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This is to certify that the
dissertation entitled

ALMOST DUAL FF-MODUL.ES

presented by

KIMBERLY ANN DYER

has been accepted towards fulfillment
of the requirements for the

 

 

 

 

Ph.D. degree in Mathematics
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Michigan State
University

 

 

ALMOST DUAL FF-MODULES
By

Kimberly A1111 Dyer

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
I\I’Iathematics

2009

ABSTRACT
ALMOST DUAL FF-MODULES
By

Kimberly Ann Dyer

In this paper we consider a subgroup, L, of a finite group of local characteristic 2.
The action of a maximal 2-loca1 parabolic subgroup containing a non—normal large
subgroup on its largest 2-reduced normal subgroup is considered in the quadratic L—
Lemma and Structure Theorem [MSS]. They show L/Op(L) E” S L2(2), S 2(2), or D27-
and obtain a 2F -offendor. The action which we are interested in can be determined
by the Malle-Guralnick—Lawther Classification of 2F-modules [GLM]. The Malle-
Guralnick—Lawther papers depend on a lC—group assumption; that is, one needs to
assume that all the simple sections of M are one of the known finite simple groups.
In this paper we explore results that do not need a lC-group assumption and therefore
do not use the classification of finite simple groups.

Let IF be a finite field with p := char IF = 2, G a finite group, and V a faithful, finite
dimensional IF G-module such that there exists an elementary abelian p—subgroup,
A S G' with TA := [V, A] and R A := [TA,A]. By considering whether or not A is
a TI—set, that is if A n A9 Q {1} for all g E G \ Ng(A), we arrive at the various
cases of the main theorem. The main theorem shows that we have IF = 2 and 4 g
|A| g 16, [In > 2 and [Al = W, or CID/(700mg) 2—: SLK(N)/Z0 where GO 2 (AG),
HG = E B 6 AG RB, N is a finite dimensional vector space over the finite field K, and
20 = {k*idV | k 6 11cm = r1}.

ACKNOWLEDGMENT

I would like to thank my advisor, Dr. Ulrich lVIeierfrankenfeld, without whose endless
support and patience this would not have been possible. I couldn’t imagine a more
supportive advisor and I’m thankful for all his help. I would also like to acknowledge
all the professors at Michigan State who have helped me along the way, especially
my committee: Dr. Jonathan Hall, Dr. Richard Hill, Dr. Christel Rotthaus, and Dr.
Clifford VVeil. To my family and friends, I appreciate all the love and support you
have given me along the way. My husband, Dr. Joshua Dyer, has my everlasting love

and appreciation for all he has done to make this and everything else possible.

iii

TABLE OF CONTENTS

Motivating Introduction .......................... 1
Almost dual FF-modules ......................... 4
The case where A is not a TI—set .................... 24
The case where A is a TI-set ....................... 37
Identifying Ln(q2) .............................. 61
Main Theorem ................................ 72

Background Lemmas ............................ 74

Definitions .................................. 76

Bibliography ................................. 79

iv

Chapter 1

Motivating Introduction

Let p be a prime and G a finite group of local characteristic p. Suppose G has a
large subgroup Q. Let Ill be a maximal p—local parabolic subgroup of G with Q _<_ M
but Q $1 III. Let Y 2 YM be the largest p—reduced normal subgroup of M. In the
Structure Theorem [MSS] the action of IV! on Y is determined. In the proof of the
Structure Theorem one runs into the following situation.

There exists a subgroup L of G such that Y S L but Y fl Op(L), M 0 L is
the unique maximal subgroup of L containing YOP(L), and L is of characteristic p.
The quadratic L-Lemma [MSS] shows L/Op(L) "2" SL2(q) where q is a power of p,
L/Op(L) ’-‘_-’ Sz(q) where q is a power of p and p = 2, or L/Op(L) ’5 D2,. where r is
an odd prime power and q = p = 2.

In this situation one tries to obtain some information about the action of III 011
Y. For this, let x E L \ M. Put D = Y 0 Op(L) and A = D“. In the Structure

Theorem, in this case, it was proved that
(a) [Y,A]C'y(A) = [y,A]Cy(A) = D for all y E Y \ D.
(b) lD/CD(A)l = lA/CA(Y)| 2 (1-

(C) lY/Dl = (1-

(<1) ID, Al S Cy(A)-

Notice that (a) implies Cy(A) = CD(A) so we have |Y/Cy(A)| = IY/CD(A)| =
lY/DllD/CplAll (g) qlD/CD(A)| (.2) IA/CA(Y)l|A/CA(Y)| = IA/C.4(Y)|2 giv-
ing |Y/Cy(A)| S lA/CA(Y)|2. Hence, A is a 2F—offender 011 Y. The Malle-
Guralnick-Lawther Classification of 2F -modules [GLM] now allows us to deter-
mine the action of M on Y. The Malls-Guralnick-Lawther papers depend 011 a
lC-group assumption; that is, one needs to assume that all the simple sections
of .M are one of the known finite simple groups. I11 this paper we would like to

explore what can be said without making a lC-group assumption.

If there exist large enough quadratically acting subgroups, then this quadratic ac-
tion can be used to determine Y and M / 011.1(Y). We consider what happens when
there is not a large enough quadratically acting subgroup. Put M = M / CA1(Y)
and H = (AM).

Let IF be a maximal subfield of the ring End ”(Y). Then Y is an lFH-module.

We assume that there is no large quadratic action on Y in the following sense:
(0) If P is a p—subgroup of M and 1 31$ (1 E P with [Y, P, a] = 0, then |P| g IIFI.

Let BO be a normal subgroup of L minimal with BO 5: Z (L) If p = 2, it can
be shown that there exists B 3 BO such that [B, Y, B] = 0 and IB/CB(Y)| 2 q. So
[Y, B/CB(Y), B/C'B(Y)] = 0 and (e) now shows that llFI Z q. A acts lF-linearly 011
Y so both [Y, A] and Cy(A) are IF-subspaces. Then (a) shows D is an lF-subspace
of Y. By (o), q = lY/Dl = |lF|dimlF(Y/D). Since |1F| Z q, we conclude that |lF| = q
and D is an lF-hyperplane of Y. Since we would prefer to work with a 1-dimensional
lF-subspace than with a hyperplane, we consider the lF-dual, V, of Y and arrive at the
following set of assumptions (where the G below is now taking the place of All / C M(Y)

from above).

Hypothesis 1.1 Let IF be a finite field with p := charlF = 2, G a finite group, and V
a faithful, finite dimensional lFG-module such that there exists an elementary abelian

p—subgroup, A S G, with TA := [V, A] and RA :2 [TA, A] such that
(2') RA is I-dimensional.

(ii) For a E A, define 45.4(0) : TA/RA —-> RA,t+ RA ——> [t,a]. Then ¢A : A -—+

I—Ion1fl.~(TA/RA,RA),a —> ¢5A(a) is onto.

(iii) Ifl 79 a E A and P is a p-subgi‘oup ofG' with [V,a, P] = 0, then |P| S llFl.

Chapter 2

Almost dual FF-modules

Definition 2.1 Put .A = A(G) = A0 and R = R(G’) = RAG. For H S G,
let A(H) = {B e A | B _<_ H}, 110 = (A(H)>, 72(H) = {RB | B e A(H)},
RH = 2139101) RB, and TH = Zoe/1(a) TB-

The goal of this paper is to prove the following theorem. To this end, we assume

that Hypothesis 1.1 holds throughout the entire document.

Theorem 2.2 Assume Hypothesis 1.]. Then Cg(A)/A is a p’-gr0up, A is a weakly
closed subgroup of G, for A are B E A, RA % RB and TA 74 TB, and one of the

following holds:
1. Each of the following holds:
(a) A E Sylp(G).
(b) IA|=l1F|2-
(c) |IF| > 2.
(d) Go/CGO(RG) g SL2(1F) 0"“ Go/CcolRG) ’5 9UP)-

(6) RC is a corresponding natural module for G0.

2. Each of the following holds:

(a) [Fl :2.
(b) 4 .<_ IAI s 16.

{c} IAflB|S2forA#BE.A.

3. GO/CGO(RG) g SLK(N)/Z0, where N is a finite dimensional vector space over
the finite field 1K. Moreover, there crisis a 1-dimensional subspace, C, of N and a
field automorphism, 0‘, ofK of order two with CK(0) = IF, K831}: HG 9:" N ®IK NU,
Z0 = {1: >1: idv I k E K,k° = k—l}, and the image ofA in PSLK(N) consists of

the identity and all trans-vections with center C.

Lemma 2.3

(a) CWT/AA) = RA.
(1)) RA is contained in every non-zero lFA-submodule of TA.

Proof. Since RA is a non-trivial p—group, CRA(A) 71$ 0. CRA(A) S RA and RA is
I-dimensional from 1.1(i) so CRA(A) = RA. Then RA S CV(A). Let v E TA \ RA.
Then there exists p E Hom(TA/RA, RA) with p(v + RA) # 0. By 1.1(ii), ch is onto
so there exists a E A witl1 ¢A(a) = p. Then 0 74 p(v + RA) 2 ¢A(a)(v + RA) = (v, a].
Hence, v E CTA(A). Then TA \ RA S CTA(A) so CTA(A) = CRA(A) = RA and (a)
is proven.

Let 0 75 W be an lFA-submodule in TA. Since A is a p—group, CW (A) 74 0. Thus by
(a), WflRA = WflC'TA(A) and 0 75 CW(A) S CTA(A) so 0 # IVDCTA(A) = WORA.

As R A is I-dimensional, RA S I’V which proves (b). E]

Definition 2.4 For H a group, let X and Y be lFH-modules. Define HomH:(X, Y) to

be the set of lF-linear maps from X to Y.

We remark that HomIF(X, Y) is an IF -space via ( f a)(a:) = f a(:r) and (0+ fl)(:r) =

a(:1:) + Mar). Also, HomIF(X, Y) is an IFH-module via (a9)(:1:) = (a(:r9-1))9.

5

Lemma 2.5

(a) qu is G-invariant.

(b) CA(TA)=1-

(c) (bA is an isomorphism of ZN0(A)-modules.
(d) dimTA/RA Z 2.

Proof. Using the remark following 2.4, gbA(a9)(a: + RA) 2 [snag] = [mg-1,a]9 =
-1
(¢A(a)(:r9 + RA))9 = (¢A(a))9(27 + RA) and (a) holds.
By 1.1(ii), qu is onto and CA(TA) is the kernel of (15A so

‘1’ I A/CA(TA) —* H01111r(TA/RA,RA).G + CA(TA) —+ @100

is an isomorphism of ZNg(A)-modules. Since R A is l-dimensional, it follows that
IA/C'A(TA)| = IlFldimTA/RA. Suppose that CA(TA) yé 1. Then there exists 1 7é a E A
with [TA, a] = 0. Since A is abelian, we get [[V, a], A] = 0 from the Three Subgroups
Lemma [Gor, 2.2.3]. We can then apply 1.1(iii) to see that IA] S llF]. Now

IA/CA<TA>I s IAI s IIFI _<. IIFIdimTA/RA-

But lA/C’A(TA)] = IIB‘IdilinTA/RA from above so IA/CA(TA)| = |A| and CA(TA) = 1
which proves (b). Then (I) = (,b A is an isomorphism and (c) is proven.

