Alt ∪ Chev(r) for some odd r by [III11 , 1.1a], so by [IA , 2.2.10], q = 2, whence p = 3 and L2 (q 2 ) ∼ = L2 (5) ∼ = A5 , and 2 ∼ M ∈ Alt ∪ Chev(5). But 5 does not divide |U6 (2)|, so M = Am , m = 5 + 3k, and then k = 1. However, no E32 group acting faithfully on A8 has more than 2 cyclic subgroups with an A5 subcomponent, contradiction. Therefore M ∈ Chev(2)∪Spor. If M ∈ Spor, then we see from the tables [IA , 5.3] that the structure of M u forces Mu ∼ = A5 , q = 2, p = 3, and M ∼ = M23 or HS. But then |M | does not divide |U6 (2)|, a contradiction. We have proved that M ∈ Chev(2). If some u ∈ De induces field or graph-field automorphisms on M , then ∼ L2 (q 2p ). Then D ≤ [D t , D t ], so by the action of t, p = 3. But M = CDt (M ) = 1, so Aut(M ) has a nonabelian 3-subgroup of exponent 3. However, for P ∈ Syl3 (Aut(L2 (q 6 ))), Ω1 (P ) is abelian, contradiction. Therefore D maps into O 2 (Aut0 (M )). Similarly if p = 3 and some u ∈ De induces graph automorphisms on M , then CM (u)(∞) ∼ = G2 (q0 ) or L± 3 (q0 ) for some q0 , or 1. In any event CM (u) 2 does not have an L2 (q ) component, contradiction. Therefore, D maps into Inndiag(M ). Next, suppose that Q := [M, Op (M )] has even order. By a Frattini argument, some 2-local subgroup P of K involves M D. The nonsolvable composition factors of the maximal 2-local subgroups of X, by the Borel-Tits theorem, are L3 (q 2 ), L2 (q), L2 (q 2 ) and U4 (q). The only one of these that admits Ep2 with more than two subgroups of order p centralizing a L2 (q 2 ) subgroup is L3 (q 2 ). Hence P in volves L3 (q 2 ), so O 2 (P ) is an extension of Eq9 by L3 (q 2 ), acting irreducibly. In particular M contains a Sylow 2-subgroup of K, so by [IA , 2.6.7], M ≤ P . This is an allowed conclusion of the lemma; moreover, D acts on M/Op p (M ) as a diagonalizable group, so the assertion about CT-systems is immediate. Hence, we may assume that Q has odd order. Suppose that Q = 1. Then there exists an odd prime r = p such that CM (Or (Q)) ≤ Op (M ), so Aut(Or (Q)) involves M and in

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particular involves a Frobenius group of order q 2 (q 2 − 1). It follows by [IG , 9.12] that mr (Or (Q)/Φ(Or (Q))) ≥ q 2 − 1. But Or (Q) ≤ K, and a Sylow r-subgroup of K has sectional rank at most 5 + 2 = 7, so q = 2. Even then, Sylow r-subgroups of K ∼ = U6 (2) are noncyclic, so r = 3 = p, a contradiction. We have proved that Q = 1, so M is quasisimple. Suppose that M is of exceptional type, but not G2 (q0 ), 3D4 (q0 ), or 2F4 (2 2 ) . Notice that En± (q0 ) ≥ F4 (q0 ) ≥ F(q0 +1)4 .8 , a Frobenius group with a Q8 complement. As the Schur multiplier of F4 (q0 ) is a 2-group [IA , 6.1.4], the minimal dimension for a nontrivial characteristic 2 projective representation of M is at least 8. This is a contradiction as M ≤ K ∼ = G2 (q0 ) or 3D4 (q0 ), then we = U6 (q). If M ∼ see from [III11 , Table 13.1], and the fact that E(CM (u)) ∼ = L2 (q 2 ) for u ∈ De , 2 2/3 2 that q0 = q or q0 = q . If q0 = q , then Φ12 (q) or Φ24 (q) divides |K|, while if q0 = q 2/3 , then Φ8 (q) divides |K|; these are all impossible by Zsigmondy’s theorem. n If M ∼ = 2F4 (2 2 ), then by [III11 , Table 13.1], q 2 = 22n+1 , an impossibility. Therefore, M is a classical group. If M contains D4 (q0 ) for any even q0 , then it contains a Frobenius group of order (q0 + 1)2 .8 and as in the F4 case above, this leads to a contradiction. Hence if M is an orthogonal group of degree at least 8, then M ∼ = Sp2n (q0 ) then n ≤ 3; and if M ∼ = L± = 2 D4 (q0 ); if M ∼ n (q0 ), then n ≤ 6 (by consideration of the nilpotence class of a Sylow 2-subgroup, see [IA , 3.3.1a]). In every case, by [IA , 4.2.2], q 2 is a power of q0 . Let V be a natural M -module and u ∈ De . Suppose that M ∼ = 2 D4 (q0 ) or M ∼ = Sp2n (q0 ) with n ≤ 3. Since 2 mp (Inndiag(M )) ≥ 2, p divides q0 − 1. Hence, with [IA , 4.8.2], q 2 = q(Mu ) = q0 , or q 2 = q02 with M ∼ = 2 D4 (q). In any case, |M | is divisible by q 4 + 1, and so M is not embeddable in U6 (q) by Zsigmondy’s theorem, a contradiction. Therefore n

M/Z(M ) ∼ = L± n (q0 ), n ≤ 6. Hence q 2 = q0 , q02 , or q03 . If q 2 = q03 , then as mp (Inndiag(M )) ≥ 2, M/Z(M ) ∼ = L6 (q0 ). As the twisted rank (5) of M exceeds that (3) of K, K does not contain M , by [III11 , 20.17]. So q 2 = q03 . If q = q0 , then again by [III11 , 20.17], M/Z(M ) ∼ = Un (q), n ≤ 6, or Ln (q), n ≤ 4. Since p divides q +1, and L2 (q 2 ) ↑p M , q(L2 (q 2 )) = q in the first case, an absurdity. If M/Z(M ) ∼ = Ln (q), then as p divides q + 1, n = 4; but then there are only two u ∈ E1 (D) such that Mu ∼ = L2 (q 2 ), contradiction. Therefore q0 = q 2 . By order considerations, U3 (q 2 ) and L4 (q 2 ) are not involved in U6 (q). Hence M/Z(M ) ∼ = L3 (q 2 ). The assertion about CT-systems again holds since D acts on M as a diagonalizable subgroup of GL3 (q 2 ). The lemma is finally proved.  Lemma 3.26. Assume (3A). Suppose that J ∈ Chev(2) is simple of level q(J) = q, and H is a group such that J ≤ H ≤ Aut0 (J). Suppose that D ∈ Ep2 (H) for some odd prime p. Let x ∈ D# and assume the following:  (a) E(CJ (x)) has a component I with I/Z(I) ∼ = Lnq (q), n ≥ 4, and mp (CH (I)) = 1; q (q); and (b) L := E(CI (D)) is a central quotient of SLn−1 # (c) For each d ∈ D , if we let Ld be the pumpup of L in CJ (d), then either Ld = L or Ld /Z(Ld ) ∼ = I/Z(I). ∼ D4 (q), or J = ∼ Lq (q), δ = 1 or 2. Then either n = 4 with J = n+δ

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∼ D4 , then either Proof. Let J be a simple algebraic group overlying J. If J = J ∼ = 2 D4 (q) or 3D4 (q). But in the last two cases no x = D4 (q), as desired, or J ∼ exists in Ip (Aut0 (J)) with a component I in CJ (x) satisfying (a), as can be seen from [IA , 4.7.3A, 4.8.2]. Thus we may assume that J = D4 . Without loss, H = JD,  so H ≤ O p (Aut0 (J)) ≤ Inndiag(J). If J is of type Am for some m, then since mp (CJx (I)) = 1 and q ≡ q (mod p), q q (q) or Ln+2 (q). So we may assume that it follows that m = n or n + 1, so J ∼ = Ln+1 J = Am for any m. Suppose that D ≤ T for some maximal torus T of J, and let Σ be the corresponding root system, and {X α }α∈Σ the set of T -root groups in J. For any d ∈ D# , let the root subsystems Σd and ΣD of Σ consist of those α such that X α is centralized by d and D, respectively. Thus Σx is of type An−1 and ΣD ⊆ Σx is of type An−2 . Since Σ is not of type Am for any m, or of type D4 , it is easy to find a root β ∈ Σ − ΣD such that the subsystem Σ1 ⊆ Σ spanned by ΣD and β is indecomposable of rank n − 1 and not of type An−1 . Moreover, as D is noncyclic there exists d ∈ D# such that β ∈ Σd , whence Σ1 ⊆ Σd . It follows that the pumpup Ld of L in CH (d) contains a subgroup M ∈ Chev(2) with untwisted root system q q (q)|2 . As M ≤ Ld ∼ (q), we have a Σ1 and of level q. But then |M |2 > |An−1 = An−1 contradiction. Thus, D ≤ T for any maximal torus T of J. By [IA , 4.10.3e], p is not a good prime for J, and so J is of exceptional Lie type with p = 3, or J = E8 with p = 5. In the latter case, since m5 (CH (I)) = 1, we see from [IA , Table 4.7.3A]   that O 5 (CH (x)) = I × Z, with Z a cyclic 5-group containing x, and I ∼ = A7q (q). Thus D lies in an E58 -subgroup E of J, whence D lies in a maximal torus of J by 3.1, contradiction. A contradiction is similarly reached in the case p = 3,  J ∼ = A6q (q); see [IA , Table 4.7.3A]). Using n ≥ 4 we see from = E7 (q) (where I ∼    [IA , Table 4.7.3A] that the only other cases are (J, I, L) = (E6q (q), A5q (q), A4q (q)) q q and (E8 (q), A8 (q), A7 (q)), and accordingly DL = D × L has 3-rank 6 or 9. In the  E6q (q) case, D lies in a maximal torus of J by Lemma 3.1, contradiction; and the E8 (q) case is impossible because then m3 (J) = 8, by [IA , 4.10.3a]. This completes the proof.  Lemma 3.27. Assume (3A), with q > 2. Suppose that J ∈ Kp and D = x, y ∈  Ep2 (Aut(J)). Suppose that CJ (D) has a component L = O p (L) such that for each d ∈ D# , L lies in a component Jd of CJ (d), and Jd ∈ Chev(2) with level q(Jd ) = q. Suppose also that mp (Jx ) ≥ 3, x ∈ Aut0 (J), and mp (CInn(J)x (Jx )) = 1. Finally, suppose that the isomorphism type of L would qualify (y, L) to be an acceptable (x, Jx )-subterminal pair [III13 , Def. 1.15]. Then J ∈ Chev(2) and q(J) = q. Proof. Since q > 2, mp (Jx ) ≥ 3, and Jx /Op (Jx ) ↑p J, we see from [III11 , 1.1ab] and [IA , 2.2.10] that J ∈ Chev(2). Since x ∈ Aut0 (J), q(J) ≤ q(Jx ) = q by [IA , 4.2.2]. Assume by way of contradiction that q(J) < q. Then as q(Jd ) = q > q(J) for all d ∈ D# , D ≤ Aut0 (J), by [IA , 4.9.1]. In particular D ∩ Inn(J) = 1, as mp (Aut0 (J)/ Inn(J)) ≤ 1. It follows from [IA , 4.8.1] that q = q(J)n for some n dividing p − 1. q (q) and Jd /Z(Jd ) ∼ Suppose first that J is a classical group. If J/Z(J) ∼ = = Lkp q p # # Lk (q ) for all d ∈ D (cf. [IA , 4.8.4]), then every d ∈ D has a preimage ud ∈ I := q (q) such that upd ∈ Sylp (Z(I)). But then the full preimage of D in I, which GLkp has noncyclic Sylow p-subgroups, contains an element u of order p, a contradiction.

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Hence, for some d ∈ D# , [IA , 4.8.1, 4.8.2] apply. In particular q = q(Jd ) = q(J)n for some n ≤ p − 1, and so q(Jx ) < q(J)p . Since mp (CInn(J)x (Jx )) = 1 by hypothesis,   Jx /Op (Jx ) ∼ = SLmq (q)/Op (SLmq (q)), where m = mp (Jx ) + 1 = mp (CJ (x)) ≥ 4, by  [IA , 4.8.2, 4.8.4]. Indeed if A ∈ Ep∗ (CJ (x)), then AJx /Op (Jx ) embeds in GLmq (q).  q (q), δ = 1 or 2. As a By definition of acceptable subterminal pair, JD ∼ = SLm−δ # result, some u ∈ D centralizes a subspace of codimension at most 2 on the natural  GLmq (q)-module. But this implies that Ju is quasisimple of level q(J). Therefore q(J) = q(Ju ) = q, a contradiction. We may therefore assume that J is of exceptional Lie type. Using [III11 , Table 13.1] and the facts that q(Jx ) > q(J), mp (Jx ) ≥ 3, we see that J ∼ = E8 (q 1/2 ) and − # ∼ p divides q + 1. Moreover, JD = A2 (q), and for all d ∈ D such that Jd > JD , ∼ − Jd ∼ = A− 3 (q) unless p = 5, in which case Jd = A4 (q). Take a σ-setup (J, σq ) for J, and a σq -invariant Borel-torus pair (B, T ). Then choose w ∈ NJ (T )/T =: W ∼ = W (E8 ) such that w2 = 1 and w ∈ O2 (P ) − Z(W ), 1+6 ∼ where P = 2+ A8 is a parabolic subgroup of [W, W ]. Let C = CP (w). Then C = (w × Q)S where Q ∼ = Q8 ∗ Q8 and S ∼ = Σ6 . Moreover, C ≤ C[W,W ] (w) ≤ P ∗ ∗ for some maximal parabolic subgroup P ≤ [W, W ], by the Borel-Tits theorem. Then P ∗ is not solvable, and so m3 (P ∗ ) = 2. We conclude that C contains a Sylow 3-subgroup of CW (w).  Let σ = σq w and identify J with O 2 (CJ (σ)). Then we may identify B with Ω1 (Op (CT (σ))), and conclude that AutJ (B) contains a copy of C. Now AutJD (B) ∼ = Σ3 and O3 (AutJD (B)) centralizes H ≤ AutJ (B) with H ∼ = Σ3 and O3 (H) conjugate to O3 (AutJD (B)) in AutJ (B). Hence, [B, O3 (H)] = D. Also, O3 (H) represents the unique class of subgroups of AutJ (B) of order 3 with nontrivial fixed points on B. On the other hand, J has a subgroup I ∼ = D8 (q 1/2 ) with B ≤ I, and there is t ∈ NI (B) of order 3 with CB (t) = b1 , b2 b3 b4 , where B = b1 , b2 , b3 , b4 and E(CI (bi )) ∼ = D6− (q 1/2 ) for each 1 ≤ i ≤ 4. Therefore D is conjugate to CB (t), so D contains some d ∈ bJ1 , whence Jd contains a copy of D6− (q 1/2 ). By order considerations, Jd does not embed in U4 (q) or U5 (q), so by [III11 , Table 13.1], Jd ∼  = D6− (q 1/2 ). But then q = q(Jd ) = q 1/2 , a final contradiction. In the remaining results in this section, p = 2 and we consider a four-group D acting on J ∈ K2 . In all but the last result, we use information about a neighborhood {Jd | d ∈ D# }, where each Jd is a component of CJ (d), to characterize the group J, and in particular to conclude that J ∈ Chev(r) for some odd r. Lemma 3.28. Suppose that J ∈ K2 and D = t, u is a four-subgroup of Aut(J). Let n = 2k + 1, k ≥ 3. Suppose that CK (D) has a component JD ∼ = Ω± n−1 (q) for some odd prime power q, that JD   CJ (tu), and that the subnormal closures of η JD in CJ (t) and CJ (u) are both isomorphic to Ωn (q). Then J/Z(J) ∼ = P Ωn+1 (q) for some sign η. Proof. Let q be a power of the prime r. Since Ωn (q) ↑2 J with n ≥ 7, we deduce from [IA , 2.2.10] and [III11 , 1.1, 1.2] that J ∈ Chev(r). Since Aut(J)/Aut0 (J) is cyclic by [IA , 2.5.12], there is v ∈ D# ∩ Aut0 (J), and then by [IA , 4.2.2], q(J) ≤ q. It follows that no element of D# is a field or graph-field automorphism, so D ≤ Aut0 (J). From [IA , 4.5.1, 4.5.2], we see that J is of type A or D, with t and u being graph automorphisms, or J is of adjoint type B. If J is of type A,

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± ∼ ± then J ∼ = A± n−1 (q), and E(CJ (tu)) must be isomorphic to Ωn−1 (q) = D(n−1)/2 (q). As n is odd, the only possibility is n = 7 with E(CJ (tu)) = J1 J2 , J1 ∼ = A± 3 (q), ± ∼ J2 = A2 (q). But then J1 is the universal version, i.e., a spin group, contrary to assumption. Thus J is of type B or D. Let V be an orthogonal space underlying J. Thus there exist  t, u  ∈ GO(V ) mapping to t and u. If  t and u  can be chosen to be commuting involutions, then multiplying them by −1 if necessary we may t)) = dim(CV ( u)) = n, and CV ( tu ) = CV ( t) ∩ CV ( u) has diassume that dim(CV ( mension n − 1. But then dim V = n + 1 and J is a quotient of Ω(V ) = Ω± n+1 (q), as claimed. Next, assume that  t cannot be chosen to be an involution. By [IA , 4.5.1], 1  2 t induces a graph automorphism on J/Z(J) ∼ = P Ω± 2n (q ). But then as n is odd, there is no v ∈ Inndiag(J) such that E(CJ (v)) has a component of level q other t and than A1 (q). As t, u ∩ Inndiag(J) = tu , this is a contradiction. Hence both  u  may be chosen to be involutions, and we need only rule out the possibility that [ t, u ] = −1. But if this occurs then u  interchanges the two eigenspaces of  t, and so no component of E(CJ (t)) is invariant under tu, contrary to assumption. The proof is complete. 

Lemma 3.29. Let J ∈ K2 and let D = x, y be a four-subgroup of Aut(J). Set Jd = E(CJ (d)) for each d ∈ D# . Suppose that for some power q of an odd prime r, Jx ∼ = Jxy ∼ = Ω5 (q), while y acts on each of Jx and Jxy as a 2-central involution,    with O r (CJx (y)) = O r (CJxy (y)) and O 2 (CJx (y))  CJ (y). Assume also that if ± J ∈ G62 , then F(J) ≤ (q 9 , BC). Then J ∼ = Ω± 6 (q) or P Ω6 (q). Proof. Suppose first that J ∈ Chev(r). Whether J is a level or nonlevel pumpup of Jx , we have J ∈ G62 and so F(J) ≤ (q 9 , BC). In the level case, J ∼ = ∼ A± (q), B (q), or C (q). If J B (q), then x and xy each act on J as involutions in = 3 3 3 3 O7 (q) with two eigenvalues equal to −1. Then y = x(xy) acts on Jx as an involution with at most two eigenvalues equal to −1. But it acts as a 2-central involution by assumption, so it acts on Jx as a reflection. Then y must have 5 eigenvalues  equal to 1, so Jy ∼ = Ω5 (q). This contradicts O 2 (CJx (y))  CJ (y), however. Thus J ∼ = C3 (q), then by [IA , 4.5.1], Jx ∼ = Ω5 (q), contradiction. = Spin5 (q) ∼ = B3 (q). If J ∼ If J ∼ (q), then the conclusion of the lemma holds unless J∼ = A± = SL± 3 4 (q), in which case again Jx ∼ = Spin5 (q), contradiction. If J is a non-level pumpup of Jx in Chev(r), then by [IA , 4.5.1, 4.9.1] and the condition F(J) ≤ (q 9 , BC), we have J ∼ = Ω5 (q 2 ) or C4 (q 1/2 ). In the first case x induces a field automorphism of order 2 on J. As y acts as a 2-central involution   on Jx , O r (CJx (y)) ∼ = SL2 (q) ∗ SL2 (q) and thus O r (CJ (y)) ∼ = SL2 (q 2 ) ∗ SL2 (q 2 ),  contradicting O 2 (CJx (y))  CJ (y). In the case J ∼ = C4 (q 1/2 ), on the other hand, y induces an inner automorphism on J and so from [IA , 4.5.1], all components of  CJ (y) have level q 1/2 , again a contradiction as O 2 (CJx (y))  CJ (y). Thus J ∈ Chev(r). The only possibility, by [IA , 2.2.10] and [III11 , 1.1ab], is q = 3, and by [IA , 4.9.1, 4.9.2] and the Borel-Tits theorem, J ∼ = L4 (4). Then both x and xy induce graph-field automorphisms on J, whence y = x(xy) induces an inner  automorphism on J corresponding to the central involution y0 of O r (CJx (y)) ∼ = SL2 (3) ∗ SL2 (3). But then y0 is 2-central in Jx ∼ = U4 (2), hence a transvection there and in J ∼ = L4 (4). Thus y0 is 2-central in J, which contradicts [III11 , Lemma 13.1]. The proof is complete. 

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Lemma 3.30. Let q be an odd prime power,  = ±1, J ∈ K2 , and let x, y be a four-subgroup of Aut(J). Suppose that E(CJ (x)) ∼ = E(CJ (xy)) is a covering group of L3 (q),  = ±1. Assume also that there is J1 × J2   CJ (y) such that J1 ∼ = SL4 (q). = J2 ∼ = SL2 (q), y ∈ J1 , and [J2 , x] = 1. Then J ∼ Proof. First suppose that J ∈ Chev(r), where q is a power of r. If x or xy induces a field or graph field automorphism on J, then so does the other and  J ∼ = A2 (q 2 ). But then as y ∈ J1 ≤ J, O r (CJ (y)) ∼ = SL2 (q 2 ), contradicting the structure of J1 and J2 . It follows that x, y ≤ Aut0 (J). Considering the structure of E(CJ (x)), we see from [IA , 4.5.1] that J ∼ = Am (q),  ± 1/2 m ≥ 3, A5 (q ), or C3 (q). Then, for O r (CJ (y)) to have a subnormal subgroup J1 ×J2 , the only possibility is J ∼ = A3 (q), the universal version since J1 J2 = J1 ×J2 . It remains to rule out J ∈ Alt∪Spor∪Chev −Chev(r). But since A2 (q) ↑2 J, the only possibility would be J ∼ = G2 (4), with x and xy inducing field automorphisms, q = 3 and  = −1, by [IA , 2.2.10, 4.9.1, 4.9.2], [III11 , 1.1ab], and the Borel-Tits theorem. But then J would have Sylow 3-subgroups P ≤ CJ (Z(P )) ∼ = SL3 (4) [IA , 4.7.3A], and so no E32 -subgroup of J would centralize an involution, contrary to the existence of y. The lemma is proved.  Lemma 3.31. Let K ∈ K2 , let U be a four-subgroup of K, and suppose that for each u ∈ U # there is a component Ju of CK (u) such that u ∈ Ju ∼ = Sp4 (q) or SL4 (q), q odd,  = ±1, and with the isomorphism type of Ju independent of u ∈ U # . Assume also that for each u ∈ U # , CLo2 (CK (u)) (Ju ) is the product of 0, 1, or 2 (solvable) components of type SL2 (q). Then K/Z(K) ∼ = P Sp6 (q) or L6 (q), respectively. Proof. If u ∈ U # ∩ Z(K), then Ju = K = Ju for all u ∈ U # , so U ≤ Z(K). But Z(Ju ) is cyclic, a contradiction. Therefore U ∩ Z(K) = 1 and U acts faithfully on K. Then without loss we can pass modulo Z(K) and assume that K is simple. By the Borel-Tits theorem, K ∈ Chev(2), and then by [III11 , 1.1ab] and [IA , 2.2.10], K ∈ Chev(r) where q is a power of the prime r. Moreover, by [IA , 6.1.4], each Ju is 2-saturated, which implies that each u has an involutory preimage in the universal version K ∗ of K. From [IA , Table 4.5.1] we see that all these hypotheses and conditions, even for a single u ∈ U # , imply that the conclusion of the lemma holds or else K/Z(K) ∼ = L5 (q). But the latter case is impossible, since L5 (q) contains no four-group all of whose involutions are the images of reflections. The lemma is proved.  Lemma 3.32. Suppose that K ∈ K2 , and x and y are commuting involutions in Aut(K). Suppose that E(CK (x)) ∼ = P Sp4 (q) for some odd prime power q > 3, and y acts as an involution y0 of K with y0 ∈ E(CK (y0 )) ∼ = SL2 (q 2 ) ∗ SL2 (q 2 ). Then 2 K∼ = P Sp4 (q ). Proof. If x is a field or graph-field automorphism, then the result is obvious. So assume otherwise. Then by [IA , 4.2.2], the structure of CK (x) forces K ∼ = d L(q 1/a ) for some positive integer a. If x is a graph automorphism, we see from ± the condition E(CK (x)) ∼ = P Sp4 (q) and [IA , 4.5.1] that K/Z(K) ∼ = A± 3 (q), A4 (q), 1/2 1/2 or D5 (q ); if x ∈ Inndiag(K) we find K/Z(K) ∼ = B3 (q) or C4 (q ). In all cases the assumed structure of CK (y0 ) is impossible, as is visible from [IA , 4.5.1]. The proof is complete. 

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Lemma 3.33. Suppose that K ∈ K2 and x, y is a four-group acting on K such that for some n ≥ 3, E(CK (x)) ∼ = E(CK (y)) ∼ = SLn (q) for some odd q, where  = ±1. Assume that E(E(CK (x)) ∩ E(CK (y))) ∼ = SLn−1 (q) is a component of CK (xy)  if (n, q) = (3, 3); if (n, q) = (3, 3), assume that O 3 (E(CK (x)) ∩ E(CK (y))) ∼ = SL2 (3)   CK (xy). Then K is a central quotient of SLn+1 (q). Proof. By [III11 , 1.1ab], K ∈ Spor ∪Alt. Suppose that K ∈ Chev(r), where q is a power of the odd prime r. Since n ≥ 3, the only possibility is K ∼ = G2 (4), with x and y inducing field automorphisms and (n, q) = (3, 3) (see [IA , 2.2.10]). But in that case xy induces an inner automorphism on K, and from the structure of CCK (x) (xy), xy acts on K/Z(K) as a 2-central involution, contradicting [III11 , Lemma 13.1]. Therefore K ∈ Chev(r). If x or y induces a field or graph-field automorphism on K, then so does the  other, K ∼ = SLn (q 2 ), xy maps into Inndiag(K), and by [IA , 4.2.2], O r (CK (xy)) is the product of Lie components of level at least q 2 , contrary to the given structure of CK (xy). Thus the image of x, y in Aut(K) lies in Aut0 (K). Now we search [IA , 4.5.1] for a group K with two (possibly equal) conjugacy classes of involutions, having Lie components in their respective centralizers isomorphic to An−2 (q) and An−1 (q) for some n ≥ 3, some odd q, and some sign . Other than Am (q) the only example is C3 (q), with x and y conjugate to t3 or t3 , and xy conjugate to t1 . However, on the natural C3 (q)-module V over the algebraic closure Fq , E(CK (x)) would preserve a decomposition V = V1 ⊕ V2 into two totally isotropic subspaces, with the representations on V1 and V2 being mutually dual natural modules. Thus E(E(CK (x)) ∩ E(CK (y))) ∼ = SL2 (q) would have two natural composition factors on V , whereas the SL2 (q) Lie component of the element t1 has only one. Therefore, K ∼ = Am (q) for some m. Then x and y act on K like involutions of GLm (q) with −1-eigenspaces Vx and Vy satisfying dim(Vx ) = dim(Vy ) = n and dim(Vx ∩ Vy ) = n − 1. The hypothesis implies that Vx ∩Vy is a full eigenspace of xy; thus CV (xy) = Vx ∩Vy . Therefore the 1-eigenspaces of x and y are disjoint, whence m = n + 1. The proof is complete.  Lemma 3.34. Let X be a K-group, q a power of the odd prime r. Let D be a four-subgroup of X, and assume that for each u ∈ D# , CX(u) has a component Ku ∼ = Sp4 (q), D ≤ Ku , and m2 (CX (Ku )) = 1. Let X0 = Ku | u ∈ D# . Then X0 ∼ = P Sp6 (q). Proof. Without loss X = X0 (note that D ≤ Ku ≤ X0 ). If O2 (X) = 1 then we set X = X/O2 (X) and conclude that X ∼ = P Sp6 (q), by induction. Let Wu = CO2 (X) (u) for each u ∈ D# . Then since Ku is a component of CX (u), [Ku , Wu ] = 1. But D ≤ Ku , so [D, Wu ] = 1. As u is arbitrary, [D, O2 (X)] = 1 and then [Ku , O2 (X)] = 1 for all u ∈ D# . As X = X0 , O2 (X) ≤ Z(X), so by L2 -balance, X is quasisimple. Since X ∼ = P Sp6 (q), whose Schur multiplier has order 2 by [IA , 6.1.4], and m2 (CX (Ku )) = 1, X ∼ = P Sp6 (q) as required. Thus, we are reduced to the case O2 (X) = 1. By L2 -balance, X = X0 = E(X), with each Ku contained in a single component Xu of X. Let u = v ∈ D# . Then CKu (v) contains L1 × L2 with L1 ∼ = L2 ∼ = SL2 (q). Since m2 (CX (Kv )) = 1 by assumption, Kv  CX (v), and so either L1 or L2 must act faithfully on Kv . Thus Ku acts nontrivially on Xv , so Xu = Xv . Thus X is quasisimple and Ku ↑2 X. Indeed if Z(X) contained an involution w, then

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w ∈ CX (Kv ) for all v ∈ D# ; but v is the unique involution of CX (Kv ), so w = v for all v ∈ D# , which is absurd. Therefore, X is simple. By [III11 , 5.11], X ∈ Spor; as each Ku ∈ Alt, X ∈ Alt; and by [III11 , 11.3], X ∈ Chev(2). Thus with [IA , 4.9.6, 2.2.10], X ∈ Chev(r). Now the conditions D ≤ X, Ku ∼ = P Sp6 (q), by examination of = Sp4 (q), m2 (CX (Ku )) = 1 force X ∼  [IA , 4.5.1]. The proof is complete. Lemma 3.35. Let K ∈ K2 and K  X. Suppose that D = x, y is a foursubgroup of K disjoint from Z(K). Let q be an odd prime power. Suppose that for each u ∈ D# , CX (u) possesses a unique subnormal SL2 (q)-subgroup Ju ∼ = SL2 (q);  moreover, u ∈ Ju and O 2 (Ju ) ≤ K. Assume if q > 3 that Ju is the unique component of CK (u), and if q = 3 that |CX (u)|2 = 3. Finally, assume that D = Ω1 (B) for some B ∼ = Z4 × Z4 , B ≤ X, and that some 3-element g ∈ X normalizes  B and satisfies B = [g, B]. Then K ∼ = L3q (q), where q = ±1 and q ≡ q (mod 4). Moreover, Jx and Jy form a standard CT- or P-pair in K, according as q ≡ 1 or −1 (mod 4). Proof. As K ∈ K2 , O2 (K) = 1. There is then no loss in assuming that Z(K) = 1. We argue first that K ∈ Chev(r), where q is a power of the prime r. Note that if q = 3, then by [III11 , 13.1], either K ∈ Chev(3) (indeed q(K) = 3) or K ∼ = M11 ; but M11 has SD16 Sylow 2-subgroups so does not contain Z4 × Z4 . If q > 3, then the existence of the components Ju forces K ∈ Spor by examination of the tables [IA , 5.3]; K ∈ Alt by [IA , 5.2.9]; and K ∈ Chev(2) by the Borel-Tits theorem. Hence K ∈ Chev(r) for some odd r, so SL2 (q) ∈ Chev(r), and since both q and r are odd, q is a power of r by [IA , 2.2.10], as asserted. Since Ju ∩ K contains Q8 , K is clearly not L2 (q) or a Ree group. Consulting [IA , 4.5.1] and using the uniqueness of Ju , as well as the fact that |CK (u)|2 = 3 when q = 3, we see that K ∼ = L± 3 (q) and Ju ≤ K. Again since Sylow 2-subgroups  of K are not semidihedral, indeed K ∼ = L3q (q). Finally, since E22 ∼ = D ≤ K, Jx and Jy form a standard CT- or P-pair in K, completing the proof.  Lemma 3.36. Let K ∈ K2 and let a, b be a four-subgroup of Aut(K). Suppose that CK (a) and CK (b) have components Ka and Kb isomorphic to E6± (q) and D6 (q)hs , respectively. Then K = Ka , Kb ∼ = E7 (q). Proof. Let K0 be the normal closure of Ka , Kb under a, b . Then by L2 balance, K0 = L2 (K0 ). Set K 0 = K0 /O2 (K0 ). It is clear from [IA , 2.2.10] and [III11 , 1.1ab] that K, and every component of K 0 , lies in Chev(r), where q is a power of the prime r. If either a or b induces a field or graph-field automorphism on K, then K ∼ = E6 (q 2 ) or D6 (q 2 ); but then b or a (respectively) must also induce a field or graph-field automorphism, contradicting the fact that Ka and Kb correspond to distinct Lie algebras. Therefore a, b ≤ Aut0 (K) and we can use [IA , 4.5.1] to deduce that K ∼ = E7 (q). If K 0 has more than one component, then |K 0 | is divisible by |Ka ||Kb |, which does not divide |K|. So K 0 is a single component, and the same argument as for K shows that K 0 ∼ = E7 (q). Hence K = K0 and the proof is complete.  Lemma 3.37. Suppose that X is a K-group, D = x, y is a four-subgroup of X, and for each u ∈ D# , Lu is a component of CX (u). Assume that for some also that odd prime power q, Ly ∼ = Lxy and Lx ∼ = E7 (q) ∼ = D8 (q)hs . Suppose # m2 (CX (u Lu )) = 1 for all u ∈ D . Then E(X) = Lu | u ∈ D# ∼ = E8 (q).

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Proof. Let Ku be the subnormal closure of Lu in X, so that  by L2 -balance, Ku is a product of 2-components of L2 (X). Without loss X = Ku | u ∈ D# D. Set X = X/O2 (X). If W := O2 (X) = 1, then by induction E(X) ∼ = E8 (q), whence L2 (CX (u)) = Lu for each u ∈ D# , according to [IA , 4.5.1]. Then if we put E = E(CLx (D)), we have 1 ≤ E ≤ Lu for all u ∈ D# , so [E, CW (u)] ≤ [E(CX (u)), O2 (CX (u))] = 1 for all such u. Consequently [E, W ] = 1, so L2 (X) = E(X) and the desired conclusion holds. We may therefore assume that W = 1. Clearly m2 (I) > 1 for all components I of Ku for all u ∈ D# and so for all components I of E(X). Hence if Ku < E(X) for some u ∈ D# , then m2 (CX (u Lu )) ≥ m2 (CX (u Ku )) > 1, contradiction. Thus Ku = E(X) for all u ∈ D# , whence E(X) is quasisimple. By [IA , 4.5.1] and [III11 , 1.1ab], E(X) ∼ = E8 (q). Finally, let X0 = Lu | u ∈ D# . The same argument applies to X0 D in place of X, and so X0 = O 2 (X0 D) ∼  = E8 (q). The proof is complete. −

Lemma 3.38. Let F ∗ (X) = L4 q (q) with q = ±1, q ≡ q (mod 4), and q > 3. − Let x, y be a four-subgroup of X such that E(CX (x)) ∼ = L3 q (q), E(CX (y)) = ∼ ∼ L1 ∗ L2 with y ∈ L1 = L2 = SL2 (q), and [x, L1 ] = 1. Then L2 x / y ∼ = P GL2 (q). −

Proof. Let K = F ∗ (X) and let X ∗ = GL4 q (q), so that X ∗ acts on K. Then we may assume that x and y act on K like x∗ = diag(−1, −1, −1, 1) ∈ X ∗ and y ∗ = diag(−1, −1, 1, 1) ∈ X ∗ . Here the SL2 (q)-subgroups L∗1 , resp. L∗2 of X ∗ corresponding to L1 and L2 are supported in the first two, resp. last two coordinates. Thus x acts on L2 like the matrix diag(−1, 1). Since q + q ≡ 2 (mod 4) by assumption, no element of Zq+q of even order is a square, and hence x induces an  outer diagonal automorphism on L2 . This implies the lemma. 4. CTP-Systems Lemma 4.1. Let L be a weak CT-system, a standard CT-pair, a weak P-system, or a standard P-pair, in a group L. Then for any α ∈ Aut(L), Lα is also a weak CT-system, standard CT-pair, weak P-system, or standard P-pair, respectively, in L, and of the same type as L. Proof. This follows directly from [III13 , Definitions 1.2, 1.5, and 1.11].



Through Lemma 4.13, as is usual when q is an odd prime power, q = ±1 and q ≡ q

(mod 4).

Lemma 4.2. Let I = Sp4 (q) or P Sp4 (q), q a power of the odd prime r. Let  t ∈ I2 (I) with O r (CI (t)) = L1 L2 , [L1 , L2 ] = 1, L1 ∼ = L2 ∼ = SL2 (q). Let ti ∈ Li  be of order 4, i = 1, 2, and set H = O r (CI (t1 t2 )). Then for both i = 1 and i = 2, {Li , H} is a standard CT-pair for I if q = 1, and a standard P-pair for I if q = −1. Proof. We use standard notation [IA , 2.10]. By [III13 , Lemma 1.8, Remark 1.9], a standard CT-pair  or P-pair in I is obtained as follows. First take the standard  CT-pair { X ±α , X ±β } in the overlying algebraic group K = C2u , where {α, β} =  Π and β is short. Set I γ = X γ , X −γ for any root γ. Let σ = σq for a CT-pair, and σ = jσq for a P-pair, where j ∈ NK (T ) inverts the maximal torus T . Then   r  O (CI α (σ)), O r (CI β (σ))

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is the desired standard CT- or P-pair in I = O r (CK (σ)). Replacing the elements t, t1 , and t2 of the lemma by conjugates in I, L1 , and L2 , respectively, as we may by Lemma 4.1, we may assume that t, t1 , t2 ∈ T with CK (t) = I α I α+2β and t = Z(I α ) or Z(I α+2β ); moreover, L1 = CI α (σ) and L2 = CI α+2β (σ). Then since we are taking σ = σq or jσq according as q ≡ 1 or −1 (mod 4), we may assume that t1 = hα (ζ) ∈ L1 ∩ T and t2 = hα+2β (ζ) ∈ L2 ∩ T 2 , where ζ ∈ Fq2 and ζ 2 = −1. From the Chevalley relations [IA , 2.4.7f] we see   that CK (t1 t2 ) = I β T . Therefore O r (CI α (σ)) = L1 and O r (CI β (σ)) = H form a standard CT-pair or P-pair, according as q ≡ 1 or −1 (mod 4). As there is u ∈ I2 (NK (L1 L2 )) interchanging L1 and L2 , and interchanging t1 and t2 , the same  holds for L2 and H, and the proof is complete. Lemma 4.3. Let K = Ω5 (q), q odd, have natural module VK = V1 ⊥ V2 ⊥ V3 , where V1 and V2 are 2-dimensional subspaces of type q and dim V3 = 1. Write Ω(V1 ⊥ V2 ) = I1 ∗ I2 with I1 ∼ = I2 ∼ = SL2 (q), and set I = Ω(V2 ⊥ V3 ). Then I1 and I form a standard CT-pair generating K if q = 1, and a standard P-pair generating K if q = −1. Proof. For a standard CT- or P-pair, by [III13 , Definitions 1.2, 1.5], we may use the isomorphism Ω5 (q) ∼ = P Sp4 (q) and simply take a standard pair for P Sp4 (q) (in reverse order). By Lemma 4.2, such a pair is {L1 , H} where L1 is either (solvable) component of CK (t) = L1 ∗L2 ∼ = SL2 (q)∗SL2 (q) for some 2-central involution  t ∈ K, and H = O r (CK (t1 t2 )), with ti ∈ Li of order 4. Clearly L1 ∗ L2 acts faithfully on [VK , t], which is therefore 4-dimensional of + type. By Witt’s lemma we may replace the given decomposition of VK by a conjugate decomposition and assume that [VK , t] = V1 ⊥ V2 . Now I2 (L1 ∗ L2 ) − Z(L1 ∗ L2 ) is a single conjugacy class, and since its elements have spinorial norm 1, it must be that [VK , t1 t2 ] is 2-dimensional of type q . Again by Witt’s lemma we  may assume that [VK , t1 t2 ] = V1 . Hence O r (CK (t1 t2 )) = Ω(V2 ⊥ V3 ). The lemma follows.  The next definition is non-standard and is introduced only for the sake of Lemma 4.5. Definition 4.4. Subgroups I ∼ = SL2 (q) and J ∼ = SL2 (q 2 ) of K = SL4 (q) ∼ = + Spin6 (q), q odd, are said to form a PP-pair if and only if the following conditions hold: (a) q ≡ −1 (mod 4) and q is a power of the prime r; (b) The natural O6+ (q)-module V , which is also a nonfaithful K-module, has a decomposition V = V1 ⊥ V2 ⊥ V3 into planes of respective signatures −1, −1, 1; and  (c) I is one of the central factors in O r (CK (V3 )) ∼ = SL2 (q) × SL2 (q), and  r J = O (CK (V1 )). Moreover, if K is any nontrivial homomorphic image of K, then the images I and J of I and J in K are also said to form a PP-pair. Lemma 4.5. Let I = Ω+ 6 (q), q ≡ −1 (mod 4), q a power of the prime r. Let {L, H} be a PP-pair for I with L ∼ = L2 (q 2 ). Let V be the natural I= SL2 (q) and H ∼ module and set VL = [V, L] and VC = CV (L), so that VC is 2-dimensional of + type.

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Let V0 be any nonsingular line in VC , and put H0 = CH (V0 ). Then H0 ∼ = L2 (q) and {L, H0 } is a standard P-pair for L, H0 = CI (V0 ) ∼ = Ω(V0⊥ ) ∼ = Ω5 (q). Proof. This is immediate from Definition 4.4 and Lemma 4.3.



Lemma 4.6. Let K = Ω6 (q), q odd,  = ±1, have natural module VK = V1 ⊥ V2 ⊥ V3 , where V1 , V2 , and V3 are 2-dimensional subspaces of types q , q , and , respectively. Write Ω(V1 ⊥ V2 ) = I1 ∗ I2 with I1 ∼ = I2 ∼ = SL2 (q), and set I = Ω(V2 ⊥ V3 ). Also if  = q , write I = I3 ∗ I4 with I3 ∼ = I4 ∼ = SL2 (q). Then the following conditions hold:  (a) If K = Ω6q (q), then any three of I1 , I2 , I3 , I4 form a weak CT-system (if q = +1) or weak P-system (if q = −1) for K; − (b) If K = Ω6 q (q) and q = +1, then I1 and I form a standard CT-pair for K; and − (c) If K = Ω6 q (q) and q = −1, then I1 and I form a PP-pair for K. Proof. In (a) we have K = Ω6q (q) ∼ = SL4q (q)/Z, where Z = ±1 ∼ = Z2 . 



Thus  = q . Let w1 and w3 be the involutions in I1 and I3 , respectively. Then w1 , w3 , and w1 w3 are all K-conjugate and are images of conjugate involutions  3 , w 1 w 3 in SL4q (q). Thus with respect to a suitable basis (orthonormal in w 1 , w 3 = diag(1, −1, −1, 1); and each Ii the unitary case), w 1 = diag(−1, −1, 1, 1) and w centralizes two basis vectors and acts faithfully on the span of the other two. It is  then clear from computing in SL4q (q) that either I1 or I2 , together with either I3  or I4 , forms a standard CT-pair or P-pair of type SL3q (q), according as q = 1 or −1. This implies (a). In (b), the proof is quite similar to that of Lemma 4.3. Now, K has a twisted

(the short fundamen = {

of type C2 with fundamental system Π α, β} root system Σ

We consider K as K/  ±1 , where K  = SU4 (q). Let L  = SL4 (q 2 ), tal root is β).  Let with respect to an ordered basis B = {v1 , v2 , v3 , v4 } of the natural L-module. 2   Lij , i = j, be the SL2 (q ) subgroup of L supported on the span of vi and vj , and  = C  (τ ) where τ (x) = ((x(q) )T )−g , and g ∈ L  fixing the other two vk ’s. Now K L maps vi to v5−i for all 1 ≤ i ≤ 4. Then L has fundamental system Π0 := {α, β, α } where α and α fold to form α

. Letting Iγ = X±γ for γ ∈ Π0 , we have that  Here Iβ = L  23 , Iα = L  12 , and {CIβ (τ ), CIα Iα (τ )} is a standard CT-pair for K. 2 ∼  34 . Thus Iβ = SL2 (q ) is one of the two isomorphic components of CL (z), Iα = L  23 ). Hence CI (τ ) ∼ where z is the involution in Z(L = SL2 (q) is one of two isoβ   14 of O r (C  (z)), where q is a power of the odd morphic SL2 (q)-components L23 , L K   12 × L  34 ∼ prime r. Likewise Iα Iα = L = SL2 (q 2 ) × SL2 (q 2 ) equals O r (CL (y)), where y = y1 y2 with yi ∈ Li of order 4, i = 1, 2. Hence CIα Iα (τ ) ∼ = SL2 (q 2 ) equals r O (CK (y1 y2 )). It follows that a standard CT-pair for K is formed by one of the   components of O r (CK (z)) and all of O r (CK (y1 y2 )). As in the proof of Lemma 4.3, we compute that [V, y1 y2 ] is 2-dimensional of type q . Replacing the decomposition V = V1 ⊥ V2 ⊥ V3 by an O(V )-conjugate decomposition, we may assume that   O r (CK (z)) = Ω(V1 ⊥ V2 ) and O r (CK (y1 y2 )) = Ω(V2 ⊥ V3 ), which implies (b). Finally, part (c) follows directly from Definition 4.4. 

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Lemma 4.7. Let K = Ωη2s (q), q odd, with |Z(K)| = 2 and s ≥ 4 (so that η = sq ). Let V be the natural K-module. Let v and t be commuting involutions in K such that dim CV (v) = dim CV (t) = 2 and CV (t) ≤ [V, v]. Let W ⊆ [V, v] be a 4-dimensional subspace of [V, v] of + type containing CV (t). Let R = CK (W ⊥ ) ∼ = Ω(W ) and J = CK (CV (v)) ∼ = Ω([V, v]), so that R ≤ J. Write R = H ∗ H ∗ with H ∼ = SL2 (q). Then there exist subgroups H = H1 , H2 , . . . , Hs−1 of K, and = H∗ ∼ an element g ∈ K, such that the following conditions hold: (a) Each Hi is K-conjugate to H; (b) g interchanges t and v by conjugation; (c) Hig = Hs−i , i = 1, . . . , s − 1; (d) H1 , . . . , Hs−2 ≤ J and J, Hs−1 = K; (e) {Hi−1 , Hi } is a standard CT-pair or standard P-pair, according as q = 1  or −1, of type SL3q (q) for all 1 < i < s; and [Hi , Hj ] = 1 for all other 1 ≤ j < i < s;  (f) Hi−1 , Hi , Hi+1 is a quotient of SL4q (q) for all 1 < i < s − 1; and (g) Z(H1 ), . . . , Z(Hs−1 ) is abelian. Proof. Write V = V1 ⊥ V2 ⊥ · · · ⊥ Vs , where each Vi is nondegenerate of dimension 2 and type q . Indeed we may assume that the orthogonal form on each Vi is given by the identity matrix, with respect to a basis Bi . For each 1 ≤ i ≤ s, let vi ∈ I2 (K) have −1-eigenspace Vi . Also for each 1 ≤ i < s, set Wi = Vi ⊥ Vi+1 , of dimension 4 and type +, and set wi = vi vi+1 . Obviously the wi ’s and vi ’s generate an elementary abelian 2-group. Without loss we may assume that v = −vs and t = −v1 . Thus [V, v] = Vs⊥ and J = CK (Vs ). For each i = 1, . . . , s − 1, Ω(Wi ) = Li1 ∗ Li2 with Li1 ∼ = Li2 ∼ = SL2 (q) and Z(Li1 ) = Z(Li2 ) = wi ; moreover, Li1 and Li2 are interchanged by a reflection on Wi , and so they are at any rate K-conjugate. Thus all the Lij are K-conjugate. For each 1 ≤ i < (s + 1)/2, let gi be an involution interchanging the bases Bi and Bs+1−i , and inducing the identity mapping on all other Vj ’s. One calculates that gi ∈ K. If i = (s + 1)/2 is an integer, we take g(s+1)/2 = 1. Then set g = g1 · · · g[(s+1)/2] , so that Vig = Vs+1−i for all i = 1, . . . , s and g 2 = 1. Then Wig − Ws−i . We may then assume that notation has been chosen so that Lgi1 = L(s−i)1 for all i = 1, . . . , s − 1. Set Hi = Li1 , i = 1, . . . , s − 1. Then all the assertions of the lemma are clear except for (f), the second statement of (d), and the first statement of (e). To complete the proof we first show that for all j, k ∈ {1, 2} and all i = 1, . . . , s − 2,   Lij , L(i+1)k ∼ = SL3q (q). Indeed, this follows directly from Lemma 4.6a, applied to the group Ω(Vi ⊥ Vi+1 ⊥  Vi+2 ) ∼ = Ω6q (q). Now as [vi , vi+1 ] = 1, (e) is obvious. Also, J, Hs−1 contains, L(s−2)1 , L(s−1)1 ,  and L(s−2)2 , which form a weak CTP-system for Ω(Vs−2 ⊥ Vs−1 ⊥ Vs ) ∼ = A3q (q), by Lemma 4.6. So  J, Hs−1 ≥ Ω(Vs⊥ ), Ω(Vs−2 ⊥ Vs−1 ⊥ Vs ) = K by [IA , 7.3.2] (or Lemma 9.1 below), proving (d). Finally, we prove (f). Note that we cannot quote [III13 , Theorem 1.10] because of the anomalous case q = 3. Fix i and consider the set T of all eight triples

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{Li−1,j1 , Li,j2 , Li+1,j3 } such that each jk equals 1 or 2. One such triple is T1 := {Hi−1 , Hi , Hi+1 }. Moreover, if we let rj be a reflection supported in Vj , i − 1 ≤ j ≤ i + 2, then ri−1 , ri , ri+2 ∼ = E23 permutes T transitively by conjugation. It  therefore suffices to find T2 ∈ T such that T2 ∼ = SL4q (q). We shall find T2 ⊥ ∼ within the subgroup I = CK (W ) = Ω(W ), where W = Wi−1 ⊥ Wi+1 . We have I∼ = D4 (q) and Z(I) = zi−1 zi+1 . There is u ∈ I of order 4 such that u2 = zi−1 zi+1 ,  u normalizes each Vj , j = i − 1, . . . , i + 2, and I0 := E(CI (u)) ∼ = SL4q (q) with zi−1 , zi , zi+1 ≤ I0 . For j = i−1, i, i+1, let Ij be the (solvable) SL2 (q)-component  of CI0 (zj ) containing zj . Then T2 := {Ii−1 , Ii , Ii+1 } ∈ T, and T2 = I0 ∼ = SL4q (q). The proof is complete.  Lemma 4.8. Suppose that the hypotheses of Lemma 4.7 hold, except that s = 3. Then there exists g ∈ K such that g interchanges t and v by conjugation, {H, H g }  is a standard CT-pair (if q = +1) or standard P-pair (if q = −1) of type A2q (q), 2 and H g = H. ∼ Lq (q), it is clear that H ≤ I = ∼ SLq (q) for some I ≤ Proof. Since K/Z(K) = 4 3 K, and then that there is a standard CT-pair or standard P-pair {H, H g } for I of the 2 asserted type such that g ∈ I and H g = H. In particular [Z(H), Z(H g )] = 1. As I has one class of involutions, every involution of the four-group E = Z(H), Z(H g ) must, like y ∈ Z(H), have a 4-dimensional commutator space on the natural  K∼ = Ω6q (q)-module. Hence the fixed-point subspaces of the three involutions of E are mutually orthogonal and 2-dimensional of type q . It follows by Witt’s lemma that {y , Z(H g )} is K-conjugate to {y , t }, and so we may assume that y g = t. 2 Since H g = H, tg = y, and the lemma is proved.  Lemma 4.9. Suppose that the hypotheses of Lemma 4.7 hold, except that K ∼ = so that Z(K) = 1. Then there exists an element g ∈ G such that (a) g interchanges t and v by conjugation; and − (b) H, H g ∼ = SL3 q (q). Moreover, t induces a non-inner but inner-diagonal automorphism on each (solvable) component of CK (CV (v)) ∼ = SL2 (q) ∗ SL2 (q). − Ω6 q (q),

Proof. Let VΩ be the natural 6-dimensional orthogonal module for K. Since t and v are involutions in K, [t, VΩ ] and [v, VΩ ] are 4-dimensional of type +. As CVΩ (t) ⊆ [v, VΩ ], [vt, VΩ ] is also 4-dimensional and so v, t # ⊆ tK . It follows that − − the preimage of v, t in Spin6 q (q) is elementary abelian. Regarding Spin6 q (q) −q as L := SL4 (q), and letting VL be the natural L-module, we see that with respect to a suitable basis {e1 , . . . , e4 } of VL (an orthonormal basis if q = 1) we have v = diag(−1, −1, 1, 1) and t = diag(1, −1, −1, 1) (with respect to an orthonormal − basis if q = 1). Because L = SL4 q (q), it follows that t induces an outer diag onal automorphism on each component of O r (CK (v)) ∼ = SL2 (q) ∗ SL2 (q), which yields the final statement of the lemma. Without loss we may assume that H is supported on the span of e1 and e2 , and centralizes e3 and e4 . Let g ∈ K fix e1 and e4 and interchange the 1-dimensional subspaces generated by e2 and e3 . Then − g interchanges t and v, and H, H g ∼  = SL3 q (q). The proof is complete. Lemma 4.10. Let q be a power of the prime r with q ≡ −1 (mod 4). Let  K = L4 (q) and let z ∈ I2 (K) with O r (CK (z)) = J1 ∗J3 and z ∈ J1 ∼ = J3 ∼ = SL2 (q).

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Then there exists J2 ≤ K such that {J1 , J2 , J3 } is a weak CT-system of type K, and such that z induces an outer diagonal automorphism on J2 . Moreover, given an involution u ∈ J1 J3 − {z}, J2 may be chosen so that [u, Z(J2 )] = 1. Proof. In matrix terms, we may assume that z = diag(1, 1, −1, −1), and that J1 and J3 are the upper and lower block diagonal SL2 (q) subgroups, respectively. Let J2 be the middle block diagonal SL2 (q) subgroup, a (solvable) component of the centralizer of diag(1, −1, −1, 1). It is then obvious that J1 , J2 , and J3 form the desired weak CT-system. Then z acts on J2 as diag(1, −1), whose determinant is not a square since q ≡ −1 (mod 4). Hence z induces an outer diagonal automorphism on J2 , as asserted. As Z(J2 ) centralizes neither J1 nor J3 , it inverts elements u1 ∈ J1 and u3 ∈ J3 of order 4, and so centralizes the involution u1 u3 . The final statement follows as J1 J3 permutes transitively its noncentral involutions.  Lemma 4.11. Let K be a K-group and x, y a four-subgroup of K. Suppose that E(CK (x)) ∼ = A3 (q) for some odd prime power q. If q > 3, assume that E(CK (y)) ∼ = A2 (q)×A1 (q), while if q = 3, assume that O 2 (CK (y)) ∼ = A2 (q)×A1 (q). Suppose that L7 , L8 , L9 , L0 are subgroups of K isomorphic to SL2 (q) and such that {L7 , L8 , L9 } and {L8 , L9 , L0 } provide weak CT systems for E(CK (x)) and E(CK (y)) (or O 2 (CK (y)) if q = 3), respectively, in accordance with the diagrams (4A)

α8

◦

α7

◦

α9

◦

and

α8

◦

α9

◦

α0

◦

Assume also that y ∈ L8 and xy ∈ L9 . Let H = L7 , L8 , L9 , L0 . Then H ∼ = A4 (q) and there is h ∈ L9 such that [L0 , Lh7 ] = 1. Proof. By assumption each Li generating H lies either in E(CH (x)) or E(CH (y)). Hence by L2 -balance, H = L2 (H). Suppose that H is a minimal counterexample to the assertion that H ∼ = A4 (q). We argue first that H ∈ K2 . Let W = O2 (H) and H = H/W . If W = 1, then by minimality H ∼ = A4 (q). We (y)) even if q = 3, so xy centralizes CW (x) and have xy ∈ L9 ≤ E(CH (x)) ∩ E(CH   #

CW (y). Therefore xy centralizes CW (d) | d ∈ x, y = W , so H is quasisimple (q), contrary to assumption. Thus W = 1. Since x, y ≤ H and H is and H ∼ A = 4 generated by the quasisimple groups E(CK (x)) and E(CK (y)), which both contain L9 , H ∈ K2 , as asserted. Thus, A3 (q) ↑2 K via x and A2 (q) ↑2 K via y. It follows from [III11 , 1.1ab] and [IA , 2.2.10] that K ∈ Chev(r), where q is a power of the prime r. As both x and y induce inner automorphisms on K, we see from [IA , 4.5.1] that the only possibility, given the structure of E(CH (x)) and E(CH (y)) (or O 2 (CH (y)) if q = 3), is K ∼ = A4 (q). It remains to prove the final assertion of the lemma. Let V = ⊕5i=1 Vi be the natural SL5 (q)-module, with each Vi 1-dimensional and x, y -invariant. Then without loss, y and L8 are supported on V1 + V2 , while xy and L9 are supported on V3 + V4 . Then L7 , whose involution commutes with both x and xy (in E(CH (x))), is, without loss, supported on V2 + V3 , fixing V1 + V4 + V5 . Finally, L0 , centralizing L8 , fixes V1 +V2 and has its support inside V3 +V4 +V5 ; this support must include V5 since the involutions of L9 and L0 commute and are unequal. Therefore V3 + V4 = V6 + V7 , where the support of L0 has the form V6 + V5 , and V7 is fixed by L0 . Let h ∈ L9 transform V6 and V7 to V4 and V3 , respectively. Then the support of Lh0 is V4 + V5 and its fixed space is V1 + V2 + V3 . Hence [Lh0 , L7 ] = 1 and the lemma follows. 

