The Classification of the Finite Simple Groups, Number 7 978-0-8218-4069-6

Table of contents :
Cover......Page 1
Title page......Page 4
Contents......Page 8
Preface......Page 10
1. Introduction......Page 12
2. The Ranking Function \FF......Page 17
3. The Set \J_{}*(), and the Four Cases......Page 18
4. The Sets \J_{}¹() and \J_{}²(), and How to Modify Stages 2 and 3a......Page 19
5. Component Pairs......Page 22
6. The Remaining Stages of Theorem \C₇*......Page 23
1. -Groups......Page 30
2. Pumpups and -Terminality......Page 31
3. Balance and Signalizers......Page 38
4. Components and Coverings......Page 39
5. Representations and Cohomology......Page 40
6. Fusion and Transfer......Page 42
7. Miscellaneous......Page 46
1. Introduction......Page 50
2. Theorem 2: The Setup and Special Cases......Page 52
3. The Cases ≅_{} or ≅, with =2......Page 56
4. The ₅ Case......Page 65
5. -Thin Configurations, Odd......Page 67
6. The General Lie Type Case, Begun......Page 74
7. The General Lie Type Case, Concluded......Page 86
8. The =2, ₂() case......Page 99
9. Theorem 3......Page 106
10. Theorem 4: The Prime ......Page 113
11. The -Uniqueness Case......Page 122
12. Theorem 5: The General Case......Page 126
13. Theorem 5: Residual Cases, =2......Page 132
14. Theorem 5: Residual Cases, >2......Page 137
15. Theorem 6......Page 154
1. Introduction......Page 156
2. Theorems 1 and 2: Most Sporadic Components......Page 158
3. Theorems 1 and 2: ₁ and Ree Components......Page 160
4. The Alternating Case: Theorem 3–Centralizers of Rank 2......Page 163
5. Theorem 4–Centralizers of Rank 1: Preliminaries......Page 168
6. Theorem 4–All Alternating Group Neighbors......Page 171
7. Theorem 4–Non-Alternating Group Neighbors......Page 178
8. Theorem 4–The Final Case: 2 Neighbors......Page 186
1. Everyday Tools......Page 192
2. Automorphisms and Coverings......Page 200
3. -Rank and -Sylow Structure of Quasisimple Groups......Page 202
4. Computations in Alternating Groups......Page 207
5. Computations in Sporadic Groups......Page 209
6. Computations in Groups of Lie Type......Page 214
7. Small Groups......Page 228
8. Embeddings......Page 233
10. Balance......Page 234
11. Subcomponents......Page 235
12. Pumpups and the Ranking Function \FF......Page 237
13. Pumpups......Page 242
14. Acceptable Subterminal Pairs, I......Page 273
15. Acceptable Subterminal Pairs, II......Page 282
16. Neighborhoods......Page 288
17. Generation......Page 299
18. Splitting Primes......Page 303
19. Double Pumpups......Page 309
20. The Double Pumpup Propositions......Page 314
21. Monotonicity of \FF, and \J_{}²()......Page 331
22. Miscellaneous......Page 348
Bibliography......Page 352
Index......Page 354
Back Cover......Page 362

Citation preview

MATHEMATICAL

Surveys and Monographs

Volume 40, Number 7

The Classification of the Finite Simple Groups, Number 7 Daniel Gorenstein Richard Lyons Ronald Solomon

The Classification of the Finite Simple Groups, Number 7 Part III, Chapters 7–11: The Generic Case, Stages 3b and 4a

MATHEMATICAL

Surveys and Monographs Volume 40, Number 7

The Classification of the Finite Simple Groups, Number 7 Part III, Chapters 7–11: The Generic Case, Stages 3b and 4a Daniel Gorenstein Richard Lyons Ronald Solomon

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 20D06, 20D08; Secondary 20D05, 20E32, 20G40. For additional information and updates on this book, visit www.ams.org/bookpages/surv-40.7 The ISBN numbers for this series of books includes ISBN ISBN ISBN ISBN ISBN ISBN ISBN

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

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. Permissions to reuse portions of AMS publication content are handled by Copyright Clearance Center’s RightsLink service. For more information, please visit: http://www.ams.org/rightslink. Send requests for translation rights and licensed reprints to [email protected]. Excluded from these provisions is material for which the author holds copyright. In such cases, requests for permission to reuse or reprint material should be addressed directly to the author(s). Copyright ownership is indicated on the copyright page, or on the lower right-hand corner of the first page of each article within proceedings volumes. 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 

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

23 22 21 20 19 18

For Rose and Lisa

Contents Preface

ix

Chapter 7. The Stages of Theorem C∗7 1. Introduction 2. The Ranking Function F 3. The Set J∗p (G), and the Four Cases 4. The Sets J1p (G) and J2p (G), and How to Modify Stages 2 and 3a 5. Component Pairs 6. The Remaining Stages of Theorem C∗7

1 1 6 7 8 11 12

Chapter 8. General Group-Theoretic Lemmas 1. p-Groups 2. Pumpups and p-Terminality 3. Balance and Signalizers 4. Components and Coverings 5. Representations and Cohomology 6. Fusion and Transfer 7. Miscellaneous

19 19 20 27 28 29 31 35

Chapter 9. Theorem C∗7 : Stage 3b 1. Introduction 2. Theorem 2: The Setup and Special Cases 3. The Cases K ∼ = He, with p = 2 = An or K ∼ 4. The F5 Case 5. p-Thin Configurations, p Odd 6. The General Lie Type Case, Begun 7. The General Lie Type Case, Concluded 8. The p = 2, SL2 (q) case 9. Theorem 3 10. Theorem 4: The Prime s 11. The s-Uniqueness Case 12. Theorem 5: The General Case 13. Theorem 5: Residual Cases, p = 2 14. Theorem 5: Residual Cases, p > 2 15. Theorem 6

39 39 41 45 54 56 63 75 88 95 102 111 115 121 126 143

Chapter 10. Theorem C∗7 , Stage 4a: Constructing a Large Alternating Subgroup G0 1. Introduction 2. Theorems 1 and 2: Most Sporadic Components

145 145 147

vii

viii

CONTENTS

3. 4. 5. 6. 7. 8.

Theorems 1 and 2: J1 and Ree Components The Alternating Case: Theorem 3–Centralizers of Rank 2 Theorem 4–Centralizers of Rank 1: Preliminaries Theorem 4–All Alternating Group Neighbors Theorem 4–Non-Alternating Group Neighbors Theorem 4–The Final Case: 2HS Neighbors

149 152 157 160 167 175

Chapter 11. Properties of K-Groups 1. Everyday Tools 2. Automorphisms and Coverings 3. p-Rank and p-Sylow Structure of Quasisimple Groups 4. Computations in Alternating Groups 5. Computations in Sporadic Groups 6. Computations in Groups of Lie Type 7. Small Groups 8. Embeddings 9. Recognition Theorems 10. Balance 11. Subcomponents 12. Pumpups and the Ranking Function F 13. Pumpups 14. Acceptable Subterminal Pairs, I 15. Acceptable Subterminal Pairs, II 16. Neighborhoods 17. Generation 18. Splitting Primes 19. Double Pumpups 20. The Double Pumpup Propositions 21. Monotonicity of F, and J2p (G) 22. Miscellaneous

181 181 189 191 196 198 203 217 222 223 223 224 226 231 262 271 277 288 292 298 303 320 337

Bibliography

341

Index

343

Preface Volumes 5, 7, and 8 of this series form a trilogy treating the Generic Case of the classification proof, of which this is the middle volume. 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 to Volume 5, to which we refer the reader. Using the Signalizer Functor Method, we have arrived by the end of Volume 5 at the existence in our K-proper simple group G, for a suitably chosen prime p, of elements x of order p whose centralizer has a generic quasi-simple component K with CG (K) of very small p-rank. Moreover, either p = 2 or K ∈ Chev(2). Significantly, also, a family of “neighboring” centralizers have semisimple p-layers. A precise statement of the initial conditions for this volume is given as Theorem 1.2 in Chapter 7 of this volume. [Note: The numbering of chapters in this volume continues that of Volume 5. In particular, Chapter 7 is the first chapter of this volume.] The principal theorem established in this volume is Theorem C∗7 : Stage 3b, which is the subject of Chapter 9 and is stated carefully in the introduction to that chapter. One of the main conclusions (Theorem 2) of this theorem is that the neighborhood N of (x, K) is vertical. The adjective “vertical” is in contrast to “diagonal”. This theorem rules out the “shadow” possibility that G = K1  x, where K is diagonally embedded in the base K1 ×· · ·×Kp of the wreathed product. Of course, this structure would violate the assumed simplicity of G, but it must be ruled out from local information. The remaining parts of Theorem C∗7 : Stage 3b address the structure of K and its “neighbors” when K is a group of Lie type. In the process we define a total ordering on the set of all groups of Lie type. This ordering is compatible with a natural partial ordering on components deriving from L-balance and the “pumpingup” process. Most of our results are proved only for those K which are maximal in the total ordering. First, we establish in Theorem 4 that, by shifting primes if need be, we may assume that p splits K, i.e., if K is defined over a field of order q, then p divides q 2 − 1. This is, of course, trivial if p = 2 and q is odd. The significance of this result is to make a good choice of the odd prime p when q is a power of 2. Once we have chosen p to split K, the last major result (Theorem 5) shows that, with few exceptions, the neighborhood N is level. Essentially, this means that the component Ly of CG (y) is defined over the same field as is K, whenever (y, Ly ) is a neighbor of (x, K). The “shadow” lurking behind this case is G = K(q p )x with x acting as a field or graph-field automorphism on E(G) = K(q p ). Again, of course, this structure violates the simplicity of G. Usually, a contradiction follows easily from the ordering we have placed on components, but sometimes a transfer argument is used to show that x ∈ [G, G], violating the assumed simplicity of G. ix

x

PREFACE

Having completed the proof of Theorem C∗7 : Stage 3b, we are now ready to attack the problem of identifying the group G0 := N(x, K, y, L) generated by K and all of the neighboring components Lu lying over a fixed 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 the Generic Case. But this will not be established until Volume 8. Moreover, in this volume, we will only identify G0 when p = 2 and K/Z(K) is either an alternating group or a sporadic group of generic type, leaving the principal case in which K is a group of Lie type as the main topic of Volume 8. The alternating and sporadic cases are the subject of Chapter 10. In view of the short list of allowable sporadic components, there is no group G0 arising when K is sporadic. The main result of Chapter 10 is then to establish that if K ∼ = An , n ≥ 9, then m2 (CG (K)) = 2 and G0 ∼ = An+4 . In view of results established in Volume 5 for the case when m2 (CG (K)) ≥ 2, it is fairly easy in this case to establish that G0 satisfies a presentation for An+4 . The more challenging case arises when m2 (CG (K)) = 1, because of the nonsimple “shadows” Σn+2 and Aut(F5 ), the latter when K ∼ = A10 . Indeed for n ≥ 11, we must construct a subgroup of G isomorphic to Σn+2 before obtaining the transfer contradiction that x ∈ [G, G]. The most delicate transfer contradiction arises in the case when n = 10 and the 2-local structure of G is that of Aut(F5 ). As a convenience for us in a couple of spots, we add one paper to the Background Results: Jack McLaughlin’s characterization of groups generated by transvections on a finite vector space of odd characteristic [McL2]. We continue the notational conventions established in Volumes 1 and 2 of this series [I1 , I2 , IG ]. We refer to the chapters of the current book as [III7 ], [III8 ], [III9 ], [III10 ] and [III11 ]. As in previous volumes, the last chapter [III11 ] collects the necessary K-group lemmas for the main chapters [III9 ] and [III10 ], and thus logically comes after [III8 ] but before [III9 ]. Both authors thank the National Science Foundation and the National Security Agency for their generous grant support which has been of great value in the preparation of this monograph and its successors. Solomon also thanks the Ohio State University Emeritus Academy and the Simons Foundation’s Collaboration Grants for Mathematicians program for their support over the past several years, which has facilitated research visits between Ohio and New Jersey, as well as the purchase of a laptop enabling virtual visits. We also thank Inna Korchagina and Gary Seitz for their help. Thanks go especially to our wives, Lisa and Rose, to whom we dedicate this volume. And our memories are still warm of our co-author and the leader of this project, Danny Gorenstein. Richard Lyons and Ronald Solomon July 2017

CHAPTER 7

The Stages of Theorem C∗7 1. Introduction This seventh volume picks up the analysis of the Generic Case where volume 5 in this series [GLS5]1 left off, and carries that analysis to the next stopping point, from which it will be completed in an upcoming volume 8. By “analysis” we mean the proof of Theorem C∗7 . In this introductory chapter we review the most important definitions from [GLS5], as well as the statement of Theorem C∗7 and the three stages – 1, 2, and 3a – of the proof of Theorem C∗7 that were carried out in [GLS5]. We then discuss the remaining stages of the proof – 3b and 4a, to be completed in this volume, and 4b and 5, which will be carried out in the next volume. 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 groups for Theorem C∗7 is as follows [III1 , p. 3];2 here q ranges over all prime powers, and  = ±1: 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} ∪ {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) = ∅ 1 The six individual chapters of [GLS5] are referenced as [III ], [III ], . . . , [III ], and [III ], 1 2 5 K the sixth and last chapter. 2 The groups U (2n ), n ≥ 3, were mistakenly omitted from the corresponding 5 A list of corrections for this and other errata is being maintained at list in [III1 ]. sites.math.rutgers.edu/∼lyons/cfsg.

1

7. THEOREMS C7 AND C∗ 7

2

is equivalent to the statement that G is of generic type [I2 , Def. 16.1, p.106] (see the formal definition (1B) below). The main feature of a p-Thin Configuration is a pair (x, K) such that x has order p, K is a p-component of CG (x), K/Op (K) ∈ Tp , and (x, K) is Ipo -terminal in G (see the formal definition 1.10 below). It is more natural to deal with p-Thin Configurations together with the cases where G has Special Odd Type (for p = 2) and Special Even Type (for p > 2). Indeed for p = 2, we have already established [GLS4, Theorem C∗2 ] that if G has a 2Thin Configuration, then G is isomorphic to a simple K-group. The corresponding assertion for p > 2 will be treated in a later monograph, together with Theorem C6 [I2 , p. 106]. At that point, we will be able to omit the hypothesis from Theorem C∗7 about p-Thin Configurations, and we will have obtained Theorem C7 [I2 , p. 106], the classification of G in the Generic Case. We combine the results from [GLS5], specifically Stages 2 and 3a of Theorem C∗7 [III4 , p. 167], [III5 , p. 274], into the following single statement. Here J1p (G) is a certain set of pairs (x, K) such that x ∈ G has order p, K is a p-component of CG (x), and (x, K) is p-terminal in G [III4 , Sec. 1]. Theorem 1.2 (Theorem C∗7 : Stages 2, 3a). Let p ∈ γ(G) and assume that G has no p-Thin Configuration. Then J1p (G) = ∅. Let (x, K) ∈ J1p (G). Then K is quasisimple, and there exists a normalized subterminal (x, K)-pair (y, L). For any such pair (y, L), the (x, K)-neighborhood of (y, L) is semisimple and nontrivial and either (a), (b), or (c) holds: (a) p = 2, K ∼ = An , and either (1) L ∼ = An−4 , CG (K) has E22 Sylow 2-subgroups, and some root foursubgroup of K centralizes some G-conjugate of K; or (2) m2 (CG (K)) = 1; (b) K ∈ Chev(r) for some r = p, and the following conditions hold: (1) mp (CG (K)) = 1; and (2) Exactly one of p and r is 2; or (c) p = 2, K is sporadic, and m2 (CG (K)) = 1. We begin by restating the relevant terminology from [GLS5], for convenience. The assumption that we are in the Generic Case is encoded in the phrase p ∈ γ(G). As always in this series, G is a K-proper finite simple group. Definition 1.3. (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)}. We call the elements of ILop (G) component pairs, or p-component pairs if the context does not make it clear what the prime p is. The principal meaning of the Generic Case is that (1A)

Lop (G) ∩ Gp = ∅ for some prime p.

1. INTRODUCTION

3

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], under the assumption that (1A) holds: (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 (1B) (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 1.4. There are two types of strong p-uniqueness subgroup for an odd prime p [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 the next volume. Definition 1.5. Let p be an odd prime. 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 with mp (M ) ≥ 4; (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. The strong p-uniqueness subgroups that we encounter will all be maximal subgroups of G. Remark 1.6. 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 volume 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)). Definition 1.7. 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

7. THEOREMS C7 AND C∗ 7

4

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 shall review the sets Gip below. However, unfortunately, a blunder in the vital lemma [IIIK , Lemma 5.1] necessitates a change in the partition that applies for odd primes p. The lemma concerns a prime p and quasisimple groups K and L in Gp . Let i and j be the unique integers such that K ∈ Gip and L ∈ Gjp . Under the assumption that K is a pumpup of L (L 3 and some  = ±1}; 3

= {L | L ∼ = A9 or A10 };

1. INTRODUCTION

5

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

(1F)

p G4p G5p G6p

= {L | L ∈ Chev ∩ Gp , mp (L) ≥ 3 and L ∈ G2p }; = {L | L ∈ Chev ∩ Gp − Alt and mp (L) = 2}; and = {L | L ∈ Spor ∩ Gp or L ∼ = An for some n ≤ 4p + 1 (but n ≤ 12 if p = 3)}.

Fortunately, this change forces no essential change in the statements or proofs of Stages 1, 2, and 3a of Theorem C∗7 . The stratification of Gp for p odd only plays a role in the proof of Stage 1, in sections 11 and 16–20 of [III3 ]. A careful reading of these sections and of the supporting lemmas in [IIIK ] shows that the proofs remain valid as they stand. This takes into account the change in the definitions of G2p and G4p , and the resulting changes in the meaning of the terms “Gp -depth,” “dp (G),”, and “wide Lie type.” Definition 1.8. Let p ∈ γ(G). (a) If K ∈ Gp , 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.) Finally, we define the terminology connected with p-Thin Configurations. For technical reasons we give a slightly different definition both of TGp -group and of p-Thin Configuration for p odd than [III1 , Definition 2.1]. This volume is the first one in which such configurations (for p odd) play any role at all, so our change does not affect the integrity of our prior work. It will be helpful in the proof of Theorem C6 . Definition 1.9. Let p be an odd prime. Then TGp = G4p ∪ G5p ∪ G6p . An element of TGp is called a TGp -group. Definition 1.10. Let p be a prime. We say that G has a p-Thin Configuration if and only if (a) G contains an Ipo -terminal Tp -pair (z, I); (b) If p = 2, then I/O2 (I) is a B2 -group; and (c) If p is odd, then d (G) lies in TGp ; moreover, (1) Every element of Lop (G) which is in Gpp dp (G) = 4 or 5; (2) If some element of Lop (G) lies in TGp ∩ Chev and has p-rank at least 3, then p = 3 and m3 (I) ≥ 2; and (3) If p = 3, J ∈ Lo3 (G) is a TG3 -group, y ∈ Aut(J) with y 3 = 1 and L3 (CJ (y)) has a 3-component J0 such that J0 /O3 (J0 ) is a B3 -group, then I/O3 (I) is itself a B3 -group, or isomorphic to G2 (8). We may also say that (z, I) affords a p-Thin Configuration in G.

6

7. THEOREMS C7 AND C∗ 7

2. The Ranking Function F The rest of the proof of Theorem C∗7 revolves around a suitably chosen prime p ∈ γ(G) and pair (x, K) ∈ Jp (G). Subject to certain conditions, and in particular only for K/Op (K) ∈ Chev, the pair will be chosen to maximize a certain invariant which we call F(K/Op (K)). In this section we introduce the function F, note its basic monotonicity property, and prepare to define the set J∗p (G) of maximal pairs which are the focus of the analysis in this volume and the next. We were unaware of the monotonicity property when we wrote [GLS5]. Now that it has been proved, it is advantageous for us to change slightly the definition of the set J1p (G) [III4 , 1.2(2b)], more precisely the definition of an indomitable component pair [III4 , 1.1]. We shall presently state the new definition of J1p (G). At the end of the next section we track this change in [III4 ] and show that the main results of that volume still hold with the altered definitions. The function F is a combination of an integer-valued function f on Lie and an ordering of the types A–G of simple Lie algebras. Definition 2.1. Let K ∈ Lie and write K = d L(q).

(2A)

Let  be the rank of the simple complex Lie algebra L, i.e., the dimension of a Cartan subalgebra of L. We define (2B)

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). Several remarks are in order. First, notice that by definition f (K) = f (K1 ) if K/Z(K) ∼ = K1 /Z(K1 ). Furthermore, since L is assumed simple, the value of f on 2 D2 (q) ∼ = A1 (q 2 ) is 2 2 f (A1 (q )) = q . Finally, in the case of ambiguous groups such as K = B2 (3)a ∼ = 2 A3 (2), f is not well-defined, but the appropriate way to view the characteristic of K (2 or 3 in this example) should always be clear from context. The following refinement F of f will be useful for comparing groups K and L of Lie type for which f (K) = f (L). Definition 2.2. Let K ∈ Chev, as in (2A). 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. 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 the six types of Lie algebras are ordered, independent of rank, as follows3 : (2C)

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

3 This ordering goes back at least to P. N. Burgoyne’s compilation of centralizers of elements of order 3 in groups in Lie(2) [GL1, Appendix, Part I].

3. THE SET J∗ p (G), AND THE FOUR CASES

7

With this definition it is straightforward to prove that if (x, K) < (x∗ , K ∗ ) in (see Section 2 of the next chapter for a discussion of such “pumpups”), with K/Op (K) and K ∗ /Op (K ∗ ) lying in Chev, then F(K/Op (K)) ≤ F(K ∗ /Op (K ∗ )). More difficult is the following monotonicity proposition, which is proved in Chapter 11 as Proposition 21.1. ILop (G)

Proposition 2.3 (Monotonicity of F). Let K, L ∈ Chev(2) and suppose that L is involved in K. Then F(L) ≤ F(K), with equality if and only if K/Z(K) ∼ = L/Z(L). 3. The Set J∗p (G), and the Four Cases For p ∈ γ(G), the set Jp (G) has several subsets of interest. In [III3 , p. 66] we defined Jp (G) = {(x, K) ∈ Jp (G) | mp (C(x, K)) > 1}  (3A) Jp (G) if Jp (G) = ∅ Jop (G) = Jp (G) if Jp (G) = ∅ In this section we introduce the subset J∗p (G), which consists of those elements which are maximal in a sense that depends on p and the isomorphism types of of K/Op (K) as (x, K) varies over Jop (G). In the case of Lie type components, the ranking function F will play the key role. Then in Proposition 3.3 below, we clarify the four cases to be analyzed in the Generic Case, according to the isomorphism type of K: two cases if K ∈ Alt, and the two cases K ∈ Spor and K ∈ Chev. Definition 3.1. Let p ∈ γ(G). For any subset X ⊆ Kp , define Jop (G)

Jop (G, X) = {(x, K) ∈ Jop (G) | K/Op (K) ∈ X}, Jp (G, X) = {(x, K) ∈ Jp (G) | K/Op (K) ∈ X}, Definition 3.2. Let p ∈ γ(G). For any (x, K) ∈ Jp (G) write K = K/Op (K). Then ⎧

⎪ (x, K) ∈ Jop (G, Alt) |K/Op (K)| is as large as possible , ⎪ ⎪ ⎪ ⎪ if Jp (G, Alt) = ∅; ⎪ ⎪ ⎪ ⎪ ⎪ ⎪Jp (G, Spor), ⎪ ⎪ ⎪ ⎨ if Jp (G, Spor) = ∅ and Jp (G, Alt) = ∅; ∗ Jp (G) = 

F(K) ≥ F(J) ∀ (x , J) ∈ J2 (G) ⎪ (x, K) ∈ J (G) ⎪ 2 ⎪ ⎪ ⎪ ⎪ if p = 2 and J2 (G, Alt ∪ Spor) = ∅; ⎪ ⎪ ⎪ 

⎪ ⎪ ⎪ (x,K) ∈ Jp (G,Chev(2)) F(K) ≥ F(J ) ∀ (x ,J) ∈ Jp (G,Chev(2)) ⎪ ⎪ ⎩ otherwise. We summarize the main properties of J∗p (G) in the following proposition. Its proof, which depends heavily on Theorem 1.2, will be given in the next section. One of its assertions, that all elements of J∗p (G) satisfy the conclusions of Theorem 1.2, can be generalized to a larger set than J∗p (G) in certain cases, and this information is part of the proposition. Proposition 3.3. Let p ∈ γ(G). Then J∗p (G) is not empty, and the conclusions of Theorem 1.2 hold for any (x, K) ∈ J∗p (G). Exactly one of the following four cases holds:

7. THEOREMS C7 AND C∗ 7

8

(a) Alternating component and root four-group. γ(G) = {2}; ∼ An for some n ≥ 9, m2 (C(x, K)) > 1, for all (x, K) ∈ J∗ (G), K = 2

and some Sylow 2-subgroup of C(x, K) is G-conjugate to a root four-subgroup of K; and Theorem 1.2 applies to all elements of J2 (G). (b) Alternating component and transposition. γ(G) = {2}; ∼ An for some n ≥ 9, for all (x, K) ∈ J2 (G), K/O2 (K) = and for all (x, K) ∈ J2 (G), m2 (C(x, K)) = 1; and Theorem 1.2 applies to (x, K) provided n is maximal. (c) Sporadic component. γ(G) = {2}; for all (x, K) ∈ J2 (G), K/O2 (K) ∈ Spor and Theorem 1.2 applies to (x, K). (d) Lie type component. Either γ(G) = {2} or 2 ∈ γ(G); for all (x, K) ∈ Jp (G), K/Op (K) ∈ Chev(r) and mp (C(x, K)) = 1, where exactly one of p and r is 2 (and r may depend on K); and if p = 2, then Theorem 1.2 applies to (x, K). 4. The Sets J1p (G) and J2p (G), and How to Modify Stages 2 and 3a [This technical section can be omitted on a first reading.] Because of the monotonicity of the ranking function F (Proposition 2.3), it becomes convenient to replace the set J1p (G), whose elements were the main characters of Theorem 1.2, with a slightly larger set J2p (G), in such a way that Theorem 1.2 remains valid for all elements of J2p (G). In this section, we introduce such a set J2p (G) and check that Theorem 1.2 can be strengthened to apply to J2p (G). One consequence of this is that J∗p (G) ⊆ J2p (G), as will be proved below. The proof depends on 21.31 in Chapter 11, which in turn depends on the monotonicity property of F. It is for this reason that in Proposition 3.3, we can assert that Theorem 1.2 applies to all elements of J∗p (G). It would be problematic, on the other hand, to prove that J∗p (G), as specified in Definition 3.2, is a subset of J1p (G). At the end of this section, we prove Proposition 3.3. The only difference between J1p (G) and J2p (G) occurs for p > 2, and concerns the “indomitability” [III4 , 1.2] of pairs (x, K) ∈ Jop (G) for which K ∈ Chev. We now give the better analogues of Definitions 1.1 and 1.2 of [III4 ], underlining the only changed features. Again we refer to Section 2 of the next chapter for the meaning of the “long pumpup” symbol 2, so that r = 2, J2p (G) = J2p (G, Chev(2)) by Theorem 1.2, and J∗p (G) is the subset of Jp (G, Chev(2)) maximizing F, hence J∗p (G) is nonempty. In this case, [III11 , 21.31] implies that J∗p (G) ⊆ J2p (G). Hence the proofs of Proposition 3.3 and Lemma 4.3 are complete.  5. Component Pairs In this section we prove a few important properties of p-component pairs. Some depend on Proposition 3.3. Corollary 5.1. Let (x, K) ∈ J∗p (G). Then one of the following holds: (a) p = 2 and K ∈ Alt; or (b) mp (C(u, I)) = 1 for every (u, I) ∈ Jp (G). Proof. If (a) fails, then we are in case (c) or (d) of Proposition 3.3, so (b) follows.  Lemma 5.2. Let (x, K) ∈ ILop (G) with K/Op (K) ∈ Gp . Then the following conditions hold: (a) There exists a long pumpup of (x, K) that is p-terminal in G; and (b) If p is odd, then mp (K) − mp (Op p (K)) > 1. Proof. If p = 2, then (a) holds by [IG , 6.10]. Suppose that p is odd. Then (b) holds by [III11 , 3.2]. By [IG , 6.22] there exists an Ipo -terminal long pumpup (x∗ , K ∗ ) of (x, K). Then K ∗ /Op (K ∗ ) ∈ Gp by [III11 , 1.3] so mp (K ∗ ) − mp (Op p (K ∗ )) ≥ 2  by (b). Then (x∗ , K ∗ ) is p-terminal by [IG , 6.27]. Lemma 5.3. Suppose that p ∈ γ(G), (x, K) ∈ ILop (G), K/Op (K) ∈ Gp and dp (K) = dp (G). If p = 2 then assume that J2 (G, Alt ∪ Spor) = ∅. Then one of the following holds:

7. THEOREMS C7 AND C∗ 7

12

(a) mp (C(x, K)) = 1; or (b) Some long pumpup (x∗ , K ∗ ) of (x, K) has a vertical pumpup. Proof. Suppose that (b) fails and choose a chain of pumpups (5A)

(x, K) = (x0 , K0 ) < (x1 , K1 ) < · · · < (xn , Kn )

such that (xn , Kn ) is p-terminal in G, and xi−1  = xi  for all i = 1, . . . , n. By assumption and [III11 , 1.2], Ki /Op (Ki ) ∈ Gp and dp (Ki /Op (Ki )) ≤ dp (K) = dp (G) for all i = 1, . . . , n. Hence equality holds by definition of dp (G), and so (xn , Kn ) ∈ Jp (G). By Proposition 3.3 and our assumption on J22 (G, Alt), Jp (G) = Jop (G), and so mp (C(xn , Kn )) = 1. Note that mp (xn  Kn ) > 1 since otherwise p = 2 and Kn /O2 (Kn ) ∼ = SL2 (q), q odd, or 2A7 , contradicting Kn /Op (Kn ) ∈ Gp . Therefore if n > 0, the last pumpup in (5A) is not diagonal, and so it is trivial. But then xn−1 ∈ C(xn , Kn ) so xn−1  = xn , contrary to choice. Consequently  n = 0 and so mp (C(x, K)) = 1. The proof is complete. From [III11 , 12.3] we get the following result, which gives an idea of the workings of the function f on which the ranking function F is based. Lemma 5.4. Let (x, K) ∈ ILp (G), let y ∈ Ip (CG (x)), and let L be a component of CK (y). Suppose that K/Op (K) ∈ Chev and write K/Op (K) = d LK (qK ) and  L/Op (L) = d LL (qL ) . Let K and L be the Lie ranks of LK and LL , respectively. Let f be defined as in (2B). Define ρ(K, L) by f (K/Op (K)) = f (L/Op (L))ρ(K,L) . Then ρ(K, L) ≥ 1, and one of the following holds: (a) y ∈ C(x, K); (b) y induces a field or graph-field automorphism on K/Op (K), and ρ(K, L) = p; a , then a ≥ 1, (c) y maps into Aut0 (K/Op (K)) and if we write qL = qK aL ≤ K , and

2 K − aL (5B) ρ(K, L) = a 1 + . aL 6. The Remaining Stages of Theorem C∗7 In order to state the remaining stages in the proof of Theorem C∗7 , we need further terminology. We recall the definitions of quasi-p-component and of subterminal (x, K)-pair [III3 , 1.3]. Definition 6.1. If X is any group, a quasi-p-component of X is a subgroup L ≤ X such that either L is a p-component of X, or p = 2 or 3 with L a solvable p-component of X and L/Op (L) ∼ = SL2 (3) or SU3 (2), respectively. Thus the extended p-layer Lop (X) of X is by definition [IG , 13.6] just the product of all the quasi-p-components of X. Definition 6.2. Let X be a group with Op (X) = 1, J a p-component of X, y ∈ Ip (NX (J)), and L a quasi-p-component of CJ (y). Then L is subterminal in X relative to y if and only if one of the following holds: (a) L is a p-component of CJ (y), and for every z ∈ Ip (CJy (L)CX (J)), either L is a p-component of CJ (z) or z centralizes J;

6. THE REMAINING STAGES OF THEOREM C∗ 7

13

(b) p ≤ 3, L is a solvable p-component of CJ (y) and L = L0 × Op (L) for some subgroup L0 such that for each z ∈ Ip (CJy (L0 )CX (J)), either L0   CJ (z) or z centralizes J. Notice that if x ∈ Ip (CX (J)) and [x, y] = 1, then L is also subterminal in X relative to xy, and indeed with respect to any element of x, y − x. Definition 6.3. Let (x, K) be a Gp -pair in G, i.e., (x, K) ∈ ILp (G) and K/Op (K) ∈ Gp . Set C = CG (x) and C = C/Op (C). Let y ∈ Ip (C) with y ∈ x and let L be a quasi-p-component of CK (y). We call (y, L) a subterminal (x, K)pair provided the following conditions hold: (a) L is subterminal in C relative to y; (b) If Lp (CK (y)) = 1, then L is a p-component of CK (y); and L is of maximal order among all p-components of CK (y) satisfying (a); and (c) None of the following cases occur: p = 2 and K ∼ = Ω5 (q), Ω6 (q) or P Ω6 (q), 2 q odd, with L2 (CK (y)) ∼ = L2 (q ); or p = 3, K ∼ = 2 D4 (2) and L ∼ = SU3 (2).  For completeness, we call (x , K) a subterminal (x, K)-pair for any power x of x with x = 1. A further refinement, the notion of level subterminal pair, incorporates a number of conditions that we wish to impose on the choice of the pair (y, L) whose neighborhood we shall analyze. Here (x, K) ∈ J∗p (G), y ∈ Ip (CG (x)) and L is a p-component of CK (y). This depends on the notion of the level q(σ) of a Steinberg endomorphism σ of a simple algebraic group K [IA , 2.1.9]. We extend this to the notion of the level q(K) of a group K ∈ Lie in the natural way. Definition 6.4. Let r be a prime and let K ∈ Lie(r) ∪ Chev(r). Let (K, σ) be a σ-setup for K, so that K is a simple algebraic group over the algebraic closure F  of F, σ is a Steinberg endomorphism of K, and K is isomorphic to O r (CK (σ)) or its commutator subgroup. Then the level q(K) is defined to be the level q(σ) of σ. Here the function q is defined on the set {σ n | n > 0} as the unique multiplicative function having positive values and such that if n > 0 and σ n normalizes a long root subgroup X of K, then q(σ n ) = |CX (σ n )|. Remark 6.5. The level is not well-defined in a few cases because of strange isomorphisms making the characteristic of K ill-defined [IA , 2.2.10]. However, the ambiguity of characteristic in these cases is always between 2 and some odd prime. The ambiguity is to be resolved by the context, namely the parity of the prime p ∈ γ(G) in the discussion at hand. Our convention is that the level of K, for any (x, K) ∈ ILp (G), or the level of a component of the centralizer of a p-element in such a group K, is to be understood as considering K ∈ Chev(r), where exactly one of the primes r, p is 2. It should be remarked that with the exception of Sections 10 and 11 of Chapter 9, the prime p will have been selected in advance and fixed. Remark 6.6. Recall that by our definition, 2 D2 (q) ∼ = A1 (q 2 ) has level q 2 , since K is to be taken to be simple. When we write K = d L(r n ), the level of K is in fact q(K) = r n , with the exception just noted (i.e., provided L is simple). The reader is warned, however, that for the Suzuki and Ree √ groups the more common notation is 2 L(q(K)2 ). Thus, 2 2 B2 (8), for the Suzuki group of order 8.7.65, for example, we write √  B2 ( 2 8), not 2  and we write F4 ( 2) , not F4 (2) , for the Tits simple group.

7. THEOREMS C7 AND C∗ 7

14

The next definition formalizes the idea that for pairs (x, K) ∈ ILop (G) with K/Z(K) ∼ = Lm (q) and p dividing q − , we always prefer to deal with those subterminal (x, K)-pairs (y, L) for which L/Op p (L) ∼ = Lm−1 (q), if such pairs exist. For p = 2 there is a similar phenomenon involving orthogonal groups and the pumpup  chain Dn (q), Bn (q), Dn+1 (q) rather than the chain Ar−2 (q), Ar−1 (q), Ar (q). We call a terminal subcomponent (y, L) of (x, K) ignorable if L and K are at the ends of such a chain, but somewhere in the group is found a chain (y  , L ) < (u, J) < (x , K  ) with the same end terms but having a middle term. Definition 6.7. Suppose that p ∈ γ(G), (x, K) ∈ J∗p (G), and (y, L) is a subterminal (x, K)-pair. We say that the quadruple (x, K, y, L) is ignorable if and only if there exists (x , K  ) ∈ ILp (G), an (x , K  )-subterminal pair (z  , J  ), and an element y  ∈ CG (x , z  ) of order p such that (a) (b) (c) (d) (e)

y  normalizes K  and J  ; K  /Op p (K  ) ∼ = K/Op p (K); CK  (y  ) has a quasi p-component L with L /Op (L ) ∼ = L/Op (L); J  ≥ L Op (J  ); and One of the following holds: (1) q is a prime power with q ≡  (mod p),  = ±1,  ≥ 4 with  ≡ −1 (mod p), K/Op (K) ∼ = A (q), L/Op (L) ∼ = A−2 (q) and J  /Op (J  ) ∼ =  A−1 (q); or (2) p = 2,  ≥ 4, q is an odd prime power,  and  are signs, K/O2 (K) ∼ =    ∼  D (q), and J /O (J ) B (q). D (q), L/O2 (L) ∼ = −1 =  2

We call (x , K  ) a replacement for (x, K) if (x, K, y, L) is ignorable. Lemma 6.8. Suppose that (x, K) ∈ J∗p (G), (y, L) is a subterminal (x, K)pair, and (x, K, y, L) is ignorable. Let (x , K  ) be a replacement for (x, K). Then (x , K  ) ∈ J∗p (G). Proof. By definition of “ignorable,” K/Op (K) ∈ Chev, so Proposition 3.3d applies. As F(K  /Op (K  )) = F(K/Op (K)), that result implies the lemma, provided we can show that (x , K  ) ∈ Jp (G). Since K/Op p (K) ∼ = K  /Op p (K  ),   K /Op (K ) is a Gp -group of the same depth as K/Op (K), namely dp (G). Therefore it suffices to show that (x , K  ) is p-terminal in G. Choose an Ipo -terminal long pumpup (z, I) of (x , K  ) [IG , 6.10]. By [III11 , 1.2], I/Op (I) ∈ Gp has level dp (G), so (z, I) ∈ Jp (G). As K/Op (K) ∈ Alt, Corollary 5.1 implies that mp (C(z, I)) = 1. But then by [III8 , 2.4], mp (C(x , K  )) = 1, whence  (x , K  ) is p-terminal in G, as desired. The proof is complete. For each p ∈ γ(G), and each (x, K) ∈ J∗p (G), we now specify a preferred subterminal (x, K)-pair (y, L) (if it exists), with which we shall work in this volume and the next. We call these subterminal (x, K)-pairs “acceptable.” It will turn out that they always exist as long as one is willing to use a replacement for (x, K) (see Definition 6.7). Definition 6.9. Let p ∈ γ(G) and (x, K) ∈ J∗p (G), and let (y, L) be a subterminal (x, K)-pair with y = x. 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. Here  = ±1.

6. THE REMAINING STAGES OF THEOREM C∗ 7

15

(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; (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 );  (q) and none of the situations (a1)–(a3) hold, y ∈ (4) K/Z(K) ∼ L = n  SL (q); and Ip (K) maximizes d := dim(CV (y)) and L ∼ = d (5) K ∼ = Sp2n (q), n ≥ 2, or Ω± 2n (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 the cyclotomic polynomial value 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) K = F4 (2 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 (see [III3 , Table 2.1]); the triple (K, L) is one of the following: (F4 (q) (q > 2), C3 (q)), (E6 (q), A5 (q)), (E7 (q),D6 (q)), (E8 (q),D7 (q)), all with m0 = 1 or 2; (E8 (q), E6 (q)) with m0 = 3; or (E8 (q), E6− (q)) with m0 = 6; (c) p = 2, K ∼ = An−4 ; = An , n ≥ 9, y ∈ K is a root involution, and L ∼ (d) p = 2, K ∈ Spor ∩ G2 , y ∈ K, and if possible, y is 2-central in K;  → (e) 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 a preimage of y in K holds: ± ∼ (1) K ∼ = L± 3 (q), Sp4 (q) or SL4 (q), and y ∈ L = SL2 (q); ± ± ∼ ∼ (2) K = Ω5 (q), Ω6 (q), or P Ω6 (q), y ∈ L = SL2 (q), and y has 4 eigenvalues −1 on V ; (3) K ∼ = SL3 (q); = P Ω6 (q), q ≡  (mod 8), y ∈ K, L ∼ ∼ ∼ (4) K = Spin7 (q) and y ∈ L = SL2 (q); 2 ∼ (5) K ∼ = Spin− 8 (q), y ∈ K, and y ∈ L = SL2 (q ) with Z(K) = Z(L); ±  ∼ ∼ (6) K/Z(K) = L2n (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) = P Sp2n (q), n ≥ 3, and y ∈ L = Sp2n−2 (q); (9) K is an orthogonal group, spin group or half-spin group, of dimension at least 7, but not Spin7 (q) or Spin− ∈ 8 (q), y ∈ I2 (K) is such that y  is an involution maximizing the dimension d of the −1-eigenspace K

16

7. THEOREMS C7 AND C∗ 7

∼ Ω± (q) or Spin± (q) (unless d = 4, in which case y ∈ L ∼ of y; L = = d d 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); (f) 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)). ∼ P Ω (q) with q ≡  (mod 8), the definition in Note that if p = 2 and K = 6 (e2) above does not apply, because the (y, L) specified there is not a subterminal (x, K)-pair. Acceptable subterminal pairs will be analyzed in great detail in this volume and the next. In volume 5, by contrast, the notion of “normalized” subterminal pair was the critical idea, imposing mild restrictions, and the main results of that volume – Theorem C∗7 , Stages 2 and 3a – apply only to normalized subterminal pairs. Our definition of acceptable subterminal pair is strictly stronger than that of normalized subterminal pair, and so the results of volume 5 apply to acceptable subterminal pairs. Proposition 6.10. Let (x, K) ∈ J∗p (G) and let (y, L) be an acceptable subterminal (x, K)-pair. Then (y, L) is a normalized subterminal (x, K)-pair. The proof is given in [III11 , Lemma 15.1]. Definition 6.9 depends on the concepts of allowability and admissibility, and ultimately on the notion of a p-source. We recall the essential features of these concepts, but for a complete discussion of this unfortunately technical framework we refer the reader to [III3 , pp. 43–57]. First, for every (x, K) ∈ J∗p (G) (indeed in J2p (G)), certain Ep3 -subgroups of CG (x), are singled out and called (x, K) p-sources. These exist, regardless of (x, K) ∈ J∗p (G). As a general rule, given (x, K) ∈ J∗p (G) and an (x, K) p-source A, a subterminal (x, K)-pair (y, L) is called an (A, x, K)-admissible pair if CA (y) contains a subgroup E∼ = Ep2 such that Lp (CL (E)) = 1. This condition is of basic importance for the proof of Stages 1 and 2 of Theorem C∗7 . However, there are further conditions and exceptions to this rule; when p = 2 or 3 and CL (E) has quasi-p-components instead of p-components, for some E ≤ A with E ∼ = Ep2 , the definition of admissible pair is more technical. In any case, given (x, K) ∈ J∗p (G) and an (x, K) p-source A, there always exist (A, x, K)-admissible pairs [IIIK , 10.1]. A neighbor of (A, x, K) arises from such an admissible pair (y, L) and an element u ∈ x, y# , and is the subnormal closure Lu  of O p (L) in CG (u). This neighbor is called semisimple if and only if (1) L and all the p-components of Lu are quasisimple; or (2) p = 2, L ∼ = SL2 (3) and either (6A) (a) Lu is a component of CG (u); or (b) Lu = O2 (L) ≤ O2 (CG (u)).

6. THE REMAINING STAGES OF THEOREM C∗ 7

17

Theorem C∗7 : Stage 2, as modified to apply to J2p (G), as discussed above, implies that for any (x, K) ∈ J∗p (G) and any (x, K) p-source A, every neighbor of (A, x, K) is semisimple (and in particular K is quasisimple)–with a single exceptional configuration. That exception is p = 2, K/O2 (K) ∼ = Sp4 (q), q odd; and even then there exists an (x, K) 2-source A such that every neighbor of (A, x, K) is semisimple. Such an A is called allowable. We recall the definitions of neighborhood and related ideas. Definition 6.11. Let (x, K) ∈ J∗p (G) and let (y, L) be a normalized level subterminal (x, K)-pair. For each u ∈ x, y# , let Lu be the subnormal closure of  O p (L) in CG (u). (Thus K = Lx .) The (x, K)-neighborhood of (y, L), specified by x, y and L, is the set  #

N(x, K, y, L) = Lu u ∈ x, y . Thus

   N(x, K, y, L)  = Lu u ∈ x, y# . 

The (x, K)-neighborhood of (y, L) is semisimple iff [O p (L), Op (CG (u))] = 1 for # all u ∈ x, y ;  trivial iff Lu = O p (L) for all u ∈ x, y − x, with L  LO2 (CG (u)) if p = 2 and L ∼ = SL2 (3); and nontrivial otherwise; vertical iff it is nontrivial, and for no u ∈ x, y − x is Lu a diagonal pumpup of L; moreover, if p = 2, L ∼ = SL2 (3), and O2 (L) ≤ O2 (CG (u)) for some u ∈ x, y − x, then L  LO2 (CG (u)); level iff either K ∈ Chev, or K ∈ Chev(r) for some prime r, and for all u ∈ x, y − x, Ku ∈ Chev(r) and q(K) = q(Ku ). Finally, for K ∈ Chev and p ∈ γ(G), we use the following notion of “splitting”. Definition 6.12. 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 (c) p divides q(K)2 − 1 if r = 2. In Chapter 9 we shall prove the following theorem. C∗7 ,

Theorem 6.13 (Theorem C∗7 : Stage 3b). Under the assumptions of Theorem J∗p (G) = ∅ for any p ∈ γ(G), and the following conditions hold: (a) For any p ∈ γ(G) and (x, K) ∈ J∗p (G) there exists (x , K  ) ∈ J∗p (G), and an acceptable subterminal (x , K  )-pair (y, L), such that K  /Z(K  ) ∼ = K/Z(K); (b) For any odd p ∈ γ(G) and (x, K) ∈ J∗p (G), mp (K) ≥ 3; (c) If p ∈ γ(G) and (x, K) ∈ J∗p (G) with K ∈ Chev, then there exists a prime p0 ∈ γ(G) and (x0 , K0 ) ∈ J∗p0 (G) such that K0 ∈ Chev and p0 splits K0 ; and (d) Suppose that p ∈ γ(G), (x, K) ∈ J∗p (G), and (y, L) is any acceptable subterminal (x, K)-pair. Let N := N(x, K, y, L) be the (y, L)-neighborhood of (x, K). Then the following conditions hold: (1) N is vertical;

18

7. THEOREMS C7 AND C∗ 7

(2) If K ∈ Chev and p splits K, then N is level except possibly in any of the following situations: (i) p = 2 and K ∼ = 3D4 (q) or G2 (q) for some q; or n (ii) p = 2 and K ∼ = 2 G2 (3 2 ) for some odd n > 1; (iii) p = 2 and K ∼ = P Sp4 (q), or a non-universal version of A± 3 (q) for some q; (3) If N is level, p splits K, and 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: (i) L ∼ = A−2 (q); or (ii) p = 2,  = 4, and Lu /O2 (Z(Lu )) ∼ = B3 (q)u ∼ = Spin7 (q) for # some u ∈ x, y ; # (iii) p = 2,  = 2, and for any u ∈ x, y , Lu ∼ = L, A2 (q), or − A2 (q). Next, Theorem C∗7 : Stage 4 is divided into two pieces, one for the cases K ∈ Alt ∪ Spor, and one for the case K ∈ Chev. Both pieces concern a quadruple (x, K, y, L) supplied by Theorem C∗7 : Stage 3b, and its neighborhood N. The setup is this: (1) (x, K) ∈ J∗p (G) for some p ∈ γ(G); (2) (y, L) is an acceptable subterminal (x, K)-pair; (6B) (3) If K ∈ Chev, then p splits K; (4) N = N(x, K, y, L); and (5) G0 = N. Theorem 6.14 (Theorem C∗7 : Stage 4a). Under the assumptions of Theorem let p, (x, K, y, L), N, and G0 be as in (6B). Suppose that K ∈ Alt ∪ Spor ∪ n { G2 (3 2 ) | n ≥ 3}. Then p = 2, K ∼ = An for some n ≥ 9, a Sylow 2-subgroup E of C(x, K) is G0 -conjugate to a root four-subgroup of K, G0 ∼ = An+4 ∈ K(7)∗ , ΓE,1 (G) ≤ NG (G0 ), and CG (G0 ) has odd order. C∗7 , 2

Theorem 6.15 (Theorem C∗7 : Stage 4b). Under the assumptions of Theorem let p, (x, K, y, L), N, and G0 be as in (6B), and assume that K ∈ Chev. Then, if necessary after replacing (x, K, y, L) by another quadruple still satisfying (6B123), and setting D = x, y, we have G0 ∈ Chev, G0 /Z(G0 ) ∈ K(7)∗ , and ΓD,1 (G) ≤ NG (G0 ). Moreover, if we put M = NG (G0 ), then C∗7 ,



G0 Op (M )/Op (M ) = F ∗ (M/Op (M )) and G0 = O p (M (∞) ). In Chapter 10 we prove Theorem C∗7 : Stage 4a. This volume then concludes with a chapter of supporting K-group lemmas. In the next volume we shall prove Theorem C∗7 : Stage 4b, and then complete the proof of Theorem C∗7 with its final stage: C∗7 ,

Theorem 6.16 (Theorem C∗7 : Stage 5). Under the assumptions of Theorem let G0 be as in the conclusion of Theorem 6.14 or 6.15. Then G0 = G.

CHAPTER 8

General Group-Theoretic Lemmas In this chapter we recall some important preparatory results from previous volumes, and present some new ones. These will be used both in this volume and the succeeding one. In the preface we noted that this chapter logically precedes Chapter 11. However, that statement is slightly incorrect; there are three lemmas in this chapter (Lemmas 1.2, 2.5, and 2.18) which depend on results from Chapter 11. None of the three are used in Chapter 11, however, so they logically belong after Chapter 11.

1. p-Groups Lemma 1.1 (Thompson Replacement Theorem). Let P be a p-group for some prime. Suppose that A and B are abelian subgroups of P of maximal order. If B normalizes A but A does not normalize B, then there exists C ≤ P , abelian of maximal order, such that [A, C] = 1 and [A, C, C] = 1. 

Proof. This is [G1, 8.2.5].

Lemma 1.2. Let X be a K-group and p an odd prime. Let K1 , . . . , Ks be those p-components of X of p-rank at least 3. Let Q ∈ Sylp (X). If s > 1 and Op (Z(X)) = 1, then Q is connected. Proof. Without loss Op (X) = 1. Suppose that s > 1 and the conclusion fails, so that mp (C  ≤ 2 for some u ∈ Ip (Q). If u does not normalize some Ki  Q (u)) u , and L0 = E(CL (u)), then L0 is a homomorphic image of and we set L = Ki Ki and CQ (u) ≥ u × (Q ∩ L0 ) with Q ∩ L0 ∈ Sylp (L0 ). Thus mp (L0 ) = 1. But then mp (Ki ) = 1 by [IG , 15.12(iv)], contrary to hypothesis. Therefore u normalizes Ki for each 1 ≤ i ≤ s, as well as Qi := Q ∩ Ki . By [IG , 10.11], each Qi has a subgroup Ui ∼ = Ep2 with Ui  Q, 1 ≤ i ≤ s; if Z(Ki ) is noncyclic, we may take Ui ≤ Z(Ki ). Set U = U1 · · · Us Ω1 (Z(X)). If u ∈ U , then as U is abelian, mp (U ) = 2 and U = U1 ; but then [u, K2 ] = 1, a contradiction as mp (K2 ) ≥ 3. Therefore u ∈ U , so mp (CU (u)) = 1 and CU (u) = CUi (u) = Ω1 (Z(X)) for all i = 1, . . . , s. As X is a K-group, if Ui ≤ Z(Ki ), then [u, Ui ] = 1 by [III11 , 2.2], a contradiction. Thus by our choice Z(Ki ) is cyclic for each i. But then mp (U/CU (u)) ≥ s ≥ 2 and u centralizes U/CU (u). As U is abelian,  mp (CU (u)) ≥ 2, a final contradiction. The proof is complete. Lemma 1.3. Let P be a p-group, p odd. Let a be the largest rank of a normal  2 a abelian subgroup of P , and let m = mp (P ) be the rank of P . Then m ≤ a + . 4 19

20

8. GENERAL GROUP-THEORETIC LEMMAS

Proof. Let A be a normal elementary abelian subgroup of P of largest order. By [IG , 10.16], A = Ω1 (CP (A)). Then the chain P > CP (A) > 1 gives mp (P ) ≤ mp (Aut(A)) + mp (A) = a + mp (GLa (p)) and the result follows from [IA , Table 3.3.1].  Lemma 1.4 (Aschbacher). Let P be a 2-group and A a set of isomorphic cyclic subgroups of P of order at least 4. Assume that A is closed under P -conjugation and has the property that for all A, A ∈ A, A = A or A ∩ A = 1. Then A is abelian. Proof. Assume false and let P be a minimal counterexample. There  P exist and A = a ∈ A and B = b ∈ A such that z = [a, b] =  1. Let X = A   Y = B P . By minimality X and Y are abelian and P = X, Y  = XY . In particular z ∈ X ∩ Y ≤ Z(P ) and P has class two. In particular z |A| = [a|A| , b] = 1. Since A∩B = 1 we may assume without loss that z ∈ A. Then Ab = az  = A so b b b b2 b2 = az 2 AA = A×A so AA = A z = A×z and | z | = |A|. But A = a contains (az 2 )|A|/2 = a|A|/2 ∈ A, so Ab = A. Hence z 2 ∈ A. As |A| > 2, 1 = z 2 ∈ A∩z, a contradiction to the structure of AAb . The proof is complete.  2

Lemma 1.5 (A. MacWilliams). If T is a 2-group of sectional rank at least 5 (that is, with subgroups T0 and T1  T0 such that T0 /T1 ∼ = E25 ), then T is connected and possesses a normal subgroup isomorphic to E23 . Proof. This follows directly from [III2 , 2.8].



Lemma 1.6. Let R = Q1 ∗ Q2 be the central product of quaternion groups Q1 and Q2 . Let y ∈ I2 (R). Then CR (y) contains a dihedral group D of order at least 8 such that Z(R) ≤ D. Proof. We have y = y1 y2 where yi ∈ Qi has order 4, i = 1, 2. We may therefore replace each Qi by a quaternion subgroup of Qi of order 8, and assume that |Q1 | = |Q2 | = 8. Write Qi = yi , wi , i = 1, 2. Then D = y1 , w1 w2  ∼ = D8 and D ≤ CR (y), as required.  Lemma 1.7. Suppose that T is a 2-group of maximal class and |T | ≥ 8. Let M be a maximal subgroup of T and x an involution in T − M . Then the image of x in Aut(M ) lies outside [Aut(M ), Aut(M )]. Proof. If M is cyclic, then Aut(M ) is abelian and the assertion is clear. Otherwise, as T is dihedral, semidihedral, or quaternion, M is dihedral or quaternion and in particular has maximal class. But the quotient T /Φ(M ) = M x /Φ(M ) is of maximal class and therefore x acts nontrivially on the four-group M/Φ(M ). Therefore the image of x in Aut(M/Φ(M )) lies outside [Aut(M/Φ(M )), Aut(M/Φ(M ))]. The assertion then follows a fortiori .  2. Pumpups and p-Terminality Lemma 2.1 (The Bp -property). For any p-subgroup Q of any K-group X, [Lp (CX (Q)), Op (CX (Q))] ≤ Op (X). Proof. See [III2 , 2.1].



2. PUMPUPS AND p-TERMINALITY

21

Let p be a prime. There are three closely related notions of “pumpups” of quasisimple groups of orders divisible by p. One applies to p-component pairs (x, K) and (x∗ , K ∗ ) ∈ ILop (G) or ILp (G), and is based on the fundamental relation (x, K) < (x∗ , K ∗ ) ⇐⇒ x∗ ∈ C(x, K) and K ∗ is a p-component of the subnormal closure of Lp (CK (x∗ )) in CG (x∗ ). We say that (x∗ , K ∗ ) is a pumpup of (x, K) (in G). We say that it is  K/Op p (K); a a vertical pumpup if and only if, in addition, K ∗ /Op p (K ∗ ) ∼ = trivial pumpup if and only if K ∗ = (K ∗ ∩ K)Op (K ∗ ); and a diagonal pumpup if and only if the subnormal closure of Lp (CK (x∗ )) in CG (x∗ ) is the product of p p-components transitively permuted by x. Any pumpup of component pairs must be of one of these three types, by Lp -balance. The transitive extension of < is written 1; or (iii) p = 2 and K ∼ = P Sp4 (q), or a non-universal version of A± 3 (q) for some q. (3) If K ∼ = A (q),  = ±1, p splits K, and N is level, 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 one of the following situations: (i) L ∼ = A−2 (q); or (ii) p = 2,  = 4, q ≡ − (mod 4), and for some u ∈ x, y# , Lu /O2 (Z(Lu )) ∼ = B3 (q)u ∼ = Spin7 (q); or #  ∼ (iii) p = 2, K = A2 (q), and for any u ∈ x, y , Lu ∼ = L, A2 (q), or − A2 (q). The proof of this theorem is divided into six subsidiary theorems. In all of them we assume that (1A)

the assumptions of Theorem C∗7 hold.

Theorem 1. Assume (1A). Then J∗p (G) = ∅ for any p ∈ γ(G). Moreover, for any p ∈ γ(G) and (x, K) ∈ J∗p (G), there exists (x , K  ) ∈ J∗p (G) and an acceptable subterminal (x , K  )-pair (y, L) such that K/Op p (K) ∼ = K  /Op p (K  ). 39

9. THEOREM C∗ 7 : STAGE 3B

40

Proof. The first statement holds by Proposition 3.3, and the second by the  preliminary lemma [III11 , 14.1]. In the next theorems, we assume that (1) p ∈ γ(G); (2) (x, K) ∈ J∗p (G); (1B) (3) (y, L) is an acceptable subterminal (x, K)-pair; and (4) N = N(x, K, y, L). Theorem 2. Assume (1A) and (1B). Then N is vertical. Theorem 3. Assume (1A) and (1B), with p odd. Then mp (K) − mp (Z(K)) ≥ 3. Moreover, dp (G) = 5. Theorem 2 is proved in Sections 2–8, and Theorem 3 in Section 9. Then in Sections 10 and 11, we prove Theorem 4. Assume (1A) and that there exists a prime p ∈ γ(G) and a pair (x, K) ∈ J∗p (G) such that K ∈ Chev. Then there exists a prime p0 ∈ γ(G) and a pair (x0 , K0 ) ∈ J∗p0 (G) such that K0 ∈ Chev and p0 splits K0 . In Sections 12–14, we prove that in almost all instances when K ∈ Chev and p splits K, the neighborhood (x, K, y, L) must be level. Theorem 5. Assume (1A) and (1B), and that K ∈ Chev and p splits K. Then N is level unless possibly p = 2 and one of the following exceptional conditions holds: (a) K ∼ = 3D4 (q) or G2 (q) for some odd q; n (b) K ∼ = 2 G2 (3 2 ) for some odd n > 1; or (c) K ∼ = P Sp4 (q) or a non-universal version of L± 4 (q) for some odd q. The exceptions in Theorem 5 will be ruled out in the next monograph. The final theorem of this chapter is proved in Section 15. Theorem 6. Assume (1A) and (1B), with K ∼ = A (q) for some rank , some sign  and some prime power q. Assume also that p splits K and that N is level. Then one of the following holds: #

(a) For every u ∈ x, y , the pumpup Lu of L in CG (u) either is trivial or satisfies Lu /Z(Lu ) ∼ = K/Z(K); (b) L ∼ = A−2 (q); # (c) p = 2, K ∼ = A4 (q), and for some u ∈ x, y , Lu ∼ = B3 (q)u ∼ = Spin7 (q). Here  = ±1 and q ≡ − (mod 4); or # (d) p = 2, K ∼ = A2 (q), A− = A2 (q), and for any u ∈ x, y , Lu ∼ 2 (q), or A1 (q). Observe that conclusions (a), (b), (c), (d1), (d2), and (d3) of Theorem C∗7 : Stage 3b follow from Theorems 1, 3, 4, 2, 5, and 6, respectively. We make the important observation, justified by [III7 , Proposition 3.3], that (1C) If (1B) holds, then Theorem 1.2 of Chapter 7 applies to (x, K).

2. THEOREM 2: THE SETUP AND SPECIAL CASES

41

2. Theorem 2: The Setup and Special Cases In the next seven sections, we prove Theorem 2. Our focus, justified by Theorem 1, is on acceptable subterminal pairs. Thus we fix p ∈ γ(G), (x, K) ∈ J∗p (G), and an acceptable subterminal (x, K)(2A) pair (y, L). As usual, for each u ∈ x, y# , 

Lu is by definition the subnormal closure of O p (L) in CG (u). Unless p = 2 with K ∼ = SL± 4 (3), Sp4 (3), or Spin7 (3), and m2 (C(x, K)) = 1, in  ∼ which cases L = SL2 (3), we have that O p (L) = L is a p-component and Lu  its pumpup in CG (u). In the three exceptional cases, O 2 (L) = O2 (L). (See [III11 , 15.10].) In volume 5 of this series [GLS5], the following assertion was proved (see Theorem C∗7 , Stages 2 and 3a [III4 , p.167], [III5 , p. 274, (1D)]): (2B)

(2C)

The (y, L) neighborhood of (x, K) is semisimple and nontrivial.

We prove Theorem 2 by contradiction, thus assuming through Section 8 that (2D)

The (y, L)-neighborhood of (x, K) is not vertical.

By definition [III7 , 6.11], we obtain immediately: # Proposition 2.1. For some u ∈ x, y , one of the following holds: (a) Lu is the product of p components cycled by x; or (b) p = 2, L ∼ = SL2 (3), O2 (L) ≤ O2 (CG (u)), but L   LO2 (CG (u)). Moreover using [III7 , Proposition 3.3] we have Proposition 2.2. mp (C(x, K)) = 1. Proof. Let U ∈ Syl2 (C(x, K)) and suppose that the proposition fails. Then by [III7 , 3.3], p = 2, K ∼ = An , n ≥ 9, U ∼ = E22 , and for some g ∈ G, V := U g is a root four-subgroup of K. Since NK (V ) is transitive on V # , we may assume by [III7 , Def. 6.9c] that y = xg . But then by [III5 , Theorem 1d, p. 274], the subnormal closure of L in CG (u) is quasisimple for all u ∈ x, y# , contradicting Proposition 2.1.  In this section we treat a number of further special cases. Certain possibilities for K have already been eliminated in the course of the proof of Theorem C7 : Stage 1. Indeed, by [III3 , Theorem 9, p. 67], we obtain 3

Proposition 2.3. If p = 2, then K is not isomorphic to L3 (q), G2 (q) or n D4 (q) for any odd q and sign , or 2 G2 (3 2 ) for any odd n > 1, or J1 . Moreover, [III11 , Lemma 14.4b] quickly yields

Lemma 2.4. If K is of Lie type, then K and L have the same level, with the following three exceptions: n 2 ∼ (a) p = 3, K ∼ = A− 3 (q), q = 2 ,  = ±1, q ≡  (mod 3), and L = A1 (q ); n 2 n/2 2 (b) p ≥ 3, K ∼ = F4 (2 2 ), n > 1, q = 2 , p divides q ± 1, and L ∼ = A1 (q 2 ) − 2 or, for p = 3, A2 (q ); or (c) p ≥ 3, K ∼ = A1 (q 2 ). = D4 (q), q = 2n , p divides q 2 + 1, and L ∼

9. THEOREM C∗ 7 : STAGE 3B

42

∼ Spin− (q), x ∈ L ∼ Proof. Otherwise by [III11 , 14.4b], p = 2 with K = = 8 n 2 2 SL2 (q ), and x = Z(K) = Z(L); or p = 2 with K ∼ = G2 (3 2 ) for some n > 1; or p = 2 and K ∼ = 3D4 (q) for some odd q. The last two are ruled out by Proposition 2.3, so assume that K ∼ = Spin− 8 (q). By Proposition 2.1, Lu is a diagonal pumpup # of L, for some u ∈ x, y . In particular x interchanges the components of Lu so  x ∈ Lu . But x ∈ L ≤ Lu , a contradiction. Next we dispose of case (b) of Proposition 2.1, in which L and Lu are solvable. #

Proposition 2.5. Proposition 2.1b fails for every u ∈ x, y . ∼ SL2 (3), and Lu = O2 (L) ≤ Proof. Under the assumptions p = 2, L = O2 (CG (u)), we must prove that L  LO2 (CG (u)). Set V = O2 (CG (u)). Since O2 (L) ≤ V , certainly u = x, so x, u = x, y. Since L∼ = Sp4 (3), SL4 (3), or Spin7 (3), with = SL2 (3), we know by [III11 , 15.10] that K ∼ x, y ≤ K. We first consider the cases ∼ Sp4 (3) or SL (3). (2E) K= 4

∼ SL2 (3) and {Z(L), Z(L1 )} = {u, ux}; ∼ L1 = Then O 2 (CK (y)) = L × L1 , where L = moreover, L and L1 are K-conjugate, and in particular u and ux are K-conjugate. If x is G-conjugate to either u or ux, then it is conjugate to both and so V ∼ = O2 (CG (x)) ≤ CG (K), whence m2 (V ) = 1 by Proposition 2.2. But an element of L of order 3 acts nontrivially on O2 (L) and hence on V , so the only possibility is that V ∼ = Q8 , whence V = O2 (L) and the assertion of the proposition holds. We may therefore assume that x is weakly closed in x, y with respect to G, whence NG (x, y) ≤ CG (x). It suffices to prove that (2F)

x centralizes V .

For then as L   CG (x, u), V  CG (u) and V ≤ CG (x, u), we have  L   LV . Thus for any Sylow 3-subgroup S of L, [V, S] = [V, S, S] ≤ L[V,S] , so [V, S] ≤ L ∩ V = O2 (L). Hence [V, S] = O2 (L) and S centralizes V /O2 (L). As L = SO2 (L), L  LV , as desired. Set V0 = V O2 (L1 ) = V O2 (LL1 ) and V1 = CV0 (x) = CV (x)O2 (LL1 ). We shall show that V0 = V1 , which will imply (2F) and the lemma. To do this we shall show that x, y char V1 , from which it will follow that NV0 (V1 ) ≤ NV0 (x, y) ≤ CV0 (x) = V1 by the weak closure property proved above, yielding the desired equality. Clearly (2G)

x, y = x, u = Z(LL1 ) ≤ Ω1 (Z(V1 )) ∩ Φ(V1 ),

and we shall prove equality to complete the proof. Set V2 = O2 (CG (x, u)). Since V = O2 (CG (u)), we have CV (x) ≤ V2 , and it is also clear that O2 (LL1 ) = O2 (CK (u)) ≤ V2 . Thus V1 ≤ V2 . Set Q = O2 (CG (x)) and note that C(x, K) ≤ CG (x, u), so CV2 (K) ≤ O2 (C(x, K)) = Q. On the other hand, by [III11 , 6.11], O2 (CAutG (K) (u)) contains the image of O2 (LL1 ) as a normal subgroup with quotient of exponent 2, and therefore Φ(V2 /O2 (LL1 )Q) = 1. In particular Φ(V1 ) ≤ Φ(V2 ) ≤ O2 (LL1 )Q. Consequently Ω1 (Z(V1 )) ∩ Φ(V1 ) ≤ Ω1 (CO2 (LL1 )Q (O2 (LL1 ))) = Z(LL1 ), the last step since m2 (Q) = 1 and [Q, LL1 ] = 1. This establishes equality in (2G) and completes the consideration of the case (2E).

2. THEOREM 2: THE SETUP AND SPECIAL CASES

43

 Finally, suppose that K ∼ = Spin7 (3). Then again y ∈ K, but now O 3 (CK (y)) = LL0 L1 with L ∼ = L0 ∼ = L1 . We argue that

(2H)

x, y# ⊆ xG .

Notice that by [III11 , 6.7a], all involutions of K − Z(K) are conjugate, so it suffices to show that xG ∩ K ⊆ Z(K). Let S ∈ Syl2 (CG (x)). By [III11 , 6.8a], [Ω1 (Z(S)), K] = 1, and as m2 (C(x, K)) = 1, Ω1 (Z(S)) = Z(K) = x. Then S ∈ Syl2 (G). By the Z ∗ -Theorem there exist z ∈ I2 (S) and g ∈ G such that z g = x = z. If z ∈ K then (2H) holds, so assume that z ∈ K. Then as m2 (C(x, K)) = 1, [III11 , 6.8b] implies that CK (z) has a quasisimple component + J ∼ = Spin± n (3) for some n ≥ 4 (but not Spin4 (3)), with Z(J) = x. Since g g g z = x, J ≤ L2 (CG (x, x )) lies in a component I of CK (x) by L2 -balance, and xg  = Z(J g ) ≤ I. Moreover xg = x as z = x. If I = K, then I ≤ C(x, K), so x, xg  ≤ C(x, K), contradicting m2 (C(x, K)) = 1. Thus I = K and xg ∈ K, so (2H) holds. As a result CG (u) ∼ = CG (x). In particular m2 (O2 (CG (u))) = m2 (O2 (CG (x))) = 1. Since O2 (L) ≤ O2 (CG (u)) by assumption, and O2 (L) admits nontrivial action by an element of L of order 3, O2 (CG (u)) must be quaternion, and indeed isomorphic to Q8 . Therefore O2 (CG (u)) = O2 (L), so LO2 (CG (u)) = L and the proof is complete.  By Propositions 2.1 and 2.5, and Lp -balance, for some u ∈ x, y − x, Lu is a product of p components of CG (u) cycled (2I) by x, with L = Lp (CLu (x)) a diagonal of Lu . In particular, L is nonsolvable. We fix u and write H = Lu = H1 · · · Hp and H = H/Op (H),

(2J)

where H1 , . . . , Hp are the components of H. Since x does not normalize H1 , we have x ∈ Lp (CG (u))Op p (CG (u)), and hence by Lp -balance and solvable L2 -balance, (1) x ∈ Lp (CCG (x) (x, u)) = Lp (CCG (x) (x, y)) = Lp (CCG (x) (y)); and (2K) (2) If p = 2, then x ∈ Lo2 (CCG (x) (y)). (2L)

For the balance of the proof of Theorem 2, except in the proof of Proposition 6.1, we shall assume that (2I), (2J) and (2K) hold.

One further condition is easily established. Let Q ∈ Sylp (C(x, K)). Then CQ (y) centralizes both L and u, so CQ (y) normalizes H, in which L is a diagonal subgroup. Therefore by [IG , 6.19], CQ (y) = x CQ (y H). As mp (Q) = 1 and [x, H] = 1, we have CQ (y) = x, that is, (2M)

x ∈ Sylp (C(x, K) ∩ CG (y))

We can quickly dispose of the following cases. Proposition 2.6. If p = 2, then K/Z(K) is not isomorphic to M c, Ly, O  N , or Co3 .

44

9. THEOREM C∗ 7 : STAGE 3B

Proof. Suppose false. We use [III11 , 15.11, 3.12] to see that according to the isomorphism type of K, we have H1 /O2 (H1 ) ∼ = 2A8 , 2A11 , 4L3 (4) or 2Sp6 (2), and in all cases y is a Sylow 2-center in K. Since m2 (CG (K)) = 1, also K  CG (x) and so y is 2-central in CG (x). Thus we may choose T ∈ Syl2 (CG (x)) with x, y ≤ Z(T ). But by inspection of the tables [IA , 5.3], every involution of Z(T ) acts on K as an inner automorphism, and so Ω1 (Z(T )) = x, y. We shall now establish the following: (1) u = y; (2) x ∈ [CG (u), CG (u)]; (2N) (3) y is 2-central in G; and (4) xy is G-conjugate to x but not to y. From (2N4) we shall then have that y is weakly closed in Z(T ). Since x ∈ [CG (y), CG (y)] by (2N1, 2), condition (2N3) will allow us to apply [III8 , 6.3] and conclude that x ∈ [G, G]. This will contradict the simplicity of G and complete the proof of the proposition. To establish (2N), observe first that since y ∈ L = L2 (CH1 H2 (x)), the elements x and xy are conjugate by an element of H1 , by [IG , 3.27(v)]. On the other hand, x cannot be conjugate to u since L2 (CG (x)) has no component isomorphic to H1 . This proves (2N4), and also that u = xy, whence u = y and (2N1) holds. Next, suppose that (2N2) fails, so that x ∈ [CG (u), CG (u)]. Then as H1 = C (u) H1x , if we consider the permutation action of x on the set H1 G of CG (u)conjugates of H1 , we see that x cannot act as a transposition, so there exists C (u) J ∈ H1 G − {H1 , H2 } such that J = J x . Then L0 = L2 (CJJ x (x)) satisfies L0 /Z(L0 ) ∼ = H1 /Z(H1 ), and L0 ≤ CCG (x) (L) as L ≤ H1 H2 . Thus the image of L0 in Aut(K) lies in CAut(K) (L). However, by [III11 , 5.1c], CAut(K) (L) ∼ = Z(L) is cyclic. Therefore L0 ≤ CG (K), a contradiction as m2 (CG (K)) = 1. Thus (2N2) holds. Finally as x, y = Ω1 (Z(T )), we have T ≤ Ty for some Ty ∈ Syl2 (CG (y)), and as H1x = H2 = H1 , certainly x ∈ Z(Ty ). But Z(Ty ) ≤ CTy (x) = T and so Ω1 (Z(Ty )) = y. A Sylow 2-center of G must be contained in Z(Ty ), so y is 2-central in G, whence (2N3) holds and the proof is complete.  Corollary 2.7. If p = 2, then d2 (G) ≥ 3 and d2 (G) = 7. Proof. The only G2 -groups of G2 -depth 1 or 2 are 2An , n ≥ 12, M c, Ly, and O  N ; and those of depth 7 are L3 (q), q > 3, q odd,  = ±1 (see [III7 , (1E)]). By [III7 , Theorem 1.2] and Propositions 2.3 and 2.6, K/O2 (K) is of none of these isomorphism types. However, (x, K) ∈ J2 (G), so by definition of this term [III7 , 1.8c],  K/O2 (K) has G2 -depth d2 (G). The result follows. Since the existence of 2-components of type 2An , n ≥ 7, or of type XL3 (4), with X of exponent 4, would imply that d2 (G) ≤ 2 by [III3 , Lemma 5.6a], we also have the following corollary. Corollary 2.8. If p = 2 and (g, J) ∈ ILo2 (G), then J/O2 (J) ∼ = 2An for any n ≥ 7; and if J/O2 2 (J) ∼ = L3 (4), then Z(J/O2 (J)) is trivial or elementary abelian.

3. THE CASES K ∼ = An OR K ∼ = He, WITH p = 2

45

3. The Cases K ∼ = An or K ∼ = He, with p = 2 We continue the proof of Theorem 2, assuming as stipulated in the previous section that (2I), (2J) and (2K) hold, so that (3A)

for some u ∈ x, y − x, the pumpup H of L in CG (u) is diagonal,

being the product of of p components H1 , H2 , . . . , Hp cycled by x. Also, L is a component of CH (x). By Proposition 2.2, (3B)

m2 (C(x, K)) = 1.

In this section we treat the cases in which p = 2, and either K ∼ = He or K ∼ = An for some n. As K is a G2 -group, n ≥ 9 in the latter case (see [I2 , 13.1]). By definition of an acceptable subterminal pair [III7 , 6.9], and with (3B) when K ∼ = He, one of the following holds in these cases, after replacing y by xy if necessary: (1) K ∼ = He, L ∼ = (2 × 2)L3 (4), and y ∈ L is a non-2-central involu(3C) tion of K; (2) K ∼ = An−4 and y is a root involution of K. = An , n ≥ 9, L ∼ Proposition 3.1. Suppose that p = 2, and K ∼ = An or He. Then (3A) does not occur. We establish the proposition by contradiction in a sequence of lemmas, so we assume that (3C1) or (3C2) holds, along with (3A). We know that L is isomorphic to a homomorphic image of H1 by [IG , 3.27(v)]. But by [IA , 6.1.4], the Sylow 2subgroup of the Schur multiplier of L/Z(L) is isomorphic to Z4 ×Z4 or Z2 according as (3C1) or (3C2) holds. Thus if H 1 ∼ = L, then Z(H 1 ) has exponent 4 with ∼ ∼ H 1 /Z(H 1 ) = L3 (4), or H 1 = 2An−4 . By Corollary 2.8, this can only happen if K∼ = A9 or A10 . Thus in any case, one of the following holds: (1) H = H 1 × H 2 with H 1 ∼ = H2 ∼ = L; or ∼ (3D) (2) K = An , n = 9 or 10, and H = H1 H2 with H 1 ∼ = H2 ∼ = 2An−4 and Z(H) = Z(H 1 ) = Z(H 2 ). We first introduce some notation that will be needed for the analysis. In view of [III11 , 4.3] and the fact that m2 (CG (K)) = 1 (3B), we have (1) E = O2 (CK (L)) ∼ = E22 and y ∈ E x − x; (2) NK (L) contains an element v of order 3 such that E v ∼ = A4 , with v ∈ CK (L) unless K ∼ = He; (3E) (3) E x contains every involution in CG (LE x); and (4) There is w ∈ K (with w = 1 if K ∼ = He) such that w2 = 1 and ∗ w E = E ≤ L, and if K ∼ = An then v w ∈ L. We fix the notation E, v, E ∗ and w of (3E). We first consider the general case of (3D) and assume through Lemma 3.16 that (3D1) holds. x

Since L is a diagonal of H 1 ×H 1 , with E ∗ ≤ L, it follows in this case from [IG , 10.31] that all involutions of E ∗ x − E ∗ are H-conjugate to x. But w conjugates E ∗ to E and centralizes x. Thus (3F)

assuming (3D1), all involutions of E x − E are G-conjugate to x.

We use (3F) to prove

46

9. THEOREM C∗ 7 : STAGE 3B

Lemma 3.2. Assume that (3D1) holds. Then (a) E centralizes H; (b) u ∈ E; and (c) H1 and H2 are components of CG (e) for each e ∈ E # . ∼ E23 centralizes L, E x leaves H invariant and some Proof. Since E x = four-subgroup F of E x leaves both H1 and H2 invariant. Then E = F × x, and since F centralizes the projection of L on H i , i = 1, 2, it follows that F = CEx (H). In particular u ∈ F . −1 For any f ∈ F # , H1 H2   CG (f, u). If f = xg for some g ∈ G, then ug ∈ −1 −1 I2 (CG (x)) and (H1 H2 )g   CG (x, ug ). Since m2 (CG (K)) = 1 but m2 (Hi ) > 1, −1 −1 i = 1, 2, L2 -balance would then imply (H1 H2 )g   CK (ug ), an impossibility ∼ as centralizers of involutory automorphisms of K = An or He have at most one component [IA , 5.2.8, Table 5.3p]. Therefore f is not G-conjugate to x, and (3F) implies that F ≤ E. Therefore F = E, proving (a) and (b). Now v permutes E # transitively and normalizes L, so for each e ∈ E # , the pumpup He of  L in CG (e) is isomorphic to H. However, [e, H] = 1 and so He contains LH = H. Therefore He = H, and (c) holds as well.  Lemma 3.3. H1 and H2 are the only components of CG (E) isomorphic to L. Moreover, H  NG (E). Proof. Lemma 3.2c implies that H1 and H2 are components of CG (E). Suppose that the first assertion fails, so that the product J of all components of CG (E) isomorphic to L and distinct from H1 or H2 is nontrivial. Then J centralizes H, J is x-invariant and Ex ∩ J ≤ CEx (H) = E, so Ex ∩ J ≤ Z(J). If Z(J) = 1, then x centralizes some z ∈ I2 (J) and z ∈ / Ex. But z ∈ J centralizes L ≤ H and hence (3E3) is contradicted. The only other possibility is that L ∼ = (2 × 2)L3 (4), in which case clearly x centralizes some z ∈ I2 (J − Z(J)) if x moves some component of J, and the same follows from [III11 , 2.6] if x leaves each component of J invariant. Again z ∈ E x and we reach the same contradiction. This proves the first  statement of the lemma, and the second follows immediately as H = H1 H2 . We shall be examining pumpups of H1 , eventually contradicting the following lemma. Lemma 3.4. H1 is not terminal in G. Proof. If H1 were terminal in G, then as m2 (H1 ) > 1, H1 would be standard in G by [II3 , Theorem PU4 ]. But H1 and the conjugate H1x = H2 commute  elementwise, so by definition H1 is not standard in G. To analyze pumpups of (u, H1 ), we first prove: Lemma 3.5. Let t ∈ I2 (CG (H1 E)) and let J be the pumpup of H1 in CG (t). Assume either that H1 ∼ = 22 L3 (4) and J ∼ = An−4 and J ∼ = He, or that H1 ∼ = Ar ∼ with r ≥ 11 and r ≥ n. Then J = K and m2 (C(t, J)) = 1. Proof. By [IG , 6.10], (t, J) 22 by assumption, this is a contradiction and the proof is complete.  In dealing with the cases K ∼ = An , n = 9, 10, or 11, we shall also need the following sharpening of Lemma 3.2. ∼ An , n = 9, 10, or 11, then m2 (CG (HE)) ≥ 4 and Lemma 3.8. If K = m2 (CG (H1 E)) ≥ 6. Proof. In these cases L ∼ = An−4 has dihedral Sylow 2-subgroups. Since H2 × CG (HE) ≤ CG (H1 E), the second assertion is immediate from the first, so it suffices to prove that m2 (CG (HE)) ≥ 4. Recall from (3E) that v ∈ CK (L) has order 3 and acts nontrivially on E, while w ∈ K is an involution with E ∗ = E w ≤ L and v w ∈ L. Set v ∗ = v w . Now L ∗ is a diagonal of H = H 1 × H 2 , and Hi is quasisimple. Hence if E i denotes the ∗ projection of E on H i , i = 1, 2, then there exists B ∗ ≤ H such that B ∗ ∼ = E24 , ∗ ∗ ∗ B = E 1 × E 2 , and E ∗ ≤ B ∗ = [B ∗ , v ∗ ]. Set B = (B ∗ )w . Since v ∗ = v w and w2 = 1, we conclude that B ∼ = E24 and E ≤ B = [B, v]. Then B v ≤ NG (E). Hence B v normalizes H by Lemma 3.3. But B v = O 2 (B v) so B v normalizes H1 and H2 . As v centralizes L, it follows that v centralizes H, whence B = [B, v] does as well. Therefore m2 (CG (HE)) ≥ m2 (B) = 4 and the proof is complete.  Now let us aim to prove that H1 is terminal in G. We first work under the following hypothesis, which is quite general in view of Lemma 3.6: (3H)

H1 is a component of CG (t) for each t ∈ I2 (H2 E).

We set X = CG (H1 ), so that H2 E ≤ X. We first prove Lemma 3.9. Assume (3H). Then H2 is a component of X. Proof. Conjugating by x, we see that it is enough to show that H1 is a component of CG (H2 ). By (3H), H1   CG (t) for any involution t ∈ H2 . Intersecting  with CG (H2 ) we get H1   CG (H2 ), as required. Now we use [III8 , 2.16b] and Lemma 3.7 to prove: Lemma 3.10. Assume (3H). Then H1 is terminal in G. Proof. We let X1 = H2 CX (H2 ), the subgroup of X inducing inner automorphisms on H2 . Also let R ∈ Syl2 (X), R1 ∈ Syl2 (X1 ) and A ∈ Syl2 (H2 E) with A ≤ R1 ≤ R. By (3H), H1 is a component of CG (a) for all a ∈ I2 (A). We shall apply [III8 , Lemma 2.16b] to deduce the terminality of H1 . Thus we need only show that A is (H1 , p)-terminating in CG (H1 ) = X. We verify the definition [III8 , 2.15]; for A, using the sequence A, H2 E, Y , X1 , X, where Y = H2 EΩ1 (Z(R1 )). By Lemma 3.7 any 2-group of rank at least 2 and order at least 8 detects pumpups of H1 in G, so the proof will be complete once we

3. THE CASES K ∼ = An OR K ∼ = He, WITH p = 2

49

verify the following statements. (1) For any u ∈ I2 (Y ), m2 (CH2 E (ug )) ≥ 3 for some g ∈ X; (2) For any u ∈ I2 (X1 ) − Y , m2 (CY (ug )) ≥ 3 for some g ∈ X; and (3I) (3) For any u ∈ I2 (X) − X1 , m2 (CX1 (ug )) ≥ 3 for some g ∈ X, or m2 (CX1 (ug )) ≥ 2 with |CX1 (ug )|2 ≥ 8. By [III11 , 3.8, 2.6ab], m2 (CH2 (u)) ≥ 2 for every involution u inducing an inner automorphism on H2 ; moreover, strict inequality holds unless H2 /O2 (H2 ) ∼ = An−4 with n − 4 = 5, 6 or 7. Thus (3I1) and (3I2) hold except possibly in these last three cases. Even in those cases, since E ≤ A ≤ R1 we have CH2 E (u) = CH2 (u)E for any involution u ∈ Y , and so m2 (CH2 E (u)) > m2 (CH2 (u)) and (3I1) holds. Likewise C := CY (u) contains CH2 (u) × (Ω1 (Z(R1 )) ∩ CX (H2 )) for any u ∈ I2 (R1 ) (and we may assume that u ∈ I2 (R1 ) by Sylow’s Theorem). Therefore m2 (C) ≥ 3, and (3I2) is proved. Finally to prove (3I3), let u ∈ I2 (X). Suppose first that H2u = H2 , so that u centralizes a diagonal I of H2 H2u , and I ≤ X1 . Thus m2 (CX1 (u)) ≥ m2 (I), so (3I3) holds as long as m2 (I) ≥ 3. Again as I is a homomorphic image of H2 by [IG , 3.27], we have m2 (I) ≥ 3, as desired, by [III11 , 2.6,3.7] unless K ∼ = An with n = 9, 10 or 11. But in these three exceptional cases H2 /O2 (H2 ) ∼ = An−4 and so m2 (Aut(H2 )) ≤ 3 by [IA , 5.2.1, 5.2.7a, 5.2.10b]. However, by Lemma 3.8, m2 (CX (H2 )) ≥ 4, and so CX (H2 H2u ) has even order. Thus u centralizes an involution b ∈ CX (H2 H2u ), whence b ∈ X1 and m2 (CX1 (u)) ≥ m2 (b I) ≥ 3, as required. We may therefore assume that u normalizes H2 , but u induces a non-inner automorphism on H2 , and m2 (CX1 (u)) ≤ 2 by the terms of (3I3). Again m2 (CH2 (u)) ≥ 3, a contradiction, by [III11 , 2.6c, 3.7], unless H2 /O2 (H2 ) ∼ = An−4 with n − 4 = 5, 6, or 7. Now u normalizes CG (H1 H2 ), and hence normalizes a Sylow 2-subgroup R2 of it. By Lemma 3.8, m2 (R2 ) ≥ 4. In particular u R2 is not of maximal class and so |CR2 (u)| > 2. But then CX1 (u) contains CH2 (u)CR2 (u), of 2-rank at least 2 and order at least 8. This completes the proof of (3I3) and the lemma.  This immediately yields: Lemma 3.11. We have K ∼ = A9 or A10 . Moreover, for some t ∈ I2 (H2 E), the pumpup J of H1 in CG (t) satisfies one of the following: (a) K ∼ = A9 and J/Z(J) ∼ = M12 or J2 ; or (b) J ∼ K. = In either case some element of E # induces an inner automorphism on J. Proof. By Lemma 3.4 and the preceding lemma, (3H) does not hold. Then by Lemma 3.6, K ∼ = A9 or A10 , either (a) or (b) holds, and the final assertion holds.  We fix t and J as in Lemma 3.11, and argue in a similar fashion. Although H1 is not terminal, we shall see that J is terminal. Since K ∼ = A9 or A10 , we have d2 (G) = 8 as (x, K) ∈ J2 (G), and so Lo2 (G) ⊆ C2 ∪ T2 ∪ {A9 , A10 }.

9. THEOREM C∗ 7 : STAGE 3B

50

By L2 -balance and [III11 , 13.23bc, 1.6] this restricts the possible pumpups of J as follows. For any a, a1 ∈ I2 (CG (J)) such that J is a component of CG (a) and [a, a1 ] = 1, the subnormal closure J1 of J in CG (a1 ) is either a diagonal (3J) or trivial pumpup of (a, J), or else J1 /O2 2 (J1 ) ∼ = Suz, with (C(a, J) ∩ CG (a1 ))/(CG (a) ∩ C(a1 , J1 )) ∼ = Z2 . We again set X = CG (H1 ) and let R ∈ Syl2 (X) with Et ≤ R. By Lemma 3.8, m2 (X) ≥ 6, so R is connected by [III8 , 1.5]. In particular as H1 is a component of CG (e) for all e ∈ E # , (3K)

the pumpup of H1 in CG (a) is semisimple for all a ∈ I2 (R)

by [III8 , Lemma 2.12b]. Next, in CX (H2 E) = CG (HE), we may choose a subgroup B0 ∼ = E24 by Lemma 3.8, and we may take E ≤ B0 . Observe that as t ∈ H2 E, we have m2 (CH2 (t)) = 2, whence B0 lies in some B ∼ = E26 such that [B, t H1 ] = 1. As B ≤ CG (t) and B centralizes H1 , B leaves J invariant. But by [III11 , 1.16], m2 (CAut(J) (H1 )) = 2, so m2 (CB (J)) ≥ 4. Choose Q ∈ Syl2 (CG (J)) with t ∈ Q and Q invariant under B. Thus (3L)

m2 (Q) = m2 (CG (J)) ≥ 4.

We use these facts in conjunction with [III8 , Lemma 2.21] to prove Lemma 3.12. J is terminal in G. Proof. If the lemma fails, then the previous paragraph and (3J) permit us to apply [III8 , Lemma 2.21], and conclude that there is a ∈ I2 (Z(Q)) and a nontrivial pumpup (t, J) < (a, I); the other alternative of that lemma, that Q is dihedral or semidihedral, is ruled out by (3L). We fix a and set I1 = II t . By (3J), in fact, I1 /Z(I1 ) ∼ = J/Z(J) × J/Z(J) or Suz, with J/Z(J) ∼ = M12 or J2 in the latter case. In any case, Q normalizes I1 and |Q/CQ (I1 )| ≤ 2, by [IG , 6.19] or (3J) according as I1 is a diagonal or vertical pumpup of J. Therefore Q = CQ (I1 ) t. On the other hand E normalizes Q, and some e ∈ E # induces an inner automorphism on J. Thus e ∈ JQ so e induces an inner automorphism on Q and in particular [e, Z(Q)] = 1. Write E = e, f  and C = CQ (I1 ) ∩ CQ (I1 )f . Then C  Q and |Q : C| ≤ 4, so C = 1 by (3L). Therefore f centralizes some element z ∈ C ∩ Z(Q) ≤ CZ(Q) (I1 ). By our choice [E, z] = 1 and clearly the subnormal closure I ∗ of J in CG (z) contains I1 . Thus (3M) the pumpup of J in CG (z) is nontrivial. The possibilities for I ∗ are given in the previous paragraph and the only choice is that I ∗ = I1 . In particular E ≤ CG (z) and E normalizes I1 . We have the vertical pumpup (f, H1 ) < (t, J) and the further pumpup (t, J) < (z, I). Moreover [E, z] = 1 and  [z,H1 ] ≤ [z, J]= 1. The subnormal closure of H1 in CG (z) then contains J = H1J and I1 = J I1 . Hence (f, H1 ) < (z, I). By Lemma 3.6, applied to (z, I) instead of (t, J), we see that I = I1 , with I ∼ = H1 , I∼ = M12 or J2 . =K∼ = A9 or A10 , or I/Z(I) ∼ But there are no involvements among the groups A9 , M12 and J2 , as can be seen by Lagrange’s Theorem. Therefore comparing the possibilities for I with those for J in Lemma 3.11, we deduce that I/Z(I) ∼ = J/Z(J). Since I = I1 , the pumpup (t, J) < (a, I) is therefore trivial, a contradiction. The proof is complete. 

3. THE CASES K ∼ = An OR K ∼ = He, WITH p = 2

51

With Lemma 3.12 we can eliminate the alternative K ∼ = J in Lemma 3.11. For ∼ J, then J ∈ G8 , whence (t, J) ∈ J2 (G). Now m2 (CG (K)) = 1 by Proposition if K = 2 2 2.2, so by definition of J22 (G), m2 (CG (J)) = 1 as well. However, this contradicts (3L). Thus we have proved: Lemma 3.13. We have K ∼ = M12 or J2 . = A9 and J/Z(J) ∼ We use the techniques in [IG , Section 18] to finish off these two residual cases. We know that J is standard in G by Lemma 3.12 and [II3 , Theorem PU4 ]. Let M be a maximal subgroup of G containing NG (J), and continue to let Q ∈ Syl2 (CG (J)), so that m2 (Q) ≥ 4 by (3L) and ΓQ,1 (G) ≤ NG (J) by [IG , 18.7]. Now Q ≤ M , so O2 (M ) ≤ ΓQ,1 (G) ≤ NG (J) by [IG , 11.23], whence [O2 (M ), J] = 1. Hence by L2 -balance, the subnormal closure J ∗ of J in M is a product of components of M , and satisfies the condition that J   CJ ∗ (a) for all a ∈ I2 (Q). But since J/Z(J) ∼ = M12 or J2 , with m2 (Q) > 1, we conclude with [III11 , 1.16] that J = J ∗ . Thus J   M and as J is standard, J  M and M = NG (J). Since J is standard and m2 (Q) ≥ 2, M is a J-preuniqueness subgroup and satisfies the 2-Component Preuniqueness Hypothesis [II3 , 1.1, (1A)]. By [II3 , Theorem PU2 ], therefore, there is E22 ∼ = U ≤ Q and h ∈ G − M such that U h ≤ M . We choose g ∈ G − M such that V = Qg ∩ M has maximal rank, and subject to this (3N) so that the order of V is maximal. Replacing g by a suitable element of gM , we may assume without loss that V, Q is a 2-group. Since U h exists, m2 (V ) ≥ 2, and by [IG , 18.8], ΓV,1 (J) < J. It follows therefore from [III11 , 5.7a] that ∼ E22 acts faithfully on J, with E(CJ (V )) ∼ V = = A5 . In particular, m2 (V ) = 2, so as m2 (Q) ≥ 4, ∼ Q. (3O) V = By (3O), [IG , 18.12] and our choice of V , we immediately conclude: Lemma 3.14. The following conditions hold: (a) For any b ∈ G − M , Qb ∩ J = 1; and (b) For any v ∈ V # , [CJ (v), V ] has odd order. Now we can reduce to a single case. Lemma 3.15. We have J ∼ = M12 . ∼ A5 . Thus I ≤ ΓV,1 (G) ≤ M g . If I centralizes Proof. Set I = L2 (CJ (V )) = g J , then as Q is a Sylow 2-subgroup of CG (J g ), we have I ∩ Qgw = 1 for some w ∈ CG (J g ). As I ≤ J, Lemma 3.14a implies that gw ∈ M . Hence J = J gw = J g , whence g ∈ M , contradiction. Therefore [I, J g ] = 1, and as I   CG (V CZ(Q) (V )) with V CZ(Q) (V ) ≤ M g and J g  M g , L2 -balance implies that I ≤ J g . Then by [IA , Tables 5.3bg], CAut(J g ) (I) has Sylow 2-subgroups which are elementary abelian of order at most 4. Suppose now that Z(J) = 1, so that J ∼ = 2M12 or 2J2 . As |Out(J)| = 2 by [IA , Tables 5.3bg], some v ∈ V # induces a (nontrivial) inner automorphism on J. But v centralizes I and by [IA , Tables 5.3bg], CJ (I) has Z4 or Q8 -Sylow 2-subgroups, respectively, each containing Z(J). It follows that v centralizes an g

9. THEOREM C∗ 7 : STAGE 3B

52

  element u of order 4 in CJ (I) with u2 = Z(J). Then u ∈ CG (v) ≤ M g , so u leaves J g invariant. But u centralizes I, so by the last sentence of the preceding paragraph, u2 centralizes J g , and then g ∈ NG (J) = M by [IG , 18.6], as J is standard in G. This is a contradiction, and we conclude that Z(J) = 1, whence J∼ = M12 or J2 . If J ∼ = J2 , then J is outer generated for 2 by [III11 , 5.7b]. But then by [IG , Prop. 18.11] there is b ∈ G − M such that Qb ≤ M , so by our choice of V ,  V ∼ = Q, contradicting (3O). Hence J ∼ = J2 and the proof is complete. The group M12 is not outer generated for 2, so we require a slight variation of the preceding analysis to force V ∼ = Q. Indeed, as before, some v ∈ V # induces an inner automorphism on J, and if we let u be the projection of v on J, then CJ (v) = CJ (u) = u × I w ∼ = Z2 × Σ5 , where w is an involution (see [IA , Table 5.3b]). Write V = v, v  , so that v  acts on CJ (v). By [IA , Table 5.3b], v  induces a noninner automorphism on J and CJ (u) v   /I ∼ = D8 . Hence [CJ (v), V ] has even order, contradicting Lemma 3.14b. This completes the proof of Proposition 3.1, assuming that (3D1) holds. That is, we have shown: Lemma 3.16. (3D2) holds. In this situation we let w ∈ K ∼ = A9 or A10 be an involution which is the product of four disjoint transpositions. Recall from (3C2) that y is a root involution of K. We choose w ∈ yL, as we may, writing w = yy ∗ with y ∗ an involution of L. We also set z = Z(H1 ). Note that w lies in a (regular) Q8 -subgroup of CK (w), so w ∈ [CG (w), CG (w)]. Lemma 3.17. The following conditions hold: (a) C(x, K) = x × O2 (CG (x)); (b) w ∈ xG ; (c) z = y = u; and (d) For any component I of CG (y) other than H1 and H2 , x normalizes I and CI (x) is solvable. Proof. Since y ∈ K, (2M) implies that x ∈ Syl2 (C(x, K)), which in turn implies (a). Then CG (x)/K has Sylow 2-subgroups of order at most 4 and containing the image of x as a central involution. By [IG , 15.12], CG (x)/K has a normal 2-complement; in particular x ∈ [CG (x), CG (x)]. As w ∈ [CG (w), CG (w)], (b) follows. Now CG (L x)/C(x, K) embeds in CAut(K) (L) ∼ = Σ4 , and so CG (L x) is solvable. Since L ≤ H1 H2 , the group CG (H1 H2 x) is solvable. In particular x, which interchanges H1 and H2 , must normalize all other components of E(CG (u)), by [IG , 3.27], and (d) holds with u in place of y. It therefore remains to prove (c). Since y ∗ ∈ L, we have y ∗ = aax for some a ∈ H1 , by [III8 , 4.3]. Since y ∗ is an involution while H1 ∼ = SL2 (5) or SL2 (9), a must have order 4 with a2 = z. Then (3P)

xa = a2 aax x = a2 y ∗ x = xy ∗ z.

Now CG (x, u) normalizes L, hence CG (x, u) normalizes H1 H2 and then centralizes z. But CK (u) = (L × E) b where L b ∼ = Σ5 or Σ6 , and E b ∼ = D8 , so O2 (Z(CG (x, u))) = x, y. Therefore z ∈ x, y. As H1 = H1x , z = x. Suppose that z = xy. Then by (3P), xa = yy ∗ = w, contradicting (b). Thus, z = y. Finally  u ∈ Cx,y (H1 ) = z, completing the proof of (c) and the lemma.

3. THE CASES K ∼ = An OR K ∼ = He, WITH p = 2

53

  Recall that the four-group E = O2 (CK (L)) normalizes LL2 (CG (u)) = H1 H2 , as does x, and so by [IG , 6.19], if we put F = CEx (H1 H2 ), then E x = F × x. We next prove Lemma 3.18. The following conditions hold: (a) F = E; (b) y is G-conjugate to w; and (c) H1 H2  CG (y). Proof. If F = E, then H1 and H2 are components of CG (E). By (3E2), NK (E)∩CK (L) contains a subgroup E v ∼ = A4 . Since [v, L] = 1, v must normalize H1 H2 . Therefore v normalizes E ∩ Z(H1 H2 ) = y, contradicting E v ∼ = A4 . This proves (a). Write E = y, y  . As F contains y but not x, F = y, y  x. Consequently y  and x have the same action on H1 H2 , and then as in (3P) we have (y  )a = zy ∗ y  = y ∗ yy  . Since y ∗ is a root involution of L and yy  ∈ E # , y ∗ yy  is K-conjugate to w, so y  is as well, and (b) follows. Finally suppose that (c) fails, so that CG (y) has a third component I ∼ = H1 with y ∈ I. By Lemma 3.17d, x normalizes but does not centralize I, so by [III11 , 6.3a], there is g ∈ I such that xg = xy. But y and y  are K-conjugate, so x, xy and xy  are all G-conjugate, and in particular CG (xy  ) has a unique nonsolvable composition factor, namely K. However, xy  ∈ F so xy  centralizes H1 H2 , and so H1 H2 / y ∼ = An−4 × An−4 must be involved in K. As this is impossible by the structure of 3- and 5-subgroups, we have a contradiction and the lemma follows.  Now we can prove Lemma 3.19. n = 10 and some G-conjugate of y acts on K as a transposition. Proof. By Lemma 3.18b we may write y = wg for some g ∈ G. We set x = xg and consider the subgroup CK g (y) = CK (w)g , whose structure is given by [III11 , 4.4]. Thus, CK g (y) ∼ = CK (w) contains a normal subgroup Rt such that (1) R = [t, R] ∼ = Q8 ∗ Q8 , t3 = 1 and Z(R) = y; and (2) A Sylow 2-subgroup of CAut(K g ) (R) is generated by the image (3Q) of y and an element b, and either b = 1 or n = 10 with b a transposition in Aut(K g ) ∼ = Σ10 . 

Moreover, since |C(x, K)|2 = 2 by Lemma 3.17, CK g (y) is solvable and CK g (y) has no subnormal subgroup isomorphic to SL2 (3). As O 2 (CG (x )) = K g × O2 (CG (x )), CG (x , y) is solvable and has no subnormal SL2 (3)-subgroup. In particular x normalizes every component of CG (y), so x normalizes H1 and H2 , with CHi (x ) ∼  SL2 (3), i = 1, 2. By [III11 , 6.3cd], CAut(Hi ) (x ) therefore has a = normal 2-complement. Now Rt = O 2 (Rt) normalizes H1 and H2 by Lemma 3.18c, and as R t centralizes x , the image of R t in Aut(Hi ), i = 1, 2, has a normal 2-complement by what we just saw. But [t, R] = R, so [R, Hi ] = 1, i = 1, 2. Finally let Q ∈ Syl2 (CH1 H2 (x )). Then [Q, R] = 1 and the image of Q in Aut(K g ) therefore is elementary abelian of order 2 or 4 by (3Q2). In view of Lemma 3.17a, |Q x  | ≤ 8. This forces x ∈ Q, for otherwise x would lie in a Q8 ∗ Q8 subgroup Q0 of H1 H2 and so |Q| ≥ |CQ0 (x )| ≥ 16, a contradiction. Consequently Q acts faithfully on K g and |Q| ≤ 4. But |Q| > 2 as Sylow 2-subgroups of H1 H2 are not of maximal class (see [IG , 10.24]), so Q = y, s for some involution s, and

9. THEOREM C∗ 7 : STAGE 3B

54

(3Q2) implies that n = 10 and s may be chosen to act as a transposition on K g . Since all involutions of H1 H2 − y are H1 H2 -conjugate, s has a G-conjugate in L −1 and hence s is conjugate to y. Then sg is an involution satisfying the conclusion of the lemma.  Now we can complete the proof of Proposition 3.1. By Lemma 3.19 there exists a G-conjugate x of x such that I = L2 (CG (x , y)) ∼ = A8 , and Sylow 2-subgroups of CCG (x ) (I) are four-groups in view of Lemma 3.17a. By L2 -balance I has a pumpup in CG (y), which must be a product of 2-components other than H1 and H2 , since I is not involved in H1 H2 . Therefore H1 H2 ≤ CCG (y) (I). Lemma 3.18c implies that x normalizes H1 H2 , and we see that x  CH1 H2 (x ) ≤ CCG (x ) (I). Thus Sylow 2-subgroups of x  CH1 H2 (x ) are embeddable in four-groups, whence H1 H2 x  has Sylow 2-subgroups of maximal class by [IG , 10.24]. This is absurd since H1 ∼ = H2 ∼ = SL2 (9), and the proposition is at last proved. 4. The F5 Case Again we assume that (2A), (2I), (2J) and (2K) hold. Here we complete the analysis when p = 2 and K ∈ Spor by showing: Proposition 4.1. If p = 2, then K ∼  F5 . = We prove Proposition 4.1 by contradiction in a sequence of lemmas, assuming that K ∼ = F5 . Since (y, L) is acceptable, y∈L∼ = 2HS, by [III7 , Definition 6.9]. Since x interchanges H1 and H2 with diagonal L, the Schur multiplier of HS is Z2 by [IA , 6.1.4], it follows with the aid of [IG , 3.27(vi)] that (1) H = H1 × H2 with H1 ∼ = H2 ∼ = L; and (4A) (2) If we set yi  = Z(Hi ), i = 1, 2, and Y = y1 , y2 , then Y ∼ = E22 and y = y1 y2 . Since H1x = H2 we must have y = Nx,y (H1 ), and then since H1 ≤ CG (u), we have (4B)

u = y.

The proof of Lemma 3.4 can be repeated, and we again conclude that (4C)

H1 is not terminal in G.

As in the previous section this will lead to a contradiction. We expand Y to Q ∈ Syl2 (C(u, H1 )), and Q to S ∈ Syl2 (CG (H1 )). Then m2 (S) ≥ m2 (Q) ≥ m2 (H2 ) = 5 by [IA , 5.6.1], and so by [III8 , 1.5] Q and S are connected. With [III8 , Lemma 2.17], this quickly implies: Lemma 4.2. There is v ∈ Y # such that H1 is not a component of CG (v). Proof. By [III11 , Lemma 17.9], E22 detects pumpups of 2HS, and we have just observed that S is connected. If H1   CG (v) for all v ∈ Y # , then since Y ∼ = E22 , [III8 , 2.17] would apply, yielding the terminality of H1 . However, this would contradict (4C), so the lemma follows. 

4. THE F5 CASE

55

Fix v as in Lemma 4.2 and let Hv be the subnormal closure of H1 in CG (v), so that Hv = H1 . Set H v = Hv /O2 (Hv ). Lemma 4.3. Hv = H1 O2 (Hv ). Moreover, [H1 , O2 (CG (v))] = 1. Proof. Assume first that Hv = H1 O2 (Hv ). Since u ∈ Y ≤ H1 H2 ≤ L2 (CG (Y )) ≤ L2 (CG (v)), u normalizes every 2-component of CG (v). Thus (v, Hv ) is a vertical pumpup of (u, H1 ). By [III11 , 13.23d], H v = Hv /O2 (Hv ) ∼ = F5 , and so by [IA , Table 5.3w], m2 (CAut(H v ) (H 1 )) = 1. But as Q is connected and m2 (Q) > 2, there is E ∈ E3 (Q) such that Y ≤ E. Then m2 (E/CE (H v )) ≤ 1, whence m2 (C(v, Hv )) ≥ m2 (CE (H v )) ≥ 2. By [IG , 6.11], there exists a 2-terminal (a, J) ∈ ILo2 (G) such that (v, Hv ) 1. Since J/O2 (J) ∼ = F5 ∼ = K, (a, J) ∈ J2 (G). However, this contradicts [III7 , 5.1], so Hv = H1 O2 (Hv ). Since Hv = H1 , the second assertion of the lemma follows immediately.  Next, let R ∈ Syl2 (CK (u)). Since u = y by (4B), and y ∈ K is a non-2central involution of K by [IA , Table 5.3w], there exists g ∈ NK (R) − R such that g 2 ∈ NK (R), and we set u0 = ug , U = u, u0  and z = uu0 . Then U ≤ Z(R). By [III11 , 5.6a], U = Z(R) ≤ L. Set Wv = O2 (CG (v)). Since z is a diagonal involution of H1 × H2 ≤ CG (v), and [H1 , Wv ] = 1 by Lemma 4.3, we have (4D)

[z, Wv ] = 1.

However, we contradict this by proving: Lemma 4.4. For any W ∈ IG (R; 2 ), we have [z, W ] = 1. Proof. By [III11 , 5.6b], we have IL (R ∩ L; 2 ) = 1. Since L is a diagonal of H1 H2 , it follows that IH1 H2 (R; 2 ) = 1. Now set W0 = [CW (u), z], so that W0 = [W0 , z] by [IG , 4.3(i)], and W0 ∈ IG (R; 2 ) since U ≤ Z(R). But z ∈ L ≤ H1 H2   CG (u) and so W0 = [z, W0 ] ∈ IH1 H2 (R; 2 ). Therefore, W0 = 1. We have shown that [z, CW (u)] = 1 for every W ∈ IG (R; 2 ), and conjugating this statement by g ∈ NK (R) ∩ CK (z), we get also that [z, CW (u0 )] = 1 for all such W . As U # = {u, u0 , z}, z then centralizes W = CW (u), CW (u0 ), CW (z) by [IG , 11.23], completing the proof.



Since Lemma 4.4 contradicts (4D), the proof of Proposition 4.1 is complete. As an immediate consequence of Propositions 2.3, 2.6, 3.1 and 4.1, and the definition of the sets Gi2 , 1 ≤ i ≤ 8 [III7 , (1E)], we have K/O2 (K) ∈ Gi2 for i = 1, 2, or 6. Then Corollary 2.7 has the following immediate consequence: Corollary 4.5. If p = 2, then d2 (G) = 6. That is, K ∈ Chev(r) for some odd r, but K ∼ = L± 3 (q) for any odd q.

9. THEOREM C∗ 7 : STAGE 3B

56

By [III7 , Theorem 1.2], we know that if p > 2, then K ∈ Chev(2). Consequently we have completed the proof of Theorem 2 in the case K ∈ Chev. Corollary 4.6. K ∈ Chev.

5. p-Thin Configurations, p Odd This section is preliminary to the proof of Theorem 2 in the case K ∈ Chev. It is devoted to a proof of the following proposition, which asserts the existence of a pThin Configuration (see Definition 1.10) for odd primes p. The proposition assumes the existence of an Ipo -terminal p-component pair (z, I) with I/Op (I) ∈ Tp ∩ Chev, and assumes one further condition relating I to our (x, K) ∈ J∗p (G) and acceptable subterminal (x, K)-pair (y, L). (See (2A).) As p-Thin Configurations in G are forbidden by the hypothesis of Theorem C∗7 , an immediate corollary is that such Ipo -terminal p-components cannot exist. Proposition 5.1. Suppose that p is an odd prime and (z, I) ∈ ILop (G) has the following properties: (a) I/Op (I) ∈ Tp ∩ Chev; (b) L/Op (L) Ω1 (PK ), then Ω1 (PK ) < Ω1 (Z(PC )) ≤ Ω1 (Z(P )) = Ω1 (PK ) x , 0 contains 31+2 while Ω1 (P ) is abelian, a so x ∈ Z(PC ). Therefore P = PC . But M contradiction. Therefore Ω1 (Z(PC )) = Ω1 (PK ). It follows that 3 divides |O3 3 (M0 )|. 0 ) = Ω1 (PK ) with CP (P ) ≤ P , we Since P normalizes M0 , and CΩ1 (Z(P )) (M C  we have in fact have Ω1 (PK ) ∩ O3 3 (M0 ) = 1. From the isomorphism type of M |PK ∩ O3 3 (M0 )| = 3. By its definition, C  NG (Ω1 (PK )) and in particular NK (Ω1 (PK )) normalizes C. Now by [III11 , 7.8], AutK (Ω1 (PK )) ∼ = D8 , necessarily acting irreducibly on Ω1 (PK ). Therefore as m3 (P ) = 4, there are exactly two NK (PK )-conjugates M0 , M0g of M0 , with g ∈ NK (Ω1 (P )), g 2 normalizing M0 and centralizing Ω1 (PK ), and Ω1 (PK ) = y × y g  where y = P ∩ O3 3 (M0 ). Both M0 and M0g are xinvariant, so Ω1 (P ) = (P ∩ M0 ) × (P ∩ M0g ), both factors being of order 9. Thus |CΩ1 (P ) (g)| ≤ 32 . On the other hand, g interchanges two independent elements of Ω1 (PK ) of order 3, while since g ∈ K  Cx , [Ω1 (P ), g] ≤ Ω1 (P ) ∩ K = Ω1 (PK ). Consequently g inverts only a subgroup of Ω1 (P ) of order 3, so |CΩ1 (P ) (g)| = 33 , contradicting what we just saw. This contradiction completes the proof of the lemma.  ∼ Lemma 5.5. M = SL31 (q1 ) for some q1 > 2, q1 ≡ 1 (mod 3), 1 = ±1. ∼ Proof. Suppose false. Then by Lemmas 5.3 and 5.4, M = G2 (q2 ) or 3D4 (q2 ) for some q2 > 2, q2 ≡ 0 (mod 3). In particular in view of [IA , Table 4.7.3A, 4.9.1], Mx := L3 (CM (x)) = 1. But Mx   L3 (CG (x, z)), and from the structure of K, therefore, Mx = L3 (CK (z)) is a single 3-component isomorphic mod 3 -core 2 to L2 (q), L− 3 (q), or L2 (q ). Let f ∈ Ω1 (P ) − x PK , so that f induces a field automorphism of order 3 on K. Then f ∈ P = CR (x) acts on M and Mx , inducing a field automorphism of order 3 on Mx by [IA , 4.2.3]. By [III11 , 6.25], f induces ∼ a field automorphism of order 3 on M = G2 (q2 ) or a graph automorphism of order 3 ∼  3 on M = D4 (q2 ). Set Mf = L3 (CM (f )), so that by [IA , Table 4.7.3A, 4.9.1], Mf /O3 (Mf ) ∼ = G2 (q3 ), G2 (q2 ) or L32 (q2 ), where q33 = q2 ≡ 2 (mod 3). Also set 1/3 ), n = 3 or 4, or B2 (q 1/3 ). Note that as K ∈ G3 , Kf = L3 (CK (f )) ∼ = A− n (q ∼ ∼  B2 (2) . K=  Sp4 (8), and hence Kf = ∗ ∗ Let Kf and Mf be the subnormal closures of Kf and Mf , respectively, in CG (f ); these have the usual structures by L3 -balance, with respect to the actions of x and z. By [III11 , 13.36], Kf∗ /O3 (Kf∗ ) is simple or the direct product of three copies of Kf , and in either case O3 3 (Kf∗ ) = O3 (Kf∗ ). But Kf∗ ∩ Mf∗ ≥ Kf ∩ Mf ≥

9. THEOREM C∗ 7 : STAGE 3B

60

1/3 Mxf := O 2 (CMx (f )), and Mxf /O3 (Mxf ) ∼ ), or L2 (q 2/3 ). = L2 (q 1/3 ), L− 3 (q ∗ ∗ Therefore Mf and Kf are the subnormal closures of Mxf in CG (f ), so Mf∗ = Kf∗ . Given the structures of Kf   CKf∗ (x) and Mf   CMf∗ (z), [III11 , 13.39] implies that Mf∗ ∈ G3 ∩ Chev with m3 (Mf∗ ) > 2, so that d3 (Mf∗ /O3 (Mf∗ )) ≤ 4. This contradicts the fact that d3 (G) = 5, completing the proof.  

The next step is: Lemma 5.6. Suppose that R ≤ X ≤ Cz and J1 , . . . , Jm are 3-components of X such that Ji ∼ = SL3i (ri ), ri ≡ i (mod 3), i = ±1. Then m ≤ 3, and x normalizes each Ji , inducing a nontrivial inner-diagonal automorphism on Ji /O3 (Ji ). Proof. If x did not normalize some Ji , or centralized some Ji /O3 (Ji ), then L3 (CX (x)) would have a 3-component J such that J/O3 3 (J) ∼ = L3i (ri ) had 3rank 2. But the only non-3-solvable composition factor of CG (x, z) = CCx (z) is isomorphic to L2 (q), L2 (q 2 ), or L− 3 (q), of 3-rank 1. Thus x normalizes each Ji . Similarly if x induces a field automorphism on Ji /O3 (Ji ), then x centralizes a 31+2 -subgroup of Ji , contrary to the fact that Ω1 (P ) is abelian. It remains to show that m ≤ 3. For each i = 1, . . . , m there exists by [IG , 10.11] an x-invariant E32 -subgroup Ui of Ji ∩R. Then U := U1 · · · Um is elementary abelian of rank greater than m, and x centralizes U = U/Z1 · · · Zm , where Zi = O3 3 (Ji )∩R. Clearly m3 (U ) = m, and so m3 (CU (x)) ≥ m. If m > 3, then as m3 (CG (x)) = 4 we must have x ∈ U ; but then m3 (CG (x)) ≥ m3 (U ) > m ≥ 4, a contradiction. This proves the lemma.  Now we can prove Lemma 5.7. For any 3-component J of Cz , Z(J/O3 (J)) = 1. Proof. Otherwise, let J ∗ be the product of all 3-components J of Cz such that Z(J/O3 (J)) = 1, so that 1 = J ∗  Cz and z ∈ J ∗ . Consequently Z(R)∩J ∗ = 1 and so m3 (Z(R)) ≥ 2. Now Ω1 (Z(R)) ≤ Ω1 (Z(P )) = Ω1 (PK ) x. One consequence is that Ω1 (Z(R)) ∩ PK = 1, whence by our choice of z (5I), z ∈ PK . Another consequence, since x ∈ Z(R), is that Ω1 (Z(R)) = z × v with v ∈ J ∗ and Ω1 (Z(P )) = x, v, z. . But K, having abelian Sylow  is nonsimple, J ∗ centralizes M Notice that as M . Therefore x ∈ J ∗ . 3-subgroups, does not involve M Now there is a 2-element t ∈ NK (PK ) such that t inverts Ω1 (PK ) elementwise, and of course [t, x] = 1. In particular t normalizes Ω1 (Z(P )) = x, v, z, inverts z, and normalizes J ∗ . As x = CΩ1 (Z(P )) (t) does not lie in J ∗ , t must invert Ω1 (Z(P )) ∩ J ∗ , and so t inverts v. Therefore Ω1 (Z(R)) = z, v is the inverted set of t on Ω1 (Z(P )), i.e., Ω1 (Z(R)) = Ω1 (PK ).

(5M)

. The argument is then similar to that in the Consequently Ω1 (PK ) centralizes M last paragraph of the proof of Lemma 5.4. Let N = NG (Ω1 (Z(R))) and M0 = 0 = M  and M0 is a 3-component of N . L3 (N ∩ M ), so that M Set M0∗ = M0N and Z = R ∩ O3 3 (M0∗ ). We claim that Z = Ω1 (Z(R)).

(5N) M0∗

has at most three 3-components by Lemma 5.6 applied to X = Namely, CG (Ω1 (Z(R))). Thus m3 (Z) ≤ 3. Also by [III11 , 7.8] there exists D0 ≤ NK (Ω1 (P ))

5. p-THIN CONFIGURATIONS, p ODD

61

such that D0 ∼ = D8 acts irreducibly on [D0 , Ω1 (P )] = Ω1 (PK ) = Ω1 (Z(R)). Obviously Z ∩ Ω1 (Z(R)) = 1, so since D0 acts irreducibly, Ω1 (Z(R)) ≤ Z. If R did not normalize every 3-component of M0∗ , there would be exactly three such 3components permuted transitively by R; but then Z would be a quotient of a rank 1 free module for R/NR (M0 ) ∼ = Z3 , and hence could not contain the trivial rank 2 module Ω1 (Z(R)). Therefore R normalizes every 3-component of M0∗ , whence Z ≤ Ω1 (Z(R)) and (5N) follows. As Ω1 (Z(R)) = z, y with y ∈ J ∗ , we must have z = R ∩ O3 3 (M0 ). By the irreducible action of D0 , there is an involution u ∈ D0 ≤ K inverting z and acting as a reflection on Ω1 (P ). We argue that u normalizes M0 . If not, then z ∈ Syl3 (O3 3 (M0 )) ∩ Syl3 (O3 3 (M0u )). But as noted above, M0∗ has at most three 3-components, and M0∗  N . Hence every element of D0 would carry some conjugate of M0 containing z in its 3-solvable radical to another, and so D0 ≤ NG (z), which is absurd. Thus u normalizes M0 , as asserted. 0 ) by [III11 , 1.8]. On the other hand by As u inverts z, u inverts Outdiag(M  Lemma 5.6, x maps into Inndiag(M0 ). Thus as [x, u] = 1, x induces an inner 0 , corresponding to some x ∈ M0 ∩ R − O3 3 (M0 ) with (x )3 ∈ automorphism on M O3 3 (M0 ). If x has order 9, then by [III11 , 1.18b], L3 (CM0 (x)) = L3 (CM0 (x )) = 1. g Conjugating by some involution g ∈ D0 such that z = z, we find that M0 = g   M0 , so since [x, g] = 1, L3 (C M D0 (x)) has more than one 3-component. Hence 0

so does L3 (CM D0  (x)), which in turn is subnormal in the single 3-component

L3 (CK (z)), a contradiction. Therefore x has order 3. Consequently x centralizes g the E34 -subgroup E := z, x  × z, x  , and then x ∈ E as m3 (Cx ) = 4; indeed as E ≤ R, E = Ω1 (P ). Thus, g acts freely on Ω1 (P ). But g, by choice, induces a  reflection on Ω1 (P ), contradiction. The lemma is proved. Now let Kz = L3 (CK (z)) = L3 (CG (x, z)) and let J be the subnormal closure of Kz in Cz . Recall that Kz /O3 (Kz ) ∼ = L2 (q b ) or L− 3 (q), where b = 1 or 2. In particular, Kz /O3 (Kz ) is 3-saturated [IA , 6.1.4]. We identify J/O3 (J) by using L3 -balance, Lemma 5.7 and the fact that d3 (G) = 5. With these and [III11 , 13.33] we immediately obtain: b Lemma 5.8. We have Kz /O3 (Kz ) ∼ = L2 (q b ). Moreover, J/O3 (J) ∼ = SL3 (q b ), or J/O3 (J) ∼ = SU6 (2) with q b = 8.