Suppose that TA/R A is 1-dimensional. Let a E A with [V, a] S RA so TA =
[V, a] + RA.

2
[V,a,a]=(va—v)a—(va—v)=va —va’—va+v=O

as p = 2 so [V,a] S CV(a). We also have [RA,a] = 0 since ORA (A) = RA. Then
TA 2 [V, a] + RA S CV(a), but TA S C'V(a) since CA(TA) = 1 from (b). This is a

6

contradiction so ((1) is proven. E]

Lemma 2.6 Let U S TA be an lF-subspace. Then A/CA(U) 2 ((U + RA)/RA)* as

N0(A) fl 00(RA) fl Ng(U)-modules, where * denotes the dual space of an lF-module.

Proof. A: Hom(TA/RA, RA) ——> Hom((U + RA)/RA,RA). The first map is an
isomorphism by 2.5 and the second map is onto. Hence, the map A -—> H0m((U +
RA)/RA,RA) is onto and its kernel is C A(U ) so we have the result by the first

isomorphism theorem. Cl
Lemma 2.7 For f 6 IF and a e A define fa := (gs/fl ftp/1(a)). Then

(a) (15,4 is lF—linear.

(b) fa is the unique element in A with [t, fa] = f[t, a] for allt E TA.

(0) A is an lF-space.

{d} Ng(A) acts lF-linearly on A.

(e) A ”-34 H0111F(TA/RA,RA) as lFN0(A)-modules.

Proof. (a) This holds by the definition of fa.

(b) Recall ¢A(a)(t + RA) = [t,a] SO (1514(f0»)(t + RA) = [tifal- A180 f(¢A(a)(t +
RA» = fit, a] and ¢A(fa) = fab/1(a) by definition so It. fa] = fit. al-

(c) Note that an lF-space is an lF-module. Consider the map it? x A —> A, (f, a) _i
fa- We have f(a+a) = ¢A—1(f¢A(a+&)) = ¢A“1(f¢A(a)+f¢A(a)) = ¢A‘1(f¢A(a))+
iii-lathe» = fa+f5~ Also, (ma = alumna» = ¢:,I(f¢A(¢;1(f¢A(a)))) =
mildew» = f(fa)-

(d) Let g E Ng(A). We have ¢A((fa)9) 2 5__(3)

371a)

(ab/100))” (f ¢A(a)) =

f(gbA(a))g 2 5(:a) fgbA (a) 2 7:(a) ¢A(fa9). Then we get (fa)9 = f(a9) since (23A is a
bijection.
(c) This now follows from part (d) and 2.5(c). Cl

7

Lemma 2.8

(a) Let U Q TA. Then CA(U) is an lF-subspace of A. IfU is an lF-subspace of TA
with RA S U, then dimU + dim CA(U) = dim TA and U = CTA(CA(U)). Also,
CA(U) = CAUFU), andforany lF-subspace, Y, ofTA, |A/CA(Y)] = IY+RA/RA|.

(b) Let X Q A. Then CT4(X) is an F-subspace of TA containing RA. IfX is an
lF-subspace of A, then dimX + dim 0ij (X) = dim TA and X = CA(CTA (X)).

(e) Let K S 00(RA). Then CA([TA,K]) = CA(K).
(d) Let K S C0(RA). Then [TA, K] + RA 2 CTA(CA(K))-
((3) Let If S CG(RA). Then CTA([A, [YD/RA = CTA/RA(I{)'

Proof. (a) Since A is an lF-space from 2.7, C A(U ) is an lF-subspace of A. If U is an
lF-subspace of TA with RA S U, then 2.6 gives A/CA(U) T——’ ((U + RA)/RA)*. Since
dimA 2'36) dimTA/RA, we have dimTA/RA —dim CA(U) = dim((U+RA)/RA) so
we get dimU +dimCA(U) = dim TA. Also, CA(U) S CA(CTA(CA(U))) _<. CA(U)
and dimU = dim TA—dimCA(U) = dim CTA(CA(U)) so we have U = CTA(CA(U))~
CTA(CA(U)) is an lF-subspace so IFU Q CTA(CA(U)) = U. Hence, CA(U) Q
CA(1FU)§ CA(U)°

Let Y be an lF-subspace of TA and put U = Y + RA. Then dim U + dim C A(U ) =
dimTA gives us that |U||CA(U)| = lTAl = [TA/RAHRAI = IAHRA]. Hence, |Y +
RA/RAl = IU/RAI = lA/CAWH = lA/CA(Y + RAll = lA/CA(Y)|-

(b) Define p : TA —> I—Iomp(X, RA) by p(t)(:L) = [t, 2:]. Then Ker p = CTA(X) and
Im p S X*. Hence, dim TA/CTA(X) S dim X“. By (a) applied with U 2' CTA (X),
dimX* = dimX S dimCA(CTA(X)) = dimTA—dim CTA(X) = dimTA/CTA(X) S
dimX*. Thus, dimX + dim CTA(X) = dim TA and since X S CA(CTA(X)), X =

CA(CTA(X))-

(e) Let X S A. Since [TA,X,K] S [R A,K] = 0, the Three Subgroups Lemma
shows that [X , K, TA] = 0 if and only if [K,TA,X] = 0. Since A acts faithfully on
TA, we conclude that [X , If] 2 0 if and only if [K , TA, X ] = 0.

(d) We ham CAN) (2) CAUTAaKl) SO CTA(CA(K)) = CTA(CA(lTA:Kl)) =
[TA, K ] + R A.

(e) v + RA E CTA/RA(K) if and only if [v,K] E RA if and only if [v,K, A] = O
by 2.3(a). Since K E 00(RA), [v,A, K] = 0 so the Three Subgroups Lemma gives
[v, K, A] = 0 if and only if [A, K , v] = 0 which holds if and only if v E CT A([A, K ])
Then CTA([A,K])/RA = CTA/RA(K)' [:1

Lemma 2.9 Let v E V and define s = 3., : A X A ——> RA, (a, b) —+ [v,a,b]. Then 3 is
symplectic and lF-bilinear. {Here symplectic means s(a, a) = 0 for all a E A, and we

will also see that s is alternating; that is s(a, b) = —s(b, a), for all a, b E A).

Proof. By definition, s(a,a) = [v,a, a] = O which means 3 is symplectic. Also 8 is
alternating since s(a, b) = [v,a,b] = —[v,b,a] = -s(b, a) by the Three Subgroups
Lemma [Gor, 2.2.3]. By 2.7(b), s is lF-linear in the second coordinate. Since 3 is

alternating, it is also lF-lincar in the first coordinate. El

Lemma 2.10 Let u E V and let 3 be as defined in 2.9. Let U be an lF-subspace of
TA with RA S U and put D = CA(U). Then

(a) {a E A I [v,a] E U} 2 Di = CA([v,D]) = CA([v, D] + RA) is an lF-subspace of
A.

(b) [v, D] + U is an lF-subspace of TA.
(c) [v.01 + U = CTAlCDflv, DD)-

(d) {a E A I [v,a] E RA} 2 rad sv = CA([v,A]) = CA([v,A] + RA) is an lF-subspace
ofA.

(e) ([v,A] + RA)/RA is an lF-subspace ofTA/RA.

9

(f) [v,/11+ RA = (Imelda, A1».
Proof. (a) Let a E A.

Claim: The following are equivalent.

1. [v,a] E U.

2. [v,a,CA(U)] = O.

3. 3,,(e, v) = 0 for all b e CA(U).

4. a e C'A(U)i.

5. s.U(b,a) = 0 for all I) E CA(U).

6. [v,CA(U),a] = 0.

Proof of Claim. 1 4:) 2 since U = CTA(CA(U)) from 2.8; 2 :> 3 by definition; 3 => 4
again by definition; 4 => 5 since 3 is alternating; 5 => 6 by definition; 6 2? 2 by the
Three Subgroups Lemma and since A is abelian. Then the claim is proven. El

Now {a E A ] [v,a] E U} = DJ- by 1 <2) 4 of the claim. And Di = CA([v,D])
by 4 (It) 6 of the claim. Also, CA([v,D]) = CA([v, D] + RA) since CA(RA) 2&3 A.
So {a E A I [v,a] E U} = DJ- 2 CA([-v, D]) = CA([v, D] + RA) is an lF-subspace of A
from 2.7. Then (a) holds.

(b) [V/U,A,A] = [TA/U, A] = RA/U = 0, so A acts quadratically 011 V/U. Let
a : D —> ([v, D] + U) / U, a ——> [v, a] + U. or is a l1o111omorphism by quadratic action;

it’s onto and its kernel is {d E D | [v, d] E U} so

Kiwi + U)/U| = WM 6 D I [ed] 6 U}| ‘3 ID/CD<1v,D1+ RA)!

2.

00

(a) lD/CDUFlvi D] + RAM.

By the last statement in 2.8(a) applied with Y = lF['U, D] + U we get lA/CA(lF[v, D] +
U)| = l(lF[v,D]+U)/RA|. Also, CA(U) = D implies CA(lF[v,D]+U) = CA(lF[v,D])fl

10

CA(U) = CD(lF[v, D]) = CD(lF[v, D] + RA). Then

lA/DllD/CDWI’U, 0] + RAll = lA/CDUFl'U: Dl + RAll = IA/C'AUFlv, 0] + U)|

=|(1F[v,Dl+ U)/RA| = IUFIMDl + U)/U||U/RA|-

Since IA/CA(U)] = IU/RA], we must have ]D/CD(lF[v,D]+RA)| = |(lF[v,D]+U)/U|.
Hence,

lllv.Dl + U)/Ul = I(IFlv.Dl + U)/Ul
and (b) holds.
(0) Since [v,D] + U is an lF—subspace of TA from (b), 2.8(a) gives [v,D] + U =
CTA(CA([U, D]+U)). Since CA(]’U, D]+U) = CD([U, D]), [v, D]+U = CTA(CD(]U, D])).