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Lemma 4.12. Let K = L6 (q). Let Li , −2 ≤ i ≤ 2, be an SL2 (q) subgroup of K. Suppose that for both  = 1 and  = −1, {L0 , L , L2 } is a weak CT-system for H := L0 , L , L2 ∼ = SL4 (q), and [H , L−2 ] = 1. Then there exists x ∈ L0 such that {L−2 , L−1 , L0 , Lx1 , L2 } is a weak CT-system for K. Proof. By assumption, all of the Curtis-Tits relations hold for L−2 , L−1 , L0 , and L2 , as well as for L−2 , L0 , L1 , and L2 , according to the A5 -diagram where the roots are numbered −2, −1, 0, 1, 2 in sequence. Indeed since L0 centralizes L±2 , the same statements hold with Lx1 in place of L1 , for any x ∈ L0 . Thus, it suffices to find x ∈ L0 such that [L−1 , Lx1 ] = 1. Write Z(Li ) = z i for each i, and let zi be the classical involution of SL6 (q) mapping on z i . Let V be the natural SL6 (q)-module, and write Vi = [V, zi ] for each i. Then V = V−2 ⊕ (V0 + V1 + V2 ) (and similarly with all signs changed), and the two summands are the supports of L−2 and H1 respectively. It suffices to show that V1x and V−1 are both z1x - and z−1 -invariant, and disjoint, for some x ∈ L0 . Writing Vi,j for Vi ∩ Vj , our hypotheses on H±1 yield V1 = V0,1 ⊕ V1,2 , and V−1 = V0,−1 ⊕ V−1,−2 , each summand being 1-dimensional. Under the action of z1 , V0 decomposes into eigenspaces V0,1 ⊕CV0 (z1 ), and a similar statement holds for z−1 . Replacing z1 by z1x for suitable x ∈ L0 we may assume that V0,1 = CV0 (z−1 ) and CV0 (z1 ) = V0,−1 . We claim that then [z1 , z−1 ] = 1, which will complete the proof. The actions of z1 and z−1 on V0 obviously commute. From the given presentation for H1 , [z1 , z2 ] = 1, and [z−1 , z2 ] = 1 as well by hypothesis. So z1 and z−1 leave V2 invariant, and as z−1 acts trivially, their actions on V2 commute. Similarly they have commuting actions on V−2 . As V = V−2 ⊕ V0 ⊕ V2 , [z1 , z−1 ] = 1, completing the proof.  Lemma 4.13. Let K = SL4 (q), where q is a power of the prime r and q ≡ −1 (mod 4). Let J1 and J2 be SL2 (q)-subgroups of K, and write zi = Z(Ji ), i = 1, 2. Assume that on the natural K-module V , Ji is supported on [V, zi ], which is 2dimensional, i = 1, 2. If [z1 , z2 ] = 1 and z1 induces a nontrivial automorphism on J2 , then J1 , J2 ∼ = SL3 (q), supported on V on the subspace [V, z1 ] + [V, z2 ]. Proof. Since [z1 , z2 ] = 1, V = ([V, z1 ]+[V, z2 ])⊕CV (z1 , z2 ) and our hypotheses imply that both direct summands are J1 , J2 -invariant; the decomposition is identical to V = [V, J1 , J2 ]⊕CV (J1 , J2 ). If CV (z1 , z2 = 0, then since dim V = 4 and dim[V, z1 ] = dim[V, z2 ] = 2, z1 centralizes [V, z2 ] = [V, J2 ]. Hence J2 = [z1 , J2 ] centralizes J2 , which is absurd. Likewise if dim CV (z1 , z2 ) = 2, then z1 inverts [V, z2 ] = [V, J2 ], leading to the same contradiction. Therefore dim CV (z1 , z2 ) = 1. As [z1 , z2 ] = 1, it is then immediate that J1 and J2 form a standard CT-pair of  type SL3 (q), and the result follows. In the final two results of this section, q is even. Lemma 4.14. Let K = D4− (q), with q > 2 a power of 2, and let p be a prime divisor of q − 1. Let V be the natural Fq K-module, let B ∈ Ep∗ (K), and write B = b1 , b2 , b3 where b1 , b2 , b3 are all K-conjugate and dim[V, bi ] = 2 for each i. Let L1 = E(CK (b1 , b2 b3 )), L2 = E(CK (b1 b2 , b3 )), and L3 = E(CK (b2 , b3 )). Then {L1 , L2 , L3 } is a weak CT-system for K. Proof. We have V = V0 ⊥ V1 ⊥ V2 ⊥ V3 where V0 = CV (B) and Vi = [V, bi ], i = 1, 2, 3 (see [IA , 4.8.1]). Using [IA , 4.8.2] we see that L1 ∼ = L2 (q) is supported

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∼ ∼ L2 (q) is supported on V1 ⊥ V2 , and L3 = CK (V2 ⊥ V3 ) = on V2 ⊥ V3 , L2 = Ω(V0 ⊥ V1 ) ∼ (q). In particular [L , L ] = 1. Set L = E(C (b b b )), so that = Ω− 1 3 12 K 1 2 3 4 L1 , L2 ≤ L12 ∼ = SL3 (q) by [IA , 4.8.2] as p divides q−1. Indeed the eigenvalues of bi on Vi are ζ and ζ −1 for some pth root of unity ζ ∈ F× q , independent of i as the bi are + − ± all K-conjugate. Write Vi = Vi + Vi where Vi is the ζ ±1 -eigenspace. Then L12  acts faithfully on V + := 3i=1 Vi+ , L1 centralizes V1+ and normalizes [V + , b2 b3 ] = V2+ + V3+ , and L2 centralizes V3+ and normalizes [V + , b1 b2 ] = V1+ + V2+ . It follows immediately that L1 and L2 form a standard CT-pair of type A2 (q). Finally, set L23 = E(CK (b3 )), so that L2 , L3 ≤ L23 ∼ = Ω(V3⊥ ). Following [BHS, Remark 2.4], we see that L2 and L3 form a standard CT-pair of type 2 D3 (q) = 2 A3 (q). The lemma follows.  Lemma 4.15. Let K = Sp2n (q), q = 2a > 2, n ≥ 3. Let  = ±1 and let p be a prime divisor of q − . Let B ∈ Ep∗ (K). Then according as  = 1 or −1, there exists a weak CT-system or weak P-system {J1 , . . . , Jn } for K such that each Ji is B-invariant. Proof. Let K = Cn and use standard notation [IA , 2.10]. a fun Let Π be damental system of T -roots and for each α ∈ Π set K α = X α , X −α . Then {K α | α ∈ Π} is a weak CT-system for which each K α is T -invariant. If  = +1,  let σ = σq and Kα = O 2 (CK α (σq )); then {Kα | α ∈ Π} is a weak CT-system for K each of whose elements is normalized by Ω1 (Op (CT (σq ))) ∼ = Epn . But by [IA , 4.10.3ac], Ep∗ (K) = Epn (K) is a single orbit under the conjugation action of K. Hence the lemma holds if  = 1. If  = −1, we use σ = σq i instead, where i ∈ NK (T ) inverts T elementwise. Then i interchanges X α and X −α for each α ∈ Π, so i and then σ normalize  each K α . This time the groups Kα = O 2 (CK α )(σ) form a weak P-system, by [III13 , Lemma 1.8]. And they are again invariant under CT (σ), which this time contains a homocyclic abelian subgroup of rank n and exponent q + 1. Hence, using  [IA , 4.10.3] as above, we obtain the assertion of the lemma.

5. Representations Lemma 5.1. In SL4 (3), there is a single GL4 (3)-conjugacy class of subgroups isomorphic to A6 , and they are absolutely irreducible. Proof. This follows immediately by applying the Steinberg tensor product theorem to SL2 (9).  Lemma 5.2 (Gilman-Griess). Suppose that A is a homocyclic abelian 3-group of rank 4, and X ∼ = A6 acts nontrivially on A. Then A is elementary abelian. Proof. The proof is taken from [GiGr1]. Let A be a minimal counterexample; then A has exponent 9. Let W be a 3 × 3 permutation matrix corresponding to a 3-cycle. For any a ∈ X of order 3, a lies in a Σ4 -subgroup Σ, and A = [A, O2 (Σ)] ⊕ CA (O2 (Σ)), with a ∈ [Σ, Σ] centralizing CA (O2 (Σ)) andactingfreely W 0 on [A, O2 (Σ)]. In particular modulo 9, a has the (4 × 4) matrix α := with 0 1 respect to some basis of A, and so Tr(a) = 1.

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Now fix a = (123) ∈ X and let b = (456). Working modulo 3, we know from Lemma 5.1 that A is the core of the natural permutation module and so we may take as a basis B3 of A/3A the elements a1 − a4 , a2 − a4 , a3 − a4 , and a4 − a5 , where X naturally permutes {a1 , . . . , a6 } and a1 + · · · + a6 = 0. We see that  modulo  3, a I v has matrix α (the reduction of α modulo 3), and b has matrix β := t . Here v 1 v is the 3 × 1 column of 1’s. of a is still We pull B3 back to a basis B9 of A such that modulo 9, the matrix  B fv α, while the matrix of b (now modulo 9 again) is of the form β = , where gv t h B = I + 3(cI + dW + eW 2 ), and c, d, e, f, g, h are integers. Here we are using the fact that ab = ba to determine β. An easy calculation then shows that Tr(α − I)β ≡ −3 (mod 9). However, αβ is the matrix of ab, which has order 3, so as shown above, Tr(αβ) ≡ 1 ≡ Tr(β) (mod 9). Thus Tr(α − 1)β ≡ 0 (mod 9). This contradiction completes the proof.  Lemma 5.3. Suppose X ∼ = A6 , A7 , M11 , or U3 (4) and X ≤ GL4 (p) for some odd prime p. Then p = 3, X ∼ = A6 , and X is absolutely irreducible. Proof. If X ∼ = M11 , then X has Frobenius subgroups of order 9.8 and 11.5, so the result follows in that case from [IG , 9.12]. If X ∼ = U3 (4), then X has a Frobenius subgroup KC with K special of order 22+4 , |C| = 3, and K/ z extraspecial for all z ∈ Z(K)# . An application of Clifford’s Theorem shows that X does not lie in GLn (p) for any n < 12 and odd p. If X ∼ = A6 or A7 , then the minimum degree of an ordinary character of X is at least 5, so as A6 ≤ A7 , p ∈ {3, 5}. But 7 does not divide |GL4 (p)| for these values, so X ∼ = A6 . Suppose that p = 5. Let V = Ep4 support the action of X. Let Z ≤ Y ≤ X with Z ∼ = A4 and Y ∼ = A5 . From the ordinary representation theory of Z we see that CV (Z) = 1, so that as Y -module, V is a quotient of the natural permutation module, hence has a trivial quotient. Dualizing, CV ∗ (Y ) = 1, so as X-module, V ∗ is a quotient of the natural X-permutation module. But that module has no 4-dimensional constituent, contradiction. Thus p = 3, and absolute irreducibility holds by Lemma 1.9.  Lemma 5.4. Let K = L3 (q), q = 2n ≡  (mod 3). Then the minimum degree of a faithful representation in characteristic 2 is 8. Proof. Since K contains U3 (2) [IA , 6.5.3], a Frobenius group of order 9.8, the minimum degree is at least 8. Let K be the universal algebraic group A2 over F2 . Then the tensor product of the natural K-module and its dual has a trivial composition factor, and is acted on trivially by Z(K). Hence a submodule or quotient of the tensor product gives a faithful representation of degree at most 8. The lemma follows.  Lemma 5.5. Suppose that X = K r ≤ GL4 (p) for some prime p ≥ 5, with K∼ = SL± 4 (3) or Sp4 (3). Then r is not a reflection in GL4 (p). Proof. Let C = CK (r) and C o = [C, C]. Then C stabilizes the 1-dimensional center and 3-dimensional axis of r, so C o centralizes the center of r and acts faithfully on the axis of r. In particular −1 ∈ C o and C o ≤ SL3 (p). If K ∼ = Sp4 (3),

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it follows that m2 (C o ) ≤ 1 and C o ∼ = [CZ(K)/Z(K), CZ(K)/Z(K)]. But from [IA , 4.5.2], CZ(K)/Z(K) contains A4 , a contradiction. Hence K ∼ = SL± 4 (3). ± o Sp (3), SL (9), L (9), or L (3) [I As C ≤ SL3 (p), E(CK (r)) ∼ = 4 2 2 A , 6.5.3 and 3 + Ω (3) or SL (3) × SL (3), by [I , 4.5.2]. But in both p. 331]. Then CK (r) ∼ = 4 2 2 A  cases, −1 ∈ C o , a contradiction. The proof is complete. Lemma 5.6. Let K ∈ Chev(2) but K ∈  Chev(r) ∪ Alt for any odd r. If K ≤ GL4 (p) for some odd prime p, then Z(K) has odd order. Proof. With [IA , 6.1.4, 2.2.10] we see that if K satisfies the hypotheses but 3 Z(K) has even order, then K/Z(K) ∼ = L3 (4), 2B2 (2 2 ), G2 (4), Sp6 (2), U6 (2), D4 (2), − F4 (2), or E6 (2). If O2 (Z(K)) is noncyclic, then K is reducible and embeds in SL2 (p)×SL2 (p), so all involutions of K lie in the center, contradicting [IIIK , 4.17]. Thus O2 (Z(K)) is cyclic, and K ≤ SL4 (p) so m2 (K) ≤ 3. Now m2 (2Sp6 (2)) = m2 (Co3 ) = 4, and the last 5 groups listed all contain Sp6 (2), so they cannot 3 3 occur. Likewise m2 (22B2 (2 2 )) > m2 (2B2 (2 2 )) = 3 by [IA , 6.4.1a], ruling out that possibility. And SL3 (4) ≤ G2 (4) with m2 (2SL3 (4)) = 5 by [IA , 6.4.1a]. So the only possible remaining case is K ∼ = 4L3 (4). In this case m2 (K) = 3, so K ≤ SL3 (p), hence K is irreducible. As K ≤ SL2 (p2 ), K is absolutely irreducible, so p ≡ 1 (mod |Z(K)|). In particular p > 3. But L3 (4) contains a Frobenius subgroup of order 9.8 and so K cannot be represented in fewer than 8 dimensions, by Clifford’s theorem. The proof is complete.  Lemma 5.7. Let n ≥ 3 and let q be an odd prime power. Let K ∈ Chev be of level q and let V be a nontrivial Fq K-module. Then the following conditions hold: (a) If K ∼ = SLn (q) or Ω± n (q), with n = 4 in the orthogonal case, then dim V ≥ n; and (b) If K ∼ = Spin± n (q), n = 4, then either dim V ≥ n or n ≤ 6 with K faithful on V . In the latter case, one of the following holds: (1) K ∼ = Spin3 (q) and dim V = 2; or (2) K ∼ = Spin5 (q) or Spin± 6 (q), and dim V = 4. Proof. Let Z be the kernel in K of the action on V , and set K = K/Z. First assume that K ∼ = SLn (q), n ≥ 3. If n = 3, then K contains A4 , while if n = 4, then K contains the central or direct product of two Q8 subgroups. These subgroups force dim V ≥ n. For n ≥ 5, let M = EP be a monomial subgroup of K, with E the diagonal subgroup of M and P ∼ = An . As V |E is nontrivial, V |E must contain a nontrivial orbit of P on the dual group E ∗ , whence dim(V ) ≥ n, as claimed. Next suppose that K ∼ = Ω± n (q). Again, if n = 3, then K contains A4 . If n is odd and n ≥ 5 then we may assume that K ≥ M = EP and the above argument gives the desired result. If n = 2m with m ≥ 3, then we have K ≥ Ω3 (q) × Ω2m−3 (q) and 3(2m − 3) ≥ 2m, so by the odd case dim(V ) ≥ 3 + (2m − 3) = 2m. Thus (a) holds. Finally, suppose that K ∼ = Spin± n (q), n ≥ 3. We may assume that V is irreducible, so Z(K) is cyclic. By (a), we may assume that K ∼ = Spin± n (q) or HSpinn (q). If n = 2k + 1, then the preimage of E, above, in K is an extraspecial 2-subgroup of width k (use [IA , 6.2.1d] and the irreducibility of P on E/Z(K)). Hence, a faithful Fq K-module has dimension at least 2k ≥ n + 1 for k ≥ 3. As Spin2k+1 (q) embeds in both Spin2k+2 (q) and HSpin2k+2 (q), we are done for n ≥ 7. For n ≤ 6, if Z = 1, then dim(V ) ≥ n by (a). Hence we may assume V is faithful

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for K. As Z(K) has even order, dim V is even. Then the listed values in (b) are clear.  Lemma 5.8. Let K ∼ = Ωn (q 2 ), n ≥ 3, q odd,  = ±1. Let V be an Fq Kmodule such that the kernel of the K-action is contained in Z(K). Then either dimFq (V ) ≥ 2n, or n = 3 and dimFq (V ) ≥ 4.  = Spinn (q 2 ). Using Steinberg’s tensor product theorem, we Proof. Let K see that Fq2 ⊗Fq V has an absolutely irreducible composition factor of the form (q) C = W1 ⊗ W2 , where the Wi are Fq2 K-modules with high weights restricted (q) to the range [0, q − 1], W1 is nontrivial, and Wi ∼  Wi if Wi is nontrivial. If = W2 is trivial then since V (q) = V by assumption, V also has a composition factor (q) C (q) = W1 ∼ = C. So we may assume that W2 is nontrivial. Suppose first that n ≥ 5. Since n2 > 2n, it follows from the preceding lemma that dim(V ) ≥ 2n, as claimed, unless n ≤ 6 and W1 or W2 is a faithful module  of dimension 4. But then, as V is a K-module, both W1 and W2 must be for K faithful, whence dim(V ) ≥ 42 > 2n. Thus we may assume that dim W1 < n ≤ 4. If n = 3, then Spin3 (q 2 ) ∼ = SL2 (q 2 ) and dim(W ) ≥ 2. The result follows as in the previous paragraph. Finally, assume that n = 4. If  = +1, then K ∼ = SL2 (q 2 ) ∗ SL2 (q 2 ) and now dim(W1 ) ≥ 4, and we are done as before. Finally, if n = 4 and  = −1, then K ∼ = L2 (q 4 ). One can again use Steinberg’s tensor product theorem, or simply observe that as |K| is divisible by a primitive prime divisor of q 8 − 1, it follows that dim(V ) ≥ 8, as claimed, completing the proof.  Lemma 5.9. Let K ∼ = SLn (q) with q odd,  = ±1, n ≥ 3, n = 4. Suppose that ± ∼ ρ : K → K1 = Ωm (q) is a nontrivial homomorphism. Then m ≥ 2n. 

Proof. As Ωm (q) ≤ Ωm (q) for all m > m and all signs ,  , we may assume by way of contradiction that m = 2n − 1. As n ≥ 3 and q is odd, we may choose a primitive prime divisor s of q n − n . (A primitive prime divisor of q n + 1 is just a primitive prime divisor of q 2n − 1.) As K1 ∼ = Bn−1 (q), it follows immediately that if n is odd, then s does not divide |K1 |. Hence n is even, so n ≥ 6. Note also that if n = 6, then  = 1, since otherwise, |K| is divisible by (q 3 + 1)2 but |K1 | is not. Suppose that  = 1. Let P = U L be a maximal parabolic subgroup of K with L ∼ = SLn−1 (q). By the Borel-Tits Theorem, ρ(P ) is contained in a maximal parabolic subgroup P1 of K1 . Let P1 = U1 L1 be the Levi decomposition. Write L1 = J1 × J2 with J1 = 1 or J1 ∼ = SLk (q), 2 ≤ k ≤ n − 1, and J2 = 1 or J2 ∼ = ±  Ω (q), 2 ≤  ≤ 2n − 3. As ρ(L ) maps isomorphically into L1 , ρ(L ) must project nontrivially into either J1 or J2 . As n − 1 is odd, ρ(L ) cannot project nontrivially into J2 , and it follows therefore that ρ(L ) maps isomorphically onto J1 ∼ = SLn−1 (q). We then conclude that ρ(U ) ≤ U1 . Let V1 = [U1 , U1 ] = Φ(U1 ) = Z(U1 ). Then, as J1 -module, V1 is the exterior square of the natural module for J1 , and hence also for ρ(L ). Suppose first that ρ(U ) ∩ V1 = 1. Then, as |U1 | = |V1 ||U |, it follows that U1 = V1 ρ(U ) = Φ(U1 )ρ(U ) = ρ(U ), a contradiction. Hence V0 := ρ(U ) ∩ V1 is a nontrivial ρ(L )-submodule of V1 , and hence is normal in U1 ρ(L ) = U1 J1 . But V1 is an irreducible J1 -module, whence V1 ≤ ρ(U1 ). Hence dim(V1 ) = (n−1)(n−2) ≤ 2 n − 1 = dim(U1 ), and so n ≤ 4, contradicting the fact that n ≥ 6. Therefore  = −1. We argue by induction on n, having reached a contradiction in the first paragraph if n = 6. This time let P = U L be a parabolic subgroup

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∼ SUn−2 (q). Again by the Borel-Tits theorem, ρ(P ) ≤ P1 for some of K with L = parabolic subgroup P1 = U1 L1 of K1 , with ρ(U ) ≤ U1 . Now U is extraspecial, so U1 is nonabelian. This rules out the possibility L1 ∼ = Ω2n−3 (q). Thus L1 ∼ = ± SLk (q) × Ω (q) with 2 ≤ k ≤ n − 1, 1 ≤  ≤ 2n − 5, and 2k +  = 2n − 1. By induction ρ(L ) projects trivially on the second direct factor, so we obtain a nontrivial homomorphism from SUn−2 (q) to SLn−1 (q). Now a primitive prime divisor of q n−3 + 1 shows that such a homomorphism cannot exist, since n ≥ 8. The proof is complete.  The next lemma clearly has analogues for other classical groups. + n Lemma 5.10. Let K = Ω± m (q), m ≥ 6, or Ω4 (q), q = 2 . Let V be a natural Fq K-module. Then considered as an F2 K-module, V is irreducible.

Proof. Let W be this F2 K-module. Then Fq ⊗F2 W is the direct sum of the n (distinct) Gal(Fq /F2 )-conjugates of V . As the summands are absolutely irreducible  and no proper subsum can be written over F2 , the result follows. We include in this section a well-known result about permutation groups. Lemma 5.11. A finite primitive permutation group containing a 3-cycle contains the alternating group. Proof. For completeness, we prove this elementary fact. Let G ≤ ΣΩ be primitive and contain a 3-cycle. For each Ψ ⊆ Ω with |Ψ| ≥ 3 let AΨ be the pointwise stabilizer of Ω − Ψ in AΩ . Let S = {Ψ ⊆ Ω | |Ψ| ≥ 3 and AΨ ≤ G}. Since G contains a 3-cycle, S = ∅. Choose Ψ0 ∈ S with |Ψ0 | maximal. If Ψ0 ∩ Ψg0 = ∅ for some g ∈ G, then Ψ0 ∪ Ψg0 ∈ S so Ψg0 = Ψ0 . Thus Ψ0 is a block for the action of G.  By primitivity Ω = Ψ0 ∈ S, completing the proof.

6. Computations in Groups of Lie Type Lemma 6.1. Let K ∈ Lie. Let X be a root subgroup in K, if K is untwisted, or a long root subgroup in K, if K is a Steinberg variation. Then CAut(K) (X) ≤ Aut0 (K). Proof. We may assume that K is the adjoint version. We use standard notation [IA , 2.10] and freely use the results in [IA , 2.5]. We may take X = Xα for some root α. Then Xα ∼ = F+ q where q = q(K), and ΦK acts on Xα as Galois automorphisms, with kernel ΦK ∩Aut0 (K). In particular CXα (ΦK ) = 1. On the other hand, X ≤ X for some root subgroup X of the overlying algebraic group K. As X is a TIset in Aut0 (K), elements of AutAut0 (K) (X) arise from AutAut0 (K) (X) and therefore are trivial or fixed-point-free on X. Hence no element of (ΦK /ΦK ∩ Aut0 (K))# has the same action on Xα as any element of AutAut0 (K) (X). By our assumptions Aut(K) = Aut0 (K)ΦK , so the lemma follows.  Through Lemma 6.22 we compute in groups of odd characteristic. ∼ L (q), q odd,  = ±1, n ≥ 4, and let x ∈ Lemma 6.2. Suppose that K = n I2 (Aut(K)) with L := E(CK (x)) ∼ = SLn−1 (q). Then CAut(K) (L) ∼ = Zq− .

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Proof. Without loss, x ∈ H := P GLn (q) is the image of the reflection x  = diag(1, 1, . . . , 1, −1); in the unitary case we take this with respect to an orthonormal basis of the un = GLn (q) and let L  be the component of C  (x) derlying vector space. Let H H  cen ∼  of CInndiag(K) (L) in H mapping on L. Then L = L and so the preimage C  Thus C  stabilizes the commutator and fixed subspace of L  on the tralizes L. ∼  x natural module, so [C, ] = 1. So CInndiag(K) (L) = CInndiag(K) (L x ) = Zq− , by [IA , 4.5.1]. Consequently CAut(K) (L) centralizes x = Ω1 (O2 (CInndiag(K) (L))), so CAut(K) (L) = CAut(K) (L x ). It remains to show that CAut(K) (L x ) ≤ Inndiag(K). Now L contains a root subgroup of K, and a long root subgroup in case  = −1, so CAut(K) (L x ) ≤ Aut0 (K) by Lemma 6.1. Now, Aut0 (K) = Inndiag(K) γ where γ is the transverseinverse automorphism. As n − 1 ≥ 3, γ induces a nontrivial graph automorphism on L. But by [IA , 4.5.1], CInndiag(K) (x) induces Inndiag(L) on L. Therefore CAut0 (K) (L) ≤ Inndiag(K), as required.  Lemma 6.3. Let K = E7 (q) or D8 (q), q an odd prime power, with Dynkin diagram Δ. Let H0 be an A4 (q)-subgroup of K corresponding to an A4 -subdiagram of Δ, i.e., generated by four fundamental SL2 (q)-subgroups of K. Let C = CK (H0 ). Then the following conditions hold: (a) If q > 3, then C (∞) = E(C) ∼ = A2 (q) or A3 (q), according as K ∼ = E7 (q) or D8 (q); (b) If q = 3, then O 2 (C) ∼ = A2 (q) or A3 (q), respectively. Proof. With standard notation [IA , Sec. 2.10] we use a σ-setup (K, σ) of K with σ = σq . The fundamental system Π corresponds to the nodes of Δ and has a subset Π0 generating an Σ0u of Σ. Without loss we may take K to  A4 -subsystem be universal. Set H 0 = X α | α ∈ Σ0 ∼ = A4 . We may assume that H0 = CH 0 (σ). Note that H0 is determined up to K-conjugacy, independent of the A4 subdiagram on which it is based. This is because the corresponding statement is easily seen to be true for A4 subdiagram subgroups of A5 (q) ∼ = SL6 (q). As one can slide an A4 -subdiagram of Δ or its extended diagram to any other, the resulting A4 (q) subgroups are all conjugate.  For each α ∈ Σ, let Lα = X α , X −α ∼ = Au1 , Lα = CLα (σ) ∼ = SL2 (q), and ∼ zα = Z(Lα ) = Z(Lα ) = Z2 . We claim that CK (H0 ) = CK (H 0 ). Let g ∈ CK (H0 ). Then for each α ∈ Π0 , [g, zα ] = 1. Hence g acts on CK (zα ), which has a unique A1 component containing zα , namely L  α . (It is for this reason that we took K to be universal.) Therefore g normalizes Lα | α ∈ Π0 = H 0 and maps into Aut0 (H 0 ) ∩ C(H0 ). But by [IA , 7.1.3], this intersection is trivial, so g ∈ CK (H 0 ), proving the claim. Set o C = CK (H 0 ) and C 1 = CT . Then C 1 is generated by T and the T -root groups in K that commute with H 0 , i.e., corresponding to roots that are orthogonal to Σ0 . Using explicit descriptions of the root systems of type E7 and D8 , it is straightforward to o check that the root system of C 1 is of type A2 or A3 , according as K = E7 or D8 . o As for C 1 /C 1 , it is covered by N 1 = NC 1 (T ) by a Frattini argument, and is isomorphic to N 1 /N 1,0 , where N 1,0 := NC o1 (T ). Let N = NK (T ), so that ∼ W . Then W1 := N 1 /T centralizes W0 = N (T )/T . Let W1,0 = N 1,0 /T , N /T = H0

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o o ∼ W1 /W1,0 ≤ CW (W0 )/W1,0 . In both the part of W in C 1 . We see that C 1 /C 1 = cases of the lemma this is a 2-group. (For example if K = E7 , then W ∼ = Z2 ×Sp6 (2), ∼ , and W . Using [I , 4.8.2] we see that in Sp (2), the centralizer Σ Σ W0 ∼ = 5 1,0 = 3 A 6 of an element of order 5 has order 2a · 3 · 5, so |CW (W0 )|/|W1,0 | is a power of 2, as o asserted.) Thus C 1 /C 1 is a 2-group. It follows that

O 2 (CC 1 (σ)) = O 2 (CC o1 (σ)) = O 2 (CC o T (σ)). For q > 3, we deduce that CC 1 (σ)(∞) = CC o (σ)(∞) ∼ = A2 (q) or A3 (q), as claimed. For q = 3, σ induces the cubing mapping on T , so CC o T /C o (σ) is a 2-group, whence O 2 (CC 1 (σ)) = O 2 (CC o (σ)) ∼  = A2 (3) or A3 (3). The proof is complete. Lemma 6.4. Let K = Spinn (q),  = ±1, n ≥ 5, q an odd prime power. For each node α of the twisted Dynkin diagram Δ of K, let zα = Z(Kα ), where Kα is the corresponding SL2 (q)- or SL2 (q 2 )-subgroup of K. Then the following conditions hold: (a) If n is odd, or n is even but  = −1, let α be the short root node; then K/ zα ∼ = Ωn (q), i.e., zα is the “spin involution” of Z(K); moreover, pro+ vided n ≥ 9, Mα := E(CK (Kα )) ∼ = Spin+ n−3 (q) or Spinn−4 (q) according as n is odd or even, with zα ∈ Mα ; (b) If n is even and  = +, let α and β be any end nodes of Δ interchanged by some symmetry of Δ of order 2, and let z = zα zβ . Then K/ z ∼ = Ω+ n (q); + ∼ provided n ≥ 10, Mα := E(CK (Kα Kβ )) = Spinn−4 (q) with z ∈ Mα . Proof. We know that K is the universal version of Ωn (q). If n is odd, or if n is even with  = −1, then Z(K) is cyclic, see [IA , 6.1.4], and so the unique involution of Z(K) is the spin involution. In the case that n is odd, we can use Chevalley generators [IA , p. 69] to see that Kα Z(K)/Z(K) induces Ω3 (q) on a 3-dimensional subspace W of the natural K-module, and centralizes the orthogonal complement W ⊥ . Thus by [IA , 6.2.1], Z(K) = Z(Kα ) with Kα ∼ = Spin3 (q) ∼ = SL2 (q), so zα is the spin involution of K. Moreover, as Kα is irreducible on W , it follows that E(CK (Kα )) = Ω(W ⊥ ) if n ≥ 8. Note that the roots β ∈ α⊥ for which the root group Xβ commutes with Kα form a subsystem of type D(n−3)/2 , so Ω(W ⊥ ) contains and then equals a copy of Ω+ n−3 (q). Hence, (a) holds in this case. Similar arguments establish (a) and (b) when n is even; one uses Chevalley generators for Ω+ n (q) as on [IA , p. 70], and for + 2 (q) in Ω (q ) as on [IA , p. 71].  the case  = −1, the embedding of Ω− n n Lemma 6.5. Let K be a quotient of Ω± m (q) for some m ≥ 7, where q is a power of the odd prime r. Suppose that t ∈ I2 (Aut(K)) and CK (t) has a subnormal r subgroup F ∼ = Ω± k (q), where m − k = 3 or 4. Then O (CK (t)) = F M where [F, M ] = 1 and M ∼ = Ω± m−k (q). Proof. We have K = Ω± m (q)/Z with Z a 2-group, so without loss we may assume that Z = 1. From the isomorphism type of F , t does not induce a field or graph-field automorphism on K, so t ∈ Aut0 (K). Then the result follows directly  from [IA , 4.5.1, 4.5.3]. Lemma 6.6. Let n = b + c with b, c integers ≥ 4. Let K = Ω± n (q), q a power of the odd prime r, let V be the natural K-module, and let V = W ⊥ X be a

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decomposition of V with dim W = b. If b = 4, assume that W is of + type. Let  ± KW be the stabilizer in K of W , and set Y = O r (KW ) ∼ = Ω± b (q) × Ωc (q). Then CAut(K) (Y ) ∼ = Z2 . Proof. As b ≥ 4, with W of + type in case of equality, Y contains a long root group, so CAut(K) (Y ) ≤ Aut0 (K) by Lemma 6.1. Since K = Ω± n (q) (and not P Ω+ (q)), GO(V ) maps onto Aut (K). Now Y acts absolutely irreducibly on W 0 8 and on X, so CGO(V ) (Y ) = (Yb × Uc ) ∩ GO(V ) where Yb and Uc are the scalar subgroups of GO(W ) and GO(X), respectively. As any element of GO(V ) scales the bilinear form on V by a constant, CGO(V ) (Y ) is generated by scalars and an involution whose −1-eigenspace is W . The result follows.  We remark that if c is changed to 1 in the previous lemma, the same proof is valid, with X interpreted as W ⊥ (on which the trivial group is absolutely irreducible). Therefore we also have the following lemma: Lemma 6.7. Let K = Ω± n (q) = Ω(V ) with q odd and n ≥ 6. Let t ∈ O(V ) be a reflection and set F = E(CK (t)). Then t maps onto CAut(K) (F ). Lemma 6.8. Let K = Ω± n (q), n ≥ 5, q a power of the odd prime r. Let t ∈  I2 (Aut(K)) and suppose that O r (CK (t)) has a subnormal subgroup F ∼ = Ω± n−2 (q). Then (a) CAut(K) (F ) is an r  -group; and (b) Suppose, if n = 6, that F ∼ = Ω+ 4 (q). Then CAut(K) (F ) is a dihedral group. ± Proof. Since K = Ω± n (q), Aut0 (K) is a homomorphic image of GOn (q). From [IA , 4.5.1, 4.5.3], t is determined uniquely up to Inndiag(K)-conjugacy by the given isomorphism type of F , and t is the image of an involution τ ∈ On± (q). (Note that K admits no triality when n = 8.) Let V be the natural K-module; replacing τ by −τ if necessary we may assume that V = [V, τ ] ⊥ CV (τ ) with dim CV (τ ) = 2 and F supported on [V, τ ]. Since n ≥ 5, F is absolutely irreducible on [V, τ ] and so CGO(V ) (F ) induces scalars on [V, τ ]. Passing to Aut0 (K) ∼ = P GO(V ), we see that CP GO(V ) (F ) ∼ = O(CV (τ )) ∼ = O2± (q) ∼ = D2(q∓1) . Under the hypothesis of (b), F contains a root subgroup of K, and a long root subgroup if K is a Steinberg group, so by Lemma 6.1, CAut(K) (F ) ≤ Aut0 (K). Finally, if (a) fails, then CAut(K) (F ) contains a field automorphism of order r by [IA , 4.9.1], contradicting [IA , 4.2.3]. This completes the proof.  − Lemma 6.9. Let K = Ω− 6 (3) or P Ω6 (3), let t ∈ I2 (K) − Z(K), and let F = + O (CK (t)) ∼ = Ω4 (3). Then CAut(K) (F ) ∼ = D8 . 3

Proof. Continuing the proof of Lemma 6.8b, we have that CV (τ ) is 2-dimensional of − type, so CAut(K) (F ) ∼  = O2− (3) ∼ = D8 . Lemma 6.10. Suppose that q is an odd prime power and n ≥ 9. Let k = n − 1 or n − 2 according as n is odd or even. Let K = Spinn (q), with z ∈ I2 (Z(K)) and η K/ z ∼ = Spink (q) be a subgroup of K whose image in Ωn (q) = Ωn (q), and let L ∼ stabilizes and acts faithfully on a nondegenerate k-space on a natural K/Z(K)module V , with , η = ±1. Then CAut(K) (L) is cyclic or dihedral. Proof. The spin involution z ∈ Z(K) is unique as n = 8. Then Aut(K) normalizes z so Aut(K) ∼ = Aut(K/ z ). Moreover, by assumption z ∈ L. As L is

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∼ CAut(K/z) (L/ z ). Our hypothesis is that L quasisimple, therefore CAut(K) (L) = acts naturally on the nondegenerate space VL := [V, L], and trivially on VL⊥ . The result then follows from Lemmas 6.8 and 6.7.  Lemma 6.11. Let X = On (q), n ≥ 7, q odd,  = ±1, with natural module V . Let z ∈ I2 (X) and let D be an elementary abelian 2-subgroup of X. Let V0 be a D-invariant nondegenerate subspace of V on which D acts as scalars. If z has an eigenspace W on V with dim W > dim V0⊥ , then there exists x ∈ Ωn (q) such that [D, z x ] = 1. In particular this conclusion holds if dim V0 > n/2. Proof. Since z has an eigenspace of dimension at least 3, it commutes with a reflection, and also with an element of SOn (q) of nontrivial spinorial norm. Hence it suffices to find x ∈ On (q) such that [D, z x ] = 1. Since dim W > dim V0⊥ , W has a subspace isometric to V0⊥ , and so replacing z by an O(V )-conjugate, we may assume that V0⊥ < W . But then z acts as scalars on W , as does D on V0 . Thus D acts as scalars on W ⊥ . It follows that [D, z] = 1, as required. Since some eigenspace of z has cardinality at least n/2, the final statement follows from the first statement.  Lemma 6.12. Let K = Bn (q), q odd, n ≥ 3. Let D = −x, u ∈ E22 (K), where x is a reflection and dim[u, V ] = 2. Then for any t ∈ I2 (Aut(K)), there is a ∈ K such that [ta , D] = 1. Proof. If t arises from an involution of O(V ), the conclusion follows from Lemma 6.11, as D centralizes ([x, V ] + [u, V ])⊥ and dim V ≥ 7. The only other possibility is that t is a field automorphism of K, which we now assume. Our assumptions imply that either [x, V ] ⊥ [u, V ] or [x, V ] ⊆ [u, V ]. Now in CK (t) ∼ = SOn (q 1/2 ) we may choose a four-group D∗ = −x∗ , u∗ such that on the natural CK (t)-module V ∗ , x∗ is a reflection, dim[u∗ , V ∗ ] = 2 = dim[u, V ], and either [x∗ , V ∗ ] ⊥ [u∗ , V ∗ ] or [x∗ , V ∗ ] ⊆ [u∗ , V ∗ ], copying the corresponding relation between [x, V ] and [u, V ]. We may regard V ∗ as an Fq1/2 -form in V . Now the conjugacy class of any involution w ∈ K is uniquely determined by dim[w, V ], since w has trivial spinorial norm. So x and x∗ are K-conjugate, and replacing t by a conjugate we may assume that x = x∗ . Then using Witt’s Lemma in O(x⊥ ) we see that u and u∗ are conjugate in CK (x). Hence D and D∗ are K-conjugate and the result follows.  Lemma 6.13. Let K = Ωn (q) have natural module V , where n ≥ 8,  = ±1, n is even, and q is odd. Let z ∈ I2 (K) be such that [z, V ] is 4-dimensional of + type. (Note that if K = Ω+ 8 (q), this condition is independent of the natural module V that is chosen.) Then there exists a four-subgroup E ≤ K such that the following conditions hold: (a) E # ⊆ z K ; (b) dim[E, V ] = 6; (c) For any τ ∈ I2 (Aut(K)), either there is g ∈ K such that [τ, E g ] = 1, or η else E(CK (τ )) ∼ = SLn/2 (q) for some η = ±1; and (d) In any case, for any τ ∈ I2 (Aut(K)), [τ, z g ] = 1 for some g ∈ K. Proof. We first set L = Ωn/2 (q 2 ) and construct an embedding φ : L → K. Let VL be a natural Fq2 L-module. We regard VL as an n-dimensional Fq -space VLo , and define (v, w)VLo = TrFq2 /Fq ((v, w)VL ) to obtain a nondegenerate orthogonal

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space of type . (If δ and δ o are the determinants of the forms on VL and VLo , respectively, then δ o = NFq2 /Fq (δ), which implies that the forms have the same type.) We can therefore identify V with VLo , yielding the embedding φ. Note that   this construction yields the components of centralizers of involutions tm/2 and tm/2 in [IA , 4.5.1], which are ΓK -conjugate, as well as the component of the centralizer  . of the involution γ(m+1)/2 2 Within L is an Ω3 (q ) subgroup L0 centralizing a subspace of codimension 3 in the natural L-module VL such that L0 has a subgroup A ∼ = A4 permuting three orthogonal unit vectors in VL . We set E = φ(O2 (A)). Then the support VE of E on V decomposes VE = V1 ⊥ V2 ⊥ V3 into its O2 (A)-homogeneous constituents; dimFq Vi = 2 and the determinant of the form on Vi is a square, for each i. Then each involution of E is supported on the sum of two of the Vi ’s. These conditions characterize E up to K-conjugacy, and imply (a) and (b). To prove (c), let τ ∈ I2 (Aut(K)). Then one of the following holds: (1) τ arises from an involution t ∈ O(V ); η (2) τ arises from an element t∗ ∈ GO(V ) with CK (t∗ ) ∼ = GLn/2 (q); (6A) (3) τ is a field or graph-field automorphism; (4) E(CK (τ )) ∼ = Ωn/2 (q 2 ), arising from the construction in the first paragraph. We have V = V1 ⊥ V2 ⊥ V3 ⊥ V4 , where V4 = CV (E). Suppose that (6A1) holds. It suffices to show that some eigenspace Vt of t is isometric (hence K conjugate) to a subspace W of the form W = 4i=1 (W ∩ Vi ), for then the corresponding conjugate of τ centralizes E. We illustrate the argument supposing that dim Vt = 5. Take W = V1 + V2 + Fq w for suitable w ∈ V3 ; as dim V3 > 1 we can choose w with any desired nonzero value of (w, w), and we choose it so that the signature of W equals that of Vt . If (6A2) holds, then there is nothing to prove. If (6A3) holds, then  = ±1, and 1/2 Kτ := E(CK (τ )) ∼ ) acts naturally on a Fq1/2 -form Vτ ⊆ V . Constructing = Ω± n (q a four-group Eτ ≤ Kτ just as we constructed E ≤ K, we see that as a subgroup of K, Eτ satisfies the conditions that characterize E up to K-conjugacy. So τ centralizes the K-conjugate Eτ of E. If (6A4) holds, then a similar argument yields the same desired conclusion, and (c) is proved. η It remains to prove (d) when Kτ := E(CK (τ )) ∼ = SLn/2 (q). Let u ∈ I2 (Kτ ) have 2-dimensional support on the natural Kτ -module. Then t has 4-dimensional support on the natural K-module, and u ∈ [K, K] as Kτ is perfect, so u ∈ z K , as required. The proof is complete.  Lemma 6.14. Let K = SUn (q), q a power of the odd prime r, n ≥ 6, and let V be the natural Fq2 K-module. Let t be an involutory graph automorphism of K such  that Kt := O r (CK (t)) ∼ = Ω± n (q). Then as an Fq Kt -module, V is the direct sum of two natural Fq Kt -modules. Proof. By [IA , 4.5.1], t is determined up to Inndiag(K)-conjugacy by the isomorphism type of Kt . The latter is in turn determined by a symmetric bilinear form on V1 := Fnq , i.e., by a symmetric matrix M over Fq . We may equally well regard M as defining a hermitian form on V2 := Fnq2 , whose isometry group is then GUn (q 2 ), the subgroup of GLn (q 2 ) consisting of those A satisfying Ah M A = M , where Ah is the hermitian transpose of A. Let t2 be the graph automorphism of

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SU (V2 ) taking A → A(q) , obtained by raising each entry of A to the qth power. Then by the first sentence of the proof, there is an isomorphism SU (V2 ) → K  taking O r (CSU(V2 ) (t2 )) to Kt . In this way, we can view elements of V and V1 as column vectors, while the module action of K on the natural module V , as well as the module action of Kt on its natural module V1 , is given by matrix multiplication.  It follows that as Fq Kt -modules, V ∼ = V1 ⊗Fq Fq2 ∼ = V1 ⊕ V1 , as desired. The next lemma helps us survive a brush with the exotic Solomon 2-fusion systems. Lemma 6.15. Let K = Spin7 (q), where q = r a , r an odd prime. Let U ∈ Sylr (K). Then the following conditions hold: (a) There is a parabolic subgroup P of K containing U with Levi decomposition P = J(U )M and the following properties: (1) J(U ) = CU (J(U )) ∼ = Eq5 is the product of four long root subgroups and one short root subgroup of U ; (2) Z(K) ≤ [M, M ] ∼ = Spin5 (q) and J(U ) is a natural (irreducible) [M, M ]/Z(K) ∼ Ω = 5 (q)-module; and (3) Long root subgroups of K are not Aut(K)-conjugate to short root subgroups of K; (b) There is P1 ≤ K such that (1) P1 is a parabolic subgroup, U ≤ P1 < K, and P1 ≥ J1 × J2 , where J1 ∼ = J2 ∼ = SL2 (q) and Z(K) ≤ J1 ; (2) If we set Q1 = Or (P1 ), then Z(Q1 ) = Z(U ) = CQ1 (J1 J2 ) ∼ = Eq , and 6 E ; and J1 J2 is irreducible on Q1 /Z(Q1 ) ∼ = q (3) Z(J2 ) centralizes Z(U ) and inverts Q1 /Z(U ) elementwise; and (c) P1 is uniquely determined by condition (b1). Proof. Let V be a 7-dimensional orthogonal space, considered as the natural K/Z(K) ∼ = Ω7 (q)-module. Let W be a 1-dimensional singular subspace of V and let P be the stabilizer in K of W . As discussed in [IA , 3.2.3], P is a parabolic subgroup. Replacing P by a conjugate we may assume that U ≤ P . Let P = QM be its Levi decomposition. By Lemma 2.3, Q = J(U ). Now M is the centralizer in K of some 2-dimensional nondegenerate subspace of V , and " the assertions of (a2) B (q), Q = follow from [IA , 3.2.3, 6.2.1b]. Since [M, M ] ∼ = 2 α∈S XS , where S is the set of all positive roots containing α1 with a positive coefficient, and α1 is the long end-node fundamental root. One checks easily that S consists of four long roots and one short root. Thus (a1) and (a2) hold. In (a3), let X = Xα0 be the highest root subgroup of U , so that X = Z(U ). Thus NAut(K) (X) covers Out(K) by a Frattini argument. If (a3) is false, therefore, there exists g ∈ K such that X g = Xα for some short positive root α. Writing g = unv with u, v ∈ U and n ∈ N , we see that Xα = Xαnv0 = Xβv , where β, the image of α0 under n, is again a long positive root. But by the commutator formula, and the uniqueness of expression of elements of U as products of root elements, this forces α = β, which is impossible. This proves (a3). Likewise if we let P1 be a parabolic subgroup containing U and stabilizing a 2-dimensional totally singular subspace W1 of V , we see with [IA , 6.2.1b] that P1 satisfies (b1) and is the unique such subgroup of K, so that (c) holds. The remaining parts of (b) follow without difficulty from the discussion in [IA , 3.2.3]. Note that

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the Fq [J1 J2 ]-module Q1 /Z(U ) is the tensor product of absolutely irreducible J1 and J2 -modules of dimensions 3 and 2 respectively, so it is irreducible.  Lemma 6.16. Let X = Aut(K), where K = P Sp4 (q) or G2 (q), q = r a , r an  odd prime. Let y ∈ K be a 2-central involution and set J = O r (CK (y)). Then J∼ = SL2 (q) ∗ SL2 (q) and OutX (J) is abelian. Proof. The first assertion is immediate from [IA , 4.5.1]. In both cases Out(K)/ Outdiag(K) is cyclic (see [IA , 2.5.12]) so it suffices to set X0 = Inndiag(K) and argue that OutX0 (J) ≤ Z(OutX (J)). Now OutInn(K) (J) ∼ = Z2 , and this completes the proof in the case K = G2 (q) as Outdiag(K) = 1 in that case. Assuming then that K ∼ = P Sp4 (q), OutInndiag(K) (J) is a four-group generated by two central subgroups of order 2, viz., OutInn(K) (J), which interchanges the two (solvable) components of J, and the stabilizer of one of those components. This completes the proof.  In the next lemma, W ∗ is the subgroup of W described in [IA , 4.1.14a]. Lemma 6.17. Let K = D4 (q)u , q odd, with extended Dynkin diagram Δ. Let Li , i = 0, 1, 2, 3, 4 be root SL2 (q)-subgroups of K, with L0 corresponding to the center node of Δ; L1 , L2 , L3 corresponding to the other fundamental nodes of Δ; and L4 corresponding to the low root. Then the following conditions hold: (a) The subgroup ΓK ∼ = Σ3 of Aut(K) centralizes L0 and L4 and permutes L1 , L2 , L3 faithfully; (b) The subgroup W ∗ of the Weyl group acts as automorphisms of Δ and permutes L1 , L2 , L3 , L4 transitively; (c) If we let zi be the central involution of Li , then z1 , z2 , z3 , z4 all commute, and the only additional relation among them is z1 z2 z3 z4 = 1. Moreover, Z(K) = z1 z2 , z1 z3 . Proof. We use standard notation [IA , 2.10] and the Chevalley relations [IA , 2.4.8]. Using [IA , 1.15.2, 1.15.5, 2.5.8c], we identify ΓK with the isometry group of the fundamental system Π of K, which in turn is identified with Σ3 via its natural action on {α1 , α2 , α3 }. We find that each γ ∈ ΓK centralizes Xα0 and X−α0 , and satisfies γ(x±αi (t)) = x±αγ(i) (t) for all i = 1, 2, 3 and all t ∈ Fq . Thus γ centralizes Xα0 as well as the obvious diagonal subgroup of Xα1 Xα2 Xα3 ,  = ±1. Since α1 , α2 , α3 are mutually perpendicular, the elements n0 := nα0 and n1 := nα1 nα2 nα3 map to involutions of N/H that generate a Weyl group of type G2 , and we see that some element of that Weyl group conjugates X±α0 to X±α4 . Thus γ centralizes L4 , proving (a). Part (b) is immediate from [IA , 4.1.14a]. In (c), the αi , i = 1, . . . , 4, are mutually orthogonal so the corresponding Li ’s, and in particular the zi ’s, commute elementwise. The group E := z1 , z2 , z3 , z4 is then W ∗ -invariant. But ΓK W ∗ ∼ = Σ4 permutes the zi naturally, so E is a quotient of the natural Σ4 permutation module, which itself is uniserial. Moreover, since α1 , α2 , α3 are fundamental roots and K is the universal version, z1 , z2 , and z3 generate their direct product. Hence |E| = 8 or 16. One calculates that for any root group Xα , the four zi ’s, 1 ≤ i ≤ 4, have identical actions on Xα . Therefore their images in K/Z(K) are identical. As Z(K) is a four-group, |E| = 8. This immediately implies the assertions of (c). 

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Lemma 6.18. Let K = Ω± n (3), n ≥ 5, and let τ ∈ I2 (O(V )) where V is the natural module for K, with τ = −1. If neither τ nor −τ is a reflection, then τ inverts an element of I5 (K). In particular if n ≥ 6, t ∈ I2 (Aut(K)), and CK (t) has an Ω− 4 (3) component, then t inverts an element of I5 (K). Proof. The second statement is a consequence of the first; the fact that, in that case, t is induced by an involution of O(V ) can be verified from [IA , 4.5.1]. To prove the first statement, suppose first that n = 5. Let T = {t ∈ I2 (SO5 (3)) | −t is a reflection}. As there are two orbits of SO(V ) on nonsingular vectors in V , T is the union of two conjugacy classes, one in Ω(V ), one outside. But also there are four classes of involutions in SO(V ) ∼ = Inndiag(B2 (3)), two in Ω(V ) and two outside (see [IA , 4.5.1]). Hence it suffices to find involutions t1 ∈ Ω(V ) and t2 ∈ SO(V ) − Ω(V ) that invert an element of I5 (K). Both t1 and t2 can be found in a subgroup L × z ∼ = A6 × Z2 of SO(V ) with z ∈ T − Ω(V ). For the general case n ≥ 6, write the eigenspace decomposition of V under τ as V = V1 ⊥ V2 with dim V1 ≥ dim V2 . As ±τ is not a reflection, dim V2 ≥ 2, and as n ≥ 6, dim V1 ≥ 3. Let X1 and X2 be nondegenerate subspaces of V1 and  V2 , respectively, with dimensions 3 and 2, respectively. Set H = O 3 (CK (X1 ⊥ ⊥ ∼ X2 ) ) = Ω5 (3). Then one of τ and −τ acts on H as an involution of determinant 1 whose negative is not a reflection. By the previous paragraph, τ inverts an element  of I5 (H), completing the proof. Lemma 6.19. Let K = L3 (q) for some odd prime power q. Let α ∈ Aut(K) − Inn(K) be an involution. Then α inverts an element g ∈ K of odd prime order ≥ 5. Proof. By [IA , 4.5.1, 4.9.1], CK (α) ∼ = P GL2 (q), L3 (q0 ), or U3 (q0 ), the latter two possible only if q = q02 for some integer q0 and  = 1. Moreover, in every case all involutions in the coset Kα are conjugate. Let p be a primitive prime divisor of q 3 − 1 or q 6 − 1, according as  = 1 or −1. Then p > 3, a Sylow p-subgroup P of K is cyclic, CK (P ) is cyclic of odd order (q 2 + q + 1)/ gcd(3, q − ), and AutK (P ) ∼ = Z3 (see [IA , 6.5.3]). Hence by a Frattini argument we may take P to be α-invariant. Thus α inverts P unless p divides |CK (α)|, which by our choice of p can occur only if  = 1 and CK (α) ∼ = U3 (q0 ). In this last case comparison of |K| and |CK (α)| shows that α inverts an element of CK (P ) whose order is a primitive prime divisor s of q03 − 1, and hence also s ≥ 5. The proof is complete.  Lemma 6.20. Suppose K ∼ = P Sp4 (q), q odd, and x ∈ I2 (Inndiag(K)) with E := E(CK (x)) ∼ = L2 (q 2 ). Then Inndiag(E) ≤ AutAut(K) (E). Proof. Let C0 = CK (x), C1 = CInndiag(K) (x), C2 = CAut(K) (x), and C 2 = C2 /C0 . Then by [IA , 4.5.1], C1 = C0 × x and C0 = E f where f 2 = 1 and f induces a nontrivial field automorphismon E. As x is determined up to K-conjugacy by the structure of E, C 2 = x × φ ∼ = Out(K), where φ is a field automorphism corresponding to a generator of Aut(Fq ). Let X ∈ Syl2 (CAut(K) (E)). Since  x ∈ Syl2 (CInndiag(K) (E)), X ∼ = X = x × X 1 where X 1 ≤ φ . If X > x , it follows that X contains an involutory field automorphism y. Thus E embeds

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in CK (y)(∞) ∼ = P Sp4 (q 1/2 ) which contradicts Lagrange’s theorem (use a primitive prime divisor of q 4 − 1). Therefore X = x . In exactly the same way, if AutAut(K) (E) contains Inndiag(E), then there exists an involutory field automorphism z ∈ Aut(K) inducing an outer diagonal automorphism on E. As CE (z) contains a cyclic subgroup of order (q 2 + 1)/2, and |CK (z)| = |Sp4 (q 1/2 )|, we again contradict Lagrange’s theorem.  Lemma 6.21. Suppose that K ∈ Chev(r), r odd, and x ∈ I2 (Inndiag(K)) with CK (x) having a component L ∼ = SL2 (r a ) such that Z(L) ≤ Z(K). Then either a 3 a/3 ∼ q(K) = r or K = D4 (r ). Proof. Without loss, K is simple. The result then holds by examination of the table [IA , 4.5.1].  ∼ Lemma 6.22. Let K = Ω+ 6 (3) and t ∈ I2 (Aut(K)) with E := E(CK (t)) = Ω5 (3). Then NK (E) is the unique maximal subgroup of K containing E. Proof. Let V be the natural K-module. Then N := NK (E) is the stabilizer in K of a nonsingular 1-subspace W ⊆ V , while E is the stabilizer in K of a nonzero vector in W . Let Ω = K/N . Then |Ω| = 117 = 1 + 36 + 80 gives the orbit lengths of N on Ω. No subsum containing 1 divides 117, so N is maximal in K. Similarly, the E-orbits on Ψ = K/E are given by |Ψ| = 234 = 1 + 1 + 72 + 80 + 80, implying the result.  For the rest of this section we compute in groups of characteristic 2. Lemma 6.23. Let K = Ln (q), where q is a power of 2,  = ±1, and n > 3. Let L be a quasisimple subgroup of K such that L/Z(L) ∼ = Ln−1 (q). Then CAut(K) (L) has odd order. Proof. Otherwise there is an involution γ ∈ Aut(K) such that [γ, L] = 1. If γ ∈ Inndiag(K), then as q is a power of 2, γ ∈ Inn(K) is induced by an involution g ∈ K with [g, L] = 1. By the Borel-Tits theorem there is a proper parabolic subgroup M of K containing L g , with g ∈ O2 (M ). The only possibility is that  = 1 and M is an end-node parabolic subgroup. But in that case O2 (M ) is a natural L-module and contains no involutions fixed by L. Thus, γ ∈ Inndiag(K). If γ is  a field or graph-field automorphism, then  = 1 and L embeds in O 2 (CK (γ)) ∼ = 1/2 L± ). But |L|2 ≥ q (n−1)(n−2)/2 ≥ q n(n−1)/4 = |CK (γ)|2 , since n > 3, so n (q equality holds, n = 4, and L and CK (γ) share a Sylow 2-subgroup S. But then applying [IA , 3.3.1b] to S in L and CK (γ), we find that S has class 2, and also class 3, a contradiction. Thus, γ is a graph automorphism. By [IA , 4.9.2beg], CK (γ) 2 embeds in C[n/2] (q), and hence L does as well. Since |C[n/2] (q)|2 = q [n/2] , a similar comparison of Sylow 2-subgroups gives n = 4, and |L3 (q)| divides |Sp4 (q)|. As L is quasisimple, L/Z(L) ∼  U3 (2). Therefore there exists a prime divisor p > 3 of = q 2 + q + 1. Then ordp q = 3 or 6, so p does not divide |Sp4 (q)|, a contradiction. The proof is complete.  Lemma 6.24. Let K ∈ Chev(2), let p be an odd prime, and let q = ±1 be such that q ≡ q (mod p). Let x ∈ Ip (K). Suppose that CK (x) possesses a component J, and x ∈ B for some B ∈ Ep (NK (J)) such that mp (B) = mp (K). Then in the following cases, CAut(K) (JB) has odd order: 

(a) (K, J) = (E7 (q) or D7q (q), D6 (q));

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(b) (K, J) = (E6q (q) or D6 (q) or A6q (q), A5q (q)); or  (c) (K, J) = (D5q (q), D4 (q)). Proof. Suppose false and let t ∈ I2 (CAut(K) (JB)). If t ∈ Aut0 (K) then by [IA , 4.2.3], t does not map into Aut0 (J), a contradiction. Thus t ∈ Aut0 (K). In all cases of the lemma, Lemma 2.2 implies that CK (B) has odd order, indeed CK (B) = T where K is an algebraic group overlying K and T is a certain maximal torus of K, with NK (B) = NK (T ) inducing the Weyl group W (K) on B. As Outdiag(K) has odd order, we may assume that t induces a graph automorphism on K. In particular K ∼  E7 (q). = According as K = E6 , Dn (n ≥ 5), or A6 , AutKΓ (B) ∼ = Z2 × W (E6 ), W (Cn ), K or Z2 × W (A6 ). Therefore |AutAut0 (K) (B) : AutInndiag(K) (B)| = 2 = |Aut0 (K) : Inndiag(K)|, so CAut0 (K) (B) = CInndiag(K) (B), which has odd order. The proof is complete.  Lemma 6.25. Let K = Ω(V ), where V is an 8-dimensional orthogonal space − type over F2n for some n. Let V− be a 4-dimensional nondegenerate subspace V of − type, and set V+ = V−⊥ , K− = CK (V+ ) ∼ = Ω(V− ), and K+ = CK (V− ) Ω(V+ ). Then CAut(K) (K− ) ∼ = O(V+ ) ∼ = O4+ (2n ).

of of ∼ =

Proof. Let C = CAut(K) (K− ) and C0 = CO(V ) (K− ). Evidently C0 stabilizes V− = [V, K− ] and V+ = CV (K− ), so C0 ∼ = O(V+ ) and |C0 : K+ | = 2. Also as K+ contains a long root group, CC (K+ ) = CAut(K) (K− K+ ) = CO(V ) (K− K+ ) = 1, using Lemma 6.1. Thus C embeds in Aut(K+ ), and the image of C0 is Aut0 (K+ ) ∼ = O(V+ ). We wish to prove that C = C0 . On the one hand, C/K+ embeds in Out(K + ), in which the image of C0 /K+ (i.e., of Aut0 (K+ )) is a direct factor, with a cyclic complement of images of field automorphisms. On the other hand, K+ = CK (K− ), so C/K+ embeds in Out(K), which is cyclic. Thus C/K+ = C0 /K+ × Φ, where Φ is a cyclic group of odd order consisting of images of field automorphisms. We must show that Φ = 1. Otherwise, by the structure of Aut(K+ ), there is f ∈ C − K+ of odd prime order r, which is then a field automorphism of K by [IA , 4.9.1]. Now there is x ∈ CK+ (f ) ≥ O4+ (2) of order 3 such that K− is a component of CK (x). By [IA , 4.2.3], f induces a field automorphism of order r on K− . This contradicts f ∈ C and completes the proof.  Lemma 6.26. Let K = D5− (q), q even, and let H ∼ = SL3 (q) be a subgroup of K generated by two SL2 (q)-subgroups of K corresponding to fundamental roots. Then  O 2 (CK (H)) = CK (H)(∞) ∼ = L2 (q 2 ). Proof. The twisted root system of K is of type B4 , say with fundamental system {a1 −a2 , a2 −a3 , a3 −a4 , a4 }. Conjugating by an element of W we may assume that H is generated by root groups for a1 − a2 , a2 − a3 , and their negatives. The  short root L2 (q 2 ) subgroup C0 corresponding to ±a4 then lies in C := O 2 (CK (H)). If q ≡ 1 (mod 3), then since CK (Z(H)) contains HC0 , it follows from [IA , 4.8.2]   that O 2 (CK (Z(H))) = HC0 , whence C = O 2 (CHC0 (H)) = C0 , as desired. If q ≡ 1 (mod 3), let p be a primitive prime divisor of q 3 − 1 and x ∈ Ip (H). Then by [IA , 4.10.3a, 4.8.2], mp (H) = 1 and on the natural K-module V , [V, x] is 6  ∼ dimensional of + type. Thus C = O 2 (CK (H)) ≤ O 2 (CK (x)) ∼ = Ω− 4 (q) = C0 , so  again C = C0 . The proof is complete.