Now return to the element f ∈ Ω1 (P ) − PK x, which induces a nontrivial field automorphism on K, on Kz , and hence on J/O3 (J) (unless J/O3 (J) ∼ = U6 (2)). As before let Kf = L3 (CK (f )), and Jf = L3 (CJ (f )). Thus Kf /O3 (Kf ) ∼ = 3 A− (r), n = 3 or 4, or B (r), where r = q. A straightforward examination of the 2 n embeddings of Kf and Jf in CG (f ) allows us to prove: Lemma 5.9. We have r b = 2, K ∼ = L4 (8), Kz ∼ = SU3 (8) = L2 (8), and J/O3 (J) ∼ or SU6 (2). Proof. It suffices to prove that r b = 2. For then the case K ∼ = Sp4 (8) is impossible as K ∈ G3 but Sp4 (8) ∈ C3 ; as Kz ∼ = L2 (8), the structures of K and J follow. b Suppose on the contrary that r b = 2. Then we have Jf /O3 (Jf ) ∼ = SL3 (r b ),   by Lemma 5.8. Note that Kf = O 3 (CK (f )) and Jf = O 3 (CJ (f )) both contain

9. THEOREM C∗ 7 : STAGE 3B

62 

O 3 (CKz (f )), which modulo 3 -core is isomorphic to L2 (r b ) and in particular is simple. It follows from this and L3 -balance that the subnormal closures of Kf and Jf in CG (f ) coincide. We call this common subnormal closure W . Thus by L3 balance, W is a single 3-component, and Kf and Jf are 3-components of CW (x) and CW (z), respectively. Now by [III11 , 13.37], we deduce that W/O3 (W ) ∈ G3 ∩ Chev with m3 (W ) > 2. Consequently (f, W ) has Gp -depth at most 4, so d3 (G) ≤ 4, a contradiction as usual. The proof is complete.  Lemma 5.10. The following conditions hold: (a) Ω1 (P ) ∩ C(z, J) = z ∈ Syl3 (O3 3 (J)); and (b) z ∈ PK . Proof. Clearly z (P ∩O3 3 (J)) ≤ C(z, J)∩Ω1 (P ) ≤ CΩ1 (P ) (Kz ) = x, z, the equality visible in Cx . Then since x acts nontrivially on J/O3 (J), C(z, J)∩Ω1 (P ) = z and (a) follows. To prove (b), suppose that z ∈ PK . We argue first that J/O3 (J) ∼ = SU3 (8). Otherwise J/O3 (J) ∼ = SU6 (2). Since Kz ∼ = L2 (8) is a component of CJ (x), x maps to a nontrivial element of Outdiag(J/O3 (J)). As z ∈ PK , there exists an involution g ∈ NK (PK ) such that g inverts z and centralizes x. Then g normalizes Kz , so g normalizes J. As g normalizes J it has equivalent actions on Outdiag(J/O3 (J)) and Z(J/O3 (J)) by [III11 , 1.8]. But this is a contradiction, since g centralizes x and inverts z. Thus J/O3 (J) ∼ = SU3 (8). We have z × Kz ≤ K so Ω1 (PK ) = z, y with y ∈ Kz . But by (a) and [III11 , 6.20], all subgroups in E1 (Ω1 (PK )) − {z} are fused in J. Combining this with the action of AutK (Ω1 (PK )) ∼ = D8 on Ω1 (PK ), we see that E1 (Ω1 (PK )) is completely fused in G. Choose z ∗ ∈ Ω1 (PK ) such that E(CK (z ∗ )) ∼ = L2 (64). By the preceding paragraph, z ∗ , like z, is 3-central in G. Now Ω1 (PK ) ≤ Z(P ), so we may expand P to R∗ ∈ Syl3 (CG (z ∗ )) ⊆ Syl3 (G). Then choose g ∈ G such that Q∗ := Qg ≤ R∗ and set H ∗ = H g and a∗ = ag . Then P , Q∗ , R∗ , and z ∗ satisfy (5G) and (5H), so by our choice in (5I), r b = 641/3 = 22 . This contradicts Lemma 5.9, so the lemma follows.  We now examine G-fusion in P and R. As K ∼ = L4 (8) and m3 (Cx ) ≥ 4, and CP (K) = x by (2M), we have P = (PK × x) f  with PK ∼ = Z9 × Z9 , f induces a field automorphism of order 3 on K, and f 3 = 1. Thus f induces a power map on PK . Set S = NR (P ), R0 = Ω1 (P ) ∼ = E34 , R1 = PK x , ∼ E33 . R2 = Ω1 (PK ) x = R0 ∩ R1 = Ω1 (R1 ) = Thus, P = R0 R1 . Lemma 5.11. The following conditions hold:   (a) R2 = z G ∩ P  NG (R0 ); G G (b) | x ∩ P | = | x ∩ R2 | = |S : P | = 3; and (c) NG (P ) ∩ NG (R2 ) acts irreducibly on R2 . Proof. Suppose first that some v ∈ R0 − R2 is G-conjugate to z; we derive a contradiction. By L3 -balance, the subnormal closure V of CK (v) ∼ = L4 (2) in CG (v) is the product of one or three 3-components permuted transitively by x. On the other hand, CK (v) ∼ = CK (z) and so if we put V = V /O3 (V ), then Z(V ) = 1 by

6. THE GENERAL LIE TYPE CASE, BEGUN

63

Lemma 5.7. Since L4 (2) is 3-saturated [IA , 6.1.4], V must be a single 3-component. Since Z(V ) = 1, we conclude from [III11 , 13.34] that V ∈ G3 ∩Chev with m3 (V ) > 2. But this gives the usual contradiction to (5A1), namely that d3 (G) ≤ 4. Therefore  G  z ∩ P ≤ R2 . Since AutK (PK ) ∼ = D8 leaves invariant only the  proper  subspaces Ω1 (PK ) and x of R2 , and z lies in neither by Lemma 5.10b, z G ∩ P = R2 , and (a) follows. As R0 = Ω1 (P ) is weakly closed in P , NG (R0 ) controls G-fusion of x in P , by [III8 , 6.4]. Since R2  NG (R0 ), the first equality in (b) follows. Since x ∈ R2 we have P ∈ Syl3 (CG (R2 )), so by a Frattini argument NG (P ) controls this fusion as well. Now A := AutNG (P ) (R2 ) ≤ GL3 (3) is not transitive on E1 (R2 ) (x and z are not conjugate), so A is a {2, 3}-group. But CA (x) is a 2-group since R2 ≤ Z(P ). As Ω1 (PK ) and x are the irreducible constituents of AutK (R2 ) on R2 , they therefore must be the irreducible constituents of a Sylow 2-subgroup of A G on R2 . It follows that | x ∩ R2 | = |NG (P ) : P |3 = |A|3 = 3 or 9. But if |A|3 = 9, then by [III8 , 7.4], A fuses E1 (R2 ) − E1 (Ω1 (PK )), and in particular fuses x and z, which is absurd. Thus |A|3 = 3 and (b) holds as well. Indeed by [III8 , 7.4], A acts irreducibly on R2 , proving (c). The proof is complete.  The preceding lemma leads directly to a contradiction. Namely as |[f, R1 ]| ≥ |[f, PK ]| = 9, R1 is the unique abelian maximal subgroup of P , so R1 char P . Thus NG (P ) ∩ NG (R2 ) normalizes Ω1 (R1 ) ∩ R2 = Ω1 (PK ) < R2 . This contradicts Lemma 5.11c, completing the verification of Definition 1.10c3. It remains to check Definition 1.10c2. Assume that there is some (a, H) ∈  is a TGp -group. We know from part (c1) of the ILop (G) such that mp (H) ≥ 3 and H  ∈ Chev, H  is a G4p -group, whence dp (G) = definition that dp (G) = 4 or 5. But as H 4 and (5A2) holds. In particular, p = 3. As m3 (L) > 1 and L/O3 (L) 1. Depending on the isomorphism type of Ht , this will contradict either [III7 , 3.3] or [III4 , Prop. 20.3].

64

9. THEOREM C∗ 7 : STAGE 3B

Recall that K ∈ Chev by Corollary 4.6. Hence L and the components of Lu lie in Chev. Also by [III11 , 14.5d] and Proposition 2.3, (6B)

If p = 2, then L/O2 (L) ∼  L2 (q) for any odd q. =

It is convenient to suspend the notation of (2I)–(2K) for the remainder of this section and (6C) the next section. To begin, we note two lemmas. The first is a “connecting” criterion for a pumpup to be trivial. For this lemma only we permit L/O2 (L) ∼ = SL2 (q), q odd. Lemma 6.2. Let (t, J) ∈ ILop (G) and let (x, K, y, L) be as in (2A). Assume that L/Op (L) p.

Indeed, once (6O) is established, we can compare (u, L1 , x, L) ∈ H with (a, H1 , w, H ∗ ) ∈ H∗ : mp (C(u, Lu )) ≥ mp (u) = 1 = mp (X) = mp (C(a, H)) by hypothesis. Thus (6O) and the maximality of (a, H1 , w, H ∗ ) ∈ H∗ will imply the desired conclusion mp (C(a, H1 )) > p. To prove (6O), first note that as t centralizes L0 and y, x, with u ∈ y, x, it follows that t leaves Lu invariant. Then since t = x, there is t ∈ t x with t leaving each Li invariant, 1 ≤ i ≤ p. But t has the same action on K as t, so we can replace t by t and assume without loss that t normalizes each Li . By (6N3) and Lp -balance, L0 = Lp (CK (t, y)) = Lp (CL (t)). Hence if we set Ii = Lp (CLi (t)), i = 1, . . . , p, then considering the projections of L and L0 on Li , we have Ii /Op (Ii ) ∼ = L0 /Op (L0 ), 1 ≤ i ≤ p, and x cycles the Ii with diagonal L0 . Set I = I1 · · · Ip .

6. THE GENERAL LIE TYPE CASE, BEGUN

71

t = Ct /Op (Ct ). Let N be the subnormal closure of K0 in Set Ct = CG (t) and C  is quasisimple or N  is the product of p p-components Ct . By Lp -balance, either N  cycled by x, and in either case K0 is a component of CN (x). Furthermore, u ∈ x, y normalizes  K0 and centralizes t, so u normalizes N . We have L0 ≤ N and I ≤ Ct , and I = LI0 since L0 is a diagonal of I. Since N   Ct , we conclude that I ≤ N . We claim that N must be the product of p p-components N1 , N2 , . . . , Np cycled  would be quasisimple and x-invariant, and I would be the by x. Indeed, otherwise N product of p components of CN (u) cycled by x. But then mp (N ) ≥ mp (I) = p ≥ 3  ∈ Gp ∩ Chev. Since mp (N ) ≥ 3 while mp (K) = 2, the and by [III11 , 1.13a], N  by definition of the sets Gip Gp -depth of K/Op (K) is strictly greater than that of N [III7 , (1F)]. But K/Op (K) has (minimal) Gp -depth dp (G) since (x, K) ∈ J∗p (G) ⊆ Jp (G) [III3 , Definitions 5.1, 5.9]. This is a contradiction, so our claim is proved. As K0 /Op (K0 ) is p-saturated by (6N1), it follows by [IG , 3.27] that  =N 1 × · · · × N p . N The group I is a product of p components of CN (u) cycled by x. As CN (u)  thus has more than one p-component, clearly u cannot cycle the components of N and consequently u leaves each Ni invariant, 1 ≤ i ≤ p. Now the desired conclusion follows quickly. First, mp (CK0 (u)) = 2 by (6N4) #  0 is a diagonal of N , K  0 projects as u ∈ x, y and x centralizes K0 . Since K i and so mp (CN (u)) = 2 for each i, whence mp (CN t (u)) = 2p + 1. onto each N i In particular mp (CG (u)) ≥ 2p + 1. If CG (u) contains more than p components of p-rank 1, then as each such component is simple, clearly mp (C(u, L1 )) > p, as required. We can therefore assume that the Li , 1 ≤ i ≤ p, are the only components of CG (u) of p-rank 1, whence Lu  CG (u). Hence some maximal subgroup M of a Sylow p-subgroup of CG (u) normalizes L1 . Furthermore mp (Aut(L1 )) ≤ 2 by [III11 , 7.2], so mp (C(u, L1 )) ≥ mp (M ) − 2 ≥ mp (CG (u)) − 3 ≥ 2p + 1 − 3 > p. 

The proof is complete. For the general analysis of the case in which H 1 is simple (6M), let R ∈ Sylp (X1 ).

Then a ∈ Z(R). Furthermore, for any t ∈ Ip (Z(R)), t normalizes H 1 , so H1 has a trivial pumpup Ht in CG (t) by Lemma 6.9. As R ≤ C(a, H1 ) ∩ CG (t), it follows that R ≤ C(t, Ht ). We choose t among all elements of Ip (Z(R)) so that a Sylow p-subgroup (6P) Rt of C(t, Ht ) containing R is as large as possible. We fix such R ∈ Sylp (X1 ), t ∈ Ip (Z(R)) and Rt ∈ Sylp (C(t, Ht )), and we put t = Ct /Op (Ct ). Also set N1 = Lp (CH (t)), so that N 1 = H 1 Ct = CG (t) and C 1 1 = H  t . Since R centralizes N 1 , R centralizes some and N R1 ∈ Sylp (N1 ), and we fix R1 . We can assume without loss that Rt was chosen so that [Rt , R1 ] = 1.

9. THEOREM C∗ 7 : STAGE 3B

72

Notice that Z(Rt ) ≤ CRt (a N1 /Op (N1 )) = CRt (a) ∩ C(a, H1 ) = R, so Z(Rt ) ≤ Z(R). In particular for any t∗ ∈ Ω1 (Z(Rt ))# , the pumpup Ht∗ of H1 in CG (t∗ ) is trivial by Lemma 6.9, as noted above. Our maximal choice of t yields: Lemma 6.13. Assume (6J). Assuming that H 1 is simple, and with the notation just introduced, for any t∗ ∈ Ip (Z(Rt )), the pumpup of Ht in CG (t∗ ) is Ht∗ , and Rt ∈ Sylp (C(t∗ , Ht∗ )).     Proof. We have Ht = N1Ht and H1 = N1H1 , so the pumpups of Ht and H1 in CG (t∗ ) coincide, proving the first statement. Finally Rt centralizes  t and Rt ≤ CG (t∗ ), so Rt ≤ C(t∗ , Ht∗ ). Our maximal choice of t implies that H  Rt ∈ Sylp (C(t∗ , Ht∗ )), completing the proof. In view of the preceding lemma and the definition of p-terminality [IG , 6.26], in order to show that (t, Ht ) is p-terminal in G, it will suffice to prove that (6Q)

Ht has a trivial pumpup in CG (b) for every b ∈ Ip (Rt ).

Thus again by Lemma 6.2 it will be enough to show that every b ∈ Ip (Rt ) centralizes an Ep2 -subgroup E of C(t, Ht ) such that Ht (6R) has a trivial pumpup in CG (e) for each e ∈ E # . To find such a subgroup E, we begin by choosing t1 ∈ Ip (R1 ), so that by our setup, t1 centralizes Rt and t1 ∈ N1 ≤ H1 . By Lemma 6.10, for each i = 2, . . . , p, the pumpup Ji of Hi in CG (t1 ) is trivial. Set J 1 = J2 · · · Jp and R0 = R ∩ H 1 ; since t1 centralizes R ≤ Rt , R0 is a Sylow p-subgroup both of J 1 and of H 1 . t = Ct /Op (Ct ). Now the components Ji of J1 Also set Ct1 = CG (t1 ) and C 1 1 1 are all simple as each H i is simple, 2 ≤ i ≤ p. Hence whether p = 2 or p is odd, we have mp (R0 ) = mp (J 1 ) ≥ 2. We let P ∗ be the subgroup of Rt inducing inner automorphisms on J1 , and with this notation we immediately obtain Lemma 6.14. Assume (6J). Then Ht has a trivial pumpup in CG (b) for every b ∈ Ip (P ∗ ). 1 , we see that b centralizes an Ep2 Proof. Indeed, from the structure of J subgroup E of R0 . But R0 ∈ Sylp (H 1 ), so by Lemma 6.10, the pumpup of H1 in   t and H 1 , CG (e) is trivial for each e ∈ E # . As usual O p (CH1 (t, e)) covers both H so the pumpup of Ht in CG (e) is also trivial for each e ∈ E # . Lemma 6.2 yields  that Ht has a trivial pumpup in CG (b), as asserted. Now we can attain our objective. Lemma 6.15. Under the hypothesis (6J), there exists an element t ∈ Ip (X1 ) which is p-central in X1 such that (a, H1 ) has a trivial pumpup (t, Ht ) in CG (t), and (t, Ht ) is p-terminal in G. Proof. If H 1 is not simple, then z and J1 as in Lemma 6.11 satisfy our requirements. So assume that H 1 is simple and continue the above analysis, begun with Lemma 6.12 and including the choice of t in (6P), and t1 ∈ R1 . As noted above we need only prove (6Q), which reduces to showing (6R). Accordingly choose any b ∈ Ip (Rt ) ⊆ CG (t1 ). Recall that for each i = 2, . . . , p, Ji is the (trivial) pumpup of Hi in CG (t1 ), J 1 = J2 · · · Jp , and R0 = R ∩ H 1 = R ∩ J 1 ∈ Sylp (J 1 ). First consider the case that b leaves invariant each component

6. THE GENERAL LIE TYPE CASE, BEGUN

73

Ji of J1 , i = 2, . . . , p. Since each Ji is simple, it is clear when p is odd that mp (CR0 (b)) ≥ p − 1, so that (6S)

mp (CR0 (b)) ≥ 2.

The same inequality holds if p = 2 by [III11 , 7.11], because of (6F). As R0 ≤ P ∗ , Lemma 6.14 implies that (6R) is satisfied, as desired. The remaining case to consider is  that in  which b fails to normalize some pb component, say J2 , of J 1 . Set M = J2 , the product of p p-components of t ) and hence on J1 . CG (t1 ). Clearly M induces inner automorphisms on Lp (C 1  b ∩ M , which is But R0 ∩ J2 ∈ Sylp (J2 ) and b ∈ Rt , so Rt ∩ M contains R0 , a b-invariant Sylow p-subgroup of M . As b cycles the simple components of M CRt ∩M (b) is a Sylow p-subgroup of CM (b), isomorphic to a Sylow p-subgroup of the simple group J2 ∼ = H 1 . Furthermore, Rt ∩ M ≤ P ∗ , so the pumpup of Ht in CG (e) is trivial for all e ∈ Ip (Rt ∩ M ) by Lemma 6.14. Hence if mp (J2 ) ≥ 2, then b centralizes an Ep2 -subgroup E of Rt ∩ M and (6R) holds, as desired. We can therefore assume that mp (J2 ) = 1. As J2 ∼ = H 1 , (6F) yields p > 2. 1 Likewise  if J ≤ M , say Jp ≤ M , then b centralizes an Ep2 -subgroup E of (Rt ∩ b M ) × Rt ∩ Jp . Clearly then E ≤ P ∗ , so again (6R) holds. Thus we can also assume that J 1 ≤ M. ) ≤ P ∗ , we may similarly assume that CR (M ) = 1. Then as (Rt ∩ M ) × CRt (M t ∗ Let Q be the subgroup of Rt leaving each p-component of M invariant. Then Rt = Q∗ b and one of the following holds, in view of [III11 , 7.3]: (1) Q∗1 := Ω1 (Q∗ ) is elementary abelian and mp (Q∗ ) > p; (6T) (2) Q∗ = Rt ∩ M is homocyclic abelian of rank p and Rt ∼ = Zpc  Zp for some c.

However, by Lemma 6.12, mp (R) > p. As R ≤ Rt by construction, mp (Rt ) > p. This contradicts (6T2), so (6T1) holds. As b has order p, b centralizes an Ep2 subgroup E of Q∗1 . But E leaves each p-component of M invariant and J 1 ≤ M , so [E, Ω1 (Rt ∩ J 1 )] = 1. Now the pumpup of J1 is trivial in CG (e) for all e ∈ Ω1 (Rt ∩ J 1 )# by Lemma 6.10; for all e ∈ E # by Lemma 6.2; and finally for e = b, again by Lemma 6.2. The proof is complete.  We now reach the goal of this section. Proposition 6.16. Assume that Proposition 6.1 fails. Choose any quadruple (a, H1 , w, H ∗ ) ∈ H∗ , and adopt the notation of (6G). Then there is a ∈ Ip (X1 ) such that the pumpup of H1 in CG (a ) is vertical, and one of the following holds: (a) H 1 is not simple, and mp (CX1 (a )) ≥ 3; or (b) H 1 is simple, and a is p-central in aH 1 . Proof. Otherwise (6J) holds and we continue the above analysis. Let t and Ht be as in the previous lemma. Since t is p-central in X1 = C(a, H1 ) and (t, Ht ) is a trivial pumpup of (a, H1 ), C(t, Ht ) contains a Sylow p-subgroup of X1 . Hence

74

9. THEOREM C∗ 7 : STAGE 3B

with the previous lemma, and as a H 1 ≤ X1 , (1) mp (C(t, Ht )) ≥ mp (aH 1 ) ≥ p ≥ 2; and (6U) (2) (t, Ht ) is p-terminal in G. First assume that p > 2. We then claim that the following hold: (1) dp (G) > 1; (2) If mp (Ht ) = 2, or if p = 3 with Ht /O3 (Ht ) ∼ = Sp6 (2), then (6V) dp (G) > 3; and (3) Ht /Op (Ht ) ∈ Chev, or p = 3 with Ht /O3 (Ht ) ∼ = A11 . Indeed since (x, K) ∈ Jp (G), dp (G) = dp (K/Op (K)) by [III7 , Definition 1.8]. Now K ∈ Chev by Corollary 4.6 so the Gp -depth of K/Op (K) is at least 2 by definition [III7 , (1F)], proving (6V1). The prime p is odd, with mp (CG (K)) = 1 by [III7 , Theorem 1.2], and L/Op (L) 2. The proof is complete.  Note that by Lemma 7.4, Ji /Op (Ji ) ∈ Chev(2) for all i = 1, . . . , r. Using [III11 , 20.4], we can argue much as in the case p = 2 to obtain:

7. THE GENERAL LIE TYPE CASE, CONCLUDED

83

Lemma 7.13. F(Jr /Op (Jr )) > F(K). Proof. Assume false, again apply Lemma 7.8, and now use [III11 , 20.4]. This time, assuming that the lemma fails, we deduce that the configuration K/Op (K) >Kp L/Op (L) 1. Let P1 ∈ Syl3 (NG (P )); thus CP1 (x) = P , P1 /P ∼ = R0 , Ω1 (Z(P1 )) = R0 , and G G E1 (Z) − E1 (R0 ) ⊆ x . As u ∈ x , u ∈ R0 . Expand P1 to P ∗ ∈ Syl3 (G). We argue that P ∗ may be chosen so that (7V)

u ∈ Z(P ∗ ).

Clearly Ω1 (Z(P ∗ )) ≤ Ω1 (Z(P1 )) = R0 . Choose v ∈ Ω1 (Z(P ∗ ))# ; to prove (7V), G we may assume that v ∈ u , so v = Ω1 (Z(P ∗ )). Set J = L3 (CK (v)), so ∼ that J = L2 (q) (see [III11 , 7.9d]) and R0 ∩ J = 1. Note that because of the field automorphism induced by a, q > 2. Let Jv be the subnormal closure of J in CG (v). Then by L3 -balance, Jv  L3 (CG (v)). As Z(P ∗ ) is cyclic and P ∗ ∈ Syl3 (CG (v)), it follows by [III8 , 4.1] that no component of CG (v)/O3 (CG (v)) is simple. Also since J is simple and 3-saturated, it follows that Jv is a single x-invariant 3-component. Then by [III11 , 13.28], Jv ∼ = SL3 (q), q ≡  (mod 3),  = ±1. We argue that ∗ v ∈ Jv . Namely, let Z = O3 3 (Jv ) ∩ P ∗ . Then Z ∗ ≤ Ω1 (CP ∗ (x J) = x, v. Also Z ∗ ≤ Φ(P ∗ ∩ Jv ), while x ∈ Φ(P ). Thus, Z ∗ is not conjugate to x, so v = Z ∗ ≤ Jv , proving our assertion. Now, Jv contains v × (R0 ∩ J) = R0 , and all elements of Jv − v of order 3 are Jv -conjugate by [III11 , 1.18a]. Since NK (R0 ) acts irreducibly on R0 , we conclude that all elements of R0# are G-conjugate. But G this is a contradiction as v ∈ u . This establishes (7V). Another application of ∗ [III8 , 4.1] proves that Z(P ) is noncyclic, whence R0 = Ω1 (Z(P ∗ )). By (7T3), Jr /O3 (Jr ) ∼ = G2 (q 2 ). Note that as m3 (K) = 2, d3 (G) = 5. By [III4 , 20.3], m3 (C(ar , Jr )) = 1. Also by [III11 , 3.16], CG (ar )/C(ar , Jr ) has a cyclic Sylow 3-center. Since Z(P ∗ ) is noncyclic, the Sylow 3-center of CG (ar ) is noncyclic as well, and so ar is 3-central in G. On the other hand m3 (Aut(G2 (q 2 ))) ≤ 3, and

7. THE GENERAL LIE TYPE CASE, CONCLUDED

85

so m3 (G) = m3 (CG (ar )) ≤ 4.

(7W)

However, a, x acts on L and hence on Lu , with x cycling the three components of Lu and a inducing a field automorphism – in particular a noninner automorphism – on L = CLu (x). Therefore some b ∈ a, x−x normalizes each component of Lu , but lies outside Lu u. Hence m3 (CG (u)) ≥ m3 (Lu u, b) ≥ 5. This contradicts (7W) and completes the proof of the lemma.  We next prove Lemma 7.14. We have (ar , Jr ) ∈ Jp (G). Proof. Recall that for any quasisimple group H, with universal covering group  we write H,  − mp (Z(H)).  m  p (H) = mp (H) Since both K/Op (K) and Jr /Op (Jr ) lie in Gp ∩ Chev (see Lemmas 7.4 and 7.5), we need only prove that the Gp -depth of Jr /Op (Jr ) is no greater than that of K/Op (K). In view of oddness of p and the definition of the sets Gip [III7 , (1F)], what we need to check is the following. (1) If m  p (K) ≥ 5, then m  p (Jr ) ≥ 5; and (7X) (2) If mp (K) ≥ 3, then mp (Jr ) ≥ 3. Assume then that m  p (K) ≥ 5 > m  p (Jr ) or mp (K) ≥ 3 > mp (Jr ). By [III11 , 15.2a], mp (L) > 1. Set b = 5 or 3 in the respective cases, and m = mp (L). We have L/Op (L) 2, if 0 ≤ i < k (3) p = 3, J i ∼ = 3 D4 (qi ), qi > 2, if k ≤ i ≤ r We choose a p-central element a ∈ J0 of order p, let J0a = Lp (CJ0 (a)) and choose R ∈ Sylp (J0a ). Thus J0a /Op (J0a ) ∼ = SU5 (q02 ) or SL30 (q0 ) according as p = 5 or 3. Replacing terms in the chain (7E) by conjugates if necessary, and using the fact that the chain is diagonal-free, we may assume that [a, ai ] = 1 and indeed R ≤ Ji for each i = 0, . . . , r. Set Jia = Lp (CJi (a)) for each i. By induction on i we see that R ≤ Jia for each i. Moreover, Jia /Op (Jia ) ∼ = SU5 (qi2 ) or SL3i (qi ), according as p = 5 or 3, for all i = 0, . . . , r.  = C/Op (C), and let M be the subnormal closure of Set C = CG (a) and C R in C. Since [R, ai ] = 1 for each i, M is ai -invariant for each i. The subnormal closure of R in CC (ai ) = CCG (ai ) (a) is Jia , and so M is the subnormal closure in C of Jia . Thus by Lp -balance, for each i = 0, . . . , r, M (which is independent of i) is a single a-invariant p-component or the product of p ai -conjugate p-components of C, and in any case Jia is a p-component of CM (ai ). As a ∈ R ≤ Jia ≤ M we have a ∈ Z(M ), and it follows from the oddness of p and [III11 , 11.2] that Jia = Lp (CM (ai )). We consider the structure of Cr = CG (ar ) and C r = Cr /Op (Cr ). We have √ mp (C(ar , Jr )) = 1 by Lemma 7.5a, with Jr /Op (Jr ) ∼ = E8 (qr ), 2 F4 ( qr ), G2 (qr ) 3 or D4 (qr ) by (7AA). Moreover, a ∈ Jr and Lp (CJr (a)) has a p-component Jra = Lp (CM (ar )) with a ∈ Jra . Thus Jra /Op (Jra ) ∼ = SU5 (qr2 ) or SL3r (qr ). We conclude from [IA , Tables 4.7.3AB] that Y := CAut(Jr /Op (Jr )) (Jra /Op (Jra )) ∼ 3D4 (qr ), in which case these has Sylow p-subgroups of order p, except if Jr = Sylow p-subgroups are isomorphic to Ep2 and meet the commutator subgroup of Aut(Jr /Op (Jr )) in the image of a. Thus Y has a normal p-complement in all cases. Since mp (C(ar , Jr )) = 1, one of the following must hold:

7. THE GENERAL LIE TYPE CASE, CONCLUDED

(7BB)

87

(1) p = 5 and m5 (CCG (ar ) (Jra /O3 (Jra ))) = 2; or (2) p = 3 and CCG (ar ) (Jra /O3 (Jra )) has a normal 3-complement and has 3-rank at most 3.

Recalling that ar normalizes M , we next set N = ar  C(a, M ). As Jra ≤ CM (ar ), N centralizes Jra /Op (Jra ). Therefore from (7BB) we deduce that one of the following holds: (1) p = 5 with m5 (CN (ar )) ≤ 2; or (7CC) (2) p = 3 with m3 (CN (ar )) ≤ 3, and CN (ar ) has a normal 3complement. We shall contradict this by remembering where J0 came from. Recall that J0 = H1 is one of p p-components H1 , . . . , Hp of CG (a0 ) cycled by an element w of order p. We use this to establish that one of the following holds, which will contradict (7CC) and complete the proof of the lemma. (7DD)

(1) p = 5 and m5 (CN (ar )) > 2; (2) p = 3 and m3 (CN (ar )) > 3; or (3) p = 3 and Lo3 (CN (ar )) = 1.

Let Hja = Lp (CHj (a)) for each j = 1, . . . , p. The subnormal closures M2 , . . . , Mp of H2a , . . . , Hpa in C have the usual structure by Lp -balance, and Hja is a p-component of Lp (CMj (a0 )). But as J0a = Lp (CM (a0 )), we have Hja ≤ M , and so Hja ≤ N for j = 2, . . . , p. Now a ∈ H1 , so Hja covers Hj /Op (Hj ) for j = 2, . . . , p. Therefore for p = 5, m5 (N ) ≥ m5 (H2 H3 H4 H5 ) ≥ 4m  5 (H2 ) = 12, as m5 (Hj ) = m5 (SU5 (q0 )) = 4. Taking a Sylow 5-subgroup S of N containing ar we have m5 (S) ≥ 12 > [5(5+4)/4] and so m5 (CS (ar )) > 2 by [III2 , 7.2], which is (7DD1). On the other hand if p = 3, we have L/O3 (L) 1, by [III11 , 15.8a]. But the latter possibility is ruled out by Proposition 2.3, so (b) holds. Similarly (a) holds; the only alternatives, by 3 1/3 ). But (8A), (8B) and [III11 , 15.8b], are that K/Z(K) ∼ = L± 3 (q), G2 (q) or D4 (q all these possibilities are ruled out by Proposition 2.3. Note that by definition of acceptable subterminal pair, y induces an inner automorphism on K in the situation of (c). Thus if (c) fails, then by [III11 , 15.8c], some involution z ∈ Z(K) satisfies z ∈ L2 (CK (y)). By (8A), z = x; but x ∈ L2 (CK (y)) by (8B), a contradiction. Thus (c) holds and the proof of the lemma is complete.  As L ∼ = SL2 (q) and L is a diagonal of H1 H2 in (2J), we must have (8C)

H/O2 (H) ∼ = H1 /O2 (H1 ) × H2 /O2 (H2 )

where, as in (2I), H is the subnormal closure of L in CG (u), and u ∈ x, y. In particular Z ∗ (H) has noncyclic Sylow 2-subgroups. But τ ∈ CG (x, y) ≤ CG (u). Hence H τ is an x-invariant product of components of CG (u) with diagonal Lτ . Since [Lτ , L] = 1, it follows that the components of H τ are distinct from H1 and H2 . So HH τ is the product of four distinct components. Moreover, if it happens that m2 (Z ∗ (HH τ )) = 2, then replacing τ by τ x if necessary, we may assume that Z ∗ (Hi ) and Z ∗ (Hiτ ) have a common Sylow 2-subgroup i = 1, 2. Thus if m2 (Z ∗ (HH τ )) = 2, then (8D)

HH τ /O2 (HH τ ) ∼ = (SL2 (q) ∗ SL2 (q)) × (SL2 (q) ∗ SL2 (q)).

8. THE p = 2, SL2 (q) CASE

89

For convenience we define

 S = SL2 (q1 ) q1 odd, q1 ≥ q , and S∗ = S ∪ {Spinn (q1 ) | n ≥ 5, q1 odd} ∪ {HSpin+ 4m (q1 ) | m ≥ 2, q1 odd} + (Here HSpin+ 4m (q1 ) is a half-spin group, that is, a covering group of P Ω4m (q1 ) + by a center of order 2, but not isomorphic to Ω4m (q1 ) unless m = 2.) We shall study pumpups of elements (b, J) ∈ ILo2 (G) with J/O2 (J) ∈ S, and the following two lemmas will be critical for our argument.

Lemma 8.3. Suppose that (z, X) ∈ ILo2 (G) is 2-terminal in G. X/O2 (X) ∈ S. If X/O2 (X) ∈ G62 ∪ G72 , then m2 (C(z, X)) = 1.

Then

Proof. We have d2 (G) = 6 by Corollary 4.5. If X/O2 (X) ∈ G62 , then (z, X) ∈ J2 (G), so m2 (C(z, X)) = 1 [III7 , Corollary 5.1]. If X/O2 (X) ∈ G72 , the desired conclusion holds by [III4 , Prop. 20.3]. The case X/O2 (X) ∈ S is impossible since (z, X) would then provide a 2-Thin Configuration, which is forbidden  by the hypothesis of Theorem C∗7 . In the next lemma we use the notation Chev(odd) := ∪r>2 Chev(r), the union over all odd primes. Lemma 8.4. Suppose that (v1 , K1 ) r ≥ 4. product of two copies of J i . But then C ∗ ∼  ∗ contains a subgroup isomorphic to H0 ,  Moreover, M ≥ M = J/O2 (J), and so M as described in (8F2). But then t, t, I1 and the r1 2-components of M ∗ satisfy the conditions (1) and (2) of (8F). As r1 > r, this violates the fact that our original choice satisfied (8F3), proving the lemma.  We next make some observations about the 2-groups R0 and R. Lemma 8.8. The following conditions hold: (a) R0 is the commuting product of r − s quaternion groups and has sectional 2-rank at least 2(r − s), whence so does R; (b) If r − s ≥ 2, then R is neither cyclic nor of maximal class; and (c) Suppose that r − s ≥ 2 and m2 (Z ∩ R) ≥ 2. Then R and all of its 2-overgroups are connected, unless possibly r − s = 2, R0 is the direct product of two quaternion groups, and Z ∩ R ≤ R0 . Proof. As R0 ∈ Syl2 (M0 ), Ri := R0 ∩ Mi ∈ Syl2 (Mi ) is a quaternion group for all s < i ≤ r. Then (a) and (b) are obvious. By MacWilliams’ Theorem [III8 , 1.5] if R has sectional 2-rank at least 5, then R contains a normal E23 subgroup and hence is connected. Hence in proving (c) we may assume that r − s = 2, R has sectional 2-rank exactly 4, and m2 (Z ∩ R) ≥ 2. If Z ∩ R  R0 , then (Z ∩ R)R0 /Φ(R0 ) has rank at least 5, contradiction. If Z ∩ R ≤ R0 , then R0 is the direct product of two quaternion groups, proving (c).  Lemma 8.9. Suppose that there is a four-group D ≤ R such that the pumpup of I in CG (d) is trivial for all d ∈ D# . Then R is not connected, and the following conditions hold: (a) r − s ≤ 2; (b) If r − s = 2 ≤ m2 (Z ∩ R), then R0 is the direct product of two quaternion groups and Z ∩ R ≤ R0 ; (c) I ∼  L± = 3 (h) or SL2 (h) for any h > 3; and (d) b ∈ IC(t, I). Proof. By our choice of notation, (b, J1 ) < (t, I). It follows from Lemma 8.4 that whenever (v1 , K1 ) < (v2 , K2 ), both being long pumpups of (t, I), we have (vi , Ki ) ∈ Chev(odd). Assuming that there exists a four-group E ≤ C(v1 , K1 ) ∩ CG (v2 ) such that the pumpup of (v1 , K1 ) in CG (e) is trivial for any e ∈ E # it follows by [III11 , 17.2] (with E in the role of V there) that the pumpup (v1 , K1 ) < (v2 , K2 ) is trivial. By an inductive argument, Iu is a trivial pumpup of I for all u lying in four-groups that are connected (8G) in R to the given four-group D. Suppose that R and all its 2-overgroups are connected. In particular Iu is a trivial pumpup of I for all u ∈ I2 (R). Choose z ∈ Ω1 (Z(R)) to maximize |C(z, Iz )|2 .

92

9. THEOREM C∗ 7 : STAGE 3B

Then R ≤ C(z, Iz ), and we expand R to R∗ ∈ Syl2 (C(z, Iz )). Then R∗ is also connected, and as in the previous paragraph we conclude that all pumpups of (z, Iz ) are trivial. Let z0 ∈ Ω1 (Z(R∗ )). Then [z0 , t] = 1, so z0 ∈ R. Then the maximal choice of |C(z, Iz )|2 yields that |C(z0 , Iz0 )|2 ≤ |R∗ |. Thus R∗ ∈ Syl2 (C(z0 , Iz0 )). We have verified by the definition that (z, Iz ) is 2-terminal in G. Moreover by Lemma 8.4, Iz /O2 (Iz ) ∈ S ∪ G62 ∪ G72 . However, m2 (C(z, Iz )) ≥ m2 (R) > 1, which contradicts Lemma 8.3. Thus, either R or one of its 2-overgroups is not connected. By Lemma 8.8c, (a) and (b) follow directly. If I/O2 (I) ∼ = SL2 (h) or L± 3 (h), h > 3, then as (t, I) is a pumpup of (b, J1 ), it follows that s = 1 and so r − s ≥ 3, a contradiction proving (c). 1 = E(C b (b)) is a central quotient of I,  whence I ∈ S, If I = I b , then M II contradicting Lemma 8.7. Finally, suppose that b induces an outer automorphism on I/O2 (I). Then by [III11 , 13.9], I/O2 (I) ∈ S∗ and s ≤ 2. Moreover, if s = 2, then E(CI (b)) = M1 M2 2 = M 1 M 1 ∗ M 2 ∼ with M = SL2 (q1 ) ∗ SL2 (q2 ), and Z ∩ I ≤ Z ∗ (I) with |Z ∩ I| = 2. As r − s ≤ 2, we have r = 4 and s = 2. Moreover, Z ≤ C(t, I) and so Z ≤ R. By (b), we have m2 (Z) = 2 and Z ≤ R0 = Q1 × Q2 , with Qi quaternion. Now, by (8F2),  ≥ H0 ∼ M = (SL2 (q) ∗ SL2 (q)) × (SL2 (q) ∗ SL2 (q)) . But then, it follows from the structure of M1 M2 that  = (M 1 ∗ M 2 ) × (M 3 ∗ M 4 ), M whence R0 is not a direct product of quaternion groups, a contradiction. Thus (d) holds and the proof is complete.  Combining Lemmas 8.9 and 8.6, we can rule out the important case I ∈ S. Lemma 8.10. The following conditions hold: (a) I/O2 (I) ∈ S; (b) b normalizes I; and (c) For any involution u ∈ CR (b), (u, Iu ) is a trivial or vertical pumpup of (t, I). Proof. Suppose that I/O2 (I) ∈ S. Then by Lemma 8.7, all components of M ∗ lie in S, whence Z ≤ Z ∗ (Ct ). Let v be any involution of Z. Set Cv = CG (v) v = Cv /O2 (Cv ). Now v ∈ O2 2 (Ct ) ≤ C(t, L2 (Ct )), so Lemma 8.6a applies and C and yields that (8F) holds with (v, Iv ) in place of (t, I). Set Kv = L2 (CJ (v)).  v , E(C v )] lie in S. In Since I/O2 (I) ∈ S, (8F5) implies that all components of [K ∗ particular Z ≤ Z (Cv ). Replacing t by v, we may assume from the outset that t ∈ Z. But then t ∈ Z ∗ (Cv ), so by L2 -balance all 2-components of Ct have trivial pumpups in Cv . As this holds for all v ∈ Z # and m2 (Z) > 1 by (8F2), any foursubgroup of Z can play the role of D in Lemma 8.9. That lemma, however, implies  SL2 (h) for any h > 3, a contradiction. This proves (a). that I/O2 (I) ∼ = By (a), (b, J1 ) < (t, I) is a vertical pumpup, so (b) holds and M1 ≤ I. Consequently in (c), (u, Iu ) is also a pumpup of (b, J1 ). Since I/O2 (I) ∈ S, certainly Iu /O2 (Iu ) ∈ S, and therefore (u, Iu ) is a vertical pumpup of (b, J1 ), as 1 and lies in Iu . Since [t, M 1 ] = 1, J1 /O2 (J1 ) ∈ S. Therefore L2 (CM1 (u)) covers M  t normalizes Iu and the proof is complete.

8. THE p = 2, SL2 (q) CASE

93

Now we sharpen our choice of the configuration (b, t, I, J1 , . . . , Jr ) satisfying (8F). We assume, as we may, that we have chosen our configuration so that in addition (1) |I/O2 (I)| is as large as possible; and (8H) (2) Subject to (1), |C(t, I)|2 is as large as possible. Lemma 8.11. The following conclusions hold: (a) Let z ∈ I2 (Z ∩ R). Then Iz is a trivial pumpup of I; and (b) b ∈ IC(t, I). Proof. As Z ≤ M , Z ≤ CCt (b). Hence by Lemma 8.10c, Iz is a trivial or vertical pumpup of I. In the vertical case, however, Lemma 8.6a shows that our maximal choice of |I/O2 (I)| in (8H) is violated. This proves (a). Suppose next that b ∈ IC(t, I). Then using [III11 , 13.9] again, we see that I/O2 (I) ∈ S∗ and Z ∩ I ≤ Z ∗ (I). Hence Z ≤ R. Then as m2 (Z) ≥ 2, any foursubgroup of Z can play the role of D in Lemma 8.9, in view of (a). Thus (b) follows from Lemma 8.9.  Now we can prove the following crucial result. Lemma 8.12. Suppose that all pumpups of (t, I) are trivial. Then (t, I) is 2-terminal in G. Proof. By definition of 2-terminality [IG , 6.26], it is enough to show that for any z ∈ Ω1 (Z(R))# , we have R ∈ Syl2 (C(z, Iz )). By Lemma 8.11b, b ∈ IC(t, I). Now our choice of R guarantees that b ∈ RCS (R). But RCS (R) centralizes z ∈ Ω1 (Z(R)) and contains a Sylow 2-subgroup of M . Hence z normalizes M1 , . . . , Mr and then centralizes M/O2 (M ), by [III11 , 6.3e]. Then by Lemma 8.6a, (b, z, Iz , J1 , . . . , Jr ) satisfies (8F), and the desired conclusion follows from the maximal choice in (8H).  Now choose, as we may by [IG , Theorem 6.10], a 2-terminal long pumpup (t∗ , I ∗ ) of (t, I). By Lemmas 8.4 and 8.3, m2 (C(t∗ , I ∗ )) = 1 and I ∗ /O2 (I ∗ ) ∈ G62 ∪ G72 . Likewise with Lemma 8.10, I/O2 (I) ∈ G62 ∪ G72 . The possible isomorphism types of I/O2 (I) are restricted by additional con1 is ditions. First, b induces a nontrivial inner automorphism on I/O2 (I), and M subnormal in CI(b) (Lemma 8.11b). Moreover, with Lemma 8.2ac, Either I/O2 (I) ∼ = L± 3 (h) for some odd h > 3, or F(I/O2 (I)) ≤ F(K) = (8I) (q 9 , A) or (q 4 , BC), according as K/Z(K) ∼ = L± 4 (q) or P Sp4 (q). 6 ∗ ∗ For, unless I/O2 (I) ∼ = L± 3 (h), h > 3, we have I/O2 (I) ∈ G2 , and so I /O2 (I ) ∈ ∗ ∗ ∗ whence (t , I ) ∈ J2 (G). But (x, K) ∈ J2 (G). By the definition of this term and by the pumpup-monotonicity of F [III7 , 3.2], [III11 , 12.3e],

G62 ,

F(K) ≥ F(I ∗ /O2 (I ∗ )) ≥ F(I/O2 (I)). Thus (8I) holds, and so we have verified the hypotheses of [III11 , 12.7]. That lemma now yields that one of the following holds: (1) Z ≤ R with s = 2; or (2) s = 1; or (8J) (3) s = 2 and I/Z ∗ (I) ∼ = P Sp4 (q1 ) or G2 (q1 ) for some q1 ≥ q, L4 (q), or U4 (q).

9. THEOREM C∗ 7 : STAGE 3B

94

In particular, r − s ≥ 2, and so m2 (C(t, I)) > 1. This quickly implies: Lemma 8.13. The following conclusions hold: (a) (t, I) has a nontrivial pumpup in G; (b) Z  R; and (c) s = 2. Proof. As m2 (C(t, I)) > 1, (t, I) is not 2-terminal in G, by Lemma 8.3. Then Lemma 8.12 implies (a). Suppose that Z ≤ R. We argue that (8K)

R ≥ ZR0 ≥ Z1 × (Q1 ∗ Q2 )

with |Z1 | = 2 and with Qi quaternion. Namely, Iz is a trivial pumpup of I for all z ∈ Z # by Lemma 8.11. If m2 (Z) ≥ 3 then by Lemma 8.9ab, Z ≤ R0 , which is the direct product of two quaternion groups. This is absurd, so there is H0 ≤ J with H0 /O2 (H0 ) the direct product of two copies of SL2 (q) ∗ SL2 (q). Let Z0 = S ∩ Z ∗ (H0 ) ∼ = E22 . We have Z0 ≤ I, for otherwise Z0 ≤ Z(R) and so R is connected as m2 (Z0 ) ≥ 2, against Lemma 8.9 with D = Z0 . As a consequence, H0 ∩ R ≥ ZR0 as in (8K). But then Lemma 8.9ab applied with Z = D implies that Z ≤ R0 and R0 is the direct product of quaternion groups, which contradicts (8K). Thus (b) holds. Finally, suppose that s = 1. As Z  R, M1 ∩ C(t, I) ≤ O2 (M1 ). Hence ∗ ∗ ∗ J = J1 × J , where J = J 2 . . . J r . It follows that either m2 (Z) ≥ 3 or H 0 ≤ J . ∗ In either case, as J ≤ C(t, I), m2 (C(t, I)) ≥ 2, and m2 (Z ∩ R) ≥ 2. Then by Lemma 8.9, r − s ≤ 2 so r ≤ 3, a contradiction proving (c).  We can now complete the proof of Proposition 8.1 as follows. Let I2v (R) be the set of all involutions u ∈ R for which Iu is a vertical pumpup of I. We argue that (8L)

I2v (R) = ∅.

Suppose the contrary. We have R0 ≤ CR (b), so that m2 (CR (b)) ≥ m2 (R0 ) ≥ 2, as r − s ≥ 2. By Lemma 8.10c, for every involution u ∈ CR (b), (u, Iu ) is either a trivial or vertical pumpup of (t, I), hence trivial by our assumption. Then by Lemma 8.9 (taking D to be a four-subgroup of CR (b)), we see that R is not connected, so R has sectional rank at most 4. Now, as M3 M4 ≤ C(t, I), it follows 4 ) = Z(M 4 ) = t. Moreover, b = b1 t for some b1 ∈ I. Hence 3 M 3 M that CR (M R = CR (b). Applying Lemma 8.10c again, we conclude that for every involution u ∈ R, (u, Iu ) is a trivial pumpup of (t, I). This contradicts Lemma 8.13a, proving (8L). We now choose u ∈ I2v (R) with |CR (u)| maximal. By Lemma 8.13bc, I/Z ∗ (I) is as in (8J3), and in particular I/O2 (I) ∈ G62 . Moreover, (u, Iu ) has a 2-terminal long pumpup (v, I1 ) by [IG , 6.10], and then F(Iu /O2 (Iu )) ≤ F(I1 /O2 (I1 )) ≤ F(K)  Iu /O2 (Iu )) since (x, K) ∈ J∗2 (G). Then by [III11 , 12.6b] the only possibilities for (I, satisfy CAut(Iu /O2 (Iu )) (I ∩ Iu /I ∩ O2 (Iu )) ∼ = Z2 . Hence if we set Ru = CR (u), then Ru = t × R1 , where R1 = C(u, Iu ) ∩ R. Furthermore by [III11 , 12.6a], the pumpup of Iu in CG (u ) is trivial for all u ∈ I2 (R1 ), which implies that Iu /O2 (Iu ) ∼ = Iu /O2 (Iu ). By Lemma 8.8b, R is neither cyclic nor of maximal class, so it follows by [III2 , 1.16] and the maximal choice of u that u ∈ Z(R). Thus Ru = R and |R/R1 | = 2. Hence any involution in a quaternion subgroup of R lies in R1 . Such a quaternion subgroup is S ∩ Mr , and

9. THEOREM 3

95

so for the involution z := zr ∈ Z ∩ Mr , (z, Iz ) is a nontrivial pumpup of (t, I). However, this contradicts Lemma 8.11a, completing the proof of Proposition 8.1. Taken together, Propositions 2.3, 2.6, 3.1, 4.1, 6.1, and 8.1 imply Theorem 2.

9. Theorem 3 In this section we establish Theorem 3. Thus we prove Proposition 9.1. Suppose that p ∈ γ(G) and (x, K) ∈ J∗p (G). If p is odd, then mp (K) − mp (Z(K)) ≥ 3 and dp (G) = 5. We assume false and argue to a contradiction in a sequence of lemmas. By [III7 , Theorem 1.2], K ∈ Chev(2). Since (x, K) ∈ J∗p (G) ⊆ Jp (G), K/Op (K) ∈ Gp has Gp -depth dp (G). As mp (K) − mp (Z(K)) ≤ 2, it follows from [III11 , 7.6] that mp (K) = 2. Then the definition of the sets Gip [III7 , (1F)] implies that dp (K) = 5.

(9A)

Hence for any (x0 , K0 ) ∈ ILop (G) such that K0 /Op (K0 ) ∈ Gp , we have dp (K0 /Op (K0 )) ≥ 5. In particular: Lemma 9.2. For any (x0 , K0 ) ∈ ILop (G) with K0 ∈ Chev ∩ Gp , we have mp (K0 ) = 2. Proof. The conclusion holds by [III7 , (1F)] if dp (K0 ) = 5, so assume that dp (K0 ) > 5. Then K0 ∈ Alt ∪ Spor by the same reference. By definition of Gp the  only choice is p = 3 and K0 /O3 (K0 ) ∼ = A8 , so that m3 (K0 ) = 2, as required. Choose an acceptable subterminal (x, K)-pair (y, L), let D = x, y, and for each u ∈ D# , let Lu be the subnormal closure of L in CG (u). By [III7 , Theorem 1.2] and Theorem 2, each Lu is a trivial or vertical pumpup of L, and Lu0 is a vertical pumpup for at least one u0 ∈ D − x. Thus (9B)

[Lu0 , x] = Lu0 .

Moreover, mp (C(x, K)) = 1. Lemma 9.3. We have mp (CG (x)) = 4. Moreover, there exists A ∈ Ep4 (CK (x)) and f ∈ A# such that the following conditions hold: (a) D ≤ A; (b) A = x × AK × f  with AK = A ∩ K ∼ = Ep2 and f inducing a field or quasifield automorphism of order p on K; (c) E(CK (f )) = 1, and one of the following holds: (1) E(CL (f )) = 1; or 3 (2) p = 3, K ∼ = SU3 (8) and CL (f ) ∼ = SU3 (2); = 2F4 (2 2 ), L ∼ (d) One of the following holds: (1) f induces a field automorphism of order p on L; or (2) p = 3, K ∼ = SL3 (q), q ≡  (mod 3), and CK (f ) ∼ = G2 (q) = 3D4 (q), L ∼ with [f, L] = 1. Proof. This is immediate from [III11 , 15.4]abcde.



9. THEOREM C∗ 7 : STAGE 3B

96

Lemma 9.4. After a suitable replacement of f by an element of f x, one of the following holds: (a) f induces a field automorphism on L and a nontrivial field or graph automorphism on Lu0 ∈ Chev(2), with Lu0 ∼ = 3D4 (q), q > 2, and p = 3 in the graph automorphism case; moreover, Cf,x (Lu0 ) = 1; (b) [f, Lu0 ] = 1. In either case, [x, CLu0 (f )(∞) ] = 1. Proof. By [III11 , 15.4f], Lu ∈ Chev(2) for all u ∈ D# . First suppose that Cf,x (Lu0 ) = 1. By (9B), Cf,x (Lu0 ) = f   = x for some  f . Without loss we may take f  ∈ f x and replace f by f  , so that [f, Lu0 ] = 1, which is (b). Moreover, since u0 was chosen so that [x, Lu0 ] = 1, the final statement of the lemma holds in this case. Therefore we may assume that (9C)

Cf,x (Lu0 ) = 1.

Next, suppose that Lemma 9.3d2 holds. Note that then since d3 (G) = 5, the n only possible vertical pumpups Lu0 of L are Lu0 ∼ = G2 (q), 3D4 (q), 2F4 (2 2 ), or SL3 (q 3 ), by [III11 , 13.32a]. If m3 (CAut(Lu0 ) (L)) ≤ 1, then since [L, f, x] = 1, we conclude that Cf,x (Lu0 ) = 1, contradiction. Therefore we may assume that m3 (CAut(Lu0 ) (L)) > 1. Hence by [III11 , 13.32b], Lu0 ∼ = 3D4 (q) and Autf,x (Lu0 ) is generated by Z(L), which is the image of x, and a quasi-field automorphism centralizing G2 (q). It follows from [IA , 4.7.3A] that CLu0 (f ) ∼ = G2 (q). By Lemma ∼ 9.3d2, CK (f ) = G2 (q) as well, and L ≤ CK (f ) ∩ CLu0 (f ). Thus (a) holds in this case, and we need to rule out the possibility that [x, CLu0 (f )] = 1. But if this were the case then CLu0 (f ) would lie in the subnormal closure of L in CG (f, x). However, the component CK (f )   CG (f, x) must be the subnormal closure of L in CG (f, x) and so CK (f ) = CLu0 (f ) would be centralized by x, u0  = D. As E(CK (D)) = E(CK (y)) = L < CK (f ) = E(CK (f )) this is a contradiction, so the lemma holds if 9.3d2 holds. On the other hand, suppose that Lemma 9.3d1 holds. We know that Lu0 ∈  Chev(2), and if Lu0 ∈ Gp , then mp (Lu0 ) = 2 by Lemma 9.2. We also have L ∼ = L2 (8), for otherwise CL (f ) ∼ = L2 (2) by Lemma 9.3d1, contradicting Lemma 9.3c. As L is a component of CLu0 (x), we may apply [III11 , 1.10] with Lu0 in the role of K there, and conclude directly that (a) holds. It remains to prove the final statement of the lemma. If f induces a field automorphism on Lu0 , then the desired statement holds by (9C) and [IA , 7.1.4] unless CLu0 (f ) is solvable, i.e., Lu0 ∼ = L2 (8), 5 2 ∼ 2 (S)U3 (8), or B2 (2 ), with p = 3, 3, 5, respectively. But since L =  L2 (8) is a component of CLu0 (x) and x acts nontrivially on Lu0 , these exceptions are not possible. By (a), the only remaining case is that f induces a graph automorphism on Lu0 ∼ = 3D4 (q) with p = 3. But in this case as well, by [III11 , 2.5] this time, 3 O (CAut(Lu0 ) (CLu0 (f )(∞) )) is usually generated by the image of f , in which case the desired statement follows from the fact that Cf,x (Lu0 ) = 1. The exceptional case is Lu0 ∼ = 3D4 (2). But this cannot occur; as Lu0 is a vertical pumpup of L, it would force L ∼ = L2 (8), again a contradiction. This completes the proof of the lemma.  Write Kf = E(CK (f )) = [CK (f ), CK (f )] and Lf = [CL (f ), CL (f )].

9. THEOREM 3

97

Thus either Lf = E(CL (f )) or the exceptional configuration of Lemma 9.3c2 holds, in which case Lf is an extension of 31+2 by an involution inverting O3 (Lf )/Φ(O3 (Lf )). In this exceptional case, O3 (Lf ) = L∞ f , the last term of the lower central series of Lf . Let (9D)

J = the subnormal closure of Kf in CG (f ).

Then J is D-invariant, and since Lf is a nonabelian subgroup of the quasisimple group Kf = [CK (f ), CK (f )], J is the subnormal closure of Lf in CG (f ). But as L ≤ Lu0 , we have Lf ≤ [CLu0 (f ), CLu0 (f )], with the latter group quasisimple in either case of Lemma 9.4, by [IA , 4.9.1,4.7.3A] and Lemma 9.3. Therefore [CLu0 (f ), CLu0 (f )] ≤ J, whence [x, J] = J by Lemma 9.4. Let CG (f ) = CG (f )/Op (CG (f )). Lemma 9.5. The following conditions hold: (a) K f   CJ (x); and (b) Either J ∈ Chev(2) ∩ Gp , or J ∼ = A11 with p = 3 and Kf ∼ = A8 . Proof. By Lp -balance and (9D), (a) holds; moreover, either J is quasisimple or J is the product J = J1 · · · Jp of p-components cycled by x, with Kf a diagonal of this product. In the latter case, since D normalizes J, ND (J1 ) is generated by some 1 = u ∈ D, and u normalizes Ji , i = 1, . . . , p. Then u centralizes the projections Pi of Lf onto Ji Op (CG (f ))/Op (CG (f )), i = 1, . . . , p. Whether Lf = E(Lf ) or O3 (Lf ) = L∞ f , it follows that Sylow p-subgroups of Pi lie in the subnormal closure of Lf in CG (u), which is Lu . Hence mp (Lu ) ≥ p ≥ 3. We have Lu ∈ Chev(2) by [III11 , 15.4f] again, and now by [III11 , 16.12], Lu /Op (Lu ) ∈ Gp . But then mp (Lu ) ≤ 2 by Lemma 9.2, a contradiction. Therefore J is a single p-component. Since mp (Kf ) = 2 and [x, J] = J, [III11 , 13.42] applies with K and Kf in the roles of H and L there. We conclude that (b) holds, unless possibly p = 3 and J/O3 (J) ∼ = A14+3k , k ≥ 0. But in this last case, (f, J) ∈ ILo3 (G) with J ∈ G13 , contradicting the fact that dp (G) = 5. The proof is complete.  The next two lemmas tackle the general case. Lemma 9.6. If J ∈ Gp ∩ Chev(2), then x induces a field automorphism on J of order p, or else p = 3 with K ∼ = 3D4 (q), q > 2. Proof. Assume that it is not the case that p = 3 and K ∼ = 3D4 (q). Thus f induces a field automorphism on K. By Lemma 9.2, mp (J) = 2. Then by [III11 , 7.4ac], J is simple and Outdiag(J) is a p -group. Thus if the image of x in Aut(J) lies in Inndiag(J), it must lie in Inn(J). In that case the simplicity of J and the facts that mp (J) = 2 = mp (Kf ) imply that Op (K f ) = 1, and x acts on J like an element of Op (K f ). In particular Sylow p-subgroups of J are nonabelian. Therefore by [III11 , 7.4b], p = 3 and J ∼ = G2 (q) √ n or 3D4 (q) or 2 F4 ( q) for some q > 2, and by [IA , 4.7.3A], K f ∼ (2 ). As = SL± 3 Kf = E(CK (f )) and f induces a field automorphism on K, however, this forces 3n K∼ = SL± 3 (2 ) ∈ G3 , a contradiction as (x, K) ∈ Jp (G). We conclude that x maps to an element of Aut(J) − Inndiag(J).

9. THEOREM C∗ 7 : STAGE 3B

98

As p is odd and mp (J) = 2, x induces a field automorphism on J, or p = 3, J ∼ = 3D4 (q), and x induces a graph automorphism on J. But in the latter case we must have CJ (x) ∼ = G2 (q), i.e., x induces a quasifield automorphism of J. For otherwise by [IA , 4.7.3A], Kf ∼ = L3 (q), and as in the previous paragraph this leads  3 ∼ to the contradiction K = L3 (q ) ∈ G3 . If CJ (x) ∼ = G2 (q), so K ∼ = G2 (q), then Kf ∼ = G2 (q 3 ), and q > 2 as G2 (8) ∈ G3 . 12 16 5 Then f (K) = q < q = f (J), and J ∈ G2 . But this is impossible as (x, K) ∈ J∗3 (G) and (f, J) ∈ ILo3 (G). The proof is complete.  Now we deal with the general case, proving: Lemma 9.7. We have p = 3 and either K ∼ = 3D4 (q), q > 2, or J ∼ = A11 . Proof. Suppose false. Then J ∈ Gp ∩ Chev(2) by Lemma 9.5, so x induces a field automorphism on J by Lemma 9.6, and mp (J) ≤ 2 by Lemma 9.2. Recalling that mp (C(x, K)) = 1 and Kf = E(CK (f )), we have Kf = Lp (CG (x, f )) = Lp (CJ (x)). Since K ∼  3D4 (q) with p = 3, f induces a field automorphism on K. We have = y ∈ Kf ≤ J with L = E(CK (y)), and we set M = Lp (CJ (y)). Note that as mp (J) ≤ 2, M is a single p-component. We have Lf ≤ CKf (y) ≤ CJ (y), so 



p ∞ O p (L∞ f ) ≤ O (CJ (y) ) = Lp (CJ (y)), 

the last equation in view of [IA , 4.2.2]. Thus O p (L∞ f ) lies in L ∩ M and is either quasisimple or a nonabelian p-group (the latter only when Lemma 9.3c2 holds). It follows that the subnormal closure Ly of L in CG (y) is the same as the subnormal closure of M in CG (y). We consider the action of x, f  on Ly , and apply [III11 , 1.10] with Ly and L in the roles of K and J there, and x and f in the roles  y = Ly /Op (Ly ): of u and v. We conclude that one of the following holds, with L y ; (1) f induces a field or graph-field automorphism of order p on L 3 ∼  (9E) (2) p = 3, Ly = D4 (q) for some q > 2, and f induces a graph  y of order 3. automorphism on L  y ∈ Gip for i < 5, contradicting Indeed, otherwise [III11 , 1.10] would yield that L (9A), or p = 3 with L ∼ = L2 (8), in which case E(CL (f )) = 1, contradicting Lemma 9.3c. On the other hand, as y induces an inner automorphism on J ∈ Chev(2), x induces a nontrivial field or graph-field automorphism on M , by [IA , 4.2.3]. Hence  y . Thus by [III11 , 1.9], f does not induce a field or graph-field automorphism on L  ∼ (9E2) must hold. In particular, M = G2 (q) or L3 (q) for some q. Then the levels of Lf and L are, in turn, q 1/3 and q. In particular q > 2 so Ly ∈ G3 . As y induces an inner automorphism on K, the level of K is at most that of L, i.e., q. Moreover by [III11 , 7.4g], since m3 (K) = 2, the untwisted Lie rank of K is at most 4, with  y ) ≥ F(K). But K of type A4 or 3 D4 (or type 2 F4 and level q 1/2 ). Therefore F(L ∗ 3  y ) = F(K) and K ∼  (x, K) ∈ J3 (G), so F(L = D4 (q), as required. ∼ 3D4 (q), which we assume through Lemma Consider now the case that K = 9.10. The argument is similar to the above, but a bit more subtle. By [III11 , 15.5],

9. THEOREM 3

99

L∼ = SL3 (q) with y = Z(L). We may assume that f was chosen to centralize L, and then that AK ≤ L. Lemma 9.8. Suppose that K ∼ = 3D4 (q). Moreover, = 3D4 (q), q > 2. Then Lu0 ∼ Ly = L. Proof. If possible take u0 = y. If m3 (Lu0 ) > 2, then by [III11 , 13.32c] and [III7 , (1F)], Lu0 /O3 (Lu0 ) ∈ Gi3 for some i < 5, contradicting d3 (G) = 5. Therefore √ m3 (Lu0 ) = 2, and by [III11 , 13.32a], Lu0 ∼ = 3D4 (q), G2 (q), 2 F4 ( q), or SL3 (q 3 ). Notice by [III11 , 7.10c1] that CCG (u0 ) (x) = CG (D) = CCG (x) (u0 ) contains a 3element w ∈ K such that w L/Z(L) ∼ = P GL3 (q). Then w normalizes L, so w normalizes Lu0 . Therefore AutCG (u0 ) (Lu0 ) contains a 3-element inducing an outer diagonal automorphism on the image P SL3 (q) of L. Hence Aut(Lu0 ) contains such √ a 3-element. By [IA , 4.7.3A], this forces Lu0 ∼ = G2 (q) or 2 F4 ( q). We next show that ∼ SL (q 3 ). (9F) Lu = 0

3

∼ SL (q 3 ), then y ∈ Z(L) ≤ Lu so by our choice, y = u0 . Choose If Lu0 = 3 0 P ∈ Syl3 (CG (x)) with y ∈ Z(P ) and expand P to R ∈ Syl3 (CG (y)). Then by [III11 , 7.10c1], Ω1 (Z(P )) = x, y = D. As x induces a nontrivial field automorphism on Lu0 , x ∈ Z(R), so Ω1 (Z(R)) = y. Hence R ∈ Syl3 (G), and NR (P ) fuses E1 (D) − {y}. In particular, y is weakly closed in D with respect to G. It suffices to show that (9G)

x ∈ [NG (y), NG (y)],

for then by [III8 , 6.3], x ∈ [G, G], contradicting the simplicity of G. But since CG (D) has the unique 3-component L, x induces a field automorphism of order 3 on precisely one 3-component of CG (y) isomorphic to Ly , namely Ly itself. Now (9G) follows from [III8 , 6.10], applied with Aut0 (Ly ) playing the role of N1 there; this completes the proof that Lu0 ∼ = 3D4 (q). Finally, if Ly > L, then by choice, u0 = y and y ∈ L ≤ Ly . Therefore  y ∈ Z(Ly ), which is absurd as Z(3D4 (q)) = 1. The proof is complete. Lemma 9.9. Suppose that K ∼ = 3D4 (q), q > 2. Then the following conditions hold: (a) CK (f ) ∼ = CLu0 (f ) ∼ = G2 (q); ∼ (b) L ≤ E(CG (f )) = K ∼ = 3D4 (q); and (c) x induces a graph automorphism on E(CG (f )) with fixed point subgroup CK (f ). Proof. By our choice of f , CK (f ) ∼ = G2 (q) and [f, D] = 1. In particular [f, u0 ] = [f, L] = 1 and so f normalizes Lu0 . Note that since L is a component of CLu0 (x), the action of x on Lu0 is that of Z(L). Since [f, L] = 1, it follows that if f induces an inner automorphism on Lu0 , then f, x, y maps to CInn(Lu0 ) (L) and hence m3 (C(u0 , Lu0 )) > 1. But then taking an I3o -terminal long pumpup (w, J) of (u0 , Lu0 ) [IG , 6.22], we have m3 (C(w, J)) ≥ 1 and F(J/O3 (J)) ≥ F(Lu0 ), with at least one of these inequalities strict, by [III8 , 2.4] and [III11 , 12.3e]. As (x, K) ∈ J∗3 (G) equality of F’s must hold, and this implies that (w, J) ∈ J∗3 (G). This contradicts [III7 , Corollary 5.1], however. Therefore f induces an outer automorphism on Lu0 . Again as [f, L] = 1 it follows by [III11 , 7.10c] that CLu0 (f ) ∼ = G2 (q), as asserted in (a).

100

9. THEOREM C∗ 7 : STAGE 3B

Let Kf = CK (f ) and M = CLu0 (f ). Then L ≤ Kf ∩ M , so Kf and M have the same subnormal closure K ∗ in CG (f ). Assuming that K ∗ is a single 3-component, K ∗ /O3 (K ∗ ) is then a proper pumpup of E(CK ∗ (x)) = Kf , so K ∗ /O3 (K ∗ ) ∼ = G2 (q 3 ) or 3D4 (q). (If K ∗ /O3 (K ∗ ) ∼ = D4 (q), then 5 = d3 ≤ ∗ ∗ d3 (K /O3 (K )) = 4, contradiction.) In the first case, however, L ≤ L3 (CK ∗ (y)) with L3 (CK ∗ (y))/O3 (L3 (CK ∗ (y))) ∼ = SL3 (q 3 ) so Ly > L, contradicting Lemma ∗ ∗ 9.8. Thus K /O3 (K ) ∼ = K. Hence (f, K ∗ ) ∈ J∗3 (G), so by [III7 , Theorem 1.2], ∗ K is quasisimple and m3 (C(f, K ∗ )) = 1. Therefore K ∗ = E(CG (f )). In proving (b), it remains to show that K ∗ is not the product of three 3components cycled both by x and by u0 . But if that were the case, then there would exist u ∈ D−x normalizing the three 3-components and with L3 (CK ∗ /O3 (K ∗ ) (u)) the product of three copies of SL3 (q). By L3 -balance and [III11 , 1.13c], Lu would then either be the product of three copies of a cover of L cycled by x, or a single component isomorphic to E6 (q) or SL9 (q). As the first alternative contradicts Theorem 2, while the second contradicts d3 (G) = 5, (b) must hold. Finally, (c) is immediate from [III11 , 7.10c] and the fact that E(CG (x, f )) = E(CK (f )) ∼ =  G2 (q). The proof is complete. We can now reduce to a single case. Lemma 9.10. We have J ∼ = A11 . ∼ 3D4 (q), and continue the above argument, Proof. Assume false, so that K = ∗ writing K = E(CG (f )) ∼ = K. Fix v ∈ AK − y and set M = L3 (CK (v)) and M ∗ = L3 (CK ∗ (v)). ∼ L2 (q 3 ) ∼ By [III11 , 7.10], M = E(CK (v)) = = M ∗ = E(CK ∗ (v)), with f and x inducing field automorphisms of order 3 on M and M ∗ , respectively. Moreover,   (O 2 (CG (x, f, v)))(∞) = (O 2 (CCG (x) (f, v)))(∞) has a unique subgroup isomorphic to L2 (q). It follows that M ∩ M ∗ ∼ = L2 (q). Let N be the subnormal closure of M in CG (v) and set N = N/O3 (N ); it follows that N is also the subnormal closure of M ∗ in CG (v). By L3 -balance, M (resp. M ∗ ) is a component of E(CN (x)) (resp. E(CN (f ))). As M = M ∗ , x, f  acts faithfully on N . If x induces a nontrivial field or graph-field automorphism on N , then since f induces a nontrivial field automorphism on the component M of CN (x), we have a contradiction to [III11 , 1.9]. Then since q > 2 and d3 (G) = 5 we can apply [III11 , 1.10] to conclude that N ∼ = 3D4 (q1 ) for some q1 , with f inducing a graph automorphism on N . But then the component M ∗ of CN (f ) is isomorphic to G2 (q1 ) ∗ ∼ 3 or A± 2 (q1 ) by [III11 , 7.10c]. As M = L2 (q ), this is a contradiction. The lemma follows.  Expand f, x to S ∈ Syl3 (CG (f )) and let t ∈ I3 (Z(S) ∩ J). We next prove Lemma 9.11. The following conditions hold: (a) K ∼ = L2 (64); = L4 (2), and L ∼ = L4 (8), Kf ∼ (b) S ∩ C(f, J) = f ; and (c) f t ⊆ f NG (S) . ∼ L4 (2), we have K ∼ Proof. Since Kf = = L4 (8), and then L ∼ = L2 (64) as (y, L) is an acceptable subterminal (x, K)-pair [III7 , Def. 6.9]. Since m3 (CG (x)) = 4, but m3 (CJ (u)) = 3 for all u ∈ I3 (Aut(J)) ⊆ Inn(J), m3 (C(f, J)) = 1. In particular S ∼ = (Z3  Z3 ) × Z3k for some k. Therefore CG (f )

9. THEOREM 3

101

contains no Z9 × Z9 × Z3 subgroup. But CG (x) does contain such a subgroup, so f is not 3-central in G. We have Ω1 (Z(S)) = t, f  with t ∈ J being the product of three 3-cycles. As Z(S) ∩ [S, S] = t and S ∈ Syl3 (G), (c) holds. In particular f  is not characteristic in S, so f  = Ω1 (Φ(Z(S))). This implies that k = 1, and (b) holds. The proof is complete.  Lemma 9.12. We have x ∈ J, and in J ∼ = A11 , x is a 3-cycle. Moreover, I3 (f (S ∩ J)) ⊆ f G . Proof. Let R ∈ Syl3 (CG (x)) with A ≤ R and set RK = R ∩ K, so that RK ∼ = Z9 × Z9 and AK = Ω1 (RK ). Then AK = x1  × x2 , where Li := E(CK (xi )) ∼ = L2 (64) and xi ∈ L3−i , i = 1, 2. Let Ii = CLi (f ) ∼ = A5 , i = 1, 2. Then AK ≤ I1 , I2  ≤ L3 (CG (f )) = J by L3 -balance, and xi ∈ I 3−i is a 3-cycle in J, i = 1, 2. Notice that the field automorphism f induces a nontrivial power map on RK , but is not G-conjugate to an element of the abelian group RK × x ∼ = Z9 × Z9 × Z3 . Therefore f AK ⊆ f G ∩ A ⊆ A − AK x. By Lemma 9.11c, f t ∈ f G as well. Therefore we have f a ∈ f G for the representatives a = x1 , x1 x2 , t of the J-conjugacy classes of elements of order 3. Letting S0 = S ∩ J, we conclude that I3 (f S0 ) ⊆ f G , and in particular (A ∩ J)f ⊆ f G . It follows that x ∈ J, and as E(CG (f, x)) ∼  = L4 (2), x is a 3-cycle in J. We now complete the proof of Proposition 9.1. Let x1 and x2 be as in the proof of Lemma 9.12, and write x0 = x. Set z = x0 x1 x2 , C = CG (z), and I = L3 (C). As the product of three disjoint cycles in J, z is 3-central in J. Furthermore, Z(S) = t × f  and f is not 3-central in G, so by Lemma 9.11c, t is a Sylow 3center in G. The same then holds for z. In particular, no component of I/O3 (I) is simple, by [III8 , 4.1]. There exists u ∈ J of order 3 such that xui = xi+1 for all i, with subscripts modulo 3. Set Ii = L3 (CI (xi )), i = 0, 1, 2. We show that Ii f  ∼ = Aut(L2 (8)) for each i = 0, 1, 2. Indeed by L3 -balance, I0 = L3 (CG (x0 , z)) = E(CK (z)) = E(CK (x1 x2 )) ∼ = L2 (8), with x1 x−1 2 ∈ I0 and f inducing a nontrivial field automorphism on I0 . Our assertion follows for all i by conjugation by u ∈ J ≤ CG (f ). We claim that I0 lies in a single 3-component H0 of I. Otherwise by L3 -balance, I0 is a diagonal of three 3-components H1 , H2 , H3 of I, cycled by x0 . However, I0 /O3 (I0 ) ∼ = L2 (8) is 3-saturated, so H1 /O3 (H1 ) is simple, contradicting what we saw above. This proves the claim. Then  H0 is x0 -invariant, and it is f -invariant since I0 is f -invariant. Let u ∗ H = H0 . If H ∗ > H0 , then H ∗ has three f -invariant components with CH0 (f ) involving L2 (2). Hence CG (z, f ) involves L2 (2) × L2 (2) × L2 (2). However, z is 3-central in J ∼ = A11 and so |CAut(J) (z)|2 = 4, a contradiction. So H0 is u-invariant. Therefore H0 is xi -invariant with L3 (CH0 /O3 (H0 ) (xi )) ∼ = L2 (8) for each i = 0, 1, 2. We saw above that Z(H0 /O3 (H0 )) = 1. But then by [III11 , 22.3], we reach the absurd conclusion that CH0 (f )/O3 (CH0 (f )) is not involved in CG (f, z)/O3 (CG (f, z)). This completes the proofs of Proposition 9.1 and Theorem 3. We are now in a position to prove the following corollary, which will be used frequently to exploit the maximality inherent in the condition (x, K) ∈ J∗p (G).

102

9. THEOREM C∗ 7 : STAGE 3B

Corollary 9.13. Let (x, K) ∈ J∗p (G) and (v, J) ∈ ILop (G). Suppose that K ∈ Chev(r) for some r = p and J ∈ Chev(r) ∩ Gp with dp (K) ≥ dp (J). Then the following conditions hold: (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 ∈ Chev(r) ∩ Gp , dp (I) = dp (K), I/Op (I) is a vertical pumpup of some covering group of J/Op (J), and F(J) < F(I) ≤ F(K). Proof. We have J ∈ Gp ∩ Chev. In particular if p = 2, then J ∈ Alt. Let (w, I) be any long pumpup of (v, J). Note that if p = 2, then since J ∈ Alt,  2Ar for any r = 9, 10, or 11. Hence by [III11 , 1.2], I/Op (I) ∈ Gp with I/O2 (I) ∼ = dp (I) ≤ dp (J) = dp (K) = dp (G). Therefore dp (I) = dp (G). Choosing (w, I) to be p-terminal, as we may by [III7 , Lemma 5.2], we conclude that (w, I) ∈ Jp (G). Thus mp (C(w, I)) = 1 by [III7 , Cor. 5.1]. Since J/Op (J) ∈ Chev(r) by assumption, I/Op (I) ∈ Chev(r) by [III11 , 1.5, 1.4], and Theorem 3. In particular if p is odd, then I/Op (I) ∈ Chev(2). Since (x, K) ∈ J∗p (G), it follows then by [III7 , Definition 3.2] that F(I) ≤ F(K). In view of [III11 , 12.4], we therefore have (9H)

F(J) ≤ F(I) ≤ F(K).

This proves (a). Next assume that F(J) = F(K). Then since J/Op (J) L. Then Lu can play the role of L0 in [III11 , 18.1h], with the automorphism ξ there being conjugation by x. We conclude that [M/Op (M ), x] = 1, a contradiction completing the proof of the lemma.  For the next two sections, we fix x, K, y, L, s, Q, and I as in Lemma 10.5, and let J be the end of the (10G) pivot (s, Q, I), i.e., J is the subnormal closure of I in CG (Q). As (s, Q, I) is nonsingular, J > IO{s,p} (J). We let J = J/Os (J) and q(K) = 2n . The next result is a crucial consequence of the existence of this pivot (and the denial of Theorem 4). It is the main result of this section. Proposition 10.6. G contains a strong s-uniqueness subgroup of component type. We assume the proposition false and argue to a contradiction in the remainder of this section. With notation as in (10G), Lemma 10.1c implies that Los (G)∩Gs = ∅, indeed that (t, J1 ) ∈ ILos (G) for any s-component J1 of J, and J1 /Os (J1 ) ∈ Gs ∩ Chev(2). As G contains no strong s-uniqueness subgroup of component type, the definition of γ(G) gives (10H)

s ∈ γ(G).

Writing Q = t , we can therefore compare (s, t, J1 ) with our choice (p, x, K) satisfying (10C). We first address the case in which J = J1 is a single s-component of CG (Q) = CG (t). Lemma 10.7. If J is a single s-component, then F(J/Os (J)) ≥ F(K); and if equality holds, then s splits J/Os (J). Proof. We know that s splits I, so s divides 22n − 1 by Lemma 10.5b. As J/Os (J) ∈ Chev(2) and I is a component of CJ (x) with xp = 1, one of the following holds, by [IA , 4.2.2, 4.9.1, 4.9.2] and [III11 , 13.40].  q(J)  = 2n , and r(J) ≥ r(I) + 2; (1) s splits J,  q(J)  = 2np and r(J) ≥ r(I); (2) s splits J,  ≥ kr(I); (3) J has level 2n/k , k ≥ 3, and r(J) (10I) n/2   (4) J has level 2 , and r(J) ≥ 2r(I) + 1; or  = 8 = 2r(I); s = 5|q 2 + 1, J ∼ (5) J has level 2n/2 , and r(J) = E8 (q), 2 ∼ SU5 (q ), where q = 2n/2 . and I = Now observe that q(I) = q(K) = 2n , and r(I) ≥ r(K) − 2, by Lemma 10.5bd.  ≥ r(K) and q(J)  = q(K), so f (J)  ≥ If (10I1) holds, then it follows that r(J)   f (K). If f (J) > f (K), then F(J) > F(K) and the assertion of the lemma holds. So  = f (K). Then r(I) = r(K) − 2, whence τ (K) = τ (I) by Lemma assume that f (J)

10. THEOREM 4: THE PRIME s

107

 so τ (K) ≤ τ (J),  and again F(K) ≤ F(J).  10.5d. But by [IA , 4.2.2], τ (I) ≤ τ (J),  As s splits J by (10I1), the lemma holds in this case. Notice that the conditions in Lemma 10.5d imply that (10J)

pr(I)2 ≥ 3r(I)2 > r(K)2 and 2r(I)(r(I) + 1) ≥ r(K)2 .

2 2 2  = 2npr(J) ≥ 2npr(I) > 2nr(K) = f (K) by (10J), Now if (10I2) holds, then f (J)  > F(K). If (10I3) or (10I4) holds, then a similar calculation shows that so F(J)  > F(K), again using (10J). Finally, suppose that (10I5) holds. Then f (J)  = F(J) 232n > f (K), as desired, unless possibly r(K) ≥ 6. In this last case as r(I) = 4, we have r(K) = 6 and τ (K) = τ (I) = A by Lemma 10.5d. As I ↑s K, we must have 2 K∼ = A− 6 (q ). But then since mp (K) ≥ 3, it follows from [IA , 4.10.3] that p splits K, a contradiction. The proof is complete. 

We can now eliminate the quasisimple case. Lemma 10.8. J is not quasisimple. Proof. Suppose that J is quasisimple. We first use Lemma 10.1b to compare  We obtain the middle dp (G), the Gp -depth of K, with ds (G) and the Gs -depth of J. inequality in  ≤ dp (K) = dp (G). ds (G) ≤ ds (J)

(10K)

The first inequality is by definition of ds (G), and the final equality because (x, K) ∈ J∗p (G). By our minimal choice (10C), we must have dp (G) ≤ ds (G), so equality holds throughout (10K):  = dp (G). ds (G) = ds (J)

(10L)

Now choose (x∗ , K ∗ ) ∈ ILos (G) such that (1) K ∗ /Os (K ∗ ) ∈ Gs ∩ Chev(2); (2) J is involved in K ∗ and ds (K ∗ /Os (K ∗ )) = ds (G); (10M) (3) Subject to (1) and (2), F(K ∗ /Os (K ∗ )) is maximal; (4) Subject to (1), (2) and (3), (x∗ , K ∗ ) is s-terminal in G, if possible. We can make such a choice because (t, J) itself satisfies (10M1, 2), by (10L) and the fact that ms (IQ) ≥ 4 (because (s, Q, I) is a pivot). Now by [IG , 6.22] and [III8 , 2.5], there exists an s-terminal long pumpup of (x∗ , K ∗ ), and it still satisfies the conditions (10M1, 2, 3), by [III11 , 13.24,21.1]. Therefore (x∗ , K ∗ ) is s-terminal in G. With (10M2), the definition of Js (G) yields (x∗ , K ∗ ) ∈ Js (G). Now choose (x∗∗ , K ∗∗ ) ∈ J∗s (G). By Lemma 10.7, our choice of K ∗ , and the definition of J∗s (G), (10N)

 ≤ F(K ∗ /Os (K ∗ )) ≤ F(K ∗∗ /Os (K ∗∗ )) F(K) ≤ F(J)

However, by (10C), F(K ∗∗ /Os (K ∗∗ )) ≤ F(K), so equality holds throughout (10N). The first equality implies by Lemma 10.7 that  J)  ∼ s splits J. The second equality implies [III11 , 21.1] that J/Z( = K ∗ /Os s (K ∗ ), ∗ ∗ ∗ ∗ so s splits K /Os (K ). The third inequality implies that (x , K ) has the required

9. THEOREM C∗ 7 : STAGE 3B

108

maximality to put (x∗ , K ∗ ) ∈ J∗s (G) [III7 , Definition 3.2]. Together with (10H), this contradicts (10B) and completes the proof of the lemma.  Lemmas 10.1a and 10.8 yield that J = J 1 · · · Jp where J1 , . . . , Jp are s-components cycled by x, and I = E(CJ (x)) is a diagonal of J. The analysis in this case is similar to the preceding, but requires a preliminary argument; also the case s = 3 is slightly exceptional. Note that Ji = Ji Os (J)/Os (J) ∼ = Ji /Os (Ji ) for each i. We have J1 /Z(J1 ) ∼ = I/Z(I), so J1 and I have the same level and untwisted  Lie rank, and s splits J1 by Lemma 10.5a. Furthermore, as s is odd and ms (I) ≥ 3,  s (J1 ) ≥ 3, by [III11 , 7.6, with J1 ∈ Chev(2), we have ms (I/Z(I)) ≥ 3 and then m 3.1]. As J1 centralizes J2 · · · Jp Q we certainly have ms (C(t, J1 )) ≥ 3. (Indeed ms (CCG (Q) (J1 )) ≥ 3p − 2, but we do not need that fact.) We choose (a, N ) ∈ ILos (G) such that (1) N/Os s (N ) ∼ = J1 /Z(J1 ); (10O) (2) |CG (a) : NCG (a) (N )| ≥ 3; (3) Subject to (1) and (2), ms (C(a, N )) is maximal. Conditions (10O1, 2) are satisfied with a = Q and N = J1 , so the choice (10O) is permissible, and moreover (10P)

ms (C(a, N )) ≥ ms (C(t, J1 )) ≥ 3.

Again by (10D3), we have ms (I) ≥ 3 or ds (I) = 2 according as dp (G) = 4 or dp (G) = 2, and in the latter case ds (N/Os (N )) = ds (J1 ) = ds (I) = 2 by [III11 , 13.24]. Letting d be the Gs -depth of N/Os (N ), we have ds (G) ≤ ds (N/Os (N )) ≤ dp (G), so our minimal choice of dp (G) implies that (10Q)

ds (G) = ds (N/Os (N )) = dp (G),

as in (10L). We set C0 = C(a, N ) and argue next that There exists a0 ∈ Is (C0 ) and E ∈ E3s (C0 ) such that (10R) (1) [a0 , E] = 1; and (2) (a, N ) has a vertical pumpup (a0 , N0 ) in CG (a0 ). Indeed, if (10R) fails, then the hypotheses of [III8 , 2.18] hold with s, a, N in the roles of p, x, and K there. Since N/Os (N ) ∈ Chev(2) ∩ Gs with ds (G) = ds (N/Os (N )) = dp (G), Es2 detects pumpups of N in G, by [III11 , 17.10], as required by [III8 , 2.18]. We conclude that there is (z, M ) ∈ ILos (G) such that either (z, M ) is a diagonal pumpup of (a, N ) and ms (C(z, M )) > ms (C(a, N )), or (z, M ) is a trivial pumpup of (a, N ) and (z, M ) is s-terminal in G. In the first case conditions (10O1, 2) are satisfied by (z, M ) and so our maximal choice of ms (C(a, N )) is violated. In the second case, since M is a trivial pumpup of N , and ms (C(a, N )) > 1, also ms (C(z, M )) > 1. This contradicts [III7 , Cor. 5.1] and establishes (10R).

10. THEOREM 4: THE PRIME s

109

0 = C0 /Os (C0 ). We choose, with [IG , 6.22] and Set C0 = CG (a0 ) and C [III8 , 2.5], a long pumpup (10S)

(a, N ) < (a0 , N0 ) < (a1 , N1 ) < · · · < (ak , Nk )

such that (ak , Nk ) is s-terminal in G. By [III11 , 13.25], Ni /Os (Ni ) ∈ (G2s ∪ G4s ) ∩ Chev(2), and ds (Ni /Os (Ni )) ≤ ds (N/Os (N )) for all i = 1, . . . , k. Hence ds (Ni /Os (Ni )) ≤ ds (G) by (10Q). Thus (ak , Nk ) ∈ Js (G). By [III7 , Cor. 5.1], (10T)

ms (C(ak , Nk )) = 1.

By (10R) and [III8 , 2.20], there is a sequence (10U)

N/Os (N ) = N 0 0 and . If such a long pumpup existed, then one would exist that is m

9. THEOREM C∗ 7 : STAGE 3B

110

3-terminal, by [IG , 6.22, 6.27]. But then f (I) > f (K) and so the maximality (10C) would be violated. In particular Mz /O3 (Mz ) is the direct product of v = 1 or 3 copies of E6η (2n )u . Let Qz ∈ Syl3 (C(z, Mz )). By [IG , 8.7], elements of E∗ (R) normalize the 3-components of Mz . Therefore m3 (Qz ) ≥ m3 (R) − vm3 (Aut(M/O3 (M ))) ≥ 35 − 3 · 7 = 14. In particular by [III8 , 1.3], Qz has a normal subgroup E ∼ = E36 . By [III11 , 17.10], E32 detects pumpups of M/O3 (M ). Now CE (ak ) is noncyclic, so there is z1 ∈ E # such that [z1 , ak ] = 1 and the pumpup H of M in CNk (z1 ) is nontrivial. As Nk /O3 (Nk ) ∼ = E8 (2n ), the only possibilities are that H/O3 (H) ∼ = n E7 (2 ) or H covers Nk /O3 (Nk ), by [IA , Table 4.7.3A]. Now the pumpup H1 of H in CG (z1 ) satisfies H1 /O3 (H1 ) ∼ = E7 (2n ) or E8 (2n ), by (10W). If the latter holds, then since m3 (CAut(E8 (2n )) (E6η (2n )u )) = 2 by [IA , Table 4.7.3A], m3 (C(z1 , H1 ) ∩ E) ≥ 6 − 2 > 1. Hence a 3-terminal long pumpup (z2 , H2 ) of H1 satisfies H2 /O3 (H2 ) ∼ = E8 (2n ) and m3 (C(z2 , H2 )) > 1, contradicting [III4 , 20.3]. However, if H1 /O3 (H1 ) ∼ = E7 (2n ), then ak centralizes H1 /O3 (H1 ), as does E1 = CE (H1 /O3 (H1 )), and |E : E1 | ≤ 3 by [IG , Table 4.7.3A]. Thus m3 (E1 ) ≥ 5, and ak normalizes E1 since it normalizes E. So CE1 (ak ) is noncyclic, and we may apply [III11 , 17.10] again and conclude that for some z2 ∈ E1# , CG (z2 ) has a 3component H2 with H2 /O3 (H2 ) ∼ = E8 (2n ) and m3 (C(z2 , H2 )) ≥ m3 (E1 ) − 1 > 1. This leads to a contradiction just as in the previous paragraph, by taking a 3terminal long pumpup of (z2 , H2 ). Thus (10V) indeed holds. We are now in a position to imitate the proof of Lemma 10.8. Choose (x∗ , K ∗ ) ∈ o ILs (G) such that (10X)

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

K ∗ /Os (K ∗ ) ∈ Chev ∩ Gs ; I/Os (I) is involved in K ∗ and ds (K ∗ /Os (K ∗ )) = ds (G); Subject to (1) and (2), F(K ∗ /Os (K ∗ )) is maximal; and Subject to (1), (2) and (3), (x∗ , K ∗ ) is s-terminal, if possible.

Since (ak , Nk ) itself satisfies (10M1, 2), such a choice is possible. As in the proof of Lemma 10.8, [IG , 6.22] implies that (x∗ , K ∗ ) is s-terminal in G, and condition (10X2) and [III11 , 13.25] yield that (x∗ , K ∗ ) ∈ Js (G). Choose (x∗∗ , K ∗∗ ) ∈ J∗s (G). By (10V), our choice of K ∗ and the definition of J∗s (G) we have (10Y)

F(K) ≤ F(Nk /Os (Nk )) ≤ F(K ∗ /Os (K ∗ )) ≤ F(K ∗∗ /Os (K ∗∗ )).

But as before F(K ∗∗ /Os (K ∗∗ )) ≤ F(K) by (10C), so equality holds throughout (10Y), and as before the three equalities imply in turn that s splits Nk /Os (Nk ) by (10V), that Nk /Os s (Nk ) ∼ = K ∗ /Z(K ∗ ) by [III11 , 21.1] and that (x∗ , K ∗ ) satisfies the maximality requirement to put (x∗ , K ∗ ) ∈ J∗s (G). As s splits K ∗ , we have again contradicted (10B). This completes the proof of Proposition 10.6.

11. THE s-UNIQUENESS CASE

111

11. The s-Uniqueness Case We continue the assumptions and notation of the preceding section. In particular by Lemmas 10.1c and 10.5 and Proposition 10.6, Los (G) ∩ Gs = ∅ but G contains a strong s-uniqueness subgroup of component type, which we shall call M . In this section we contradict the fact that p ∈ γ(G) by proving: Proposition 11.1. M is a strong p-uniqueness subgroup of G of component type. We shall verify that M satisfies the definition [III1 , Definition 1.1] of such a subgroup. The requirements are: (1) M is a maximal subgroup of G and is a p-component preuniqueness subgroup of G, i.e., M has a p-component K such that CG (u) ≤ M for all u ∈ CM (K/Op (K)) of order p; (11A) (2) M is almost strongly p-embedded in G; and (3) Either Op (M ) = 1, or M has a component K ∈ Chev(2) such that mp (CG (K)) ≤ 1. Fix R ∈ Syls (M ). Since M is a strong s-uniqueness subgroup of component type, M is a maximal subgroup of G and is almost strongly s-embedded in G. In particular (see [I2 , 8.1– 8.4]), (11B)

ΓR,2 (G) ≤ M ; in particular R ∈ Syls (G) and NG (R) ≤ M .

We continue to fix an acceptable subterminal (x, K)-pair (y, L), a nonsingular pivot (s, Q, I), and its end J, as in (10G) and Lemma 10.5. Replacing M by a suitable G-conjugate and setting R0 = CR (x) we may assume without loss that R0 ∈ Syls (CG (x)). By (10D), ms (CG (x)) ≥ ms (IQ) ≥ 4 with I ≤ K, so (11C)

ms (R0 ) ≥ 4 and ms (R0 ∩ K) ≥ 3.

We first prove Lemma 11.2. We have CG (x) ≤ M . Proof. As s splits K and ms (R0 ) ≥ 4, K = ΓR0 ,2 (K) ≤ ΓR,2 (G) ≤ M by [III11 , 17.5] and (11B). Then since mp (C(x, K)) = 1 < mp (K), we have K  CG (x), and a Frattini argument and (11B) yield CG (x) ≤ KNG (R0 ∩ K) ≤ KΓR,2 (G) ≤ M.



As in the previous section, we write C = CG (Q). We know from Lemma 10.1c that I is a component of CJ (x), and J is either a single x-invariant s-component of C or the product of p s-components of C cycled by x. We next prove Lemma 11.3. We have J ≤ M .

9. THEOREM C∗ 7 : STAGE 3B

112

Proof. Let R∗ ∈ Syls (C) with R∗ containing a Sylow s-subgroup R1 of C ∩ CG (x). Then R1 ≤ CG (x) ≤ M by the preceding lemma, so ms (R∗ ∩ M ) ≥ ms (R1 ) ≥ ms (IQ) ≥ 4 (see (10D3)). But then using (11B) we see that R∗ ≤ M and then ΓR∗ ,2 (G) ≤ M . In particular Os (J) ≤ M . Let J0 be any s-component of J, and R0∗ = NR∗ (J0 ). Then IQ ≤ R0∗ so ms (R0∗ ) ≥ 4. As the components of J lie in Chev(2), we find again from [III11 , 17.5] that J0 /Os (J0 ) is covered by  ΓR0∗ ,2 (G) ≤ ΓR∗ ,2 (G) and so J ≤ M , as required. Now let 

J



N be the subnormal closure of I in M .   and K = I K , N is the subnormal closure in M both of J and of

Since J = I K. As J is a product of s-components of CM (Q), N is a product of s-components of M by Ls -balance. Since p divides |I|, each s-component of J has order divisible by p, and so the same holds for all the s-components of N/Os (N ). Thus Op p (N ) ≤ Os s (N ) as Sol(N/Os s (N )) = 1. Likewise since K is a p-component of CM (x), N is a product of p-components of M permuted transitively by x, by Lp -balance. But then Op p (N ) is the unique maximal characteristic subgroup of N , whence (1) Op p (N ) = Os s (N ), and therefore (11D) (2) N = N1 · · · Nr , where r = 1 or p, and each Ni is both a pcomponent and an s-component of N . Set M = M/Op (M ) and fix this notation. We next prove Lemma 11.4. The following conditions hold: (a) N is simple; (b) K < N ; (c) N ∈ Gp ∩ Chev(2); and (d) N = F ∗ (M ). Proof. By (10G), IOp (J) < J, so x acts nontrivially on J/Op (J) and hence on J. Consequently x acts nontrivially on N and so K < N , proving (b). Set V = CM (N ), so that Op (M ) ≤ V but x ∈ V . Moreover V , like N , is x-invariant, and we argue that V is a p -group. Otherwise CV (x) would contain an element a of order p, whence a ∈ CG (x) and [a, K] is a p -group as a centralizes K ≤ N . Hence a ∈ CCG (x) (K). But mp (CG (K)) = 1 and so x ∈ a ≤ V , a contradiction. Therefore V is a p -group, and so CF ∗ (M ) (N ) is as well. But N is a product of components of M , so CF ∗ (M ) (N ) ≤ Op (M ) = 1. Consequently N = F ∗ (M ), proving (d), and each N i is simple. If N = N 1 is a single component, then (a) holds. Furthermore K 1. Hence M is of strongly closed type, and the definition of this  term [III1 , Definition 1.2(4)] completes the proof. We use Lemma 11.5 to pull p-local subgroups of G into M . We consider first p-elements of M centralizing a “good” Es2 -subgroup. Definition 11.6.

 D = d ∈ Ip (M ) there exists Ed ≤ CM (d), Ed ∼ = Es2 , ΓEd ,1 (G) ≤ M . d = Cd /Os (Cd ). Then one Lemma 11.7. Let d ∈ D and set Cd = CG (d) and C of the following holds: (a) Cd ≤ M ; or d ) is simple; moreover either (b) F ∗ (C (1) CM (d) is s-solvable; or d ) ∼ M (d)) ∼ (2) s = 5 with F ∗ (C = A10 or F i22 , and F ∗ (C = A5 × A5 or D4 (2), respectively. Proof. Using Lemma 11.5 we apply [III11 , 17.6], with the roles of p, X, and Γ there played by s, Cd Ed , and CM (d)Ed . The upshot is that either (a) or (b) holds, or else CM (d) has an s-component H such that H/Os (H) ∈ Chev(2). But N ∈ Chev(2) is the only non-s-solvable composition factor of M , and so H must be an s-component of CN (d), whence by [IA , Cor. 4.9.6], H/Os (H) ∈ Chev(2), a contradiction. The lemma is proved.  We next prove: Lemma 11.8. Let B ∈ Ekp (M ), k = 2 or 3, and suppose that ΓB,∗−1 (G) ≤ M . If g ∈ G and B g ≤ M , then g ∈ M . That is, B ∈ U(G, M, p). Proof. Set Γ = ΓB g ,∗−1 (M ). By assumption Γ ≤ M g . Since N ∈ Chev(2), and K  CN (x) with p not splitting K and K/Op (K) ∈ Gp , it follows by [III11 , 17.4a] that N = ΓB g ,∗−1 (N ), and hence N Op (M ) ≤ Γ ≤ M g . But then ms (M ∩ M g ) ≥ ms (N Op (M )) ≥ ms (I) > 1. However, by (11B) we have ΓR,2 (G) ≤ M (R ∈ Syls (M )) and so g ∈ M by  [IG , 17.10]. The proof is complete.

9. THEOREM C∗ 7 : STAGE 3B

114

As an application, we prove: Lemma 11.9. x ∈ N . Proof. Suppose false. As N = F ∗ (M ), we may identify M with a subgroup of Aut(N ). If x ∈ Inndiag(N ) − N , then with [IA , 4.2.2], we see that p splits the component K of CN (x), so p splits K, contrary to our assumption in this section. So x ∈ Aut(N ) − Inndiag(N ). Again as p does not split K, p > 3. Thus x is a field automorphism of N . In particular, x ∈ [M, M ]. However, by [III11 , 18.3], there exists a subgroup B ∼ = Ep2 of M such that B = x × d where d ∈ B ∩ N , CN (d) contains a subgroup Ed ∼ = Es2 with ms (CN (Ed )) ≥ 3, and such that CN (d) has a Lie component which, if s = 5, is neither s-solvable nor isomorphic to A5 or D4 (2). Then ΓEd ,1 (G) ≤ M since ΓoR,1 (G) ≤ M , so d ∈ D; moreover CG (d) does not satisfy either of the conditions in Lemma 11.7b. Thus CG (d) ≤ M . As p does not split K, p does not divide | Outdiag(N )|, so by [III11 , 2.3], the subgroups xdj , 0 ≤ j < p, are all N conjugate. As CG (x) ≤ M by Lemma 11.2, we have altogether ΓB,1 (G) ≤ M . Thus by Lemma 11.8, B ∈ U(G, M, p). But then by [III8 , 6.1h], B ≤ [M, M ]. As x ∈ [M, M ] this is a contradiction and the lemma is proved.  Next, let (11E)

A be an (x, K) p-source.

By [III3 , 2.6e] and the fact that mp (K) ≥ 3, we have that A ≤ K ≤ N . Now x ∈ N and p does not split K, so Z(K) is a p -group; thus (11F)

mp (N ) ≥ mp (A × x) = 4.

Using D as in the proof of Lemma 11.9, we can now prove: Lemma 11.10. ΓA,2 (G) ≤ M . Proof. We apply [III11 , 18.2] to obtain the existence of some A0 ∈ E2 (A) such ∼ that for each d ∈ A# 0 , CN (d) contains a subgroup Ed = Es2 with ms (CN (Ed )) ≥ 3, and CN (d) has a Lie component which is non-s-solvable and not isomorphic (if s = 5) to either A5 or D4 (2). Then ΓEa ,1 (G) ≤ M as ΓoR,1 (G) ≤ M so A# 0 ⊆ D. # But for any d ∈ A0 , Lemma 11.7b again cannot hold by our conditions. Thus by Lemma 11.7a, CG (d) ≤ M for all d ∈ A# 0 . Now let D ≤ A with |D| = p2 ; we must show that NG (D) ≤ M . Then D ∩ A0 = 1 as mp (A) = 3, and so CG (D) ≤ M . Thus ΓA,−1 (G) ≤ M , so A ∈ U(G, M, p) by Lemma 11.8. Returning to D, we have A ≤ CG (D), and we expand A to W ∈ Sylp (CG (D)). Then W ≤ M , and so for any g ∈ NG (W ) we have Ag ≤ W ≤ M and then g ∈ M as A ∈ U(G, M, p). So NG (W ) ≤ M and a Frattini  argument ends the proof: NG (D) = CG (D)NNG (D) (W ) ≤ M . We now can complete the proofs of Proposition 11.1 and Theorem 4. First we verify all the conditions (1)–(7) of [III4 , (15A)]. Conditions (1) and (2), that (x, K) ∈ J1p (G) and A is an (x, K) p-source, hold by the hypothesis of Theorem C∗7 : Stage 3b and (11E). Conditions (3,5,6,7) are Lemmas 11.2, 11.4ad, 11.10 and 11.8 respectively. Finally CCM (x) (K) centralizes not only K but also the quasisimple

12. THEOREM 5: THE GENERAL CASE

115

group K  CG (x), so mp (CCM (x) (K)) ≤ mp (CG (K)) = 1. Thus in M0 := CM (K), mp (CM0 (x)) = 1, whence x is p-central in M0 . It follows that any Sylow p-subgroup of C(x, K) is a Sylow p-subgroup of CM (K), which is condition (4). Since [III4 , (15A)] holds, we may apply [III4 , Prop. 17.5], obtaining ΓoP,1 (G) ≤ M for any P ∈ Sylp (M ). However, since N ∈ Chev(2), K is a component of CN (x), and p does not split K, we have mp (CM (u)) ≥ 3 for all u ∈ Ip (M ) by [III11 , 17.4b]. Therefore by definition of ΓoP,1 (G) [III4 , Definition 2.2], CG (u) ≤ M for all such u. By [IG , Prop. 17.11], M is strongly p-embedded in G. As F ∗ (M ) = N , M is a p-component uniqueness subgroup relative to N [I2 , Definition 6.1]. Moreover CG (x) ≤ M and so mp (M ) ≥ 4 as (x, K) ∈ ILop (G). If Op (M ) = 1, then N ∼ = N ∈ Chev(2) and M = NG (N ) by the maximality of M , whence CG (N ) = 1 as N is simple. Thus all the conditions (11A) are verified, and Proposition 11.1 is proved. But p ∈ γ(G) by the hypothesis of Theorem C∗7 , so G has no strong p-uniqueness subgroup of component type by definition of γ(G) (1B2b). This contradiction completes the proof of Theorem 4. 12. Theorem 5: The General Case In this section we make the major reduction in the proof of Theorem 5, whose proof will be completed in the next two sections. We prove the following proposition, in which (12A)

r(H) is the untwisted Lie rank of the group H ∈ Chev.

The setup is as follows: (1) p ∈ γ(G); (2) (x, K) ∈ J∗p (G); (3) K ∈ Chev(s) with p splitting K; (12B) (4) (y, L) is an acceptable subterminal (x, K)-pair, and D = x, y; and (5) N is the (x, K)-neighborhood of (y, L). Proposition 12.1. Under the conditions (12B), one of the following holds: (a) N is level; (b) p = 2 and r(K) ≤ 3; 2 ∼ (c) p = 2, K ∼ = Spin− 8 (q), and L = SL2 (q ); 3 3 (d) p = 2, K ∼ = SL2 (q ); = D4 (q), and L ∼ (e) p = 2, K/Z(K) ∼ = L6 (q) and L ∼ = SL4 (q),  = ±1; − n 2 n (f) p > 2, L ∼ = L4 (q), p divides q − , K/Z(K) ∼ = L− 6 (2 ) or D4 (2 ), and − p for some u ∈ x, y − x, Lu ∼ = L4 (q ); n (g) p > 2, K ∼ (2 ), p divides q − , L ∼ = L− = SL− 7 5 (q), and for some u ∈ − p x, y − x, Lu ∼ = SL5 (q ); (h) p > 2, Op (K) = 1, K is p-saturated, mp (K) = 5 and for some u ∈ x, y − x, x induces a nontrivial field automorphism on Lu ; (i) p > 2, K ∼ = SL3 (q) or = L4 (q) or Sp6 (q), q > 2, p divides q − , L ∼ Sp4 (q), respectively, and for some u ∈ x, y − x, Lu ∼ = SL3 (q p ) or p Sp4 (q ), respectively; (j) p = 3, K ∼ = L6 (2n ) with = E6 (2n ) or A6 (2n ),  = (−1)n , and L/Z(L) ∼  3n Lu /Z(Lu ) ∼ = L6 (2 ) for some u ∈ x, y − x;

9. THEOREM C∗ 7 : STAGE 3B

116

∼ L6 (4n ), L ∼ (k) p = 3, K/Z(K) = = L4 (4n ), and Lu ∼ = L8 (2n ),  = ±1, p n divides 2 + , for some u ∈ x, y − x; (l) p = 5 or 7, K/Z(K) ∼ = Lp (2n ), L ∼ = SLp−2 (2n ), 2n ≡  (mod p), and for  some u ∈ x, y − x, Lu ∼ = SLp−2 (2pn ) or (with p = 5 and  = 1 only) Aη5 (2n/2 ), η = (−1)1+(n/4) ; or (m) p = 5, K/Z(K) ∼ = = SU5 (q), and Lu ∼ = U6 (q), q ≡ −1 (mod 5), L ∼ 1/2 E8 (q ). We assume that the alternative conclusions of the proposition all fail, and we argue to a contradiction. Let the level of K be q(K) = q = sb , where s is prime. Thus exactly one of s and p equals 2, by [III7 , Theorem 1.2]. If q(L) = q, then by [III11 , 14.4], (b), (c) or (d) of the proposition holds, or p > 2 with mp (K) = 2, contradicting Theorem 3. Therefore we may assume that (12C)

q(L) = q(K) = q = sb .

Since (a) fails, we may choose u ∈ D − x such that the pumpup Lu of L in CG (u) satisfies Lu ∈ Chev(s) with q1 := q(Lu ) = q, or Lu ∈ Chev(s). In particular, L < Lu . By Theorem 2, Lu is quasisimple. We set r = r(K). Lemma 12.2. We have r ≥ 4. Proof. If p = 2, the lemma holds as (b) of Proposition 12.1 is assumed to fail. So assume that p > 2 and r < 4. Then s = 2. Since mp (K) > 2 by Theorem 3, we must have r = 3 by [IA , 4.10.3], and indeed the only possibilities are K ∼ = A3 (q) and C3 (q), with  = ±1 and p dividing q − . As K ∈ Gp , q > 2. In these cases L ∼ = SL3 (q) or Sp4 (q) by definition of acceptable subterminal pair [III7 , 6.9], so K and L are as in (i). Clearly L ∈ Alt ∪ Chev(s ) for any odd s . Also mp (L) > 1, so Lu ∈ Chev(2) by [III11 , 1.1ab]. Now, dp (G) = dp (K) = 4. If mp (Lu ) ≥ 3, then dp (Lu ) ≤ dp (K), so F(Lu ) ≤ F(K) by Corollary 9.13. Therefore in these two cases for L, x acts as a field automorphism on Lu , by [III11 , 11.7], and Proposition 12.1i holds, contrary to n assumption, unless possibly L ∼ = SU3 (2n ), n odd, and Lu ∼ = 2F4 (2 2 ). n We eliminate this 2F4 (2 2 ) configuration to complete the proof of the lemma. Note that |Lu |3 = |L|3 , and with [IA , 4.2.3], AutAut(Lu ) (L) is an extension of Inn(L) by a cyclic group of field automorphisms. In particular, AutCG (u) (L) does not contain an outer diagonal automorphism. However, as K ∼ = U4 (2n ), NK (L) n contains a copy of GU3 (2 ), and so AutCG (u) (L) covers P GU3 (2n )/U3 (2n ) ∼ = Z3 . This contradiction completes the proof.  We postpone the proof of the next lemma, which is of a different character, to the end of this section. Lemma 12.3. Assume (12B). Suppose that the conclusions of Proposition 12.1 all fail. Then Lu ∈ Chev(s) for all u ∈ x, y# . Notice also that by [III11 , 15.2a], (12D)

If p > 2, then mp (L) > 1.

12. THEOREM 5: THE GENERAL CASE

117

We write r(L) = r − δ, so that δ is a nonnegative integer, and next prove Lemma 12.4. One of the following holds: (a) δ ≤ 1; or (b) δ = 2 and K/Z(K) ∼ = Lr+1 (q). Moreover either q− ≡ r+1 ≡ 0 (mod p), or p is odd with q ≡ − (mod p). Proof. Assume that the lemma is false. Since p splits K and (b), (c), and (d) of the proposition are assumed to fail, [III11 , 15.12] implies that p = 2, K ∼ = − + ∼ HSpin+ 4k (q) or Spin4k (q), k ≥ 3, and x, y ≤ L = Spin4(k−1) (q), so that δ = 2. But then given the isomorphism type of L, it follows from [III11 , 13.11] that q(Lu ) = q, a contradiction. The lemma is proved.  Returning to the proof of the proposition, we know by [III11 , 13.40] that one of the following holds: (1) q1 = q p and r(Lu ) = r(L); (2) q1 = q 1/a and r(Lu ) = ar(L) + c, for some integers a ≥ 2 and c ≥ 1; or (12E) (3) q1 = q 1/a and r(Lu ) = ar(L) for some integer a ≥ 2. Moreover if r(L) ≥ 3, then p ∈ {2, 5}, and if p = 5, then L ∼ = SU5 (q), q ≡ −1 (mod 5). Let (x∗ , K ∗ ) be an Ipo -terminal long pumpup of (u, Lu ); the existence of (x∗ , K ∗ ) follows from [IG , 6.10]. We next prove Lemma 12.5. dp (Lu ) = dp (K ∗ ) = dp (K), K ∗ /Op (K ∗ ) ∈ Chev(s), and (x , K ∗ ) ∈ Jp (G). ∗

Proof. We claim first that (12F) Lu /Op (Lu ) ∈ Chev(s) ∩ Gp , and Lu is unambiguously of characteristic s. Otherwise, we quote [III11 , 16.11], knowing that conclusions (a) and (b) of that lemma fail. Of the remaining alternative conclusions in [III11 , 16.11], (c) yields Proposition 12.1c or 12.1d, (d) contradicts Theorem 3, (e) yields Proposition 12.1f, (f) contradicts [III3 , 5.6] as d2 (G) ≥ 6 (when p = 2) by (12B2, 3), and (g) contradicts Lemma 12.2. As we are assuming that Proposition 12.1 fails, this proves the claim. As a consequence, by [III11 , 1.2], it follows that the long pumpup K ∗ /Op (K ∗ ) of Lu /Op (Lu ) lies in Gp . This last conclusion also makes use of [III3 , 5.6] and the fact that d2 (G) ≥ 6 to rule out the possibility 2An ∈ Lo2 (G), 9 ≤ n ≤ 11. As the characteristic of Lu is unambiguous, all the terms in a chain of pumpups from Lu to K ∗ are in Chev(s), in view of [III11 , 1.1a], unless along the way there occurs a pumpup I 2, then I ∈ Chev(p) or |I|p = p, whence the same holds for Lu , contradicting the unambiguousness of Lu and the fact (12D) that mp (L) > 1. Thus, (12G)

K ∗ ∈ Chev(s) unambiguously.

9. THEOREM C∗ 7 : STAGE 3B

118

As (x∗ , K ∗ ) is Ipo -terminal, it is p-terminal, by [III7 , 2.5]. It remains to prove that dp (Lu ) = dp (K ∗ ) = dp (G). But (x, K) ∈ Jp (G), so we have dp (G) = dp (K) by definition of Jp (G) [III3 , 5.9]. Thus with [III11 , 1.2], dp (Lu ) ≥ dp (K ∗ ) ≥ dp (K) and we need to prove equality. If p = 2, then since r ≥ 4, we have d2 (G) = 6 and m2 (L) > 2 by [III11 , 15.9], whence Lu ∈ G62 . Thus (x∗ , K ∗ ) ∈ J2 (G) in this case. Therefore we may assume that p > 2, whence mp (K) > 2 by Theorem 3, and  p (K) ≥ 5 or not. For the lemma to fail we can so dp (K) = 2 or 4 according as m only have dp (K) = 2 and dp (Lu ) ≥ 4, or dp (K) = 4 and dp (Lu ) ≥ 5. But then by [III11 , 16.13], one of the conclusions (fghijlm) of Proposition 12.1 holds, contrary to assumption. The lemma is proved.  Lemma 12.6. We have f (K) < f (Lu ) ≤ f (K ∗ /Op (K ∗ )). Proof. As Lu /Op (Lu ) 2 as Sp6 (2) ∈ G3 ), L4 q (q), or L5 q (q). Let z = b0 b1 . By 2 the structure of O (W ) and [III11 , 22.2], z ∼G β. Therefore a Sylow 3-subgroup P1 of CG (z) is isomorphic to P , so Z(P1 ) = z, m3 (P1 ) = 4, and |P1 | = 36 . On the other hand, by [III7 , Theorem 1.2], K1 pumps up to a component Kz of CG (z). As Z(P1 ) = z, z ∈ Kz . Now K1 is simple with m3 (K1 ) = 2, so m3 (Kz ) ≥ 3. Now K1 ∈ Chev(3) so by [III11 , 1.1ab], Kz ∈ Spor ∪ Chev(3). As z ∈ Kz with m3 (Kz ) = 3, Kz ∈ Alt, by [IA , 6.1.4]. Clearly K1 ∈ Chev(r) for any odd r, so q 6 Kz ∈ Chev(2). As m3 (SL± 3 (q)) ≤ 2, while |SL6 (q)|3 > 3 , we have a contradiction to P1 ∈ Syl3 (CG (z)). This contradiction completes the proof of Proposition 12.8 and with it the proofs of Lemma 12.3 and Proposition 12.1. 

13. Theorem 5: Residual Cases, p = 2 In this section we assume the setup (12B), and we prove the following result. Proposition 13.1. Assume (12B), and suppose that p = 2. Then one of the following holds: (a) (b) (c) (d)

For every u ∈ D# , Lu is a level pumpup of L; K∼ = 3D4 (q) or G2 (q) for some q; n K∼ = 2 G2 (3 2 ) for some odd n > 1; or K∼ = P Sp4 (q), or a non-universal version of A± 3 (q) for some q.

Cases (b–d) will be ruled out in monograph 8. Throughout this section we assume that p = 2. We assume that the proposition is false, and we fix a counterexample u ∈ D# . Thus in view of Proposition 12.1, one of the following holds:

9. THEOREM C∗ 7 : STAGE 3B

122

± u u (1) K ∼ = A± 2 (q) (q > 3), A3 (q) , B2 (q) , B3 (q), or C3 (q); − 2 ∼ ∼ (13A) (2) K = Spin8 (q), L = SL2 (q ), and x = Z(K) = Z(L); or (3) K/Z(K) ∼ = L4 (q), for some  = ±1. = L6 (q), and L/Z(L) ∼ We rule these out case by case.

Lemma 13.2. Condition (13A2) does not occur. Proof. Suppose that it does occur, and choose u ∈ D − x such that Lu > L. Note that CK (D) has subnormal subgroups L, L1 , and L2 with x ∈ L ∼ = SL2 (q 2 ), ∼ ∼ u ∈ L1 = SL2 (q), and ux ∈ L2 = SL2 (q) (see [IA , Tables 4.5.1, 4.5.2]. Since x ∈ Lu − Z(Lu ), ux acts nontrivially on Lu , whence by (solvable) L2 -balance,   O 2 (L2 ) ≤ Lu . Thus both L and O 2 (L2 ) are subnormal in CLu (x). Given the isomorphism types of L and L2 , however, this is impossible by [III11 , 13.6]. The lemma follows.  Lemma 13.3. Condition (13A3) does not occur. Proof. Suppose that (13A3) occurs. By definition of acceptable subterminal pair [III7 , 6.9], and given the structure of L, y induces an inner automorphism  on K; and letting s be the characteristic of K, we have O s (CK (y)) = LL1 with  1/2 L ∼ ), = SL4 (q) and L1 ∼ = SL2 (q). By [III11 , 13.15a], Lu ∼ = SL4 (q 2 ) or SL± 8 (q and in either case Ω1 (Z(Lu )) = Ω1 (Z(L)) ≤ D. It follows that Ω1 (Z(Lu )) = Ω1 (Z(L)) = CD (Lu ) = u. 1/2 Suppose that Lu ∼ ). Then with [IA , Table 4.5.1], CAut(Lu ) (L) is = SL± 8 (q 2 supersolvable. It follows that [O (L1 ), Lu ] = 1. In particular as [x, Lu ] = 1, x∈ / u L1 . This implies that LL1 is a central product and consequently Z(K) has odd order. Now choose an involution v ∈ L − Z(L). Note that there is g ∈ K such that v g = u and ug = v. Also v ∈ L ≤ Lu and as an element of Lu , v has four eigenvalues equal to −1, so putting J = E(CLu (v)), we have J = J1 × J2 1/2 with J1 ∼ ) and v ≤ Z(J1 ). As u = Ω1 (Z(Lu )) it follows that = J2 ∼ = SL± 4 (q ∗ vu ≤ Z(J2 ). Let J = E(CG (u, v)), so that J is a product of components of J ∗ . Also g normalizes J ∗ , and as g interchanges u and v it follows that Ω1 (Z(J1g )) = u, whence J1g is a component of J ∗ distinct from both J1 and J2 . Thus [J1g , J1 J2 ] = 1. As CAut(Lu ) (J1 J2 ) is solvable, [J1g , Lu ] = 1, i.e., J1g ≤ C(u, Lu ). In particular m2 (C(u, Lu )) > 1. Hence by Corollary 9.13c, there exists a pumpup I of some covering group of Lu such that (13B)

F(Lu ) < F(I) ≤ F(K).

However, this is impossible by [III11 , 12.5a]. The remaining case is Lu ∼ = SL4 (q 2 ), with x inducing a field or graph-field automorphism of Lu , according as  = 1 or −1. In particular x does not induce an inner automorphism on Lu , so x ∈ L2 (CG (u)), whence by L2 -balance (13C)

x ∈ L2 (CG (D)).