If U = RA, then D = A and (d), (e), and (f) follow from (a), (b), and (c)

respectively. [:1

Lemma 2.11 Let v E V and let X be an lF-space of [v,A] + RA with [v,A] + RA =-
X 69 RA. Define q :2 qu : A -—> RA so that q(a) is the unique element of RA such
that [v, a] -— q(a) E X. Then for all a, b E A,

q(ab) = q(a) + 8v(a, 1)) + (10))

Proof. Let x(a) = [v, a] — q(a). [v, a] : va — v so v“ = v + :1:(a) + q(a) with :1:(a) E X
and q(a) E RA. Also for any b E A, [2:(a), b] = [v, a, b] — [q(a),b] = [v,a,b] = sv(a, b)

since q(a) E RA implies q(a)b = q(a). Then a:(a)b = :L‘(a) + sv(a, b). Hence,

vab = he)” = (v + eta) + q(a))b
= vb + to)” + em)” = v + 1:(b)+ q(b) + me) + s(a, b) + q(a)

= v + (13(0) + 115(0)) + (0(0) + Sula, 1)) + (1(0))-

11

It follows that 33(ab) = a;(a) + :I:(b) and q(ab) = q(a) + sv(a, b) + q(b). C]

We remark that q from the previous lemma does not need to be a quadratic form.

I11 particular, we do not know whether q( fa) = f 2q(a) for all f E IF and a E A.
Lemma 2.12 Let 1 7é a E A and put Aa = CA([V, a]). Then

(a) Au 2 lFa.

(b) [V,Aa] = [V, a] = CTA(a) is a hyperplane of TA and RA S [V, Aa].

(c) AG is quadratic on V.

(d) Aa E Sylp(CG(lV,al))-

(6) Aa S] NG(lV,al)-

(f) CGUVvall/Aa 2'8 a P’-gr0ui).

Proof. From 2.3(b) we have RA S [V, a] and by 2.5(b), TA 74 CTA (a). Thus
(*) RA S [V,a] S CTA(a) < TA

since [V,a, a] = 0.

Observe that [V, a] is an lF-subspace of V. Thus, [Aal = |CA([V,a])| 21—3—53)
IT A/ [V, a]| = |IF|n, where n is a positive integer, as TA / [V, a] is an lF-space. Also,
[V, a,Aa] = 0 so |Aa| S |lF| by 1.1(iii). Hence, [Au] = [IF] and A0 = lFa. Then (a) is
proven.

We have just shown |IF| = ITA/ [V, a]| and thus [V, a] is a hyperplane of TA. [V, a] S
CTA (a) < TA and since [V, a] is a hyperplane of TA, [V, a] = CTA(a). Let 1 7é b E Aa.
Then [V, a] S CT A (b) = [V, b] and again, since [V, a] is already a hyperplane of TA,
we conclude that [V, a] = [V, b] = [V,Aa]. Since RA S [V,a] for any 1 79 a E A,
R A S [V, Aa]. So (b) is proven.

I2

Then [V,Aa,Aa] = [V, a, A0,] = [V,Aa,a] : [V, a, a] = 0 from (b). So Aa is
quadratic on V and (c) is proven.

Let P be any p—subgroup of G with Aa S P. Since [V, a, Aa] = 0, Aa S Cp([V, a]).
If Ag, # Cp([V, (1]), then lAal < le([V,a])| S ]lF| from 1.1(iii) which contradicts
(a). So Cp([V,a]) 2 Au for any p—subgroup, P, of G with A0, S P. Thus, Aa E
Sylp(CC,v([V, a])) and (d) is proven.

We have

[V, <Al’G“V’“”>1 = w. AaNG‘W) = <1v, alNG<IVel>> = iv, a]

by (b). [V, a, Aa] = 0 so [V, a, Aflvcflv’ab] = 0 which means [V, (AQ’GGV’aDH = [V, a] S

CV(<AbVG([V’aD)). Therefore, (A£VG([V’a]) ) is quadratic 011 V. Hence, (AfGfiVfll)

) is

_ . . _ NG(]V,a]) T .
a p—group so (d) implies that Aa — (Aa ). Thus Aa _<_l Ag([V,a]) and (e) 18
proven. 1

Au S 1VG([V, (1]) and Aa E Sylp(Cg([V, a])) imply that Cg([V, a])/Aa is a p’—group

and (f) is proven. Cl
Lemma 2.13 C0(TA) is a p'-group.

Proof. Let P be a p—subgroup of Cc(TA) and let 1 # a E A. Then P S Cg(TA) S
Cg([V,a]). By 2.12(f), Cg([V, a])/Aa is a p'-group. So P S Aa S A which implies
P S CA(TA) making P = 1 by 2.5. E]

Lemma 2.14 Let P be a p-subgroup ofG' with A S P S N0(TA) fl 00(RA). Then
CP(TA/RA) = A and CG(V/TA) r1 Ce'(TA/RA) D CG(RA) = A-

Proof. Extend gbA to (I) : Cp(TA/RA) —> Hom(TA/RA, RA),s —> (t + RA ——> [t, 5]) so
that <IJ|A = gbA. Since qu is onto, CP(TA/RA) = (Ker <I>)A. This gives CP(TA/RA) =
CCP(TA/RA)(TA)A' CC;(TA) is ap’-group from 2.13 so CCP(TA/RA)(TA) = 1. Hence,

CP(TA/RA) = A-

13

We have Cg(V/TA) fl Cg(TA/RA) fl C0(RA) is a p—group by A.1, so using what

we just proved,

CG(V/TA)RCG(TA/RAIDCGWA) = CoGw/TAmCG(TA/RAmCGmA)(TA/RA) = A-

El
Lemma 2.15 Let A, B E A with TA = TB and RA = R3. Then A = B.

P7‘00f. A = CG(V/TA) fl CG(TA/RA) n Com/1) = CG(V/TB) D CG(TB/RB) fl
Cg(RB) = B from 2.14. D

Lemma 2.16 CC;(A)/A is a p’-group.

Proof. Let P be a p—subgroup of CC;(A) with A S P S C0(A). Now P centralizes
A and so it also normalizes A. P normalizes V as V is an lFG-module. Then P
normalizes [V, A] = TA and [TA, A] = R A- Since R A is l-dimensional and CR A (P) 75

0 as both are p—groups, we have C R A(P) = R A- Hence, P centralizes R A- Therefore,
P S CG(A) D Com/1) S CG(HO“1(TA/ RA: RAD D 00(RA) S CG(TA/RA)

from 2.7(e). Then P = Cp(TA/RA) = A by 2.14 and CC;(A)/A is a p’-group. Cl

Definition 2.17 Let H be a group and let X S Y S H. We say that X is weakly
closed in Y with respect to H if for all h E H with X h S Y, we have X h = X. That
is, ifX is the only H-conjugate of X contained in Y. We say that X is a weakly
closed subgroup of H if there erists a Sylow p-subgroup, P, of H such that X is weakly

closed in P with respect to H.

Lemma 2.18 Let R be a p-subgroup of a finite group H. Then the following are

equivalent:

(a) R is a weakly closed subgroup of H.

14

(b) Any p-subgroup ofH contains at most one conjugate of R.
(e) Let h e H with [12,12’1‘] g n on”. Then R = R".
((1) UR S S S X S H and S is a p-group, then R S X.

Proof. (0. => b) Every p—subgroup of H is contained in a Sylow p—subgroup of H which
contains only one conjugate of R by the definition of weakly closed.

(b => c) Since [12, Rh] 3 R r1 Rh, we have R g Namh) and Rh 5 NCO?) so RR"
is a subgroup, in fact it’s a p-group. We have R S RRh and Rh S RRh. Then (b)
gives R = R”.

(6 => d) Let a: E X and R S S S X S H. If R S S, then conjugating by :1: we
get R‘” S S since S S X. So we have [R,R‘”] S R O R‘”. Then R = R3 so R S X.
So we see that R S S S X => R S X. Consider R S NS(R) S NS(NS(R)). Then
R S NS(N3(R)) S NS(R) which gives N3(R) = N5(NS(R)) = S and R S S.

((1 => 0.) Let S = X E Sylp(H) so RSS. Thus, S E Sylp(NH(R)). Assume R" S
S for some h E H. Conjugating S E Sylp(NH(R)) by h we get, Sh E Sylp(NH(Rh)).
Also, RSS}‘_1 by (d) so Rh SS. Then S E Sylp(NH(Rh)). Hence, S = Sht for some
t E N0(Rh). Then we have R S SSS(ht) S H again from (d) so R = Rm = Rh. Cl

Lemma 2.19 A is a weakly closed subgroup of C.

Proof. Otherwise 2.18 implies that there exists B E .A with [A,B] S A O B and
A S B. By 2.16, C0(A)/A is a p’-group so B S Cg(A). Hence, [A,B] S 1. Then
[V, [A, B]] S 0. Since B normalizes A, B normalizes TA and TA 0 TB. Similarly, A
normalizes TA (1 TB-

As U and AB are p—groups, CU(AB) S 0. So we have 0 S CU(AB) S CTA(A) =
RA by 2.3. By symmetry, CU(AB) S RB and since RA is l-dimensional, CU(AB) 2
RA = R3.

Now if TA = TB, 2.15 gives a contradiction to A S B, so TA S TB. Let 1 S a E
A O B. 2.12(b) states that [V, a] is a hyperplane of TA. Then [V,a] S [V, A D B] S

15

U < TA since TA S TB. We see that [V, a] = [V, A (I B] = U is a hyperplane of TA
and similarly of TB. So U = [V, a] S CV(a) for all a E Am B; hence U S CV(A OB).
Then A O B S CA(U) = A0, and

Aa = CAIleal) S CABUVMII S 00([V,a]).

By 2.12((1), CC;([V, a])/Aa is ap’-group so CAB([V, (1]) = Aa and Similarly CABUV, a]) =
Ba-

Now A O B S Aa 2 Ba S A O B and we have
A D B = Aa = CAB(lVa (II) = CAB(U)

which has order |lF| from 2.12(a). Also, [U, AB] S [TA,A][TB, B] = RARB = RA =
R3.

Define r : AB ——> Hom(U/RA,RA),l —> (u + RA —> [u,l]). Restricted to A, r
is onto from 1.1 so AB = A(Ker (7’)). We get AB = A(CAB(U)) = A(A O B) = A
which contradicts A S B. C]

Lemma 2.20 If H S C, then H acts transitively on A(H) and A(H) = CH for any
C e A(H).

Proof. Let A E A(H). For any C E A(H), (A,Ch) is a 2-group for some 11. E H by
Sylow’s theorem. Since A is a weakly closed subgroup of G by 2.19, 2.18(b) gives
A = C9. Then H acts transitively on A(H) and A(H) = CH. Cl

Notation: From 110w 011 let A S B E .A and put L := (A,B), R := Op(L),

E:= (AnR)(Bnn), ZzzAflB, U:= TAflTB, and W :2 RA+(TAflTB)+RB.

Lemma 2.21 Let L be a finite group and A be an elementary abelian weakly closed
subgroup ofL where B E Ai’ and L = (A, B). Let R = Op(L), E = (A flOp(L))(B fl
0,)(L)), and C 2: CL(AI:). Then

16

(f) C/Op(L) is an abelian p’-group.