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Lemma 6.27. Let K = 2 D4 (q), q = 2n . Let p be a prime divisor of q + 1, and let y ∈ Ip (K) with L := E(CK (y)) ∼ = L4 (q). Then CAut(K) (L y ) ∼ = Zq+1 . Proof. We use [IA , 4.8.1, 4.8.2]. Let V be the natural Fq K-module. Then V = V0 ⊥ V1 ⊥ V2 ⊥ V3 where dim Vi = 2 for each i, V0 is of − type, and V1 , V2 , V3 are of + type. Since L ∼ = L4 (q), the only possibility is that dim CV (y) = 6, and so without loss V0 = [V, y]. Now CO(V ) (L y ) centralizes CV (y), on which L is absolutely irreducible, so CO(V ) (L y ) ∼ = CO(V0 ) (y) ∼ = Zq+1 . Clearly L contains a long root subgroup of K, so CAut(K) (L) ≤ Aut0 (K) ∼ = O(V ) by Lemma 6.1. This completes the proof.  Lemma 6.28. Let p be an odd prime, and let K and L be as in [III15 , Table 15.1]. Let y ∈ Ip (Aut(K)) with L a component of CK (y). Let C = CAut(K) (y L). Then Ip (C) ⊆ Inndiag(K), and in cases (a), (b), and (e) of the table, C is psolvable. Proof. In every case as mp (CInn(K) (L)) = 1, we have L  CK (y), and hence by [IA , 4.2.3], no element of Ip (C) induces a nontrivial field or graph-field automorphism on K. If t ∈ Ip (C) induces a graph automorphism on K, then K ∼ = D4 (q)  and L ∼ = A3q (q) as in case (j) of the table. But by [IA , 4.7.3A], t cannot then normalize L, and in particular does not centralize y, contradiction. This proves the first assertion. Finally, in cases (a), (b), and (e) (and most other cases, to be sure), CInndiag(K) (L) has p-rank 1, so by the first assertion, y is the unique subgroup of C of order p, whence C is p-solvable, as asserted.  Lemma 6.29. Let K = Sp8 (q), q = 2n , and suppose that H = H1 × H2 ≤ K with H1 ∼ = H2 ∼ = Sp4 (q). Then CAut(K) (H) = 1. Proof. Let Di ≤ Hi with Di ∼ = E32 , i = 1, 2. Then since CK (Di )(∞) ≥ ∼ = Sp4 (q) , it follows (see [IA , 4.8.2]) that the natural K-module has a decomposition V = V1 ⊥ V2 with Vi = CV (H3−i ). Then CK (H) must normalize each Vi and induce scalar mappings on it, since Hi is absolutely irreducible on Vi . But all scalar mappings in Sp(Vi ) are trivial as q is even. Thus, CK (H) = 1. As Outdiag(K) = ΓK = 1 [IA , 2.5.12], CAut0 (K) (H) = 1. Clearly H1 contains a root subgroup so CAut(K) (H) ≤ Aut0 (K) by Lemma 6.1. The proof is complete.  (∞) H3−i

Lemma 6.30. Let K = Sp6 (q), q = 2a > 2, and let B ∈ Ep∗ (K) for some prime divisor p of q − , where  = ±1. Let K ∗ , K3 , and K4 be B-invariant subgroups of K such that (a) K ∗ /Z(K ∗ ) ∼ = K4 ∼ = L2 (q); = L3 (q) and K3 ∼ ∗ (b) K4 ≤ K ; (c) K3 , K4 ∼ = Sp4 (q); and (d) Ki is a component of CK (Bi ) for some hyperplane Bi of B, i = 3, 4. Let J = E(CK ∗ (B ∩ K3 )). Then J ∼ = L2 (q) and [J, K3 ] = 1. Proof. Let V be the natural Fq K-module. With [IA , 4.8.1], there is a unique decomposition V = V1 ⊥ V2 ⊥ V3 into B-invariant nondegenerate planes, and we may write B = b1 , b2 , b3 with each bi supported on Vi , i = 1, 2, 3, and with all three bi ’s having the same eigenvalues on V . Then NK (B) contains a subgroup W ∼ = Z2  Σ3 with the base group normalizing each bi , and = AutK (B) ∼ a complement permuting {b1 , b2 , b3 } faithfully. Using [IA , 4.8.2] we see that B3

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must be W -conjugate to either b1 , b2 or b1 , bη2 b3 for some η = ±1, as must B4 . If B3 and B4 are both of the first type, then K3 K4 = K3 × K4 , while if they  are both of the second type, then up to W -conjugacy, B3 = b1 b2 , b3 and B4 = b−1 1 b2 , b3 or B4 = b1 , b2 b3 , in which cases K3 K4 = K3 × K4 (recall that q is even), or  ∼ K3 , K4 = E(CK (b1 b2 b±1 3 )) = SL3 (q), respectively. Also, if B3 = b1 , b2 and B4 = b1 b2 , b3 , then again K3 K4 = K3 × K4 . Hence, by assumption (c), up to W -conjugacy and interchanging B3 and B4 , we have that B3 = b1 , b2 and B4 = b1 , b2 b3 , and K3 , K4 = CK (V1 ). Next, consider the B-invariant subgroup K ∗ . We argue that CB (K ∗ ) = 1. Otherwise, since mp (B) = mp (K), CK (K ∗ ) is a p -group and B = B 0 × b0 , with b0 inducing a field automorphism on K ∗ and B 0 ≤ K ∗ . But then some p-element of AutK ∗ (B) acts quadratically on B, contradicting AutK (B) ∼ = Z2  Σ3 . This proves our claim. Let b ∈ CB (K ∗ )# . Then K ∗ ≤ CK (b), and with [IA , 4.8.2] we get that b =  ±1 ∗ ∗ . In particular, no W -conjugate b1 b2 b±1 3 in K . By (b), K4 ≤ K and  of K3±1lies . as B4 = b1 , b2 b3 , we must have b = b1 (b2 b3 ) ∼ Now B ∩ K3 = b3 and so J = E(CK ∗ (b3 )) = E(CK ∗ (b1 b±1 2 )) = L2 (q). In particular, J is supported on V1 ⊥ V2 , while K3 is supported on V3 . Hence, [J, K3 ] = 1, completing the proof.  The next three lemmas address the following setup from Chapter 16. G0 = d L(q) ∈ Lie(2); p is a prime divisor of q − q , q = ±1, and Op (G0 ) = 1; G0  M , M = M/Op (M ), and G0 = F ∗ (M ) ∈ Gp ;  Op (M )O p (M ) ≤ H  M and H is the full preimage in M of Inndiag(G0 ); ± (5) G0 is not isomorphic to L± n (2), n ≤ 7, Cn (2) or Dn (2), n ≤ 4, D5− (2), or F4 (2); (6) B ∈ Ep∗ (H) and mp (B) ≥ 4; and  (7) b ∈ B # , Lb = O 2 (CG0 (b)), and we write OE,p (CM (b)) for the full preimage in CM (b) of Op (CM (b)/Lb ). (1) (2) (3) (4)

(6B)

Lemma 6.31. Suppose that (6B) holds with Op (M ) = 1 and Lp (Lb ) = 1. Let B ≤ Pb ∈ Sylp (CM (b)) with Pb ≤ P ∈ Sylp (H). Then either Pb is abelian with B = Ω1 (Pb ) and 3 < mp (B) ≤ p, or Pb = P and one of the following holds:  (a) G0 ∼ = Lpq (q), with p ≥ 5, and P = A w with A abelian, Ω1 (A) = B, 3 < mp (A) ≤ p, and wp ∈ A with CA (w) = b ; or (b) p = 3, q = 2, and G0 ∼ = U8 (2), U9 (2), or 2E6 (2), with b 3-central in G0 in all cases.  In particular, p > 3, except for the cases listed in (b). Moreover, if G0 ∼ = Lnq (q) and p > 3, then n ≤ p. Proof. Suppose first that ∼ Lnq (q) with q > 2, and (n, q − )p = pm ≥ 1. (6C) G0 =  a Sylow p-subgroup of  be the full preimage of H in GLn (q), denote by B Let H  and fix a preimage b ∈ B  of b. As n > 3, and with the full preimage of B in H,

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 is diagonalizable and bp = ζI for some pm -th Lemma 2.8a applied to H × Zp , B root ζ of 1. We may assume that b = diag(η1 , . . . , ηn ) with ηip = ζ for all i. Hence, b has at most p distinct eigenvalues. On the other hand, if some eigenspace for b has dimension k, with 2 ≤ k < n, then as q ≥ 4, it follows that CG (b) has a 0  component isomorphic to SLkq (q), whence Lp (Lb ) = 1, a contradiction. Hence, the ηi are distinct, whence n ≤ p. Moreover, if n < p or b ∈ Z(P ), then we have Pb abelian of rank at most p. On the other hand, if n = p and Pb = P , then (a) holds. Thus if (6C) holds, then the lemma holds. Suppose next that G0 is a classical group with q > 2, but not as in the preceding paragraph, and let V be the natural G0 -module. We use [IA , 4.8.1, 4.8.2] without further comment. Now P acts naturally on V , and the minimal support of any element of P of order p on V is 2-dimensional. Then if dim(V ) < 2p, then P is abelian of rank less than p, and the lemma holds. On the other hand, suppose that dim(V ) ≥ 2p. If ζ = 1 is an eigenvalue for b with multiplicity k > 1, then as q > 2, Lp (CG0 (b)) has a component isomorphic to 2 SL± k (q) or SLk (q ) (as q ≡ ±1 (mod p)), contrary to assumption. Hence, the 1eigenspace of b has dimension at least 2p − (p − 1) = p + 1 ≥ 4. Again as q > 2, Lp (CG0 (b)) is nontrivial as it has a normal subgroup isomorphic to the derived group of the isometry group of this eigenspace, a contradiction. Thus, if G0 is a classical group, then we may assume that G0 = d L(2), whence p = 3. Again let V be the natural L-module. For the moment exclude the case G0 ∼ = U3m (2). Again P acts naturally on V . Let ω be a primitive cube root of unity. If ω or ω −1 has multiplicity k as an eigenvalue of b, then L3 (CG0 (b)) has a component isomorphic to SUk (2) (if k ≥ 4) or SLk (4) (if k ≥ 2 and G0 ∼ = Ln (2)), contrary to assumption. Hence ω and ω −1 occur as eigenvalues of b with multiplicity at most 3, and indeed at most 1 if G0 ∼ = Ln (2). As for the 1-eigenspace CV (b), if G0 ∼ = Ln (2) or Sp2n (2), then as L3 (CG0 (b)) = 1, dim CV (b) ≤ 2. Thus, if G0 ∼ = Ln (2) or Sp2n (2), then n ≤ 4. But this contradicts (6B5). Similarly, if G0 ∼ = Ω± 2n (2), then dim CV (b) ≤ 4, with CV (b) of Witt index 2 in case of equality. Hence n ≤ 4 or G0 ∼ = Ω− 10 (2), again contradicting (6B5). Hence we may assume G0 ∼ = Un (2). If n ≡ 0 (mod 3), then dim CV (b) ≤ 3, so n ≤ 9. But n ≥ 8 by (6B5), so n = 8. Moreover, two eigenspaces have dimension 3 and the other has dimension 2, so b is 3-central, and (b) holds. For the last classical case, assume that G0 ∼ = U3m (2), so m ≥ 3 by (6B5). Then ∼  = SU3m (2) or GU3m (2). In H = U3m (2) or P GU3m (2) and accordingly we set H  of B is elementary abelian. Let b be a preimage of b in either case the preimage B  Then no eigenspace of b can be larger than 3-dimensional, so every eigenspace B. is 3-dimensional, m = 3 and b is 3-central. This completes the proof in the classical case. Finally, suppose that G0 is an exceptional group. If p = 3, or if G0 = E8 (q) with p = 5, then as G0 ∼ = F4 (2), the data in [IA , Tables 4.7.3A,B] show that Lp (CG0 (b)) = 1, contradicting our assumption, unless G0 = 2E6 (2), in which case conclusion (b) holds. In particular we may assume that p ≥ 5, with strict inequality if G0 = E8 (q). Thus, q > 2. Moreover, using [IA , 4.10.2] we conclude that P is

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abelian of rank less than p, whence the lemma holds, except in the following cases:  (6D) G0 = E6 (q), p = 5  q − ; G0 = E7 (q), p = 5 or 7; G0 = E8 (q), p = 7. Hence it remains only to deal with these cases. By Lemma 2.34, there is X ≤ G0 such that b ∈ B ≤ NG0 (X) and X ∼ = As (q) where, according to the three possibilities for G0 in (6D), s = 5, 7, or 8. But then s+1 > p, and the first paragraph   of the proof implies that Lp (CX (b)) = 1. As Lp (CX (b)) ≤ O p (O 2 (CG0 (b)))(∞) = Lp (CG0 (b)) by [IA , 4.2.2], and Lp (CG0 (b)) = 1, we have a contradiction completing the proof.  Lemma 6.32. Suppose that (6B) holds with Op (M ) = 1. Then there exists Tb∗ ≤ CM (b) such that if we set Cb1 = Lb Tb∗ , the following conditions hold: (a) Lb is a commuting product of Lie components in Lie(2); (b) Tb∗ is an abelian group inducing inner-diagonal automorphisms on each Lie component of Lb ; (c) Cb1 ≤ H, |CH (b) : Cb1 | = 1 or p, and Cb1  CM (b); (d) B ≤ OE,p (CM (b)) ∩ Cb1 ; and (e) One of the following holds: (1) Cb1 = CH (b); or (2) G0 ∼ = Lpm (q) with p|q − , or G0 ∼ = E6 (q) with p = 3 dividing q − ;  moreover, if G0 ∼ = Lp (q), then b is p-central in H; in any case, CG0 (b) has nonabelian Sylow p-subgroups. Proof. Conclusions (a), (b), and (c) follow from [IA , 4.2.2, 4.9.3]. By hypothesis, B ≤ H and, by [IA , 4.2.2d], if (p, |Outdiag(G0 )|) = 1, then CH (b) = Cb1 , whence (d) and (e) hold. So, we may assume that either G0 ∼ = Lpm (q) with p|q −   ∼ or G0 = E6 (q) with p = 3 dividing q − . In the latter case, inspection of [IA , Table 4.7.3A] shows that either CH (b) = Cb1 or CH (b)/Lb is a 3-group, in either of which case (d) and (e) hold, or b is of type t1,6 . Note that if b were of type t1,6 or t3 , then m3 (CInndiag(G0 ) (b)) = 4 < m3 (G0 ), contradicting b ∈ B ∈ E3∗ (H). Likewise if b is of type t1,6 , as b ∈ B ∈ E3∗ (G0 ), no element of B can induce a triality on Lb ∼ = D4 (q), and so B ≤ Lb CG0 (b Lb ), with CG0 (b Lb ) abelian, proving (d); and (e) holds as Sylow 3-subgroups of Lb are nonabelian.  the full preimage Suppose next that G0 ∼ = Lpm (q) with p|q − . Then, with H    as in the of H in GLn (q) and B a Sylow p-subgroup of the preimage of B in H, 1  is diagonalizable, whence B ≤ C and again (d) holds. proof of Lemma 6.31, B b Moreover, if Cb1 = CH (b), then some p-element of CH (b) permutes eigenspaces so that CH (b) has a nonabelian Sylow p-subgroup. Last, let G0 ∼ = Lp (q). Again, as discussed in Lemma 6.31, Cb1 = CH (b) unless b is p-central and then CG0 (b) has nonabelian Sylow p-subgroups, as claimed, completing the proof in all cases.  ∼ U8 (2) or U9 (2), and with p = 3. Let z Lemma 6.33. Assume (6B) with G0 =  be a 2-central involution of G0 . Let R = O3 (M ) and G1 = RO 3 (M ). Then there exist subgroups Uz ≤ G0 and Lz such that the following conditions hold: (a) O3 (CM (z)) = R × Uz and CG1 (z) = (RUz )Lz . Moreover, either Lz ∼ = GUn−2 (2) or (with n = 9) Lz ∼ = SU7 (2); (b) Uz is extraspecial, Z(Uz ) = z , and Uz / z is a natural F22 Lz -module; (c) For any nontrivial normal subgroup Q  Lz , [Q, Uz ] = Uz ; (d) There exists B0 ∈ E33 (CG1 (z)) such that L3 (CG0 (b)) = 1 for all b ∈ B0# ;

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(e) There exist b0 , b1 ∈ B0# such that E(CL (b0 , b1 )) ∼ = U4 (2) and z

E(CLz (b0 )) ∼ = U5 (2).

Proof. Since G0 is simple, we may pass to M/R and assume that R = 1, whence F ∗ (M ) = G0 by (6B). Now z is a transvection, and by a matrix calculation, CAut(G0 ) (z) = CInndiag(G0 ) (z) g = Uz L∗z g where Uz ∼ = 21+2(n−2) , ∗ ∼ Lz = GUn−2 (2), Uz / z is a natural F22 Lz -module, and g is an involution inducing a graph automorphism on both G0 and L∗z . Thus (a) and (b) hold; note that P GU8 (2) ∼ = U8 (2) but |P GU9 (2) : U9 (2)| = 3, which accounts for the two possible structures of Lz in (a). If N is any minimal normal subgroup of Lz , then either N ∼ = Z3 acting fixed-point-freely on Uz / z , or N ∼ = SUn−2 (2) = Z(GUn−2 (2)) ∼ acting naturally on Uz / z . In either case, (c) holds. Finally, let B1 ∈ E3∗ (Lz ). As a subgroup of Lz ∼ = GUn−2 (2) or SU7 (2), B1 is diagonal with respect to an orthonormal basis of the natural module, and m3 (B1 ) ≥ 6, by (a). Let B0 be the subgroup of B1 defined by the equality of the first four diagonal entries. Thus m3 (B0 ) = m3 (B1 ) − 3 ≥ 3, and L3 (CG0 (B0 )) contains U4 (2). Similarly there exists B0∗ ∈ E2 (B1 ), defined by the equality of the first five diagonal entries, with E(CLz (B0∗ )) ∼ = U5 (2), which implies (e) and completes the proof.  Lemma 6.34. Let K = SUn (2), let V be the natural F22 K-module and let t ∈ I2 (K) with dimF22 [V, t] = r. Then CK (t) contains commuting subgroups L ∼ = SUn−2r (2) and M ∼ = SUr (2) such that [V, L] is a natural F22 L-module and [V, M ] is the direct sum of two natural F22 M -modules. Proof. Let U = [V, t], let v1 , v2 ∈ V , and let ui = [vi , t] ∈ U , i = 1, 2. Evaluating the hermitian form on V , we have (u1 , u2 ) = (v1 + v1t , v2 + v2t ) = 2(v1 , v2 ) + 2(v1 , v2t ) = 0. Thus U is totally isotropic. We may write V = V1 ⊥ V2 where V1 + U = U ⊥ , V1 and V2 are nondegenerate, and U is a maximal totally isotropic subspace of V2 . We then take L = CK (V2 ) and M ≤ CK (V1 ), and we are reduced to the case V = V2 . In particular n is even. Now P := NK (U ) is a parabolic subgroup of type A n2 −1 (4), with t ∈ O2 (P ). Let W be a maximal totally isotropic subspace of V with W ∩ U = 0. As the form defines a perfect pairing of W and U , N := [NP (W ), NP (W )] ∼ = SL n2 (4) has U and W as dual-conjugate natural modules. For v, w ∈ W , A(v, w) := (v, wt ) = (v, w + wt ) defines a nondegenerate sesquilinear form on W , since the mapping w → w + wt is onto U . Moreover, A(w, v) = (w, v t ) = (wt , v) = A(v, w)2 , so A is hermitian. Now NP (W ) induces GL n2 (4) on W . Suppose that x ∈ NP (W ) and x preserves A. Then for all v, w ∈ W , (wx , v tx ) = (w, v t ) = A(w, v) = A(wx , v x ) = (wx , v xt ). As v and w are arbitrary, this implies that xt = tx on W ; obviously they commute on U as t|U = idU . Hence CN (t) contains SU (W ; A) ∼ = SU n2 (2), acting naturally on both W and U . The lemma is proved.  Lemma 6.35. Let K ∈ Chev(2) be of type A or E, with untwisted root system Σ of rank at least 3 and having lowest root α∗ with respect to a fundamental system Π in Σ. Let L and M be subsystem subgroups for {±α∗ } and Σ ∩ α∗⊥ , respectively. Let p be a prime divisor of q(K)2 − 1 and x ∈ Ip (L). Then (a) M  CK (x); and

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(b) If y ∈ Ip (Aut(K)) and CK (y) has a component centrally isomorphic to M , then the image of x in Aut(K) lies in y Inn(K) . Proof. Notice first that if K/Z(K) ∼ = Ln (q),  = ±1, then replacing x and L by conjugates we may assume that L is the upper block-diagonal SL2 (q), which has a unique class of elements of order p, and M is the lower block-diagonal SLn−2 (q). Then the assertions of the lemma are elementary (see [IA , 4.8.2, 4.8.4, 4.9.1]). So we can assume that K has type E. Let (K, σ) be a σ-setup for K. Let (B, T ) be a σ-invariant Borel-torus pair, identify Σ with the set of T -roots, and let L and M be subsystem subgroups corresponding to {±α∗ } and Σ ∩ α∗⊥ . Thus (L, σ|L ) and (M , σ|M ) are σ-setups for L and M , respectively. We have x ∈ L, and to prove (a) we show that M  CK (x). Replacing x by an L-conjugate we may assume that x ∈ T . Then to prove (a), we need only show that for any root α ∈ Σ, α(x) = 1 if and only if α ∈ α∗⊥ . But there is a unique β ∈ Π joined to α∗ in the extended Dynkin diagram, and its coefficient in the Π-expression for α∗ is −2. Hence for any α ∈ Σ, α ≡ kβ (mod α∗⊥ ), where k ∈ {0, −1, −2}. As [x, M ] = 1 and x has order p, β(x) is a primitive pth root of unity. Then α(x) = β(x)k = 1 if and only if k = 0, i.e., α ∈ Σ ∩ α∗⊥ , as desired. Hence, (a) holds. Note that according as K ∼ = E6 (q),  = ±1, E7 (q), or E8 (q), we  ∼ have M = A5 (q) (universal if and only if K is universal), D6 (q), or E7 (q). Using [IA , 4.10.3a, 4.7.3A] we see that mp (CInndiag(K) (M )) = 1. In proving (b), we may assume that K is of adjoint type. Let y ∈ Ip (Aut(K)) have a component J  E(CK (y)) with J/Z(J) ∼ = M/Z(M ). From the structure of J, y is obviously neither a field nor a graph-field automorphism of K, so y ∈  O p (Aut0 (K)) ≤ Inndiag(K). Hence we may regard y ∈ K, indeed up to conjugacy we may take y ∈ T . Let Σy be the the largest component of the set of roots in Σ annihilating y. As J/Z(J) ∼ = M/Z(M ), Σy is isometric to Σ ∩ α∗⊥ . Note also that RΣy ∩ Σ is a root subsystem of Σ of corank 1 containing Σy , so it equals Σy . (The root system of D5 does not contain that of A5 , nor does E6 contain D6 ; the Weyl groups do not embed.) Therefore by Lemma 1.1, Σy is NK (T )-conjugate to Σ∩α∗⊥ . It follows that J is Inndiag(K)-conjugate to M . As mp (CInndiag(K) (M )) = 1 it follows that y is conjugate to the image of x in Inndiag(K), and then in Inn(K), by [IA , 4.2.2j]. The lemma is proved.  Lemma 6.36. Let K = D4 (q), q = 2n > 2. Let r and p be prime divisors of q − 1 and q + 1, respectively. Let h ∈ Ir (K) and D = x, y ∈ Ep2 (CK (h)), and put J = E(CK (h)). Suppose that Lp (CJ (x)) ∼ = Lp (CJ (y)) ∼ = Ω− 4 (q), with + ∼ x ∈ Lp (CJ (y)) and y ∈ Lp (CJ (x)). Then J = Ω6 (q). Proof. Let V be a natural orthogonal module for K. Then V = [V, h]⊕CV (h) with [V, h] an orthogonal space of even dimension 2k > 0 and, since r divides q − 1, of maximal Witt index. Let K0 = CK (CV (h)). Then, any component of E(CK0 (h)) is isomorphic to SLk (q) for some j, 2 ≤ j ≤ k. Moreover, either CV (h) = {0} or CV (h) is an orthogonal space of maximal Witt index and dimension 8 − 2k. Let K1 = CK ([V, h]). Then E(CK1 (h)) ∼ = Ω+ 8−2k (q) = 1 if k > 2. ∼ ∼ Suppose that k = 2. Then E(CK1 (h)) ∼ = Ω+ 4 (q) = L2 (q)×L2 (q). As E(CJ (y)) = − 2 Ω4 (q) ∼ = L2 (q ), [E(CJ (y)), E(CK1 (h))] = 1. Since x ∈ E(CJ (y)), this contradicts the structure of E(CJ (x)). Hence, k = 2. If k = 1, then E(CK0 (h)) = 1 and J = E(CK1 (h)) ∼ = Ω+ 6 (q), as claimed. Thus we may assume that k = 3 or 4,

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whence E(CK1 (h)) = 1 and J = E(CK0 (h)) ∼ = SL3 (q) or SL4 (q). Given the structure of Lp (CJ (x)), we must have J ∼ = SL4 (q) ∼ = Ω+ 6 (q), as claimed, completing the proof.  Lemma 6.37. Let K = D4 (q), q = 2n > 2. Let r be a prime divisor of q − 1, and let a4 ∈ Ir (K) be such that E(CK (a4 )) ∼ = L4 (q). Then there exist a1 , a2 , a3 ∈ Ir (E(CK (a4 ))) such that if we put A0 = a1 , a2 , a3 , a4 , then A0 is elementary abelian and NK (A0 ) contains an element cycling {a1 , a2 , a3 , a4 } by conjugation. Proof. By Lemma 2.24, there is a natural K-module V such that dim(CV (a4 )) = 6 with E(CK (a4 )) supported on U := CV (a4 ). Moreover, if U4 := [V, a4 ], then U = U1 ⊥ U2 ⊥ U3 with Ui isometric to U4 for all i. In particular, we may choose ai ∈ Ω(Ui ) ≤ E(CK (a4 )) with ai of order r, so that A0 := a1 , a2 , a3 , a4 is elementary abelian. By Lemma 2.40, there is N ≤ K with N∼ = Σ4 permuting {U1 , U2 , U3 , U4 } naturally; and indeed, we may choose a 4-cycle in N cycling {a1 , a2 , a3 , a4 }, after a suitable change in the choice of ai in ai , for i = 1, 2, 3. The proof is complete.  Lemma 6.38. Let K = D4 (2) and let V be an 8-dimensional orthogonal F2 space which is a natural F2 K-module. Let X be the set of all totally singular 4dimensional subspaces of V . Let X ∈ X, set P = NK (X), and let P = O2 (P )L be a Levi decomposition of P . Then L normalizes some XL ∈ X such that X ∩XL = 0. Moreover, if M ≤ P with M ∼ = L(∼ = L4 (2)), and M normalizes some XM ∈ X with XM ∩ X = 0, then M is P -conjugate to L. Proof. In standard notation for K, we may take L = Xα | α ∈ Σ0 for some A3 -subsystem Σ0 of the root system Σ of K. Note that NO(V ) (P ) ≤ Ω(V ). Let n ∈ N map to −1 ∈ W . Then Ln = L but n ∈ P = NK (P ). Thus we can take XL = X n . Since L ∼ = L4 (2) is irreducible on X, X ∩ XL = 0. Conversely, let X and XM be as in the lemma. Then the bilinear form on V gives a perfect pairing between X and XM . Hence there exist bases {b1 , b2 , b3 , b4 } of X and {b5 , b6 , b7 , b8 } of XM such that (bi , bj ) = 1 if i + j = 9, and (bi , bj ) = 0 otherwise. Together with the fact that the bi are singular, these equations determine the quadratic form on V . But one can similarly find bases {a1 , a2 , a3 , a4 } and {a5 , a6 , a7 , a8 } of X and XL , respectively, satisfying analogous conditions, and then there is g ∈ O(V ) such that agi = bi for each i = 1, . . . , 8. Then g ∈ NO(V ) (X) = NΩ(V ) (X) = P and XM = XLg , completing the proof.  Lemma 6.39. Let K = U6 (q), q = 2n , and let P be a parabolic subgroup of K with E(P/O2 (P )) ∼ = A2 (q 2 ). Then O2 (P ) is abelian and the following conditions hold: (a) If X ≤ P and X covers E(P/O2 (P )), then either E(X) ∼ = A2 (q 2 ) or O2 (P ) ≤ X; and (b) Let x ∈ Ip (NAut(K) (P )), where p is a prime dividing q + 1, and set Px = Lp (CP (x)). If Px = 1, then [Px , CO2 (P ) (x)] = 1. Proof. The subgroup P is the stabilizer of maximal totally isotropic subspace of the natural Fq2 K-module. A matrix calculation shows that O2 (P ) is elementary abelian and, as a module for E := E(P/O2 (P )), O2 (P ) is an Fq -form of the tensor product U := V ⊗Fq2 V (q) , where V is the natural Fq2 E-module and V (q) its twist under the involution in Gal(Fq2 /Fq ). By the Steinberg tensor product theorem,

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E acts irreducibly on O2 (P ). Hence if O2 (P ) ≤ X, then O2 (P ) ∩ X = 1 and X embeds in P/O2 (P ), proving (a). Let x be as in (b). It is clearly enough to show that Lp (CP (x)), if nontrivial,  acts nontrivially on CU (x). Now, O p (NInndiag(K) (O2 (P ))) ∼ = L3 (q 2 ) or (if p = 3) 2 P GL3 (q ). Hence if x ∈ Inndiag(K), then x acts diagonalizably on E, say as diag(α1 , α2 , α3 ), where αip = 1 for i = 1, 2, 3. As p divides q + 1, the eigenvalues of x on O2 (P ) are the nine products αi αjq = αi αj−1 . If Lp (CE (x)) = 1, then at least two αi are equal, say α1 = α2 , and the α1 -eigenspace V1 of x on V is at least (q) 2-dimensional and is a natural module for Lp (CE (x)). Then CU (x) ⊇ V1 ⊗ V1 , on which Lp (CE (x)) ∼ = SL(V1 ) acts nontrivially, even irreducibly, by the Steinberg tensor product theorem. Thus (b) holds in this case. If x ∈ Inndiag(K), then x is a field automorphism of order p, arising, as p is odd, from an element of Ip (Gal(Fq2 /F2 )), and CK (x) = P SU (V1 ) or P GU (V1 ), where V1 is an Fq2/p -form in V . Then it is clear (for instance by examining a matrix representation of P ) that P1 := CP (x) is the stabilizer of a maximal totally isotropic subspace of V1 , and P1 is a parabolic subgroup of E(CK (x)) with O2 (P1 ) ≤ O2 (P ). Thus CO2 (P ) (x) = O2 (P1 ) and E(P1 /O2 (P1 )), which can be identified with E(CE (x)), acts nontrivially on O2 (P1 ) by the Borel-Tits theorem. The proof is complete.  Lemma 6.40. Let K = D5 (2) and S ∈ Syl2 (K). Then S has exactly three subgroups isomorphic to E210 . They are normal and self-centralizing in S, and two of them are conjugate in NAut(K) (S). Moreover, S splits over each of them, with complements isomorphic to Sylow 2-subgroups of L5 (2). Finally the product of any two of the three E210 -subgroups equals J(S). Proof. Let P1 and P2 be the two L5 (2)-type parabolic subgroups of K containing S. Thus for i = 1, 2, Pi is a semidirect product Ai Li with E210 ∼ = Ai  Pi and Li ∼ = L5 (2). A straightforward matrix calculation shows that Ai is isomorphic to the additive group of skew 5 × 5 matrices M over F2 , and A ∈ Li acts via M → AM AT . Thus, Ai ∼ =Li Vi ∧ Vi , where Vi is a natural F2 Li -module. By [IA , 3.3.1], m2 (S) = 10 = m2 (A1 ), so Ai ∈ A(S), i = 1, 2, and thus A1 A2 ≤ J(S). We claim first that J(S) = A1 A2 . For this, we take the positive root system corresponding to S to be all ai ± aj , 1 ≤ i < j ≤ 5, where {a1 , . . . , a5 } is an orthonormal basis in Euclidean space R5 . We may then assume that A1 is generated by all root groups Xai +aj , 1 ≤ i < j ≤ 5, and A2 is the image of A1 under the graph automorphism of K normalizing S and corresponding to the orthogonal reflection a5 → −a5 of R5 . Thus A1 ∩ A2 is the product of the root groups Xai +aj , 1 ≤ i < j ≤ 4. We define a linear functional λ on R5 by choosing N >> 0 and minuscule positive numbers i , 1 ≤ i ≤ 5, which are linearly independent over Q. Then we set λ(ai ) = N − i + i , 1 ≤ i ≤ 4, set λ(a5 ) = N − 8 + 5 , and extend linearly, and order Z5 by stipulating that v > w if and only if λ(v − w) > 0. Then the roots whose corresponding root groups generate A1 A2 , namely, all ai + aj and ai − a5 , are larger than all other roots. As the only abelian subsets of roots of maximum cardinality (10) consist of such roots, it follows by Mal’cev’s argument (see [IA , 3.3.4] and its proof) that any element of A(S) lies in A1 A2 , so J(S) = A1 A2 , as claimed. Identify A1 with V1 ∧V1 , and let V1 have basis {v1 , . . . , v5 }. Then R := A2 ∩L1 ∼ = E24 and P := NL1 (R) is the stabilizer in L1 of a hyperplane X of V1 ; we may assume

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that X is spanned by v1 , . . . , v4 . (If P were, instead, the stabilizer of a vector in V1 , then |CV1 ∧V1 (R)| = 24 , whereas A1 ∩ A2 ≤ CA1 (R) with |A1 ∩ A2 | = 26 .) Moreover, R = O2 (P ) and A2 = (A1 ∩ A2 )R. Now A1 ∩ A2 is identified with X ∧ X ⊆ V1 ∧ V1 . We may label the elements of R as rv , v ∈ X, where rv maps v5 to v5 + v. Thus for v, w ∈ X, (v5 ∧ w)rv = v5 ∧ w + v ∧ w. In particular, for v ∈ X # , CV1 ∧V1 (rv ) is spanned by X ∧ X and v5 ∧ v. Now let A ∈ A(S) − {A1 , A2 }. Since A = A1 , the previous sentence implies that A ∩ A1 = A1 ∩ A2 , so AA1 = A1 A2 . For each w ∈ X # , write tw for the element of A1 corresponding to v5 ∧ w. It follows that A contains an element of the form qw = rw tcww , where cw = 0 or 1. Note that for w, w ∈ X # , [tw , rw ] =: qw,w is the element of A1 ∩ A2 corresponding to w ∧ w . Hence as A is abelian, and A1 A2 has class two, cw +cw . 1 = [qw , qw ] = [tw , rw ]cw [rw , tw ]cw = qw,w  Therefore cw +cw ≡ 0 (mod 2) for all w = w ∈ X # , whence c := cw is independent of w. If c = 0, then R ≤ A so A = A2 , contradiction. Thus c = 1, and A is uniquely determined and exists. Moreover, A ∩ Ai = A1 ∩ A2 for i = 1, 2, so AAi = J(S) for i = 1, 2. Note that CS (A) ≤ CS (A1 ∩ A2 ) = A1 A2 , and the irreducible action of NL1 (A1 A2 ) ∼ = E24 L4 (2) on A1 A2 /A implies that CS (A) = A. Finally, S ∩ L1 is a complement to A in S. Namely, it has order |S : A1 | = |S : A|, and S ∩ L1 ∩ A ≤ S ∩ L1 ∩ A1 A2 . In S := S/A1 ∩ A2 , S ∩ L1 ∩ A1 A2 = A2 , which is disjoint from A. Hence S ∩ L1 ∩ A ≤ L1 ∩ A1 = 1. This completes the proof.  7. Outer Automorphisms, Covering Groups, and Envelopes Lemma 7.1. Let K ∈ Chev(s) for some prime s. Then (a) Out(K)/ Outdiag(K) is abelian unless K ∼ = D4 (q) for some q; and (b) If s = 2, then [Out(K), Out(K)] is cyclic. Proof. This follows directly from [IA , 2.5.12]: Out(K)/ Outdiag(K) is the product of ΦK and ΓK , which commute with each other elementwise; ΦK is cyclic, and ΓK is abelian unless K ∼ = Σ3 . Finally if = D4 (q) for some q, in which case ΓK ∼ s = 2, then Outdiag(K) is cyclic, and in fact if K ∼ = D4 (2n ), then Outdiag(K) = 1. The lemma follows.  ∼ L2 (q), Bn (q), or Cn (q), q odd, n ≥ 2. Lemma 7.2. Suppose that F ∗ (X) = K = Then the following conditions hold: (a) Out(K) = Outdiag(K) × Φ, where Outdiag(K) ∼ = Aut(Fq ) is = Z2 and Φ ∼ the image of a (cyclic) group of field automorphisms of K; and (b) Every involution in Aut(K) − Inn(K) maps into Outdiag(K) or Φ. Proof. Part (a) is a direct consequence of [IA , 2.5.12]. In (b), we may assume that Φ has even order, so Φ contains the image of a field automorphism f of order 2. By [IA , 4.9.1], all involutions in Inndiag(K)f are field automorphisms and hence Inndiag(K)-conjugate to f . But since Out(K) is abelian, all such involutions have image in Φ, as asserted.  Lemma 7.3. There is no group X such that F ∗ (X) ∼ = SL2 (q 2 ) for some odd ∗ ∗ prime power q and OutX (F (X)) ≥ Ω1 (O2 (Out(F (X)))). Proof. See [IVK , 1.9].

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Lemma 7.4. Let K = D2n (q)hs , where q is an odd prime power and n ≥ 3. Then Out(K) is abelian and contains the image of no graph automorphism of order 2.  ∼   = D2n (q)u . Then Z(K) for some 1 = Z ≤ Proof. Let K = E22 and K ∼ = K/Z   Since n > 2 by assumption, there is a unique spin involution zspin ∈ Z(K), Z(K). + ∼   i.e., such that K/ zspin = Ω2n (q), and Out(K) is the stabilizer of Z in Out(K). By  = Outdiag(K)Φ   Γ  with Outdiag(K)Φ   centralizing [IA , 6.3.1a, 2.5.12], Out(K) K K K   and Γ  ∼ Z(K), K = Z2 interchanging the two involutions of Z(K) − zspin . Hence   , which is abelian. The proof is complete. Out(K) ∼  = Outdiag(K)Φ K Lemma 7.5. Le K = Dn (q), q odd, n ≥ 5,  = ±1, with Z(K) cyclic and K not a homomorphic image of Ω2n (q). Let X = K t with t2 = 1 and t acting as a reflection on K/Z(K). Write z = Ω1 (Z(T )). Then tz ∈ tK .  be the spin involution. Thus,  = Spin2n (q) and let z ∈ K Proof. Let K  = 4, then as t induces a graph automorphism on K,  z maps to z. If |Z(K)| ∼    [Z(K), t] = z and the result follows. So assume that Z(K) =  z , whence K = K.  (q). There is y ∈ K such that in X, t and ty are commuting Let X = X/Z(K) ≤ O2n reflections. By [IA , 6.2.1], tty has order 4. But tty has order 2 so t, ty ∼ = D8 with center z . The lemma follows.  Lemma 7.6. Let K = Sp2n (q), q odd. Relative to a fixed root system Σ of type Cn , let K1 , . . . , Kn be the long root SL2 (q) subgroups of K, and write Z(Ki ) = zi . Then Z(K) = z1 · · · zn . Proof. We may write Σ = {±ai ± aj , ±2ai | 1 ≤ i, j ≤ n, i = j} and take Ki = X2ai , X−2ai . Set z = z1 · · · zn . Then zi = h2ai (−1), i = 1, . . . , n, and zi inverts the root subgroups X±ai ±aj , j = i, and centralizes all other root subgroups. As a result z centralizes all root subgroups and so lies in Z(K). But Z(K) ∼ = Z2 . It remains to show that z = 1. Using the Chevalley relations we see that z = ha1 −a2 (−1)ha3 −a4 (−1) · · · han−1 −an (−1) or ha1 −a2 (−1)ha3 −a4 (−1) · · · han−2 −an−1 (−1)h2an (−1) according as n is even or odd. As K is the universal version, hα1 (−1), . . . , hαn (−1) are independent for a fundamental system {α1 , . . . , αn }, and so z = 1. The proof is complete.  Lemma 7.7. Suppose that F ∗ (X) = E(X) = K ∗ K t , where t ∈ I2 (X), K ∼ = SL2 (q) for some odd q > 3, and Z(K) = Z(K t ). Let x ∈ I2 (X) induce a nontrivial field automorphism on K and centralize t. Then the following conditions hold: (a) There are at most two KK t -classes of involutions in the coset KK t x; and (b) Any involution of CX (K) − Z(K) induces a nontrivial field automorphism on K t . Proof. By [III11 , 1.17], any involution in KK t x is sheared to Z(K) in K. Using [IA , 4.9.1d] we see that in KK t x /Z(K), there are four conjugacy classes of involutions outside KK t /Z(K), represented by the images of x, xy, xy t , and

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xyy t for a suitable y ∈ K. Not all of these pull back to involutions of KK t x ; for example xg and xg t do not, where g ∈ CK (x) has order 4. This implies (a). For (b), let w ∈ I2 (CX (K)) − Z(K); we may assume X = KK t x, w, t . We have F ∗ (CX (K)) = K t , so if w induces an inner automorphism on K t , then w ∈ I2 (K t ) ⊆ Z(K), contrary to assumption. Thus we may assume, by Lemma 7.2, that w induces an outer diagonal automorphism on K t . Set X0 = NX (K). As |X : X0 | = 2, X0  X and we may replace w by a conjugate andassume that x, w, t is a 2-group. Set X = X/K1 K2 . As Out(K) is abelian, X = w, t × x ∼ = D8 × Z2 . and similarly for K t . Hence, putting C = We have CInndiag(K) (x)   ≤ Inn(K) CX (x), we see that C ∩ w, wt

= 1, so that |X : C| ≥ 4. As x  X, and the

t

coset xKK contains at most two conjugacy classes of involutions by (a), it follows  that |X : C| ≤ 2, a contradiction completing the proof. Lemma 7.8. Suppose that K ∈ Chev(2), p is an odd prime, mp (K) ≥ 3, and K/Op (K) ∈ Gp . Then either K ∈ Lie(2) or p = 3 with K/O2 (K) ∼ = F4 (2) or E6− (2). 1

Proof. Note that the nonsimple groups B2 (2), G2 (2), 2F4 (2 2 ) all have odd rank 2, so they do not arise here. Suppose that K ∈ Lie and let K = K/O2 (K). Since K/Z(K) ∼ = A6 , it follows from [IA , 6.1.4] that if the lemma fails, then K ∼ = 3 2 2 D4 (2), U6 (2), C3 (2), U4 (2), L4 (2), G2 (4), L3 (4), or B2 (2 ). As mp (K) ≥ 3, p = 3 in every case, and the last four of these are not possible [IA , 4.10.3a]. The previous  four lie in C3 and so also are not possible [I2 , 12.1]. The lemma follows. 8. p-Structure of Quasisimple K-groups  Lemma 8.1. Suppose that K ∼ = E6q (q)a with q = 2n ≡ q (mod 3), q = ±1, but q ≡ q (mod 9). Then m3 (K) = 5.

Proof. Our assumptions yield |K|3 = 39 . By [IA , 4.10.3a], m3 (K) ≥ 5. Let B ∈ E3∗ (K). By Lemma 2.2, NK (B) is an extension of CK (B) by W (E6 ). As |W (E6 )|3 = 34 , m3 (K) = m3 (B) ≤ 5, completing the proof.  Lemma 8.2. Suppose that K ∈ Spor, P ∈ Sylp (K), p > 3, mp (P ) > 1, and P is abelian. Then |P | = p2 . Proof. This is easily checked from the tables [IA , 5.3].



Lemma 8.3. Let K ∈ Chev(2) be simple and K ≤ H ≤ Aut(K). Let p ≥ 5 be a prime and let P ∈ Sylp (H). Then P ∼ = p1+2 . ∼ p1+2 . Write K = d L(q). Notice that as p ≥ 5 and Proof. Suppose that P =  q mp (K) ≤ 2, K ∼ = Lkp (q), q ≡ q (mod p), for any k or q = 2n , so Z(K), the Schur multiplier of K [IA , 6.1.4], and Outdiag(K) are p -groups. We may assume that K is simple and identify K with Inn(K). If P ≤ K, then P = (P ∩K) x , and again as p ≥ 5, x is a field automorphism of order p. Hence mp (K) = 2 > 1 = mp (CK (x)). 1 But q p ≡ q (mod p), so by [IA , 4.10.3a], mp (K) = mp (CK (x)), contradiction. Therefore, P ≤ K. Let C = CK (Z(P )). Since the Schur multiplier of K is a  p -group, it follows from [IA , 4.2.2] that C = O 2 (C)T with T abelian. Thus

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Z(P ) ≤ O 2 (C), so O 2 (C), as the product of groups in Chev(2), must have a q (q). Again this leads to a contradiction component J with J/Z(J) of the form Lkp as p ≥ 5 and mp (K) ≤ 2. The proof is complete.  Lemma 8.4. Let K = L3 (4) and P ∈ Syl3 (K). Then NK (P ) ∼ = U3 (2). Proof. Clearly U3 (2) embeds in L3 (4), and we compute that |CK (x)| = 32 for each x ∈ I3 (K). Hence by Sylow’s theorem, NK (P ) = P Q with Q8 ≤ Q and |Q|/8 ≡ 1 (mod 3). If |Q| > 8, then Q does not embed in GL2 (3) so CK (P ) = 1, a contradiction. The lemma follows.  Lemma 8.5. Let G0 ∈ Chev(2), q, p, and q be as in [III15 , Table 15.1], with Z(G0 ) = 1. Let G0 ≤ X ≤ Inndiag(G0 ). Let D ≤ Inndiag(G0 ) with D ∼ = Ep2 , CG0 (D) having a component L as in the table, and mp (CX (D)) ≥ mp (B) as given in the table. Then the following conditions hold: (a) mp (CX (D)) = mp (X); and (b) If b is a field automorphism of G0 of order p, then mp (CX (b)) = mp (X). 

Proof. Let H = O p (CG0 (b)). Suppose first that p does not divide |Outdiag(G0 )|. Then since q 1/p ≡ q (mod p), it follows from [IA , 4.10.3a] (using the notation of that lemma) that mp (H) = nm0 = mp (G0 ), as desired. Here m0 = ordp q = ordp q 1/p . Indeed that same lemma implies that if p divides |Outdiag(G0 )| and (b) fails, then mp (H) = nm0 − 1 and mp (X) = nm0 . We have   G0 ∼ = E6q (q) and p = 3, or G0 ∼ = Lmq (q) and p divides m, with q ≡ q = ±1 (mod p) in both cases. Since q is a pth power, q ≡ q (mod p2 ). In particular in the E6 cases, we consider a σ-setup (K, τ ) for CG0 (b), together with a suitable τ -invariant maximal torus T on which τ acts as the q q 1/3 power mapping. Then (K, τ 3 ) is a σsetup for G0 , on which τ induces the same field automorphism as some generator of b . Then CK (τ 3 ) ∼ = Inndiag(G0 ), while CT (τ ) and CT (τ 3 ) have homocyclic abelian Sylow 3-subgroups A and A∗ of rank 6 and of exponent (q 1/3 − q )3 and (q − q )3 , respectively. As |Outdiag(G0 )| = 3, we see that Ω1 (A) = Ω1 (A∗ ) ≤ Φ(A∗ ) ≤ G0 .  Hence m3 (CG0 (b)) = 6, as required. Finally in the cases G0 ∼ = Lmq (q), let H be the exponent-p subgroup of the full diagonal subgroup of X. Then H is obviously centralized by a conjugate of b, and contains a p-subgroup of X of maximal rank, so conclusion (b) holds in this case.  To prove (a), suppose first that G0 ∼ = Lmq (q). If D is diagonalizable, then the result is clear. If D is not diagonalizable, then D is the image of a nonabelian  q (q), p-subgroup of GLmq (q) centralized by L, so p divides m, L is involved in SLm/p and mp (L) ≤ (m/p) − 1. Thus (m/p) − 1 ≥ m − 4, which is impossible. So assume G0 does not have this form. In many cases either mp (L) = mp (Inndiag(G0 )) − 2 and p does not divide ||Outdiag(L)||, or p is a good prime for G0 , both of which imply the conclusion of (a) (see [IA , 4.10.3f]). So we may assume that these conditions both fail. The only remaining case is G0 = E7 (q), p = 3. But then by assumption, m3 (CX (D)) ≥  m3 (B) = 7 = m3 (X), completing the proof. Lemma 8.6. Let p be an odd prime, and let K ∈ Chev with mp (K) > 1. Let α ∈ Ip (Aut(K)) − Inndiag(K) with mp (CK (α)) = 1. Then K ∈ Chev(p) − Chev(2). Proof. If α is not a field automorphism, then p = 3 and α is a graph automorphism or graph-field automorphism. Hence K ∼ = D4 (q) or 3D4 (q). If

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q = 3n , then the desired conclusion holds, while otherwise, by [IA , 4.7.3A, 4.9.1],  CK (α) ∼ = 3D4 (q), G2 (q), or L3q (q), and in all cases m3 (CK (α)) = 2 by [IA , 4.10.3a], contradiction. Thus we may assume that α is a field automorphism. Since mp (CK (α)) = 1, the Schur multiplier of CK (α) is a p -group. This easily implies the same for K, 1 in view of [IA , 6.1.4]. Hence by [IA , 4.10.3a] and the fact that q p ≡ q (mod p), mp (K) = mp (CK (α)) = 1, a contradiction. The lemma is proved.  In the next three lemmas, q is a power of the odd prime r, X is a group with F ∗ (X) = K ∈ Chev of (8A) level q, and T ∈ Syl2 (X). We use the conventional q = ±1 with q ≡ q (mod 4). We shall use, often without reference, the material in [IA , Section 4.10]. In particular, in all cases considered, K contains the set S(T ∩ K) of all fundamental (long root) SL2 (q)subgroups J such that J ∩ T ∈ Syl2 (J). The elements of S(T ∩ K) commute elementwise and their product is called S(T ∩ K). Except for the case K = G2 (3n ), which has an automorphism interchanging long and short root SL2 (q)’s, S(T ∩ K) and S(T ∩ K) are invariant under NAut(K) (T ∩ K). In any case they are invariant under CAut(K) (T ∩ K). Lemma 8.7. Assume (8A). Then 



O 2 (AutCX (T ∩K) (K)) ≤ O 2 (CAut(K) (T ∩ K)) ≤ CAut0 (K) (S(T ∩ K)). Proof. The first inclusion is obvious. Let T0 = T ∩ K and choose any J ∈ S(T0 ). Thus J contains a long root subgroup U . Also J is the unique element of S(T0 ) containing T ∩ J, since distinct elements of S(T0 ) commute elementwise but T ∩ J is not abelian. Thus CAut(K) (T0 ) normalizes J. By [III11 , 1.17], CAut(J) (T ∩  J) has odd order, so O 2 (CAut(K) (T0 )) ≤ CAut(K) (J) ≤ CAut(K) (U ) ≤ Aut0 (K),  the last inclusion by Lemma 6.1. As J was arbitrary in S(T0 ), O 2 (CAut(K) (T0 )) centralizes S(T ∩ K). The proof is complete.  In the next lemma, by “classical involution” in K ∈ Chev(r), r odd, is meant an involution z in a long root SL2 (q)-subgroup of K. Lemma 8.8. Assume (8A). Let Z = Z(T ). Then the following conditions hold: (a) Suppose that K is a non-universal version of L± 4 (q). Then we have |Ω1 (Z∩  K) : Ω1 (Z(K))| = 2, and O r (CK (Ω1 (Z))) ∼ = SL2 (q) ∗ SL2 (q); (b) Suppose that K ∼ = L± 4 (q) and |Z| > 2. Then there is t ∈ I2 (Z) such that t ∈ [Aut(K), Aut(K)]) and E(CK (t)) ∼ = P Sp4 (q); (c) If K ∼ = P Sp4 (q) or G2 (q), then |Z| = 2 and Z is generated by a classical involution;  (d) If K ∼ = P Sp4 (q), then |Z| = 2 and O r (CK (Z)) ∼ = SL2 (q) ∗ SL2 (q); o (e) If K ∼ = Sp4 (q) ∗ SL2 (q); = P Sp6 (q), then |Z| = 2 and Lr (CK (Z)) ∼  − − (f) If K ∼ = L8 q (q), then |Ω1 (Z)| = 2 and and O r (CK (Ω1 (Z))) ∼ = SL4 q (q) ∗ − SL4 q (q);  − − (g) If K ∼ = L6 q (q), then |Ω1 (Z)| = 2 and O r (CK (Ω1 (Z))) ∼ = SL4 q (q) ∗ SL2 (q); and ± ∼ ∼ (h) If K ∼ = L± 5 (q), then Ω1 (Z) = Z2 and E(CK (Ω1 (Z))) = SL4 (q).

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Proof. By Lemma 8.7, we may assume that AutX (K) ≤ Aut0 (K), and we know that [Z, S(T ∩ K)] = 1. ±  ∼  Suppose that K is a non-universal version of L± 4 (q). If K = GL4 (q), then K contains (S1 ∗ H1 ) × (S2 ∗ H2 ) with Hi cyclic, Si Hi isomorphic to a subgroup of index 2 in GL± (Si ). As K is non-universal, it follows that 2 (q), and Hi = CGL± 2 (q) Z(S1 ) and Z(S2 ) are identified in K. We may assume that S(T ∩ K) is the image L1 ∗ L2 of S1 S2 , so [Z, L1 L2 ] = 1. Then Z ∩ K is contained in the image of H1 × H2 , and it follows that Ω1 (Z ∩ K) = Z(K) × Z0 , where Z0 = Z(S(T ∩ K)) has order 2. As r is odd and H1 H2 is an r  -group, it follows easily that   ∼ SL2 (q) ∗ SL2 (q), O r (CK (Ω1 (Z))) = O r (CK (Z0 )) = L1 L2 = proving (a). Now suppose that Z(K) = 1, but |Z| > 2. Again let Li be the image in K of Si . Then Z centralizes L1 ∗ L2 . But an element of T ∩ K inverts the cyclic group CK ∗ (L1 L2 ), where K ∗ = Inndiag(K). Hence, Z ∩ K ∗ = Ω1 (Z ∩ K ∗ ) =: z . First, suppose that Z is cyclic and let v ∈ Z with v 2 = z. There exists y ∈ L1 with y 2 = z. Thus t := yv ∈ I2 (X − K ∗ ), centralizing L2 . It follows using [IA , Table 4.5.1] and [IA , Proposition 4.9.1] that E(CK (t)) ∼ = P Sp4 (q). But then as L2 ≤ E(CK (t)), we have L1 ∗ L2 ≤ E(CK (t)), a contradiction. Hence, Z is not cyclic, and we may choose t ∈ I2 (Z)−{z}. Again, t must centralize L1 ∗L2 , whence E(CK (t)) ∼ = P Sp4 (q) and t ∈ [Aut(K), Aut(K)], proving (b). We remark concerning K = SL± 4 (q), for later use, that S1 and S2 are interchanged by an element of T ∩ K. Therefore Z ∩ S1 S2 ≤ Z(K). Hence if X ≤ Inndiag(K), then (8B)

CInndiag(K) (T ) = 1.