Hence Z(K) has odd order, for otherwise LL1 = L × L1 and x ∈ D = Z(LL1 ), contradicting (13C). We argue that (13D)

x ∈ [CG (u), CG (u)].

For this it suffices, in view of the fact that x maps to an element of Out(Lu ) outside the commutator subgroup [IA , 2.5.12], to show that Lu  CG (u), indeed to show that there is no component J = Lu of CG (u) with J ∼ = Lu . Suppose then that

13. THEOREM 5: RESIDUAL CASES, p = 2

123

  such a component J exists. Let J ∗ = J x and let J0 = E(CJ ∗ (x)). Then by [IA , 4.2.2e, 4.9.1], J0 is a nontrivial product of components of level at least q 2 , or J0 ∼ = SL± 4 (q). But J0 is a product of components of CG (u, x) = CG (D). We have Lo2 (CK (D)) = LL1 with L ≤ Lu ≤ CG (J0 ) and L1 /O2 (L1 ) ∼ = SL2 (q), so the only possibility is that J0 ≤ C(x, K). As m2 (C(x, K)) = 1 it then follows that x ∈ J0 ≤ L2 (CG (D)), contradicting (13C). This establishes (13D). Let S ∈ Syl2 (CG (x)) with D ≤ S. As Z(K) has odd order, Ω1 (Z(S)) = x, u, by [III11 , 3.11e]. Since x induces a field or graph-field automorphism on Lu ∼ = SL4 (q 2 ), x is CG (u)-conjugate to xu. From the structures of Lu and E(CG (x)) it is obvious that x ∈ uG , so u is weakly closed in Ω1 (Z(S)). Hence by (13D) and [III8 , 6.3], x ∈ [G, G], contradicting the simplicity of G. The proof is complete.  We are therefore left with the cases (13A1), in which K has untwisted Lie rank at most 3. To start, an argument analogous to that of Lemma 13.3 rules out (P )Sp6 (q): Lemma 13.4. We have K ∼  C3 (q). =  Proof. Suppose that K ∼ = C3 (q). By [III7 , 6.9], O s (CK (y)) = LL1 with L∼ = Sp4 (q) and L1 ∼ = SL2 (q). By [III11 , 13.15b], Lu ∼ = Sp4 (q 2 ) or Sp8 (q 1/2 ), and in either case Ω1 (Z(Lu )) = Ω1 (Z(L)) = CD (Lu ) = u. If Lu ∼ = Sp8 (q 1/2 ), we repeat the second paragraph of the proof of Lemma 13.3, with Sp8 (q 1/2 ) and Sp4 (q 1/2 ) in place of SL8 (q 1/2 ) and SL4 (q 1/2 ). We again obtain (13B), which again is impossible, this time by [III11 , 12.5b]. Thus Lu ∼ = Sp4 (q 2 ), with x inducing a field automorphism of Lu . We repeat the third paragraph of the proof of Lemma 13.3, again with symplectic groups replacing linear groups, to establish (13D). Again using [III11 , 3.11d], we see that in a Sylow 2-subgroup S of CG (x), u is weakly closed in Ω1 (Z(S)) = D, and again [III8 , 6.3] provides the contradiction x ∈ [G, G] that completes the proof. 

We next prove Lemma 13.5. We have K ∼  B3 (q). = Proof. First suppose that K ∼ = Ω7 (q) or 3Ω7 (3), so that by [III7 , 6.9], y ∈ K  and L ∼ (3). Here  = Ω6q (q) or 3Ω− q = ±1 and q ≡ q (mod 4). With the help of 6 Corollary 9.13, we see that the nonlevel pumpup L < Lu satisfies F(L) < F(Lu ) ≤ F(K). However, this is impossible by [III11 , 12.5c]. Therefore, K ∼ = Spin7 (q) or 3Spin7 (3). Then by [III7 , 6.9] and [IA , Table  4.5.2], O s (CK (D)) = L1 L2 L3 with L = L1 ∼ = L2 ∼ = L3 ∼ = SL2 (q), and the centers of the three Li ’s are the three subgroups of D of order 2. We choose notation so that u ∈ L2 and xu ∈ L3 . Since both x and xu act nontrivially on Lu , it follows  by (solvable) L2 -balance that O 2 (L1 L3 ) ≤ Lu . In particular u = x(xu) ∈ Lu . If q > 3, then we apply [III11 , 13.4] to Lu . Note that Lu ∼ = 3 D4 (t) for any t, the latter group having trivial Schur multiplier by [IA , 6.1.4]. The conclusion from [III11 , 13.4] is that Lu has level q, contrary to our assumption that Lu is not a level pumpup of L. Therefore q = 3, and Lu is not of the form d L(3) for any d and L. In this case we apply [III11 , 13.1] to restrict the isomorphism type of Lu . Notice that as Z(L1 ) = x = u, the alternative conclusion (b) of that lemma fails; so does

9. THEOREM C∗ 7 : STAGE 3B

124

∼ M11 , as the Schur multiplier of M11 is trivial. We the alternative conclusion Lu = conclude that Lu ∈ Chev(2), but Lu ∈ Chev(r)∪Alt for any odd r; also x centralizes no Sylow 2-subgroup of Lu . In particular m2 (Lu ) > 1. On the other hand by [III11 , 3.10a], if we let T ∈ Syl2 (CG (x)), then Ω1 (Z(T )) = O2 (Z(K)) = x, the latter since m2 (C(x, K)) = 1. Therefore x char T so T ∈ Syl2 (G). As all involutions of K − Z(K) are conjugate [III11 , 6.7a], we may assume that T was chosen so that x, u  T . Set Tu = CT (u). Since x does not centralize any Sylow 2-subgroup of Lu , we have Tu ∈ Syl2 (CG (u)). But as |T : Tu | = 2, u is 2-central in G, hence conjugate to x. However, such conjugacy is impossible as CG (x) has no component isomorphic to Lu , since Lu ∼ = K and  m2 (Lu ) > 1 = m2 (C(x, K)). This contradiction completes the proof. Lemma 13.6. We have K/O2 (K) ∼  SL4 (q),  = ±1, or Sp4 (q). = Proof. Suppose that K has one of these isomorphism types and let T ∈ Syl2 (CG (x)). With [III11 , 3.11a], we have x = Ω1 (O2 (Z(K))) = Ω1 (Z(T )) as m2 (C(x, K)) = 1. In particular x char T so T ∈ Syl2 (G). Let x, v1  be a normal four-subgroup of T with v1 ∈ K. Thus v1 has two eigenvalues −1 on the natural  K-module; E(CK (v1 )) (if q > 3) and O 3 (CK (v1 )) (if q = 3) equal L1 × L2 with L1 ∼ = L2 ∼ = SL2 (q); and x = v1 v2 where vi  = Z(Li ). We have v1T = {v1 , v2 }. Let V = v1 , v2  = x, v2 . By [III7 , 6.9], either y is K-conjugate to v1 , or E(CK (y)) ∼ = L3 (q). First suppose that v1 (and then v2 ) is 2-central in G. By [IG , 16.21], NG (V )/CG (V ) is transitive on V # . Thus, Lo2 (CG (V )) has three 2-components or quasi-2-components. The transitive action of NG (V ) implies that these three components are isomorphic, so C(x, K) has a normal subgroup L3 isomorphic to L1 and L2 . We conclude easily that Li ≤ C(vi , E(CG (vi ))) for i = 1, 2, and (13E)



O 2 (Li ) pumps up trivially in CG (vi ), i = 1, 2.

We argue next that (13E) holds even if v1 and v2 are not 2-central in G. For in that case, TV := CT (V ) satisfies |T : TV | = 2, so TV ∈ Syl2 (CG (vi )) for each i, i.e.,  x is 2-central in CG (vi ). Let Ji be the subnormal closure of O 2 (Li ) in CG (vi ), and  assume that Ji is a nontrivial pumpup of O 2 (Li ). If q > 3 then by [III11 , 13.2], x centralizes no Sylow 2-subgroup of Ji , a contradiction as [x, TV ∩ Ji ] = 1. If q = 3 then we obtain the same contradiction from [III11 , 13.3]. This contradiction establishes (13E). We next claim that the situation ∼ SL (q) and L = ∼ SL (q) (13F) K/O2 (K) = 4

3

cannot occur. If it does, then replacing v1 and v2 by conjugates we may assume that [V, D] = 1, v1 ∈ L, and L1 = E(CL (v1 )). Then by [III11 , 13.14], since L < Lu is a nonlevel pumpup and f (Lu ) ≤ f (K) = q 9 if Lu /O2 (Lu ) ∈ G62 (by Corollary 9.13a), we have Lu ∼ = SL3 (q 2 ), unless possibly q = 3,  = −1, and Lu /Z(Lu ) ∼ = G2 (4). In the former case let L∗1 = E(CLu (v1 )) ∼ = SL2 (q 2 ). Then ∗ ∗ ∗ L1 ≤ L1 and as the normal closure of L1 in L1 is L1 , (13E) is contradicted. In the latter case, since G2 (4) has a parabolic subgroup whose unipotent radical has the form 22+8 , T must have sectional rank at least 8. However, since m2 (CT (K)) = 1 and CT (K) ≥ Z(K) ∼ = Z4 (as  = −1), CT (K) is cyclic. As Aut(K) has sectional 2-rank at most 6 by [III11 , 6.12c], T has sectional 2-rank at most 7, contradiction. This shows that (13F) cannot occur.

13. THEOREM 5: RESIDUAL CASES, p = 2

125

It follows that D and V are conjugate, and without loss we take D = V and u = v1 . The pumpup of L1 in CG (u) being trivial, we must have L = L2 and Z(L) = xu. But then CCG (u) (v2 ) = CG (V ) has L2 as a component, so CLu (v2 ) has the subnormal subgroup L2 with v2 ∈ L2 ∼ = SL2 (q). If q > 3, then by [III11 , 13.4], therefore, Lu has level q or Lu ∼ = 3D4 (q 1/3 ). As Lu is not a level pumpup of L, Lu ∼ = 3D4 (q 1/3 ). But then O 2 (CLu (D)) is the 1/3 central product of SL2 (q ) and SL2 (q). On the other hand CG (D) = CCG (x) (D) does not contain two subnormal SL2 -subgroups with the same central involution, contradiction. Therefore q = 3. Next we observe that x is 2-central in CG (u). For otherwise, as in the first paragraph of the proof, NG (V ) = NG (D) is transitive on D# , and it follows easily that Lu ∼ = K, a level pumpup of L. We apply [III11 , 13.1]. Using the facts that CD (Lu ) = u, x is 2-central in CG (u), and Lu is not in Chev(3) of level 3, we conclude that the only possibility is that Lu ∼ = M11 . As ux ∈ Lu ∩ Z(TV ), TV normalizes Lu , and then as M11 is complete [IA , Table 5.3a] and TV ∈ Syl2 (CG (u)), we have TV = (TV ∩ Lu ) × CTV (Lu ). Let t ∈ T be such that ut = ux, and set T0 = CTV (Lu , Ltu )  TV . Since u ∈ Ltu we have T0 ∩ D = 1. But D = Ω1 (Z(TV )) since CΩ1 (Z(TV )) (T ) = x is cyclic. Therefore T0 = 1. It follows that TV = (TV ∩ Lu ) × (TV ∩ Ltu ) and T ∼ = SD16  Z2 . In particular T has no normal Z4 -subgroup, so |CT (K)| = 2. This rules out K ∼ = SU4 (3) as well as Sp4 (3); in the latter case |T | = |CG (x)|2 ≤ 2|Aut(K)|2 = 28 , contradicting |T | = 29 . Therefore K∼ = SL4 (3). This final case can be ruled out as follows. Let T1 = T ∩ K. Thus |T : T1 | = 2 so by the Thompson Transfer Lemma, every involution of T is G-conjugate into I2 (T1 ) = {x} ∪ (uG ∩ T1 ), and hence to either x or u. Let Tu = CTV (Lu ) ∼ = SD16 . For any involution v ∈ Tu , v ∈ xG since [v, Lu ] = 1 and Lu is not embeddable in CG (x). Hence v ∈ uG and so Lu is the unique normal subgroup of CG (v) of its isomorphism type. Therefore v ∈ uNG (Lu ) , so all involutions of CG (Lu ) are NG (Lu )-conjugate, hence 2-central, hence CG (Lu )-conjugate. We now have TV = (T ∩ Lu ) × Tu , with all involutions in either direct factor lying in uG , and all diagonal involutions lying in xG . On the other hand there exists E ∼ = E23 with D ≤ E ≤ T1 , so that E ≤ TV , and all involutions of E − x lie in uG and hence lie  in either T ∩ Lu or Tu . This is impossible, and the lemma is finally proved. In view of Lemmas 13.2–13.6, the only remaining case in (13A) to be ruled out is K ∼ = L± 3 (q). Since K ∈ G2 , q > 3. Lemma 13.7. We have K ∼  A2 (q), q > 3, q odd,  = ±1. = ∼ SL2 (q). Proof. Suppose that K ∼ = A2 (q), q > 3, so that d2 (G) = 7 and L = Since (x, K) ∈ J2 (G), the definitions of this set and of the sets Gi2 imply that for any (t, I) ∈ ILo2 (G), either I/O2 (I) ∈ C2 ∪ T2 or I/O2 (I) ∼ = L3 (q0 ) for some odd prime power q0 > 3, or I/O2 (I) ∼ = A9 or A10 . Consequently by [III11 , 13.5],

126

9. THEOREM C∗ 7 : STAGE 3B

if v ∈ I2 (CG (t)) and CI (v)/O2 (CI (v)) has a subnormal SL2 (q)-subgroup (with q as in the statement of the lemma), then (13G)

I/O2 (I) ∼ = SL2 (q), SL2 (q 2 ), or L± 3 (q)

or else I/O2 (I) ∼ = 2An , n ≥ 7, or [X]L3 (4) with center of exponent 4. But since (x, K) ∈ J∗2 (G), we have d2 (G) = 7, so the last two possibilities cannot occur by [III3 , 5.6]. Thus (13G) is the full list of possibilities. We may assume that notation is chosen so that y = Z(L). Since L < Lu is a nonlevel pumpup, it follows from (13G) that Lu ∼ = SL2 (q 2 ), so that x induces a field automorphism on Lu . Then CD (Lu ) = Z(L) = y so u = y. As x acts nontrivially on Ly , x is Ly -conjugate to xy, by [III11 , 6.3a]. Let S be a Sylow 2-subgroup of CG (x) containing y in its center; then as m2 (C(x, K)) = 1 and Z(K) has odd order, [III11 , 6.4c] implies that D = Ω1 (Z(S)). Note that x ∈ y G . For otherwise all three involutions of D would be conjugate in G, and then in NG (S) ≤ NG (D); so each involution of D would lie in a rank 1 component of CG (D), whereas xy clearly does not lie in such a component in CG (D) = CCG (x) (y), contradiction. Thus y is weakly closed in Z(S) with respect to G, whence by the simplicity of G and [III8 , 6.3], x ∈ [CG (y), CG (y)]. As x maps to a nontrivial element of the abelian group Out(Ly ), it follows from [III8 , 6.8] that there is g ∈ CG (y) such that Lgy = Ly and x induces a nontrivial field automorphism on Lgy . Hence CG (D) = CG (y, x) has a component J = L, J ∼ = L such that u = Z(J) = Z(L). Thus JL ≤ CG (x). As y ∈ C(x, K), JL maps isomorphically to its image in CG (x)/C(x, K) ≤ Aut(K). But this is absurd as Aut(K)(∞) = Inn(K) ∼ = L3 (q) has 2-rank 2 while m2 (JL) = 3. The proof is complete.  By the case division (13A), Lemmas 13.2–13.7 complete the proof of Proposition 13.1.

14. Theorem 5: Residual Cases, p > 2 We complete the proof of Theorem 5 in this section. Thus we prove Proposition 14.1. Assume (12B), and suppose that p > 2. Then for every u ∈ D# , Lu is a level pumpup of L. Throughout this section we assume that the proposition is false. Then one of the cases (f)–(l) of Proposition 12.1 holds. In all but three of these cases, x induces a nontrivial field automorphism on Ld for some d ∈ D − x. We deal with these cases first, listing them according to the

14. THEOREM 5: RESIDUAL CASES, p > 2

127

pairs (K/Z(K), L): − (1) (L− 6 (q), SL4 (q)); − (2) (L7 (q), SL− 5 (q)); (3) (L4 (q), SL3 (q)); (4) (L6 (q), SL5 (q)), p = 3; (5) (Lp (q), SLp−2 (q)), p = 5 or 7; (6) (Sp6 (q), Sp4 (q)), q > 2; (14A) (7) (Sp10 (q), Sp8 (q)); or − (8) (Ω− 8 (q), L4 (q));  (9) (Ω10 (q), Ω+ 8 (q)); − (q), Ω (10) (Ω− 12 10 (q));  (11) (E6 (q), (S)L6 (q)), p = 3; or (12) (L7 (q), SL6 (q)), p = 3. In (14A), q is a power of 2, q 2 ≡ 1 (mod p),  = ±1, and q ≡  (mod p). Thus, the following result accomplishes the bulk of Proposition 14.1. (The three residual cases from 12.1 are taken up afterwards, in Lemmas 14.20, 14.25 and 14.29) Lemma 14.2. Assume (K/Z(K), L) are as in (14A). Then x does not induce a nontrivial field automorphism on Ld for any d ∈ x, y# . #

We assume false, fix d ∈ x, y such that x induces a field automorphism of order p on Ld , and argue to a contradiction in a sequence of lemmas. We first rule out an easy case. Lemma 14.3. It is not the case that K/Z(K) ∼ = Lp+1 (q) with p = 3 or 5, or  ∼ that K/Z(K) = L7 (q) with p = 3. Proof. Suppose that K and p have one of these forms. We will use [III8 , 6.3] to show that x ∈ [G, G]. Let C = CG (x) and Px ∈ Sylp (C) with D ≤ Px . We have L ∼ = SLkp (q) with  p ∼ k = 1 or 2. Let u = D ∩ L. As Ld = SL (q ) for some d ∈ D# , we see kp

that d = u ∈ Z(Lu ). Since CK (D) is a quotient of GLp (q) by a p -group, while mp (C(x, K)) = 1, we have Zx := Ω1 (Z(Px )) = x × Z(Px ∩ L). If k = 1, it follows that Zx = D, while if K/Z(K) ∼ = L7 (q), we have that Zx = x, u, v where o v ∈ L and L3 (CK (v)) = L1 × L2 ≤ L with L1 ∼ = L2 ∼ = SL3 (q). As x induces a nontrivial field automorphism on Ld , x is not p-central. Moreover, as Zx ∩ L ≤ Lu and Lu ∼ = SLkp (q) is the universal version, the subgroups of order p in Zx −(Zx ∩L) are all fused in CG (u). In particular, if k = 1, then u is p-central in G and u is weakly closed in Zx . As x induces a nontrivial field automorphism on Lu , its image in Aut(Lu ) lies outside [Aut(Lu ), Aut(Lu )]. If there exists a second component L∗u ∼ = Lu of NG (u) on which x induces a field automorphism, then CG (D) contains at least two components isomorphic to L, whereas E(CG (D)) = E(CK (u)) = L is a single component, a contradiction. Hence with [III8 , 6.9], x ∈ [NG (u), NG (u)]. If k = 1, this contradicts the simplicity of G, by [III8 , 6.3]. Hence, K/Z(K) ∼ = L7 (q). Let Z = Zx ∩ L. As L1 L2 ≤ L ≤ Lu , we have that E(CLu (Z)) = K1 × K2 with L i ≤ Ki ∼ = SL3 (q p ). Suppose that v = ug for some v ∈ Z − u and g ∈ G. Then

9. THEOREM C∗ 7 : STAGE 3B

128

L1 L2 = Lo3 (CG (x, v)) and K1 K2  E(CG (u, v)). Let Lv = Lgu . If [K1 K2 , Lv ] = 1, then CLv (u) has no component, which is impossible. Hence, Ki ≤ Lv for some i, and Ki is a component of E(CLv (u)). As x induces a field automorphism on Ki , it induces a field automorphism on Lv . But then CG (x, v) has a component isomorphic to L, a contradiction. Hence, u is weakly closed in Zx with respect to G. It follows that u is 3-central in G, and again we may invoke [III8 , 6.3] to obtain a final contradiction.  Lemma 14.4. We have x ∈ Sylp (C(x, K)). Proof. It follows from the definition of acceptable subterminal pair [III7 , 6.9] that y ∈ K in all cases of (14A). Let Q ∈ Sylp (C(x, K)). Then Q is cyclic and Ω1 (Q) = x, which induces nontrivial field automorphisms on Ld . Therefore Q/ x acts faithfully on E(CLd (x)) = L. But [Q, L] ≤ [Q, K] = 1 so Q = x, as desired.  We adopt the following notation. If J is a quasisimple subgroup of G of Lie type, we let NG (J)0 be the set of all g ∈ NG (J) inducing an inner-diagonal or graph automorphism on J. Again let C = CG (x), and set C0 := C ∩ NG (K)0 . Choose B ∈ Ep (C0 ) of maximal rank such that D ≤ A ≤ B for some A ∈ Ep∗ (Kx). Lemma 14.5. Either B ≤ x, K, or K/Z(K) ∼ = E6 (q) with p = 3 and ∼ KB/x = Inndiag(K). Proof. If B ≤ KC(x, K), then by Lemma 14.4, B ≤ x, K, as claimed. If not, then there exists b ∈ B inducing an outer diagonal automorphism on K. If K/Z(K) ∼ = Lp (q) and = E6 (q), then the lemma holds. If not, then K/Z(K) ∼  ∼ ∼ L/Z(L) = Lp−2 (q) with p ∈ {5, 7}. As AutC (K) ≥ KB/Z(KB) = P GLp (q), and B ≥ A, we have E(CK (b)) ∼ = SLp−1 (q) for some b ∈ B # . Therefore, (x, K, y, L) is ignorable, contrary to the definition of acceptable subterminal pair [III7 , 6.9]. This completes the proof.  When the Schur multiplier or outer-diagonal automorphism group of K/Z(K) or L/Z(L) has order divisible by p, there are extra complications. The exceptional triples (K/Z(K), L, p) requiring special attention are easily determined from the list (14A) (recall that q ≡  (mod p)) and [IA , 2.5.12,6.1.4]. Taking Lemma 14.3 into account, we thus have a case division into the following three cases: (1) | Outdiag(K)|, |Z(K)|, and | Outdiag(L)| are p -groups; (2) (K/Z(K), L, p) = (Lp (q), SLp−2 (q), 5 or 7), with p dividing (14B) | Outdiag(K)|; (3) (K/Z(K), L, p) = (E6 (q), A5 (q), 3), with p = 3 dividing both | Outdiag(K)| and | Outdiag(L)|. The fact that x induces a nontrivial field automorphism on Ld yields information about C. This is partly because of the following important fact [III11 , 6.16a]:

(14C)

Let δ, δ  ∈ E1 (Inn(K)) and suppose that CK (δ) and CK (δ  ) each have a component isomorphic to L. If δ and δ  are Aut(K)-conjugate, then they are K-conjugate.

14. THEOREM 5: RESIDUAL CASES, p > 2

129

Lemma 14.6. The following conditions hold: ∼ SL (q) as in (14B2); (a) If x ∈ Z(K), then K = p (b) If f ∈ C is a p-element inducing a field automorphism of order p on K, then f p  = x; (c) Let Px ∈ Sylp (C). Then Ω1 (Px ) ≤ C0 . Moreover if (14B1) holds, then Ω1 (Px ) ≤ K x; and (d) B is of maximal rank in C. ∼ Proof. If x ∈ Z(K) and (a) fails, then we are in case (14B3), with K = E6 (q)u . Then by [III7 , 6.9] and [IA , Table 4.7.3A], L ∼ = SL6 (q) with Z(L) ≤ Z(K) = x. Thus x ∈ Z(L) ≤ Ld , contrary to our assumption that x induces a nontrivial field automorphism on Ld , proving (a). Let Px ∈ Sylp (C), and suppose f ∈ Px and f induces a field automorphism of K of order p. By Lemma 14.4, f p ∈ x. Then by [III11 , 6.16b], there is w ∈ f K such that w centralizes D, whence w induces a field automorphism of order p on L by [IA , 4.2.3]. As w ∈ CG (d) and w normalizes L, w normalizes Ld , and by the action of w on L, we see that w induces a field automorphism of order p2 on Ld . As wp ∈ x, it follows that f p  = wp  = x, proving (b). Now with [IA , 2.5.12,4.9.1] and from our list (14A), any element of C − C0 of order p must induce a field automorphism of K of order p, which contradicts (b). Thus Ω1 (Px ) ≤ C0 . The second statement of (c) also follows since in case (14B1), Outdiag(K) is a p -group. Finally, since Ω1 (Px ) ≤ C0 and B was chosen in Ep∗ (C0 ), (d) is obvious. The proof is complete.  We set B1 = [B, NK (B)], Bd = B ∩ Ld , and BD = B ∩ L. Lemma 14.7. The following conditions hold: (a) If (14B3) does not hold, then B1 = B ∩ K, AutC (B) acts indecomposably on B1 and irreducibly on B/ x, with B1 ∼ = B/ x if x ∈ K; (b) Let s ∈ AutC (B) be of order p. If s acts quadratically on B1 , then (14B3) holds. In no case does s induce a transvection on B1 ; and (c) If (14B3) does not hold, then B = D × Bd and Bd = BD . Proof. If (14B3) holds, we only have to prove that no element of AutC (B) induces a transvection on B1 . But p = 3 and K ∼ = E6 (q) in this case, so B1 /B1 ∩ x ∼ = AutK (B), the = E35 can be identified with the natural module for SO5 (3) ∼ Weyl group of type E6 (see [St1, p. 201]). However, no orthogonal group of odd characteristic contains transvections (this fact is elementary, but also follows from [III8 , 5.1]), so the lemma holds in this case. Next assume that (14B2) holds. Since (y, L) is not ignorable, B contains no element z such that E(CK (z)) ∼ = SLp−1 (q). As x ∈ Sylp (C(x, K)) and B ≤ C0 , it follows, if K is simple, that B = B1 × x with B1 ∼ = V , where V is the irreducible core of the natural permutation module for AutK (B) ∼ = Σp . Note that an element of AutK (B) has a single Jordan block on the natural permutation module, and hence a single Jordan block on V , of size p − 2 ≥ 3. If x ∈ K, then B1 is the trace-0 submodule of the permutation module for AutK (B), and hence is indecomposable. Thus (a) holds in this case. Since B1 involves V , (b) follows. Again since (y, L) is not ignorable, every element of CB (L) of order p is congruent

130

9. THEOREM C∗ 7 : STAGE 3B

to a power of y modulo CB (K) = x, i.e., D = CB (L). Clearly B induces innerdiagonal automorphisms on L, and as Outdiag(L) is a p -group in this case, and Op (L) = 1, B = (B ∩ L) × D = BD × D. As L ≤ Ld , BD ≤ Bd . But x induces a field automorphism on Ld and [d, Ld ] = 1, with Op (Ld ) = 1, so D ∩ Bd = 1. It follows that Bd = BD , proving (c).  In the remaining cases, (14B1) holds. As B ≤ C0 and O p (C0 /Op (C0 )) is the direct product of the images of K and x, we have B = (B ∩ K) × x. Moreover by [III11 , 6.17b], NK (B) is irreducible on B ∩ K so B1 = B ∩ K and (a) holds. Likewise by [III11 , 6.17c], B = BD × D. As at the end of the previous paragraph, this implies that BD = Bd , so (c) holds. Finally, no element of order p in AutK (B)  acts quadratically on B, by [III11 , 6.17d]. The proof is complete. Next, let (14D) Pd be an x-invariant Sylow p-subgroup of CLd (Bd ). Lemma 14.8. Assume that if (14B3) holds, then K and L are simple. Then the following conditions hold: (a) x induces a power mapping of order p on Pd ; (b) Either Pd is homocyclic abelian, or (14B3) holds with Pd isomorphic to a maximal subgroup of a homocyclic abelian group; and (c) Unless (14B3) holds, Bd = Ω1 (Pd ) = [Pd , x]. Proof. Assume that (14B3) does not hold. Then L is p-saturated and Z(L) and Outdiag(L) are p -groups. Moreover, as L is a classical group, p is a good prime for L and so BD represents the unique conjugacy class of maximal elementary abelian p-subgroups of L, by [IA , 4.10.3f]. Also BD = Bd by Lemma 14.7c. Since L = E(CLd (x)) and x induces a field automorphism of order p on Ld by our assumption that Lemma 14.2 fails, Ld is also p-saturated with Z(Ld ) and Outdiag(Ld ) being p -groups. Thus Bd is also maximal elementary abelian in Ld . As Bd ≤ Z(Pd ) by definition of Pd , Bd = Ω1 (Pd ). It follows from [III11 , 6.18] that Pd is homocyclic abelian and x induces a power map of order p on Pd . Hence (a) and (b) hold, and it follows immediately that [Pd , x] = Bd , proving (c). We may therefore assume that p = 3 and L ∼ = L6 (q 3 ). Notice = L6 (q), with Ld ∼ ∼ that CLd (x) = Inndiag(L) by [III11 , 2.4], as a result of which m3 (AutB (L)) = m3 (Inndiag(L)) = 5 and so m3 (Bd ) = 5 = m3 (Ld ). Indeed (see [IA , 4.10.2]) Bd lies in a homocyclic abelian Sylow 3-subgroup of a maximal torus of Inndiag(Ld ) of 3-exponent at least 9, and of rank 5, while | Outdiag(Ld )| = 3. Thus (b) holds,  and again (a) follows from [III11 , 6.18]. The proof is complete. Lemma 14.9. If (14B3) does not hold, then the following conditions hold: (a) Op (AutG (B)) = 1 and CB (Op (AutG (B))) = B1 ; (b) x ∈ Z(K); (c) d ∈ B1  NG (B) and [Pd , B1 ] = 1; (d) C/K has abelian Sylow p-subgroups; and (e) If K/Z(K) ∼ = Lp−2 (q), p = 5 or 7, then p2 does not divide = Lp (q) with L ∼ q − . Proof. Suppose that Op (AutG (B)) = 1. For any s ∈ Pd − CPd (x), the image of s in AutG (B) is a transvection on B = Bd × D moving x. By McLaughlin’s

14. THEOREM 5: RESIDUAL CASES, p > 2

131

Theorem [III8 , 5.1], the subgroup A0 of AutG (B) generated by all its transvections decomposes as A0 = A1 × · · · × Am for some m ≥ 1, and B decomposes as B = CB (A0 ) × B 1 × · · · × B m , such that for each i, [Ai , B j ] = 1 for all i = j, and Ai induces either SL(B i ) or Sp(B i ) (with respect to some symplectic form) on B i . Since AutC (B) is either irreducible on B1 with B = B1 × x, or indecomposable on B = B1 and irreducible on B/ x, it follows easily that m = 1, and AutG (B) contains Sp(B) in its action on B. But then AutG (B) is transitive on B # , contrary to the fact that d ∈ xG . Thus, Op (AG (B)) = 1, and (a) follows. In particular, B is not a simple AutG (B)-module. If x ∈ K, then by Lemma 14.7a, B = B1 is indecomposable as AutC (B)-module with B/x simple. As x is not AutG (B)invariant, this is impossible. Hence x ∈ Z(K), proving (b). Then x and B1 = B ∩ K are the unique proper AutC (B)-submodules of B, whence B is indecomposable as AutG (B)-module with B1 = CB (Op (AutG (B))) the unique proper submodule. Moreover, (e) follows. We turn to (c). Suppose that Pd  CG (B1 ). As Pd centralizes d × Bd , and Bd = BD ≤ B1 , it follows that d ∈ B1 . Thus z = xi d ∈ B1 for some 1 ≤ i < p. Suppose first that z W ∩ Bd = ∅, where W = AutK (B). Let w ∈ W with z w = z1 ∈ Bd . Now, [Pd , xj ] = Bd for all j with 1 ≤ j < p, and so (xj )Pd = xj Bd for all such j. Then dw = (x−i z)w = x−i z w = x−i z1 ∈ x−i Bd ⊆ (x−i )Pd , a contradiction. Thus, if (c) fails, then z W ∩ Bd = ∅. ∼ L (q) with n ∈ {4, 6}, and m(B1 ) = By [III11 , 6.27], this implies that K/Z(K) = n m = n − 1. Moreover, m = p by Lemma 14.3. Now, let A1 := NG (B)/NG (B) ∩  denote the projection map from NG (B) to A1 . As CG (B1 ) and let H → H ∼  acts irreducibly on B1 , we see that B1 is a simple W ≤ NG (B) and W = W A1 -module. In particular, Op (A1 ) = 1. If s ∈ Pd with 1 = s, then as s centralizes Bd , a hyperplane of B1 , it follows that s induces a transvection on B1 . Again, by McLaughlin’s Theorem [III8 , 5.1], we conclude that A1 ≥ Sp(B1 ) or SL(B1 ) in its action on B1 . As m = m(B1 ) = 3 or 5, we see that A1 ≥ SL(B1 ). Let A0 ≤ A1 with A0 ∼ = SL(B1 ). As B1 is a hyperplane of B, [B, A0 ] ≤ B1 . Hence, xA0 ≤ xB1   C0 and W  acts and so |xA0 | ≤ pm . Also, C0 := CA0 (x) = NC (B)/CG (B). As W ∼   absolutely irreducibly on B1 , it follows that |CC0 (W )| ≤ p − 1. As W = Σn , it follows that |A0 | ≤ pm |W | · 2(p − 1) with |W | = 24 or 720, according as m = 3 or 5. On the other hand, |A0 | = |SL(B1 )| = |SL(m, p)| = pm(m−1)/2 (pm − 1) . . . (p2 − 1). If m = 3, then p ≥ 5 and 53 − 1 > 48. If m = 5, then 35 · (35 − 1) > 1440 = 2|W |. Thus, we have a contradiction in both cases. This proves that Pd ≤ CG (B1 ). As [Pd , x] = 1, it follows that B1 = CB (Pd ) and so d ∈ B1 , completing the proof of (c). Finally consider (d). Since x ∈ Sylp (C(x, K)) it is enough to show that Out(K) has cyclic Sylow p-subgroups. But if this fails, then Out(K) contains the image of a field automorphism of order p, and p divides | Outdiag(K)| as well. This implies that (14B2) holds, so that K ∼ = SLp−2 (q), K contains no = Lp (q). As L ∼  element of order p centralizing an SLp−1 (q) subgroup. Thus q ≡  (mod p2 ). But since K admits a field automorphism of order p, q = q0p where q0 ≡  (mod p), and  so q ≡  (mod p2 ). This contradiction completes the proof of the lemma.

9. THEOREM C∗ 7 : STAGE 3B

132

Lemma 14.10. Assume that (14B3) does not hold. Set R = Op (AutG (B)). Then the following conclusions hold: (a) R ∼ = B1 as NK (B)-module, with B = B1 × x; (b) NG (x) = CG (x); and (c) xR = xNG (B) = xB1 . Proof. Since (14B1) or (14B2) holds, and x ∈ K, B = B1 × x. Then [B, R] ≤ B1 , so the map fx : R → B1 defined by fx (r) = [r, x] for all r ∈ R is welldefined. As B1 = CB (R), fx is a group homomorphism. Moreover, for w ∈ NK (B), we have that fx (r)w = [r w , xw ] = [r w , x] = fx (r w ). Thus, fx is a homomorphism of NK (B)-modules with kernel CR (x) = CR (B) = 1. As B1 is a simple NK (B)module and R = 1, it follows that fx is an isomorphism of NK (B)-modules, proving (a). As xr = x[x, r], it follows that xR = xB1 ⊆ xNG (B) . For any b ∈ B1# , b = [r, x] for some r ∈ R and so b is a commutator in a Sylow p-subgroup of CG (b). However, by Lemma 14.9bd, x is not a commutator in a Sylow p-subgroup of CG (x). Thus xG ∩ B1 = ∅, so xNG (B) ⊆ B − B1 = x# B1 . Therefore, (c) will follow from (b). It remains to prove (b). Let g ∈ NG (x); we must prove that g ∈ C. Now g normalizes K. Thus, replacing g by a suitable element of gK, and using (14C), we may assume that g normalizes d. Hence g normalizes L = E(CK (d)) and then Ld . But the image of x in Aut(Ld ) lies outside [Aut(Ld ), Aut(Ld )], so [g, x] = 1. The lemma is proved.  Now, still assuming that (14B3) does not hold, set N1 = NG (B1 ) and C1 = CG (B1 ). Since d ∈ B1 by Lemma 14.9c and in view of Lemma 14.9a, C1 ≤ Cd := CG (d) and NG (B) ≤ N1 . Of course B ≤ C1 and we expand B to U ∈ Sylp (C1 ), and set NU = NN1 (U ), so that N1 = NU C1 by a Frattini argument. As U ≤ C1 ≤ Cd and B1 ∩ Ld ≤ Z(Ld ), U acts on Ld . Let U1 be the preimage in U of Inndiag(Ld ) and set V = CU (Ld ), so that V  U . Notice that since (14B1) or (14B2) holds, Z(Ld ) and Outdiag(Ld ) are p -groups. Thus U1 = Pd × V induces inner automorphisms on Ld . Lemma 14.11. Assume that (14B3) does not hold. Then the following conditions hold: (a) V is a cyclic TI-set in NU ; (b) V  NU ; (c) U1 is homocyclic abelian with Ω1 (U1 ) = B1 ; and (d) U1 is the unique largest abelian subgroup of U . Proof. As N1 = C1 NU and NG (B) ≤ N1 ,  NU   N1   NG (B)  d = B1 . = d ≥ d Since d ∈ V but B1 ≤ V , V is not normal in NU , proving (b).

14. THEOREM 5: RESIDUAL CASES, p > 2

133

Since (14B1) or (14B2) holds, B1 = d × Bd and so C(d, Ld )  C1 , whence V ∈ Sylp (C(d, Ld )). Now we show that V is cyclic. By choice, x ∈ B ≤ U , so V = CU (Ld ) is x-invariant. Since mp (CC (L)) = 2, we have mp (CC(d,Ld ) (x)) = 1 and in particular CV (x) is cyclic, of course containing d. As [U, B1 ] = 1, it follows from [III11 , 6.18a] that the image of Ω1 (U ) in Aut(Ld ) is abelian. Setting U ∗ = [Ω1 (U ), Ω1 (U )], we conclude that U ∗ ≤ U ∩ C(d, Ld ) = V . But U ∗ is NU -invariant, while N1 and NU act irreducibly on B1 , and B1 ∩ V = d < B1 . Therefore d ∈ U ∗ . On the other hand if V were noncyclic we could use [IG , 10.11] to choose V0  V x, V0 ∈ Ep2 (V ). As CV (x) is cyclic and contains d, it would follow that [x, V0 ] = d, whence d ∈ U ∗ , a contradiction. Thus, V is cyclic. Moreover if g ∈ NU with V ∩ V g = 1, then d = Ω1 (V ) is g-invariant. As U ∩ Ld ≥ B1 ∩ Ld is noncyclic while U ∩ C(d, Ld ) is cyclic, g must normalize Ld , hence g normalizes C(d, Ld ) ∩ U = V . This completes the proof of (a). Since Pd and V are abelian, so is U1 = Pd × V . Moreover the image of U/U1 in Out(Ld ) is cyclic, generated by the image of a field automorphism. As Pd is homocyclic abelian, U/U1 induces power mappings on U1 . Let k = mp (Pd ) ≥ 2. These facts quickly imply that U1 is the largest abelian subgroup of U . For if A is an abelian subgroup of U and pa = |AU1 /U1 | > 1, then |U1 : U1 ∩ A| ≥ |U1 : CU1 (A)| = pak > |AU1 : U1 | = |A : U1 ∩ A|, and hence |A| < |U1 |. Thus (d) holds. Furthermore, U1 is characteristic in U and so U1 is invariant under NU , which covers N1 /C1 and so is irreducible on B1 = Ω1 (U1 ). Hence U1 is homocyclic, proving (c) and the lemma.  Lemma 14.12. Assume that (14B3) does not hold. Let P ∈ Sylp (N1 ) with U ≤ P . Then the following conditions hold: (a) P ∈ Sylp (G); (b) U1 is the largest abelian subgroup of P ; and (c) x ∈ [N1 , N1 ]. Proof. We first show that (14E)

U Op (N1 )  N1 and U1 Op (N1 )  N1 .

Since U1 char U by Lemma 14.11d, and Op (N1 ) = Op (C1 ) with U ∈ Sylp (C1 ), it suffices to show that C1 has a normal p-complement. Thus by [III8 , 6.13] it is enough to show that NC1 (Z(Ja (U ))) has a normal p-complement, where Ja (U ) is the subgroup of U generated by all its abelian subgroups of largest order. But Ja (U ) = U1 is abelian. Hence it suffices to show that NC1 (U1 ) has a normal pcomplement. By Frobenius’ Theorem it suffices to show that if U ∗ ≤ U and g is a p -element of NC1 (U ∗ ) ∩ NC1 (U1 ), then [g, U ∗ ] = 1. Replacing U ∗ by U ∗ U1 we may assume that U1 ≤ U ∗ . Then U1 = Ja (U ∗ ) is g-invariant and as B1 = Ω1 (U1 ), [g, U1 ] = 1. But U1 = CU ∗ (U1 ) so [g, U ∗ ] = 1, proving (14E). Next we argue that (14F)

x induces a power map on U1 .

We know that U1 = Pd × V , and x induces, say, mth and nth power maps on Pd and V , respectively. Choose g ∈ N1 ∩ C such that dg  = d. Then dg ∈ B1 . Letting N1 = N1 /Op (N1 ), we have U 1  N1 and V ∩ V g = 1. Hence V g ≤ U 1 and V g embeds in U 1 /V ∼ = U1 /V ∼ = Pd . As g ∈ C, x induces the mth power mapping on Pd , V g , V g , and V . So m = n, proving (14F).

9. THEOREM C∗ 7 : STAGE 3B

134

Since U1 is homocyclic and [Pd , x] = Ω1 (Pd ) = Bd , it follows that [V, x] = Ω1 (V ) and so [U1 , x] = Ω1 (U1 ) = B1 . Thus xU1 = xB1 = xNG (B) , the second equality by Lemma 14.10. This implies that B = Ω1 (U ), and it follows that NU = U1 CNU (x). Using (14E) we see that N1 = Op (N1 )NU , and as U ≤ P , we have P ≤ NU and P = U1 CP (x). Now if (b) fails, then by the Thompson Replacement Theorem [III8 , 1.1], there is an abelian subgroup A ≤ P (of maximal order) acting quadratically and nontrivially on U1 . If [A, B1 ] = 1, then A ≤ C1 ∩ P = U , so A ≤ Ja (U ) = U1 , contradiction. Thus [A, B1 ] = 1. As P = U1 CP (x), there is a ∈ CP (x) acting quadratically and nontrivially on B1 , and hence on B. This contradicts Lemma 14.7b, however, so (b) is proved. As NG (U1 ) ≤ NG (B1 ) = N1 and P ∈ Sylp (N1 ), we conclude that P ∈ Sylp (G), proving (a). Finally to prove (c), let N 1 = N1 /Op (N1 ) = N U . If x ∈ [N1 , N1 ], then x = [N 1 , N 1 ] = [U 1 C 1 , U 1 C 1 ] with U 1  N 1 and C 1 = CN 1 (x), so x ∈ U 1 [C 1 , C 1 ]. Thus x ∈ CU 1 (x)[C 1 , C 1 ] = 1 (U 1 )[C 1 , C 1 ]. But U 1  N 1 , NK (B1 ) ≤ N1 ∩ K is irreducible on Ω1 (U 1 ) = B 1 , and U1 is homocyclic abelian, so 1 (U 1 ) = [N1 ∩ K, 1 (U 1 )] ≤ N1 ∩ K. Therefore x ∈ N1 ∩ K[C 1 , C 1 ], so x ∈ COp (N1 ) (x)K[C, C] and thus x ∈ K[C, C]. However, C/K has abelian Sylow p-subgroups by Lemma 14.9d, and x ∈ K by Lemma 14.9b. As x ∈ Z(C) it follows from Burnside’s Theorem [IG , 15.12(i)] that xK ∈ [C/K, C/K]. This contradiction completes the proof.  Now we can prove Lemma 14.13. Case (14B3) holds, i.e., p = 3 and K ∼ = E6 (q). Proof. Assume false and continue the above argument. As NG (U1 ) ≤ NG (Ω1 (U1 )) = NG (B1 ) = N1 , x ∈ [NG (U1 ), NG (U1 )] by Lemma 14.12c. But U1 is a weakly closed abelian subgroup of P ∈ Sylp (G) by Lemma 14.12ab. Therefore by the Hall-Wielandt Theorem [IG , 15.27], NG (U1 ) controls p-transfer in G. Thus x ∈ [G, G], contradicting the simplicity of G and completing the proof.  ∼ E  (q), For the remainder of the proof of Lemma 14.2, we have p = 3 and K = 6 the adjoint version (see Lemma 14.6a). We let W = NK (B)/CK (B), isomorphic to the Weyl group W (E6 ) ∼ = SO5 (3). Then B1 = [B, W ] ≤ B ∩ K with B1 a 5dimensional simple W -module which we can regard as the natural SO5 (3)-module. Again let Bd = B ∩ Ld . Set B2 = [B ∩ L, NL (B)], so that B2 ≤ Bd ∩ B1 . Now, Ld ∼ = L6 (q 3 ). As B2 ≤ Bd it follows from [III11 , 6.22] that Td := CLd (Bd ) is a maximal torus of Ld , with Bd = Ω1 (Pd ), where Pd ∈ Syl3 (Td ). Recall that D = x, d ≤ B. In fact, K contains a product J ∼ = A1 (q) × A5 (q)  ∼ [GL1, p. 34]. As L = E(CK (d)) = A5 (q) is maximal with respect to inclusion among all centralizers of elements of order 3 in K [IA , 4.7.3A], the projection of d on K is conjugate to a subgroup d  of the A1 (q) component of J. Since d is inverted in A1 (q), we conclude that d ∈ x (B1 − B2 ). Lemma 14.14. The following conclusions hold: (a) d ∈ B1 = d × B2 and dW ∩ B2 = ∅; and (b) If d1 = xi y with y ∈ dW ⊆ B1 and d1 ∈ dG , then i = 0.

14. THEOREM 5: RESIDUAL CASES, p > 2

135

Proof. As L ∼ = L6 (q), B2 is a 4-dimensional simple NL (B ∩L)-module. Moreover, B2 ≤ B1 = [B, NK (B)]. Also, B2 ≤ Bd = Ω1 (Pd ). We see that Pd ≥ Hd with Hd homocyclic abelian of rank 4 and exponent 3s+1 , where 3s = (q − )3 , and [Pd , x] = [Hd , x] = Ω1 (Hd ) = s (Pd ) ≤ Bd . As Td = CLd (Bd ) and NLd (Bd ) = Td NL (Bd ), it follows that NL (Bd ) acts on [Pd , x]. Now, Bd is indecomposable as NL (Bd )-module with unique simple submodule B2 . It follows that [Pd , x] = B2 . Thus xi B2 ⊆ (xi )NG (B) for all i ∈ {1, 2}, and so dG ∩ (xi )B2 = ∅. Let y = xj d1 ∈ B1 with E(CK (y)) ∼ = E(CK (d)) and with d1 ∈ dG ∩ B1 x. Since B1 is the natural SO5 (3)-module, the hyperplane B2 of B1 contains representatives of all W -orbits on B1 , so y W ∩ B2 = ∅. Let g ∈ NK (B) with y g ∈ B2 . Then (d1 )g = (x−j y)g = x−j y g ∈ x−j B2 . As dG = (x−j )G , it follows that j = 0. In particular, d ∈ B1 x ∩ K = B1 . As d ∈ B2 , B1 = B2 × d, proving (a). Then for any y ∈ dW , we have y ∈ B1 ≤ B ∩ K and E(CK (y)) ∼ = E(CK (d)), and so if y = xj d1 with d1 ∈ dG , then j = 0, proving (b).  Recall that (Cd )0 is the subgroup of Cd := CG (d) inducing inner-diagonal automorphisms on Ld ; and that KB/ x ∼ = E6 (q)a or Inndiag(E6 (q)). Thus, by [IA , 4.10.3], m3 (B) ≤ 7. On the other hand, CLd (x) ∼ = P GL6 (q) and so m3 (Bd ) = 5 and m3 (d × Bd ) = 6. As x ∈ d, Bd , it follows that (14G)

m3 (B) = 7.

We set B0 = B ∩ (Cd )0 ; then d × Bd = B0 . In particular, B0 ≥ d × B2 = B1 . Set N1 = NG (B0 ) and C1 = CG (B0 ) ≤ Cd . As [B, W ] = B1 ≤ B0 , we see that NK (B) ≤ N1 . We choose U ∈ Syl3 (C1 ) with B ≤ U , and we let U1 = U ∩ (Cd )0 , the subgroup of U inducing inner-diagonal automorphisms on Ld . Thus the 3-group U/U1 injects into Out(Ld )/ Outdiag(Ld ) and so U/U1 is cyclic. Set V = CU (Ld ) ≤ U1 . Lemma 14.15. The following conclusions hold: (a) V is cyclic; and (b) dG ∩ B = dW ⊆ B1 . Proof. We have Ld U1 /V ∼ = L6 (q 3 ) or P GL6 (q 3 ), and x acts on Ld U1 /V as a field automorphism of order 3. It follows that Ω1 (U/V ) ≤ V B/V and so Ω1 (U ) = Ω1 (V x) × Bd . In particular Bd ∩ Φ(Ω1 (U )) = 1. Using NK (B) ≤ N1 and a Frattini argument, we have dW ⊆ dN1 = dC1 NN1 (U) = dNN1 (U) , since C1 = CG (B0 ) ≤ CG (d). But dW ∩ B2 = ∅ by Lemma 14.14a, whence for some t ∈ NG (U ), dt ∈ B2 ≤ Bd . As Bd ∩ Φ(Ω1 (U )) = 1, it follows that d ∈ Φ(Ω1 (U )). Thus, d ∈ Φ(Ω1 (V x)). As Φ(Ω1 (V x)) ≤ V , it follows that d, x ∩ Φ(Ω1 (V x)) = 1. Now, x CV (x) ≤ CCG (x) (L); but m3 (C(x, K)) = 1 = m3 (CAut(K) (L)), so m3 (x CV (x)) = 2. Hence, CΦ(Ω1 (V x)) (x) = 1, whence Ω1 (V x) is elementary abelian. As m3 (x CV (x)) = 2, Ω1 (V ) = d and V is cyclic, proving (a).

9. THEOREM C∗ 7 : STAGE 3B

136

Suppose that d1 = dg ∈ B and let L1 = (Ld )g . As CG (d1 , L1 ) has cyclic Sylow 3-subgroups by (a), it follows that L1 = E(CG (d1 )) ∼ = L6 (q 3 ). Then, E :=  E(CK (d1 )) = E(CL1 (x)). By [III11 , 6.26], either E ∼ = L6 (q) or E is not defined over Fq and E ∼ = L3 (q 3 ). It follows by [IA , Table 4.7.3A] that E ∼ = L6 (q) and that the images of d1 and d in Aut(K) are conjugate. Consequently d1 ∈ xi {dh | h ∈ NK (B)}  for some i, 0 ≤ i ≤ 2. Then by Lemma 14.14b, d1 ∈ dW , proving (b). Lemma 14.16. xG ∩ B = xNG (B)∩CG (d) . Proof. Let Px ∈ Syl3 (C) with B ≤ Px . We have Ω1 (Px ) ≤ x × K or x × K ∗ with K ∗ ∼ = Inndiag(K). Hence by [IA , 4.10.3c], B = J(Px ) is the unique elementary abelian subgroup of Px of maximal rank. Therefore B is weakly closed in Px , and so xG ∩ B = xNG (B) by [IG , 16.9]. Now let g ∈ NG (B). Then by Lemma −1 14.14b, dg = dw for some w ∈ NK (B). Then dwg = d and xwg = xg . Thus NG (B) x = xNG (B)∩CG (d) , and the lemma is proved.  Lemma 14.17. U1 = Ja (U ) and U1 is abelian. Proof. Suppose that A is an abelian subgroup of U with A  U1 . As U/U1 is cyclic, so is U1 A/U1 . Let |U1 A/U1 | = 3a . By the action of U1 A/U1 on U1 /V , we see that |U1 /V : CU1 /V (A)| ≥ 34a . Hence, (14H)

|U1 : U1 ∩ A| ≥ |U1 /CU1 (A)| ≥ 34a , and so |A| ≤ |U1 |/33a < |U1 |/3.

As U ∈ Syl3 (C1 ) and NK (B) ≤ N1 with NK (B) ∩ C1 = CK (B), a Frattini argument applied to C1 NK (B) shows that AutNG (U) (B0 ) ≥ AutNK (B) (B0 ).  N (U)   N (B)  ≥ d K > d. Suppose that U1 is not abelian. Then In particular, d G  3 Ld U1 /V ∼ P GL (q ). Let P = U ∩ = d 1 Ld and let h ∈ U1 − V Pd . As U1 /V is abelian, 6 [Pd , h] ≤ V ∩ Pd = 1. Also, Vh := CV (h) has index 3 in V . Then V × Pd and Uh := Vh × Pd , h are abelian 3-subgroups of index 3 in U1 , with U1 = (V × Pd )Uh . By (14H), U1 contains every abelian subgroup of U of order at least |U1 |/3. Hence, U1 = Ja (U ) and d = [Ja (U ), Ja (U )] is normalized by W , a contradiction. Thus, U1 is abelian and, by (14H), U1 is the unique abelian subgroup of U of  maximum order. Hence, U1 = Ja (U ). Lemma 14.18. The following conditions hold: (a) C1 = U O3 (C1 ); (b) C1 NK (B) is a subgroup of N1 containing a Sylow 3-subgroup P of N1 such that U  P and U1  P ; (c) P/U1 = U/U1 × Q where Q ∼ = Z3  Z3 ; and (d) B0 = Ω1 (U1 ). Proof. Part (a) is equivalent to the assertion that C1 has a normal 3-complement, and since U ∈ Syl3 (C1 ), it suffices by [III8 , 6.13] to show that NC1 (Z(Ja (U ))) has a normal 3-complement. By Lemma 14.17 it suffices to show that NC1 (U1 ) has a normal 3-complement. Now, m3 (U1 ) = m3 (U1 V /V ) + m3 (V ) ≤ m3 (Inndiag(Ld )) + 1 = 6. On the other hand B ≤ U with U/U1 cyclic and x ∈ B − U1 . As m3 (B) = 7 by (14G), it follows that B0 = Ω1 (U1 ), proving (d). Thus, Ω1 (U1 ) ≤ Z(NC1 (U1 )). The proof of (a) may be completed exactly as that of (14E) above.

14. THEOREM 5: RESIDUAL CASES, p > 2

137

As for (b), notice that AutK (B) acts faithfully on B1 and on B0 . Hence NK (B) ≤ N1 and |C1 NK (B)|3 = |AutK (B)|3 |C1 |3 = 34 |U |. On the other hand, dN1 ⊆ B0 as d ∈ B1 , so by Lemma 14.15b, |N1 | = |N1 ∩ CG (d)||dW | = |N1 ∩ CG (d)||W : CW (d)| = |N1 ∩ CG (d)||SO(5, 3) : Σ6 |. Hence, |N1 |3 = 32 |N1 ∩ CG (d)|3 . Now, |CG (d)|3 = 32 |U |, with B0 normal in a Sylow 3-subgroup of CG (d), and so |N1 |3 = 34 |U | = C1 |NK (B)|3 . Thus, C1 NK (B) contains a Sylow 3-subgroup P of N1 with U  P . Then by Lemma 14.17, U1 = Ja (U )  P , completing the proof of (b). As U  P , P/U1 lies in an extension of the cyclic group U/U1 by 

O 3 (NK (B)U1 /U1 ). The latter group is isomorphic to B2 (3)a ; hence it is perfect, has Z3  Z3 Sylow 3subgroups and a Schur multiplier which is a 3 -group [IA , 6.1.4]. Therefore U/U1 ≤  Z(P/U1 ) and U/U1 is a direct factor of P/U1 , proving (c). Next, set N 1 = N1 /O3 (C1 ), so that U1 ∼ = U 1  N 1 by Lemma 14.18a. By 1 = the structure of Cd , U1 ∈ Syl3 (CCd (U1 )) and so U1 ∈ Syl3 (CG (U1 ). Set N ∼ ∼  N1 /U1 O3 (C1 ) = N 1 /U 1 = AutN1 (U1 ). Let N = NG (U1 ). Since B0 = Ω1 (U1 ), N ≤ N1 and P ∈ Syl3 (N ). Lemma 14.19. The following conclusions hold: (a) U1 is a weakly closed abelian subgroup of P ; (b) P ∈ Syl3 (G); and (c) O 3 (N ) = N . Proof. Suppose that (a) fails and choose any g ∈ G such that A := U1g ≤ P with A = U1 . Let exp(U1 ) = 3s and |U/U1 |3 = 3t . Since U/U1 embeds in Out(Ld )/ Outdiag(Ld ), Aut(Ld ) contains a field automorphism of order 3t , and so s > t. As x ∈ U − U1 , t > 0 and so s ≥ 2. According as Ld U1 /V ∼ = P GL6 (q 3 ) or L6 (q 3 ), U1 /V is isomorphic to a homocyclic group of exponent 3s and rank 5, or to a maximal subgroup of such a group. In either case [NLd (U1 ), Ld ∩ U1 ] contains a homocyclic abelian subgroup U2 of rank 4 and exponent 3s with Ω1 (U2 ) ≤ B1 . Since NK (B) is irreducible on B1 , and B0 /B1 is cyclic, s−1 (U1 ) = B0 = Ω1 (U1 ) or B1 according as U1 is or is not homocyclic. Correspondingly, U1 is homocyclic of rank 6 and exponent 3s or U1 is isomorphic to a maximal subgroup of such a group.  = 1, then B g ≤ s−1 (A) ≤ By Lemma 14.18, P ∼ = Z3t × (Z3  Z3 ). If s−1 (A) 1 g U1 , so B1 ≤ Ω1 (U1 ) = B0 . But d ∈ B1 so by Lemma 14.15b and the irreducibility of AutG (B1 ) on B1 , B1g = B1 . Thus g ∈ NG (B1 ) = NK (B)NCd (B1 ), again by Lemma 14.15b. But U 1  N 1 with NK (B) ≤ N1 . Hence we may write g = g1 g2 with g1 ∈ NN1 (U1 ) and g2 ∈ NCd (B1 ). Thus A = U1g2 . However, U1 is weakly closed  = 1. in a Sylow 3-subgroup of Cd , so A = U1 , a contradiction. Therefore s−1 (A)  Since t < s, the only possibility is that s = 2, t = 1, and A contains an element  × Q,  |A|  ≤ 27, with x  ≤A  in case of equality. In the of order 9. As P = U ∈U latter case, choose a ∈ A ∩ U1 x; as U1 is abelian, we get |U1 : A ∩ U1 | ≥ |U1 : CU1 (a)| = |U1 : CU1 (x)| ≥ 34 > |A : A ∩ U1 |,

9. THEOREM C∗ 7 : STAGE 3B

138

∼ Z9 , whence = so |A| < |U1 |, a contradiction. Hence we are reduced to the case A A ∩ B1 is a hyperplane of B1 . But then if g ∈ A − (A ∩ U1 ), we have that g acts as a transvection on B1 , contradicting Lemma 14.7b and completing the proof of (a). As U1 is weakly closed in P , NG (P ) ≤ N . As P ∈ Syl3 (N ), it follows that P ∈ Syl3 (G), proving (b). As G = O 3 (G) and U1 is a weakly closed abelian subgroup of P , it follows from the Hall-Wielandt Theorem that N = O 3 (N ), completing the proof of the lemma.  Now we complete the proof of Lemma 14.2. By the action of x on Ld U1 /V , x is a power map on U1 /V . There exists w ∈ NK (B) ≤ CN1 (x) with d = dw . Consequently V w ∩ V = 1 and so V w maps isomorphically into U 1 /V . Hence, x  is a power map on V w . As [w, x] = 1, it follows that x  is a power map on U 1 , whence , N  ]. These conditions imply that  ). Moreover as N = O 3 (N ), x  ∈ [N x  ∈ Z(N  and U  is a direct factor of ∈U x  ∈ [P, P], by [IG , 15.12]. However, since 1 = x    P by Lemma 14.18c, x  ∈ [P , P ]. This contradiction completes the proof of Lemma 14.2. One of the three remaining configurations to be ruled out for Proposition 14.1 is the following (from Proposition 12.1j): (1) p = 3, K/Z(K) ∼ = L6 (4n ), L ∼ = L4 (4n ), and Lu ∼ = L8 (2n ), n+1  = (−1) ; and (14I) (2) P ∈ Syl3 (CG (x)) and d ∈ B = J(P ), the subgroup of P generated by all elementary abelian subgroups of maximal rank. Lemma 14.20. The configuration (14I) does not occur. We assume that Lemma 14.20 is false and argue to a contradiction in a short sequence of lemmas. The argument is in part taken from work of Finkelstein and Frohardt [FinFr1]. Lemma 14.21. Assume (14I). Then B ∼ = E35 . Also, if Q ∈ Syl3 (CG (d)) with B ≤ Q, then B = J(Q). Proof. We first argue that (14J)

m3 (C(d, Ld )) = 1.

Notice that m3 (Ld ) = m3 (L8 (2n )) = 4, as  = (−1)n+1 . Therefore d3 (Ld ) = 4 = d3 (K). Hence if (d, Ld ) were 3-terminal in G, we would have (d, Ld ) ∈ J3 (G) and so m3 (C(d, Ld )) = 1 by [III4 , 20.3], as desired. So we assume that (d, Ld ) is not 3-terminal in G and argue to a contradiction. By [IG , 6.22] and [III8 , 2.5], there is a 3-terminal long pumpup (v, J) of (d, Ld ), and a chain from (d, Ld ) to (v, J) must contain at least one vertical pumpup (v0 , J0 ) < (v1 , J1 ), with J0 /O3 (J0 ) a covering group of Ld , by [III8 , 2.4]. Clearly J/O3 (J) ∈ Chev(2). Then F(Ld ) = F(J0 ) < F(J1 ) ≤ F(J) ≤ F(K), the last by [III7 , Definition 3.2] as (v, J) ∈ J3 (G) and (x, K) ∈ J∗3 (G). However, there is no J1 satisfying these inequalities, by [III11 , 12.5h]. This contradiction proves (14J). In particular Ld = L3 (CG (d)). Next suppose that there is f ∈ I3 (CG (x)) inducing a nontrivial field automorphism on K. Then without loss we may take f to centralize y, and by [IA , 4.2.3], f induces a nontrivial field automorphism first on L and then on Ld . Set C = CG (f ) and let J be the subnormal closure of Kf := L3 (CK (f )) in C. Let Lf = L3 (CL (f ))

14. THEOREM 5: RESIDUAL CASES, p > 2

139

    K L and Ldf = L3 (CLd (f )). Since Kf = Lf f and Ldf = Lf df , J is also the subnormal closure of Lf , and of Ldf , in CG (f ). Now J is x, y-invariant and by L3 -balance, J is the product of one or three x, y-conjugate components with Kf and Ldf being 3-components of CJ (x) and CJ (d), respectively. If J has three 3components, one of which is J1 , then Kf /O3 3 (Kf ) ∼ = J1 /O3 3 (J1 ) ∼ = Ldf /O3 3 (Ldf ), which is not the case. Therefore J is a single 3-component, simultaneously a pumpup of Kf and of Ldf . But then by [III11 , 6.24], m3 (CJf  (d Ldf /O3 (Ldf ))) ≥ 3. On the other hand CAut(Ld ) (Ldf ) is generated by the image of f , by [IA , 7.1.4]. Thus, CCG (d) (Ldf ) has Sylow 3-subgroup R × f , where R ∈ Syl3 (C(d, Ld )). As R is cyclic by (14J), m3 (CG (d Ldf /O3 (Ldf ))) = 2, a contradiction. Thus we have proved that every element of CG (x) of order 3 induces an innerdiagonal automorphism on K. A similar argument shows that every element of CG (d) of order 3 induces an inner-diagonal automorphism on Ld . Let E ≤ P be an elementary abelian subgroup of P of maximal rank m. If m ≥ 6, then as m3 (C(x, K)) = 1, m3 (AutE (K)) ≥ 5 and so some b ∈ E # satisfies E(CK (b)) ∼ = SL5 (4n ). But then (x, K, y, L) is ignorable by definition [III7 , 6.7], which contradicts the fact that (y, L) is an acceptable subterminal (x, K)-pair. Consequently m ≤ 5. As m3 (SL6 (4n )) = 5 it follows that E ≤ K x. In fact either x ∈ K ∼ = SL6 (4n ) or K ∼ = L6 (4n ), and in the latter case, m3 (K) = 4, whence n 4 ≡ 1 (mod 9). In either case, AutK (E) ∼ = Σ6 and E = Ω1 (CP (E ∩ K)) is the unique subgroup of its isomorphism type in P . Therefore E = J(P ) = B. Finally if B ≤ Q ∈ Syl3 (CG (d)), then in view of (14J), Ω1 (Q) = d × Ω1 (Q ∩ Ld ) ∼  = E32 × (Z3  Z3 ), so B = J(Q) as well. The lemma is proved. Let A = AutG (B) ≤ Aut(B) ∼ = GL5 (3), Ax = AutNG (x) (B) and Ad = AutNG (d) (B). Therefore, Ad has the normal subgroup AutLd (B) ∼ = Z2  Σ4 acting monomially  on B ∩ Ld ∼ = E34 and centralizing d. Now CAut(Ld ) (B) covers O 3 (Out(Ld )), which is a cyclic group generated by the image of a field automorphism. Hence, |Ad |3 = 3. Likewise Ax has the normal subgroup AutK (B) ∼ = Σ6 , acting irreducibly on (B ∩ K)/Z(K), and, if Z(K) = 1, indecomposably on B. Lemma 14.22. Assume (14I). Then the following conditions hold: (a) The Ad -orbits on E1 (B ∩ Ld ) all have length divisible by 4, and the Ad orbits on E1 (B) − E1 (B ∩ Ld ) − {d} all have length divisible by 8; and (b) The Ax -orbits on E1 (B) − {x} all have length divisible by 5.   Proof. In (a), the orbit lengths on E1 (B ∩ Ld ) are 2i−1 4i , 1 ≤ i ≤ 4, all of which are divisible by 4. Since every element of B ∩ Ld is conjugate to its  inverse under AutLd (B), the orbit lengths on E1 (B) − E1 (B ∩ Ld ) − {d} are 2i 4i , 1 ≤ i ≤ 4, proving (a). Part (b) is obvious.  Lemma 14.23. Assume (14I). Then the following conditions hold: (a) A is irreducible on B; (b) O2 (A) = 1; and (c) |Ax |/|Ad | = 15/8 or 15/4.

9. THEOREM C∗ 7 : STAGE 3B

140

Proof. We have CB (Ad ) = d and CB (Ax ) = x, with Ad and Ax irreducible on B/ d and B/ x respectively. Obviously CG (d) ∼ = CG (x) so d and x are not A-conjugate. Suppose that (a) fails. Then the only proper A-invariant subgroup of B must be B ∩ Ld = B ∩ K ∼ = E34 , with K ∼ = L6 (4n ). Let Ω be the A-orbit of d. Then Ω ⊆ E1 (B) − E1 (B ∩ Ld ) so by Lemma 14.22, |Ω| ≡ 1 (mod 8) and |Ω| ≡ 0 (mod 5) (since x ∈ Ω). Thus |Ω| = 25 or 65. As 9 divides |Ax | and |A|, 9 therefore divides |Ad |. But we saw above that |Ad |3 = 3, a contradiction. Thus (a) holds. Let Z = Z(AutLd (B)). Thus Z = z where z is an involution, [z, B ∩ Ld ] = B ∩ Ld , and [z, d] = 1. Then CA (z) normalizes CB (z) = d so CA (z) = Ad . In particular CA (z) is a {2, 3}-group. Moreover by the structure of Aut(Ld ), O2 (CA (z)) = 1. In particular CO2 (A) (z) = 1 so O2 (A) is abelian. By irreducibility of A, O3 (A) = 1 so O2 (A) is a {2, 3} -group. Let v ∈ O2 (A). Then v = zz a for some a ∈ A. Since z and z a each invert a hyperplane of B, |B : CB (v)| ≤ 32 . But then as v is a {2, 3} -element, v centralizes B/CB (v) and then B, whence v = 1. Thus, O2 (A) = 1, so (b) holds. By irreducibility z ∈ Z(A). Hence by the Z ∗ -Theorem there is a ∈ A such that a [z, z ] = 1 and z = z a . Consequently z a normalizes CB (z) = d and [B, z] = B∩Ld , and then as z = z a , z a inverts d. Now AutAut(Ld ) (B) ∼ = Z2  Σ4 ∼ = AutLd (B), and ∼ it follows that Ad = Z2 × (Z2  Σ4 ). In particular, some element of A inverts B elementwise, and |Ad | = 28 3. On the other hand AutAut(K) (B) = z0  × AutK (B) with z0 inducing a graph or field automorphism on K, and inverting B ∩ K elementwise. It follows that  |Ax | = 6!r with r = 2 or 4. As a result, (c) holds and the lemma is proved. G

G

Let Ξ = x ∩ E1 (B) and Υ = d ∩ E1 (B). Lemma 14.24. Assume (14I). Then the following conditions hold: (a) |Υ|/|Ξ| = 15/8 or 15/4; (b) |Υ| ≡ 1 (mod 4); and (c) |Ξ| ≡ 1 (mod 5). Proof. Since B = J(P ) = J(Q) (see Lemma 14.21), NG (B) is transitive on both Ξ and Υ. Then part (a) is immediate from Lemma 14.23c, while (b) and (c) are immediate from Lemma 14.22.  Now we can reach a contradiction completing the proof of Lemma 14.20. For some integer n > 0 dividing |Aut(B)| = |GL5 (3)|, |Υ| = 15n, |Ξ| = 8n or 4n. In particular 19n ≤ |E1 (B)| = 121 so n ≤ 6. Since |Υ| = 15n ≡ 1 (mod 4), n ≡ 3 (mod 4). So n = 3, whence |Ξ| = 24 or 12. This contradicts |Ξ| ≡ 1 (mod 5), however. The proof of Lemma 14.20 is complete. The second configuration to be ruled out comes from Proposition 12.1l: η p = 5, K/Z(K) ∼ = SL3 (q 2 ), and Ld ∼ = A5 (q), q = 4n , = L5 (q 2 ), L ∼ (14K) n+1 , for some d ∈ D − x. η = (−1) Lemma 14.25. The configuration (14K) does not occur. We assume that (14K) does occur, and fix d as in that condition. We choose E ∈ E54 (K x) to contain D = x, d. Then E ≤ H x for some Cartan subgroup H of K. Set Cx = CG (x), Cd = CG (d), and Nx = NG (x). We argue to a contradiction

14. THEOREM 5: RESIDUAL CASES, p > 2

141

by investigating AutG (E), which in particular has the subgroups AutNx (E) and AutCd (E). We set A = AutG (E). Lemma 14.26. The following conditions hold: (a) (b) (c) (d) (e)

E = Ω1 (O5 (H x)) = D × (L ∩ E) with L ∩ E ∼ = E52 ; AutK (E) ∼ = Σ5 with AutK (E)  AutNx (E); E = (E ∩ Ld ) × d; CE (AutK (E)(∞) ) = x; and A acts absolutely irreducibly on E.

Proof. Let Px ∈ Syl5 (Cx ) contain a Sylow 5-subgroup of H, and set Rx = Px ∩ K. If x ∈ K, then Ω1 (J(Rx )) = Ω1 (O5 (H)) ∼ = E54 admits AutK (H) ∼ = Σ5 faithfully, so (b) holds in this case; also m5 (K) = 4 in this case [IA , 4.10.3] and m5 (L) = 2, so (a) holds as well. On the other hand if x ∈ K, then since (y, L) is an acceptable subterminal (x, K)-pair, it is not ignorable; thus there is no y  ∈ K of order 5 with E(CK (y  )) ∼ = SL4 (q 2 ). It follows that Ω1 (O5 (H)) ∼ = E53 , so E = x × Ω1 (O5 (H)) in this case, and again (a) and (b) follow easily. In either case, the 3-dimensional module E/ x for AutK (E)(∞) ∼ = A5 is the core of the permutation module and so is absolutely irreducible. This implies (d). It also implies the extension E of E to the algebraic closure of F5 has an irreducible constituent of dimension at least 3. Thus if E is irreducible for A, it will be absolutely irreducible and (e) will hold. We claim that E induces inner automorphisms on Ld . Otherwise, since Outdiag(Ld ) is a 5 -group, some e ∈ E # induces a field automorphism of order 5 on Ld . Since [x, e] = 1 and Ld is a component of CLd (x) by L5 -balance, it follows from [IA , 4.2.3] that e induces a nontrivial field automorphism on L. But by definition E ≤ K x, and no element of AutK (L) is a nontrivial field automorphism η (of L). This proves the claim. Note that the Schur multiplier of Ld /Z(Ld ) ∼ = L6 (q)  is a 5 -group, and Sylow 5-subgroups of Ld are homocyclic of rank 3 as 5 | q + η. It follows that ELd = Ld × U , where U is an elementary abelian 5-group containing d. Then U ≤ CCx (Ld ). If |U | > 5, then there exists u ∈ U # such that E(CK (u)) ∼ = SL4 (q 2 ), and so (y, L) is ignorable, a contradiction. Hence |U | = 5 and m5 (E) = m5 (ELd ) = 4. In particular C(d, Ld ) × Ld has an abelian Sylow 5subgroup Pd with Ω1 (Pd ) = E. Hence (c) holds; moreover AutLd (E) is irreducible on E ∩Ld , indeed it contains A4 , by [III11 , 6.21]. In particular x is not AutLd (E)invariant. For later use, we note also that a Sylow p-subgroup Sd of Cd containing Pd is generated by Pd and a field automorphism, and as a result, E ≤ Z(Sd ). To complete the proof we must show that E is irreducible for A. Suppose false, and let V be a minimal nonzero submodule of E. Then either V = x or K ∼ = L5 (q 2 ) with V = E ∩ K. The first case is impossible as x is not even AutLd (E)invariant. As a result, V = E ∩ K = E ∩ Ld ∼ = L5 (q 2 ). As noted = E53 and K ∼ 2 above, O (AutLd (V )) is irreducible on V = E ∩ Ld , so AutNG (E) (V ) ∩ SL(V ) > AutNK (E) (V ) ∼ = Σ5 . From order considerations and [IA , 6.5.3] we conclude that AutNG (E) (V ) ≥ SL(V ). In particular |AutNG (E) (V )|5 = 53 . However, as noted above, E ≤ Z(Sd ) for Sd ∈ Syl5 (Cd ). It follows that the orbit of d under a Sylow 5-subgroup of AutNG (E) (V ) is regular, so |E1 (E) − E1 (V ) ∩ dG | ≥ 53 . As G G |E1 (E) − E1 (V )| = 53 , we have E1 (E) − E1 (V ) ⊆ d . But x ∈ V so x ∈ d .  As E(Cx ) = K ∼ = Ld , this is absurd, and the lemma follows.

142

9. THEOREM C∗ 7 : STAGE 3B

Lemma 14.27. The following conditions hold: (a) m2 ([A, A]) ≤ 3; and (b) Sol(A) = Z(A). Proof. Since [A, A] embeds in SL(E) and any E24 subgroup of GL(E) contains a reflection, by Thompson’s Dihedral Lemma, (a) follows. As for (b), first suppose that A preserves some direct decomposition E = E1 × · · · × Er ,

r > 1.

∼ A5 normalBy irreducibility, r divides 4 = dimF5 (E). In particular AutK (E)(∞) = izes each Ei , and |Ei | ≤ 52 . As AutK (E)(∞) contains A4 , it must centralize each Ei . This is a contradiction. As a result every normal abelian subgroup Na of A lies in Z(A), as A preserves the decomposition of E into Na -eigenspaces. By absolute irreducibility, therefore, |Na | is cyclic of order dividing 5 − 1 = 4. Suppose that Sol(A) > Z(A) and let N be minimal such that Z(A) < N  A and N/Z(A) is elementary abelian. By the previous paragraph N is not abelian, so N has nilpotence class 2. By minimality N/Z(A) is a p-group for some p; then Op (N ) ∩ Z(A) = 1 by the previous paragraph, so p = 2. Any characteristic abelian subgroup of N , being normal in A, is cyclic by the previous paragraph. Thus N is of symplectic type. Let f ∈ AutK (E) be of order 5 and set N0 = [N, f ], an extraspecial group. As N0 ≤ [A, A], m2 (N0 ) ≤ 3 by (a), and then as N0 = [N0 , f ] with f of order 5, N0 ∼ = Q8 ∗ D8 . Hence N0 is absolutely irreducible on E, whence CA (N0 ) = Z(A). Thus N = Z(A)N0 . Now by [III11 , 22.5], N0 is AutK (E)(∞) -invariant and m2 (N0 AutK (E)(∞) ) = 4, contradicting (a). The proof is complete.  Lemma 14.28. E(A) = AutK (E)(∞) . Proof. Since Sol(A) = Z(A) by Lemma 14.27b, F ∗ (A) = Z(A)E(A) and in particular E(A) = 1. Let J be a component of E(A). Suppose first that AutK (E)(∞) ∼ = A5 does not normalize J, and let f ∈ AutK (E)(∞) be of order 5. Then the f -orbit of J consists of five distinct  components of A. Choosing any  2-element a ∈ J − Z(J) we see that B := af  is an abelian 2-group on which f acts nontrivially, so m2 ([f, B] ≥ 4, contradicting Lemma 14.27a. Thus f normalizes all components of E(A). Consequently AutK (E)(∞) normalizes each such component and induces inner automorphisms on it, by the Schreier property. As F ∗ (A) = Z(A)E(A), AutK (E)(∞) ≤ E(A). Now let J = J1 · · · Jr be the product of all components of E(A) on which AutK (E)(∞) projects nontrivially. Since J ≤ E(A) ≤ SL(E), and since CA (d) is a 5 -group, we may quote [III11 , 6.14], with AutK (E)(∞) and J in the roles of Y and X there, to conclude that J = J1 = AutK (E)(∞) , completing the proof.  Lemmas 14.26 and 14.28 give an immediate contradiction. Namely, as AutK (E)(∞)  AutG (E), AutG (E) normalizes CE (AutK (E)(∞) ) = x, which contradicts the irreducibility of AutG (E) on E. This contradiction completes the proof of Lemma 14.25. The final configuration to be ruled out is from Proposition 12.1m: p = 5, K/Z(K) ∼ = SU5 (q), and Ld ∼ = U6 (q), L ∼ = E8 (q 1/2 ), for some (14L) d ∈ D − x, with q ≡ −1 (mod 5).

15. THEOREM 6

143

Lemma 14.29. The configuration (14L) does not occur. Proof. Suppose that it does occur. Let CD = CG (D). Then as D = Z(L) × x, AutCD (L) = AutCG (x) (L) contains Inndiag(L). However, by [IA , 4.2.2] and the (corrected) Table 4.7.3B of [IA ], AutCD (L) ∩ Inndiag(L) ≤ AutCG (d) (L) ∩ Inndiag(L) ≤ AutInndiag(Ld ) (L) = Inn(L), 

a contradiction.

Now Lemmas 14.2, 14.20, 14.25, and 14.29 imply Proposition 14.1. In turn, Propositions 13.1 and 14.1 complete the proof of Theorem 5. 15. Theorem 6 In this section we complete the proofs of Theorem 6 and Theorem C∗7 : Stage 3b, by proving Proposition 15.1. Under the conditions (12B), if N is level and K ∼ = Am (q), then one of the following holds: (a) For any u ∈ x, y# , Lu ∼ = K/Z(K); = L or Lu /Z(Lu ) ∼  ∼ (b) L = Am−2 (q); (c) p = 2, m = 4, q ≡ − (mod 4), and for some u ∈ x, y# , Lu ∼ = B3 (q)u ∼ = Spin7 (q); or # (d) p = 2, K ∼ = A2 (q), and for any u ∈ x, y , Lu ∼ = L, A2 (q), or A− 2 (q). We begin the proof of the proposition by choosing any u ∈ x, y − x such that Lu is a vertical pumpup of L. We assume that (b) and (c) fail, and argue that ∼  Lu /Z(Lu ) ∼ = K/Z(K) or that Lu ∼ = A− 2 (q) with K = A2 (q); thus as u is arbitrary, (a) or (d) holds. Since (b) fails and as p splits K, the definition of acceptable subterminal pair [III7 , 6.9] implies that (15A)

K∼ = Am (q) and L ∼ = Am−1 (q), with p dividing q − .

Moreover L ∼ = SLm (q). Since N is level by assumption, Lu ∼ = d Lr (q) with r ≥ m − 1. Suppose that r = m − 1. Then in view of [IA , 4.2.2] applied to Lu and its automorphism induced by x, either p = 2 or Lu is of exceptional Lie type, with p a bad prime for Lu in the latter case. Then from [IA , Tables 4.5.2, 4.7.3AB], we have ± (Lu , p, L) = (B3 (q), 2, A3 (q)), (E7 (q), 2, A± 7 (q)) or (E8 (q), 3, A8 (q)).

∼ and in neither of the last two cases is L the universal version. Therefore p = 2, Lu = B3 (q), and L ∼ = SL4 (q). But as K ∼ = A4 (q), AutCG (x,y) (L) contains P GL4 (q). Hence from [IA , Table 4.5.1], applied to Lu , q ≡ − (mod 4). Thus (c) holds, contrary to assumption. This contradiction shows that (15B)

2

2

r ≥ m, whence f (K) = q m ≤ q r = f (Lu ).

We will use the maximality of F(K) that is contained by definition in the assumption (x, K) ∈ J∗p (G). For that we need the following lemma.

9. THEOREM C∗ 7 : STAGE 3B

144

Lemma 15.2. Lu ∈ Gdp , where d = dp (G). Proof. Since (x, K) ∈ J∗p (G), dp (G) = dp (K). If p = 2, the combination ∼ ± K∼ = A± m (q), m > 2, and Lu = A2 (q) is impossible as r ≥ m, so by the definition i of the sets G2 [III7 , (1E)], d2 (K) = 6 = d2 (Lu ), as required. On the other hand if p > 2, then by Theorem 3, mp (K) ≥ 3. By definition of the sets Gip [III7 , (1F)], the conclusion of the lemma holds unless (15C)

 p (Lu ), or mp (Lu ) = 2. m  p (K) ≥ 5 > m

Assume, for a contradiction, that (15C) holds. Since mp (K) ≥ 3, certainly mp (L) ≥ 2 by [III11 , 15.2a]. We then apply [III11 , 16.4] with Lu and NCG (u) (Lu ) in the roles of I0 and C there, to obtain a contradiction. The assumptions of [III11 , 16.4] hold because Lu is a vertical and level pumpup of L; by the structures of K and L; and since NCG (u) (Lu ) ∩ K contains a copy of GLm (q) whose commutator subgroup is L. The lemma is proved.  Lemma 15.3. Let u ∈ x, y# be such that Lu > L. Then Lu ∼ = Aηm (q) for some sign η depending on u. Proof. Let (x∗ , K ∗ ) be an Ipo -terminal long pumpup of (u, Lu ) [IG , 6.10]. Using Lemma 15.2 and [III11 , 1.2] we get (x∗ , K ∗ ) ∈ Jp (G). (Note that if p = 2, then Lu is not involved in A11 , as Lu ∈ Gd2 ⊆ Chev, while L± 3 (3) and P Sp4 (3) lie in C2 .) Moreover if p > 2, then Lu is clearly unambiguously of characteristic 2, so K ∗ /Op (K ∗ ) ∈ Chev(2). Then by [III7 , Definition 3.2] and the fact that (x, K) ∈ J∗p (G), it follows that F(K ∗ ) ≤ F(K). But F(K ∗ ) ≥ F(Lu ) by [III11 , 12.4], and therefore (15D)

F(Lu ) ≤ F(K).

Now Lu and K have the same level since N is level, and their respective untwisted Lie ranks satisfy r ≥ m. Therefore (15D) implies first that r = m, and then by the definition of F [III7 , 2.2], the type of the Lie algebra Lu underlying Lu cannot exceed that of K in the ordering [III7 , (2C)]. But Am underlies K, and so  Lu = Am . The lemma follows. Lemma 15.4. Let u and η be as in the preceding lemma, and  as in (15A). If m ≥ 3 or p ≥ 3, then η =  and K/Z(K) ∼ = Lu /Z(Lu ). η Proof. On the one hand x acts on Lu with L  E(CLu (x)), so L ∼ = Am−1 (q) and p divides q − η. However, L ∼ = Am−1 (q), and p splits K by assumption. If p is odd, this implies that η ≡ q ≡  (mod p), and so η =  as desired. If p = 2 and m ≥ 3, then as m − 1 ≥ 2, the groups Am−1 (q) and A− m−1 (q) are not isomorphic, so again η = , as required. 

Now Proposition 15.1 follows at once. Since K ∈ Gp , we have m > 1 [I2 , 12.1, 13.1]. If m = 2 = p, then either condition (d) or condition (a) of the proposition holds, according as η =  or η = , where η is as in Lemma 15.3. Otherwise condition (a) of the proposition holds by Lemma 15.4. With Proposition 15.1, the proof of Theorem 6 is complete. As observed at the beginning of this chapter, Theorem C∗7 : Stage 3b now follows, as it is a consequence of Theorems 1–6.

CHAPTER 10

Theorem C∗7 , Stage 4a: Constructing a Large Alternating Subgroup G0 1. Introduction Having established Theorem C∗7 : Stage 3a in [III5 ], we selected in the previous chapter a prime p ∈ γ(G), a pair (x, K) ∈ J∗p (G), and an acceptable subterminal (x, K)-pair (y, L). In the case K ∈ Chev, for example, we developed finer information about the structure of the neighborhood N = N(x, K, y, L), possibly changing the prime p and even the pair (x, K) in order to achieve the condition that p splits K if K ∈ Chev. The next step for us is to construct a simple or quasisimple K-subgroup G0 of G. The isomorphism type of G0 of course depends heavily on the prime p and the particular isomorphism types of the terminal pair (x, K) and its neighbors (u, Lu ) in the neighborhood N. Thus, we divide the analysis according to these isomorphism types. We shall prove: Theorem C∗7 : Stage 4. Under the assumptions of Theorem C∗7 , there exists a prime p ∈ γ(G), a pair (x, K) ∈ J∗p (G), and an acceptable subterminal (x, K)-pair (y, L) satisfying the conclusions of Theorem C∗7 : Stages 3a, 3b, such that if we set G0 = N(x, K, y, L), then CG (G0 ) is a p -group and one of the following holds: ∼ An , and G0 = ∼ An+4 for some n ≥ 9. Moreover for any root (a) p = 2, K = four-group E ≤ G0 , ΓE,1 (G) ≤ NG (G0 ); or (b) K ∈ Chev(r) and G0 ∈ Chev(r) for some prime r = p such that 2 ∈ {p, r}, and ΓD,1 (G) ≤ NG (G0 ) where D = x, y. The cases (a) and (b) being quite different, we treat them separately. We shall handle the non-Lie type case in this chapter. The Lie type case will be the principal subject of the following volume. Stages 3a and 3b of Theorem C∗7 provide us with a prime p ∈ γ(G) and a quadruple (x, K, y, L) such that (x, K) ∈ J∗p (G), (1A) (y, L) is an acceptable subterminal (x, K)-pair, and if K ∈ Chev, then p splits K, and the various other conclusions of those stages hold [III7 , Theorems 1.2, 6.13]. Important consequences of these stages, for the sets Jp (G) and J∗p (G), include the four-part case division [III7 , Proposition 3.3], which the reader may review at this point. For any quadruple (x, K, y, L) as in (1A), which we shall typically choose and fix, we use the following notation: 145

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

146

(1B)

(1) D = x, y, and for each u ∈ D# , Lu is the subnormal closure of L in CG (u); thus, K = Lx ; # (2) N = N(x, K, y, L) = {L }; and u | u ∈ D # (3) G0 = G0 (N) = N = Lu | u ∈ D .

By Theorem C∗7 : Stages 1, 2, 3a and 3b, N is semisimple, nontrivial, and vertical, i.e., (1) For each u ∈ D# , Lu is quasisimple, or else p = 2, L ∼ = SL2 (3), (1C) and L  LO2 (CG (u)); and (2) For some u ∈ D − x, Lu > L. In this chapter we shall prove: Theorem C∗7 : Stage 4a. Assume that for some p ∈ γ(G) and (x, K) ∈ J∗p (G), n we have K ∈ Chev or K ∼ = 2 G2 (3 2 ) for some odd n ≥ 3 (with p = 2 in the latter case). Then the following conditions hold: (a) p = 2 and K ∼ = An for some n ≥ 9; (b) For any choice of acceptable subterminal (x, K)-pair (y, L), we have G0 ∼ = An+4 (notation as in (1B)); (c) A Sylow 2-subgroup E of CG (K) is G0 -conjugate to a root four-subgroup of K; (d) ΓE,1 (G) ≤ NG (G0 ); and (e) CG (G0 ) has odd order. The proof is divided into the following four subtheorems. Theorem 1. Let p ∈ γ(G) and (x, K) ∈ J∗p (G). Then K ∈ Spor.

In the course of proving Theorem 1, we need to rule out the possibility K ∼ = J1 . n The argument in this case effectively rules out the possibilities K ∼ = 2 G2 (3 2 ) for any n ≥ 3 at the same time, and so as a side benefit for the next volume we shall also obtain: n Theorem 2. Let p ∈ γ(G) and (x, K) ∈ J∗p (G). Then K ∼  2 G2 (3 2 ) for any = odd n ≥ 3. This explains the incongruous appearance of the Ree groups in Theorem C∗7 : Stage 4a. Theorems 1 and 2 are proved rather quickly, while the following result takes more work. It was originally proved by Aschbacher [A22]. Theorem 3. Let p ∈ γ(G) and (x, K) ∈ J∗p (G), and suppose that K ∈ Alt. Then p = 2. If m2 (C(x, K)) > 1, then K ∼ = An for some n ≥ 9, and a Sylow 2subgroup E of CG (K) is a G-conjugate to a root four-subgroup of K. Moreover, for any acceptable subterminal (x, K)-pair (y, L), we have G0 ∼ = An+4 , with notation as in (1B). Finally, ΓE,1 (G) ≤ NG (G0 ), and CG (G0 ) has odd order. Theorem 4. Let p ∈ γ(G) and (x, K) ∈ J∗p (G), and suppose that K ∈ Alt. Then p = 2 and m2 (C(x, K)) > 1. Clearly Theorems 1, 2, 3, and 4 imply Theorem C∗7 : Stage 4a. Note also that by [III7 , Prop. 3.3], the assumptions on K in Theorems 1, 2, 3, and 4 imply in each case that (1D)

p = 2,

which condition shall be in force throughout this chapter. (In Theorem 3, if p = 3, then K ∼  A8 as m3 (CG (x)) ≥ 4.) =

2. THEOREMS 1 AND 2: MOST SPORADIC COMPONENTS

147

2. Theorems 1 and 2: Most Sporadic Components In this short section we consider the case K ∈ Spor and rule out all but one case, proving: Proposition 2.1. Assume (1A), (1B) and (1C). If K ∈ Spor, then K ∼ = J1 . We assume the proposition to be false. By (1D), p = 2. Then as (x, K) ∈ J∗2 (G) by (1A), we have K/O2 (K) ∈ G2 by definition of J∗2 (G) [III7 , 3.2]. Hence by the definition of G2 [I2 , 13.1], (2A) K/Z(K) ∼ = O  N , M c, Ly, He, Co3 , or F5 . By [III11 , Lemma 5.1a], Z(K) has odd order. Moreover by [III7 , Prop. 3.3], m2 (CG (K)) = 1. We choose an acceptable subterminal (x, K)-pair (y, L), so that by [III7 , Definition 6.9], (2B)

In the six cases of (2A), L/O2 (Z(L)) is isomorphic, respectively, to 4L3 (4), 2A8 , 2A11 , 22 L3 (4), 2Sp6 (2), or 2HS.

By [III11 , Lemma 5.1c], every involution in CAut(K) (L) is the image of an involution in Z(L). Since the definition [III7 , 6.9] of acceptable subterminal pair specifies y ∈ K in the sporadic case, we have (2C)

y ∈ Z(L).

Using (2C) we first show Lemma 2.2. L is a component of CG (y). Proof. Suppose not, so that L < Ly and Ly is quasisimple. Given the isomorphism type of L  E(CLy (x)), [III11 , 5.1d] implies that only the cases K/Z(K) ∼  Co3 . By the same = M c, Ly and Co3 are possible. Suppose first that K ∼ = lemma, L ∼ = 2Am , m = 10 or m ≥ 12. Note from the = 2An , n ∈ {8, 11}, and Ly ∼ isomorphism type of K and the fact that (x, K) ∈ J∗2 (G) that d2 (G) = d2 (K) > 1. But this is impossible by [III3 , Lemma 5.6ab]. Thus K ∼ = (2 × 2)D4 (2), again by = Co3 and L ∼ = 2Sp6 (2), whence Ly ∼ [III11 , 5.1d]. We shall reach a transfer contradiction. As E(CLy (x)) = 1, x induces a non-inner automorphism on Ly by the Borel-Tits theorem. Moreover, a Sylow 2-subgroup of C(y, Ly )∩CG (x) embeds in CCG (x) (L), which lies in y×C(x, K) by [III11 , 5.1c]. As m2 (C(x, K)) = 1, the only component of CCG (y) (x) = CCG (x) (y) isomorphic to L is L itself, and so CCG (y) (x) must normalize L. But Out(Ly ) ∼ = Σ3 by [IA , 2.5.12]. It follows that a Sylow 2-subgroup of C(x, K), already known to be of rank 1, equals x. Then CE(CG (y)) (Ly ) x, y has Sylow 2-subgroups of maximal class by [IG , 10.24], whence Ly  CG (y). This implies in turn that x ∈ [CG (y), CG (y)]. Let T ∈ Syl2 (CG (x)) with x, y ≤ Z(T ). Now Out(Co3 ) = 1 [IA , Table 5.3j], and Sylow 2-centers of Co3 are of order 2 by [III11 , 3.12]. Therefore T = x × (T ∩ K) and Z(T ∩ K) = y. As |T | = 211 < |Ly |2 , y ∈ xG and T ∈ Syl2 (G). Since Z(T ) = x, y, xy ∈ xNG (T ) , whence y is weakly closed in Z(T ) with respect  to G. By [III8 , 6.3], x ∈ [G, G], contradicting the simplicity of G. The next lemma gives a symmetry between x and xy.

148

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

Lemma 2.3. Lxy ∼ = K and m2 (C(xy, Lxy )) = 1. Proof. Let (z, Kz ) be a 2-terminal long pumpup of (xy, Lxy ) [IG , 6.10]. Since N is a vertical neighborhood and Ly = L, L < Lxy . Then if Lxy ∼  K, by = [III11 , 5.1d], Lxy ∼ = M c or Ly), or Lxy ∼ = 2An , n = 10 or n ≥ 12 (with K ∼ = 22 D4 (2) (with K ∼ = Co3 ). Then y ∈ Z(L) ≤ Z(Lxy ), so CG (Lxy ) ≥ xy, y = x, y, which is absurd. Thus Lxy ∼ = Lxy . = K. Again by [III11 , 5.1g], Kz /O2 (Kz ) ∼ Thus (z, Kz ) ∈ J2 (G), so m2 (C(z, Kz )) = 1 by [III7 , 5.1]. This implies that  m2 (C(xy, Lxy )) = 1 by [III8 , 2.4], completing the proof. Lemma 2.4. L is the unique component of CG (y) of its isomorphism type. Proof. Set Q := C(y, L) and C := CAut(K) (L). Then x ∈ Q and by [III11 , 5.1c], C ∼ = Z(L), unless L ∼ = 2HS, in which case C ∼ = Z4 . Note that CQ (x) normalizes K and centralizes L, whence |CQ (x)/C(x, K)| ≤ 4. Recall that m2 (C(x, K)) = 1. Now suppose that the lemma fails and that L1  E(CG (y)) with L1 ∼ = L, L1 = L. Suppose that x does not normalize L1 and set L∗1 = L1 Lx1 . Then CL∗1 (x) ≤ CQ (x) but as x is the unique involution of C(x, K), CL∗1 (x)∩C(x, K) has odd order. Thus CL∗1 (x) has Sylow 2-subgroups of order at most 4. But CL∗1 (x) contains a diagonal component L0 isomorphic to some central quotient of L1 , so |L1 /Z(L1 )|2 ≤ 4, a contradiction. Therefore, x normalizes L1 . Next, if x ∈ L1 , then as in the previous paragraph C(x, K) ∩ L1 has odd order, whence Sylow 2-subgroups of CQ (x) embed in C and hence have order at most 4. But given the possible isomorphism types for L1 , this contradicts [III11 , 5.1e]. Therefore x ∈ L1 . But then CL1 (x) contains an abelian 2-subgroup of rank at least 2 + m2 (Z(L)), by [III11 , 5.1f]. Since m2 (CQ (x)) ≤ m2 (C(x, K)) + m2 (C) = 1 + m2 (Z(L)), this is a contradiction and the proof of the lemma is complete.  Now let V = {v ∈ y K ∩ L | v ∈ Z(L)}. For any v ∈ y G , we let L(v) denote the unique component of CG (v) isomorphic to L. In particular, if v ∈ V, then L(v) = E(CK (v)). Lemma 2.5. V is a nonempty set. Moreover, for any v ∈ V, L(v) is the unique component of CG (v) of its isomorphism type. Proof. The first statement is clear if I2 (K) = y K , as in K = O  N , M c, or Ly. If K ∼ = Co3 , then L contains a Sylow 2-subgroup of K with Z(L) = y. Hence v ∈ V exists by the Z ∗ -Theorem [IG , 15.3]. Thus we may assume that K ∼ = He or F5 . Let T ∈ Syl2 (CK (y)), Z = T ∩ Z(L), T ≤ S ∈ Syl2 (K). In these cases we see from [IA , Tables 5.3pw] that T ∈ Syl2 (NK (L)), L = E(CK (z)) for all z ∈ Z # , and y is not 2-central in K. Therefore there exists s ∈ NS (T ) − T with Z s ≤ Z(T ), Z s ∩ Z = 1. Let v = y s . Then v ∈ Z(T ) − Z. As v ∈ Z(T ), v ∈ L by [III11 , 3.13], whence v ∈ V. Thus V = ∅ in all cases. The final assertion of the lemma has already been discussed above.  Lemma 2.6. One of the following holds: (a) y G ∩ K = y K and y G ∩ Lxy = y Lxy ; or (b) CG (y) normalizes K, and K ∼ = Co3 or F5 .

3. THEOREMS 1 AND 2: J1 AND REE COMPONENTS

149

Proof. If K ∼ = M c, Ly, or O  N , then I2 (K) = y K and (a) is clear, using the symmetry between x and xy. Suppose next that K ∼ = He and that (a) fails, say for K. Thus there is v ∈ (∞) G y ∩ K with v 2-central in K. Let C1 = CK (v). As C1 = C1 and L(v)  CG (v), C1 ≤ L(v)Q1 , where Q1 = C(v, L(v)), by the Schreier Property. Let CG (v) = CG (v)/Z(L(v)), so that C1 ≤ L(v) × Q1 . By the structure of C1 and of L(v), the projection of C 1 into L(v) is either trivial or isomorphic to L3 (2), since L3 (4) contains no 2-local subgroup involving L3 (2). In either case, as L(v) ∩ C1  C1 , it follows that L(v)∩C1 = v and Q1 ∩C1 ≥ O2 (C1 ). Now CAut(K) (O2 (CK (v))) = v by [III11 , 5.4]. As x normalizes L(v) and CL(v) (x) ≤ CCG (x) (Q1 ), it follows that CL(v) (x) ≤ C(x, K) × v. Then m2 (CG (x, K)) = 1 implies that m2 (CL(v) (x)) ≤ 2. Thus, by [III11 , 2.6ab], x induces an outer automorphism on L(v). But then, as x is the unique involution of CG (x, K), it follows that v ∈ Syl2 (CL(v) (x)), whence Sylow 2-subgroups of L(v) x are of maximal class, which is absurd. Thus (a) holds in this case. Finally, if K ∼ = Co3 or F5 , we argue that (b) holds. We claim that V ∪ {y} is invariant under CG (y). Note that CK (y)/ y ∼ = Aut(L) in both cases [III11 , 5.3a]. Hence, as L  CG (y), it follows that CG (y) = CK (y)C(y, L). As V = y K ∩ (L − Z(L)), it follows that V ∪ {y} is CG (y)-invariant. As V is non-empty, K = L, L(v) | v ∈ V by [III11 , 5.3b]. Hence CG (y) ≤ NG (K), as claimed.  Now we can prove Lemma 2.7. Let v ∈ V. Then L(v) ≤ CK (y). Proof. First suppose that Lemma 2.6a holds. As L ≤ K ∩ Lxy , we have y K ∩ L = y G ∩ L = y Lxy ∩ L. So v ∈ y K ∩ y Lxy . This implies that L(v) ≤ K ∩ Lxy ≤ CG (x) ∩ CG (xy) ≤ CG (y), whence the desired conclusion. If Lemma 2.6b holds, let Q ∈ Syl2 (C(x, K)). Then as x is the unique involution of Q, Q maps isomorphically into CAut(Lxy ) (L), whence Q is cyclic by [III11 , 5.3c]. Thus CG (K) is 2-nilpotent and so K = NG (K)(∞) . As v ∈ V ⊆ y K , Lemma 2.6b gives CG (v) ≤ NG (K), and so CG (v)(∞) ≤ K, whence CG (v)(∞) = CK (v)(∞) = L(v). But also v ∈ L ≤ Lxy , and CLxy (v)(∞) ≤ CG (v)(∞) = L(v), so from the structure of centralizers of involutions in Lxy ∼ = Co3 or F5 , we must have CLxy (v)(∞) = L(v). Thus, L(v) ≤ K ∩ Lxy ≤ CG (y). The lemma follows.  But now for any v ∈ V, L(v) ≤ CK (y)(∞) = L by Lemma 2.7 and so [v, L] = 1. This contradicts the definition of V and completes the proof of Proposition 2.1. 3. Theorems 1 and 2: J1 and Ree Components In this section we complete the proofs of Theorems 1 and 2 by proving√ Proposition 3.1. K ∼  J1 or 2 G2 (q) for any odd power q = 3n/2 of 3 with = n ≥ 3. ∼ J1 or 2 G2 (q), q = 3n/2 > 3, and we use We suppose on the contrary that K = the notation x, K, y, L, D = x, y from (1A) and (1B). Thus m2 (C(x, K)) = 1 by [III7 , 5.1]. As K is quasisimple, C(x, K) = CG (x K). We also let T ∈ Syl2 (CG (x)), Q = T ∩ C(x, K), and R = T ∩ K.

150

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

∼ Z2 , R ∼ Lemma 3.2. We have Q = = E23 , AutK (R) ∼ = F7.3 , the Frobenius group of order 21, and T = R × Q ∼ = E24 . Proof. By [III11 , 5.2], |Out(K)| has odd order and the Schur multiplier of K has odd order, so T = R × Q. By [III11 , 7.14], R ∼ = E23 and AutK (R) ∼ = F7.3 . If |Q| > 2, then as m2 (Q) = 1, x = Ω1 (Φ(Q)) = Ω1 (Φ(T )) so x  NG (T ). However, as T is abelian, [III8 , 6.3a] implies that xG ∩ T = xNG (T ) so x is weakly closed in T , contradicting the Z ∗ -Theorem [IG , 15.3]. Therefore |Q| = 2 as asserted.  Lemma 3.3. We have xR ⊆ xG , and one of the following conditions holds: (a) NG (T )/CG (T ) ∼ = GL(3, 2) and R is the unique minimal normal subgroup of NG (T ) contained in T ; or (b) NG (T )/CG (T ) ∼ = E23 .F7.3 and a Sylow 2-subgroup H of O 2 (NG (T )) is homocyclic abelian of rank 3 and exponent 4, with Ω1 (H) = R. Moreover NG (T ) = O 2 (NG (T )) x. Proof. If T ∈ Syl2 (G), then by Thompson’s Transfer Lemma, all involutions of T are conjugate into R and hence to one another. Thus |xNG (T ) | = 15 and so |AutG (T )| = 15.21. But Aut(T ) ∼ = GL(4, 2) has no subgroup of this cardinality [III11 , 22.4], a contradiction. Hence T ∈ Syl2 (G) and therefore |xNG (T ) | = 8 and |AutG (T )| = 168. Moreover xNG (T ) = T − R, since the only alternative, xNG (T ) = {x} ∪ R# , is not permuted transitively by any subgroup of Aut(T ). If O2 (AutG (T )) = 1, then AutG (T ) ∼ = GL(3, 2) and (a) holds. Otherwise, AutG (T ) is solvable with the structure given in (b). Let S ∈ Syl2 (NG (T )). Then AutS (T ) ∼ = E23 and by a Frattini argument, there exists a 7-element w ∈ NG (S) ∩ NG (T ) inducing an automorphism of order 7 on S and acting transitively on R# . It follows from [III8 , 7.6] that S = Hx with H = [S, w] of order 64, and H is either elementary abelian or homocyclic abelian of exponent 4. Note that since H is abelian and |CG (x)|2 = 24 < |H|, xG ∩ S ⊆ Hx. In particular S ∈ Syl2 (G), for otherwise x ∈ [G, G] by the Thompson Transfer Lemma, contradicting the simplicity of G. Suppose that H ∼ = E26 . Then x acts freely on H, so I2 (Hx) = xH = xNG (T ) . Then NG (S) normalizes H = J(S), so normalizes I2 (Hx) = T , whence S ∈  Syl2 (G), a contradiction. Hence H has exponent 4 and (b) holds. Now consider L := E(CK (y)). Write v = D ∩ K and consider the quasisimple pumpup Lv of L in CG (v). Lemma 3.4. The following conditions hold: (a) L < Lv ; (b) K ∼ = A5 ; = J1 and L ∼ (c) Lv /Z(Lv ) ∼ U = 3 (4), L3 (4), L2 (16) or M12 , with x inducing a non-inner automorphism on Lv , or Lv ∼ = J2 ; and (d) CLv (x) has no subgroup isomorphic to Σ5 . Proof. Suppose that L = Lv , so that L   CG (v). Let N = NG (T ) and N = N/O2 (N ) = N/O2 (CG (T )). Let Y ∈ Syl3 (N ∩L). Then Y ≤ N ∩L   CN (v) so Y ≤ N ∩ L   CN (v) = CN (v). But then Y T /T = N ∩ LT /T   CN (v)T /T ,

3. THEOREMS 1 AND 2: J1 AND REE COMPONENTS

151

whence Y ≤ O3 (CN (v))T /T . But in both cases of Lemma 3.3, CN (v)T /T has a normal A4 -subgroup and Sylow 3-subgroup of order 3, so O3 (CN (v)T /T ) = 1, a contradiction. This proves (a). Next we show that  K. (3A) Lv ∼ = Suppose that Lv ∼ = K, choose an x-invariant Sylow 2-subgroup Q of C(v, Lv ), and expand x Q to S ∈ Syl2 (CG (v)). Then x CQ (x) = x, v so x Q is of maximal class by [IG , 10.24]. Hence, S = (S ∩ Lv ) × Q with [S,  S] = [Q, Q]  cyclic. On the other hand, as |xNG (T ) | = 8 is a power of 2, x−1 xNG (T ) = x−1 xS = [x, S] is cyclic, which is absurd. This proves (3A), and [III11 , 13.10], with Lv and x here in the roles of K and y there, yields (b) and (c). As T ∈ Syl2 (CG (x)) is abelian, (d) is clear. The proof is complete.  We next set C = CG (v) and prove: Lemma 3.5. x induces an inner automorphism on Lv ∼ = J2 . Proof. By Lemma 3.3, xG ∩T = T −R and v G ∩T = v K ∩T = R# . Therefore v is 2-central in G and CG (x) controls the G-fusion of v in T . Hence by [III8 , 6.3], x ∈ [C, C]. On the other hand, again (3B)

x, v ∈ Syl2 (CG (x, L)),

and so Sylow 2-subgroups of CG (L) – and hence Sylow 2-subgroups of C(v, Lv ) – are cyclic or of maximal class, by [IG , 10.24]. Consequently by Lemma 3.4c, Lv  C. As the outer automorphism group of Lv is 2-nilpotent with abelian Sylow 2-subgroup, by [III11 , 2.1], x ∈ [C, C] induces an inner automorphism on Lv . Then  Lv ∼ = J2 by Lemma 3.3. We can now derive a final contradiction completing the proof of Proposition 3.1. As Lv ∼ = J2 , O2 (CLv (L)) ∼ = E22 by [IA , Table 5.3g]. Therefore CG (x, L) has 2-rank at least 3, contradicting (3B). This contradiction completes the proof of Proposition 3.1. Propositions 2.1 and 3.1 together imply Theorems 1 and 2. These theorems have consequences for the isomorphism types of J/Z(J) for any (z, J) ∈ ILo2 (G). We combine these with some more basic restrictions in the following corollary. Corollary 3.6. Suppose that p = 2. Then d2 (G) ≥ 3. Moreover, suppose that (z, J) ∈ ILo2 (G) and set J = J/O2 (J). Then the following hold: (a) J ∼  2Am for any m ≥ 7; = ∼ L3 (4), then Z(J)  J)  =  is elementary abelian; (b) If J/Z( J (c) If J ∈ Spor ∩ G2 , then J ∼ = 1 , He, or Co3 ; ∼ (d) If K = An , n = 9 or 10, and J ∈ G2 , then J ∼ = A9 or A10 ; (e) If K ∼ = SL2 (q) for some odd q, then q ∈ FM9; = An , n = 9 or 10, and J ∼ (f) If K ∼ = Am for any m > n. = An , n ≥ 11, and J (G) = ∅, then J ∼ 2

Proof. Let d = d2 (G). By [III3 , Theorem 1, p. 61], J ∼ = 2Am for any m ≥ 12, so d > 1. Let (w, I) be a 2-terminal long pumpup of (z, J); this exists by

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

152

[IG , 6.10]. Set I = I/O2 (I). By the hypothesis of Theorem C∗7 , G possesses no  I)  ∼ 2-Thin Configuration, so I ∼  2Am for any 7 ≤ m ≤ 11, and if I/Z( = = L3 (4),  ∼   then Z(I) is elementary abelian. If J = M c, Ly, or O N , then by [III11 , 13.17a],  By the definition of J∗ (G), (w, I) ∈ J∗ (G), but this contradicts Theorem 1. I ∼ = J. 2 2 Therefore d ≥ 3, and a similar argument shows that it is not possible that J ∼ = F5 . Thus, (a) and (c) hold, by definition of G2 ∩ Spor and [III7 , (1E)]. Moreover  not elementary abelian, then by [III11 , 13.17ab],  J)  ∼ if J/Z( = L3 (4) with Z(J)  ∼  I = O N , again contradicting Theorem 1. Thus (b) holds. In (d) and (e), d = d2 (K) = 8 as K ∈ J∗2 (G). This immediately implies (d). In (e), again the lack of 2-Thin Configurations applies to give I ∼ = SL2 (q ∗ ) for any odd q ∗ . Thus by [III11 , 13.8], q ∈ FM9, as required. Finally in (f), I ∼ = Ar for some r ≥ m > n by [III11 , 1.7]. By assumption, J2 (G) = ∅ so (w, I), like (x, K), lies in Jo2 (G). But (x, K) ∈ J22 (G) and as r > n, we have a contradiction by the definition of J22 (G) [III7 , 4.2]. The proof is complete.  4. The Alternating Case: Theorem 3–Centralizers of Rank 2 In the next five sections we fix (x, K) as in (1A), (1B) and (1C), and we complete the analysis of the case K ∈ Alt. Thus (a) or (b) of [III7 , Prop. 3.3] holds, and in particular p = 2. ∼  2An , n ≥ 12, by Corollary 3.6, we have Since K ∈ G2 , but K = K∼ = An , n ≥ 9, according to [IA , 6.1.4] and the definition of G2 [I2 , 13.1]. In this section we deal with Case (a) of [III7 , Prop. 3.3], that is, we assume that m2 (CG (K)) ≥ 2. We prove Theorem 3 by constructing a suitable subgroup G0 ∼ = An+4 . In accordance with [III7 , Prop. 3.3a], [II3 , Theorem PU4 ], and the hypotheses of Theorem 3, we assume the following setup: (1) x ∈ I2 (G); (2) K is a standard component of CG (x) with K ∼ = An , n ≥ 9; (4A) (3) C = CG (K), E ∈ Syl2 (C), E is a four-group and E is Gconjugate to a root four-subgroup of K. This can be tightened as follows. Proposition 4.1. Under the hypotheses (4A), the following conditions hold: (a) C/O2 (C) ∼ = A4 ; (b) There exists g ∈ G interchanging the four-group E with a root foursubgroup E1 of K by conjugation. Proof. We fix a root four-subgroup E1 of K and an element g ∈ G such that E g = E1 . We set N = NG (E), N1 = NG (E1 ), and K1 = K g ≤ N1

4. THE ALTERNATING CASE: THEOREM 3–CENTRALIZERS OF RANK 2

153

and fix this notation. Since E ∈ Syl2 (C), CC (E) has a normal 2-complement, so K = E(CG (E))  N, and therefore K1 = E(CG (E1 ))  N1 . There are root A4 - and An−4 -subgroups A1 and J of K such that (4B)

A1 J = A1 × J, and E1 ≤ A1 .

Since n ≥ 9, J = E(CK (E1 )) and so J = E(CG (EE1 )). As E and E1 are conjugate, E lies in a conjugate A2 of A1 , and since Out(K) is a 2-group, A2 ≤ KCG (K) = K × C with A2 ∩ K = E ∩ K = 1. Thus C contains a subgroup isomorphic to A4 . Let N 1 = N1 /CN1 (K1 ) ≤ Aut(K1 ) ∼ = Σn , and set X = J × C ≤ CG (E1 ) ≤ N1 and D = CC (K1 ), so that C ∼ C/D embeds in B = CAut(K1 ) (J). But = J = E(CG (EE1 )) is a component of CK1 (E), and in particular J is a root An−4 subgroup of K 1 . Thus B and hence C is embeddable in Σ4 by [III11 , 4.2a]. Since C contains A4 and |C|2 = 4 the only possibility is that C/O2 (C) ∼ = A4 , proving (a). Moreover the image of C/O2 (C) in Aut(K1 ) centralizes J, so is the image of a root A4 -subgroup A of K1 . Then C ≤ ACG (K1 E1 ). But CG (E1 K1 ) ∼ = CG (EK) has a normal 2-complement, so E ≤ O 2 (C) ≤ O 2 (A), whence E is a root four-subgroup of K1 . As E1 is a root four-subgroup of K, E1g is a root four-subgroup of K g = K1 , and so E = E1gh for some h ∈ K1 . Then E gh = E1h = E1 , so E and E1 are interchanged by gh, proving (b).  Constructing a subgroup of G isomorphic to An+4 is now fairly straightforward. We shall prove the following result. Proposition 4.2. Under the hypotheses and notation of (4A), KE ≤ G0 ∼ = An+4 , ΓE,1 (G) ≤ NG (G0 ), and CG (G0 ) has odd order. We first set up some notation. We are aiming for the target group G∗ = AΩ ∼ = An+4 , where Ω = {1, 2, . . . , n + 4}. We let E ∗ and E1∗ be the (disjoint) root foursubgroups of G∗ supported on Ψ = {1, 2, 3, 4} and Ψ1 = {n + 1, n + 2, n + 3, n + 4}, respectively, and set W ∗ = NG∗ (E ∗ ), W1∗ = NG∗ (E1∗ ), K ∗ = E(W ∗ ) and K1∗ = E(W1∗ ). Thus K ∗ and K1∗ are the root An -subgroups of G∗ supported on Ω − Ψ and Ω − Ψ1 , respectively. We also set Y ∗ = W ∗ ∩ W1∗ , so that Y ∗ = J ∗ (S1∗ × S ∗ ) where J ∗  Y ∗ , J ∗ ∼ = An−4 , J ∗ ∩ (S1∗ × S ∗ ) = 1, S1∗ ∼ = S∗ ∼ = Σ4 , S ∗ ≤ K ∗ , S1∗ ≤ K1∗ , and [K ∗ , O 2 (S1∗ )] = [K1∗ , O 2 (S ∗ )] = 1. Then we may take S ∗ and S1∗ so that in the action of (S1∗ × S ∗ )/O 2 (S1∗ × S ∗ ) on J ∗ , involutions s∗1 ∈ S1∗ − O 2 (S1∗ ) and s∗ ∈ S ∗ − O 2 (S ∗ ) act as identical transpositions. Consequently CY ∗ (J ∗ ) is the subgroup of S1∗ × S ∗ of index 2 not containing either S1∗ or S ∗ . In G, on the other hand, we let C, N , N1 , K1 , A1 and J be as in (4A)–(4B). We fix an element g interchanging E and E1 , by Proposition 4.1, and we will presently refine the choice of g. There is a subgroup S1 of K such that E1  S1 ∼ = Σ4 , all involutions of S1 are root involutions and NK (E1 ) = S1 J, where J is the root An−4 subgroup of K supported on the fixed points of E1 . We fix an involution s1 ∈ S1 − O2 (S1 ), so that s1 induces transpositions on both the support of E1 and the support of J. Note that A1 = O 2 (S1 ). Lemma 4.3. Set A = Ag1 . Then [A, K] = 1, Ag = A1 and the following conditions hold: (a) N = NG (K) = NG (A) > KC;

154

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

(b) g normalizes J; and (c) Every root four-subgroup of K lying in J is also a root four-subgroup of K1 . Remark 4.4. Assertion (c) is trivial by [III11 , 4.2b] unless n − 4 = 6. Proof. Set H = O 2 (N ∩ N1 ) = O 2 (NN (E1 )). Then H ≤ K × C, and the projection of H into K is O 2 (NK (E1 )) = A1 × J. Hence H = CH (K) × A1 × J. Since g interchanges E and E1 , g normalizes H and J = E(H), proving (b). Moreover A = Ag1 , like A1 , is a direct factor of H containing E, whence H = A1 × J × A × O2 (H), in view of Proposition 4.1a. But then A acts on K and centralizes A1 J, so with [III11 , 4.2a], [A, K] = 1. Moreover g 2 normalizes E and E1 and so normalizes K, whence g 2 normalizes A1 = CK (O 2 (CK (E1 ))). Thus Ag = A1 . Now C = O2 (C) × A, and NG (A) ≥ KC. If equality holds, then A  [NG (A), NG (A)]. But A1 ≤ [NK (A1 ), NK (A1 )], a contradiction. Hence NG (A) > KC. We have NG (K) = (NG (K)∩NG (E1 ))K ≤ NG (A) as A = CK1 (O 2 (CK1 (E1 ))). As NG (A) ≤ NG (E) ≤ NG (K), (a) follows. Finally let F and F1 be subgroups of J which are root four-subgroups of K and K1 , respectively. Thus F and F1 are G-conjugate to E1 and E, so are both conjugate to E. We prove that F and F1 are J-conjugate, which will imply (c). In any case the involutions in F and F1 are root involutions of J, so replacing F1 by a J-conjugate we may assume that F ∩ F1 = 1. But by (a) and Proposition 4.1, E = O2 (CG (e)) for all e ∈ E # . Consequently F = O2 (CG (F ∩ F1 )) = F1 , as required.  We now refine our choice of g. For any u ∈ J, if we put g  = ug, then we see that g  still interchanges A and A1 (since J centralizes them). Since s1 acts on J like a transposition, and g like an element of Σn−4 , we may replace g by such a suitably modified element g  and assume that g acts on J like an element of s1 . This implies that g 2 centralizes J. But g 2 normalizes E1 and E, so g 2 normalizes NK (E1 ) = S1 J and A1 . Then the action of g 2 on S1 J/A1 ∼ = Σn must be trivial since g 2 centralizes A1 J/A1 ∼ = J. Consequently g 2 normalizes S1 . Now set S = S1g and s = sg1 , so that g interchanges S and S1 , A = O 2 (S), and (4C) s ∈ S. Since the action of g on J is that of an element of s1  ∼ = S1 J/J, s and s1 have identical actions on J. Moreover [s, A1 ] = [s1 , A]g = 1 so s acts on K as a transposition. Hence CS (K) < S, and so CS (K) = A and SK/A ∼ = Σn . Lemma 4.5. There exist subgroups W ≤ N and W1 ≤ N1 and isomorphisms φ : W → W ∗ , φ1 : W1 → W1∗ and φY : Y → Y ∗ , where Y := W ∩ W1 , such that φ|Y = φ1 |Y = φY . ∼ An and A ∼ Proof. Set W = KS = (K × A) s. We have K = = A4 . Also s acts as a transposition on both A and J, and s centralizes A1 , and these facts determine the action of s on KA up to an inner automorphism (normalizing A1 ).