{g} Moreover, if L/C E” SL201), Sz(q) with q a prime power ofp larger than 2, 0r
L/C E“ D2,. with r an odd prime, then C = E = Op(L).

Proof. (a) Notice that AR is a p—group. Since A is a weakly closed subgroup of L, 2.18
shows that R normalizes A. Similarly, R normalizes B. Thus, [R, A] S AOR S E S R
and [R,B] S BflR S E S R. So [R,L] S E. Since E S R, [E,L] S [R,L] S E and
then [R, L] S E S L.

(b) )E’ = [.210 R Bn 1%] g [21 mm [[2, a} 3 Ana. SinceAflB 3 2a), Ané s it
so AflB S Z(L)OE

(0) Since Op(L)A
by 2.18(d) as A< S0p(L )A_ S Op(L )A S C. A is normal in any p—subgroup so

is a p~group and A is a weakly closed subgroup of L, A SOp(L)A

0p(L) S NC;(A). Thus, Op(L )_ < C.

(d) If J E A(L), then A n C normalizes J O C by definition of C. Let E :=
(D O C | D E A(L)). E is a p—group since each D is a conjugate of A. Thus
E S 013(L), yielding

A
V

C ..

Anopui) < AnégAnEgAnOpw).

So A O 0,,(L) = A O C. This, along with the definition of E, gives us A O E S
AflOp(L) S E. Hence, ADC: AflOp(L) = ADE.

17

(e) Notice that C S L and A S C by definition. We have [A, C] S A 0 C S E and
similarly, [[3, C] S E 0 C S E. Thus, [L,C] S E S Op(L).

(f) As [L,C] S E S Op(L), we have C/OI,(L) S Z(L/Op(L)). Since 0p(L/Op(L)) =
1, C/OI,(L) is an abelian p’—group.

(g) In addition, suppose now that L/C g SL2(q),Sz(q) or Dgr. Let A S S E
Sylp(L) and let L = L/ C For notational simplicity let S = S. Define M = NflS)
and let NI := .M be the inverse image of TV in C. Since [C , S] (S) Op(L), S is normal
in CS so S is the unique Sylow p—snbgroup of CS. As S SET since CS S Ni, S is the
unique Sylow p—subgroup of NI. Hence, S 2 0pm?!) S A}. Since A is a weakly closed
subgroup of L, A S [1.71.

Consider the L/ C E’ SL2(q) case. we may assume S := {( it] ) ] s E lF} and is
a Sylow p—subgroup of SL2(q). Let A :2 A/C S S E Sylp(L) and NSL2(q)(S) cor‘
respond to (i A91 ). We have(:‘)‘91 “(11(1))“:19‘ ):(,\—12a(1))'80 NSL2(
acts irreducibly on S. Since A S Ail, A = S.

For L/C 53’ Sz(q), it follows easily from [Huppert, Chapter X1 Section 3] that
Z (S) = 91(S), H acts irreducibly on Z (S), and M acts irreducibly on S/Z(S).
Since A is elementary abelian, A S 91(S) = Z (S) Since A S M and M is irreducible
on Z(S), we get A = Z(S).

For L/C ”:11 Dgr with r an odd prime, IA] 2 2. Since r is prime, A is a maximal
subgroup, so A = N.

Define L = L/E and for X S L let X = XE/E. From [L,C] (S) E we have
[L, C] = 1, so C S Z(L). Also, A F‘I C S E from (d), so A H C =1.

Case 1: Suppose that L/C ’-‘.:’ D2,. Then ID] = 2 = IA] = IA/(A fl C)| (g)
IA/(A O E)] = IAE/EI = |A|. Hence, L = (A, E) is also a dihedral group of order 23
for some 3.

Since A and D are weakly closed subgroups of L, they are conjugate in L. If s

is even, then (a,b)/((ob)2) g D4 = C2 x Cg which contradicts that A and E are

18

conjugate. So it follows that s = l—Q—l is odd. Then C S Z (i) = 1 so C = 1 and

C = E. Also, E (S) Opal) (S) C so C = E = R and we are done in this case.

Case 2: Suppose now that 13/ C T—_’ S L2(q) or S 2((1) with q > 2. RI acts irreducibly
on 250 A g [A, M10. Hence, A = [A, NIX/106‘) (3-) [A, mam). So A = [A, M] g
[3. Similarly, I? S 1:. Thus, I: = i’.

[Griess, Table 1] gives the orders of the Schur multipliers. Since the Schur mul-
tipliers of SL201) and Sz(q) are p—groups, we conclude that C is a p—group and so
6* = 1}.

Claim: There exists a complement, R, to R in S.

Proof of Claim. Consider Z/R = Z/C 9-1 SL2(q). W’e have R S Sylp(1~L). Also

A A

SC = AC since (as proved above) X = S, so AR = S. Let P = A.
A n I“: = (xi/E) n (fr/E) = 1.

So R = A is a complement to R in S.

Now consider i/R C—‘i Sz(q). Then [17! acts irreducibly on S/AC and IS/ACI = q.
AC/A S Z(S/A) and [C, 1:] = 1. Also, S/AC is elementary abelian so S’ S AC.
Then t* : S/AC >< S/AC —> AC/A, (ELT, (ST) —+ [51,, E)]/1 is a symplectic map.

Assume S/A is not abelian so (S/A)’ 7g 1. Choose H S (S/A)’ with l(S//1)’/H| =
2. t : S/AC x S/AC —> (S/A)’/H is a symplectic form over IFg. If radt = S/AC,
then S / A is abelian. If radt = 1, then t is non-degenerate and 11:1 -invariant on S / AC
So S / AC has even dimension over 1F2. This is a contradiction since (1 is an odd power
of 2 for the Suzuki groups, so we conclude that S / A is abelian.

(A?) x [$21,111] =

Let P = [S, 1171]. Since [CI/S is a p’—group, we get S/A = CS/A
03/21“?! ) x R/A from [Gor, 5.2.2.3]. Since A71 acts irreducibly on Z (S) and on

__ .7 .. ~ , .. ~ ~ (e) ~ .
S/Z(S), we have Cgflll) = 1. Thus, C3011) S R. Also, [R, L] S E so R S C301 )
and R = C3011). Then S 2 RR and S/A = AR/A x R/A gives RflR S(/1R)F113 S

19

A. So R D P S R (’1 A = 1 since A = [A, 1171] S P. This completes the proof of the
claim that there exists a complement to R in S. E]

Now that the claim is proven, Gaschiitz’ Theorem [Asch, 10.4] implies that there
exists a complement I? to R in L. Then, L = R x 1?, L = L’ = [L, L]
[R, R][R, 1?][133 1%] = [112R] since R = C is in the center of L. So L S 11’ S

C] bwlA

and we have L = 11’ and R =1. Thus C/B = R/B = 1 so C = 11:} = R.
Lemma 2.22 TL = TA + TB = [V, L] = [V, 0])(L)] + U.

Proof. We have .
TL = [V,Ll = [V,/4] + [V, B] = TA + TB-

Let Y = [V, OP(L)]. we have A S SA and B S SB where SA and SB are in
Sylp(L). So {if = 33 for :r e L = sAomL). Then 541 = 5/1“” for z e OP(L). S0
A1 S S Al 2 S 3- However, B. S S B so A1 = B since A is a weakly closed subgroup
of G from 2.19. Then TL = TA + T}, 3 TA + [TAJ] 3 TA + Y. TA 3 TL and
Y S [V, L] = TL so TL = TA + Y. Then A.2 gives Y = [Y, L] = [Y, A] + [Y, B]. Thus,
TL = TA+[Y, A]+[Y, B] and [Y, A] S [V, A] = TA so TL = TA+[Y, B]. T3 = TLflTB

since T B S TL so
TB = TLnTB = (TA+ [Y,B]) nTB = TAnTB + [Y,B]
as [Y, B] S TB. Similarly TA 2 [Y, A] + TA (1 TB. Hence,
[V,L] = TL = [Y,A]+(TAnTB)+[Y,B] = [Y,L]+U = Y+U = [V,01’(L)]+U. [:1

Lemma 2.23 RA # RB.

Proof. Assume not and choose B such that R A = R B with A # B and, in addition,
L = (A, B) minimal. By 2.19, any Sylow p—subgroup contains only one conjugate of

A so L is not a p—group.

2O

By the quadratic L-Lemma [MSS], we have L/Op(L) ’5 SL201), Sz(q) or 027-,
where q is a power of p and r is an odd prime. Then by 2.21(e), [L, C] S Op(L)
where C = CL(AL). So CL(AL)/OP(L) S Z(L/Op(L)) = 1 as L/Op(L) is simple.
Hence, CL(AL) = R.

If T A 2 TB, then by 2.15, A = B, yielding a contradiction, so TA 75 TB. Notice
2.21(d)
that [V, Z] = [V,AF‘I B] S U and Z S Z(L). Suppose that Z 7E 1. Then 2.12(b)

implies that [V, Z] contains a hyperplane of TA and TB. Since U is also a hyperplane
of TA and TB, U = [V, Z]. Since [U, L] S [TA,A][TB,B] = RARE = RA = R3, A.1
says L/CL(U) is a p-group. So 07’(L) S CL(U) by the definition of 0P(L). Hence,

(*) [U, O”(L)] = 0.

Thus [V, Z, Op (L) = O, and the Three Subgroups Lemma gives [V, 01) (L), Z] = 0. Let
a E Z. Then [V, a] = U from 2.12 and [V, a] = [V, Z]. [V, a, a] = 0 gives [V, Z, a] = 0.
Hence, [V, Z, Z] = 0. Then

[v, L, Z] (232) [[v, OP(L)] + U, Z] = ([v, 0%)] + [v, Z], Z] = 0.

Since A S L, we conclude [V, A, Z] = 0. Thus, Z S A centralizes TA and 2.5 yields a

contradiction making Z = 1.

E’ = [AflR,BflR] S AflB = Z = 1 so E is abelian. Recall that R = E by
2.21. Then

E = (A n 12% = (A n R)[A n 12, L] 3 (An R)[E, L] = (A n R)[E, A][E, B]

= (A n R)[E, B] s E.

Thus, EflB = [E,B](AflBflR) = [E,B]Z = [E, B] and similarly EflA = [E, A], so
E = (A n R)[E, B] = (A n E)]E, B] = [E, AME, B] = [E, L]. Hence, E = [E,OP(L)]

21

by A.4 and we have R S OP(L).