∼ P Sp4 (q) or G2 (q). Then Aut0 (K) = Inndiag(K) or Inn(K), Suppose that K =  respectively. From [IA , Table 4.5.1], Z ∩ K = z with z 2 = 1 and O r (CK (z)) =: L1 ∗ L2 ∼ = SL2 (q) ∗ SL2 (q). In particular Z = z if K ∼ = G2 (q). If K ∼ = P Sp4 (q), then COutdiag(K) (z) induces Outdiag(L1 ) on L1 , and (c) and (d) follow. A similar argument applies to P Sp6 (q), proving (e). − Suppose that K ∼ = Ln q (q) for n = 6 or 8. From [IA , Table 4.5.1], we see that  − − K contains an involution z with O r (CK (z)) =: L = L1 ∗ L2 ∼ = SL4 q (q) ∗ SLr q (q) with r = n − 4. From (8B) and the fact that CInndiag(K) (L) is cyclic of order q + q , we see that Ω1 (Z)∩Inndiag(K) = z . As the group ΓK ∼ = Z2 induces an element of Aut(L1 ) − Inndiag(L1 ), but AutK (L1 ) ≤ Inndiag(L1 ), it follows that Ω1 (Z) = z , proving (f) and (g). A very similar argument handles the case K ∼ = L± 5 (q), proving (h), and hence all parts of the lemma.  Lemma 8.9. Assume (8A). Let Z = Z(T ). Then the following conditions hold: (a) If K ∼ = D4 (q) and K is simple, then Z ≤ K, |Z| = 2 and Z is generated by a classical involution, i.e., an involution with a 4-dimensional −1eigenspace of type + on some (or any) natural K-module; (b) If K ∼ = Ω+ 2a (q), a ≥ 2, then CAut(K) (T ∩ K) has odd order; (c) If K ∼ = Ω2a +s (q), s odd, s < 2a ≥ 4, then there is z ∈ I2 (Z) such that  ∼ O r (CK (z)) = Kz Mz with z ∈ Kz ∼ = Ω+ 2a (q) and Mz = Ωs (q) (by convention Ω1 (q) = 1), and z maps to a Sylow 2-subgroup of CAut(K) (Kz Mz ); (d) If K ∼ = Ω7 (q), then Z = Z(T ∩ K) = z, y , where z is as in part (c),  E(CK (y)) ∼ = Ω6q (q), and on the natural K-module the support of y contains the support of z;

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(e) If K ∼ = Spin± n (q), n ≥ 5, or HSpin4n (q), then CAut(K) (T ∩ K) has odd order; (f) If K ∼ = E7 (q)a , then Z is cyclic; and (g) If K ∼ = Spin9 (q). = F4 (q), then |Z| = 2 and CK (Z) ∼ Proof. As in Lemma 8.8, we may assume that AutX (K) ≤ Aut0 (K), and we know that [Z, S(T ∩ K)] = 1. If K ∼ = D4 (q) with K simple, then S(T ∩K) = {S1 , S2 , S3 , S4 }. Also S(T ∩K) = S1 ∗ S2 ∗ S3 ∗ S4 , and by [IA , 4.5.1], CAut0 (K) (S(T ∩ K)) = Z(Si ) for all i = 1, . . . , 4. This proves (a). To prove (b), we proceed by induction on a. If a = 2 then K = S(T ∩ K)  is centralized by O 2 (CAut(K) (T ∩ K)), as desired. If a > 2, then K contains a subgroup K0 = K1 × K2 , where K1 ∼ = K2 ∼ = Ω2a−1 (q) with T ∩ K0 ∈ Syl2 (K0 ), r and K0 = O (NK (Z(K0 ))). Therefore K0 is CAut(K) (T ∩ K)-invariant, and by  induction O 2 (CAut(K) (T ∩ K)) centralizes K0 . Suppose that there is an involution z ∈ CAut(K) (T ∩ K). Then z ∈ Aut0 (K) by Lemma 8.7, so z has a preimage z, a 2-element acting on a natural K-module V as an isometry or similarity, with z2 acting as a scalar on V . Now V = [V, K1 ] ⊕ [V, K2 ], with [V, Ki ] a natural Ki module, and thus absolutely irreducible for Ki . It follows that z acts as scalars z is the image of Z(K1 ) in Aut(K). Therefore Z(K1 ) or on each [V, Ki ], and so  Z(K2 ) lies in Z(T ). But K1 and K2 are T -conjugate, so Z(K1 ) and Z(K2 ) are T -conjugate, a contradiction. Thus (b) holds. Next, let K ∼ = Ωn (q) with n = 2a + s, s < 2a ≥ 4, and s odd. We may choose an orthogonal decomposition of the natural module V as V = U ⊥ W with U a 2a dimensional space of + type. Then there exists z ∈ I2 (T ∩ K) with [V, z] = U and  ∼ CV (z) = W . We have O r (CK (z)) = Kz Mz with z ∈ Kz ∼ = Ω+ 2a (q) and Mz = Ωs (q), where Mz = 1 if s = 1. By [IA , Table 4.5.1], CAut0 (K) (Kz Mz ) = z . Clearly Kz contains a long root subgroup of K, so CAut(K) (Kz Mz ) = z , by Lemma 6.1. It remains to show that |CX (z)|2 = |X|2 , whence we may assume that z ∈ Z. Clearly, CX (z) covers X/K, and so we may assume that X = K. In the notation of [IA , Proposition 4.10.6], T ≤ R := NK (S(T )), where S(T ) = J1 × · · · × Jm with Ji ∼ = Ω+ 4 (q), and n = 4m + r with r = 1 or 3. Replacing T by a conjugate, we may assume that the support of z equals the support of J := J1 × · · · × J2a−2 , whence J ≤ Kz . Note that if R1 is the kernel of the permutation action of R on {J1 , . . . , Jm }, then R1 centralizes Z(J1 ) × · · · × Z(Jm ) ≥ z . Now, NK (S(T )) contains Σ ∼ = Σm transitively permuting {J1 , . . . , Jm }, and we may assume that T ∩ Σ ∈ Syl2 (Σ). Then T ∩ Σ is a direct product of factors corresponding to the dyadic decomposition of m = 2a−2 + k, where s = 4k + r. As s < 2a , k < 2a−2 . Therefore T ∩ Σ = T1 × T2 , where T1 acts on J, while T2 normalizes Ji for each "2a−2 i ≤ 2a−2 . Thus z = i=1 zi , where zi = Z(Ji ). Hence, T ∩ Σ centralizes z. Thus T = (T ∩ R1 )(T ∩ Σ) centralizes z, and z ∈ Z, completing the proof of (c). Now, we specialize (c) to the case K ∼ = Ω7 (q), n = 22 + 3, and z as in (c). ∼ By (c), z ∈ Z and CX (z) = (Kz × Mz )Hz Fz , where Kz ∼ = Ω+ 4 (q), Mz = Ω3 (q), + (Kz × Mz )Hz is isomorphic to a subgroup of O4 (q) × SO3 (q), and Fz acts faithfully as field automorphisms on both Kz and Mz . In particular, there is an involution w ∈ CK (z) such that Mz w ∼ = SO3 (q) ∼ = P GL2 (q). It follows that Z = z, z1 , where z1 = Z((T ∩ Mz ) w ). Let y = zz1 and let V be the natural K-module.

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Then dim([V, y]) = 6 and [V, z] ⊆ [V, y]. Moreover, E(CK (y)) ∼ = Ω6q (q), as claimed, completing the proof of (d). ∼ ± Next, let K ∼ = Spin± n (q), n ≥ 5. Then K = K/ z = Ωn (q) for some z ∈ Z(K). + Note that if K ∼ = Spin8 (q), ΓK ∼ = Σ3 acts faithfully on Z(K) [IA , 2.5.12], so CAut(K) (T ∩ K) ≤ Inndiag(K), using Lemma 8.7. Thus in any case, CAut(K) (T ∩ K) ≤ P GOn± (q). Suppose that x ∈ I2 (CAut0 (K) (T ∩ K)). If x is induced by an element ξ ∈ I2 (K), then by [IA , 6.2.1e], ξ ∼K ξz, so by [III8 , 6.12], ξ is not 2-central in K, contradiction. Hence 

(8C)

x is not induced by an involution in K.

∼ Now, x centralizes S(K) = J1 · · · Jm , where m = [n/4] or (n − 4)/4, and Ji = Spin+ (q) for each i = 1, . . . , m. Moreover, if V is the natural K-module, then 4 dim[V, J i ] = 4. Write Z(J i ) = z i , where zi ∈ Z(Ji ) ∼ = E22 . For any j = 1, . . . , m, let yj = z1 · · · zj . Suppose that n > 4m. Then by induction, x centralizes S(T ∩ K). As T  centralizes y m , T normalizes O r (CK (ym )) = S(T ∩ K) ∗ L, where L = 1 if n ≡ 1 or 2 ∼ 2 (mod 4), L ∼ = Spin3 (q) ∼ = SL2 (q) if n ≡ 3 (mod 4), and L ∼ = Spin− 4 (q) = SL2 (q ) if n ≡ 0 (mod 4). In particular T ∩ L ∈ Syl2 (L) so by [III11 , 1.17], x centralizes L. If n ≡ 2 (mod 4), then we see from [IA , 4.5.1] that there is only one involution in CAut0 (K) (S(T ∩ K)L). Therefore x is induced by an element of ym Z(K) ≤ K,  contradicting (8C). So n ≡ 2 (mod 4). Since O r (CK (x)) contains S(T ∩ K) and in view of (8C), the only possibility, by [IA , 4.5.1], is that x acts on K as a reflection. But then one calculates that |CK (x)|2 = |Spinn−1 (q)|2 < |Spinn (q)|2 = |K|2 , a contradiction. Therefore n = 4m, m > 1, and K ∼ = Spin+ 4m (q). ∼ HSpin4m (q) in this paragraph. To complete the proof of (e) we also allow K = In that case as well we have subgroups J1 , . . . , Jm , which are isomorphic images of their counterparts in Spin+ 4m (q). In either case, if T is transitive on S(T ), then n is a power of 2, and the result follows from (b). So we may assume that T is not transitive on S(T ), and that {J1 , . . . , Jr } is an orbit of T , with r < m.  Then T centralizes y r and by induction, x centralizes O r (CK (yr )) = L ∗ M , where + + L∼ = Spin4r (q) and M ∼ = Spinn−4r (q). But then by [IA , 4.5.1], CAut0 (K) (LM ) is the image of yr . Therefore again x is induced by an element of yr Z(K) ≤ K, contradicting (8C) and completing the proof of (e). If K ∼ = E7 (q)a , there exists z ∈ I2 (T ∩ K) with CT ∩K (z) ∈ Syl2 (CK (z)) r and O (CK (z)) = J ∗ L, where J ∼ = SL2 (q) and L ∼ = HSpin12 (q). Using (e), we see that [Ω1 (Z), JL] = 1. But then as CAut0 (K) (JL) = z by [IA , 4.5.1], it follows by Lemma 8.7 that Ω1 (Z) = z , whence Z is cyclic, proving (f). Likewise if K ∼ = Spin9 (q). Using (e), we see = F4 (q), there exists z ∈ T with CK (z) ∼ that Ω1 (Z) = z . As CAut(K) (CK (z)) = z by Lemma 8.7 and [IA , 4.5.1], this completes the proof of (g), and hence of all parts of the lemma.  Lemma 8.10. Let X = Σ2n , n ≥ 5. Let S ∈ Syl2 (X). Then the following conditions hold: (a) J(S) = T1 × · · · × T[n/2] × Z where |Z| ≤ 2 and Ti ∼ = D8 for each i = 1, . . . , [n/2]; moreover, m2 (S) = n;

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(b) Z(T1 ), . . . , Z(T[n/2] ) are generated by pairwise disjoint root involutions; (c) |Z| = 2 if and only if n is odd; (d) If n is odd, then J(S ∩ A2n ) ∼ = T1 × · · · × T[n/2] and [J(S ∩ A2n ), J(S ∩ A2n )] = [J(S), J(S)]; and (e) J(S ∩ A2n ) ∼ = E2n if n is even.  Proof. Let X act naturally on Ω, with |Ω| = 2n. Let n = i ci 2i be " the binary expansion of n and let I = {i | ci = 1}. Then S is a direct product I Si , pairwise disjoint supports on Ω. By where Si ∈ Syl2 (Σ2i+1 ) and the Si ’s have " [IG , 10.30(i)], J(S) is the direct product I J(Si ). Hence in proving (a), (b), and (c), we may assume that n = 2i . If i = 0 the lemma is trivial, as it is for i = 1. For i ≥ 2, there is a bisection Ω = Ω0 ∪ Ω1 and there are Sylow 2subgroups Sij of ΣΩj , j = 0, 1, such that Si = (Si0 × Si1 ) t ∼ = Si0  Z2 . Here t is an involution interchanging Ωi0 and Ωi1 . We show that J(Si ) ≤ Si0 × Si1 , so that J(Si ) = J(Si0 ) × J(Si1 ), and then (a), (b), and (c) follow directly. If J(Si ) ≤ Si0 × Si1 , then, for a suitable choice of t, there is A ∈ E∗ (Si ) with t ∈ A. Then m2 (Si0 Si1 ) ≤ m2 (Si ) = m2 (A) = 1 + m2 (CSi0 Si1 (t)) = 1 + m2 (Si0 ). Consequently m2 (Si1 ) ≤ 1, which is absurd as i ≥ 2. Thus (a), (b), and (c) hold. If n is odd, then it is clear from (b) and (c) that every element of E∗ (S) contains a transposition, so m2 (An ) = m2 (S) − 1. On the other hand, S has a central transposition so any element of E∗ (S ∩ A2n ) lies in an element of E∗ (S). Hence J(S ∩ A2n ) = J(S) ∩ A2n , which implies (d). If n is even, then J(S) = T1 × · · · × Tn/2 contains E = E1 × · · · × En/2 ∈ E∗ (S) ∩ E(A2n ), where each Ei is a root four-subgroup of S. Conversely, for any " F ∈ E∗ (S ∩ A2n ), F is the direct product n/2 i=1 (F ∩ Ti ), with each F ∩ Ti ≤ A2n .  Thus F ∩ Ti is a root four-group and (e) follows as well. Lemma 8.11. Let X = EΣ, where E = F ∗ (X) is an elementary abelian 2group, Σ ∼ = Σn or An , n ≥ 5, and as F2 Σ-module, E is either the natural permutation module or its trace-0 submodule. Let S ∈ Syl2 (X). Then the following conditions hold: (a) If E is the natural permutation module and Σ ∼ = Σn , then S is isomorphic to a Sylow 2-subgroup of Σ2n , with a permutation basis of E mapping to {(12), (34), . . . , (2n − 1, 2n)}; moreover, m2 (E) = n = m2 (S); (b) If E is the trace-0 module and Σ ∼ = Σn , then S is isomorphic to a Sylow 2-subgroup of A2n , with a basis of E mapping to {(12)(34), (34)(56), . . . , (2n − 3, 2n − 2)(2n − 1, 2n)}; moreover, if n is odd, then m2 (E) = n − 1 = m2 (S); (c) If E is the trace-0 module and Σ ∼ = An with n even, then m2 (E) = n − 1 = m2 (S) and J(S) = E, F for some (indeed any) F ∈ E∗ (S) − {E}. Moreover, [J(S), J(S)] lies in E and contains ti + tj for some 1 ≤ i < j ≤ n. Here ti , tj are part of a permutation basis for the natural permutation module E ∗ for Σ, which we regard as containing E. Proof. Let Σn and Σ2n act naturally on {1, . . . , n} and {1, . . . , 2n}, respectively. For each σ ∈ Σn define σ  ∈ Σ2n by σ  (2k − δ) = 2σ(k) − δ, k = 1, . . . , n,

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δ = 0, 1. Then σ → σ  is an injective homomorphism; let Σ be its image. Part (a) follows easily, as does (b) with the help of Lemma 8.10d. Let X = EΣ be as in (c). Then X has index 2 in a subgroup X ∗ = EΣ∗ of A2n , where Σ ≤ Σ∗ ∼ = Σn and |A2n : X ∗ | is odd. Let S ≤ S ∗ ∈ Syl2 (X ∗ ), and ∗ define X = X ∗ /E, which we identify with Σ∗ . Since n is even, Lemma 8.10 implies that m2 (X ∗ ) = n and J(S ∗ ), as a subgroup of A2n , is the direct product of n/2 pairwise disjoint root four-groups. Therefore J(S ∗ ) is generated by n/2 pairwise disjoint transpositions. In particular J(S ∗ ) ≤ X, so m2 (X) = n − 1 = m2 (E). Set J = J(S ∗ ) ∩ X ≤ S. Then EJ, considered as a subgroup of EΣ = X, is generated by elementary abelian subgroups of rank n − 1 (E included), so EJ ≤ J(S). We argue that (8D)

EJ = J(S).

Let F ∈ E∗ (S) − {E}. Let O1 , . . . , Or be the nontrivial orbits of F , as a subgroup of X ∼ = An . Suppose first that no Oi has length 2. We show that this leads to a contradiction. Because of this assumption we can embed F in an elementary ∗ abelian 2-subgroup F = F 1 × · · · × F r ≤ X, where each F i is a regular elementary ∗ abelian subgroup on Oi . Then CE (F ) = CE (F ), and as F has maximal rank it ∗ follows that F = F . Say |Oi | = 2ai , i = 1, . . . , r. For ease of computation let E ∗ be the natural permutation module for Σ∗ and Σ, and consider E ∗ to contain E. Then r r # # ai = m2 (F ) = m2 (E/CE (F )) ≥ m2 (E ∗ /CE ∗ (F )) − 1 = −1 + (2ai − 1) i=1

i=1

with strict inequality unless CE ∗ (F ) ≤ E, i.e., unless O1 , . . . , Or cover {1, . . . , n}. As each ai ≥ 2 the only solution is that r = 1, a1 = 2, and equality holds, whence n = 4. This contradicts our hypothesis that n ≥ 5. Therefore F has an orbit of length 2, as a subgroup of X ∼ = An . Let g be the ∗ transposition on that orbit and set F = F g , an elementary abelian 2-subgroup ∗ ∗ ∗ ∗ of X ∼ = Σn . Clearly F > F but CE (F ) = CE (F ). It follows that CE (F )F , ∗ considered as a subgroup of the semidirect product EΣ , is an elementary abelian ∗ subgroup of S ∗ of maximal rank. Hence F = J(S ∗ ) so F = J. This proves (8D). To complete the proof of (c) we must show that ti +tj ∈ [J(S), J(S)] for some i = j. Choose an involution h = (ij)(kl) ∈ F . Since n > 4, there is a letter m ∈ {i, j, k, l}.  Then [ti + tm , h] = ti + tj , as required.

9. Generation Lemma 9.1. Let V be a linear, unitary, symplectic or orthogonal space over a finite field F with V = V0 ⊥ V1 ⊥ V2 , where each Vi is a nonzero nondegenerate subspace of V . Let C(V ) denote the isometry group of V , and C(V )o = [C(V ), C(V )]. Suppose that the following conditions hold: (a) dim(V ) ≥ 4 if V is a unitary space and |F| = 2; (b) dim(V0 ) ≥ 4 if V is an orthogonal space and |F| is even; (c) dim(V0 ) ≥ 2 if V is an orthogonal space and |F| is odd; and (d) C(V ) = O4+ (3). Then C(V )o = C(V0 ⊥ V1 )o , C(V0 ⊥ V2 )o .

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Proof. This is [IA , 7.3.2]. Here we interpret C(V0 ⊥ Vi ), i = 1, 2, as CC(V ) (V3−i ) except in the linear case, in which case we interpret it as CGL(V ) (V3−i )∩ NGL(V ) (V0 + Vi ).  A similar special case we shall need is: Lemma 9.2. Let K = Ω+ 6 (q) ∈ K, q even. Let V be the natural (6-dimensional) Fq K-module, and suppose V = V0 ⊥ V1 ⊥ V2 where V0 is 2-dimensional of + type. If V1 is of + type, assume that q > 2. Then K = CK (V1 ), CK (V2 ) . Proof. Since V is of + type, V1 and V2 are both of + type or both of − type. First suppose that they are of + type, so that q > 2 by assumption. Let p be a prime divisor of q − 1. For each i = 0, 1, 2, CK (Vi⊥ ) ∼ = Ω(Vi ) ∼ = Zq−1 contains an element bi of order p. Set B = b0 , b1 , b2 ∼ = Ep3 . Then CK (Vi ) =  O 2 (CK (bi )) ∼ = Ω(Vi⊥ ) ∼ = L2 (q) × L2 (q). Now K ∼ = SL4 (q) and we let W be the natural 4-dimensional Fq K-module. Then B is diagonalizable and preserves a unique frame {Fq wi | 1 ≤ i ≤ 4} in W . Let Wij = Fq wi + Fq wj for each i = j, and let Lij be the SL2 (q) subgroup of K supported on Wij and centralizing Wk , where {i, j, k, } = {1, 2, 3, 4}. As CK (Vi ) ∼ = L2 (q) × L2 (q), we must have (up to a permutation of indices) CK (V1 ) = L12 × L34 and CK (V2 ) ≥ L23 . Then CK (V1 ), CK (V2 ) ≥ L12 , L23 , L34 = K, as desired. Now assume that V1 and V2 are of − type, and let p be a prime divisor of q + 1. Let bi ∈ Ip (Ω(Vi )), i = 1, 2, so that [b1 , b2 ] = 1, and set Ki = Ω(Vi⊥ ) = E(CK (bi )) ∼ = L2 (q 2 ), i = 1, 2. Thus bi ∈ K3−i . Set X = K1 , K2 ; then mp (X) = 2 and we must prove that X = K. Suppose that X < K. Note that if q = 2, then K∼ = A8 , and in the natural permutation representation of K on an 8-element set Ψ, K1 and K2 are root A5 -subgroups whose supports cover Ψ, so X = K. We may thus assume that q > 2. By Lp -balance, X = Lp (X). As m2 (X) ≥ m2 (K1 ) ≥ 4 but ms (K) ≤ 3 for all odd primes s, O2 (X) ≤ Z(X). Also since q 2 + 1 divides |X|, X is not embeddable in a parabolic subgroup of K, so O2 (X) = 1 by the Borel-Tits theorem. Therefore X = E(X). As ms (K) = 1 for a prime divisor s of q 2 + 1, X is simple. Using the facts that q > 2 and mp (X) > 1, and [III11 , 1.1ab], we see that X ∈ Chev(2). Let r(X) be the untwisted Lie rank of X. As b1 , b2 ≤ X, r(X) > 1 and q(X) = q 2/n for some n such that the extended Dynkin diagram of X contains n totally disconnected nodes, by [IA , 4.2.2]. On the other hand, the nilpotence class of a Sylow 2-subgroup of K is 3, so for X it is at most 3. It follows easily from the Chevalley commutator formula that r(X) ≤ 3, whence n = 1 or 2. 2 If n = 1, then |X|2 ≤ |S|2 = q 6 forces X ∼ = A± 2 (q ), so |X|2 = |S|2 and O2 (X) = 1 by [IA , 2.6.7], contradiction. Thus n = 2 and q(X) = q. If r(X) = 3 or X ∼ = G2 (q) − then |X|2 ≥ q 6 = |S|2 , again a contradiction. As mp (X) = 2, X ∼ A (q) or B2 (q); = 2 L (q) or 1, a final contradiction.  but then by [IA , 4.8.2], E(CX (bi )) ∼ = 2 Lemma 9.3. Let q be an odd prime power, q > 3, and set F = Fq . Let V = Fv0 ⊥ Fv1 ⊥ · · · ⊥ Fvm , m ≥ 3, be a nondegenerate F-orthogonal space with (vi , vi ) = 1 for all 1 ≤ i ≤ m, but with (v0 , v0 ) a nonsquare in F. Let W1 = Fv0 ⊥ Fv1 and W2 = Fv1⊥ . Set Ω(Wi ) = CΩ(V ) (Wi⊥ ), i = 1, 2. Then Ω(V ) = Ω(W1 ), Ω(W2 ) .

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∼ Ω− (q) ∼ Proof. Suppose first that m = 3, whence Ω(V ) = = L2 (q 2 ) and 4 Ω(W2 ) ∼ = Ω3 (q) ∼ = L2 (q). As q > 3, NΩ(V ) (Ω(W2 )) is maximal in Ω(V ) and is the unique proper overgroup of Ω(W2 ). Hence if the result fails then Ω(W1 ) normalizes Ω(W2 ), whence Ω(W1 ) normalizes W2⊥ = Fv1 . But then Ω(W1 ) is an elementary abelian 2-group, so q = 3, contrary to assumption. Hence the lemma holds for m = 3. For m > 3 we let W3 = W2 ∩ Vm⊥ and V  = Vm⊥ and inductively obtain Ω(W1 ), Ω(W2 ) ≥ Ω(W1 ), Ω(W3 ) ≥ Ω(V  ). Hence, Ω(W1 ), Ω(W2 ) ≥ Ω(V  ), Ω(W2 ) = Ω(V ) by Lemma 9.1, completing the proof.



Lemma 9.4. Let K = F4 (q) or E6 (q)u , q a power of the odd prime r,  = ±1. Let c ∈ I2 (K) with O 2 (CK (c)) = J ∗ I, c ∈ J ∼ = SL2 (q) and c ∈ I ∼ = Sp6 (q) or SL6 (q). Let D be a four-subgroup of I such that c ∈ D and CJI (D) has a subnormal subgroup H with Z(H) = D c and H being the central product of four SL2 (q) subgroups. Then the following conditions hold: (a) For each d ∈ D# there exists Ed  CI (d) such that Ed ∼ = Sp4 (q) or SL4 (q), according to the isomorphism type of I;  (b) I = Ed | d ∈ D# ; and (c) II (D; r) = {1}. Proof. Our assumptions imply that CI (D) is the central product of three components or solvable components isomorphic to SL2 (q). Let I = I/Z(I). We give the proof for the I ∼ = L6 (q) case; the P Sp6 (q) case is similar but easier. Obviously |CI (D)| is divisible by r, proving (c). According to [IA , 4.5.1], for any  d ∈ D# , Id := O r (CI (d)) has Lie components A4 (q), A1 (q)A3 (q), A2 (q)A2 (q), or   A2 (q 2 ). As O r (CI (D)) = O r (CId (D/ d )) has three SL2 (q) Lie components, the onlypossibility is Id = Jd Ed with Jd ∼ = A1 (q) and Ed ∼ = A3 (q), proving (a). Then # by Lemma 9.1. The proof is complete.  I = Ed | d ∈ D Lemma 9.5. Let K = D4 (q), q = 2n > 2. Let p be a prime divisor of q + 1 and D ∈ Ep2 (K). Suppose that on some natural K-module, D is supported on a nondegenerate 4-dimensional subspace. Let I ∈ Chev(2) with K ≤ I and suppose that ΓD,1 (I) ≤ K. Then I = K. 

Proof. Let Γ = Γ2D,1 (I). First observe that, since K = D4 (q) with q = 2n > 2 and K ≤ I, it follows from [IA , 7.3.3] that I = Γ.  Let x ∈ Ip (I). By [IA , 4.2.2], Qx := O 2 (CI (x)) is the product of the Lie components of CI (x). We denote by Hx the product of all Lie components of CI (x) which are either solvable or p -groups; and by Lx the product of the remaining Lie components of CI (X). Thus, Qx = Hx Lx with [Hx , Lx ] = 1. Note that by assumption, any component of Lx , for x ∈ D# , is a component of CK (x), and 1 hence not B2 (2) , G2 (2) , or 2F4 (2 2 ) . Thus for x ∈ D# , Lx = Lp (CI (x)). Now, let V be the natural K-module such that D is supported on a 4-dimensional subspace U of V . Then U ⊥ is isometric to U , and there is a K-conjugate A of D supported on U ⊥ . In particular, Lx = Lp (CI (x)) ≤ K for all x ∈ D# ∪ A# .

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On the other hand, Mx := Lp (CK (x)) satisfies Mx = O p (O 2 (CK (x))(∞) ) ≤   O p (O 2 (CI (x))(∞) ) = Lx , so Mx = Lx for all x ∈ D# ∪ A# . Let e ∈ D# . Then, A ≤ Lp (CK (U )) = Lp (CK (D)) ∼ = Ω+ 4 (q), and by Lp balance, Lp (CK (D)) ≤ Le . In particular, [A, He ] = 1. Choose a ∈ A# with [V, a] = U ⊥ . Then D ≤ La . Let L0 = Lp (CI (e, a )). Then, L0 ≤ La , and indeed, either L0 = 1 or L0 is a product of components of La isomorphic to L2 (q).   Since [He , a] = 1, He ≤ O 2 (CQa (e)) = O 2 (CHa ×La (e)) = Ha × L0 . But also He  CI (e), so He  Ha × L0 , and hence, He ≤ Ha . On the other hand, D ≤ La  and [Ha , La ] = 1, whence Ha ≤ Qe = He Le . As before, Ha ≤ O 2 (CHe Le (a)) = He × L0 , and so Ha ≤ He . Thus, He = Ha . As e was chosen arbitrarily in D# , it follows that Hd = Ha for all d ∈ D# . Hence, Ha  CI (d) for all d ∈ D# , and  so Ha  ΓD,1 (I) = I. Thus, Ha = 1, and so Qd = Ld for all d ∈ D# . Then,  I = Γ = ΓD,1 (I) ≤ K, and so I = K, as claimed. Lemma 9.6. Let K = Dn± (q), where q is a power of 2 and n ≥ 4. Let p be have an odd prime divisor of q − ,  = ±1. Let B ∈ Ep∗ (K)  and let B1 ∈ E1 (B) minimal support on the natural K-module. Then K = E(CK (B1 ))NK (B) . Proof. AutK (B) ∼ = W (Dn ) or W (Cn−1 ) preserves the frame B = {B1 , B2 , . . . , Bk } of all cyclic subgroups of B with minimal support on the natural K-module; indeed it is transitive on that frame. Here k = n or n − 1, and the  minimal support of each Bi is 2-dimensional as p divides q 2 − 1. In any case E(CK (B1 ))NK (B) ≥ E(CK (B1 )), E(CK (B2 )) = K by Lemma 9.1, as required.  In the next several lemmas we consider the following setup. (1) p is an odd prime, X is a K-group, Op (X) = 1, and L   X with L simple; (9A)   (2) D ≤ X, D ∼ = Ep2 and H ≤ X with ΓD,1 ( LD ) ≤ H < LD . By [IA , 3.28(ii)], LD = L and of course CD (L) = 1.  be the Lemma 9.7. Assume (9A) with L ∼ = Alt(Ω) ∼ = An for some n. Let D image of D in Inn(L), which we identify with L. For each d ∈ D# write d for the  and let Ωd be the set of fixed points of d on Ω. Write Ld for image of d in D, the subgroup of CL (d) supported on and inducing Alt(Ωd ) on Ωd . Then one of the following conclusions holds: (a) Ω = Ω0 ∪ Ω1 , with Ω0 = Ωd for all d ∈ D# , and Ω1 is a union of regular  on Ω; moreover, H  L0 ×L1 ∼ orbits of D = Ar ×Akp2 , = Alt(Ω0 )×Alt(Ω1 ) ∼ with k ≥ 1 and r ≥ 1; or (b) |Ωd | = 0 or p for all d ∈ D# and Soc(H) = L1 × · · · × Lr , where d1 , . . . , dr are the subgroups of D with |Ωdi | = p, and Li = Ldi . Also, 2 ≤ r ≤ p + 1, and H involves S ∼ = Ar permuting {L1 , . . . , Lr } naturally. Proof. Let Γ = ΓD,1 (L),  on Ω, and let Ω1 and suppose that Γ ≤ H < L. Let Ω0 be the fixed point set of D  be the union of all regular D-orbits on Ω. Then the complement Ω1 = ∪d∈D# Ωd .

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By [IA , 7.5.3], AΩ1 ≤ Γ. Assume first that Ω0 = ∅. Then by [IA , 7.5.1], AΩ1 ≤ Γ. Since Γ < L, AΩ1 = 1. If Ω0 = Ω1 , then for some d ∈ D# , there is g ∈ CL (d) such that Ωg1 meets both Ω1 − Ω0 and Ω1 nontrivially. Then Γ contains AΩ1 , AgΩ1 , AΩ1 = L by [IA , 7.5.2], contradiction. Therefore Ω0 = Ω1 , so Γ  L0 ×L1 as in (a). If some g ∈ H does not normalize L0 ×L1 , then Ωgi ∈ {Ω0 , Ω1 } for both i = 0 and i = 1, and again by [IA , 7.5.2], H ≥ L0 L1 , Lg0 Lg1 = L, a contradiction. Thus (a) holds if Ω0 = ∅. We may then assume that Ω0 = ∅. Assume next that Ω1 = ∅. For any α ∈ Ω1 , there is d ∈ D# such that αd = α. Hence d has a p-cycle containing α, as well as (at least p) p-cycles covering Ω1 . Thus CL (d) has an element g interchanging a p-cycle of d containing α with a p-cycle supported in Ω1 , while also  fixing the support of some p-cycle of d supported in Ω1 . It follows by [IA , 7.5.2]  first that Γ ≥ AΩ1 , AgΩ1 ≥ AΩ1 ∪{α} , and then since α ∈ Ω1 was arbitrary, Γ = L, contradiction.  of length Now we may assume also that Ω1 = ∅. Thus, Ω is a union of D-orbits  has no fixed points on Ω. Suppose p. If D = x, y , then Ωx ∩ Ωy = ∅, since D  O in Ω − Ωx , there exists that x can be chosen with |Ωx | > p. For any D-orbit y ∈ D − x with O ⊆ Ωy . Then xy has no fixed points on Ωx ∪ Ωy , and so CL (xy) has a section inducing the symmetric group Σk on the set of k D-orbits contained in Ω − Ωxy , with k ≥ 3. It follows that the normal closure of Lx in CL (xy) is L(xy) , the alternating group on Ω − Ωxy . As O ∪ Ωx ⊆ Ω − Ωxy and O is an arbitrary D-orbit on Ω − Ωx , it follows by [IA , 7.5.2] that Γ = L, contrary to assumption.  Hence |Ωd | ≤ p for all d ∈ D# . Let Ω = O1 ∪ · · · ∪ Or as a union of D orbits of length p. On each Oi , the kernel of the D-action is a group di of  fixes or order p. Then Oi = Ωdi and Li := Ldi ≤ Γ. As each element of D cycles each Oi , the system of imprimitivity Ω = O1 ∪ · · · ∪ Or is Γ-invariant. As L1 ≤ H < L and L1 contains 3-cycles, H is either intransitive or imprimitive on Ω, by Lemma 5.11. Now, CL (d1 ) induces Σr−1 on {O2 , . . . , Or }, and similarly for CL (dr ) on {O1 , . . . , Or−1 }. Thus, if r > 2, then Γ induces Σr on {O1 , . . . , Or }, so H is transitive on Ω, hence imprimitive. The doubly transitive action of Σr implies that H preserves {O1 , . . . , Or }, so Soc(H) = L1 × · · · × Lr . On the other hand, if r = 2, then whether H is intransitive or transitive but imprimitive, we clearly have Soc(H) = L1 × L2 . As D has exactly p + 1 subgroups of order p, r ≤ p + 1. Thus conclusion (b) holds, completing the proof.  Lemma 9.8. Assume (9A), with L ∈ Spor. Suppose that H contains a Sylow p subgroup of L. Suppose also that J is subnormal in H with J = O p (J) ∈ Chev(2). If mp (L) ≤ 2, then p = 5, L ∼ = F i22 , J ∼ = D4 (2) and H ∼ = Aut(D4 (2)). Proof. As L ∈ Spor, |Out(L)| ≤ 2, so D induces inner automorphisms on L. As ΓD,1 (L) ≤ H < L, we must have that mp (L) = 2 and then, by [IA , 7.5.5], one of the following holds: (a) (b) (c) (d)

p = 3 and L ∼ = M11 or M12 ; or p = 5 and L ∼ = M c, HS, Ru, or F i22 ; or p = 7 and L ∼ = He, O  N , or F i24 ; or p = 11 and L ∼ = J4 .

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∼ Aut(D4 (2)) is a maximal If L ∼ = F i22 with p = 5, then by [IA , 7.6.2], ΓD,1 (L) = ∼ subgroup of L. We conclude that J ∼ D (2) and H = 4 = Aut(D4 (2)), the desired conclusion. If L ∼ = M11 or M12 , then from the list of maximal subgroups of L [IA , 5.3ab], H is solvable, a contradiction. In all the remaining cases we have p ≥ 5 and P ∼ = p1+2 . Without loss we may replace D by its image in L, and assume D ≤ L. Thus we may assume D ≤ P ; let z = Z(P ) ≤ D. In these cases, mr (L) ≤ 2 for all primes r ≥ 5, so J has at most 2 conjugates in H and in particular is P -invariant. As Z(P ) is cyclic, z ∈ J  H. By [IA , Table 5.3], CL (z) is solvable, so [z, J] = 1. Thus, P is a Sylow p-subgroup of AutH (J). But Inn(J) ≤ AutH (J) ≤ Aut(J), and this contradicts Lemma 8.3, completing the proof.  Lemma 9.9. If K ∈ K and |K| has just three distinct prime divisors, then K∼ = U4 (2). = A5 , A6 , L2 (7), L2 (8), L2 (17), L3 (3), U3 (3), or P Sp4 (3) ∼ Proof. Clearly if K ∈ Alt ∪ Spor, the result holds. If K ∈ Chev, it follows in a straightforward way from Zsigmondy’s theorem and the fact the the only solution of pa − q b = 1 with p, q primes and a > 1 < b is 9 − 8 = 1.  Lemma 9.10. Assume (9A), with L ∈ Chev(r), r = p. Then r = 2 and one of the following holds: (a) p = 3, L ∼ = L2 (8), Sp4 (2) , G2 (2) , L3 (4), or U4 (2), and H is 3-constrained; (b) p = 3, L ∼ = A6 ; = L3 (4) and H ∼ ∼ (c) p = 3, L = Sp6 (2) and H ∼ = O6− (2); ∼ (d) p = 3, L = Sp8 (2) and H ∼ = O8+ (2); 1 2  ∼ (e) p = 3, L = F4 (2 2 ) and H ∼ = Aut(L3 (3)); or 5 1 (f) p = 5, L ∼ = 2B2 (2 2 ) or 2F4 (2 2 ) , and H is 5-constrained. Proof. That p and L must be as asserted is immediate from [IA , 7.3.4]. Suppose that L ∼ = P Sp4 (3), but H is not 3-constrained. All 3-local = U4 (2) ∼ subgroups of L are solvable, so if we put H = H/O3 (H), then O3 (H) = 1 and F ∗ (H) is simple; by Lemma 9.9, F ∗ (H) ∼ = A5 or A6 . Therefore |H|3 ≤ 32 . On the other hand, by [IA , 7.3.4], we may assume that D ≤ L, whence D ≤ U4 (2) is diagonalizable. Hence, D ≤ E ≤ Γ ≤ H for some E ∼ = E33 , contradiction. In cases (c) and (d), H ≥ H0 ∼ = O6− (2) or O8+ (2). We see from [IA , 4.8.1] that H0 controls L-fusion in a Sylow 3-subgroup. Since ΓD,1 (L) ≤ H, it follows that Dg ∩ H = 1 for all g ∈ L − H. Therefore |L : H| ≡ 1 (mod 9). As |L : H| divides |L : H0 | = 22 7 or 23 17, this implies H = H0 , as asserted. In case (e), suppose that H > H0 ∼ = Aut(L3 (3)). By the Borel-Tits theorem and [IA , 7.6.2], H0 normalizes no nilpotent subgroup of L, so H ≤ N0 = NL (L0 ) for some simple L0 > F ∗ (H0 ).  L2 (5a ) By order considerations, L0 ∈ Chev(r) for some r ∈ {3, 13}. Clearly L0 ∼ = a a so r = 2. As all 2-local subgroups of L are of order 2 .3 or 2 .5, q(L0 ) = 2 and L0 has twisted rank at most 2. But the few groups with these properties have orders not divisible by 13, contradiction. Thus H = H0 in case (e), as required. 5 In the remaining cases, L (or Aut(L), if L ∼ = L2 (8) or 2B2 (2 2 )) has a strongly p-embedded Sylow p-normalizer NL (Q), and mp (Aut(L)) = 2 = mp (Q). We may assume that D ≤ Q, so D ∩ Z(Q) = 1 and Q ≤ Γ. We may also assume that Q  H, so there is g ∈ H such that H0 := Q, Qg > NH (Q). This rules out L∼ = G2 (2) ∼ = L2 (9) and L ∼ = U3 (3), for which Qg permutes Sylp (L)−{Q} = Sp4 (2) ∼

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transitively and so H0 ≥ L, contradiction. It also rules out L ∼ = L2 (8) and 2B2 (2 2 ), 1 by [IA , 6.5.1, 6.5.4]. Finally, if L ∼ = L3 (4) or 2F4 (2 2 ) , we may assume D = Q, so as Q  H, NH (Q) is strongly p-embedded in H. By [IA , 7.6.1, 7.6.2], the only possibilities are E(H) ∼ = L3 (4), or E(H) ∼ = L2 (52 ) = A6 or M11 with p = 3 and L ∼ 1 2  ∼ ∼ 2 with p = 5 and L = F4 (2 ) . By [IA , 6.5.3], H = A6 in the first case, and by Lagrange’s theorem, the second case is impossible. In the third case, the centralizer of a 2-central involution in L contains Z5 so NL (Q) contains an involution z with CQ (z) = 1. Hence z ∈ H so z induces a nontrivial field automorphism on H, whence CH (z) is nonsolvable. But CL (z) is solvable, contradiction. The proof is complete.  5

Lemma 9.11. Assume (9A) with p > 3. Suppose that L has an abelian Sylow p-subgroup A, and Lp (H) = 1. Then one of the following holds: (a) L ∼ = L2 (pm ), m ≥ 2; 1 (b) p = 5 and L ∼ = 2F4 (2 2 ) ; or 5 (c) p = 5 and L ∼ = 2B2 (2 2 ) and DL has a nonabelian Sylow 5-subgroup. Proof. If L ∈ Chev(p), then as L ∩ A is an abelian Sylow p-subgroup of L and p > 3, it follows that L ∼ = L2 (pm ). Since ΓD,1 (L) < L, we must have m ≥ 2, whence (a) holds. If L ∈ Alt, then by Lemma 9.7, Lp (H) = 1, as p > 3. If L ∈ Chev(r), 1 5 r = p, then by Lemma 9.10, p = 5 and L ∼ = 2F4 (2 2 ) or 2B2 (2 2 ). But in the latter case, AutLD (L) does not have abelian Sylow 5-subgroups. Hence (b) or (c) holds. Finally, if L ∈ Spor, then since p > 3, we must have |L| = p2 by Lemma 8.2. In particular, L has a strongly p-embedded subgroup, whence by [IA , 7.6.1], p = 5 and L∼ = Aut(D4 (2)) is a maximal subgroup of = F i22 . However, in this case, ΓD,1 (L) ∼ L by [IA , 7.6.2], a contradiction to L5 (H) = 1, proving the lemma.  Lemma 9.12. Let K ∈ Chev(r) and let p be a prime distinct from r. Let D∼ = Ep2 act on K and suppose that KD is a p-component of CK (D). If KD is a component of CK (d) for all d ∈ D# , then [K, D] = 1. Proof. Since KD is a p-component of CK (D), KD ∈ Chev(r) by [IA , 4.9.6].  In particular IK (D; r) = {1}. Now by [IA , 4.2.2], O r (CK (d)) normalizes every  component of CK (d), for any d ∈ D# . In particular KD is normalized by ΓrD,1 (K),  which equals K by [IA , 7.3.1]. Hence KD = K, proving the lemma. Lemma 9.13. Let K = F ∗ (X) ∈ Chev(2) be simple, and X ≤ Inndiag(K) and B ∈ Ep (X), where p is an odd prime. Let b ∈ B # and let L be a p-component of CK (b). Set B0 = CB (L). If mp (B0 ) > 1, then there exists b0 ∈ B0# and a component L0 of CK (b0 ) such that L < L0 . Proof. Clearly L is a component of CK (B0 ). Since [B0 , K] = K and B0 is noncyclic, it follows from Lemma 9.12 that L is not a component of CK (b0 ) for some b0 ∈ B0# . By the Bp -property and Lp -balance, there then must exist p components L1 , . . . , Lp of CK (b0 ) cycled by b and with L = E(CL1 ···Lp (b)). But as p is odd this  is impossible, by [III11 , 1.13b]. The proof is complete. Lemma 9.14. Let K = SLn (q), with q a power of 2,  = ±1, and n ∈ {m − 1, m} where m ≥ 6. Let p be an odd prime divisor of q − . Let D = x, v ∈ Ep2 (Inndiag(K)) act on K as a diagonalizable subgroup of GLn (q). Assume that

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∼ SL (q) and E(CK (x)) ≥ E(CK (v)) ∼ E(CK (x)) = = SLm−δ (q), δ = 2 or 3. Set m−2 K1 = E(CK (x)), E(CK (v)) . Then either K1 = K or n = m and K1 ∼ = SLm−1 (q). Proof. This is a straightforward application of Lemma 9.1. Since D is diagonalizable we may assume that it is induced by a diagonal subgroup of GLn (q), and thus Kx := E(CK (x)) and Kv := E(CK (v)) are block-diagonal subgroups. Let V be the natural K-module, with basis {ei }ni=1 (orthonormal if  = −1). Then we may assume that Kx is supported on the span of e1 , . . . , em−2 and fixes the remaining one or two ei ’s. Since Kx ≥ Kv , and m−3 ≥ 3, we may assume that Kv is supported on ej , ej+1 , . . . , ek and fixes the remaining ei ’s, where j ≤ m − 2 and k = m − 1 or m. Choose g ∈ Kv fixing e1 , . . . , em−3 and interchanging em−2 with em−1 . Then Kx , Kv ≥ Kx , Kxg ∼ = SLm−1 (q). If Kv fixes em or if n = m − 1, then Kx , Kv is supported on e1 , . . . , em−1 and Kx , Kv = Kx , Kxg . Otherwise  n = m and there is h ∈ Kv interchanging em−1 and em , and we get Kx , Kv ≥ Kx , Kxg , Kxgh = K. The lemma is proved.  Lemma 9.15. Suppose that J ∈ Chev(r) and D ∈ Ep2 (Aut(J)), where p and r are distinct primes. Suppose that for some h ∈ Ip (Aut(J)), CJ (h) has a p-component I with I/Op (I) ∈ Gp . Then J = ΓD,1 (J). Proof. Since h and I exist, J ∈ Gp by [III11 , 1.2]. On the other hand, if J = ΓD,1 (J), then J is one of the groups listed in [IA , 7.3.4]. With the definitions of Cp and Tp [I2 , 12.1, 13.1], we see that J ∈ Cp ∪ Tp , contradiction, unless possibly p = 3 and J ∼ = Sp8 (2). But in that case, with [IA , 4.8.2], we find that I ∼ = A6 , Sp6 (2), or U4 (2), all of which lie in C3 . Thus I ∈ G3 , contrary to assumption, and the lemma follows.  Lemma 9.16. Suppose that X ∈ Kp and G0 ≤ X with G0 ∈ Chev(r) for some prime r distinct from p. Suppose that D ≤ Aut(X) with D ∼ = Ep2 . Assume that for some d ∈ D# , E(CX (d)) has a component K ≤ G0 with K/Op (K) ∈ Gp . If p is odd, assume that mp (Aut(K/Op (K))) ≥ 3. Assume also that ΓD,1 (X) normalizes G0 . Then G0 = X. Proof. If X ∈ Chev(r), then by Lemma 9.15, ΓD,1 (X) = X so G0  X, and thus G0 = X. So we may assume that X ∈ Chev(r). On the other hand, since K ≤ G0 , K is a component of CG0 (d), and since G0 ∈ Chev(r), also K ∈ Chev(r). Thus, K is a component of CX (d) with K/Op (K) ∈ Chev(r) ∩ Gp . Suppose that X ∈ Chev(q) for some q = r. Then as Chev(p)∩Gp = ∅, p ∈ {q, r}. But K ∈ Chev(q) ∩ Chev(r). As K/Op (K) ∈ Gp , mp (K) > 1. Now, the groups of ambiguous characteristic are listed in [IA , 2.2.10], and as p ∈ {q, r} it follows easily that mp (K) = 1, contradiction. If X ∈ Spor, then by [III11 , 1.1b], the condition K ∈ Chev(r) ∩ Gp is never met. Thus we may assume that X ∈ Alt. Then K ∈ Chev(r) ∩ Alt implies that K/O2 (K) ∈ Chev(2), and as K ∈ Gp , p > 2; indeed the only possibility is that p = 3 and K ∼ = A8 . But then mp (Aut(K)) = 2, contradicting our hypothesis. The lemma follows.  ∼ An , n ≥ 13. Let Lemma 9.17. Suppose that X ∈ K2 and G0 ≤ X with G0 = D ≤ G0 be a root four-group and assume that for some d ∈ D# , Gd := E(CG0 (d)) is a component of CX (d) and D ∈ Syl2 (CX (Gd )). Then G0 = X.

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Proof. As X ∈ K2 and G0 is simple, Z(X) × D centralizes Gd with Z(X) a 2group. Since D ∈ Syl2 (CX (Gd )), Z(X) = 1 and X is simple. We have Gd ∼ = An−4 . As d ∈ X and n − 4 ≥ 9, we see from [III11 , 1.6ef] that X ∼ = Am for some m ≥ n. Since Gd is a component of CX (d), CX (Gd ) contains a copy of Am−n+4 supported on the support of d. In particular m ≡ n (mod 4) and |Am−n+4 |2 ≤ |D| = 4.  Therefore m = n and G0 = X. The lemma is proved. Lemma 9.18. There are no groups X, K, P , D, and R with the following properties: K  X = KP with K ∈ K3 , P ∈ Syl3 (X), z = Z(K) = Z(P ) ∼ = Z3 , and |P | = 39 or 310 . Moreover, z D ≤ R ≤ P with D ∼ = E32 , R ∼ = 31+6 , and E1 (D) ⊆ U(X, NX (R), 3). Proof. If such a configuration exists, then m3 (K/ z ) ≥ m3 (R/ z ) = 6 but |K|3 ≤ 310 . As Z(K) ∼ = Z3 , it is impossible that K ∈ Alt ∪ Spor [IA , 6.1.4, 5.6.1], and it is impossible that K ∈ Chev(3) [IA , 6.1.4, Table 3.3.1]. So K ∈ Chev(r), r = 3. As ΓD,1 (K) ≤ NK (R) < K but Z(K) = 1, it follows from [IA , 7.3.4] and [IA , 6.1.4] that K ∼ = SL3 (4), which is absurd as m3 (K) ≥ 6. The proof is complete.  10. Pumpups Lemma 10.1. Suppose that K ∈ Chev(r), x ∈ K has prime order s = r, and L is a Lie component of CK (x). Suppose that f ∈ Aut(K), [f, x] = 1, and f induces a nontrivial field automorphism on L. Suppose also that K is s-saturated. Then f ∈ Inndiag(K). Proof. Suppose that the result is false. Passing first modulo Or (Z(K)), we may assume that K ∈ Lie(r). The saturation hypothesis implies that the kernel N of the universal covering Ku → K is an s -group. Since f ∈ Inndiag(K), f centralizes not only N but also the unique preimage of x in Ku of order s. We may therefore assume that K is the universal version. Let (K, σ) be a σ-setup for K with K universal. Then Inndiag(K) is induced by elements of CK/Z(K) (σ). Consequently f acts on K like an element g ∈ C = CK (x) such that [g, σ] ∈ Z(K) ∩ C ≤ Z(C). By Steinberg’s connectedness theorem [IA , 4.1.3], C is connected and reductive. Therefore g, which maps into CC/Z(C) (σ), induces an inner-diagonal automorphism on L. This contradicts our assumption on f and completes the proof.  2

In the next lemma, f is the ranking function f (L) = q(L)r(L) , where q(L) and r(L) are the level and untwisted Lie rank, respectively, of L ∈ Chev. Lemma 10.2. Let J, K ∈ Chev(2) and suppose that J ↑p K for some prime p ≥ 7. Suppose that max(7, q(K)) < q(J), and suppose that the untwisted Lie rank of K is at least 6. Suppose also that f (K) ≤ q(J)16 . Then J ∼ = L2 (q(J)). Proof. By [IA , 4.2.2, 4.9.1], and as p > 3, J ↑p K via an inner-diagonal automorphism. Moreover, if we let q = q(J), then q(K) = q 1/k and q 16 ≥ f (K) ≥ q 

2

/k

≥ qj

2

k

for some k ≥ 2 such that the untwisted Dynkin diagram of K, of rank , contains a subdiagram consisting of k isometric connected components of rank j. (Note

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that as p ≥ 7, every inner-diagonal automorphism of K is of parabolic type.) Our assumption on f (K) yields j 2 k ≤ 2 /k ≤ 16. As k ≥ 2, j ≤ 2. Clearly also jk < , just by counting nodes of the Dynkin diagram. By assumption  ≥ 6, so k ≥ 3. If k = 3 then  ≤ 6 and so j = 1. If k = 4 then  ≤ 8 and so j = 1. If k > 4 then again j = 1. Thus in any case, j = 1 and  J∼ = A1 (q). Lemma 10.3. Suppose that p is an odd prime and K = d L(q) ∈ Chev(2) is unambiguously of characteristic 2, with L = Am for any m. Let K ↑p J via x for some J. Then J ∈ Chev(2) and either q(J) = q(K) or x is a field or graph-field automorphism of J. Proof. As K is unambiguously of characteristic 2, using [III11 , 1.1ab] and [IA , 2.2.10], we deduce from [IA , 4.9.6] that J ∈ Chev(2). If x is a graph automorphism of J, then p = 3 and q(J) = q(K) by [IA , 4.7.3A]. So by [IA , 4.9.1], we may assume that x ∈ Inndiag(J) and q(J) = q(K). Let Δ be the untwisted Dynkin diagram of J, and Δ∗ the extended diagram. Then by [IA , 4.2.2], q(K) = q(J)d for some integer d. Moreover, by deleting some nodes from Δ, or by deleting a single node from Δ∗ marked with the prime p, we must be able to obtain d pairwise disjoint copies of the Dynkin diagram of L. As p > 2, it is straightforward to check from the diagrams that this can only happen for d > 1  if L is of type Am for some m. Lemma 10.4. Let G0 , L and p > 2 be as in [III15 , Table 15.1]. Let u ∈ Ip (Aut(G0 )) and let J be a component of CG0 (u). Assume that (a) mp (CAut(G0 ) (J)) ≥ 2; and (b) L ↑p J. 



q Then equality holds in (a). Moreover, either (G0 , J) = (Ln+2 (q), SLnq (q)), or q q (G0 , J) = (E6 (q), SL5 (q) or D4 (q)) with p = 3.