4. THE ALTERNATING CASE: THEOREM 3–CENTRALIZERS OF RANK 2

155

These data determine the isomorphism type of W , so there is an isomorphism φ : W → W ∗ carrying A, A1 to A∗ , A∗1 , respectively. Next we have K1 = E(CG (E1 )) = K g and we set W1 = S1 K1 = W g = (K1 × A1 ) s1  , We similarly obtain an isomorphism ψ : W1 → W1∗ carrying A and A1 to A∗ and A∗1 , respectively. Now Y ∗ = W ∗ ∩ W1∗ = NW ∗ (E1∗ ), which is a split extension of J ∗ by S1∗ × S ∗ ∼ = Σ4 × Σ4 . Also Y = W ∩ W1 = NW (E1 ) = NW1 (E) is mapped to Y ∗ by both φ and ψ; and Y is g-invariant so JS1 S ≤ Y as S = S1g . Since JS1 ≤ K and S ∩ K = 1, while J ∩ S1 = 1 within K, we have |JS1 S| ≥ |Y ∗ |, and consequently both φ and ψ map Y onto Y ∗ . We shall choose a suitable automorphism γ of W1∗ mapping Y ∗ to itself and set φ1 = γψ and φY = φ|Y . The only requirement that we need to satisfy is then that φ|Y = φ1 |Y , that is, that γψ|Y = φ|Y . Setting α = (φ|Y )(ψ|Y )−1 , we have α ∈ Aut(Y ∗ ), and we are reduced to showing that there exists γ ∈ Aut(W1∗ ) such that γ|Y ∗ = α. Now α is not arbitrary; for one thing, since φ and ψ both carry E and E1 to E ∗ and E1∗ , α stabilizes both E ∗ and E1∗ . Furthermore, any root four-subgroup B ∗ of G∗ lying in J ∗ corresponds under φ and ψ to root four-subgroups of K and K1 , respectively. By Lemma 4.3c, B ∗ and α(B ∗ ) are J ∗ -conjugate. Now the extendability of α to γ ∈ Aut(W1 ) is immediate from [III11 , 4.2c], and the proof is complete.  Now let β = φ ∪ φ1 , so that β is a bijection from W ∪ W1 to W ∗ ∪ W1∗ which is multiplicative on W as well as on W1 . The image of β is the set of all permutations in G∗ which stabilize either Ψ or Ψ1 . Among these are the permutations (123) and (12)(i i + 1) for each i = 3, 4, . . . , n + 3, since n ≥ 9, and we define     y1 = β −1 (123) and yi = β −1 (12)(i + 1 i + 2) , i = 2, 3, . . . , n + 2. Thus y2 ∈ E. We wish to verify that these elements satisfy the following relations, which define An+4 by [III11 , 9.2]: (1) y1 has order 3 and y2 , . . . , yn+2 are involutions; (2) (y1 y2 )3 = 1; (4D) (3) (y1 yi )2 = 1 for all i = 3, . . . , n + 2; (4) (yi yi+1 )3 = 1 for all i = 2, . . . , n + 1; and (5) (yi yj )2 = 1 for all i and j such that 2 ≤ i and i + 2 ≤ j ≤ n + 2. These relations are clearly satisfied by the images β(yi ) ∈ G∗ . Moreover,  among the elements yi , the only one not lying in W is theny3 = β −1 (12)(45) , and the only one not lying in W1 is yn−1 = β −1 (12)(n n + 1) . Therefore any pair (yi , yj ), with the single exception of the pair (y3 , yn−1 ), lies in either W or W1 ; since β|W and β|W1 are isomorphisms, the desired relations between yi and yj are consequences of the corresponding relations in either W ∗ or W1∗ . Lemma 4.6. The yi satisfy the relations (4D). Proof. The previous discussion shows that it remains only to check that (4E)

[y3 , yn−1 ] = 1.

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

156

Set X = CG (yn+2 ). The relations (4D) which we already know include that y2 , . . . , yn centralize yn+2 and so lie in X. We shall verify (4E) in X. By (4D1, 4), all the yi except y1 are G-conjugate. Thus X ∼ = CG (y2 ), with y2 ∈ E # , and so (∞) X = E(X) ∼ = An . More precisely, (4F)

y2 , y3 , . . . , yn are all E(X)-conjugate.

Indeed in W the elements y4 , . . . , yn are all conjugate via 3-cycles of K which are disjoint from yn+2 and hence lie in O 2 (CK (yn+2 )) = E(CK (yn+2 )) ≤ E(X) (recall that n ≥ 9). In W1 the elements y2 , y3 , y4 are similarly conjugate by 3-cycles in O 2 (CK1 (yn+2 )) ≤ E(X). Set L = E(X) y2 , so that by (4F), L contains y2 , . . . , yn . Since yn+2 is the involution (12)(n + 3 n + 4) in W , yn+2 induces a transposition on K, so E(CG (y2 , yn+2 )) ∼ = An−2 . Consequently y2 acts on E(X) as a transposition; by (4F), so do y3 , . . . , yn , and L ∼ = Σn . On the other hand from W1 we have that y2 y3 and y3 y4 are of order 3, so they are 3-cycles in E(X). Since y2 and y4 are distinct commuting transpositions in L, the support of y3 (in a set underlying X) lies in the union of the supports of y2 and y4 . However, y2 yn−1 and y4 yn−1 are both of order 2 so yn−1 is a transposition of L disjoint from both y2 and y4 . Therefore yn−1 is disjoint from y3 and so  [y3 , yn−1 ] = 1, completing the proof of (4E) and the lemma. Now we can prove Lemma 4.7. Set G1 = y1 , y2 , . . . , yn+2  and M = NG (G1 ). Then the following conditions hold: (a) KE ≤ G1 ; (b) G1 ∼ = An+4 ; (c) ΓE,1 (G) ≤ M ; and (d) CG (G1 ) has odd order. Proof. It is clear from our definition of β that KE = yi | 1 ≤ i ≤ n + 2, i = 3 ≤ G1 , so (a) holds. Part (b) holds by Lemma 4.6 and [III11 , 9.2]. To prove (c) we first show (4G)

NG (K) ≤ M.

By (4A2, 3), K  CG (x) and CG (K x) has Sylow 2-subgroups of order 4, so CG (K x) has a normal 2-complement. Thus K = CG (x)(∞) . By Proposition 4.1, all involutions in E are G-conjugate to x, and since E ≤ CG (K), we have K = CG (e)(∞) for each e ∈ E # . For each u ∈ xG , set Ku = CG (u)(∞) ∼ = K. Let R(G1 ) be the set of all root involutions of G1 . Then R(G1 ) ⊆ xG . Thus if u ∈ R(G1 ), then E(CG1 (u)) ∼ = An ∼ = K so E(CG1 (u)) = Ku . But as n ≥ 7, [III11 , 17.1] then implies that G1 = Ku , K | u ∈ R(G1 ) ∩ K. Since NG (K) stabilizes the set R(G1 ) ∩ K of root involutions of K, this yields NG (K) ≤ NG (G1 ) = M , proving (4G). Since K  CG (e) for each e ∈ E # and K = CG (E)(∞)  NG (E), we have   ΓE,1 (G) = CG (e), NG (E) | e ∈ E # ≤ NG (K) ≤ M, and (c) holds as well.

5. THEOREM 4–CENTRALIZERS OF RANK 1: PRELIMINARIES

157

Finally, as CG (G1 ) ≤ CG (K) with E ∈ Syl2 (CG (K)) and CE (G1 ) = 1, CG (G1 ) has odd order, proving (d).  Recall that D = x, y, with y ∈ K a root involution by [III7 , Def. 6.9]; and G0 = N(x, K, y, L) = K, Ly , Lxy . It remains to show: Lemma 4.8. G0 = G1 . Proof. Both x and y are root involutions in G1 ∼ = An+4 with n ≥ 9. It follows that Ly ∼ =K ∼ = An and by [III11 , 17.1], G1 = K, Ly . Thus, we need only show that Lxy ≤ G1 . In fact we show that Lxy = L. Now E ≤ CG (xy) with [L, E] = 1, so E normalizes Lxy . Moreover, for each e ∈ E # , K  CG (e) so L = E(CK (xy)) is a component of CLxy (e). Hence by [III11 , 1.16], either Lxy ∼ = An+4k , k ≥ −1, or n = 5 with K/Z(K) ∼ = M12 or J2 ; and then [III11 , 17.8] yields that for a root involution z ∈ L, Lxy ≤ L, CG (z). But z ∈ xG1 so CG (z) ≤ ΓE,1 (G)G1 ≤ M , where again we put M = NG (G1 ). Hence Lxy ≤ M . But L = E(CG1 (xy))  CM (xy), so  L = Lxy . The proof is complete. Lemmas 4.7 and 4.8 immediately imply Proposition 4.2, which in turn immediately implies Theorem 3. 5. Theorem 4–Centralizers of Rank 1: Preliminaries For the remaining configuration with a standard alternating component [III7 , Prop. 3.3b], namely m2 (C(x, K)) = 1, no simple groups actually arise. However, some groups with a subgroup of index 2 do arise, namely Aut(F5 ) and Σn+2 . Moreover, for small values of n (n ≤ 12), the possible existence of neighbors which are not themselves alternating groups is troublesome. It therefore will take some effort to rule out this case, that is, to prove Theorem 4. The proof will occupy the next four sections. In this section we prove some generalities about these situations; in the subsequent three sections we deal with the specific cases. Again in accordance with [III7 , 3.3b] and Theorem C∗7 : Stage 3b, we assume the following hypotheses and notation, including notation for an acceptable subterminal pair (y, L): (1) (x, K) ∈ J∗2 (G) with K ∼ = An , n ≥ 9; (2) Q ∈ Syl2 (C(x, K)) and m2 (Q) = 1; (3) y is a root involution of K and D = x, y ∼ = E22 ; (5A) (4) L = E(CK (y)) ∼ = Am , m = n − 4; (5) For each u ∈ D − x, the pumpup Lu of L in CG (u) is quasisimple; (6) For some u ∈ D − x, Lu > L. We shall use the following additional notation: (1) V = y, y   is the root four-subgroup of K containing y; (2) t ∈ CK (L) has order 3; (3) B = Ω1 (V Q) = V × x ≤ Z(V × Q); (5B) (4) For each b ∈ B − x, Lb is the pumpup of L in CG (b); (5) E is a root four-subgroup of K with E ≤ L, and e ∈ E is an involution; and (6) h ∈ K and h interchanges E and V as well as e and y. Thus [V, L] = 1 and indeed L = E(CK (v)) for all v ∈ V # . Moreover CK (L) = t V ∼ = A4 .

158

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

Since t permutes V # transitively, every b ∈ B − x is CG (L)-conjugate to an element of D − x, and so the pumpup Lb is quasisimple by (5A5). Similarly we have Lemma 5.1. Let u ∈ I2 (CCG (x) (L)). Then L ≤ E(CG (u)). Proof. Clearly u normalizes B = V × x, so B1 := CB (u) is noncyclic. We have just seen that L ≤ Lb ≤ E(CG (b)) for all b ∈ B # and in particular for all b ∈ B1# . Then by [III8 , 2.13], L ≤ E(CG (u)).  Our original choice of (x, K) limits the possible structures of 2-components of centralizers of arbitrary involutions. In particular it limits the possibilities for Lu , for any u ∈ B − x. Lemma 5.2. Assume (5A). Then for any u ∈ B − x such that Lu > L, one of the following holds: (a) Lu ∼ = K; (b) Lu /O2 (Z(Lu )) ∼ = An−2 ; (c) n = 12, m = 8 and Lu ∼ = L4 (4), HS or 2HS; (d) n = 10, m = 6 and Lu ∼ = Sp4 (4), L± 5 (2), P Sp4 (3), L4 (3), 2HS, or ∼ Lu /O2 (Z(Lu )) = U4 (3) or 2U4 (3); (e) n = 9, m = 5 and Lu /O2 (Z(Lu )) ∼ = (X)L3 (4) with X = 1, 2 or 2 × 2, or Lu ∼ = U3 (4), L2 (24 ), L2 (52 ), M12 , 2M12 , J2 or 2J2 . Proof. As noted above, we may replace u by a t-conjugate and assume that u ∈ D − x. In view of the preceding lemma, Corollary 3.6c, [III11 , 13.19], and [IA , 6.1.4], since Lu is a vertical pumpup of L ∼ = Am , m = n − 4, the only possibility other than the desired conclusions is that Lu ∼ = An+2k , k > 0. In that  case, however, Lu ∈ G2 , contradicting Corollary 3.6df. Of course Lu is quasisimple; the only reason for mentioning O2 (Lu ) above is that A7 , U4 (3), and L3 (4) have nontrivial coverings with kernels of odd order. Lemma 5.3. Assume (5A) and let u ∈ D − x with Lu > L. Let w be an involution in CG (BL). Then the following conditions hold: (a) w ∈ B; and (b) If w centralizes Lu , then Lu ≤ Lw . Moreover, either Lu = Lw or else Lu ∼ = An−2 or A8 or P Sp4 (3), with Lw ∼ = An or Lw /Z(Lw ) ∼ = HS or ± L4 (3), respectively (and n = 10 in the last two cases). Proof. The assumption [w, BL] = 1 implies that w lies in CG (x) and maps into CAut(K) (L) ∼ = Σ4 . But V maps onto the root four-subgroup of this Σ4 and since [w, B] = 1, the image of w is the same as that of an involution b ∈ V . Thus w ∈ V C(x, K), and since x is the unique involution of C(x, K), w ∈ B. Now assume that w centralizes Lu . As L ≤ Lu , the pumpup Lw of L in CG (w) is also the pumpup of Lu in CG (w), so that Lu ≤ Lw . Now Lemma 5.2 applies to w and Lw as well as to u and Lu , with the same values for m and n. Thus Lw has one of the isomorphism types given in that lemma, but Lw also contains Lu . Hence by [III11 , 13.20], either Lw = Lu or Lw and Lu satisfy one of the exceptional conclusions of (b), as desired.  Since [B, L] = 1, B normalizes Lu . We next prove

5. THEOREM 4–CENTRALIZERS OF RANK 1: PRELIMINARIES

159

Lemma 5.4. Assume (5A), let u ∈ D −x with Lu > L, and set B0 = CB (Lu ). Write D# = {x, u, v}. Then one of the following holds: (a) (b) (c) (d) (e)

B0 B0 B0 B0 B0

= V and t normalizes Lu ; = u and x, y   acts faithfully on Lu ; is noncyclic, Lu ∼ = L4 (3) or U4 (3); = P Sp4 (3) and Lv /Z(Lv ) ∼ is noncyclic, Lu ∼ = A8 and Lv ∼ = HS or 2HS; or is noncyclic, Lu ∼ = An−2 and Lv ∼ = K.

Proof. We have B ∼ = E23 . Since Lu > L, x ∈ B0 . Suppose first that B0 is noncyclic. Since t normalizes B, B0 ∩ B0t then contains an involution w = bt , b ∈ B0 . Since t centralizes L, it follows that Lw = Ltb . Now the preceding lemma implies that both Lw and Lb contain Lu . If Lw ∼ = Lu ∼ = Lb , then it follows that Lw = Lu = Lb , whence t normalizes Lu . But then B0 is t-invariant, and as x ∈ B0 , we get B0 = [B, t] = V , and (a) holds. So, as Lw = Ltb ∼ = Lb , we may assume that  Lu for both v  = w and v  = b. The preceding lemma implies that Lu is as in Lv  ∼ =  Lu , v  and u (c–e), with Lv having the structure asserted for Lv there. Since Lv ∼ = are not CG (L x)-conjugate and so they are not t-conjugate. But v  ∈ B − x is t-conjugate to some element of D − x, which must then be v. Hence Lv ∼ = Lv , and one of (c–e) holds. Finally if B0 is cyclic, then obviously B0 = u, and since u ∈ x, y − x we  have B = x, y   × B0 , and (b) holds. Lemma 5.5. Let u ∈ B − x be an involution such that Lu > L. Let ξ ∈ Aut(Lu ) be the automorphism of Lu induced by x, and set Cξ = CAut(Lu ) (L ξ). Then one of the following holds: (a) Cξ ∼ = Z2 ; (b) Lu /Z(Lu ) ∼ = E22 ; = HS, M12 or J2 , and Cξ ∼ ∼ (c) Lu = K or Lu /Z(Lu ) ∼ = L± 4 (3), and Cξ is isomorphic to a subgroup of D8 . Proof. This is immediate from Lemma 5.2 and [III11 , 13.21].



We prove the next two lemmas together. Lemma 5.6. Q is cyclic of order at most 4. Moreover, if Q = x, then for any u ∈ B − x such that Lu > L, we have Lu /Z(Lu ) ∼ = L4 (3) or U4 (3). Lemma 5.7. If u ∈ B − x and Lu ∼ = K, then m2 (C(u, Lu )) = 1. Proof. Since m2 (Q) = 1 and [x, Lu ] = 1, Q acts faithfully on Lu . Let B1 = Q × V . Then B1 ≤ CG (u) and B1 normalizes Lu . Therefore we have embeddings Q → B1 /CB1 (Lu ) → CAut(Lu ) (L ξ) = Cξ where ξ and Cξ are as in Lemma 5.5. It follows from that lemma that either Q = x, or Lu ∼ = L± = K or Lu /Z(Lu ) ∼ 4 (3), and Q maps isomorphically to a  subgroup Q of Cξ , whence Q is cyclic of order at most 4, as claimed. In order to complete the proofs of Lemmas 5.6 and 5.7, we may henceforth assume that  =Q  is cyclic. Hence, Lu ∼ = CCξ (Q) = K. Moreover, if Q = x, then B1 /CB1 (Lu ) ∼ B1 /CB1 (Lu ) is cyclic; but m2 (B) = 3 so (5C)

m2 (C(u, Lu )) > 1.

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

160

Thus the failure of either lemma leads to (5C), and we complete their proofs by arguing from this last condition to a contradiction. By [IG , 6.10] there exists a 2-terminal pair (u∗ , L∗ ) such that (u, Lu ) L, Lu ∼ = K and Lu /O2 (Lu ) ∼ = An−2 . We proceed by contradiction, assuming that ∼K = ∼ An or Lu /O2 (Lu ) ∼ (6A) For all u ∈ D − x, Lu = L or Lu = = An−2 . This configuration does arise in the nonsimple group Σn+2 , and we shall eventually rely on a transfer argument to contradict the simplicity of G. We continue the notation (5B). In particular V is a root four-subgroup of CK (L), D = x, y with y ∈ D ∩ V , and B = V × x. We let B ∗ ∈ Syl2 (CCG (x) (L)) with B ≤ B ∗ . We choose B ∗ , as we may, so that y ∈ Z(B ∗ ). By Lemma 5.6, Q = x ∈ Syl2 (CG (K)), and so B / x embeds in CAut(K) (L) ∼ = Σ4 , whence |B ∗ | ≤ 24 . We also recall that E is a root four-subgroup of L and e ∈ E is an involution. The next lemma is the key to our analysis. ∗

Lemma 6.2. Ly /O2 (Ly ) ∼ = An−2 . Proof. Suppose first that Ly ∼ = An . Then by Lemma 5.8, y ∈ Syl2 (C(y, Ly )), and so |CG (y)/O2 (CG (y))Ly | ≤ 4. But then y ∈ [CG (y), CG (y)], contradicting Lemma 5.9. Hence we may assume, by way of contradiction, that Ly = L.

6. THEOREM 4–ALL ALTERNATING GROUP NEIGHBORS

161

Since NK (L) permutes V # transitively, (6B)

Ly1 = L for all y1 ∈ V # .

Let u = xy, so that uV # ⊆ uK . Since N is vertical and Ly = L, we have Lu > L. Suppose that Lu ∼ = Σn−2 with L = E(CLu (x)), and so x = CLu x (L). = An−2 . Then Lu x ∼ In particular, setting Vu := CB (Lu ), we see that B = x × Vu . As Vu is a foursubgroup of B, V ∩Vu contains an involution y1 . As y1 ∈ Vu , L < Ly1 , contradicting (6B). In view of Hypothesis (6A), Lu ∼ = An . Let t1 ∈ I3 (CLu (L)). If x ∈ Lu , then x is a root involution in Lu and so x ∼Lu e ∼K y, a contradiction. Hence, y ∈ Lu and, as y G ∩ B = V # , we have B ∩ Lu = V . Hence, [x, t1 ] = y1 for some y1 ∈ V # . Now, t1 ∈ CG (e) and L1 := E(CK (e)) is a component of CG (e) with V < L1 . Since t1 normalizes V , t1 normalizes L1 . But x centralizes L1 , so y1 = [x, t1 ] centralizes L1 , a contradiction completing the proof of the lemma.  We set By = CB (Ly ) ∼ = E22 . Now we can precisely determine the structure of CG (x)/O2 (CG (x)). Lemma 6.3. CG (x) = (O2 (CG (x)) × x × K)τ , where τ is an involution acting on K as a transposition, Tx := By τ  ∈ Syl2 (C(y, Ly ) ∩ CG (x)), with Tx ∼ = D8 , and Kτ  ∼ = Σn . Proof. We have that CG (y) ≥ By × Ly x, where B = By × x, and Ly x ∼ = Σn−2 . Let T be an x-invariant Sylow 2-subgroup of C(y, Ly ) containing By . Since E(CG (x, y)) = E(CCG (x) (y)) = L, Ly is CG (x, y)-invariant, so CT (x)×x×CLy (x) contains a Sylow 2-subgroup of CG (x, y), say S. As y ∈ [S, S] and y ∈ CT (x), it follows that y ∈ [CT (x), CT (x)]. On the other hand, it is clear in CG (x) that |CS (L) : B| ≤ 2. It follows that CT (x) is dihedral of order 8, and CT (x) × x = CS (L). Let τ be an involution in CT (x) − By . As τ centralizes L, it follows that τ acts as a transposition on K, and Kτ  ∼ = Σn , as claimed. Moreover, CT (x) = By τ  = Tx , and the lemma follows directly.  In the next five lemmas, we will determine the precise structure of CG (y)/O2 (CG (y)). Lemma 6.4. Ly  CG (y). Proof. Suppose not. The normal closure Ey of Ly in CG (y) is a product of isomorphic commuting components, which we may write as L0 ∗ Ly with L0 ∩ Ly ≤ Z(Ly ) (and Z(Ly ) is trivial, unless possibly n = 9 and |Z(Ly )| = 3). Now, both Ly and L0 are x-invariant and C(y, Ly )-invariant. Moreover, since CK (L y) is a 2-group and L ≤ Ly , Lemma 6.3 gives C(y, Ly ) ∩ CG (x) = O2 (CG (x))Tx . Now CL0 (x) has even order and is normal in C(y, Ly )∩CG (x). But then y ∈ CL0 (x), whence y ∈ Z(L0 ), a contradiction. Hence Ly is the unique component of CG (y)  isomorphic to An−2 .

162

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

We set M = CK (L). Thus M = V t is a root A4 subgroup of K, and V = O2 (M ). First, we determine the structure of CG (M ). Lemma 6.5. There exists s ∈ I2 (CG (x)) such that if we set Ls := E(CK (s)) and let Ks be the pumpup of Ls in CG (s), then M ≤ Ls ∼ = An−2 and Ks ∼ = An . ∼ ∼ Moreover, Ly = An−2 (i.e., Ly =  3A7 ). Proof. From Lemma 6.3, there is an involution τ ∈ C(y, Ly ) ∩ CG (x), with [τ, B] = y, and with τ acting as a transposition on K. Let Lτ = E(CK (τ )) ∼ = An−2 . Then L ≤ Lτ . Let Kτ be the normal closure of Lτ in E(CG (τ )) (see Lemma 5.1). As L ≤ Lτ ≤ Kτ , it follows that Ly ≤ Kτ . As Ly = Lτ , we conclude that Kτ is a nontrivial pumpup both of Lτ and of Ly . y Suppose that Kτ = K1 × K1x with K1 ∼ = An−2 . Then K1 = K1x , and so xy normalizes both K1 and K1x . As E(CKτ (x, xy)) = L, it follows that E(CK1 (xy)) ∼ = E(CK1x (xy)). Let K ∗ = E(CKτ (xy)) ∼ = An−4 ∼ = An−4 × An−4 . ∗ ∗ ∗ ∗ Then L = E(CK (x)), and so K = [K , L] ≤ Lxy , with K   CLxy (τ ). However, by (6A), Lxy ∈ Alt, so E(CLxy (τ )) is trivial or simple by [III11 , 4.1], a contradiction. It follows that Kτ is a vertical pumpup of Lτ ∼ = An−2 , n ≥ 9. As Kτ is also a pumpup of Ly , it follows that O2 (Ly ) = 1, proving the second statement of the lemma. Now we look more carefully in CG (y). By Lemma 6.3, CG (y) ≥ Hy := By τ  × Ly x. As before, we let Tx = By τ . Then CHy (x) = x × Tx × Lτ1 , where τ1 may be chosen in τ K ∩ CG (Tx ). In particular, as τ and τ1 are transpositions in Kτ  with disjoint supports, τ τ1 ∈ y K . On the other hand, as τ1 ∈ Hy and τ1 acts as a transposition on Ly , there exists x1 ∈ xTx with xg1 = τ1 for some g ∈ Ly . As τ and τ y play the same roles in this regard, we may assume that either x1 ∈ xBy or x1 = xτ . But, in the latter case, τ1 = xg1 = τ xg , and so xg = τ τ1 ∈ y K , a contradiction. Hence, x1 ∈ xBy ⊆ B, and so τ is G-conjugate into B, whence Kτ , being a vertical pumpup of Lτ , must be isomorphic to K, as a consequence of (6A). Finally, if we choose s ∈ τ K with M ≤ Ls := E(CK (s)), then the lemma holds for this choice of s.  Now define CM := CG (M ). Fix s as in Lemma 6.5. Lemma 6.6. One of the following conclusions holds: (a) CM = (O2 (CM ) × Ly )x; or (b) CM = O2 (CM )(x × Ls). Proof. Of course, CM ≤ CG (V ) ≤ CG (y). Let Qy = C(y, Ly ). By Lemma 6.4, Qy  CG (y). Now, CG (x) ∩ CM = O2 (CG (x))(x × Ls). As L < Ly and CQy (x) ∩ CM  CG (x) ∩ CM , it follows that CQy (x) ∩ CM ≤ O2 (CG (x)). Now Qy ∩ CM  CM , and x acts on Qy ∩ CM with CQy ∩CM (x) of odd order. Hence Qy ∩ CM is an odd order normal subgroup of CM . On the other hand, O2 (CM ) normalizes Ly and centralizes L, so O2 (CM ) ≤ Qy . It follows that Qy ∩ CM = O2 (CM ). Thus CM /O2 (CM ) maps injectively into Ly x. As x × Ls is a maximal subgroup of Ly x contained in CM , either (a) or (b) holds, as claimed. 

6. THEOREM 4–ALL ALTERNATING GROUP NEIGHBORS

163

Lemma 6.7. CM = (O2 (CM ) × Ly )x, and NG (M ) = O2 (CM )(M τ  × Ly x), where M τ  = CKτ  (L) ∼ = Σ4 . Proof. In Ks , Ls = E(CKs (x)) is a root An−2 -subgroup, and M is in turn a root A4 subgroup of Ls . As n − 2 > 6, it follows from [III11 , 4.2b] that M is a root A4 -subgroup of Ks . Let L0 = E(CKs (M )) ∼ = An−4 . As x acts non-trivially on Ks centralizing M , it follows that [L0 , x] = 1. If case (b) of the previous lemma holds, then as Ly  CG (y), (CM )(∞) = L, whence [(CM )(∞) , x] = 1. But then [L0 , x] = 1, a contradiction. Hence case (a) holds. Now, as [M, Ly ] = 1, it follows that V = By ≤ Tx ∈ Syl2 (C(y, Ly ) ∩ CG (x)). Thus M Tx = M τ  ≤ CG (Ly ) ∩ CG (x), with M Tx ∼ = Σ4 ∼ = Aut(M ). It follows that M Tx × Ly x covers NG (M )/O2 (CM ), and the lemma holds.  Finally, we can determine CG (y). Lemma 6.8. CG (y) = O2 (CG (y))(Tx × Ly x). Proof. Let M1 be a K-conjugate of M contained in L. Let CM1 = CG (M1 ) and L1 = E(CK (M1 )), and let KM1 be the subnormal closure of L1 in CM1 . Then, it follows from Lemma 6.7 that CM1 = (O2 (CM1 ) × KM1 )x, ∼ Σn−2 . We shall compute CG (M1 y) = with KM1 ∼ = An−2 and KM1 x = CG (M1 × y) two ways. First, we use (6C). In L1 , y is a root involution, and L1 is in turn a root An−4 -subgroup of KM1 ∼ = An−2 . Hence by [III11 , 4.2b], y is a root involution of KM1 . Hence, we have (6C)

CG (M1 y) = O2 (CM1 )(T1 × E1 x), with y ∈ T1 ∼ = Σn−6 . = D8 , E1 ∼ = An−6 , and E1 x ∼ Now we consider CG (M1 y) as a subgroup of Cy = CG (y), noting that M1 < L < Ly , each group a root alternating subgroup of the next, with Ly ∼ = An−2 . We argue that

(6D)

M1 is a root A4 subgroup of Ly . By [III11 , 4.2b], this is immediate unless L ∼ = A6 . Suppose then that (6E) fails, so that n = 10. We again set Qy = C(y, Ly ). Since Ly  Cy and Ly x ∼ = Σ8 = Aut(Ly ), we have Cy = (Qy ×Ly )x. Thus, as M1 ≤ Ly ∼ = A8 , CLy x (M1 ) is a 2-group since (6E) fails, so O 2 (CG (y M1 )) = O 2 (Qy ). But by (6D), O 2 (Qy ) = O2 (CM1 ) × F1 , with F1 ∼ = A4 . Let z1  = CF1 (x). As T1 × z1 , x ∈ Syl2 (CCG (M1 y) (x)), it follows that z1 is 2-central in a Sylow 2-subgroup of CQy (x). However, by Lemma 6.3, Tx ∈ Syl2 (CQy (x)) with y = Z(Tx ). As z1 = y, this is a contradiction. Thus we have established that M1 is a root A4 subgroup of Ly . Then (6E)

CG (M1 y) ≥ Qy × Ey x, ∼ Σn−6 . As Qy ≥ Tx with Tx ∼ with Ey ∼ = An−6 and Ey x = = D8 , it follows by comparison with the previous calculation that Qy = O2 (CG (y))Tx , proving the lemma. 

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

164

Before proceeding to the general case, it is convenient to dispose of the cases n = 9 and n = 10. Lemma 6.9. We have n ≥ 10. Proof. Suppose on the contrary that n = 9. We claim that x is 2-central in G. Let S ∈ Syl2 (CG (x)) and J = J(S). As CG (x)/O2 (CG (x)) ∼ = Σ9 × Z2 D by Lemma 6.3, we have J = x × U1 × U2 with Ui ∼ , and S = Js with = 8 U1s = U2 and s2 = 1 (see [III11 , 3.15]). We may choose notation so that y ∈ Z(U1 ). As Ly x ∼ = Σ7 , it follows from the previous lemma that J ∈ Syl2 (CG (y)). By the Krull-Schmidt Theorem [IG , 3.22], NG (J) permutes the set {y, y s }, and we conclude that S ∈ Syl2 (NG (J)), whence S ∈ Syl2 (G), as claimed. Now, Z(S) = x, yy s  with yy s  = Z(S) ∩ [S, S]. By the Thompson Transfer Lemma [IG , 15.16], x has an extremal conjugate z in S1 := (U1 × U2 )s, whence z ∈ Z(S) ∩ S1 = yy s . But, by Burnside’s Lemma, NG (S) controls G-fusion in Z(S), and yy s   NG (S), a contradiction, completing the proof.  Lemma 6.10. We have n ≥ 11. Proof. Suppose on the contrary that n = 10. By Lemma 6.8, CG (y)/O2 (CG (y)) ∼ = D8 × Σ8 . We claim that y is 2-central in G. Let P ∈ Syl2 (CG (y)) and J = J(P ). Then we may write J = U0 × U1 × U2 , with U0 = Tx , Ui ∼ = D8 for all i, and P = Jw with U1w = U2 and w2 = 1. Let yi be the involution in Z(Ui ), so that y0 = y. On the one hand, NG (J) controls fusion in Z(J) by [IG , 16.9], so y G ∩ Z(J) = y NG (J) ⊆ {y0 , y1 , y2 } by the Krull-Schmidt Theorem. On the other hand, Z(P ) = {1, y, y1 y2 , yy1 y2 }. Thus y G ∩ Z(P ) = {y}. Hence P ∈ Syl2 (G), as claimed. Next, let S ∈ Syl2 (CG (x)) with τ ∈ S. Then, in view of Lemma 6.3, S = x × S1 τ , where S1 τ  ∈ Syl2 (Kτ ), with Kτ  ∼ = Σ10 and τ a transposition (see [III11 , 3.15]). Thus, S = x, τ  × (R × Rv )v, G with R ∼ = D8 and with Z(R) ∈ y . Then m2 (S) = 6 = m2 (P ). Let S < S ∗ ∈ Syl2 (G). Then J(S) ≤ J(S ∗ ) ∼ = D8 × D8 × D8 , and S = CS ∗ (x). It follows that we may write J(S ∗ ) = R0 × R × Rv with x ∈ R0 . Then {Z(R0 ), Z(R), Z(Rv )} is G-conjugate to {Z(Ui ) | i = 0, 1, 2}, by Sylow’s Theorem and the Krull-Schmidt Theorem. As Z(R) ∈ yG and Z(Tx ) = y, it follows easily that all three centers lie in yG , i.e., yi ∈ y G , i = 0, 1, 2. Let r0  = Z(R0 ) ∈ yG . We claim that

(6F)

{r0 } = y G ∩ Z(S).

Set Z(R) = r and z = rr v . Then Z(S) = x, r0 , z with Z(J(S ∗ )) = r0 , r, r v . As y G ∩ Z(J(S ∗ )) = {r0 , r, r v } and y G ∩ {x, xr0 } = ∅, it will suffice ∗ to argue that xz ∈ y G , since xr0 z ∈ (xz)J(S ) . Suppose on the contrary that xz ∈ y G . Then by Lemma 6.8, CG (xz) = O2 (CG (xz))(T × K ∗ x∗ ), ∼ A8 . We compute C := CG (x, xz) = with T ∼ = D8 , xz ∈ Z(T ), and K ∗ = CG (x, z) in CG (x). As z is a 2-central involution in K, we see that O 2 (C) = O2 (CG (x)) × H, where H = Qu, with Q ∼ = Q8 ∗ Q8 , z = Z(Q), and u3 = 1. Now, looking in CG (xz), we conclude that H ≤ O 2 (CG (xz)) = O2 (CG (xz)) × K ∗ .

6. THEOREM 4–ALL ALTERNATING GROUP NEIGHBORS

165

In particular Q ≤ K ∗ . As xz ∈ Z(T ), we conclude that T × Q ≤ C with xz, z = [T × Q, T × Q]. However, x ∈ [CG (x), CG (x)], whence x ∈ [C, C], a contradiction. Thus xz ∈ y G , and so r0 is isolated in Z(S), proving (6F). It follows by [III8 , 6.3] that x ∈ [CG (r0 ), CG (r0 )]. As r0 ∈ y G , C0 := CG (r0 ) =  O2 (C0 )(T0 × K0 )x0 , with T0 ∼ = D8 and K0 x0  ∼ = Σ8 . Then x ∈ O 2 ([C0 , C0 ]) = / y G = r0G , x induces an inner automorphism of order 2 on r0  × K0 . Since x ∈ , whence C K0 ∼ A = 8 G (x, r0 ) is solvable. On the other hand, r0 , z ≤ Z(S), and x ∈ r0 , z; it follows that r0 acts on K either as a transposition, in which case CG (x, r0 ) contains a copy of A8 , or as a product of five disjoint transpositions, in which case CG (x, r0 ) involves A5 . In either case, this contradicts the previous paragraph and completes the proof.  We remark that the preceding argument would also handle the case n = 11. However, our general argument below works for all n ≥ 11. Our goal is to construct a subgroup G1 of G with G1 ∼ = Σn+2 . We continue in the spirit of Lemma 6.3, but change notation slightly, setting e1 := τ . We let Ω = {1, 2, . . . , n}, and fix an isomorphism φ : Ke1  → ΣΩ with φ(e1 ) = (1, 2) and φ(M ) the root A4 on {1, 2, 3, 4}. We let ei be the K-conjugate of e1 with φ(ei ) = (i, i + 1), 1 ≤ i ≤ n − 1. We set M ∗ = M e1 . Then by Lemma 6.7, CG (M ∗ ) = O2 (CG (M ∗ ))Ly x, with Ly x ∼ = Σn−2 . Lemma 6.11. The following conditions hold: (a) ei ∈ xG , 1 ≤ i ≤ n − 1; (b) xe1 e5 ∈ xG ; and (c) If a, b ∈ xG and [a, b] = 1, then the action of a on E(CG (b)) ∼ = An is not the action of a root involution. Proof. The automorphism induced by xe1 e5 on K is K-conjugate to that induced by y. Since we know that x ∈ y G , to prove (b) it suffices to show that xy ∈ xG . Suppose false, so that xg = xy for some g ∈ G. Then x, xg  is a four-group and E(CG (x, xg )) = E(CG (x, y)) = L ∼ = An−4 . Set u = g −1 and u conjugate by u; thus x, x  is a four-group and E(CG (x, xu )) ∼ = An−4 . Hence there is h ∈ K such that xuh = xg . But now y uh = (xxg )uh = xuh xh = xg x = y, so x and xg = xy are CG (y)-conjugate. On the other hand, with Lemma 6.8, y∈ / [x, CG (y)], so such a conjugation is impossible. Thus, (b) holds. Conjugating in (c), it suffices to show that for b = x, no element a ∈ xG ∩CG (x) acts on K like y. But the only involutions in CG (x) with this action are y and xy, which we have just seen are not in xG . For (a), let C = CG (M ∗ ) and C = C/O2 (C) ∼ = Σn−2 . Then E(CC (x)) ∼ = An−4 and, calculating in CG (x), E(CC (x, e5 )) ∼ = An−6 . Hence either e5 ∈ xC or xe5 ∈ xC . Suppose the latter case holds. As e1 ∈ M ∗ , xe1 e5 ∈ (xe1 )C . But xe1 ∈ (xe5 )K , whence xe1 e5 ∈ xG , contradicting (b). Thus e5 ∈ xC . As ei ∈ eK 5 for 1 ≤ i ≤ n − 1, the proof is complete.  We now construct a subgroup G1 of G with G1 ∼ = Σn+2 . First, we set K1 = E(CG (e1 )) ∼ = An , and set K1∗ = K1 x. We set en+1 = x. We let Ω1 denote the natural permutation domain for K1∗ . For s ∈ K1∗ , we let S(s) denote the support of s on Ω1 .

166

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

Lemma 6.12. There exists en ∈ xK1 such that G1 := e1 , e2 , . . . , en , en+1  ∼ = Σn+2 . Moreover, setting H := [G1 , G1 ], we have that N := NG (H) = (X × H)x, with X := O2 (N ). Proof. We claim first that ei ∈ K1∗ for 3 ≤ i ≤ n − 1. First, [e1 , ei ] = 1 and ei acts on O 2 (CK1 (x)) = O 2 (CK (e1 )) ∼ = An−2 as a transposition. As ei ∈ xG , Lemma 6.11c implies that ei acts as a transposition on K1 . Thus either ei or e1 ei lies in K1∗ O2 (CG (e1 )). But e1 ei ∈ y G = xG , so ei ∈ K1∗ O2 (CG (e1 )). Let X1 = O2 (CG (e1 )). Then e∗i := xi ei ∈ K1∗ for some xi ∈ X1 . As S(e∗i )∩S(x) = ∅, we have [x, e∗i ] = 1. Since also [x, ei ] = 1, we conclude that [x, xi ] = 1. Note that K is locally 1-balanced with respect to e1 , whence xi ∈ CO2 (CG (x)) (e1 ) = CO2 (CG (x)) (ei ). As ei inverts xi , we have xi = 1, as claimed. Now there is a unique transposition en ∈ K1∗ such that (ei en )2 = 1 for 3 ≤ i ≤ n − 2 and (en−1 en )3 = 1 = (en en+1 )3 . We set G1 := e1 , e2 , . . . , en , en+1 . In order to prove that G1 ∼ = Σn+2 , we check the defining relations [III11 , 9.1]: for all i ≤ j, the first, second, or third power of ei ej is 1, according as i = j, i < j − 1, or i = j − 1 [I1 , (11.1)]. In view of the relations already visible in CG (e1 ) and CG (x), it will suffice to prove that [e2 , en ] = 1. Let K4 := E(CG (e4 )) and K4∗ = K4 x. Set f := e3 e4 e5 e4 e3 . Then f ∈ CG (x) and f acts as the transposition (36) on K, commuting with e4 . We claim that e1 , e2 , f, e6 , e7 , . . . , en , en+1  ≤ K4∗ ∼ = Σn . and {e1 , e2 , f, e6 , . . . , en−1 } is a set of standard generators for a subgroup of CK4∗ (en+1 ) isomorphic to Σn−2 acting on the fixed point set of en+1 . For each i = 3, . . . , n + 1, ei ∈ K1∗ and ei acts as a transposition on K1∗ . Therefore for all 3 ≤ i < j ≤ n + 1, ei ej ∈ K1 . Therefore for all 6 ≤ i < j ≤ n + 1, ei ej ∈ E(E(CG (e1 )) ∩ CG (e4 )) ≤ E(CG (e4 )) = K4 , by L2 -balance. As en+1 = x ∈ K4∗ by definition, we have ei = (ei en+1 )en+1 ∈ K4∗ for all 6 ≤ i ≤ n + 1. Likewise e1 ej ∈ K for all 1 ≤ j ≤ n − 1, and so e1 f ∈ K. Thus by L2 -balance e1 e2 , e1 e3 , e1 f, e1 e6  ≤ E(CK (e4 )) ≤ K4 . It follows that e1 , e2 , f, e6 , . . . , en+1  ≤ K4∗ , and our claim follows. For w ∈ K4∗ , let M (w) denote the support of w on the natural permutation domain for K4∗ . As en−1 , en  ∼ = Σ3 ∼ = en , en+1 , en and en+1 are K4 -conjugate to en−1 , so they act as a transpositions of K4∗ . Moreover en−1 = x = en+1 . Thus M (en ) ⊆ M (en−1 ) ∪ M (en+1 ). As [e2 , en−1 ] = 1 = [e2 , en+1 ] and en−1 = e2 = en+1 , we conclude that M (e2 ) and M (en ) are disjoint sets. Hence [e2 , en ] = 1, as claimed. Thus, G1 ∼ = Σn+2 . Now set H = [G1 , G1 ], X = CG (H). Then CX (x) ≤ Yx = O2 (CG (x)). Hence |X| is odd, and N = (X × H)x, completing the proof of the lemma.  Lemma 6.13. For any root involution y1 ∈ H, CG (y1 ) ≤ N .

7. THEOREM 4–NON-ALTERNATING GROUP NEIGHBORS

167

Proof. Using H-conjugacy we may assume that y1 = y = e1 e3 . Let w = e5 e7 . Then w acts as a root involution on Ly ∼ = An−2 . As n ≥ 11, Ly is locally balanced with respect to w. Moreover, Ly is quasisimple and so by conjugacy L2 (CG (w)) = E(CG (w)). As y ∈ E(CG (w)), O2 (CG (w)) = O2 (CG (w)) ∩ CG (y) ≤ O2 (CG (y)). By symmetry between y and w, O2 (CG (w)) = O2 (CG (y)). Hence O2 (CG (y)) centralizes E(CG (y)), E(CG (w)) = E(CG1 (y)), E(CG1 (w)) = H, so O2 (CG (y)) ≤ N . Now it follows from Lemma 6.8 that CG (y) ≤ N , as claimed.  Lemma 6.14. |G : N | is odd. Proof. Let S ∈ Syl2 (N ). By [III11 , 3.15], J(S) = Z × U1 × · · · × Um , where m = [n/4], |Z| ≤ 2 and each Ui ∼ = D8 with Z(Ui ) generated by a root involution. By the Krull-Schmidt Theorem [IG , 3.22], NG (S) permutes the set {Z(U1 ), . . . , Z(Um )}. But by Lemma 6.13, CG (Z(Ui )) ≤ N for all i, and clearly Z(U1 ), . . . , Z(Um ) are all N -conjugate. Therefore NG (S) ≤ N and the lemma follows.  Lemma 6.15. xG ∩ H = ∅. Proof. Any involution t ∈ H is the product of disjoint root involutions, each in the center of a D8 subgroup of G1 = H x. So t ∈ [CG1 (t), CG1 (t)], whence / [CG (x), CG (x)], so t ∈ xG .  t ∈ [CG (t), CG (t)]. But x ∈ Now we immediately reach our final contradiction. By Lemmas 6.14 and 6.15 and the Thompson Transfer Lemma [IG , 15.16], x ∈ [G, G], contradicting the simplicity of G. This completes the proof of Proposition 6.1. 7. Theorem 4–Non-Alternating Group Neighbors We now begin to address some of the particular exceptional configurations of Lemma 5.2. We continue to assume (5A), and shall prove Proposition 7.1. Assume (5A). Then Q = x. Moreover, for any u ∈ D−x with Lu > L, one of the following holds: (a) Lu ∼ = K; (b) Lu /O2 (Z(Lu )) ∼ = An−2 ; or (c) m = 6 and Lu ∼ = 2HS. We also continue the notation (5B), so that B = Ω1 (V × Q) = V × x, and for a fixed u we shall use the notation B0 = CB (Lu ), as in Lemma 5.4. We choose S ∈ Syl2 (CG (x)) with B < S, and set R = S ∩ K. The remaining possibilities from Lemma 5.2 will be ruled out in a few batches. We first rule out the possibilities (7A) Lu /O2 (Lu ) ∼ = U4 (3) or 2U4 (3) for some u ∈ x, y − x, deriving a contradiction from this assumption in the next five results. The first lemma will also apply in the case Lu ∼ = L4 (3). In both these cases, n = 10.

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

168

Lemma 7.2. Suppose that u ∈ D − x and Lu /O2 (Lu ) ∼ = L± 4 (3) or 2U4 (3). Then B0 = u. Proof. Suppose false, so that B0 = V by Lemma 5.4. Recall that CK (L) = V t ∼ = A4 . Now Lu is a component of CG (V ). As [t, L] = 1 and t normalizes V , t normalizes and then centralizes Lu , since by Lemma 5.5, CAut(Lu ) (x L) is a 2-group. Set H = CG (t) and Kt = E(CK (t)). Then Kt ∼ = A7 and L ≤ Kt , and Kt is a component of CH (x). Set H = H/O2 (H), so that by L2 -balance, L ≤ K t ≤ E(H) and then Lu ≤ E(H). As [Lu , x] = 1, x ∈ O2 (H). But from CG (x), Q ∈ Syl2 (CG (x, t, Kt )), whence O2 (H) = 1 and F ∗ (H) = E(H). Again by L2 balance, the subnormal closure I of K t in H is a single x-invariant component or x, and in either the product of two components covering A7 and interchanged by   L  u ≤ KtLu , Lu ≤ I case K t ∼ is a component of C (x). But since L = L A = 7 u I as well. Therefore the diagonal case is impossible, so I is a single component. Now Q ∈ Syl2 (CH (x, K t )) and m2 (Q) = 1, so m2 (CH (K t )) = 1. Hence by [III11 , 13.18], I∼ = A9 or He. But neither of these groups involves L± 4 (3), by comparison of orders. This contradiction establishes the lemma.  For the next four lemmas we choose u satisfying (7A). Lemma 7.3. We have u = y, and either y ∈ Lu or O2 (Lu ) = 1. Proof. By Lemma 7.2, B0 = u, and so B/B0 ∼ = E22 . As B/B0 embeds in CAut(Lu ) (L), [III11 , 6.10b] applies. It yields that for any involution e ∈ L = E(CLu (x)), the involutions x and xez are Lu -conjugate, where z is a generator of O2 (Z(Lu )) (possibly z = 1). Since B normalizes Lu , z centralizes B and so z ∈ B by Lemma 5.3a. Thus z ∈ B0 , so z ∈ u. Now e is a root involution of L ∼ = A6 , hence a root involution of K, and hence CG (x)-conjugate to y, whence xe and xy are CG (x)-conjugate. If z = 1, then x ∼G xy, whence u = y, as claimed. Suppose then that z = 1. Suppose for a contradiction that z = xy. Then x is Lu -conjugate to xez = xexy = ey. The involution ey, being the product of two disjoint root involutions of K, lies in the center of a subgroup R = O 2 (R) ∼ = (Q8 ∗ Q8 )Z3 of K by [III11 , 4.4]. However, x lies in the center of no such subgroup of CG (x); indeed x O 2 (CG (x))/KO2 (CG (x)) ∼ = x O 2 (C(x, K)/O2 (CG (x))) has 2-rank 1. This contradiction completes the proof.  Now set J = E(CLy (V ))   E(CG (V )) (as y ∈ V ). Lemma 7.4. The following conditions hold: (a) (b) (c) (d)

[t, J] = 1; J∼ = P Sp4 (3) ∼ = E(CLy (xy  )), where V = y, y  ; Ly ∼ = 2U4 (3), and y ∈ Ly ; and There exists an involution w ∈ Ly ∩ CG (x) with I := E(CK (w)) ∼ = A8 , and with w being Ly -conjugate to z, a 2-central involution of K. Moreover, either CQ (w) = x, or I has a trivial pumpup Iw in CG (w), and u CQ (u) ∈ Syl2 (C(w, Iw )).

7. THEOREM 4–NON-ALTERNATING GROUP NEIGHBORS

169

Proof. We first prove (b). By Lemmas 7.2 and 7.3, B0 = y, whence x, y   acts faithfully on Ly , centralizing L. By [III11 , 6.10d], E(CLy (y  )) ∼ = E(CLy (xy  )) ∼ = P Sp4 (3). Of course E(CLy (y  )) = E(CLy (V )) = J, completing the proof of (b). Now t ∈ NG (V ) centralizes the subgroup L of J, so t normalizes J. But by [III11 , 13.21], CAut(J) (L) is a 2-group. Thus [t, J] = 1, proving (a). By [III11 , 6.6a], CJ (x) contains an involution w such that L w ∼ = Σ6 . Since w ∈ J, we have [w, V t] = 1. Considering w ∈ CG (x) we therefore see that w acts on K like an odd element of Σ10 fixing the support of V pointwise. Replacing w by a suitable involution of Lw we may further assume that w acts on K like a transposition of Σ10 . Set C = CG (w), C = C/O2 (C), I = E(CK (w)) and Iw = the subnormal (7B) closure of I in C. By [III11 , 13.19], L2 -balance, and Corollary 3.6d, ∼ L4 (4), A8 , A10 , HS, 2HS, or A8 × A8 , Iw = (7C) the possibility of 2A8 components being removed by Corollary 3.6. In any case, unless I w = I, CQ (w) acts faithfully on I w , centralizing I, whence CQ (w) = x by [III11 , 13.21]. If CQ (w) > x, then I = I w and x char w CQ (w) ∈ Syl2 (C(w, Iw )∩CG (x)), so w CQ (w) ∈ Syl2 (C(w, Iw )). Suppose that Ly ∼ = U4 (3). Of course w ∈ J ≤ Ly , and by [III11 , 6.12a], w is Ly -conjugate to an involution w ∈ L. But w is K-conjugate to y, so w ∈ y G . Let Lw be the component of CG (w) that is isomorphic to Ly . By (7C), Iw = Lw . But Q, w contains a Sylow 2-subgroup of CG (x, wI). If CQ (w) = x, it follows that x contains a Sylow 2-subgroup of CLw (x), which is clearly impossible. Otherwise, m2 (Lw ) > 2 = m2 (w CQ (w)), a contradiction. Hence O2 (Z(Ly )) = 1. By Lemma 7.3, y ∈ Ly , proving (c). The argument of the previous paragraph shows that w ∈ (w )Ly , and so w is Ly -conjugate to z := yw , a 2-central involution of K, proving (d).  We continue the notation (7B), and (7C) still holds. In particular, by [IA , 2.5.12, Tables 5.3, 5.2.1] (7D)

Out(I w ) is a 2-group.

Also let R be a Sylow 2-subgroup of CC (I) containing x. Lemma 7.5. The following conditions hold: (a) R has sectional 2-rank at most 3; (b) Iw  CG (w); and (c) Iw /O2 (Iw ) ∼ = 2HS. Proof. If x, w is a Sylow 2-subgroup of CG (x, w, I), then CR (x) = x, w, whence R has maximal class by [IG , 10.24], and (a) holds. Otherwise, w×CQ (w) ∈ Syl2 (C(w, I)) and as m2 (Q) = 1, (a) follows. If (b) were false, we could find a 2-component I0 of CG (w) isomorphic to a 2-component of Iw but with [I 0 , I w ] = 1. Then R would contain an isomorphic copy of a Sylow 2-subgroup of Iw , which is contrary to the structure of R in (a). This proves (b).

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

170

Using Lemma 7.4, set w = z g , where z is 2-central in K. By [III11 , 4.4], if we set H1 = O 2 (CK (z)), then H1 = S1 U1 , with z = Z(S1 ), S1 ∼ = Q8 ∗ Q8 , |U1 | = 3 and S1 = [S1 , U1 ]. Set H = H1g , S = S1g and U = U1g , so that similar properties hold for H. Because of (b) we may consider the action of H = O 2 (H) on I w . By (7D) and as H = O 2 (H), H induces inner automorphisms on I w , so that H ≤ I w W , where W = CC (I w ). Since I ≤ I w , (a) implies that Sylow 2-subgroups of W have sectional rank less than that of S, and so S ∩ I w = 1. But S ∩ I w is normal in H and so contains w. Thus w ∈ I w , so the components of I w are not simple. By  (7C), I w ∼ = 2HS, completing the proof of the lemma. Now we can prove Proposition 7.6. Lu /Z(Lu ) ∼  U4 (3) for any u ∈ D − x. = Proof. Assume false and continue the above analysis. Consider the structure of CG (w, y). On the one hand w ∈ J ≤ Ly and so O 2 (CLy (w)) is the central product M1 M2 of two copies of SL2 (3) interchanged by an element h ∈ CLy (w) [III11 , 6.12b]. By solvable balance [IG , 13.8], for each i we have either O2 (Mi ) ≤ Iw or [O2 (M i ), I w ] = 1. Whichever of these two conclusions holds for i = 1 must hold for i = 2, since h normalizes Iw by Lemma 7.5b. But O2 (M1 M2 ) ∼ = Q8 ∗ Q8 has sectional 2-rank 4 so cannot centralize I w by Lemma 7.5a. Thus O2 (M1 M2 ) ≤ I w . Therefore a Sylow 3-subgroup of M 1 M 2 acts faithfully on I w . On the other hand y is a root involution of I = E(CIw (x)) ∼ = A8 , and since ∼ I w = 2HS, [III11 , 5.9] implies that y is a 2-central involution of Iw and so m3 (CAut(I w ) (y)) = 1 by [IA , Table 5.3w]. As a Sylow 3-subgroup of M 1 M 2 has 3-rank 2 and acts faithfully on I w , centralizing y, we have a contradiction. This completes the proof.  Next we rule out the possibility (7E) Lu ∼ = L4 (3) for some u ∈ D − x. As before, we assume that Lu ∼ = L4 (3) for some u ∈ D − x, we choose u = y if possible, and we argue to a contradiction in a series of lemmas. By Lemma 7.2 we may apply [III11 , 6.10c], obtaining CLu (x) = (Z × L) w 2 ∼ ∼ where Z = Z4 , w = 1, Z w = D8 , and L w ∼ = Σ6 . Lemma 7.7. Q = x, and E(CL (y  )) ∼ = E(CL (xy  )) ∼ = P Sp4 (3).

(7F)

u

u

Proof. In the notation of Lemma 5.5, (Q × V )/B0 embeds in Cξ , which is isomorphic to a subgroup of D8 . It follows that Q = x, as claimed. Also, as B0 = u, x, y   acts faithfully on Lu by Lemma 5.4, and the final assertion holds  by [III11 , 6.10d]. On the other hand, for some root involution w1 ∈ K acting as a transposition on the support of L, CQ×K (u) = x × (V × L)w1  contains a subgroup of index at most 2 in a Sylow 2-subgroup of CG (x, u), and Lw1  ∼ = Σ6 , with V w1  ∼ = D8 and Z(V w1 ) = y. By (7F), CG (x, u, L) contains Z ∼ = Z4 , so B ∈ Syl2 (CG (x, u, L)). Therefore, we must have x × V ∗ ∈ Syl2 (CG (x, u, L)) with V < V ∗ ∼ = D8 . We may assume that Z < x×V ∗ , whence

7. THEOREM 4–NON-ALTERNATING GROUP NEIGHBORS

171

y ∈ Ω1 (Z) = Z(V ∗ ) ≤ Lu and u = xy. We then have proved (a) and (b) of the next lemma. Lemma 7.8. The following conditions hold: (a) B0 = xy; (b) E(CLxy (y  )) ∼ = E(CLxy (xy  )) ∼ = P Sp4 (3); (c) xy ∈ Syl2 (C(xy, Lxy )); (d) y ∈ Z, and Z y   ∼ = D8 centralizes x and maps onto a Sylow 2-subgroup of CAut(K) (L); and (e) Ly ∼ = L4 (3). Proof. Assertion (e) holds because we were to have chosen u = y if possible. Thus it remains to prove (c) and (d), and we have seen already that y ∈ Z < x × V ∗ . Thus Z y   ≤ CG (x). But as Q = x is disjoint from y, y  , Z y   embeds in CAut(K) (L) and hence is isomorphic to D8 , proving (d). Now let R be an x, y  -invariant Sylow 2-subgroup of C(xy, Lxy ) and set R0 = CR (x). Then Z y   ∩ R0 = 1 and so |R0 | = |Z y   R0 |/8 ≤ |CCG (x) (L)|2 /8 ≤ 24 /8 = 2, so that R0 = xy. Now if R > R0 , it would follow that x is R-conjugate to x(xy) = y, contradicting Lemma 5.9 and completing the proof.  Lemma 7.9. Ly ∼ = P Sp4 (3) and Ly  CG (y). Proof. There exists an involution g ∈ NK (V ) such that y g = y  . We have by g, L ≤ L1  L = E(CG (V x)) ≤ E(CLxy (y  )) ∼ = P Sp4 (3), and conjugating   E(CG (xy  , y)) ≤ E(CG (y)) with L1 ∼ = P Sp4 (3). Thus L1 ≤ LE(CG (y)) = Ly ; indeed L1 is a component of CLy (xy  ). As a pumpup of L ∼ = A6 containing L1 , Ly is isomorphic to P Sp4 (3), as asserted, or to A10 , U5 (2), L4 (3), U4 (3) or 2U4 (3), by Lemma 5.2 and Lagrange’s Theorem. But the last three of these have been ruled out by Lemma 7.7 and Proposition 7.6. And neither A10 nor U5 (2) is a pumpup of P Sp4 (3), by [IA , 5.2.9, 4.9.1, 4.9.2, 3.1.4]. Thus Ly ∼ = P Sp4 (3). Suppose that Ly  CG (y). Then CE(CG (y)) (Ly ) contains an x-invariant direct product X of copies of Ly . Using [III11 , 6.6b], we get m2 (CG (Ly x)) ≥ m2 (CXy (x)) = 1 + m2 (CX (x)) ≥ 1 + 3. But m2 (CCG (x) (Ly )) ≤ m2 (B) = 3, a contradiction. The lemma is proved.



Lemma 7.10. There exists an involution v ∈ CG (LD) ∩ CG (Ly ) such that K v ∼ = P Sp4 (3). = Σ10 and CLxy (v) ∼ Proof. By Lemma 7.8d, any involution v  ∈ Z y   − V acts on K as a transposition, and centralizes x, y = D as well as L. By Lemma 7.9 and [III11 , 6.6a], v  , x maps into CAut(Ly ) (L) ∼ = Z2 . We define v by v = Cv ,x (Ly ). As [x, Ly ] = 1, v = v  or v  x. In either case the first two asserted properties of v hold; it remains to check that CLxy (v) ∼ = P Sp4 (3). But v Z ∼ = D8 embeds in  CAut(Lxy ) (L) and so the lemma follows from [III11 , 6.10d]. We are now ready to rule out (7E).  L4 (3) for any u ∈ x, y − x. Proposition 7.11. Lu ∼ = Proof. Assume false and continue the above analysis. Let J be the subnormal closure of L in CG (v), and set J = J/O2 (J). By L2 -balance and Lemma

172

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

7.10, J is a 2-component of CG (v), D acts on J, and CJ (x), CJ (y), and CJ (xy) have components isomorphic to A8 , P Sp4 (3), and P Sp4 (3), respectively. But by [III11 , 6.13], no such K-group J exists. This contradiction completes the proof of Proposition 7.11.  As a corollary, we have: Lemma 7.12. Q = x and x ∈ y G . Proof. By Proposition 7.11 and Lemma 5.6, Q = x. Then by Lemma 5.9,  x ∈ y G . The last special case of this section is ∼ 2M12 or 2J2 for some u ∈ D − x. Lu = In this case, n = 9. We let z be the involution in Z(Lu ). Proposition 7.13. Lu ∼ = 2M12 or 2J2 for any u ∈ D − x. Proof. Suppose false. There exists an involution s ∈ CK (y) = CK (u) such that L s ∼ = Σ5 . Then s normalizes L so s normalizes Lu , and so CAut(Lu ) (x) contains a copy of L s ∼ = Σ5 . Consequently, we see from [IA , Tables 5.3bg] that x induces an inner automorphism on Lu , corresponding to an element g ∈ Lu of order 4 with g 2 = z. Setting h = xg −1 , we see that [h, Lu ] = 1 and h2 = z. Furthermore, z ∈ CB (s) = D. Indeed if some involution of CG (x L) acts on K as a transposition (let us call this the “involution case”), then CG (x L)/O2 (CG (x)) ∼ = Z2 × Σ4 and the only square in x, y, h lies in K, so z = y. On the other hand if no involution of CG (x L) acts in this way on K, then a Sylow 2-subgroup of CG (x L) lies in a crown product of Σ4 and Z4 . In this case no involution in K is a square in CG (x L), so z = h2 = xy. Whether z = y or xy, g is real in Lu by [IA , Tables 5.3bg], so x and xz are Lu -conjugate. As y ∈ xG by Lemma 7.12, we have z = xy, so z = y and the involution case holds. As x and xy are conjugate in this case, we must also have y = u. Moreover, if T is a Sylow 2-subgroup of CG (Ly ) containing h, then CT (h) ≤ CT (x, y, L, h) = h, whence T is cyclic or of maximal class. In particular, Ly  CG (y). Let e ∈ L be an involution. By [III11 , 5.7c], e is 2-central in Ly . We set C = CG (e) and W = O 2 (CLy (e)) h and reach a contradiction by considering the embedding of W in C. Both e and y are root involutions of K so e = y j and y = ej for some involution j ∈ K. Set H = Ljy , so that H  CG (e) and H ∼ = Ly with e ∈ Z(H). Set C = C/CC (H) ≤ Aut(H). We have e ∈ Ly − Z(Ly ), and conjugating by j we get y ∈ H − Z(H). Thus y = 1, so that h ∈ C is an element of order 4.  If Ly ∼ = 2J2 , then by [IA , Table 5.3g], y ∈ O 5 (W ), but on the other hand  CC (h) contains no element of order 5. Therefore y ∈ O 5 (W ) = 1, a contradiction. Thus Ly ∼ = 2M12 . Set W0 = O2 (O 2 (CLy (e))). By [IA , Table 5.3b], W0 is special with Z(W0 ) = y, e, and an element w ∈ W of order 3 acts without fixed points on W0 /Z(W0 ). Since y = 1, it follows that W 0 is nonabelian, with CW 0 (w) = y. In particular W 0 must contain a w-invariant Q8 subgroup, so W contains a copy of SL2 (3). On the other hand by construction [W , h] = 1. However, CH (h) contains  no SL2 (3)-subgroup by [III11 , 5.8], a contradiction completing the proof.

7. THEOREM 4–NON-ALTERNATING GROUP NEIGHBORS

173

These exceptions having been ruled out, we have the following consequences. Recall that m = n − 4 and L ∼ = Am . Lemma 7.14. Assume (5A). Let u ∈ D − x and suppose that Lu > L. Then one of the following holds: (a) Lu ∼ =K∼ = An or Lu /O2 (Lu ) ∼ = An−2 ; (b) m = 6 with Lu ∼ = 2HS; (c) m = 8 with Lu ∼ = L4 (4), HS or 2HS; (d) m = 6 with Lu ∼ = Sp4 (4), L± 5 (2), or P Sp4 (3); (e) m = 5 with Lu /Z(Lu ) ∼ = L3 (4) or U3 (4), and with O2 (Z(Lu )) elementary abelian; (f) m = 5 with Lu ∼ = L2 (24 ), L2 (52 ), M12 or J2 . Moreover, if we let ξ be the image of x in Aut(Lu ), then CAut(Lu ) (L ξ) is a 2-group. Proof. These are the only possibilities for Lu , by Lemma 5.2 and Propositions 7.6, 7.11 and 7.13. The final assertion holds by Lemma 5.5.  Now we begin the proof of Proposition 7.1 by contradiction. Fix u ∈ D − x such that the conclusion of the proposition fails for Lu . Thus Lu satisfies one of Lemma 7.14cdef, and in particular n = 9, 10 or 12. Again set B0 = CB (Lu ). We quickly get the following refinement of Lemma 5.4. Lemma 7.15. One of the following holds: (a) B0 = V and [t, Lu ] = 1; or (b) B0 = u, with Lu ∼ = M12 or J2 . Proof. Indeed, alternative (c) of Lemma 5.4 cannot occur by Propositions 7.6 and 7.11, and alternatives (d) and (e) would contradict the fact that one of Lemma 7.14cdef holds. If Lemma 5.4a holds, then B0 = V and t normalizes Lu , with t centralizing both x and the component L of E(CLu (x)). Then the final assertion of Lemma 7.14 immediately implies that [t, Lu ] = 1, so (a) holds. On the other hand if Lemma 5.4b holds, then the four-group x, y   acts faithfully on Lu and centralizes L, so Lemma 5.5 implies that Lu /Z(Lu ) ∼ = M12 or J2 . Then (b) holds by Proposition 7.13.  We rule out the second possibility: Lemma 7.16. B0 = V and [t, Lu ] = 1. Proof. Suppose false, so that B0 = u, Lu ∼ = A5 = M12 or J2 , with L ∼ and n = 9. Let e ∈ L be an involution. By [III11 , 5.7c], e is 2-central in Lu , and we set W = F ∗ (CLu (e)) = O2 (CLu (e)), so that W is extra-special of order 21+4 and the only involution in CAut(Lu ) (W ) is e [III11 , 5.7d]. Now CL×B (Lu ) = CB (Lu ) = B0 = u. Thus there exists v ∈ V # such that [v, Lu ] = 1, and then [ve, Lu ] ≥ [ve, L] = [e, L] = 1 and so v does not act on Lu like e. Therefore [v, W ] = 1, so e ∈ [v, W ]. ∗ ∼  Let J = E(CK (e)) = A5 and let Je be the pumpup of J in CG (e) and Je = CG (e) Je . Then v ∈ J ≤ Je∗  CG (e), and W ≤ CG (e), so e ∈ [v, W ] ≤ Je∗ .

174

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

Choosing an involution g ∈ K interchanging e and y, we  g interchanges J  see that CG (y) ∗ we conclude that and L and so g interchanges Je and Ly . Setting Ly = Ly ∗ y ∈ Ly . In particular Ly is not simple. Since m = 5, Lemma 7.14 implies that Ly /Z(Ly ) ∼ = L3 (4) and O2 (Z(Ly )) is elementary abelian. In particular, y as well as u is a counterexample to our proposition. Set By = Ω1 (CB (Ly )), the analogue of B0 but for y instead of u. Lemma 7.15 applies to y as well as to u and implies that [V t , Ly ] = 1. Choose an involution z ∈ Z(Ly ); since x normalizes Ly we may choose z to centralize x. Then z ∈ CG (x) normalizes K and centralizes LV t, so it centralizes K. Consequently z = x,  which is absurd as [x, Ly ] = 1. This contradiction proves the lemma. Lemma 7.17. Lu /O2 (Z(Lu )) ∼  L3 (4). = Proof. Suppose false, so that n = 9 and V = B0 centralizes Lu , by the preceding lemma. Thus u = y, and L = CLy (x). Let V1 ∈ Syl2 (L). Choose an involution s ∈ K normalizing V ×L, centralizing y as well as an involution of V1 , and such that Ls ∼ = Σ5 . We may even choose s to invert t. Then s ∈ y K and x, s induces a four-group of outer automorphisms of Ly . We further choose an involution w ∈ CK (s) such that V w = V1 . Set t1 = tw ∈ L, so that [t1 , V ] = 1 and s inverts t1 . By [III11 , 6.15a], H1 := [x, CLy (V1 )] is homocyclic of exponent 4 and rank 2, H1 is inverted elementwise by x, Ω1 (H1 ) = V1 and [H1 , t1 ] = H1 . As s ∈ NCG (x) (V1 ), s normalizes H1 . As w interchanges V and V1 , we have K1   CG (V1 ), where K1 := Lw y , and with L1 := E(CK (V1 )) = CK1 (x) ∼ = A5 . Moreover, V ∈ Syl2 (L1 ). Set H = H1w , an x, s, t-invariant homocyclic subgroup of exponent 4 in K1 with Ω1 (H) = V and [H, t] = H. Notice that Ly is the unique component of CG (y) of its isomorphism type. For otherwise, C(y, Ly ) would contain an x-invariant product M of components isomorphic to Ly , and then CM (x) would contain a subgroup R ∼ = E22 or Q8 , with R ∩ D = R ∩ x, y = 1. But R × D ≤ CCG (x) (L x), contradicting the fact that a Sylow 2-subgroup of the latter group embeds in x × CAut(K) (L) and thus in Z2 × D8 . Thus, H normalizes Ly and then H = [H, t] centralizes Ly , by Lemma 7.16. We now consider CG (s). As s normalizes H, with s2 = 1 and CV (s) = y, we have CH (s) = h ∼ = Z4 , with h2 = y. We may choose g ∈ K interchanging y and s. Then y induces an outer automorphism on Lgy , the unique component of CG (s) isomorphic to L3 (4). But then h normalizes Lgy , and since h2 = y we have a contradiction to the fact that Out(L3 (4)) has no element of order 4 [III11 , 6.15b]. This proves the lemma.  We now complete the proof of Proposition 7.1 by examining C := CG (t). Set C = C/O2 (C) and J = E(CK (t)) ∼ = An−3 = Am+1 ,

8. THEOREM 4–THE FINAL CASE: 2HS NEIGHBORS

175

so that J is a component of CG (t, x) = CC (x). By L2 -balance J has a subnormal closure I in C which is either a single x-invariant 2-component or the product of two 2-components interchanged by x, and in either case J is a component of CI (x). Furthermore, it is clear in K that L ≤ J, and therefore L ≤ I ∩ Lu . Since Lu ≤ C by Lemma 7.12, and Lu is quasisimple, it follows that Lu ≤ I, so that Lu ≤ I.

(7G)

But the isomorphism type of Lu is given in Lemma 7.14, so Lu /Z(Lu ) is not embeddable in J by Lagrange’s Theorem. Hence I has a nonsolvable composition factor not isomorphic to J, so (7H)

I is a single 2-component and J < I.

Moreover since Q = x by Lemma 5.6, we have CC (J x) = CCG (x) (J t) = W t, x where W = O2 (CG (x)). Of course J  CCG (x) (t) = CC (x) and CC (x) (t)/CC (x) (J t) ∼ = J or Σm+1 , whence G

G

CC (J x) = CC (J x) = W x, ∼ with CC (x)/CC (J x) = J or Σm+1 . In particular x ∈ Syl2 (CC (J)), and since [x, I] = 1 and J ≤ I, this implies that CC (I) has odd order. Consequently I = F ∗ (C) is simple, so that we may regard C ≤ Aut(I). However, [III11 , 13.22] implies that the only simple K-group I with an involutory automorphism x, a subgroup J satisfying (7G) and (7H) with J ∼ = Am+1 and |CInn(I)x (J)|2 = 2, and a subgroup Lu with one of the isomorphism types specified in Lemma 7.14cdef for the appropriate value of m, is I = U4 (3), with Lu /O2 (Lu ) ∼ = L3 (4) and m = 5. This contradicts Lemma 7.17 and completes the proof of Proposition 7.1. 8. Theorem 4–The Final Case: 2HS Neighbors In this section we continue the assumptions and notation (5A) and rule out the third possibility of Proposition 7.1, that ∼ 2HS for some u ∈ D − x, y. (8A) n = 10, m = 6 and Lu = Proposition 8.1. Assume (5A). Then the case (8A) of Proposition 7.1 does not occur. Thus, the assumption (8A) will be in force throughout this section. We shall reach a contradiction to the simplicity of G, but this will require some effort, because of the existence of the group Aut(F5 ), in which an outer involution x satisfies CF5 (x) ∼ = A10 , and F ∗ (CF5 (y)) ∼ = 2HS for a root involution y ∈ CF5 (x) (see [III11 , 5.5]). We fix u as in (8A) (we shall see in a moment that u = y) and an x-invariant Sylow 2-subgroup W of C(u, Lu ) = CCG (u) (Lu ). Lemma 8.2. The following conditions hold: (a) CG (x)/O2 (CG (x)) ∼ = Z2 × Σ10 ; (b) u = y; (c) y ∈ W ∼ = Z4 ; and (d) CG (y) = O2 (CG (y))Ly W y  , with y  inverting W , Ly y  / y ∼ = Aut(HS) and CLy (y  ) ∼ = Z2 × A8 .

176

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

∼ A6 Proof. We know that Q = x by Lemma 5.6. We have x ∈ CG (u) and L = is a component of CLu (x), so by [IA , Table 5.3m],x induces an inner automorphism  of Lu , corresponding to an element s ∈ Lu with s2 = Z(Lu ). Thus x = sw with w ∈ W and w2 = s2 . Set U = s W , so that x ∈ U ≥ s, w ∼ = Z4 × Z2 , and L ≤ CLu (s) ≤ CG (U ). Thus U ≤ CCG (x) (L). In particular as s∩x = 1, the image of s in Aut(K) ∼ = Σ10 is a 4-cycle disjoint from L. Furthermore, CLu (s) contains a subgroup L∗ ∼ = Σ6 , by [IA , Table 5.3m]. Fix an involution v ∈ L∗ − L. Then v ∈ CG (x), and the image v of v in Aut(K) centralizes the image of s, and induces an outer automorphism on L. Therefore v induces an outer automorphism on K. Now CG (x) = O2 (CG (x))K v x, and so (a) holds. As CG (x)/K has elementary abelian Sylow 2-subgroups, u = w2 ∈ K, whence u = D ∩ K = y, which is (b). Now CAut(K) (L) ∼ = Σ4 , so by (a), Sylow 2-subgroups of CCG (x) (L) are embeddable in Z2 × Σ4 . As s ∈ Z(U ) has order 4, while s CW (w) = CU (w) = CU (x) embeds in Z2 ×D8 , we have CU(x)∼ = Z2 ×Z4 and CW (w) = w ∼ = Z4 . In particular as u ∈ Z(W ), u = Z(Lu ) = w2 . By [IA , Table 5.3m], there is an involution r ∈ Lu such that sr = s−1 and L s ∼ = P GL2 (9). If W > w then there exists t ∈ W such that wt = w−1 , whence xrt = (sw)rt = sr wt = suwu = sw = x. Thus rt ∈ CG (x)∩NG (L) so AutCG (x) (L) contains Aut(L). However, AutCG (x) (L) = AutAut(K) (L) ∼ = Σ6 is a proper subgroup of Aut(L), a contradiction. Thus W = w, and (c) follows. Since W is cyclic, C(y, Ly ) has a normal 2-complement by Burnside’s Theorem. Moreover, as W is cyclic, x, y   acts faithfully on Ly , centralizing L and with x inducing an inner automorphism. Thus with [IA , Table 5.3m], y  induces a noninner automorphism on Ly , so since Out(Ly ) ∼ = Z2 , CG (y) = O2 (CG (y))W Ly y  .  ∼ Moreover CLy (y ) contains L = A6 and so CLy (y  ) ∼ = Z2 × A8 by [IA , Table 5.3m]. It remains only to prove that y  inverts W . But CLy (y  ) ∼ = Z2 × A8 contains no element of order 4 centralizing L, so y  inverts s. As y  centralizes x = sw, y  inverts W . The lemma is proved.  We keep the notation x = sw, s2 = w2 = y, s ∈ Ly , w = W of the preceding proof. We choose any involution e ∈ L. We begin the fusion analysis with the following elementary facts. Lemma 8.3. The following conditions hold: (a) x and xy are conjugate in G; (b) x, y and ye represent distinct conjugacy classes in G; and (c) For every involution z ∈ Ly − y, the pair {z, zy} consists of one Gconjugate of y and one of ye. Proof. From [IA , Table 5.3m], or the last paragraph of the preceding proof, we see that s is Ly -conjugate to s−1 = sy. Therefore x = sw is Ly -conjugate to syw = swy = xy, proving (a). Since E(CG (y)) ∼ = 2HS while K ∼ = A10 is a component of CG (x), clearly x and y are not G-conjugate. Also by Lemma 8.2a, x is not a square in CG (x), hence not in G. But ye ∈ y L ≤ K ∼ = A10 and every involution of K is clearly a square in K. Thus x is not conjugate to ye.

8. THEOREM 4–THE FINAL CASE: 2HS NEIGHBORS

177

Let z be any involution commuting with y. Then z acts on Ly , so surveying all the possibilities in [IA , Table 5.3m] we see that CLy (z)/O2 (CLy (z)) has order divisible by 5. Therefore O 2 (CG (y, z)/O2 (CG (y, z))) does as well. Were y and ye conjugate, it would follow that O 2 (CG (ye, x)/O2 (CG (ye, x))) had order divisible by 5. However, O 2 (CG (ye, x)/O2 (CG (ye, x))) is involved in CK (ye), by Lemma 8.2a, and CK (ye) is a {2, 3}-group, contradiction. Thus (b) holds. Finally by [IA , Table 5.3m], all involutions of Ly − y are 2-central, and all are conjugate modulo y. Therefore it suffices to prove (c) for some involution of Ly − y, and for z = e the assertion is clear.  We choose Sylow 2-subgroups P ≤ T of CG (y) and G, respectively. Thus P = CT (y). We may suppose that P was chosen so that D ≤ P , and thus, w ∈ P . Notice that W ∈ Syl2 (O2 2 (CG (y))). We set W = {W g | g ∈ G, W g ≤ T }, and A = W, so that A is the weak closure of W in T . We also set N = NG (A) and N = N/O2 (N ). Since A is weakly closed in T by definition, NG (T ) ≤ N and N controls G-fusion in Z(A), the latter because of [IG , 16.9]. We verify that W satisfies the conditions of [III8 , Lemma 1.4]. Clearly W is closed under T -conjugation, and consists of cyclic subgroups of order 4. Next, if T0 is any 2-subgroup of G containing W , then NT0 (Ω1 (W )) = NT0 (W ). (Indeed NT0 (Ω1 (W )) = CT0 (y) normalizes W = T0 ∩ O2 2 (CG (y)).) This statement may be conjugated by any element of G and yields that NT (Ω1 (W ∗ )) = NT (W ∗ ) for any W ∗ ∈ W. Finally, if two distinct elements of W had a nontrivial intersection, then conjugating by an appropriate element of G we would obtain a conjugate W ∗ = W of W such that y ∈ W ∩ W ∗ and W, W ∗  is a 2-group. But W ∗ = W g for some g ∈ G, and then g ∈ CG (y). As W ∈ Syl2 (O2 2 (CG (y))), it follows that W = W ∗ , a contradiction. Thus [III8 , Lemma 1.4] applies, yielding that A is abelian, so that A ≤ CG (y), A is the weak closure of W in P , and N (8B) controls G-fusion in A. We also let Z = Ω1 (A), let Z0 be the subgroup of A generated by the conjugates of ye which it contains, and set A0 = {a ∈ A | a2 ∈ Z0 }. Thus Z0  N and hence A0  N .  We also set N = N /A0 , put R = O2 (N ) ∩ [N , N ], let R be the preimage of R in T and next prove Lemma 8.4. The following conditions hold: (a) A ∼ = Z4 × Z4 × Z4 × Z4 and A ∈ Syl2 (CG (A)); (b) N ∩ CG (y) is a split extension (AF )M with F ∼ = Z2 inverting A, and M∼ = L3 (2); A F ∼ (c) N = (RAF )M , a split extension, with R = E25 having as its M composition factors one natural and two trivial modules;

178

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

(d) |W| = |T : P | = 23 , and T is transitive on W; (e) Each involution in Z − Z0 lies in an element of W, and is N -conjugate to y. Proof. First, W ∼ = Z4 lies in CCG (x) (L). But (8C) CG (x)/O2 (CG (x)) ∼ = Σ10 × Z2 , with L mapping on a root A6 subgroup of Σ10 . It follows that we may choose the isomorphism (8C) in such a way that the image of w corresponds to a 4-cycle in Σ10 (disjoint from the support of the image of L). In Σ10 such a 4-cycle clearly has a distinct conjugate with which it commutes, and therefore W is not weakly closed in some Sylow 2-subgroup of G, hence not weakly closed in T . Thus A > W . Choose W0 ∈ W with W0 = W . As with W above, we have (8D)

[W0 , CG (Ω1 (W0 ))] ≤ Ω1 (W0 )O2 (CG (Ω1 (W0 ))).

Moreover, W0 ≤ A ≤ CT (W ) = T ∩ Ly W = P0 W , where P0 = T ∩ Ly ∈ Syl2 (Ly ). Let z0 be the involution in W0 . Then z0 is a square in Ly W so z0 ∈ Ly . Consequently by Lemma 8.3, z0 and z0 y are not conjugate, so that z0 is not conjugate to any element of the coset z0 W other than itself. Therefore if we put C = CG (y) and  = C/O2 2 (C), we see that CL W (z0 ) = CL (z0 )W covers C  ( C y y Ly z0 ). It follows by 0 , C  ( (8D) that [W z0 )] ≤  z0  O2 (C  ( z0 )); but the second factor is trivial (see Ly

Ly

0 , C  ( [IA , Table 5.3m]) and so [W z0 , whence Ly z0 )] ≤  0  C  ( W Ly z0 ).  be the weak closure of W 0 in P with respect to L  y and let B be the complete Let B  By  0 = N  (B). preimage of B in P . Since P ≤ T it follows that B ≤ A. Also set N C ∼ 0 , B  = C  (B),  [III11 , 5.10], B = Z4 × Z4 × Z4 contains exactly 7 conjugates of W C  F by M , where F ∼  M ∼ 0 is a split extension of B and N = Z2 inverts B, = L3 (2) acts   naturally on Ω1 (B), and M splits over W .  = B,  and as W ≤ B it follows that B = A. This  ≤ C  (B) Since A is abelian, A C  immediately yields (a). A Frattini argument implies that the image of N ∩ C in C  is precisely N0 . Then all assertions of (b) follow as well, except for the assertion about the action of F on A, which will be verified below. 0 has 7 conjugates under N 0 , and W contains preimages of these as Since W well as W itself, we have |W| ≥ 8. Furthermore, we know that distinct elements of W have distinct subgroups of order 2, which lie in Ω1 (A) = Z and are G-conjugate to y. But by Lemma 8.3c, the involution z0 y of Z is not G-conjugate to y. Since an element of N of order 7 has orbits of lengths 1, 7 and 7 on Z # , it must be that Z contains exactly 8 G-conjugates of y, and so |W| = 8. But N is transitive on W, with the stabilizer of W equalling N ∩ C, so |N : N ∩ C| = 8. Therefore the Sylow 2-subgroup T of N is transitive on W, and as P = T ∩ C ∈ Syl2 (C), it follows that |T : P | = 8, proving (d) as well as (e). , and are permuted The seven elements of W − {W } are all conjugate under M    by F , which centralizes M modulo A. Since a 7-cycle in Σ7 is self-centralizing, all elements of W − {W } must be normalized by F, and F either centralizes them all  F inverts them all, and as it also inverts W or inverts them all. Since F inverts B, by Lemma 8.2d, F inverts A, completing the proof of (b).

8. THEOREM 4–THE FINAL CASE: 2HS NEIGHBORS

179

It remains to prove (c). Because of (b), we see that CN (Z) = AF . Therefore N /AF acts faithfully on Z so embeds in Aut(Z) ∼ = L4 (2), permuting the eight elements of W transitively and with the image of M being the stabilizer of y. Thus |N /AF |2 = |L4 (2)|2 , so N /AF embeds as a parabolic subgroup of Aut(Z), the only possibility being NAut(Z) (Z0 ). Thus N /AF is the split extension of a subgroup R1 /AF ∼ = E23 centralizing Z0 by the image of M ∼ = L3 (2), acting naturally. Here 1 is an extension of A F ∼ R1 satisfies AF ≤ R1  N . Then R = E22 , on which M 1 ∼ must act trivially, by R1 /AF , a natural M -module. It follows that R = E25 , completing the proof.  By a transfer argument, we can now prove: Lemma 8.5. Let T0 = T ∩ [N, N ]. Then the following conditions hold: (a) w ∈ T0 ; and (b) A ≤ R, |T : T0 | = 2, and T = T0 F . Proof. We have [T, T ] ≤ T0 . Recall that P0 = P ∩ Ly ∈ Syl2 (Ly ). Now (N ∩ Ly )/ y ∼ = (Z4 × Z4 × Z4 )L3 (2) is a perfect subgroup of Ly / y of odd index, so by [IG , 15.12(iii)], N ∩ Ly is perfect and so (8E)

P0 ≤ T ∩ [N, N ] = T0 .

We shall compute VG→T /T0 (x), first proving (8F)

For any extremal G-conjugate x of x in T , we have T0 x = T0 w.

Indeed, let x = xg be such an extremal G-conjugate. Then by Sylow’s Theorem we may assume that g is chosen so that CT (x)g ≤ CT (x ) ≤ T . In particular W g ≤ T , and since T is transitive on W by Lemma 8.4d, there exists h ∈ T such that W gh = W . Set x = xgh . As [T, T ] ≤ T0 , T0 x = T0 x so it is enough to prove the assertion for x . Replacing x by x we may therefore assume that x ∈ CG (W ). Therefore x ∈ CT (W ) = T ∩ Ly W = P0 W , as P0 = T ∩ Ly . But by Lemma 8.3bc, P0 contains no G-conjugate of x, and so x ∈ P0 W − P0 = P0 w. Thus (8E) implies (8F). As T ∈ Syl2 (G), we therefore have τG→T /T0 (x) = T0 w, by [III8 , 6.5]. Therefore the simplicity of G implies that w ∈ T0 , proving (a). Consequently A ≤ [N , N ] ∩ O2 (N ) = R, so A ≤ R. Then in Lemma 8.4c we have a split extension N = (RF )M . Hence by [III8 , 7.5a], [N , N ] = RM and (b) follows.  We are now able to reach a final contradiction completing the proof of Proposi is an indecomposable M -module, with socle A ∼ tion 8.1. We now know that R = Z2   and R/A a natural M -module permuting W regularly. Therefore by [III8 , 7.5b],  ≤ C  (   ∼ 2 A R t) = [R, t] = E2 for all involutions t ∈ M . Choose any such involution t. Then t inverts some element v ∈ M of order 3. We have ∼ [R, ∼ E22 . =R v × [R,  v], with A ≤R v := C  (  v] = R v) = R

  Since  t normalizes both direct factors and acts freely on R, t acts nontrivially on both factors. In particular  (8G) C  ( t) = A. Rv

180

10. THEOREM C∗ 7 : STAGE 4A: A LARGE ALTERNATING SUBGROUP

  Define N0 to be the complete inverse image of N0 := CR (v) v, t and choose u ∈ R such that u ∈ CR (v)−A. Also set N0 = N0 Z/Z. Then N0 ∩ A = W ×N0 ∩ A0 is   a four-group, and N0 ∩ R = (N0 ∩ A) u has order 8. Since u ∈ O2 (N ) − A, u does not normalize W . Therefore it does not normalize W . But W u ∈ W, so W = W u are the two subgroups of N0 ∩ A other than N0 ∩ A0 . It follows that N0 ∩ R ∼ = D8 ∼ with Z(N0 ∩ R) = N0 ∩ A0 . As N0 ∩ A = E22 and N0 ∩ R are normal subgroups of N0 , t induces an inner automorphism on N0 ∩ R. In particular t centralizes v . This the Frattini quotient N0 ∩ R/N0 ∩ A0 of N0 ∩ R. But then t centralizes R contradicts (8G), and the proof of Proposition 8.1 is complete. Now Theorem 4 is immediate. Indeed if Theorem 4 fails, then p = 2 and there is (x, K) ∈ J∗2 (G) with K ∼ = An , n ≥ 9, and m2 (C(x, K)) = 1. Fix a root involution y ∈ K and let L = E(CK (y)). Then (y, L) is an acceptable subterminal (x, K)-pair by definition, and so by Theorem C∗7 , Stage 3a, the conditions (5A) hold. Then Proposition 7.1 gives the possible isomorphism types of the nontrivial pumpups Lu for any u ∈ x, y − x. But Proposition 6.1 shows that the possibility (c) of Proposition 7.1 must occur for some such u and Lu . Then Proposition 8.1 yields a final contradiction. This completes the proof of Theorem 4 and with it the proof of Theorem C∗7 , Stage 4a.

CHAPTER 11

Properties of K-Groups 1. Everyday Tools Lemma 1.1. Suppose that I, J ∈ Kp and I 2, and Li = A1 (q), i = 1, 2, 3. Remark 1.14. Part (b) asserts that in a K-group X, the following situation, permitted in principle by Lp -balance, actually never occurs: (1A)

(x, K) < (y, L) is a diagonal pumpup, where x, y ∈ X are commuting elements of odd prime order p, K and L are p-components of CX (x) and CX (y) respectively, and [y, K/Op (K)] = 1.

A K-group-free proof that (1A) is impossible in a simple group would be interesting. Proof. Since U6 (2) and D4 (2) are excluded from (c), (a) is a consequence of (c) and the definition of Gp [I2 , 12.1, 13.1]. If K ∈ Alt ∪ Spor ∪ Chev(p), then CK (x) does not have as many as three components, by [IA , 5.2.2d], [IA , Tables 5.3], and [IA , 4.9.1,4.9.2], respectively. Hence K ∈ Chev(r) − Alt − Chev(p) for some r = p. Without loss, K is the adjoint version. Note also that since L has p conjugates, mp (K/Z(K)) ≥ p. We have that x ∈ Inndiag(K), since otherwise, with the help of [IA , 4.9.1, Table 4.7.3A] and the oddness of p, CK (x) has at most one p-component.

1. EVERYDAY TOOLS

185

  If K is a classical group and its natural vector space is V , write K = K/Z( K)  where K ≤ Isom(V ), the group of isometries of V . Let x  ∈ GL(V ) be a p-element of least order inducing the automorphism x on K. Then according as x p = 1 or not, the structure of E(CK (x)) is given by [IA , 4.8.1,4.8.2] or [IA , 4.8.4]. In the p = 1. Let V0 = CV ( x). By latter case, CK (x) has at most one component, so x [IA , 4.8.1,4.8.2], E(CK (x)) is the product of the images of L0 := E(Isom(V0 )) and at most (p−1)/dT linear or unitary groups. Here dT is a natural number depending on the type T (linear, unitary, symplectic, orthogonal) of the space V , as well as on the multiplicative behavior of ±q modulo p, where q is the level of K. In particular (1B)

dT = 1 if and only if V is a linear or unitary space, and accordingly p divides q − 1 or q + 1.

Let n and n0 be the numbers of components of E(CK (x)) and L0 , respectively. Thus n0 ≤ 1 unless V is an orthogonal space and V0 is of + type and dimension 4, in which case n0 = 2. We have p−1 n ≤ n0 + . dT If dT > 1, then n ≤ 2 + (p − 1)/2 = (p + 3)/2, with equality only in the orthogonal case discussed above. Thus since n ≥ p, we have n = p = 3, V is an orthogonal space, dim(V0 ) = 4. The components of E(CK (x)) are of the same type as those of Isom(V0 ), i.e., A1 (q). But the component of E(CK (x)) supported on ± ⊥ V0⊥ is SL± m/2 (q), where m = dim V0 ; therefore m = 4 and K = D4 (q). Moreover |V0⊥ so Isom(V0⊥ ) must be, like V0 , of + type. Thus K = D4 (q), Isom(V0⊥ ) contains x as in (c3). Note also in this case that two of the three components have the same support on V , so the cycling element y cannot lie in Isom(V ); it must therefore induce a graph or graph-field automorphism on K. Thus by [IA , Table 4.7.3A], E(CK (y)) ∼ = G2 (q) or A± 2 (q); in particular E(CK (y)) is not a homomorphic image of L1 and so (b) holds in this case. Therefore we may assume dT = 1, whence K ∼ = Lm (q) and p divides q − , x) are supported on the p eigenspaces of x ; as  = ±1. The components of CK ( these components are cycled by y, the eigenspaces must be of equal dimension, and (c1) holds. Moreover the diagonal Ly , a cover of Lk (q), acts homogeneously on V , with p (isomorphic) nontrivial irreducible constituents. In particular it is not irreducible on its support on V . Using [IA , 4.9.1,4.8.2,4.8.4], however, we see that the only elements of order p with an Ak−1 (q) component J in their centralizers have J acting irreducibly on its support on V . This implies (b) in this case, and completes the proof in the classical case. Suppose next that K is of exceptional type. If p = 3, or p = 5 with K = E8 (q), we see from [IA , Tables 4.7.3AB] that K can only be E6 (q), q ≡  (mod 3), with p = 3 and (c2) holding. Moreover, the same table shows that Z(L1 L2 L3 ) has order at most 3, so Ly ∼ = A2 (q)a . Again the table, together with [IA , 4.9.1], shows that no component of the centralizer of any automorphism of order 3 is isomorphic to Ly , proving (b) in this case. Finally suppose that p > 3, and p > 5 if K = E8 (q). Then p exceeds all the coefficients of the high root on the extended Dynkin diagram, and so by [IA , 4.2.2], x is of parabolic type. Therefore by deleting nodes from the Dynkin diagram of

186

11. PROPERTIES OF K-GROUPS

K, we must be able to reach a disjoint union of p isomorphic subdiagrams. But as p ≥ 5, and even p ≥ 7 when K = E8 , this is easily seen to be impossible. This contradiction completes the proof of the lemma.  Lemma 1.15. Let p and r be distinct primes and let K ∈ Chev(r). Let E be an elementary abelian p-subgroup of Aut(K) and J a Lie component of CK (E). Assume that (a) E is noncyclic; (b) J is a nonsolvable Lie component of CK (e) for some e ∈ E # ; (c) If p = 2, then J/Z(J) ∼  L2 (q) for any odd q. = Then there exists e1 ∈ E # such that CK (e1 ) has a Lie component J1 satisfying [J1 , J1 ] > [J, J]. Proof. Let L = [J, J], a component of both E(CK (e)) and E(CK (E)). Since E is noncyclic, and r divides the order of J ≤ CK (E), Seitz’s Lemma [IA , 7.3.1]   implies that ΓrE,1 (K) = K. But by [IA , 4.2.2], for each e1 ∈ E # , O r (CG (e1 )) normalizes all Lie components of CK (e1 ). It follows that (1C)

for some e1 ∈ E # , L is not a component of CK (e1 ). 



For otherwise, for each e1 ∈ E # , O r (CK (e1 )) would normalize J. As ΓrE,1 (K) is  generated by such subgroups O r (CK (e1 )), we would have J  K, contradicting the quasisimplicity of K. With e1 as in (1C), let L1 be the subnormal closure of L in CK (e1 ). Since CK (e1 )(∞) is the product of the commutator subgroups of the Lie components of CK (e1 ), it follows that either L1 = [J1 , J1 ] > [J, J] for some Lie component J1 of CK (e1 ), or L1 is the diagonal of p components of CK (e1 ) cycled by e. In the former case we are done, so assume the latter. If p is odd, then Lemma 1.13b is contradicted. If p = 2, we assume without loss that K is universal, and we consult [IA , Table 4.5.2]. The only cases of permuted components not of type A1 are (K, L1 , e1 ) = (SL2n (q), SLn (q)2 , tn ), (Sp4n (q), Sp2n (q)2 , tn ), + + (q), Dn+ (q)2 , tn ), (D2n (q), Dn− (q)2 , tn ), (D2n ± (q), Bn−1 (q)2 , γn ), and (D2n−1

all for n ≥ 3. In every case, the automorphism of K induced by e1 is determined up to K-conjugacy by having a component in its centralizer with a central quotient of the given isomorphism type. Thus L and the components of L1 must be conjugate in Aut(K). However, in the first two cases the centers of the components of L1 are not contained in Z(K), while the diagonal L satisfies Z(L) ≤ Z(K). In the last three cases, the components of L1 are universal, while the diagonal L is not universal. These contradictions completes the proof.  Lemma 1.16. Suppose that K ∈ Kp , E ≤ Aut(K) with E ∼ = Ep2 , and I ≤ K is a p-component of CK (e) for all e ∈ E # . Then (a) − (d) hold: (a) E ∩ Inn(K) = 1; (b) One of the following holds:

1. EVERYDAY TOOLS

187

∼ An , and E acts (1) K/Z(K) ∼ = An+kp2 , k > 0, n ≥ 5, IZ(K)/Z(K) = on K/Z(K) like a subgroup of Σn+kp2 with k regular orbits and n fixed points; or (2) K ∈ Spor, and the triple (p, K/Z(K), IZ(K)/Z(K)) is one of the following: (2, M12 or J2 , A5 ) (with I ∼ = A5 ), (2, Co1 , G2 (4)), (2, Suz, L3 (4)), 3 2 2 (2, He, 2 L3 (4)), (2, Ru, B2 (2 2 )), (3, Suz, A6 ), or (3, O  N, A6 ); (c) If (b1) holds with kp2 = 4, then E  CAut(K) (I) ∼ = Σ4 ; (d) If (b2) holds, then E contains all elements of CAut(K) (I) of order p; in particular mp (CAut(K) (I)) = 2. Proof. Without loss, K is simple. First suppose that K ∈ Chev. If some element e ∈ E # is a field or graph-field automorphism, then by [IA , 7.1.4], CAut(K) (I) is cyclic, which is impossible as [E, I] = 1. Therefore E ≤ Aut0 (K). Say K ∈ Chev(r). If r = p, then Outdiag(K) is a p -group. So the image of E in Out(K) consists of images of graph automorphisms, and hence is cyclic [IA , 2.5.12]. But then there is 1 = e ∈ E ∩ Inn(K), and we reach the contradiction I ≤ Lp (CK (e)) = 1 by  the Borel-Tits theorem. Therefore r = p. In this case for each e ∈ E # , O r (CK (e)) is the commuting product of its Lie components, and every p-component of CK (e) is the last term of the derived series of one of these Lie components. Thus    I  O r (CK (e)) | e ∈ E # . But since E centralizes a Sylow r-subgroup of I, which is nontrivial, the group on the right equals K by [IA , 7.3.1], i.e., I  K, which is absurd. Therefore K ∈ Alt ∪ Spor. Suppose that K ∈ Alt, so K ∼ = Am . If m = 6 then CK (e) is solvable for all e ∈ Aut(K), so m = 6. Thus Aut(K) ∼ = Σm , so (a) is clear. As I is a p-component of CK (e) for each e ∈ E # , the elements of E # must all have the same support and the same set of (at least 5) fixed points on n letters. Thus every nontrivial orbit of E is regular, and (b1) and (c) follow easily. Suppose that K ∈ Spor. Again |Out(K)| ≤ 2 so (a) is trivial. Just the condition that there exists e ∈ Ip (Aut(K)) and a component I ≤ E(CK (e)) such that mp (CAut(K) (e I)) > 1 yields, upon examination of the tables in [IA , 5.3], the eight examples of (b2) together with seven other examples that we shall presently rule out: (p, K, I) = (3, Co1 , [3 × 3]U4 (3)), (5, Co1 , A5 ), (2, HS, A6 ), (2, F i22 , Sp6 (2)), (2, F i23 , [2 × 2]U6 (2)), (2, F i24 , [2 × 2]U6 (2)), (2, F2 , F4 (2)). In the eight examples of (b2), E contains some Ee ∈ Sylp (CAut(K) (e I)). This certainly implies that E contains a Sylow subgroup of CK (I). Hence (d) holds if p > 2, or if Out(K) = 1, or if Ee ∩ Inn(K) ∼ = Z2 for some e. These conditions cover five cases; to complete the proof of (d) we then must consider only cases K ∼ = J2 , Suz, and He. In these cases E22 ∼ = E ≤ K, and so if (d) failed there would exist an involution t ∈ Aut(K) − Inn(K) centralizing I. However, I is not embeddable in CK (t) in these three cases, as CK (t)(∞) ∼ = L2 (7); M12 or J2 ; and 3A7 , by [IA , Tables 5.3]. This completes the proof of (d). It remains to rule out the seven triples listed in the previous paragraph. We refer without comment to [IA , Tables 5.3]. In the HS case, m2 (CInn(K) (I)) = 1 so there would have to exist an involution t ∈ Aut(K) − Inn(K) such that CK (t) has a component isomorphic to I; but no such involution exists. In the F i24 and F i22 cases, similarly, there would have to exist an involution t ∈ Inn(K) such that CK (t)

188

11. PROPERTIES OF K-GROUPS

has a component isomorphic to I, and no such t exists. In the F i23 case, E would have to equal O2 (CK (2B)) ∼ = E22 , with I = O 2 (CK (2B)) ∼ = [22 ]U6 (2). However, I contains a subgroup t × J with t of order 3 and J/E ∼ = U4 (2). Let C = CK (t); then F ∗ (C), isomorphic to Z3 × B3 (3) or E36 , is not embeddable in I. On the other  hand using [IA , 7.3.3], F ∗ (C) ≤ Γ3E,1 (K) ≤ I, a contradiction. In the F2 case, again E = O2 (CK (e)) and I ∼ = F4 (2), and all involutions of E would have to be conjugate, hence conjugate in NK (E). But then as Out(F4 (2)) is a 3 -group, NK (E) would be isomorphic to A4 ×F4 (2). On the other hand no element of order 3 has a centralizer containing a copy of F4 (2), contradiction. In the Co1 case with p = 3, again all elements of E ∼ = E32 would have to be conjugate, but they are not, according to the notes to [IA , Table 5.3l]. Finally consider Co1 with p = 5; again all elements of E ∼ = E52 would have to be conjugate. However, for P ∈ Syl5 (K), J(P ) ∼ = E53 and fusion in J(P ) is controlled by NK (J(P )), the product of power mappings and SO3 (5) acting naturally. But one checks easily that in a 3-dimensional orthogonal space over a field of odd characteristic, any hyperplane contains some pair of nonisometric 1-spaces, so the requisite fusion could not occur. This completes the proof of the lemma.  ∼ SL2 (q), q odd, and u ∈ I2 (Aut(K)). Let R be Lemma 1.17. Suppose that K = a u-invariant Sylow 2-subgroup of K. Then [u, R] = 1 and there is x ∈ R of order 4 such that xu = xz, where z = Z(K). Proof. Work in the semidirect product R u. Let Ru = CR (u). If u induces a field automorphism, then Ru is a quaternion group which by order considerations has index 2 in R, and we can take x to be any element of order 4 in R − Ru . If u induces an outer diagonal automorphism, then u acts nontrivially on the four-group  R = R/Φ(R) = a, b where b generates a u-invariant cyclic maximal subgroup of R. Then u inverts uua , which maps on b, and some power of which may be taken as x. Finally if u induces an inner automorphism on K, it must correspond to some element y ∈ R of order 4. There exists a Q8 -subgroup Q ≤ R with y ∈ Q, and we can take x ∈ Q − y, as required.  Lemma 1.18. Let K = SL3 (q), q ≡  (mod 3),  = ±1. Let K = K/Z(K). Then the following conditions hold: (a) All elements of K − Z(K) of order 3 are K-conjugate; (b) If x ∈ I3 (K) and x has a preimage x ∈ K of order 9, then CK (x) has a Lie component isomorphic to SL2 (q); (c) Sylow 3-subgroups of K, and of any group X such that Inn(K) ≤ X ≤ Aut(K), have centers of order 3; Proof. In (a), if x and y are two such elements, then x and y have eigenvalues 1, ω, and ω 2 , where ω 3 = 1 = ω. Hence x and y are diagonalizable, hence conjugate in P GL3 (q) ∼ = Inndiag(K). By [IA , 4.2.2j], x and y are K-conjugate. In (b), let α be an eigenvalue of x and ω = α3 . Then x3 = ω so the eigenvalues of x are α, αω a , and αω b . If these were all different, then det x = α3 = 1, contradiction. Hence two eigenvalues of x are equal and the result follows. In (c), a Sylow 3-subgroup of O 2 (Aut(K)) ≤ P ΓL (q) has the form H w F , where H may be taken to consist of images of diagonal matrices and is homocyclic of exponent (q−)3 and rank 2, and w ∈ L3 (q) is a permutation matrix of order 3. One computes that |CH (w)| = 3. Here F is a group of field automorphisms centralizing

2. AUTOMORPHISMS AND COVERINGS

189

w and acting faithfully on H by power maps. It follows easily that H0 := H ∩ L3 (q) satisfies CHwF (H0 ) = H, whence for any P such that H0 w ≤ P ≤ H w F , Z(P ) ≤ CH (w) ∼ = Z3 . Moreover, a Sylow 3-subgroup of K, being nonabelian, is absolutely irreducible in the natural 3-dimensional representation of K; hence its centralizer in K consists of scalars, i.e., equals Z(K). This completes the proof of (c) and the lemma.  2. Automorphisms and Coverings Lemma 2.1. Let K be a simple K-group. Then the following conditions hold: (a) Either Out(K) has a normal 2-complement or K ∼ = D4 (q), q odd, in which case Out(K)/ Outdiag(K) has a normal 2-complement and Outdiag(K) ∼ = E22 . (b) If Out(K) has a nonabelian Sylow 2-subgroup, then K ∈ Chev(r) for some odd r, and | Outdiag(K)| ≡ 0 (mod 4). Proof. If K ∈ Alt ∪ Spor, then |Out(K)| divides 4 [IA , 5.2.1,Tables 5.3]. So assume that K ∈ Chev(r) for some prime r. Then by [IA , 2.5.12], we have a semidirect product decomposition (2A)

Out(K) = Outdiag(K)(ΓK ΦK ).

Here [ΓK , ΦK ] = 1, ΦK is cyclic, and indeed either ΓK ΦK is abelian or ΓK ΦK = ΓK × ΦK with ΓK ∼ = Σ3 . In any event, ΓK ΦK has a normal 2-complement. Thus if Outdiag(K) has odd order, then both (a) and (b) hold; in particular we are done if r = 2. So assume that r > 2. If K ∼ = E22 and again both parts of the lemma hold, = D4 (q), then Outdiag(K) ∼ so assume that K does not have this form, whence |ΓK | = 1 or 2. In particular ΓK ΦK is abelian. We next prove (b). Suppose that | Outdiag(K)|2 ≤ 2. Then the decomposition (2A) gives a direct product decomposition of Sylow 2-subgroups. Thus (b) holds. To prove that Out(K) has a normal 2-complement it now suffices to show that for T ∈ Syl2 (Outdiag(K)), we have [T, O2 (ΓK ΦK )] = 1. This is obvious if T is cyclic, and the only other case is K ∼ = D2n (q), with Outdiag(K) a four-group. But  O2 (ΓK ΦK ) ≤ ΦK , and [T, ΦK ] = 1 by [IA , 2.5.12h]. The proof is complete. Lemma 2.2. Let p be an odd prime and K ∈ Kp . If x ∈ Ip (Aut(K)), then [x, Ω1 (Z(K))] = 1. Proof. This is obvious if x ∈ Inn(K), so we may assume that x ∈ Inn(K), whence p divides |Out(K)|. As p is odd, K ∈ Chev(r) for some r. If r = p, then either Z(K) = 1 or Out(K) is a p -group [IA , 6.3.1], while if r = p, then Z(K) has  cyclic Sylow p-subgroups [IA , 6.1.4]. The result follows. Lemma 2.3. Let K ∈ Chev and let p be a prime such that p does not divide |Outdiag(K)|. Let x be a field or graph-field automorphism of K of order p. Then any element of x Inn(K) of order p is Inn(K)-conjugate to x. Proof. Let y ∈ x Inn(K) have order p. Let Ku be the universal version of K, and ξ the automorphism of Ku corresponding to x. Then Z(Ku ) ∼ = Outdiag(K) is a p -group, so there is η ∈ Ku ξ of order p mapping on y. We may therefore assume that K is universal, and must prove that x and y are K-conjugate. This is

190

11. PROPERTIES OF K-GROUPS

well-known: let (K, σ) be a σ-setup of K. Then there is a Steinberg endomorphism τ of K such that τ p = σ and τ induces x on CK (σ) ∼ = K. Since y ∈ x Inn(K), there p is τ1 ∈ τ (K σ) such that τ1 induces y on K, and τ1 = σ = τ p (this is argued in the second paragraph of the proof of [IA , 4.9.1d]). Then by Lang’s Theorem, τ1 and τ are K-conjugate, and the preceding equations imply that the conjugating element  lies in CK (σ). But as K is universal, CK (σ) = K, completing the proof. Lemma 2.4. Suppose that p is an odd prime and x is a field automorphism of order p of K ∈ Chev. Assume that p2 does not divide | Outdiag(K)|. Then  O p (CInndiag(K) (x)) ≤ Inn(K). Proof. Let r be the characteristic of K. Since x exists, K ∈ Lie(r). Let C =    O p (CInndiag(K) (x)) and K0 = O r (CK (x)). By [IA , 4.9.1], C ∼ = O p (Inndiag(K0 )) and we identify these two groups. Obviously Inn(K0 ) ≤ K. If the lemma fails, then | Outdiag(K)|p = p and C covers a Sylow p-subgroup Pd of Outdiag(K). Choose a p-element g ∈ C of minimal order mapping on an element of Outdiag(K) of order p. Thus g ∈ C − Inn(K0 ). We claim that g has order p. Indeed it suffices to show that Inndiag(K0 ) contains an element of order p outside Inn(K0 ). For K0 ∼ = E6 (q) and p = 3 dividing q − , this is evident from [IA , Table 4.7.3A]. As p is odd, the only other case is K0 ∼ = Lnp (q), p dividing q − , with − either p2 not dividing q −  or p not dividing n. Since L− np (q) and P GLnp (q) share + 2 + 2 Sylow p-subgroups with Lnp (q ) and P GLnp (q ), respectively, the claim reduces to the case  = 1. If p2 does not divide q − , then we can take g to be the image of a diagonal matrix of GLnp (q) − SLnp (q) of order p. If p does not divide n, we can take g to have minimal polynomial tp − ω, where ω generates F× q . This proves the claim. Now x and g are commuting elements of order p, and so the elements xg i ∈ Inndiag(K)x, 0 ≤ i < p, have order p. Hence by [IA , 4.9.1], they are all Inndiag(K)conjugate. Hence in the group H = Outdiag(K) x, which embeds in Out(K), the images of the xg i are all fused. But Outdiag(K) is abelian with Sylow p-subgroup of order p, so H has a normal p-complement and a Sylow p-subgroup of order p2 . Hence this fusion cannot occur in H. This contradiction completes the proof.  ∼ 3D4 (q), q ≡ 0 (mod 3), q > 2, and x ∈ Aut(K) Lemma 2.5. Suppose that K =  is a graph automorphism of order 3. Let M = CK (x)(∞) . Then O 3 (CAut(K) (M )) = x. 

Proof. Since q > 2, M = O r (CK (x)) where q is a power of the prime r. By the table [IA , Table 4.7.3A], CInndiag(K) (M ) = 1 and CInndiag(K) (x) ∼ = Inndiag(M ). As a σ-setup for K we can take (K, σq γ), where γ = γ 1 or γ 2 according to the conjugacy class of x (notation as in [IA , p. 215]). Thus σq and x−1 have the same action on K. Let K γ = CK (γ) ∼ = G2 or Aa2 . Then (K γ , σq ) is a σ-setup for M , and so σr  / σq  induces field automorphisms on M corresponding to the elements of Gal(Fq /Fr ). As Aut(K) = Inndiag(K)ΦK with ΦK = σr |K , and σq3 |K = 1, it follows that CAut(K) (M ) = σq |K , which is equivalent to the assertion of the lemma.  Lemma 2.6. Suppose that K ∼ = 22 L3 (4). Then (a) Every involution of K/Z(K) is the image of an involution of K;

3. p-RANK AND p-SYLOW STRUCTURE OF QUASISIMPLE GROUPS

191

(b) Every involution of K lies in an elementary abelian 2-subgroup of K of rank m2 (K) = 6; and (c) If X = K g with g 2 = 1, then m2 (CX (g)) ≥ 4. Proof. Parts (a) and (b) hold by [IA , 6.4.1, 6.4.4a] and the fact that K/Z(K) has one class of involutions. In (c), suppose first that g induces a graph-field automorphism–i.e. a unitary automorphism–on K. Then g maps to Z(Out(K)) and so by [IA , Table 6.3.1], [g, Z(K)] = 1. Then CK/Z(K) (g) ∼ = U3 (2) has Q8 Sylow 2-subgroups. Let T be the preimage in K of such a Sylow subgroup. Then [T, g] ≤ Z(K) so [T, g, T ] = 1 and hence [T, T, g] = 1 by the Three Subgroups Lemma. Consequently [Ω1 (T ), g] = 1. But by [IA , 6.4.1], Ω1 (T ) splits over Z(K) and hence m2 (CK (g)) ≥ 3, as desired. On the other hand if g induces a graph or field automorphism on K, then with [IA , Table 6.3.1], we see that g acts nontrivially on Z(K). In these cases CK/Z(K) (g) ∼ = L2 (4) or L3 (2) and in particular g centralizes a four-group U/Z(K), U ≤ K. As g acts freely on Z(K), m2 (CU (g)) ≥ 3, so  m2 (CK (g)) ≥ 3 and the lemma is proved. 3. p-Rank and p-Sylow Structure of Quasisimple Groups Lemma 3.1. Let K ∈ Chev(2) and suppose that mp (K/Z(K)) ≥ 3 for some odd prime p. Then mp (K) − mp (Z(K)) ≥ 3. Proof. We may assume that Z(K) is a nontrivial p-group, and then by [IA , 6.1.4], K ∈ Lie(2) (note that K ∼  3A6 as mp (K/Z(K)) ≥ 3). Thus K is a quotient = of SLn (q) for some sign  and some n such that n ≡ q −  ≡ 0 (mod p), or p = 3 and K is the universal version of E6 (q), q ≡  (mod 3). In either case we need to show that mp (K) ≥ 4. We quote [IA , 4.10.3] to get m3 (K) = 6 in the E6 case; mp (K) ≥ p − 1 in the linear and unitary cases; and if K ∼ = SL3 (q), then  m3 (K/Z(K)) = 2. The lemma follows. Lemma 3.2. Suppose that p is an odd prime and K ∈ Gp . Then mp (K) − mp (Z(K)) ≥ 2. Proof. If mp (K) = 1 then by definition [I2 , 12.1,13.1], K ∈ Cp ∪ Tp , contradiction. Therefore mp (K) ≥ 2. In particular if Z(K) = 1 then the conclusion holds. So assume that Z(K) = 1. If K ∈ Alt, then the only possibilities are K∼ = 3A6 or 3A7 , but these lie in T3 by definition. The only sporadic group in Gp with Z(K) = 1 is K = 3O  N with p = 3; but then 31+4 ≤ K so the result holds. Finally, if K ∈ Chev(r) for some r, then r = p so by [IA , 6.1.4], Z(K) is cyclic and K/Z(K) ∼ = Lkp (q), q ≡  (mod p),  = ±1, or E6 (q) with p = 3. We may assume that K is the universal version, and then mp (K)−mp (Z(K)) = mp (K)−1 = kp−2 or 5, respectively, by [IA , 4.10.3]. Hence mp (K) − mp (Z(K)) ≥ 2 unless p = 3 and  K/Z(K) ∼ = L3 (q). But then by definition K ∈ T3 , a final contradiction. Lemma 3.3. Suppose that K ∈ Alt∪Spor and p is an odd prime. If mp (K) ≥ 3, then mp (CK (x)) ≥ 3 for all x ∈ Ip (K). Proof. We may assume that K ∈ Kp . Suppose first that K ∈ Alt. If Z(K) = ∼ An for some n. As x lies in the 1, then mp (K) = 2 by [IA , 5.2.3]. Therefore K = product of groups generated by [n/p] disjoint p-cycles, mp (CK (x) = [n/p] = mp (K) by [IA , 5.2.10]. If K ∈ Spor, then the result follows quickly from the information

192

11. PROPERTIES OF K-GROUPS

in [IA , 5.6.1, Tables 5.3]. Note that if p = 3 and m3 (K) ≥ 4, then by Konvisser’s theorem [IG , 10.17], a Sylow 3-subgroup of K has a normal subgroup A ∼ = E34 , which immediately implies that m3 (CK (x)) ≥ 3.  Lemma 3.4. Let K ∈ Chev(r) ∩ Gp where p and r are distinct primes and p is  be a universal covering group of K. Then the following conditions hold: odd. Let K   (a) Suppose that mp (K)−mp (Op (K)) ≥ 5 but mp (K)−m p (Z(K)) ≤ 4. Then  ∼ p = 3 and K = L6 (q), q > 2, q ≡  (mod 3); and  − (b) Conversely, if K/Z(K) ∼ = L6 (q), q > 2, q ≡  (mod 3), then m3 (K)  = 4. m3 ((Z(K))) ∼ SL (q), m3 (K)  =  = Proof. Part (b) holds by [IA , 6.1.4, 4.10.3], which yield K 6  ∈ Kp be any p-saturated covering group of K/Z(K). 5 and m3 (Z(K)) = 1. Let K  ∼  p (K) Part (a) is trivial if K/O = K/Z(K). Otherwise by [IA , 6.1.4], we may take    ∼ K = SLnp (q), or E6 (q), with p = 3 in the latter case, and with q ≡  (mod p) and  = 1 in either case. Consequently mp (K)  − 1 ≤ 4 so mp (K)  ≤ 5. This mp (Z(K))  p = p with np = 3, 5, rules out the E6 -case and gives np − 1 ≤ 5. Thus |Z(K)| or 6, whence mp (K) = mp (K/Z(K)) = np − δ, δ = 1 or 2. Then np − 1 ≥ 5 by assumption, so np = 6 and the lemma follows.  Lemma 3.5. Suppose that p is an odd prime, I, J ∈ Kp and I 2. The only other possibility is K ≤ GL4 (3), which is absurd by order considerations. The proof is complete. 

In the next few lemmas we use the following setup: (1) K = F ∗ (X) ∈ K2 ; (3A) (2) T ∈ Syl2 (X); and (3) q = r a is a power of the odd prime r. Lemma 3.10. Assume (3A). In any of the following situations, Z(T ) ≤ Z(K): (a) K ∼ = Spinn (q), n odd, n ≥ 5; (b) K ∼ = 2Sp6 (2); (c) K ∼ = HSpin+ 12 (q); or (d) K ∼ = Ω+ 2n (q), n ≥ 3. Proof. We use [IA , Table 4.5.1, 4.9.1] freely. First consider (d). We argue ∼ inductively, beginning with the non-quasisimple group K = Ω+ 4 (q) = SL2 (q) ∗ SL2 (q), for which the assertion CX (T ) = Z(K) follows immediately from Lemma 1.17. The inductive step follows from the fact that K contains J ∼ = Ω+ 2n−1  Z2 . and o ∗ o L2 (J) contains F (CK (Z(L2 (J)))).