Let a E A \ R. By minimality of L, L = (a, B). Thus, TL = (T6) = (Tlga’m) =
[T3,a]+TB and TA = TAflTL = [T3,a]+(TBflTA) = [T3,a]+U. RA S CTA(a) =
[V,a] < TA by 2.12(b). If U = RA, then TA = [T3,a] + U S [V,a] < TA by
2.12(b) again, a contradiction. This implies that U 74 R A- Observe that [TA,a] =
[T3, a, a] + [U, a] = [U, a]. However, 2.3 gives [T1], (1] 75 0 and hence [U, a] 74 0.

On the other hand, [U, E] = [U, R] S [U, OP(L)] (L) 0 and so for any d E A we have
(1 E E if and only if [U, d]— — 0. Thus, AflR= CA(U) and IU/RA] 2 _8__(a) IA/CA(U)I =
|A/ (A n R)| = IAR/ R] by the second isomorphism theorem. If L/ R ¥ D2,, then
L/R is simple so L/R = OWL/R) and L = OP(L)R. Since R S OP(L), we have
L = OWL). Then [U, L] = [U,O”(L)] = 0 since OP(L) S CL(U), and therefore
[U,a] = O, a contradiction. Hence, L/R 9:" Dgr. Thus, IAR/RI = 2 and lU/RA] = 2.
Since lU/RAI is an lF—space, IF] = 2 and therefore W] = 4. Now let 1 # a E A 0 E.
We have [V, a] = CTA(0) from 2.12(b). Then |[V, a]] =2 ICTA(a)| = [TA/RA] = |A|
and so |[V,a] /U|= 1711' Since R: E, IA/A (1 E]: |U/RAI = 2. We also have

anEl = T] = 'f‘.

[[V, a,] B n E] S [TA,B n E] S TA 0 [V, B]: U since
E S C S NL(A). Define D 2: CBnE([V,a]/RA). Since B (1 E centralizes U,
(B (1 E)/D embeds into HomF([V, a]/U,U/RA). So |(B F1 E)/D| S |[V, a]/U] =
Lat—l. Since IB 0 E] = Jig], it must be the case that D # 1. As Z = 1, D S A
so IADI > IA]. [V,a,A] S RA and [V,a,D] S R3 2 RA so [[V,a],AD] S RA.
Then IAD/CAD([Va“l)l S |H0m(lV,alRA/RA,RA)| = IlvwalRA/RAI = IAI/2 and
since MD] > |A| we have |CAD([V,0.])] 2 4. Observe that AD is a p—group and
[V, a,CAD([V,a])] = 0. So we can apply 1.1(iii) yielding ICAD([V, a])l S [IF] = 2, a

contradiction. Therefore, RA 75 R3. [:1
Lemma 2.24
(a) NG(A) = NG(RA)°

(b) CL(A(L)) = CL(R(L))-

22

(C) 0p(L) _<_ CL(RL) S CL(A(L))-

Proof. (a) NG(A) g Ng(RA). Let t e Nam/1). We have RA] = RA‘ = RA so
A1 = A from 2.23. Then N0(RA) S Ng(A).

(b) CL(~A(L)) = ODeA(L) NL(D) = f) NL(RD) = CL(R(L)) from (a)-

(c) Op(L) S CL(AL) by 2.21(c) so Op(L) normalizes each D E A(L) and so
centralizes each RD. Thus Op(L) S CL(RL). Also, CL(RL) = CL(E7Z(L)) S

Come» ‘9 CL(«4(L))- n

Lemma 2.25 If RA S TB, then also RB S TA.

Proof. Observe that CB(RA) = CB(RA + RB), so by 2.8(a), IB/CB(RA)] = IRA +
RB/RBI = WI and CTB(CB(RA)) = RA + RB-

We have CB(RA) S N0(RA) = N0(A) S Ng(TA) by 2.24. Suppose that RB S
TA- Then (RA + RB) fl TA 2 RA n TA = RA. Also,

[TA 0 TB, CB(RA)l S [TAa CB(RA)l 0 [T3, CB(RA)l S TA 0 [713,3] S TA 0 RB = 0

SO TAOTB S CTB(CB(RA))OTA = (RA+RB)OTA = RA- NOW [TA, CB(RA)l S TA
and [TA,CB(RA)] S [V, B] 2 TB so

(*) [TA»CB(RA)l S TA 0 TB S RA-

(*)
Let P = ACB(RA). Then 2.14 implies Cp(TA/RA) = A, and thus CB(RA) S

Cp(TA/RA) = A. By 2.5(d), [B] 2 HF]2 and since IB/CB(RA)| 2 [IF], we have
CB(RA)741. Pick 1 7g 1) E CB(RA). Then by 2.12(b), RB S [V, b] S [V, CB(RA)] S
[V, Al = TA. [:1

23

Chapter 3

The case where A is not a TI-set

Lemma 3.1 Suppose that RA S TB. Then RL = RA + RB, E = CL(RL), L/E ’5’
SL2(IF), TA # TB, and one of the following holds:

(o) |1F| > 2, IA] : |1r|2, and |AnB| = [IF].
(1)) [IF] = 2, IAnBI = 2, and |A| g 24.

Proof. From 2.25 we have that RB S TA. It follows that both A and B normalize
RA +123. Then RL = (1251) 3 RA + R3 5 RL gives 12L = RA + RB as RA +123 is
L—invariant. So R L is 2-dimensional.

By 38(8), lA/CA(RB)| = IRA + RB/RAI = IIFI- Then ACL(RLl/CL(RL) =
{(19 )l* 6 IF} and BCL(RLl/CL(RL) = {(51' )l* E 11“}, SO L/CL(RL) 9—“ SL201“).
Now 2.21 and 2.24(c) give CL(RL) S CL(AL) = E = R S CL(RL). Hence, E =
CL(RL) and L/E a SL2(IF). |AI 2 Ir]? by 2.5(d) so IA 0 El = |An CL(RL)| =
ICA(RB)| 2 IF]-

Since Z S Z(L) by 2.21(b), Z S L. Define L = L/Z.

1o Chooseb E (BflE)\Z. Then 024(5) 2 ADE and IIA,b]Z/ZI = IA/CA(E)I = IIFI.

Proof of (1°). Suppose a E A \ E with [(1,1)] 6 Z. I5,(a,B)] = [5, aIIh, B] = 1

since B is abelian and Ia, b] E Z. Then 5 is centralized by L z: (a, B). Since

24

L/E z SL2(T) 9.: LE/E, LE = L. In particular, there exists 1 e L with AE = B‘E.
Since A is a weakly closed subgroup in C, A 2 Bl. Then b E bZ = blZ S Bl = A
and b E A n B = Z, which is a contradiction. Thus, CA(5) S A n E. [A H E, B F!
E] S [A, E] (1 [E, B] S Z so [A flE,b] S Z and we have A HE S CA(5). Hence,
A 0 E = C A(h). Consider the commutator map which sends a to [a, b]Z; this gives
|[A,b]Z/Z|= IA/CA(5)|= |A/AnE|=|1F|. [3

2° TA 74 TB.

Proof of (2°). Suppose that TA 2 T3. Then [V, L,A] S [TA, A]+ [T3, A] = [TA,A] =
RA S RL so [V,L,L] S RA + R3 S RL and hence [V,L’] S RL and

M [V, L’, E] = 0

since E = CL(RL). Assume that E # Z. By 2.21(b), Z S E so B 0 E 75 Z and
there exists (7 E B F] E \ Z. Thus, (1°) implies that [A,b] 75 1. Let 1 =75 0. E [A,b].
[A,b] g A since o e E = CL(AL) g NL(A). [A,b] g [L,L] = L’ so o e An L’. By
(*), [V, a, E] = 0 so E S Cg([V,a]). E = R is a p—group so by 2.12, E S Aa S A.
Thus b E Z which contradicts (1°). Hence, E]: Z.

We already have that IA n E] 2 IlFI, so Z = E 74 1. Then 1 76 [TA, Z] S RA and
since RA is l-dimensional, R A 2 [TA, A 0 B] 2 [T3, A O B] = R3, a contradiction to
2.23. El

30 OLE—(A) = A m E and C—E(OP(L)) = 1.

Proof of (3°). 014(5) = A (1 E sé A from (1°). So CA6) 79 A and 5 ¢ CW(A) so
Cbn—E(A) = 1. E = (A (1 E)(B (1 E) from 2.21. Since A is abelian, A H E S C-E(A)

 

 

and we conclude that CE(A) = (A H E)((B n E) (1 CE(A)) = A H E and similarly
CE(B) = B F) E. Also,

 

(IE-(L)=CE(A)flC-§(B)=AflEflBflE=AnB=Z=1.

25

Hence, CE(OP(L)) = 1. ‘ El
4° Z # 1.

Proof of (4°). Suppose that Z = 1 and let 1 aé b E B ['1 E. Then (1°) and Z = 1
imply C1402) = CA(b) = A n E. From 2.8(d) we have [TA,l)] + RA = CTA(CA(b)) =
CTA (A n E). By 2.8(b),

ICTAM n E)l = Int/(ME)! = ITHA/(AnEh “—3) In?

Also, AflE S E = CL(RL) so AflE centralizes RL and we have RL S CTA (AflE).
Then, as they have the same order, R L = CTA(A (1 E). So we have [TA, b] + RA 2
CTA(AF\ E) = R3. Now [TA,b] + RA 2 RL for all I) 6 EH E, so [TA,BflE] S RL

and hence also
(**) [TA, El = [TAaA fl EllTA,B n E] S RARL S RL-

Let n = dim A. Since IA/AflEI = IIFI, we have IAflEI = IlFI"_1. E = (AflE)(BflE)
and z = 1 give |E| = ITEM-1).

Let a E A D E and put Y = CTA(a) '2 [V,a]. Then Y is a hyperplane of
TA from 2.12. RL S TA and [RL,E] = [RL,CL(RL)] = 0, so RL S Y. Then
dimY/RL=(n+1)—1—2=n—2and

(* * a [x12] 3 [TA E1 (2‘) 12o.

Consider the map ’7 : E —+ Hom(Y/RL, RL) which takes 6 —+ (y + RL —> [y, e]). This
gives lE/CE(Y)| s Irl<dim<Y/RL>><dimRL> = W“) and so ICE(Y)| 2 IIFIQ- By
2.12, CE(Y) S Aa and [Au] 2 [IF] which is a contradiction. Hence, Z ¢ 1. El

By (4°) we can choose 1 75 z E Z. Then [V, z] S TA (1 T3 = U. From (2°)
and 2.12 we get that U = [V, 2:] is a hyperplane of TA and T3 and U = CTA(z).

26

Z S CZ([V,Zl) S AZ (1 Bz S Z so AZ (1 B; = Z. However, AZ is the unique Sylow
p—subgroup of CC;([V,z]) so AZ = 8,2 and Z = A2 = le. Then [Z] = IlFI.