Proof. By (a) there is a noncyclic elementary abelian p-group U ≤ Aut(G0 ) such that u ∈ U and J is a component of CG0 (U ). We first claim that Ip (CAut(G0 ) (J)) ⊆ Inndiag(G0 ). Otherwise, there is v ∈ Ip (Aut(G0 )) such that [v, J] = 1 and v is either a field, graph-field, or graph automorphism. In the last  case, G0 ∼ = G2 (q) = D4 (q) as in case (e) of the table, p = 3, and J1 := O 2 (CG0 (v)) ∼ q q or L3 (q). By [IA , 4.7.3A], m3 (CAut(G0 ) (J1 )) = 1. Hence SL3 (q) = L ↑3 J ↑3 J1 . This forces J1 = G2 (q) and then J = L, contrary to L ↑3 J. In the field and graph-field cases, q(J1 ) = q 1/p and J1 and G0 have the same untwisted Lie rank r(J1 ) = r(G0 ) = r, say. Moreover, by [IA , 4.9.1], mp (CAut(G0 ) (J1 )) = 1 so again L ↑p J ↑p J1 . Since q(L) = q, it follows from [IA , 4.2.2] that r(L) ≤ r(J1 )/p = r(G0 )/p. In every case of the table, however, this inequality is false. Thus, our claim holds. It now follows by [III11 , 1.15] that (10A)

L ↑ J ↑ J1 ↑ G 0 p

p

p

for some J1 ∈ Chev(2). Note that if q(J) > q, then q(J) = q p and L ↑p J via a field or graph-field automorphism. In that case q(J1 ) = q p or q and we get r(L) = r(J) ≤ r(G0 )/p, giving the same contradiction as before. A similar argument

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reaches the same contradiction if q(J1 ) > q. Consequently by [IA , 4.2.2], q(J1 ) = q. Again if q(J) < q, then q(J) = q 1/p and we reach a similar contradiction. So we have q(L) = q(J) = q(J1 ) = q(G0 ) = q. In particular, L ↑p J via an element of Ip (Aut0 (J)). Suppose that the untwisted Lie ranks in (10A) are strictly increasing from left to right, whence r(L) + 3 ≤ r(G0 ). This only occurs in cases (b), (c), (d), and (f) − of the table. In case (c), however, since p divides q − q , J1 /Z(J1 ) ∼ = Lm q (q) with m ≤ n+2−2 = n and similarly we get n−3 = r(L) ≤ r(J)−2 ≤ r(J1 )−4 ≤ r(G0 )− 6 = n − 5, contradiction. In case (d), r(G0 ) − r(L) = 3 so r(J1 ) = 5 = r(J) + 1. − By [III11 , Table 13.1], the only possibility is J1 ∼ = L6 q (q), and then as in case (c), r(J) = 3, a contradiction. In cases (b) and (f) we have one of the desired   conclusions. Notice that in case (f), J1 ∼ = A5q (q) or D5q (q) by [III11 , Table 13.1], yielding the claimed possibilities for J. It remains to consider the possibility that for one of the pumpups in (10A), the untwisted Lie ranks of the two groups are equal. By [IA , 4.2.2], p is the mark on an end node of the Dynkin diagram of the larger group. The only possibility is the pumpup L9 (q) ↑3 E8 (q). But then G0 , as a long pumpup of E8 (q) of level q, must be E8 (q), so we are in case (n) of the table. Also as J1 is of type A, so are J and L. But L ∼ = D6 (q) by the table, a contradiction completing the proof of the lemma.  Lemma 10.5. Suppose that K ∈ Chev(2) is simple, mp (Inndiag(K)) ≥ 4, and there is x ∈ Ip (Inndiag(K)) such that CK (x) has a p-component L with L/Op (L) ∈ Gp . Then K is not isomorphic to any of the following groups: L± n (2), n n ≤ 7, Cn (2) or Dn± (2), n ≤ 4, D5− (2), G2 (q), F4 (2), or 2F4 (2 2 ) . Proof. Suppose that K is isomorphic to one of the stated groups. The condition mp (Inndiag(K)) ≥ 4 already rules out most possibilities, forcing p = 3 and − K∼ = L− n (2), 5 ≤ n ≤ 7, C4 (2), D4 (2), D5 (2), or F4 (2) (see [IA , 4.10.3a]). Accordingly, using [IA , 4.8.2, 4.7.3A], we find that L ∼ = SUm (2), m ≤ 6 or L2 (8); Cm (2), m ≤ 3, or U4 (2); U4 (2); D4 (2), U4 (2) or U5 (2); or C3 (2). None of these groups lies  in G3 by definition [I2 , Section 13], so the lemma follows. ∼ H2 = ∼ Lemma 10.6. Suppose that K ∈ K and H1 × H2 × H3 ≤ K, where H1 = H3 ∼ = L2 (q), q = 2n , n > 1. Let p be a prime divisor of q 2 − 1. Assume that for any  i = j and any xi ∈ Ip (Hi ), Hj Hk = O p (E(CK (xi ))), where {i, j, k} = {1, 2, 3}. p Assume also that for B0 ∈ E3 (H1 H2 H3 ), AutK (B0 ) is a {2, 3}-group. Then K ∈ Chev(2) and q(K) = q, or else f (K) > q 9 . Proof. If K ∈ Alt, then by [IA , 5.2.8a], CK (xi ) has at most one component, contradiction. If K ∈ Spor, then the conditions that L2 (q) ↑p K, and mr (K) ≥ 3 for all r dividing |L2 (q)|, narrow down the possibilities to p = 5 and K/Z(K) ∼ = Co1 (see [IA , 5.3, 5.6.1]). But then AutK (B0 ) is nonsolvable for any B0 ∈ E53 (K) [IA , 5.3l], a contradiction. Hence, K ∈ Chev(s) for some s. Suppose that s > 2. Then s = 5 and q = 4, or s = 3 and q = 8, as L2 (q) ∈ Chev(s). As x1 ∈ H1 induces 1 an inner automorphism on K, the case s = 3, L2 (8) ∼ = 2 G2 (3 2 ) is impossible by [IA , 4.2.2], so s = 5. By the Borel-Tits theorem, p = s, so p = 3. By the structure of E(CK (x1 )) and [IA , 4.7.3A], K is not of exceptional type, or of type D4 . In particular, K is a classical group. Then it follows from [IA , 4.8.2] that E(CK (x1 )) m has at most one component not isomorphic to SL± n (5 ) for some m and n, a

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contradiction. We have proved that K ∈ Chev(2). We assume that q(K) = q and show that f (K) > q 9 to complete the proof. Obviously x1 , x2 , x3 ≤ K. Thus, q(K) = q 1/a for some a > 1, by [IA , 4.2.2]. Suppose that K is a classical group. Let V be a natural module for a covering  of K (as K ∈ Chev(2), a proper covering group is not needed unless group K possibly K is of linear or unitary type). Using [IA , 4.8.2, 4.8.4] and the fact that q(K) = q, we see that neither H2 nor H3 is supported on CV (x1 ). Hence as xi ∈ Hi , i = 1, 2, 3, CV (x1 )  CV (x3 ) and by symmetry, CV (x3 )  CV (x1 ), contradiction. Thus, K is of exceptional type. By [IA , 4.7.3AB], p is a good prime for K. By [IA , 4.2.2], the untwisted Dynkin diagram of K must contain at least 2a totally disconnected nodes. As 2a ≥ 4, K ∼ = E7 (q 1/a ) or E8 (q 1/a ), and a ≤ 3. Then 72 /3 9 f (K) ≥ q > q . The proof is complete.  Lemma 10.7. Suppose that L, K ∈ Kp and L 2. Then L4 (q) is unambiguously of characteristic 2 [IA , 2.2.10]. Moreover, mp (L4 (q)) = 3, so by [III11 , 1.1ab], K ∈ Chev(2). So suppose that q = 2, whence p = 3 and U4 (2) ↑3 K. A look at the tables [IA , 5.3] shows that if K ∈ Spor, then K ∼ = Co2 . By [IA , 2.2.10], U4 (2) ∈ Alt ∪ Chev(r) for any r ≥ 5. So K ∈ Chev(2) ∪ Chev(3), and we may assume the latter. Then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], since U4 (2) ∼ = B2 (3)a , K ∼ = B2 (33 )a , finishing the proof.  Lemma 10.10. Suppose that I ∼ = L4 (q) where q is a power of 2 and  = ±1. Suppose that p is a prime, p > 3, q ≡ − (mod p), and q ≡  (mod 3). Let J ∈ K and suppose that I 1 by [IA , 4.10.3a]. Also, since q ≡ ±1 (mod p) and p > 3, we have q > 2. Hence I ∈ Alt ∪ Chev(r) for any r > 2. Now by [III11 , 1.1ab], J ∈ Chev(2) − Chev(r) for any r > 2. As mp (I) > 1, also mp (J) > 1. Note that if q(J) = 2, then as q > 2, it follows from [IA , 4.2.2] that J has untwisted Lie rank at least 6. These last two conditions, combined with the definition of C3 -groups [I2 , 12.1], imply that J/O3 (J) ∈ C3 . Hence if the lemma fails, then J/O3 (J) ∈ T3 , whence by definition [I2 , 13.1], J/O3 3 (J) ∼ = A6 , L3 (q1 ), or G2 (8). In any case m3 (J) = 2. Now I is p-saturated, so since I 2 and D ∼ = Ep2 for some prime p dividing  (X) = 1. Let D ≤ J ≤ X with J ∼ q + 1. Let X be a K-group with Op = L4 (q), #  and for each u ∈ D set Ju = Lp (CJ (u)). Assume that J0 ≤ X with J0 = 1 or J0 ∼ = L2 (q)×L2 (q), and that Lp (CX (u)) = J0 Ju for each u ∈ D# , with [J0 , Ju ] = 1. Assume also that if q = 4, then m5 (CX (u)) ≤ 5 for all u ∈ D# . Then J0  Lp (X), J  X and [J0 , J] = 1. Proof. By assumption and Lp -balance, J0 × Ju ≤ Lp (X) for each u ∈ D# , so J0 × J ≤ Lp (X) by [IA , 7.3.3] applied to the action of D on J. Suppose that

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J projects nontrivially on k > 1 components J1 , . . . , Jk of Lp (X). Then as J is simple, J1 involves J, but also each nontrivial Ju , u ∈ D# , projects nontrivially on J1 , . . . , Jk . Hence by Lp -balance, Ju is a homomorphic image of J1 , a contradiction as J1 involves J. Therefore J lies in a single component X1 of X. For some u ∈ D# , L2 (q 2 ) ∼ = Ju ↑p X1 . By [III11 , 1.1a] and inspection of the tables [IA , 5.3], and as q > 2, X1 ∈ Chev(2). Similarly, each component J1 of J0 lies in a single D-invariant component X0 of X. For otherwise, by Lp -balance, J1 would project nontrivially on exactly p components Y1 , . . . , Yp of X, and each u ∈ D# would cycle Y1 , . . . Yp by conjugation. As D ∼ = L2 (q), q > 2, so = Ep2 this is impossible. Assuming that J0 = 1, we have J1 ∼ as in the previous paragraph, X0 ∈ Chev(2), or else q = 4 and p = 5, with X0 ∼ = As , s ≥ 30, by [III11 , 1.16]. The latter case is impossible as it implies m5 (CX (u)) > 5 for every u ∈ D# , contrary to hypothesis. Thus X0 ∈ Chev(2). Then by Lemma 9.12, X0 = J1 . It follows that J0  E(X), and as J does not embed in J0 , we have [J0 , X1 ] = 1. We shall show that 

(10B)



2 # (1) O 2 (C X1 (u)) = Ju O (CX1 (D))  for every u ∈ D ; and  (2) X1 = O 2 (CX1 (u)) | u ∈ D# .

 # Together  J imply that X1 = Ju | u ∈ D CX1 (D) = JCX1 (D), whence  X these X1 = D 1 = D = J. Furthermore, writing E(X) = J0 JI with I a product of components of X, we have Lp (CX (u)) ≥ J0 Ju Lp (I) for any u ∈ D# , so Lp (I) = 1 and then I ≤ Op (X) = 1. Hence, E(X) = J0 J, which will complete the proof. To prove (10B), choose u1 , u2 ∈ D# such that Jui ∼ = L2 (q i ), i = 1, 2 (see [IA , 4.8.2]). If X1 /Z(X1 ) is not p-saturated, then X1 /Z(X1 ) ∼ = Lkp (q1 ), or E6 (q1 ) with p = 3, and with  = ±1 and p dividing q −  in either case. But then by [IA , 4.8.2, 4.8.4, 4.7.3A], since D ≤ J ≤ X1 , {q, q 2 } = {q(Ju1 ), q(Ju2 )} ⊆ {q1 , q1p }, an impossibility. Hence, X1 /Z(X1 ) is p-saturated.  In (10B1), [IA , 4.2.2] then gives CX1 (u) = O 2 (CX1 (u))Tu for any u ∈ D# , where Tu is abelian of odd order and induces inner-diagonal automorphisms on  each Lie component of O 2 (CX1 (u)). If Ju = 1, then in J we see that D = u ×  (D ∩ Ju ), whence CX1 (D) covers CX1 (u)/O 2 (CX1 (u)) by [IA , 4.2.2j], as claimed in (10B1). If Ju = 1, on the other hand, then as [J0 , X1 ] = 1 we have Lp (CX1 (u)) =  1. Then O 2 (CX1 (u)) is a p -group, and as CX1 (u) induces only inner-diagonal  automorphisms on it, [D, O 2 (CX1 (u))] = 1 and D ≤ Tu . Hence D ≤ Z(CX1 (u)) in this case, completing the proof of (10B1). Finally, suppose that (10B2) fails. Then X1 is one of the groups listed in [IA , 7.3.3], but X1 /Z(X1 ) is p-saturated, as seen above. Each group in [IA , 7.3.3] is ruled out by this or by the facts that E(CX1 (u2 )) ∼ = L2 (q 2 ) with q > 2, and u2 has order a divisor of q + 1. (The order |X1 | is only divisible by p|L2 (q 2 )| if  L2 (42 ), by [IA , 4.8.2].) The lemma X1 = C4 (2) and q = 4, but then E(CX1 (u2 )) ∼ = is proved. 

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Lemma 10.13. Suppose q = 2n ≡  (mod 3),  = ±1. If A5 (q) ↑3 K and m  3 (K) ≤ 4, then K ∼ = A5 (q 3 ). Proof. Suppose that K ∈ K3 and J ↑3 K via x, with J ∼ = A5 (q). By [IA , 2.2.10] and [III11 , 1.1ab], K ∈ Chev(2). If Z(K) = 1, or K is not 3-saturated, then η as m  3 (K) ≤ 4, we see by [IA , 6.1.4, 4.10.3ab] that K ∼ = A5 (q  ), q  ≡ η (mod 3). It is then clear that x is a field automorphism, as desired. So assume that Z(K) = 1 and K is 3-saturated. If x ∈ Inn(K), then since m  3 (J) = 4, m3 (K) > 4, contradiction. Since K is 3-saturated, Outdiag(K) is a 3 -group, so x ∈ Inndiag(K). If x is a  graph or graph-field automorphism, then O 2 (CK (x)) ∼ = 3D4 (q 1/3 ), G2 (q), or A2 (q), contradiction. Thus x is a field automorphism and the proof is complete.  Lemma 10.14. Suppose that q is a power of 2,  = ±1, and p is a prime divisor of q − . Let J = A5 (q) and suppose that J ↑p K via x, with K ∈ Chev(2) and 2 q(K) = q. If K ∈ Gp and K has Gp -depth 2, assume that F(K) ≤ (q 6 , E). Then the following conditions hold: (a) K/Z(K) ∼ = L7 (q), D6 (q), or E6 (q); (b) Suppose that H is a root SL2 (q)-subgroup of J and b ∈ Ip (H), so that CJ (b) has a component I ∼ = A3 (q). If we consider J as a subgroup of K, then I is properly contained in a component of CK (b); and (c) CAut(K) (J x ) has odd order, and if K ∼ = D6 (q), then CAut(K) (J) has odd order. Proof. Since q(K) = q, x is not a field or graph-field automorphism, and from [IA , 4.7.3A] it is clear that x is not a graph automorphism. Thus x ∈ Inndiag(K). Write K = d L(q) and let Δ be the Dynkin diagram of L. If x is of equal-rank type as in [IA , 4.2.2], then p = 3 and K ∼ = E6 (q). Thus in any case L has rank  p (K) > 4,  ≥ 6. Together with Lemma 10.13 and [IA , 4.10.3a], this implies that m 2 2 so K ∈ G2p . Hence by assumption, (q  , L) ≤ (q 6 , E), and (a) follows. In (b), H is a (long) root SL2 (q)-subgroup of K. From the Dynkin diagram it follows that CK (b) contains a Lie component ∼ A (q)u , D4 (q), or A (q), I0 = 4

5

according to the isomorphism type of K. Thus if I < I0 , then (b) holds. Other wise, since I ≤ O 2 (CK (b)), a product of Lie components, there would exist a Lie component I1 containing a copy of A3 (q), and [I0 , I1 ] = 1. In every case, however, this leads to mp (I0 I1 ) ≥ mp (I0 ) − mp (Z(I1 )) + mp (I1 ) > mp (K), a contradiction as I0 I1 ≤ K. Hence (b) holds. Suppose now that CAut(K) (J) contains an involution t. By Lemma 6.1, t ∈ Aut0 (K). Suppose that t ∈ ΓK ∪ ΦK and t is a graph automorphism. Then by [IA , 4.9.2], CK (t) ∼ = Sp6 (q), Sp10 (q), or F4 (q), according to the isomorphism type of K. But unless CK (t) ∼ = Sp10 (2), |J| does not divide |CK (t)| (use a primitive prime divisor of q 5 −  or q 3 −  to see this). And if CK (t) ∼ = Sp10 (2) and J ∼ = U6 (2), then |CK (t)|3 = |J|3 but m3 (CK (t)) = 5 = m3 (J) = 4. Thus J does not embed in CK (t), contradiction. Hence if t is a graph automorphism, then t ∈ ΓK ∪ΦK . By [IA , 4.9.2e] and the Borel-Tits theorem, we may then assume that F ∗ (CK (t)) = O2 (CK (t)) and CK (t) lies in a t-invariant parabolic subgroup P of K. Since A5 (q) is not involved in D5 (q), by similar considerations to the above, P must be of type A5 (q) (with  = 1), A5 (q), or A5 (q), according to the isomorphism type of K.

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If K ∼ = A6 (q), however, O2 (P ) is a natural A5 (q)-module and so some element of A5 (q) of odd order acts without fixed points on O2 (P ). Thus, t must be a graph automorphism; but no graph automorphism of A6 (q) normalizes an A5 (q)-parabolic subgroup, contradiction. If K ∼ = D6 (q), then for the same reason t cannot be a graph automorphism, so t ∈ O2 (P ). But again O2 (P ) has no fixed points as an A5 (q)-module, contradiction. (A fixed point would correspond to an A5 (q)-invariant symmetric bilinear form on the natural module, but the only such form is the 0 form.) Thus K ∼ = E6 (q) and we may assume that t centralizes J x . If t induces a (nontrivial) graph automorphism on K, then it induces such an automorphism on P (∞) /O2 (P (∞) ) ∼ = A5 (q), contrary to assumption. So t induces an inner automorphism on K, and we may assume that t ∈ CK (x). Then by [IA , 4.2.2],  t ∈ O 2 (CK (x)) =: L, a central product of Lie components of CK (x). If x is of equal rank type, then p = 3 and the diagram Δx for L is a product of A2 diagrams. But J is a component of CK (x). Hence, x is of parabolic type and Δx is an A5 diagram. Then t ∈ L = J, whence t ∈ Z(J). But |Z(J)| is odd, a final contradiction. 

Lemma 10.15. Suppose that p is an odd prime dividing q 2 − 1, where q = 2n . If K ∈ Kp ∩ Chev(2), D4 (q) ↑p K, and q(K) = q, then m  p (K) ≥ 5. Proof. Say D4 (q) ↑p K via x ∈ Aut(K). Since q(K) = q, x is not a field or graph-field automorphism, and by the isomorphism type of D4 (q), and [IA , 4.7.3A], x is not a graph automorphism. So x ∈ Inndiag(K). By [IA , 4.2.2], the Dynkin diagram or extended Dynkin diagram of K contains a D4 diagram.  If K ∼ = E6q (q) with p = 3 and q ≡ q (mod 3), q = ±1, then the universal  p (K) = 5. version Ku satisfies m3 (Ku ) = 6 and m3 (Z(Ku )) = 1, so m Otherwise, p does not divide |Outdiag(K)| or the order of the Schur multiplier of K, so x ∈ Inn(K) and m  p (K) = mp (K) ≥ mp (x E(CK (x))) ≥ 1 + mp (D4 (q)) = 5.



Lemma 10.16. Suppose q = 2n ≡  (mod p), p an odd prime,  = ±1. If K and K is involved in E6 (q), then K ∼ = E6 (q).

D5 (q) ↑p

Proof. Suppose that J = D5 (q) ↑p K via x ∈ Aut(K). Then x is not a graph or graph-field automorphism, as J is not of type 3 D4 , G2 , or A2 [IA , 4.7.3A]. If x is a field automorphism, then K = D5 (q p ) is involved in E6 (q), so q 20p ≤ q 36 , contradiction. Thus x ∈ Inndiag(K). If q(K) < q, then using [IA , 4.2.2], we see that the only possibility is K ∼ = D10 (q 1/2 ), and again Lagrange’s theorem prevents  K from being involved in E6 (q). Thus q(K) = q, and the only possibilities by Lagrange’s theorem are K = E6± (q) and D6± (q). By considering primitive prime  divisors of q 5 ± 1, we see that the only possibility is K = E6 (q), as asserted.

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Lemma 10.17. Let q = 2n and let p be a prime divisor of q 2 −1. Suppose  = ±1 and q ≡  (mod p). Let K ∈ Kp with K/Z(K) ∼ = L− n (q), n = 6 or 7. Suppose p that X ∈ Kp and K ↑p X. Let B ∈ E∗ (X), and suppose that mp (B) = 4 and the subgroup of AutX (B) generated by all its reflections is isomorphic to W (C4 ). Then X/Z(X) ∼ = L− n+2 (q). Proof. As usual, we use [III11 , 1.1ab] and [IA , 2.2.10] to get X ∈ Chev(2). Say K ↑p X via x ∈ Ip (Aut(X)). If x is a field automorphism, then X/Z(X) ∼ = p p (q ), and since q ≡ q ≡  (mod p), m (X) = 3 by [I , 4.10.3a], contrary to L− p A n assumption. It is clear from the isomorphism type of K/Z(K) and [IA , 4.7.3A, 4.9.1] that x is not a graph or graph-field automorphism. Therefore, x ∈ Inndiag(X). If X is a classical group, then with [IA , 4.8.2, 4.8.4], we have X/Z(X) ∼ = L− m (q), m ≥ n + 2, m ≡ n (mod 2). Since mp (X) = 4, m = n + 2, so the lemma holds in this case. Thus, assume that X is of exceptional type, so that the untwisted Lie rank is at most 8. As the untwisted Lie rank of K is more than half that of X, it follows from [IA , 4.2.2] that q(X) = q. Note also that as mp (X) = 4, we cannot have p = 3 with X ∼ = E6 (q). So x ∈ Inn(X), and Z(X) is a p -group. As mp (X) = 4, it follows that K is terminal in X, so K appears in [III11 , Table 13.1]. The only choice is n = 6 and X ∼ = W (F4 ) by Lemma = E6− (q). But then AutX (B) ∼ 2.6, contrary to assumption. The proof is complete.  Lemma 10.18. Let q be a power of 2,  = ±1, and p an odd prime such that q ≡  (mod p). Let m ≥ 5 and L, J ∈ Kp , and suppose that L ↑p J, where L is a central quotient of SLm−1 (q). If q = 2, assume that m ≥ 7. Let K = Lm (q). Assume that if J ∈ Gp , then the Gp -depth of J is at least as great as that of K, and in case of equality, F(J) ≤ F(K). Suppose further that I ↑p J for some I ∈ Kp with I/Z(I) ∼ = Lm−2 (q). Then ∼ J/Z(J) = K. Proof. Given the isomorphism type of L, J must lie in Chev(2) by [IA , 2.2.10] and [III11 , 1.1ab]. Let J have untwisted Lie rank  and type τ = A, BC, . . . . Let L ↑p J via α ∈ Aut(J). As m ≥ 5, the structure of L shows by [IA , 4.7.3A] that α is not a graph automorphism. If α is a field or graph-field automorphism, then no I can exist as assumed. Therefore α ∈ Inndiag(J). Suppose first that F(J) ≤ F(K). 2

2

If q(J) = q, then (q  , τ ) ≤ (q (m−1) , A). Since L ↑p J, m − 2 ≤ . If strict inequality holds, then  = m − 1 and τ = A, and the conclusion of the lemma follows. Otherwise  = m − 2 so α is an equal-rank automorphism. As p is odd, the only possibility is p = 3, m = 10, and J ∼ = E8 (q). But in that case the group I in our hypothesis cannot exist, by [IA , 4.7.3A]. Still assuming that F(J) ≤ F(K), suppose now that q(J) = q (= q(L)). It follows from [IA , 4.2.2] that for some d ≥ 2, q(J) = q 1/d and  ≥ d(m − 2). But 2 2 then f (J) ≥ q d(m−2) > q (m−1) = f (K) as m ≥ 5, a contradiction. Therefore F(J) > F(K). As a result, our assumptions imply that either J/Op (J) ∈ Gp , or else the Gp depth of J/Op (J) is strictly larger than that of K. If q > 2, then as J is a proper pumpup of L with L/Z(L) ∼ = Lm−1 (q), m ≥ 5, it is clear with the help of

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[IA , 4.2.2] and the definition of Gp [IA , 12.1, §13] that J/Op (J) ∈ Gp . If q = 2, then L/Z(L) ∼ = Un (2), n ≥ 6, and p = 3, and we again see that J/Op (J) ∈ Gp . Also mp (L) ≥ 3, so mp (J) ≥ 3. Thus the only possibility is that J/Op (J) has Gp -depth 4 and K/Op (K) has Gp -depth 2, see [III12 , (1I)]. By [III11 , 1.2], L/Op (L) then has Gp -depth 4 as well. Since dp (K) = 2, m ≥ 6, with strict inequality if p = 3 [III12 , (1I)]. The condition on L implies that m − 1 ≤ 6, with p = 3 in case of equality. Hence, m = 6 or 7 according as p = 3 or p = 3. Suppose that p does not divide |Outdiag(J)|. Then Z(J) = 1 and J is psaturated, by [IA , 6.1.4]. Hence 4 ≥ mp (J) ≥ 1 + mp (L) − mp (Z(L)). The only possibility is p = 5, m = 6, L/Z(L) ∼ = L5 (q), and m5 (J) = 4. If J is a classi cal group, then from [IA , 4.8.1] we see that J contains O 5 (GL5 (q)), contrary to m5 (J) = 4. Thus J is of exceptional type. If q(J) = q, then by [IA , 4.2.2] and the fact that L/Z(L) ∼ = A4 (q), we must have J ∼ = E8 (q 1/2 ). But then by [IA , 4.7.3B], there is no I ↑5 J as in our hypothesis. And if q(J) = q, then as q ≡  (mod 5) and m5 (J) = 4, we have J ∼ = F4 (q) or E6− (q). But in neither case does J have an element of order 5 with an A4 (q) component in its centralizer, by [III11 , Table 13.1], a contradiction. Suppose finally that p divides |Outdiag(J)|. Thus J is of type A or of type E6 . In the latter case p = 3 and by [IA , 4.10.3a], the universal version of J has 3-rank 6 with |Z(J)| = 3, whence J ∈ G23 , contradiction. In the former case as J ∈ G4p we must have p = 3 and m = 7 with J ∼ = A40 (q0 ). Here = A50 (q0 ), or m = 6 with J ∼ q0 is a power of 2, 0 = ±1, and q0 ≡ 0 (mod p). But L ∼ = A5 (q) or A4 (q) in these , 4.2.2], L cannot be a component of CJ (α), a final cases, respectively, so by [IA contradiction.  Lemma 10.19. Let q be a power of 2, p an odd prime, and q = ±1 such q (q). Set that q ≡ q (mod p). Let n ≥ 5 with n ≡ −1 (mod p), and L = SLn−1 q   K = Ln+1 (q) and suppose that K/Op (K) ∈ Gp . Then L/Op (L) ∈ Gp . Moreover, if I is a level vertical pumpup of L, then I/Op (I) ∈ Gp and one of the following holds: (a) The Gp -depth of I/Op (I) is less than that of K/Op (K); (b) F(I) > F(K) and the Gp -depths of I/Op (I) and K/Op (K) coincide; ∼ Lnq (q) or K/Z(K); or (c) I/Z(I) = −n (d) I ∼ = Sp2n−2 (q), or Ω q (q), or En−1 (q) (n = 7, 8, 9), or E − (q) (n = 7). 2n−2

6

Proof. If q = 2, then p = 3 and q = −1; as K/O3 (K) ∈ G3 , we must have n + 1 ≥ 7. But p divides n + 1, so n + 1 ≥ 9, n − 1 ≥ 7, and thus L/O3 (L) ∈ Gp in this case. If q > 2, then L/Op (L) ∈ Gp since we are assuming that n − 1 ≥ 4 > 3. Thus the first assertion of the lemma holds, and by [III11 , 1.3], I/Op (I) ∈ Gp as well. Consequently, as mp (I) ≥ 3, I ∈ Spor. Then by [IA , 2.2.10] and [III11 , 1.1a], I ∈ Chev(2). Suppose that (a), (b), and (c) fail. Since n ≥ 5, mp (K) ≥ 4, and indeed K/Op (K) ∈ G2p if n > 5 or p > 3, while K/Op (K) ∈ G4p if n = 5 and p = 3 (see [IA , 4.10.3a] and [III12 , (1I)]). Accordingly, as I ∈ Chev(2), and (a) fails, we have I/Op (I) ∈ G2p ∪G4p , or I/Op (I) ∈ G4p . (Notice that mp (L) ≥ 3 as n ≥ 5, so that L/Op (L) and hence I/Op (I) have Gp -depth at least 4.)

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Let  be the untwisted Lie rank of I, so that f (I) = q  . Assume first that 2  ≥ n. We have F(K) = (q n , A), so F(I) > F(K) (otherwise (c) holds, contrary to assumption). Since (b) fails, we must have I/Op (I) ∈ G4p and K/Op (K) ∈ G2p . Thus n > 5 or p > 3. Moreover, mp (I) ≤ 5, with strict inequality unless p = 3  and I/Z(I) ∼ = L6q (q). But in the latter case, (c) must hold, contradiction. Thus mp (I) ≤ 4. As I is a level pumpup of L, the pumpup is not by a field or graph-field automorphism. It follows that mp (L) ≤ 3. Hence n = 5 and p > 3. This is a contradiction as p divides n + 1. We conclude that  < n. If  ≤ n − 2, then since L has untwisted Lie rank n − 2, I must be a pumpup of L via an inner-diagonal automorphism of equal-rank type [IA , 4.2.2]. As p is odd,  the only possibility is L ∼ = E8 (q), with p = 3. But in that case, = A8q (q) and I ∼ L is not the universal version [IA , 4.7.3A], contrary to our assumption. Therefore  = n − 1. The untwisted Dynkin diagram of I must contain an An−2 -subdiagram [IA , 4.2.2] and it follows easily that either (c) or (d) holds. Notice the absence of exceptional Schur multipliers [IA , 6.1.4] in (c) and (d), since we know that if q = 2, then n + 1 ≥ 9. The proof is complete.  Lemma 10.20. Suppose that q is a power of 2, p is an odd prime,  = ±1 and q ≡  (mod p). Suppose that J ↑p I is a pumpup with I, J ∈ Chev and q(I) = m q(J) = q. Assume that J ∼ = Ω2m (q) with m ≥ 4. Then mp (I) ≥ 5 and one of the following holds: m+1 (a) I ∼ = Ω2m+2 (q); (b) m ≤ 7 and I ∼ = E6 (q), E7 (q), or E8 (q); or (m+2)2 , A). (c) F(I) > (q Proof. Since p is an odd prime and because of the isomorphism type of J, the (level) pumpup must be via an inner-diagonal automorphism α of I. Using [IA , 4.2.2] we see that α must be of parabolic type. Thus, the untwisted Dynkin diagram Δ of I properly contains a Dm subdiagram, so Δ must be a Dn or En diagram for some n. As mp (J) = m ≥ 4 by [IA , 4.10.3a], mp (I) ≥ 5 unless possibly α induces an outer diagonal automorphism on I; but even in that case, p = 3, I∼ = E6 (q), and m3 (I) ≥ 5, proving the first assertion. If Δ is of type Dn , then clearly n ≥ m + 1, and mp (I) ≥ m + 1 as well, so either (a) or (c) holds. On the other hand, if Δ is of type En , then as mp (I) ≥ 5, I∼  E6− (q) by [IA , 4.10.3a]. Also m < n ≤ 8 as α is of parabolic type, so (b) holds. = The proof is complete.  Lemma 10.21. Suppose that K ∼  = Dn (q),  = ±1, n ≥ 4, q even, but K ∼ = D4 (2). Let p be a prime divisor of q 2 − 1 with q ≡ q = ±1 (mod p), and set q (q). Let J ∈ Chev(2) ∩ Kp with L ↑p J and q(J) = q. Then J ∈ Gp . If L = Dn−1 2 either dp (J) > dp (K) or F(J) ≤ (q n , D), then one of the following holds: (a) J ∼ = K; or (b) n = 4 and J ∼ = A± 4 (q). Proof. If J ∈ Cp ∪ Tp , then using [IA , 4.2.2] and the definitions of these sets [I2 , 12.1], we conclude that L cannot have the assumed form, contradiction. Thus, J ∈ Gp . Say L ↑p J via x. Suppose that dp (J) > dp (K). If dp (K) = 4, then dp (J) = 5 so − mp (J) = 2. Hence mp (L) ≤ 2, whence n = 4 and L = A3 q (q). Thus mp (L) = 2,

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so x induces a non-inner automorphism on J. Because A3 q (q)  E(CJ (x)), x does not induce an outer diagonal automorphism on J. As q(J) = q(K) = q(L), x does not induce a field or graph-field automorphism on J. Thus x induces a graph automorphism on J, and in particular p = 3. But then the structure of L is inconsistent with the structure of CJ (x) as found in [IA , 4.7.3A]. We have 2 proved that dp (J) ≤ dp (K), so F(J) ≤ (q n , D) by assumption. Consequently the untwisted rank of J is at most n, and if it equals n, then J is a group of type A or D. As the untwisted Dynkin diagram Dn−1 of L is a subdiagram of the extended untwisted diagram of J, J cannot be of type A or F , unless n = 4 and J ∼ = A± 4 (q), an allowed conclusion. Likewise if J is of type D, then since L ↑p K and L ↑p J, we must have J ∼ = K. Thus we are done unless possibly J has untwisted rank n − 1, ± (q) or Bn−1 (q). But then as p is odd, the extended diagram for J with J = En−1 does not yield the Dn−1 diagram when a node is suppressed whose coefficient in  the lowest root is p. Hence by [IA , 4.2.2], L ↑p J gives a final contradiction. Lemma 10.22. Suppose that L ↑p K, where p is an odd prime dividing 22n − 1. (a) If L ∼ = E7 (2n ) and K embeds in E8 (2n ), then K ∼ = E8 (2n ); and n n ∼ (b) If L = D6 (2 ) and K embeds in E7 (2 ), then K ∼ = E7 (2n ). Proof. Say L ↑p K via x ∈ Aut(K). If x ∈ Inndiag(K), then by [IA , 4.9.1, 4.9.2], K ∼ = E7 (2np ) or D6 (2np ). As K embeds in E8 (2n ) or E7 (2n ), 263np ≤ 2120n 30np or 2 ≤ 263n , both of which are impossible. Therefore x ∈ Inndiag(K). We apply [IA , 4.2.2]. In (a), the untwisted extended Dynkin diagram Δ of K has the untwisted diagram Δ0 of L as a subdiagram and is not itself E7 , so K = E8 (q) for some q. As Δ does not contain two disjoint copies of Δ0 , q = 2n , as claimed. In (b), we can make a similar argument unless Δ is the extended Dm diagram for ± n/k some m ≥ 7. But in the latter case, we would have m ≥ 6k, with K ∼ (2 ) = Dm n and k ≤ 2. As K embeds in E7 (2 ), comparing Sylow 2-subgroups would yield 63n ≥ m(m − 1)n/k ≥ 6k(6k − 1)n/k, ± n whence k = 1. Hence K = Dm (2 ). In particular D7± (2n ) would embed in E7 (2n ). But this would contradict Lemma 16.2. The proof is complete. 

Lemma 10.23. Suppose that O3 (X) = 1, |X|3 ≤ 36 , and m3 (X) ≤ 4. Let K be a component of X and x ∈ I3 (NX (K)) an element such that CK (x) has a subnormal subgroup R ∼ = 31+2 , and R ≤ I ∼ = SU3 (2) for some I   CX (x). Assume that Z(I) ≤ Z(X). Then K ∼ = SU3 (8) and x induces a field automorphism on K. Proof. This is a minor strengthening of [III11 , Lemma 13.29], whose proof reduces us to obtaining a contradiction if K ∼ = 3J3 or 3O  N . Since R  I, I ≤ NX (K). Hence I embeds in H := O3,2 (CAut(K) (x)). However, by [IA , 5.3hs],  |H|2 ≤ 4 < |I|2 , a contradiction. Lemma 10.24. Suppose that U4 (2) ↑2 K via x. Then K ∼ = L4 (4), L± 5 (3), or ± P Ω (3) with k ≥ 6. Moreover, in the first three cases, all Ω5 (9), or K/Z(K) ∼ = k involutions in Kx are Inndiag(K)-conjugate. Proof. First, inspection of the tables [IA , 5.3] shows that K ∈ Spor. Since U4 (2) is not in Alt ∪ Chev(r) for any r ∈ {2, 3} by [IA , 2.2.10], neither is K, and so K ∈ Chev(2) ∪ Chev(3). If K ∈ Chev(2), then by the Borel-Tits theorem and

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∼ L4 (4), and the final [IA , 4.9.1, 4.9.2], x is a graph-field automorphism of K = assertion holds in this case. We may therefore assume that K ∈ Chev(3), and regard U4 (2) ∼ = Ω5 (3). If x is a field or graph-field automorphism, the only possibility is K ∼ = Ω5 (9), and the final assertion holds in that case by [IA , 4.9.1]. Suppose then that x ∈ Aut0 (K). Clearly 1 K∼  2 G2 (3 2 ) , so by [IA , 4.2.2], q(K) = 3. We then survey the possibilities for K = ± ∼ in the table [IA , 4.5.1]. Suppose that K ∼ = A± n (3). If n = 3 then K = P Ω6 (3), so suppose that n = 3. We see that the only possibility is n = 4 and all involutions in the coset Inndiag(K)x are Inndiag(K)-conjugate, as desired. If K ∼  A± = n (3), then the table shows that K must be Bn (3), n ≥ 3, or Dn± (3), n ≥ 4, as asserted. (If K ∼ = Cn (3), then K has a subcomponent C2 (3) = B2 (3), but it is the universal  version, whereas Ω5 (3) is simple.) Lemma 10.25. Suppose that K ∈ Chev(2) is a classical group and K  X = KB, where B is an elementary abelian 3-group inducing inner-diagonal automorphisms on K and containing every element of I3 (CX (B)). Suppose that b ∈ B # , J is a component of CK (b), J ∼ = U4 (2) or D4 (2), and CB (J) = b . Suppose further that AutCAut(K) (b) (J) contains O6− (2) or O8+ (2), according to the isomorphism type of J. Then K ∼ = Un (2), n ≤ 6, Dn± (2) or Cn (2), n ≤ 4, or D5− (2). Proof. By assumption B induces inner-diagonal automorphisms on K. As q(J) = 2 it follows from [IA , 4.2.2, 4.8.4] that q(K) = 2 and [IA , 4.8.2] applies. If J ∼ = D4 (2), we conclude that K is an orthogonal group, and as CB (J) = b , K∼  Ln (2), = D5− (2) is the only possibility. Next, assume that J ∼ = U4 (2). Then K ∼ = ∼ and if K = Un (2), the condition CB (J) = b forces n ≤ 6. Note that Z(K) is a 3 -group since CB (J) = b and Z(J) is a 3 -group. We are reduced to the cases K∼ = Dn± (2) and Cn (2), and the condition CB (J) = b yields m3 (B) = 4 and then n ≤ 4, as desired, or K ∼ = D5 (2). In this final case, let V be the natural 10-dimensional F2 K-module. Then b ∈ K and dim[V, b] = 8. Hence CAut(K) (b) = CO(V ) (b) = CO([V,b]) (b) × O(CV (b)) ∼ = GU4 (2) × Z2 For the last isomorphism, see [GL1, p. 95]. Thus AutCAut(K) (b) (J) ∼  = U4 (2) ∼ = − O6 (2), contrary to hypothesis. The proof is complete.  Lemma 10.26. Let X be a K-group and t ∈ I2 (X) with CX (t) having a component H ∼ = A8 . Suppose that H ≤ I for some component I of X. Assume also that D ≤ X with D ∼ = A5 for two u ∈ E1 (D), and = E32 , D normalizes H, E(CX (u)) ∼ E(CX (u)) = 1 for the other two u ∈ E1 (D). If m3 (CX (D)) ≤ 3, then H = I. ∼ Proof. Since t and D normalize H, they normalize I. By [III11 , 1.6], I = A8+2n , n ≥ 0, L4 (4), HS, or 2HS. In the last three cases, it follows from [IA , 4.8.2, 5.3] that E(CX (u)) contains a copy of A5 for all u ∈ D# , contrary to hypothesis. Thus, I ∼ = A8+2n and we assume for a contradiction that n > 0. Let Ω be a set of size 8 + 2n on which I acts naturally. As H = E(CI (t)), Ω = X ∪ Y , where X is the support of H of size 8, and Y is the fixed point set of H. Now, D ≤ I × CX (I), and indeed, since D normalizes H, we have D = O 2 (D) ≤ H × CI (H) × CX (I). For u ∈ D, we shall speak of the action of u on Ω, meaning the action of the projection of u into I. Our assumptions imply that I ≤ CX (u) for all u ∈ D# . If u ∈ D# centralizes H, then H ≤ E(CX (u)), contrary to assumption. Hence,

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D projects onto a Sylow 3-subgroup of H. If u ∈ D projects to a 3-cycle of H, then E(CH (u)) = E(CX (u)), whence u acts without fixed points on Y . If k is the number of D-orbits on Y , then m3 (CI (D)) ≥ 2 + k, whence k = 1 and |Y | = 3 or 9. But, |Y | = 2n, a contradiction.  ± Lemma 10.27. If K ∈ ↑2 (Sp6 (2)), then K/Z(K) ∼ = Sp6 (4), L± 6 (2), L7 (2), or F i22 .

Ω± 8 (2),

Proof. By [IA , 4.9.5, 5.2.8], K ∈ Chev(2) ∪ Spor. Then the result follows  directly from [IA , 4.9.2, 5.3] and the Borel-Tits theorem. Lemma 10.28. Let K ∈ K2 and x ∈ I2 (K). Then there is no component L of CK (x) such that x ∈ L ∼ = 2D4 (2). Proof. Suppose false. As x ∈ L, we may factor away Z(K) and assume that K is simple. By [IA , 4.9.5, 5.2.8, 5.3], K ∈ Chev(2), but this violates the Borel-Tits theorem.  Lemma 10.29. Suppose K ∼ = E6− (2), Z(K) has odd order, and K ↑3 K ∗ ∈ K3 . Then the Schur multiplier of K ∗ has odd order. Proof. Clearly K ∗ ∈ Alt ∪ Spor [IA , 5.3]. Thus K ∗ ∈ Chev(2). But E6− (2) is the largest group in Chev(2) with Schur multiplier of even order, by [IA , 6.1.4], and the result follows.  Lemma 10.30. Let {J, K} ⊆ Gp ∩ Chev(r) for some prime r such that exactly one of p and r is 2. In the following cases there is no p-pumpup I ∈ Chev(r) of any covering group of J such that F(J) < F(I) ≤ F(K): ± 2 (a) p = 2, (J, K) = (SL± 8 (q), L6 (q )); 2 (b) p = 2, (J, K) = (P Sp8 (q), Sp± 6 (q )); q (c) p = 2, (J, K) = (Ω6 (q), Ω7 (q)) and I is a nonlevel pumpup of J; (d) p = 2, (J, K) = (L3 (q 2 ), L± 4 (q)); 2 (e) p = 2, (J, K) = (L5 (q), L± 4 (q )); 2 (f) p = 2, (J, K) = (P Sp4 (q ) or G2 (q 2 ), L± 4 (q)); (g) p = 5, (J, K) = (A4 (q), D4 (q)), q ≡  (mod p),  = ±1; (h) p = 7, (J, K) = (A6 (q), E6 (q)), q ≡  (mod p),  = ±1; and 2 (i) p = 3, (J, K) = (L8 (q), L± 6 (q )). Proof. Suppose by way of contradiction that J is a covering group of J, J ↑p I via x ∈ Aut(I), and (10C)

F(J) < F(I) ≤ F(K).

We set ρ(K, I) = logf (I) (f (K)) and define ρ(K, J) and ρ(I, J) similarly. Thus (10D)

ρ(K, J) = ρ(K, I)ρ(I, J) ≥ ρ(I, J).

We have 50 50 18 9 9 18 9 , , , , , , 1, 1, 49 16 9 8 16 8 49 in the respective cases (a)–(i). In particular ρ(K, J) = logf (J) (f (K)) =

(10E)

ρ(I, J) ≤ ρ(K, J) < 2,

and indeed (10F)

ρ(I, J) ≤ ρ(K, J) < ( + 1)2 /2 ,

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where  is the untwisted Lie rank of J. If x is a field or graph-field automorphism of I, then ρ(I, J) = p, contradicting (10E). Next suppose that x is a graph ± automorphism of I. By [IA , 4.5.1], either (b) holds with I ∼ = A± 7 (q) or E6 (q), ± ± ± 2 2 ± 2 or (f) holds with I ∼ = A3 (q ), A4 (q ), or Dn (q ), n ≥ 4, or D5 (q). In every case, ρ(I, J) ≥ min(2, ( + 1)2 /2 ), contrary to (10E) and (10F). Thus we may

Then with assume that x ∈ Inndiag(I). Suppose that I is a nonlevel pumpup of J. [IA , 4.5.1] and [IA , 4.2.2], the untwisted Lie rank of I is at least k times that of J, where k is the degree of the field of definition of J over that of I. Since k ≥ 2, it again follows that ρ(I, J) ≥ 2, a contradiction.

In particular (c) does not We may now assume that I is a level pumpup of J. occur. By (10F) and [IA , 4.2.2], the untwisted Lie rank of I must be , and x is of equal-rank type. In particular the possibilities for I and J, if any, are to be found

we see from those tables in [IA , 4.5.1, 4.7.3A, 4.7.3B]. As I is a level pumpup of J, that no such x and I exist. This completes the proof.  Lemma 10.31. Let {J, K} ⊆ G2 ∩ Chev(r) for some odd prime r. Let I ∈ Chev(r) be a vertical 2-pumpup of some covering group of J such that F(J) < F(I) ≤ F(K). Then the following conditions hold: 2 (a) If (J, K) = (SL4 (q), L± 4 (q )) and J is a component of CI (x) for some ∼ x ∈ I2 (J), then I/Z(I) = L5 (q); 2 ∼ (b) If (J, K) = (D4± (q), L± 4 (q )), then I = B4 (q); ± ± (c) If (J, K) = (L3 (q), L4 (q)) and I is a nonlevel pumpup of J, then I ∼ = L3 (q 2 ); − ± (d) If (J, K) = (L± 4 (q), D4 (q)) and I is not a pumpup of SL4 (q), then I is a ± quotient of Ω7 (q) or Ω8 (q); ∼ (e) If K ∼ = Spin± 2k (q) for some k ≥ 4 and odd q, J = Spin2k−1 (q), I is a level pumpup of J, and J is a component of CI (Z(J)), then k = 5 and I∼ = F4 (q); and ± (f) If (J, K) = (Dn−2 (q) or Cn−2 (q), L± n (q)), n even, n ≥ 8, and I is a ± ∼ (q), but J ∼  HSpin2n−4 (q). pumpup of J, then I = Bn−2 (q) and J ∼ = Dn−2 = Proof. We continue the notation and argument of Lemma 10.30, with p = 2. Here in (a) and (b), we have ρ(K, J) = 2, 18/16, respectively. Moreover, in (b),  = 4, and ρ(K, J) < ( + 1)2 /2 . The argument of Lemma 10.30 shows that x ∈ Inndiag(I) in case (b), and furthermore that I is a level pumpup of J and x is of equal-rank type, with I of untwisted rank 4. Then from [IA , 4.5.1] we see that I∼ = B4 (q), as desired. In case (a), x ∈ J ≤ I by assumption, with x ∈ Z(I). Since J is the universal 2 version, we see from [IA , 4.5.1, 4.5.2] that I ∼ = A± m (q), m ≥ 4. Thus ρ(I, J) = m /9. ± But ρ(I, J) ≤ ρ(K, J) = 2, so I ∼ = A4 (q) and (a) follows. In (c), we are done if x induces a field or graph-field automorphism on I. From [IA , 4.5.1] and the structure of J, x does not induce a graph automorphism on I, so x induces an inner-diagonal automorphism on I. Then since I is a nonlevel pumpup of J, [IA , 4.2.2] implies that the centralizer of x in a simple algebraic group overlying I must have at least two components of type A2 . The only possibility, by 1/2 ); but then f (I) = q 25/2 > q 9 = f (K), [IA , Tables 4.3.1, 4.5.1], is that I ∼ = A± 5 (q a contradiction. In (d) we again have ρ(K, J) = 16/9 = ( + 1)2 /2 < 2. As in (b) and Lemma 10.30, we see that x does not induce a field or graph-field automorphism on I. If

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∼ A± (q), x induces a graph automorphism on I, then from [IA , 4.5.1] we get I = 5 contradicting F(I) ≤ F(K). So x induces an inner-diagonal automorphism on I. As ρ(K, J) < 2 we see as in Lemma 10.30 that I is a level pumpup of J. Thus I has untwisted Lie rank 3 or 4. From [IA , 4.5.1] and the fact that I is not a pumpup of ± ∼ SL± 4 (q), we get that I = B3 (q) or D4 (q), but in neither case is I a full spin group, whence (d) follows. In (e), similar arguments show that x does not induce a field or graph-field automorphism on I; since J is a component of CI (Z(J)), [IA , 4.5.1] shows that I = F4 (q) is the only possibility, with k = 5. Finally in (f), ρ(K, J) = (n − 1)2 /(n − 2)2 = ( + 1)2 /2 ≤ 49/36 < 2, so as usual, x ∈ Aut0 (I). If x induces a graph automorphism of I, then from [IA , 4.5.1], I∼ = A± 2n−5 (q), and 2n − 5 > n − 1 as n ≥ 8, so F(I) > F(K), contradiction. So we may assume that x induces an inner-diagonal automorphism of I. As in the proof of Lemma 10.30, I is a level pumpup of J. If I has untwisted rank n − 1, then as F(I) ≤ F(K), I ∼ = A± n−1 (q); but then it is not the case that J ↑2 I, by [IA , 4.5.1]. Therefore I has untwisted rank n − 2 and x is of equal-rank type. As n ≥ 8, the ± (q). As J is a component of CI (x), only possibility is I = Bn−2 (q) and J ∼ = Dn−2 ± ± J∼ Ω (q) or Spin (q). This completes the proof.  = 2n−4 2n−4 n

Lemma 10.32. For n odd and n ≥ 3, ↑2 (2 G2 (3 2 )) = {G2 (3n )}. Proof. Suppose that K ∈ K2 , x ∈ I2 (Aut(K)), and CK (x) has a component n isomorphic to 2 G2 (3 2 ). Inspection of [IA , 5.3] shows that K ∈ Spor; and by [IA , 5.2.8], K ∈ Alt; and then by [IA , 2.2.10], K ∈ Chev(3), as n ≥ 3. Then  [IA , Table 4.5.1] forces K ∼ = G2 (3n ), as asserted. Lemma 10.33. Suppose that K ∈ K2 with Z(K) nontrivial and cyclic. Let  x ∈ I2 (Aut(K)), and suppose that O 2 (F ∗ (CK (x))) = I x where either x = 1 or x ∈ K is an involution inducing the automorphism x, and I is the central product of n = 1 (or 2) subgroups I1 (and I2 ) such that each Ii is a component or subnormal Q8 -subgroup, and I2 (Ii ) ⊆ Z(K). If n = 2, assume that at most one Ii is isomorphic to Q8 . Then one of the following holds: (a) n = 1, K ∼ = SL2 (q) for some odd q, x is a field automorphism, and  1 I1 ∼ = O 2 (SL2 (q 2 )); ∼ (b) n = 1, K = Spin5 (q), x is an inner-diagonal but noninner automorphism  of K, and I1 ∼ = O 2 (SL2 (q)) or SL2 (q 2 ); (c) n = 2 and x ∈ [Aut(K), Aut(K)]; or (d) K ∼ = L3 (4) with Z(K) of exponent 4. = 2An for some n ≥ 7, or K/Z(K) ∼ In cases (a) and (b), Out(K) is abelian. Proof. We argue first that Either CK (x) has a component of 2-rank 1, or O2 (CK (x)) = 1, or CK (x) (10G) is 2-constrained with |CK (x)|2 ≤ 25 . Indeed, if CK (x) has no such component or core, then n = 1 and I = F ∗ (CK (x)) = I1 x with I1 ∼ = Q8 . Then CK (x)/F ∗ (CK (x)) acts faithfully on the chain I > Z(I) ≥ Z(I1 ) > 1 so |CK (x)|2 ≤ 2|I| ≤ 25 , proving (10G). Now (10G), together with the condition Z(K) = 1, implies that K ∈ Spor, by inspection of [IA , Table 5.3]. If K ∈ Alt, then K ∼ = 2An for some n ≥ 5. Obviously

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∼ SL2 (9) satisfies (a); otherwise (d) 2A5 does not satisfy the hypotheses, and 2A6 = holds. So we may assume that K ∈ Chev(r) − Alt for some r. Suppose next that r is odd and n = 2. By conclusion (c), we may assume that x ∈ [Aut(K), Aut(K)], so as x is an involution, x ∈ Inndiag(K). We consult [IA , Tables 4.5.1–4.5.3]. In cases where CK (x) has a single Lie component L, the tables show that CCK (x) (L) is abelian, and so n = 1, contradiction. Thus  CK (x) must have at least two Lie components, and as O 2 (F ∗ (CK (x))) = I1 I2 , it has exactly two Lie components. The only candidates from [IA , Table 4.5.1] are conjugates of the element t2 ∈ Aut(B3 (q)), and conjugates of the element t2 ∈ Aut(D4 (q)). (Note that if K ∼ = D4 (q), then since |Z(K)| = 2, the cosets of Inn(K) called s and s in [IA , Table 4.5.1] lie outside [Aut(K), Aut(K)].) If x = t2 ∈ Aut(B3 (q)), then x ∈ Inn(K) by [IA , Table 4.5.1]; but Out(K) is abelian in this case and so (c) holds. If x = t2 ∈ Aut(D4 (q)), then by [IA , Table 4.5.3] the Lie components of CK (x) are simple and so do not contain Z(K), contrary to assumption. Thus the lemma holds if r is odd and n = 2. Now suppose that r is odd and n = 1. If x induces a field or graph-field automorphism on K, then (a) holds by [IA , 4.2.3]. Otherwise we may again use the tables [IA , 4.5.1, 4.5.2]. As n = 1, CK (x) has a unique Lie component L, and  L is universal of type A1 . Moreover, since I1 = O 2 (F ∗ (CK (x))), O2 (CCK (x) (L)) = Ω1 (Z(K)) x1 or Ω1 (Z(K)) according as x is or is not an inner automorphism.  Notice that if K = A± 3 (q) and x is conjugate to t2 in the tables, then as L is universal, K is universal, and then from [IA , Table 4.5.2], |O2 (CCK (x) (L))|2 ≥ 8 or 4 according as x is or is not an inner automorphism, contradicting what we just saw. −   In the cases (K, x) = (A± 3 (q), g2 ) and (D4 (q), t2 ), L is not universal. Moreover, if K = B2 (q) and x is inner, then x is induced by an element of order 4, not by an involution as we have hypothesized. The only surviving cases are for K = B2 , with x being an outer diagonal automorphism and L ∼ = A1 (q) or A1 (q 2 ). Therefore (b) holds. We remark that it is clear from [IA , 2.5.12] that in cases (a) and (b), Out(K) is the direct product of Outdiag(K) and a cyclic group of field automorphisms, and in particular Out(K) is abelian. Suppose finally that K ∈ Chev(2) − Alt − ∪r>2 Chev(r). Note that if O2 (CK (x)) = 1, then by [IA , 7.7.1c], K/Z(K) ∼ = L3 (4) and CK/Z(K) (x) ∼ = U3 (2), 2 ∗ contradicting the given structure of O (F (CK (x))). Consequently O2 (CK (x)) =  1, so F ∗ (CK (x)) = O 2 (F ∗ (CK (x))). In most cases, (10H)

some 2-central involution z of K/Z(K) splits and is stable.

If (10H) holds, then by definition [IA , 5.5.1], some preimage z ∈ I2 (K) of z lies in CK (x). We may take z to be 2-central in CK (x), and then z centralizes F ∗ (CK (x)) by its given structure, so z ∈ Z(F ∗ (CK (x))). But our hypothesis implies that all central involutions of F ∗ (CK (x)) lie in Ω1 (Z(K)) x , so we must have z ≡ x (mod Z(K)), i.e., x acts on K like z. Consequently CK (x) contains a Sylow 2subgroup of K, so CK (x) is 2-constrained by Tits’s lemma [IA , 2.6.7] and |K|2 ≤ 25 by (10G). But then the Schur multiplier of K has odd order by [IA , 6.1.4], a contradiction. We may therefore assume that (10H) fails, whence by [IA , 6.4.2], 3 K/Z(K) ∼ = 4L3 (4) (in the last case, conclusion = Sp6 (2), 2B2 (2 2 ), or D4 (2), or K ∼ (d) of the lemma holds).

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If Ii is a component for some i = 1, 2, then by the Borel-Tits Theorem and ∼ D4 (2) and Ii ∼ [IA , 4.9.2], K = = Sp6 (2), so m2 (Ii ) > 1, contrary to assumption. Therefore n = 1 and I1 ∼ = Q8 , so |CK (x)|2 ≤ 25 by (10G). Consequently |CK/Z(K) (x)|2 ≤ 25 . 3 3 This rules out K/Z(K) ∼ = 2B2 (2 2 ), since |2B2 (2 2 )|2 = 26 and all involutions of 3 2 B2 (2 2 ) are 2-central. If K ∼ = Sp6 (2), then Out(K) = 1 so x is an involution. Let T ∈ Syl2 (K) with CT (x) ∈ Syl2 (CK (x)), and let U be a normal four-group in T . By the Thompson Transfer Lemma, x has an extremal conjugate in CT (U ). Replacing x by that conjugate we may assume that U ≤ CT (x). Since F ∗ (CK (x)) is a 2-group and [U, CT (x)/Z(K)] = 1, U ≤ Ω1 (F ∗ (CK (x))) = x Z(K). Therefore x acts on K as an element of U , whence |CK (x)|2 ≥ |K|2 /2 > 25 , a contradiction. Therefore K ∼ = D4 (2). Let T be an x-invariant Sylow 2-subgroup of K. There are three maximal parabolic subgroups Pi , 1 ≤ i ≤ 3, of K containing T , with Levi factors Li ∼ = A3 (2) and unipotent radicals Qi . We claim that x normalizes Qi for some i such that Qi ∼ = E27 . This will imply that m2 (CK (x)) ≥ 5. But since Sylow 2-subgroups of CK (x) are of order at most 25 and contain Q8 , this will be a  → K be a universal covering, so that Z(K)  contradiction. To prove the claim, let K    is a four-group [IA , 6.1.4]. Let Pi and Qi be the preimages of Pi and Qi in K. Since i , Q  i ]| ≤ 2, and Li acts absolutely irreducibly on Qi /Z(K) ∼ = E26 , we must have |[Q  the commutator quotient of Qi is elementary abelian. Now x normalizes some Pi , so i is not elementary abelian. As a triality automorphism we may assume that that Q  we can number the involutions cycles the Pi as well as the involutions of Z(K), i , Q  i ] = yi for all i. We may  of Z(K) as yi , 1 ≤ i ≤ 3, in such a way that [Q  assume that the kernel of the covering K → K is y1 . As x ∈ Aut(K), x lifts  fixing y1 and hence normalizing Q 1 . But Q 1 / y1 is to an automorphism of K elementary abelian, so our claim is proved. The lemma follows.  Lemma 10.34. In (a), (b) of Lemma 10.33, K does not involve G2 (q1 ), 3D4 (q1 ), n or G2 (3 2 ) for any odd q1 or n. 2

Proof. With [IA , 4.8.10], we see that as q1 is odd, G2 (q1 ) involves the Froben 1 nius group F of order 8.7. Likewise for any odd n, 2 G2 (3 2 ) ≥ 2 G2 (3 2 ) ∼ = L2 (8) contains a copy of F . As G2 (q1 ) ≤ 3D4 (q1 ), it suffices to show that K does not involve F . In Lemma 10.33a, K has quaternion Sylow 2-subgroups so K has sectional 2-rank 2. Thus K cannot involve F in this case. In Lemma 10.33b, K is involved in I := Spin5 (q) ∼ = Sp4 (q). If our lemma is false, then there are subgroups H ≤ I and H0  H such that H/H0 ∼ = F . Taking H to be minimal we see that H is 2-closed by a Frattini argument, and then H = Q y with y a 7-element and Q = O2 (H). Minimality then implies that Q is elementary abelian of order 8 or special, with |Q/Z(Q)| = 8 and y centralizing Z(Q) in the latter case. If Q is special then for any hyperplane Q0 of Z(Q), Q/Q0 must be extraspecial by the action of y; but this is absurd as |Q/Z(Q)| is an odd power of 2. Thus, Q ∼ = E23 . By Clifford’s Theorem any representation of H in odd characteristic has degree at least 7. This  contradicts H ≤ I ∼ = Sp4 (q) and completes the proof. Lemma 10.35. Let K  X = K t with K ∼ = Sp4 (q), q odd, t2 = 1, Z(K) = z , and X/ z ∼ = Inndiag(K). Then t ∼K tz.