194

11. PROPERTIES OF K-GROUPS

Let K = K/Z(K). In (c), an involution z ∈ Z(T ) satisfies F ∗ (CK (z)) ≤ ∼ + Lo2 (CK (z)) = I∗J with I ∼ = Ω+ 8 (q), J = Ω4 (q), and Z(I) = Z(J) = z. Then Z(T ) centralizes I and J by (d) and Lemma 1.17. Elements of CAut(K) (z)−Inndiag(K) do not induce algebraic automorphisms on I and J, and it follows that CAut(K) (IJ) = z. On the other hand if z ∈ K maps on z, then z is an involution and is sheared in K to Z(K) by [IA , 6.2.1]. By [III8 , 6.12], Z(T ) = Z(K), as asserted. In (a), suppose by way of contradiction that z ∈ Z(T ) but the image z of z in X = X/Z(K) is nontrivial. Without loss we may assume that z is an involution. If 1 z induces a field automorphism on K, then CK (z) embeds in Inndiag(Ωn (q 2 )), and using the order formula of type Bn we see that |CK (z)|2 < |K|2 , a contradiction. So z acts as an involution of Inndiag(K) ∼ = SOn (q). One easily finds a K-conjugate z 1 of z such that [z, z 1 ] = 1 and zz 1 has a 2-dimensional eigenspace for −1 on the natural K-module. Let z1 be the corresponding K-conjugate of z. Then (zz1 )2 = Z(K) by [IA , 6.2.1]. As z, z1  has order 8 and z and z1 have the same order it follows that [z, z1 ] = 1. Thus z is sheared to the involution in Z(K), and [III8 , 6.12] yields a contradiction. This proves (a), and (b) is an immediate consequence since  2Sp6 (2) embeds in Spin7 (3) with odd index [IA , 6.2.2]. Lemma 3.11. Assume (3A). Let Z = Ω1 (Z(T )). Then the following hold: (a) If K ∼ = SL± 4 (q) or Sp4 (q), then Z ≤ Z(K); (b) If K is a nonuniversal version of A± 3 (q), then Z ≥ z Z(K) where z ∈  I2 (K) is such that O r (CK (z)) ∼ = SL2 (q) ∗ SL2 (q). If Z > z Z(K), then for any involution u ∈ Z − z Z(K), E(CK (u)) ∼ = P Sp4 (q), and the image of u in Out(K) lies outside [Out(K),Out(K)]; ± r ∼ (c) If K ∼ = L± 5 (q), then Z(T ) is cyclic and O (CK (Z)) = SL4 (q); (d) If K ∼ = Z2 ; = Sp6 (q), then Z/Z(K) = Ω1 (Z(T /Z(K))) ∼ ± r ∼ ∼ (e) If K ∼ = L± 6 (q), then Z = Z2 and O (CK (Z)) = SL4 (q) ∗ SL2 (q); ± ± r ∼ ∼ ∼ (f) If K = L8 (q), then Z = Z2 and O (CK (Z) = SL4 (q) ∗ SL± 4 (q); and (q) and z ∈ I (K) with E(C (z)) having a D4 (q) (g) If K ∼ = Ω9 (q) or Ω± 2 K 10 component, then z is 2-central in X. Proof. Let u ∈ Z − Z(K). If u does not induce an inner-diagonal automorphism on K, then it induces a graph, field, or graph-field automorphism on K, and CK (u) is given by [IA , Table 4.5.1, 4.9.1]. In every case but one, a routine calculation shows that |CK (u)|2 < |K|2 , a contradiction. The exception is in (b), with u a graph automorphism and E(CK (u)) ∼ = Ω5 (q) ∼ = P Sp4 (q). By [IA , 2.5.12], the image of u in Out(K) lies outside [Out(K), Out(K)] in this exceptional case, as asserted. Therefore for the rest of the proof we may assume that u induces an inner-diagonal automorphism on K, and indeed that the image of X in Aut(K) lies in Inndiag(K). r ∼ Suppose that K ∼ = SL± 4 (q) or Sp4 (q). Let z ∈ I2 (K) with E = O (CK (z)) = SL2 (q) × SL2 (q). Replacing z by a suitable conjugate, we may assume that u centralizes a Sylow 2-subgroup of CK (z). By Lemma 1.17, [u, E] = 1. This quickly yields u ∈ Z(E), since u is an involution. But E embeds in an SL2 (q)  Z2 subgroup of K, so u is on the diagonal of Z(E), i.e., u ∈ Z(K), contradiction. Thus (a) holds. In (b), we may take K = SL± 4 (q)/Z0 with 1 = Z0 ≤ Z(K). A similar argument shows that u is the image of an element of GL± 4 (q) whose eigenspaces are the supports of the two Lie components of E on the natural SL± 4 (q)-module. As u

3. p-RANK AND p-SYLOW STRUCTURE OF QUASISIMPLE GROUPS

195

is an involution, u is in Z(E)/Z0 . Moreover, since Z0 = 1, Z(E)/Z0 ≤ Z/Z0 , and (b) follows. The proofs of (c), (d), (e), (f), and (g) follow a similar pattern; we illustrate  with (d). This time let z ∈ I2 (K) with E = O r (CK (z)) ∼ = Sp4 (q) × SL2 (q); we may assume that u centralizes a Sylow 2-subgroup of E. By (a) and Lemma 1.17, [u, E] = 1, As the components of E are absolutely irreducible on their supports, this implies that u ∈ Z(E). Moreover, we calculate using [IA , Table 4.5.1] that  |CInndiag(K) (z)|2 = | Inndiag(K)|2 , so Z(E) ≤ Z, and (d) follows. Lemma 3.12. Assume (3A) with K ∼ = HS, Co3 , M c, Ly, or O  N . Then Z(T ) ∼ = Z2 . Proof. The result for Co3 follows from the analogous result for the 2-central involution centralizer 2Sp6 (2) (Lemma 3.10b) and the fact that Out(Co3 ) = 1 [IA , Table 5.3j]. For M c and Ly, it follows from the result for 2An and 2Σn , n ∈ {8, 11} [IA , 5.2.8b]. For HS, choose z ∈ I2 (Z(T ∩ K)). From [IA , Table 5.3m], we have F := F ∗ (CK (z)) = O2 (CK (z)) and Z(T ∩ K) ≤ Z(F ) ∼ = Z4 . But as CK (t) ∼ = Z3 × Σ5 for any t ∈ I3 (CK (z)), Z(F ) is inverted in CK (z). Thus |Z(T ∩ K)| = 2, and it is evident from [IA , Table 5.3m] that no involution of T − K lies in Z(T ) or even centralizes the Z4 × Z4 × Z4 subgroup A = Ja (T ). But if x ∈ Z(T ) − A, then x2 ∈ Z(T ∩ K) ∩ A ≤ Ω1 (A) and so xA contains an involution, a contradiction. Finally for O  N , a similar argument shows that Z(T ) ∼ = Z2 , using the information in [IA , Table 5.3s] to obtain that Z(T ∩ K) ∼ = Z2 and K has a 2-local subgroup NK (Ja (T )) with Ja (T ) ∼ = Z4 × Z4 × Z4 and no outer involution of  Aut(K) centralizing Ja (T ). The proof is complete. Lemma 3.13. Assume (3A), with K ∼ = L3 (4) or HS. Then Ω1 (Z(T )) ≤ K. Proof. For K/Z(K) ∼ = HS this follows from Lemma 3.12. Suppose false for ∼ K = L3 (4) and let z ∈ I2 (X) − K with [z, T ∩ K] = 1. Then z induces an outer automorphism on K and |CK/Z(K) (z)|2 = |K/Z(K)|2 . However, by [IA , 4.9.1,4.9.2], CK/Z(K) (z) ∼ = L3 (2), U3 (2), or L2 (4). Therefore |CK/Z(K) (z)|2 < |K/Z(K)|2 , a contradiction completing the proof.  Lemma 3.14. Let X be a group and K a component of X such that K ∼ = SL3 (q), where  = ±1 and q ≡  (mod 3). Let P ∈ Syl3 (X). Then [Z(P ), K] = 1. Proof. Suppose false and let X be a minimal counterexample. Since P ∩ K ≤ Z(K), Z(P ) normalizes K. Then X = KZ(P ). Set X = X/CX (K). By Lemma 1.18, Z(P ) ≤ K, so X = K. Hence by minimality, X = K, and Lemma 1.18 finishes the proof.  Lemma 3.15. Let X = Σn , acting naturally on Ω = {1, . . . , n}, and let S ∈ Syl2 (X). Then J(S) = Z × U1 × · · · × U[n/4] where Z = 1 (n ≡ 0, 1 (mod 4)) or Z is generated by a transposition (otherwise); each Ui is isomorphic to D8 and has support on Ω of cardinality 4; and the supports of the Ui are pairwise disjoint. i=[n/4] Moreover, S = J(S)T where J(S) ∩ T = 1, T faithfully permutes {Ui }i=1 , and T is isomorphic to a Sylow 2-subgroup of Σ[n/4] . Proof. Write n = 4m+k, 0 ≤ k < 4; then n!2 = (4m)!2 [(k +2)/2] so the proof quickly reduces to the case n = 4m. Then n!2 = 8m m!2 so S may be taken within an S4 Sm subgroup. Then S = JT with J ∩T = 1, T ∈ Syl2 (Σm ), and J = U1 ×· · ·×Um

196

11. PROPERTIES OF K-GROUPS

as described in the statement of the lemma. It remains only to prove that J = J(S). Let A be an elementary abelian subgroup of S of maximal rank r. Let O1 , . . . , Os be the A-orbits on n = 4m letters, of lengths n1 , . . . , ns . Then by maximality A = A1 × · · · × As where each Ai acts regularly on Oi , i = 1, . . . , s, and fixes Oj pointwise for all j = i. Then r = m2 (A) = log2 (n1 ) + · · · + log2 (ns ). As n1 + · · · + ns = 4m is fixed, the maximality of r is easily seen to imply that each ni equals 2 or 4, so that r = 2m. Moreover, since each Ai permutes the supports of U1 , . . . , Ur , each Oi must be contained in one of these supports. This implies that A ≤ J. As J is generated by two elementary abelian subgroups each of rank 2m, J = J(S) and the proof is complete.  Lemma 3.16. Let K = G2 (q) and let K ∗ be any group such that Inn(K) ≤ K ≤ Aut(K). If q is not a power of 3, then Sylow 3-subgroups of K ∗ have cyclic centers. ∗

Proof. As Outdiag(K) = ΓK = 1, we have K ∗ = Inn(K)Φ, where Φ is a group of field automorphisms of K. We may choose Φ to normalize the centralizer L = CG (x) ∼ = SL3 (q), q ≡  (mod 3),  = ±1, where x has order 3 and is 3-central in K [IA , Table 4.7.3A]. Then Φ faithfully induces a group of field automorphisms on L [IA , 4.2.3]. Thus CLΦ (L) = Z(L). Let P ∈ Syl3 (NK ∗ (L)) ⊆ Syl3 (K ∗ ). Then  Z(P ) ≤ CLΦ (L) = Z(L) by Lemma 3.14, completing the proof. 4. Computations in Alternating Groups Lemma 4.1. Suppose that K ∼ = An for some n. If x ∈ I2 (Aut(K)), then E(CK (x)) is trivial or simple. Proof. This is immediate from [IA , 5.2.9].



Let K = An , n ≥ 4, acting naturally on a set Ω of cardinality n. Let 4 ≤ k ≤ n. By a root Ak -subgroup of K we mean the (pointwise) stabilizer of some Ω1 ⊆ Ω with |Ω1 | = n − k. By a root involution or root four-subgroup of K we mean an involution or four-subgroup of a root A4 -subgroup of K. For n = 6, since Aut(K) ∼ = Σn , these notions depend only on K and not on Ω. However, for n = 6, as Aut(A6 ) > Σ6 , these definitions depend on the choice of Ω, except the notion of root involution (because A6 has a unique class of involutions). In the next result, these notions should be understood to be defined with respect to a fixed but unspecified faithful action on n letters. Lemma 4.2. Let K = An , n ≥ 8. Then the following conditions hold: (a) If L is a root Ak -subgroup of K, 4 ≤ k ≤ n − 1, then CK (L) ∼ = An−k and CAut(K) (L) ∼ = Σn−k ; (b) If L ∼ = Am , 4 ≤ m < n, and φ : L → K is an embedding whose image is a root Am -subgroup of K, then φ carries root involutions to root involutions. Moreover, φ carries root four-groups to root four-groups unless possibly m = 6; (c) Suppose that n ≥ 13 and let E and E1 be root four-subgroups of K with disjoint supports. Let N = NK (E) and M = N ∩ NK (E1 ), so that J := E(M ) ∼ = An−8 . Then any automorphism of M carrying root foursubgroups of J to root four-subgroups of J extends to an automorphism of N;

4. COMPUTATIONS IN ALTERNATING GROUPS

197

(d) If t ∈ I2 (K), then either t is a root involution of K or t ∈ O 2 (CK (t)); and (e) Let Σ be a partition of {1, . . . , n} into [n/2] subsets of size 2 (and one singleton, if n is odd). Let T be the subgroup of K stabilizing every element of Σ. Then T is an elementary abelian 2-group weakly closed in a Sylow 2-subgroup of K, and containing representatives of all conjugacy classes of involutions in K. ∼ Σn . In (b), L and φ(L) are Proof. Part (a) is routine since Aut(K) = Aut(K) = Σn -conjugate, so we may assume that L = φ(L), i.e., φ ∈ Aut(L). If m = 6 then φ is conjugation by an element of Σm , so φ carries root four-groups to root four-groups and in particular root involutions to root involutions. If m = 6 then all involutions in L are root involutions, so (b) follows. In (c), let ψ : M → M be an automorphism carrying root four-subgroups to root four-subgroups. It is enough to show that ψ lies in the Σn−8 -subgroup of Aut(M ). This is obvious if n − 8 = 6, and if n − 8 = 6 it still holds since Aut(A6 )/Σ6 interchanges the two conjugacy classes of four-subgroups of A6 . In (d), t is the product of 2k disjoint transpositions, k ≥ 2, and there is no loss in assuming that t = (12)(34) · · · (n − 1 n) with n = 4k, k > 1. Then (12)(34) = [(12)(56), (153)(264)] ∈ O 2 (CK (t)). Similarly any product of two of t’s transpositions lies in O 2 (CK (t)), which quickly implies (d). In (e), let z be the involution which is the product of all the transpositions supported on some 2-element set in Σ. Then CK (z) = T Σ where T  CK (z), Σ∼ = Σ[n/2] permutes the 2-element parts of Σ naturally, and T , the stabilizer of all parts of Σ, consists of all products of even numbers of the transpositions constituting z. Obviously T contains representatives of all conjugacy classes of involutions of K, and z G ∩ T = {z}, so CK (z) = NK (T ). To complete the proof it suffices to prove that T is weakly closed in NK (T ), for then NK (T ) will contain a Sylow 2-subgroup of K. Suppose then that g ∈ K and T g ≤ NK (T ). If Σg = Σ then clearly T g = T , as desired, so we assume by way of contradiction that Σg = Σ. Let {a, b} be a part of Σg that is not a part of Σ, so that there are parts {a, a } and {b, b } of Σ, where a, b, a , b are all distinct. Since n ≥ 8 there is a part {c, d} of Σg disjoint from {a, a , b}. Then h := (ab)(cd) ∈ T g so h preserves Σ. But {a, a }h = {b, a }, not a part of Σ, a contradiction. The proof of the lemma is complete.  Lemma 4.3. Suppose that K ∼ = An , n ≥ 9, or He. Let x ∈ I2 (Aut(K)) and let I be a component of CK (x). Assume that I ∼ = An−4 or 22 L3 (4), respectively. Then the following conditions hold: ∼ E22 ; (a) E := O2 (CK (I)) = (b) NK (I) contains an element v of order 3 such that E v ∼ = A4 , and if K∼ = An then [v, I] = 1; (c) E contains every involution in CAut(K) (IE); and (d) If K ∼ = An , there is w ∈ I2 (K) such that (E v)w ≤ I. Proof. For K ∼ = An , let X be a root A4 -subgroup of K and I be the root An−4 -subgroup whose support on n letters is disjoint from that of X. Then E = O2 (X) and we take any v ∈ I3 (X). Then the various statements are easily checked; in (d), take w to be a product of four disjoint transpositions such that w maps the

198

11. PROPERTIES OF K-GROUPS

support of X into the support of I. For K ∼ = He the various assertions are contained  in the table [IA , Table 5.3p]. Lemma 4.4. Let K = An , n = 9 or 10, and let z ∈ K be the product of four disjoint transpositions. Set C = CK (z). Then C has a normal subgroup R t such that (a) R = [t, R] ∼ = Q8 ∗ Q8 , t3 = 1, and Z(R) = z; (b) |C : R t | = 2 or 4 according as n = 9 or 10; and (c) CΣn (R) = z, z  , where according as n = 9 or 10, z  = 1 or z  is the (unique) transposition disjoint from z. Moreover, C has no subnormal subgroup isomorphic to SL2 (3). Proof. Let L and L∗ be the root A8 subgroup of K and the root Σ8 -subgroup of Aut(K) ∼ = Σn , respectively, supported on the support of z. Then CL (z) and CL∗ (z) are the kernels of the (surjective) restriction mappings from C and CΣn (z), respectively, to the symmetric group on the fixed points of z. Moreover CL (z) has a normal subgroup R t of index 2 with the structure given in (a), and with CL∗ (R) = z (this can be easily seen for instance using the isomorphisms L ∼ = L4 (2), L∗ ∼ = Aut(L4 (2))). As |C : CL (z)| = (n − 8)!, (b) and (c) follow. The final statement holds as t ∈ Syl3 (C) and the subnormal closure of t in C contains  [t, R] = R of order 25 . This completes the proof. Lemma 4.5. Let K = Σ11 and K0 = A11 ≤ K. Then the following conditions hold: (a) If z ∈ K is the product of three pairwise disjoint 3-cycles, then |CK (z)|2 = 4 and Sylow 2-subgroups of CK (z) are four-groups; and (b) K does not involve SU3 (2). Proof. Let z ∈ J ≤ K with J ∼ = Σ9 . Then J × Z ≤ K with |Z| = 2 so for (a), it suffices to show that |CJ (z)|2 = 2. But this holds since any 3 -subgroup of CJ (z) must permute faithfully the set of three orbits of z on 9 letters. Finally if K involved SU3 (2), then the normalizer of z would involve Q8 , and so CK (z) would contain an element of order 4, contradicting (a). The lemma follows. 

5. Computations in Sporadic Groups Two sets of sporadic groups play a particular role in this volume. One is Spor ∩ G2 = {J1 , Co3 , He, M c, Ly, O  N, F5 }, the set of sporadic groups arising as possible standard components in G. The other is {M12 , J2 }, because of these groups’ bad generational properties for p = 2 (see Lemma 1.16). In addition, 2HS makes an appearance as a component in the centralizer of an involution in F5 . Lemma 5.1. Let K ∈ G2 ∩ Spor − {J1 }. Then the following conditions hold: (a) K is simple; (b) There exists y ∈ I2 (K) such that L := E(CK (y)) is a single component which is standard in K, y ∈ L, and the pair (K, L) is one of the following: (Co3 , 2Sp6 (2)), (M c, 2A8 ), (He, 22 L3 (4)), (Ly, 2A11 ), (O  N, 4L3 (4)), or (F5 , 2HS); (c) In (b), we have CAut(K) (L) ∼ = Z(L) except in the case K ∼ = F5 , in which case CK (L) ∼ = Z(L) ∼ = Z2 but CAut(K) (L) ∼ = Z4 ;

5. COMPUTATIONS IN SPORADIC GROUPS

199

∼ K or (L, J) is one of the following: (d) If J ∈ K2 and L ↑2 J, then either J = (2Sp6 (2), 22 D4 (2)) or (2Ak , 2Ak+2m ), k = 8 or 11, m ≥ 1; (e) In (b), if u ∈ I2 (Aut(L)), then |CL (u)|2 > 4; and (f) In (b), if u ∈ I2 (L), then m2 (CL (u)) − m2 (Z(L)) ≥ 2. (g) If J ∈ K2 and K 0. Finally if J ∈ Spor, the tables [IA , 5.3] show that the only possibility in each case is J ∼ = K. Part (g) follows in a similar way. In (e), since Sylow 2-subgroups of L are evidently not of maximal class, the result follows for u ∈ Inn(L) by [IG , 10.24] applied to L. For noninner automorphisms, notice that it suffices to prove the result for L/Z(L), by [IG , 9.16] applied to a Sylow 2-subgroup of L u. If L/Z(L) ∼ = An , note that every involution of Σn (n ≥ 8) centralizes an E23 subgroup of the form (ab)(cd), (cd)(ef ), (ef )(gh). If L/Z(L) ∼ = L3 (4), = HS, then |CL/Z(L) (u)|2 ≥ 8 by [IA , Table 5.3m]. If L/Z(L) ∼ then CL/Z(L) (u) ∼ = L3 (2), U3 (2), or L2 (4), by [IA , 4.9.1,4.9.2]. Although |L2 (4)|2 = 4, in this case CL (u) covers CL/Z(L) (u) and contains an involution of Z(L), so |CL (u)|2 > 4 even in this case. Finally Out(Co3 ) = 1, completing the proof of (e). In (f), suppose first that L/Z(L) ∼ = L3 (4). Then u is conjugate to a root involution and the preimage U of a root subgroup of L/Z(L) in L is abelian by the discussion in [IA , Case 2, p. 323]. As every involution of L/Z(L) splits over Z(L) [IA , 6.4.1], m2 (CL (u)) ≥ m2 (U ) = 2 + m2 (Z(L)), as required. In all the other cases m2 (Z(L)) = 1 and we claim that it suffices to show that m2 (L) ≥ 3. Namely, let S ∈ Syl2 (L). As S is not of maximal class, there is U  S with U ∼ = E22 . By the Thompson Transfer Lemma u has a conjugate centralizing U , so we may assume that [U, u] = 1. If u ∈ U then m2 (CL (u)) ≥ m2 (U u) = 3 = m2 (Z(L)) + 2, as required, while if u ∈ U then by [IG , 10.20(i)], m2 (CL (u)) = m2 (L) ≥ 3, as required. That m2 (L) ≥ 3 follows from [IA , 6.4.4c] for L ∼ = 2Sp6 (2); from the embeddings A8 ≤ 2HS and J1 ≤ O  N [IA , 5.3ms]; and from E24 subgroups of 2An ,  n ≥ 8, which exist by [IA , 5.2.4g]. This completes the proof of the lemma. Lemma 5.2. Let K = J1 or 2 G2 (3 2 ) for some odd n. Then Out(K) and the Schur multiplier of K have odd order. n

Proof. This information is contained in [IA , Table 5.3f,6.1.4,2.5.12].



Lemma 5.3. Let K = Co3 or F5 , and let z ∈ I2 (K) be 2-central or non-2central, respectively. Let L = E(CK (z)) ∼ = 2Sp6 (2) or 2HS, respectively. Then the following conditions hold:

200

11. PROPERTIES OF K-GROUPS

∼ (a) CK (z)/  z = AutK (L) = Aut(L);  (b) K = L, Lh | h ∈ K, z h ∈ L − Z(L) ; and (c) CAut(K) (L) ∼ = Z2 or Z4 according as K = Co3 or F5 . Proof. In this proof we assume that K is a K-group, i.e., all simple sections of K lie in K. Parts (a) and (c) follow directly from the information in the tables [IA , 5.3],  together with the fact that Sp6 (2) is complete [IA , 2.5.12]. In (b), let K0 = L, Lh | h ∈ K, z h ∈ L − Z(L) and assume that K0 < K. Then for any h ∈ K such that z h ∈ L, we have that Lh is a component of E(CK0 (z h )), hence by L2 -balance, Lh ≤ L2 (K0 ). Thus K0 = L2 (K0 ) and L is a component of E(CK0 (z)). Since |L/Z(L)|22 > |K|2 in both cases, with K0 generated by copies of L, K0 is a single 2-component. We claim that z h ∈ L − Z(L) for some h ∈ K. Note that Z(L) = z in both cases. If K ∼ = Co3 , then L contains a Sylow 2-subgroup of K, so our claim follows from the Z ∗ -Theorem. If K ∼ = F5 , then as no involution of Aut(HS) − HS centralizes a Sylow 2-subgroup of HS (see [IA , Table 5.3m]) and HS has Sylow 2-center of order 2 (Lemma 3.12), Ω1 (Z(T )) ≤ L for any T ∈ Syl2 (CK (z)). But z is not 2-central in K and we can take h ∈ NS (T ) − T , where T < S ≤ K with |S : T | = 2. This establishes the claim. As Z(L) = z, Lh and L are distinct modulo O2 (K0 ). Therefore L ↑2 K0 / O2 (K0 ). But then as z ∈ L, the Borel-Tits Theorem implies that K0 /O2 (K0 ) ∈ Chev(2), so K0 /O2 (K0 ) ∈ Spor, given the isomorphism types of L. From [IA , Tables 5.3] it is then clear that the only possibility is K0 /O2 (K0 ) ∼ = K, so K0 = K. Thus (b) holds and the proof is complete.  Lemma 5.4. Let X = Aut(K), K ∼ = He, and let z ∈ K be a 2-central involution of X. Set Q = O2 (CK (z)). Then CX (Q) = z. Proof. By [IA , Table 5.3q] CK (Q) = z and Q ∼ = D8 ∗ D8 ∗ D8 . Suppose that g ∈ CX (Q) − K. Then g 2 = 1 or z and in either case, gQ contains an involution t such that CQ (t) contains D8 ∗ D8 ∗ Z4 . But by [IA , Table 5.3q], Sylow 2-subgroups  of CK (t) have order 24 , contradiction. The lemma follows. Lemma 5.5. Let K = F5 and let u ∈ I2 (Aut(K)). If u ∈ Inn(K) and CK (u) involves a nonabelian simple group other than A5 , then F ∗ (CK (u)) ∼ = 2HS. Proof. This is immediate upon inspection of [IA , Table 5.3t].



Lemma 5.6. Let K = F5 . (a) Let y ∈ K be an involution such that L := E(CK (y)) ∼ = 2HS. Let R ∈ Syl2 (CK (y)). Then Z(R) ≤ L and Z(R) ∼ = E22 ; and (b) With R as in (a), we have IL (R ∩ L; 2 ) = {1}. Proof. Let C = CK (y) and C = C/ y ∼ = Aut(HS). By [IA , Table 5.3m], L has an E24 Σ6 -subgroup, so Z := Z(R ∩ L) ∼ = Z2 ; also the preimage of Z in L is a four-group. Therefore it suffices to prove that Z(R) = Z and that IL (R ∩ L; 2 ) = 1. Let H = CC (Z) and H 0 = H ∩ L = QΣ with Q = F ∗ (H) ∼ = D8 ∗ D8 ∗ Z4 and Σ ∼ = Σ5 . Suppose that Z(R) > Z, so that |Z(R)| = 4. Then F ∗ (H) = O2 (H) = QZ(R) ∼ = Q × Z2 . Let Y = Z(O2 (H)) ∼ = E22 and choose t ∈ Y − Z. Thus t induces an outer automorphism on L, so by [IA , Table 5.3m], |CL (t)|2 = 27 .

5. COMPUTATIONS IN SPORADIC GROUPS

201

But also t is centralized by QO 2 (H 0 ), a subgroup of index 2 in CL (Z). Hence |CL (t)|2 ≥ |L|2 /2 = 29 , a contradiction. This proves (a). If (b) is false, then there exists W ∈ IL (R ∩ L; 2 ) which is an elementary abelian p-group, of order dividing 32 , 52 , 7, or 11. If, say |W | = 52 , then taking any w ∈ W # we have 210 = |R ∩ L| ≤ |Aut(W )|2 |CL (W )|2 ≤ |Aut(W )|2 |CL (w)|2 ≤ 25+2 , a contradiction. A similar argument rules out the other cases as well. The proof is complete.  Lemma 5.7. Let K ∈ K2 with K/Z(K) ∼ = M12 or J2 . (a) Let E be an elementary abelian 2-subgroup of Aut(K) with m2 (E) ≥ 2. If ΓE,1 (K) < K, then E ∼ = E22 and E(CK (E)) ∼ = A5 ; (b) If K = J2 , then K is outer generated for p = 2; (c) If t ∈ I2 (Aut(K)) with E(CK/Z(K) (t)) ∼ = A5 , then involutions of E(CK (t)) ∼ = A5 are 2-central in K; and (d) If Z(K) = 1 and t ∈ I2 (K) is 2-central, then Q := O2 (CK (t)) ∼ = Q8 ∗ Q8 or Q8 ∗ D8 , and CAut(K) (Q) = t. Proof. We first prove (a) and (b), using freely the information in [IA , Tables 5.3bg]. Suppose then that ΓE,1 (K) < K. First consider the case in which some e ∈ E # induces an inner automorphism on K corresponding to a 2-central involution f of K/Z(K). Then there is an involution f ∈ K mapping on f , and f is 2-central in K. Let C = CK (f ). Then C/Z(K) = CK/Z(K) (f ) is maximal in K/Z(K). It follows that ΓE,1 (K) = C. Let e1 ∈ E − e. Then CK (e1 ) ≤ C, which implies that e1 too acts on K as a 2-central involution f 1 ∈ K/Z(K).  But then we similarly  get C = CK (f1 ) for a preimage f1 of f 1 , and so f , f 1 is a four-subgroup of   Z(C/Z(K)). However, Z(C/Z(K)) = f , a contradiction. Suppose next that K/Z(K) ∼ = J2 and some e ∈ E # induces an outer automor∼ phism on K. Then E(CK (e)) = L2 (7) lies in a unique maximal subgroup M of K, with M (∞) = E(CK (e)). Therefore ΓE,1 (K) ≤ M , and in particular CK (e1 ) cannot involve A5 for any e1 ∈ E # . But E ∩ Inn(K) = 1 and for any e1 ∈ E ∩ Inn(K), CK (e1 ) involves A5 , a contradiction. This proves (b). We also deduce in (a) that E(CK (e)) ∼ = A5 for all e ∈ E # . If K/Z(K) ∼ = J2 we must have E ≤ Inn(K), and using L2 -balance and the structure of all maximal subgroups of K we see that ΓE,1 (K)/Z(K) ∼ = E22 × A5 , so (a) holds in this case. Suppose that K/Z(K) ∼ = M12 . Then in a similar fashion a maximal subgroup M of K containing ΓE,1 (K) must be isomorphic to Z2 × Σ5 , and again (a) follows. In (c), the tables [IA , 5.3bg] show that E(CK (t)) ∼ = A5 , so the involutions of E(CK (t)) are 2-central as they split over Z(K). Finally (d) is immediate from the  tables [IA , Tables 5.3bg]. Lemma 5.8. Let K = M12 or 2M12 . Then in Aut(K), no element of order 4 centralizes an SL2 (3)-subgroup. Proof. Suppose on the contrary that u ∈ Aut(K) has order 4, H ≤ Aut(K) with H ∼ = SL2 (3), and [u, H] = 1. Then H ≤ O 2 (Aut(K)) = K. Only 2-central involutions of K centralize a nonabelian 2-group, so z = Z(O2 (H)) is 2-central in K and Aut(K), and u H ≤ C := CAut(K) (z). Now C ∩K = QΣ with Q ∼ = Q8 ∗Q8 ,

202

11. PROPERTIES OF K-GROUPS

Σ∼ = Σ3 , and Q = [Q, O3 (Σ)]. It follows that O2 (H) ≤ Q and QH covers O3 (Σ). As [u, H] = 1, we have u ≤ O2 (C) = u Q. Now there are exactly two Q8 -subgroups of Q invariant under a 3-element w ∈ H, and one of them is O2 (H); hence the other one, say Q1 , is also u-invariant. As [u, Q1 ] is w-invariant, we get [u, Q1 ] = 1. But then choosing t ∈ O2 (H) of order 4, we see that ut is an involution of Aut(K) − K and [ut, Q1 ] = 1. But by [IA , Table 5.3b], no outer involution of Aut(K) centralizes a nonabelian 2-subgroup of K. This contradiction completes the proof.  Lemma 5.9. Let K = HS or 2HS. If t ∈ I2 (Aut(K)) and L := E(CK (t)) ∼ = A8 , then all involutions of L are 2-central in K. Proof. It suffices to prove the assertion for K = 2HS. In that case, from [IA , Table 5.3m] we see that any involution x ∈ K − Z(K) maps to a 2-central involution of K/Z(K) and is weakly closed in the coset xZ(K); thus x is 2-central in K. The lemma follows as L ∩ Z(K) = 1.  Lemma 5.10. Suppose that K = HS and P ∈ Syl2 (Aut(K)). Let z ∈ Ω1 (Z(P )) and C = CK (z). Then the following conditions hold: (a) There is a unique subgroup Z ∼ = Z4 such that Z  C; and (b) Let B be the weak closure of Z in P with respect to Aut(K). Then (1) B ∼ = Z4 × Z4 × Z4 contains exactly 7 conjugates of Z; (2) NAut(K) (B) = B(M × F ), a split extension, where F ∼ = Z2 inverts B elementwise and M ∼ L (2) acts naturally on Ω (B); and = 3 1  (3) Let K = 2HS be the universal covering group of K. Then M splits  over Z(K). Proof. From [IA , Table 5.3m], Z := Z(O2 (C)) ∼ = Z4 , and C/O2 (C) ∼ = Σ5 acts ∼ 4 irreducibly on O2 (C)/Z = E2 . This yields (a). Note that if g ∈ K and Z ∩ Z g = 1, then g ∈ CK (Ω1 (Z)) = C so Z = Z g . Therefore [III8 , 1.4] applies, and implies in (b) that B is abelian. By [IA , Table 5.3m] there exists N ≤ K such that N is a split extension of O2 (N ) ∼ = Z4 × Z4 × Z4 by L ∼ = L3 (2), and F ∗ (N ) = O2 (N ). We may choose N to contain P ∩ K. Then, z ∈ O2 (N ) and we may choose x ∈ N of order 3 with z ∈ CO2 (N ) (x) ∼ = Z4 . As x ∈ C, [x, Z] = 1, and as CK (x) has D8 Sylow 2-subgroups, it follows that Z ≤ O2 (N ). Hence, O2 (N ) ≤ B and B contains 7 conjugates of Z, one for each involution of O2 (N ). Since N/O2 (N ) acts faithfully on O2 (N ) and B is abelian, B = O2 (N ). Let B0 = Ω1 (B). Then NK (B0 ) = CK (B0 )N = N because N contains a Sylow 2-subgroup of K, and no element of K of odd prime order centralizes an E23 -subgroup of K. Let N ∗ = NAut(K) (B) = NAut(K) (B0 ). Since B is the weak closure of Z in P , a Frattini argument gives |N ∗ : N | = |Aut(K) : K| = 2, and so N ∗ = C0 L, where C0 = CAut(K) (B0 ) and |C0 : B| = 2. Let Q ∈ Syl7 (L) and F = CC0 (Q) ∼ = Z2 . Then F induces an outer automorphism on K, and it is clear from [IA , Table 5.3m] that CK (F ) ∼ = Σ8 . Then CN (F ) = NCK (F ) (B0 ) must be the split extension of E8 by L3 (2), so N ∗ = B(M × F ) as asserted, and CB (F ) = B0 . As HomM (B/B0 , B0 ) ∼ = Z2 , F must invert B elementwise. Finally from [IA , Table 5.3m], for any involution t ∈ K that does not split over  Z(K), t is a direct factor of CK (t). However, every involution of M is central in  whence M a D8 -subgroup of M . Therefore the involutions of M split over Z(K),  as its universal covering group is SL2 (7).  itself splits over Z(K)

6. COMPUTATIONS IN GROUPS OF LIE TYPE

203

Lemma 5.11. Let K ∈ Spor and x ∈ I2 (Aut(K)). Suppose that CK (x) has a component L ∈ Chev(r) for some prime r > 2. Then L ∼ = L2 (5), L2 (7), L2 (9), or L2 (17). Accordingly K/Z(K) ∼ = M12 , J1 , or J2 ; J2 ; HS; or J3 . Proof. This follows directly from the tables [IA , Tables 5.3].



Lemma 5.12. Let K ∈ Spor ∩ Kp and x ∈ Ip (Aut(K)), where p is an odd prime. Suppose that CK (x) has a p-component L such that L ∈ Chev − Chev(p). Then (K/Z(K), p, L/Z(L)) is one of the following triples: (M23 or HS, 3, L2 (4)), (M24 or He, 3, L2 (7)), (Suz, 5, L2 (9)), (F2 , 7, L3 (4)), or (F1 , 13, L3 (3)). Proof. This too follows directly from the tables [IA , 5.3].



6. Computations in Groups of Lie Type Lemma 6.1. Let K ∈ K2 ∩ Chev(r) for some odd r, with K/Z(K) ∼  L2 (q) or = n G2 (3 2 ) for any n. Let z be a classical involution of K, i.e., z ∈ L   CK (z) for some L ∼ = SL2 (r n ) for some n, with L containing long root subgroups of K. Let  α ∈ Aut(K) be a field or graph-field automorphism of order 2 with O r (CK (α)) ∼  = n 2 G2 (3 2 ) for any n. Then for some g ∈ K, [z g , α] = 1 and α induces a field automorphism on Lg of order 2. 2

Proof. Let (K, σ) be a σ-setup for K. Then there is a Steinberg automorphism τ of K such that τ 2 = σ and the image of τ  in Aut(K) is α. Then  n O r (CK (τ )) ∼ = L2 (q) or 2 G2 (3 2 ) for any q or n, by assumption. Choose a τ invariant Borel-torus pair and adopt standard notation for K (with respect to σ) [IA , 2.10]. Then B, a τ -invariant Borel subgroup of K, has a unique highest root subgroup Xα , and J = Xα , Xαw0  is a K-conjugate of L, where w0 is the long element of the Weyl group N /T . As τ centralizes w0 , τ normalizes J, and as τ acts as a field automorphism of order 2 on Xα ∼ = F+ r n , τ acts as a field automorphism of order 2 on J. Since J is K-conjugate to L the proof is complete.  Lemma 6.2. If K ∼ = SL2 (q) for some odd q and x ∈ I2 (Aut(K)), then E(CK (x)) is trivial or isomorphic to SL2 (r), with r 2 = q. Moreover x induces an inner, outer diagonal, or field automorphism on K. Proof. This well-known fact is also contained in [IA , Table 4.5.1,4.9.1].



Lemma 6.3. Suppose that K ∼ = SL2 (q) for some odd q, and K  K g where g is an involution. (a) If g acts nontrivially on K, then there is h ∈ K of order 4 such that Z(K) = [g, h]. (b) g induces an inner-diagonal automorphism or a field automorphism on K. (c) If g induces an inner-diagonal automorphism on K, then CK (g) ∼ = Zq±1 , CInn(K) (g) ∼ = Dq±1 , CInndiag(K) (g) ∼ = D2(q±1) , and CAut(K) (g) has a normal 2-complement; 1 (d) If g induces a field automorphism on K, then CK (g) ∼ = SL2 (q 2 ). (e) Let T ∈ Syl2 (K). Then CAut(K) (T ) has odd order. Proof. These facts are well-known. For (a), see [IIIK , 2.1c]; for (b) and (d), see [IA , 4.9.1] and the discussion preceding it; for (c), see [IIIK , 2.1b] and  [IIK , 4.1b]; and (e) holds by Lemma 1.17.

11. PROPERTIES OF K-GROUPS

204

Lemma 6.4. Assume that K = L3 (q), q odd,  = ±1. Set M = Aut(K) and M0 = Inndiag(K), and let S ∈ Syl2 (M ). Then the following hold: (a) If (q − )2 = 2m > 2, then S ∩ K ∼ = Z2m  Z2 and m−1 (S) ≤ K; (b) If S/S ∩ K is non-cyclic, then K ∼ = L3 (q) and q is a square; (c) Let Z = Z(S ∩ K). Then Z = CS (S ∩ K) ∼ = Z(q−)2 , and in particular Ω1 (Z) ∼ = Z2 ; (d) K has one class of involutions; (e) Let Z be as in (c). Let Ω1 (Z) = z and let Kz be the normal SL2 (q)subgroup of CK (z). Then z is the unique involution of CM (Kz ); (f) K has one class of four-subgroups, and for any four-subgroup V ≤ K, AutK (V ) = Aut(V ) ∼ = Σ3 ; (g) Let z and Kz be as in (e). Then CAut(K) (Kz ) is 2-nilpotent; its Sylow 2subgroups lie in Aut0 (K) and are quaternion or cyclic of order 2(q − )2 , according as q ≡  or − (mod 4). On the other hand, Sylow 2-subgroups of CInn(K) (Kz ) are cyclic of order (q − )2 . (h) In (g), suppose that  = −q and let T0 ∈ Syl2 (CAut(K) (Kz )), so that T0 ∼ = Z4 . Then there is a 2-element g ∈ NK (Kz ) such that z ∈ [T0 , g]. Proof. We first prove (a) and (b). Suppose that K ∼ = U3 (q). By [IA , 2.5.12], Out(K) has cyclic Sylow 2-subgroups, so S/S ∩ M is cyclic. Suppose further that ν2 (q + 1) = m > 1. Then q is not a square and so by [IA , 2.5.12], |S/S ∩ M | = 2. As m > 1, m−1 (S) ≤ K, as claimed. Thus we may assume that K ∼ = L3 (q). If q is not a square, then again by [IA , 2.5.12], |S/S ∩ M | = 2, and in particular, S/S ∩ M is cyclic, proving (b). Now suppose that m−1 (S)  K for some m > 1. Then Out(K)/ Outdiag(K) contains m a cyclic group of field automorphisms of order 2m . Hence q = r 2 for some odd prime power r. Since k+1

r2 k+1

it follows that ν2 (r 2 follows that

k

k

− 1 = (r 2 − 1)(r 2 + 1), k

− 1) = 1 + ν2 (r 2 − 1) for all k ≥ 1. Since ν2 (r 2 − 1) ≥ 3, it m

ν2 (q − 1) ≥ ν2 (r 2 − 1) ≥ 3 + (m − 1) = m + 2 > m, completing the proof of (a). Part (d) is immediate from [IA , Table 4.5.1]. Next let z ∈ I2 (K), so that by [IA , Table 4.5.1], CK (z) has a unique normal SL2 (q)-subgroup Kz , and CM0 (Kz ) ∼ = Zq− . We verify (g). As Out(K) is 2-nilpotent (see Lemma 2.1) and CM0 (Kz ) has cyclic Sylow 2-subgroups, CM (Kz ) is 2-nilpotent, proving the first statement. Without loss z is the image of diag(−1, −1, 1). Let γ be the graph automorphism of K induced by the transpose-inverse mapping on SL3 (q). Thus Aut0 (K) = M0 γ. Now γ centralizes z, inverts CM0 (Kz ), and acts on Kz like an element t ∈ L such that t2 = z. It follows that CAut0 (K) (L) has a Sylow 2-subgroup T0 with the Z4 or quaternion structure asserted in the lemma. Moreover as CM (z)/CAut0 (K) (z) faithfully induces field automorphisms on Kz , CM (Kz ) ≤ Aut0 (K) and (g) follows. Then (e) is an immediate consequence. Moreover, in (h), CM0 (z) ∼ = GL2 (q), and  we choose h ∈ CM0 (z) corresponding to a 2-element of GL2 (q) with determinant −1. We can take T0 = γt, and then compute [γt, h] = z, proving (h). ∼  = SL3 (q) has a subgroup L In (f), note that K = GL2 (q) which is the centralizer of an involution and has odd index. Obviously GL2 (q) has a unique class of

6. COMPUTATIONS IN GROUPS OF LIE TYPE

205

 and K each have a unique class of ordered pairs non-central involutions and so K of unequal but commuting involutions, which implies (f). We also have that S is semidihedral or Zq−  Z2 according as q ≡ − or  (mod 4). It remains to prove (c). Using Lemma 1.17 and the proof of (g) we have CS (S ∩ K) ≤ CS (Kz ) ∼ = Z4 or Q2(q−)2 according as q ≡ − or  (mod 4), and Z = Z(S ∩ K) is a cyclic subgroup of CS (S ∩ K), with CS (S ∩ Kz ) = Z or Z g, where g = γt, t ∈ Kz of order 4. We need to show, therefore, that [g, S ∩ K] = 1. If Z g is quaternion, this is clear as Z ≤ S ∩ K. So assume that q ≡ − (mod 4), whence S ∩ K is semidihedral. Now γ, the transpose inverse mapping, centralizes a four-subgroup of S ∩ K, while the element t ∈ Kz of order 4 centralizes no such four-group, by the structure of S ∩ K. Hence [g, S ∩ K] = 1. The proof is complete.  Lemma 6.5. Let K = P Sp4 (q), q odd. Then the following conditions hold: (a) There is a unique Aut(K)-conjugacy class of E24 -subgroups E ≤ K; (b) Let A ≤ K with A ∼ = E24 . Then CAut(K) (A) = A × F where F is a cyclic group of field automorphisms of K. Moreover AutK (A) ∼ = A5 if q ≡ ±3 (mod 8), and AutK (A) ∼ = Σ5 otherwise; (c) Let y, u ∈ I2 (K) with y 2-central and u ∈ y K . Then y may be chosen in A, and |A ∩ y K | = 5, A ∩ O 2 (CK (y)) ∩ y K = {y}, and |A ∩ uK | = 10. Proof. We identify K = P Sp4 (q) with Ω5 (q) and Inndiag(K) with SO5 (q). Let V be the natural SO5 (q)-module. Let D be the set of all decompositions V =⊥5i=1 Vi of V into five orthogonal and mutually isometric nondegenerate 1spaces. Fix such an element of D. By Witt’s Lemma, SO5 (q) is transitive on D. The simultaneous stabilizer A in SO5 (q) of all the Vi obviously is isomorphic to (O1 (q)5 ) ∩ SO5 (q) ∼ = E24 , and since the Vi are all isometric the spinor norms of the elements of A# are all trivial; so A ≤ K. Similarly the stabilizer N of the decomposition in SO5 (q) is an extension of A by Σ5 , for which A is the heart of the permutation module. Since the Vi are the irreducible constituents of V as an A-module, and the Vi are inequivalent A-modules, CSO5 (q) (A) = A. Let ΦK ≤ Aut(K) be a “standard” group of field automorphisms of K (see  [IA , 2.5.10]). Then K0 := O r (CK (ΦK )) ∼ = P Sp4 (r), where q is a power of the prime r. The previous paragraph applies to K0 as well as K, and so K0 contains an Inndiag(K)-conjugate of A. Replacing A by an Inndiag(K)-conjugate we may assume that [A, ΦK ] = 1. Then CAut(K) (A) = CInndiag(K) (A)ΦK = CSO5 (q) (A)ΦK = A × ΦK . To complete the proof of (a) and (b) it remains to determine the structure of AutK (A); we know that AutInndiag(K) (A) ∼ = Σ5 . Since | Outdiag(K)| = 2, certainly AutK (A) ≥ A5 . Let S be a Sylow 2-subgroup of the preimage in K of A5 , so that |S| = 26 . Then S = AB where B ∈ Syl2 (A5 ) is a four-group acting freely on A. So A = J(S)  NK (S). Hence if |K|2 > 26 , then |NK (A)|2 > 26 and so AutK (A) ∼ = Σ5 . Of course if |K|2 = 26 , which is the case if q ≡ ±3 (mod 8), then AutK (A) ∼ = A5 . Thus (a) and (b) are proved. Finally as A is the heart of the permutation module for A5 ≤ AutK (A), the NK (A)-orbits on A are of length 5 and 10. Now some 2-central involution y of NK (A) must be 2-central in K, and on the other hand y lies in A = CK (A) in the orbit of odd length on A# . Moreover y has four eigenvalues −1, so y pulls back to an involution in Spin5 (q). On the other hand A contains 10 elements

206

11. PROPERTIES OF K-GROUPS

with two eigenvalues −1, and these pull back to elements of order 4 in Spin5 (q). Let u ∈ A be such an involution. Therefore y and u are not K-conjugate, and so they represent the two classes of involutions in K (see [IA , Table 4.5.1]). We have O 2 (CK (y)) ∼ = SL2 (q) ∗ SL2 (q), with preimage SL2 (q) × SL2 (q) in Spin5 (q). Therefore y is weakly closed in O 2 (CK (y)) with respect to K. Thus (c) holds and the proof is complete.  Lemma 6.6. Let K = P Sp4 (3) and t ∈ I2 (Aut(K)). Then the following conditions hold: (a) If E(CK (t)) ∼ = A6 , then CK (t) ∼ = Σ6 and CAut(K) (E(CK (t))) = t; and (b) m2 (CK (t)) ≥ 3. Proof. In (a), regard K as U4 (2). Thus Aut(K) = K γ where γ is a graph automorphism of order 2. Using the Borel-Tits Theorem and [IA , 4.9.2b1] we conclude that either CK (t) is 2-constrained or CK (t) ∼ = Sp4 (2) ∼ = Σ6 , and CK (CK (t)) has odd order in the latter case. As K ∈ Chev(3), the Borel-Tits Theorem quickly yields CK (CK (t)) = 1. Thus (a) holds. Part (b) may be quickly read off the table  [IA , Table 4.5.1]; note that Aut(K) = Inndiag(K). Lemma 6.7. Let K = Spin7 (q), q odd. Then the following conditions hold: (a) All involutions of K − Z(K) are K-conjugate; and (b) m2 (K) = 4. Proof. Let K = K/Z(K) ∼ = Ω7 (q) have natural module V . Then for any z ∈ K such that z is an involution, z is an involution if and only if the −1eigenspace Vz of z on V is 4-dimensional (see [IA , 6.2.1]). As z ∈ Ω7 (q) has trivial spinor norm, the determinant of the orthogonal form on Vz must be a square and so z is determined up to conjugacy. By [IA , 6.2.1] the coset zZ(K) is fused in K, so z is determined up to conjugacy, proving (a). Now for any elementary abelian subgroup # E of K, E ⊆ z K and m2 (E) = 1 + m2 (E). Let χ be the complex character of E corresponding to its representation on V , so that χ(1) = 7 and χ(z) = −1 # for each z ∈ E . Since (χ, 1E ) ≥ 0 we get |E| ≤ 8. On the other hand, using a frame in V consisting of seven isometric 1-dimensional subspaces, an example of such an E is generated by (−1, −1, −1, −1, 1, 1, 1), (−1, −1, 1, 1, −1, −1, 1), and (−1, 1, −1, 1, −1, 1, −1). The proof is complete.  Lemma 6.8. Let K = Spin7 (3). Then the following conditions hold: (a) For any S ∈ Syl2 (K), CAut(K) (S) has odd order; and (b) For any involution t ∈ Aut(K) not induced by an involution of K, CK (t) has a quasisimple component J with Z(K) ≤ J. Proof. Part (a) holds by Lemma 3.10a. In part (b), note that Out(K) ∼ = Outdiag(K) ∼ = SO7 (3). The two eigenspaces V+1 , V−1 of t = Z2 , so Aut(K) ∼  on the natural O7 (3)-module are orthogonal, and so O 3 (CK (t)) contains a normal subgroup isomorphic to Spin(V+1 ) (as long as dim(V+1 ) ≥ 3), which contains Z(K), by [IA , 6.2.1bc]. A similar statement holds for Spin(V−1 ). If either V+1 or V−1 has dimension at least 5, therefore, then the conclusion of (b) holds. Since t ∈ SO7 (3) the only remaining case is dim(V−1 ) = 4, in which case the conclusion of (b) holds + ∼ ∼ provided Spin(V−1 ) ∼ = Spin− 4 (3) = SL2 (9). But if Spin(V−1 ) = Spin4 (3), then t acts as an involution of K − Z(K), contrary to assumption. This completes the proof. 

6. COMPUTATIONS IN GROUPS OF LIE TYPE

207

Lemma 6.9. Let K = Spin± n (q), q odd, n ≥ 5, and write Z(K) = z. Let x ∈ I2 (Aut(K)) and assume that one of the following holds: (a) x is an involution in P SOn± (q) ≤ Aut(K) which is the image of an involution in On (q); or (b) n is odd and x ∈ Inndiag(K). Then there is y ∈ K such that y x = yz, and x centralizes no Sylow 2-subgroup of K. Proof. If n is odd then Inndiag(K) = SOn (q) ∼ = P SOn (q), so (b) implies (a); thus it suffices to show that (a) implies the conclusion. Let V be the natural On± (q)module, so that the eigenspaces of x give an orthogonal decomposition V = V1 ⊥ V2 . As n ≥ 5 there are isometric nondegenerate 1-spaces U1 ≤ V1 and U2 ≤ V2 . Let y ∈ O(V ) invert U1 + U2 and centralize (U1 + U2 )⊥ . Then y ∈ Ω(V ) and we choose y ∈ K mapping to y. In the semidirect product K x, the elements x and xy are then conjugate, so they have the same order, while y has order 4 by [IA , 6.2.1d]. Therefore x, y is nonabelian, so y x = y −1 = yz. This implies that x centralizes  no Sylow 2-subgroup of K, by [III8 , 6.12]. The proof is complete. Lemma 6.10. Suppose that K ∼ = L± 4 (3) or 2U4 (3). Let t ∈ I2 (Aut(K)) and A with m suppose that L := E(CK (t)) ∼ = 6 2 (CAut(K) (L)) ≥ 2. Let u ∈ I2 (L). Then the following conditions hold: ∼ L4 (3), then t induces an inner automorphism on K; regarding (a) If K = t ∈ K, we have t ∼K u; (b) If K/Z(K) ∼ = P Ω− = U4 (3) and Z(K) = z, then t acts on K/Z(K) ∼ 6 (3) − as an involution  t ∈ O6 (3) whose eigenspaces on the natural module are of dimensions 2 and 4; moreover,  t ∼K  tuz; (c) In either case there is w ∈ I2 (CK (t)) such that L w ∼ = Σ6 . Moreover if ∼ K∼ L (3), then C (t) = (Z × L) w with Z Z and Z w ∼ = 4 = 4 = D8 ; and K (d) In either case, for any involution t ∈ CAut(K) (L t) such that t = t, we have E(CK (t )) ∼ = P Sp4 (3). Proof. We first argue that t ∈ Inndiag(K). We have Aut(K) = Inndiag(K) γ where γ is a graph automorphism of K of order 2. If t ∈ Inndiag(K), therefore, t is a graph automorphism and as m2 (CAut(K) (L)) ≥ 2, m2 (CInndiag(K) (L)) ≥ 1. However, by [IA , Table 4.5.1], if t is a graph automorphism then CInndiag(K) (L) = 1, contradiction. Thus, t ∈ Inndiag(K). By [IA , Table 4.5.1], t is then uniquely determined up to K-conjugacy by the structure of L, and m2 (CInndiag(K) (L)) = 1. Notice that the conclusions when K ∼ = U4 (3) are immediate consequences of the conclusions when K ∼ = L4 (3) or 2U4 (3), = 2U4 (3). Thus we assume that K ∼ i.e., K ∼ = Ω6 (3),  = ±1. Let K0 be the group generated by all involutions in Inndiag(K); then t ∈ K0 ∼ = P SO6 (3). Let V be the natural O6 (3)-module. Then t acts on K as an element  t ∈ I2 (SO6 (3)) whose +1 and −1 eigenspaces V+ , V− are of dimension 4 and 2 respectively, with V+ , the support of L, being of − type, i.e., the determinant of the symmetric form on V+ is −1. On the other hand the determinant of the symmetric form on V is −. Thus the spinorial norm of  t is , so t ∈ Inn(K) if  = 1 and t ∈ Inn(K) if  = −1. Since u ∈ L ≤ K, u has eigenspaces

208

11. PROPERTIES OF K-GROUPS

of dimension 4 and 2, and u has spinorial norm 1. Therefore  t ∼K u if  = 1; if  = −1 we calculate that  tuz has spinorial norm −1 and eigenspaces of the same dimensions as those of  t. This yields (a) and (b). In (c), let w ∈ K be conjugate to u, but with −1-eigenspace W satisfying dim(W ∩ V+ ) = dim(W ∩ V− ) = 1. Then w induces a graph automorphism on Ω(V+ ), i.e., a field automorphism on 2 D2 (3) ∼ = A1 (32 ), so that L w ∼ = Σ6 . Likewise if  = 1, w inverts Ω(V− ) ∼ = Z4 , and (c) follows. In (d), it follows from the first paragraph that t induces a graph automorphism on K. Also if E(CK (t )) ∼ = A6 , then m2 (CAut(K) (E(CK (t )))) = 1, so E(CK (t )) =  L. Therefore E(CK (t )) ∼ = A6 . Since E(CK (t )) ≥ L ∼ = A6 , the only possibility  from [IA , Table 4.5.1] is E(CK (t )) ∼ = P Sp4 (3). Lemma 6.11. Let K = Sp4 (3) or SL± 4 (3), and let Inn(K) ≤ X ≤ Aut(K). Let z be an involution of K whose image in K/Z(K) is 2-central. Then O2 (CX (z))/ O2 (CInn(K) (z)) is elementary abelian. Proof. This is trivial for K = Sp4 (3) and SL4 (3), for which (see [IA , 2.5.12]) Out(K) is an elementary abelian 2-group. For K = SU4 (3), Out(K) ∼ = D8 . It therefore suffices to show that for the preimage X ∼ = P GU4 (3) in Aut(K) of the Z4 -subgroup of Out(K), we have |O2 (CX (z)) : O2 (CInn(K) (z))| ≤ 2. Suppose for a contradiction that this fails, whence O2 (CX (z)) covers Outdiag(K). Let P ∈ Syl3 (CX (z)). We conclude that AutCX (z) (P ) = AutCInn(K) (z) (P ). However, by using the embedding GU2 (3)  Z2 ≤ GU4 (3) (which has odd index), one sees that AutCX (z) (P ) ∼ = D8 while AutCInn(K) (z) (P ) ∼ = E22 . This contradiction completes the proof of the lemma.  Lemma 6.12. Let K = U4 (3). Then the following conditions hold: (a) K has one class of involutions; (b) If t ∈ I2 (K), then O 2 (CK (t)) = L1 ∗ L2 with L1 ∼ = L2 , and L1 = SL2 (3) ∼ and L2 are interchanged by an element of CK (t); and (c) Aut(K) has sectional 2-rank at most 6. Proof. Parts (a) and (b) are well-known and are contained in the information 4.5.1]. In fact CK (t) := CK (t)/ t is a semidirect product (L1 × in [IA , Table  L2 ) a, b where a and b are commuting involutions, a interchanges L1 and L2 ,       and L1 b ∼ = L2 b ∼ = P GL2 (3). Let T ∈ Syl2 (CK (t)). Thus T = J a, b with J = T ∩L1 L2 ∼ = E24 and a, b acting freely on J. In particular J = J(T ). Since the preimage of J in T is not elementary abelian, m2 (T ) ≤ 4. Moreover as t ∈ Φ(T ), it follows that T /Φ(T ) ∼ = T /Φ(T ) ∼ = E23 . Thus T has no E25 quotient. Suppose that T has a section S/R ∼ = E2m , m ≥ 5. By the previous paragraph, S < T and R > 1. But |T | = 27 so |S| = 26 and |R| = 2, and then R = Φ(S)  T . As Z(T ) = t, R = t and so m2 (T ) ≥ 5, contrary to what we saw above. Therefore T has sectional 2-rank at most 4. Since Out(K) ∼ = D8 , (c) follows and the proof is complete.  Lemma 6.13. Let K ∈ K2 and let t, t  be a four-subgroup of Aut(K). If E(CK (t)) and E(CK (t )) each have a component isomorphic to P Sp4 (3), then E(CK (tt )) does not have a component isomorphic to A8 .

6. COMPUTATIONS IN GROUPS OF LIE TYPE

209

Proof. Suppose false. Since P Sp4 (3) ∈ Alt, K ∈ Alt. Therefore by Lemma 1.6, K ∼ = L4 (4), HS, or 2HS. By [IA , 4.9.1, 4.9.2, Table 5.3m] we see that it is impossible then for E(CK (t)) to have a P Sp4 (3) subcomponent. The proof is complete.  Lemma 6.14. Let V be a 4-dimensional vector space over F5 . Let Y ≤ X ≤ SL(V ) with X = E(X) and Y ∼ = A5 projecting injectively into every component of X. Assume that Y fixes a decomposition V = V1 ⊕ V2 with V1 a natural Y ∼ = Ω3 (5)module, and Y  NX (V2 ). Suppose further that for some 0 = v ∈ V , NX (v) is a 5 -group. Then X = Y . Proof. We have |X| dividing |SL(V )| = 28 32 56 .13.31, and m2 (X) ≤ m2 (SL(V )) = 3. As Y projects nontrivially into each component of X, X has at most two components. Suppose that X has two components. Then each component, having a nonabelian centralizer, embeds in SL2 (5). As Y is irreducible on V1 , X must be irreducible on V , and so X ∼ = SL2 (5) ∗ SL2 (5), acting on V via the tensor product of the natural representations. Thus X = Ω(V ), with respect to a suitable symmetric bilinear form on V . As CX (v) contains a Sylow 5-subgroup of X if v ∈ V is singular, and contains Ω(v ⊥ ) if v is nonsingular, this case is ruled out by hypothesis. Hence, X is quasisimple. From the order of X, we see that X ∈ Spor, and if X ∈ Alt, then Y = X ∼ = A6 , as Y ≤ X. Suppose that = A5 , as claimed, or X ∼ ∼ X = A6 . Then X contains a subgroup H ∼ = A4 with t ∈ I3 (H) having CV (t) = 0. But H is a Frobenius group so CV (t) = 0, contradiction. Now we may assume that X ∈ Chev(r) − Alt. If r ∈ {3, 13, 31} then by order considerations, we have X ∼ = L2 (31). But by Clifford’s theorem, a Borel subgroup of X then requires at least 15 dimensions for a faithful F5 -representation. Suppose that X ∈ Chev(2) − Alt. Given that |X|3 ≤ 32 , m2 (X) ≤ 3 and a Y ≤ X, we must have X/Z(X) ∼ = L± m (2 ) with m ≤ 3 and a ≤ 3. As 7 divides the orders of L2 (8) and L3 (2), the only possibility is X ∼ = U3 (4). But the restriction of X ≤ SL(V ) to T ∈ Syl2 (X) must have a summand V0 on which the kernel Z has order 2, with T /Z extraspecial of order 21+4 . But then V = V0 , whereas Z acts nontrivially on V , a contradiction. Assume finally that X ∈ Chev(5) − Chev(2) and Y = X. If V2 is X-invariant, then as Y  NX (V2 ), Y = X, contradiction. Hence X is irreducible either on V or on V1 . As NX (v) is a 5 -group, a Sylow 5-subgroup P of X has a regular orbit on E1 (V ). As |E1 (V )| = 156, |P | ≤ 53 and if equality holds, then |X : NX (v)| = |P |. This fact, together with Y < X and and order considerations from X ≤ SL(V ), yield X ∼ = L2 (52 ) or L3 (5). In the first case, X has a unique representation on V by the Steinberg tensor product theorem, and so X = Ω− 4 (5). The same argument ∼ as for Ω+ 4 (5) above shows that 5 divides |NX (v)|, contradiction. Finally, if X1 = 3 3 L3 (5), then |P | = 5 and so |X1 : NX1 (v)| = 5 , as we saw above. But L3 (5) has  no subgroup of this index [IA , 6.5.3]. The proof is complete. Lemma 6.15. Suppose that K ∼ = L3 (4). Then the following conditions hold: (a) Let t ∈ I2 (Aut(K)) be a graph automorphism, and let S be a t-invariant Sylow 2-subgroup of K. Then NK (S) = SH where |H| = 3 and [t, H] = 1. Moreover, [t, S] ∼ = Z4 × Z4 is inverted by t and C[t,S] (H) = 1. (b) Out(K) ∼ = Z2 × Σ3 .

11. PROPERTIES OF K-GROUPS

210

Proof. By [IA , 4.9.2bg], |CK (t)|2 = 4 and t is determined up to K-conjugacy. Now Z(S) = Φ(S) is a four-group. If [t, Z(S)] = 1, then CS (t) covers CS/Φ(S) (t). But |S/Φ(S)| = 24 , so |CS (t)| ≥ 23 , a contradiction. Therefore [t, Z(S)] = 1 and so Z(S) ∈ Syl2 (CK (t)). As S = CK (Z(S)), the pair (t, S) is determined up to Kconjugacy. As a result we can choose S to be the upper triangular unipotent group, and t = t1 t2 where t1 is the transpose-inverse automorphism and t2 is conjugation by ⎡ ⎤ 0 0 1 ⎣0 1 0⎦ . 1 0 0 The assertions of (a) are then easily checked, and (b) follows from [IA , 2.5.12].  Lemma 6.16. Let p be an odd prime and let K ∈ Chev(2) be of level q, where q ≡  (mod p) for some  = ±1. Let u ∈ Ip (K) and g ∈ Aut(K), and let L be a component of Lp (CK (u)). Suppose that one of the following cases holds: −   (K/Z(K), L) = (L− n (q), SLn−2 (q)), n = 6, 7; (Ln (q), SLn−1 (q)), n = 4, 6, p = 3; − +    (Spn (q), Spn−2 (q)); (Ω− 4n (q), Ω4n−2 (q)); (Ω4n+2 (q), Ω4n (q)); (E6 (q), (S)L6 (q)), p =   3; (Lp (q), SLp−2 (q)), p = 5, 7. In the last case assume that u is the image of an element of SLp (q) with p − 2 eigenvalues equal to 1. Suppose also that mp (K) ≥ 3. Let v ∈ Ip (K) be such that E(CK (v)) has a component isomorphic to L. Then the following conditions hold: (a) u and ug  are Inn(K)-conjugate; (b) The images of u and v in Inn(K) are Inn(K)-conjugate, unless possibly (K/Z(K), L) = (Lp (q), SLp−2 (q)), with p = 5 or 7; and (c) If g induces a field automorphism on K of order p, then g centralizes some K-conjugate of u. Proof. We first prove (b), for which we may assume that Z(K) = 1. E6 (q),

u and v are Inndiag(K)-conjugate and then K-conjugate, by [IA , For K = Table 4.7.3A and 4.2.2j]. For (K, L) = (Ln (q), SLn−1 (q), u and v are congruent modulo scalars to diagonal matrices centralizing a (nonsingular) hyperplane on the natural module, so u and v are Inn(K)-conjugate. In the other cases (except the last), let V be the natural K-module; the centralizers of u and v are then given by [IA , 4.8.2]. Since CK (v) has a component isomorphic to L, it follows that CV (u) and CV (v) must be isometric. (This is not necessarily true for Ω+ 8 (q), but K is not of that isomorphism type, by assumption.) Moreover, in each case mp (CK (L)) = 1; [V, u] is a minimum-dimensional nontrivial submodule of V for u, and similarly for the isometric subspace [V, v]. We have mp (Isom([V, u])) = 1. Now K is transitive on subspaces of V isometric [V, u]; say [V, v] = [V, u]k for some k ∈ K. Then by  to  −1

Sylow’s theorem, u and uk are conjugate subgroups of the p-rank 1 subgroup CK (L), and (b) follows in these cases. In view of (b) we need only prove (a) in the last case for (K, L), and in general under the assumption ug  ≤ Z(K) u. In the last case for (K, L), ug  also is the image of an element with p−2 eigenvalues equal to 1, as g acts as a the product of a semilinear transformation of V with either 1 or the transpose-inverse automorphism. Now the isometry types of all the eigenspaces of u are uniquely determined and u and ug are diagonalizable; K is transitive on frames in V (orthogonal frames

6. COMPUTATIONS IN GROUPS OF LIE TYPE

211

if  = −1); hence u and ug  are K-conjugate. Assume then in general that ug  ≤ Z(K) u. If Z(K) is a p -group, then ug  = u, so we need only consider the case (K, L) = (E6 (q)u , SL6 (q)) with p = 3. Let K = K/Z(K). In this case u is inverted in K, by [IA , Table 4.7.3A], so NInn(K) (u permutes the three elements of E1 (u Z(K)) − {Z(K)} either transitively or in orbits of lengths 1 and 2. In any case, each orbit is stable under g, which proves (a) in this case. For (c), write L = Lη (q), η = ±1, and q0 = q 1/p . Set K0 = E(CK (g)), of the same type as K but of level q(K0 ) = q 1/p . We have q0 ≡ q ≡  (mod p), so in all cases there exists v ∈ K0 such that E0 := E(CK0 (v)) ∼ = Lη (q0 ). (This follows easily from [IA , 4.8.2] if K is a classical group, and from [IA , Table 4.7.3A] if K ∼ = E6 (q).)   If (K, L) = (Lp (q), SLp−2 (q)), we may assume that v is chosen (like u) to have p − 2 eigenvalues equal to 1. Now [g, v] = 1, and we complete the proof by showing that u and v are K-conjugate. Let Kv = E(CK (v)). Then g normalizes Kv and by Lp -balance, E0 is a component of CKv (g). Now the components of Kv are of level at least q by [IA , 4.2.2], while E0 has level q0 . Hence E0 lies in a single component Kv0 of Kv , which must be of level q and g-invariant, and on which g induces a field or graphfield automorphism of order p. Since L ∼  3D4 (q) by assumption, g induces a field = automorphism on Kv0 . Then E0 = E(CKv0 (g)), and since E0 ∼ = Lη (q0 ), Kv0 ∼ = L. By the observation made above in the proof of (a), this isomorphism implies that u and v are K-conjugate, except possibly in the case (K, L) = (Lp (q), Lp−2 (q)). But in this final case we chose v to have p − 2 eigenvalues equal to 1, so it is obvious that u and v are K-conjugate. The proof is complete.  Lemma 6.17. Let p, K, q, , u, and L be as in Lemma 6.16, and suppose that Outdiag(K) and Outdiag(L) are p -groups. Let A ∈ Ep∗ (CK (u)). Then the following conditions hold: (a) (b) (c) (d)

A ∈ Ep∗ (K); NK (A) is irreducible on A; A = (A ∩ L) × u; and No element of AutK (A) of order p acts quadratically on A.

∼ E6 (q) with p = 3, or Lp (q). In Proof. Since Outdiag(K) is a p -group, K = particular, K is a classical group. Hence by [IA , 4.10.3f], K has a unique conjugacy class of maximal elementary abelian p-subgroups, which implies (a). In every case L is a level subcomponent of K and so it follows from [IA , 4.8.2, 4.8.4] that A induces only inner-diagonal automorphisms on L. But | Outdiag(L)| is a p -group, as is the Schur multiplier of L/Z(L), and so A = (A ∩ L) × CA (L). In every case L is a terminal subcomponent of K, so mp (CA (L)) = 1 and (c) holds. Let V be the natural K-module. For the moment exclude the cases K/Z(K) ∼ = Ln (q). Now m = mp (A) ≥ 3, and we may write V = V0 ⊥ V1 ⊥ · · · ⊥ Vm (with respect to the zero form in the linear group case) and A = A1 × · · · × Am . Here for all 1 ≤ i ≤ m, Ai ∼ = Zp centralizes all Vj for all j = i, Vi is a minimal classical vector space over Fq (Fq2 in the unitary case) admitting an isometry of order p, and mp (Isom(Vi )) = 1 (cf. [IA , 4.8.2]). Indeed as p divides q 2 − 1 by assumption, dim(Vi ) = 2 for all 1 ≤ i ≤ m. This implies that each Ai is inverted by an involution of Isom(Vi ), yielding an elementary abelian 2-subgroup E ≤ AutIsom(V ) (A). As m ≥ 3, and V1 , . . . , Vm are isometric, AutK (A) contains a subgroup E0 ≤ E whose only irreducible constituents on A are the Ai . As AutK (A) also permutes the Ai

212

11. PROPERTIES OF K-GROUPS

transitively, AutK (A) is irreducible on A, which is (b). Moreover, as NK (A) must permute the Vi , any element of AutK (A) of order p must cycle some set of p linearly independent elements of A, and so cannot act quadratically on A. Thus (d) holds in this case. It remains to prove (b) and (d) in the case K/Z(K) ∼ = Ln (q). By assumption p does not divide n(n − 1). As a result A is of rank n − 1 and is the trace-0 submodule of the natural permutation module for AutK (A) ∼ = Σn . Since p does not divide n, AutK (A) is irreducible on A, so (b) holds. Moreover if g ∈ AutK (A) has order p, then since p < n − 1, g has a free summand on the trace 0 submodule of the natural permutation module, whence g does not act quadratically on A. The lemma is proved.  Lemma 6.18. Let p be an odd prime and let K ∈ Chev(2) be simple of level  q = q(K) with p dividing q 2 − 1. Set K ∗ = O p (Inndiag K) and let X = K ∗ x ≤ Aut(K) with x a field automorphism of order p. Let A be an elementary abelian p-subgroup of K ∗ of largest rank. Assume that A ≤ K and [A, x] = 1. Suppose also that either K is a p-saturated classical group or K ∼ = E6 (q) with p = 3 dividing q − . Then the following conditions hold: (a) C := CK ∗ (A) is abelian and Op (C) is homocyclic; and (b) x induces a power mapping (of order p) on Op (C). Proof. First assume that K is a classical group and V is its natural module,  = [Isom(V ), Isom(V )] by a p -group. Let so that K ∗ = K is a quotient of K  be a preimage of A in K  with A ∼ A = A. To prove (a) it suffices to prove it for   C := CK (A) in place of C. Let Vmin be a minimal subspace of V such that p divides  yields V = V0 ⊥ V1 ⊥ · · · ⊥ Vm | Isom(V0 )|. Decomposing V under the action of A  and each Vi ∼  is a linear V where V0 = CV (A) as a classical space. (If K = min group, “⊥” should be read “⊕”.) Then it follows easily from [IA , 4.10.2, 4.8.2], the fact that K is p-saturated and simple, and the oddness of p that Isom(Vmin ) has a cyclic Sylow p-subgroup Pmin , CIsom(Vmin ) (Pmin ) is cyclic, and Isom(V ) contains a subgroup M = M0 MW ∼ = Isom(Vmin )  Σm centralizing V0 and with p not dividing | Isom(V ) : M |. Let P ∈ Sylp (M ), so that P = P0 PW where P0 is homocyclic abelian and a Sylow subgroup of the (abelian) base group M0 of the wreath product M , while PW is a Sylow subgroup of the complement MW ∼ = Σm .    is In particular P0  M . Since A is of maximal rank, A = Ω1 (P0 ); furthermore, A   the permutation module for PW , so CP (A) = P0 . Now | Isom(V ) : K| is relatively  Moreover, C  stabilizes each Vi and so lies in M , whence prime to p, so P ≤ K.  ≤ M0 . As M0 is abelian and P0 is homocyclic abelian, (a) follows. C If p divides q − 1, then one checks that A = Ω1 (Op (H)), where H is a Cartan subgroup of K, and that CU (A) = 1, where U is unipotent and U H is a Borel subgroup of K. Let N = NK (A). Then N is a monomial subgroup of K and N x = NKx (A). Now H is the direct product of copies of F× q , on which some field automorphism g of K of order p acts naturally. In particular g acts on H, and thus on P0 , as a power mapping of order p. Replacing x by a power of itself we may assume that g ∈ Kx. Then CKx (A) = H g, whence x ∈ Hg. Hence x, like g, induces a power mapping on H and in particular on P0 . Hence (b) holds in this case.

6. COMPUTATIONS IN GROUPS OF LIE TYPE

213

Suppose that p divides q + 1. Let J be the adjoint untwisted group of the same type as K but over Fq2 . Also let J ∗ = Inndiag(J). Then for some field or graph-field automorphism g of J of order 2, and for some field automorphism y of order p, [g, y] = 1, 

O 2 (CJ (g)) and CJ (g) may be identified with K and K ∗ , and y induces a field automorphism of K of order p. Moreover as p is odd, one computes that |J|p = |K ∗ |p . Hence a Sylow p-subgroup of K is also one of J, and as (b) holds for J, it also holds for K. In the same way we reduce the proof when K = E6 (q) to the case  = 1. Suppose finally that K = E6 (q), q ≡ 1 (mod 3). By [IA , 4.10.3c], and as q is a cube by assumption, A ∼ = E36 is the exponent 3 subgroup of a Cartan subgroup of K and of a Cartan subgroup H ∗ of K ∗ , and of course the latter is homocyclic abelian. Then as in the classical case, CK ∗ x (A) = H ∗ g where g is a field automorphism inducing a power mapping on H ∗ of order p, and we may assume that x ∈ H ∗ g. This quickly yields both (a) and (b), and the proof is complete.  Lemma 6.19. Let K ∈ Lie(2), let U ∈ Syl2 (K) and let x ∈ I2 (Z(U )). Set C =   O 2 (CK (x)). Then there is a parabolic subgroup P containing U such that O 2 (P ) = C. Let C = C/O2 (P ) and assume that C = 1. Then up to an automorphism of K, one of the following holds: (a) x is a long root involution, and (K, C) is one of the following, with  = ±1: (An (q), An−2 (q)), n ≥ 3; (Cn (q), Cn−1 (q)), n ≥ 2;  (q)), n ≥ 4; (3D4 (q), A1 (q 3 )); (G2 (q), A1 (q)), (Dn (q), A1 (q)Dn−2 n n 2 (F4 (q), C3 (q)), ( F4 (2 2 ), 2B2 (2 2 )), (E6 (q), A5 (q)), (E7 (q), D6 (q)), (E8 (q), E7 (q)); (b) x is a short root involution, and (K, C) = (Cn (q), A1 (q)Cn−2 (q)), n ≥ 3; (c) x is the product of a high long root involution and a high short root involution, and (K, C) = (Cn (q), Cn−2 (q)), n ≥ 3, or (F4 (q), C2 (q)). Proof. By Tits’s Lemma [IA , 2.6.7], C ≤ P for some parabolic subgroup P = K containing U , and CH = P where H is some Cartan subgroup of P . We may assume that K has twisted rank at least 2, for otherwise P is 2-closed, contrary n to assumption. Unless K ∼ = 2F4 (2 2 ) we quote [IA , 3.3.1] and find that Z(U ) is the unique root subgroup of greatest height in U , except if K ∼ = Cn (q), Bn (q), or F4 (q), in which case Z(U ) is the product of the highest root subgroup and the highest short root subgroup in U . 8 n In K ∼ = 2F4 (2 2 ), using the notation of [IA , 2.4.6], we have U = i=1 Xi , and we can deduce from the commutator formula [IA , 2.4.5d] that Z(U ) = Y5 = Φ(X5 ). For example, [y5 (t), x8 (u)] = [x5 (t0 )2 , x8 (u)] = [x5 (t0 ), x8 (u)]2 [x5 (t0 ), x8 (u), x5 (t0 )] = y7 (t0 u)2 [y7 (t0 u), x5 (t0 )] = [x7 (t1 )2 , x5 (t0 )] = x6 (t1 t0 )2 = 1 (mod [X6 , X7 ] = 1), = t and t1+2θ = t0 u, in the notation of [IA , 2.4.5d]. Similar calculations where t1+2θ 0 1 with other root elements replacing x8 (u) show that Y5 ≤ Z(U ). Likewise if z ∈ 8 Z(U ) and we write z = i=1 zi with zi ∈ Xi for each i, then we consider [z, x] = 1 for various x ∈ U to conclude that z ∈ Y5 . From x ∈ X3 and x ∈ X7 we get that z1 = 1; from x ∈ X8 , that z2 = 1 and then z3 = 1; from x ∈ X2 , that z8 = 1; from

214

11. PROPERTIES OF K-GROUPS

x ∈ X1 and x ∈ X5 , that z7 = 1 and then z6 = 1; from x ∈ X1 and x ∈ X8 , that z6 = z4 = 1, and then from x = x5 (u), that z5 ∈ Y5 , as desired. We consider Y5 as the high root group in this case. Thus in every case x lies in the high root group, as in (a); or x lies in the product of the high root group and the highest short root group, as in (b) or (c), with the Dynkin diagram of K having a double bond. Now C = U (N ∩ C)U by the Bruhat decomposition. Suppose that x lies in the high root group Xα . Then for n ∈ N ∩ C, the image n of n in N/H ∼ =W fixes α , so n is the product of fundamental reflections fixing α . It follows that C is generated by U and the negative root subgroups Xβ that it contains; moreover, for negative β, Xβ ≤ C if and only if nβ ∈ C if and only if β ⊥ α . This yields (a) (cf. [IA , 3.2.6]). On the other hand if x lies in the high short root group Xαs , then we consider the dual root system Σ∗ . There is a version K ∗ of Σ∗ (q) such that K ∼ = K ∗ via an ∗ isomorphism carrying Xαs to a long root subgroup of K , and together with (a), this yields (b). Finally consider (c), for the cases K ∼ = Cn (q) and F4 (q). As in the proof of (a), C is generated by U and those negative root groups orthogonal to both α and αs . A straightforward calculation shows then that P is of type Cn−2 or C2 , respectively. The proof is complete.  Lemma 6.20. Let K = SU3 (8). and x ∈ I3 (Aut(K)) with L3 (CK (x)) ∼ = L2 (8). Let y ∈ L3 (CK (x)) be of order 3. Then the coset yZ(K) is completely fused in K. Proof. This is an immediate consequence of Lemma 1.18.



Lemma 6.21. Suppose that K = L2n (q) where q = 2m . Let p be a prime divisor of q + 1 and P an elementary abelian p-subgroup of K of maximal rank. Then AutK (P ) contains Z2  Σn . Proof. Since p is odd, it does not divide q − 1. Hence the universal version K ∗ = SL2n (q) has a p -center, and it suffices to prove the result for K ∗ . Clearly K ∗ ≥ L2 (q)  Σn ≥ D  Σn where D is dihedral of order 2p, so the lemma follows.  Lemma 6.22. Let q be a power of 2 and let K = L6 (q 3 ), q ≡  (mod 3),  = ±1.  Let x be a field automorphism of K of order 3 and set L = O 2 (CK (x)). Let A be an elementary abelian 3-subgroup of CK (x) of maximal rank and set A0 = [NL (A), A]. Then T := CK (A0 ) is a maximal torus of K, and if P ∈ Syl3 (T ), then A = Ω1 (P ). Proof. Set L∗ = CK (x). By Lemma 2.4, L∗ ∼ = E35 is the = P GL6 (q). Thus A ∼  image of a diagonalizable subgroup H ≤ GL6 (q), and AutL (A) ∼ = AutL∗ (A) ∼ = Σ6 . Then A0 , the (4-dimensional) core of the natural AutL (A)-module, is the image of H0 = H ∩ SL6 (q). Since H0 contains diag(ω, ω −1 , 1, 1, 1, 1) (where ω 3 = 1 = ω) and all its Σ6 -conjugates, T = CK (A0 ) is the image of a diagonalizable subgroup of SL6 (q), and so is a maximal torus of K. Finally A ≤ T and as A has maximal  rank, A = Ω1 (P ), completing the proof. Lemma 6.23. Assume that K = F ∗ (X) ∼ = E8 (q) and q = r m with r prime and r = 3. Write η = ±1 ≡ q (mod 3). Let P ∈ Syl3 (X). Then |Z(P )| = 3 and  η η O r (CK (Z(P )) is the central product K1 K2 with K1 ∼ = E6 (q))u , K2 ∼ = SL3 (q), and Z(K1 ) = Z(K2 ) = Z(P ).

6. COMPUTATIONS IN GROUPS OF LIE TYPE

215

Proof. Let t = t7 ∈ K, in the notation of [IA , Table 4.7.3A]. Let Q ∈ Syl3 (CK (t)) and Q ≤ P ∈ Syl3 (K). The information in the table will complete the proof once we show that Q = P and t = Z(P ). For this it is sufficient to show that t = Z(Q), as then NP (Q) ≤ NP (Z(Q)) = Q, so Q = P . Let C = CK (t).  We have O r (C) = K1 K2 with K1 and K2 of the claimed isomorphism types, η η and with C/K1 ∼ = P GL3 (q) and C/K2 ∼ = Inndiag(E6 (q)). By [IA , Table 4.7.3A] applied to K1 , there is a Q-invariant subgroup J ≤ K1 such that J = J1 J2 J3 ; Ji ∼ = SL3 (q) for i = 1, 2, 3; the subgroups Z(Ji ) are distinct for i = 1, 2, 3 and  are permuted transitively by Q; and O 3 (CAut(K) (J)) = Z(J) ∼ = E32 . By Lemma  3.14, Z(Q) ≤ O 3 (CC (JK2 )) = Z(JK2 ). As Q permutes the Z(Ji ) transitively,  Z(Q) ∼ = Z3 , as required. The proof is complete. Lemma 6.24. Let J ∈ K3 and suppose that for some n and some x, y ≤ Aut(J) with x, y ∼ = L8 (2n ), with m3 (H) = 4, = E32 , CJ (y) has a component H ∼ n+1 , and CJ (x) has a component I such that I/Z(I) ∼  = (−1) = L6 (4n ). Then m3 (CJ (H)) > 1 and Z(J) = 1. Proof. Obviously y does not induce a graph-field automorphism on J. Suppose that y induces a field automorphism on J. Then q(J) = 8n , and so by [IA , 4.2.2, Table 4.7.3A], x does not induce an inner-diagonal or graph automorphism on J, as q(I) < 8n . Hence x induces a field or graph-field automorphism on J, whence J/Z(J) ∼ = L6 (82n ), which is absurd. Given the structure of = L8 (8n ) ∼ CJ (y) it is clear from [IA , Table 4.7.3A] that y does not induce a graph automorphism on J, and that J is not of exceptional Lie type. Hence J is a classical group and y ∈ Inndiag(J). By [IA , 4.8.2, 4.8.4] and the fact that  = (−1)n+1 , so that 2n ≡  (mod 3), it must be that J/Z(J) ∼ = Lm (2n ) for some m. As 2n ≡  (mod 3) and J ∈ K3 , Z(J) = 1. Then again by [IA , 4.8.2] and the structure of I, m ≥ 12,  whence m3 (CJ (H)) > 1 as asserted. Lemma 6.25. Suppose that K = G2 (q) or 3D4 (q), q > 3, q ≡ 0 (mod 3), and x ∈ I3 (K) with E(CK (x)) ∼ = SL2 (q) or SL2 (q 3 ), respectively. If y ∈ I3 (Aut(K)), y centralizes x, and y induces a nontrivial field automorphism on E(CK (x)), then according as K = G2 (q) or 3D4 (q), y induces a field automorphism or graph automorphism on K. Proof. By [IA , 2.5.12], Out(K) is cyclic in both cases, and any non-inner automorphism of K of order 3 is, respectively, a field or graph automorphism. Thus it suffices to show that y ∈ Inndiag(K)(= Inn(K)). But by [IA , Table 4.7.3A], no inner automorphism of CK (x) induces a nontrivial field automorphism on E(CK (x)). The result follows.  Lemma 6.26. Let L = L6 (q 3 ), q = 2n ≡  (mod 3),  = ±1. Let v ∈ I3 (Aut(L)). Let E = E(CL (v)). If some component of E is defined over Fq , then E ∼ = L6 (q). Moreover, E ∼ = L3 (q 3 ). Proof. If v ∈ Inndiag(L), then v induces a field automorphism of order 3 on L and E ∼ = L6 (q). If v ∈ Inndiag(L), then by [IA , 4.8.2, 4.8.4], the components of E are isomorphic to Ld (q1 ) with q1 ≥ q 3 , confirming the first assertion of the lemma. Suppose then that E ∼ = L3 (q 3 ). Let X = GL6 (q 3 ). We may regard v as an element ˆ ˆ ∼ ˆ be the pullback of E in CX (v). As E/Z( E) of X of order 3 or 9. Let E = L3 (q 3 ), v must have a 3-dimensional eigenspace W on the natural 6-dimensional module

216

11. PROPERTIES OF K-GROUPS

∼ SL (3, q) with ˆ normal in the stabilizer in X of W . But then E ˆ = for X with E ˆ ∩ Z(X) = 1, whence E ∼ Z(E) = SL (3, q), a contradiction completing the proof.  Lemma 6.27. Let K = K(V ) be a quasisimple classical linear group with dim(V ) ≥ 5, and dim(V ) ≥ 7 if V is an orthogonal space. Let y ∈ K be a semisimple element of odd order with CV (y) of co-dimension 2. Then y K ∩ E(CK (y)) = ∅, unless K ∼ = U5 (2). Proof. As y is semisimple, V = [V, y] ⊕ CV (y) with both subspaces nonsingular and with y ∈ I([V, y]) , where I([V, y]) is the isometry group of [V, y]. As dim(CV (y)) ≥ 3, CV (y) contains a subspace W isometric to [V, y] and so unless  O 2 (CK (y)) ∼ = SU3 (2), E(CK (y)) contains I(W ) , which contains a K-conjugate of y.  Lemma 6.28. Suppose that K ∈ K3 and x, y ∈ I3 (Aut(K)). Assume that n CK (y) has a component Ky ∼ = SL3 (q), G2 (q), 3D4 (q), or 2F4 (2 2 ) for some  = ±1, q ≡  (mod 3), q > 2, and odd n > 1. Assume also that Lo3 (CK (x)) = 1. Then  K∼ = SL3 (q  ), q  ≡  (mod 3),  = ±1. Proof. Let q be a power of the prime r. The assumption q > 2 guarantees that Ky ∈ Chev(s)∪Alt for any s = r, so K ∈ Chev(s)∪Alt. Moreover as Ky ∈ Chev(3), Lemma 5.12 implies immediately that K ∈ Spor. Thus, K ∈ Chev(r). If K is of exceptional type, including D4 (r n ), then the assumption Lo3 (CK (x)) = 1 yields, by inspection of the table [IA , Table 4.7.3A], that K = D4 (2) or G2 (2), whence q = 2, a contradiction. Thus K is a classical group and so Ky ∼ = SL3 (q), q > 2. If y induces a field automorphism on K, then the desired conclusion holds, so we may assume that y induces an inner-diagonal automorphism on K. Let V be the natural module for K. Thus dim(V ) > 3, from the structure of Ky . Now Inndiag(K) is a section of Isom(V ), and we let y ∈ Isom(V ) correspond to y. Likewise x is not a field or graph-field automorphism of K, for if it were then Lo3 (CK (x)) = 1 would imply K ∼ = A1 (23 ) and y could not exist. If x were a graph automorphism, then K ∼ = D4 (2), but then q = 2, a contradiction. So x ∈ Inndiag(K). Let x  ∈ Isom(V ) correspond to x. Suppose first that V is an orthogonal or symplectic space. Then from [IA , 4.8.2], Ky is supported on [V, y] which has dimension 6, and K has level q. Then either dim(CV ( x)) ≥ 4 or dim([V, x ]) ≥ 4, and as q > 2, and using [IA , 4.8.2] again, we see that Lo3 (CK (x)) = 1, contradiction. Therefore V is a linear or unitary group. Write K/Z(K) ∼ = Lηn (s). If s ≡ −η (mod 3), then we cannot have s = q and η =  by assumption, so from [IA , 4.8.2], q = s2 and dim[V, y] = 6. Again either dim[V, x ] = 2m ≥ 4, in which case L3 (CK (x)) contains SLm (q), or dim(CV ( x)) ≥ x). Thus this 4, in which case L3 (CK (x)) has a component supported on CV ( case cannot occur so s ≡ η (mod 3). If dim(V ) > 6, then it follows easily from [IA , 4.8.2,4.8.4] that L3 (CK (x)) = 1, so we have dim(V ) ≤ 6. In particular by [IA , 4.8.4], y is diagonalizable on V , and so by [IA , 4.8.2], s = q. Now as dim(V ) ≥ 4, it follows that L3 (CK (x)) has a component containing SL2 (q). This final contradiction completes the proof. 

7. SMALL GROUPS

217

7. Small Groups Lemma 7.1. Suppose that K ∈ Kp and P ∈ Sylp (K). If Ω1 (P ) ≤ Z(K), then p = 2 and K ∼ = SL2 (q), q odd, or 2A7 . Proof. By [IIK , 2.13], p = 2 and m2 (K) = 1. Then by [IG , 15.3], K has a unique involution z and quaternion Sylow 2-subgroups, so K/ z ∼ = L2 (q) or A7 (see [IA , 5.6.3]). The result follows with [IA , 6.1.4].  Lemma 7.2. If p is odd and K ∈ Kp with mp (K) = 1, then K is simple with mp (Aut(K)) ≤ 2. Moreover, for any p-subgroup P of Aut(K) properly containing a Sylow p-subgroup of K, P splits over P ∩ K, and Ω1 (P ) ∼ = Ep2 . Proof. This is contained in [IIK , 5.1abc].



Lemma 7.3. Suppose that X = M P where M = F ∗ (X) is the direct product of p copies of a component of p-rank 1, p is an odd prime, and P ∈ Sylp (X) permutes the components J1 , . . . , Jp of J transitively. Then one of the following holds: (a) Ω1 (NP (J1 )) is elementary abelian of rank mp (P ) > p; or (b) mp (P ) = p and P ∼ = Zpa  Zp , where pa = |J1 |p . Proof. Let X0 = NX (J1 ) = ∩pi=1 NX (Ji ), so that M = E(X0 ) and X0 embeds in Aut(J1 ) × · · · × Aut(Jp ). By Lemma 7.2, a Sylow p-subgroup of this latter group splits over its intersection with Inn(J1 )×· · ·×Inn(Jp ), so X0 = M P0 with P0 ∩M = 1 and P0 ≤ P . By Lemma 7.2, the projection of Ω1 (P ) on any Aut(Ji ) is abelian. Since CX (M ) = 1, Ω1 (P ) is abelian. Now if P0 = 1 then Ω1 (P ) > Ω1 (P ∩M ) ∼ = Epp , so (a) holds. Suppose then that P0 = 1. Then P ∩M is homocyclic of rank p, and in the series P ∩ M > Φ(P ∩ M ) > Φ(Φ(P ∩ M )) > · · · 1, P/P ∩ M acts freely on each successive quotient. Hence by [IG , 9.22,9.25(iii)], P/Φ(P ∩ M ) splits over P ∩ M/Φ(P ∩ M ), say with complement P1 /Φ(P ∩ M ). Then applying a similar argument to P1 and continuing this process we find that P = (P ∩ M )P0 for some P0 of order p. But  then P ∼ = Zpa  Zp where pa = |J1 |p . The lemma is proved. Lemma 7.4. Let K ∈ Chev(2) ∩ Gp and suppose that mp (K) = 2 for some odd prime p. Let P ∈ Sylp (K) and Q ∈ Sylp (Aut(K)). Then the following conditions hold: (a) K is simple, so we may regard P as a subgroup of Q; (b) Either P is homocyclic abelian or else p = 3 with K ∼ = G2 (q) or 3D4 (q) √ 2 n or F4 ( q) for some q = 2 > 2; (c) Out(K) has cyclic Sylow p-subgroups, and | Outdiag(K)| is not divisible by p; in particular mp (Aut(K)) ≤ 3; (d) Unless p = 3 and K ∼ = 3D4 (q), every element of Q − P of order p is a field automorphism of K; (e) If p = 3 and K ∼ = 3D4 (q), then there exist quasifield automorphisms f of order 3 in Q − P , and for any such f , CK (f ) ∼ = G2 (q); (f) If P is homocyclic abelian then Ω1 (Q) is abelian; and (g) If p = 3 and K has untwisted Lie rank at least 4, then K has untwisted Lie rank exactly 4, and K is isomorphic to A− 4 (q), q ≡  (mod 3),  = ±1, n 3 D4 (q), or 2F4 (2 2 ) for some n.

11. PROPERTIES OF K-GROUPS

218

Proof. First suppose that (7A)

√ p = 3 with K ∼ = G2 (q), 3D4 (q), or 2 F4 ( q).

As K ∈ G3 , q > 2 by the definition of G3 . By [IA , 6.1.4], Z(K) is a 3 -group and hence is trivial as K ∈ K3 . Thus (a) holds, and (b) is trivial. By [IA , 2.5.12], Outdiag(K) = 1 in these cases, and Out(K) is cyclic, generated by the image of the Frobenius mapping x → x2 in all cases. With [IA , 4.9.1], (d) holds. In the case K = 3D4 (q), the outer automorphism of order 3 is a graph automorphism whose structure is given in [IA , Table 4.7.3A]. By definition a quasifield automorphism in this case is a graph automorphism γ with CK (γ) ∼ = G2 (q), so (c) and (e) hold. In this case (f) is vacuous as P is not abelian, by [IA , 4.10.3c]; and (g) is obvious. Thus we may assume that (7A) fails. Recall that the groups A2 (q), q ≡  (mod 3),  = ±1, lie in T3 by definition, and so K cannot have such an isomorphism type when p = 3, as K ∈ G3 by assumption. Suppose that K is not simple, so that Z(K) is a nontrivial p-group. Thus by [IA , 6.1.4], K ∼ = Akp−1 (q), q ≡  (mod p), or E6 (q) with p = 3. But then using [IA , 4.10.3b] we see that mp (K) ≥ 3, with the exception of the A2 (q) case that was ruled out in the previous paragraph. So K is simple, proving (a). Then Outdiag(K) ∼ = Z(Ku ) is a p -group, where Ku is the universal version of K. In the notation of [IA , 2.5.12], the group ΦK ΓK /ΦK is also a p -group, for otherwise p = 3 with K ∼ = D4 (q) or 3D4 (q); the former is impossible as m3 (D4 (q)) = 4 > 2 and the latter as we are assuming (7A) fails. Thus Sylow p-subgroups of Out(K) lie in the image of ΦK . This implies (c) and (d). To prove (b) we use [IA , 4.10.2bc]. As there, let m0 be the least integer such that p divides Φm0 (q). Write K = d L(q) and let fK be the polynomial such that fK (q) = |d L(q)u |2 for all powers q of 2. Then Φm0 divides fK with multiplicity mp (K) = 2; moreover, in order to prove (b) it suffices to check that Φpa m0 does not divide fK for any a > 0. One can then check all the order formulas for groups of Lie type. As an example, suppose that K = E6− (q). Then fK = Φ41 Φ62 Φ23 Φ24 Φ36 Φ8 Φ10 Φ12 Φ18 , so m0 = 3 or 4. This implies that p > 3, since 3 divides Φ1 (q)Φ2 (q). But then by inspection Φpa m0 is not a factor of fK , as desired. As another example take K = Am (q). Then Φm0 has multiplicity [(m + 1)/m0 ] − [1/m0 ] in fK . This multiplicity equals 2 if and only if m0 = 1 with m = 2, or m > 2 with 2m0 ≤ m + 1 < 3m0 . In the second case pm0 ≥ 3m0 so Φpm0 does not divide fK ; thus (b) holds in this case. In the first case, fK = Φ21 Φ2 Φ3 , but p > 3 by the assumption that K ∈ T3 , so again Φpm0 = Φp is not a divisor of fK , and (b) holds. In (f), Q/P is cyclic by (c), and so if Ω1 (Q) > P then with (d), Ω1 (Q) = f, Ω1 (P ) with f a field automorphism and Ω1 (P ) ∼ = Ep2 . Without loss we may assume that CP (f ) ∈ Sylp (CK (f )). Let q0 = q 1/p ; then q ≡ q0 (mod p), so |CP (f )| = |CK (f )|p = |Φm0 (q0 )|2p . But |Φm0 (q0 )|p = |Φm0 (q)|p /p, and by the same argument as for (b), CP (f ) is homocyclic of rank 2. Therefore CP (f ) = Φ(P ), and f centralizes first Ω1 (P ) and consequently P/Φ(P ). It follows that P f  has nilpotence class 2, so as p is odd, Ω1 (f  P ) = f  Ω1 (P ) ∼ = Ep3 , proving (f). Finally in (g), since p = 3, we have m0 = 1 or 2. Apart from the groups in (7A) d 2 3 and the groups A± 4 (q), the order formula for K = L(q) is divisible by (q − 1) , as K has untwisted rank at least 4. But then m3 (K) ≥ 3, contradiction. So we may

7. SMALL GROUPS

219

assume K = A4 (q),  = ±1. Now using [IA , 4.10.2], we see that m3 (K) = 2 if and only if q ≡ − (mod 3). The proof of the lemma is complete.  Lemma 7.5. Suppose that K ∈ Chev(2)∩K3 and m3 (K) = 2. Then K/Z(K) ∼ = − n ∼ L− (q), or L (q),  = ±1, q = 2 ≡  (mod 3), or K C (q), G (q) or = 2 2 4 5 n for some q = 2n , or K ∼ = 2F4 (2 2 ) for some odd n.

L3 (q), 3 D4 (q)

Proof. This follows easily from [IA , 4.10.3].



Lemma 7.6. Suppose that p is an odd prime, K ∈ Chev(2) ∩ Gp , and mp (K) − mp (Z(K)) = 2. Then mp (K) = 2. Moreover, K ∈ G5p or p = 3 with K ∼ = L4 (2) ∼ = A8 . Proof. By definition of G5p we need only prove mp (K) = 2, and this holds by [IIIK , 4.2].  Lemma 7.7. Suppose that p is an odd prime, K ∈ Kp , and mp (K)−mp (Z(K)) ≤ 1. Then K ∈ Cp ∪ Tp . Proof. If Z(K) = 1, then the definition of Tp [I2 , 13.1] implies the result. So assume that p divides |Z(K)|. For K ∈ Alt ∪ Spor, we must have p = 3 and mp (Z(K)) = 1 [IA , 6.1.4], so mp (K) = 2. Then with [IA , 5.6.1] we see that K/Z(K) ∼ = A6 , A7 , or M22 , and indeed K ∈ T3 by definition. Now assume that K ∈ Chev(r) − Alt for some prime r. If r = p then K ∈ Cp by definition [I2 , 12.1], so assume that r = p. By [IA , 6.1.4], K is a quotient of SLn (q), n ≡ q −  ≡ 0 (mod p),  = ±1, or p = 3 with K = E6 (q)u . From the assumption mp (K) = 2 and [IA , 4.10.3] we see that p = 3 and K is a quotient of  SL3 (q). But then by definition K ∈ T3 . The proof is complete. Lemma 7.8. Let K ∈ K3 with K ∼ = A− n (q), n = 3 or 4, or B2 (q), with q ≡  (mod 3),  = ±1. Let P ∈ Syl3 (K). Then P is homocyclic abelian of rank 2, and AutK (P ) ∼ = AutK (Ω1 (P )) ∼ = D8 . Proof. Using [IA , 4.10.2bc] as in the proof of Lemma 7.4 we see that m3 (K) = 2 with P homocyclic abelian in all cases. In particular, P ∈ Syl3 (CK (Ω1 (P ))), so AutK (Ω1 (P )) = AutNK (P ) (Ω1 (P )) by a Frattini argument. Thus the restriction mapping AutK (P ) → AutK (Ω1 (P )) is onto, so it is an isomorphism by [IG , 11.6].  = SL− (q) or K  = Sp4 (q) of K Let V be the natural module for the covering K n+1 ∼    and P ∈ Syl3 (K). Thus P = P . By [IA , 4.8.1], we may write P =  a1 ,  a2  and ai , V ] for i = 1, 2 and V0 = CV (P). Then V1 and V = V0 ⊕ V1 ⊕ V2 where Vi = [ a1  V2 are isometric and K contains SL(V1 )  Z2 stabilizing the set {V1 , V2 }. As   is inverted in SL(V1 ), we conclude that AutK (Ω1 (P )) contains D8 . Furthermore, a2  are not conjugate to the other two subgroups of P of order 3, whose  a1  and  commutators on V are 4-dimensional. But if D8 < AutK (Ω1 (P )) ≤ GL2 (3), then such fusion would have to occur. Thus AutK (Ω1 (P )) ∼ = D8 , and the lemma is proved.  Lemma 7.9. Let K = P SL4 (q),  = ±1, q ≡ − (mod 3). Let P ∈ Syl3 (K) and P ∗ ∈ Syl3 (Aut(K)) with P ≤ P ∗ . Then the following conditions hold: (a) P is homocyclic abelian of rank 2 and AutK (P ) ∼ = D8 ; (b) P ∗ = P φ where φ induces a field automorphism on K and φ faithfully induces power mappings on P ;

220

11. PROPERTIES OF K-GROUPS

(c) Ω1 (P ∗ ) = Ω1 (P ) × Ω1 (φ);   (d) The subgroups E of P of order 3 satisfy O 3 (CK (E)) = P O 3 (HE ) where HE ∼ = A1 (q) or A1 (q 2 ) (two E’s of each type); and (e) AutAut(K) (P ∗ ) = AutKP ∗ (P ∗ ). Proof. By Lemma 7.8, (a) holds. Now Ω1 (P ) has two irreducible constituents  on the natural K-module. The kernels E of these constituents satisfy O 3 (CK (E)) =  P O 3 (HE ), HE ∼ = A1 (q), HE ≤ CK (E), while the other two subgroups E of P of  order 3 satisfy O 3 (CK (E)) = P HE , HE ∼ = A1 (q 2 ), by [IA , 4.8.2]. Thus (d) holds. Moreover, since AutAut(K) (Ω1 (P )) has two orbits on E1 (P ), AutAut(K) (Ω1 (P )) ∼ =  D8 and in particular O 3 (NAut(K) (P )) centralizes Ω1 (P ). Next, a Sylow 3-subgroup of Out(K) is generated by the image of a field automorphism φ, and we may take φ ∈ P ∗ , whence φ normalizes P ∗ ∩ K = P . By the previous paragraph, φ centralizes Ω1 (P ), so φ acts on each HE , E ∈ E1 (P ), inducing a field automorphism of order |φ| on it by [IA , 4.2.3]. Now the four cyclic groups P ∩HE are each φ-invariant, and they cover the four subgroups of P/Φ(P ) of order 3, as can be seen from the action of AutK (P ) ∼ = D8 . The φ-invariance implies that φ induces a power mapping on P . Thus (b) holds, and (c) is a consequence of (b). Finally, the image of φ in Out(K) lies in Z(Out(K)). Therefore the kernel of the restriction mapping AutAut(K) (P ∗ ) → AutK (P ) is a 3-group. This yields (e), completing the proof of the lemma.  Lemma 7.10. Let K = 3D4 (q). Then the following conditions hold: (a) Out(K) is cyclic; (b) mp (K) ≤ 2 for every prime p not dividing q; (c) If q > 2 and q ≡  (mod 3),  = ±1, x ∈ I3 (Aut(K)), and L = CK (x)(∞) , then exactly one of the following holds: (1) x is a Sylow 3-center in any subgroup of Aut(K) containing Inn(K), L∼ = P GL3 (q); = SL3 (q), and AutK (L) ∼ 3 (q ), and elements of CAut(K) (x) of order 3 (2) x ∈ Inn(K), L ∼ SL = 2 induce nontrivial field automorphisms on L if and only if they are not in Inn(K); (3) x ∈ Inn(K), and L ∼ = G2 (q) or L3 (q). Proof. Part (a) is immediate from [IA , 2.5.12bc], and part (b) from [IA , 4.10.3]. From [IA , Table 4.7.3A] we see that there are four conjugacy classes of elements of Aut(K) of order 3, with centralizer structure as given in (c1,2,3). It remains only to prove that if L ∼ = SL3 (q), then x is a Sylow 3-center in any group X with K ≤ X ≤ Aut(K). By [IA , Table 4.7.3A], x ∈ Syl3 (CInn(K) (L)) and AutK (L) ∼ = P GL3 (q). Let P ∈ Syl3 (K). Then by Lemma 1.18, Z(P/ x) has order 3 so lies in the image  P SL3 (q), and then Z(P ) = x. Thus (c1) holds for the case O 3 (X) = K. Assume then that 3 divides |X : K|. We have Aut(K) = KΦ with Φ ∼ = Aut(Fq3 ), and Φ has an element γ1 of order 3 centralizing L. Let Q ∈ Syl3 (X); we may assume that Q = P (Q∩Φ). Let Q0 = P γ1 . If Z(Q) = x, then as Q/P is cyclic, Z(Q0 ) = x. Thus, we may assume that Q = Q0 . By Lemma 1.18 again, Z(Q0 ) ≤ CQ0 (L) = x, γ1 , the last by [IA , Table 4.7.3A]. However, γ1 ∈ Z(Q0 ) since CK (γ1 ) ∼ = G2 (q)  and |G2 (q)|3 < |K|3 . The proof is complete.

7. SMALL GROUPS

221

Lemma 7.11. Suppose that K ∈ K∩Chev(odd), K is simple, and x ∈ I2 (Aut(K)) with m2 (CK (x)) = 1. Then K ∼ = L2 (q) for some odd q. Proof. If x ∈ Inn(K) then this is obvious. If x is a field or graph-field automorphism, then O 2 (CK (x)) ∈ Lie and so O 2 (CK (x)) ∼ = SL2 (q) for some odd q; but then K ∼ = SL2 (q 2 ), contradicting simplicity. This rules out G2 (q) and 3D4 (q). If K x has sectional 2-rank at least 5 then a Sylow 2-subgroup of K x is connected, so a Sylow 2-subgroup of CKx (x) cannot be of 2-rank 2. Thus K x has sectional 2-rank at most 4. This rules out types Dn± , n ≥ 4, and their overgroups, n ∼  2 G2 (3 2 ). as well as A± n , n ≥ 5, Bn and Cn , n ≥ 3. Since |Out(K)| is even, K = ± The remaining cases, An (q), 2 ≤ n ≤ 4, and B2 (q), are ruled out by inspection of [IA , Table 4.5.1].  Lemma 7.12. Suppose that K ∈ K2 , m2 (K) > 1, and v ∈ I2 (Inn(K)). Then m2 (CK (v)) ≥ 2. Proof. Let T ∈ Syl2 (K), K = K/Z(K), and x ∈ T be such that v is conjugation by x. If T has maximal class, then by [IG , 15.17(ii)], K has one class of involutions, and the conclusion holds. If T does not have maximal class, then by [IG , 10.11], T has a normal four-subgroup U . Let C = CT (U ). Then |T : C| ≤ 2 so by the Thompson Transfer Lemma [IG , 15.16] and the simplicity of K, v has a K-conjugate in C. The lemma follows.  Lemma 7.13. Suppose K ∈ K2 and m2 (K) = 2. If Z(K) = 1, then K ∼ = Sp4 (q) for some odd prime power q. Proof. If K ∈ Spor, then as Z(K) has even order, m2 (K) > 2 by [IA , 5.6.1]. If K ∈ Alt, then by [IA , 6.1.4], K ∼ = 2An , n ≥ 5, and by [IA , 5.2.10d], m2 (K) ≡ 1 (mod 3). Thus we may assume K ∈ Chev(r) for some prime r. Suppose that r is odd. n Let q be any power of r. Of course m2 (A1 (q)) = 1, and m2 (2 G2 (3 2 )) > 2 as K ≥ Z2 × L2 (3n ) in that case. By consideration of Dynkin diagrams, every other choice for K, with the exception of K ∼ = B2 (q), contains a copy of A2 (q) or A− 2 (q) − − (q) = SU (q). So K or A3 (q), and even A3 (q) = P SU4 (q) contains the version A− 3 2 ± ± (q), and regardless of the version of A (q), m (Z(K)×A (q)) = contains Z(K)×A± 2 2 2 2 m2 (Z(K)) + 2 > 2, contrary to assumption. So K ∼ = B2 (q) and by [IA , 6.1.4], K must be the universal version Sp4 (q). We are reduced to showing that the case K ∈ Chev(2), K ∈ Chev(r) ∪ Alt for all odd r, is impossible. If K ∼ = 2Sp6 (2), then by [IA , 6.2.2], there is an embedding K ≤ K ∗ = Spin7 (3), and we calculate that |K ∗ : K| is odd, so m2 (K) = m2 (K ∗ ) ≥ 3 by the previous paragraph. Thus K ∼  2Sp6 (2), and by [IA , 6.4.1], therefore, every = 2-central involution of K = K/Z(K) is the image of an involution of K. Suppose for the moment that Z(K) is elementary abelian. As m2 (K) = 2, we must have |Z(K)| = 2 and there exists no four-subgroup of K consisting of 2central involutions. In particular K must be defined over F2 , whence by [IA , 6.1.4], K∼ = D4 (2), U6 (2), or 2 E6 (2). (Note that because of the graph-field automorphism, F4 (2) has a Sylow 2-center which is a four-group.) But the corresponding groups K contain, respectively, K 0 ∼ = Sp6 (2), Sp6 (2), and F4 (2), as the centralizer of a graph automorphism of order 2. In all cases the full preimage K0 satisfies m2 (K) ≥ m2 (K0 ) > 2 by the cases already covered.

222

11. PROPERTIES OF K-GROUPS

∼ L3 (4), Thus we may assume that Z(K) is not elementary abelian, whence K = in which every involution is 2-central. Again if Z(K) is noncyclic, then m2 (K) > 2, so we may assume that Z(K) is cyclic, whence Z(K) ∼ = Z4 by [IA , 6.1.4]. From an A1 (4)-type parabolic subgroup of K we choose subgroups T, A ≤ K such that T ∼ = Z , and T is minimal A-invariant. Then T = [T, A], and as T /Ω (Z(K)) E24 , A ∼ = 5 1 is elementary abelian, T is either extraspecial of order 25 or abelian. In either case m2 (K) ≥ m2 (Z(K)T ) > 2 (note that Z4 ∗ D8 ∗ Q8 contains Q8 ∗ Q8 , of 2-rank 3). The proof is complete.  Lemma 7.14. Let K be a simple K-group and let T ∈ Syl2 (K). Then T ∼ = E23 n 2  ∼ 2 if and only if K ∼ J or K G (3 ) . Moreover, in all cases except K ∼ = 1 = = 2 1 2  ∼ 2 G2 (3 ) = L2 (8), T = CK (T ) and AutK (T ) is a Frobenius group of order 21. Proof. The first statement is (a corrected version of) [IA , 5.6.4]. For the second statement, let t ∈ I2 (K) and C = CK (t) and T ∈ Syl2 (C). Then C = t×L2 (q), q > 3, q ≡ ±3 (mod 8), so CK (T ) = T and CAutK (S) (t) ∼ = Z3 . Moreover K has one class of involutions, so NK (T ) is transitive on T # and AutK (T ) ∼ = F7.3 , as asserted. 

8. Embeddings Lemma 8.1. The following conditions hold: (a) W (E8 ) has no subgroup isomorphic to A10 ; and (b) W (E7 ) has no subgroup isomorphic to A9 . Proof. In (a), we show in fact that H := W (E8 ) has no element of order 21. Set H1 = O 2 (H/O2 (H)) ∼ = D4 (2); it is enough to show that H1 has no element of order 21. Let x ∈ H1 be of order 7 and let V be a natural H1 -module. Then dim[x, V ] = 6 and since x induces an isometry of V of order 7, [x, V ] is of + type. Since V is of + type, so is the 2-dimensional space [x, V ]⊥ . Thus O 2 (CH1 (x)) ∼ = ∼ O 2 (CΩ+ (2) (x)). Since Ω+ (2) A has no element of order 21, (a) follows. = 8 6 6 We prove (b) by observing that A9 contains a copy of the Frobenius group U3 (2) of order 9.8, but W (E7 ) ∼ = Z2 × Sp6 (2) does not. Since Z(U3 (2)) = 1 it is enough to show that U3 (2) does not embed in Sp6 (2), and this in turn is true since the minimum dimension of a faithful representation of U3 (2) in characteristic 2 is at least 8, by [IG , 9.12].  Lemma 8.2. Let I = Sp4 (2m ) and J = L3 (2n ). Then I is not involved in J. Proof. Suppose false and let H ≤ J be minimal subject to H/H0 ∼ = I for some H0  H. By a Frattini argument, H0 is nilpotent. If |H0 | is even, then H lies in a proper parabolic subgroup of J by the Borel-Tits Theorem, so I is involved in L2 (2n ), which is absurd. Thus H0 has odd order. Let z be root involution in H. Then CH (z) covers CH/H0 (z) and in particular involves L2 (2m ). However, CJ (z) is a 2-group, a contradiction. The lemma follows. 

10. BALANCE

223

9. Recognition Theorems Proposition 9.1. For any n ≥ 2, let X = {t1 , . . . , tn−1 } and let R be the set of words comprised of yi2 , 1 ≤ i < n − 1, (yi yi+1 )3 , 1 ≤ i < n − 2, and (yi yj )2 for all other pairs of indices with 1 ≤ i < j < n. Then gp X | R ∼ = Σn . Proof. This is [Hu1, Beispiel I.19.7].



Proposition 9.2. Let n ≥ 11. Let X = {y1 , . . . , yn } and let R be the set of words comprised of y13 , yi2 , 2 ≤ i ≤ n, (y1 y2 )3 , (yi yi+1 )3 , 2 ≤ i < n, and (yi yj )2 = 1 for all other pairs of indices with 1 ≤ i < j ≤ n. Then gp X | R ∼ = An+2 . Proof. This is [Hu1, Beispiel I.19.8].



10. Balance Lemma 10.1. Let K ∈ Chev(3) have level q(K) = 3. Then K is locally balanced with respect to p = 2. Proof. We may assume that K is simple, and we let K ≤ H ≤ Aut(K) and z ∈ I2 (H) and assume by way of contradiction that W := O2 (CH (z)) = 1. Since  q(K) = 3, z ∈ Aut0 (K). Also [W, O 3 (CInndiag(K) (z)] = 1. Hence from [IA , Table 4.5.1] we see that W ∩ Inndiag(K) = 1. The only possibility is that W ∼ = Z3 3 induces graph automorphisms on K, with K ∼ D (3) or D (3). Consequently = 4 4  z ∈ Inndiag(K). By [IA , 4.9.2eg], CK (W ) embeds in G2 (3). Hence O 3 (CK (z)) embeds in an involution centralizer in G2 (3), and hence in SL2 (3) ∗ SL2 (3). But there is no such involution in K, by [IA , Table 4.5.1], a contradiction proving the lemma.  Lemma 10.2 (F3 -balance). Let K ∈ Chev(3) with q(K) = 3, and suppose that K is a component of a group X with O2 (X) = 1. Let x ∈ I2 (K) and suppose that L   CX (x) with L ∼ = SL2 (3). If O2 (L) ≤ K, then L ≤ K. Proof. Write L = QR with |R| = 3 and Q ∼ = Q8 . Since L normalizes Q, L normalizes K. We claim that R induces inner automorphisms on K. As q(K) = 3 the only alternative is for K ∼ = D4 (3) or 3D4 (3), with R inducing graph automorphisms on K. In either case there is a unique R-invariant conjugacy class of involutions in Inn(K), namely, 2-central involutions. According to the isomorphism type of K, we have that CK (x) has a normal subgroup N which is the central product of four copies of SL2 (3), permuted in orbits of length 1 and 3 by R; or N ∼ = SL2 (3) ∗ SL2 (33 ) with R inducing an inner, respectively field automorphism on the components of N . In either case there is an R-invariant B of CG (x) such that B ∼ = E33 and  B  3-subgroup 1+2 ∼ ∼ 3 BR = Z3  Z3 . Therefore R is nonabelian. However, this contradicts =   the fact that LCK (x) is the commuting product of conjugates of L, so its Sylow 3-subgroups are abelian. This proves the claim. We may assume that R ≤ K. As O2 (K) ≤ O2 (X) = 1, KR = K × R1 and R is a diagonal subgroup of R0 × R1 , where R0 ∼ = R1 ∼ = Z3 and R0 ≤ K acts on K like R. Moreover, R0 and R1 are x-invariant so both are centralized by x. Now L = QR   CX (x), so L   CKR (x). Projecting onto K, we get

224

11. PROPERTIES OF K-GROUPS

QR0   CK (x)   CX (x). Thus L and QR0 are both subnormal SL2 (3)-subgroups of CX (x). On the other hand [O2 (CX (x)), K] = 1 by Lemma 10.1, and as x ∈ K and O2 (X) = 1, [O2 (CX (x)), CF ∗ (X) (K)] = 1. Therefore O2 (CX (x)) = 1, and it follows by [IG , 13.5] that L = QR0 ≤ K, completing the proof.  Lemma 10.3. Let K ∈ Chev(2) and let p and r be (not necessarily distinct)  odd primes. Let x ∈ Aut(K) be of order r, and W = O p (CK (x)(∞) ). Then W  centralizes every W -invariant p -subgroup of CK (x). Proof. We have K = L where L ∈ Lie(2), and then CK (x)(∞) = CL (x)(∞) , which by [IA , 4.2.2] is the commuting product of the last terms of the derived series of all the Lie components of CL (x). These derived subgroups are quasisimple and so CK (x)(∞) = E(CK (x)). Moreover, W  CK (x) is the product of those components which are not p -groups, so W is perfect. If X is any W -invariant p -subgroup of CK (x), then [W, X] ≤ W ∩ X ≤ Op (W ) ≤ Z(W ), so [W, X] = 1 by [IG , 3.8]. The proof is complete.  11. Subcomponents Lemma 11.1. Let K = A (q),  ≥ 4, or D (q),  ≥ 4,  = ±1, with q odd in either case. Let V be a four-subgroup of Aut(K) such that L2 (CK (V )) ∼ = A−2 (q)   (q) for some sign  , respectively. Then there exists v ∈ V # such that one or D−1 of the following holds: (a) K ∼ = A (q) and L2 (CK (v)) ∼ = A−1 (q); or  (b) K ∼ = D (q) and L2 (CK (v)) ∼ = B−1 (q). Proof. Write q = r a , where r is prime, and let KV = L2 (CK (V )) = E(CK (V )). For every v ∈ V # , let Kv be the subnormal closure of KV in CK (v), so that by L2 -balance and the B2 -property, Kv is a product of components of E(CK (v)). r # Thus  by [IA , 4.2.2]. If Kv = KV for all v ∈ V , then KV    Kv  O (CK (v)) O r (CK (v)) | v ∈ V # = ΓrV,1 (K). Hence KV  K by [IA , 7.3.1], which is of course absurd. Thus for some v ∈ V # , which we fix, Kv > KV . It is enough to show that Kv has the isomorphism type specified in (a) or (b), for then we see with [IA , Table 4.5.1] that CK (v, Kv ) is solvable, so Kv = L2 (CK (v)) as needed. By L2 -balance, if Kv is not a single component then it is the product of two isomorphic components. Let v and V be the untwisted Lie ranks of any component of Kv and of KV , respectively. We use the facts that (1) KV is a component of CKv (V / v), and in particular (11A) (2) v ≥ V , by [IA , 4.2.2, 4.9.1]. In (a), therefore, v ≥ −2 ≥ 2. If Kv is not a single component, 2 we get 2v ≤ ,  = 4; but |A4 (q)| is not divisible by |A± 2 (q)| , contradiction. Thus Kv is quasisimple. From [IA , Table 4.5.1, 4.9.1] we see that the only possibilities for Kv , as a component of CK (v) of such untwisted rank and not isomorphic to KV , are Kv ∼ = A−1 (q), A (q 1/2 ) ( = 1), B2 (q) ( = 4), C3 (q) ( = 5), and A± 3 (q) ( = 5). In the last three cases, Kv does not have an automorphism of order 2 centralizing a component A−2 (q) [IA , Table 4.5.1, 4.9.1], contradiction. The case Kv ∼ = A (q 1/2 ) leads to a similar contradiction; in order for KV to have level q, we

11. SUBCOMPONENTS

225

would have to have KV ∼ = A(−1)/2 (q) [IA , Table 4.5.1]; but ( − 1)/2 <  − 2 as  ≥ 4. Thus the lemma holds in case (a). A similar but easier argument applies in case (b).  Lemma 11.2. Let p be an odd prime, K ∈ Kp , and x ∈ Ip (Aut(K)). Suppose that J is a p-component of Lp (CK (x)) such that J ∩ Z(K) = 1. Then J = Lp (CK (x)). Proof. If K ∈ Alt, then K ∼ = 3An , n = 6 or 7, and L3 (CK (x)) = 1. If K ∈ Spor, the result follows from inspection of the tables [IA , 5.3]. If K ∈ Chev(p), then Lp (CK (x)) consists of at most one p-component, regardless of the structure of Z(K), by the Borel-Tits Theorem and [IA , 4.9.1,4.9.2]. So we may assume that K ∈ Chev(r), r = p. Then since Z(K) = 1, [IA , 6.1.4] gives K ∼ = E6 (q)u , p = 3, q ≡  (mod 3). = SLkp (q), q ≡  (mod p),  = ±1, or K ∼ If x ∈ Inndiag(K), then x, having order p, is a field automorphism and CK (x) has at most one p-component by [IA , 4.9.1]. So suppose that x ∈ Inndiag(K). In the SLkp (q) case, if x acts like an element of GLkp (q) of order p, then by [IA , 4.8.2], every p-component of CK (x) has fixed points on the natural module, so meets the fixed-point-free subgroup Z(K) trivially. If x acts otherwise, then by [IA , 4.8.4], CK (x) has at most one p-component. Finally in the E6 case, we see from [IA , Table 4.7.3A] that if CK (x) has more than one 3-component, then all 3-components of CK (x) are the universal version, even if K itself is the adjoint version. This implies the result.  Lemma 11.3. Let K ∈ Chev(2) and x ∈ I2 (Aut(K)), and let L be a component of CK (x). Then O2 (Z(L)) ≤ Z(K). Proof. Assume false and form H = K x. Let y ∈ O2 (Z(L))# . By L2 balance, L ≤ L2 (CH (x, y)) ≤ L2 (CH (y)) = 1, the last by the Borel-Tits Theorem. This contradiction completes the proof.  Lemma 11.4. Suppose that K ∈ Chev − Chev(p) for some prime p, and f is a field or graph-field automorphism of K of order p. Let K0 be the Lie component of CK (f ) and suppose that K0 has level q. Let x ∈ K0 be of order p and let J0 be a Lie component of CK0 (x) of the same level q and of order divisible by p. Then there exists a component J of CK (x) such that f induces a field or graph-field automorphism on J and J0 is a Lie component of CJ (f ). Moreover the level of J is q p . Proof. Let (K, σ) be a σ-setup for K. Then by [IA , 2.5.17], there exists a Steinberg endomorphism τ of K such that σ ∈ τ , | τ  : σ | = p, and (K, τ ) is   a σ-setup for K0 . Without loss K0 = O r (CK (τ )) and K = O r (CK (σ)), where K ∈ Chev(r). Moreover, τ  / σ induces the group f  on K. In particular x ∈ K, and we set C = CK (x). By [IA , 4.2.2], there is a simple component J of  C which is τ -invariant (since J0 has level q) and such that J0 = O r (CJ (τ )). Set  J = O r (CJ (σ)); then J has level q p ; f , like τ  / σ, induces field or graphfield automorphisms on J, again by [IA , 2.5.17]; and J0 is a Lie component of  CJ (τ ) = CJ (f ). The lemma follows. Lemma 11.5. Let K ∈ Spor and x ∈ I2 (Aut(K)). Suppose that CK (x) has a component L ∈ Alt. Then L ∼ = A5 , A6 , 3A7 , A8 , 2A8 , A10 , or 2A11 . Accordingly K/Z(K) ∼ = M12 , J1 , or J2 ; HS; He; HS; M cL; F5 ; or Ly.