C3(U) is a p—subgroup in CC;(U) so by 2.12, C3(U) S Aa = Z = CA(U) S
C3(U). Therefore, C3(U) 2: Z. [TA/U] 2 IF] and IT3/UI = [F] as U is a hyperplane
of TA and T3, so ITL/UI = IlFIQ.

Let U0 be an lF—subspace of U with U = U0 EB RL. Put A0 = CA(U0), B0 =
C3(U0) and L0 = (A0, BO). Since RL S U, we have CL(U) S CL(RL) = E and so
CL(U) = C3(U) = Z. Notice that IUD, L0] = O by construction. CL0(U) S CL(U) =
Z S CL0(U) so LO/Z acts faithfully on U. CL0(RL) = CL0(RL+U0) = CL0(U) = Z
so LO/Z acts faithfully on RL. By 2.8(b), IAOI = [TA/(U0 + RA)I = ITA/UIIU/(Uo +
R/lll = ITA/Ul|(U0+RA+RB)/(U0+RA)| = [Fl2- Hence, le/CL0(RL)I = le/Zl =
IIFI. Thus A0 acts as (1(1) ) and B0 acts as ((1); ) on RL. Then A0 induces SL2(IF)
on RL and LO/Z 2’ SL2(IF).

5° The commutator map, TL/U —> R1,, t+ U —> [t,z] is a well defined isomor-

phism.

Proof of (5°). TL/CTL(2) E“ [TL,z] = [TA,2] + [T3,z]. Also, 1 75 z E A OB so
by 2.5, 1 74 [TA,z] S R A- Since R A is l-dimensional, [T A,z] = RA and similarly
[Tar-”l = RB. Then TL/CTL(Z) '5 [Tad = [TA,zl + [Taft] = RA + RB = 3L-
Now TL / U and T L / CTL(z) '5 RL are both 2-dimensional so U S CTL(z) S TL gives

U = CTL(Z) = Cv(z) OTL and TL/U 3 RL- \/
6° [CV(.:), L] S U, [U, L] S RL, and [Cv(z),OP(L)] S RL.

Proof of (6°). Let V0 S V be maximal with [V0,z] S U0. 0 : V/Cv(z) —> U,
o —> [22,13] is an isomorphism. VO/Cv(z) is the inverse image of U0 under a, so
VO/Cv(z) 9-1 U0 as Lo-modules. Then, as L0 centralizes U0, L0 also centralizes
VO/Cv(z). We have [Cv(z),L] S Cv(z) (1 [V, L] S CV(z) (1 TL 2 U. Also, [U, L] S
[TA,A] fl [T3,B] S RL so [U/RL,L0] = 1. L0 stabilizes the series RL S U <

27

CV(::) S V0 and so also stabilizes the series 1 S U/RL S Cv(z)/RL S VO/RL. Then
by A.1, [VO/RL,OP(L0)] =1. Hence, [V0,OP(L0)] S RL. RL S U S CV(z) so OP(L)
centralizes CV(z)/RL. Then, [Cv(z),OP(L)] S RL. El

Suppose that IlFI > p. Since LO/Z E’ SL2(IF) and 07’(SL2(1F)) = SL2(IF), we have
0P(L0/Z) = LO/Z. By A.3, 0P(L0/Z) = OP(L0)Z/Z. Then LO/Z = OP(L0)Z/Z
and L0 = 07’(L0)Z. So A0 = (OP(L0)flA0)Z and since IAO/ZI = IIFI, A0 S Z. Then
A0 S E since Z = CL0(RL) = CL(RL)F1L0 = EflLO. So there exists a E OP(L0)flA0
with a 9% E. Since a E Op(L0), [V0,a] S RL. Since p = 2, [V0,a] S C3L(a).
CLO(RL) = Z and a g2 Z so [R3,a] S 0. Hence, RA S C3L(a) < R3. R3 is
2-dimensional and R A is l-dimensional so C 3 L(a) = R A-

[V,z] = U0 69 RL so [V/U0,z] = (R3 + U0)/U0 3 RL g U/UO. Then RL ”:3
lV/U0,Zl 9-“ (V/Uol/CV/UO(Z) g (V/Uol/(Vo/UO) g V/Vo- We know RL/lRLtal
1-di1110118i01131 SO (V/V0)/lV/V ,a] 9‘3 (V/Vo)/(([V,a] + Vol/V0) "—“-’ V/(thal + V0) is

y—u

S

1-di1nensional. We have shown [V0, a] S CRL (a) = R A so [[V, a] + V0,a] S R A- we
have also shown V/(IV,a] + Vb) is 1-dimensional so V = IFU + [V, a] + V0 for some
21 E V\([V, a]+V0). Then [V, a] S IFIU, a]+RA. Hence, [V, a] is at most 2-dimensional.
[V, a] is a hyperplane in T A so it follows that TA / R A is at most 2-dimensional and 2.5

gives that TA / R A is exactly 2-dimensional so IA] 2 IlFI2 and (a) holds in this case.

Suppose now that IIFI = p. If dim TA/RA = 2, then dimA = 2, so IAI = 4,
IA FIB I = 2, and (b) holds. So now we may assume that dim TA / R A > 2. From (1°),
we have IA/Afl EI = IlFI so IA 0 E] > IIFI. However, IAfl Z] = IZI = IIFI so E 75 Z.

Let D be a normal subgroup of L in E minimal with respect to D S Z. By
(3°), CE/Z(OP(L)) = 1 so [D,OP(L)] # 1. If [D,OP(L)l S Z, then D/Z S
CD/Z(OP(L)) = 1, but D S Z so [D, OP(L)] S Z. The minimality of D implies

(>1: * Mr) D = [D, OP(L)] S OP(L).

Let 1 e L. If [D,A] 3 Z, then [D,A‘] 3 Z, but L = (AL) so [D,L] 3 Z. Then

28

[D,OP(L)] S Z which is a contradiction. Hence, [D,A] S Z and therefore there
exists a E (An D) \ Z. U/CU(a) E” [U,a] S RA so IU/CU(a)I S IIFI. Now a E D S
E = CL(RL) SO RL S CU(a). Then

0

( >
[Co(a.).L] s RLSCUW

so L normalizes CU(a). Since a E D\Z, (aL) S Z, but (aL) S D since D is normal in
L and so by minimality, D 2 (oz). Then [CU(a), D] = [CU(a), (aL)] = [CU(a), a]L =
1.

Define X/CU(a) = CV/CU(a)(Z)° Observe that Cv(z) S X so CV(2) = CX(z).
CU(a) S U = [V, 2:] so IX, 2] = CU(a). Also, CU(a) = IX, 2] g X/CX(z) L—‘J X/CV(z)
as L—modules. 1 = [CU(a),D] 2.: [X/CV(::),D] and so IX, D] S Cv(z). Hence,
ix, D.OP<L>1 s [Ci/(o), 01’(L)] ((2)122 s [rooms Now dim(TL/Rn/(TA/Ro) =
dim TL/TA = dim(T3 + TA)/TA = dim T3/(TA (1 T3) = 1 since TA 0 T3 is a hyper-
plane of T3. So T A is a hyperplane of TL. [T A: a] S R A S R L so TA / R L is centralized
by 0., making TA/RL S CTL/RL(a). Let IV = [TL,a] + RL. We have dim W/RL =
dimlTL/RLa 0] = dim(TL/RLl/CTL/RL(G) S dim(TL/RLl/(TA/RL) = 1- SO W/RL

(5°)

is at most l-dimensional and 1V is at most 3-diirielisi011al. Also, 1 = [RL,E] g

[TL/U, E] gives [11,3] s U. 121, s U so we have 12L 3 W s U. Then (W, L] g
(60) .. .. s

[U, L] S RL S IV and so IV is L—invariant and W = ([TL, aIL)+RL = [TL, DI+RL.

Thus, IX, 0P(L), D] g [V, L, D] = [TL, D] 5 W.

The Three Subgroups Lemma gives [X , [D, Op(L)]] S l/V. Then IX, D] S [V since
D (*:*) [D,OP(L)]. Thus, [X, a] S [V Observe that [U, a] S [TA,a] S RA and
[U,a] aé 1 since (1 ¢ Z. Therefore [U,a] 2 RA as RA is l—dimensional. So U/CU(a)
is l-diinensional. Then V/X 91 (V/CU(a))/(X/CU(a)) = (V/CU(a))/CV/CU(a)(a) E”
[V/CU(a),a] = [V, aI/CU(a) is 1-dimensional since IV, a] and [V,zI = U are both

hyperplanes of TA and U / CU(a) is 1-dimensional. So we have V/X is l-dimensional

29

and 'W at most 3-dimensional. Since X/l’V S CV/W(a)’ (V/l'V)/(CV/W(a)) is a
quotient of (V / W) / (X / W) and therefore at most 1-dimensional. Hence, [V/ l/‘V, a]
and therefore, ([V, a] + l’V)/W is at most 1-dimensional. Then [V, a] is at most 4-

dimensional. So IAI S 241 and (b) holds in this case. B

Lemma 3.2 Let A S S E Sylp(C) ands E S\A. Then dimIA, s] 2 2. In particular,
if dimA S 3, then A E Sylp(G).

Proof. Suppose dimIA, s] < 2. Since A is a weakly closed subgroup of G, A S S. By

2.16, C5(A) = A; thus [A, s] 71$ 1. Pick 1 S a E [A, 3]. Since [A, s] is an lF-subspace of

2.8(e) 2.12
A, [A,s] 2 Fa 2 Au. Then UTA/1321(8) = CTA([A,s])/RA = CTA(Aa)/RA =
[V,a]/RA, Hence, [V,a,s] S RA. Since A induces HomIF([V,a]/RA,RA) on [V, a],
s E C5([V,a])A = AaA = A by 2.12. This is a contradiction to the choice of 3.
Therefore, dimIA, s] _>_ 2.

Assume dimA S 3 and suppose S S A. S \A has an element of order 2, so choose

.3 E S \ A with 32 E A. Then [A, s] S CA(s) S A. So
2 * dimIA, s] = dim A/CA(s) + dimIA, s] S dimA S 3.

Then dimIA, s] < 2 which is a contradiction and therefore S = A. El

Lemma 3.3 Suppose that RA S T3, IlFI > 2, IAI = IlFIQ, and IA (1 BI = IIFI. Then

one of the following holds:

(a) A F) B S C, R3 = [V, A F) B] is 2-dimensional, Co/CG0(R0) '5 SL2(IF), and RC

is the corresponding natural module.

(1)) RC is 4-dimensional, CO/CGO(RG) ’5 ”4+0”, and RC is the corresponding nat-

ural module.

Proof.

30

1° A is a Sylow p-subgroup ofG' and both N0(A)/A and Ng(A)/Cg(TA/RA) are
p’ - groups.