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Proof. By [IA , 4.5.1, 4.5.2], CK (t)/ z < CK/z (t z / z ). This implies the lemma.  Lemma 10.36. Let K be the central product of two components K1 , K2 . Let x ∈ I2 (Aut(K)). Suppose for each i = 1, 2, x normalizes Ki and [x, Ki ] = 1, and suppose further that Ki and the image of x in Aut(Ki ) satisfy the hypotheses of Lemma 10.33 but not conclusion (c) or (d). Suppose further that Z(K1 ) = Z(K2 ) has order 2. Form the semidirect product K x and let d be the number of Kconjugacy classes of involutions y in the coset xK such that CK (y) ∼ = CK (x). If d > 1, then d = 2 and each Ki satisfies Lemma 10.33a. Proof. Consider X = K x and X = X/Z(K). Now K1 is of one of the types given in Lemma 10.33ab. By [III11 , 1.17] and Lemma 10.35, x ∼K xz; indeed for any involution x ∈ xK such that CK1 (x) ∼ = CK1 (x ), x ∼K x z. Therefore it suffices to prove the corresponding statement about K-classes of involutions in xK. If K1 ∼ = Spin5 (qi ), then x induces an inner-diagonal but noninner automorphism on K i . For these groups |Outdiag(K i )| = 2, so X/K 3−i ∼ = Inndiag(K i ). By [IA , 4.5.1], the image of x in Aut(K i ) is determined up to conjugacy by its order, the fact that it is noninner, and the isomorphism type of CK i (x). On the other hand, if K 1 ∼ = L2 (q) for some odd q, then K 1 x contains two classes of involutions, both represented by field automorphisms (and fused by any noninner automorphism in P GL2 (q)). The classes are represented by x and xu, for any particular choice of involution u ∈ O 2 (CK1 (x)). Now if K1 is a spin group of dimension 5 as above, then we argue that d = 1. The assertions above about K 1 x ∼ = X/K2 show that d is the number of conjugacy classes of involutions y ∈ K2 x such that CK2 (y) ∼ = CK2 (x). If K2 is also a spin group, then the same assertions applied to K2 x then imply that d = 1. If K2 ∼ = SL2 (q), then either d = 1 or there is an involution xv, with v an involution of O 2 (CK 2 (x)). But this implies that v ∈ CK2 (x). As K2 has a unique involution, v has order 4, so xv has order 4, a contradiction. Thus d = 1 in this case as well. Finally if Ki ∼ = SL2 (qi ), i = 1, 2, then similar considerations show that d = 2, with the second class represented by xv1 v2 , where vi ∈ CKi (x) is an element of order 4, i = 1, 2. The proof is complete.  ∼ L± (9) or K/Z(K) Lemma 10.37. For any K ∈ ↑2 (A6 ) ∩ Chev(3), either K = 3 is a projective orthogonal group of dimension at least 4 over F3 or F9 . Moreover, ± if K ∼ = Sp4 (4), L± 5 (2), or L3 (9), then A6 ↑2 K only via an outer automorphism. Proof. Let K ∈ Ch3 . Thus K ∈ Chev(3) and A6 ↑2 K, via x ∈ I2 (Aut(K)), say. If x ∈ Aut0 (K), then by [IA , 4.9.1], x is a field automorphism, so K ∼ = L2 (34 ) ∼ = − Ω4 (9), as required. If x ∈ Aut0 (K), then q(K) = 3 or 9 by [IA , 4.2.2]. Excluding groups of type B and D , as well as A3 = D3 and C2 = B2 , we survey [IA , 4.5.1] for a subcomponent L ∼ = A1 (9)a and find only K ∼ = L± 3 (9). This proves the first assertion. The second assertion holds by the Borel-Tits theorem and the fact that if ∼  J ↑2 L± 3 (9) via an inner automorphism, then J = SL2 (9), not L2 (9). Lemma 10.38. Let K = d L(q) ∈ G62 , q odd, q > 3, and let x ∈ I2 (Aut(K)) be such that CK (x) has a component J ∼ = SL2 (q) with Z(J) ≤ Z(K). If F(K) ≤ (q 9 , A), then either K ∼ (q) with x a graph automorphism, or K ∼ = SL± = Spin5 (q). 4 In the latter case J = E(CK (x)).

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∼ A± (q), A± (q), B2 (q), or Proof. The condition on F(K) implies that K = 2 3 G2 (q). Since Z(K) has even order, K ∼ = B2 (q) is an allowed conclusion, while ∼ ± K∼ = A± 2 (q) and G2 (q) are impossible [IA , 6.1.4]; so assume that K = A3 (q). Then  the lemma follows from the information in [IA , 4.5.1, 4.5.2]. Lemma 10.39. Suppose that q is an odd prime power, q > 3, K ∈ K2 , and 3 x ∈ I2 (K) − Z(K) with x ∈ L2 (CK (x)) ∼ = SL2 (q). Then K ∼ = L± 3 (q) or D4 (3). Proof. Since x ∈ I2 (K) − Z(K), we may factor out Z(K) and assume that K is simple. Then from [IA , 5.2.8, 5.3] we see that K ∈ Alt ∪ Spor, so K ∈ Chev(r) for some r. By the Borel-Tits theorem, r > 2, and then by [IA , 2.2.10], q is a power  of r. As E(CK (x)) ∼ = SL2 (q), the result follows by inspection of [IA , 4.5.1]. Lemma 10.40. Let K ∈ K2 with K = F ∗ (X). Let y ∈ I2 (X) and suppose that CK (y) has a component L with y ∈ L ∼ = SL2 (q 2 ) for some odd prime power q. 2 Then O (CK (L y )) has no subnormal subgroup containing y and isomorphic to SL2 (q) (if q > 3) or Q8 (if q = 3). Proof. We can of course have y ∈ Z(K), but then the conclusion follows as K = F ∗ (X). So assume that [y, K] = 1. By assumption y ∈ K, and there is no loss in assuming that X = K. We quote [III11 , 13.4] to conclude that K ∈ Chev, and either q(K) = q 2 or K ∼ = 3D4 (q 2/3 ). In the latter case, the conclusion of the lemma holds by the information in [IA , 4.5.1]. By [IA , 4.2.2] any component M = L of CK (y) lies in Chev and has level q(M ) ≥ q 2 , which yields the desired assertion if q > 3. Suppose then that q = 3. Then K has level 9 and we need to derive a contradiction from the assumed existence of a Q8 -subgroup H ≤ O2 (CK (L)). But then as the Lie components of CK (y) have level at least 9, we get H ≤ C :=  CCK (y) (O 3 (CK (y))). However, by [IA , 4.5.1], C/Z(K) is cyclic, so C is abelian, a final contradiction.  Lemma 10.41. Let q be a power of the odd prime r, q > 3, and let J = SL2 (q 2 ). Let K ∈ K2 and suppose that x ∈ I2 (K) − Z(K) and x ∈ L ≤ K with L ∼ = J and L   CK (x). Then one of the following holds: 2 2 2 ∼ (a) K ∼ = L± 3 (q ) or G2 (q ), or K/Z(K) = P Sp4 (q ); 6 9 (b) K ∈ G2 and f (K) > q . Proof. Since q > 3, q 2 ≥ 25, so [IA , 2.2.10] and [III11 , 1.1ab] imply that K ∈ Chev(r). b We may assume that Z(K) = 1. Because x ∈ L, clearly K ∼  L2 (r b ) or 2 G2 (3 2 ) = ± − for any b. Thus if K ∈ G2 , then K ∼ = L± 3 (3), L4 (3), Ω6 (3), or G2 (3), none of which 2 2 have an SL2 (q ) subcomponent, q ≥ 25, by [IA , 4.5.1]. Therefore K ∈ G2 . b If K ∈ G62 , it follows that K ∈ G72 , so K ∼ = L± 3 (r ) for some b. But then K has b ∼ one class of involutions and SL2 (r ) = E(CK (x)), so r b = q 2 and (a) holds. With [IA , 4.2.2], we may now assume that K = d L(q 2/a ) ∈ G62 for some positive integer a. We may also assume that (b) fails, whence f (K) ≤ q 9 , so 22 ≤ 9a where  is the rank of L. Obviously  > 1. If a = 1, then  = 2 and (a) follows immediately. Otherwise by [III11 , 13.4], a = 3 and K ∼ = 3D4 (q 2/3 ). But then 22 = 32 > 9a, a contradiction. The lemma follows.  Lemma 10.42. Let K = d L(q) ∈ G62 , q odd, q > 3, and let x ∈ I2 (K) be such that two of the components of CK (x) have center equal to x , are isomorphic to

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SL2 (q), and are interchanged by an involution t of Aut(K). If F(K) ≤ (q 9 , A), n then either K is a quotient of Ω± n (q) for n = 5 or 6, or q = 3 for some odd n and K∼ = G2 (q), with t being a graph-field automorphism of K. Proof. First suppose that K is simple. Since x ∈ K, we see from [IA , 4.5.1] that K ∼ = P Ω± n (q), n ≥ 5, or G2 (q). In the latter case the two components of CK (x) are long- and short-root A1 (q)’s, and so they are not conjugate in Aut(K) except by a graph-field automorphism when q = 3n ; and for such an automorphism to be an involution (even in Out(K)) it is necessary that n be odd. In the case K∼ = P Ω± n (q), the condition on F(K) forces n ≤ 6, as claimed. Now for arbitrary K, the assumptions go over to K/Z(K), and so in view of ∼ [IA , 6.1.4] it remains to rule out K ∼ = Spin± n (q), n = 5 or 6, i.e., K = Sp4 (q) or ± + ∼ ∼ SL4 (q). But in these cases it is clear that E(CK (x)) = Spin4 (q) = SL2 (q) ×  SL2 (q), contrary to assumption. The lemma follows. Lemma 10.43. Suppose that q > 3 is an odd prime power. Let K ∈ K2 and let x, y be commuting involutions in Aut(K) with the following properties: (a) E(CK (x)) has a component L0 ∼ = SL2 (q) and E(CK (y)) has a component 2 (q ) with L ≤ M ; M1 ∼ SL = 2 0 1 (b) AutAut(K) (M1 ) contains P GL2 (q 2 ); and (c) Z(L0 ) ≤ Z(K). Then K ∈ G62 . If F(K) ≤ (q 9 , A), then K ∼ = SL± 4 (q) and the image of x in Out(K) lies outside [Out(K), Out(K)]. Proof. By [III11 , 5.11], K ∈ Spor. Since q > 3, q 2 > 9 and so SL2 (q 2 ) ∈ Alt ∪ Chev(2); therefore K ∈ Alt ∪ Chev(2). By [IA , 2.2.10], K ∈ Chev(r) where ± q is a power of the prime r. Again since q 2 > 9, K/Z(K) ∼ = L± 3 (3), L4 (3), or  L2 (q0 ) for any q0 , and G2 (3). The existence of L0 and M1 implies that K/Z(K) ∼ = so K ∈ C2 ∪ T2 , i.e., K ∈ G2 . Another consequence is that neither x nor y induces a field automorphism on K. If K ∼ = L± 3 (q0 ), then using [IA , 4.5.1] we see from M1 2 that q0 = q , but from L0 that q0 = q, contradiction. Therefore d(K) = 7 and so K ∈ G62 . Since x does not induce a field automorphism on K, the level of K is q 1/a for some positive integer a. But then the structure of M1 implies, by inspection of [IA , 4.5.2], that a < 2. (Note that we cannot have K = D4 with y of class t2 in the table, for then the component of CK (y) would be the adjoint version.) Hence K has level q. If K has untwisted rank 2, then K = Sp4 (q); but then from Lemma 6.20, AutAut(K) (M1 ) does not contain P GL2 (q 2 ), contrary to assumption. Therefore if F(K) ≤ (q 9 , A), then K has untwisted rank 3, whence K ∼ = A± 3 (q). Since Z(L0 ) ≤ Z(K) it follows by inspection of [IA , 4.5.1] that K ∼ = SL± 4 (q), and that x induces a graph automorphism on K. This implies the final assertion, with  the help of [IA , 2.5.12]. Lemma 10.44. Suppose q is a power of the odd prime r, with q > 3. Let K ∈ K2 possess an involution z such that E(CK (z)) has components K1 , K2 with z ∈ K1 ∩ K2 and Ki /O2 (Ki ) ∼ = SL2 (q), i = 1, 2. Then K is unambiguously in Chev(r), i.e., K ∈ Chev(r) − Alt − ∪s=r Chev(s). Proof. Using the fact that K1 ↑2 K, and inspection of the tables in [IA , 5.3], we see that K ∈ Spor. Likewise as Ki ∼ = An for any n, while z ∈ Z(K), we have K ∈ Alt [IA , 5.2.8]. Thus, K ∈ Chev(s) for some s. By the Borel-Tits theorem,

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and as q > 3, s is odd. As r and s are both odd, s = r by [IA , 2.2.10], completing the proof.  Lemma 10.45. Suppose that K ∈ K2 and x ∈ I2 (K − Z(K)) with L ∼ = Sp4 (q) n or SL± (q) for some component L of C (x) and some q = r , r an odd prime. K 4 Then K is unambiguously in Chev(r). Proof. This follows from [III11 , 1.1ab] and [IA , 2.2.10].



a Lemma 10.46. Suppose that K ∈ K2 with K/Z(K) ∼ = P Ω± n (r ), r an odd prime, n ≥ 6. Then ↑2 (K) ⊆ Chev(r).

Proof. If K ↑2 J, then J ∈ Chev(r) as a consequence of [III11 , 1.1ab] and  [IA , 2.2.10]. Lemma 10.47. Suppose that q is an odd prime power, q > 3, K ∈ K2 , and x ∈ I2 (K − Z(K)) with L2 (CK (x)) = L1 L2 , where L1 ∼ = SL2 (q) and [L1 , L2 ] = 1. Assume that L2 /O2 (L2 ) ∼ = 2A7 or SL2 (q0 ) for some odd q0 . Suppose also that x ∈ L2 and Z(L1 ) = v = x , with vx ∈ Z(K). Then K ∼ = SL± 4 (q) or Sp4 (q), K and v ∈ x . Proof. If K ∈ Alt then Li /O2 (Li ) ∈ Alt with vx ∈ L1 ∩ L2 by [IA , 5.2.8], contradicting the fact that L1 L2 /O2 (L2 ) is a direct product. By inspection of the tables [IA , 5.3], K ∈ Spor. Thus K ∈ Chev(r) for some r, and r > 2 by the Borel-Tits Theorem. Thus and q and q0 are both powers of r, by [IA , 2.2.10]. Now by [IA , Tables 4.5.1–4.5.3], the assertions of the lemma hold. Note that as Z(K) = 1, K ∼  G2 (q) or 3D4 (q1 ) for any q1 . Also, as L1 L2 = L1 × L2 with = Z(K) ∩ L1 = Z(K) ∩ L2 = 1, if K ∼ = Spin± n (q) or HSpinn (q), then n = 5 or 6 as  q > 3 and CK (x) has no other components than L1 and L2 . Lemma 10.48. Suppose that L ↑2 K, where L and K are elements of Chev of the same odd level. If L ∼ = L± 3 (q), then K has untwisted rank at least 3. Proof. Suppose that K has untwisted rank at most 2, and let Δ be its extended diagram. Then by [IA , 4.2.2], Δ must have a node marked “2” which, when deleted, leaves an A2 -diagram. But it is easily checked that there are no such diagrams Δ. The proof is complete.  Lemma 10.49. Suppose that q is odd and L3 (q) ↑2 K,  = ±1, and K has an involution t such that t ∈ L ∼ = SL2 (q 2 ) for some component L of CK (t). Then 2 ∼ K = L3 (q ). Proof. By [III11 , 13.4], either K ∼ = d L(q 2 ) for some d and L, or K ∼ = 2/3 D4 (q ). By assumption there exists y ∈ I2 (Aut(K)) such that CK (y) has a component L with L/O2 (L) ∼ = L3 (q). Using [IA , 4.2.2, 4.5.1] we see that y cannot induce either an inner-diagonal or graph automorphism on K, so y induces a field  or graph-field automorphism. Then the result follows by [IA , 4.9.1].

3

Lemma 10.50. Suppose L3 (q) = J ↑2 I via x ∈ Aut(I) for some I ∈ K2 , some  = ±1, and some odd q. If I ∈ Chev(r) for some odd prime r, assume that F(I) ≤ (q 9 , A). Then one of the following holds: (a) x ∼I xz for all z ∈ I2 (J); or (b) I ∼ = SL4 (q).

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Proof. By [III11 , 1.1a], [IA , 2.2.10], and the tables [IA , 5.3], I ∈ Alt ∪ Spor, ∼ G2 (2) . But in that case by the Borel-Tits theorem and and if I ∈ Chev(2), then J = [IA , 4.9.1, 4.9.2], x induces a field automorphism on I ∼ = G2 (4), and (a) holds. Thus, we may assume that I ∈ Chev(r) − Chev(2) for some odd r. Again if x ∈ Aut0 (I) then I ∼ = L3 (q 2 ) and (a) holds by [IA , 4.9.1], so assume that x ∈ Aut0 (I). We consult [IA , 4.5.1]. If I is not of type A , then the only possibility is I ∼ = C3 (q), 1/a whence F(I) = (q 9 , BC) > (q 9 , A). Thus by [IA , 4.2.2], I ∼ (q ) for some = A±  √ 2 /a 9 integer a ≥ 1, and so q = √f (I) ≤ q , whence  ≤ 3 a. If a > 1, then by [IA , 4.5.1], a = 2 and 5 =  ≤ 3 2, a contradiction. Thus a = 1, so I ∼ = SL± 4 (q)/Z ± for some Z ≤ Z(SL4 (q)). Moreover, x acts as a reflection on the natural SL± 4 (q)module. For any z ∈ I2 (J), it follows that −xz is also a reflection, so x ∼I (−z)x. If −1 ∈ Z, then (a) holds, and otherwise, (b) holds. The proof is complete.  Lemma 10.51. Suppose that J, K ∈ K2 and J ↑2 K ∈ Chev(s), s odd. Let rK be the untwisted Lie rank of K. Then the following conditions hold: (a) If J ∼ = SL2 (q), q > 3, then one of the following holds: (1) K has level q; (2) K ∼ = SL2 (q 2 ); 1 1 (3) K has level q 2 , and either rK ≥ 3 or K ∼ = Sp4 (q 2 ); or 1 1 (4) K has level q 3 , and K ∼ = 3D4 (q 3 ); (b) If J ∼ SL (q), q > 3, and f (K) ≤ q 4 , then one of the following holds: = 2 1 (1) K ∼ = Sp4 (q 2 ) with Z(J) = Z(K) and f (K) = q 2 , and J ↑2 K via a non-inner automorphism of K; or (2) K ∼ = L± = SL2 (q 2 ), or K/Z(K) ∼ 3 (q), P Sp4 (q), or G2 (q); ± ∼ (c) If J = A2 (q), then F(J) < F(K); (d) If J ∼ = C2 (q), then the following conditions hold: (1) F(J) < F(K); 1 (2) If f (K) ≤ q 8 , then K ∼ = C4 (q 2 ) or C2 (q 2 ), with f (K) = q 8 ; moreover if J is universal, then so is K, and J ↑2 K via a non-inner automorphism of K. Proof. Say that J ↑2 K via x ∈ I2 (Aut(K)). If x is a field or graph-field automorphism, then K ∼ = SL2 (q 2 ), SL2 (q 2 ), A3 (q 2 ), 2 or C2 (q ), in the respective parts, and the appropriate conclusions follow easily. Thus, we may assume that x ∈ Aut0 (K). We use [IA , 4.5.1, 4.5.2]. If J ∼ = SL2 (q) and K has level q, then (a) holds, and in (b), rK ≤ 2 as f (K) ≤ q 4 . This leads to the final three groups in (b2). If J ∼ = SL2 (q) and K has level different from q, then [IA , 4.5.1, 4.5.2] give the only possibilities as spin or half-spin groups of level q 1/2 and of dimension at least 5, all arising from the isomorphism 1/2 ∼ ) = SL2 (q); and the one extra possibility (a4). Thus (a3) or (a4) holds. Spin− 4 (q Note that if rK ≥ 3 with K of level q 1/2 , then f (K) ≥ q 9/2 > q 4 , and likewise if (a4) holds, then f (K) = q 16/3 > q 4 . Therefore the first statement of (b1) holds, and the remaining assertions of (b1) are contained in [IA , 4.5.1, 4.5.2]. Hence (a) and (b) hold. We have F(J) < F(K) by [III11 , 12.4]. So it remains to establish (d2). We have J∼ = C2 (q) and f (K) ≤ q 8 . Again we use [IA , 4.5.1]. If x is a graph automorphism, 1/2 the only possibilities are K ∼ ), for both of which f (K) ≥ q 9 , = A± 3 (q) and D5 (q contradiction. Thus x is an inner-diagonal automorphism. Using [IA , 4.2.2] and examination of Dynkin diagrams, we see that the only possibilities for K of level q

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satisfy rK ≥ 3, so f (K) ≥ q 9 , again a contradiction; and the only other possibility is K∼ = Sp8 (q 1/2 ). In this last case the remaining statements of (d2) hold by [IA , 4.5.1, 4.5.2].  Lemma 10.52. Let K ∈ K2 and x ∈ I2 (K) − Z(K). Suppose that there is L   CK (x) such that x ∈ L ∼ = Sp4 (q), q a power of the odd prime r. Then K ∈ G62 9 and F(K) > (q , A). Proof. By [III11 , 1.1ab] and [IA , 2.2.10], K ∈ Chev(r). As x ∈ K, [IA , 4.2.2] implies that K ∼ = d L(q 1/a ) for some positive integer a. Moreover, since L ∼ = B2 (q), the untwisted diagram of L has a double bond. It follows that K ∈ G62 . From [IA , 4.5.1] we see that the only possibilities are for a = 1 with the rank of L at least 3, in which case F(K) ≥ (q 9 , A), or for a = 2 with L = C4 . But in the latter case the involutory automorphism x must be outer by [IA , 4.5.1], a final contradiction. The proof is complete.  Lemma 10.53. Let K be a 2-component with K/O2 (K) ∈ K2 and let t ∈ I2 (K) − Z ∗ (K). Suppose that CK (t) has a normal subgroup which is a central product K1 ∗ K2 , where K1 ∼ = SL2 (q), K2 ∼ = SL4 (q) for some odd prime power q, and K1 ∩ K2 = t . Suppose also that K is embeddable in HSpin12 (q). Then K/Z(K) ∼ = L6 (q).  L6 (q). Since Proof. First suppose that K/O2 2 (K) ∼ = L6 (q) but K/Z(K) ∼ = ∼ t ∈ K1 ∩ K2 , it follows that K/O2 (K) = L6 (q). As O2 (K) ≤ Z(K) and K is a 2-component, there exists an odd prime s and an s-subgroup R  K such that CK (R) ≤ O2 (K). We choose s and R so that R is minimal. Then R is of class at most 2 and exponent s. We have K ≤ J ∼ = HSpin12 (q). If s divides q, then by the Borel-Tits theorem, K lies in a parabolic subgroup P of J with R ≤ Os (P ). Given the isomorphism type  of K, I := O s (P )/Os (P ) ∼ = A5 (q), and so Os (P ) ∼ = V ∧ V where V is the natural I-module. But V ∧ V is irreducible for I, so F ∗ (K) = R = Os (P ). Examining V ∧ V , we see that [K2 , CR (t)] = 1. But K2  CK (t) by assumption, and so [K2 , CR (t)] ≤ K2 ∩ R ≤ K2 ∩ O2 (K) = 1, a contradiction. Thus, s = q. Now s is a good prime for J = HSpin12 (q) and so Z(R) lies in a maximal torus of the overlying algebraic group D6 . Consequently, by a Frattini argument, AutJ (Z(R)) is involved in the Weyl group 25 Σ6 of type D6 , and so AutK (Z(R)) is as well. Thus AutK (Z(R)) cannot involve L6 (q), so Z(R) ≤ Z(K). Choose any 1 = z ∈ Z(R). Then by [IA , 4.2.2], K lies in the product of the Lie components of CJ (z). Using [IA , 4.5.1] we see that the only possibility is that CJ (z)(∞) ∼ = A5 (q), a single Lie component. But as K is perfect, this forces K = CJ (z)(∞) to be quasisimple, contrary to assumption. By the previous paragraphs, it is enough to show that K/O2 (K) ∼ = L6 (q). Let K = K/O2 (K). Then K is simple, and t ∈ I2 (K) with t ∈ Z(Ki ), K i ∼ = Ki a component of CK (t). By Lemma 10.45, K ∈ Chev(r), where q is a power of r. As K 1 ∗ K 2 is a product of components of CK (t), it follows from [IA , 4.5.1] that r either K ∼ = L6 (q), as claimed, or K ∼ = Ω+ 10 (q). But in the latter case, O (CK (t)) = K 0 ∗ K 1 ∗ K 2 with K 0 and K 1 interchanged in CK (t), contrary to the assumption  that K1 ∗ K2  CK (t), completing the proof. Lemma 10.54. Suppose K ∈ K2 and D = u, v is a four-subgroup of Aut(K). Let KD be a component of E(CK (D)), and for each t ∈ D# let Kt be a component

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of E(CK (t)) containing KD . Suppose that q is a power of an odd prime r, and that KD ∼ = A3 (q), Ku ∼ = D5 (q)u , Kv ∼ = D4± (q), and Kuv ∼ = A3 (q) or Ku , with  = ±1. Then one of the following holds: η (a) K ∼ = D6 (q), η = ±1, and E(CK (uv)) is the central product of A3 (q)u and η A3 (q); (b) K ∈ Chev(r) and F(K) > F(E6 (q)). Proof. From [III11 , 1.1ab] and [IA , 2.2.10], we have that K ∈ Chev(r). If some element of D# induces a field or graph-field automorphism on K, then K has level q 2 . Some element t ∈ D# must however lie in Aut0 (K), as Aut(K)/Aut0 (K) is cyclic [IA , 2.5.12]. As CK (t) is assumed to have a component of level less than q 2 , however, this is impossible. Therefore D ≤ Aut0 (K), and the assumed components Kt , t ∈ D# , must appear in [IA , 4.5.1]. It follows easily that K must be an orthogonal group of level q and dimension at least 11. Since F(D7 (q)) > F(B6 (q)) > F(E6 (q)), (b) holds unless K ∼ = D6± (q) or B5 (q), which we now assume. If K ∼ = P SO11 (q). From the component = B5 (q), then D ≤ Inndiag(K) ∼ structures, u must act on K as a reflection on the natural module, and v as an involution with eigenspaces of dimensions 3 and 8. Then uv has eigenspaces of dimensions 2 and 9, or 4 and 7, and in either case no component Kuv of the assumed type exists. η Therefore K ∼ = D6 (q), η = ±1. On the natural module, u has eigenspaces of dimension 2 and 10 and v has eigenspaces of dimension 4 and 8, but D has no eigenspace of dimension larger than 6. It follows that the 2-dimensional eigenspace of u and the 4-dimensional eigenspace of v are orthogonal, and the lemma follows easily.  Lemma 10.55. Let K be a quasisimple K-group, t ∈ I2 (Aut(K)), and J a component of E(CK (t)) such that J ∼ = A± 5 (q) for some odd q, and 1 = O2 (Z(J)) ≤ Z(K). Assume that K ∈ Chev has level q. Then either K ∼ = D6 (q) or C6 (q), or K has untwisted Lie rank at least 7. Proof. Since K has level q, t is not a field or graph-field automorphism, so K and t can be found in [IA , 4.5.1, 4.5.2]. Other than the desired isomorphism types for K, among groups of untwisted Lie rank at most 6, this yields either ∼ ± K ∼ = A± 6 (q) or K = E6 (q), for both of which Z(K) has odd order, contradicting  1 = O2 (Z(J)) ≤ Z(K). The lemma follows. Lemma 10.56. Suppose that K ∈ K2 , x ∈ I2 (Aut(K)), and L is a component of CK (x) with L ∼ = D6 (q), q odd. Then the following conditions hold: (a) If L u / u ∼ = E7 (q) = D6 (q)hs for some u ∈ Z(K) with u2 = 1, then K ∼ 2 or D6 (q ); (b) If K/Z(K) ∼ = P Ω± n (q) for some n and CK (x, L) has a component or solvable 2-component, then n ≥ 15; and + 2 (c) If L u / u ∼ = Ω+ 12 (q) or P Ω12 (q) for some u ∈ Z(K) with u = 1, and ± ± ∼ K has level q, then K/Z(K) = P Ωn (q), n > 12, or A11 (q); in the first case, L u / u ∼ = Ω+ 12 (q). Proof. From [III11 , 1.1ab] and [IA , 2.2.10], we get that K ∈ Chev(r), where q is a power of r. If x induces a field or graph-field automorphism on K, then K∼ = D6 (q 2 ) [IA , 4.9.1] and all parts of the lemma hold. Otherwise K and L appear

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in [IA , Table 4.5.1]. If x induces a graph automorphism on K, the only possibility + is K ∼ = A± 11 (q), with L a quotient of Ω12 (q), and again all assertions of the lemma are clear. We may therefore assume that x ∈ Inndiag(K). If K ∼ = D6 (q)hs , so all parts = E7 (q) then from [IA , Table 4.5.1], L/L ∩ Z(K) ∼ + of the lemma obviously hold. Another possibility is K ∼ (q 1/2 ). In that case = D12 + by [IA , Table 4.5.2], L is a quotient of Ω12 (q) so the hypotheses of (a) do not hold; |CK (L)| ≤ 2 by [IA , Table 4.5.1] so the hypotheses of (b) do not hold; and K does not have level q as in (c). The only remaining possibilities for K, by [IA , Table 4.5.1], are Bn (q), n ≥ 6, and Dn± (q), n ≥ 7. In these cases, by the table LZ(K)/Z(K) ∼ = Ω+ 12 (q), so (c) and ± (q) or D (q), then C (L) is supersolvable by the table, (a) hold. Finally, if K ∼ B = 6 K 7 so CK (x, L) has no component or solvable 2-component; thus (b) holds, completing the proof.  Lemma 10.57. Suppose that K is a quasisimple K-group, K  X, y ∈ I2 (K) − Z(K), and CX (y) has a normal subgroup K1 L with [K1 , L] = 1, L ∼ = Sp2n (q), K1 ∼ = SL2 (q), q odd, n ≥ 2, and Z(K1 ) = Z(L) = y . Then K ∼ = P Sp2n+2 (q), or n = 3 and K ∼ = P Sp2n+2 (q). = F4 (q). If F(K) ≤ F(P Sp2n+2 (q)), then K ∼ Proof. Write q = r a where r is an odd prime and a is an integer. If n = 2, then by Lemma 10.45, K ∈ Chev(r). If n > 2, then as L ↑2 K, we deduce from [III11 , 1.1ab] and [IA , 2.2.10] that K ∈ Chev(r). By L2 -balance, L ≤ K, and indeed K1 ≤ K unless possibly q = 3. Even if q = 3, we get K1 ≤ K provided K has level 3, by solvable L2 -balance and F3 -balance [III11 , 10.2]. Since y ∈ K − Z(K) we search [IA , 4.5.1] for an inner automorphism of K/Z(K) with a component isomorphic to (universal) Sp2n (q). The only results are K/Z(K) ∼ = F4 (q), n = 3. By the first paragraph = P Sp2m (q), m > n, and K ∼ K1 ≤ K as well, and from [IA , 4.5.1] this forces K/Z(K) ∼ = P Sp2n+2 (q) or F4 (q). If Z(K) = 1, then by [IA , 6.1.4] and [IA , 4.5.2], LK1 = L × K1 , contrary to assumption. Thus Z(K) = 1. As F(C4 (q)) < F(F4 (q)), the final statement holds, and the proof is complete.  Lemma 10.58. Suppose that L ↑2 K via x ∈ Inn(K) with L ∼ = Sp2n (q), q odd, n ≥ 2. Suppose that F(K) ≤ F(Sp2n+2 (q)). Then the following conditions hold: (a) If Z(L) ≤ Z(K), then n = 2 and K ∼ = Spin7 (q); and (b) If Z(L) ≤ Z(K), then K ∼ = Sp2n+2 (q) or P Sp2n+2 (q), and x acts on K as an element of I2 (K). Proof. As x ∈ Inn(K), K ∈ Chev(2) by the Borel-Tits theorem. As L ↑2 K, K ∈ Spor by inspection of the tables [IA , 5.3]. By [III11 , 1.1a] and [IA , 2.2.10], K ∈ Chev(r), where q is a power of r. Since x induces an inner automorphism on K, the (untwisted) Dynkin diagram of K has a Cn -subdiagram, by [IA , 4.2.2]. Using [IA , 4.5.1] and the inner condition we see that K ∼ = Bk (q), k > n = 2; Ck (q), k > n, 2 or F4 (q), n = 3. The inequality f (K) ≤ f (Sp2n+2 (q)) = q (n+1) forces K ∼ = B3 (q) or Cn+1 (q). In the former case, [IA , 4.5.1] shows that K must be universal with Z(L) ≤ Z(K), whence the conclusion (a) holds; while in the latter case, [IA , 4.5.1, 4.5.2] yields the conclusion (b). The proof is complete.  Lemma 10.59. Let K ∈ K2 and t ∈ I2 (Aut(K)). Let q be a power of the odd prime r. Suppose that CK (t) has a subnormal subgroup R × I0 such that R ∼ =  r π ∼ O r (Ω+ (q)) and I (Ω (q)) for some even n ≥ 8 and some sign π, with O = 0 n−4 4

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Z(I0 ) of even order. Assume also that Ωηn−2 (q) ↑2 K for some sign η. If q = 3 assume that K ∈ Chev(3) has level 3. Then K ∼ = Ωπn (q). Proof. By Lemma 10.46, K ∈ Chev(r), where q is a power of r. If q > 3, then CK (t) has at least 3 components of level q; by [IA , 4.9.1], t ∈ Aut0 (K) and then the conclusion follows from the information in [IA , 4.5.1]. Note in this regard that Z(K) = 1 since Z(I0 ) = 1 and R × I0 is a direct product. If q = 3, then we are given that K ∈ Chev(3) has level 3 and is a pumpup of Ω± n−2 (3). These facts, and the structure of RI0 , again yield the conclusion of the lemma, by examination  of [IA , 4.5.1]. Lemma 10.60. Let K ∈ K2 and t ∈ I2 (Aut(K)). Suppose that CK (t) has a subnormal subgroup HJ, where J is quasisimple, [H, J] = 1 and for some odd prime  ± ∼ power q = r m , H ∼  Ω− = O r (Ω+ 4 (q)) and Z(H) ≤ Z(K). Then J = 4 (q) or Ωn (q) for any n ≥ 5. Proof. Suppose false. If K ∈ Chev(s), s odd, then by [IA , 4.2.2], HJ = O 2 (HJ) ≤ LT , where L is the product of Lie components, L  LT , and T is an abelian s -group. Given the isomorphism type of J, we must have s = r by [IA , 2.2.10]. Thus L has at least 3 Lie components, and Z(H) ≤ Z(K). But these conditions never occur for K ∈ Chev(s), as is shown by [IA , 4.5.1]. Suppose that K ∈ Chev(2). Choosing an involution x ∈ H − Z(K) we have J ≤ E(CK (t, x)) ≤ L2 (CK (x)) by L2 -balance. But L2 (CK (x)) = 1 by the BorelTits theorem, contradiction. If K ∈ Alt, then K ∼ = 2An for some n, t acts as the product of k transpositions for some k, and so O 2 (CK (t)) lies in the central product A ∗ B, where B ∼ = 2An−2k (if n − 2k ≥ 4; otherwise just An−2k ) and F ∗ (A) is a 2-group. Hence n − 2k ≥ 5 and J∼ = 2An−2k . But then J does not have the given isomorphism type, contradiction. Finally if K ∈ Spor, then J ∈ Chev(2) by [III11 , 1.1b]. Hence q = 3, and by inspection of [IA , 5.3], CK (t) does not have the alleged structure. The proof is complete.  Lemma 10.61. Let K ∈ K2 . Let x ∈ I2 (Aut(K)), let J be a component of CK (x), and suppose that J ∼ = Ω± 2m (q), q odd, m ≥ 3, with 1 = Z(J) ≤ Z(K). Then 6 (m+1)2 , D). K ∈ G2 and F(K) > (q Proof. By Lemma 10.46, K ∈ Chev(r), where q is a power of r. Since m ≥ 3, 6 7 J ∈ G62 or else J ∼ = Ω± 6 (3); in any case, as J ↑2 K, K ∈ G2 ∩ Chev(r) ⊆ G2 ∪ G2 . If ± ± K ∈ G72 , then K ∼ = L3 (q1 ) for some q1 , so J/Z(J) ∼ = L3 (q) or L2 (q), a contradiction. Thus, K ∈ G62 . 2 Suppose F(K) ≤ (q (m+1) , D). If x induces a field or graph-field automorphism 2 2 on K, then f (K) = (q 2 )m > q (m+1) , contradiction. If x induces a graph auto(2m−1)2 > morphism, then from [IA , 4.5.1] we get K/Z(K) ∼ = L± 2m (q), so f (K) = q 2 (m+1) , contradiction. Thus x ∈ Inndiag(K). Since 1 = Z(J) ≤ Z(K), it is imq ∼ possible by [IA , 4.5.1] that K be nonclassical, A±  (q1 ), or B (q1 ), unless J = A3 (q) 2 1/2 49/2 (m+1) >q , contradic(m = 3) and K ∼ = A7 (q ); but in that case, f (K) = q tion. If K ∼ = C (q1 ), the only possibility by [IA , 4.5.1] is K ∼ = C4 (q), m = 3 and  = 4; but then again F(K) > F(J). Finally suppose K ∼ = D± (q1 ). If q1 = q, then the conditions  ≤ m + 1 and 1 = Z(J) ≤ Z(K) again force m = 3 and K∼ = D4 (q). Then from [IA , 4.5.3] we get J ∼ = Spin± 6 (q), a contradiction. If q1 = q

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then K ∼ = D2m (q 1/2 ) and f (K) = q 2m > q (m+1) , a final contradiction. The proof is complete.  2

2

Lemma 10.62. Suppose that K ∼ = Bn−1 (q), n ≥ 4, and J is = Dn± (q), q odd, L ∼ a nontrivial level pumpup of L. If J contains a subgroup X which is the product of ± (q), then F(J) > F(K). two components isomorphic to Dn−1 2

Proof. Let  be the untwisted Lie rank of J. If the lemma fails, then q  = 2 f (J) ≤ f (K) = q n , so  ≤ n; likewise if  = n then J is of type A or D. Using [IA , 4.5.1] we see that J ∼ = F4 (q) (with n = 5) or Dn (q). In either case we compute  2 |X|r ≥ |Dn−1 (q)|r ≥ |J|r , where r is the prime dividing q. Hence equality holds throughout, and by Tits’s lemma [IA , 2.6.7], X lies in a proper parabolic subgroup  of J, forcing Or (X) = 1, a contradiction. Lemma 10.63. Let K = Ω± n (q) for some n ≥ 8 and some odd prime power q = r m , and let J = Ωn−2 (q),  = ±1. Suppose that I ∈ K2 , J ↑2 I, and if I ∈ Chev, then F(I) ≤ F(K). Then I is a quotient of Ωηn (q) or Ωηn−1 (q) for some sign η. Proof. By Lemma 10.46, I ∈ Chev(r). Suppose that t ∈ I2 (Aut(I)) and J ↑2 I via t. Let k be the untwisted Lie rank of K. Thus k = [n/2] ≥ 4 and the untwisted Lie rank of J is k − 1. If t is a field or graph-field automorphism of I, then I has level q 2 and untwisted 2 2 Lie rank k − 1, so f (I) = q 2(k−1) > q k = f (K), contradiction. Thus t ∈ Aut0 (I). Suppose that I has level q0 = q. Then from [IA , 4.5.1], (I, k) = (A7 (q 1/2 ), 4), ± ± (D2k−2 (q 1/2 ), k) or (D2k−1 (q 1/2 ), k). In each case f (I) > f (K), contradiction. Thus we may assume that I has level q. Hence the untwisted rank of I is at most k, by [IA , 4.2.2]. If I = A (q), then t must be a graph automorphism and 2 2 2 n − 2 =  + 1, so f (I) = q  = q (n−3) > q k = f (K), contradiction. Otherwise, if I is not of type B or D, then again from [IA , 4.5.1], n = 8 and I ∼ = C4 (q), so (q). Since J is not a spin group, F(I) > F(K), contradiction. Thus I/Z(I) ∼ = P Ω± m (q). The condition F(I) ≤ F(K) then immediately implies I is a quotient of Ω± m that m ≤ n, which yields the lemma.  In the next several lemmas we deal with the following setup.

(10I)

(1) K ∈ K2 ∩ Chev(r) for some odd r, x ∈ I2 (Aut(K)), and L is a component of CK (x); (2) K and L have the same level q.

We introduce additional notation:

(10J)

(1) K ∗ ∈ K2 and K ∗ → K is a 2-saturated covering of K, with kernel Y ; (2) L∗ is the component of CK ∗ (x) mapping on L, and Z is the kernel of the covering L∗ → L.

Notice that x induces an automorphism of the universal version K u of K, and hence of the 2-saturated group K ∗ ∼ = K u /O2 (K u ). Thus CK ∗ (x) makes sense in (10J2).

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In this situation we know that (1) L∗ /O2 (L∗ ) ↑2 K ∗ ; (2) Z = L∗ ∩ Y ; and (10K) (3) x does not induce a field or graph-field automorphism on K, so x and K ∗ are to be found in [IA , 4.5.2]. Indeed (1) and (2) are obvious, and (3) is as well, since L, L∗ , K, and K ∗ all have the same level. Lemma 10.64. Assume (10I), (10J) and (10K). Suppose that L ∼ = Spin± m (q) or HSpinm (q) for some m ≥ 7. Then (a) One of the following holds: ± ∼ (1) L ∼ = Spin± m (q), and K = Spinn (q) or HSpinn (q) with n > m; moreover, L ∩ Z(K) contains an involution zL such that L/zL ∼ = ± Ω± m (q) or P Ωm (q); (2) L ∼ = HSpin8 (q) ∼ = Ω+ 8 (q); (3) (K, L) is one of the pairs (F4 (q), Spin9 (q)), (E6 (q), Spin10 (q)), a (E7 (q)u , Spin+ 12 (q)), (E7 (q) , HSpin12 (q)), (E8 (q), HSpin16 (q)); (b) If x is induced by an involution ξ ∈ L and L/ ξ ∼ = Ω± m (q), then either + ∼ (a3) holds or L = Spin8 (q); and 2 ∼ (c) If (K, L) = (E7 (q)u , Spin+ 12 (q)), then O (CK (x)) = LL1 where L1 = ∼ SL2 (q), [L, L1 ] = 1, Z(L1 ) ≤ Z(L), and L/Z(K) = L/Z(L1 ) ∼ = HSpin12 (q). Proof. First consider the case L ∼ = Spin± m (q), which is 2-saturated by [IA , ∗ ∗ 6.1.4]. Then L = L ↑2 K . We show that (a1) or (a3) holds. Indeed from [IA , 4.5.2], K ∗ is as in (a1) or (a3), so to prove (a) in this case we need to show that if K ∗ is a spin group, then K = K ∗ or K is a half-spin group. However, the only alternative is that K is a quotient of Ω± n (q), in which case the tables [IA , 4.5.1, 4.5.3] show that L cannot be the universal version, i.e., a spin group, which is a contradiction. Moreover, if L ∩ Z(K) contains no involution zL such that L/ zL ∼ = (q) as in (a1), then C (x) has the component LZ(K)/Z(K), which is a Ω± K/Z(K) n spin or half-spin group, and is isomorphic to L if L ∼ (q). However, K/Z(K) = Spin+ 8 has no such involutory automorphism, by [IA , 4.5.1]. Thus (a1) holds, proving (a) in this case. Under the assumptions of (b), if (a3) fails and L ∼  Spin+ = 8 (q), then ± zL is the unique involution of Z(L) such that L/ zL ∼ = Ωm (q). Hence zL = ξ. But zL ∈ Z(K) while ξ ∈ Z(K), a contradiction. So (b) holds in this case. In (c), all the assertions follow from the information in [IA , 4.5.2] except the isomorphism L/Z(K) ∼ = HSpin12 (q), which is supplied in [IA , 4.5.1]. Consider next the case L ∼ = HSpinm (q), so that m ≡ 0 (mod 4) and L/Z(L) ∼ = + P Ωm (q). In (a), we may assume m ≥ 12. Then |Z| = 1 or 2 according as L∗ = L ∗ ∗ or L∗ ∼ = Spin+ m (q). If L = L, then L ↑2 K and we see from [IA , 4.5.2] that ∼ (a3) holds with K = E8 (q). So assume that L∗ ∼ = Spin+ m (q). As m ≥ 12, there ∗ ∗ is a unique involution zL∗ ∈ Z(L ) such that L / zL∗ ∼ = Ω+ 2m (q), and zL∗ ∈ Z. Thus zL∗ ∈ Y . If K ∼ = E7 (q), then |Y | ≤ 2, whence Z = Y = Z(K ∗ ) and so K is the adjoint version with m = 12, and (a3) again holds. Otherwise we see from [IA , 4.5.1] that K ∼ = D± (q) or B (q), and L∗ /L∗ ∩ Z(K ∗ ) has version “c,” which means exactly that zL∗ ∈ Z(K ∗ ). Then Z zL∗ is a four-subgroup of L∗ ∩Z(K ∗ ) and so Z(L∗ ) = Z(K ∗ ). Consequently K ∼ = D+ (q) with  even. Letting + c ∗ K = K / zL∗ ∼ = Ω2 (q), we conclude that the central involution y = −1 of K c

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∼ Ω+ (q), whose unique central involution is not lies in L∗ / zL∗ . But L∗ / zL∗ = 2m y, but rather acts on the natural Fq K c -module by inverting only the support of L∗ . This contradiction completes the proof of (a). The hypothesis of (b) cannot ∼ + hold as the only proper quotient of L is P Ω+ m (q) = Ωm (q), so (b) holds vacuously, as does (c). The proof of the lemma is complete.  Lemma 10.65. Assume (10I), (10J) and (10K). Suppose that L is a quotient of Ω± m (q) for some m ≥ 6. Then one of the following holds: (a) K is a quotient of Ωn (q) for some n > m and some sign ; (b) K is a quotient of SL± m (q), and if Z(L) has even order, then K is 2saturated; or (c) K ∈ G62 and F(K) > F(Ωr (q)), where r = 2[(m + 2)/2] is the least even number strictly larger than m. Proof. In the row of [IA , 4.5.1] corresponding to x and L, the version of L must be either “c” or “a” if m > 6; at any rate the version cannot be “u”. This rules out the exceptional types K = F4 (q) or E (q),  = 6, 7, 8, so K is a classical group. Suppose first that K ∗ ∼ = SL±  (q) for some . Then from [IA , 4.5.1], m =  and LZ(K)/Z(K) is the adjoint version, so Z(L) ≤ Z(K). Also from [IA , 4.5.2], |Z(L∗ )| ≤ 2. Consequently if Z(L) has even order, then |Z(L∗ )| = 2 and Z has odd order. But Z = L∗ ∩ Y ≤ Z(K ∗ ) with Z(K ∗ ) cyclic, whence Y has odd order and K is 2-saturated. Thus (b) holds. If K ∼ = C (q), then from [IA , 4.5.1], m = 6,  = 4, and so (c) holds. It remains to consider the cases K = B (q) and D± (q). Then K ∗ is a spin group and from [IA , 4.5.2], L∗ is also a spin group. Since L is not a spin group, Z = 1 and so Y = 1. If CZ(K ∗ ) (x) is cyclic then Y contains the unique involution of CZ(K ∗ ) (x), so K ∼ = K ∗ /Y satisfies (a). We are reduced to the case that CZ(K ∗ ) (x) is a four-group, that is, K ∗ = D2k (q), k ≥ 2 and Y = 1 but Y does not contain any involution y such that K ∗ / y = Ω+ 4k (q). In particular x is not a graph automorphism. Thus x ∈ Inndiag(K). Also, k = 2, so k ≥ 3 and K∼ = D2k (q)hs . Our conditions imply that in the notation of [IA , 4.5.2, 4.5.3], x is conjugate in its action on K to ti or ti , 1 ≤ i ≤ m/2. But then by these tables, since + ∗ L is a quotient of Ω± m (q), Y must contain the kernel of the covering K → Ω4k (q). This contradicts what we saw above, and completes the proof.  Lemma 10.66. In Lemma 10.64a1, write q = r a with r prime. Then we have  that O r (CK/Z(K) (L)) is either trivial or an orthogonal group. In particular if x  induces an inner automorphism on K corresponding to ξ ∈ I2 (K), and O r (CK (L))  has a subnormal subgroup I ∼ = SL2 (q) with ξ ∈ I, then O r (CK (L)) has a second subnormal SL2 (q) subgroup. Proof. The isomorphism type of L shows that x is Aut0 (K)-conjugate to one of the following involutions in [IA , 4.5.1]: ti , ti (1 ≤ i < m/2); tm/2 , tm/2 + (K = D2 (q)); or γi (1 ≤ i ≤ (m + 1)/2). It follows that there is a natural Fq K-module and an involution ξ ∈ O(V ) such that x acts on K like ξ. Writing V = V1 ⊥ V2 , where V1 and V2 are the two eigenspaces of ξ, we see that we  may assume that L = CΩ(V ) (V2 ). Since CΩ(V1 ) (L) consists of scalars, O r (CK (L))  preserves V1 and V2 and equals O r (CΩ(V ) (V1 )), which is trivial or isomorphic to Ω(V1 ), as claimed. The lemma follows. 

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Lemma 10.67. In Lemma 10.64a1, suppose that L ↑2 K via ξ ∈ I2 (K), and ± ∼ ξ ∈ L ∼ = Spin+ 8 (q). Suppose also that K/Z(K) = P Ωn (q) with n ≤ 12. If   O 2 (O 2 (CK (ξ))) = L(M1 × M2 ) with Mi ∼ = O 2 (SL2 (q)), i = 1, 2, and [L, M1 M2 ] = + ∼ 1, then K = Spin12 (q) or HSpin12 (q). In the latter case Z(K) is the diagonal subgroup of Z(M1 ) × Z(M2 ). Proof. Given the structure of K/Z(K) and L, we have K ∼ = Spin± n (q), 9 ≤ n ≤ 12, or HSpin12 (q), by [IA , 6.1.4, 4.5.2, 4.5.3]. Since CK (L) is not an r  group, n = 9 or 10, by Lemmas 6.7 and 6.8. If K ∼ = Spin11 (q) or Spin− 12 (q) but − (∞) ∼ ∼ ∼ not Spin11 (3), then CK (L t ) = Spin3 (q) = SL2 (q) or Spin4 (q) = SL2 (q 2 ), contradicting the existence of M1 M2 . If K ∼ = Spin11 (3) then it follows from [IA , 4.5.1] that O 2 (CK (L)) ∼ = Spin3 (3) ∼ = SL2 (3), again contradicting the existence of M1 M2 . Thus K ∼ = Spin+ 12 (q) or HSpin12 (q), as asserted, and by [IA , 4.5.2], 2 M1 M2 = O (X) where X ∼ = Spin+ 4 (q) and the images of L and X in K/Z(K) have orthogonal supports in the natural representation of K/ zK ∼ = Ω+ 12 (q). If + i be the K ∼ = HSpin12 (q), then consider the covering Spin12 (q) → K and let M SL2 (q)-preimage of Mi , i = 1, 2. By [IA , 6.2.1b], the diagonal involution z ∈ 2 ) is the spin involution of Spin+ (q). But the spin involution maps 1 ) × Z(M Z(M 12 on the involution of Z(K), which therefore lies on the diagonal of M1 M2 . The proof is complete.  Lemma 10.68. Suppose that K ∈ K2 , x ∈ I2 (K), and CK (x) has a component L such that x ∈ L ∼ = Spinn (q) for some odd n ≥ 7 and some odd prime power q. Then either L = K or else n = 9 with K ∼ = F4 (q). Proof. Suppose that L = K. By Lemma 10.46, K ∈ Chev(r), where q is a power of r. By [IA , 6.1.4], x ∈ Syl2 (Z(L)) and L is the universal version. Since L = K, there is no loss in passing to K/Z(K); but then [IA , 4.5.1] shows that n = 9  and K ∼ = F4 (q). In the next lemmas we assume that (10L)

p is a prime and I, J ∈ Kp with I ↑ J via some x ∈ Ip (Aut(J)). p

We may identify I with I0 /Op (I0 ) for some component I0 of CJ (x). With this point of view, we say that (10M)

I is anchored in J if and only if I0 ∩ Op (J) = 1.

Lemma 10.69. Assume that (10L) holds with p = 2. Then the following conditions hold: (a) If I ∼ = 2An , n ≥ 7, J/Z(J) ∼ = L3 (4) with = SL2 (q), q odd, q > 3, then J ∼ 2 6 ∼ Z(J) of exponent 4, J = SL2 (q ), J ∼ (q), or J ∈ G ; = L± 2 3 2 6 ∼ (b) If I ∼ = L± 3 (q), q odd, q > 3, then either J = L3 (q ) or J ∈ G2 ; ∼ (c) If I = Sp2n (q), q odd, n ≥ 2, with I anchored in J, then J ∈ G62 . Moreover, if F(J) ≤ F(P Sp2n+2 (q)), then n = 2 and J ∼ = Sp4 (q 2 ), SL± 4 (q), 1 Sp8 (q 2 ), or Spin7 (q), with x inducing, respectively, a field automorphism, a graph automorphism, an outer diagonal automorphism, or an innerdiagonal automorphism on J; and (d) If I ∼ = Ω+ 8 (q), q odd, and I is anchored in J, then x ∈ Inn(J).

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Proof. In (a), we may assume by [III11 , 13.5] that J ∈ G2 with d2 (J) ≤ 6. Since J ∼ = 2An and J ∈ G62 are allowed conclusions, we may assume that 1 < d2 (J) < 6. Thus J ∼ = An , n ≥ 11, or M c, Ly, O  N , F5 , J1 , Co3 , or He. But then it is clear from [IA , 5.3] and [III11 , 1.6] that J is not a pumpup of I. Thus (a) holds. In (b) and (c), using [III11 , 1.1ab] and [IA , 2.2.10], we get J ∈ Chev(r), where q is a power of r, except possibly if I ∼ = Sp4 (3). In the latter case, Lemma 10.24 yields the same conclusion since J/Z(J) ∼ = L4 (4) would force Z(J) = 1. In (b), we may assume that d2 (J) = 6, whence d2 (J) = 7, i.e., J ∼ = L3 (r a ) for 2 some sign  and some integer a. Then by [IA , 4.5.1, 4.9.1], J ∼ = L3 (q ), as desired. In (c), we have I ∈ G62 , so J ∈ Gd2 , d ≤ 6. Since J ∈ Chev(r), d2 (J) = 6. 2 Suppose then that F(J) ≤ F(P Sp2n+2 (q)) = (q (n+1) , BC). Using [IA , 4.9.1, 4.5.1, u 4.5.2], we find that J ∼ = Sp2n (q 2 ) (with x a field automorphism), A± 2n−1 (q) (with 1/2 x a graph automorphism), Sp4n (q ) (with x an outer diagonal automorphism), 2 2 or Spin7 (q) (with n = 2 and x ∈ Inndiag(J)). Accordingly f (J) = q 2n , q (2n−1) , 2 2 q 2n , or q 9 , with n = 2 in the last case. As f (J) ≤ q (n+1) and n ≥ 2, we conclude that n = 2, which gives the five groups in the conclusion of (c). Finally in (d), examination of [IA , 4.5.1] shows that if x ∈ Inndiag(J), then 1 J ∼ = D8+ (q 2 ) and x is conjugate to t4  or t4  , which automorphisms are indeed non-inner. The proof is complete.  Lemma 10.70. Assume that (10L) holds with p = 2. Suppose that (a) I ∼ = SL2 (q), q odd, q > 3, and I is anchored in J; (b) AutAut(J) (I) contains Inndiag(I); and (c) If J ∈ G62 , then F(J) ≤ (q 9 , A). Then one of the following holds: (a) J ∼ = L3 (4) with Z(J) of exponent 4; = 2An , n ≥ 7, or J/Z(J) ∼ (b) J ∈ Chev has level q; or η η (c) J ∼ = SL2 (q 2 ), SL4 (q 1/2 ) or D4 (q 1/2 ), η = ±1. Moreover, in all cases except J ∼ = D4+ (q 1/2 ), E(CJ (x)) has a unique component isomorphic to SL2 (q). Proof. Hypothesis (a) rules out the possibility J ∈ Spor, as can be seen in [IA , 5.3]. If J ∈ Alt then as 1 = Z(I) ≤ Z(J), J ∼ = 2An by [IA , 6.1.4], and n ≥ 7 since q > 3, so conclusion (a) holds in this case. If J ∈ Chev(2)−Alt−∪r>2 Chev(r), then by the Borel-Tits theorem and [IA , 4.9.1, 4.9.2], J/Z(J) ∼ = L2 (42 ), L± 3 (4), ± Sp4 (4), or L5 (2). Then as Z(J) = 1, J/Z(J) ∼ = L3 (4) by [IA , 6.1.4]. Since elements of I of order 4 map to involutions of IZ(J)/Z(J), Z(J) must have exponent 4, in view of [IA , 6.4.1], and again (a) holds. We may therefore assume that J ∈ Chev(r) − Alt for some odd r. If x induces a field or graph-field automorphism on J, then conclusion (c) holds. Otherwise CJ (x) can be examined in [IA , 4.5.1, 4.5.2], and assuming that conclusion (b) fails, we see that either conclusion (c) holds or else J ∼ = Bm (q 1/2 ), 2 ≤ m ≤ 4,  x conjugate to t2 . But in that case OutAut(J) (I) does not contain Outdiag(I), contradicting hypothesis (b). The lemma follows.  Lemma 10.71. Let X = E(X) be a K-group with O2 (X) = 1. Let x ∈ I2 (Aut(X)) and suppose that CX (x) has components L1 ∼ = L2 ∼ = SL2 (q), q > 3, q

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 odd, with Z(L1 ) = Z(L2 ) ≤ Z(X) and X = (L1 L2 )X . Let X1 be any component of X. Then one of the following holds: (a) X1 ∈ Chev and X1 has level q; (b) X1 ∼ = SL2 (q 2 ) or Sp4 (q 1/2 ); (c) X1 /Z(X1 ) ∼ = An , n ≥ 7, or L3 (4), with Z(X1 ) of exponent 2 or 4, respectively; (d) X1 ∈ G62 and f (X1 ) > q 4 . Proof. Let Ji be the subnormal closure of Li in X. By L2 -balance, either Ji is the direct product of two copies of Li or Ji is a single x-invariant component of E(X). In any case every component of E(X) is conjugate to a component of J1 or J2 , and so by symmetry it suffices to prove that the components of J1 satisfy one of the four conditions (a)–(d). This is obvious if J1 is not quasisimple, so assume that J1 is quasisimple. We assume that J1 does not satisfy one of (a)–(d). In particular by (b), x does not induce a field or graph-field automorphism on J1 , so x ∈ Aut0 (J1 ). By Lemma 10.69a, and in view of (a), (b) and (c), J1 ∈ G62 , as the Schur multiplier of L± 3 (q) has odd order [IA , 6.1.4]. Thus by (d), f (J1 ) ≤ q 4 . Using [IA , Table 4.5.1] we see that since (a) fails, q(J1 ) = q 1/2 . Let  be the untwisted Lie rank of J1 . Obviously, 1/2  > 1. But also 2 /2 = logq (f (J1 )) ≤ 4 so  = 2. As J1 ∈ G62 , J1 ∼  A± ). As = 2 (q n 1/2 2 ∼ ∼ Z(L1 ) ≤ Z(J1 ), J1 =  G2 (q ) or G2 (3 2 ) for any odd n. Hence, J1 = Sp4 (q 1/2 ) and (b) holds, contradiction. The proof is complete.  11. Small Groups Lemma 11.1. Let p be an odd prime and K ∈ Kp with mp (K) = 1. If Aut(K) − Inn(K) contains an element a of order p, then K ∈ Chev and a is a field automorphism. Proof. This is [GL1, (7-13)(2)].