226

11. PROPERTIES OF K-GROUPS

Proof. This follows directly from the tables [IA , 5.3].



Lemma 11.6. ↓2 (U4 (3)) = {A6 , U3 (3), P Sp4 (3)}; and  1/2 ↓2 (L± ), L2 (q), SL2 (q)},  = ±1, for any odd q. 3 (q)) ⊆ {L3 (q Proof. This is immediate from [IA , Table 4.5.1, 4.9.1].



Lemma 11.7. Suppose that L 1, then x is a field automorphism; and (b) If L/Z(L) ∼ = L3 (2n ) with p dividing 2n − , then either x is a field auton morphism, or mp (L∗ ) ≥ 3 with f (L∗ ) > 29n , or p = 3 with L∗ ∼ = 2F4 (2 2 ). Proof. If x is a field automorphism, then there is nothing to prove. If x is a graph-field automorphism, then as p is odd, L ∼ = 3D4 (q) for some q, contradiction. If x is a graph automorphism, then p = 3, and by [IA , Table 4.7.3A], L and L∗ have the same level, contrary to hypothesis. Thus we may assume that x ∈ Inndiag(L∗ ). Let Δ and Δ∗ be the Dynkin diagrams of L and L∗ , respectively. By [IA , 4.2.2], the disjoint union of some number a ≥ 2 of copies of Δ can be obtained either as some subdiagram of Δ∗ , or as the subdiagram of the extended diagram of Δ∗ obtained by deleting a node marked with the prime p. Note in (a) that Δ∗ does not have two double bonds, so two copies of Δ must be obtained from the extension of Δ∗ , which must be of type Cn . However, all nodes of that extension are marked with 1 or 2, not with p, contradiction. Thus (a) holds. In (b), let  be the untwisted Lie rank of L∗ . Then 2a ≤ , and L∗ has level 2 n/a 2 , so f (L∗ ) = 2n /a ≥ 24an . Hence if a > 2, then f (L∗ ) > 29n as desired. If a = 2, we still get f (L∗ ) > 29n unless  = 4. But then the two copies of Δ must be obtained from the extended diagram of Δ∗ , and the only possibility is p = 3 and n L∗ ∼ = 2F4 (2 2 ) with n odd. It remains to show that mp (L∗ ) ≥ 3 except in this case. Suppose by way of contradiction that mp (L∗ ) = 2. If the Schur multiplier of ∗ L or Outdiag(L∗ ) has order divisible by p, then by [IA , 6.1.4, 4.10.3], we must ∗ have L∗ /Z(L∗ ) ∼ = L± 3 (q) for some q, which is absurd by the structure of Δ and Δ . ∗ ∗  Therefore x ∈ Inn(L ), and Z(L ) is a p -group so x corresponds to an element of L∗ of order p. We have mp (L) = 2 as p divides q − , so x must act as an element of Z(L), and in particular p = 3. If L∗ is a classical group then using [IA , 4.8.2, 4.8.4] and the fact that m3 (L∗ ) = 2, we see that CL∗ (x) has no non-level component. If L∗ is of exceptional Lie type, then using [IA , Table 4.7.3A] we see n  that L∗ ∼ = 2F4 (2 2 ). The proof is complete.

12. Pumpups and the Ranking Function F In this section we begin to study the behavior of the ranking function F under (single) pumpups. Later, in Sections 19 and 20, we shall make a more detailed study, using F to study long pumpups. If I, J ∈ Kp and I g(x) ≥ 1. Proof. Parts (a) and (b) are clear. As for (c), note that for fixed 0 ≤ x ≤ 1, the function b + (x/b) is strictly increasing in b for b ≥ 1. Writing a = b2 , we thus have  x 2 (12A) ag(a−1 x) = b + > (1 + x)2 = g(x), as required. b  Recall that for any K ∈ Chev, with K = d L(q) , we have defined   2 f (K) := q r(K) and F(K) = f (K), τ (K) , where r(K) is the rank of the simple Lie algebra L, i.e., the untwisted Lie rank of K, and τ (K) ∈ {A, BC, D, E, F, G} is the type of L. In the case of “ambiguous” K ∈ Chev(r) ∩ Chev(s) for distinct primes r and s, one of r and s is always 2 (see [IA , 2.2.10]). Whenever we use the function f (K), it will be clear from context whether to consider ambiguous K’s as lying in Chev(2) or Chev(r) for some odd r. Recall also that we say that F(K) < F(K1 ) if and only if either f (K) < f (K1 ), or f (K) = f (K1 ) and τ (K) < τ (K1 ) in the ordering A < D < E < BC < F < G. Definition 12.2. Let K1 , K ∈ Chev(r) for some prime r = p. We define ρ(K, K1 ) = logf (K1 ) f (K). The definition of ρ will be used mostly when (x, K) is a pumpup or long pumpup of (x1 , K1 ) for some (x, K) and (x1 , K1 ) ∈ ILop (G); but this context is not required for the definition. It is obvious that ρ is multiplicative in the sense that for any K, K1 , K2 ∈ Chev(r), (12B)

ρ(K2 , K) = ρ(K2 , K1 )ρ(K1 , K).

Note also that if K = d L(q) and K1 = d1 L1 (q) with 1 = r(K1 ) ≤ r(K) = , and if we let k be the nonnegative integer r(K) − r(K1 ), (12C) ρ(K, K1 ) = (r(K)/r(K1 ))2 = [1 + (k/r(K1 ))]2 = g(kr(K1 )−1 ), with k ≥ 0. The next result gives basic information about ρ(K, K1 ) when K, K1 ∈ Kp for some prime p and K1 ↑p K. Lemma 12.3. Let K ∈ Chev(r) and let x ∈ Aut(K) be of prime order p = r. Let K1 be a p-component of CK (x). Let K have level q(K) = q and untwisted Lie rank r(K) = . Similarly let K1 have level q(K1 ) = q1 and untwisted Lie rank r(K1 ) = 1 . Let a be such that q1 = q a , and set k =  − a1 . Then the following conditions hold: (a) ρ(K, K1 ) ≥ 1; (b) (Algebraic type (a, k)) If x ∈ Aut0 (K), then the following conditions hold: (1) a is a positive integer and a1 ≤ , so that k ≥ 0;

228

11. PROPERTIES OF K-GROUPS

(2) ρ(K, K1 ) = 2 /a21 = ag(ka−1 1 −1 ) ≥ a; (3) If k > 0 but ρ(K, K1 ) ≤ 4, then ρ(K, K1 ) ≥ g(k1 −1 ) ≥ g(1 −1 ), with ρ(K, K1 ) > g(k1 −1 ) if a > 1; and (4) If k = 0, then ρ(K, K1 ) = a; (c) (Field type) If x ∈ Aut0 (K), then a = 1/p,  = 1 , and ρ(K, K1 ) = p; (d) If ρ(K, K1 ) = 1, then 1 =  and q1 = q, x ∈ Inndiag(K), and (p, K1 , K) is one of the following triples: (2, D± (q), B (q)),  ≥ 3, (2, B4 (q), F4 (q)),   (2, A± 7 (q), E7 (q)), (2, D8 (q), E8 (q)), (3, A2 (q), G2 (q)) or (3, A8 (q), E8 (q)), with  = ±1 ≡ q (mod 3) in the last two cases; and (e) F(K1 ) < F(K). Proof. For x ∈ Aut0 (K), the structure of CK (x) is described in [IA , 4.2.2]. From parts (e,f,g) of that result, assertion (b1) of our lemma holds if x ∈ Inndiag(K). On the other hand if x is a graph automorphism, then p = 2 or 3 and (b1) holds by inspection of the tables [IA , Tables 4.5.1, 4.7.3A]. By definition, logq (f (K)) = 2 and logq (f (K1 )) = a21 , so

k 2 (k + a1 )2 ρ(K, K1 ) = 2 = = ag , a1 a21 a1 proving (b2). Then (b4) is immediate. In (b3), first assume that k−1 1 > 1. Then by (b2) and monotonicity of g, ρ(K, K1 ) > ag(a−1 ), whence ρ(K, K1 ) > g(1) = 4 by Lemma 12.1c, so (b3) holds in this case. On the other hand if k−1 1 ≤ 1, then (b3) is immediate from (b2) and Lemma 12.1ac. For x ∈ Aut0 (K), x is a field or graph-field automorphism [IA , 4.9.1], and we have L = L1 ,  = 1 , and a = 1/p. Then ρ(K, K1 ) =

2 = p, a21

and (c) holds. Then (b) and (c) imply (a). In (e), if ρ(K, K1 ) > 1 then f (K1 ) < f (K) and so F(K1 ) < F(K). So in the proofs of (d) and (e) we may assume by (a) that ρ(K, K1 ) = 1, whence x ∈ Aut0 (K) by (c). Then by (b3), if k > 0, we have 1 ≥ g(−1 1 ) > 1, a contradiction, so k = 0. By (b4), 1 = a. That is, K1 and K have the same untwisted rank, and hence the same level q. If x is a graph automorphism then L = Am , Dm or E6 with p = 2, or L = D4 with p = 3. In these cases K1 and K have distinct untwisted ranks, by [IA , Tables 4.5.1, 4.7.3A]. This is a contradiction, and so x ∈ Inndiag(K). By [IA , 4.2.2], the Dynkin diagram of K1 is obtained from the extended Dynkin diagram of K by deleting a node corresponding to a fundamental root whose coefficient in the highest root is p. Since the Dynkin diagram of K1 is connected, the deleted node is an end node not connected to the lowest root in the extended diagram. The list of possibilities in (d) follows easily from this on inspection of [IA , Tables 4.5.1, 4.7.3A]. In all cases we have not only 1 = , but also q1 = q and τ (K1 ) < τ (K). Hence f (K1 ) = f (K) and so F(K1 ) < F(K). Thus (d) and (e) hold and the proof is complete.  The main point for this section is the following fundamental consequence. (The assumption that r = p is not actually necessary, but in all our applications, it will hold.)

12. PUMPUPS AND THE RANKING FUNCTION F

229

Lemma 12.4. Let K, K1 ∈ Chev(r) ∩ Kp , r = p, and suppose that K1 3, assume that F(I) ≤ (q , A) or (q , BC). Then one of the following holds: (a) Z(L) ≤ Z(I) and s = 2; (b) s = 1; or   b (c) s = 2, and I/Z(I) ∼ = P Sp4 (q  ), L± 4 (q), or G2 (q ) for some q = r , b ≥ a. Proof. We assume, as we may, that q1 ≥ q2 ≥ · · · and that s ≥ 2. Write 2 I = d L(q0 ) and let  be the rank of L, so that F(I) = (q0 , τ (I)). Since s ≥ 2,  ≥ 2 n  2 G2 (3 2 ) for any n. and L = A2 ; also since s = 0, I ∼ = First suppose that q0 ≥ q. Then our assumption on F(I) implies that  ≤ 3, with equality possible only if τ (I) = A and q0 = q. If  = 3, then (c) follows with I/Z(I) ∼ = L± 4 (q). If  = 2, then by the previous paragraph I = B2 (q0 ) or G2 (q0 ), and (c) again holds (q0 = q1 in all these cases by [IA , Table 4.5.1]). Thus we may assume that q0 < q ≤ qi for all i ≥ 1. As b ∈ Inn(I), qi = q0ni for all i = 1, 2, . . . , with ni > 1, by [IA , 4.2.2]. An examination of [IA , Table 4.5.1] √ shows that the condition s ≥ 2 cannot be achieved. (Note that if I = D4 ( q) and b is conjugate to t2 , then b is not an inner automorphism, contrary to assumption.) The lemma follows.  13. Pumpups Lemma 13.1. Let X be a K-group with O2 (X) = 1, let x ∈ I2 (X), and let L  be a quasi-2-component of CX (x). Set L0 = O 2 (L) and let K be the subnormal closure of L0 in X. Then one of the following holds: (a) K ≤ O2 (X); (b) K ∼ = 2An , n ≥ 6, and Z(K) is the unique subgroup of L of order 2; (c) K ∼ = M11 ; (d) K ∈ Chev(3) and K has level q(K) = 3; or (e) K ∈ Chev(2), but K ∈ Alt ∪ Chev(r) for any odd r, and x does not centralize any Sylow 2-subgroup of K. Proof. Assume that (a) fails. By solvable L2 -balance [IG , 13.8], L0 lies in a component of X, which must be K. Then as L0  L, L normalizes K. We may pass to KL x /O2 (KL x) and assume that X = KL x. There exists a 3-element t ∈ L such that L = O2 (L)L0 t, and we may pass to KL0 t, x and assume that X = K t, x, with t3 ∈ O2 (CX (x)). Since L   CX (x), t centralizes

232

11. PROPERTIES OF K-GROUPS

every t-invariant abelian 2-subgroup of CX (x), and consequently [t, Z(K)] = 1 by the A × B-lemma. Thus [O 2 (X), Z(K)] = 1. Now Z(K) is a 2-group, and so in X := X/Z(K) we have O 2 (CX (x)) = O 2 (CX (x)). As L   O 2 (CX (x)) it follows that L   CX (x). Hence we may assume that either Z(K) ∩ L = 1 or Z(K) = 1.  Furthermore, as O 2 (L) = L0 ≤ K, KL/K has odd order. But also KL  KL x = X, so O2 (KL) = 1 and hence F ∗ (KL) = K. Suppose that K ∈ Alt ∪ Spor. Then |Out(K)| ≤ 2 by [IA , 5.2.1,Tables 5.3], and since L = O 2 (L), we have L ≤ K. For K ∈ Spor but K ∼  M11 , we use the condition that L ≤ O2 23 (CK (x)) = to narrow the possibilities in [IA , Tables 5.3] for K/Z(K) and the conjugacy class of x to (M12 , 2B), (M22 , 2A), (J4 , 2A), (F i22 , 2C), (F i23 , 2C), or (F i24 , 2B); in particular O2 (CK (x)) = 1, so L ∼ = SL2 (3). Let R = O2 (CK (x)), 2a = |R| and b 2 = |CCK (x)/R (t)|2 . Then as [t, R] ∼ = Q8 , |CR (t)| = 2a−2 , so |CK (t)|2 ≥ 2a+b−2 . In every case, however, we see from the tables [IA , 5.3] that |CK (u)|2 < 2a+b−2 for every u ∈ K of order 3, a contradiction. For example, in K = F i24 , a = 13 and 2b = |U4 (3)|2 = 27 , but the largest Sylow 2-subgroup of the centralizer of any element u ∈ K of order 3 occurs for u in class 3A: |CK (u)|2 = 212 , and 12 < 13 + 7 − 2. In this way we see that (c) holds if K ∈ Spor. Suppose that K ∈ Alt. Let X = X/Z(K). Then L   CK (x) with L ∼ = SL2 (3) or A4 , the latter possible only if Z(K) = 1. We may assume that x acts on K like an element of Σn (otherwise K x ∼ = P GL2 (9) and CK (x) is a 3 -group). Say x is the product of r > 0 disjoint transpositions, with s ≥ 0 fixed points, so that K∼ = A2r+s , as in [IA , 5.2.8a]. Indeed, by that lemma, and since t ∈ O2 ,2,3 (CK (x)) with [t, O2 2 (CK (x))] having a Sylow 2-subgroup which is either E22 or Q8 , we see that either r = 3 or s = 4. In both cases, O2 (L) = [t, O2 (L)] is a four-group, so Z(K) = 1, whence K ∼ = 2An [IA , 6.4.1]; moreover n = 2r + s ≥ 6, so (b) holds. Thus we may assume that K ∈ Chev(r) − Alt for some prime r. Suppose that r is odd. If x induces a field or graph-field automorphism on K, and C := CK (x) is nonsolvable, then by [IA , 4.9.1,7.1.4c], CInn(K) ([C, C]) = 1, a contradiction since [C, C, L0 ] = 1. Thus if x induces a field or graph-field automorphism on K, C is solvable, forcing K/Z(K) = L2 (9) ∈ Alt, contradiction. Therefore we may assume in this case that x induces an inner-diagonal or graph automorphism on K. If r > 3, or if r = 3 with K having level 3n , n > 1, then  Kx := O r (CK (x)) is the commuting product of quasisimple groups and so L0 , which lies in Sol(CK (x)), maps into CInn(K) (Kx ). But by [IA , Table 4.5.1], Sylow 2-subgroups of CInn(K) (Kx ) are either cyclic or dihedral, and in the latter case they contain the image of x in Aut(K). In either case there is no SL2 (3) or A4 section of CAut(K) (x Kx ) with Sylow 2-subgroup contained in Inn(K), contradicting the 1 existence of L. Thus, r n ≤ 3. If r n < 3 then K ∼ = 2 G2 (3 2 ) , for which Z(K) = 1 and Sylow 2-subgroups of Aut(K) are abelian, contradicting the existence of L0 . Thus r n = 3 and (d) holds. We may now assume that K ∈ Chev(2) − Chev(s) − Alt for all odd s. To prove (e) we assume that x centralizes some T ∈ Syl2 (K), and argue to a contradiction. The actions of graph, graph-field, and graph-field involutions on

13. PUMPUPS

233

T /Φ(T ) are all nontrivial, so x induces an inner automorphism on K, corresponding  to an involution of Z(T )/Z(K). Let C = CK (x) and C0 = O 2 (C). Then T ≤ C, so by [IA , 2.6.7] there exists a parabolic subgroup Q ≤ K such that if we put  Q0 = O 2 (Q), then C0 = Q0 and C ≤ Q. Moreover, O2 (CX (x)) ≤ O2 (X) = 1 by the Borel-Tits Theorem and so O2 (L0 ) = 1. Thus L0 ∼ = Q8 with L0 ≤ K, and L = L0 t with t3 = 1. As ∗ O2 (X) = 1, F (X) = K x. Thus t acts nontrivially on O2 (Q). Hence,  0 ≤ O2 (Q) and indeed L0 =  L [t, O2 (Q)]. Since L   CX (x) we conclude that tQ0 O2 (Q)/O2 (Q) is an elementary abelian 3-group. Suppose next that (13A)

t normalizes T . Then in CX (x) we see that [t, T ] = L0 ∼ = Q8 . If t ∈ K then t lies in a Cartan subgroup by (13A), and then t centralizes all but one root subgroup of T . However, the roots corresponding to CK (t) form a subsystem of the root system of K, and as that system is indecomposable there is no subsystem missing exactly one root (and its negative). Therefore t ∈ K. If t induces a field automorphism on K, then t acts nontrivially on each root subgroup of T . Again as [t, T ] = L0 the only possibility is K/Z(K) ∼ = L2 (8); but then Z(K) = 1 by [IA , 6.4.1], and so [t, T ] is a four-group, contradiction. Finally if t induces a graph or graph-field automorphism on K, we have either K ∼ = 3D4 (q), and O2 (Q)/Z(K) is special of = D4 (q) or K ∼ 1+8 order q . In the D4 (q) case t cycles each of two sets of three root groups mapping injectively to the Frattini quotient of O2 (Q)/Z(K), so |[t, T ]| ≥ 24 , a contradiction. 3 3 ∼ In  D4 (q) case, t induces a field automorphism on Q0 /O2 (Q0 ) = L2 (q ), so  Qthe t O2 (Q0 )/O2 (Q0 ) is not a 3-group, contradiction. Therefore t does not normalize T . In particular t does not centralize Q0 /O2 (Q0 ) or map into a Cartan subgroup of t Q0 /O2 (Q0 ). Consequently t Q0 /O2 (Q0 ) has a normal elementary abelian 3-subgroup not contained in Z(t Q0 /O2 (Q0 )). With Lemma 6.19 and the fact that K ∈ Chev(r) ∪ Alt for any odd r, we conclude that K ∼ = C3 (2). Then by Lemma = Dn (2), n ≥ 4, or Cn (2), n ≥ 3. Suppose that K ∼ 6.19, Q0 /O2 (Q0 ) ∼ = A1 (2) or A1 (2) × A1 (2). In particular |O2 (Q0 )/Z(K)| ≥ 27 and so |CO2 (Q0 )/Z(K) (t)| ≥ 25 . However, using [IA , 4.8.2] we see that any automorphism τ of K of order 3 must satisfy |CK/Z(K) (τ )|2 ≤ 24 , so |CO2 (Q0 )/Z(K) (t)| ≤ 24 , a contradiction. Similarly, if K ∼ = D4 (2), then |O2 (Q0 )/Z(K)| = 29 so |CO2 (Q0 )/Z(K) (t)| ≥ 7 2 , but from [IA , Table 4.7.3A], |CK/Z(K) (t)|2 ≤ 26 , contradiction. These two cases aside, we have |Out(K)| ≤ 2 and Q0 is a parabolic subgroup of K, with O3 (Q0 /O2 (Q0 )) ∼ = Z3 generated by an element t such that m2 ([t, O2 (Q0 )/Φ(O2 (Q0 ))]) = 4(n − 2) > 2. As this commutator is a four-group, we have a contradiction. The proof is complete.  Lemma 13.2. Let K ∈ K2 and x ∈ I2 (Aut(K)). Suppose that CK (x) has a component I ∼ = SL2 (q) for some odd q > 3, or I ∼ = 2A7 . Assume that Z(I) ≤ Z(K). Then there is y ∈ K such that y x = yz, where z = Z(I). Moreover, x does not centralize any Sylow 2-subgroup of K. Proof. If y exists as asserted, then the lemma holds, by [III8 , 6.12]. Thus it suffices to produce y. By Lemmas 5.11 and 11.5, K ∈ Spor. Suppose that K ∈ Alt. Then since Z2 ∼ = Σn as = Z(I) ≤ Z(K), K ∼ = 2An for some n ≥ 7. Writing x ∈ Aut(K) ∼

234

11. PROPERTIES OF K-GROUPS

x = (13) or x = (12)(34) · · · and letting y ∈ K map on (13)(24), we find by using [IA , 5.2.4df] that y x = yz, as required. Therefore we may assume that K ∈ Chev. Suppose that K ∈ Chev(2) − Alt. Then I ∼ = SL2 (q) ∈ Chev(2) by [IA , 4.9.6], so q = 5, 7, or 9, by [IA , 2.2.10]. By the Borel-Tits Theorem, x is a field, graph-field, or graph automorphism. Us± ing [IA , 4.9.1,4.9.2] we find that K/Z(K) ∼ = L2 (16), L± 3 (4), Sp4 (4), L4 (2), or ± L5 (2). As Z(K) = 1 and L4 (2) ∈ Alt, the only possibilities by [IA , 6.1.4] are K/Z(K) ∼ = U4 (2) ∼ = P Sp4 (3) and K/Z(K) ∼ = L3 (4). In the first case, K ∼ = Spin5 (3) and Aut(K) ∼ = SO5 (3), so the desired conclusion follows from Lemma 6.9. So assume that K/Z(K) ∼ = SL2 (5) or SL2 (7), x is a graph = L3 (4). According as I ∼  automorphism or a field automorphism. Let K = (4 × 4)L3 (4), the 2-saturated covering group of K in K2 . Then as x inverts Outdiag(K), x acts nontrivially on  (see [IA , 6.3.1]). As I is 2-saturated it has a preimage I ∼  Ω1 (Z(K)) = I, I ≤ K.  But  is the unique minimal normal subgroup of K  x, whence K ∼ Then Z(I) = K. then either y ∈ Ω1 (Z(K)) − Z(I) satisfies y x = yz, as required. Now we may assume that K ∈ Chev(r) − Chev(2), for some odd r. If x is a field automorphism, then the result follows from Lemma 6.3a. No graph-field automorphism has a centralizer with an SL2 (q) component. So we may assume that x ∈ Aut0 (K). We consult [IA , Tables 4.5.1,4.5.2] for such automorphisms x, i.e., such that in the universal version Ku and adjoint versions Ka we have corresponding components of CKu (x) and CKa (x) isomorphic ± ∼ to SL2 (q) and L2 (q), respectively. Keeping in mind that Spin± 6 (q) = SL4 (q) and ∼ Spin5 (q) = Sp4 (q) we see that in almost all cases, Ku is a spin group Spin(V ) of dimension at least 5 and x arises from an element of the corresponding orthogonal group O(V ); the desired result holds in these cases by Lemma 6.9. The only remaining cases are for K = D4+ (q), with x = t2 , t2 , γ2∗ , or γ2∗∗ . But these automorphisms are triality-conjugate to t2 and γ2 , for which the previous argument holds. This completes the proof.  Lemma 13.3. Let K ∈ K2 and x ∈ I2 (Aut(K)). Suppose that CK (x) has a subnormal subgroup I ∼ = SL2 (3), with Z(I) ≤ Z(K). Then x does not centralize any Sylow 2-subgroup of K. Proof. Lemma 13.1 restricts the possibilities for K, and in particular implies that we may assume that K ∈ Chev(2). Since Z(K) = 1, we only have the possibilities K = 2An , n ≥ 6, and K = d L(3) to consider. If K ∼ = 2An then the argument in the proof of Lemma 13.2 shows that if x ∈ Σn , then there exists a 2-element y ∈ K such that y x = yz, where z = Z(K). Hence x cannot centralize a Sylow 2-subgroup of K in this case. If x ∈ Σn , then n = 6 and x K/Z(K) ∈ P GL2 (9), with dihedral Sylow 2-subgroups, so x centralizes no Sylow 2-subgroup even of K/Z(K). Suppose then that K ∈ Chev(3) is of level 3. By assumption I ≤ K. From [IA , Table 4.5.1] we see, just as in the proof of Lemma 13.2, that since I = A1 (3) is the universal version in CKu (x) but the adjoint version in CKa (x) (where Ku and Ka are the universal and adjoint versions, respectively) the only possibilities occur for K = Spin± n (3) and x acting as an Aut(K)-conjugate of an involution of

13. PUMPUPS

235

SOn± (3). In those cases x centralizes no Sylow 2-subgroup of K, by Lemma 6.9. The proof is complete.  Lemma 13.4. Let K ∈ K2 and x ∈ I2 (K) − Z(K). Suppose that there is L   CK (x) such that x ∈ L ∼ = SL2 (q), q odd, q > 3. Write q = r a for an odd prime r and some positive integer a. Then K ∈ Chev(r) and one of the following holds: (a) K ∼ = d L(q) for some d and L; (b) K ∼ = 3 D4 (q 1/3 ) and L is uniquely determined (as a subgroup) by K and x. In both cases f (K) ≥ q 4 and K ∈ Gi2 , i = 6 or 7. Proof. Without loss we may pass to K/Z(K) and assume that K is simple. If K ∈ Chev(r), then as x ∈ L induces an inner automorphism on K, we see from [IA , Table 4.5.1] that the only way L can be a nonadjoint version of A1 (q) 1/2 is for (a) or (b) to hold. (Note that in the case K ∼ ), that table records = A± 3 (q that Z(L) = 1.) If K ∈ Chev(p) for some prime p, then as E(CK (x)) = 1, p > 2 by the Borel-Tits Theorem [IA , 3.1.4]. But then L2 (q) ∈ Chev(p) and so p = r by [IA , 2.2.10]. If K ∈ Alt, then L ∼ = Am for some m, which is absurd as Z(L) = 1. Likewise the possibility K ∈ Spor is ruled out by inspection of centralizers of inner automorphisms in [IA , 5.3]. It is obvious that K ∼  A1 (r b ) for any b. Thus in (a), the rank of L is at least 2. = We find that in (a) and (b), f (K) ≥ q 4 and f (K) = q 16/3 respectively, so f (K) ≥ q 4 in all cases. Finally suppose that K ∈ G62 ∪ G72 . Then K ∈ C2 ∪ T2 , and we conclude from the definition of these sets [I2 , 12.1,13.1] that K ∼ = A1 (r b ) for some b ≥ 0, ± ∼ which we have seen is impossible, or K/Z(K) = L3 (3), L± 4 (3), or G2 (3). In those cases again [IA , Table 4.5.1] shows that no A1 (3b ) component of the centralizer of an inner involution exists, contrary to assumption. The proof is complete.  Lemma 13.5. Suppose that K ∈ C2 ∪ T2 ∪ G72 ∪ G82 and x ∈ I2 (Aut(K)). Suppose that CK (x) has a 2-component I such that I/O2 (I) ∼ = SL2 (q) for some odd q > 3. Then one of the following holds: (a) K ∼ = SL2 (q 2 ) or L± 3 (q); or (b) K/Z(K) ∼ = An , n ≥ 7, or L3 (4), with Z(K) of exponent divisible by 2 or 4, respectively. Proof. Let z be an involution of I. By Lemma 5.11, K ∈ Spor. If K ∈ Alt, then since Z ∗ (I) > O2 (I), we must have K ∼ = 2An for some n, and then L ∼ = 2Am , m ≤ n − 2. As q > 3, m ≥ 5, so n ≥ 7 and (b) holds. If K ∈ Chev(2), then q = 5, ± 2 7, or 9, and with [IA , 4.9.1,4.9.2] we see that K/Z(K) ∼ = L± 3 (4), L2 (4 ), L4 (2), or Sp4 (4). But by Lemma 11.3, z ∈ Z(K), so K has a nontrivial exceptional Schur multiplier in [IA , 6.1.4], which narrows the list to L3 (4) and L± 4 (2). But 2L4 (2) ∼ = 2A8 , and 2U4 (2) ∼ = Sp4 (3) ∈ G62 , the latter case therefore not satisfying our hypothesis. If K/Z(K) ∼ = L3 (4) and the exponent of Z(K) is not divisible by 4, then all involutions of K/Z(K) are 2-central and pull back to involutions in K [IA , 6.4.1], and so K contains no quaternion group whose center lies in Z(K). This contradicts the existence of I. We may therefore assume that K ∈ Chev(r)−Chev(2)−Alt for some odd prime r. If K ∈ C2 , then K ∼ = L3 (3), G2 (3), or a quotient of L± 4 (3) by a nonidentity central

236

11. PROPERTIES OF K-GROUPS

subgroup. Using [IA , Tables 4.5.1,4.5.2] we see that there is no A1 (q) component L in the first two cases, and in the last case L ∼ = L2 (9), not SL2 (9). Therefore K ∈ C2 . If K ∈ T2 , then K ∼ = SL2 (q1 ) for some q1 , and then obviously q1 = q 2 , and (a) holds. Since K ∈ Alt, we are reduced by our assumptions to the case ∼ K∼ = L± 3 (q1 ) for some q1 . Since L = SL2 (q), this forces q1 = q, so (a) again holds and the proof is complete.  Lemma 13.6. Let K ∈ K2 with K   X and O2 (X) = 1. Suppose that a ∈ I2 (X), and CK (a) has a subnormal subgroup K1 such that a ∈ K1 and  K1 /O2 (K1 ) ∼ = SL2 (q 2 ), q odd. If Y   CX (a) with Y ∼ = SL2 (q), then [O 2 (Y ), K] = 1. 

Proof. Suppose that [O 2 (Y ), K] = 1. By Lemma 13.1 and L2 -balance, 2 O (Y ) ≤ K. Clearly, a ∈ Z(K). By Lemma 5.11, K ∈ Spor. If K ∈ Alt then since Z ∗ (K1 ) contains the involution a, we have a ∈ Z(K), contradiction. If K ∈ Chev(2) then Lemma 11.3 again yields the contradiction a ∈ Z(K). So K ∈ Chev(r) − Chev(2) for some odd r. By L2 -balance and Lemmas 13.1 and 10.2, Y ≤ K. Now searching [IA , Tables  4.5.1,4.5.2] for an involution a with O r (CK (a)) having subnormal SL2 (q 2 ) and SL2 (q) subgroups, we find only K ∼ = B3 (q), a = t2 , or K = D4− (q), a = t2 ; and in both cases Z(K1 ) ≤ Z(K). Therefore again a ∈ Z(K), a final contradiction. The proof is complete.  Lemma 13.7. Let q be an odd prime power, q > 3, and define S to be the set of all groups in K2 isomorphic to SL2 (q1 ), q1 odd, q1 ≥ q. Suppose that K1 ∈ S and H ∈ S, with K1 ↑2 H. Then H ∈ (G2 − G82 ) ∪ {2An | n ≥ 7} ∪ {(X)L3 (4) | X has exponent 4}. Proof. By definition S ⊆ T2 . By Lemma 1.1c, it follows that H ∈ T2 ∪ G2 . If H ∈ G82 , then H ∼ = An−2k = An , n = 9 or 10, so K1 ↑2 H implies that K1 ∼ for some k > 0, contradicting K1 ∈ S. And if H ∈ T2 but H is not of one of the asserted isomorphism types, then H ∼ = A7 ; thus H has = L2 (r) for some odd r or H ∼ dihedral Sylow 2-subgroups and cannot contain a copy of K1 with its quaternion Sylow 2-subgroup. The proof is complete.  Lemma 13.8. Suppose that K ∼ = SL2 (r a ), r odd, and r a ∈ FM9. If I ∈ ↑2 (K), then I ∈ Chev(r). Moreover either I ∼ = SL2 (r 2a ) or I ∈ Gd2 , d = 6 or 7. Proof. Since r a ∈ {5, 7, 9}, and r a is odd, K ∈ Alt ∪ Chev(2), and thus I ∈ Alt∪Chev(2). By Lemma 5.12, I ∈ Spor. As there are no isomorphisms between Chevalley groups of different odd characteristics, I ∈ Chev(r). Since K ↑2 I, I ∼ = L2 (r n ) for any n. Suppose that I ∈ C2 − Chev(2). Then r = 3 and I/Z(I) ∼ = L3 (3), L± 4 (3), or G2 (3), and from [IA , Table 4.5.2] the only possibility is r a = 9, which is assumed not to be the case. Likewise if I ∈ T2 , then it is immediate from the definition of T2 that I ∼ = SL2 (q) for some odd q, and q = r 2a is the only possibility. Finally, assume that I ∈ G2 . As I ∈ Alt ∪ Spor, the definition of the sets Gd2 shows that  I ∈ G62 or G72 , proving the lemma. Lemma 13.9. Suppose K ∈ Chev(r) ∩ K2 for some odd prime r, and x ∈ I2 (Aut(K)) − Inn(K), with CK (x) having a component J1 /O2 (J1 ) ∼ = SL2 (q1 ) for

13. PUMPUPS

237

some q1 ≥ 5. Let J1 , . . . , Js be all the components of CK (x) isomorphic modulo core to some SL2 (q), q ≥ 5. Then the following conditions hold: (a) s ≤ 2, and if equality holds then J1 J2 = J1 ∗ J2 ; (b) O2 (Z(J1 · · · Js )) ≤ Z(K); (c) K is isomorphic to one of the following groups: SL2 (q 2 ), q ≥ 5, q odd; + Spin± n (q), n ≥ 5, q odd; or a half-spin group HSpin4m (q), q odd. Proof. If x ∈ Aut0 (K), then x is a field or graph-field automorphism and K ∼ = SL2 (q 2 ) [IA , 4.9.1,4.9.2]. So assume that x ∈ Aut0 (K) − Inn(K), in which case the structure of CK (x) is tabulated in [IA , Tables Tables 4.5.1,4.5.2]. The rest of the proof makes detailed use of these tables, which we use without further comment. We first observe that as x ∈ Inn(K), when Ka is the adjoint version of K, then CKa (x) has no universal A1 -type component, i.e., no SL2 component. This is equivalent to assertion (b). Let Ku be the universal version of K. Next we see that there are just three cases in which CKu (x) has more than one universal component of type A1 : these +  are K of type A± 3 , or B3 with x of class γ2 or t2 , or K of type D4 with x of     class t2 , t2 , t2 . In all these cases CKu (x) is the central product of exactly two (universal) components of type A1 . This yields (a). In the A± 3 and B3 cases, Z(Ku ) is cyclic so for CK (x) to have universal components we must have K = Ku , + so that K ∼ = Spin± 6 (q) or Spin7 (q). In the D4 case we must have Z(K) = 1, and so because of triality-induced automorphisms, K is isomorphic to Spin+ 8 (q) or a half-spin group. Thus (c) holds in these multiple-component cases. It remains to check (c) in those cases where CKu (x) has a single SL2 -component.  Again K is of type A, B or D. Namely K is of type A± 3 with t of class t2 , type   B2 with x of class t1 or t1 or t2 , depending on the residue of q modulo 4, or type ± Dm , m ≥ 4 (but not D4+ (q)), with x of class t2 . In the first two cases again, Z(Ku ) is cyclic and so K = Ku ∼ = Spin5 (q) or Spin± 6 (q). In the last case, now using [IA , Table 4.5.3] as well, we see that K cannot be a quotient of Ω± 2m (q), so K must be a spin or half-spin group, the latter existing only when m is even.  Lemma 13.10. Suppose that K ∈ K2 , y ∈ I2 (Aut(K)), L   CK (y), and L∼ = L2 (q), q odd. Assume also that (a) CK (y) has elementary abelian Sylow 2-subgroups; and n  J1 (q = 5). (b) K ∼ = 2 G2 (3 2 ) (q = 3n ) for any odd n, and K ∼ = ± ∼ Then q = 5. Moreover K/Z(K) = L3 (4) or M12 , or K ∼ = L2 (24 ) or J2 . In all but the J2 case, y ∈ Inn(K). Proof. First suppose that K ∈ Chev(r) for some odd r. By assumption (b), n K ∼ = 2 G2 (3 2 ) for any n ≥ 1. Note that as CK (y) has elementary abelian Sylow 2subgroups, r = ±3 (mod 8) and every component of CK (y) is of the form L2 (r a ) for some odd a. In particular, q is not a square, and every component of CK/Z(K) (y) likewise has the form L2 (r a ). Notice that y is neither a field nor a graph-field automorphism, since if it were, then K/Z(K) ∼ = L2 (q 2 ), and according as Z(K) = 1 or Z(K) = 1, CK (y) would contain P GL2 (q) or SL2 (q), and would have nonabelian Sylow 2-subgroups, contradiction. Therefore the pair K/Z(K), y must appear in [IA , Table 4.5.1]. Keeping in mind that the “c” version of D2 (q) is the central product SL2 (q) ∗ SL2 (q), and that 2 D2 (q) ∼ = L2 (q 2 ), and that B2 (q) ∼ = C2 (q) we find the only possibilities to be K = B2 (q) with y conjugate to t1 or t1 in the table,

238

11. PROPERTIES OF K-GROUPS

± K = A± 2 (q) with y conjugate to γ1 , and K = A3 (q) with y conjugate to γ2 . In the ± K = A2 (q) case, | Inndiag(K) : K| = 3 and so CK (y), like CInndiag(K) (y), contains u P GL2 (q), with nonabelian Sylow 2-subgroups, a contradiction. In the K = A± 3 (q) case, [IA , Table 4.5.2] shows that CK (y) has nonabelian Sylow 2-subgroups. In the ± a K = A± 3 (q) and K = Ω6 (q) cases, we have |Z(K)| ≤ 2, so a Sylow 2-subgroup of CInn(K) (y) must have an abelian subgroup of index 2, and a similar argument yields a contradiction. We have L × L1  CInn(K) (y) with L ∼ = L1 ∼ = L2 (q). Moreover, CInndiag(K) (y) contains an involution interchanging the two components of CK (y), and CInndiag(K) (y) covers the four-group Outdiag(L)×Outdiag(L1 ). Hence, a Sylow 2-subgroup of CInndiag(K) (y) has no abelian subgroup of index 4. As y inverts Outdiag(K), |CInndiag(K) (y) : CInn(K) (y)| ≤ 2. Therefore a Sylow 2-subgroup of CInn(K) (y) has no abelian subgroup of index 2, a contradiction. Consider finally the K = B2 (q) case. If K is the universal version then L ∼ = SL2 (q) by [IA , Table 4.5.2], contradicting the fact that CK (y) has abelian Sylow 2-subgroups. So assume that K is the adjoint version, whence CInndiag(K) (y) ∼ = P GL2 (q)×D2(q−) ,  = ±1, has nonabelian Sylow 2-subgroups, by [IA , Table 4.5.1]. If y is non-inner then y covers Outdiag(K), so CInndiag(K) (y) = y × CInn(K) (y) must have abelian Sylow 2-subgroups, contradiction. Thus y ∈ Inn(K), so by [IA , Table 4.5.1], q ≡  (mod 4). Then the dihedral direct factor D2(q−) has nonabelian Sylow 2-subgroups, so CInndiag(K) (y) has no subgroup of index 2 with abelian Sylow 2-subgroups, again a contradiction. Therefore K ∈ Alt ∪ Spor ∪ Chev(2). If K ∈ Spor, the assertions follow directly  J1 . So K, and therefore from the tables [IA , 5.3] and our assumption that K ∼ = L, lies in Alt ∪ Chev(2). As L ∼ = L2 (q), q ≡ ±3 (mod 8), we must have q = 5. If K ∼ = 2Am for some m, contradiction; while if K ∼ = An , then = 2An , then L ∼ n = 7 + 2k, k ≥ 0, and CK (y) involves Σ5 , with nonabelian Sylow 2-subgroups, contradiction. Therefore K ∈ Chev(2). By the Borel-Tits Theorem and [IA , 4.9.1, 4.9.2], K ∼ = L3 (4). and y ∈ Inn(K). The proof is = L2 (42 ) or U3 (4) or K/Z(K) ∼ complete. 

Lemma 13.11. Suppose that I ↑2 J via x, where x ∈ I ∼ = Spin4k (q), k ≥ 2. Then J ∈ Chev and q(J) = q. Proof. Since k ≥ 2, I is clearly unambiguously in Chev(r), where q is a power of the prime r. Therefore J ∈ Chev(r), and the pair (I, J) can be found in [IA , Table 4.5.1]. We have x ∈ I ≤ J. In particular x induces an inner automorphism on J. Examination of [IA , Table 4.5.1] shows that if q(J) = q, then x does not induce an inner automorphism on J, contradiction. The lemma is proved.  Lemma 13.12. Let K ∈ K2 and let x ∈ I2 (Aut(K)). Suppose that among the components of E(CK (x)) are two isomorphic covering groups of Am for some m ≥ 5. Then K ∈ G2 , m = 5 or 6, and accordingly K ∈ Chev(5) or K ∈ Chev(3). Proof. If K ∈ Alt then E(CK (x)) consists of at most one component, so K ∈ Alt. The same holds if K ∈ Chev(2), by the Borel-Tits Theorem and the structure of non-inner involutory automorphisms [IA , 4.9.1,4.9.2]. The hypotheses do not hold for any K ∈ Spor, by inspection of the tables [IA , 5.3]. Therefore K ∈ Chev(r) − Chev(2) − Alt for some odd r, and by [IA , 4.9.6], Am ∈ Chev(r). Therefore m = 5 or 6, and correspondingly r = 5 or 3. It remains to show that K ∈ G2 . But if K ∈ Chev(r) − G2 , then by [I2 , 12.1,13.1], K/Z(K) ∼ = L2 (r m ), or

13. PUMPUPS

239

± r = 3 with K/Z(K) ∼ = L± 3 (3), L4 (3), or G2 (3). Using [IA , Table 4.5.1] we see that in no case can E(CK (E)) have the assumed structure. The proof is complete. 

Lemma 13.13. Let K ∈ K2 and let x, y be a four-subgroup of Aut(K). Let K0 be a component of CK (x) and let K1 and K2 be distinct components of CK (y). Assume that K1x = K2 and that K0 = E(CK1 K2 (x)). Suppose further that K0 /O2 (Z(K0 )) ∼ = Am , m ≥ 5. Then (a) m = 5 or 6; (b) K/Z(K) ∼ = P Ωr (q) for some r ≥ 5 and sign . Moreover, q = 5 or 9, or ∼ K/Z(K) = P Ω+ 8 (3). Proof. First, K ∈ Chev(2) ∪ Alt ∪ Spor, since centralizers of involutory automorphisms of groups in these sets have at most one component (see [IA , 5.2.9] for Alt, [IA , Tables 5.3] for Spor, and the Borel-Tits theorem and [IA , 4.9.1,4.9.2] for Chev(2)). Thus K ∈ Chev(p) for some odd p. As K0 ∈ Chev(p) ∩ Alt, we immediately have m = 5 or 6. Furthermore, K0 /O2 (K0 ) ∼ = L2 (pa ) for pa = 5 or 9. Thus (a) holds, and to prove (b) it remains to show that K/Z(K) ∼ = P Ωr (pa ) for some r, or P Ω+ 8 (3). Consulting the table [IA , Table 4.5.1] for groups and an involution centralizer containing an adjoint A1 (pa ) component, as well as a second involution centralizer (perhaps the same as the first) with two (isomorphic) A1 (pa ) components, we see immediately that K cannot be of exceptional type, or of type Cn , n ≥ 3. Furthermore if K ∼ = Bn (q), n ≥ 2, or Dn (q), n ≥ 4, then with one exception the pair (K1 , K2 ) of components of CK (x) arises as an entry D2 (q) in the table, so that q = 5 or 9. The exception is K ∼ = D4 (q), where K1 and K2 can be of type A1 (q 2 )    (and x of class t2 , t2 , or t2 ), so q = 3. It remains to consider the case K ∼ = An (q). Again from [IA , Table 4.5.1], the structure of CK (x) forces n = 3 and x of class t2 or γ2 . As A3 (q) = D3 (q),  K/Z(K) ∼ = P Ω6 (q), q = 5 or 9, and the lemma follows. Lemma 13.14. Suppose that q is odd and J ↑2 K, where J = L3 (q),  = ±1. Then one of the following holds: (a) (b) (c) (d)

K ∈ Chev and K has level q(K) = q; K∼ = L3 (q 2 ); K ∈ G62 and f (K) > q 9 ; or  = −1 and q = 3, and K ∼ = G2 (4).

Proof. Since J ∈ Alt, K ∈ Alt. Let q = r a with r prime. If K ∈ Chev(r), then using [IA , Tables 5.3] and [IA , 2.2.10] we see that J ∼ = U3 (3) ∼ = [G2 (2), G2 (2)], with K ∈ Chev(2), indeed K ∼ = G2 (4) with the help of [IA , 4.9.1,4.9.2] and the Borel-Tits Theorem. Thus (d) holds in this case. We may then assume that K ∈ Chev(r). Say J ↑2 K via x ∈ I2 (Aut(K)). If x is a field or graph-field automorphism, then (b) holds, by [IA , 4.9.1]. By inspection of [IA , Table 4.5.1], x cannot be a graph automorphism. Thus, assume x ∈ Inndiag(K). By [IA , 4.2.2], q(K) = q 1/b for some integer b, such that b disjoint copies of the A2 Dynkin diagram can be obtained by deleting a node from the Dynkin diagram or extended Dynkin diagram of L, according as x is of parabolic or equal-rank type. Moreover, the deleted node must be marked as having coefficient 1 or 2, respectively, in the highest root. By (a) we may assume that b > 1. The

240

11. PROPERTIES OF K-GROUPS

1/2 only possibility is b = 2, K ∼ ), in which case (c) holds with f (K) = q 25/2 . = A± 5 (q The proof is complete. 

Lemma 13.15. Suppose that L ↑2 K, and suppose that if L and K both lie in Chev(r) for some odd r, then the levels q(L) and q(K) are not equal. Then the following conditions hold: 1/2 ) with  = 1 or (a) If L ∼ = SL± = SL4 (q), q odd,  = ±1, then K ∼ 8 (q 2 K∼ = SL4 (q ), and in either case Ω1 (Z(K)) = Ω1 (Z(L)); (b) If L ∼ = Sp4 (q), q odd, then K ∼ = Sp8 (q 1/2 ) or K ∼ = Sp4 (q 2 ), and in either case Ω1 (Z(K)) = Ω1 (Z(L)). Proof. Say L ↑2 K via x ∈ I2 (Aut(K)). If x is a field or graph-field automorphism, then we see from [IA , 4.9.1] that K ∼ = SL4 (q 2 ) or Sp4 (q 2 ) in (a) or (b), respectively, and it is clear that Ω1 (Z(L)) = Ω1 (Z(K)) in these cases. So we may assume that x is an inner-diagonal or graph automorphism, and use the tables [IA , Tables 4.5.1,4.5.2]. In our situation L = SL4 (q) or Sp4 (q) is a non-level subcomponent and is the universal version. Inspection of the tables shows that 1/2 ) and K = Sp8 (q 1/2 ) are possible. Since K/Ω1 (Z(K)) does only K = SL± 8 (q not satisfy the hypotheses, it must be that Ω1 (Z(L)) ≤ Ω1 (Z(K)). The lemma follows.  Lemma 13.16. Suppose that p is a prime, I, K ∈ Kp , and I ↑p K. Let D = Chev − Chev(p) − Alt. If I ∈ D, then K ∈ D unless (p, K, I) = (3, M24 , L3 (2)), (2, J3 , L2 (17)), (3, He, L3 (2)), (7, F2 , L3 (4)) or (13, F1 , L3 (3)). If p > 2, then in all the exceptional cases, mp (I) = 1. Proof. For p > 2, this is immediate from Lemma 5.12. Suppose that p = 2. If K ∈ Chev(2) ∪ Alt, then I ∈ Chev(2) ∪ Alt by [IA , 4.9.6,5.2.9], contrary to assumption. Thus if K ∈ Chev, then K ∈ D, as asserted. We may therefore assume that K ∈ Spor. One checks the tables [IA , 5.3] to see that the only example  is K ∼ = J3 . Lemma 13.17. Suppose that K ∈ K2 and I ∈ ↑2 (K). (a) If K ∈ Spor ∩ G2 , then K ∼ = J1 and I ∼ = O  N ; and ∼ (b) If K/Z(K) = L3 (4) with Z(K) not elementary abelian, then I ∼ = O N ∼ and Z(K) = Z4 . Proof. By definition Spor∩G2 = {J1 , Co3 , He, M c, Ly, O  N, F5 }. Since K ↑2 I and K ∈ Chev ∪ Alt, I ∈ Chev ∪ Alt [IA , 5.2.6,5.2.8,4.9.6]. Thus I ∈ Spor in (a). In (b), we have L3 (4) ∈ Chev(r) ∪ Alt for any odd r, so we must have I ∈ Chev(2) ∪ Spor. As Z(K) = 1, we can choose z of order 2 in Z(K). Then if K ↑2 I via x, we have 1 = L2 (CI (x, z)) ≤ L2 (CI (z)) = 1 by L2 -balance and the Borel-Tits Theorem. This contradiction shows that I ∈ Spor. Now I ∈ Spor in both cases, and inspection of the tables [IA , 5.3] yields the lemma.  Lemma 13.18. Suppose that J, K ∈ K2 , K ↑2 J via x, K ∼ = A7 , and m2 (CInn(J)x (K x)) = 1. Then J ∼ = A9 or He. Proof. As K ∈ Chev, J ∈ Chev. If J ∈ Spor then the tables [IA , 5.3] show that J ∼ = He is the only possibility. So we may assume that K ∈ Alt, whence J ∼ = A7+2k , k > 0. If k > 1 then CInn(J) (K) contains a four-group so m2 (CInn(J)x (K x) > 1, a contradiction. Thus k = 1, completing the proof. 

13. PUMPUPS

241

In the next three lemmas we assume the following situation. (1) J, K ∈ K2 , and K ↑2 J via t ∈ I2 (Aut(J)); (2) K ∼ = An for some n ≥ 5;  F5 ; (3) J ∼ (13B) = = XL3 (4) with X of exponent 4, and J ∼ (4) If J ∈ G2 , then d2 (J) ≥ d2 (An+4 ); and (5) If n = 6, then AutCJ (t) (K) ∼ = A6 or Σ6 . Lemma 13.19. Assume (13B). Then one of the following holds: (a) J ∼ = An+2k for some k > 0; (b) n = 8 and J ∼ = L4 (4), HS, or 2HS; ∼ (c) n = 6 and J ∼ = Sp4 (4), L± 5 (2), P Sp4 (3), L4 (3), 2HS, or J = U4 (3) or 2U4 (3); or (d) n = 5 and J ∼ = M12 , 2M12 , J2 , 2J2 , or U3 (4), L2 (16), L2 (52 ), or J/Ω1 (O2 (Z(J))) ∼ = L3 (4). Proof. We use Lemma 1.6. It implies that if J ∈ Alt ∪ Spor ∪ Chev(2), then J is as in this lemma, unless (n, J) = (10, F5 ), (7, He), (6, HS), or (5, J1 ). By assumption J ∼  F5 . Also d2 (A11 ) > d2 (He), and d2 (A9 ) > d2 (J1 ), so the second = and fourth possibilities are ruled out. If J ∼ = HS with n = 6, then AutCJ (t) (K) = Aut(A6 ) ≤ Σ6 , contrary to hypothesis. Thus we may assume that J ∈ Chev(r) for some odd r, and Lemma 1.6 implies that we may assume that r = n = 5 or r = 3, n = 6. In particular if J ∈ G2 , then d2 (J) ≥ d2 (An+4 ) = 8, a contradiction as J ∈ Chev. So J ∈ C2 ∪ T2 . If r = 3, then n = 6 and K ∼ = L2 (9). The only choices, in view of [IA , Table 4.5.1,4.9.1,4.9.2], are J ∼ = L± 4 (3), 2U4 (3), P Sp4 (3), as asserted, and L2 (81). In the last case, however, AutCJ (t) (K) ∼ = P GL2 (9), contrary to assumption. Similarly, if r = 5 then n = 5, K ∼ = L2 (5), and the only choice is J ∼ = L2 (52 ). The proof is complete.  Lemma 13.20. Assume (13B). Assume also that another quasisimple group J0 satisfies the same conditions as J (for the same value of n), and furthermore J ↑2 J0 . Then (J/Z(J), J0 /Z(J0 )) is one of the following pairs: (An+2k , An+2 ),  > k > 0; (A8 , HS) (with n = 6); or (P Sp4 (3), L± 4 (3)) (with n = 6). Proof. If both J and J0 lie in Alt, then the first conclusion holds. Suppose next that J ∈ Alt but J0 ∈ Alt. By Lemma 13.19, the only possibility is for K∼ = A6 and J ∼ = A8 , and for J0 to appear in both Lemma 13.19b and 13.19c. Thus J0 ∼ = 2HS with J ∼ = A8 and the lemma holds in this case. We may therefore assume that J ∈ Alt, whence J0 ∈ Alt, and J and J0 both appear in the same part ((b), (c), or (d)) of Lemma 13.19. By considering the possibilities for J0 case by case, and using [IA , Table 4.5.1, 4.9.1, 4.9.2, Tables 5.3] to determine the possible subcomponents J of J0 , we see that the only choice is ∼  J0 ∼ = L± 4 (3) or 2U4 (3), with J = P Sp4 (3), as required. Lemma 13.21. Assume (13B). Let ξ ∈ I2 (Aut(J)) with K0   CJ (x) and K0 /O2 (Z(K0 )) ∼ = K. Set C = CAut(J) (K0 ξ). Then one of the following holds: (a) |C| = 2; (b) C ∼ = A6 ), or M12 or J2 (K0 ∼ = = HS or U4 (3) (K0 ∼ = E22 , with J/Z(J) ∼ A5 ); (c) C is embeddable in D8 , with J ∼ = L4 (3) (K0 ∼ = A6 ); or (d) C is embeddable in Σ2k , with J ∼ = An+2k .

242

11. PROPERTIES OF K-GROUPS

Proof. We use Lemma 13.19. If J ∼ = An+2k , k > 0, then from [IA , 5.2.8a] we see that K0 is a root An subgroup of J, so CAut(J) (K0 ) ∼ = Σ2k and (d) holds. If J ∈ Spor and |C| > 2, then C is a four-group, by Notes 1, 1, and 2 in [IA , Tables 5.3bgm], respectively, so (b) holds. If J ∈ Chev(2), we argue that |C| = 2. Now ξ is a field, graph-field, or graph automorphism of J. In the first two cases, |C| = 2 by [IA , 7.1.4]. In the graph automorphism case, C ∩ Inndiag(J) = 1 by [IA , 4.9.2bc]. This implies that |C| = 2 except possibly in the cases J ∼ = L3 (4). In the latter case since = U3 (4) or J/Z(J) ∼ Out(J)/ Outdiag(J) is elementary abelian, it suffices to show that ξ is the unique involution in CAut(J) (K0 ). But the centralizer of any outer involution not in the Inndiag(J)-coset of ξ is a 5 -group, by [IA , 4.9.1], so we are done in this case. If J∼ = U3 (4), then let (J, σ) be a σ-setup of J. Then with standard notation we may assume that σ = τ σ4 , where σ4 is a field automorphism of J fixing the field of four elements only. Then ξ acts like σ4 on J, and σ2  covers Out(J) ∼ = Z4 , with σ2 acting on K0 and J as a field automorphism fixing the field of two elements. Therefore CAutJ (ξ)/ ξ ∼ = Σ5 and so |C| = 2 in this case as well. 2 (5 ), then |C| = 2 by [IA , 7.1.4]. Finally, assume that n = 6 and If J ∼ L = 2 ± J/Z(J) ∼ L (3). One possibility is that ξ is a graph automorphism of J. But = 4 in that case ξ covers Out(J)/ Outdiag(J), and it follows from [IA , Table 4.5.1] that |C| = 2. We may therefore assume that ξ ∈ Inndiag(J). We identify Aut(J) with P GO6± (3); then ξ acts as an involution of O6± (3) whose −1-eigenspace is 4dimensional of − type. Therefore CAut(J) (K0 ) ∼ = O2∓ (3), which proves that C and J are as in (b) or (c). This completes the proof.  Lemma 13.22. Suppose that the following conditions hold: (a) J ≤ L ≤ I, where I is simple; (b) J ∼ = Am+1 for m = 5, 6, or 8; (c) J ↑2 I via x ∈ I2 (Aut(I)), and |CInn(I)x (J)|2 = 2; (d) AutCI (x) (J) ∼ = J or Σm+1 ; and (e) L and m are as follows: for m = 5, L ∼ = M12 , J2 , L2 (24 ), L2 (52 ), or ∼ U3 (4), or L/Z(L) = L3 (4) with O2 (Z(L)) elementary abelian; for m = 6, ∼ L∼ = Sp4 (4), L± 5 (2), or P Sp4 (3); for m = 8, L = L4 (4), HS, or 2HS. Then m = 5, I ∼ = U4 (3), and L ∼ = L3 (4). Proof. The condition J ≤ L and Lagrange’s Theorem rule out all cases in (e) except (m, L) = (5, M12 ), (5, J2 ), (5, XL3 (4)), (6, L5 (2)), (8, L4 (4)). If m = 8 then J ∼ = A11 , which does not contain L4 (4), by = A9 , so condition (c) yields I ∼ Lagrange’s Theorem. Likewise if m = 6 then J ∼ = A7 , so I ∼ = A9 or He, neither of whose orders are divisible by |L5 (2)|. So m = 5. By 3-fusion J ∼ = A6 is not contained in J2 , since Sylow 3-subgroups of J2 are isomorphic to 31+2 and contain a weakly closed subgroup of order 3, whereas the E32 -Sylow subgroups of A6 have no such weakly closed subgroup. Since J ∼ = A6 or Σ6 by (d), I ∼ = A 8 , L± = A6 and AutCI (x) (J) ∼ 5 (2), or Sp4 (4), or I ∈ Chev(3), by Lemma 1.6. In the last case, using conditions (c) and (d) and [IA , Table 4.5.1], we get I ∼ = P Sp4 (3) or L± 4 (3). Lagrange’s Theorem applied to L ≤ I leaves only the following possibilities for L ≤ I: M12 ≤ U5 (2), L3 (4) ≤ A8 , XL3 (4) ≤ L5 (2), L3 (4) ≤ U4 (3), 2L3 (4) ≤ U4 (3). Both M12 and L3 (4) contain Frobenius groups of order 9.8, so by [IG , 9.12] they do not embed in L5 (2) or U5 (2). Furthermore, L3 (4) does not embed in A8 ,

13. PUMPUPS

243

as they are nonisomorphic of equal orders. If XL3 (4) ≤ L5 (2) with X a nontrivial power of 2, then by the Borel-Tits Theorem, XL3 (4) would embed in a parabolic subgroup of L5 (2), and then L4 (2) ∼ = A8 would have to involve L3 (4), which is again impossible. By Lemma 6.12, centralizers of all involutions of U4 (3) are solvable, so an embedding 2L3 (4) ≤ U4 (3) is impossible. The only remaining possibility is  L3 (4) ≤ U4 (3), as asserted in the lemma. Lemma 13.23. The following conditions hold: (a) If n ≥ 9, then ↑2 (An ) = {An+2k | k > 0} or {An+2k | k > 0} ∪ {F5 }, the latter if and only if n = 10. (b) ↑2 (J2 ) = {Suz}. (c) ↑2 (M12 ) = {Suz, Co3 }. (d) ↑2 (2HS) = {F5 }. (e) ↑2 (F5 ) = ∅. (f) ↑2 (E7 (q)u ) = {E7 (q 2 ), E8 (q)} for any odd q. Proof. Part (a) follows from Lemma 1.6. Suppose that K ↑p J. Inspection of [IA , Tables 5.3] establishes (b), (c), (d), and (e). Finally in (f) it is clear from [IA , 4.9.1,4.9.2,2,5.2.8,Tables 5.3] and the Borel-Tits theorem that J ∈ Alt ∪ Spor ∪ Chev(2), and the assertion then follows from [IA , Table 4.5.1,4.9.1]. The proof is complete.  Lemma 13.24. Suppose that K ∈ Gip for some odd prime p. If K ↑p J, then J ∈ Gjp for some j ≤ i. Moreover if i = 2, then j = 2. Proof. The first statement holds by Lemma 1.2. If the second statement fails, then j = 1, so J ∼ = Am for = An and K ∈ Chev with mp (K) > 2. As K ↑p J, K ∼ some m. But then since K ∈ Alt ∩ Chev, it follows from [IA , 2.2.10] that m ≤ 8,  whence mp (K) ≤ 2, a contradiction completing the proof. Lemma 13.25. Suppose that K ∈ G2p ∪ G4p for some odd prime p, and K ↑p J. Then J ∈ G2p ∪ G4p . Moreover, if K ∈ Chev(2), then J ∈ Chev(2). Proof. By Lemma 1.2, J ∈ ∪4i=1 Gip . But G1p ∪ G3p ⊆ Alt. So if the lemma fails then J ∈ Alt, and so K ∈ Alt. Thus K ∈ Alt ∩ Chev, and mp (K) ≥ 3, by definition of G2p and G4p . There are no such groups K, by [IA , 2.2.10], so the first assertion follows. Similarly, if K ∈ Chev(2), then as mp (K) ≥ 3 and K ∈ Gp , K ∈ Chev(r) for any odd prime r (see [IA , 2.2.10]). Thus J ∈ Chev(r) and the lemma follows.  Lemma 13.26. The following conditions hold: (a) ↑3 (3A6 ) = {M24 , J2 , Ru, 3O  N }, ↑3 (3A7 ) = {He}, and ↑3 (3M22 ) = {J4 }; and (b) ↑3 (L) = ∅ for L ∈ {M24 , J4 , He, 3O  N, Ru}. Proof. Suppose that L ↑3 K with K, L ∈ K3 . If K ∈ Spor, then the assertions follow from [IA , Tables 5.3]. But if L ∈ Spor, then K ∈ Spor. So we may assume that L ∼ = 3A6 or 3A7 , and K ∈ Spor. If K ∈ Alt, then by [IA , 5.2.8], 3 divides |Z(K)|, so K ∼ = L, which is absurd. Thus we may assume that K ∈ Chev(r), with r = 2 or 3 and L ∼ = 3A6 . If r = 2, then O2 (K) ≤ O3 (K) = 1 as K ∈ K3 . It follows easily that K ∈ Lie(2), whence L is the commutator subgroup of a group in Lie(2), by [IA , 4.2.2]. But 3A6 = 3Sp4 (2) ∈ Lie(2), contradiction. If r = 3, then

244

11. PROPERTIES OF K-GROUPS

L must be the fixed points of a graph, field, or graph-field automorphism of order  3, so K ∼ = L2 (9), contradiction. The proof is complete. = L2 (36 ) and then L ∼ Lemma 13.27. The following conditions hold: (a) ↑3 (B3 ) ∩ T3 ⊆ B3 ∪ {G2 (8), J2 }; (b) ↑3 (3A6 ) ∩ Chev = ∅; (c) Define G5,6 = G53 ∪ G63 . Then ↑3 (B3 ) ∩ G5,6 = {G2 (q) | q > 2, q = 8, q = 3 3 √ a 3 3 } ∪ { D4 (q), 2 F4 ( q) | q > 2, q = 3a } ∪ {M24 , J4 , He, Ru, 3O  N }. All these groups are 3-saturated and contain 31+2 . In all of them except 3O  N , which has 31+4 Sylow 3-subgroups, the centralizer of a 3-central element of order 3 has a component J such that J/O3 (J) ∈ B3 ; (d) If A− 3 (q) ↓3 J ∈ K3 via an inner automorphism, q ≡  (mod 3),  = ±1, q > 2, then J ∈ {L2 (q), L2 (q 2 )}; (e) If A− 4 (q) ↓3 J ∈ K3 via an inner automorphism, q ≡  (mod 3),  = 2 ±1, q > 2, then J ∈ {L− 3 (q), L2 (q )}; (f) If B2 (q) ↓3 J ∈ K3 via an inner automorphism, q ≡ 0 (mod 3), q > 2, then J ∼ = L2 (q); and (g) Let L ∈ B3 and let L 2, q = 3a } ∪  {M24 , J2 , J4 , He, Ru, 3O N }. Proof. Part (b) is immediate from Lemma 13.26a. In (a), suppose that L ∈ B3 and L ↑3 K with K ∈ T3 − B3 . Since Z(L) = 1, m3 (L) > 1 and Sylow 3subgroups of L are nonabelian. Hence the same holds for Sylow 3-subgroups of K; also Z(K) = 1 as K ∈ B3 . From the definition of T3 and these conditions we then get K ∼ = L3 (q), q ≡  (mod 9), M12 , J2 , or G2 (8). In the first case, all components of centralizers of elements of I3 (Aut(K)) are SL2 (q) or L3 (q 1/3 ); in the M12 case, such centralizers have no components at all. Hence, (a) follows. In (c), let L ∈ B3 and K ∈ G5,6 3 with L ↑3 K, say via x ∈ I3 (Aut(K)). We have Z(L) = 1 and L ∈ T3 by definition of B3 . If K ∈ Alt, then L ∈ Alt so L ∼ = 3A6 or 3A7 . Then K ∼ = An for n ≥ 9, so we reach the contradiction Z(L) = 1. Thus  ∼ K ∈ Spor ∪Chev, and since K ∈ G5,6 3 , m3 (K) = 2 or K = 3O N . So we may assume that m3 (K) = 2. As in (a), K has nonabelian Sylow 3-subgroups. If K ∈ Spor, then the assertion of (c) follows directly from this last fact and the list of sporadic groups in G3 [I2 , p. 103]. So we may assume that K ∈ Chev ∩ G3 ⊆ Chev − Chev(3). Note that as L ∈ B3 , L contains a 31+2 -subgroup. (This can be found within a monomial subgroup for L ∼ = G2 (8), and in all other = SL3 (q), within SU3 (8) for L ∼ cases because for P ∈ Syl3 (L), |P | = 33 , |Z(P )| = 3, and NL (P ) acts irreducibly on P/Z(P ).) Therefore K contains such a subgroup, so it has one of the isomorphism types claimed in (c), by [IA , 4.10.3c]–and by the fact that K ∈ G3 , which gives the restrictions on q. The remaining statements in (c) about properties of K follow directly from [IA , Tables 4.7.3A,5.3]. Parts (d), (e), and (f) follow directly from [IA , 4.8.1, 4.8.2]. The proof of (g) is similar to that of (c). By Lemma 1.1c, K ∈ T3 ∪ G3 . We have Z(L) = 1 and as in the proof of (c), L contains a 31+2 subgroup, so K does as well. If K ∼ = An−3k for some k, contrary to Z(L) = 1. = An , then we must have L ∼ If K ∈ Spor, then as K ∈ T3 ∪ G3 , K is one of the groups in the conclusion or else K ∼ = M12 , M22 , M23 , or HS. But K cannot be isomorphic to M12 since, as above, it has no components in centralizers of automorphisms of order 3; and the

13. PUMPUPS

245

other three groups have Sylow 3-groups of order only 9. Thus again we may assume that K ∈ Chev − Chev(3), and we may assume that K ∈ T3 ∪ G53 − {3O  N }, so m3 (K) = 2. Again we appeal to [IA , 4.10.3c] to complete the proof.  Lemma 13.28. Let K ∈ K3 , with K a component of the K-group X. Let x ∈ I3 (NX (K)) with I := L3 (CK (x)) satisfying I/O3 (I) ∼ = L2 (q), q ≥ 8, q ≡  (mod 3),  = ±1. Suppose that m3 (CX (x)) ≤ 4, that Z(K) = 1, and that Sylow 3-centers of X are cyclic. Then K ∼ = SL3 (q). ∼ L2 (q), K ∈ K3 , and Z(K) = 1 rule Proof. The simultaneous conditions I = out the possibility K ∈ Alt ∪ Spor (see [IA , 6.1.4,Tables 5.3]). If K ∈ Chev(3), then 3 I ∈ Chev(3), and the only possibility is q = 8, K ∼ = 2 G2 (3 2 ); but then Z(K) = 1 by [IA , 6.1.4], contradiction. Thus K ∈ Chev(r), r = 3, and as Z(K) = 1 we η have K/Z(K) ∼ = L3k (q0 ) with q0 ≡ η (mod 3), η = ±1. (The only other case with η Z(K) = 1, namely K ∼ = E6 (q0 ), gives the wrong structure for I = L3 (CK (x)) by [IA , Table 4.7.3A].) Using the structure of CK (x) and [IA , 4.8.2,4.8.4] we see that either the conclusion of the lemma holds, or K/Z(K) ∼ = L6 (q 1/3 ). But in the latter case, we let P ∈ Syl3 (X) be such that NP (K) ∈ Syl3 (NX (K)). Then P ∩ K contains P0 = P1 × P2 with index 3, where P1 and P2 each centralize a 3-dimensional subspace of the natural SL6 (q 1/3 )-module, and so P1 and P2 are normal in NP (K). Thus Z(P ) ∩ K ≥ Z(P1 ) × Z(P2 ) and it follows easily that  Z(P ) ∩ K P is noncyclic, contrary to assumption. The lemma follows. Lemma 13.29. 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 I ∼ = SL3 (2n ) for some n, where 2n ≡  (mod 3) and  = ±1. Assume that Z(I) ≤ Z(X). Then x induces a possibly trivial field automorphism on K. Proof. We have Z(I) ≤ K ∩ Z(X) ≤ Z(K). Also we may assume that x acts nontrivially on K, whence m3 (x I/Z(I)) = 3. If K ∈ Chev(3), then by [IA , 6.1.4], K/Z(K) ∼ = G2 (3), U4 (3), or Ω7 (3), so |K|3 > 36 , contradiction. Thus, K ∈ Chev(3). If K ∈ Chev(r), r = 3, then by [IA , 6.1.4] and the conditions |K|3 ≤ 36 and η m3 (K) ≤ 4, K ∼ = SL3 (r m ) for some m and η = ±1 with r m ≡ η (mod 3). As m3 (Inndiag(K)) = m3 (P GLη3 (r m )) = 2 but m3 (x I/Z(I)) = 3, x induces a field automorphism on K, the desired conclusion. If K ∈ Alt, then K ∼ = 3A6 or 3A7 and the condition m3 (Aut(K)) = 2 leads to a contradiction. Finally assume that K ∈ Spor. Then m3 (K/Z(K)) ≥ m3 (x I/Z(I)) = 3. Using this condition and our other hypotheses, and [IA , 6.1.4] and [IA , Tables 5.3], we see that K ∼ = 3J3 or 3ON . If 2n > 2, then I is a component of CK (x), but no component of that isomorphism type exists. Thus I ∼ = SU3 (2) ≤ O3,2 (CK (x)). But from [IA , Tables 5.3hs], in both cases |O3,2 (CK (x))| ≤ 2, a contradiction as |I|2 = 8. The proof is complete.  Lemma 13.30. Suppose that p is an odd prime, I, J ∈ Kp ∩ Chev(2), and I 2, and L ↑3 K. Then the following conditions hold: n (a) If m3 (K) = 2, or K ∈ C3 ∪ T3 , then K ∼ = G2 (q), 3D4 (q), 2F4 (2 2 ), or SL3 (q 3 ); (b) In (a), if m3 (CAut(K) (L)) > 1, then K ∼ = 3D4 (q), and CAut(K) (L) = Z(L), f  where f is a quasi-field automorphism of K and CK (f ) ∼ = G2 (q); (c) If m3 (K) > 2, then K ∈ Chev(2) ∩ G3 . Proof. Since L ∈ Alt, K ∈ Alt; also by inspection of the tables [IA , 5.3], K ∈  Spor. By [IA , 2.2.10], L ∈ Chev(r) for any r > 2, so K ∈ Chev(r); hence K ∈ Chev(2). Also, L ∈ T3 by definition, so by Lemma 1.1, K ∈ C3 . Let x ∈ I3 (Aut(K)) with CK (x) having a 3-component Lx such that Lx /O3 (Lx ) ∼ = L. If x is a field automorphism, then K ∼ = SL3 (23n ). By [IA , 7.1.4], m3 (CAut(K) (L)) = 1 in this case. As m3 (K) = 2 in this case, all parts of the lemma hold. Also, x cannot be a graph-field automorphism as then L would be isomorphic to 3D4 (q), contradiction. Thus we may assume that x ∈ Aut0 (K). In (a), suppose that K ∈ T3 . It follows easily from [IA , 4.10.3, 5.6.1] that m3 (K) = 2. Thus in (a), m3 (K) = 2. Since  = (−1)n , L contains 31+2 , so a Sylow

13. PUMPUPS

247

3-subgroup of K has more than one elementary abelian 3-subgroup of maximal rank. Then [IA , 4.10.3c] implies that K is of one of the isomorphism types listed in (a), with the level being determined by [IA , Table 4.7.3A]. n In (b), m3 (CAut(K) (L)) = 1 in the G2 (q) and 2F4 (2 2 ) cases by [IA , Table 4.7.3A]. This reference also implies that x = Z(L) ∈ Syl3 (CInn(K) (L)) in the 3 D4 (2n ) case. Then (b) follows since Out(K) ∼ = Aut(F23n ) acts naturally on L, with kernel of order 3 generated by a quasi-field automorphism. Finally in (c), K ∈ T3 since otherwise m3 (K) ≤ 2, as noted above. As K ∈ C3 , we have K ∈ G3 , completing the proof.  Lemma 13.33. Suppose that L, K ∈ K3 and L ↑3 K, where L ∼ = L2 (q) or q ≥ 8, q ≡  (mod 3),  = ±1. Suppose also that Z(K) = 1. Then one of the following holds: (a) K ∼ = L2 (q); = SL3 (q) with L ∼ ∼ (2) with L (b) K ∼ SU = L2 (8); or = 6 (c) K ∈ G3 ∩ Chev with m3 (K) > 2.

SL− 3 (q),

Proof. Our assumptions give L ∈ Alt ∪ Chev(3), so K ∈ Alt ∪ Chev(3). Moreover by inspection of [IA , Tables 5.3] and the facts that L ↑3 K and Z(K) = 1, K ∈ Spor. Then as Z(K) = 1, either K ∼ = E6± (q0 ) for some q0 , in which case η (c) holds by definition of G3 and [IA , 4.10.3], or K/Z(K) ∼ = L3k (q0 ), ±1 = η ≡ q0 (mod 3), in which case we use [IA , 4.8.2,4.8.4,4.9.1] to determine the possible subcomponents of K. If k > 1, then again by definition of G3 and [IA , 4.10.3], either (b) or (c) holds, so we may assume that k = 1. But then the only possible L  is L ∼ = SL3 (q). This completes the proof. = L2 (q) with K ∼ Lemma 13.34. Suppose that K ∈ K3 , L4 (2) ↑3 K, and Z(K) = 1. Then K ∈ G3 ∩ Chev and m3 (K) > 2. Proof. By definition of C3 and T3 , L4 (2) ∈ C3 ∪ T3 , so L4 (2) ∈ G3 . Then by Lemma 1.1cd, K ∈ G3 . As L4 (2) ↑3 K, we see from [IA , Tables 5.3] that K ∈ Spor, and then as L4 (2) ∈ Chev(r) for any odd r, K ∈ Chev(2) ∪ Alt. But if K/Z(K) ∼ = An , then obviously n > 8, so Z(K) = 1 by [IA , 6.1.4]. Thus K ∈ Chev(2). Suppose finally that m3 (K) = 2. Then Lemma 7.4a implies that Z(K) = 1, a contradiction.  Thus m3 (K) > m3 (L4 (2)) = 2, and the lemma follows. Lemma 13.35. ↑3 (F4 (2)) = {F4 (8)}. Proof. Suppose that F4 (2) ↑3 K via x. By Lemma 1.1ab and [IA , 2.2.10], K ∈ Chev(2). From [IA , 4.8.2, 4.8.4, Table 4.7.3A], x ∈ Inndiag(K), so x is a field automorphism and the lemma is proved.  ∼ A− (q), Lemma 13.36. Suppose that K, L ∈ K3 and L 2. Proof. Since Kz is a component, q > 2. Consequently Kx ∈ G3 , by definition, and as Kx ↑3 K we have K ∈ G3 by Lemma 1.2. Moreover Kz is unambiguously in Chev(r), where q is a power of the prime r, so K ∈ Chev(r), with the help of Lemma 13.16. It remains to show that m3 (K) > 2. By the structure of Kx and [IA , Table 4.7.3A], x does not induce a graph or graph-field automorphism on K. If x induces a field automorphism on K, then 3 3 K∼ = A− n (q ) or B2 (q ), and Kz cannot have the asserted structure, by [IA , 4.8.2, 4.8.4, 4.9.1]. Thus x induces an inner-diagonal automorphism on K. If K is simple and 3-saturated, then Inndiag(K) = Inn(K), so x induces an inner automorphism on K and m3 (K) ≥ m3 (x Kx ) > m3 (Kx ) = 2. So by [IA ,   6.1.4], we may assume that K/Z(K) ∼ = E6 (q  ) or L3m (q  ), q  ≡  (mod 3),  = ±1.  We then have m3 (K) ≥ 4 and are done, by [IA , 4.10.3], unless K/Z(K) ∼ = L3 (q  ). But then again by [IA , 4.8.2, 4.8.4, 4.9.1], Kx could not have the hypothesized structure, a final contradiction.  Lemma 13.38. If q is a power of a prime other than 3, and q > 2, then ↑3 (G2 (q)) = {G2 (q 3 ), 3D4 (q), D4 (q)}. Proof. Write q = r a with r prime, r = 3. Let K ∈ ↑3 (G2 (q)). Since q > 2, G2 (q) is unambiguously in Chev(r). Consequently K ∈ Alt ∪ Chev(s) for any prime s = r. By Lemma 13.16, K ∈ Chev(r). If K were a classical group, then G2 (q) would be as well, by [IA , 4.8.2, 4.8.4]. So K is of exceptional type. According as G2 (q) ↑3 K via an inner-diagonal-graph automorphism or not, we quote [IA , Table 4.7.3A] or [IA , 4.9.1] to get K ∼ = D4 (q) or 3D4 (q), or G2 (q 3 ). This completes the proof.  Lemma 13.39. Suppose that K, L, M ∈ K3 and L ∼ = A− n (q), n = 3, 4, or B2 (q),  q ≡  (mod 3),  = ±1 (but not B2 (2) ). Assume that M ∼ = G2 (q1 ) or L31 (q1 ), q1 ≡ 1 (mod 3), 1 = ±1. Suppose that L ↑3 K and M ↑3 K. Then K ∈ G3 ∩ Chev and m3 (K) − m3 (Z(K)) ≥ 3. Proof. Notice that K ∈ Chev(3), since otherwise L ∈ Chev(3), which is not the case. Also M ∈ Alt so K ∈ Alt; and the facts that L ↑3 K and M ↑3 K, together with the tables [IA , 5.3], show that K ∈ Spor. Thus K ∈ Chev(r), r = 3.

13. PUMPUPS

249

∼ G2 (q1 ), Suppose first that M ∼ = G2 (q1 ), so that q1 > 2. By Lemma 13.38, K = G2 (q13 ), 3D4 (q1 ), or D4 (q1 ). In the D4 (q1 ) case, m3 (K) = 4 by [IA , 4.10.3], and as q1 > 2, Z(K) = 1 by [IA , 6.1.4] and K ∈ K3 −C3 −T3 = G3 by the definition of these sets [I2 , 12.1,13.1]. Thus the desired conclusions hold. In the other three cases, [IA , Table 4.7.3A,4.9.1] show that no component of CK (x) for any x ∈ Aut(K) of order 3 is isomorphic to L, contradiction. Therefore we may assume that M ∼ = L31 (q1 ), so again q1 > 2. In particular  1 M ∼ = K or M ↑3 K via a graph, graph= SL3 (q1 ), by the congruence on q1 . If M ∼ field, or field automorphism, then the possibilities are K ∼ = D4 (q1 ), 3D4 (q1 ), or 1 a L3 (q1 ), a = 1 or 3, and similar arguments to those in the previous paragraph give the desired conclusions. So we may assume that M ↑3 K via an inner-diagonal automorphism. Consequently m3 (Inndiag(K)) ≥ 1 + m3 (M ) = 3. In particular K/Z(K) ∼ = A± 2 (q0 ) for any q0 (see [IA , 4.10.3]). If K is of exceptional Lie type, then [IA , Table 4.7.3A] shows that the condition M ↑3 K forces K ∼ = E6± (q0 ) for some q0 . But then we can again check the definitions to get K ∈ G3 , and use [IA , 4.10.3] to get m3 (K) − m3 (Z(K)) ≥ 4. Finally assume that K is a classical group. Using [IA , 4.8.2, 4.8.4] we see that the only way to have an adjoint component such as M in the centralizer of an 0 (q0 ), q0 ≡ 0 (mod 3), inner-diagonal automorphism of order 3 is for K/Z(K) ∼ = L3n n ≡ 0 (mod 3). In particular the Lie rank of K is at least 8, and we check as in the previous paragraph that the desired conclusions hold. The proof is complete.  In the next lemma, r(X) is the untwisted Lie rank of X ∈ Chev. Lemma 13.40. Suppose that L ↑p I with L, I ∈ Chev of respective levels q and q b , where b = 1 and p is relatively prime to q. Then one of the following holds: (a) b = p and r(I) = r(L); (b) b = 1/a and r(I) = ar(L) + c for some integers a ≥ 2 and c ≥ 1; or (c) b = 1/a and r(I) = ar(L) for some integer a ≥ 2. Moreover if r(L) ≥ 3, then p ∈ {2, 5}; and if p = 5, then L ∼ = SU5 (q), q ≡ −1 (mod 5). Proof. With the exception of the last sentence of (c), this follows from parts (b1) and (c) of Lemma 12.3. With a as in that lemma, we have b = 1/a. Moreover, the situation of (c) can only arise from an equal-rank automorphism of I of order p, so for p odd, I must be of exceptional type with p = 3, or I ∼ = E8 (q) with p = 5. The tables [IA , Tables 4.7.3AB] then show that the only possibility with r(L) ≥ 3 and p odd is I ∼ = E8 (q 1/2 ), p = 5, L ∼ = SU5 (q), and q ≡ −1 (mod 5). The proof is complete.  In the next lemma we assume that (13C)

p is a prime and I, J ∈ Kp with I ↑ p J via some x ∈ Ip (Aut(J)).

We may identify I with I0 /Op (I0 ), where I0 is a component of CJ (x). With this point of view, we say that I is anchored in J if and only if I0 ∩ Op (J) = 1. η Lemma 13.41. Assume (13C) with p = 3, I ∼ = E6 (q)u or E7 (q) or E8 (q), η = ±1, q ≡ η (mod 3). Let k be the untwisted Lie rank of I. Then the following conditions hold: η (a) J ∼ = Ek (q 3 ) or E (q) with k <  ≤ 8; and η (b) If I is anchored in J, then k = 6 and J ∼ = E6 (q 3 )u .

(13D)

250

11. PROPERTIES OF K-GROUPS

Proof. Write q = r m where r is prime. Clearly I ∈ Chev(s) ∪ Alt for any s = r, so by Lemma 13.16, J ∈ Chev(r). If x induces a field automorphism on J, then both (a) and (b) hold. If x induces a graph or graph-field automorphism n on J, then I ∼ = 3D4 (r n ), G2 (r n ), or A± 2 (r ) for some n, which is absurd. Thus we may assume that x induces an inner-diagonal automorphism on J. By [IA , 4.2.2] the untwisted Dynkin diagram or extended Dynkin diagram of J has a subdiagram of type E6 . Therefore J is of type Ek , k = 6, 7, 8, and the proof is completed by  inspection of [IA , Table 4.7.3A]. Lemma 13.42. Let p be an odd prime and L, J ∈ Kp with L ↑p J. Assume that there exists H ∈ Chev(2) ∩ G5p and a field or quasifield automorphism f of H such that L = E(CH (f )). Then J ∈ Chev(2) ∩ Gp , or p = 3 and J ∼ = A11+3k , k ≥ 0. Proof. Since H ∈ G5p , mp (H) = 2, and hence mp (L) = 2. Of course L ∈ Chev(2) as H ∈ Chev(2). Certain isomorphism types cannot occur for L, namely 1 A6 , 3A6 , G2 (2) , 3D4 (2), all for p = 3, and 2F4 (2 2 ) , for p = 5. For H ↓p L would imply in these cases that H ∈ Chev(2) ∩ G5p ; the groups Sp4 (8), G2 (8), 3D4 (2) are 5  SL3 (q) or P SL3 (q) for p = 3 since the not in G3 , and 2F4 (2 2 ) ∈ G5 . Likewise L ∼ =  3 groups (P )SL3 (q ) are not G3 -groups. Note that the fixed points of a quasifield automorphism of 3D4 (q) of order 3 are isomorphic to G2 (q). If J ∈ Spor, then the conditions (13E)

L ↑p J, mp (L) = 2, L ∈ Chev(2)

and [IA , Tables 5.3] yield that (J, L, p) is one of the triples (M24 , 3A6 , 3), (J2 , 3A6 , 3), ((3)J3 , A6 , 3), ((3)Suz, A6 , 3), (Ru, 3A6 , 3), ((3)O  N, (3)A6 , 3); in all of these the structure of L is impossible, as shown above.  A6 for If J ∈ Alt, then J ∼ = = An , n ≥ 8. As L ↑p J and L ∈ Chev(2), but L ∼ p = 3, while mp (L) = 2, L ∼ = A8 with p = 3. Therefore J ∼ = A8+3k , k > 0, as asserted. If J ∈ Chev − Chev(2), then L ∈ Chev(2) ∩ Chev(r) for some odd r. Since mp (L) is exactly 2 the only choice is L ∼ = G2 (2) ∼ = U3 (3) with p = 3, an impossible structure as shown above. Thus J ∈ Chev(2). If J ∈ Cp ∪ Tp , then the conditions (13E) similarly lead to structures for L that have been ruled out above. Thus J ∈ Gp and the proof is complete.  Although the tables [IA , Tables 4.5.1, 4.5.2, 4.7.3AB] and the results [IA , 3.1.4, 4.8.2, 4.8.4, 4.9.1, 4.9.2] give much information about p-layers of centralizers of elements of order p in groups of Lie type, they do not deal with CK (x) if K ∈ Lie(r) is of exceptional Lie type and x ∈ Ip (Inndiag(K)), where p = r is an odd prime that is good for K. Table 13.1 deals with this case when r = 2, listing all the isomorphism types of “terminal Lie components” L of CK (x) in this situation for any odd prime p, good or bad. The data in the table for bad primes p, and for K = 3D4 (q) for p = 3, are simply taken from [IA , Tables 4.7.3AB]. Unusual behavior is indicated in the the “terminal L’s” column by an asterisk; the correcting details appear in the “*” column. To verify the table, we fix K ∈ Lie(2) of exceptional type (including 3D4 (q)) and of level q(K) = q. We also fix an odd prime p that is good for K. In particular, p = 3 if K = E6± (q). Therefore p does not divide |Z(K)| or | Outdiag(K)|.

13. PUMPUPS

251

Table 13.1. Terminal Subcomponents in Exceptional Lie-Type Groups, p > 2 K

Φm0 (q)

mp

terminal L’s

∗

8 4 4 2 2 2

± ∗ A± 7 (q) , D7 (q), E7 (q) E6± (q) 2 ∗ D6− (q), A− 3 (q )? ± A4 (q) − D4 (q), A1 (q 4 )? 3 2 D4 (q), A− 2 (q )?

A± 8 (q) if p = 3

E8 (q)

Φ1 (±q) Φ3 (±q) Φ4 (q) Φ5 (±q) Φ8 (q) Φ12 (q)

E7 (q)

Φ1 (±q) Φ3 (±q) Φ4 (q)

7 3 2

± A± 6 (q), D6 (q), E6 (q) 3 A± 5 (q), D4 (q) − D4 (q)

E6 (q)

Φ1 (q) Φ1 (q), p = 3 Φ2 (q) Φ3 (q) Φ6 (q) Φ4 (q)

6 6∗ 4 3 2 2

A5 (q), D5 (q)∗ D4 (q), A2 (q 3 ) A5 (q), D4− (q) 3 D4 (q) − A2 (q), A2 (q 2 ) A− 3 (q)

F4 (q)

Φ1 (±q) Φ3 (±q) Φ4 (q)

4 2 2

C3 (q) A± 2 (q) C2 (q)

G2 (q)

Φ1 (±q)

2

A1 (q)∗

A± 2 (q) if p = 3

Φ1 (±q) Φ3 (±q)

2 2

A1 (q)∗ , A1 (q 3 ) A± 2 (q)

A± 2 (q) if p = 3

Φ1 (q 2 )

2

A1 (q 2 ), 2 B2 (q)

Φ2 (q 2 ) Φ4 (q 2 )

2 2

A1 (q 2 )∗ 2 B2 (q)

3

D4 (q)

2

F4 (q) n+ 12

q=2 n>0

†

3

2 A− 4 (q ) if p = 5

D4 (q)† for Inndiag(K)

2 A− 2 (q ) if p = 3

Note: D4 (q) if and only if p = 3, q ≡ 1 (mod 9)

The data in the table depend on the arithmetic of the prime p and the level q, specifically on the multiplicative order (13F)

m0 = ordFp (q).

Thus m0 is the least positive integer such that p divides the cyclotomic polynomial value Φm0 (q). We extend the meaning of “terminality” in K to Lie components as follows. Let K = d0 L(q) be as in Table 13.1. Let x ∈ Inndiag(K) be of prime order p > 2, and let L be a Lie component of CK (x) of order divisible by p. (By our earlier

11. PROPERTIES OF K-GROUPS

252 

remark, O p (Inndiag(K)) = Inn(K) and Z(K) is a p -group so we may regard x ∈ K.) We say that (x, L) is terminal in K if and only if L is a Lie component of CK (y) for all y ∈ CK (x L) of order p. We define T Cp (K) to be the set of all pairs (x, L) such that x ∈ Ip (K) and L (of order divisible by p) is a terminal Lie component of CK (x). Note that if such a pair is to exist, it is necessary that mp (K) > 1. We shall prove: Lemma 13.43. Let K ∈ Lie(2) be of exceptional Lie type and level q, and let p be an odd prime with mp (K) > 1. Let m0 be the least integer such that p divides Φm0 (q). Then depending on K and m0 , the isomorphism types of terminal Lie components, and the p-rank mp = mp (Inndiag(K)), are as listed in Table 13.1. The three question marks for E8 (q) mean that if m0 = 4, 8 or 12, no proof will be 2 4 given of the existence of a terminal Lie component isomorphic to A− 3 (q ), A1 (q ) − 2 or A2 (q ). We remark that no assertion about conjugacy of elements of order p is being made. The proof of Lemma 13.43 will occupy us through Lemma 13.55. We fix K ∈ Lie(2) and an odd prime p and first prove: Lemma 13.44. Let x ∈ K be of order p, and let L be a Lie component of CK (x) of order divisible by p. Then (x, L) ∈ T Cp (K) if and only if mp (CK (x L)) = 1. Proof. The backward implication is trivial. Suppose that the forward implication is false and let E ≤ CK (x, L) be an elementary abelian p-group of rank mp (CK (x L)) > 1. Then for every y ∈ E # , L   CK (y), since (x, L) is termi  nal, indeed L  O 2 (CK (y)). Therefore L  Γ2E,1 (K). But E centralizes a Sylow  2-subgroup of L, so Γ2E,1 (K) = K by [IA , 7.3.1]. Thus L  K, which is absurd. The proof is complete.  Now let x ∈ Ip (K). Since we may assume by [IA , Tables 4.7.3AB] that p is good, (13G) p > 3, and if K ∼ = E8 (q), then p > 5. As Outdiag(K) and the center of the universal version of K are p -groups, we may assume without loss that K and K are universal, and, as already remarked, that x ∈ K. Moreover, by [IA , 4.8.5], (13H)

x is of parabolic type.

We use the calculus of [IA , 4.2.2]. Let (K, σ) be a σ-setup of K, and let n be the rank of K, i.e., the untwisted rank of K. Let (B 0 , T 0 ) n

be a standard σ-invariant Borel torus pair. Except in the case K = 2F4 (2 2 ), we may take (13I)

σ = σq γ,

where σq induces the qth power mapping on T 0 and on each T 0 -root subgroup of B 0 , and γ is a graph automorphism of K of order d0 ≤ 3 normalizing B 0 and T 0 .

13. PUMPUPS

253

n

The case K = 2F4 (2 2 ) is the only case in which σ interchanges long and short root subgroups of K. Using [IA , 4.2.2], we have the following. Lemma 13.45. Let (x, L) ∈ T Cp (K). Then CK (x) has a minimal σ-invariant  product L = L1 · · · La of components such that L = O 2 (CL (σ)), and then (L1 /Z(L1 ), σ a ) is a σ-setup for L/Z(L). The level of L is q(σ a ) = q(σ)a = q a , and the Dynkin diagram of L is obtained from that of K by deleting some nodes. Lemma 13.46. Let e ∈ Ip (K) and let I be a Lie component of CK (e) of order divisible by p. If either K or I is not quasisimple, then Table 13.1 is valid for K and p. 1

Proof. Note that if p = 5 and K = F4 (2) or 2F4 (2 2 ), then by [IA , 4.8.6, 4.8.7], the table is valid for K and p. So we may assume that these cases do not occur. If I is solvable, then since p > 3 and p divides |I|, we must have p = 5 1 1 and I ∼ = 2F4 (2 2 ), = 2B2 (2 2 ), so σ interchanges long and short roots, whence K ∼ 1 contradiction. Next assume that I is B2 (2), G2 (2), or 2F4 (2 2 ). By Dynkin diagram considerations (Lemma 13.45), the only choice is B2 (2), whence p = 5. As K is of exceptional type, the double bond forces K ∼ = F4 (2), contradiction. Finally if K is 1 not quasisimple, then as mp (K) ≥ 2 and p > 3, the only possibility is K ∼ = 2F4 (2 2 ), a final contradiction.  We can therefore (13J)

ignore the cases K = F4 (2) or 2F4 (2 2 ) , with p = 5, 1

and assume for any (x, L) ∈ T Cp (K) that L is quasisimple. We next prove: Lemma 13.47. The following conditions hold: (a) T Cp (K) = ∅; and (b) If e ∈ Ip (K) and J is a p-component of CK (e), then there exists (e∗ , J ∗ ) ∈ T Cp (K) such that J ≤ J ∗ . Proof. We first claim that there exists e ∈ Ip (K) and a Lie component I of CK (e) of order divisible by p. If false, then we pick any E ∈ Ep2 (K). As p does not divide | Outdiag(K)|, it follows from [IA , 4.2.2] that for any e ∈ E # , E induces inner-diagonal automorphisms on every Lie component of CK (e), hence   on O 2 (CK (e)). But O 2 (CK (e)) is a p -group since no I exists. It follows that   [E, O 2 (CK (e))] = 1 and hence as e was arbitrary, Γ2E,1 (K) ≤ CK (E) < K. As  p > 3, and by (13J), we have Γ2E,1 (K) = K by [IA , 7.3.3], a contradiction. This proves the claim. By Lemma 13.46, I is quasisimple. Hence it suffices to prove (b). Notice in (b) that as every p-component of CK (e) lies in a Lie component, J is itself a Lie component of CK (e), again by Lemma 13.46. Since K is universal, all Lie components of all centralizers CK (y), y ∈ Ip (K), are normal in CK (y). Among all pairs (e∗ , J ∗ ) such that e∗ ∈ Ip (K), J ≤ J ∗ , and J ∗ is a Lie component of CK (e∗ ) choose one such that J ∗ is maximal with respect to inclusion. Then as J is quasisimple, so is J ∗ . By the previous paragraph, Lp -balance, and

11. PROPERTIES OF K-GROUPS

254

the maximality of J ∗ , J ∗ is a component of CK (y) for all y ∈ Ip (CK (J ∗ e∗ )), and then for all y ∈ Ip (CK (J ∗ )). Hence (e∗ , J ∗ ) ∈ T Cp (K), completing the proof.  We begin the verification of Table 13.1 by getting the awkward case K ∼ = n F4 (2 2 ) out of the way. In this case q = 2n/2 , n odd, and σ interchanges fundamental long and short 2 root subgroups X α and X αs , with x±α (t)σ = x±αs (t) and x±αs (t)σ = x±α (tq ). Thus σ 2 = σq2 .

2

Lemma 13.48. Table 13.1 is correct for K = 2 F4 (q). Proof. We have σ 2 = σq2 . By (13J), q 2 > 2. Let (x, L) ∈ T Cp (K), and let  K1 = CK (σ 2 ) ∼ = F4 (q 2 ) and L1 = O 2 (CK1 (x)). Then σ normalizes L1 , interchanging long and short root groups, and L  CL1 (σ). As x is of parabolic type, the only possibilities are L1 ∼ = A1 (q 2 ) × A1 (q 2 ), with σ interchanging the components, and 2 L1 ∼ = B2 (q ), with σ acting as a graph-field automorphism. Thus L ∼ = A1 (q 2 ) or n 2 B2 (2 2 ). From the commutator formula [IA , 2.4.5d], orthogonal root groups commute, n n so K contains both A1 (q 2 ) × A1 (q 2 ) and 2B2 (2 2 ) × 2B2 (2 2 ). It follows that both n 2 2 2 A1 (q ) and B2 (2 2 ) occur for p dividing q −1, since neither embeds in the other. If p divides q 2 + 1 or q 4 + 1, then every parabolic subgroup has p-rank at most 1. This implies that m2,p (K) ≤ 1 by the Borel-Tits theorem. Hence if y is of order p and I is a Lie component of CK (y), then p divides |I|. This fact rules out L ∼ = A1 (q 2 ) n 4 2 2 if p divides q + 1, and B2 (2 2 ) if p divides q + 1. As T Cp (K) = ∅, the lemma follows.  Consider any other row in Table 13.1. The row is determined by K and applies to all odd primes p with a given value (or pair of values) of m0 as in (13F). (Note that Φn (−q) = Φ2n (q) for all odd n.) Recall our Borel-torus pair (B 0 , T 0 ). Lemma 13.49. Suppose that the group L = d L1 (q a ) ∈ Chev(2) belongs in the row determined by K and m0 . Then (a) mp (Inndiag(L)) ≤ r(L); and r(K) − ϕ(m0 ) ar(L) ≤ . (b) mp (K) − 1 ≤ mp (Inndiag(L)) ≤ ϕ(m0 ) ϕ(m0 ) Proof. Part (a) is immediate from [IA , 4.10.3]. For (b), set C = CK (x). Given our restrictions on p (13G), and by [IA , 4.1.3,4.1.8], C is connected and reductive, and, as already noted, x is of parabolic type. By Lemma 13.44, mp (CK (L)) = 1. By [IA , 4.1.4], we may choose a σ-invariant maximal torus T with x ∈ T ≤ C. g By [IA , 4.10.3e] we may assume that mp (CT (σ)) = mp (K). Then T = T 0 for some g such that in Aut(K), σ −1 gσg −1 = w0 ∈ N0 , where N0 is a complement to T 0 in NK (T 0 ). (In standard notation one may take N0 to be generated by the elements nα (1) as α ranges over the root system of K.) The actions of σ on T and C are then g −1

transported by g −1 to the actions of gσg −1 = σw0 on T 0 and C . Set w = γw0 , where γ is as in (13I), and σ  = σw0 = σq γw0 = σq w. −1 −1 We replace (x, L) by the conjugate pair (xg , Lg ), so that C is replaced by g −1

C and T by T 0 . After this replacement, we see by Lang’s Theorem that (K, σ  ) is a σ-setup for an isomorphic copy of K. Moreover, (13K) σ  and w each permute {L1 , . . . , La } transitively.

13. PUMPUPS

255

Note that as x is of parabolic type, and K is simply connected, L1 · · · La is also simply connected. Let T L = CT 0 (L1 · · · La ), so that x ∈ T L . We next argue that (13L)

x ∈ (T L )o .

Suppose (13L) fails. The connected group T 0 L1 · · · La equals T L L1 · · · La = (T L )o L1 · · · La . Thus T L = (T L )o Z where Z = Z(L1 · · · La ) Hence as (13L) fails, then p divides |Z|. Since x is of parabolic type, The Dynkin diagram of L1 · · · La is a proper subdiagram of the (unex(13M) tended) Dynkin diagram of K. The only possibilities, given (13G), are p = 5 or 7 with L1 = Ap−1 and a = 1. Moreover K ∼ = En with n ≥ 6 if p = 5, and n ≥ 7 if p = 7. Then |Z| = p, o T L L1 = (T L ) × L1 , and L1 is simply connected. Now CT o (σq w) × CZ(L1 ) (σq w) is L contained in CK (L), but mp (CK (L)) = 1 by Lemma 13.44. Therefore x = Z(L1 ),  and O p (CK (L)) = x = Z(L). Moreover C = C 1 × L1 where C 1 = (CC (L1 ))o is σq w-invariant, and it follows that CC 1 (σq w) is a p -group. Thus mp (CK (x)) = mp (Inndiag(L)) = p − 1. However, any elementary abelian subgroup of L of order pp−1 is contained in an elementary abelian p-subgroup of K of rank mp (K), by [IA , 4.10.3e]. Thus mp (K) = mp (CK (x)) = p − 1. As L is a linear or unitary group and x ∈ Z(L), q ≡ ±1 (mod p). Hence by [IA , 4.10.3a], p−1 = mp (K) is the multiplicity of q ±1 as a polynomial divisor of the order of K. Since K = En , the only possibility is that p = 5, K = E6 , K ∼ = E6± (q) ∓ 5 ∼ and accordingly p divides q ± 1. But then L = SL5 (q). However, |q ± 1| does not divide |E6± (q)|, as a consequence of Zsigmondy’s Theorem [I1 , 1.1]. Thus |L| does not divide |K|, a contradiction. This establishes (13L). The actions of w on T 0 , and then on T 0 ∩ L1 · · · La and on (T L )o , are all given by integer matrices. Hence in the actions of w on the exponent-p subgroups of these tori, all primitive m0 th roots of unity in Fp have the same nonzero multiplicity as eigenvalues. Exactly one of these roots of unity gives rise to fixed points of σ  = σq w. Hence dim((T L )o ) ≥ ϕ(m0 ) and ar(L) = dim(T 0 ∩ L1 · · · La ) ≥ ϕ(m0 )mp (Inndiag(L)), proving the middle inequality of (b). Moreover, (T L )o ∩ L1 · · · La is finite, so ar(L) + ϕ(m0 ) ≤ dim T 0 = r(K), proving the right-hand inequality of (b). Finally, let E ∈ Ep∗ (CK (x)), so that mp (E) = mp (K) by [IA , 4.10.3f]. Since CK (x) is connected, E induces inner automorphisms on L1 · · · La and so E maps into Inndiag(L), with kernel x as mp (CK (L)) = 1. The left-hand inequality of (b) follows, completing the proof.  Let n → n∗ be the permutation of the positive integers interchanging m and 2m for all odd m, and fixing all other positive integers. That is, Φn∗ (q) = Φn (−q). Also, for any X ∈ Lie with σ-setup (X, σ), define X ∗ ∈ Lie to have σ-setup (X, σι), where ι ∈ Aut0 (X) inverts elementwise a maximal torus of some σ-invariant Borel∗ torus pair of X. Ignoring differences in isogeny classes, we thus have A± n (q) = n ±(−1) ± ∗ A∓ (q) for n ≥ 4, E6± (q)∗ = E6∓ (q), and X ∗ = X n (q) for n ≥ 2, Dn (q) = Dn for all other types. Lemma 13.50. Suppose that L appears in the row of Table 13.1 corresponding to K and m0 . Then in the row corresponding to K ∗ and m∗0 , a group M appears in which L∗ or L is embeddable, according as a is odd or even.

11. PROPERTIES OF K-GROUPS

256

Proof. Let w, a, σ  , T 0 , T L , and L1 · · · La be as in Lemma 13.49. Let ι ∈ NKΓ (T 0 ) be an element inverting T 0 elementwise and centralizing σ  . Now (K, σ  ) K

and (L1 , (σ  )a |L1 ) are σ-setups for K and L, respectively. Hence (K, σ  ι) is a σsetup for K ∗ . Moreover all the prime divisors p∗ of |C(T L )o (σ  ι)| satisfy ordp∗ (q) = 

(ordp (q))∗ for our prime divisor p of |C(T L )o (σ  )|. Let L0 = O 2 (CL1 ···La (σ  ι)). Then according as a is even or odd, L0 is isogenous to L or L∗ . Finally, by Lemma 13.47b, there exists (e∗ , J ∗ ) ∈ T Cp∗ (K) for some prime p∗ as above such that L0 embeds in J ∗ . This completes the proof.  Now, through Lemma 13.54, we assume that K ∼  3D4 (q), and we verify the = rows of Table 13.1, one m0 at a time. We write Δ(X) for the untwisted Dynkin diagram of the finite or algebraic group X. The p-rank of K is given by [IA , 4.10.3], from which it also follows that the table lists all m0 for which mp (K) ≥ 2, i.e., such that Φm0 (q), as a polynomial, occurs at least twice in a factorization of the order formula for |K|2 . Lemma 13.51. Table 13.1 is correct for m0 = 1 and m0 = 2. Proof. As ϕ(m0 ) = 1, Lemma 13.49 gives mp (K) − 1 ≤ mp (Inndiag(L)) ≤ r(L) ≤ (r(K) − 1)/a. Given the values of mp (K) and r(K), this implies that a = 1. Suppose first that mp (K) = r(K). Then r(L) = r(K) − 1. As x is of parabolic type, Δ(L) is obtained by removing an end node α from Δ(K). One consequence is that if K = F4 (q) or G2 (q), then L is uniquely determined up to isomorphism, so it must appear in the table, by Lemma 13.47. A second consequence is that if K = E7 (q) or E8 (q), then as mp (Inndiag(L)) = r(K) − 1 = r(L), the only possibilities for L are those in the table. The same holds if K = E6 (q) with m0 = 1. To see that the listed groups actually occur in the cases K = En (q), with m0 = 1, one need only take x in a Cartan subgroup H of K, acting nontrivially on the root groups X±α but centralizing X±β for all the other nodes β of the Dynkin diagram of K. For the groups K = 2 E6 (q), E7 (q) and E8 (q) with m0 = 2, use the σ-setup (K, σ  ) of K, where σ  = σq ι and ι ∈ Aut0 (K) inverts the standard maximal torus T 0 . Let  J = O 2 (CK ((σ  )2 )) ∼ = En (q 2 ) for the appropriate value of n. Then mp (K) = mp (J)  and K = CJ (σ ), with σ  inducing a field or graph-field automorphism of order 2 on J. By the case m0 = 1, and [IA , 4.10.3e], there exist pairs (e, I ∗ ) ∈ T Cp (J) with I ∗ of any of the isomorphism types listed in the table for m0 = 1 (but of level  q 2 ). Letting I = O 2 (CI ∗ (σ  )), we have (e, I) ∈ T Cp (K). Hence all the listed L’s for m0 = 2 do occur. Next, consider (K, m0 ) = (2 E6 (q), 1). In this case several conditions, namely, |L| divides |K|, Δ(L) ⊆ Δ(K), mp (L x) ≤ mp (K), and mp (Inndiag(L)) ≥ mp (K) − 1 = 3, − narrow down the possibilities for L (for m0 = 1) to A− 5 (q), D4 (q), and A3 (q). The first two of these actually occur in centralizers of appropriate elements of order p in a Cartan subgroup, but we must rule out L ∼ = A3 (q). There exists a graph automorphism δ of K such that Kδ := CK (δ) ∼ = F4 (q), and the order formulae show

13. PUMPUPS

257

that δ centralizes some P ∈ Sylp (K). Choose, by way of contradiction, (e, L) ∈ T Cp (K) with L ∼ = A3 (q). We may replace (e, L) by a conjugate to obtain CP (e) ∈ Sylp (CK (e)); then δ normalizes e and L, centralizing P ∩ L ∈ Sylp (L). We have δ 2 = 1, but CAut0 (L) (P ∩ L) has odd order, so [δ, L] = 1. Hence (e, L) ∈ T Cp (Kδ ), so A3 (q) belongs in the table for F4 (q), a contradiction. This completes the proof for K ∼ = E6 (q) and m0 = 2, the only possibilities for = 2 E6 (q). Similarly for K ∼ L are A5 (q), D4− (q), and A− 3 (q). The last possibility is ruled out using the graph automorphism δ as above. By Lemma 13.50, there are (e1 , L1 ) and (e2 , L2 ) ∈ T Cp (K) such that A5 (q) and D4− (q) embed in L1 and L2 respectively. Since neither A5 (q) nor D4− (q) embeds in the other (by their orders and Zsigmondy’s theorem),  L1 ∼ = A5 (q) and L2 ∼ = D4− (q). The proof is complete. Lemma 13.52. All rows of Table 13.1 with m0 = 3, 5, 6, or 10 are correct. Proof. Suppose that m0 has one of these values, and let (x, L) ∈ T Cp (K). Suppose first that m0 = 3 or 6, so that by Lemma 13.49, (13N)

2(mp (K) − 1) ≤ 2mp (Inndiag(L)) ≤ ar(L) ≤ r(K) − 2 and mp (K) − 1 ≤ r(L).

For the moment, exclude K = E6 (q) and 2 E6 (q) if m0 = 6 and 3, respectively. The inequalities (13N), together with the condition (13M) and [IA , 4.10.3], restrict the isomorphism type of L, the only possibilities being (K, L) = (F4 (q), A2 (q)), (E6 (q), 3D4 (q)), (E7 (q), A5 (q)), (E7 (q), 3D4 (q)), and (E8 (q), E6 (q)) for m0 = 3; and the corresponding groups (K ∗ , L∗ ) for m0 = 6. Since T Cp (K) = ∅, these all occur, with a possible exception if K = E7 (q). But in that case for m0 = 3, the extended Dynkin diagram shows that K contains A2 (q)A5 (q), so it contains Zp × A5 (q); also K ≥ E6 (q) ≥ Zp × 3D4 (q). Since neither A5 (q) nor 3D4 (q) embeds in the other, Lemma 13.47b implies that both must occur in the table. A similar argument applies if m0 = 6. Similarly, if m0 = 5 or 10, then K = E8 , and the inequalities of Lemma 13.49b, and (13M), leave only the respective possibilities L ∼ = A4 (q) or A− 4 (q), which must occur as T Cp (K) = ∅. Now assume that m0 = 6 and K ∼ = E6 (q), so that mp (L) = 1 and by (13N), 2 ar(L) ≤ 4. The only L’s satisfying these conditions are A1 (q 3 ), A− 2 (q), A2 (q ), ± ± and D4 (q). We rule out D4 (q) as follows. If such an L existed, it would contain a parabolic subgroup of type A± 3 (q), and so by the Borel-Tits theorem K would contain a parabolic subgroup Q containing Zp × I, where I ∼ = A± 3 (q). In turn,  Q ≤ Q for some σ -invariant parabolic subgroup Q ≤ K. But Q would then be of type D4 , D5 , or An X, where 3 ≤ n ≤ 5 and X = A1 or 1. Taking fixed points of σ  , which has level q, we see that CQ (I) would be a p -group, contradiction. Thus L∼  D4± (q). = We use a graph automorphism δ of K of order 2 such that K δ := CK (δ) ∼ = F4 and δ centralizes some P ∈ Sylp (K), which is homocyclic of rank 2 [IA , 4.10.1]. Let Kδ = CK (δ). By [IA , 4.10.3], P lies in a maximal torus, in both K δ and K. Let P0 (resp. P1 ) be the set of all subgroups R of P of order p such that CK (R) (resp. CKδ (R)) has even order. As δ has order 2, it is clear that P0 = P1 . From the Borel-Tits theorem we see that m2,p (K) = 1 = m2,p (Kδ ). It follows that every Lie component L of CK (e) (resp. CKδ (e)) must have order divisible by p, for all e ∈ Ip (K) (resp. Ip (Kδ )); consequently, (e, L) ∈ T Cp (K) (resp. T Cp (Kδ )). Now if

258

11. PROPERTIES OF K-GROUPS 

(e, L) ∈ T Cp (K) with e ∈ P , then O 2 (CL (δ)) must be nontrivial and a terminal Lie component of CKδ (e). This implies, for one thing, that L cannot be A1 (q 3 ). For if it were, then CL (δ) would be A1 (q 3 ) or an elementary abelian 2-group, and not A− 2 (q), as required. Since L ∼ = A− 2 (q) for all (e, L) ∈ T Cp (Kδ ), the Wielandt-type formula [IA , 7.4.1] ∼ for Kδ = F4 (q) reads q 24 = |Kδ |2 = (q 3 )|P1 | , so |P0 | = |P1 | = 8. Applying the same formula to K, we see that 36 = 3a + 6b, where a and b are, respectively, the  2 number of e ∈ E1 (P ) such that O 2 (CK (e)) ∼ = A− 2 (q) and A2 (q ), respectively. But a + b = |P0 | = 8, so a = b = 4. In particular a and b are positive, so the L’s in 2 this row are A− 2 (q) and A2 (q ). Essentially the same argument handles the case K ∼ = 2 E6 (q), m0 = 3. The proof is complete.  Lemma 13.53. All rows of Table 13.1 with m0 = 4 are correct. Proof. Suppose that m0 = 4, and let (x, L) ∈ T Cp (K). Suppose first that K = F4 (q). By Lemma 13.49b, ar(L) = 2. If a = 2 then by (13K), L1 and L2 are interchanged in Aut0 (K) = Inn(K), so they are both A1 ’s corresponding to two fundamental roots of the same length. But no such product of A1 ’s exists. Therefore a = 1. The conditions that p divides |L| and (13M) holds yield only the possibility (K, L) = (F4 (q), B2 (q)). Suppose next that K = E6± (q). As in the previous proof let P ≤ K with P ∼ = = Ep2 and choose a graph automorphism δ of K such that P ≤ Kδ := CK (δ) ∼  F4 (q). Then δ induces an automorphism of L of order at most 2 with O 2 (CL (δ)) ∼ =  O p (CKδ (x)) ∼ = B2 (q). Thus with (13M), δ induces a graph automorphism of 2 L ∼ (q) or A± = A± 3 4 (q). Again let P = {P1 ≤ P | |P1 | = p, O (CKδ (P1 )) = 1}; |P| no proper parabolic subgroup of Kδ has p-rank 2 so by [IA , 7.4.1], |B2 (q)|2 = |K0 |2 = q 24 . Thus |P| = 6. No proper parabolic subgroup of K has p-rank 2, so   # for any y ∈ P  with O 2 (CK (y)) = 1, also O 2 (CK0 (y)) = 1 and so y ∈ P. Thus q 36 = |K|2 = P |CK (P1 )|2 . As |P| = 6, the only choice is L ∼ = A± 3 (q). ∼ Now suppose that K = E6 (q). If L = A3 (q), then there is L0 ≤ L such that L0 ∼ = A2 (q) and x × L0 normalizes a nontrivial 2-subgroup of L. However, it is easily seen that no proper parabolic subgroup of K contains Zp × A2 (q), so the Borel-Tits Theorem is contradicted. Hence L ∼ = A− 3 (q), completing the case − K = E6 (q). As a result, if K = E6 (q), then Lemma 13.50 implies that the only choice is L ∼ = A3 (q). Now suppose that K ∼ = E7 (q), so that mp (K) = 2 and ar(L) ≤ 5. Then there exist subgroups H1 ∼ = E6 (q) and H2 ∼ = 2 E6 (q) of K, and |K|p = |H1 |p = |H2 |p so we may assume the Hi chosen so that H1 ∩ H2 ≥ P ∼ = Ep2 . Let P1 = {P1 ≤ P | |P1 | =  − 2 ∼ p, O 2 (CH1 (P1 )) ∼ A (q)} and P = {P ≤ P | |P = 3 2 1 1 | = p, O (CH2 (P1 )) = A3 (q)}. ± By the E6 (q) analysis, |Pi | = 6 and |CHi (P )| is odd, i = 1, 2. Define b = |P1 ∩P2 | ≤ 6 6. Then by [IA , 7.4.1], and as |A± 3 (q)|2 = q ,  $ % |CK (P1 )|2 /|CK (P )|2 . q 63 ≥ q 6(12−2b) |CK (P )|2 P1 ∩P2

Note that K ≥ D6 (q)A1 (q) ≥ D4 (q)A1 (q)3 and mp (D4 (q)) = 2. Therefore, |CK (P )|2 ≥ q 3 . Let P1 ∈ P1 ∩ P2 ; change notation so that x ∈ P1 . Then L1 ≥ A3 and as ar(L) ≤ 5, a = 1. Since mp (L) = 1, L contains copies of A3 (q) and A− 3 (q). Since mp (L) = 1, the only possibilities are L1 = D4 or A5 with L ∼ = D4− (q) or

13. PUMPUPS

259

− 10 14 A± or |A± 5 (q). Then |L : CL (P )|2 = |D4 (q)|2 /|D2 (q)|2 = q 5 (q)|2 /|A1 (q)|2 = q . Our inequality becomes q 63 ≥ q 6(12−2b) q 10b |CK (P )|2 ≥ q 72−2b+3 . Therefore b = 6, equality holds, and L = D4− (q) is the only entry. Finally suppose that K ∼ = E8 (q). Then mp (K) = 4 and ar(L) = 6, by Lemma 13.49b. As mp (Inndiag(L)) = 3, we get only the possibilities L ∼ = D6− (q) and − + − 2 L∼ = A− 3 (q ). We have Zp × D6 (q) ≤ D8 (q) ≤ K, and D6 (q) is not embeddable in − 2 − A3 (q ), so D6 (q) belongs in the table, by Lemma 13.47b. The proof is complete. 