2.14
Proof of (1°). A E Sylp(C) by 3.2. A S CC;(TA/RA) so Nc(A)/C0(TA/RA) and
Ng(A)/A are p’-groups. [I]

Let ’P be the set of 1-dimensional subspaces of TA / R A- Observe that R L / R A E ’P.

2° Let 1 # z E Z. Then E = Z and R3 = [V, 2].

2.24(c)
Proof of (2°). E is a p—group and E S Op(L) S CL(A(L)) S NC;(A), so E S A

by (1°). 132 (AnE)(BnE) _<_ E(BnA) g Ans: ZsoE= 2. [714,4 = RA and

[T3, 2:] = R3 so [T3, z] = R3 and therefore,
(*l RL = [TLiZl = [TLiZl S [V,Z]-

RL and [V, Z] are both 2-dimensiona‘l so RL 2 [V, Z] = [V,z]. Then, since Z is
2.12(e)
l-dimensional, Z = Aa S Ng(RL) so NC;(RL) = Ng(Z). [:1

3° N0(A) has orbits of length 1, IlFI — 1, and 1 on P. Also, Ng(A) S Ng(Z).

Proofof (3°). Let 1 7E z E Z. [R3, Z] = 080 RL S CTL(Z) S TL. Now TL/CTL(z) g
[TL, 2] (2 R L as L-modules. Since A is 2-dimensional, T A is 3-dimensional. T L /T A =
(T3 + TA)/TA ’5 T3/(T3 0 TA) and since TA 0 T3 is a hyperplane of T3, TL/TA is
l-dimensional. Hence, TL is 4-dimensional and TL / R L is 2-dimensional. Then both
TL/CTL(Z) and TL/RL are 2-dimensional, so CTL(Z) = R3 and TL/RL ’5 RL.
Let p : TL/RL -—> RL be an L—isomorphism. Then CRL (A) 9—” CTL/RL(A), but
CRL(A) = RA and CTL/RL(A) = TA/RL making TA/RL ’.‘—:’ RA as NL(A)-modules.
Claim: TA/RA E’ RL as lFH-modules where H = NL(A) fl NL(B).
Proof of Claim. R3 = R4 EB R3 so RL/RA E’H (RA + R3)/RA g3 R3/(RA 0
R3) 9111 R3. Then RL = R4 69 R3 2H TA/RL EB RL/RA. NL(A)/A is a p’-group

by (1°). Masehke’s theorem [Asch, 12.9] gives the existence of Y/RA S TA/RA

31

with TA/RA = RL/RA EB Y/RA. Y/RA E’H (TA/RA)/(RL/RA) 2H TA/RL. Then
RL 23 TA/RL EB RL/RA 2’3 Y/RA EB RL/RA = TA/RA. So the claim is proven.
We now calculate the orbits of H on RL instead of on TA/RA. Let K =

{(3 A91 ) I 0 # A 6 IF}. The lengths of the orbits of H on R3 are the same as

the lengths of orbits of K on the 1-subspaces of IF x IF.

Observe that IF(1, 0) and IF(O, 1) are fixed points of K. Let 2:, y 6 IF\ {0}. Since

1 = A2 for some

0 75 A 6 IF. Hence, H/Z corresponds to (3)31 )(1,1) = (/\,)\"1) = /\$_1(:1:,y).

Thus, K has three orbits whose lengths are 1, 1, and IF — 1. Then H has three orbits

the characteristic of IF is 2, each element in IF is a square so cry—

on TA / R A whose lengths are 1, 1, and IF — 1.

A/Z corresponds to {(19 )I* 6 IF} and NL(A)/Z corresponds to {(i‘ A91 ) |* 6
IF} so N L(A) = HA. A acts trivially on TA/ R A so the orbits of N L(A) are the same
as the orbits of H on P, namely orbits of length 1, 1, and IIF I — 1 on P. An orbit of
Ng(A) and NC;(A) fl N0(RL) is a union of orbits of NL(A) so the possible lengths
of such orbits are:

1) 1, 1, IIFI —1

2) IIFI + 1

3) 1i IIFI

4) 2, IIFI — 1.

Since Ng(A)/A is a p'—group by (1°), the orbits of RL/RA have p'-length so
options 3 and 4 are not viable.

Suppose for a contradiction that Ng(A) is transitive on P. Since IIF I > 2, N L(A)
has exactly 2 fixed points: RL/RA and Ri/RA for some 9 E Ng(A) with RL/RA 75
syn/1. Then NL(A)9"1 fixes Iii—l/RA and RL/RA. so NL(A)9“1 g NG(A) n
NG(RL)-

From (1°), N0(RL) flNG(A) does not have an orbit of length IIFI on P but it fixes

RL/RA so it does have an orbit of length 1. This implies that N0(RL)F1 Ng(A) has

32

orbits of length 1, IIF I -— 1, and 1 on P. Notice that this means that N0(RL) flNg(A)
and NL(A) have the same orbits. So Ng(RL) fl Ng(A) fixes RL/RA and Ri/RA.
Since we proved that NL(A)9—1 S N0(A)flNg(RL), this means that NL(A)9_1 fixes
RL/RA and Ri/RA. Since the fixed points of NL(A)9“1 are RL/RA and Ri—l/RA,
this gives R% = R%_1. Hence, R312 2 RL.

Notice that gA/A has odd order from (1°) so (g)A = (g2)A. Thus R9 = R3.
This is a contradiction to the choice of g. So option 2 is not viable and the orbits of
Ng(A) and Ng(A) fl NG(RL) must be of length 1, 1, and IIFI — 1.

Let Y/RA and RL/RA be the fixed points of Ng(A) on P. Let Z := CA(Y).

[R3, Z] = 0 so Z S C A(RL) S A. Since A is 2-dimensional, Z is 1-dimensional,
and A S CA(RL), we have Z = CA(RL). Ng(A) normalizes CA(RL) so it normalizes
Z and we have NC;(A) S Ng(Z). Z = CA(Y) and since Y is a fixed point of N0(A),

~

New) S NG(Z)- 5

4° Let 1 S t E A. Then one of the following holds:
(a) t 9} Z and A is the unique Sylow p-subgroup of Cg(t).
(o) t e z and op’(cG(t)) = cam.) = L.

(e) t E Z and there crisis A S B E A with t E A 0 B. Moreover, for any such B,
Z = A n B and L := (A, [3) = op’(cG(t)) = cG(t)0.

Proof of (4°). Notice that Sylp(C(;(t)) = A(Cg(t)). If {A} = A(CG(t)), then (a)
holds. Suppose from now on that {A} S A(CG(t)). Then there exists B E A(Cg(t))
with A S B. So tB is a p—group and B is a Sylow p—subgroup of Cg(t). Thus t E B
and t E A (1 B.

1ft e Z, then t 6 AanB and RA + R3 +123 5 [V,t] by 2.3. [V,t] = 121,
from (2°). Therefore, R B S RL so R B = R Al for some I E L. Thus B = A1 by 2.23
and A(Cg(t)) 2 AL. Then Cg(t)0 = L. OPI(H) = (Sylp(H)) for any finite group so
Op’(Cg(t)) = (A(C(;(t))) and (b) holds.

33

So suppose that t ¢ Z and put L = (A, B) Then RL 2 RA + RB S [V, t] S
TA (1 T B' So L fulfills all the assumptions L does. Then we can apply (b) to get
Op’(Cg(t)) :2 Cg(t)0 = L. we can also apply (3°)3 to see that RZ/RA is a fixed
point of Ng(A) on P. Hence, RI. :2 Y and CL(RL)3=1E (2:0) A n B applied to L

gives 2 = CA(Y) = 0,,(1213) = A r1 B. So (c) holds. (3

5° Let g E G. Then ZZg = Z x Z9 E A(L) and IZG ADI = 1 for all D E A
where Z0 /\D := {Z9 I g E G,Z9 S D}.

(4 1(b )L. Also ,3.1

Proof of(5° ). For 1 S 2 E Z we have 0p(Cg(z)) S Op (CG (2 ))

gives L/E ’.—_—\_’ SL2(IF) so Op(L) = E. Then Op(Cg(z)) S Op(L) = E 3S1 Cg(RL) (12:)
C0([V,z]). Hence, Op(Cg(z)) S Op(Cg([V,z])) = A; since A; is the unique Sylow
p—subgroup of Cg([V, 25]) from 2.12(f). AZ S Cg(z) so AZ S Op(Cg(z)). Hence,
Oprszll = Az-

Recall that Ng(A) S NC;(Z) by (3°). Let h E NC;(A). 2:" E Zh = Z so 2 is not
conjugate in N0(A) to any involution in A \ Z. By IGor, 7.1.1], since A is an abelian
Sylow p—subgroup of G, 2 is not conjugate in C to any involution in A \ Z. Then
zGflA C_I Z, ZGAA= {Z}, and IZGADI=1 for any DE A.

Let 2" E Cg(z). A is a Sylow p—subgroup of C0(z) so 2" E Ak for some I: E Cg(z).
Then zhk_1 E A and zhk_1 E Z since 2G Q A = {Z}. So 2" E Zk = Z, Z S Cg(z),
and 20 flCg(z )2 {29le E Cg(.. )} C Z.

9 E G and let 1 S c E Z9 and 2 := c9 1. Recall Z = A flB = CA(Y). Since Z
corresponds to a different fixed point in TA / R A than Z does, Z S Z. Then 2 and c
are not conjugate in G. Notice that (z, c) 9:” D2,, where n is even, otherwise :5 and c
would be conjugate by Sylow’s Theorem. Since n is even and (zc) is a cyclic group
of order n, there exists t E (zc) with ItI— — 2. z (and 0 send each element to its inverse
in Dgn so t E Z((z, c)). Then t E OPI(CG (2:))(2 L. Pick X E Sylp(L) with t E X.
Then Z S Op(L) S X. We can then apply (4°) to X in the place of A.

Suppose for a contradiction that (4°)(b) holds. Since t E Z((z, e)), c E Opl(Cg(t))

34

= L. Hence, [z,c] = 1 and t E zc. So c E tz E Z. Therefore, 2 = c9.1 E CG DA g Z.
This is a contradiction to 27. E Z and Z 0 Z = 1.