Lemma 11.2. Suppose that K ∈ K3 , m3 (K) = 2, and Z(K) = 1. Then ∼ SL (q), q ≡  (mod 3),  = ±1, or 3A6 , 3A7 , or 3M22 . K= 3 Proof. If K ∈ Spor ∪Alt, then as Z(K) = 1, the result follows from [IA , 6.1.4,  5.6.1]. If K ∈ Chev(r) − Alt and r = 3, then by [IA , 6.1.4], K/Z(K) ∼ = L3kq (q) or q E6 (q), q = ±1, q ≡ q (mod 3), and then as m3 (K) = 2, [IA , 4.10.3a] yields the desired result. Finally if K ∈ Chev(3) − Alt, then by [IA , 6.1.4, 6.4.4], the result holds unless possibly K ∼ = 3G2 (3). We complete the proof by showing that m3 (3G2 (3)) > 2. Namely, let E be the preimage in K of the product of two long root subgroups corresponding to roots inclined at π/3. The discussion of K on [IA , p. 319] shows that the preimages of the long root subgroups commute and  have exponent 3. Hence E ∼ = E33 , completing the proof. Lemma 11.3. Suppose that X is a K-group, O2 (X) = 1, T ∈ Syl2 (X), and T is dihedral or semidihedral with T ∩ Z(X) = 1. Then one of the following holds: (a) X = T ; or (b) There exists X1  X such that X1 ∼ = SL2 (r) for some odd prime power r. Moreover, T is semidihedral and every involution of T − Z(T ) induces an outer diagonal automorphism on X1 .

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Proof. Let X1 = O 2 (X) and let T1 be the cyclic maximal subgroup of T . Then no involution of T − T1 is X-conjugate into T1 , so by the Thompson Transfer Lemma, applied to every involution of T − T1 , T ∩ X1 has a unique involution. If T ∩ X1 is cyclic then X1 = 1 by Burnside’s Theorem, and (a) holds. So assume that T ∩ X1 is quaternion. Then Z(T ) = Z(X), and we set X = X/Z(T ), so that O2 (X) = 1 and T is dihedral. We may assume that T < X. If F ∗ (X) is solvable, then X must have a normal four-group and then X ∼ = Σ4 , so X1 ∼ = SL2 (3) and (b) ∼ holds. Therefore E(X) = 1, so with [IA , 5.6.3], E(X) = L2 (r) or A7 for some odd r > 3. As T is dihedral, E(X) = F ∗ (X). Now Out(E(X)) is abelian so X 1 ≤ E(X), whence equality holds and X1 ∼ = SL2 (r) or 2A7 . Let t be any noncentral involution of T . If X1 ∼ = Aut(A7 ) ∼ = Σ7 has 2-rank 3, contradiction. So = 2A7 then X 1 t ∼ ∼ X1 = SL2 (r). Similarly we reach a contradiction if t induces a field automorphism on X 1 . But T 1 < T so the only possibility is that X ∼ = P GL2 (r), and the lemma follows.  Lemma 11.4. Let X be a K-group and x ∈ I2 (X). Assume that CX (x) has a Sylow 2-subgroup x, z ∼ = E22 for some z ∈ Z ∗ (X). Set X = X/O2 (X). Then either X is a 2-group of maximal class or for some odd q, O 2 (X) ∼ = SL2 (q) and X/ z ∼ = P GL2 (q). Proof. We may assume that O2 (X) = 1, so that z ∈ Z(X). Expand x, z to T ∈ Syl2 (X). Then by [IG , 10.24], T has maximal class, and as x2 = 1, T is dihedral or semidihedral. Then Lemma 11.3 applies to X and yields the lemma.  Lemma 11.5. Let K ∈ Chev(p) for some odd p. Suppose that K  X, x ∈ I2 (X), and CX (x) has abelian Sylow 2-subgroups with m2 (CX (x)) ≤ 2. Then K∼ = A1 (pn ) for some n. Proof. Suppose that K ∼  A1 (pn ). If K ∼ = 2 G2 (3 2 ) ∼ = L2 (8), then m2 (CK (x)) = = m2 (K) = 3, contradiction. So we may assume that K ∈ Lie(p). If x induces a field or graph-field automorphism on K, then x ∈ K, so CK (x) has cyclic Sylow 2-subgroups and a normal 2-complement. But CK (x) ∈ Lie(p), which is a contradiction. Thus x acts on K as an element of Aut0 (K). Since K ∼ = A1 (pn ), CK (x) has a Lie component Kx , by [IA , 4.5.1], and as Kx has abelian Sylow 2-subgroups, Kx ∼ = L2 (pn ) for some pn ≡ ±3 (mod 8). But then again m2 (Kx x ) = 3, a contradiction. The proof is complete.  1

Lemma 11.6. Suppose that X = SL2 (q) (q > 3, q odd) or Q8 . Suppose that ∼ X and Y acts on X. Then [Z(Y ), X] = 1. Y = Proof. If X = Q8 , then Aut(X) ∼ = Σ4 , which does not contain Q8 . If X = SL2 (q), q > 3, then Aut(X)(∞) ∼ (q), which similarly has dihedral Sylow 2L = 2 subgroups and so does not contain a quaternion group. This implies the result.  ∼ SL2 (q), q odd, α2 = 1, and Lemma 11.7. Suppose that X = K α where K = K α /Z(K) ∼ = P GL2 (q). Then the following conditions hold: (a) α inverts a cyclic subgroup of K of order q + q , where q = ±1 and q ≡ q (mod 4); and (b) I2 (X − Z(K)) = αK .

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Proof. From [IA , 4.5.2], we see that |CK (α)|2 = 2. Hence a Sylow 2-subgroup T of X is of maximal class, and as it contains noncentral involutions and a quaternion maximal subgroup, T must be semidihedral. Hence all noncentral involutions of T are T -conjugate, which, with Sylow’s theorem, implies (b). From [IA , 4.5.1], we see that C := CX/Z(K) (α) contains a subgroup D ∼ = D2(q+q ) . Since K/Z(K) does not contain a Zq+q -subgroup, the preimage of C ∩ K/Z(K) in K is a semidirect product E β , where E is a cyclic group of order (q + q )/2 inverted by β of order 4. As |CK (α)|2 = 2, α inverts β. Then αβ ∈  I2 (X − Z(K)) and αβ inverts E. So (b) implies (a), completing the proof. Lemma 11.8. If K = SL2 (q), q odd, u ∈ I2 (Aut(K)), and Z(K) is complemented in a Sylow 2-subgroup of CK (u), then u ∈ Inndiag(K) − Inn(K). Proof. Let S ∈ Syl2 (CK (u)). As Z(K) = Ω1 (S), S = Z(K) ∼ = Z2 . But if u is a field automorphism, then CK (u) ∼ = SL2 (q 1/2 ), while if u ∈ Inn(K), then u acts on K like some v ∈ K of order 4 and v ∈ S. Thus |S| > 2, a contradiction in either case. The lemma follows.  Lemma 11.9. Let X = Ω+ 4 (3) and D = t, u ≤ Aut(X) with t and u acting like distinct commuting reflections on X. Then     O2 (X) ≤ O 3 (CX (t)), O 3 (CX (u)) . Proof. Let Q = O2 (X) ∼ = Q8 ∗ Q8 , and X = X/Z(Q). Also let Xt = 3 O (CX (t)) and Xu = O (CX (u)). We have CX (t) ∼ = CX (u) = Z(Q) × Xu ∼ = Z2 × A4 . It suffices to show that Q ≤ X t , X u . Now t interchanges the two SL2 (3) central factors of Q, so elements of X t and X u of order 3 are fixed-pointfree on Q. Therefore if the result fails, then O2 (X t ) = O2 (X u ) and then X t = X u (note that a Sylow 3-subgroup of X normalizes no four-subgroup of Q). But then [u, CX (t)] ≤ Z(Q) and so [u, O 2 (CX (t))] = 1, whence [u, CX (t)] = 1 and CX (t) = CX (u). As t and u are reflections this implies t = u, a contradiction that completes the proof.  3

Lemma 11.10. Suppose that F ∗ (X) = K1 K2 , where [K1 , K2 ] = 1, Ki ∼ = SL2 (qi ) for some odd qi , i = 1, 2, and Z(K1 ) = Z(K2 ). Then the following conditions hold: (a) Let f ∈ I2 (X) induce a nontrivial field automorphism on both K1 and K2 . 1 Then CF ∗ (X) (f ) = I1 I2 d , where Ii = CK (f ) ∼ = SL2 (q 2 ), Ii d /Z(Ki ) ∼ = i

1 2

i

2

P GL2 (qi ), i = 1, 2, and d = 1; and (b) All involutions of F ∗ (X) − Z(K1 ) are F ∗ (X)-conjugate. Proof. Let (K, σ) be a σ-setup for I1 = CK1 (f ). Then we may identify CK (σ 2 ) with K1 . By Lang’s Theorem there is d1 ∈ K such that df1 = dσ1 = d1 z, 2 where z = Z(K1 ). Then dσ1 = d1 so d1 ∈ K1 . Moreover, by [IA , 2.5.4, 2.5.8], 1 ∼ P GL2 (q 2 ). As K1 has quaternion Sylow 2-subgroups, so does I1 d1 , I1 d1 / z = i and replacing d1 by a suitable element of the coset d1 I1 we may assume that d21 = z. A similar element d2 ∈ K2 exists with respect to I2 . Then (a) holds with d = d1 d2 . In (b), any involution of K1 K2 has the form x1 x2 with xi ∈ Ki of order 4, i = 1, 2;  the assertion holds as xi is uniquely determined up to Ki -conjugacy, i = 1, 2.

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Lemma 11.11. Let K = L3 (q), q odd, let Inn(K) ≤ X ≤ Aut(K), and let − T ∈ Syl2 (X). Also let K − = L− ∈ Syl2 (K − ). Then T has no normal 3 (q) and T − subgroup isomorphic to T . Proof. Suppose false and regard T −  T and K = Inn(K). If q ≡ −1 (mod 4), then K admits no field automorphisms. In that case let 2n = (q + 1)2 ≥ 4. Then T has a subgroup T0 = T ∩K ∼ = Z2n Z2 . = SD2n+2 and |T : T0 | ≤ 2, while T − ∼ 2n+1 − n+3 = |T | ≤ |T | ≤ 2 . As n ≥ 2, we have n = 2 and T − = T . Therefore 2 − ∼ Therefore T = Z4  Z2 has a maximal subgroup T0 ∼ = SD16 . Let J be the abelian maximal subgroup of T − . Then Ω1 (J) = Φ(J) ≤ Φ(T ) ≤ T0 so Ω1 (J) is a normal four-subgroup of T0 . But T0 has no such subgroup, contradiction. Suppose then that q ≡ 1 (mod 4) and let 2n = (q − 1)2 ≥ 4. Then T − is semidihedral of order 2n+2 , and T1 := T ∩K  T with T1 ∼ = Z2n Z2 . Moreover, T /T1 embeds in Out(K) and hence (as |Outdiag(K)| = 1 or 3) in Out(K)/ Outdiag(K) ∼ = Γ K × ΦK ∼ = Z2 × Zn [IA , 2.5.12]. Let J1 be the abelian maximal subgroup of T1 ; then T ≤ T1 (ΓK × ΦK ) with ΓK × ΦK faithfully inducing power maps on J1 . If T − ∩ J1 (ΓK × ΦK ) contains an element x ∈ J1 , then Ω1 (J1 ) ≤ [x, J1 ] ≤ [T − , T ] ≤ T − , so Ω1 (J1 ) is a normal four-subgroup of the semidihedral group T − , a contradiction. Thus T − ∩ J1 (ΓK × ΦK ) ≤ J1 . As |T1 : J1 | = 2, it follows that |T − : T − ∩ J1 | ≤ 2, so T − ∩ J1 is the cyclic maximal subgroup of T − . Thus T − ∩ J1 ∼ = Z2n+1 , contradicting the fact that J1 has exponent 2n . The proof is complete.  Lemma 11.12. Suppose that F ∗ (X)/Z(F ∗ (X)) ∼ = G2 (q) or 3D4 (q) for some q, and Op (X) = 1 for some prime p not dividing q. Then X splits over F ∗ (X). Proof. Let K = F ∗ (X). By hypothesis, Or (K) = 1, where r is the prime dividing q. Then by [IA , 6.1.4], K is simple, and so it suffices to show that Aut(K)  splits over K. But this holds by [IA , 2.5.12], since Outdiag(K) = 1. Lemma 11.13. Let K = G2 (q) or 3D4 (q), q odd. Let x ∈ I2 (K). Then F (CK (x)) = F ∗ (CAut(K) (x)). Proof. Let F = F ∗ (CAut(K) (x)) and F0 = E(CK (x)). If q > 3, then F0 ∼ = ∼ SL2 (33 ). In any SL2 (q) ∗ SL2 (q k ), where k = 1 or 3. If K = 3D4 (3), then F0 = case, F = F0 CF (F0 ). Now Aut(K) = Inn(K)ΦK ΓK , where ΦK ΓK is cyclic and acts faithfully on F0 , for an appropriate choice of x, unless K ∼ = G2 (3). Hence in general, CAut(K) (F0 ) ≤ Inn(K), whence F ≤ K, as desired. In the exceptional case O 2 (CK (x)) ∼ = SL2 (3) ∗ SL2 (3), with the two central factors interchanged by a  generator of CAut(K) (x)/CK (x) ∼ = Z2 , so the result holds in this case as well. ∗

Lemma 11.14. Let K be a nonuniversal version of C2 (q) or A3 (q), q a power  of the odd prime r,  = ±1. Let z ∈ I2 (K) be such that z ∈ O r (CK (z)) ∼ = SL2 (q) ∗  SL2 (q). Let R = CAut(K) (O r (CK (z))). Then according to the isomorphism type of K, R ∼ = Z2 or R is dihedral of order 2(q − ), with cyclic maximal subgroup  CInndiag(K) (O r (CK (z))). Moreover, in the latter case, every involution w ∈ R other than the image of z satisfies E(CK (w)) ∼ = P Sp4 (q). 

Proof. As O r (CK (z)) contains a long root subgroup, R ≤ Aut0 (K) by Lemma 6.1. Let IK = Inndiag(K).  Suppose first that K ∼ = P Sp4 (q). Then by [IA , 4.5.1], CIK (O r (CK (z))) is generated by the image of z, and Aut0 (K) = IK , so R ∼ = Z2 .

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Now suppose K/Z(K) ∼ = L4 (q). We may assume that z = diag(−1, −1, 1, 1). 

Then R0 := CIK (O r (CK (z))) induces scalar mappings on each eigenspace of z, so it is a cyclic group of order q −  consisting of images of diagonal matrices. Let γ be the transpose-inverse (graph) automorphism of K, so that Aut0 (K) = IK γ . Then  O r (CK (z)) has a noncentral involution t centralized by γ such that γt centralizes  O r (CK (z)). It follows that we may take R = R0 γt . Since γ inverts R0 and  t ∈ O r (CK (z)) centralizes R0 , R is dihedral, as claimed. Finally, for any involution  g ∈ R − R0 , g is a graph automorphism of K, so by [IA , 4.5.1], O r (CK (g)) ∼ =    r r r P Sp4 (q) or P Ω± 4 (q). But since z ∈ O (CK (g)) and O (CK (z)) ≤ O (CK (g)), the   only possibility is O r (CK (g)) ∼ = P Sp4 (q). The proof is complete. Lemma 11.15. Let K = P Sp4 (q), q = r n odd, r prime. Then the following conditions hold: (a) K has two conjugacy classes of involutions, represented say by y1 and y2 , where y1 is 2-central; (b) CK (y1 ) = LLt t = L, t with [L, Lt ] = 1, L ∼ = SL2 (q), t ∈ y1K , ty1 ∈ y2K , t and CAut(K) (LL ) = y1 ;  (c) K2 := O r (CK (y2 )) ∼ = L2 (q) and AutK (K2 ) = Inndiag(K2 ); (d) If q ≡ 1 (mod 8), then there exists an involution u ∈ K such that y1u = t  and u centralizes O r (CK (y1 t)); (e) Preimages of y1 and y2 in Sp4 (q) are of orders 2 and 4, respectively; and (f) I2 (E(CK (y1 ))) − {y1 } ⊆ y2K . Proof. Parts (a), (c), and (e), and all but the last three assertions of (b), follow  = Sp4 (q), C  (y1 ) ∼ directly from the tables [IA , 4.5.1, 4.5.2]. In K = SL2 (q)  Z2 , so K  is an involution. Then a preimage of t may be selected so that a preimage  t in K ty1 has order 4, so (e) implies that t ∈ y1K and ty1 ∈ y2K . Moreover, the preimage of every non-central involution u of E(CK (y1 )) has order 4, and hence u ∈ y2K , proving (f). The final assertion of (b) holds by Lemma 11.14. Finally to prove (d), suppose q ≡ 1 (mod 8). Let L12 = E(CLLt (ty1 ))  CK (ty1 ). We identify K with Ω5 (q) and L12 with the centralizer of a hyperbolic plane inverted by ty1 in the natural module. This shows that S := CCK (ty1 ) (L12 ) is dihedral of order q − 1. As |S|2 ≥ 8, while t and y1 are commuting non-2-central involutions of S, there exists an involution u ∈ S interchanging t and y1 . Thus (d) holds and the lemma is proved.  Lemma 11.16. Let K = P Sp4 (q), q odd. Let y ∈ K be a 2-central involution, so that CK (y) = LLt t , L ∼ = SL2 (q), [L, Lt ] = 1, t2 = 1. Then CAut(K) (t, y ) ∩ NAut(K) (L) covers Out(K). Proof. Set C = CAut(K) (y), C0 = C ∩ Inn(K) = LLt t , and C = C/y . From [IA , 4.5.1], K has two classes of involutions, with nonisomorphic centralizers,  CK (ty), so CAut(K) (t, y ) = so C covers Out(K). By Lemma 11.15bc, CK (t) ∼ = NAut(K) (t, y ) ∩ CAut(K) (t) = NAut(K) (t, y /y ). Thus it suffices to show that  t C ( t )∩N (L) covers C/C 0 . Since C 0 = LL t ∼ = L2 (q)Z2 , C 0 acts transitively C

C

t

by conjugation on the set of all pairs (M , u) such that M = L or L , and u is an

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460 t

involution of LL t. As such pairs are permuted by C, it follows that the stabilizer of (L, t) covers C/C 0 . But this is the assertion of the lemma.  Lemma 11.17. Let T ∈ Syl2 (X) and K   X. Let q be an odd prime power. Then the following conditions hold: (a) If K ∼ = Sp4 (q) or SL2 (q), then [Z(T ), K] = 1; (b) If K ∼ = Z2 ; and = P Sp4 (q), then Z(T )/CZ(T ) (K) ∼ (c) If K ∼ = P Sp4 (q) and CX (K) = 1, then Z(T ) ≤ [T, T, T ] ≤ O 2 (CK (Z(T ))). Proof. From [III11 , 6.3a], (a) holds if K ∼ = SL2 (q). As Z(T ) ≤ NT (K), we may assume that X = KNT (K). Let X = X/CX (K) and S = T ∩ K. If K ∼ = P Sp4 (q), then by Lemma 8.8c, |Z(T )| = 2, proving (b). If K = Sp4 (q), we  get Z(T ) ≤ Z(S) and have O r (CK (Z(T ))) = L × Lt with q = r a , L ∼ = SL2 (q), and t2 = 1. Hence Z(S) = Z(K) so (a) follows. For (c), we assume that K = P Sp4 (q) and X = Aut(K). By (b), |Z(T )| = 2. Then Out(K) is abelian, and |CK (Z(T )) : O 2 (CK (Z(T )))| = 2, so [T, T, T ] ≤ O 2 (CK (Z(T ))). To complete the proof it suffices to show that [S, S, S] = 1. Let K1 and K2 be the subnormal SL2 (q)-subgroups of CK (Z(T )). Then since K1 and K2 are interchanged by an element of S, [S, S] acts nontrivially on Ki for each i. By (a), [S, S] ≤ Z(S), so [S, S, S] = 1 and the lemma is proved.  ∼ L (q), q = r a , r an odd prime,  = ±1. Let Lemma 11.18. Suppose that K = 4 x ∈ Aut(K) be a 2-element with E(CK (x)) ∼ = P Sp4 (q). Then x2 = 1. Proof. Let t be the involution in x . Now t is conjugate to the involution γ ∈ ΓK if  = 1, and to the involution γ ∈ ΦK if  = −1; for otherwise, we see from [IA , 4.5.1, 4.9.1] that |CK (t)|r ≤ q 3 < |P Sp4 (q)|r , contradiction. It follows that E(CK (t)) = E(CK (x)) contains a long root group, so x ∈ CAut(K) (E(CK (t))) ≤ Aut0 (K) by Lemma 6.1. But by [IA , 4.5.1], CAut0 (K) (t E(CK (t)) = t has order 2, completing the proof.  Lemma 11.19. Let K = F ∗ (X) = L4 (q),  = ±1, q = r a , r an odd prime. Let T ∈ Syl2 (X) and let z be an involution in Z(T ) ∩ K. Then the following conditions hold:  (a) O r (CK (z)) = L1 Lt1 where t ∈ I2 (CK (z)) ∩ z K , Lt1 = L1 ∼ = SL2 (q); (b) If X = Aut(K), then CCX (z) (L1 Lt1 ) is dihedral of order 2(q − ), with cyclic maximal subgroup CInndiag(K) (L1 Lt1 ). For every involution u ∈ CCX (z) (L1 Lt1 ) − {z}, u induces a graph automorphism on K and E(CK (u)) ∼ = P Sp4 (q); (c) If y = vw with v and w of order 4, v ∈ L1 , and w ∈ CCX (z) (L1 Lt1 ) maps into Inndiag(K) (this only occurs if q ≡  (mod 4)), then y is an involution and E(CK (y)) ∼ = L3 (q); (d) If u ∈ I2 (Aut(K)) centralizes z and induces a nontrivial field automorphism on both L1 and Lt1 , then the image of u in Out(K) lies outside [Out(K), Out(K)]. Proof. In (a), we may take z and t to be images of diag(−1, −1, 1, 1) and the permutation matrix corresponding to the permutation (13)(24). Then (a) and (b) follow from Lemma 11.14 and [IA , 4.5.1]. In (d), there exists u ∈ ΦK inducing the

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same field automorphism as u on L1 . Then u u−1 ∈ CAut(K) (L1 ) ≤ Aut0 (K) by Lemma 6.1. Hence u ∈ Aut0 (K)u so (d) follows from [IA , 2.5.12]. Finally in (c), w has a preimage diag(i, i, 1, 1) in GL4 (q), where i ∈ Fq2 and 2 i = −1. It follows that u = vw has a preimage u0 centralizing the support of Lt1 on the natural module, and acting as a non-scalar involution on the support of L1 . Therefore the eigenvalues of u0 are −1, 1, 1, 1, and (c) follows. The proof is complete.  Lemma 11.20. Let K  X with K ∼ = SL± 4 (q), q odd. If t ∈ I2 (X) and Y := o CK,t (t), then either Ω1 (O2 (K)) ≤ L2 (Y ) or else m2 (Y ) ≥ 3. Proof. If t ∈ K, then t has two eigenvalues −1 on the natural K-module, and so Ω1 (Z(K))) ≤ Lo2 (Y ) ∼ = SL2 (q) × SL2 (q), as desired. So we may assume that t ∈ K and m2 (Y ) ≤ 2, and argue to a contradiction. As t × Ω1 (Z(K)) ≤ Z(Y ), Ω1 (Z(K)) must then be the unique subgroup of Y ∩ K = CK (t) of order 2. Let q be  a power of the prime r, and let Y o = O r (CK (t)) ≤ Y ∩ K. The structure of Y o is given in [IA , 4.5.2], and as Y o contains at most one involution, the only possibility is Y o ∼ = SL2 (q 2 ). But then Ω1 (Z(K)) = Z(Y o ), and as Y o ≤ L2 (Y ) ≤ Lo2 (Y ) the proof is complete.  Lemma 11.21. Let K ∼ = L4q (q), q odd, q ≡ q = ±1 (mod 4), and let y be a  2-central involution of K. Write q = r n with r prime, and let L = O r (CK (y)) = L1 ∗ L2 with L1 ∼ = L2 ∼ = SL2 (q). Then I2 (L) ⊆ y K .  = SLq (q), so that K ∼   Proof. Let K K). Then y has a preimage = K/Z( 4  which is an involution with exactly two eigenvalues equal to −1. Let j be y ∈ K × an element of order 4 in F× q or Fq 2 , according as q = ±1. i ∼ Let t ∈ I2 (L), so that t = t1 t2 with ti of order 4, i = 1, 2. Let L = SL2 (q)  The be the subgroup generated by all r-elements of the full preimage of Li in K.   unique preimage ti of ti in Li has eigenvalues j and −j. Then one preimage of t in  is   K t = j t1  t is K-conjugate t2 , an involution with the same eigenvalues as y. Hence  K  to y, whence t ∈ y . The proof is complete. 

Lemma 11.22. Let K = SL6 (q), q odd, and u ∈ I2 (Aut(K)). Then CK (u) is not an extension of a group of 2-rank at most 1 by a subgroup of SL2 (q). Proof. Suppose by way of contradiction that N  CK (u) with m2 (N ) = 1 and CK (u)/N embeddable in SL2 (q). Then any component or solvable component  of O r (CK (u)) (where q = r a , r an odd prime) must be of rank 1. In particular  if u ∈ Aut0 (K), then O r (CK (u)) must be a product of A1 -type components, and [IA , 4.5.1] shows that there is no such u. Otherwise u is a field or graph-field  1/2 automorphism, and O r (CK (u)) ∼ ), which is not of rank 1. The proof is = SL± 6 (q complete.  Lemma 11.23. Suppose that K ∼ = D4 (q), q = r a odd, and K  X, and let  ∼ x ∈ I2 (X). If z = Z(K) = Z2 and O r (CK (x)) is the product of two SL2 (q 2 ) components each containing z, then x is K-conjugate to xz, and the image of x in Aut(K) lies outside [Aut(K), Aut(K)].  be the universal version of K, so that Proof. Let K = K/Z(K) and let K there are homomorphisms  →K→K K

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 so with respective kernels  y and z of order 2. Let z be a preimage of z in K,  that Z(K) =  y , z .  and K of order 3. Then τ acts nonLet τ be a graph automorphism of K   = trivially on each of the four-groups Z(K) and on Outdiag(K) = Outdiag(K) Outdiag(K). Hence there is no loss in assuming that K = Ω(V ), where V is an 8-dimensional orthogonal space of + type over Fq . Then Out(K) ∼ = D8 × ΦK where ΦK is a group of field automorphisms and the D8 direct factor is the product of Outdiag(K) by γ , where γ is a graph automorphism of order 2. It follows that [Out(K), Out(K)] = [Outdiag(K), γ], which is the subgroup of order 2 referred to in [IA , 4.5.1] as containing the coset “d”. From [IA , 4.5.1, 4.9.1] there are just three conjugacy classes Cd , Cs , Cs of in volutory automorphisms t of K such that O r (CK (t)) is the product of two A1 (q 2 )components, and they are inner-diagonal automorphisms, mapping to distinct nontrivial cosets in Outdiag(K). Thus τ cycles Cd , Cs and Cs , where the subscript indicates the coset of Inn(K) in which these automorphisms lie. We will show that x maps to an element of Cs or Cs , which in view of the previous paragraph will imply that the image of x in Aut(K) lies outside [Aut(K), Aut(K)]. Indeed elements of Cd are induced by involutions ξ of SO(V ) both of whose eigenspaces on V  − 2 2 ∼ ∼ are 4-dimensional of − type, so O r (CK (ξ)) = Ω− 4 (q) × Ω4 (q) = L2 (q ) × L2 (q ) =  r O (CK (x)). Thus, x cannot map to an element of Cd . This proves the last assertion of the lemma. Letting ξ be as in the previous paragraph, we see by Witt’s Lemma that ξ is O(V )-conjugate to ξz, and since CK (ξ) contains elements with arbitrary combinations of determinant and spinorial norm, ξ is K-conjugate to ξz. Hence in the  ξ , ξ is sheared to either z or zy. Applying a suitable power of τ we see group K  to z or zy (in the that for (at least) one of x0 = x, x0 = xξ, x0 is sheared in K  x0 ). Therefore x0 is K-conjugate to x0 z. One of the two possibilities for group K x0 lies in Cs , the other in Cs , since ξ maps to an element of Cd . But the graph automorphism γ of order 2 interchanges Cs and Cs and fixes  y . Therefore both possibilities for x0 are sheared in K to z, and in particular x ∼K xz. The proof is complete.  Lemma 11.24. Suppose that K ∼ = Sp4n (q), q odd, and K  X. Let x ∈ I2 (X) with E(CK (x)) ∼ = Sp2n (q 2 ). Then x is K-conjugate to xz, where z = Z(K). Proof. From [IA , 4.5.1, 4.9.1], Inndiag(K) = Inn(K) ξ , where ξ is the automorphism of K induced by x. Comparing [IA , 4.5.1] and [IA , 4.5.2], we conclude that CK/Z(K) (ξ) contains an element inducing a nontrivial field automorphism on L, while CK (ξ) contains no such element. This implies that x ∼K xz, as required.  n Lemma 11.25. Let K = P Ω+ 12 (q) for some odd prime power q = r . Then the following conditions hold:

(a) There is a unique conjugacy class of 2-central involutions u ∈ Aut(K). Moreover, CK (u) has a component I ∼ = D4 (q); also, u is the image of an involution v ∈ Ω+ (q) whose −1-eigenspace on the natural module is 12 8-dimensional of +-type;  (b) O r (CK (u)) = IJ1 J2 with J1 ∼ = J2 ∼ = SL2 (q); and r (c) O (CAut(K) (IJ1 J2 )) = 1.

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Proof. The involution v specified in (a) clearly has trivial spinorial norm  = Ω+ (q). Let u be the image of v in K. As and determinant, so lies in K 12 + + ∼  = 2, (b) holds. The uniqueness of u, CO+ (q) (v) = O8 (q) × O4 (q), and |Z(K)| 12 up to conjugacy, is immediate from [IA , 4.5.1, 4.9.1], so (a) holds. By [IA , 4.5.1], CInndiag(K) (IJ1 J2 ) is a 2-group. As |Aut0 (K) : Inndiag(K)| = 2 and I contains a root group, Lemma 6.1 yields that CAut(K) (IJ1 J2 ) = CAut0 (K) (IJ1 J2 ), which is a 2-group. The proof is complete.  Lemma 11.26. Let t ∈ I2 (Aut(K)), where K = Spinn (q), q odd, n ≥ 5, n odd. Then Z(K) lies in a component or solvable component of CK (t). Proof. Let V be the natural K/Z(K) ∼ = Ωn (q)-module. Since n is odd, |Z(K)| = 2 and Inndiag(K) ∼ = SOn (q). If t ∈ Inndiag(K), then t acts on K t has an eigenspace like some involution  t ∈ SOn (q) = SO(V ), and as n ≥ 5,  V0 on V of dimension at least 3. Then by [IA , 6.2.1bc], CK (t) has a normal subgroup L ∼ = Spin(V0 ), and L is the required component or solvable component. If t ∈ Inndiag(K), then by [IA , 2.5.12, 4.9.1], t is a field automorphism. Thus there exists a σ-setup (K, σq ) of K with K universal, and σq1/2 acts on K as a conjugate  of t. Then CK (t) ∼ = CK (q 1/2 ) ∼ = Spinn (q 1/2 ), completing the proof. Lemma 11.27. Let K = L4 (q),  = ±1, q = 2n > 2, q ≡  (mod 3). Then none of the following groups J embeds in K: J = L2 (q) × L2 (q) × L2 (q), G2 (q), L3 (q), 1 1 D4 (q 3 ), or 3D4 (q 3 ). Proof. Let p be a prime divisor of q +  (which exists as q > 2). Then mp (L2 (q)) = 1 but mp (K) = 2 by [IA , 4.10.3a], ruling out J = L2 (q)3 . Lemma 5.4 1 1 1 rules out L3 (q) as well as D4 (q 3 ) and 3D4 (q 3 ), which contain L3 (q 3 ) [IA , 4.7.3A]. Finally if J ∼ = G2 (q), then |J|2 = q 6 = |K|2 so by [IA , 2.6.7], if J ≤ K, then J lies in a parabolic subgroup of K, which is clearly not the case. The lemma is proved.  Lemma 11.28. m2,3 (O8+ (2)) < 4. Proof. By [IA , 4.10.3], O8+ (2) has a unique class of E34 -subgroups, and they are easily seen to be self-centralizing in O8+ (2) (see for example [IA , 4.8.2]). Suppose E ∈ E34 (O8+ (2)) normalizes the nontrivial 2-group T . Then T = [T, E] ≤ Ω+ 8 (2), and (2) by the Borel-Tits theorem. However, so E lies in a parabolic subgroup of Ω+ 8 every such parabolic subgroup has 3-rank at most 3, contradiction. The lemma follows.  Lemma 11.29. Suppose that X is a solvable subgroup of O(V ), where V is an 8-dimensional orthogonal space over F2 of + type. Suppose that there is E ≤ X with E ∼ = E34 . Then one of the following holds: (a) E  X; or (b) There exists F ∈ E1 (E) such that F  X and |[F, V ]| = 4. Proof. Let Q = F (X). By Lemma 11.28, O2 (Q) = 1. No p-local subgroup of O(V ), p > 3, contains E, so Q = O3 (X). Let E ≤ P ∈ Syl3 (X). If Q ≤ E the conclusion holds as X is 3-constrained, so assume that Q ≤ E. As |O(V )|3 = 35 , P ∈ Syl3 (O(V )). Thus P ∼ = Z3 × (Z3  Z3 ) and Z := Z(P ) = Z1 × Z2 ≤ E ∩ O3 (X) with dim[Z1 , V ] = 2, dim[Z2 , V ] = 6, and [Z1 , V ] ∩ [Z2 , V ] = 0. Hence Z1 is weakly

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closed in Z. We may assume that Z1  X, for otherwise (b) holds. Therefore Z  X, so Ω1 (Z(Q)) > Z. As Q ≤ E, it follows from the structure of P that Q = Z x for some x ∈ P − E, and then that Q  P , a contradiction. The proof is complete.  Lemma 11.30. Suppose that K ∼ = G2 (4) or 2G2 (4), K  X, and x ∈ I2 (X). Then m2 (CKx (x)) ≥ 3. Proof. Let K = K/Z(K) ∼ = G2 (4). There is an x-invariant parabolic subgroup H ≤ K such that for Q = O2 (H), H/Q ∼ = GL2 (4) acts naturally on # ∼ Z := Z(Q) = E24 . Then H is transitive on Z . Consequently the full preimage Z of Z in K is elementary abelian of order 24 |Z(K)|. Then m2 (CZx (x)) ≥ 3, as required.  12. Subcomponents Lemma 12.1. Suppose that K ∈ Chev(r), x ∈ Ip (Aut0 (K)), where p and r are distinct primes, and L is a component of CK (x). Then the following conditions hold: (a) q(L) ≥ q(K); and (b) If q(L) = 3, then q(K) = 3. Proof. By [IA , 4.2.2], q(L) = q(K)n for some positive integer n. This implies (a). In (b), if q(K) = 3, the only possibility is that K is a Ree group, namely 1 K = 2 G2 (3 2 ) . But then CK (x) is solvable, so L does not exist, contradiction. The proof is complete.  Lemma 12.2. Let H ∈ Lie(r) for some odd r, but with H ∼  A1 (q), G2 (q), =  or D4 (q) for any odd q. Let c ∈ I2 (H) and suppose that c ∈ J  O r (CH (c)) for some J ∼ = SL2 (r n ) and some n. Then q(H) = r n and J is a fundamental n SL2 (r )-subgroup of H. 3

 be the universal version of H and Proof. As J   CH (c), c ∈ Z(H). Let H r  ∼    J the full preimage of J in H. Then J0 = O (J) = J and we let  c be the preimage c) maps into CH (c), J0   CH ( c). Therefore we are reduced to of c in J0 . As CH ( the case in which H is universal.  Given that 3D4 (q) is excluded, and that SL2 (r n ) ∼ = J  O r (CH (c)), we see from [IA , 4.5.1, 4.5.2] that the only possibilities for c are those involutions of H mapping to the automorphisms t given in the table in [IA , 4.5.5]. In each case, by [IA , 4.5.5bc], and since G2 (q) is excluded, there is a unique normal fundamental  SL2 (q) subgroup in CH (t) containing c. The lemma follows. In the next lemmas we study the following groups:

(12A)

(1) H ∈ Chev(r) for some odd r and H ∼ = Ln (q), n ≥ 5, P Sp2n (q),  n ≥ 3, F4 (q), E6 (q), E7 (q), or E8 (q); and (2) C = {c ∈ I2 (H) | c is in a fundamental SL2 (q)-subgroup of H}.

Lemma 12.3. Assume (12A). Then C is an H-conjugacy class, and c ∈ C if and only if c ∈ J0 (c)  CH (c) for some subgroup J0 (c) ∼ = SL2 (q) depending on c. Moreover, given c ∈ C, the subgroup J0 (c) is unique and is a fundamental SL2 (q) subgroup of H. Finally, C is Aut(H)-invariant.

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Proof. From [IA , 4.5.1, 4.5.2], and given the possibilities for H, there exists a unique H-conjugacy class C0 of involutions z such that z ∈ J0  CH (z) for some J0 ∼ = SL2 (q). By Lemma 12.2, J0 is a fundamental SL2 (q) subgroup in H. As there is a unique class of fundamental SL2 (q) subgroups in H, by [IA , 3.2.7], C = C0 , C is Aut(H)-invariant, and the lemma follows.  Lemma 12.4. Assume (12A). Suppose that Inn(H) ≤ X ≤ Aut(H). Let b ∈ I2 (X) and suppose that b ∈ J  CX (b) with J ∼ = SL2 (q) (if q > 3) or J ∼ = Q8 (if q = 3). Let c ∈ C, and suppose also that CX (b)/O2 (CX (b)) is involved in CX (c)/O2 (CX (c)). Then b ∈ C. Proof. If q > 3, then as Out(H) is solvable, b ∈ J ≤ Inn(H) and J  CInn(H) (b), whence b ∈ C by Lemma 12.3. Hence, we may assume q = 3. If b ∈ Inn(H), then Out(H) has a nonabelian Sylow 2-subgroup, whence H ∼ = A± n (3). But then a Sylow 2-subgroup of Out(H) is dihedral, a contradiction. Hence, q = 3 and b ∈ Inn(H).  Assume that b ∈ C; we argue to a contradiction. Set Hb = O 3 (CX (b)) =  O 3 (CInn(H) (b)). Since b ∈ C, and given (12A1), it follows from [IA , 4.5.1] that no Lie component of Hb is isomorphic to SL2 (3). Hence Hb is perfect. As [J, Hb ] ≤ O2 (Hb ) ≤ Z(Hb ), the Three Subgroups lemma gives [J, Hb ] = 1. But by [IA , 4.5.1], Y := CInndiag(H) (Hb ) is cyclic of order at most 4, and Out(H)/ Outdiag(H) is generated by a graph automorphism. As |J| = 8, we see from the table that we − ∼ must have H ∼ = A± n (3) or E6 (3), with Y = Z4 and some element t ∈ J in the Inndiag(H)-coset of a graph automorphism. Finally, in these cases, O2 (CX (b)) = O2 (CX (c)) = 1 by [IA , 7.7.8a], so by − (∞) assumption CX (c)(∞) ∼ . In the E6 and A+ = A± n n−2 (3) or A5 (3) involves CX (b) − − + cases, this means that A5 (3) involves D5 (3), or An−2 (3) involves A(n−1)/2 (9), both of which are absurd by order considerations. Hence we are in the A− n (3) case, − − (∞) ∼ whence A− A (3) involves C (b) (3)A (3), 3 ≤ i ≤ n/2, or A− = X n−2 n−1 (3). i−1 n−i Consideration of orders shows that only the first of these is possible. But in that case a graph automorphism of H of order 2 induces (nontrivial) graph automorphisms on both components of CX (b)(∞) , on which t must therefore act nontrivially. As t ∈ J, this is a contradiction and the proof is complete.  We introduce the following additional notation:  (12B) For any c ∈ C, J0 (c) as is in (12A), and J(c) = O 2 (J0 (c)). Lemma 12.5. Assume (12A) and (12B). Let c, c ∈ C. Then the following conditions are equivalent: (a) [J0 (c), J0 (c )] = 1; (b) [J(c), J(c )] = 1; and (c) [T, T  ] = 1 for some T ∈ Syl2 (J(c)) and some T  ∈ Syl2 (J(c )). Proof. Obviously (a) =⇒ (b) =⇒ (c). Suppose that (c) holds. As J0 (c)   CH (c), T  normalizes and then centralizes J0 (c), by Lemma 11.17a. Sim  ilarly, [T, J0 (c )] = 1. But J(c) ≤ CH (T  ) ≤ CH (c ) so J(c) = T J(c) centralizes  J0 (c ). Thus (a) follows unless q = 3. In that case, as O 3 (CH (c)) = J0 (c)K with  [J0 (c), K] = 1, J0 (c ) ≤ O 3 (CH (J(c))) = K ≤ CH (J0 (c)), so (a) holds in any case. The proof is complete. 

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Lemma 12.6. Assume (12A) and (12B). Suppose that Inn(H) ≤ X ≤ Aut(H), and let c ∈ C. Then CH (c) acts transitively on the set of all X-conjugates ch of c such that [J(c), J(c)h ] = 1.   Proof. Let T ∈ Syl2 (J(c)) and set T = T g |, g ∈ H, [T g , T ] = 1 . By Sylow’s theorem and [IA , 4.10.6], NH (T ) permutes T transitively, since the groups in (12A) give rise to doubly transitive groups in [IA , Table 4.10.6], acting on S(P ) for P ∈ Syl2 (H). Now each T g lies in a unique J(c0 ), c0 ∈ C, namely for Z(T g ) = c0 . Hence by Lemma 12.5, NH (T ) permutes transitively the set of all c0 ∈ C such that  [J(c), J(c0 )] = 1. By Lemma 12.3, cX = cH , completing the proof. Lemma 12.7. Assume (12A). Set X = Aut(H), and let x ∈ I2 (X) and y ∈ C. Then one of the following holds (a) [x, y g ] = 1 for some g ∈ H; or (b) H ∼ = P Sp4n (q) and E(CH (x)) ∼ = P Sp2n (q 2 ). Moreover, in case (b), x lies outside a subgroup of X of index 2, and there exists z ∈ I2 (Hx) such that z ∈ xX . Proof. If x ∈ Aut0 (H) then (a) holds by [III11 , 6.1]. Now assume that x ∈ Aut0 (H). ± If H = A± n (q), then y is the image of an involution of SLn+1 (q) with two eigenvalues −1 on the natural module V . If x is the image of an involution of  GL± n+1 (q), then it is obvious that (a) holds. If x ∈ Inndiag(H) but no preimage x of x is an involution, then n is odd and x is conjugate to the element t(n+1)/2 of x) ∼ [IA , 4.5.1]; thus, CGL n+1 (q) ( = GL(n+1)/2 (q 2 ), and a reflection in GL(n+1)/2 (q 2 ) has two eigenvalues −1 on V , i.e., is conjugate to a preimage of y, as desired. If x is a graph automorphism, then CH (x) is an symplectic group or orthogonal group of dimension n ≥ 3, and these groups obviously contain involutions with two eigenvalues equal to −1. If H ∼ = P Sp2n (q), then again y is characterized up to conjugacy as having two eigenvalues −1 on the natural module V , and again any image x of an involution of Sp2n (q) clearly centralizes a conjugate of y. The involutions x = tn/2 and x = tn/2 of [IA , 4.5.1] have centralizers isomorphic to GLn (q) and GUn (q), respectively; in the former case V = W ⊕ W ∗ where W is the natural GLn (q)-module, and in the latter case V is just the natural GUn (q)-module regarded as a module over Fq rather than Fq2 . In either case a reflection in CH (x) is the desired conjugate of y for (a). The only remaining class of involutions is that of x = tn/2 , for which (b) and the final assertions of the lemma hold on the basis of the information in [IA , 4.5.1]. In the remaining cases we freely use the information in [IA , 4.5.1]. In H = F4 (q), E6± (q), and E8 (q), there are two classes of inner involutions; one is represented by y and the other consists of 2-central involutions, so (a) holds provided x ∈ Inn(H). Suppose next that x is a graph automorphism of H = E6± (q), so that Hx :=  O r (CH (x)) ∼ = F4 (q) or C4 (q). In either case Hx contains an involution z such that   z ∈ O r (CH (z, x )) = O r (CHx (z)) ∼ = A1 (q) ∗ C3 (q). If z ∈ y H , then (a) holds. r r Otherwise z ∈ Hz := O (CH (z)) ∼ = Spin± 10 (q). But O (CHz (x)) has a component isomorphic to C3 (q), and there is no such involutory automorphism x of Spin± 10 (q), contradiction.

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The last exceptional case to consider is H = E7 (q). But in this case y is centralized by a Sylow 2-subgroup of Inndiag(H), so (a) holds trivially. The proof is complete.  Lemma 12.8. Let K = A1 (q), Bn (q), Cn (q) or E7 (q), where q is a power of r for some odd prime r. Let α ∈ Aut(K) be a field automorphism of order 2, and let  K0 = O r (CK (α)). Then Inndiag(K0 ) ∼ = CInndiag(K) (α) ≤ Inn(K). Proof. Let Ku be the universal version of K, with center Z(K) = z ∼ = Z2 .  Let K0 = O r (CK (α)). By [IA , 4.9.1b], CInndiag(K) (α) ∼ = Inndiag(K0 ), whereas the image of CKu (α) in K is isomorphic to K0 . In the semidirect product Ku α , the involutions α and αz are Ku -conjugate by [IA , 4.9.1]. Thus there is g ∈ Ku such that [α, g] = z. Consequently the image of g in K lies in CK (α) but not in the image of CKu (α). Thus CK/Z(K) (α) > CK (α)/Z(K). Since | Inndiag(K0 ) : K0 | = 2, the result follows.  Lemma 12.9. Let K ∈ Chev(r), r odd. Assume that the underlying Lie algebra has two root lengths, and let z ∈ I2 (K) be a short root involution. Then the following conditions hold: (a) If K ∼ = F4 (q), then CK (z) ∼ = B4 (q)u ; ∼ (b) If K = C3 (q), then z ∈ L  CK (z) with L ∼ = C2 (q)u . Proof. From [IA , Table 4.5.1] there exists an involution z ∈ K in either case (and a group L in case (b)) satisfying the desired conclusions. It remains only to check that z is a short root involution. This follows from Lemma 6.4a applied to CK (z) or L, respectively. Note that B2 = C2 and that CK (z) and L (in (a) and (b), respectively) are subsystem subgroups, so a short root group in either of them is a short root group in K.  ∼ L (q), q a power of the odd prime r,  = ±1, Lemma 12.10. Suppose that K = n  n ≥ 5. Let x ∈ I2 (K) be a classical involution, so that O r (CK (x)) = K1 ∗ K2 with K1 ∼ = SL2 (q) and K2 ∼ = SLn−2 (q). Then CAut(K) (K2 )/K1 ∼ = Zq− . Proof. Let C = CAut(K) (K2 ) and C0 = C ∩ Inndiag(K). As K2 contains a long root group, C = CAut0 (K) (K2 ) by Lemma 6.1. Clearly C0 normalizes the support of K2 , as well as the subspace of fixed points of K2 , on the natural (projective) K-module. Therefore K1  C0 and indeed K1 = E(C0 ) or O 2 (C0 ) according as q > 3 or q = 3. In any case, K1 char C0 so K1  C. Without loss we may assume that K1 and K2 are the images in K of block-diagonal subgroups of SLn (q). Writing Aut0 (K) = Inndiag(K)ΓK as in [IA , 2.5.12], we see that as n − 2 ≥ 3, every nonidentity element of ΓK maps to an element of Aut(K2 ) outside Inndiag(K2 ). As NInndiag(K) (K2 ) maps to Inndiag(K2 ), it follows that C0 = C. Now CGL n (q) (K2 ) ∼ = GL2 (q) × Zq− , and factoring out K1 as well as   Z(GLn (q)) ∼ = Zq− shows that C ∼ = Zq− , as asserted. Lemma 12.11. Let K = Sp2n (q) or Ω+ 2n (q), q a power of the odd prime r, and n ≥ 6, n even. Suppose that x, y1 is a four-group acting on K such that   L = O r (CK (x)) ∼ = SLn (q),  = ±1, y1 ∈ L, and O r (CL (y1 )) = JI, where J∼ = SL2 (q), I ∼ = SLn−2 (q), and [J, I] = 1. Then there is a unique normal subgroup H of CK (y1 ) such that H ∼ = Sp4 (q) or Ω+ 4 (q), respectively. Moreover, [H, I] = 1.

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Proof. By [IA , Table 4.5.1, 4.9.1], there is a unique Sp2n (q)-class (resp.,  of involutions t with O r (CK (t)) ∼ = L. According as  = 1 or −1, the natural Fq K-module, restricted to L, is either the direct sum of two totally isotropic subspaces which are natural Fq L-modules, or a natural Fq2 L-module considered as an Fq L-module. Likewise, by [IA , Table 4.5.2], and as n is even, there are two L-classes y1L and (y1 z)L with centralizers as described, where z is the involution in Z(L) and Z(K). In any case, it follows that y1 has nondegenerate eigenspaces U1 and U2 of dimension 4 and 2n − 4 respectively on V , with V = U1 ⊥ U2 . If K ∼ = Sp4 (q) and = Sp2n (q), it now follows immediately that J ≤ H = Sp(U1 ) ∼ I ≤ Sp(U2 ). Thus, [H, I] = 1, and as 2n − 4 ≥ 8, H  CK (y1 ) is unique of its isomorphism type. In the orthogonal case, the same argument holds with the  additional observation that, since J ≤ Ω(U1 ), we must have Ω(U1 ) ∼ = Ω+ 4 (q). + O2n (q)-class)

Lemma 12.12. Suppose H ∼ = Cn (3), n odd, n ≥ 5, and x ∈ I2 (Aut(H)) with E := E(CK (x)) ∼ = Un (3) ∼ = SUn (3). Let y ∈ E be an involution such that y ∈ E(CE (y)) ∼ = SUn−1 (3). Then E(CH (y)) ∼ = Cn−1 (3). ∼ Sp2n (3). Since Proof. Since n is odd, there is no loss in assuming that H =  y = 1, either the conclusion of the lemma holds or M := O 3 (CH (y)) is the product of two Lie components Sp2i (3)Sp2j (3) with i ≤ j < n − 1 and i + j = n, by [IA , 4.5.2]. On the other hand, E(CM (x)) ∼ = SUn−1 (3). As i ≤ j < n − 1, however,  CM (x) can have no such component, by [IA , 4.5.1]. The lemma follows. 2

− Lemma 12.13. Let K  X with K ∼ = Sp8 (q), Spin7 (q), Ω+ 8 (q), Spin8 (q), or Spin9 (q), q a power of the odd prime r. Let x ∈ I2 (X) and suppose that CKx (x) has a normal subgroup N such that I2 (N ) = {x}. Then CKx (x)/N does not embed in GL2 (q).

Proof. Suppose false. Then m2 (CKx (x)) ≤ m2 (N ) + m2 (GL2 (q)) = 1 + 2 =  3. Moreover, C := x O r (CK (x)) is an extension of a group of 2-rank 1 by a subgroup of SL2 (q). Consequently, C has sectional 2-rank at most 4. So CK (x) has at most two components and solvable components; if there are two, then one embeds, modulo center, in P SL2 (q), and each has sectional 2-rank 2. In particular, x is not a field or graph-field automorphism, so x ∈ Aut0 (K) and hence x and CK (x) appear in [IA , 4.5.2, 4.5.3]. Notice also that our hypotheses imply that if x induces a noninner automorphism on K, then CN (x) has odd order and so CK (x)/O2 (CK (x)) embeds in GL2 (q). However, for the listed groups K, no x and CK (x) exist satisfying all these conditions, as can be seen from [IA , 4.5.2, 4.5.3], so the lemma holds.  Lemma 12.14. Let K ∈ Chev(r), r odd, and let x ∈ I2 (Aut(K)) − Inn(K). 1 Assume that K ∼ = Sp4 (q 2 ), Sp8 (q 2 ), or SL± 4 (q), and in the last case that x induces  a graph or graph-field automorphism on K. Then O r (CK (x)) is not embeddable in SL2 (q).  Proof. Let H = O r (CK (x)). Using [IA , 4.5.1, 4.9.1] we see that if K ∼ = 1 Sp4 (q 2 ), then H/Z(H) ∼ = P Sp4 (q), L2 (q 4 ), or L2 (q 2 ); if K ∼ = Sp8 (q 2 ), then 1 1 ± ∼ ∼ 2 H/Z(H) ∼ = P Sp8 (q 4 ), P Sp4 (q), or L± 4 (q ); and if K = SL4 (q), then H/Z(H) = − 21 2 L4 (q ), P Sp4 (q), L2 (q) × L2 (q), or L2 (q ). In every case |H/Z(H)|r > q =  |SL2 (q)|r and so H is not embeddable in SL2 (q).

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Lemma 12.15. Let K ∈ K2 and let u ∈ I2 (Aut(K)). Let L be a component of CK (u). Let v ∈ I2 (L) and suppose that I is a component of CL (v). Then the following conditions hold: (a) If K ∼ = E6 (q), q odd,  = ±1, L ∼ = A5 (q), and v ∈ I ∼ = A3 (q), then  u ∼ E(CK (v)) = D5 (q) ; (b) If K ∼ = Dn± (q), q odd, n ≥ 5, L ∼ = An−1 (q) for some sign , and v ∈ I ∼ =  Ar (q) for some r < n − 1, then the subnormal closure of I in CK (v) is  isomorphic to Dr+1 (q) for some sign  ; and  (c) If K ∼ = D6 (q), and v ∈ I ∼ = D4 (q)u , then O r (CK (v)) ∼ = = E7 (q), q odd, L ∼  O r (CK (u)). Proof. In every case, v ∈ Z(L), so v ∈ Z(K); we may thus pass to K/Z(K) and assume that Z(K) = 1. Let J be the subnormal closure of I in CK (v). Since v ∈ I ≤ J, I is an anchored subcomponent of CJ (u). In (a), we see from [IA , 4.5.1] that I is the universal version, and also if the desired statement fails, then J ∼  = A5 (q). But then v ∈ I ≤ J, so CJ/Z(J) (u) ∼ = ± SL4 (q), a contradiction. Similarly in (c), I is the universal version and if the desired statement fails, then as v ∈ I ≤ J, Z(J) = 1, so J ∼ = A± 7 (q). But then, with [IA , 4.5.2], J is a proper homomorphic image of the universal version, which implies that the D4 (q) component of CJ (u) cannot be the universal version, a contradiction. Thus it remains to prove (b). With [IA , 4.5.1], u is unique up to conjugacy in Aut(K). Let (K, σ) be a σ-setup for K with K adjoint. Then K = Dn = P SO(V ) where V is a 2n-dimensional orthogonal space over the algebraic closure Fq of Fq , and u acts on V as an element of SO(V ) of order 4 whose eigenspaces W1 , W2 are a pair of opposite maximal totally singular subspaces of V . Thus the component L of CK (u) isomorphic to An−1 has contragredient natural representations on W1 and W2 , and (L, σ|L ) is a σ-setup for L. We conclude that v ∈ K is an involution each of whose eigenspaces is evenly split between W1 and W2 . This implies that the component J of CK (v) containing J is isomorphic to Dr+1 , whence the desired conclusion. The proof is complete.  Lemma 12.16. Suppose that K ∈ Chev(2), p is an odd prime, and a, f ∈ Ep2 (Aut(K)), with a ∈ Inndiag(K). Let L be a component of CK (a) such that q(L) = q(K), and assume that f induces a nontrivial field automorphism on L. Then f ∈ Inndiag(K). Proof. This follows immediately from [IA , 4.2.2e3].



Lemma 12.17. Suppose that q is a power of 2,  = ±1, and p is an odd prime dividing q − . Let K = Ln (q) with n ≥ 5 and let x ∈ Ip (Inndiag(K)) be such that  J := O 2 (CK (x)) ∼ = SLn−1 (q). Let V be a natural Fq -module for J and let b ∈ Ip (J)   with dim[V, b] = 2. Then O 2 (CK (b)) ∼ = SLn−2 (q) and O 2 (CJ (b)) ∼ = SLn−3 (q). Proof. Since q ≡  (mod p), b is diagonalizable in J and K; we may assume b = diag(ω, ω −1 , 1, . . . , 1) in J where ω is a primitive pth root of unity. As J =  O 2 (CK (x)) we may assume that J is a block-diagonal subgroup of K. Then the eigenvalues of b as an element of K are those as an element of J, with one extra 1.  As b is also diagonalizable in K (and ω = ω −1 ), the result follows.