Lemma 13.54. All rows of Table 13.1 with m0 = 8 or 12 are correct. Proof. In both cases we have K = E8 (q), mp (K) = 2, mp (L) = 1, ϕ(m0 ) = 4, and ar(L) = 4, by Lemma 13.49b. As mp (L) = 1, it follows easily that L ∼ = D4− (q) − 4 3 2 or A1 (q ) for m0 = 8, and L ∼ = D4 (q) or A2 (q ) for m0 = 12. Also, no proper parabolic subgroup of K has p-rank 2. Hence, every Lie component of CK (y), for any y ∈ Ip (K), has order divisible by p and p-rank 1. Suppose that m0 = 12. A parabolic subgroup Q0 of the form q 1+56 E7 (q) has an element y0 ∈ Ip (Q0 ). Then O2 (Q1 )/Z(O2 (Q1 )) is a direct sum of the fixed points of y with one or more summands of order q 12 . It follows that |CO2 (Q1 ) (y0 )|2 ≥ q 9 . Hence CK (y0 ) has a Lie component L ∼ = 3D4 (q). If m0 = 8, we have y × D4− (q) ≤ D4− (q) × D4− (q) ≤ D8+ (q) ≤ K for some y ∈ Ip (K), so CK (y) has a Lie component L ∼ = D4− (q). The proof is complete.  Finally, we show: Lemma 13.55. Table 13.1 is correct for K ∼ = 3D4 (q). Proof. Let (x, L) ∈ T Cp (K). Since Φ4 (q) does not divide |K|, a = 2. Suppose that m0 = 1 or 2. By Lemma 13.49b, ar(L) ≤ 3. For m0 = 1, a Cartan subgroup contains a Sylow p-subgroup of K, so L is a subsystem subgroup and as p > 3, only a long root L ∼ = A1 (q) and a short root A1 (q 3 ) are possible. They both occur, since there are a long root A1 (q) and short root A1 (q 3 ) that centralize each other. Using Lemma 13.50 we see that these are the only possibilities for m0 = 2 as well. They both occur for m0 = 2, since in K1 = 3D4 (q 2 ), A1 (q 2 ) and A1 (q 6 ) both occur in elements of T Cp (K1 ). If m0 = 3 or 6, then ar(L) = 2, so a = 1 and r(L) = 2, and the only possibilities are L ∼ = A2 (q) and A− 2 (q), respectively. As T Cp (K) = ∅, the lemma follows.  Now Lemmas 13.48–13.55 complete the verification of Table 13.1. Lemma 13.56. Let I ∈ Chev(2) be of level q = q(I) and let p be an odd prime such that mp (I) ≥ 2. Then one of the following holds: There exists x ∈ I of order p and a p-component of Lp (CI (x)) of level q; D4 (2), or L± p = 3, q = 2, and I ∼ = B2 (2) , G2 (2) , 3√ 4 (2); 1 2  I∼ = F4 (2 2 ) with p = 3 or 5, and q = 2; n I ∼ = 2F4 (2 2 ) for some n > 1, and there is x ∈ I of order p and a pcomponent of Lp (CI (x)) of level q 2 ; (e) p = 3, I ∼ = L3 (q), q ≡  (mod 3), q ≡  (mod 9), q > 2; or (f) I ∼ = D4 (q), p divides q 2 + 1, and there is x ∈ I of order p and a pcomponent of Lp (CI (x)) isomorphic to L2 (q 2 ).

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

260

11. PROPERTIES OF K-GROUPS

Proof. Assume that (a) fails. Since mp (I) > 1, we may assume in view of (c) 1 that I ∼  2F4 (2 2 ) . Assume first that I is not a classical group. Then there exist =  elements x ∈ Inndiag(I) of order p such that O 2 (CI (x)) has a Lie component J as listed in Table 13.1. In every line of that table – i.e., regardless of the prime p – such a Lie component J of CI (x) exists of order divisible by p and of level q, or else (d) holds. We choose J if possible to be nonsolvable. Note also that in  the one case in which O p (Inndiag(I)) = Inn(I) or p divides |Z(I)|, namely p = 3 and K ∼ = E6 (q) with q ≡  (mod 3), we see from [IA , Table 4.7.3A] that x may be chosen in I with J ∼ = A5 (q), giving the conclusion (a). Thus we may assume that p divides neither |Z(I)| nor | Outdiag(I)|. Since (a) fails, the Lie component J chosen 1 1 above must be solvable. The case J ∼ = 2F4 (2 2 ) , so = 2B2 (2 2 ) is impossible as I ∼ − J ∼ = A1 (2) or A2 (2). In particular q = 2 and p = 3. By our choice of J, all Lie components of level 2 of CI (x), for all x ∈ I of order 3, are isomorphic to A1 (2) or  3 ∼ A− 2 (2). Now [IA , Table 4.7.3A] shows that I = G2 (2) or D4 (2), so that (b) holds. This completes the proof in the non-classical case. Assume then that I is a classical group.  ∼ Suppose first that I = An (q) with p dividing q − . Let ω be a primitive pth root of unity in Fq2 and let x ∈ I be the image of an element of SLn+1 (q) having eigenvalues ω, ω −1 , 1, . . . , 1 on the natural module. Then CI (x) has a subnormal subgroup J ∼ = SLn−1 (q). As (a) fails, J is solvable, whence one of the following ∼ ∼ ± holds: I = A2 (q), I ∼ = A− 4 (2) with p = 3, or I = A3 (2) with p = 3. In the last case (b) holds. In the second case replace x by an element with 4 eigenvalues ω and one eigenvalue ω −1 , contradicting our assumption that (a) fails as then CI (x) has a SU4 (2) component. Consider then the case I ∼ = A2 (q). If p = 3 we replace x by an element with two eigenvalues ω and one eigenvalue ω −2 to obtain a component L2 (q) of Lp (CI (x)). If p = 3 then we can replace x by an element of Z(I) of order 3 (if Z(I) = 1) or by an element with eigenvalues α, α, α−2 , with α ∈ Fq2 of multiplicative order 9 (if Z(I) = 1 and q ≡  (mod 9)). Otherwise if p = 3, then (e) holds. We may therefore assume that if I is of type A, then q ≡  (mod p). Consequently if V is a natural I-module, then I is a quotient of the commutator subgroup of Isom(V ) by a central p -group. Set H = Isom(V ) and choose an element x ∈ H of order p such that CV (x) has maximal dimension d. Then since mp (H) ≥ 2, 2d ≥ dim(V ). Moreover, CH (x) has  a normal subgroup isomorphic to Isom(CV (x)), so O p (CI (x)) has a corresponding normal subgroup L, which is a classical group based on CV (x). Since (a) fails, either L is solvable or L is not of level q. If L has a normal subgroup isomorphic to A− 2 (2), then V is a unitary space and H has level q = 2 and p = 3 = q + 1, the case considered in the previous paragraph. The only other solvable possibility for L is that L is the direct product of copies of A1 (2), again with p = 3. As 2d ≥ dim(V ) this can only arise from I = L4 (2), U4 (2), or B2 (2) , so (b) holds if L is solvable. Therefore L is not of level q. As L is a classical group over the same field as I, and of the same type – linear, unitary, symplectic, or orthogonal – it must be 2 ∼ ∼  that L ∼ = Ω− 4 (q) = A1 (q ). Then I = D (q),  ≥ 4, so d = 4,  = 4, and  = 1. In 2 particular p divides q + 1, and the lemma is proved.  Lemma 13.57. Let p and q be distinct odd primes. Suppose that K, L ∈ Kp and L ↑p K, with L/Oq (L) ∈ Chev(2) ∩ Gq . Then the following conditions hold:

13. PUMPUPS

261

(a) K/Oq (K) ∈ Gq ; (b) If mq (L) ≥ 3, then K ∈ Chev(2); and (c) If K ∈ Chev(2) and K and L have the same level s, with p not dividing s2 − 1, then r(K) ≥ r(L) + 2, where r(X) denotes the untwisted Lie rank of the group X ∈ Chev. Proof. Suppose first that K ∈ Spor. Using the conditions L ↑p K and L/Oq (L) ∈ Chev(2) we see from the tables [IA , 5.3] that L/Z(L) ∼ = L2 (r), r = 5, 7, 8, 9, L3 (4), or U4 (2). In all cases L/Oq (L) ∈ Cq ∪ Tq , contrary to hypothesis. Likewise if K ∈ Alt, then L ∈ Alt and the conditions (13O) imply that L/Z(L) ∼ = An , n = 5, 6, or 8. As L/Oq (L) ∈ Gq we must have q = 3 and L/Z(L) ∼ = A8 . As L ∈ Kp , p = 5 or 7, and K/Z(K) ∼ = A8+kp for some k > 0. This implies that K/Oq (K) ∈ Gq . Moreover, mq (L) = 2. Thus all conclusions hold if K ∈ Alt. We may therefore assume that K ∈ Chev(r) for some r. If r > 2, then L/Z(L) ∈ Chev(2) ∩ Chev(r), so L/Z(L) may be found in [IA , 2.2.10]. Thus either mq (L) = 1 or L/Z(L) ∼ = U3 (3) or P Sp4 (3) with q = 3. In any case L/Oq (L) ∈ Gq , a contradiction. Hence K ∈ Chev(2). Since L ↑p K, mp (Aut(K)) > 1. Suppose that (a) fails, so that (13O)

K/Oq (K) ∈ Cq ∪ Tq . If mq (K) = 1 then mq (L) = 1, whence L/Oq (L) ∈ Gq , contrary to assumption. So mq (K) > 1 as well. By definition of Cq and Tq [I2 , 12.1,13.1], and the facts that mp (Aut(K)) > 1 and p ∈ {2, q} [IA , 4.10.3], (K/Oq q (K), q, p) is one of the following triples: (L3 (2m ), 3, p), with  = (−1)m ; (G2 (8), 3, 7); (D4 (2), 3, 5); (3D4 (2), 3, 7); 1 1 5 (F4 (2), 3, 5 or 7); (2F4 (2 2 ) , 3, 5); (Sp4 (8), 3, 7); (2F4 (2 2 ) , 5, 3); or (2F4 (2 2 ), 5, p).  m m  m/p In case K/O3 3 (K) ∼ ) and = L3 (2 ), L/O3 3 (L) is clearly either L2 (2 ) or L3 (2 in either case L/O3 (L) ∈ T3 , a contradiction. In the succeeding cases with p = 3, we have the following respective possibilities for L/O3 3 (L), using [IA , 4.8.2] and 1 Table 13.1: L2 (8); L2 (4); L3 (2); A6 or L3 (2); no L if K ∼ = 2F4 (2 2 ) ; L2 (8). In all cases L/O3 (L) ∈ C3 ∪ T3 , contradiction. Similarly, if p = 5 we must have 5 1 5 K∼ = 2F4 (2 2 ) , L2 (32), or 2B2 (2 2 ); again we get the contradiction = 2F4 (2 2 ) and L ∼ L ∈ C5 ∪ T5 . This proves (a). It remains to prove (c). Say that L ↑p K via x ∈ Ip (Aut(K)). Since K and L have the same level, x cannot be a field or graph-field automorphism. If x is a graph automorphism then p = 3 with K ∼ = G2 (s) or A± = D4 (s) or 3D4 (s), and L ∼ 2 (s) by [IA , Table 4.7.3A], and so (c) holds in this case. We may therefore assume that x ∈ Inndiag(K). If x is not of parabolic type, then p = 3 or 5, with K ∼ = E8 (s) in the latter case. Since p does not divide s2 − 1, we have p = 5, and the assertion of (c) follows directly from [IA , Table 4.7.3B]. So we may assume that x is of parabolic n n type. Likewise Table 13.1 shows that K ∼  2B2 (2 2 ) , and if K ∼ = 2F4 (2 2 ) , then = L is of type A1 , A2 , or B2 , implying (c) in that case. So we may assume that n n K ∼  2B2 (2 2 ) or 2F4 (2 2 ) . Let (K, σ) be a σ-setup for K. We may assume in = addition that K and K are adjoint versions, so that x ∈ CK (σ). Now in (c), assume that r(L) ≥ r(K) − 1; we show that p divides s2 − 1. As x is of parabolic type, r(L) = r(K) − 1 (see [IA , 4.2.2]). Then CK (x)o = T x L where L is a simple algebraic group, (L, σ) is a σ-setup for L, and T x is a torus of dimension r(K) − r(L) = 1 containing x. Let T be a maximal torus of CK (x) invariant under

262

11. PROPERTIES OF K-GROUPS

σ. Then σ acts on T as the product φs w where φs is the sth power mapping and w lies in a complement W ∗ = W ΓK to T in NAut0 (K) (T ), where W is a copy of the Weyl group of K and ΓK induces graph automorphisms on K. Now the actions of w on T , T ∩ L, and T x are mod p-reductions of rational representations of w. Since T x is 1-dimensional it follows that xw = x±1 . Hence x = xσ = x±s , so p divides s ∓ 1, completing the proof. (The argument of the last paragraph could be replaced by an invocation of [IA , 4.8.2] for the classical case for K, and Table 13.1 for the exceptional cases.) 

14. Acceptable Subterminal Pairs, I Lemma 14.1. Under the assumptions of Theorem C∗7 , suppose that p ∈ γ(G) and (x, K) ∈ J∗p (G). Then there exists (x , K  ) ∈ J∗p (G) and an acceptable subterminal (x , K  )-pair (y, L) such that K/Op p (K) ∼ = K  /Op p (K  ). We separate the cases p = 2 and p > 2, proving Lemma 14.1 as Lemmas 14.2 and 14.3. Lemma 14.2. Lemma 14.1 holds if p > 2. Proof. Since p > 2 and (x, K) ∈ J∗p (G) ⊆ ILop (G), K ∈ Chev(2) ∩ Gp by Theorem C∗7 : Stage 3a and mp (CG (x)) ≥ 4 by definition of ILop (G). It is routine that if K is a classical group with K/Op (K) ∈ Gp , then there is y ∈ Ip (K) such that CK (y) has a component L satisfying the conditions of [III7 , Def. 6.9a], except possibly for the ignorability condition. Likewise, using Table 13.1, if K/Op (K) ∈ Gp is of exceptional type, then y and L exist satisfying [III7 , Def. 6.9b]. The only two issues for this lemma are therefore ignorability and the (A, x, K)-admissibility of (y, L) for some allowable p-source A. We make three general remarks. First, as p > 2, any p-source A is allowable by Theorem C∗7 : Stage 2 [III4 , p. 167] (see also [III5 , 1.1]). Second, for any subterminal (x, K)-pair (y, L) as in [III7 , Def. 6.9ab], y induces an inner-diagonal automorphism on K. Third, the terms of our lemma allow us, if it turns out that (x, K, y, L) is ignorable, to start over, using a replacement (x , K  ) for (x, K). Indeed by [III7 , 6.7], we will have K  /Op p (K  ) ∼ = K/Op p (K) ∼ = Lpn (q), and we will choose (y  , L ) so that L /Z(L ) ∼ = Lpn−1 (q), by the terms of [III7 , Def. 6.9]. This will yield a non-ignorable (x , K  , y  , L ). Recall the definition and existence statement of an (x, K) p-source A [III3 , 2.4, 2.5]. The conditions are that A ≤ K x, A ∼ = Ep3 , and the embedding of A in K x is specified in [III3 , 2.4ab, 2.1]. It suffices to find a subterminal (x, K)-pair (y, L) as in [III7 , Def. 6.9ab], and an (x, K) p-source A, such that the following conditions hold: (1) [y, A] = 1; and (14A) (2) Lp (CL (E)) = 1 for some E ∈ E2 (A). For then [x, y , A] = 1 by (14A1), and as a result of (14A2), y will be nonsolvably (A, x, K)-admissible with respect to L [III3 , Definition 1.6]. (Note that the exceptional configuration of that definition does not occur since by definition of acceptability, if K ∼ = L2 (82 ).) Hence by [III3 , Definition = L4 (8) and p = 3, then L ∼ 1.10], (y, L) will be an (A, x, K)-admissible pair.

14. ACCEPTABLE SUBTERMINAL PAIRS, I

263

To verify that such a pair (y, L) exists, suppose first that mp (K) ≤ 2. As K/Op (K) ∈ Gp , mp (K) > 1, so mp (K) = 2. Choose y ∈ K as in the second paragraph of the proof, let B ∈ Ep2 (K) with y ∈ B, and set A = x × B. If p = 3 and K ∼ = 3D4 (q), q > 2, then choose B to contain some y  such that  E(CK (y )) ∼ = L2 (q 3 ), as we may by [IA , Table 4.7.3A]. Then A is by definition an (x, K) p-source. Since mp (K) = 2 and K/Op (K) ∈ Gp , K/Z(K) ∼ = Lpn (q) for any n > 0 and q ≡  (mod p) (where  = ±1). Therefore (x, K, y, L) is not ignorable. Then (14A) clearly holds with E = x, y. We may now assume that mp (K) ≥ 3. Thus the situations (a3), (b1), (b2), (b3) of [III7 , Def. 6.9] do not occur. First consider the case in which K is a classical group.  = K/Op (K). The conditions (14A) will be satisfied with respect to K if Let K  so and only if the images of y, L, and E, and A satisfy them with respect to K,  → K where K  without loss we assume that Op (K) = 1. Then K has a covering K   ∼  acts on the natural module V , viz., K = SLn (q), Sp2n (q), or Ω2n (q). By definition [III3 , 2.1ad, 2.4a], an (x, K) p-source A is the image in K of an Ep3 -subgroup ≤K  which maximizes dim CV (A).  A Suppose that (14B)

K/Z(K) ∼  Ln (q), q ≡  =

(mod p).

Let d be the minimal dimension of the support of a faithful classical representation  =  Zp → Isom(V ) (see [IA , 4.8.1, 4.8.2, p. 220]). Then d > 1 by (14B), A y1  × y3  and V = V0 ⊥ V1 ⊥ V2 ⊥ V3 with Vi = [V, yi ] being d-dimensional,  y2  ×  i = 1, 2, 3. The only choice for y satisfying (a4) or (a5) of [III7 , Def. 6.9] is for y  to be K-conjugate to y1 . So we may take y = y1 ∈ A. Let Lo = O 2 (CK (y)), y ) = V0 ⊥ V2 ⊥ V3 of dimension dy , say. Then a classical group on the space CV ( dy = 2d + dim V0 ≥ 4. If equality holds and V is an orthogonal space, then V0 = 0 and dim V = 6, contradicting [III7 , Def. 6.9a5]. It follows that L = [Lo , Lo ] = Lp (CK (y)). Because of (14B), (x, K, y, L) is certainly not ignorable. It remains  =  only to verify (14A2). For that we let E y1 , y2 , and let E be the image of 2   E in K. Then O (CK (E)) is the classical group on the space V0 ⊥ V3 , and to show that Lp (CL (E)) = 1 it suffices to show that Isom(V0 ⊥ V3 ) is not psolvable. Since dim V3 ≥ 2, and dim V0 ⊥ V3 ≥ 4 in the orthogonal case, this is clear unless q(K) = 2, p = 3, and V0 ⊥ V3 is a 4-dimensional orthogonal space of + type, a 3-dimensional unitary space, or a 2-dimensional space. The first possibility does not occur since it would imply p = 3 and K/O3 (K) ∼ = D4 (2) ∈ G3 ; the second possibility does not occur since V would then be a unitary space and (14B) would be contradicted. If dim(V0 ⊥ V3 ) = 2 then V0 = 0 and dim(V ) = 6. But K/Z(K) ∼  U6 (2) or U4 (2) ∼  O6+ (2) ∼ = = O6− (2) as K/Op (K) ∈ Gp ; K ∼ = = L4 (2) as ∼ m3 (K) ≥ 3; and so K = L6 (2). In this final case we make the different choice  to obtain L3 (CL (E)) ∼  y1 , y2 y3  of E = L2 (4). This completes the proof in the classical case if (14B) holds. Now assume that (14C)

K/Z(K) ∼ = Ln (q), q ≡ 

(mod p).

264

11. PROPERTIES OF K-GROUPS

Among all (x , K  ) ∈ J∗p (G) such that K  /Z(K  ) ∼ = K/Z(K) we choose one if possible such that for some y ∈ Ip (CG (x )), CK  (y) has a Lie component L with L∼ = SLn−1 (q). If this is possible, rename (x , K  ) as (x, K). If this is not possible, fix y ∈ Ip (K) with n − 2 eigenvalues equal to 1, so that CK (y) has a component L∼ = SLn−2 (q). We claim first that L ∼ = SLn−1 (q) unless possibly n is divisible by p, L ∼ = SLn−2 (q), and (14D) (x, K, y, L) is not ignorable. Namely, if p does not divide n, we choose y ∈ K to have exactly n − 1 equal eigenvalues ω; as ω n = 1, such an element y exists. If p does divide n and some (x , K  ) was available, we have replaced (x, K) by it and L ∼ = SLn−1 (q). If p   does divide n and no (x , K ) was available, then (x, K, y, L) is not ignorable, and L∼ = SLn−2 (q), as claimed. We claim next that L is quasisimple. Assume false. Notice that n ≥ 4, since mp (K) ≥ 3. Therefore if L is not quasisimple then q = 2; either L ∼ = SL− 3 (2) or ∼ ∼ L = SL2 (2). But then K = U4 (2) or U5 (2), all with p = 3. Therefore K ∈ C3 , a contradiction proving the claim. ≤K  is such that dim(CV (A))  is minLet A be any (x, K) p-source. Since A  is diagonalizable, and therefore replacing y by a imized, it follows easily that A K-conjugate if necessary, we may assume that [y, A] = 1. Again it remains to  Let {V1 , . . . , Vn } be a check (14A2). Let y be a preimage of y of order p in K. frame for V (an orthogonal frame in the case  = −1), stabilized by both y and  We may assume that A  was chosen to centralize V1 + · · · + Vn−4 and to act A.  induces the exponent p subgroup of on Vn−3 + · · · + Vn . Since mp (L4 (q)) = 3, A a full diagonal subgroup of SL(X), where X = Vn−3 + · · · + Vn . We may also assume that y was chosen so that its eigenspace W of dimension n − 2 or n − 1 is W = V1 + · · · + Vn−2 or W = V1 + · · · + Vn−1 . Then L = SL (W ), central to be generated by izing the remaining two or one Vi ’s. In most cases we take E e1 = diag(1, 1, ω, ω −1 ) and e2 = diag(ω, ω, ω −1 , ω −1 ) (describing the action of these   contains SL (Vn−3 + Vn−2 ), which is nonsolvelements on X). Then O 2 (CL (E)) able provided q > 2. So assume that q = 2, whence p = 3 and  = −1. Since  to be generated by e1 and e3 = diag(1, ω, ω −1 , 1) K ∈ G3 , n ≥ 7. We then take E and find that L3 (CL (E)) contains SL− (V1 + · · · + V4 ) ∼ = U4 (2), a nontrivial group. This completes the consideration of the case in which K is a classical group. Now suppose that K is of exceptional Lie type with mp (K) ≥ 3. If mp (K) = 3, we see from Table 13.1 that K ∼ = E6 (q) or E7 (q) and p divides Φ3 (q) for some sign . Since clearly p = 3, q = −2. With [IA , 4.10.3], K has an abelian Sylow p-subgroup P . We may take A = Ω1 (P ) as our p-source. By Table 13.1, there is y ∈ A# and a component L ∼ = 3D4 (q) or A5 (q) of Lp (CK (y)), as in [III7 , Def. 6.9b4]. Moreover, y ∈ E ∈ E2 (A) for some E such that Lp (CL (E)) ∼ =  A− A2 (q) ∼ = 2 (2), so (14A) holds. We may therefore assume that (14E)

mp (K) ≥ 4. A p-source A is then defined as a p-source of an “inductive associate” K0 ≤ K, where (K, K0 ) is (F4 (q), C3 (q)), (E6 (q), A5 (q)), or (E7 (q), D6 (q)) (with ordp (q) = 1

14. ACCEPTABLE SUBTERMINAL PAIRS, I

265

or 2); (E8 (q), D7 (q)) (with ordp (q) = 1 or 2 or 4); (E8 (q), E6 (q)) (with ordp (q) = 3); or (E8 (q), E6− (q)) (with ordp (q) = 6). In all cases there exists L ≤ K and y ∈ Ip (K) such that (y, L) is a subterminal (x, K)-pair, with L = K0 except for the triple (E8 (q), D7 (q), 4), in which case L ∼ = D6− (q). Even in this last case, A may be chosen in L, and so in all cases, [y, A] ≤ [y, L] = 1. It remains therefore to check that Lp (CL (E)) = 1 for some E ∈ E2 (A). In the cases where K ∼ = E8 (q) and  (q), as noted in the analysis of the case m (K) = 3, we have q = −2 L ∼ E = 6 p  and Lp (CL (E)) ∼ A (q) for some E ∈ E (A). In the other cases L is a classical = 2 2 group, A was chosen to have maximal fixed points on the natural module, and we choose E ∈ E2 (A) with maximal fixed points on the natural module. Then, for example, for the triple (E8 (q), D7 (q), 4), we have L ∼ = D6− (q) embedded naturally − in K0 ∼ = D7 (q) and Lp (CL (E)) ∼ = D2 (q). In general Lp (CL (E)) = 1, and so we are done, but there is one exceptional case, namely K ∼ = E6− (2) and p = 3,  with K0 = L ∼ = U6 (2); our general construction fails since O 2 (CL (E)) ∼ = A− 2 (2). In this case, notice, however, that A contains E1 = y1 , y2 , where y1 and y2 are elements of orthogonal root SL2 (2)-subgroups of K0 and hence of K. We set L1 = L3 (CK (y1 )) ∼ =L∼ = U6 (2) and argue that the conditions (14A) hold for y1 , L1 , and E1 (and A). As y1 ∈ A, [y1 , A] = 1. As y2 lies in a root SL2 (2)-subgroup of L1 , L3 (CL1 (E1 )) = L3 (CL1 (y1 )) ∼ = U4 (2) = 1. This completes the proof of the lemma.  Lemma 14.3. Lemma 14.1 holds if p = 2. Proof. Recall from [III5 , Def. 1.1ff.] that every (x, K) 2-source is allowable unless possibly K ∼ = Sp4 (q) for some odd q. We let A be an allowable 2-source, to be specified more precisely in certain cases. If K ∈ Alt − Chev, ∼ then K = An , n ≥ 9, by [III3 , Theorem C∗7 :Stage 1 (c)], the fact that K/O2 (K) ∈ G2 , and [IA , 6.1.4]. Let E be a root four-group in An and A = E × x. Then by [III3 , Def. 2.4b, Table 2.2], A is an (x, K) 2-source. Let y ∈ E # and L = E(CK (y)) ∼ = An−4 . Then (y, L) is a subterminal (x, K)-pair and since [y, A] = 1 and L = E(CL (A)), (y, L) is an (A, x, K)-admissible pair. Hence (y, L) is acceptable, by [III7 , Def. 6.9]. If K ∈ Spor, then as K/O2 (K) ∈ G2 , K/O2 (K) ∼ = J1 , M c, Ly, Co3 , He, O  N , or F5 . Choose y ∈ I2 (K) such that E(CK (y)) = 1, if possible with y being 2-central. Set L = E(CK (y)). In every case, (y, L) is a subterminal (x, K)-pair. We choose a fourgroup E ≤ K containing y and set A = E × x. As long as E can chosen to be a 2-internal source [III3 , Table 2.2], A will be an (x, K) 2-source by [III3 , Def. 2.4b], and (y, L) will be an (A, x, K)-admissible pair because CA (L) contains the fourgroup x, y. Except in the case K ∼ = F5 , it is clear from [III3 , Table 2.2] that E can be so chosen; when K ∼ = He or O  N we have to be careful to choose E ≤ Z(L) or E ≤ L, respectively. If K ∼ = A12 and we = F5 , then there exists H ≤ K with H ∼ want E to be conjugate to a root four-subgroup of H. It is sufficient to show that y is conjugate to a root involution t ∈ H. But this is clear because CH (t) ∼ = A8 , while the only class of involutions in K whose centralizers contain A8 is y K .

266

11. PROPERTIES OF K-GROUPS

We may now assume that K ∈ Chev. As in the proof of [IIIK , 10.1], the embedding of A in CG (x) is given by [III3 , Table 2.2, Definition 2.4], and we begin with the case that K is of “wide” Lie type as indicated in [III3 , Table 2.2], but not E8 (q). Repeating the third paragraph of [IIIK , p. 426], we find a subterminal (x, K)-pair (y, L) with y ∈ K, such that (y, L) is an (A, x, K)-admissible pair for some (allowable) 2-source A, and such that y and L are of one of the types listed in [III7 , Def. 6.9]. For these cases, therefore, the only condition remaining to be checked is that (x, K, y, L) is not ignorable. Assuming that (x, K, y, L) is ignorable we replace (x, K) by another pair, as we did in the previous lemma, and find an involution y ∈ CG (x) and component L ± of L2 (CK (y)) such that (K/Z(K), L) = (L± 2n (q), SL2n−1 (q)) or (Dm (q), Bm−1 (q)), with n ≥ 3 or m > 6, respectively. Here y is a reflection inducing an inner-diagonal, resp. graph, automorphism on K. As the new (x, K, y, L) is not ignorable, we need only check that (y, L) is an (A, x, K)-admissible pair for some 2-source  A 2 2 2 4 2 or , + − , + − of (x, K). Now with respect to some standard basis, A = −  4 4 4 8 4 − , + − , + − in the linear and orthogonal cases, respectively; as y acts on K as a reflection we may suppose y and A to have been chosen so that y = −1 . Assuming  case that zspin = 1 in K, we have [y, A] = 1. Taking  2 in6 the orthogonal 4 12 we have [y, E] = 1 and L2 (CL (E)) = 1. Thus (y, L) is E = − , − or − , − ± (q), m > 6, an (A, x, K)-admissible pair, as desired. On the other hand if K ∼ = Dm and zspin = 1, we change our point of view and may assume that y acts as +12 −1 . Then L contains a Spin+ 12 (q) subgroup H with A ≤ H. Choosing E ≤ A as before we again have L2 (CL (E)) = 1, so again (y, L) is an (A, x, K)-admissible pair. To complete the proof in the wide cases we need only treat the case K = E8 (q). In this case, [III7 , Def. 6.9] forces us to choose y ∈ I2 (K) such that L = E(CK (y)) ∼ = D8 (q)hs , whereas A = Z(L1 × L2 × L3 ), where the Li are root SL2 (q) subgroups. Set zi  = Z(Li ), i = 1, 2, E = z1 , z2 , and z = z1 z2 . Using [IA , Tables   4.5.1,4.5.2], we have O r (CK (z1 )) = L1 ∗ J1 , J1 ∼ = E7 (q)u , and then O r (CK (E)) = (L1 × L2 ) ∗ J2 , where J2 ∼ = D6 (q)u and Z(J2 ) = E. If z ∈ z1K , then similarly  r calculating O (CK (E)) using E = z, z1 , we would obtain O r (CK (E)) = (L1 × L0 ) ∗ J1 with L0 ∼ = SL2 (q) and z ∈ L0 . Hence L1 × L0 = COr (CK (E)) (J1 ) = L1 × L2 so z ∈ L2 , contradiction. We conclude that z ∈ z1K . As K has only two classes of involutions, z ∈ y K . We then may choose y = z and obtain J2 ≤ E(CL (E)), so E(CL (E)) = 1, as needed. This completes the proof in the wide cases. Consider then the narrow Lie type cases for (x, K). We follow the proof of [III3 , 10.1] in the last paragraph on [III3 , p. 426] where possible. If K∼ = Sp4 (q), then by [III3 , Theorem C∗7 :Stage 1], |C(x, K)|2 > 2, and A is as given in [III3 , Definition 2.4]; in particular A ∩ K ∼ = E22 . Choosing y ∈ A ∩ K − O2 2 (K), we may take a quasi-2-component L ∼ = SL2 (q) with y ∈ L, as stipulated in [III7 , Def. 6.9c1]; (x, K, y, L) is trivially not ignorable, and y is (A ∩ K, x, K)-admissible with respect to L and L, as argued in [III3 , 10.1]. Thus by [III3 , Definition 1.10], (y, L) is an (A, x, K)-admissible pair, and the lemma holds in this case. This case aside, we have by [III3 , Definition 2.4] that A = E or A = E × x, according as m2 (E) = 3 or 2, where E is a 2-internal source as specified in

14. ACCEPTABLE SUBTERMINAL PAIRS, I

267

[III3 , Table 2.2]. In most cases where m2 (E) = 2, we choose a subterminal (x, K)pair (y, L) with y ∈ E # such that y is (x, y , x, K)-admissible with respect to L and 2 ∼ L, as follows. If K/Z(K) ∼ = L± 3 (q), q > 3, we choose y = − and y ∈ L = SL2 (q); if ± K is a quotient of Ωn (q), n = 5 or 6, q > 3, we choose y = −4 and y ∈ L ∼ = SL2 (q), if such a pair (y, L) is indeed (x, K)-subterminal; if K/Z(K) ∼ = L5 (q), we choose n y = −4 , y ∈ L ∼ = SL4 (q); and if K ∼ = G2 (q), q > 3, 3D4 (q), or 2 G2 (3 2 ), n ≥ 3, then K has one class of involutions [IA , Table 4.5.1] and we choose any y ∈ E # , giving L ∼ = SL2 (q), SL2 (q 3 ), or L2 (q), respectively. Since L is a component of E(CK (y)) = E(CK (x, y), y is (x, y , x, K)-admissible with respect to L and L, and so the pair (y, L) is an (A, x, K)-admissible pair. The only remaining case is K ∼ = P Ω6 (q), q ≡  (mod 8), for which there is no subterminal (x, K)-pair with L∼ = SL2 (q), we choose a subterminal = SL2 (q); instead of using y = −4 and L ∼ (x, K)-pair (y  , L ) with y  ∈ CK (y L) and L ≤ L ∼ = SL3 (q). That such a pair  is available is evident when we regard K ∼ = L4 (q). Since L is a component of E(CL (y)) = E(CL (x, y), y  is (x, y , x, K)-admissible with respect to L and L; therefore the pair (y  , L ) is an (A, x, K)-admissible pair. This completes the case m2 (E) = 2. We may then assume that m2 (E) = 3. The only such narrow cases occur for ± classical groups K. Suppose first that K/Z(K) ∼ (q), n ≥ 3, = A2n−1 (q) or Dm m ≥ 4. Then there exists a subterminal (x, K)-pair (y, L) such that L ∼ = A± 2n−3 (q) ± ± or A2n−2 (q) in the first case, Dm−1 (q) or Bm−1 (q) in the second case. In either case we choose the latter if possible. Then by the usual replacement argument we are assured that (x, K, y, L) is not ignorable. Call the cases (K/Z(K), L/Z(L)) = (A2n−1 (q), A2n−2 (q)), n ≥ 3, and ± (Dm (q), Bm−1 (q)), m ≥ 4 “unusual,” and disregard them for the moment. Fix a standard basis of the natural K-module and obtain A from [III3 , Table 2.2]. We generally choose y ∈ K − Z(K), as specified in [III7 , Def. 6.9], to be an involution acting on the natural module as −a with a as large as possible, and we take L to be a quasi-2-component of CK (y) supported on that −1-eigenspace. The one exception is if K ∼ = Spin− 8 (q), 2 when we take L to be the SL2 (q ) component supported on the +1-eigenspace of y. Note that we are using the four-dimensional module as the natural module for ± ∼ Spin± 6 (q) = SL4 (q). We also choose a suitable four-group E≤A and check that (y, L) is a subterminal (x, K)-pair such that y is (E, x, K)-admissible with respect to L and some I. Hence (y, L) is an (A, pair. Specifi  2x, K)-admissible 2 2 and note that y ∈ E, (q), we choose E = − , + − cally, when K/O2 (K) ∼ = SL± 4 as in [IIIK , p. 427]. Then we can choose L so that y ∈ L ∼ = SL2 (q) and [E, L] = 1, so y is (E, x, K)-admissible with respect to L and L (if q = 3, solvable admissibility holds by [III3 , Definition 1.7 (6a)]). When K/O2 (K) ∼ = L6 (q) or P Sp6(q), y = −4  and L ∼ = SL4 (q) or Sp4 (q), respectively. In this case we choose E = −2 , +2 −2 ; then y is (E, x, K)-admissible with respect to L and I ∼ = SL2 (q) as in the proof of [IIIK , 10.1], using [III3 , Definition 1.7 (6b)] if q = 3. Next, if K/Z(K) ∼ = P Ω± n (q), − 7 ≤ n ≤ 12, but K ∼ = Spin8 (q), then the proof of [IIIK , 10.1] shows that y is (E, x, K)-admissible for some E ∈ E2 (A), as desired. The choices are as follows. If

268

11. PROPERTIES OF K-GROUPS

  4 4 4  8 according as K/Z(K) ∼ = P Ω± n (q), n > 8, we take E = − , zspin or − , + − zspin is nontrivial or trivial ([III3 , Definition 1.7 (6b)] comes into play here and in the next case); if m2 (Z(K)∩A) = 2 we take E ≤ Z(K); in  the remaining  cases with 4 2 6 or − according as (q), n = 7 or 8, we take E = − , z , − K/Z(K) ∼ = P Ω± spin n zspin is nontrivial or trivial. Finally we return to the (narrow) “unusual” cases, in which y acts on K like a reflection, and K is a 2n-dimensional linear or unitary group, n ≥ 3, or a 2n-dimensional orthogonal group, n ≥ 4. Then L is the corresponding 2n − 1dimensional group. We need to check in every narrow unusual case that (y, L) may be chosen so that y is (E, x, K)-admissible with respect to L and some I, for some E ∈ E2 (A). The following E’s and y’s fit the bill. If K/O2 (K) ∼ = L6 (q), we take 2 2 2 5 E = − , + − and y = − and see that (y, L) is (E, x, K)-admissible, using condition (6a) of [III3 , Definition 1.7] if q = 3. If K/Z(K) ∼ = P Ω± n (q), n > 8, with zspin = 1, then by replacing y by a K-conjugate we may assume that y centralizes of −8 . Then y centralizes the E that we used above, viz.,   8 the −1-eigenspace ± ∼ − , zspin , and L2 (CL (E)) contains Spin+ 8 (q), as required. If K/Z(K) = P Ωn (q), n > 8, with zspin 1, then y may be taken to centralize the +1-eigenspace of −4 ,   = and using E = −4 , +4 −4 as before, we see that CL (E) contains Ω+ 4 (q) supported on the −1-eigenspace of +4 −4 , and y is then (E, x, K)-admissible; when q = 3, condition (6b) of [III3 , Definition 1.7] applies. If K ∼ = Spin+ 8 (q), wetake the natural  module annihilated by the group CZ(K) (y) of order 2. We let A = −4 , Z(K) , take y to centralize the −1-eigenspace of −4 , and set E = CA (y). Then E is a four-group and y is (E, x, K)-admissible; when q = 3 condition (6a) of [III3 , Definition 1.7] is satisfied. If K ∼ y to centralize the −1-eigenspace of = Spin− 8 (q) then we can take  4 − , which implies that y centralizes E = −4 , zspin and Lo2 (CL (E)) ∼ = SL2 (q). In with I nonsolvable for all e ∈ E # . If this case, [III3 , Definition 1.7] is satisfied  2 2 2 e ± + ∼ K = Ω8 (q) or P Ω8 (q), we take E = − , + − . Every nonzero element of Fq is the length of some vector in the −1-eigenspace of −2 , so we may take y to have its −1-eigenspace inverted by −2 . Then [y, E] = 1 and Lo2 (CL (E)) contains Ω± 4 (q); again it is easily checked that y is (E, x, K)-admissible. This completes the proof of the lemma.  Now we observe some general properties of acceptable subterminal pairs. Lemma 14.4. Let p ∈ γ(G) and (x, K) ∈ J∗p (G) ∩ Chev. Let (y, L) be an acceptable subterminal (x, K)-pair. Then the following conditions hold: (a) (x, K, y, L) is not ignorable; (b) We have q(K) ≤ q(L). Moreover, either q(K) = q(L) or one of the following holds: n (1) p = 2, K ∼ = 2 G2 (3 2 ) for some n ≥ 3 and L ∼ = L2 (3n ); − 2 (2) p = 2, K ∼ = Spin8 (q) and L ∼ = SL2 (q ) with Z(K) = Z(L); (3) p = 2, K ∼ = SL2 (q 3 ); = 3D4 (q) and L ∼ 2 ∼ (4) p = 3, K/Z(K) ∼ = A− 3 (q), q ≡  (mod 3) and L = SL2 (q ); n 2 2n n (5) K ∼ = F4 (2 2 ), n ≥ 3, p divides 2 − 1, and L ∼ = SU3 (2 ) or L2 (2n ) according as p = 3 or p > 3; (6) p > 2, K ∼ = L2 (q 2 ); = D4 (q), p divides q 2 + 1, and L ∼ (c) If p > 2, then y induces an inner-diagonal automorphism on K; (d) If p > 2 and p does not divide | Outdiag(K)|, then y acts on K like an element of A for some A ∈ Ep∗ (K); and

14. ACCEPTABLE SUBTERMINAL PAIRS, I

269

(e) mp (CK (L)) − mp (Z(K)) ≤ 1. Proof. Part (a) is immediate from [III7 , Defs. 6.7,6.9]. In (b), note that the only parts of [III7 , Def. 6.9] in which K ∈ Chev and 2 ∼ q(L) = q(K) are (a3), (a5) (with L/Z(L) ∼ = Ω− 4 (q) = L2 (q )), (b2), (e5), (f1), and 3 D (q). These give rise to the configurations in parts (4), (6), (5), (2), (f3) with K ∼ = 4 (1), and (3) of (b) of the current lemma, and in every case q(L) > q(K). Thus (b) holds. Note that in case (a5) of [III7 , Def. 6.9], d = 4 is the maximum dimension of the fixed subspace of an element of Ωη2n (q) of order p, while mp (K) > 1 as K ∈ Gp . This forces 2n = 8, η = +1, and p | q 2 + 1, as claimed in (b6). We next prove (c). If y induces a field or graph-field automorphism on K, then of course y ∈ K and CK (y) is given by [IA , 4.9.1]. This contradicts the conditions specified in [III7 , Def. 6.9ab]. Similarly if y induces a graph automorphism on K, then p = 3, again y ∈ K ∼ = D4 (q) or 3D4 (q), and CK (y) is given by [IA , Table 4.7.3A], again contradicting [III7 , Def. 6.9ab]. So (c) holds. In (d), it follows from [IA , 6.1.4] that the Schur multiplier of K/Z(K) is a p group. By (c), y induces an inner automorphism on K. Therefore by [IA , 4.10.3], if p is a good prime for K, then the conclusion of (d) holds. Also if mp (K) = 2 then (d) holds trivially. We may therefore assume that K is of exceptional Lie type and p = 3, or K ∼ = E8 (q) with p = 5. In particular the order m0 of q modulo p divides 4. Hence we are in one of the cases (b5,b6) of [III7 , 6.9], and the result follows directly from the values of mp (K) and mp (L) given by [IA , 4.10.2]. In (e), note first that mp (CKy (y L)) − mp (Z(K)) ≤ 1. ∼ Ep2 ≤ CKy (y L) with CE (K) = 1. Then For otherwise there would exist E = L would be a component of CK (e) for all e ∈ E # , by subterminality, contradicting Lemma 1.16. In particular, supposing that (e) fails, we have y ∈ LZ(K), Z(K) = 1 and y shears to a nonidentity element z ∈ Z(K). For p > 2, this is impossible. Namely, if K/Z(K) ∼ = SLn−1 (q) = Ln (q) with p dividing n ≥ 5 and q − , then L ∼  or SLn−2 (q), and such shearing is impossible as y and yz have different sets of eigenvalues. If K ∼ = E6 (q) with p = 3, then we see directly from [IA , Table 4.7.3A] that m3 (CK (L)) − m3 (Z(K)) = 1. Thus, p = 2. If K is of exceptional Lie type, then as Z(K) = 1, K ∼ = E7 (q)u u ∼ by [IA , 6.1.4], and L = D6 (q) , and it is immediate from [IA , Table 4.5.1] that the desired conclusion holds. Hence we may assume that K is a classical group. As y ∈ LZ(K), we are in one of the following cases of [III7 , Definition 6.9]: (e3), (e6) (with L ∼ = SL2n−1 (q)), (e9), or (e10). In cases (e3), (e6), and (e10), CAut(K) (L) has cyclic Sylow 2-subgroups, so our result holds in these cases. Thus we may assume that K is an orthogonal group, spin group or half-spin group. As in [III7 , 6.9],   be the corresponding orthogonal group On± (q), n ≥ 7, then y ∈ K if we let K is an involution with maximal dimension d for its −1-eigenspace (subject to the condition, of course, that y is an involution), and d > 4. If K = Ω± n (q), then d = n − 1 or n − 2, according as n is odd or even. As n ≥ 7, y does not shear + to an involution in Z(K). So suppose that K ∼ = Spin± n (q) or HSpinn (q), whence + ∼ d is a multiple of 4 and L = Spind (q). But then y ∈ LZ(K), and the proof is complete.  Lemma 14.5. Let p ∈ γ(G) and (x, K) ∈ J∗p (G) ∩ Chev. Let (y, L) be an acceptable subterminal (x, K)-pair. Then the following conditions hold:

270

11. PROPERTIES OF K-GROUPS

∼ Ω± (q) or P Ω± (q), n ≥ 7, then L ∼ (a) If p = 2 and K = = Ω± n n n−a (q), where a = 1 or 2; moreover if a = 2 then n is even; (b) If p = 2 and K/Z(K) ∼ = P Ω± n (q), n ≥ 9, but K is not as in (a), then + L∼ = Spind (q) where d = 4[(n − 1)/4]; (c) If p = 2 and y induces a graph automorphism on K, then the untwisted Lie ranks of K and L differ by 1; n (d) If p = 2 and L ∼ = 2 G2 (3 2 ), where r = 3n , = L2 (r) for some odd r, then K ∼ n ≥ 3; 1 2 2  (e) If p = 2, then L ∼ = P Sp4 (3), L± 3 (3), G2 (3 ) , or G2 (3);  (f) If p = 3, then L/Z(L) ∼  L2 (8), B2 (2) , U4 (2), U5 (2), 3D4 (2), G2 (2) , = 1 G2 (8), 2F4 (2 2 ) , or F4 (2); (g) If p = 3 and L ∼ = L6 (2) or 2 D4 (2); = L4 (2), then K ∼ ∼  (h) If p = 2 and L/O2 (L) = SL2 (q), and K has untwisted Lie rank at least 1/2 4, then either K ∼ ); = 3D4 (q 1/3 ) or K ∼ = Spin− 8 (q n 5 2  ∼ ∼ (i) If p > 2, then L =  F4 (2 2 ) for any n, and L =  2B2 (2 2 ); (j) If p = 3 and L/Z(L) ∼ = L4 (q), G2 (q), = L3 (q), q ≡  (mod 3), then K ∼ n 3 2 n D4 (q), or F4 (2 2 ) where q = 2 ; and Z(L) = 1; (k) If p = 3 and L ∼ = Sp2n+2 (q) or n = 3 and q > 2 = Sp2n (q), then either K ∼ ∼ with K = F4 (q); and (l) If p = 2, then L ∼  B2 (q)a for any odd q. = Proof. We refer to Definition [III7 , 6.9]. First assume that p = 2. Since (x, K) ∈ J∗2 (G), m2 (Z(K)) = 1 by Theorem + C∗7 : Stage 3a, and in particular K ∼  Spin+ = 4k (q), k ≥ 2. We shall regard HSpin8 (q) + as Ω8 (q). Parts (a) and (b) have been observed in the proof of Lemma 14.4e. In (c), the only case of the definition in which y induces a graph automorphism on K is (e10), with K ∼ = Bn−1 (q). Parts (d), (e), (h), and (l) are = Dn (q) and L ∼ immediate by inspection of the definition and the facts that K/O2 (K) ∈ G2 and m2 (Z(K)) = 1. Now assume that p > 2. In (f), suppose that L is of one of the listed types. If K is of exceptional type, then as p = 3, case (b3) of the definition must apply. But then m3 (K) = 2, so K ∼ = G2 (q) or 3D4 (q), and case (b1) applies, contradiction. Therefore K is a classical group, so L is as well. From the various cases (a1)–(a5), we see that the isomorphism type of L forces K/Z(K) ∼ = U5 (2), U6 (2), U3 (8), Sp6 (2), or D4 (2), and so K ∈ G3 , a contradiction. Hence, (f) holds. Likewise (g), (i), (j), 5 and (k) are easily checked against the definition; notice in (i) that if L ∼ = 2B2 (2 2 ), 5 2 ∗ then K ∼ = F4 (2 2 ); as (x, K) ∈ Jp (G), mp (C(x, K)) = 1 so mp (Aut(K)) ≥ 3; thus  p = 5 and so K ∈ C5 , a contradiction. Lemma 14.6. Let p ∈ γ(G) and (x, K) ∈ J∗p (G) ∩ Chev. Let (y, L) be an acceptable subterminal (x, K)-pair. Then the following conditions hold: (a) If p = 3 and L has untwisted Lie rank at least 2, and ρ(K, L) ≥ 3, then − (K, L) = (3D4 (q), A2 (q)) or (A− 4 (q), A2 (q)), where  = ±1 and q ≡  (mod 3); (b) If p > 2 and m  p (K) ≥ 5, then mp (L) ≥ 4; and (c) If p = 3, m  p (K) ≥ 5, and L ∈ Chev(2) is a linear or unitary group of dimension n ≤ 6, then n = 6. Proof. Let K have level q and untwisted rank , and let L have level q a and untwisted rank k. Then a ≥ 1 by Lemma 14.4b. Then ρ(K, L) = 2 /ak2 .

15. ACCEPTABLE SUBTERMINAL PAIRS, II

271

In (a), we see directly from [III7 , Def. 6.9] that  − k ≤ 2; note in case (b3) of the definition that p cannot be 3 since (b1) and (b2) do not hold. Therefore 3 ≤ ρ(K, L) ≤ 2 /a( − 2)2 . Since k ≥ 2 by assumption, this yields  = 4, k = 2, and a = 1. As p = 3, examination of [III7 , Def. 6.9] yields the two asserted examples only. Suppose that m  p (K) ≥ 5. If p does not divide | Outdiag(K)|, then there is no loss in assuming K universal; Z(K) is a p -group. By Lemma 14.4de, y induces an inner automorphism on K, with y ∈ B ∈ Ep∗ (K) for some B, and mp (CK (L)) = 1. As Z(K) is a p -group, B maps into Inndiag(L) by [IA , 4.2.2]. But mp (B) ≥ 5 by hypothesis, so if mp (L) < 4, then some element of B induces an outer diagonal automorphism on L, forcing mp (L) = 3. Since p divides | Outdiag(L)|, the only possibility is p = 5 and L ∼ = L5 (q), q ≡  (mod 5),  = ±1. From [III7 , Def. 6.9] we see that K is not of type E8 in this case, so p = 5 is a good prime for K, and thus as K is universal, so is L, a contradiction. Hence if (b) fails, then p divides | Outdiag(K)|. If p = 3 and K ∼ = E6 (q), q ≡  (mod 3), then  L = A5 (q) from the definition, so m3 (L) ≥ 4, as claimed. The only other possibility is K/Z(K) ∼ = SLk (q), k = n = An (q), with p dividing n + 1 and q − . Then L ∼ or n − 1, so if (b) fails then n − 2 ≤ 3, n ≤ 5. But then m  p (K) ≤ 4, contrary to assumption. Hence (b) holds. Finally suppose that n ≤ 5 in (c). By (b), mp (L) ≥ 4 so n = 5 and L/Z(L) ∼ =  L5 (q), q ≡  (mod 3). It follows easily from [III7 , Def. 6.9] that K/Z(K) ∼ = Ln (q) for some n. As m  3 (K) ≥ 5, n ≥ 7, so as q ≡  (mod 3), (y, L) is not a subterminal pair, contradiction. The proof is complete.  15. Acceptable Subterminal Pairs, II For most of the results of this section we assume the following setup: (1) p ∈ γ(G), (x, K) ∈ J∗p (G), and (y, L) is an acceptable subterminal (x, K)-pair; (15A) (2) mp (C(x, K)) = 1; and (3) If p > 2 then K/Op (K) ∈ Chev(2). Notice that as (x, K) ∈ J∗p (G), in particular we have K/Op (K) ∈ Gp . Lemma 15.1. Assume (15A). Then (y, L) is a normalized subterminal (x, K)pair. Proof. This is a matter of comparing [III5 , Def. 1.3], defining “normalized,” with [III7 , Def. 6.9]. Both definitions require (y, L) to be an (A, x, K)-admissible pair for some allowable (x, K) p-source A. Condition (a1) of the “normalized” definition is implied by conditions (b6) and (f3) of the “acceptable” definition. Conditions (a2) and (a3) of the “normalized” definition are implied by conditions (e1,e2,e3) and (e10) of the “acceptable” definition. Condition (a4) of the “normalized” definition follows from the condition that (x, K, y, L) not be ignorable, in the “acceptable” definition. Finally condition (b) of the “normalized” definition is easily seen by inspection of the tables [IA , 5.3] to follow from condition (d) of the “acceptable” definition, in view of the fact that G2 ∩ Spor = {J1 , Co3 , M c, Ly, He, O  N, F5 }.  Lemma 15.2. Assume (15A), with p odd. If mp (L) = 1, then the following conditions hold:

272

11. PROPERTIES OF K-GROUPS

(a) mp (K) = 2; (b) If |L|p = p, then mp (Aut(K)) ≤ 2; and (c) If K admits a field or quasifield automorphism of order p, then there is t ∈ CG (x) inducing such an automorphism such that [t, y] = 1, K0 := Lp (CK (t)) is simple and p-saturated of p-rank 2, L0 := Lp (CK0 (y)) is a nontrivial subgroup of L, and mp (CK0 (y)) = 2. Proof. Consider the various cases of [III7 , Definition 6.9]. In case (a1), K/Z(K) ∼ = Ln (q), p divides gcd(n, q − ), so if mp (K) > 2 then  n > 4. But L contains SLn−2 (q) so mp (L) > 1, contradiction. Likewise in case (a2), mp (L) = 1 forces n = 3, so mp (K) = 2. In case (a3), mp (K) = 2 by [IA , 4.10.3]. In cases (a4) and (a5), the support of y on the natural module is minimal, so mp (L) + 1 = mp (K) and (a) holds. In all the cases of (b) of [III7 , Def. 6.9], it follows from [IA , 4.10.3] that mp (K) = 2 or mp (L) ≥ 2, so the proof of (a) is complete. Assume that (b) fails. Observe that by Lemma 7.4, Outdiag(K) and Z(K) are p -groups. Hence, as mp (K/Z(K)) = 2, either there exists a field automorphism f ∈ Aut(K) of order p, or p = 3 with K ∼ = 3D4 (q). In the latter case, however, by definition of acceptable subterminal pair, L ∼ = SL3 (q), q ≡  (mod 3), and so m3 (L) > 1, contradiction. As L3 (q) ∈ G3 , we conclude from [IA , 4.10.3] that all Ep2 -subgroups of K are conjugate. But it follows quickly from [IA , 4.10.2] that mp (CK (f )) > 1. Hence every Ep2 -subgroup of K is centralized by a K-conjugate of f . Consequently we may choose f so that [f, y] = 1. Then by [IA , 4.2.3], f induces a field automorphism on L. This implies that |L|p > p. Indeed if m = ordp (q), where q p = q(L), then |L|p ≥ |Φm (q p )| = |Φm (q)||Φmp (q)| ≥ p · p = p2 (see [GL1, p. 112]). This completes the proof of (b). In (c), if K ∼ = 3D4 (q) and p = 3, then as observed above, m3 (L) > 1, contrary to assumption. This case aside, any quasifield automorphism of order p is a field n automorphism. Similarly the cases K ∼ = G2 (q) or 2F4 (2 2 ) with p = 3 lead to the contradiction m3 (L) > 1, by [III7 , Definition 6.9]. Then p is a good prime for K; otherwise K is of exceptional Lie type with p = 3, or K ∼ = E8 (q) with  L3 (q) p = 5, and in all cases mp (K) ≥ 4, contradiction. Note also that K ∼ = with p = 3, q ≡  (mod 3), since K ∈ Gp . As a result, and because mp (K) = 2, it follows from [IA , 6.1.4] that K and K0 are simple and p-saturated. Moreover, by [IA , 4.10.3f], any two Ep2 -subgroups of K are K-conjugate. It follows that t may be chosen to centralize y. Now by [IA , 4.10.3a] and Fermat’s Little Theorem, mp (K0 ) = mp (K) = 2, so mp (CK0 (y)) = 2. It remains to show that Lp (CK0 (y)) is nontrivial. For this it suffices to show that Lp (CL (t)) = 1. But by [IA , 4.2.3], t induces a field automorphism of order p on L. As mp (L) = 1, it follows that 5 Lp (CL (t)) = 1 unless p = 3 with L ∼ = L2 (8), or p = 5 with L ∼ = 2B2 (2 2 ). But these are impossible by Lemma 14.5fi. The proof is complete.  Lemma 15.3. Suppose (x, K) ∈ ILop (G), and K/Op (K) ∈ Chev(r) ∩ Gp with mp (C(x, K)) = 1, where exactly one of r and p is 2. Let (y, L) be an acceptable subterminal (x, K)-pair and suppose that L/Op (L) 2. Then the following conditions hold: (a) L ∼ = SL3 (q),  = (−1)n ; and (b) If X = K a with a of order 3 and F ∗ (X) = K < X, then there exists f ∈ X inducing a quasifield automorphism of K of order 3, such that [f, L] = 1 and CK (f ) ∼ = G2 (q). Proof. Since (y, L) is an acceptable subterminal (x, K)-pair, (a) follows from [III7 , Def. 6.9]. Now by [IA , 2.5.12], Out(K) is cyclic and its subgroup of order 3 is the image of γ, where γ is a graph automorphism of order 3. Then by [IA , Table 4.7.3A], there is f ∈ X of order 3 inducing a quasifield automorphism and with  CK (f ) ∼ = G2 (q). Let g be a 3-central element of CK (f ). Then O 2 (CK (g)) ∼ = L,  and we let z = Z(L). Then O 2 (CK (z)) contains a copy of L. It is then evident     from [IA , Table 4.7.3A] that O 2 (CK (z)) ∼ = L, [O 2 (CK (z)), f ] = 1 and z = y h −1 for some h ∈ K. Replacing f by f h , we get (b).  Lemma 15.6. Assume (15A), with p = 3 and m3 (L) = 1. Then K ∼ = A− 3 (q), or B2 (q), where q ≡  (mod 3),  = ±1, and q > 2.

A− 4 (q),

Proof. By Lemma 15.2a, m3 (K) = 2, and then by Lemma 7.5, either the n lemma holds or K ∼ = A2 (q), q ≡  (mod 3),  = ±1, G2 (q), 3D4 (q), or 2F4 (2 2 ) for some q or n. In the first case, K ∈ T3 , contradicting the assumption K ∈ G3 . In the other three cases by [III7 , Definition 6.9b1,b2], L ∼ = A2 (q), q ≡  (mod 3), and so m3 (L) = 2. But m3 (L) = 1 by assumption, contradiction. Note that as m3 (Aut(K)) ≥ 3, K admits a nontrivial field automorphism, so q > 2. The proof is complete.  Lemma 15.7. Assume (15A). Suppose that p is odd, mp (K) ≥ 3 and L ∼ = A0 (q0 ) with 0 > 1,  = ±1 and q0 ≡  (mod p). Let  = r(K). Then  ≤ 0 + 2. Moreover, if equality holds, then (K, L) = (A0 +2 (q0 ), A0 (q0 )u ), and 0 ≡ −3 (mod p). Proof. Without loss we may assume that  ≥ 3. Since mp (K) ≥ 3 it follows from Lemma 14.4 and [IA , 4.10.3] that q(K) = q0 . If K is of exceptional Lie type, then since mp (K) ≥ 3, K ∼ = En (q0 ) or F4 (q0 ), and as q0 ≡  (mod p), it is immediate from [III7 , Definition 6.9] that  = 0 + 1. So we may assume that K is a classical group. If K is a symplectic or orthogonal group, then as q0 ≡  (mod p), the minimal dimension support among elements of Ip (K) on the natural module is 2, whence from [III7 , Definition 6.9],  = 0 + 1 again. If K ∼ = A−  (q0 ), then in a ∼  similar way L ∼ = A− −2 (q0 ), a contradiction as 0 > 1. Therefore K = A (q0 ), and again by [III7 , Definition 6.9],  − 0 ≤ 2, with the desired conditions in case of equality. The proof is complete.  Lemma 15.8. Assume (15A), with p = 2 and K ∈ Chev. Suppose that L/O2 2 (L) ∼ = L2 (q) for some odd q, or L/O2 (L) ∼ = SL2 (3). Then the following conditions hold:

15. ACCEPTABLE SUBTERMINAL PAIRS, II

275

∼ 2 G2 (3 n2 ) for some odd n > 1; (a) If L/O2 (L) ∼ = L2 (q), then q = 3n and K = (b) If L/O2 (L) ∼ = SL2 (q), then either some involution of Z(K) lies in ± ± Lo2 (CK (y)) or else K ∼ = L± 3 (q), L4 (q), Ω6 (q), P Sp4 (q), G2 (q), or 3 D4 (q 1/3 ); and (c) If K/Z(K) ∼ = L± 4 (q) or P Sp4 (q) and y induces an inner automorphism on K, but no involution of Z(K) lies in Lo2 (CK (y)), then K/O2 (K) ∼ = τ SL± 4 (q) or Sp4 (q), and L2 (CK (y)) = LL for some involution τ ∈ CK (y), with Z(L) = Z(Lτ ) and L = Lτ . Proof. Part (a) is immediate from Lemma 14.5d. In (b), if K is of none of the listed isomorphism types, then by [III7 , Def. 6.9ef], K ∼ = Spin7 (q) or Spin− 8 (q) ∼ (q) with support on and y acts on K like an element of K/Z(K) = Ω7 (q) or Ω− 8 the natural module having dimension 4 and type +. In these cases, as well as the o cases K ∼ = Spin± n (q), n = 5 or 6, two of the Lie components of L2 (CK (y)) generate + ∼ a subgroup isomorphic to Spin4 (q) = SL2 (q) × SL2 (q) and containing Z(K), so Z(K) ≤ Lo2 (CK (y)). This proves (b) and the first statement of (c). Note in (c) that q > 3, as proper quotients of SL± 4 (3) and Sp4 (3) are not in G2 . By [IA , Table 4.5.1], it remains only to prove that an element τ ∈ CK (y) interchanging the components of E(CK (y)) can be taken to be an involution. Regarding K as a quotient of Ω± n (q), n = 5 or 6, we may take τ to be a conjugate of y with 3-dimensional −1-eigenspace on the support of y on the natural K-module. The proof is complete.  Lemma 15.9. Assume (15A), with p = 2, K ∈ Chev, and r(K) ≥ 4. Then either m2 (L) > 2 or K ∼ = Spin− 8 (q) for some odd q. Proof. If K/Z(K) ∼ = Ln (q), q odd, then n ≥ 5. Hence any acceptable subterminal (x, K)-pair (y, L) contains a central quotient of SL4 (q) and so has 2-rank at least 3. If K/Z(K) ∼ = P Sp2n (q), then n ≥ 4, so any acceptable subterminal (x, K)-pair (y, L) contains Sp6 (q), whence m2 (L) ≥ 3. If K/Z(K) ∼ = P Ωn (q), then n ≥ 8. If n > 8 then L, being acceptable, must contain a central quotient of Spin+ 8 (q) and hence must have 2-rank at least 3. If n = 8 and  = + then K is a quotient of Ω+ 8 (q), so as L is acceptable, L contains  a central quotient of Ω6q (q), which has 2-rank at least 3. If n = 8 and  = − then a similar argument is valid if Z(K) = 1, while if Z(K) = 1 then K ∼ = Spin− 8 (q), which is an exception in the lemma. If K ∼ = C3 (q)u or B4 (q)u ; if K ∼ = E6 (q) then L is a 2-saturated = F4 (q) then L ∼  ∼ version of A5 (q); if K = E7 (q) or E8 (q) then L contains some version of E6 (q), D6 (q), or A6 (q). In all cases these conditions imply that m2 (L) ≥ 3, completing the proof of the lemma.  Lemma 15.10. Assume (15A) with p = 2, K quasisimple, Z(K) cyclic, and L∼ = SL2 (3). Then the following hold. (a) y ∈ K ∼ = Sp4 (3) or SL± 4 (3) or Spin7 (3); and (b) Moreover, in the first two cases, O 2 (CK (y)) = L × L1 with L1 being Kconjugate to L; and {Z(L), Z(L1 )} = {y, yz} where z ∈ I2 (Z(K)). Proof. First suppose that K ∈ Chev. As m2 (L) = 1, Lemma 15.9 yields that 1 K has untwisted Lie rank at most 3. Also q(K) ≤ q(L) = 3. As 2 G2 (3 2 ) ∼ = L2 (8) ∈ ± ± C2 , q(K) = 3. Thus K ∼ = A2 (3), B2 (3), G2 (3), A3 (3) or B3 (3). Note that C2 ± contains A± 2 (3), G2 (3), and the non-universal versions of B2 (3) and A3 (3). Also,

276

11. PROPERTIES OF K-GROUPS

∼ Ω7 (3), then by Lemma 14.5a, L/Z(L) ∼ if K ∼ = B3 (3)a = = U4 (3), contradiction. Thus K is as in (a), and y ∈ K by [III7 , Definition 6.9e1,e4]. Then y has two −1 eigenvalues in the symplectic, linear and unitary cases, and (b) follows easily.  Lemma 15.11. Assume (15A), with p = 2 and K ∈ Spor. Then y ∈ K, L = E(CK (y)), and (K, L) is one of the following pairs: (J1 , A5 ), (Co3 , 2Sp6 (2)), (M c, 2A8 ), (Ly, 2A11 ), (He, 22 L3 (4)), (O  N, 4L3 (4)), or (F5 , 2HS). In all cases but K = He and F5 , y is a 2-central involution of K. Proof. Since K ∈ G2 , K is one of the listed groups. The structure of E(CK (y)) is then immediate from Definition [III7 , Def. 6.9] and the tables [IA , 5.3].  Lemma 15.12. Assume (15A). Suppose that p splits K ∈ Chev(s), where exactly one of p and s is 2. Set r = r(K) and assume that r ≥ 4; if p is odd, assume that mp (K) > 2. Write r(L) = r − δ. Then δ ≤ 1 or one of the following holds: (a) δ = 2 and K/Z(K) ∼ = Lr+1 (q). Moreover either q− ≡ r+1 ≡ 0 (mod p), or p is odd with q ≡ − (mod p). In the latter case mp (K) = [(r + 1)/2]; 2 ∼ (b) p = 2, K ∼ = Spin− 8 (q), r = 4, δ = 3, and L = SL2 (q) or SL2 (q ); 3 3 (c) p = 2, K ∼ = SL2 (q ); or = D4 (q), r = 4, δ = 3, and L ∼ (d) p = 2, K ∼ (q), k ≥ 3, δ = 2, and x, y ≤ L ∼ = HSpin4k (q) or Spin− = 4k Spin+ (q). 4(k−1) Proof. Assume that δ > 1. First suppose that p > 2 and let r = r(K) and q = q(K). Since p splits n K, p divides q 2 − 1. As mp (K) > 2, K ∼  2F4 (2 2 ) so q is an integral power of = s = 2. We consider the various cases of the definition of acceptable subterminal pair [III7 , Definition 6.9]. If (a1) or (a2) of that definition holds, then K ∼ = Ln (q)  ∼ and L = Ln−δ (q) with q ≡  (mod p) and δ = 2, so (a) of this lemma holds. If (a3) of that definition holds, then mp (K) = 2 and p > 2, contrary to assumption. If (a4) or (a5) of that definition holds, then K is a classical group and as p splits K, an orthogonally irreducible representation of Zp has dimension 2. Thus if n is the dimension of the natural K-module, then n − 2 is the dimension of the natural L-module. Therefore if K is a symplectic or orthogonal group, then δ = 1, while if K/Z(K) ∼ = Ln (q), then δ = 2. In (a4) of the definition of acceptable subterminal pair, p does not divide q − , so p divides q +  as p splits K. Then as an application of [IA , 4.10.3], mp (K) = [(r + 1)/2]. Therefore the lemma holds if p > 2 and K is a classical group. If p > 2 and K is of exceptional Lie type, then our conditions that mp (K) > 2, ordp (q) = 1 or 2, and δ > 1 yield a immediate contradiction to all parts of [III7 , Definition 6.9]. This completes the proof if p > 2. Suppose then that p = 2. Then since δ > 1, (a) obviously holds in most cases of [III7 , Definition 6.9] that apply, indeed unless (e5) or (e9) applies or (f3) applies with K ∼ = 3D4 (q). Now (e5), and (f3) imply our conclusions (b) and (c), respectively, so we need only consider (e9), i.e., K is an orthogonal group, spin group or half-spin group. We have K/Z(K) ∼ = P Ω± n (q) for some n. If K is a quotient of ± ± ∼ Ωn (q), then by Lemma 14.5a, L = Ωn−1 (q) or Ω± n−2 (q), and in either case δ = 1, contradiction. Thus by Lemma 14.5b, L ∼ = Spin+ d (q) with d = 4[(n − 1)/4]. Thus r(L) = d/2 = 2[(n − 1)/4] and [n/2] = r(K) = r(L) + δ ≥ r(L) + 2 = 2[(n + 3)/4].

16. NEIGHBORHOODS

277

Therefore n ≡ 0 (mod 4), and as Z(K) is cyclic by (15A2), K ∼ = HSpin4k (q) or + ∼ ∼ Spin− (q). The case K HSpin (q) Ω (q) is excluded, and the case K ∼ = = = 8 8 4k Spin− (q) yields conclusion (b) of our lemma, so we may assume that k ≥ 3. Then 8 L∼  = Spin± 4(k−1) (q), so (d) holds. The proof is complete. 16. Neighborhoods J∗p (G)

Given (x, K) ∈ and an acceptable subterminal (x, K)-pair (y, L), the neighborhood that they determine consists of the pumpups Lu of L in CG (u), as u varies over x, y# . In this section we study the (long) pumpups of L in G. The setup is the following: (1) (x, K) ∈ J∗p (G) for some prime p, with K/Op (K) ∈ Chev(r), r = p, r = 2 if p > 2; (16A) (2) (y, L) is an acceptable subterminal (x, K)-pair; (3) L/Op (L) 2 and r = 0. Suppose that I ∈ Tp ∩Chev. Then the following conditions hold: (a) L/Op (L) ∈ Tp ; (b) One of the following holds: (1) mp (K) = 2, mp (L) = 1, and dp (G) = 5; or (2) p = 3, m3 (K) = 3 or 2, Z(L/O3 (L)) = 1, and L/O3 (L) ∈ B3 . Proof. Since I ∈ Tp ∩ Chev, it follows from the definition of Tp that either 5 I ∈ Chev − Chev(p) with mp (I) = 1 (but with I ∼  2B2 (2 2 ) if p = 5), or p = 3 with = I∼ = 3A6 , G2 (8), SL3 (q) or P SL3 (q),  = ±1, q ≡  (mod 3). Then as p is odd and L/Op (L) 1. From our description of Tp ∩ Chev, p = 3 with I ∼ = 3A6 , G2 (8), or (P )SL3 (q). If (16B) L/O3 3 (L) ∼ = L (q), q ≡  (mod 3), 3

then (a) is obvious and (b2) holds by Lemma 14.5j. So we may assume that (16B) does not hold. Thus, as L/O3 (L) 2 and r = 1. Suppose that I1 ∈ Chev(2) and I0 ↑p I1 via h ∈ Ip (Aut0 (I1 )). Suppose that m := mp (I1 ) = mp (I0 ) = 2 or 4, and that correspondingly mp (K) > 2 or m  p (K) > 4. Then one of the following holds: (a) mp (I1 ) = 4, p = 5, I0 /O5 (I0 ) ∼ = SU5 (q 2 ) and I1 /O5 (I1 ) ∼ = E8 (q) for some q ≡ ±2 (mod 5); (b) mp (I1 ) = 2, p = 3, I0 /O3 (I0 ) ∼ = SL3 (q) and I1 /O3 (I1 ) ∼ = G2 (q), 3 D4 (q), √ 2 or F4 ( q) for some  = ±1 and q ≡  (mod 3), q > 2; (c) mp (I1 ) = 2, p = 3, I0 /O3 (I0 ) ∼ = G2 (q), and I1 /O3 (I1 ) ∼ = 3 D4 (q) for some q. Proof. If h induces a graph automorphism on I1 , then by [IA , Table 4.7.3A], p = 3 and I0 ∼ = G2 (q) or L3 (q), q ≡  (mod 3),  = ±1. Thus m = m3 (I0 ) = 2 so as I1 possesses a graph automorphism of order 3, I1 ∼ = 3D4 (q). If I0 ∼ = L3 (q),   then since L/O3 (L) 2 and r = 0. Then (a) I ∈ Chev(p ) unless p = 3, L ∼ = L4 (2), m3 (K) = 3 and I ∼ = A11+3k for some k ≥ 0; (b) Suppose that p = 3, with either m3 (I) = 2 or I ∼ = Sp6 (2). Then m3 (K) ≤ 4; (c) Suppose that I ∈ Cp . Then p = 3 and I ∼ = Sp4 (8), Sp6 (2), D4 (2), F4 (2), U5 (2) or SU6 (2) or U6 (2); and (d) If L/Op p (L) ∼ = I/Op (I), then I ∈ Chev(p) ∪ Alt unless I ∼ = A11+3k and L∼ = L4 (2) are as in (a), with p = 3. Proof. As observed at the beginning of this section, mp (Aut(K)) ≥ 3. In particular by Lemma 15.2b, |L|p > p, and so |I|p > p. Suppose that I ∈ Alt. Then by Lemma 1.1a, L ∈ Alt ∩ Chev(2). As |L|p > p, p = 3 with L/Z(L) ∼ = A6 or A8 . The first of these is impossible by Lemma 14.5f, so L/O3 (L) ∼ = A8 . As I ∈ K3 , it follows immediately that I ∼ = A11+3k , k ≥ 0, or A8 . Moreover q(K) = 2 by Lemma 14.4b. Note that Outdiag(K) is a 3 -group, since otherwise by [IA , 2.5.12], K/Z(K) ∼ = E6− (2) or Un (2) for some n; but then K has no subcomponent isomorphic to L4 (2), contradiction, by [IA , Table 4.7.3A,4.8.2,4.8.4]. Then Z(K) is a 3 -group, and by Lemma 14.4c, g induces an inner automorphism on K, so m3 (K) > m3 (L) − m3 (Z(L)) = 2. If m3 (K) > 3, then by Lemma 14.4d, m3 (CK (y)) > 3. As Out(L) is a 3 -group, m3 (CK (y L)) > 1. But this violates Lemma 14.4e. Thus m3 (K) = 3 and the conclusions of (a), (b) and (d) hold. Suppose that L ∈ Chev(p). Then L ∈ Chev(p)∩Chev(2), and since |L|p > p, we have p = 3 and L ∼ = L2 (8), G2 (2) , or U4 (2) by [IA , 2.2.10]. This contradicts Lemma 14.5f. Thus L ∈ Chev(p). Consequently by Lemma 1.1ab, I ∈ Spor ∪ Chev(p). As a consequence, (a) and (d) are completely proved. In (c) we have I ∈ Cp , whence L ∈ Cp by Lemma 1.1c. As L ∈ Chev(p) ∪ Alt, ∼ 3D4 (2) or 2F4 (2 21 ) , either p = 3 and L is one of the groups in (c), or p = 3 and L = 1 5 5 or p = 5 and L ∼ = 2F4 (2 2 ) , 2F4 (2 2 ), or 2B2 (2 2 ), by the definition of Cp [IA , 12.1]. In particular L is unambiguously in Chev(2). Since I ∈ Chev(r) for some r, it follows that I ∈ Chev(2) ∩ Cp , so the same isomorphism types are the possibilities for I. Using 14.5fi we see that L is one of the groups in (c), and p = 3. Then with [IA , Table 4.7.3A], the same holds for I, as L 2. The proof is complete.  Lemma 16.11. Assume (16A), and K has level q(K) = q. L/Op (L) 2, then mp (L) > 1; (b) I ∈ Chev(r) and q(I) = q(K); 2 (c) p = 2 and (K, L) = (3D4 (q), SL2 (q 3 )) or (Spin− 8 (q), SL2 (q )); (d) p > 2 and mp (K) = 2; (e) p = 3, L ∼ = L6 (2) or 2 D4 (2); = L4 (2), and K ∼ (f) p = 2 and I ∼ = 2An , n ≥ 7; or (g) r(K) ≤ 3. Proof. Suppose first that I ∈ Spor. By Lemma 15.3, p > 2 and L ∈ Chev(p). Thus K ∈ Chev(2), whence L ∈ Chev(2) by Lemma 1.1. If mp (L) = 1, then (d) holds by Lemma 15.2a. If mp (L) > 1, then L∼ = G2 (2) , U4 (2), or B2 (2) by [IA , 2.2.10], contradicting Lemma 14.5f. Thus we may assume that I ∈ Spor. Suppose next that I ∈ Alt, ∼ An , n = 5, 6, or 8. If O2 (L) = 1 and p = 2 then so that L ∈ Alt and L/Z(L) = (b), (c), or (f) holds, by Lemma 14.4b. If p = 2, then n = 8 as L ∈ Chev(r), r odd; thus L/O2 (L) ∼ = L2 (q) for q = 5 or 9, contradicting Lemma 14.5d. Therefore p > 2, and we may assume that mp (L) > 1 as in the previous paragraph. Then p = 3 and n = 6 or 8. But n = 6 is again ruled out by Lemma 14.5f. If n = 8, so that L ∼ = L4 (2), then (e) holds by Lemma 14.5g. Thus, we may assume that I ∈ Alt, whence I ∈ Chev(s) for some s. Choose such an s different from r, if possible. If s = r and p > 2, then as in the first paragraph we may assume that mp (L) > 1 by Lemma 15.2a, and [IA , 2.2.10] and Lemma 14.5f give a contradiction. If s = r and p = 2, then [IA , 2.2.10] and Lemma 14.5deh show that (c) or (g) holds. Therefore, we may assume that s = r and both L and I are unambiguously in Chev(r). Note that if p = 2 and I ∈ G82 , then I ∼ = An , n = 9 or 10, and L ∼ = A5 or A6 , contradicting Lemma 14.5d. Thus in any case if I ∈ Gp , then (a) holds, so we may assume that I ∈ Cp ∪ Tp . Note that if p = 2 and L/Z(L) ∼ = L2 (q) for some odd q, then (c) or (g) holds by Lemma 14.5dh. Therefore we may assume that K has untwisted Lie rank at least 4, L/Z(L) ∼ = L2 (q), q odd, if p = 2, and I/Z(I) is as in [I2 , Definition 12.1(1b,2a,3a)] or [I2 , Definition 13.1(2)]. If I/Z(I) ∼ = L/Z(L), then by Lemma 14.4b, (b), (c), or (d) holds. So we may assume that I/Z(I) ∼  L/Z(L). = If p = 2 and I/Z(I) ∼ = L3 (3), L± 4 (3), or G2 (3), then r = 3 and L is unambiguously in Chev(3) with L/O2 (L) 2, so as above, mp (L) > 1. 1 5 Suppose next that p = 5, so that I ∼ = 2F4 (2 2 ) or 2F4 (2 2 ). With [IA , 4.9.1] 5 1 and Table 13.1 we get that the only possibility is I ∼ = 2F4 (2 2 ) , = 2F4 (2 2 ) and L ∼ contradicting Lemma 14.5i. Thus p = 3. If I ∈ T3 , then I/Z(I) ∼ = G2 (8) or L3 (q), q ≡  (mod 3), and as m3 (L) > 1, the only possibility by [IA , Table 4.7.3A, 4.8.2, 4.9.1] is L/Z(L) ∼ = Lη3 (q  ), q  ≡ η (mod 3). Since K has untwisted Lie rank at least 4, Lemma 14.5j n η this time gives K ∼ = 3D4 (q  ) or 2F4 (2 2 ). As I ∈ G3 we get I/Z(I) ∼ = L3 ((q  )3 ) or  G2 (8), with q = 8. Accordingly conclusion (d) or (b) holds, as desired. Thus we may assume that I ∈ C3 , as in [I2 , Definition 12.1(2a)]; in particular I ∈ Chev(2). If q(I) > 2 then I ∼ = B2 (8) and so L ∼ = L2 (8) or B2 (2) , contradicting Lemma 14.5e. Thus q(I) = 2. Likewise if q(L) > 2 we see from [IA , 4.8.2, Table 4.7.3A] that I/Z(I) ∼ = L2 (8), again a contradiction. = 3D4 (2) or U6 (2) with L ∼ Therefore q(L) = 2. If q(K) = 2 then (b) holds, so by Lemma 14.4b, q(K) < 2, 1  forcing K ∼ = 2F4 (2 2 ) ∈ G3 , a final contradiction. Lemma 16.12. Assume (16A) with p > 2 and mp (K) = 2 and r = 0. Suppose that L/Op (L) ↑p I, I ∈ Chev(2), and mp (I) ≥ 3. Then I ∈ Gp . Proof. By Lemma 15.4, some f ∈ Ip (CG (x)) induces a field or quasifield automorphism on K; in the latter case p = 3, K ∼ = 3D4 (q) (q > 2 as K ∈ G3 ), and G (q). Accordingly we have f inducing a field or graph-field automorCK (f ) ∼ = 2  (q), q ≡  (mod p). In any case phism of order p on L, or [f, L] = 1 with L ∼ SL = 3 q(L) > 2. If I ∈ Chev(p), then L ∈ Chev(p) by Lemma 1.1. As L ∈ Chev(2) admits a field or graph-field automorphism of order p, the only choice, from [IA , 2.2.10], is L∼ = L2 (8) with p = 3, contradicting Lemma 14.5f. Thus I ∈ Chev(p). Suppose that I ∈ Cp ∪ Tp . Using the definitions of Cp and Tp [I2 , 12.1, 13.1] and [IA , 4.10.3] to calculate mp (I) ≥ 3, we obtain the following complete list of possibilities for I: I/Z(I) ∼ = U5 (2), U6 (2), Sp6 (2), D4 (2), and F4 (2), all with p = 3. Using [IA , 4.8.2, 4.8.4, Table 4.7.3A], we see that the subcomponent L of I must have level 2, contradicting the first paragraph, or L ∼ = L2 (8), contradicting Lemma 14.5f. The proof is complete.  Lemma 16.13. Assume (16A) with p > 2 and r = 0. Suppose that dp (K) = 2 or 4, and p splits K. Suppose that L/Op (L) ↑p I via x, I ∈ Gp , and I is unambiguously in Chev(2). If dp (I) > dp (K), then one of the following holds: (a) q(I) = q(K); (b) Op (K) = 1, K is p-saturated, mp (K) = 5, and x induces a nontrivial field automorphism on I; − ∼ (c) K/Z(K) ∼ = L− n (q), n = 6 or 7, L = SLn−2 (q) with p dividing q − , and − p ∼ I = SLn−2 (q ); ∼ − p (d) K ∼ = 2 D4 (q), L ∼ = L− 4 (q), p divides q − , and I = L4 (q );  n  n ∼ ∼ (e) p = 5 or 7, K/Z(K) = Lp (2 ), L = SLp−2 (2 ), and 2n ≡  (mod p); moreover, I ∼ = SLp−2 (2pn ); (f) p = 3, K ∼ = E6 (q) or A6 (q), L/Z(L) ∼ = L6 (q), q ≡  (mod 3); moreover,  3 I/Z(I) ∼ = L6 (q );

11. PROPERTIES OF K-GROUPS

286

(g) K ∼ = L4 (q) or Sp6 (q), q > 2, p divides q − , L ∼ = SL3 (q) or Sp4 (q),  p 2 1/2 respectively, and I ∼ = SL3 (q ) or F4 (q ) (if K ∼ = L4 (q)) or Sp4 (q p ) (otherwise); or (h) K/Z(K) ∼ = U6 (q), p = 5 divides q + 1, L ∼ = SU5 (q), and I ∼ = E8 (q 1/2 ). Proof. Since dp (K) = 5 by assumption, mp (K) > 2. Thus r(K) ≥ 3 by [IA , 4.10.3], mp (L) > 1 by 15.2, and q(L) = q(K) by Lemma 14.4b. In particular p splits L. We assume that (a) fails, so that (16G)

q(I) = q(L).

First suppose that dp (I) = 5 > dp (K), so that mp (I) = 2. Then mp (L) = 2. Working from the definition [III7 , Definition 6.9] and using [IA , 4.10.3] to compute p-ranks, we see that one of the following holds, with  = ±1 and p dividing q −  in every case: (1) K/Z(K) ∼ = SL3 (q); = L5 (q), p = 5, L ∼ − (2) K/Z(K) ∼ L (q), n = 6 or 7, L ∼ = n = SL− n−2 (q); − 2 ∼ ∼ (16H) (3) K = D4 (q), L = L4 (q); (4) K ∼ = SL3 (q); = L4 (q), L ∼ ∼ (5) K = Sp6 (q), L ∼ = Sp4 (q), p divides q 2 − 1. The only other case to consider is that dp (I) = 4 > dp (K), so that m  p (K) > 4  p (I) ≤ 4. Again using [III7 , Definition 6.9], and noting by [IA , 6.1.4] but m  p (L) ≤ m that the universal p-covers of K and L are quotients of their universal versions, we get the following possibilities only: (1) Op (K) = 1, K is a p-saturated classical group, and mp (K) = 5; (2) K/Z(K) ∼ = SL5 (q); = L7 (q), p = 7 divides q − , and L ∼ (16I)  (3) K/Z(K) ∼ = L6 (q); or = E6 (q), p = 3 divides q − , and L/Z(L) ∼   (4) K/Z(K) ∼ = SL6 (q). = L7 (q), p = 3 divides q − , and L ∼ It remains only to prove the assertions about I. Consider first the cases in which mp (I) = mp (L) = 2, i.e., the cases of (16H). We have that L/Op (L) ↑p I via x ∈ Ip (Aut(I)). Note that in each case of (16H) except (16H4), Z(L) is a p -group. Suppose that p divides | Outdiag(I)|. As p is odd, the only possibility then is p = 3 and I/Z(I) ∼ = L3 (q0 ) for some q0 . Since p = 3, (16H1) does not hold. If ± (16H4) holds, then I ∼ = SL3 (q 3 ), as desired. Otherwise L involves L± 4 (q) or Sp4 (q). But then L/Z(L) is not embeddable in I/Z(I) by Lemma 8.2. This contradiction shows that we may assume that Outdiag(I) is a p -group. As a result, Op (Z(I)) = 1. Hence if x is an inner-diagonal automorphism, it is induced by an element ξ ∈ Ip (I) and so 2 = mp (I) ≥ mp (ξ L), whence p = 3 and (16H4) holds. As n q(L) = q(I), Lemma 13.31 yields I ∼ = 2F4 (2 2 ) with n odd, and q(K) = 2n . Thus (g) holds. So we may assume that x is not an inner automorphism. If x is a graph automorphism, then p = 3 and q(L) = q(I), by [IA , Table 4.7.3A], contradiction. Assume then that x is a field or graph-field automorphism. Since L ∼ = 3D4 (q0 ) for any q0 , x is a field automorphism and I is as claimed in the lemma, in (c), (d), (e), and (g).

16. NEIGHBORHOODS

287

Now consider the cases of (16I), and suppose by way of contradiction that x does not induce a field automorphism on I. Suppose that (16I1) holds. Then using [III7 , Definition 6.9] we find that mp (L) = mp (K) − 1 = 4 and Op (Z(I)) = 1. Consequently x does not induce a graph or graph-field automorphism on I, so x induces an inner-diagonal automorphism on I. If p does not divide | Outdiag(I)|, then unless x acts on I like an element of Z(L), we have Zp × L embedded in I and disjoint from Z(I), so m  p (I) ≥ 5, contradiction. If p does not divide | Outdiag(I)| but x acts on I like an element of Z(L), we must have L ∼ = SL5 (q0 ) with p = 5. Using [III7 , Definition 6.9] we find that q0 = q and K ∼ = SL6 (q), with 5 dividing q − . Since  5 (I) = 4, and then by [IA , 4.8.2, 4.8.4] q(I) = q(L) and m  5 (I) ≤ 4, we have m 1/2 and Table 13.1, I ∼ E (q ). Thus, (h) holds. Therefore, we may assume that p = 8 η divides | Outdiag(I)|. As m  p (I) ≤ 4 but mp (I) ≥ mp (L) = 4, we have I ∼ = SL5 (q0 ) η with p = 5 or I/Z(I) ∼ = L6 (q0 ) with p = 3, and in either case p divides q0 − η. Since x induces an inner-diagonal automorphism on I, and L is a component of CI (x) with mp (L) = 4, we see from [IA , 4.8.2, 4.8.4] that q(I) = q0 = q(L), contradiction. If (16I2) holds, then p = 7 does not divide | Outdiag(I)|. (If it did, I/Z(I) ∼ =  7 (I) ≥ 5, contradiction.) As x does not induce a Lη7n (q0 ), 7 divides q0 − η, and so m field automorphism on I, it then induces an inner automorphism and Z7 × SL5 (q) embeds in I, so m7 (I) ≥ 5, contradiction. If (16I3) holds, then similarly, p = 3 does not divide | Outdiag(I)|, and we may assume that x induces an inner automorphism on I. If I contains Z3 × L then m  3 (I) ≥ 5, contradiction. Therefore x acts (nontrivially by hypothesis) on I as an element of Z(L) of order 3. Consequently L ∩ Z(I) is a 3 -group and m  3 (I) ≥ m3 (L) = 5, contradiction. This completes the proof of the lemma.  Lemma 16.14. Assume (16A) with p = 2, r = 0, and K/O2 (K) ∈ G62 , but with  L2 (q) for any odd q. Then the following conditions hold: L/O2 2 (L) ∼ = (a) I ∈ G2 − G82 , or I ∼ = L± 4 (3) or 2U4 (3); ∼  L/O2 2 (L), then I ∈ G2 − G82 ; and (b) If I/O2 (I) = (c) If I/O2 (I) ∼  L/O2 2 (L), then I ∈ Chev − Chev(2) − Alt. = Proof. Since K/O2 (K) ∈ G62 , K ∈ Chev(r) for some odd r, and so L ∈  L2 (r a ) for any a, L ∈ Alt. Chev(r). As L/O2 2 (L) ∼ = Suppose that L ∈ Chev(s) for some s = r. Then by [IA , 2.2.10, 6.1.4] and Lemma 14.5e, L ∼ = Sp4 (3), r = 3, and s = 2. We claim next that no long pumpup I1 of L/O2 (L) lies in Spor, and prove the claim by induction on the length of a pumpup chain in K2 : L/O2 (L) 3, u induces a field automorphism on H. Then from [IA , 4.10.3], mp (CH (u)) = mp (H) > 1, the last inequality as H ∈ Cp ∪ Tp . Thus mp (CX (u)) ≥ mp (u × CH (u)) ≥ 3, as asserted. So in proving (b), we may assume that u ∈ Ip (X) induces an inner automorphism on H. Then by [IA , 4.10.3e], there exists A ∈ Ep∗ (H) such that [u, A] = 1. Thus (b) holds if mp (u H) ≥ 3. As mp (H) > 1, we are done unless mp (H) = 2 and u ∈ H, and CX (H) = 1, i.e., F ∗ (X) = H is simple. We may then assume that either x ∈ H = X, or X = H x with x inducing a field automorphism on H. In the first case, note that mp (K) > 1 as K/Op (K) ∈ Gp , so x ∈ K as mp (H) = 2. But then x ∈ Z(K) so p splits K, contradiction. Hence, X = H x with x inducing a field automorphism on H. By [IA , 4.10.3f], any two Ep2 -subgroups of H are conjugate, and also mp (CH (f )) = 2 for any field automorphism f of H of order p,  by [IA , 4.10.3]. These conditions imply (b), so the proof is complete. Lemma 17.5. Suppose that X has a component K ∈ Chev(2) ∩ Gp for some odd prime p, and X = KP with P ∈ Sylp (X), and mp (P ) ≥ 4. Then X = ΓP,2 (X). Proof. Suppose false; thus ΓP,2 (K) < K. Let A ∈ E∗ (P ), so that mp (A) ≥ 4. By [IA , 7.3.6], ΓA,−3 (K) = K, so mp (A) = 4. Moreover p = 3 and K has level 2, for otherwise ΓP,2 (K) ≥ ΓA,−2 (K) = K. If CA (K) = 1, then write A = A0 × a with 1 = a ∈ CA (K); then ΓP,2 (K) ≥ ΓA0 ,1 (K) = K by [IA , 7.3.7]. Thus CA (K) = 1. As m3 (A) = 4 = m3 (P ) and K has level 2, and in view of [IA , 4.10.3,7.3.5], we have − ∼ K∼ = A− 4 (2), A5 (2), B4 (2), D4 (2), or F4 (2). As K ∈ G3 , K ∈ C3 and so K = B4 (2).

290

11. PROPERTIES OF K-GROUPS

But B4 (2) ≥ SL2 (2)×E33 , so there is a 2-local subgroup of K containing a subgroup B ≤ P, B ∼ = E33 . Then ΓP,2 (K) ≥ ΓB,−1 (K) = K by [IA , 7.3.1]. The proof is complete.  Lemma 17.6. Let p be an odd prime and X a K-group with P ∈ Sylp (X) and mp (P ) ≥ 2. Assume that Γ ≤ X and the following conditions hold: (a) ΓP,2 (X) ≤ Γ, and  if mp (P ) ≥ 3, then  ΓoP,1 (X) := ΓP,2 (X), CX (a) mp (CP (a)) ≥ 3 ≤ Γ; and (b) For some D ≤ P with D ∼ = Ep2 , ΓD,1 (X) ≤ Γ. Then one of the following holds: (a) Γ = X; (b) Γ has a p-component H such that H/Op (H) ∈ Chev(2); or (c) Op (X) ≤ Γ < X, and I := F ∗ (X/Op (X)) is simple. Moreover either Γ is p-solvable or else p = 5 with I ∼ = A10 or F i22 , in which cases the (unique) non-p-solvable chief factor of Γ is A5 × A5 or D4 (2), respectively. Proof. Since Op (X) ≤ ΓD,1 (X) ≤ Γ, there is no loss in passing modulo Op (X) and assuming that Op (X) = 1. We argue that for any 1 = R  P , we have NX (R) ≤ Γ. As ΓP,2 (X) ≤ Γ, this is obvious if mp (R) > 1. But if mp (R) = 1, then Ω1 (R) ≤ Z(P ), so NX (Ω1 (R)) lies in ΓD,1 (X) if mp (P ) = 2, and mp (CP (Ω1 (R))) ≥ 3 otherwise. Hence NX (R) ≤ Γ by our hypotheses. In particular we may assume that Op (X) = 1. Moreover we may assume that (17B)

for any 1 = N  X, we have N ≤ Γ,

for otherwise X = N NX (P ∩ N ) ≤ Γ as P ∩ N  P . This implies that CX (N ) = 1, as CP (N )  P and N ≤ NX (CX (N )). Similarly, N has no nontrivial decomposition N = N1 × N2 with N1 and N2 both P -invariant. Since Op p (X) = 1, E(X) = K1 × · · · × Kr where the Ki are all simple, and P permutes the Ki transitively. If r > 1, it follows that r ≥ p ≥ 3. In that case mp (CQ (u)) ≥ r ≥ 3 for all u ∈ Ip (Q), where Q = P ∩ E(X). Hence by assumption (a), Γ contains ΓQ,1 (K) and hence E(X) ≤ Γ, contrary to (17B). Therefore r = 1 and we let K = K1 = E(X). Suppose that mp (P ) ≥ 3. Since ΓP,2 (X) ≤ Γ, Γ controls X-fusion in P by [III8 , 6.14]. If in addition (17C)

mp (CX (g)) ≥ 3 for all g ∈ X of order p,

it follows that CX (g) ≤ Γ for all g ∈ Ip (P ) that are extremal in P , and then for all g ∈ Ip (P ) by control of fusion. In this case we conclude that ΓP,1 (X) ≤ Γ. Then the structure of K and Γ is given by [IA , Theorems 7.6.1, 7.6.2], as well as the maximality of ΓP,1 (X) = Γ in X. As mp (X) ≥ 3, K ∈ Chev(p), of twisted rank 1, and Γ is solvable, so conclusion (c) holds. Likewise if K ∈ Chev(p), then [IA , Theorem 7.6.3] implies that K has twisted rank 1 and Γ = ΓP,1 (X) is solvable, so (c) holds. We may therefore assume that (17C) fails and that K ∈ Chev(p). Suppose that K ∈ Spor ∪ Alt. Since (17C) fails, Lemma 3.3 implies that mp (K) = 2. If K ∈ Alt, then P ∼ = Ep2 and Γ = ΓP,1 (K) = K unless K ∼ = A2p , in which case F ∗ (Γ) ∼ = Ap × Ap , p ≥ 5. Then conclusion (b) holds if p > 5 and (c) holds if p = 5. If K ∈ Spor and ΓP,1 (X) < X, then [IA , Theorems 7.6.1, 7.6.2] again imply that

17. GENERATION

291

conclusion (c) holds. According to [IA , Theorem 7.5.5], this leaves the following sporadic cases: (p, K) = (3, M12 ), (5, HS), (5, Ru), (7, He), (7, O  N ), (7, F i24 ), with NK (D) ∼ = AGL2 (3) in the first case. In that case NK (D) is maximal in X by [IA , Table 5.3b], so conclusion (a) or (c) holds. In the remaining cases, P ∼ = HS, Z(P ) is not weakly = p1+2 and except for K ∼ closed in P with respect to K. By assumption NX (Z(P )) ≤ ΓD,1 (X) ≤ Γ. Also, Γ again controls X-fusion in P , and so NX (R) ≤ Γ for all R ≤ P such that R ∈ Z(P )X . Except for the HS case, there is then no loss in assuming that E1 (D) ⊆ Z(P )X . Using [IA , Tables 5.3prsv] to see that Op (CX (a)) = 1 for all a ∈ P # and that NK (P ) is irreducible on P/Z(P ), we may apply the argument of [IA , Proposition 7.5.6] and conclude that L := F ∗ (Γ) is simple. But NK (Z(P )) ≤ Γ, and P ∈ Sylp (K), and so by Lemma 17.3, the only possibility is that p = 7, K ∼ = F i24 , and L ∼ = He. Therefore Γ has a component isomorphic to He and conclusion (b) holds. In the case K ∼ = HS, p = 5, we have O5 (CX (a)) = 1 for all a ∈ P # , and Z(P ) is weakly closed in P with respect to K, hence with respect to Γ. If Z(P )  Γ, then Γ is solvable, while otherwise, by [IA , Proposition 7.8.2], F ∗ (Γ) ∼ = U3 (5) or F ∗ (Γ) is sporadic. Hence conclusion (b) or (c) holds. We may finally assume that K ∈ Chev −Chev(p), and Γ = X. If m := mp (P ) ≥ 3, then we choose A ∼ = Epm , A ≤ X. By assumption (a), ΓA,1 (X) ≤ Γ. Hence by [IA , 7.3.6] Γ = X unless possibly m = 3, p = 3, and K ∈ Chev(2) has level 2. Then by [IA , 7.3.5, 4.10.3], the only possibilities are that K ∼ = U4 (2) or Sp6 (2). As P Sp (3) ∈ Chev(3), that case does not arise; and if K∼ U4 (2) ∼ = = Sp6 (2), then K 4 contains L2 (2) × E32 so Γ = X by [IA , 7.3.1]. Thus we may assume that m = 2. This time, excluding the Chev(3)-groups B2 (2) ∼ = A1 (9) and G2 (2) ∼ = U3 (3) and 1 2  ∼ ∼ 2 L2 (8) = G2 (3 ) , [IA , 7.3.5, 4.10.3, 7.3.8] yield only K = L3 (4) with p = 3, or 5 1 2 B2 (2 2 ) or 2F4 (2 2 ) with p = 5. In all cases Ω1 (P ) is abelian, so Γ = ΓP,1 (X) and Γ is solvable by [IA , 7.6.2]. Hence conclusion (c) holds and the proof is complete.  We conclude this section with several results related to Lemma 1.16. Lemma 17.7. Let the assumptions be as in Lemma 1.16. Let r be a prime distinct from p, such that either r or p equals 2. Suppose that there is a group I ∗ ∈ Chev(r) such that I ∗ /Op (I ∗ ) 2 and since I is nonsolvable, n ≥ 5. This implies n + kp2 ≥ 5 + p2 ≥ 4p + 2, with strict inequality if p > 3, and the proof is complete. 

292

11. PROPERTIES OF K-GROUPS

Lemma 17.8. Let the assumptions and notation be as in Lemma 1.16, with p = 2 and I ∼ = Am for some m ≥ 5. Let t be a root involution of I. Then K = I, CK (t). Proof. By Lemma 1.16b, (K, I) = (Am+4k , Am ), or (K/Z(K), I) = (M12 , A5 ) or (J2 , A5 ). In the first case, since I is a component of CK (x) for some involution x ∈ Aut(K), I has support ΩI with |ΩI | = m in the natural permutation representation of K on Ω = {1, . . . , m+4k}. The support Ωt of t has cardinality 4 and Ωt ∪ΩI = Ω.   As  Ωt ∩ ΩI = ∅. Both |ΩI | and |Ωt | are at least 5, so I, CK (t) =  m ≥ 5, AΩI , AΩt = AΩ = K, as required. Suppose that K = J or 2J, where J = M12 or J2 . Let J = 2J. In each case for J, we see from [IA , Tables 5.3bg] that J has two classes of involutions: the 2-central  and the non-2-central involutions, which do involutions z, which split over Z(J),   then z is 2-central in J.  not split over Z(J). Moreover if z is a preimage of z in J, It follows that the involutions in I are 2-central in K. Let y ∈ I be an involution. Again by [IA , Tables 5.3bg], CK (y) is maximal in K. As obviously I ≤ CK (y), we have K = I, CK (y), and the lemma follows.  Lemma 17.9. E22 detects pumpups of 2HS. Proof. By definition [III8 , 2.10ff.], we must show that there is no quasisimple group L and four-subgroup V of Aut(L) such that for some I ≤ CL (V ) with I ∼ = 2HS, I is a component of CL (v) for all v ∈ V # . In particular m2 (CAut(L) (I)) ≥ 2. Since 2HS ↑2 L, L ∈ Spor and then from [IA , Tables 5.3], L ∼ = F5 and CAut(L) (I) has Z4 Sylow 2-subgroups. The result follows.  Lemma 17.10. Let p be an odd prime and K ∈ Chev(2) ∩ (G2p ∪ G4p ). Then Ep2 detects pumpups of K. Proof. Otherwise there would exist J ∈ Kp and an Ep2 -subgroup E ≤ Aut(J) and a common p-component K ∼ = E (q) of all CK (e), e ∈ E# . But then J and K would satisfy the hypothesis of Lemma 1.16 (in place of K and I there), contradicting our assumption that K ∈ Chev(2) ∩ (G2p ∪ G4p ). The proof is complete.  18. Splitting Primes Lemma 18.1. Let (x, K) ∈ ILop (G) with mp (C(x, K)) = 1, mp (K) > 2, and K/Op (K) ∈ Gp ∩ Chev(2), where p is an odd prime. Let X = CG (x). Assume that p does not split K. Let (y, L) be an acceptable subterminal (x, K)-pair. Let L0 ≤ G with L0 ∈ Kp and suppose that L is a component of CL0 (ξ) for some ξ ∈ Ip (Aut(L0 )). Then there exist an odd prime s = p and subgroups Q and I of K with the following properties: (a) |Q| = s and Q ≤ L; (b) I is a component of CK (Q) with p and s dividing |I|; (c) I/Os (I) ∈ Gs ∩ Chev(2); (d) ms (IQ) ≥ 4, and if dp (K/Op (K)) = 2, then ds (I/Os (I)) = 2; (e) s splits both I and K, and q(I) = q(K); (f) If s = 3, then q(K) ≤ 23 ; moreover, if K ∼ = E6 (2n ) ∼ = E6 (2), with p n n dividing Φ3 (2 ), then s divides 2 − ;

18. SPLITTING PRIMES

293

Table 18.1. Groups in the Proof of Lemma 18.1 K

a

L

H

I

E6 (q)3+ , 6− 3D4 (q)A1 (q 3 )A5 (q)

L0

I0

E6 (q), E7 (q)

A5 (q), D6 (q)

E7 (q)

3

A5 (q) A3 (q) D6 (q) A5+3n (q),E7 (q)A3+3n (q),D6 (q)

E7 (q)

6

− − − A− 5 (q) A3 (q) D6 (q) A5+3n (q),E7 (q)A3+3n (q),D6 (q)

E8 (q)

3

E6 (q) A5 (q) E7 (q)

E8 (q)

E7 (q)

E8 (q)

3

E6− (q) A− 5 (q) E7 (q)

E8 (q)

E7 (q)

E8 (q)

4

D6− (q) D4− (q) E7 (q) D8 (q), E8 (q)

D6 (q), E7 (q)

(g) One of the following holds: (1) r(K) = r(I) + 2 ≥ 6 and τ (K) = τ (I); or (2) r(K) = r(I) + 1 ≥ 4 and if τ (K) = BC, then τ (I) = BC; (h) Let M be the subnormal closure of I ∩ L in CL0 (Q). Then ξ does not centralize M/Op (M ); (i) One of the following holds: (1) There exists a component H of CL (Q) such that H ≤ I and p divides |H|; or (2) There exist Q1 ≤ L and I1 ≤ K such that conclusions (a), (b), (c), (d), and (i1) hold with Q1 and I1 in place of Q and I; moreover, the following conditions hold: (i) [Q, Q1 ] = 1; (ii) I is a component of CI1 (Q); (iii) mp (Os (CX (Q))) = 1; and   (iv) O s (I ∩ L) ≤ [I ∩ L, I ∩ L], and O s (I ∩ L) ≤ Z(I). Proof. We write K = d L(q), q = 2n , and let a be the first positive integer for which p divides Φa (q). By assumption, p does not split L, so (18A)

a ≥ 3 and p > 3.

Also by assumption (18B)

mp (K) ≥ 3.

By Lemma 14.4 and the oddness of p, q(K) = q(L). Suppose first that (18C) K is of exceptional Lie type. In this case our selection of s, Q, and I is independent of the isomorphism type of the given group L0 , but depends on p, K, and L. From Table 13.1 we see that the conditions (18A) and (18B) restrict us to the possibilities for K and a in Table 18.1. (Note: 3+ , 6− in the second column means that a = 3 or 6 according as  = + or −.) By definition of acceptable subterminal pair [III7 , Definition 6.9], L is then as in Table 18.1.

294

11. PROPERTIES OF K-GROUPS

In every row of Table 18.1 but the first, we set  = 1. If q = 2 and  = 1, we take s = 3; otherwise we take s to be any prime divisor of q −  other than 3, if one is available. If none is available, we take s = 3. Then (f) holds. We choose Q = t, where t is an element of order s in a long root A1 (q)subgroup L1 of L. Moreover, L1 is also a long root A1 (q)-subgroup of K, so putting I = E(CK (Q)) and H = E(CL (Q)), we get that I and H are as in Table 18.1, and alternative (i1) holds. Evidently r(K) = r(I) + 1, so (g2) holds. By [IA , 4.10.3], mp (K) = 3 or 4, with Op (Z(K)) = 1 by [IA , 6.1.4], so dp (K) = 4. By our choice of s and [IA , 4.10.3], ms (I) ≥ 4, so (d) holds. If s = 3, then (c) is clear from the definition of Gs [I2 , p. 102]. Indeed given the groups I in the table, we have I/Os (I) ∈ Gs unless s = 3 and I ∼ = A− 5 (2), so that − K∼ = E6 (2). But in that case there is no prime divisor p of |K| for which a = 6, a contradiction. Thus (c) holds. As (a), (b), and (e) are clear, it remains to prove (h). Notice first that L ∈ Chev(r) ∪ Alt for any odd r, and mp (L) ≥ 2 in every case (see [IA , 4.10.3]). Thus with Lemma 1.1ab it follows easily that L0 , like any pumpup of L, must lie in Chev(2). If (h) fails, then since our choice of (s, Q, I) is independent of L0 , we may choose a counterexample with L0 minimal; it then follows that L is terminal in L0 . If ξ ∈ Inndiag(L0 ), then the candidates for L0 can be read off from Table 13.1 and [IA , 4.8.2, 4.8.4], and they are listed in Table 18.1. Since Q lies in a long root A1 -subgroup of L, it lies in a long root A1 -subgroup of L0 , and this gives the possibilities for I0 := E(CL0 (Q)) in Table 18.1. Since p divides |I0 |,  I0 = O p (CL0 (Q)(∞) ) contains H and it is clear from the table that (18D)

H/Op (H) ∼ = I0 /Op (I0 ).

On the other hand if ξ ∈ Inndiag(L0 ), then since p > 3, ξ is a field automorphism of L0 . Hence by [IA , 4.2.3], ξ induces a field automorphism of order p on I0 , with  H = O p (CI0 (ξ)). Thus (18D) holds in this case as well. Now H ≤ I ∩ L, so I0 is contained in the subnormal closure M of I ∩ L in CL0 (Q). Since (h) fails, [M/Op (M ), ξ] = 1, so [I0 /Op (I0 ), ξ] = 1. Then CL0 (ξ Q) covers I0 /Op (I0 ). As H   CL (Q)   CL0 (ξ Q) and H ≤ I0 , it follows that HOp (I0 )   I0 , contradicting (18D). Thus (h) holds. This finishes the proof if (18C) holds. Suppose then that K is a classical group:  (q), or Bm (q). K∼ = Am (q), Dm For the moment we exclude the following case, which will require separate consideration: (18E)

± (q), k > 0 K∼ = D4 (q), L0 ∼ = D4+2k = D6− (q), p divides q 2 + 1, L ∼

With this case excluded we again choose (s, Q, I) independently of L0 . Let V be the natural module for K, and let d be the minimal dimension of a classical vector space of the same type as K and admitting an isometry of order p. Consider the case (18F)

± (q) or Bm (q). K∼ = Dm

Then d is even, and as p does not divide q 2 − 1, we must have d ≥ 4. As mp (K) ≥ 3 it follows that 2m = dim V ≥ 3d ≥ 12. Moreover by [III7 , Definition 6.9], L ∼ = D (q) or B (q), where 2m = 2 + d and thus  ≥ d ≥ 4.

18. SPLITTING PRIMES

295

We make some remarks in the “lowest” case L ∼ = D4 (q). Namely as mp (L) ≥ 2, −  = 1 and p divides q 2 + 1, whence K ∼ = D6 (q). As we are excluding (18E), the only possibility, by [IA , 4.8.2] and Table 13.1, is that ξ ∈ Inndiag(L0 ), so ξ is a field automorphism and L0 ∼ = D4 (q p ). Returning to the general case (18F) but still assuming (18E) does not hold, we choose s dividing q 2 − 1 avoiding s = 3 if possible; thus (f) holds. This time we choose Q = t of order s to lie in L and have (minimal) support, of dimension 2, on V . The supports of Q on V and on the natural L-module CV (y) coincide. ± Then I = E(CK (Q)) ∼ (q) or Bm−1 (q), implying (c) as well as (g2) since = Dm−1 ± m ≥ 6. Moreover H = E(CL (Q)) ∼ (q) or B−1 (q) has order divisible by p = D−1 since mp (L) ≥ 2, so (b) and (i1) hold. With [IA , 4.10.3], ms (I) ≥ m − 2 ≥ 4. Also if dp (K/Op (K)) = 2, then mp (K) ≥ 5, whence 2m ≥ 5d ≥ 20 and it is clear that ms (I) ≥ 5. Therefore (d) holds. As (a) and (e) are obvious, it remains to verify (h). Suppose that (h) fails. Let I0 be the subnormal closure of H in CL0 (Q). It suffices to verify (18D), for then the paragraph following (18D) can be repeated to obtain a contradiction. If ξ ∈ Inndiag(L0 ) then again as p > 3, ξ is a field automorphism on L0 and then, by [IA , 4.2.3], on I0 as well, with H = E(CI0 (ξ)), and (18D) is clear. If ξ ∈ Inndiag(L0 ) then we determine the possible isomorphism types of L0 from [IA , 4.8.2] and Table 13.1, and the facts that mp (L) ≥ 2 and p does not divide q 2 − 1. The main possibilities are L0 ∼ = Dn± (q) or Bn (q), n ≥ m, with natural module V0 = CV0 (L) ⊕ [L, V0 ], and with [L, V0 ] being the natural L-module. (It is here and only here that we have to exclude (18E) because the term “the natural D4 (q)-module” is not well defined, and in particular the action of Q on V0 in this case would not be uniquely determined.) In these main cases for ± L0 , Q has 2-dimensional support on V0 so I0 contains Dm−1 (q) or Bm−1 (q), and certainly (18D) holds, as desired. There is a single further possibility: K ∼ = D8 (q), L∼ = D6− (q), p divides q 2 + 1, and L0 ∼ = E8 (q). But in this case there is a subgroup L ≤ L1 ≤ L0 with L1 ∼ = D8 (q), so Q has 2-dimensional support on the natural L1 -module and again I0 contains D7± (q), yielding (18D), as desired. Now consider the case K∼ = Am (q), which is similar. We make only a few remarks about it. This time d ≥ 3 so m + 1 = dim V ≥ 3d ≥ 9 and m ≥ 8. We have L∼ = Am−d (q). We choose s just as before, and again choose Q = t of order s to have 2dimensional support on V . This Q lies in a root A1 (q) subgroup of L and of K. We have I ∼ = Am−d−2 (q) has order divisible by p since = Am−2 (q), and H ∼ m − d − 2 ≥ d. This time (g1) holds. Also ms (I) − ms (Z(I)) ≥ [(m − 2)/2] ≥ 3, proving the first part of (d). This time the main possibilities are L0 ∼ = An (q), n ≥ m − 1, with natural module V0 , and with [L, V0 ] the natural L-module. Hence I0 ∼ = An−2 (q) contains Am−3 (q) so (18D) holds. There is again a single further possibility: K ∼ = A5 (q), H ∼ = A3 (q), a = 3 or 6 according as  = 1 or = A8 (q), L ∼ −1, and L0 ∼ = E7 (q). In that case there is L ≤ L1 ≤ L0 with L1 ∼ = A6 (q), and so  I0 contains E(CL0 (Q)) ∼ = A4 (q). This implies (18D), and as remarked above, (h) follows as before. To complete the proof of the lemma it remains only to consider the case (18E). We choose s as before, so that (f) holds. Now we let Q1 be a subgroup of L ∼ =

296

11. PROPERTIES OF K-GROUPS

D4 (q) of order s with 2-dimensional support on the natural K-module V . Then Q1 has 2-dimensional support on [V, L], which is a natural L-module. We set I1 = E(CK (Q1 )) ∼ = D5± (q). As noted in the proof of the orthogonal case above, all parts of the lemma except (h), and including (i1), hold for Q1 and I1 . However, our previous proof of (h) fails for Q1 and I1 , and we choose Q of order s instead to be contained in a root A1 (q) subgroup of L and of K, with [V, Q1 ] ≤ [V, Q]. Thus dim[V, Q] = 4. Then we may assume that [Q, Q1 ] = 1 as the orthogonal group on [V, Q] has abelian Sylow s-subgroups. Set I = E(CI1 (Q)) ∼ = D4− (q) ∈ Gs ; then I is a component of CK (Q), of order divisible by p, so (b) and (c) hold. Moreover ms (I) = 3 = mp (K), so (d) holds, and (g1) is evident, as are (a) and ∼ (e). As I ∩ L ∼ = Ω+ 4 (q) = SL2 (q) × SL2 (q), (i2(iv)) holds. Furthermore, I contains a Sylow p-subgroup of CK (Q), and s divides |I|, so Os (CK (Q)) is a p -group, implying (i2(iii)). It remains as usual to check (h). Let V0 be the natural L0 module, of dimension n ≥ 12. Since Q has 4-dimensional support on [L, V ], it has 4-dimensional support on each of the three natural L-modules. Thus it has ± (q), I0 has order 4-dimensional support on V0 , so I0 := E(CL0 (Q)) ∼ = D(n−4)/2 divisible by p, and [I0 , ξ] = 1. But M contains the subnormal closure of I ∩ L in I0 , which is I0 , so [M/Op (M ), ξ] = 1. The proof of the lemma is complete.  In the next two lemmas, we assume the following setup. (1) N ∈ Chev(2) and x ∈ Ip (Aut(N )); (2) K is a p-component of CN (x), mp (CAut(N ) (K)) = 1, and K/Op (K) ∈ Gp ; (18G) (3) p is an odd prime not splitting K; (4) A is a p-internal source of K with mp (A) = 3; and (5) s is an odd prime splitting K, and if K ∼ = E6 (2) and p = E6 (q) ∼ n n divides Φ3 (2 ), then s divides 2 − . Lemma 18.2. Assume (18G). Suppose that x induces an inner-diagonal automorphism of N . Then there exists a hyperplane A0 of A and E ≤ CN (A0 ) such that the following conditions hold: (a) E ∼ = Es2 and ms (CN (E)) ≥ 3; (b) For each a ∈ A# 0 , there exists a Lie component L of CN (a) such that L is not s-solvable and furthermore if s = 5, then L ∼  A5 or D4 (2). =

Proof. Suppose first that N is a classical group. Then  /Z N =N  is the universal cover of N and Z ≤ Z(N  ). Since p does not split K, where N    Z(N ) is a p -group. Let V be the natural N -module, over, say, the finite field F of characteristic 2. Let V (p) be a minimal-dimensional classical space over the same field F and of the same type (linear, unitary, symplectic, orthogonal) as V , nondegenerate in the nonlinear cases, and admitting isometries of order p. Let d(p) = dim V (p). Thus d(p) ≥ 3 in the linear and unitary cases, and d(p) ≥ 4 with d(p) even, in the symplectic and orthogonal cases. Then by definition of p-internal source, and because mp (CN (K)) = 1 and p does not split K, we may write V = V0 ⊥ V1 ⊥ V2 ⊥ V3 ⊥ W and A = A1 × A2 × A3 ,

18. SPLITTING PRIMES

297

∼ V1 = ∼ V2 = ∼ V3 ∼ i , V ] for i = 1, 2, 3, and where V0 = = V (p), V0 = [x, V ], Vi = [A ⊥  V0 = V1 + V2 + V3 + W is a natural K-module. Thus [A, V ] = V1 ⊥ V2 ⊥ V3 . Here i and A  are isomorphic preimages of Ai and A in N  . (See [IA , 4.8.1–4.8.4] for a A discussion of the action of classical groups on their natural modules.) We take A0 = A1 × A2 , 0 ) = V0 + V3 + W . Let d = dim(W ), so that so that CV (A  6 if K is linear or unitary  (18H) dim CV (A0 ) = 2d(p) + d ≥ 8 if K is symplectic or orthogonal  be an elementary abelian s-subgroup of C  (V1 + V2 ) of maximal rank. Let E N Then  = 1, so [A0 , E] = 1 0 , E] [A  in N . Moreover where E is the (isomorphic) image of E (18I)

ms (E) ≥ 3

since dim(V0 + V3 + W ) ≥ 6 and s divides 22n − 1, where 2n is the level of K and N. a Now given a ∈ A# , choose an Es2 -subgroup Ea ≤ E such that its preimage E   in E has a fixed subspace whose dimension dEa is as large as possible. Then Ea fixes a subspace Va of V containing V0 + V3 + W with codimension at least 2. Therefore for each e ∈ Ea , CN (e) has a component of the form SL± (Va ) with dim Va ≥ 8, or Ω(Va ) with dim Va ≥ 10. All parts of the lemma now follow, completing its proof when N is a classical group. Now assume that N is an exceptional group. Since mp (N ) ≥ 4 but p does not split K, the only possibility (see Table 13.1) is N ∼ = E8 (2n ), with mp (N ) = 4, and p dividing Φm0 (2n ) for m0 = 3, 4, or 6. We shall find subgroups N0 and I such that N0 , I ≤ N , N0 and I are semisimple, [N0 , I] = 1, A0 := A ∩ N0 ∼ = Ep2 , (18J) ms (I) > 1, ms (N0 ) > 0, and I has a non-s-solvable composition factor, not isomorphic to A5 . These conditions imply the desired conclusions. Indeed for any a ∈ A# 0 , I ≤ O 2 (CN (a)), so CN (a) has a Lie component not isomorphic to A5 . Moreover if s = 5 then n is even as s splits N ; so by [IA , 4.2.2] every Lie component of CN (a) has level a power of 4. Thus (b) holds. We take Ea ∈ Es2 (I); then A0 ≤ N0 ≤ CN (I) ≤ CN (E), as desired. Also as ms (N0 ) > 0, there are elements of order s in N0 − Z(N0 ) by [IG , 16.11], so (a) holds as [I, N0 ] = 1. The groups N0 and I are found as follows. Let N1 be an inductive associate of N [III3 , Table 2.1], so that N1 ∼ = E6 (2n ), D7 (2n ), or E6− (2n ) according as m0 = 3, 4, or 6. Note that if m0 = 6, then n > 1 as p > 3. By definition of p-internal source [III3 , Definition 2.1], A is a p-internal source of N1 , which in these cases simply means that A ≤ N1 with A ∼ = Ep3 . For m0 = 4, we take N0 ∼ = D4 (2n ) to centralize a subspace of dimension 6 on the natural N1 -module, and I = E(CN1 (N0 )) ∼ = A3 (2n ). 2n Since s splits K, s divides 2 −1, so clearly ms (I) ≥ 2 and ms (N0 ) > 0, as required. The cases m0 = 3, 6 are similar, with N0 ∼ = I isomorphic to A2 (q)A2 (q) and  = 1 or −1, respectively. The proof is complete.  

11. PROPERTIES OF K-GROUPS

298

Lemma 18.3. Assume (18G). Suppose that x is a field automorphism of N . Then there exists a ∈ Ip (K) and E ≤ CN (a) such that the following conditions hold: (a) E ∼ = Es2 and ms (CN (E)) ≥ 3; (b) There exists a Lie component L of CN (a) such that L is not s-solvable and furthermore if s = 5, then L ∼  A5 or D4 (2). = Proof. The proof is similar to that of Lemma 18.2. If N is a classical group,  be its universal cover and V the natural N  -module, and take a as the then let N   -module V . Then image of  a ∈ Ip (N ) with minimal support V1 on the natural N ⊥ as p does not split K and mp (N ) ≥ 3, dim V1 ≥ 6 and as in Lemma 18.2 we find  ≤ C  (V1 ). a suitable E ≤ CN (a) as the image of some E N If N is an exceptional group and mp (N ) > 3 we can simply choose A0 as in Lemma 18.2 and any a ∈ A# 0 will satisfy our requirements. So assume that N is exceptional and mp (N ) = 3. The only cases, since p does not split N , are N ∼ = E6 (2n ) or E7 (q), p dividing Φ3 (q),  = ±1. We take commuting subgroups I ∼ = A2 (2n )A2 (2n ) (for N = E6 (2n )) or A5 (q) (for N = E7 (q)) = A2 (2n ), and N0 ∼  and we can take E ≤ N0 and a ∈ Ip (I) to reach the desired conclusion. 19. Double Pumpups In this section we begin to analyze vertical pumpup chains of length greater than one, for their effect on the function F. The reader should review Lemma 12.3 before proceeding. We fix a prime p. Lemma 19.1. Suppose that K, x ∈ Aut(K), K1 , k, and a are as in Lemma 12.3 with k = 0 and a > 1. If p = 2, assume that 1 > 1. Then a and the triple (p, K1 , K) are one of the following possibilities: ± ± 2 (a) a = 2, (p, K1 , K) = (2, Dm (q ), D2m (q)), m ≥ 3, and K1 is not 2saturated; (b) a = 2, (p, K1 , K) = (2, Cm (q 2 ), C2m (q)), m ≥ 2; n (c) a = 2, (p, K1 , K) = (3, 2A2 (2n ), 2F4 (2 2 ));  3  (d) a = 3, (p, K1 , K) = (3, A2 (q ), E6 (q)), with q ≡  (mod 3) and  = ±1; (e) a = 2, (p, K1 , K) = (5, 2A4 (q 2 ), E8 (q)), with q 2 ≡ −1 (mod 5). Proof. From Lemma 12.3, the assumption k = 0 implies that  = a1 , with a ≥ 2,

(19A) where K = L (q) and K1 = Suppose first that d

d1

L1 (q a ).

x ∈ Inndiag(K). By [IA , 4.2.2], a1 is the number of nodes in the “unfolded” Dynkin diagram Π1 of K1 , which in turn is obtained from the Dynkin diagram Π of K by deleting a nonempty set of nodes, or from the extended Dynkin diagram of K by deleting a node corresponding to a fundamental root whose coefficient in the expansion of the highest root is p. As a1 = , we must be in the latter case; furthermore as K1 is a single component, Π1 is the disjoint union of a copies of Σ1 , where K1 = d1 Σ1 (q a ). A routine inspection of all Dynkin diagrams shows that the only possibilities for (a, p, K, Σ1 ), where K is the algebraic group overlying K, and such that a > 1 and 1 > 1, are (2, 2, C2n (n > 1), Cn ), (2, 2, D2n (n > 2), Dn ),

19. DOUBLE PUMPUPS

299

(2, 3, F4 , A2 ), (3, 3, E6 , A2 ), and (2, 5, E8 , A4 ). In each case there is a unique choice for the root to be deleted from the extended diagram of K, viz., αn , αn , α2 , α3 , and α5 , respectively. Descending to the finite group K from K, we find the enumeration of corresponding automorphisms x of K and the structure of their centralizers in [IA , Tables 4.5.1-2, 4.7.3A, and 4.7.3B1 ], according as p = 2, p = 3, and p = 5, respectively. The quadruples (x, p, K, K1 ) that occur, up to conjugacy, are (tn , 2, C2n (q), Cn (q 2 )), n + − + 2 (tn or t (tn , 2, D2n (q), Dn− (q 2 )), (t2 , 3, 2F4 (2 2 ), 2A2 (2n )), n , 2, D2n (q), Dn (q )),    3 n odd, (t3 , 3, E6 (q), A2 (q )),  = ±1, q ≡  (mod 3), and (t5 , 5, E8 (q), 2A2 (q 2 )), q 2 ≡ −1 (mod 5). The case K = F4 (q) does not occur with a > 1; in the E6 case, the automorphisms t±1 3 , in the notation of that table, do satisfy the hypotheses of this lemma; and in the E8 case, the automorphism t5 satisfies the hypotheses of our lemma iff q 2 ≡ −1 (mod 5). Finally, the additional assertion of (a) is also found in [IA , Table 4.5.2], completing the proof in the case x ∈ Inndiag(K). By Lemma 12.3, we have a = 1/p if x ∈ Aut0 (K), contrary to our assumption a > 1. The only remaining case is that in which x ∈ Aut0 (K) − Inndiag(K), so that x is of graph type. But from [IA , Tables 4.5.1, 4.7.3A], however, (19A) is never satisfied in these cases. The lemma is proved.  Lemma 19.2. Suppose that K, x ∈ Aut(K), K1 , and a are as in Lemma 12.3 with a > 1, p > 2, and mp (K1 ) ≥ 2. Then K1 ∼ = A1 (q1 ),  = ±1, with p dividing q1 − . Proof. We argue first that x ∈ Inndiag(K). If x is of field type then a = 1/p. If x induces a graph automorphism then p = 3 and K ∼ = D4 (q) or 3D4 (q) and by [IA , Table 4.7.3A], a = 1. Since we are assuming a > 1 these cases cannot occur, so x ∈ Inndiag(K). Let q be the level of K, so that q1 = q a . If K is a classical group, then the conclusion of the lemma follows quickly from [IA , 4.8.2, 4.8.4]. If K is an exceptional group, we refer to Table 13.1 and, using mp (K1 ) ≥ 2 and [IA , 4.10.3], 2 2   3 deduce that (K, K1 ) = (E8 (q), A− 1 (q )) with p dividing q + 1, (E6 (q), A2 (q )) with √ p = 3 dividing q −  and hence dividing q 3 − , or (2 F4 ( q), A− 2 (q)) with p = 3 dividing q + 1. The lemma follows.  In the next two lemmas we begin to investigate double pumpups. The setup is the following: (1) p and r are distinct primes, one of which is 2; (2) Ki ∈ Chev(r) ∩ Kp , i = 0, 1, 2; (3) Ki−1 is a central quotient of a component of CKi (xi ) for some xi ∈ Aut(Ki ) of order p, for i = 1, 2; (19B) (4) Ki = di Li (qi ), i = 0, 1, 2; (5) ρi = ρ(Ki , Ki−1 ), i = 1, 2, and ρ := ρ1 ρ2 = ρ(K2 , K0 ) (see (12B)); and (6) 0 > 1. Notice that Lemma 12.3 applies to the pumpups Ki−1 ↑p Ki , 1 In

i = 1, 2.

the last line of Table 4.7.3B, “t4 ” is incorrect and should be “t5 .”

11. PROPERTIES OF K-GROUPS

300

We assume that ρ is not too large, and enumerate the possibilities for the Ki ’s. Specifically, the next two lemmas together do this enumeration whenever −1 ρ ≤ g(−1 0 ), and even when ρ ≤ g(20 ) in the presence of special hypotheses. 2 (Recall that g(x) = (1 + x) .) Lemma 19.3. Assume the setup (19B). Then ρ ≥ 1 and the following conditions hold. (a) If ρ = 1, then p = 2 and (K0 , K1 , K2 ) = (D4± (q0 ), B4 (q0 ), F4 (q0 )); (b) If 1 < ρ ≤ g(−1 0 ), then q0 = q1 = q2 and one of the following holds: (1) p = 2 and (K0 , K1 , K2 ) is one of the following: ± (i) (A± −1 (q0 ), D (q0 ), B (q0 )),  ≥ 3; (ii) (B−1 (q0 ), D± (q0 ), B (q0 )),  ≥ 3; ± (iii) (D−1 (q0 ), D± (q0 ), B (q0 )),  ≥ 5; ± (iv) (A3 (q0 ) or B3 (q0 ), B4 (q0 ), F4 (q0 )); ± (v) (A± 6 (q0 ), A7 (q0 ), E7 (q0 )); ± (vi) (A7 (q0 ) or D7± (q0 ), D8 (q0 ), E8 (q0 )); ± ± (q0 ), B−1 (q0 ), B (q0 ) or D± (q0 )),  ≥ 4; (vii) (D−1 (viii) (A± 7 (q0 ), E7 (q0 ), E8 (q0 )); ± (2) p = 3 and (K0 , K1 , K2 ) = (A± 7 (q0 ), A8 (q0 ), E8 (q0 )). Proof. By Lemma 12.3a, each ρi ≥ 1, so ρ ≥ 1. Also, ρi ≤ ρ ≤ g(−1 0 ) ≤ g(1) = 4 for i = 1, 2, by our assumptions. Lemma 12.3 implies that for each i = 1, 2, one of the following holds: (1) (Ki , Ki−1 ) is of algebraic type (ai , ki ); or (19C) (2) (Ki , Ki−1 ) is of field type, ρi = p, and i−1 = i . In the algebraic type case, for each i, according to the value of ki in (19C1), we have one of the following: −1 (1) ki > 0 and ρi ≥ g(ki −1 i−1 ) ≥ g(i−1 ); moreover if ai > 1, then −1 (19D) ρi > g(ki i−1 ); or (2) ki = 0 and ρi = ai . These properties follow from Lemma 12.3b. For assertion (a) of the lemma, since ρi ≥ 1 for each i we have ρ2 = ρ1 = 1. Thus, (K0 , K1 , K2 ) must arise from concatenating two triples (p, K0 , K1 ) and (p, K1 , K2 ) in Lemma 12.3d. Clearly the only possibility is the one stated in (a). Now we prove (b). First suppose that (19E)

for some j = 1, 2, (Kj , Kj−1 ) is of field type,

so that ρj = p and j = j−1 , by (19C). Define i by {1, 2} = {i, j}.

(19F) g(−1 0 )

by the hypothesis of (b), and in particular p ≤ g(−1 Then pρi = ρ ≤ 0 ). By assumption 0 > 1, so p ≤ 9/4. Therefore p = 2, 0 = 2 (as g(1/3) < 2), and ρi ≤ 9/8. In particular (Ki , Ki−1 ) must be of algebraic type (ai , ki ). We claim that i−1 = 2. We have just seen that 0 = 2, so suppose that i = 2. Then j = 1, so (K1 , K0 ) is of field type and 1 = 0 = 2, as claimed. If ρi > 1, then ρi ∈ Z, whence by (19D), ki > 0 and then g(−1 i−1 ) ≤ ρi ≤ 9/8. However, by the previous paragraph g(−1 ) = g(1/2) = 9/4, a contradiction. i−1

19. DOUBLE PUMPUPS

301

Therefore ρi = 1, whence (Ki , Ki−1 ) is as in Lemma 12.3d, and in particular as p = 2, that lemma gives the contradiction i−1 > 2. Hence (19E) must fail, that is, (19D) applies to both (K2 , K1 ) and (K1 , K0 ). −1 −1 As ρ2 ≥ 1, we have ρ1 ≤ ρ ≤ g(−1 0 ), so if k1 > 0, then g(k1 0 ) ≤ g(0 ). But g is increasing, so k1 = 0 or 1. We argue next that

(19G)

a1 = 1, and according as k1 = 0 or 1, ρ1 = 1 or ρ2 = 1.

Consider first the possibility k1 = 0 and a1 > 1. g(−1 0 ).

Then a1 = ρ1 ≤ ρ1 ρ2 = ρ ≤ Since 0 > 1 and a1 ∈ Z, the only possibility is that 0 = 2, a1 = 2, and ρ2 ≤ 9/8. Then Lemma 19.1 applied to the pair (K1 , K0 ) n gives the only possibilities for (p, K1 ) to be (2, C4 (q1 )), (3, 2F4 (2 2 )) and (3, E6± (q1 )). Therefore the pair (K2 , K1 ) cannot be as in Lemma 12.3d, so ρ2 > 1. As ρ2 ≤ 9/8, ρ2 is not an integer, so k2 = 0 and g(−1 1 ) ≤ 9/8, by Lemma 12.3b3. But this is absurd as 1 = 4 or 6. Thus, if k1 = 0, then a1 = ρ1 = 1. On the other hand if k1 = 1, then ρ1 ≥ g(−1 0 ) ≥ ρ = ρ1 ρ2 by (19D) and the hypothesis of (b). As ρ2 ≥ 1, equality must hold throughout. Hence ρ2 = 1 and with (19D), a1 = 1 as well. Thus (19G) holds, and according as k1 = 0 or 1, the pair (K1 , K0 ) or the pair (K2 , K1 ) is then as in Lemma 12.3d. If k1 = 1, moreover, then as a1 = 1, we have 0 = (1 − k1 )/a1 = 1 − 1. For each possibility for (K2 , K1 ) satisfying ρ2 = 1 in Lemma 12.3d, the corresponding possibilities for K0 can be read off from Tables 4.5.2 and 4.7.3A of [IA ]. This yields the triples (i–vi) in (b1) and the one triple in (b2). Notice in (b1)(vi) that D8 (q0 ) is a half-spin group, so we cannot have K0 ∼ = B7 (q). Finally if k1 = 0, then 1 = a1 0 = 0 and ρ1 = a1 = 1. Hence ρ2 = ρ ≤ −1 g(−1 0 ) = g(1 ) by the hypothesis of (b). Therefore equality holds throughout when we apply (19D1) with i = 2. Consequently k2 = a2 = 1 and then 1 = 2 − 1. As p and (K1 , K0 ) are given by Lemma 12.3d, and the Dynkin diagram of L1 has rank 1 less than that of L2 , we obtain the remaining triples in (b1), using  [IA , Tables 4.5.2, 4.7.3A]. The proof is complete. The next lemma is a continuation of the previous one, allowing somewhat larger values of ρ, but only under additional special restrictions on K0 . −1 Lemma 19.4. Assume the setup (19B). Suppose that g(−1 0 ) < ρ ≤ g(20 ) and  K0 ∼ = A0 (q0 ), the p-saturated version, with  = ±1. Assume further that 0 ≡ −3 (mod p) and q0 ≡  (mod p). Then q0 = q1 = q2 and 2 ≥ 0 + 2.

Proof. We have (19H)

ρ ≤ g(2−1 0 ) ≤ g(1) = 4.

p Again consider first the field type case (19E). Then ρj = p and qj = qj−1 . In particular p ≤ ρ, which when combined with (19H) and the hypothesis 0 > 1

11. PROPERTIES OF K-GROUPS

302

yields either p = 2 with 0 ≤ 4, or p = 3 with 0 = 2. But 0 ≡ −3 mod p, so p = 2 and 0 = 3, whence L0 = A3 , by our hypotheses. As in (19F), write {i, j} = {1, 2}. Then 2ρi = ρi ρj = ρ ≤ g(2/3) = 25/9. By Lemma 12.3bc, −1 ρi ≥ g(−1 i−1 ), so g(i−1 ) ≤ 25/18 and thus i−1 ≥ 6. As 0 = 3, we must have i = 2, whence j = 1. But then 3 = 0 = 1 ≥ 6, contradiction. Therefore (19D) applies to both (K2 , K1 ) and (K1 , K0 ). We argue next that (19I)

if kj = 0 for some j = 1, 2, then aj = 1.

Indeed otherwise, (Kj , Kj−1 ) is given by Lemma 19.1. We know that K0 is isomorphic to the p-saturated version of A± 0 (q0 ), with 0 ≡ −3 (mod p). If j = 1, then these conditions and Lemma 19.1 yield the single possibility p = 2 and K0 ∼ = D3± (q), but not the 2-saturated version, contradiction. Hence j = 2. As (K2 , K1 ) is as in Lemma 19.1, p ≤ 5. If p = 3, then by Lemma 19.1, we have L1 = A2 . But 0 > 1 so L0 = A2 , contradicting the hypothesis 0 ≡ −3 (mod p). The only possibilities are then (p, a2 , L1 , 0 ) = (2, 2, Dm or Cm , 2d + 1), d ≥ 1, m ≥ 2d + 1 ≥ 3, and (5, 2, A4 , 2). In all cases ρ2 = a2 = 2, so by assumption (19J)

2ρ1 = ρ ≤ g(2−1 0 ) ≤ 4.

Suppose that p = 2. Then 0 ≥ 3. On the other hand, g(2−1 0 ) ≥ 2 by (19J), so 0 < 5 and thus 0 = 3 and d = 1. Then by (19J), ρ1 ≤ g(2/3)/2 = 25/18. If k1 > 0, then by (19D), ρ1 ≥ g(k1 −1 0 ) ≥ g(k1 /3) ≥ 16/9 > 25/18, contradiction. So k1 = 0 and then a1 = 1 by the previous paragraph. Therefore (K1 , K0 ) is as in Lemma 12.3d, whence L1 = B3 . However, this is impossible as (K2 , K1 ) is as in Lemma 19.1. Consequently p = 5, so that K0 ∼ = A2 (q0 ), q0 ≡  (mod 5) by assumption, − 2 while K1 ∼ = A4 (q1 ), q1 = q2 ≡ −1 (mod 5). But K0 is a central quotient of a component of CK1 (x) for some x ∈ Aut(K1 ) of order 5. Hence by [IA , 4.8.2, 4.8.4] we must have q1 = q0 . Then a1 = 1 and k1 = 2, so ρ1 = g(2−1 0 ), which contradicts (19J). This contradiction establishes (19I). Suppose next that k1 k2 = 0. Then √ 2 ≥ a2 1 + 1 ≥ a2 (a1 0 + 1) + 1 ≥ a1 a2 0 + 2 ≥ a1 a2 (0 + 2), as 0 ≥ 2 and a1 a2 ≥ 1. In particular 2 − 0 ≥ 2. Now √  a1 a2 (0 + 2) 2 22 21 22 ρ = ρ1 ρ2 = · = ≥ = g(2−1 √ 0 ) ≥ ρ, a2 21 a1 20 a1 a2 20 a1 a2 0 the last by assumption. Hence equality holds throughout both displayed chains of √ inequalities. In particular a1 a2 0 + 2 = a1 a2 (0 + 2), whence a1 a2 = 1. Then q0 = q1 = q2 and the conclusion of the lemma holds. Finally suppose that k1 k2 = 0. Then for some i = 1, 2, ki = 0 and ai = 1 by (19I), so ρi = 1. Thus (1) (Ki , Ki−1 ) is as in Lemma 12.3d; and (19K) (2) Either p = 2 and 0 is odd, or p = 3 and 3 divides 0 . Indeed (19K2) follows from (19K1) and the assumption 0 ≡ −3 (mod p). Consider the subcase i = 1, i.e., k1 = 0.

20. THE DOUBLE PUMPUP PROPOSITIONS

303

Then (19K) implies that p = 2 and (K1 , K0 ) is either (B3 (q0 ), A± 3 (q0 )) or (q )). In particular q = q and  =  . If also q = q , (E7 (q0 ), A± 0 0 1 0 1 1 2 then us7 ing our hypothesis we get −1 2 2 2 2 g(−1 0 ) < ρ = ρ1 ρ2 = ρ2 = 2 /1 = 2 /0 = g((2 − 0 )0 )

so 2 − 0 > 1, and the lemma holds with conclusion (b). Thus, we may assume that q2 = q1 , i.e., a2 > 1, so q2 < q1 . Using [IA , Table 4.5.1] we see that the only 1 2 possibility is that (K0 , K1 , K2 ) = (A± 3 (q), B3 (q), D7 (q )). But then by [IA , Table 4.5.2], K1 is not the 2-saturated version, so neither is K0 , contrary to assumption. Thus it remains to consider the subcase i = 1, −1 whence i = 2, a2 = 1 by (19I), 2 = 1 and g(−1 0 ) < ρ1 ≤ g(20 ). If a1 = 1, these inequalities force 1 − 0 > 1, as in the previous paragraph, and the lemma holds with conclusion (b). So assume that a1 ≥ 2. As L0 = A0 , and (K2 , K1 ) is as in Lemma 12.3d, and in view of the hypothesis ρ ≤ g(2−1 0 ), it is impossible that p = 3, so p = 2, and then the only possibilities are that (K0 , K1 , K2 ) = ± 2 ± (A3 (q 2 ), A± 7 (q), E7 (q)) or (D3 (q ), D6 (q), B6 (q)). In these cases as well, [IA , Table 4.5.2] provides the contradiction that K0 is not the 2-saturated version. The proof is complete. 

20. The Double Pumpup Propositions The gist of the double pumpup propositions (one for p = 2, the other for p > 2) is that if (x, K) ∈ ILop (G) with K/Op (K) ∈ Gp ∩ Chev, then taking an acceptable subterminal (x, K)-pair (y, L) and pumping L up vertically twice or more yields, in general, a component I such that F(I) > F(K). The application of these two propositions comes in the proof of Theorem 2 in Chapter 9 of this volume. The proofs of the propositions depend heavily on the results of the previous section. The statements of the propositions are unfortunately not clean, and we shall introduce some conditions describing the failure of the desired inequality F(I) > F(K). Before that, we set up the situation and study the invariant ρ(K, L) for K and L as above. The setup for K and L is the following. (1) (x, K1 ) ∈ ILop (G) with K1 ∈ Chev(s), where p and s are distinct primes, one of which is 2; (2) (y, L1 ) is an acceptable (x, K1 )-subterminal pair; (3) K = K1 /Op (K1 ) ∈ Gp and L = L1 /Op (L1 ); (4) The untwisted Lie ranks of K and L are  and 0 , respectively, (20A) and the level of K is q(K) = q; (5) mp (C(x, K1 )) = 1; and (6) If p = 2, then the following conditions hold: (a) 0 > 1; (b) x ∈ Lo2 (CK1 (y)); n (c) K ∼  2 G2 (3 2 ) or 3D4 (q0 ) for any odd n or q0 , and q(L) = q. = Lemma 20.1. Assume the setup (20A). Then the following conditions hold: (a) 1 ≤ ρ(K, L) ≤ g(p−1 0 ); (b) If p = 2, then the following conditions hold:

11. PROPERTIES OF K-GROUPS

304

−1 ∼  (1) If ρ(K, L) > g(−1 0 ), then ρ(K, L) = g(20 ) and K = A0 +2 (q) with  u q ≡  (mod p), 0 ≡ −3 (mod p), and L1 ∼ = A0 (q) ; and ± ∼ (2) If K/Z(K) = P Ωn (q) for some n, then K is a quotient of Ω± n (q); and (c) If p > 2, then the following conditions hold: (1) ρ(K, L) < p2 ; and (2) One of the following holds: (i) ρ(K, L) ≤ 4; (ii) p > 3, mp (K) = 2, and (K, L) is one of (E6 (q), A− 2 (q)), (E6− (q), A2 (q)), (D4 (q), A1 (q 2 )), (3D4 (q), A1 (q)), or (2 F4 (q), A1 (q 2 )); in these cases ρ(K, L) = 9, 9, 8, 16, and 8, respectively; or (iii) For some m and  = ±1, K ∼ = A2m−1 (q), mp (K) = 2, q ≡  (mod p), m − 1 is the least positive integer such that p divides |Am−1 (q)|, L ∼ = Am−1 (q) (or A1 (q 2 ) if K ∼ = A3 (q) and p = 3), −1 and ρ(K, L) = 4g((2m−2) ) (or 9/2 if K ∼ = A3 (q) and p = 3).

Proof. By Lemma 12.3a, ρ(K, L) ≥ 1. Next observe that (c2) implies (c1). Indeed in (c2)(iii), g((2m−2)−1 ) decreases with increasing m, while equalling (5/4)2 for m = 3 and (3/2)2 for m = 2. Thus (c1) reduces to the true statements 25/4 < 9 and 9 < p2 for p > 3. Hence we must prove (a) and (b) when p = 2, and (a) and (c2) when p > 2. We next reduce to the case in which (20B)

the level of L is q.

Indeed if (20B) fails with p > 2, we have (c2)(iii) holding, or (K, L) ∼ = (D4 (q), A1 (q 2 )) 2 2 or ( F4 (q), A1 (q )) with p > 3 (in which case (c2)(ii) holds), or (K, L) ∼ = 2 (2 F4 (q), A− (q )) with p = 3 (in which case (c2)(i) holds), all by Lemma 14.4. 2 In these cases one easily computes that (a) holds. n On the other hand, if (20B) fails with p = 2, then K ∼ = 3D4 (q), 2 G2 (3 2 ) or Spin− 8 (q), by Lemma 14.4 again. These are ruled out by our special hypotheses (20A6a, 6c) for p = 2. Thus (20B) holds. As a result, since  is the untwisted Lie rank of K, we have (20C)

−1 ρ(K, L) = g(( − 0 )−1 0 ) = g(k0 ), where we have set k =  − 0 .

Now assume that p = 2. Suppose, as in (b2), that K is a covering group of P Ω± n (q), but not a quotient of Ω± n (q). Since m2 (C(x, K1 )) = 1 by (20Ae), K1 is a spin or half-spin group with cyclic center containing x. By definition of acceptable subterminal pair [III7 , Def. 6.9], and [IA , 6.2.1b], Ω1 (Z(K1 )) ≤ Lo2 (CK1 (y)), and so x ∈ Lo2 (CK1 (y)), contradicting (20A6b). Thus (b2) holds. Since (b1) implies (a) when p = 2, we may assume by (20C) that (20D)

k ≥ 2.

We go through the various cases of parts (e) and (f) of the definition of acceptable subterminal pair [III7 , 6.9]. Begin with the classical case for K. Since 0 > 1 when p = 2 by assumption, cases (e1)–(e5) of that definition do not apply. If one of (e6), (e7), (e8), or (e10) of that definition holds, then because of (20D), we can

20. THE DOUBLE PUMPUP PROPOSITIONS

305

only have the exceptional configuration in conclusion (b) of this lemma. If (e9) of the definition applies, then K1 is a quotient of an orthogonal group Ω± n (q) by (b2), and according to case (e9), L1 is an orthogonal group of dimension n − 1 or n − 2. Therefore k =  − 0 ≤ 1, contradicting (20D). n Now in the nonclassical cases [III7 , 6.9f], since K ∼ = 3D4 (q) or 2 G2 (3 2 ) by assumption, clearly (f2) must apply and yield k = 1, contradicting (20D). This completes the proof for p = 2. For p > 2, first assume that K ∼ = A (q) with q ≡  (mod p). Then by definition [III7 , 6.9a1,2], k = 1 or 2, and in the latter case, p divides  + 1 with L1 universal. −1 Then ρ(K, L) = g(k−1 0 ) ≤ g(p0 ), proving (a) in this case. Moreover as g(1) = 4, we have ρ(K, L) ≤ 4 unless k = 2 and 0 = 1, whence  = 3. But as p is odd and p divides  + 1, this is impossible. Next suppose that K is a classical group with natural module V , of dimension n, as in [III7 , 6.9a4,5]. Let d be the dimension of a minimal subspace V0 of V admitting an isometry of order p (and nonsingular in every case but the case K ∼ = A+ (q)). Let W be an orthogonal complement to V in V . Then by [III , 6.9a4,5], 0 7  L/Z(L) ∼ = L0 /Z(L0 ) where L0 = [Isom(W ), Isom(W )]. Recall by [IA , 4.8.2, p. 220] that d ≤ p − 1. If K = An−1 (q), then  = n − 1, 0 = n − d − 1 and

d ρ(K, L) = g . n−d−1 Thus (a) follows in this case. Moreover if n − d − 1 ≥ d, then ρ(K, L) ≤ ρ(1) = 4. Otherwise n − d − 1 = d − 1, since mp (L) ≥ 1 by assumption, and so conclusion (c2)(iii) holds. On the other hand, if K is a symplectic or orthogonal group, then  = n/2 while 0 = (n − d)/2, with n − d ≥ d. So ρ(K, L) = g(d/(n − d)) ≤ g(1) = 4, and as d < p, (a) holds as well. It remains to consider the case p > 2, K of exceptional Lie type. As in the case p = 2, (20C) implies that (a) holds as long as k ≤ p. We go through the cases of [III7 , 6.9b]. If (b1) holds, then p = 3 ≥ k so (a) holds, and ρ(K, L) ≤ 4, as desired. In particular if K ∼ = 3D4 (q) then we may assume that p > 3. Since q(K) = q(L), we skip (b2). In (b4)–(b6) we check easily that  ≤ 20 and k ≤ 3 ≤ p, so that (a) and (c1) of this lemma hold. It remains to check case (b3) of [III7 , 6.9], i.e., mp (K) = 2. As L is subterminal, it can be found in Table 13.1. Except for the cases listed in (c2)(ii), that table shows that  ≤ 20 , whence ρ(K, L) ≤ 4. Moreover, we have p > 3 in those cases–by the previous paragraph if K ∼ = 3D4 (q), and because p does not divide q 2 − 1 in the other cases. This completes the proof of (c2). Finally, (a) follows from (20C) if k ≤ p, which in turn follows from Table 13.1 as we are assuming that mp (K) = 2. Indeed, if k > 3 then k = 4, in one of three cases for K = E8 (q) and one case for K = E6 (q). In these last cases ordp (q) > 2, so p ≥ 5 > k. The proof of the lemma is complete.  We are almost ready to state the double pumpup propositions, one for p = 2, and one for p > 2.

11. PROPERTIES OF K-GROUPS

306

Here is the setup for the chain of length at least two: (1) p and s are distinct primes, one of which is 2; (2) I0 , I1 , . . . , Ir ∈ Kp ∩ Chev(s), where r ≥ 2; (3) For each i = 0, . . . , r − 1, Ii ∈ Kp is a covering group of Ii ; (20E) moreover, Ii = Ii for i = 0, i = r − 2, and i = r − 1; (4) For all i = 1, . . . , r, Ii−1 ↑p Ii ; and (5) I0 is a covering group of L, where K and L are as in (20A). In particular if r ≤ 3, then Ii−1 ↑p Ii for all i = 1, . . . , r. For the p = 2 case, we also need the following terminology. Definition 20.2. Suppose that J, K ∈ K2 and J ↑2 K. (a) We say that J and K share a central involution if and only if for any a ∈ Aut(K) such that CK (a) has a component J1 with J1 /O2 (J1 ) ∼ = J, we have Z(J1 ) ∩ Z(K) = 1; (b) We say that J is large (resp. very large) in K if and only if for any a ∈ Aut(K) such that CK (a) has a component J1 with J1 /O2 (J1 ) ∼ = J, every nonabelian simple section of the group CK (a J1 ) is isomorphic to L2 (q) for some odd q (resp. |CAut(K) (J1 )|2 = 2). The case p = 2 reads as follows. Proposition 20.3. Assume that p = 2 and (20A) and (20E) hold, but L/Z(L) ∼  = L2 (q) for any odd q. Then one of the following holds: (a) F(Ir ) > F(K); (b) r = 2 and (K, L, I1 , I2 ) = (A0 +2 (q), A0 (q), A0 +1 (q), A0 +2 (q)), 0 ≥ 2,   or (D0 +1 (q), D0 (q), B0 (q), D0 +1 (q)), 0 ≥ 3, where q = sa , and ,  ,  are signs; (c) For each i = 1, . . . , r, Ii−1 and Ii share a central involution and Ii−1 is large in Ii ; or (d) m2 (I0 ) > 2, and for each i = 1, . . . , r − 1, Ii−1 and Ii share a central involution, and Ii−1 is large in Ii ; moreover Is−1 is very large in Is . Proof. Assume that conclusion (a) fails, i.e., F(Ir ) ≤ F(K). Then ρ(Ir , K) ≤ 1, and indeed with the help of Lemma 12.4, we conclude that (20F)

ρ(Ii , K) ≤ 1 and F(Ii ) ≤ F(K), for all i = 1, . . . , r.

We focus on the initial subchain I0 , I1 , I2

(20G)

in (20E) of length 2. Since I0 is a covering group of L, (20H)

ρ(I2 , I0 ) = ρ(I2 , L).

In view of (20F), this implies that (20I)

ρ(I2 , I0 ) = ρ(K, L)ρ(I2 , K) ≤ ρ(K, L).

The conditions on ρ(K, L) in Lemma 20.1ab, together with (20I) and (20H), therefore imply that if we let 0 be the untwisted Lie rank of L, which is the same as that of I0 , then we have (1) ρ(I2 , I0 ) ≤ g(2−1 0 ); and ∼ ± ∼ ± u (20J) (2) If ρ(I2 , I0 ) > g(−1 0 ), then L1 = A0 (q) and K = A0 +2 (q) for some power q of s.

20. THE DOUBLE PUMPUP PROPOSITIONS

307

Accordingly we make the case division (1) ρ(I2 , I0 ) = 1; or (2) 1 < ρ(I2 , I0 ) ≤ g(−1 0 ); or (20K) −1 ∼ ± (3) g(−1 0 ) < ρ(I2 , I0 ) ≤ g(20 ), with L = A0 (q) being 2-saturated ± ∼ and K = A0 +2 (q). We shall apply Lemma 19.3 in the first two of these cases to the chain (20G), and in the third case we shall apply Lemma 19.4. Write Ii = di Li (qi ) and i = Lie rank of Li , i = 0, 1, . . . , r, and (20L) K = d L (q). Since I0 is a covering group of L, and q(L) = q by (20A), q = q0 . Observe that the notation in (20L) is consistent with that in (20J2). Consider now the first case of (20K), ρ(I2 , I0 ) = 1. By Lemma 19.3a, I2 ∼ = F4 (q), I1 ∼ = B4 (q) and I0 ∼ = D4± (q). Now F(I2 ) ≤ F(K), so either L = F4 or  ≥ 5. But L is an acceptable subterminal subcomponent of K; using [III7 , 6.9], we see that K ∼ = D5 (q). Moreover, if I2 is not the last term Ir in (20E), then the next term is I3 ∼ = F4 (q 2 ) or E6± (q), the only proper pumpups of F4 (q) (see [IA , Table 4.5.1, 4.9.1]). But in either case we calculate ρ(Ii , K) > 1 contradicting (20F). Therefore I2 = Ir , whence the chain (20E) is (20M)

(20N)

D4± (q) 1 and so 2 > 2. Thus the possibilities L2 = B2 ; D2 ; F4 ; E7 ; and E8 of (20P3) yield the respective conclusions that L = B2 , C2 or F4 ; D2 , E6 , E7 , E8 , B2 or C2 ; F4 ; E7 , B7 or C7 ; E8 , B8 or C8 .

11. PROPERTIES OF K-GROUPS

308

We rule these out in several subcases. In the case L = C0 +1 , the definition of acceptable subterminal subcomponent yields L0 = C0 , whence by (20P4), L0 = B2 = C2 and L2 = B3 . Then I1 ∼ = A3 (q). u Moreover L ∼ = C2 (q) by [IA , Table 4.5.1]. As I0 is a covering group of L, we have I0 ∼ = L. Therefore by [IA , Table 4.5.1] again, I1 is 2-saturated. So I1 f (I2 ), so I2 must be the last term in (20E). Therefore r = 2 and conclusion (c) of the lemma holds. If L = B , then by definition of acceptable subterminal pair, L0 = D unless possibly K = B (q)u and  is odd. But the former gives  = 0 , contradicting (20R), while the latter cannot happen by Lemma 20.1b2. If L = F4 , then L ∼ = C3 (q) by [III7 , Def. 6.9]. But this contradicts (20P4). Thus we may assume that L = A , D , or E . If L = L2 , then the only possibility for (I0 , I1 , I2 ) in Lemma 19.3b1 is the triple ± (D−1 (q), B−1 (q), D± (q)),  ≥ 4,

as I0 /Z(I0 ) ∼ = L/Z(L) with L an acceptable subterminal subcomponent of K. Thus conclusion (b) of the proposition will hold once we prove that r = 2. Again Lemma 12.4 shows that any vertical pumpup J of I2 satisfies F(J) > F(I2 ), so I2 must be the last term in (20E) and r = 2, as desired. Now assume that L = L2 . Since F(I2 ) ≤ F(K), it follows with (20R) that τ (I2 ) < τ (K) in our ordering of types. As τ (K) = A, D, or E, τ (I2 ) = E and so τ (I2 ) = A or D. Again the only possibility in Lemma 19.3b1 is case (vii), so ± that I0 = D−1 (q) and I2 = D± (q). Then τ (K) = E, and since L is an acceptable subterminal subcomponent of K with L/Z(L) ∼ = I0 /Z(I0 ), we see that the only possibility is  = 7, with L ∼ = B−1 (q), and using = D6 (q)hs or D6 (q)u . We have I1 ∼ [IA , Table 4.5.2] we find that I1 must be universal. Moreover any vertical pumpup J of I2 ∼ = D7± (q) must satisfy F(J) > F(E7 (q)) = F(K), so r = 2. Then I0 and I2 are universal as well. As in the L2 = C2 case above, we find that conclusion (c) holds. We have proved that we may assume that  > 2 = 0 + 1 if (20O) holds. We now complete the proof in the case (20O). Since  > 0 + 1, Lemma 20.1b1 implies that L = A0 +2 and 0 is odd; indeed L ∼ = SL0 +1 (q0 )/O2 (SL0 +1 (q0 )), which is a 2-saturated covering group  of P SL0 +1 (q0 ). By hypothesis 0 > 1. As I0 is a covering group of L, it too is obviously 2-saturated. We know by Lemma 19.3b that I0 , I1 , I2 is one of the following chains: (20S)

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



A0 (q), D0 +1 (q), B0 +1 (q); A7 (q), D8 (q), E8 (q);  A3 (q), B3 (q), B4 (q) or D4 (q); or A3 (q), B4 (q), F4 (q).

Lemma 19.3b also allows the possibility A7 (q) 2. We have ρ(K, I2 ) = ((0 + 2)/(0 + 1))2 < 2 and of course ρ(K, I3 ) ≥ 1, so ρ(I3 , I2 ) = ρ(K, I2 )/ρ(K, I3 ) < 2. By Lemma 12.3bc, it follows that q3 = q2 (= q), and similarly q = qi for every i > 2. Now by [IA , Table 4.5.1], none of the groups I2 in (20S) has A± 0 +2 (q) as a long pumpup. Thus as (0 +2)2 , we must have i = 0 + 1 for all i > 2. f (Ii ) ≤ f (K) = q In view of the previous paragraph and given the possibilities in (20S) and [IA , Table 4.5.1], we conclude that one of the following holds: (1) r = 2;  (2) I2 ∼ = D4 (q), and I3 ∼ = B4 (q). Either r = 3 or r = 4, with (20T) I4 ∼ = F4 (q) in the latter case; or (3) r = 3, I2 ∼ = B4 (q), and I3 ∼ = F4 (q). With [IA , Table 4.5.1] we see that in general, in the pumpups involved in (20S) and (20T) beyond I0 4, then mp (K) = 2, and p > 3 in the final three cases of (20X). As in the proof of Proposition 20.3, we set up further notation as follows. Let q = q(K), q0 = q(L) = q(I0 ) (as I0 is a covering group of L) and qi = q(Ii ) for each 1 ≤ i ≤ r. Let L , of rank , be the complex semisimple Lie algebra underlying the group K, while Li , of rank i , underlies Ii , 0 ≤ i ≤ r. Furthermore, for any i such that the pumpup Ii−1 1, qiai = qi−1 , and ρi = ai ; (20Y) (4) For some integer ai > 1, ki = i − ai i−1 ≥ 1, qiai = qi−1 , and −1 −1 ρi = ai g(ki a−1 i i−1 ) > max(ai , g(ki i−1 )). Like the pairs (Ii−1 , Ii ), the pair (L, K) also satisfies one of a similar set of possibilities; however, the condition that L is acceptable subterminal [III7 , 6.9] specifies in most cases the isomorphism type of L, depending on K and p. In particular if (20Y4) holds, then by Lemma 14.4, L ∼ = 2 F4 (q), = A1 (q 2 ) and K ∼ − D4 (q), or A3 (q) (with p = 3 and q ≡  (mod 3) in the last case). By Lemmas 12.3d and 19.1, (1) If (20Y1) holds for some i, then p = 3 and (Ii , Ii−1 ) is either (G2 (qi ), A2 (qi )) or (E8 (qi ), A8 (qi )), where  = ±1 and qi ≡  (mod 3); and (2) If (20Y3) holds for some i, then (p, a, Ii , Ii−1 , ρi ) equals (20Z) 2 2 b (3, 2, 2 F4 (qi ), A− 2 (qi ), 2) such that qi = 2 for some odd b;   3 (3, 3, E6 (qi ), A2 (qi ), 3) with qi ≡  (mod 3) and  = ±1; or 2 2 (5, 2, E8 (qi ), A− 4 (qi ), 2) with qi ≡ −1 (mod 5). With this notation fixed, we first prove Lemma 20.5. If for some i, Ii−1 ai + 2 if ai > 1. In any case, ρi ≥ 4. Proof. One of the cases of (20Y) applies, with i−1 = 1, but (20Y1) is impossible since it would imply that Ii = A1 (qi ) ∼ = Ii−1 . If (20Y2) holds, then i ≥ 2 and ρi = 2i ≥ 4. By (20Z2), (20Y3) cannot hold. Finally if (20Y4) holds, then ki ≥ 1 −1 2 and ρi = ai g(ki a−1  i ) ≥ ai g(ai ) = (ai + 1) /ai > ai + 2 ≥ 4, as asserted. This enables us to rule out any non-algebraic pumpups in (20E). Lemma 20.6. Every pumpup Ii−1 1, then by Lemma 12.3, ρj ≥ min(2, (3/2)2 ) = 2, a contradiction; so ρj = 1. By Lemma 12.3d, L ∼ = A2 (q), so K ∼ = 3D4 (q) and 3 Ij has Lie algebra G2 . If i ≥ 3 then we must have I2 ∼ = D4 (q) or D4 (q), so f (I3 ) > f (I2 ) = f (K), contradicting (20V). If i ≤ 2 then I2 ∼ = G2 (q 3 ). Then r = 2, 3 3 3 for otherwise I3 ∼ = D4 (q ) or D4 (q ), contradicting (20V) again. Thus conclusion (c) of Proposition 20.4 holds, contrary to our assumption that the proposition fails. The lemma is proved.  Next we deal with the case mp (K) = 2. Lemma 20.7. We have mp (K) ≥ 3. Proof. Suppose false. Since Proposition 20.4e fails, and by Lemma 13.30a, (20BB)

mp (I0 ) ≤ mp (I1 ) ≤ · · · ≤ mp (Ir ) ≤ 2.

As these pumpups are algebraic by Lemma 20.6, and r ≥ 2, we see from Lemma 13.30bc that p = 3, m3 (I1 ) = 2, and for all i = 2, . . . , r, (Ii−1 , Ii ) has one of the following forms for some power q∗ of 2: √ (20CC) (G2 (q∗ ), 3D4 (q∗ )) (Aη2 (q∗ ), G2 (q∗ ) or 3D4 (q∗ ) or 2 F4 ( q∗ )), where q∗ ≡ η (mod 3), η = ±1. Since p = 3 and m3 (K) = 2 with K ∈ G3 by hypothesis, we have K = G2 (q), − 3 D4 (q), 2 F4 (q), B2 (q), A− 3 (q), or A4 (q), q ≡  (mod 3). In the first three cases,  L∼ = A2 (q), by [III7 , 6.9], whence by the previous paragraph, the chain (20E) can only be A2 (q) 1; (c) mp (Ii ) > 1 for all i = 0, . . . , r; and (d) ρ ≤ 4. Proof. By Lemmas 20.7 and 15.2a, mp (I0 ) > 1. This immediately implies (b); with Lemma 13.30a it implies (c); and with Lemma 14.4 it implies (a), since K has p-rank 2 in all the counterexamples to (a). Finally (d) is immediate from Lemmas 20.7 and 20.1c.  We next aim to prove that (20DD)

q0 = q1 = · · · = qr .

Since pumpups of field type do not appear in (20E), by Lemma 20.6, the obstructions Ii−1 ai respectively. Here ai ≥ 2. Lemma 20.9. Suppose that (20DD) fails. Then there is a unique i = 1, . . . , r such that qi−1 = qi . Moreover, i > 1. Proof. Suppose that i = j are two indices satisfying the stated condition. As just noted, ρi ≥ 2 and ρj ≥ 2. Hence (20EE)

ρ1 · · · ρr ≥ ρi ρj ≥ 4 ≥ ρ,

the last inequality by Lemma 20.8d. Therefore by (20W), equality holds throughout (20EE). In particular ρi = ρj = 2, whence (20Y3) applies for both i and j, and ρm = 1 for all other m, 1 ≤ m ≤ r, whence (20Y1) applies for all indices m = i, m = j. Thus each pumpup in the chain (20E) is as in (20Z1) or (20Z2). This implies, however, that for each pumpup in (20E), the smaller term has associated Lie algebra of type A while the larger term does not. This can only be true if r = 1, contrary to assumption. This proves the first assertion of the lemma. Suppose that the second assertion fails, so that q0 > q1 = q2 = · · · = qr . By Lemma 20.8abc, q = q0 , 0 > 1, and mp (L) ≥ 2. As q > q1 , we have L ∼ = A0 (q) with q ≡  (mod p), by Lemma 19.2. Then since (y, L1 ) is an acceptable (x, K1 )subterminal pair,  ≤ 0 + 2 by Lemma 15.7, and one of the following holds: (1)  ≤ 0 + 1; or (20FF) (2)  = 0 + 2, (K, L) = (A0 +2 (q), A0 (q)u ), and 0 ≡ −3 (mod p). Then (19B) applies with I0 , I1 , I2 playing the roles of K0 , K1 , K2 , respectively. Also, the hypotheses of Lemma 19.3 or Lemma 19.4 hold according as (20FF1) or (20FF2) holds. These lemmas yield q0 = q1 , contrary to assumption. The proof is complete.  Now we can attain our objective. Lemma 20.10. We have q = q0 = q1 = · · · = qr .

11. PROPERTIES OF K-GROUPS

314

Proof. Let i be the unique index with qi−1 = qi , so that i > 1 by Lemma 20.9. Thus qt = qi for all i ≤ t ≤ r, and qt = q for all 0 ≤ t ≤ i − 1. Notice that mp (Ii−1 ) ≥ 2 as mp (I0 ) ≥ 2. Thus by Lemma 19.2, Ii−1 ∼ = Ai−1 (q) and p divides q − . Following the chain from Ii−1 down to I0 , and using [IA , 4.8.2, 4.8.4] and the fact that q0 = q1 = · · · = qi−1 , we find that It ∼ = At (q) for all 0 ≤ t ≤ i − 1. Thus 0 < 1 < · · · < i−1 . Moreover, with Lemma 20.8, we can apply Lemma 15.7 and conclude that  ≤ 0 + 2 and 0 ≥ 2. As i > 1, we have q0 = q1 and ρ1 ≥ (0 + 1)2 /20 . But ρi ≥ ai ≥ 2, and we get ρ1 ρi ≥ 2(0 + 1)2 /20 . However, ρ1 ρi ≤ ρ by (20W), and ρ ≤ ρ(I0 , K) = 2 /20 . We conclude that 2 ≥ 2(0 + 1)2 . As 0 + 1 ≥ 3, our inequalities force  > 0 + 2. But we saw above that   ≤ 0 + 2. This contradiction completes the proof. Before proceeding we make an obvious reduction: by inserting refining terms in the sequence I0 , I1 , . . . , Ir−1 , Ir as much as possible, we may assume that each Ii−1 is terminal in Ii , without forfeiting the conclusion of the proposition. Now the analysis focuses on the ranks and types of the Lie algebras L, L0 , L1 , . . . , Lr associated with the groups K, I0 , I1 , . . . , Ir , respectively. Recall that the associated ranks are , 0 , 1 , . . . , r , respectively and we set k =  − 0 and ki = i − i−1 , i = 1, . . . , r. In view of Lemma 20.10, our indirect assumption F(Ir ) ≤ F(K) means that r 

(20GG)

ki ≤ k

i=1

and in case of equality (20HH)

Lr ≤ L in the type ordering A < D < E < BC < F < G.

We now define decompositions k = φ + f, ki = φi + fi of the rank differences k, k1 , . . . , kr into “main” φ-terms and “error” f -terms. The φ-terms depend only on the type (and twist, if any) of the larger Lie algebra L as well as on the arithmetic of p and q, but not on the rank of L. Thus (20GG) will take the form (20II)

r  i=1

φi +

r 

fi ≤ φ + f.

i=1

Specifically, for an odd prime p, we define φL (, q, p), usually abbreviated to φL , as a positive integral invariant of the subgroup ±q ≤ F× p of the multiplicative group of the field of p elements, as follows: ⎧ φA = | q | ⎪ ⎪ ⎪ ⎨φ BCD = φB = φC = φD = | q, −1 / −1 | (20JJ) φL = φL (, q, p) = ⎪ φEF G = φE = φF = φG = ϕ(| q |) ⎪ ⎪   ⎩ φ2 F = ϕ(| q 2 |)

20. THE DOUBLE PUMPUP PROPOSITIONS

315

In the case of the classical groups, these values have familiar geometric significance. Lemma 20.11. The following conditions hold: (a) For any n ≥ 1, φA is the minimal dimension of support on the natural module among all elements of Ip (GLn (q)); and (b) φBCD is half the minimal dimension of support on the natural module among all elements of Ip (Bn (q)) (for any n ≥ 2) or Ip (Dn± (q)) (for any n ≥ 4). Proof. This is well-known. In (a), one checks that φA is the minimal value of n such that p divides |GLn−1 (q)|; similarly in (b), φBCD is the minimal value of n  such that p divides |Bn (q)| or at least one of |Dn± (q)|. Using the facts that mp (K) ≥ 3, 0 > 1 and mp (L) > 1 (Lemmas 20.7 and 20.8bc), and the definition of acceptable subterminal pairs [III7 , 6.9], we can show that the error term on the right side of (20II) is practically nothing: Lemma 20.12. We have f = 0, i.e, k = φL (, q, p), unless K ∼ = A (q) with q ≡  (mod p) and p dividing  + 1, in which case f = 1. Proof. We go through the cases of [III7 , 6.9]. In (a1) and (a2) of that definition, K/Z(K) ∼ = Ln (q) and p divides q − , so φA = 1. The definition specifies  ∼ L = SLn−2 (q) or SLn−1 (q), so k = 2 or 1, and thus f = k − φA = 1 or 0. The case f = 1 occurs only if p divides n in (a1) of the definition, which is the exception allowed in our lemma. Part (a3) of [III7 , 6.9] does not pertain to this lemma since it applies to a case in which mp (K) = 2. In (a4), again K/Z(K) ∼ = Ln (q), and L ∼ = SLd (q) where d is the maximum dimension of fixed points on the natural module of an element of Ip (K). Thus k = n − d = φA by Lemma 20.11a, so f = 0. A similar argument handles case (a5) of [III7 , 6.9], using Lemma 20.11b. Parts (b1)–(b3) of [III7 , 6.9] are again irrelevant since we know that mp (K) ≥ 3. In the remaining three parts of [III7 , 6.9], K, L, and the order m0 of q in F× p are given, and they obviously satisfy k = φL . For example in (b5), L = E8 and  L0 = D6 so k = 2, while φE = ϕ(m0 ) = ϕ(4) = 2. We next show that the error terms on the left side of (20II) are rarely negative. Again we know by Lemma 20.8c that mp (Ii ) ≥ 2 for all i = 0, . . . , r. Lemma 20.13. Suppose fi < 0 for some i = 1, . . . , r. Then p = 3 and (Ii−1 , Ii ) is either (A2 (q), G2 (q)) or (A8 (q), E8 (q)), q ≡  (mod 3),  = ±1. Proof. Say that xi ∈ Ip (Aut(Ii )) yields the subcomponent Ii−1 in its centralizer. Since qi−1 = qi , xi ∈ Aut0 (Ii ). If xi is a graph automorphism, then p = 3 and Li = D4 with Li−1 = G2 or A2 , so that ki = 2. But in this case | q | ≤ 2 = |F× 3 |, so φi = 1 and fi = 1 ≥ 0, contradiction. Therefore we may assume that xi ∈ Inndiag(Ii ). Suppose first that Ii is a classical group. Consider first the case Ii = Aii (qi ), qi ≡ i (mod p). Then with [IA , 4.8.2, 4.8.4] and the fact that qi−1 = qi , and using the terminality of Ii−1 , we conclude that Ii−1 = Aii −ki (qi ) with ki = 1 or 2. But φA = | i qi  | = 1 in this case, so fi = 0 or 1, contrary to assumption. Next, if Ii = Aii (qi ) with | i qi  | = φA > 1, we see from [IA , 4.8.2] that either Ii−1 has

316

11. PROPERTIES OF K-GROUPS

level qi−1 = qiφA = qi , contradicting Lemma 20.10, or ki = i − i−1 = φA (, q, p) and the support of Ii−1 on the natural module for Ii is the fixed-point subspace of xi . In that case fi = 0, contradiction. The cases Ii = Bi (qi ) and Di (qi ) are argued in a similar way. For example if Ii = Bi (qi ), [IA , 4.8.2] implies that Ii−1 is of one of two types. The first is that Ii−1 is supported on the natural Ii -module on the fixed points of xi , and then Ii−1 = Bi−1 (qi ) with ki = i − i−1 = φB = φi and fi = 0, using Lemma 20.11. The second only occurs if qi2 ≡ 1 (mod p), i.e., φi = 1, because qi−1 = qi , and then Ii−1 ∼ = Ai −1 (qi−1 ), so again ki = 1 and fi = 0. We may therefore assume that Ii is of exceptional Lie type (including 3D4 (qi )). We then use Table 13.1 to see the possible terminal subcomponents and the corresponding values of φi and ki . Suppose first that mp (Ii ) = 2. We have mp (Ij ) ≤ 2 for all j ≤ i, and in particular for j = i − 1. Using [IA , 4.10.3] to determine pranks, we see that p = 3 and Ii ∼ = 2 F4 (qi ), 3D4 (qi ), or G2 (qi ), with φi = 1 and  Ii−1 ∼ = A2i (qi ). As ki − φi = fi < 0, we must have ki = 0, yielding the desired G2 (qi ) exception. Beyond that we may assume that mp (Ii ) = 3, and bearing in mind that qi = qi−1 we see easily from Table 13.1 that ki ≥ φi unless p = 3, Ii = E8 (qi ) and Ii−1 = A8i (qi ), qi ≡ i (mod 3). The lemma follows.  We can strengthen the previous result: Lemma 20.14. We have fi ≥ 0 for all i = 1, . . . , r. Proof. Suppose false, so that by Lemma 20.13, p = 3 and for some i = 1, . . . , r, we have (Ii−1 , Ii ) = (A2 (q), G2 (q)) or (A8 (q), E8 (q)), q ≡ 

(mod 3).

Thus for all j < i, Lj is of type A, so the value i is unique, and indeed Ij ∼ = Aj (q), with the sequence 0 < 1 < · · · < i−1 being strictly increasing. In particular, I0 and L are of type A. Suppose that Ii ∼ = E8 (q). Then Ii has no proper pumpup of level q, so i = r. As F(K) ≥ F(Ir ), K has untwisted rank at least 9, or K ∼ = B8 (q) or E8 (q). Now if K were of type D or B, then so would be L, by [III7 , Definition 6.9], whereas L is of type A. Likewise by the same definition if K ∼ = D7 (q), = E8 (q), then L ∼  ∼ contradiction. So K = A (q),  = 9 or 10. But then  ≡ −1 (mod 3), so 0 =  − 1. The only choice is L ∼ = A8 (q), forcing r = 1, a contradiction. Suppose then that Ii ∼ = G2 (q), so that Ii−1 ∼ = A2 (q), q ≡  (mod 3). As 0 > 1, we must have i = 1 in this case. The only level pumpups of G2 (q) for p = 3 are then I2 ∼  3D4 (q) or 2 F4 (q), and inspection of = D4 (q) or 3D4 (q). Since m3 (K) ≥ 3, K ∼ = Table 13.1 shows that having a terminal 3-component isomorphic to L, K cannot be of exceptional Lie type. Hence L is a classical group. As in the previous paragraph, with [III7 , 6.9], K cannot be an orthogonal or symplectic group, and K ∼ = A (q).  ∼ Since F(Ii ) ≤ F(K),  ≥ 5. But then as q ≡  (mod 3), I0 = A0 (q), 0 = 3 or 4, a contradiction. The proof is complete.  Now we can obtain the exception (b) in the conclusion of Proposition 20.4 as the only outcome of the case f = 1. Lemma 20.15. We have f = 0 and k ≥ 2. Proof. First, if k < 2, the right side of (20II) is at most 1. As each φi ≥ 1 and fi ≥ 0, (20II) then implies r ≤ 1, contrary to assumption. So k ≥ 2.

20. THE DOUBLE PUMPUP PROPOSITIONS

317

∼ A (q) with p dividing Suppose then that f > 0, so that by Lemma 20.12, K =    + 1 and q − , and I0 ∼ = A−2 (q). The right side of (20II) then equals 2. Suppose that the left side of (20II) also equals 2. Then r =  and Lr ≤ L in the ordering of Lie algebra types; since L is of type A, so is Lr . Thus Ir ∼ = Ar (q). If q ≡ r (mod p), then the only terminal level subcomponent of Ir would be r (q), which is not a long pumpup of I0 , contradiction. Therefore r =  Ir−1 ∼ = A−2 and Ir /Z(Ir ) ∼ = A−1 (q), = K/Z(K). It follows immediately that r = 2 and Ir−1 ∼ so conclusion (b) of Proposition 20.4 holds, contrary to assumption. Therefore the left side of (20II) is at most 1. Again, as each φi ≥ 1 and fi ≥ 0, r = 1, a contradiction. The lemma follows.  With Lemmas 20.14 and 20.15, we can now drop the “error” terms fi and simply use (20KK)

r 

φi ≤ φ

i=1

in place of (20II). We complete the proof of Proposition 20.4 by considering the various possible types for L0 , arguing to a contradiction in each case. Case 1. L0 has type F or G. Since L is an acceptable subterminal pair, [III7 , Definition 6.9] shows that these types never occur. Case 2. L0 has type E. Then [III7 , Definition 6.9] shows that L has type E. Moreover by [IA , 4.8.2, 4.8.4, 4.9.2] and Table 13.1, L1 and L2 are then also of type E, so (20KK) gives the contradiction rφE ≤ φE . Case 3. L0 has type B. Then [III7 , Definition 6.9] shows that L has type B or F . Moreover by [IA , 4.8.2, 4.8.4, 4.9.2] and Table 13.1, L1 and L2 are then also of type B or F , with at most one F . As φB ≥ φF , (20KK) then yields φB + φF ≤ φL ≤ φB , a contradiction. Case 4. L0 has type D. If L ∼ = 3D4 (q), then the only possibility for K, by Table  ∼  13.1, is E6 (q) with | q | = 3 or 6 in F× p . The same table then shows that I1 = E6 (q) or E7 (q). In any case φ1 = φ so (20KK) gives φ + φ2 ≤ φ, a contradiction. Thus L∼ = D0 (q). Then L is of type D or E, and each Ii is of type D or E, by [IA , 4.8.2, 4.8.4] and Table 13.1. Let a be the number of Ii ’s of type D, 1 ≤ i ≤ r. Then (20KK) reads φ = φD or φE . aφD + (r − a)φE ≤ φ, Note that φD ≥ φE , by their definitions. This forces φ = φD and a = 0 in the above inequality, since r ≥ 2. But then Ir is of type E and K is of type D; as D < E (20KK) must be a strict inequality in this case, so 2φE < φD . Now Ir−1 must be of type E. So from Table 13.1 we see that q has order dividing 6, and then from the definitions we get 2φE ≥ φD , a contradiction. Case 5. L0 has type A. Suppose first that K is not a classical group. Keeping in mind that mp (K) ≥ 3, and that φ ≥ 2 from (20KK), we see from [III7 , Definition 6.9] that K ∼ = A5 (q), with q of order 3 or 6 according as  = 1 or = E7 (q) and I0 ∼ −1. If I1 is not a classical group, then Table 13.1 yields I1 ∼ = E7 (q), so φ1 = φ and (20KK) gives the contradiction φ2 ≤ 0. Therefore I1 is a classical group. But then φ1 ≥ min(φA , φB ) ≥ 3 while φ = φE = 2, and (20KK) again gives a contradiction. Therefore K is a classical group. Suppose that L = D ,  ≥ 4, or B ,  ≥ 2. Since I0 is of type A and 0 > 1, the only possibility is that L0 = A3 = D3 and L is of type D. But mp (L) ≥ 2, which

11. PROPERTIES OF K-GROUPS

318

implies that p divides q 2 − 1, which in turn implies by [III7 , 6.9] that L = D4 . Thus k =  − 0 = 1, contradicting Lemma 20.15. Therefore K ∼ = A (q). If L1 is also of type A, then φ1 = φA = φ and (20KK) gives the contradiction φ2 ≤ 0. So L1 is not of type A. Suppose that q 2 ≡ 1 (mod p). Then using [IA , 4.8.2] we see that I1 cannot be a classical group of type B or D, as this would force q1 = q0 . Moreover p = 3. If I1 ∼ = G2 (q) or 3D4 (q), then by Table 13.1, mp (I0 ) = 1, contradiction. If I1 ∼ = F4 (q), E7 (q), or E8 (q), then I1 has no vertical pumpup of level q relative to a prime p for which | q | > 2, contradicting the existence of I2 . Thus I1 ∼ = E6± (q). As mp (I0 ) ≥ 2, mp (I1 ) ≥ 3, and as q 2 ≡ 1 (mod p), Table 13.1 gives the contradiction I0 ∼ = 3D4 (q). 2 Therefore q ≡ 1 (mod p). It follows from [III7 , 6.9] and [IA , 4.8.2] that k ≤ 2. Therefore the right side of (20II) equals 2, and so equality holds in (20II). As F(Ir ) ≤ F(K), Lr ≤ L in the ordering of types of Lie algebras. But L is of type A, so Lr is as well, and then L1 is too. This contradicts the previous paragraph and completes the consideration of Case 5. This completes the proof of Proposition 20.4. Lemma 20.16. Suppose that K0 ≤ K1 , with Ki ∈ Chev(2), i = 0, 1. Let P0 be a parabolic subgroup of K0 , P0 < K0 , and set Q0 = O2 (P0 ). Then there exists a parabolic subgroup P1 ≤ K1 such that P0 ≤ P1 and Q0 = P0 ∩ Q1 , where Q1 = O2 (P1 ). Proof. By the Borel-Tits Theorem, P1 exists such that P0 ≤ P1 and Q0 ≤ Q1 .  Then Q1 ∩ P0  P0 so Q1 ∩ P0 ≤ O2 (P0 ) = Q0 . Inductively we obtain Corollary 20.17. Let K0 and K1 be as in Lemma 20.16. Let B0 = P00 < P10 < · · · Pr0 < K0 be a chain of parabolic subgroups of K0 , with B0 a Borel subgroup. Then there exists a chain P01 < P11 < · · · Pr1 < K1 such that if we set Qji = O2 (Pij ) for all i and j, we have Q0i = Pi0 ∩ Q1i and Pi0 ≤ Pi1 for all i = 0, . . . , r. Related to Proposition 20.4 is the following simpler lemma, relating as well to splitting primes. Lemma 20.18. Suppose that K and I = I0 , I1 , . . . , In lie in Chev(2) ∩ Gis for some odd prime s and i = 2 or 4. Assume that the following conditions hold for some odd prime p: (a) I = I0 1. Proof. If 0 = 1, then using the F -inequality and the c-inequality, we have c(K1 ) ≥ 21 . But then Table 21.2 shows that d1 L1 = A1 , contradicting Lemma 21.2(b).  Definition 21.8. By the twisted rank of a parabolic subgroup P of I ∈ Chev(2) we mean the sum of the twisted ranks of the Lie components of a Levi subgroup of P.   Lemma 21.9. Let L, M ∈ Chev(2) with twisted ranks rL , rM and untwisted ranks rL , rM , respectively. Assume that L ≤ M . Then the following conditions hold:   ≤ rM ; (a) rL (b) Any Borel subgroup of L lies in some parabolic subgroup of M of twisted   − rL ; and rank at most rM 2 (c) If L = L(q), L = ArL , DrL , or E6 , and if M is of type A, B, or D, but not 2 B2 or 3 D4 , then rL ≤ rM .  Proof. The proofs are by induction on rL .  Suppose that rL = 1. Then there is nothing to prove in (a), and, when rL = 1,  = 1 is L = 2A2 (q) and rL = 2. But then L in (c). The only other case when rL has nonabelian Sylow 2-subgroups, so rM > 1 (the only groups with rM = 1 are  = 1. So does (b), as a consequence of M = L2 (q)). Hence (a) and (c) hold if rL the Borel-Tits Theorem [IA , 3.1.3]. In general take an end-node maximal parabolic subgroup P of L, and a Levi subgroup J of P . In (c), moreover, choose P so that J is of type 2ArL −2 (q), ArL −2 (q), or 2 D4 (q), according to the type of L. Then O2 (P ) is nonabelian in (c). By the Borel-Tits Theorem [IA , 3.1.3], there is a parabolic subgroup P ∗ < M such that O2 (P ) ≤ O2 (P ∗ ) and P ≤ P ∗ . Let J ∗ be a Levi subgroup of P ∗ , and set   J0 = O 2 (J) and J0∗ = O 2 (J ∗ ). Since P is an end-node subgroup, J0 ∈ Chev(2)   and the twisted rank of O 2 (J) is rJ = rL −1. But J0 embeds in J0∗ , which in turn is ∗ ∗ the central product of groups J1 , . . . , Jk ∈ Chev(2) whose twisted ranks r1 , . . . , rk    sum to at most rM − 1. Then induction in (a) yields rL − 1 ≤ rM − 1, completing the proof of (a). Part (b) is immediate from Corollary 20.17. In (c), since O2 (P ) ≤ O2 (P ∗ ), O2 (P ∗ ) must be nonabelian. But in characteristic 2, the unipotent radicals of end-node parabolic subgroups of untwisted groups of type A, B, and D are all abelian; and the same holds for end-node parabolic subgroups of twisted groups of those types, if the untwisted rank of the parabolic subgroup is rM −1. Therefore P ∗ cannot be an end-node subgroup in the untwisted case, and P ∗ has untwisted rank at most rM − 2 in the twisted case. In all cases it follows that all the central factors J1∗ of P ∗ have untwisted rank at most rM − 2. Again since J0 /Z(J0 ) injects into one of them, induction yields rL − 2 ≤ rM − 2, and the proof is complete. 

As immediate consequences for our embedding K0 ≤ K1 we have:   Corollary 21.10 (Rank inequality). rK ≤ rK , where the symbols denote 0 1 twisted ranks.

324

11. PROPERTIES OF K-GROUPS

Corollary 21.11 (Borel embedding). B0 ≤ P for some Borel subgroup of K0   and some parabolic subgroup P of K1 of twisted rank at most rK − rK . 1 0 Lemma 21.12. Let Bi be a Borel subgroup of Ki , i = 0, 1 and let P1 be a rank one parabolic subgroup of K1 containing B1 . Assume that q0 > q1 . Then the following conditions hold: (a) If K0 and K1 are untwisted groups, then B0 is not embeddable in B1 ; (b) If K0 and K1 are untwisted groups, then B0 is not embeddable in P1 , unless possibly K0 ∼ = A2 (4) and q1 = 2; and (c) Suppose that K0 is untwisted and K1 ∼ = D−1 (q1 ). If 0 > 2, then B0 is not embeddable in P1 . Proof. In (a), choose a prime power pk dividing q0 −1 but not q1 −1 [III8 , 7.10].  0 of K0 contains Zpk × Zpk . Since 0 > 1 by Corollary 21.7, the universal version K  But Z(K0 ) is cyclic as K0 ∈ Chev(2), so K0 contains Zpk . As the odd part of the exponent of B1 divides q1 − 1, B1 contains no copy of B0 , proving (a). The proof of (b) is similar. Here the maximal odd order abelian subgroups of P1 are the direct products of two (coprime) groups, one of exponent dividing q1 − 1 and the other of exponent dividing q1 + 1; the latter group is cyclic since P1 has rank 1. Therefore with pk chosen as in the previous paragraph, and now chosen if possible so that p = 3, we conclude that if pk divides the exponent of P1 , then p divides q1 + 1 and so Sylow p-subgroups of P1 are cyclic. Thus, P1 does not contain a copy of Zpk × Zp . On the other hand unless K0 = A2 (q0 ) and p = 3, B0 even contains Zpk ×Zpk . Finally if K0 = A2 (q0 ) and p = 3, then since p = 3 was the only choice, Zsigmondy’s Theorem [IG , 1.1] implies that q0 = 26 or 22 . But in the first case B0 = L3 (64) contains Z9 × Z3 , so again B0 is not embeddable in P1 . Thus, (b) holds. Part (c) follows similarly. In this case, for any maximal odd order abelian subgroup A ≤ P1 , either A has exponent q12 − 1 and p-rank at most 2 for all prime divisors p of q1 + 1; or A has exponent q14 − 1 and has p-rank at most 1 for all p dividing either q12 + 1 or q1 + 1. Notice that as q1 is a power of 2, q1 − 1, q1 + 1, and q12 + 1 are pairwise relatively prime. Now, we choose pk as before to divide q0 − 1 but not q1 − 1. Then K0 contains the direct product of 3 copies of Zpk , and so is not embeddable in P1 .  Lemma 21.13 (Untwisted Borel Embedding). Suppose that K0 , K1 ∈ Chev(2) are simple groups of respective untwisted ranks rK0 , rK1 and levels q0 , q1 . Suppose  that K1 is untwisted. Let rK be the twisted rank of K0 . Suppose that one of the 0 following conditions holds:  (a) q0 > q1 , 2rK > rK1 , and K0 is the universal version; 0  (b) q0 > q1 and 2rK > rK1 + 1; or 0 (c) There exists a prime power pn not dividing q1 − 1 and a subgroup Q of a  . Cartan subgroup of K0 such that Q ∼ = (Zpn )m for some m > rK1 − rK 0 Then K0 is not embeddable in K1 . Proof. We observe first that each of (a) and (b) implies (c). Indeed if (a) r

0 holds, then a Cartan subgroup H of K0 contains a subgroup Zq0K−1 . Since q0 > q1 , n there is a prime power divisor p of q0 − 1 not dividing q1 − 1. Then (c) holds with   > rK 1 − rK . If (b) holds, on Q the exponent pn subgroup of H, of rank m = rK 0 0

21. MONOTONICITY OF F, AND J2p (G)

325

the other hand, then since the Schur multiplier of K0 has cyclic Sylow p subgroups  ∼ Z rK0 −1 . With pn as above, (c) for all odd primes p, H contains a subgroup H0 = q0 −1  − 1. now holds with Q the exponent pn subgroup of H0 , of rank m = rK 0 Thus we assume that (c) holds but K0 ≤ K1 , and derive a contradiction. If K1 is one of the three groups that is not an adjoint group of Lie type, then it is the commutator subgroup of the universal version, and we replace it without loss by the corresponding element of Lie(2). By Corollary 21.11 and our assumptions, a Borel subgroup B0 of K0 lies in a  parabolic subgroup P of K1 whose rank r(P ) satisfies r(P ) ≤ rK1 −rK < m. Let L 0  be a Levi subgroup of P containing a Cartan subgroup H of B0 , and set J = O 2 (L). Then J is the product of groups in Lie(2). Now J is untwisted of rank r(P ), and AutL (J) ≤ Inndiag(J). Using [IA , 4.10.3], we have mp (Inndiag(J)) ≤ r(P ) < m for all odd primes p. Therefore mp (H/CH (J)) < m as well. Replacing Q by a conjugate we may assume that Q ≤ H. It follows immediately that CH (J) contains a subgroup isomorphic to Zpn . However, CH (J) lies in a Cartan subgroup of K1 and so has exponent dividing q1 − 1. Therefore pn divides  q1 − 1, contrary to assumption. The proof is complete. To introduce the next invariants, which derive from the p-structure of Ki for certain odd primes p, we write the odd part of the order of the universal version  i of Ki as K   i |2 = Φk (qi )mik . (21G) |K k>0

Definition 21.14. e(Ki ) := max{k | mik = 0}. The value of e(Ki ) depends only on di Li ; the function e is tabulated in Table 21.2. For untwisted groups Ki , for example, e(Ki ) is simply the largest of d1 , . . . , di . The following lemma prepares for an application of Zsigmondy’s Theorem. Lemma 21.15. e(K0 )s0 = 6. Proof. Suppose on the contrary that e(K0 )s0 = 6. Since 0 > 1 by Corollary 21.7, we have e(K0 ) > 2 (see Table 21.2), and so (e(K0 ), s0 ) = (3, 2) or (6, 1), whence by the table and the solvability of U3 (2), K0 ∼ = L3 (4), L6 (2), U4 (2), Sp6 (2), D4 (2), or G2 (2) . Accordingly f (K0 ) = 28 , 225 , 29 , 29 , 216 , or 24 . Moreover as type A is the minimal type, we have, respectively, s1 21 ≤ 7, 24, 8, 9, 16, or 4. We derive a contradiction in every case. If K0 ∼ = L3 (4) or G2 (2) , then as 1 > 1 by Lemma 21.2, 1 = 2 and s1 = 1. Thus [K1 , K1 ] ∼ = A2 (2), B2 (2) , or G2 (2) . None of these groups contain U3 (2), but L3 (4) does; and none contain a subgroup of SU3 (2) of index 2 except G2 (2) , which does. So the inclusion K0 ≤ K1 is impossible in these two cases. If K0 ∼ = D4 (2) then m3 (K0 ) = 4 by [IA , 4.10.3] and so m3 (K1 ) ≥ 4. Another application of [IA , 4.10.3] yields 1 ≥ 4. Therefore s1 = 1 and 1 = 4, and f (K0 ) = f (K1 ). Since we are proving the proposition by contradiction, F(K0 ) ≥ F(K1 ), ± 3 whence K1 ∼ = A± 4 (2), D4 (2), or D4 (2). Again by [IA , 4.10.3], m5 (K0 ) = 2 > m5 (K1 ), and the inclusion K0 ≤ K1 is impossible in this case as well. In the remaining three cases m3 (K0 ) = 3 by [IA , 4.10.3] and as in the previous case we deduce that m3 (K1 ) ≥ 3 and 1 ≥ 3. This rules out K0 ∼ = U4 (2). Moreover

11. PROPERTIES OF K-GROUPS

326

∼ Sp6 (2), then u(K1 ) > u(K0 ) = 1 by the u-inequality and Table 21.1; the if K0 = table then yields 1 ≥ 4, so s1 21 ≥ 16, contradiction. The final case is K0 ∼ = L6 (2). In this case since K0 has twisted rank 5, 1 ≥ 5 by Lemma 21.9a, so s1 21 ≥ 25, a contradiction completing the proof.  Now we can prove: Lemma 21.16 (e-inequality). a ≤

e(K1 ) . e(K0 )

 0 | by Proof. Clearly Φs0 e(K0 ) (2) divides Φe(K0 ) (2s0 ), which in turn divides |K definition of e(K0 ). By Lemma 21.15 and Zsigmondy’s Theorem, we may choose a prime p dividing 2s0 e(K0 ) − 1 but not dividing 2n − 1 for any 1 ≤ n < s0 e(K0 ). Thus ordp (2) = s0 e(K0 ). Moreover, p divides |K0 | and hence divides |K1 |. It follows that p divides Φk (q1 ) for some k such that m1k = 0 (see (21G)), i.e., for some k ≤ e(K1 ). But then 2s1 k = q1k ≡ 1 (mod p), so s0 e(K0 ) divides s1 k. A  fortiori , s0 e(K0 ) ≤ s1 k ≤ s1 e(K1 ), and the lemma follows as a = s0 /s1 . In a similar way we can prove: − Lemma 21.17. Assume that neither K0 nor K1 is isomorphic to D2n (q), n ≥ 2,  1 be the universal version of K1 , assume that K1 ∼ or 3D4 (q), for any q. Let K = 3 D4 (q1 ), and write

 1 |2 = |K

(21H)

1 

(q d1i − ω1i )

i=1

with each ω1i = ±1. For each i let b1i = 1 or 2 be the multiplicative order of ω1i and set Di (K1 ) = {t ∈ Z | t > 0, t = b1i r for some r dividing d1i }, 1 ≤ i ≤ 1 . Define Di (K0 ) similarly, i = 1, . . . , 0 , with exponents d0i and orders b0i . Choose any t ∈ Di (K0 ) such that ts0 = 6, and define 

nt = i 1 ≤ i ≤ 0 and Φt (X) divides X d0i − ω0i 

mt = i 1 ≤ i ≤ 1 and b1i d1i is an integer multiple of at .  ⎧  0 )| ⎨1 if all primes dividing 2ts0 − 1 also divide |Z(K (2n − 1) zt = 0 1. The Borel embedding gives an embedding of the Cartan  i ) is subgroup of K0 in that of K1 , viz., Zqr2 −1 /Z0 in Zqr2 −1 /Z1 , where Zi ∼ = Z(K 0 1 cyclic. As r > 1 we see from the exponents of the Cartan subgroups that q02 − 1 divides q12 − 1. By [III8 , Lemma 7.11], q1 is a power of q0 , that is, a = k1 for some k ∈ Z+ . By the F -inequality, a≥

21 (0 − 1)2 2 1 1 = =1− + 2 > . 2 2 0 0 0 0 2

11. PROPERTIES OF K-GROUPS

328

Hence a = 1, i.e., q0 = q1 . But this is absurd since it implies that |K0 | = |U0 +1 (q0 )| > |U0 (q0 )| = |K1 |, contrary to K0 ≤ K1 . Thus, K1 ∼ = A1 (q1 ). But for any odd prime divisor p of q0 + 1, we know from [IA , 4.10.3] that mp (K1 ) ≤ 1 while mp (K0 ) ≥ 0 − z0 , where z0 = 1 if p divides 0 + 1, and z0 = 0 otherwise. As mp (K0 ) ≤ mp (K1 ), it follows that mp (K0 ) = 0 − 1 = 1 . Now the e- and F -inequalities give

2 2 1 0 − 1 0 e(K1 ) 1 = $ 0 % = ≤ ≤ , 0 0 e(K0 ) 2 4 2 +2 so that 0 = 3. But then 0 + 1 = 4 is not divisible by p, and so 0 = mp (K0 ) ≤  mp (K1 ) ≤ 1 , a contradiction completing the proof.  B1 (q1 ), 1 ≥ 2. Lemma 21.19. K1 ∼ = Proof. Suppose that K1 ∼ = B1 (q1 ). By the u-inequality and Table 21.1, ± 3 ∼ K0 = A ± (q ), D (q ), or D (q ). 0 0 4 0 0 0 The 3D4 (q0 ) case is impossible by the F - and e-inequalities, which give 21 /16 ≤ 21 /12 and thus 1 = 2; but B2 (q1 ) has abelian Sylow 3-subgroups while 3D4 (q0 ) has nonabelian Sylow 3-subgroups, contradicting the inclusion K0 ≤ K1 . Suppose next that (21J)

L0 = D0 , 0 ≥ 4.

Then in the untwisted case, the F - and e-inequalities give 21 21 1 ≤ = , 2 0 20 − 2 0 − 1 so that 1 ≤ [0 + 1 + (1/(0 − 1))] = 0 + 1; the rank inequality gives 0 ≤ 1 . In the twisted case the F - and e-inequalities instead give 21 /20 ≤ 1 /0 , so that 1 ≤ 0 , while the rank inequality gives 0 − 1 ≤ 1 . So  0 + 1 or 0 if K0 is untwisted 1 = 0 − 1 or 0 if K0 is twisted.   − rK = 0 or 1. By the Borel In either case, the untwisted ranks satisfy rK 1 0 embedding, a Borel subgroup B0 of K0 is embeddable in a parabolic subgroup P1 of K1 of rank 1. The maximal abelian odd-order subgroups of P1 are isomorphic −1 −1 × Zq1 ±1 , while those of B0 are isomorphic to Zq00−1 × Zq0 ±1 according as to Zq11−1 K0 is twisted or not. As 0 ≥ 4 it follows that q0 − 1 divides q1 − 1. Thus by [III8 , Lemma 7.11], q1 is an integral power of q0 , and so 1/a ∈ Z. In particular a ≤ 1, so 1 ≤ 0 by the F -inequality. Now suppose that 1 = 0 . Then a ≥ 1 by the F -inequality, so a = 1, i.e., q0 = q1 . Consequently f (K1 ) = f (K0 ). As D < B, we have f (K0 ) < f (K1 ), contrary to assumption. Therefore 1 < 0 , whence K0 is twisted and 1 = 0 − 1. −2 × Zq02 −1 in a Cartan subgroup of The Borel embedding in this case embeds Zq00−1 0 −1 2 K1 , isomorphic to Zq1 −1 . Consequently q0 − 1 divides q1 − 1, so a ≤ 1/2, again with the help of [III8 , Lemma 7.11]. The F -inequality gives (0 − 1)2 /20 ≤ 1/2, a contradiction as 0 ≥ 4. We have proved that (21J) cannot hold. Therefore,

(21K)

L0 = A0 , 0 ≥ 2.

21. MONOTONICITY OF F, AND J2p (G)

329

Suppose first that K0 = Al0 (q0 ). The F - and e-inequalities give 21 21 , ≤ 2 0 0 + 1 whence

220 2 = 2(0 − 1) + . 0 + 1 0 + 1 Since 0 + 1 ≥ 3, it follows that 1 ≤ 20 − 2. If q0 > q1 , then by Lemma 21.13b, 20 ≤ 1 + 1, a contradiction. Hence q0 ≤ q1 , i.e., a ≤ 1. Then by the F -inequality, 1 ≤ 0 . On the other hand, by the rank inequality, 0 ≤ 1 . Hence equality holds, and so f (K0 ) ≤ f (K1 ). Since A < B, it follows that F(K0 ) < F(K1 ), contrary to assumption. Hence, K0 ∼ = A− 0 (q0 ). Now, the e- and F -inequalities give 1 ≤

21 21 21 or ≤ 2 0 2(0 + 1) 20 according as 0 is even or odd. Thus 1 ≤ 0 , with strict inequality if 0 is even. In particular as 1 > 1, 0 ≥ 3. On the other hand, let p be any prime divisor of q0 + 1, and let pa be the p-part of q0 + 1. Then mp (K0 ) ∈ {0 − 1, 0 }, with mp (K0 ) = 0 if pa does not divide 0 + 1. On the other hand, mp (K1 ) ≤ 1 . Hence, 1 − 1 ≤ 0 − 1 ≤ 1 , i.e., 1 ∈ {0 − 1, 0 }. Moreover, 1 = 0 unless q0 + 1 divides 0 + 1. Suppose that q0 > q1 . If possible, choose a prime p dividing q02 − 1 = 22s0 − 1 but not dividing 2n − 1 for any n < 2s0 ; this is possible by Zsigmondy’s Theorem unless q0 = 8, in which case we choose p = 3. This exceptional case aside, p does not divide q12 − 1, and so mp (K1 ) ≤ [1 /3], by [IA , 4.10.3], whereas mp (K0 ) ≥ 0 −1 by the same result. As mp (K0 ) ≤ mp (K1 ), we see that 0 − 1 ≤ [1 /3] ≤ 1 /3 ≤ 0 /3, which is absurd as 0 ≥ 3. Thus if q0 > q1 , then q0 = 8 and q1 = 4 or 2. Therefore by [IA , 4.10.2], Sylow 3-subgroups of K1 have order 31 |Σ1 |3 , while those of K0 have order at least 90 −1 |Σ0 +1 |3 . As K0 ≤ K1 it follows that 90 −1 ≤ 31 , with strict inequality if 0 ≡ −1 (mod 3). As 0 ≥ 1 we conclude in turn that 2(0 − 1) ≤ 0 , 0 = 2, and 2(0 − 1) < 0 , again a contradiction. Hence q0 ≤ q1 , whence 0 > 1 . Thus, 1 = 0 − 1. As shown above, q0 + 1 divides 0 + 1. In particular, if 0 ≤ 3, then 0 = 2 = q0 . But then K0 ∼ = U3 (2) is not simple, contrary to assumption. Hence 0 ≥ 4. By the F -inequality, a ≥ (0 − 1)2 /20 ≥ (3/4)2 > 1/2. On the other hand, K0 contains a homocyclic abelian group of rank 0 − 1 = 1 and exponent q0 + 1. Hence, K1 likewise contains such a group. By [IA , 4.10.3], it follows that q0 + 1 divides q12 − 1. By [III8 , 7.11], 2s0 divides 2s1 and so a = s0 /s1 is the reciprocal of an integer. Therefore a = 1. Now the e-inequality yields e(K1 ) ≥ e(K0 ), that is, 2(0 − 1) ≥ 4[0 /2] + 2 ≥ 20 , a final contradiction.  = D±1 (q1 ) for any 1 ≥ 4. Lemma 21.20. K1 ∼ Proof. Suppose false and set K2 = B1 (q1 ). 2

Then K1 is isomorphic to

a subgroup of K2 . Moreover f (K1 ) = q11 = f (K2 ), so f (K2 ) ≤ f (K0 ). If F(K2 ) ≤ F(K0 ), then by Lemma 21.19, equality holds and K0 ∼ = K2 . However, this is impossible as |K0 | ≤ |K1 | < |K2 |. Therefore F(K0 ) < F(K2 ). Hence f (K2 ) = f (K0 ), so f (K0 ) = f (K1 ), and L0 has type A, D or E, by definition of F

330

11. PROPERTIES OF K-GROUPS

[III7 , Definition 2.2]. Note that by the u-inequality, u(K0 ) < u(K1 ) < 1. Hence by Table 21.1, K0 is not of type E. If L0 has type A, then F(K0 ) < F(K1 ), contrary to our assumption. Thus K0 has type D, like K1 . Since u(K0 ) < u(K1 ), we see by Table 21.1 that 1 > 0 . By the F -inequality, a > 1, so q0 > q1 . Also, by the F - and e-inequalities, 1 21 2(1 − k1 ) ≤ ≤a≤ , 20 2(0 − k0 ) 0 − k0 where ki = 1 or 0 according as Ki is untwisted or twisted. One consequence is that 1 ≤ [20 /(0 − 1)] = 0 + 1. As 1 > 0 , we must have 1 = 0 + 1. Another consequence is that k0 = 1, for otherwise 21 /20 < 1 /0 , which is absurd as 1 > 0 . Therefore K0 ∼ = D1 −1 (q0 ). A third consequence is that a ≤ 1 /(1 − 2) ≤ 2, since 1 ≥ 4.  = 1 − 1. Either K1 ∼ The twisted rank of K0 is rK = D1 (q1 ) has twisted rank 0 −   ∼ rK1 = 1 , or K1 = D1 (q1 ) with rK1 = 1 − 1. Accordingly the Borel embedding embeds a Borel subgroup B0 of K0 into a rank 1 parabolic subgroup P1 of K1 ∼ = D (q1 ) or into a Borel subgroup B1 of K1 ∼ = D− (q1 ). Thus a Cartan subgroup −1 of B0 embeds either in a maximal abelian subgroup of P1 of odd order, H0 ∼ = Zq01−1 −1 −2 which is isomorphic to Zq11−1 × Zq1 ±1 , or in a Cartan subgroup H1 ∼ × Zq12 −1 = Zq11−1 a of B1 . However, q0 = q1 with 1 < a. As q0 > q1 , there is a prime power pc dividing q0 − 1 but not dividing q1 − 1. Choose such a prime power with c maximal. Then H0 has 1 − 1 ≥ 3 direct factors isomorphic to Zpc , but Zq1 −1 has none, so B1 and P1 have at most one such direct factor. This contradicts the fact that H0 embeds  in B1 or P1 . The proof is complete. n Lemma 21.21. K1 ∼ = 2B2 (2 2 ) for any odd n and K1 ∼  3D4 (q1 ). = n  2B2 (2 2 ). Proof. By Lemma 21.2, K1 has twisted rank at least 2, whence K1 ∼ =  = 2. By the rank inSuppose that K1 ∼ = 3D4 (q1 ), so that the twisted rank is rK 1  equality, rK0 ≤ 2; by the u-inequality, u(K0 ) < u(K1 ) = 3/4. The only possibilities − are d0 L0 = A− 3 or A4 . Now, we use [IA , 4.10.3] to obtain mp (K0 ) ≥ 3 for any prime p dividing q0 + 1, but also mp (K1 ) ≤ 2 for all odd primes, a contradiction as  K0 ≤ K1 . The lemma follows.

Lemma 21.22. K1 ∼  G2 (q1 ). = Proof. Suppose that K1 ∼ = G2 (q1 ). Then m3 (K1 ) = 2 and K1 has a subgroup of order 3 that is weakly closed in a Sylow 3-subgroup of K1 . By [IA , 7.8.2] and Corollary 21.5, m3 (K0 ) ≤ 1. As 0 > 1, [IA , 4.10.3] gives K0 ∼ = 2 B2 (q0 ) or A− 2 (q0 ) with  = ±1 ≡ q0 (mod 3) in the latter case. Since A < G and B < G, and 1 = 2 = 0 in all cases, the F -inequality yields a > 1, i.e., q0 > q1 . Let H0 be a Cartan subgroup of K0 . In the case K0 ∼ = A2 (q0 ), |H0 | = (q0 − 1)2 , while in the other two cases, |H0 | = q02 − 1. In the first case H0 embeds in a Cartan subgroup H1 of K1 via the Borel embedding; in the second case H0 embeds in a rank 1 parabolic subgroup P1 of K1 . The largest odd order abelian subgroups of H1 , resp. P1 , have orders (q1 − 1)2 , resp. q12 − 1, but these are  strictly less than |H0 | in every case. This contradiction completes the proof. Lemma 21.23. K1 ∼ = 2 F4 (q1 ).

21. MONOTONICITY OF F, AND J2p (G)

331

∼ 2 F4 (q1 ), so that m3 (K0 ) ≤ m3 (K1 ) = 2 by [IA , Proof. Suppose that K1 = 4.10.3]. Consequently L0 = A2 , B2 , or G2 , or K0 ∼ = A− 0 (q0 ), 0 = 3 or 4, q0 ≡  3 ∼ (mod 3), or K0 = D4 (q0 ). The F - and c-inequalities rule out L0 = B2 and G2 . If K0 ∼ = 3D4 (q0 ), then we see from [IA , Table 4.7.3A] that for some x ∈ K0 of order 3, CK0 (x) contains L2 (q03 ), while CK1 (x) ∼ = SU3 (q12 ). Comparing 2-ranks of these centralizers yields 3 2 q0 ≤ q1 , so a ≤ 2/3. However, a ≥ 1 by the F -inequality, a contradiction. Likewise if K0 ∼ = A0 (q0 ), 0 = 3 or 4, then for a suitable x ∈ K0 of order 3, CK0 (x) contains L2 (q02 ) or L3 (q0 ), respectively. In the first case q02 ≤ q12 so a ≤ 1, contradicting the F -inequality as 0 = 3. The second case is impossible because L3 (q0 ), having no strongly embedded subgroup, is not embeddable in SU3 (q12 ). The cases K0 ∼ = A− 0 (q0 ), 0 = 3 or 4 are ruled out because mp (K0 ) ≥ 3 for any prime divisor p of q0 + 1, whereas mp (K1 ) ≤ 2 by [IA , 4.10.3], contradiction. In the case K0 ∼ = A− 2 (q0 ), the F - inequality gives a > 4 since A < F ; but a ≤ 4 by the e-inequality, a contradiction. The only remaining case is K0 ∼ = A2 (q0 ), so that again a > 4 by the F -inequality. As Cartan subgroups of K0 and K1 are of exponent q0 − 1 = q1a − 1 and q12 − 1, the Borel embedding is contradicted. The proof is complete.  = 2 E6 (q1 ). Lemma 21.24. K1 ∼ Proof. Suppose that K1 ∼ = 2 E6 (q1 ). By the u-inequality, L0 = A0 or D0 . If L0 = D0 then apart from the case K0 ∼ = 3D4 (q0 ), the F - and c-inequalities yield 13 26 36 , ≤ a ≤ 0  = 2 0 0 (0 − 1) 2 ∼ so 0 /(0 − 1) ≥ 36/26. As 0 ≥ 4 this is a contradiction. The possibility K0 = D4 (q0 ) is similarly ruled out by the F - and e-inequalities, which imply that 36/16 ≤ 18/12. Therefore L0 = A0 . Suppose that K0 ∼ = A− 0 (q0 ). First consider the cases 0 ≥ 6 (resp. 7). Then for any prime divisor p of q0 + 1, mp (K0 ) ≥ 6 (resp. 7). Note here when 0 = 6 that q0 = 2s0 ≡ −1 (mod 7). On the other hand mp (K1 ) ≤ 6 by [IA , 4.10.3], with equality possible only if p divides q1 + 1. As K0 ≤ K1 , we conclude that mp (K1 ) = 0 = 6 and every prime divisor of q0 + 1 divides q1 + 1. But a > 1 by the F -inequality, since A < E. Therefore q0 > q1 . By Zsigmondy’s Theorem applied to q02 − 1, this is possible only if q0 = 8. But then |K0 |3 = 96 |Σ7 |3 > 36 |W (E6 )|3 ≥ |K1 |3 , contradicting K0 ≤ K1 . Therefore 0 ≤ 5. The F - and e-inequalities yield

3

(21L)

18 36 , ≤a≤ 20 4[0 /2] + 2

  so 8[0 /2] + 4 ≤ 20 . This forces 0 = 5. Now rK = 4 and rK = 3, so by the Borel 1 0 embedding, a Cartan subgroup H0 of K0 embeds in a rank 1 parabolic subgroup P1 of K1 . The invariant factors of H0 are (q0 + 1)/z, q02 − 1 and q02 − 1, where z = gcd(3, q0 + 1), while the maximal odd order subgroups of P1 are the direct product of a group of exponent dividing q12 − 1 with a cyclic group of exponent dividing q14 − 1. It follows that q02 − 1 divides q12 − 1. Therefore a ≤ 1, which contradicts (21L).

11. PROPERTIES OF K-GROUPS

332

We have proved that K ∼ = A0 (q0 ). By the F - and c-inequalities, 36 13 . ≤ [(0 + 1)2 /4] 0 2 This implies that 0 > 4, contradicting the rank inequality and completing the proof.   F4 (q1 ). Lemma 21.25. K1 ∼ = Proof. Suppose that K1 ∼ = F4 (q1 ). By the rank inequality and u-inequality, L0 = A0 , 2 ≤ 0 ≤ 4; A− , 2 ≤ 0 ≤ 8; B0 , 2 ≤ 0 ≤ 4; 2 B2 ; D4± ; 3 D4 ; D5− ; or 0 − E6 . By [IA , 4.10.3], mp (K0 ) ≤ mp (K1 ) ≤ 4 for all odd primes p. This rules out − − all the cases K0 ∼ = A− 0 (q0 ), 5 ≤ 0 ≤ 8, D5 (q0 ), and E6 (q0 ), for all of which mp (K0 ) > 4 for some prime p dividing q0 + 1, with the single exception of U6 (2). If K0 ∼ = U6 (2), there is a semisplit torus H ∼ = E34 in K0 such that AutK0 (H) ∼ = Σ6 . On the other hand, K1 has a unique conjugacy class of E34 -subgroups by [IA , 4.10.3]; these are semisplit tori and AutK1 (H) ∼ = W (F4 ). As W (F4 ) is solvable, it does not contain a copy of Σ6 , so K0 ∼ = U6 (2) is impossible. We have proved that 0 ≤ 4. n The cases K0 ∼ = 3D4 (q0 ) and 2B2 (2 2 ) are impossible as the F - and e-inequalities yield 1 < a ≤ 12/12 and 4 ≤ a ≤ 12/8, respectively. Similarly, we get the analogous contradiction when K0 ∼ = A± 2 (q) or B2 . These arguments use the type inequalities d0

(21M)

A < F, D < F, and B < F.

For the cases K0 ∼ = A3 (q0 ), A4 (q0 ), and D4± (q0 ), we use the Borel embedding to reach a contradiction. We have a > 1 in all cases by the F -inequality, using (21M) when necessary. Thus q0 > q1 . The Borel embedding sends a Cartan subgroup  = 4) H0 of K0 into H1 , which is either some Cartan subgroup of K1 (when rK 0 or some maximal abelian odd order subgroup of a rank 1 parabolic subgroup P1 of  = 3). In any case H1 ∼ K1 (when rK = Zq31 −1 × Zq1 ±1 . But H0 has a direct product 0 decomposition with at least three direct factors that are cyclic of order at least q0 − 1, and just the existence of two such factors already prevents the existence of any embedding H0 ≤ H1 , a contradiction. If K0 ∼ = B0 (q), 0 = 3 or 4, then K0 has a subgroup K0∗ ∼ = A3 (q0 ) or D4 (q0 ), respectively, with f (K0∗ ) = f (K0 ) ≥ f (K1 ). Because of (21M), our assumption F(K0 ) > F(K1 ) implies that F(K0∗ ) > F(K1 ), which we have just shown leads to a contradiction. − The only remaining cases are K0 ∼ = A− 3 (q0 ) and A4 (q0 ). For them we have a > 1 as usual by the F -inequality, so q0 > q1 . Hence by Zsigmondy’s Theorem there is a prime p dividing q02 − 1 but not dividing 2n − 1 for any power 2n < q02 , unless q0 = 8, a case that we consider in a moment. Then K0 contains a Ep3 subgroup, and so mp (K1 ) ≥ 3. Consequently p divides q12 − 1, by [IA , 4.10.3]. Unless q0 = 8, we deduce that q12 ≥ q02 and so q1 ≥ q0 , a contradiction. When q0 = 8, q1 = 2 or 4, whence |K1 |3 = 36 , and using [IA , Table 4.7.3A] we see that a Sylow 3-subgroup of K1 has a maximal subgroup that is extraspecial of order 35 and exponent 3. This is inconsistent with the fact that K0 contains a copy of Z93 . The proof is complete.  Lemma 21.26. K1 ∼ = E1 (q1 ), 1 = 6 or 7.

21. MONOTONICITY OF F, AND J2p (G)

333

Proof. Suppose that K1 ∼ = E1 (q1 ), 1 = 6 or 7. Consider first the case that K0 is untwisted. Thus, by the u-inequality, K0 ∼ = A0 (q0 ), D0 (q0 ), or (for 1 = 7 only) B0 (q0 ) or E6 (q0 ). Then 1 ≥ 0 by the rank inequality, so a ≥ 1 by the F -inequality. Moreover, since A < E and D < E, a > 1 unless K1 ∼ = E7 (q1 ) and K0 ∼ = B7 (q0 ). So q0 ≥ q1 , with equality only if K1 ∼ = E7 (q1 ) and K0 ∼ = B7 (q1 ). In this latter case, however, |K1 | q 18 − 1 = q114 · 14 |K0 | q1 − 1 is not an integer, yielding a contradiction. Hence, we have q0 > q1 . We claim that 0 ≤ 3. Choose any prime power pn dividing q0 − 1 but not dividing q1 − 1 [III8 , Lemma 7.10]. Then the exponent pn subgroup Q of a Cartan subgroup of K0 satisfies Q ∼ = Zp0n /Y , where Y = 1 or Y ∼ = Zp , the latter occurring if and only if p = 0 + 1 = 5 or 7 and L0 = A0 , or p = 3 and L0 = E6 . If Y = 1 and 20 > 1 , then the conditions of Lemma 21.13c hold with 0 in the role of m there, and that lemma yields a contradiction. Thus if Y = 1, then 20 ≤ 1 ≤ 7, so 0 ≤ 3 as claimed. If Y ∼ = Zp0n−1 = Zp5n , = Z7 , then in any case Q contains a subgroup Q0 ∼ and Lemma 21.13 applies again to give a contradiction, with Q0 in place of Q, and m = 5, since 0 = 6. A similar argument applies if p = 3 and L0 = E6 . Likewise in the case Y ∼ = Z5 , Lemma 21.13 applies and yields our claim if we can replace the choice p = 5 by some other prime divisor p of q0 − 1 not dividing q1 − 1. Note that s0 is a multiple of 4 since q0 ≡ 1 (mod 5), so Zsigmondy’s Theorem provides such a prime p unless q0 = 16. But then as q1 < q0 , we can use p = 3 in this situation unless q1 = 4. We then have a = 2, so the F -inequality gives 21 ≤ 220 = 32, a contradiction. Thus our claim holds. In the residual untwisted cases, that is, L0 = A2 , A3 , B2 , or B3 , the F and e-inequalities together yield a contradiction. They also give a contradiction n n if K0 ∼ = 3D4 (q0 ), 2B2 (2 2 ), or 2F4 (2 2 ), and likewise if K0 ∼ = E6− (q0 ). It remains − − ∼ to consider the cases K0 = A0 (q0 ) and D0 (q0 ). If 0 ≤ 6, then the F - and einequalities provide the desired contradiction unless K0 ∼ = D6− (q0 ) and K1 ∼ = E7 (q1 ), in which case Lemma 21.13b gives a contradiction. Thus 0 > 6, and so for any prime divisor p of q0 + 1, mp (K1 ) ≥ mp (K0 ) ≥ 7. But mp (K1 ) ≤ 1 by [IA , 4.10.3]. Therefore mp (K0 ) = mp (K1 ) = 7 and L1 = E7 . Consequently 0 = 7 or 8. Furthermore, p divides q12 − 1, again by [IA , 4.10.3]. Unless q0 = 8, we may choose p not to divide 2n − 1 for any n < 2s0 , so the multiplicative order of 2 (mod p) is 2s0 . Therefore 2s0 divides 2s1 , so a is the reciprocal of an integer. Since both A < E and D < E, the only possibility in the F -inequality is that a = 1 and 0 = 8. In particular q0 = q1 . Then L0 = D8 is impossible because |K0 | would be divisible by Φ16 (q0 ), which has a prime divisor not dividing any of the cyclotomic divisors Φn (q0 ) of |K1 |. Hence L0 = A8 but mp (K0 ) = 7. This forces p = 3, so q0 = 2. In this case Sylow 3-subgroups of K0 and K1 coincide, and for any E ≤ K0 such that E ∼ = E37 , AutK0 (E) ∼ = Σ9 embeds in AutK0 (E) ∼ = W (E7 ). However, such an embedding is impossible by Lemma 8.1. We conclude that q0 = 8. By the F -inequality, a ≥ 72 /82 > 3/4 so q1 ≤ 8. If q1 < 8, then ord7 q1 = 3, so m7 (K1 ) = 3. But m7 (K0 ) > 3, a contradiction. Thus, q1 = 8 = q0 and a = 1. Again by the F -inequality, 0 = 8. If L0 = A8 , then Sylow 3-subgroups of K0 and K1 coincide and have a unique E37 subgroup, and we reach the same contradiction as in the previous paragraph. Thus, L0 = D8 . As in the

334

11. PROPERTIES OF K-GROUPS

previous paragraph, |K0 | but not |K1 | is divisible by Φ16 (q0 ), a final contradiction proving the lemma.  Lemma 21.27. If K1 ∼ = E8 (q1 ), then K0 ∼ = A3 (q0 ), A4 (q0 ), D4 (q0 ), A− 0 (q0 ), − 5 ≤ 0 ≤ 8, D5 (q0 ), or U10 (4). Proof. Suppose that K1 ∼ = E8 (q1 ). Suppose first that K0 is untwisted. By the rank inequality, 0 ≤ 8. Thus a ≥ 1 by the F -inequality, with strict inequality for types A and D. If 0 < 1 , then q0 > q1 and, by Lemma 21.13, K0 cannot be of rank 5, 6, or 7. If K0 ∼ = B8 (q0 ), then c(K1 ) = 36 = c(K0 ) so the F - and c-inequalities force a = 1, i.e., q0 = q1 . However, then Φ16 (q0 ) divides |K0 | but not |K1 |, a contradiction. Thus, 0 ≤ 4. The F - and c-inequalities yield contradictions if L0 = A2 , B2 , B3 , B4 , F4 , or G2 . Only the untwisted groups in the statement of the lemma remain. Now suppose that K0 is twisted, of twisted rank at most 8 by the rank inequal− ω ity. The F - and e-inequalities rule out K0 ∼ = A− 0 (q0 ), 2 ≤ 0 ≤ 4, D4 (q0 ), D4 (q0 ), n n − 2 2  B2 (2 2 ), F4 (2 2 ) , and E6 (q0 ). By [IA , 4.10.3], mp (K1 ) ≤ 8 for all odd primes p, − but the groups A− 0 (q0 ) and D0 (q0 ) with 0 ≥ 9 satisfy mp (K0 ) ≥ 9 for some prime p dividing q0 + 1, with the single exception of A− 9 (4). The twisted possibilities for − (q ) and D (q ) with 5 ≤ 0 ≤ 8, and A− K0 are now restricted to A− 0 0 9 (4). 0 0 − Finally, if K0 ∼ = D0 (q0 ), 0 = 6, 7, or 8, we have a > 1 by the F -inequality and   = 0 − 1, so 2rK > 9, and Lemma 21.13b yields a the fact that D < E. Then rK 0 0 contradiction. The proof is complete.  Lemma 21.28. K1 ∼  E8 (q1 ). = Proof. We assume that K1 ∼ = E8 (q1 ) and must rule out the various residual possibilities for K0 from Lemma 21.27. We shall use the F -, e-, and c-inequalities to restrict a and then use Lemma 21.17 for appropriate t’s. We define E, using the notation of Lemma 21.17, by E := {bi di |i = 1, . . . , 8} = {2, 8, 12, 14, 18, 20, 24, 30}. When we trace an integer t in |K1 | we set Eta = {x ∈ E | x is an integer multiple of ta}. Thus mt , in Lemma 21.17, is |Eta |. ∼ A3 (q0 ), so that Suppose that K0 = 30 64 7. However, we readily see from (21N) that E2a ⊆ {30}, so |E2a | = 1, a contradiction. Thus, K ∼ = A3 (q0 ). Similarly consider K0 ∼ = A4 (q0 ). Then 4 < a ≤ 6. We trace t = 5 through |K1 |; clearly ts0 = 6, and 20 < ta ≤ 30. Therefore E5a = {24} or {30}, with a = 24/5 or 6, respectively. We can also trace t = 4 in |K1 |; of the two possible values for a, only a = 6 yields a nonempty E4a . To obtain a contradiction from a = 6 we take a parabolic subgroup P0 of K0 of type A3 and observe that Φ4 (q0 ) divides |P0 |. Since q0 = q16 , |P0 | is divisible by Φ24 (q1 ). By the Borel-Tits Theorem P0 ≤ P1 for some

21. MONOTONICITY OF F, AND J2p (G)

335

proper parabolic subgroup P1 of K1 . But it is easy to check that Φ24 (q1 ) does not divide |P1 | for any such P1 . For K0 ∼ = D4 (q0 ) we trace t = 6 in |K1 |. In this case 4 < a ≤ 5 by the F - and e-inequalities. We have s0 = as1 > 4, and so ts0 = 6. The only possibility this time is a = 5. But then if we trace t = 2, we have mp (K0 ) = 4, and so there must be four elements of E that are integer multiples of at = 10, a contradiction. Next consider − K0 ∼ = A− 5 (q0 ) and D5 (q0 ). The F - and e-inequalities give 2.56 < a ≤ 3. Tracing t = 10 we see that the only possibility is a = 3, so q0 ≥ 8. Assume for the moment that q0 > 8, i.e., s0 = 3. Then any prime p chosen to divide q02 −1 but not 2n −1 for any n < 2s0 is not p = 3, so mp (K0 ) = 5. Then tracing t = 2, we get |E2a | ≥ mp (K0 ) = 5. But 2a = 6 and by inspection, |E2a | = 4, a contradiction. Finally if s0 = 3, then s1 = 1, and with  0 = SU6 (8) contain a homocyclic group of exponent [IA , 4.10.2], K0 = D5− (8) and K 9 and rank 5. Note on the other hand that again by [IA , 4.10.2], since q1 = 2, a Sylow 3-subgroup of K1 has a normal elementary abelian 3-subgroup (of rank 8) with nonabelian quotient of order 35 , isomorphic to a Sylow 3-subgroup of W (E8 ). Such a 3-group cannot contain Z95 , so K0 = A− 5 (8). In this case K0 contains a 3-constrained subgroup N such that F ∗ (N ) ∼ = Z3 × Z94 and N/F ∗ (N ) ∼ = Σ6 . This time, the contradiction is that there is no elementary abelian 3-subgroup of N that is normal and of index at most 35 in a Sylow 3-subgroup of N . This finishes the case 0 = 5. If K0 ∼ = A− 6 (q0 ) then 64/36 < a ≤ 30/14, as usual by the F - and e-inequalities. We take t = 14 (so that ts0 = 6) and get 224/9 < at ≤ 30, so at = 30 and a = 15/7. We then have s0 ≥ 15, so we may take t = 6, for which mp (K0 ) = 2 by [IA , 4.10.3], and find that at least two elements of E are integer multiples of at = 90/7. This is absurd. If K0 ∼ = A− 7 (q0 ) then 64/49 < a ≤ 30/14, and we trace t = 14. Then 18 < 14a ≤ 30, so a = 20/14, 24/14, or 30/14. So s0 ≥ 10 and tracing t = 6 as in the A− 6 (q0 ) case we get that at least two elements of E are integer multiples of at = 60/7, or of at = 72/7, or of 90/7, again an absurdity. If K0 ∼ = A− 8 (q0 ) then 1 < a ≤ 30/18, and we trace t = 18, concluding that a = 20/18, 24/18, or 30/18. Then s0 is divisible by either 5 or 4, so we may trace t = 14 and conclude that some element of E is an integer multiple of 14a, which clearly is not the case. ∼ The final case is K0 ∼ = A− 9 (4) = U10 (4). With [IA , 4.10.3], we have m5 (K1 ) ≥ m5 (K0 ) = 8 and then that 5 divides q12 − 1. Note that a > 64/81 > 1/2 by the F -inequality, so q1 < q02 = 16. The only choice is q1 = 4 = q0 . Then K0 contains a subgroup AΣ, where A ∼ = E58 and Σ ∼ = Σ10 acting on A as the nontrivial composition factor of the natural permutation module. By [IA , 4.10.3c], there is a unique conjugacy class in K1 of E58 subgroups, and since a semisplit torus has this isomorphism type, we conclude that AutK1 (A) ∼ = W (E8 ). But by Lemma 8.1,  W (E8 ) contains no subgroup isomorphic to Σ10 . This completes the proof. By Lemmas 21.18 through 21.26 and Lemma 21.28, we have reached a contradiction completing the proof of Proposition 21.1. We shall next derive Proposition 21.31 below from Proposition 21.1. First we need two easy lemmas.

336

11. PROPERTIES OF K-GROUPS

Lemma 21.29. Suppose that K, J ∈ Kp for some odd prime p, with K k. Thus (a) holds in this case. Finally suppose that each Z(Xi ) is noncyclic. As Xi is a K-group, p = 3 and Xi ∼ = 32 U4 (3), by [IA , 6.1.4]. By [IA , 6.4.4], m3 (Xi ) = m3 (Z(Xi )) + m3 (Xi /Z(Xi )), which implies that every 3-central element of Xi /Z(Xi ) of order 3 is the image of an element of Xi of order 3. Choose xi ∈ Xi − Z(Xi ), i = 1, . . . , k, such that xi Z(Xi ) ∈ Z(P/Z(Xi )), and set V = x1 , . . . , xk  Z. As in the previous case we argue that m3 (CV (x) x) > k, completing the proof of (a). In (b), it is enough to show that X1 ∼ = Sp4 (q); the final assertion then follows from [IA , Table 4.5.2]. Our assumptions X1 ∈ Chev and X1 /Z(X1 ) ∼ = L2 (q) imply that m2 (X1 ) > 1. Suppose that m2 (X1 ) = 2. If Z(X1 ) = 1, then by Lemma 7.13,

338

11. PROPERTIES OF K-GROUPS

X1 ∼ = Sp4 (q) for some odd q, a case we shall discuss further below. If Z(X1 ) = 1, then by our hypothesis and [IA , 5.6.3], X1 ∼ = L3 (q),  = ±1; but then by [IA , Table η 4.5.1, 4.9.1], x acts either as a 2-central involution of X1 or CX1 (x) ∼ = L3 (q 1/2 ), η = ±1, and in either case m2 (CX1 (x)) = 2, contrary to assumption. Now assume that m2 (X1 ) ≥ 3. By [III2 , 6.7], if x induces an inner automorphism on X1 , then m2 (CX1 (x)) ≥ 2, a contradiction. Thus (22A)

x induces a non-inner automorphism on X1 .

Write s2 (T ) for the sectional 2-rank of the group T , that is, the largest 2-rank of any section of T . If s2 (X1 x) ≥ 5, then by [III8 , 1.5], X1 x has a connected Sylow 2-subgroup, which immediately implies that m2 (CX1 (x)) ≥ 2, contradiction. Therefore (22B)

s2 (X1 x) ≤ 4. A± 5 (q)

each have sectional 2-rank at least 6 since they Note that B3 (q), C3 (q), and contain three mutually commuting SL2 (q)-subgroups. Hence by (22B), X1 cannot involve any of them. Moreover, the only non-inner involutory automorphisms x of G2 (q) and 3D4 (q) are field automorphisms, so m2 (CX1 (x)) > 1 in those cases; and n Out(2 G2 (3 2 )) has odd order. These considerations reduce us to the following cases: X1 ∼ = An (q), n = 3 or 4,  = ±1, or X1 ∼ = B2 (q). If x induces a field or graph-field automorphism on X1 , then clearly m2 (CX1 (x)) > 1, contradiction. Likewise we reach the same contradiction if x induces a graph automorphism or if X1 ∼ = B2 (q)a , by [IA , Tables 4.5.1, 4.5.2]. These arguments reduce the proof to the cases X1 ∼ = B2 (q)u (the desired conclu ∼ sion) and X1 = A3 (q), with x mapping to a nonidentity element of Outdiag(X1 ) in the latter case. By [IA , Tables 4.5.1, 4.5.2], we may assume that X1 ∼ = SL4 (q), with 2 ∼ E(CX1 (x)) = SL2 (q ) and Z(E(CX1 (x))) ≤ Z(X1 ). Considering X1 as Spin6 (q), we see that x acts on X1 as an involution of O6 (q) with two eigenvalues equal to −1 on the natural module V . Then x has an X1 -conjugate x whose −1-eigenspace on V is contained in the 1-eigenspace of x. This implies that xx is an involution of CX1 (x) (see [IA , 6.2.1cd]). As xx ∈ Z(X1 ), this contradicts m2 (CX1 (x)) = 1 and completes the proof.  Lemma 22.2. Let X = Σ6 and let M be a 4-dimensional irreducible F3 [K]module. Let V ≤ [X, X] be a four-group acting freely on M and write M = ⊕3i=0 Mi , where M0 = CM (V ) and the Mi are the nontrivial irreducible V -submodules of M . Then the orbits of X on E1 (M ) have cardinalities 15, 15, and 10, and they are represented, respectively, by M0 ; M1 ; and any element of E1 (M0 ⊕ M1 ) besides M0 and M1 . Proof. Considering A6 as P SL2 (9), we can use the Steinberg Tensor Product theorem to determine the A6 -irreducible modules over F3 , and there is just one of dimension as small as 4. We recognize this as the core of the natural permutation module. This representation then has two extensions to X, which are the natural permutation module and its twist under an outer automorphism of X. Without loss, then, we may assume that M is the core of the natural permutation module & ∗ M ∗ . Here M ∗ has F3 -basis {bi }6i=1 permuted by X, and in M := M ∗ /  bi , M is generated by all bi −bj , i = j. There are two choices for V , up to conjugacy; the two are handled similarly. Suppose for example that V = (12)(34), (34)(56).    Then one calculates that M0 = b1 + b2 − b3 − b4 and without loss, M1 = b1 − b2 .

22. MISCELLANEOUS

339

Both of these are easily seen to have 15 conjugates under  X. Then the two other  1-dimensional subspaces of M0 ⊕ M1 are bi + b3 + b4 , i = 1, 2, and they are centralized by the Sylow 3-subgroup Pi of X containing (i34). As the unique maximal subgroup of X containing Pi is NX (Pi ), of index 10, the lemma follows.  Lemma 22.3. Suppose that K ∈ K3 with Z(K) = 1, and E32 ∼ = x, f  ≤ Aut(K). If L3 (CK (x)) ∼ = L2 (8) and f induces a field automorphism on L3 (CK (x)), then CK (f )/O3 (CK (f )) is not involved in C := CZ3k ×Σ11 (z) where z is a 3-central element of Σ11 of order 3. Proof. One sees easily that |C|2 = 4 and m3 (C) = 4. 3 If K ∈ Chev(3) ∪ Alt ∪ Spor, then given L3 (CK (x)) ∼ = L2 (8), we have K ∼ = 2 G2 (3 2 ) or Co3 , by [IA , 2.2.10, Tables 5.3]; but then Z(K) = 1 by [IA , 6.1.4], a contradiction. A similar contradiction arises if x is a field automorphism. Thus, x ∈ Inndiag(K). It follows √ from [IA , 4.2.2] that 8 is an integer power of the level of K, so K has level 2, 8, or 8. Indeed using [IA , 4.8.2, 4.8.4, Table 4.7.3A], as well as Table 13.1 and the hypothesis Z(K) = 1, we see that the only possibilities for K/Z(K), as a pumpup of L2 (8), are U3n (8), n odd, E6− (8), and U6 (2). This last case arises from viewing a 2-dimensional unitary F23 -space as a 6-dimensional unitary F2 -space, so f acts freely on the natural module, whence CK (f ) contains Σ3 × Σ3 × Σ3 , contradicting (22C). Thus K has level 8, whence f is a field automorphism of K. By Lemma 4.5, C does not involve SU3 (2). Hence K ∼  E6− (8) by [IA , Table = ∼ 4.7.3A], and K =  SU3n (8). As Z(K) = 1, n must be divisible by 3, whence CK (f ) involves U9 (2), which by orders is not involved in A11 . The proof is complete. 

(22C)

Lemma 22.4. L4 (2) has no subgroup of order 32 .5.7. Proof. If such a subgroup H existed, it would be solvable and hence contain a subgroup of order 5.7. But groups of that order are abelian, whereas any 7-cycle  in L4 (2) ∼ = A8 is self-centralizing. This proves the lemma. Lemma 22.5. Suppose that X is a group and Q = F ∗ (X) = O2 (X) = Z(Q)E, where E is extraspecial and Z(Q) is cyclic of order at most 4. Suppose that X = QY with Y ∼ = A5 . Then m2 ([X, X]) ≥ 4. Proof. Suppose that m2 ([X, X]) ≤ 3. Let y ∈ Y of order 5. Then [Q/Φ(Q), y] is the direct sum of minimal y-invariant subgroups, and is the image of a subgroup of X of 2-rank at most 3. It follows that [Q, y] ∼ = D8 ∗ Q8 , and so Y has a unique nontrivial composition factor F on Q/Φ(Q). Thus F is a Y ∼ = SL2 (4) natural − module or a natural Y ∼ Ω (2) module. Since [Q, y] is extraspecial, F must = 4 − possess a Y -invariant quadratic form over F2 , forcing F to be the Ω4 (2)-module. But then F is a free module for a Sylow 2-subgroup of Y , hence H 1 (Y, F ) = 0 and so [Q, y] is Y -invariant. Then we may assume that X = [Q, y]Y , whence X is perfect. Now [Q, y] has exactly 5 four-subgroups and they are permuted transitively by Y . Let V be one of these four-groups; then X ≥ V N , where N = NY (V ) ∼ = A4 . ∼ 2 4 centralizes V and V [N, N ] E , contradicting m (X) ≤ 3. Then [N, N ] ∼ E = 2 = 2 2 The proof is complete. 

Bibliography Michael Aschbacher, Standard components of alternating type centralized by a 4-group, J. Algebra 319 (2008), no. 2, 595–615, DOI 10.1016/j.jalgebra.2006.09.033. MR2381797 [FinFr1] Larry Finkelstein and Daniel Frohardt, Simple groups with a standard 3-component of type An (2), with n ≥ 5, Proc. London Math. Soc. (3) 43 (1981), no. 3, 385–424, DOI 10.1112/plms/s3-43.3.385. MR635563 [GL1] Daniel Gorenstein and Richard Lyons, The local structure of finite groups of characteristic 2 type, Mem. Amer. Math. Soc. 42 (1983), no. 276, vii+731, DOI 10.1090/memo/0276. MR690900 [GLS1] Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite simple groups, Mathematical Surveys and Monographs, vol. 40, American Mathematical Society, Providence, RI, 1994. MR1303592 , Outline of Proof, ch. 2 in [GLS1], American Mathematical Society, 1994. [I2 ] , Overview, ch. 1 in [GLS1], American Mathematical Society, 1994. [I1 ] Daniel Gorenstein, Richard Lyons, and Ronald Solomon, The classification of the finite [IG ] 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 [IA ] 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 [GLS4] Miles Reid, Chapters on algebraic surfaces, Complex algebraic geometry (Park City, UT, 1993), IAS/Park City Math. Ser., vol. 3, Amer. Math. Soc., Providence, RI, 1997, pp. 3–159. MR1442522 , General Lemmas, ch. 1 in [GLS4], American Mathematical Society, 1999. [II1 ] , p-Component Uniqueness Theorems, ch. 3 in [GLS4], American Mathematical [II3 ] Society, 1999. , Properties of K-Groups, ch. 4 in [GLS4], American Mathematical Society, 1999. [IIK ] , Strongly Embedded Subgroups and Related Conditions on Involutions, ch. 2 in [II2 ] [GLS4], American Mathematical Society, 1999. , Chapter 1: Theorem C7 : General Introduction, ch. in [GLS5], American Math[III1 ] ematical Society, 2002. , Chapter 2: General Group-Theoretic Lemmas, ch. in [GLS5], American Math[III2 ] ematical Society, 2002. , Chapter 3: Theorem C∗7 : Stage 1, ch. in [GLS5], American Mathematical Soci[III3 ] ety, 2002. , Chapter 4: Theorem C∗7 : Stage 2, ch. in [GLS5], American Mathematical Soci[III4 ] ety, 2002. , Chapter 5: Theorem C∗7 : Stage 3a, ch. in [GLS5], American Mathematical So[III5 ] ciety, 2002. , Chapter 6: Properties of K-Groups, ch. in [GLS5], American Mathematical [IIIK ] Society, 2002. [GLS5] 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 [A22]

341

342

[G1] [Hu1] [McL2] [St1]

BIBLIOGRAPHY

Daniel Gorenstein, Finite groups, 2nd ed., Chelsea Publishing Co., New York, 1980. MR569209 B. Huppert, Endliche Gruppen. I (German), Die Grundlehren der Mathematischen Wissenschaften, Band 134, Springer-Verlag, Berlin-New York, 1967. MR0224703 Jack McLaughlin, Some groups generated by transvections, Arch. Math. (Basel) 18 (1967), 364–368, DOI 10.1007/BF01898827. MR0222184 Robert Steinberg, Lectures on Chevalley groups, Yale University, New Haven, Conn., 1968. Notes prepared by John Faulkner and Robert Wilson. MR0466335

Index

Gp , 3 Gip , 4 Gip -groups, 5