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13. Acceptable Subterminal Pairs In this section we assume familiarity with the concepts and terminology discussed in Chapter 12. Lemma 13.1. Let p be an odd prime, and let (x, K) ∈ J∗p (G) with K ∈ Chev(2), mp (C(x, K)) = 1, and p splitting K. Let (y, L) be an acceptable subterminal (x, K)pair. Suppose that mp (K) ≥ 3. Then y lies in an element of Ep∗ (CG (x)). Moreover,  if p = 3 and K ∼ = E6q (q), then y lies in an element of E36 (x, y L). Proof. Clearly K  CG (x). Let P0 ∈ Sylp (CG (x)) with y ∈ P0 . Let A ∈ Ep∗ (P0 ) and set C = K x, y, A . It is enough to show that y lies in an element of Ep∗ (C). Passing modulo Op (C), there is no loss in assuming that Op (C) = 1 and that P = P0 ∩ C ∈ Sylp (C). As mp (C(x, K)) = 1, CC (K) ≤ P . By [III11 , 14.4c], y induces an inner-diagonal automorphism on K. If some a ∈ Ip (C) induces a graph or graph-field automorphism on K, then so does an element of A# and we may take a ∈ A. Since mp (K) ≥ 3 by assumption, the only possibility is K ∼ = D4 (q) for some q, and of course p = 3. We may write A = a × A1 in such a way that every element of A# 1 induces an inner or field  3 automorphism on K, and x ∈ A1 . Then CK (a) ∼ D (q), G2 (q) or P GL3q (q) has = 4 3-rank 2 by [IA , 4.7.3A, 4.10.3]. Let a1 ∈ A# induce a field automorphism on K, if such an a1 exists; otherwise set a1 = 1. As m3 (CC (K)) = 1 and A ≤ CC (a), we conclude that m3 (A) = 4 + m3 (a1 ). But m3 (CK (a1 )) = 4, so m3 (C) ≥ m3 (x K a1 ) = 5 + m3 (a1 ) > m3 (A), a contradiction. Thus we may write A = Ad × af , where Ad is the preimage of Inndiag(K) in A, and af , possibly trivial, induces a field automorphism on K. Let K1 = CK (a1 ). Suppose that p does not divide |Outdiag(K)|. Then Z(K) is a p -group, by [IA , 6.1.4], so A = x (A ∩ K) a1 . In this case it follows easily from [IA , 4.10.3] that mp (K1 ) = mp (K). Hence mp (C) = mp (A) = 1 + mp (K) + mp (a1 ), whence mp (A ∩ K) = mp (K). Also y induces an inner automorphism on K, corresponding to an element of K of order p as Z(K) is a p -group. If p is a good prime for K, it follows from [IA , 4.10.3e] that y is K-conjugate to an element of A ∩ K, and so y lies in a K-conjugate of A, yielding the desired conclusion. So we may assume that p is a bad prime for K. Hence K is of exceptional Lie type. As p splits K by assumption, part (b6) of [III12 , Def. 1.15] must apply with m0 = 1 or 2. Keeping in mind, if K ∼ = E6± (q), that p = 3 does not divide |Outdiag(K)|, we see in every case that Z(L) is a p -group and mp (K) = 1 + mp (L). Hence mp (CK (y)) = mp (K). Then by [IA , 4.10.3d], y is K-conjugate to an element of A ∩ K, and the conclusion of the lemma holds. We may thus assume that  q ∼ Lpn (q) or E q (q), p divides |Outdiag(K)|, so that K/Z(K) = 6

with p = 3 in the latter case. Set C = C/CC (K). It is enough to show that g −1 [A1 , y g ] = 1 for some g ∈ C and some A1 ≤ C with A1 ∼ = A. For then, A1 normalizes Q := y, CC (K) , and hence normalizes Ω1 (Q) = x, y as CC (K) is a  ∼ g cyclic p-group. Then A2 := x, y, CA1 (y) = A1 is the required element of Ep∗ (C). Let P1 be the preimage of Inndiag(K) in P . Then mp (P ) = mp (P1 ) or mp (P1 )+ 1, and mp (P 1 ) ≤ pn − 1. If mp (A) ≥ pn + 1, then mp (P 1 ) = pn − 1, whence A ∩ P1 = J(P 1 ) is the image of a maximal diagonalizable subgroup of GLnp (q) of exponent p. But then y has a conjugate in A ∩ P1 , as desired. If mp (A) ≤ pn − 1,

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then again as y is diagonalizable, y centralizes a maximal elementary abelian psubgroup A1 of K. Then mp (CC (y)) ≥ mp (A1 x ) ≥ np − 1, as desired. Finally, suppose that mp (A) = pn. The argument just given shows that if y ∈ Inn(K), then mp (CC (y)) ≥ mp (A1 x, y ) ≥ np, as desired. So assume that y ∈ Inn(K). Let V = J(P ∩ K), the largest elementary abelian image of a diagonalizable psubgroup of K. Then some conjugate of y centralizes V and so we may assume that y ∈ V . We complete the proof in this case by showing that [A, V ] = 1. Let R = P ∩ K. Then R = HWp where H is an abelian p-group and Wp is isomorphic to a Sylow p-subgroup of the Weyl group W = W (Apn−1 ) ∼ = Σpn . Set V = Ω1 (H); we argue that [A, V ] = 1. Otherwise, by the Timmesfeld Replacement Theorem [IG , 26.23], A has a proper subgroup A∗ acting nontrivially but quadratically on V . However, as W -module, V contains either the core or the trace-0 submodule of the natural Fp W = Fp Σnp permutation module. Moreover, AutAut(K) (V ) ∼ = W . Since np > 3 (as mp (K) ≥ 3, for example), this implies that no element of W of order p acts quadratically on V , a contradiction. Thus, [A, V ] = 1, completing the proof if K ∼ = Lnp (q).  It remains to consider the case K ∼ = E6q (q), p = 3. By [III12 , Def. 1.15] and  [IA , 4.7.3A], y ∈ K and L ∼ = A5q (q), with Z(L) = Z(K). First, if a1 = 1, then from [IA , 4.2.3, 4.7.3A] we deduce that a1 centralizes an element of y K . Hence it suffices to prove the result for CC (a1 )/ a1 in place of C, and so we are reduced to the case a1 = 1. Then C ≤ Inndiag(K).   Since L x contains either SL6q (q) or L6q (q)×Z3 , with y ∈ L x , m3 (CC (y)) ≥ m3 (x, y L) ≥ 6, proving the final statement of the lemma. We may therefore assume that m3 (C) ≥ 7. Then m3 (A) ≥ 6 so by [IA , 4.10.3bc], equality holds and A = J(P ). As above, it suffices to show that y ∈ A, and so it suffices to show that m3 (CC (y)) = 6. But CK (y) contains Zq−q × L, and CInndiag(K) (y) ∼ = ∼ Zq−q × Inndiag(L). Thus we are done if m3 (L) = 5, or if C = Inndiag(L), which has rank 5. Otherwise, we have C = KCC (K) and q ∼  q (mod 9). But then by = Lemma 8.1, m3 (C) = m3 (K) = 5 < 6 = m3 (A), which is absurd. The proof of the lemma is complete.  Lemma 13.2. Let (x, K) ∈ J∗p (G) and let (y, L) be an acceptable subterminal (x, K)-pair, with p > 2, K ∈ Chev(2), mp (C(x, K)) = 1, and mp (K) ≥ 3. Let B ∈ Ep∗ (CG (x)) with y ∈ B, and with every element of B inducing an inner-diagonal or field automorphism on K. Then B ∩ L ≤ Z(L). Proof. We have L ∈ Chev(2) by [IA , 4.9.6]. First we show that B normalizes  L. Otherwise, by [IG , 8.7(iii)], p = 3 and L ∼ = SL3q (q), or 3A6 . Since mp (K) ≥ 3, we must be in case (a2) of the definition of acceptable subterminal pair [III12 , 1.15],   so K ∼ = L4q (q) and m3 (K) = 3. But m3 ( LB ) > 3, a contradiction. Thus, B normalizes L. If Z(L) ≤ Op (L), the result is trivial, so assume that p divides |Z(L)|. Again q (q) with using [III12 , 1.15] and [IA , 4.7.3A], we find that either K/Z(K) ∼ = Lkp+1    q q q u ∼ ∼ ∼ L = SLkp (q), or p = 3, K = E6 (q) , and L = SL6 (q) with Z(L) = Z(K). In particular, Z(L) is cyclic. It suffices to show that (13A)

mp (B) > mp (CKB (L)/Z(L)) + mp (Out(L)) + 1. ∼ CKB (L)/Z(L) by BL/LCBL (L), which For BL/L is an extension of LCBL (L)/L = embeds in Out(L). Hence (13A) implies that mp (B ∩ L) ≥ 2, yielding the assertion

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of the lemma as Z(L) is cyclic. In the E6 case, m3 (B) ≥ 6 while m3 (Out(L)) ≤ 2 and m3 (CKB (L)/Z(L)) ≤ 2, by [IA , 4.7.3A, 4.2.3] and the fact that m3 (C(x, K)) = 1. q So the lemma holds in this case. In the Lkp+1 (q) case, Z(K) and Outdiag(K) are p -groups and B = (B ∩ K) x, f , with f p = 1, f inducing a possibly trivial field automorphism on K. Moreover, B ∩ K is the image in K of a maximal diagonal elementary abelian p-group of rank kp, with respect to a suitable basis of the natural module. But as y ∈ B, L is a block-diagonal subgroup, and so B ∩L is a hyperplane  of B ∩ K. Thus mp (B ∩ L) ≥ 2, completing the proof. Lemma 13.3. Assume that (x, K) ∈ J∗2 (G), K ∈ Chev(2), and (y, L) is an acceptable subterminal (x, K)-pair, with p odd. Suppose that p, K, and L are as in [III15 , Table 15.3]. Then CK (L y ) is p-solvable. Proof. In most cases, a stronger statement is true: any element of Ip (CInn(K) (L y )) (if indeed this set is nonempty) is the image of a power of y. This is evident in cases (a), (b), and (e). In all other cases except possibly (f), (g), and (m) (with p = 3), Op (Z(K)) = Op (Z(L)) = 1 and mp (K) = 1 + mp (L) by [IA , 4.10.3a], which implies our stronger assertion. In cases (f) and (g), the desired assertion follows because Ip (CK (L y )) ⊆ Z(K) y . This in turn holds since oth erwise, there would exist z ∈ Ip (CK (L y )) such that L < E(CK (z)) ∼ = SL5q (q) or q L8 (q), respectively, contradicting the fact that the acceptable subterminal (x, K)pair (y, L) must by definition not be ignorable [III12 , Def. 1.15]. Finally, in case (m) with p = 3, it follows from [IA , 4.7.3A] that CInndiag(K) (L) has 3-rank 1, its subgroups of order 3 being conjugates of the image of y . This implies the desired conclusion, finishing the proof.  Lemma 13.4. Let (x, K) ∈ ILo2 (G) and suppose that for some n ≥ 7 and odd q, ∼ K = HSpinn (q) (n ≡ 0 (mod 4), n > 8) or Spin± n (q). Assume that m2 (Z(K)) = m2 (C(x, K)) = 1. If n = 7, assume that x ∈ Syl2 (C(x, K)). Then the following conditions hold: (a) There exists y ∈ I2 (KC(x, K) − C(x, K)) such that CK (y) has a component L ∼ = Spinηm (q) for some m and some sign η, x ∈ L, and L/ x ∼ = η Ωm (q); (b) In (a), if n ≥ 9, then y may be chosen so that y ∈ K and m is the largest ∼ multiple of 4 less than n. In that case L ∼ = Spin+ m (q) and L/Z(K) = + Ωm (q); moreover, either (y, L) is an acceptable subterminal (x, K)-pair or (x, K, y, L) is ignorable; (c) In (a), if y and L are chosen with m as large as possible, then m is even; and (d) In (c), if |C(x, K)|2 > 2, then m is the largest even number less than n and L ∼ = Spinηm (q), where m = 2k and η = q k . Here q = ±1 and q ≡ q (mod 4).  = Spin±  Proof. Let K n (q) be the universal version of K. For t ∈ K, write t for   and t for the image of  t modulo the spin involution z of K, (any) preimage of t in K, (q). For those t ∈ K such that t is an involution with −1-eigenspace V i.e., in Ω± t of n  z -module V ,  dimension 2k on the natural K/ t2 = z k [IA , 6.2.1d]. Thus for such t ∈ K with k even, t is an involution. Furthermore CK (t) has subnormal subgroups ± + isomorphic to Spin± mt (q) and Spinm+ (q), where mt = dim Vt and mt = dim CV (t), t

provided these dimensions are at least 3; see [IA , 6.2.1b].

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 = Suppose that K = HSpin4j (q), so that by assumption j ≥ 3. Then K + ∼  Spin4j (q) has Z(K) = E22 . By the previous paragraph there is an involution t ∈ K such that Vt has dimension 4(j − 1) and + type, and then E(CK (t)) has a ∼ + component L ∼ = Spin+ m (q), m = 4(j − 1), with Z(K) ≤ L and L/Z(K) = Ωm (q). In particular (a) holds in this case with y = t. By the previous paragraph no involution u ∈ K − Z(K) satisfies dim Vu > dim Vt , so by [III12 , Definition 1.15], either (y, L) is an acceptable subterminal (x, K)-pair or (x, K, y, L) is ignorable, proving (b) in this case.  = Spinn (q). Since m2 (C(x, K)) = We next prove (a) and (b) in the case K = K  ∼ 1, x = Ω1 (Z(K)) and K := K/ x = Ωn (q). If x ∈ Syl2 (C(x, K)), fix w ∈ C(x, K) such that w2 = x. Let m0 and m1 be the largest multiples of 4 and 2, respectively, strictly less than n. Then there is an involution t0 ∈ K such that dim Vt0 = m0 . Moreover, if m1 > m0 then m1 ≡ 2 (mod 4) and so there is t1 ∈ K such that t21 = x and the −1-eigenspace of t1 on V has dimension m1 . Then (a) follows with y = t0 if n ≥ 8. (Note that if n = 8, then as Z(K) is cyclic, K ∼ = − 2 ∼ Spin− (q) and E(C (t )) has a component isomorphic to Spin (q) SL (q ).) If = K 0 2 8 4 n = 7, then w exists by assumption and (a) holds with y = t1 w and m = 6. If n ≥ 9, then y = t0 also satisfies the conditions of (b) (see [III12 , Definition 1.15] for the final assertion of (b)). It remains to prove (c) and (d). Suppose that m is as in (c) but m is odd. By (a), y induces an inner automorphism on K and E(CK (y)) has a component isomorphic to Spin± m (q). But as m is odd, an examination of [IA , 4.5.1] shows that n cannot be even. Let m0 and m1 be as above, so that m1 = n−1. If |C(x, K)|2 > 2 then let w ∈ C(x, K) be as above. Also let t0 , t1 ∈ K be as above. If m0 = m1 we take y = t0 , while if m0 = m1 − 2 and |C(x, K)|2 > 2, we take y = t1 w, thereby achieving m = m1 . So (c) and (d) hold in these cases. The only remaining case is that m0 = m1 − 2 and x ∈ Syl2 (C(x, K)). But then if m is odd, we must have m > m0 by (b), so m = m0 + 1 = n − 2 and n is odd. As y is an involution inducing an inner automorphism on K, we have y ∈ K. The eigenspaces of y then are of dimensions 2 (for the eigenvalue −1) and n − 2. But then y has order 4 by  [IA , 6.5.2d], contradiction. The proof is complete. Lemma 13.5. Suppose that p = 2, K/Z(K) ∼ = P Sp2n (q), n ≥ 2, q odd, Sp (q). Then CAut(K) (L) ∼ y ∈ I2 (K), L   CK (y), and L ∼ = = SL2 (q). More2n−2 over, if we put IK = Inndiag(K), then NIK (L)/NInn(K) (L) maps isomorphically to Outdiag(L). Proof. As L contains a long root subgroup, CAut(K) (L) ≤ Aut0 (K) = IK by Lemma 6.1 and [IA , 2.5.12]. Let V be the natural Sp2n (q)-module, i.e., Sp(V ) ∼ = Sp2n (q) is the group of all isometries of the 2n-dimensional Fq -vector space V equipped with a nondegenerate alternating form. Let GSp(V ) be the group of all similarities of that form, i.e., linear transformations T carrying the form to a nonzero multiple γ(T ) of itself. Then the group Z of all scalar mappings lies in GSp(V ), and P GSp(V ) := GSp(V )/Z ∼ = Inndiag(Sp(V )). Indeed as T ranges over . GSp(V ), γ(T ) ranges over F× q Now NGSp(V ) (L) stabilizes each term of the decomposition V = [V, L] ⊥ CV (L). As L ∼ = Sp([V, L]) is absolutely irreducible on [V, L], CGSp(V ) (L) acts on [V, L] via scalars, and acts as similarities on CV (L). Modulo Z, we see that CIK (L) faithfully induces Sp(CV (L)) ∼ = SL2 (q) on CV (L).

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Similarly, NGSp(V ) (L) induces all possible similarities on [V, L], and Z induces all scalar mappings on [V, L], so AutIK (L) ∼ = Inndiag(Sp([V, L])) ∼ = Inndiag(L). On the other hand, NK (L) induces only Sp([V, L]) on [V, L] so AutInn(K) (L) ∼ = Inn(Sp([V, L])) ∼ = Inn(L). As Outdiag(K) ∼ = Outdiag(L) ∼ = Z2 , the lemma is proved.  14. Fusion Lemma 14.1. Let x ∈ Ip (X) for some prime p. Under either of the following conditions, x Aut(X) = x X : (a) X = SLn (q) and x is diagonalizable; or (b) Ln (q) ≤ X ≤ P GLn (q) and x is diagonalizable. Proof. Without loss, x is a diagonal matrix or the image of a diagonal matrix. Let H be the full diagonal subgroup of X. Then the graph automorphism g → (g −1 )T inverts H elementwise, and for any automorphism ϕ ∈ Aut(Fq ), the corresponding field automorphism of X induces a power mapping on H. Moreover,  CAut(X) (H) covers P GLn (q)/Ln (q). The result follows directly. Lemma 14.2. Let q be an odd prime power and let K = Ωn (q) for some n ≥ 4. Let x ∈ I2 (K). Then xAut(K) = xK . Proof. If x ∈ Z(K), the result is obvious as |Z(K)| ≤ 2. Also if n = 4, then K has cyclic center and a single class of noncentral involutions, so the assertion holds. Suppose then that n ≥ 5 and x ∈ Z(K). Let V be the natural K-module. As x ∈ K, [V, x] has dimension 2k < n and type η := kq , where q ≡ q = ±1 η η (mod 4). Then O 2 (CK (x)) = K1 × K2 ∼ = O 2 (Ω2k (q)) × O 2 (Ωn−2k (q)). Moreover, if k > 1, then x ∈ K1 , while if k = 1, then K1 has odd order and x ∈ O 2 (CK (x)). It follows that the isomorphism type of the pair (O 2 (CK (x)), x) determines the involution x up to O(V )-conjugacy. This gives the first equality in the chain xAut(K) = xO(V ) = xSO(V ) ⊆ xInndiag(K) = xK . The second inequality holds as x centralizes some reflection; the next inclusion as  P SO(V ) ≤ Inndiag(K); and the last equality is by [IA , 4.2.2j]. In the next two lemmas we use the following terminology. Definition 14.3. Let I ∈ Chev(2) with level q, and let p be a prime divisor of q 2 − 1. By a long root p-element of I we mean an element of order p in some fundamental SL2 (q) subgroup of I. Lemma 14.4. Let p be an odd prime and I ∈ Chev(2) with level q, where p divides q 2 − 1, but not I ∼ = Sp4 (q) or F4 (q). Let x be a long root p-element of I. Aut(I) I Then x = x . Proof. We use standard notation for I [IA , 2.10]. Replacing x by an Iconjugate, we may assume that x ∈ Ip (L), where L is the subsystem subgroup corresponding to {±α}, α being the highest root. Let N = NAut(I) (L). By Sylow’s N I theorem, N ≤ INAut(I) (x ), so x ⊆ x . Write Aut(I) = Inndiag(I)ΦI ΓI [IA , 2.5.10]. As L is a subsystem subgroup, ΦI ≤ N . Moreover, since Sp4 (q) and F4 (q) are excluded, ΓI is trivial unless I

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is untwisted with one root length, in which case ΓI normalizes the root subgroups corresponding to α and −α; thus, ΓI ≤ N . Finally, a Cartan subgroup of I lies in N and covers Inndiag(I)/ Inn(I), so Aut(I) ≤ Inn(I)N and x Aut(I) = x N Inn(I) =  x I , as asserted. We make a definition for use in the next lemma. Definition 14.5. For a group J ∈ Chev(2) and a subgroup A0 ≤ Out(J), we say that A0 is triality-free unless J ∼ = D4 (q) and in Out(J) = Γ × Φ, the projection of A0 on Γ contains an element of order 3. Here Γ is the image of ΓJ in Out(J) [IA , 2.5.10] and Φ ∼ = Aut(Fq ) is the isomorphic image of a subgroup of Aut(J) consisting of field automorphisms. Lemma 14.6. Suppose that p is an odd prime, I ∈ Chev(2) with level q, p divides q 2 − 1, Op (I) = 1, Ep2 ∼ = x, y ≤ I, y ∈ K = E(CI (x)), D = x, y , L = E(CK (y)), u ∈ D# , and L ≤ Lu  E(CI (u)). Assume that I, K, and L are as in [III15 , Table 15.3c d f h i j k n], with I playing the role of G0 , and that Lu is either as in that table or Lu = L. Then there exists y1 ∈ Ip (L), independent of u, such that the following conditions hold: I

I

(a) y ∈ x and y1 ∈ x ; (b) Let AutI (K) ≤ A ≤ Aut(K). If K ∼ = D4 (q) and L ∼ = A± 3 (q), assume that A AutI (K) ; and A/ Inn(K) is triality-free. Then y = y (c) Let AutI (Lu ) ≤ A ≤ Aut(Lu ). If K ∼ = D4 (q) and L ∼ = A± 3 (q), assume that A/ Inn(Lu ) is triality-free. Then one of the following holds: A Aut (L ) (1) y1 = y1 I u ; or (2) Lu = L, CAut(I) (L) has abelian Sylow p-subgroups of rank 2, and L contains an I-conjugate of D. Proof. First suppose that (14A)

Case (c), (h), (i), (j), or (k) holds, with n > 4 in case (j).

Case (i) is exceptional because I is not a classical group, but there does exist a subsystem subgroup H ≤ I in that case such that H ∼ = Sp8 (q) and long root subgroups of H are long root subgroups of I. (If they are short root subgroups of I, replace H by H γ where γ is a graph-field automorphism of I.) For uniformity set H = I in cases (c), (h), (j), and (k). Note from the table that in all these cases, including (i), Lu ∼ = K or L. Moreover, K is a component of CH (x) and L is a component of CH (D). Now, H is a classical group, with natural module V , say. Further, in each case, x is a subgroup of I of order p with minimal (2-dimensional) support on V , and CV (x) is a natural module for K, which we identify with the subgroup of H stabilizing CV (x) and centralizing [V, x]. As an element of K, y likewise has minimal support among elements of Ip (K), whether the support is considered on V H I or CV (x). In particular y ∈ x ⊆ x . Then CV (D) = CV (x, y ) is a natural module for L, of dimension at least 4. We take y1 ∈ L to have minimal support on CV (D), hence also on CV (x) and V . Then conclusion (a) follows (see [IA , 4.8.1]). We give an argument for conclusion (c) under the assumption that Lu > L; the proof of conclusion (b) is very similar. If Lu ≤ H, then Lu is a component of CH (u). Since Lu ∼ = K in these cases, and H is not D4 (q) or C2 (q), u must have minimal support on V . Thus CV (u) is a natural Lu -module. As y1 has minimal

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support on V and CV (D), it does as well on CV (u). Since we are assuming n > 4 in g case (j), Lu ∼ = D4 (q). For any g ∈ Aut(Lu ), E(CLu (y1 )) ∼ = E(CLu (y1 )). Hence by g [IA , 4.8.1], y1 has minimal support on the natural Lu -module CV (u), so by Witt’s Lemma, y1 and y1g are conjugate in the isometry group C of CV (u). As V has dimension at least 6, this conjugacy can be made in [C, C] and hence in Lu , the desired conclusion of (c). If Lu ≤ H, on the other hand, then H < I so we are in case (i). As we are assuming that Lu > L, we have Lu ∼ = Sp6 (q). In this case, y1 lies in a maximal torus T of Lu of exponent q − q and rank 3, and Out(Lu ) is covered by a subgroup of NAut(Lu ) (T ) inducing power mappings on T and therefore normalizing y1 . Hence NAut(Lu ) (y1 ) covers Out(Lu ), completing the proof of conclusion (c) if Lu > L and (14A) holds. As noted above, conclusion (b) follows similarly in these cases. Now suppose that (14A) holds, u ∈ D# , and Lu = L. We have V = [V, D] ⊕ CV (D) with dim[V, D] = 2k, where k is the minimal dimension of support of an element of Ip (H) on V . Moreover, CV (D) is a natural L-module. As mp (CH (D)) ≥ 4, it follows that CV (D) has a subspace isometric to [V, D]. Using Witt’s lemma it follows that Dg ≤ L for some isometry g of V , and then as dim CV (D) ≥ 4, g can be taken in H, hence in I. It also is easily checked, using [IA , 4.8.1], that CH (L) has abelian Sylow p-subgroups of rank 2. Thus conclusion (c) holds if H = I. In the only remaining case of (14A), I = F4 (q) and it remains to check that for P ∈ Sylp (CI (L)), P is abelian of rank 2. Let P ≤ Q ∈ Sylp (I). If p > 3, then Q is abelian of rank 4 [IA , 4.10.3], and the result is clear as mp (L) = 2 and [D, L] = 1. If p = 3, then for the same reasons, it is enough to show that P is abelian. But otherwise, choose z ∈ Ip (Z(P )) ∩ [P, P ]. From [IA , 4.7.3A], we    deduce that O 2 (CI (z)) ∼ = A2q (q) ∗ A2q (q), where q ≡ q = ±1 (mod 3). But then  L∼ = Sp4 (q) embeds in A2q (q), which is impossible by Zsigmondy’s theorem applied to q 4 − 1. This contradiction completes the proof if (14A) holds. Next, suppose that (14B)

Case (d), (f), or (n) holds,

± ∼ L±q (q) or D6 (q). Then by Lemma 6.35, so that K ∼ = L6 q (q) or E7 (q), with L = 4 ± x is a long root p-element of I ∼ = E6 q (q) or E8 (q), and as y ∈ K, the isomorphism type of L implies that y is a long root p-element of the subsystem subgroup K of I. Hence y ∈ x I . We choose y1 ∈ y K ∩ L, so that y1 is a long root pelement of L, and conclusion (a) holds. Moreover, whatever the isomorphism type of Lu , including the possibility Lu = L, y1 is a long root p-element of Lu . Hence conclusions (b) and (c) follow from Lemma 14.4. Finally, assume

(14C)

Case (j) holds with n = 4,

  that is, I ∼ = D5q (q), K ∼ = D4 (q), and L ∼ = A3q (q). In this case, the table specifies I that y ∈ x , and we again choose y1 ∈ y K ∩ L, so conclusion (a) holds. Both y and y1 have minimal support on the natural I-module. By our assumptions,  AutI (K) ≤ A and A/ Inn(K) is triality-free. But AutI (K) ∼ = O8q (q), so the projection of A/ Inn(K) on the graph automorphism group ΓK ∼ = Σ3 has even order. Hence (see Definition 14.5) that projection has order 2, so A is an extension of

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14. FUSION

477

AutI (K) by field automorphisms. Thus A is a group of semilinear transformations of the natural K-module. Let g ∈ A; then y and y g have support of the same (minimal) size on CV (x), whence y and y g are conjugate in the isometry group of CV (x). As in the case (14A), it follows that they are conjugate in K. Hence conclusion (b) holds, and the proof of conclusion (c) is similar. The lemma follows.  Lemma 14.7. Suppose that K ≤ X ≤ Aut0 (K), K ∈ Chev(2) ∩ Kp , K and the odd prime p as in [III15 , Table 15.3]. Let q = ±1 be such that q ≡ q (mod p).  Let B ∈ Ep∗ (X) and let w ∈ Ip (NX (B)) with wX ∩ B = ∅. Set Kw = O 2 (CK (w)). Then one of the following holds:  (a) K ∼ = Lpq (q), and CK (w) is a Frobenius group with kernel of order q p − q /p(q − q ) and complement of order p; q  (q), m > 1, and Kw /Z(Kw ) ∼ (b) K ∼ = Lmq (q p ); = Lpm q  3 ∼ ∼ (c) K = E6 (q), p = 3, and Kw = D4 (q) or L3q (q 3 ); or ∼ (d) K = D4 (q), q > 2, p = 3, w induces a graph automorphism on K, and  Kw ∼ = G2 (q) or L3q (q). Proof. If w ∈ Inndiag(K), then (d) holds, by [IA , 2.5.12, 4.7.3A] (q > 2 as K ∈ K3 ). So we may assume that w ∈ Inndiag(K). Likewise every element of Ep∗ (X) lies in Inndiag(K), so we may assume that X ≤ Inndiag(K). Unless p divides |Outdiag(K)| or p is a bad prime for K, wK ∩ B = ∅ by [IA , 4.10.3ef]. Suppose that p does not divide |Outdiag(K)| but p is a bad prime for K. Then − (K, p) = (F4 (q), 3), (E6 q (q), 3), (E7 (q), 3), or (E8 (q), 3 or 5). Now mp (X) = mp (K) is given by [IA , 4.10.3a], and centralizers of elements of order p by [IA , 4.7.3AB], and we see in every case that mp (CK (w)) = mp (K), whence wK ∩ B = ∅ by [IA , 4.10.3c]. Thus we may assume that p divides ||Outdiag(K)||. q (q), then by Lemma 2.13, B is the image of a diagonalizable subgroup If K ∼ = Lkp q q (q). Hence if w is the image of a diagonalizable p-element of GLkp (q), then of GLkp some K-conjugate of w centralizes B, so lies in B. Otherwise by [IA , 4.8.1, 4.8.4], (a) or (b) holds.  Finally suppose that K ∼ = E6q (q) with p = 3. Since B = J(P ) for any Sylow 3-subgroup P of X containing B, by Lemma 2.4, it suffices to prove the lemma assuming that X = Inndiag(K). Assuming that (c) fails, we deduce from [IA , 4.7.3A] that m3 (CX (w)) = 6 (= m3 (X)). Indeed, this is clear for the elements t1 , t2 , t1,6 in the table; t3 is 3-central in X; and t4 lies in a root SL2 (q) subgroup L  which in turn lies in L × M with M ∼ = P GU6 (q). This completes the proof. Lemma 14.8. Suppose that K ≤ X ≤ Aut0 (K) where K ∼ = Lkp (q), p is an odd prime, k > 1,  = ±1, and q ≡  (mod p). Let x ∈ Ip (X), set L = E(CK (x)), and assume that L/Z(L) ∼ = Lk (q p ). Then mp (CX (x)/L) ≤ 2. Moreover, if k = 2, then x ∈ Inn(K). Proof. Without loss we may assume X = Inndiag(K) = P GLkp (q), since  = GL (q). By [IA , 4.8.4], O p (Aut0 (K)) ≤ Inndiag(K). Let C = CX (x) and X kp 0 ∼  of C in X  satisfies C =C 0 ϕ ,  where C  is the full preimage C = GLk (q p ) and ϕ  p ∼     a field automorphism of C0 of order p. Then L := [C0 , C0 ] = SLk (q ) maps onto 0 ϕ  which has the cyclic  /L, L. Hence, CX (x)/L is a homomorphic image of C  of index p. 0 /L subgroup C 

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Suppose that k = 2. We have x p = ωI where ω ∈ Sylp (F× q 2 ). Then the minip mal polynomial of x  is t − ω, which is irreducible, so the characteristic polynomial is (tp − ω)2 . Therefore det x  = ω 2 . Since q = 2a , ω 2 is not a 2pth power in Fq2 , so   x does not map into L2p (q) = Inn(K). The lemma is proved. Lemma 14.9. Assume the situation of Lemma 14.7, but assume that conclusion (a) of that lemma does not hold. Then mp (CB (w)) ≥ 2 and B ∩ Kw = 1. Proof. In cases (b), (c), and (d) of Lemma 14.7, mp (B) > p, so since w normalizes B, mp (CB (w)) ≥ 2. On the other hand, (14D) mp (CAut0 (K) (w)/Kw ) ≤ 2, and m3 (CInn(K) (w)/Kw ) ≤ 1. This is by Lemma 14.8 in case (b), and by [IA , 4.7.3A] in cases (c) and (d). Hence if mp (CB (w)) ≥ 3, then B ∩ Kw = 1 and we are done. We therefore assume that mp (CB (w)) = 2 and B∩Kw = 1. In case (b), kp−1 ≤ mp (B) ≤ 2p so k = 2 and w ∈ Inn(K). But also CB (w) ≤ [B, w] ≤ Inn(K) so as mp (CInn(K) (w)/Kw ) ≤ 1, mp (CB (w)) ≤ 1, contradiction. In case (c), we again have p = 3, m3 (B) = 5 so CB (w) ≤ CInn(K) (w) and |CInn(K) (w)|/Kw | ≤ 3, so CB (w) ∩ Kw = 1. Finally in case (d), B ≤ K since B ∈ E3∗ (X), and CK (w)/Kw is cyclic, so CB (w) ∩ Kw = 1. The proof is complete.  Lemma 14.10. Let q be an odd prime power,  = ±1, and K, L ∈ K2 . In the following cases there exists a unique Inn(K)-conjugacy class of involutions t ∈ Aut(K) such that CK (t)/O2 (CK (t)) has a component isomorphic to L: (a) L = A5 (q) with |Z(L)|2 = 2; K = E6 (q) or A7 (q); (b) L = D6 (q)u ; K = E7 (q)u ; or (c) L = D6 (q)u ; D8 (q)hs . Proof. For uniqueness, by [IA , 4.2.2j], it is enough to check that in [IA , Table 4.5.1] there is a unique row for K corresponding to a component with the appropriate simple quotient. Existence, for (a) and (b), follows from the information in [IA , 4.5.1, 4.5.2] as well. Finally, existence in (c) holds by Lemma 13.4b. The proof is complete.  Lemma 14.11. Let K ∈ K2 and t ∈ I2 (K) − Z(K) and suppose that CK (t) has a component I ∼ = Spin+ 8 (q) for some odd q, with t ∈ Z(I). Then t is not weakly closed in Z(I) with respect to K. Proof. From [III11 , 1.1ab] and [IA , 2.2.10], K ∈ Chev(r), where q is a power of the odd prime r. Then as t ∈ I, it follows by [IA , 4.5.1] that K is either of type Bn (q) (n ≥ 4) or Dn± (q) (n ≥ 5), but in neither case is K of adjoint type. Indeed in the Dn± (q) case, it follows from [IA , 4.5.3] that K is not isomorphic to Ω± 2n (q), either. Hence in any event, K is either a spin group or a half-spin group.  be the universal version of K, i.e., Spin2n+1 (q) or Spin± (q), and let Let K 2n  Then K ∼  zspin be the spin involution in Z(K). where zspin ∈ Z. Since I = K/Z has no proper covering group [IA , 6.1.4], we see from [IA , 4.5.1, 4.5.2] that there  Let   such that zspin ∈ I. t be the preimage of  t in is a preimage I ∼ = I of I in K  Then by [IA , 6.2.1e],   I. t is K-conjugate to  tzspin . Taking images in K yields the desired result. 

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14. FUSION

479

Lemma 14.12. Let p be an odd prime, q a power of 2,  = ±1 with q ≡  (mod p), and K = D4 (q) or D5 (q). If K = D5 (q), let L = E(CK (x)) for some p x ∈ Ip (Inn(K)) such that L ∼ = D4 (q). If K = D4 (q), let L = K. Let B ∈ E∗ (K) p with B ∩ L ∈ E∗ (L). Set W = AutL (B ∩ L). Let V be a natural Fq [L]-module (not necessarily extendible to K). Let {b1 , b2 , b3 , b4 } be a W -invariant frame in B ∩L. Let C be the set of all members of E1 (B ∩ L) which are in some W -invariant frame in B ∩ L. Then the only elements of CAut(K) in b1 , b2 are b1 and b2 . Proof. Besides B = {bi | 1 ≤ i ≤ 4}, one checks that B + = {b1 b22 b33 b44 | i = ±1, 2 3 4 = 1} and B − = {b1 b22 b33 b44 | i = ±1, 2 3 4 = −1} are W -invariant frames. Here we assume b1 , b2 , b3 , b4 to have been chosen to be W -conjugate, and of course linearly independent. Now any W -invariant frame must be acted on diagonally by some elementary abelian subgroup E of O2 (W ) of order 8 which is normal in W ; and E determines the frame, as the set of its homogeneous components on B. As there are just three such elementary abelian subgroups of O2 (W ), the three frames mentioned above are the only W -invariant ones. Thus if L = K, then the assertion of the lemma is evident. Suppose that K∼ = D5 (q). Then there is a AutAut(K) (B)-invariant frame B0 ∪ {b5 } in B, where B0 is one of the three frames above. The elements of b1 , b2 in question are of the form b1 bi2 , 1 ≤ i < p. If B0 = B, then AutAut(K) (B) stabilizes {b1 , . . . , b5 } and the assertion is clear. If B0 = B + or B − , the arguments are essentially the same, so assume that B0 = B + . We write B + = {ci | 1 ≤ i ≤ 4}; then b1 and b2 have −1 −1 i 1+i 1−i the form,  say b1 = c1 c2 c3 c4 and b2 = c1 c2 c3 c4 . Thus b1 b2 = (c1+c2 ) −(c3 c4 ) . i Then b1 b2 is not conjugate in Aut(K) to an element of B ∪ B ∪ B ∪ {b5 } since that would entail the existence of an exponent a ∈ F× p such that a = 1 + i and ±a = 1 − i. This is impossible as it forces 0 = 2 or 0 = 2i. The proof is complete.  Lemma 14.13. Let X = D4 (q), q = 2n , and let p be a prime and  = ±1 be such that q ≡  (mod p). Let x, y, z ∈ Ip (X) with K = E(CX (x)) ∼ = E(CX (z)) ∼ = L4 (q),  X ∼ y ∈ K and E(CK (y)) = SL3 (q). Then z ∩ x, y = ∅. Proof. By [IA , 4.7.3A] there is one Aut(X)-conjugacy class of elements ξ ∈ X of order p such that E(CX (ξ)) ∼ = L4 (q); this class splits into three X-conjugacy classes, which are permuted naturally by ΓX ∼ = Σ3 . Now, with [IA , 4.8.2] we may choose a natural module V for X such that dim[V, x] = 2. Since X ∼ = Ω+ 8 (q), we may write V = V1 ⊥ V2 ⊥ V3 ⊥ V4 with V1 = [V, x] and the Vi ’s all isometric and of type . For each i = 1, . . . , 4, let xi ∈ Ip (X) fix all Vj , j = i, elementwise (and act on Vi ). We may take x = x1 . Replacing y by an X-conjugate and if necessary replacing some xj by suitable we may then assume that y = x2 x3 x4 . Then xy = x1 x2 x3 x4 and powers, x−1 y = x−1 1 x2 x3 x4 are interchanged by a transvection (graph automorphism of order 2) in O8+ (q) fixing V2 ⊥ V3 ⊥ V4 . Hence, by [IA , 4.8.2], they and x represent the three X-classes of subgroups of order p with L4 (q) components in their centralizers. The lemma follows. 

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480

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15. Balance and Signalizers

 Lemma 15.1. Let K = SL2 (q), q odd, q > 3, and let u ∈ K with u2 = Z(K). Write q = r m where r is prime. Then for any H such that Inn(K) ≤ H ≤ Aut(K), O2 (CH (u)) is the semidirect product of a cyclic group of order (q − q )2 with a cyclic group of order dividing m. Proof. Let f ∈ Aut(K) be a field automorphism of order m, so that Aut(K) = Inndiag(K) f . As CK (f ) ∼ = SL2 (r), f centralizes a K-conjugate of u and so without loss we may assume [f, u] = 1. Then CAut(K) (u) = CInndiag(K) (u) f . As CInndiag(K) (u) is cyclic of order q −q , CAut(K) (u) has a normal 2-complement equal to O2 (CK (u))O2 (f ). The result follows.  Lemma  15.2. Let K = Sp4 (q), q odd, and let y ∈ I2 (K) − Z(K) and v ∈ K with v 2 = Z(K). Then the following conditions hold: (a) K is locally balanced with respect to y ; (b) y K is the unique conjugacy class of involutions of K − Z(K); and (c) For any H such that Inn(K) ≤ H ≤ Aut(K), O2 (CH (v)) is cyclic of order dividing q − q . Proof. By [IA , 4.5.1, 4.5.2], (b) holds and CK (y) and CK (v) each have a subnormal A1 (q) subgroup Ly , Lv , respectively. Moreover, Aut(K) = Inndiag(K)ΦK , where ΦK is the cyclic group of automorphisms of the field Fq . As CK (ΦK ) ∼ = Sp4 (r), where r is the prime divisor of q, we may replace y and v by K-conjugates and assume that [y, ΦK ] = [v, ΦK ] = 1. Then ΦK faithfully induces field automorphisms on Ly and Lv , by [IA , 4.2.3]. But O2 (CH (y)) centralizes Ly and similarly for v. We conclude that O2 (CH (y)) ≤ O2 (CInndiag(K) (y)) and similarly for v. The balance assertions in (a) and (c) then follow from [IA , 4.5.1]. The proof is complete.  Lemma 15.3. Let K = D4− (q) with q a power of 2. Let p be a prime divisor of q + 1 and y ∈ Ip (K) a p-element with minimal support on the natural Fq K-module. Let L = E(CK (y)). Then the following conditions hold: (a) L centralizes every element of IK (L; p ); and (b) CAut(K) (L) ∼ = O2− (q) ∼ = D2(q+1) . Proof. Let X ∈ IK (L; p ). We have L ∼ = Ω+ 6 (q). If X has even order, then by a Frattini argument, there is a Sylow 2-subgroup T of X such that NK (T ) covers XL/X. Then by the Borel-Tits theorem, K contains a parabolic subgroup P whose Levi complement involves L. No such subgroup P exists, however, so X has odd order. Now m2 (L) ≥ 4 while mr (K) ≤ 3 for all odd primes r [IA , 3.3.1, 4.10.3a]. Hence by the Thompson Dihedral lemma and the simplicity of L, L centralizes any L-invariant r-subgroup of K. In particular [L, F (X)] = 1, which  implies L = O 2 (L) ≤ CK (X) by [IG , 3.17(iii)]. Thus (a) holds. Let K ∗ = O8− (q), which we can regard as Inn(K) γ = Aut0 (K), where γ ∈ ΦK is a graph automorphism of order 2. Clearly L contains a long root subgroup of K, so by Lemma 6.1, CAut(K) (L) ≤ K ∗ . ∼ O − (q) = ∼ This implies that CAut(K) (L) = O([V, y]) × CO([V,L]) (L) = O([V, y]) = 2 D2(q+1) , as claimed in (b). 

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Lemma 15.4. Suppose that K ∼ = D8 (q) or E7 (q), q odd. Let t ∈ Aut(K) be an involution with CK (t) having a D6 (q) component. Then K is locally balanced with respect to t . Proof. Let I be the product of all Lie components of CK (t). Since one of these components is isomorphic to D6 (q), we see from [IA , 4.5.1] that CAut0 (K) (I t ) ∼ = Z2 . Hence if K is not locally balanced with respect to t , there is a field automorphism f of K of odd prime order centralizing I and t. But then by [IA , 4.2.3], f induces a nontrivial field automorphism on the (unique) D6 (q)-component of I, a contradiction. The lemma is proved.  16. Miscellaneous Lemma 16.1. D5 (q) does not involve Aη5 (q), for any q, and signs , η. Proof. If q 3 − η has a primitive prime divisor p, then by [IA , 4.10.3], mp (D5 (q)) = 1 but mp (Aη5 (q)) = 2, which implies the desired conclusion. So assume that q 3 − η has no primitive prime divisor. By Zsigmondy’s theorem [IG , 1.1], q = 4 and η = 1, or q = 2 and η = −1. But even in those cases we find by [IA , 4.10.3] that m7 (D5 (4)) = 1 < m7 (A5 (4)); m11 (D5 (2)) = 0 < m11 (A− 5 (2)); and  |D5− (2)|3 = 35 < |U6 (2)|3 , completing the proof. Lemma 16.2. Let q = 2n . Then D7± (q) does not embed in E7 (q). Proof. By the order formulas and Zsigmondy’s theorem, |D7± (q)| is divisible by (q 4 − 1)3 but |E7 (q)| is not.  Lemma 16.3. Co2 does not involve P GU6 (2). Proof. |P GU6 (2)|3 = 37 while |Co2 |3 = 36 [IA , Table 5.3k].



Lemma 16.4. Let K = A8 , let t ∈ I2 (K) be a 2-central involution, and let Q = O2 (CK (t)) ∼ = Q8 ∗ Q8 . Then there is a four-group V ≤ Q such that I2 (V ) consists of root involutions (i.e., involutions with support of size 4 on 8 letters). Proof. We may take t = (12)(34)(56)(78) and V = (12)(34), (12)(56) . Now for any 2-element x ∈ CK (t), x lies in Q if and only if x induces an even permutation on the set of four t -orbits. This implies that V ≤ Q.  Lemma 16.5. Let K ∈ K with Z(K) of odd order. Suppose that K ≤ SL4 (7), but K contains no transvection. Then K ∼ = L2 (q), q ∈ {5, 7, 72 }. Proof. Since m2 (SL4 (7)) = 3 and Z(SL4 (7)) has even order, m2 (K) = 2. Thus K/Z(K) ∼ = L2 (s) or L± 3 (s) for some odd s, A7 , M11 , or U3 (4), by [IA , 5.6.3]. If Z(K) = 1, then K ∼ = 3A6 , 3A7 , or SL± 3 (s), s odd. Hence K embeds in SL3 (7), acting naturally on an eigenspace of Z(K) and centralizing a complement. As 5 does not divide |GL3 (7)|, and s3 divides it, we must have K ∼ = SL3 (7) containing a transvection, contradiction. Hence, K is simple. A Sylow 2-subgroup of U3 (4), and a Sylow 11-normalizer of M11 , are not embeddable in SL4 (7). If K ∼ = L2 (s), s not a power of 7, then K contains a Frobenius group of order s(s − 1)/2, so (s − 1)/2 ≤ 4 and s ≤ 9. But A6 has no characteristic 0 or 7 nontrivial representation of degree  A7 . And if s is a 4, so K does not contain a copy of A6 . Thus s < 9 and also K ∼ = power of 7, then Lagrange’s theorem and Zsigmondy’s theorem yield s = 7 or 72 .

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17. PRELIMINARY PROPERTIES OF K-GROUPS

Finally suppose that K ∼ = L± 3 (s). Let t ∈ I2 (K). As CK (t) embeds in GL2 (7)  Z2 , and |K| divides |SL4 (7)|, K ∼ = U3 (3) or L3 (7). Hence K contains a Frobenius subgroup of order 24 .3 and exponent 12 or of order 9.8, respectively. This does not  embed in SL4 (7), contradiction. The proof is complete. Lemma 16.6. Suppose that J ∈ K, Z(SL4 (7)) ≤ J ≤ SL4 (7), 7 divides |J|, and m2 (J) > 1. Then J contains transvections of SL4 (7). 



Proof. Let t ∈ I2 (J) − Z(J). Then O 7 (CJ (t)) ≤ O 7 (CSL4 (7) (t)) =: Kt , with Kt = K 1 × K 2 ∼ = SL2 (7) × SL2 (7), both components containing transvections. So we may assume that Ki ≤ J, i = 1, 2. In particular J ∼  Sp4 (7) or SL± = 4 (7). Now 1 < m2 (J) ≤ 3 and we may repeat the two paragraphs following [III15 , (6J)] with J in place of E to conclude that J ∈ Chev(s) for some odd s. The arguments in the subsequent paragraph show (together with the facts that m5 (SL4 (7)) = 1 and 7 divides |J|) that J ∼ = SU4 (3), which contradicts Lagrange’s theorem. The proof is complete.  Lemma 16.7. Suppose that X is a K-group, O3 (X) = 1, x ∈ I3 (X), H is a component of X, and I 210 , contradiction. So K ∈ Alt. If K ∈ Spor, then as |K|2 = 210 and m2 (K) = 6 (see [IA , Table 5.6.1]), K ∼ = M24 or He, so K has a nontrivial Sylow 23-subgroup or nonabelian Sylow 7-subgroup, whence K does not embed in GL10 (2), contradiction. Thus, K ∈ Chev(r) for some prime r. Suppose that r > 2. Since K embeds in GL10 (2), |K|r ≤ r 6 . Hence, if K is a classical group, then it must have a natural module V of dimension at most 4. But applying Clifford’s theorem to K2 we see that dim V ≥ 15, a contradiction. Thus K is of exceptional type, so K ∼ = G2 (r), and we reach a similar contradiction from the adjoint representation of K. Therefore, r = 2. Hence K2 is a maximal parabolic subgroup of K, whence K has twisted rank 4, and combined with |K|2 = 210 , this forces K ∼  = L5 (2), as claimed. Lemma 16.9. Let K = A8 = L4 (2) and let V be a (4-dimensional) natural F2 K-module. Then H 1 (K, V ) = 0.

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16. MISCELLANEOUS

483

Proof. Otherwise there exists an indecomposable F2 K-module V1 such that V ⊆ V1 and dim(V1 ) = 5. Let x1 , x2 ∈ K ∼ = A8 be disjoint 3-cycles. Then CV (xi ) = 0, i = 1, 2. In particular V0 := CV1 (x1 ) = CV1 (x2 ) is 1-dimensional. But then V0 is invariant under CK (x1 ), CK (x2 ) = K, contradicting indecomposability  of V1 . Lemma 16.10. Let X be a simple subgroup of SL5 (p) for some odd p. Then X∼ = G2 (3) or L± 4 (3). Proof. G2 (3) contains a Frobenius group of order 8.7, and U4 (3) contains a subgroup N ∼ = 24 A6 transitive on hyperplanes of O2 (N ). By Clifford’s theorem, X can have neither of these isomorphism types. If X ∼ = L4 (3), then X contains a Frobenius group of order 33 .13, so similarly, p = 3. But X also contains Z4 × A6 , which is not embeddable in SL5 (3) (an embedding would yield an embedding of  A6 in either SL3 (3) or SL2 (9), both of which are impossible). Lemma 16.11. Let K = 22 E6− (2) and let x ∈ I3 (Inn(K)) be such that E(CK/Z(K) (x)) ∼ = U6 (2). Then E(CK (x)) ∼ = 22 U6 (2), the full 2-covering group of U6 (2). Proof. Let X = 2F2 . By [IA , 5.3y], there is an involution t ∈ X such that E(CX (t)) = CX (t)(∞) ∼ = K, and we identify E(CX (t)) with K. Let ξ ∈ I3 (K) act on K like x and set L = E(CX (ξ)) = CX (ξ)(∞) . Since |U6 (2)| does not divide |O6− (2)|, it follows from [IA , 5.3y] that L = CX (ξ)(∞) ∼ = 2F i22 . As CL (t) has a U6 (2) composition factor, we see from [IA , 5.3t] that CL (t) ∼ = 22 U6 (2). Then CL (t) = CL (t)(∞) ≤ CK (ξ) = CK (x), and the lemma follows. 

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Bibliography [A9] [BHS] [Di2]

[FinS1]

[GiGr1]

[GL1]

[GLS1]

[I2 ] [I1 ] [IG ]

[IA ]

[GLS4]

[II3 ] [IIK ] [II2 ] [III1 ] [III2 ] [III3 ] [III4 ]

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. MR0498828 R. Blok, C. Hoffman, and S. Shpectorov, Classification of Curtis-Tits and Phan amalgams with 3-spherical diagram, Israel J. Math., to appear. Leonard Eugene Dickson, Representations of the general symmetric group as linear groups in finite and infinite fields, Trans. Amer. Math. Soc. 9 (1908), no. 2, 121–148, DOI 10.2307/1988647. MR1500805 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 Robert H. Gilman and Robert L. Griess Jr., Finite groups with standard components of Lie type over fields of characteristic two, J. Algebra 80 (1983), no. 2, 383–516, DOI 10.1016/0021-8693(83)90007-8. MR691810 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 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. , Overview, ch. 1 in [GLS1], American Mathematical Society, 1994. 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 Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 4. Part II. Chapters 1–4: Uniqueness theorems, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1998. With errata: The classification of the finite simple groups, Number 3, Part I. Chapter A. MR1675976 , p-Component Uniqueness Theorems, ch. 3 in [GLS4], American Mathematical Society, 1999. , 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. , Chapter 1: Theorem C7 : General Introduction, ch. 1 in [GLS5], American Mathematical Society, 2002. , Chapter 2: General Group-Theoretic Lemmas, ch. 2 in [GLS5], American Mathematical Society, 2002. , Chapter 3: Theorem C∗7 : Stage 1, ch. 3 in [GLS5], American Mathematical Society, 2002. , Chapter 4: Theorem C∗7 : Stage 2, ch. 4 in [GLS5], American Mathematical Society, 2002. 485

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486

[III5 ] [IIIK ] [GLS5]

[GLS6]

[IVK ] [GLS7]

[III7 ] [III8 ] [III9 ] [III10 ] [III11 ] [G1] [Grm2] [McL2] [Ph1] [Ph2] [St1] [Wag3]

[Wag1]

[Wag2]

[Wo1] [ZS]

BIBLIOGRAPHY

, Chapter 5: Theorem C∗7 : Stage 3a, ch. 5 in [GLS5], American Mathematical Society, 2002. , Chapter 6: Properties of K-Groups, ch. 6 in [GLS5], American Mathematical Society, 2002. Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 5. Part III. Chapters 1–6, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2002. The generic case, stages 1–3a. MR1923000 Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups. Number 6. Part IV, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2005. The special odd case. MR2104668 , Preliminary Properties of K-Groups, ch. 10 in [GLS6], American Mathematical Society, 2005. , The Classification of the Finite Simple Groups. Number 7. Part III, Chapters 7–11: The generic case, stages 3b–4a, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 2018. , Chapter 7: The Stages of Theorem C∗7 , ch. 7 in [GLS7], American Mathematical Society, 2018. , Chapter 8: General Group-Theoretic Lemmas, ch. 8 in [GLS7], American Mathematical Society, 2018. , Chapter 9: Theorem C∗7 : Stage 3b, ch. 9 in [GLS7], American Mathematical Society, 2018. , Chapter 10: Theorem C∗7 , Stage 4a: Constructing a Large Alternating Subgroup, ch. 10 in [GLS7], American Mathematical Society, 2018. , Chapter 11: Properties of K-Groups, ch. 11 in [GLS7], American Mathematical Society, 2018. Daniel Gorenstein, Finite groups, 2nd ed., Chelsea Publishing Co., New York, 1980. MR569209 Ralf Gramlich, Developments in finite Phan theory, Innov. Incidence Geom. 9 (2009), 123–175. MR2658896 Jack McLaughlin, Some groups generated by transvections, Arch. Math. (Basel) 18 (1967), 364–368, DOI 10.1007/BF01898827. MR0222184 Kok Wee Phan, On groups generated by three-dimensional special unitary groups. I, J. Austral. Math. Soc. Ser. A 23 (1977), no. 1, 67–77. MR0435247 Kok-wee Phan, On groups generated by three-dimensional special unitary groups. II, J. Austral. Math. Soc. Ser. A 23 (1977), no. 2, 129–146. MR0447427 Robert Steinberg, Lectures on Chevalley groups, Yale University, New Haven, Conn., 1968. Notes prepared by John Faulkner and Robert Wilson. MR0466335 Ascher Wagner, The faithful linear representations of least degree of Sn and An over a field of odd characteristic, Math. Z. 154 (1977), no. 2, 103–114, DOI 10.1007/BF01241824. MR0437631 Ascher Wagner, Determination of the finite primitive reflection groups over an arbitrary field of characteristic not 2. I, Geom. Dedicata 9 (1980), no. 2, 239–253, DOI 10.1007/BF00147439. MR578199 Ascher Wagner, Determination of the finite primitive reflection groups over an arbitrary field of characteristic not two. II, III, Geom. Dedicata 10 (1981), no. 1-4, 191–203, 475– 523, DOI 10.1007/BF01447423. MR608141 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. MR0379679 A. E. Zalesski˘i and V. N. Sereˇ zkin, Finite linear groups generated by reflections (Russian), Izv. Akad. Nauk SSSR Ser. Mat. 44 (1980), no. 6, 1279–1307, 38. MR603578

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Index

when p = 2, 30 E q (X), 94 even type, 303 restricted, 8 extendible, 13

A, 158 a, superscript, 30 acceptable subterminal pair, 7 agreeable subgroup, 25 almost strongly p-embedded subgroup, 4 types of, 4 Alperin, J., 314 Alt, 1 anchored component, 453 Aschbacher, M., x, 12 classical involution theorem, 29

F, F(K), 6 Φm (mth cyclotomic polynomial), 7 f , f (K), 6 Finkelstein, L., x, 15 G0 , 29 G0 , 2 G0 ∈ Chev, construction of, 29–290 Gi2 -groups, 5 γ(G), 4 Generic Case, 4 Gilman, R. H., x, 15, 29 Gilman-Griess theory, 14 Gorenstein, D., x Gp , 5 Gip , 5 Gip -groups, 6 Gramlich, R., xi, 12 Griess, R. L., Jr., x, 15, 29 G∗ -trick, 253

B ∗ , 157 B0 , 175 Background Results, xi Baer, R., 332 Bender, H., xi Bennett, C., 12 Blok, R., xi, 10, 12 breather, 303 brush, 393 C o , C(V )o , 15 Chev, 1 CI , CI  , CIo , 16 classical involution, 89, 294, 299 compatible, 14 Cp , 5 CT-pairs and CT-systems, 376–383 CT-system, see also weak CT-system Curtis-Tits theorem, 9 C(V ), 15

Harris, M. E., x HD , 18 Hoffman, C., xi, 10, 12 Holt, D., 318 Horn, M., 12 hs, superscript, 30

De , 29, 155 Devillers, A., 12 Dickson, L. E., xi, 323 + − (q), Dn (q), 16 Dn dp , 6 dp (G), 6 dp (K), 6 Dunlap, J., 12

ignorable, 7 ILop (G), 4 Ipo (G), 4 J∗p (G), 7 Jp (G), 6 Jr , Jr , 171

E6+ (q), E6− (q), 16 q when p is odd, 16, 155

κ, 15 K(7)∗ , 1 K(7)+ , 3 487

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488

K¨ ohl, R., xi, 12 K-proper, 1 Kp , 5 K-preuniqueness subgroup, 4 Lq2 (X), 94 long root p-element, 474 Lop (G), 4 L+ (q), L− (q), 15 Mal’cev, A. I., 337, 405 McLaughlin, J., xi milestone, 303 m  p (K), 5 M¨ uhlherr, B., 12 N, N(x, K, y, L), x, 2, 29 Natural, Mr., see also Griess, R. L., Jr. neighborhood, 2 Nickel, W., 12 P, 18 P-pairs and P-systems, 376–383 pair, 4 acceptable subterminal, 7 p-component preuniqueness subgroup, 4 p-generic, 5 Phan system, see also weak P-system Phan theorem, 13, 14 Phan theory, 10–14 Phan, K.-W., 12 Phan-able, 11 PP-pair, 377 primitive prime divisor, 147, 386 p-Thin Configuration, 1 Q(G), Qp (G), 156 Q∗ (G), 156 R, 173 ranking functions F, 6 F, F(K), 6 τ , τ (K), 6 f , f (K), 6 reflection module, 326 ρ(K, I), 436

INDEX

Stage 4b+, 3 Stage 5+, 291 Thompson, J. G., x, xi, 46, 246 torally embedded, 339 Tp , 5 triality-free, 475 u, superscript, 30 uniqueness subgroup almost strongly p-embedded subgroup strongly p-embedded type, 4 strongly closed type, 4–5 strong p-uniqueness subgroup, 4 strong p-uniqueness subgroup of component type, 4 VI , VI  , 16 Vκ,p , 16 W , 158 W ∗ , 159 W1 , 173 Wagner, A., 322 Walter, J. H., x weak CT-system, weak Curtis-Tits system, 9 weak CTP-system, 14 weak P-system extendible, 13 weak Phan system, weak P-system, 13 WH , WK , WL , WLu , 158 Wielandt, H., 359 Witzel, S., 12 W (L), 159 Wong, W., 15 Wong, W. J., x W ≡ W (L), 159 W ≡V W (L), W ≡ W (L), 322 Zalesski˘ı, A. E., 322

Sereˇ zkin, V. N., 322 Shpectorov, S., xi, 10, 12 Solomon 2-fusion systems, 79, 393 Solomon, R., 15 splitting K ∈ Chev(r), 8 Spor, 1 standard CT-pair, 9 standard Phan pair, standard P-pair, 10 Suzuki, M., 332 τ , τ (K), 6 Theorem C∗7 , 1

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