Suppose (4°)(c) holds. The setup is symmetric in Z and Z so we arrive at a
similar contradiction to the (4°)(b) case. Thus, (4°)(a) holds. Then X is the unique
Sylow p—subgroup of CGftl- z E Z S X and c E Cg(t) so c E X. Thus, [z,c] = 1
and c E 0p,(CG(t)) = L. This holds for any 0 E Z9. Then Z9 S L, Z E Z(L), and
Z D Z = 1. So (Z, Z9) 2 Z x Z9 and has order IIFI2. Therefore, ZZg E Sylp(L) =
A(L). El

In particular, A = Z x Z. By 3.1, L/E E’ SL2(IF) and by (3°), E = Z; hence
L ’5 Z x SL2(IF) by Gaschiitz’s Theorem [Asch, 10.4]. We know Z S Z(L) and
S L2(IF ) is perfect so L' E SL2(IF). Since Ng(A) normalizes Y, it also normalizes
CA(Y) = Z. So NL(A) normalizes Z. Then [A,NL(A)] = [Z x Z, NL(A)] S Z. We
have A/Z = [A/Z,NL(A)] inside SL2(IF) as Z x (:191 ) corresponds to NL(A) S
NG(A) g NG(Z) from (3°). So 2 g A = ZIA,NL(A)] and as [A,NL(A)] g 2, we
get 2 = (Z n Z)[A,NL(A)I = [A,NL(A)I g L'. Hence, 2 g L’ a: SL2(IF). Then
(ZL) = L' since A/Z has order IIFI.

Recall ZZg E A(L) from (5°) so ZZg = A1 for some l E L making Z S Alg—l.
Thus, A = Alg—lh by 2.20. Therefore, lg‘lh E N0(A) (S) Ng(Z). S0 Z = Zlg-lh
and Z = Z"'_1 = Zlg-l. Then Z9 = Z1 and we get (Z0) = (ZL) = L’. Hence,
[(20), (20)] = 1. Then Co = (A) = (A0) = (20) x (20).

Case 1: If Z S C, then Z = ZG S AG 2 (ZZ)G = (ZZ)L = AL. RC =
RAL = R3 so RC = R3 (3:) [V, Z] is 2—dimensional. As Z S C, CO = Z x (Z0) so
C0 = Z x L’ = L. Also recall from 3.1 and 2.21 that Z = CL(RL), so Z = CGO(RC,~)
and CO/CG0(RG) 9-3 L/Z 95 SL2(IF) and (a) holds.

Case 2: Suppose that Z S Z9 for some g E G. Then by (5°), B := ZgZ =
(229*)? e A. Then A n B = Z by (4°). (20) a»: SL201") S (20). Hence,
(:0 = (26‘) x (20) S SL201?) x 5L2(r) S afar).

35

Let a e (20) \ NG(A). We have ('20) = L’, (20) = L’, and L = ZL’ =
Z(ZG). Then (RS265) = (Ri<ZG)) = (R51) 2 RL giving RC = RGO = (Rio) =
(RAszIIZGI) = (12220)). Hence, R3 2 RA + RAID so RC = (RIJZGI) = RIIZG) +
Riga) = R L + R1313 which are both 2-di1nensional so RC is 4-din1ensional. Both
R L and Rim are natural SL2(IF)-modules for (Z0) = L’.

Hence, as an IF(ZG)-module, RC = V1619 V2 and as an IF(ZG)-module, RC = V169

V2, where V,- and V, are natural S L2 (IF)-modulcs. Notice that End (V1) ’3—1 IF and

(ZG
so by [Asch, 27.14(5)] , Ra E’ V1 (3'11“ J for some IF(ZG)-module, J. Theli RG E“ JED J
as IF(ZG)-modules and so J E’ V1 as IF(ZG)-modules. Hence, RG E’ V1 E’ V1 as
IF CO-modules.

The same argument also shows that the natural 52: (IF )-module for C0 is the tensor

product of two natural S L2 (IF)-modules and so RC is a natural {ZION-module. This

gives (b). [:1

36

Chapter 4

The case where A is a TI-set

Corollary 4.1 IfA is a TI—set, then RA S T3, D is a TI-set for any D E A, and
RB 0 TA = 0.

Proof. Assume R A S T3. Then by 3.1, IA (1 B I S 1 which contradicts the definition
of TI-set, so RA S T3.
Let D = Ah for some [2. E C. If D is not a TI-set, then there exists g E C\N0(D)
such that AhflAhg S 1. This implies AflAg S 1 which contradicts that A is a TI -set.
If 0 S R3 (1 TA S R 3, then R3 0 T A = R3 since R3 is 1-dimensional. But then

R3 S TA which contradicts that B is a TI-set. El
Notation 4.2

(a) H7 2 RA + (TA 0T3) + R3.

(b) A1: NAUV), B1 = N3(VV), and L1=(A1,Bl).

(C) A0 = NA(RA + RB); Bo = NB(RA + RB): and L0 = ((40.30).

If X is one of the symbols just defined, then we will sometimes write X (A, B) in place
of X, to indicate the dependence ofX on A and B. Also if C, D E A with C S D,
then X (C, D) is defined as above with C and D in place of A and B.

37

We also fix 0 S r3 E R3 and put 3 = 3TB and 51 = sIAl. (For the definition of
8’13 see 2.9). Finally, if RA S T3, pick X S [r3,A] + RA with TA 0T3 S X and

[TBiAI + RA = X EB RA, then put q = qu,X {see 2.11).
Lemma 4.3 Suppose that A is a TI—set. Then NA(R3) S NA(B).

Proof. Let a E NA(R3). Then R3 = R“ = R3a and by 2.23, B“ = B so a E NA(B)
and we have NA(R3) S NA(B). [:1

Lemma 4.4 Let 0 S r3 E R3.

(a) I'V flTA = (TA 0T3) + RA.

(b) A normalizes IV flTA.

(c) A1 centralizes IV/LVflTA, (IV flTA)/RA, and RA.
(d) A1 = {a E A I [r3,a] E IVflTA}.

(e) A0 = {a E A I Ir3,a] E R4} = rads.

Proof. (a) Clearly (TAflTB) +RA S TA. Also, R3 flTA S TAflTB and so lVflTA 2
(RA + (TA 0 TB) + RB) 0 TA = (RA + (TA 0 T3)) + (RB 0 TA) = RA + (TA 0 TB)-

(b) [WHTA,A]SRASWHTA. '

(C) [IV 0 TA, A] S R A S W 0 TA also gives us that A1 centralizes (IV r1 TA) / R A-
A1 clearly centralizes R A- Observe that A1 normalizes W and W (1 TA and that
W/(WDTA) (i) (R3+(I/VflTA))/(IVF1TA) is at most 1-dimensional so A1 centralizes
I'V/W 0 TA.

((1) Let a E A. Suppose [r3,a] E W flTA. Then (a) gives W = IFr3 + W flTA so
[W, a] = [IFr3,a] + [IV (1 TA,a] S IV 0 TA S IV since a normalizes W flTA. Hence,
a E NAUV) = A1. Suppose a E A1. Then by (c), A1 centralizes LV/W (1 TA so
[IV/11] S W PITA and we get [r3,a] E W flTA.

(e) Let a E A. Suppose [r3,a] E RA. Then [RA + R3,a] = IRA,a] + [R3,a] =

[RA,a] +IFIr3,a] S RA S RA +R3. Hence, a E NA(RA + R3) = A0. Now

38

suppose a E A0. (R A + R3) / R A is a 1-space normalized by A0 so A0 acts trivially
011 it and [r3,a] E RA. Hence, A0 = {a E A I [r3,a] E RA} and 2.10 gives

{aEAIIr3,a]ERA}=rads. CI
Lemma 4.5 Suppose that A is a TI-set and NB(A) S 1. Then

(a) NA(B) S 1-

(b) A 0 E = NA(B) = CA(RB) = CA(W) = CA(TA r1TB)-

(c) A n E is an IF-subspace of A.

(d) IA/AflEl = ITA rlTBlo

(e) [TA,BflEI =TAflT3.

(f) dimTA 0T3 2 2.

(g) dimA Z 4.

(h) PV is an L~submodule of V. In particular, RL S W and L 2 L1.

Proof. From 4.1, with A and B interchanged, we have R3 S TA. [TA,N3(A)] S
[V, B] = TB and [TA, N3(A)] = [V,/1, N3(A)] S TA 80 we get

(*) ITAa NBfAll S TA 0 TB
and
(**) [TA n T3, N3(A)] 5 TA r1 R3 = 0.

1° ThenlSNA(B)=AflE.

Proof of (1°). Let a: E CA(N3(A)). Then 1 S N3(A) = N3(A)“: S B (1 B3. As B
is a TI-set this gives B 2 B93. Hence, 1 S CA(N3(A)) S NA(B). Then NA(B) S 1

and (a) is proven. In particular, the setup is symmetric in A and B.

39

From (*) and (**) we have [TA,NB(A),N3(A)] S [TA (1 T3,N3(A)I S 0. So
N 3(A) acts quadratically 011 TA and therefore on TA / R A- Then, by 2.5(c), N 3(A)
acts quadratically on A“ and therefore also 011 A. Since N 3(A) acts quadrati-
cally on A, we have [A,N3(A)] S CA(N3(A)) S NA(B). So [NA(B)N3(A),A] S
NA(B) S NA(B)N3(A). Thus, A normalizes NA(B)N3(A), and similarly B nor-
malizes NA(B)N3(A). Then NA(B)N3(A) S (A, B) = L. NA(B)N3(A) is ap—group
so it’s in OP(L). So also NA(B) S Op(L). Then A D Op(L) S NA(B) S A 0 Op(L).
So A (1 Op(L) = NA(B). Also, A F) Op(L) S A H E S A H Op(L) so NA(B) = A 0 E
and symmetrically N 3(A) = B 0 E. El

(**) also gives us that N 3(A) centralizes R A, R 3, and TA 0T 3 and therefore W.
Thus, [[V, N3(A)] = 0 and so N3(A) S C3(IV) S C3(RA) S N3(RA) 4S3 N3(A).
Then using the symmetry ofA and B we have ADE = NA(B) = CA(R3) = CA(W).
So (b) is proven except for the last equality.

We have E = (A F] Op(L))(B fl Op(L)) = (A (1 E)(B n E). Since A is a TI-set, we
have AflB = 1. Hence, E is abelian. Then CA(BflE) S AflE and CA(BF1E) Z AflE
so CA(B n E) = A n E. By 2.7(d) N3(A), and therefore B 0 E, acts IF-linearly 011
A. Therefore, C A(B n E) = A 0 E is an IF-subspace of A. Then (c) is proven. Also,

2.8(d) gives
(***) ITA,Bf1E]+RA=CTA(CA(BDE))=CTA(AHE).

2° TAflT3=ITA,BflE].
Proof of (2°). RA S CTA(AF1 E) so

(**) *IHI (*)and(b)
(TAOTB)+RA S CTA(AOE)(=)[TA,BOE]+RA S (TAOTB)+RA.

So equality holds everywhere. Then T A (1 T3 2 [TA, B F] E] + (TA (1 T3 (1 R A) 2
[TA, B D E]. El So (e) is proven.

40

3° IV 2 [TA + T3, E] = [T3, E] is a L-submodule and RL S IV.

Proof of (3°). [TL,E] = [TA + T3, E] = [TA,A H E] + [TA,B 0 E] + [T3,A n E] +
[T3, B (1 E] 2 RA + (TA 0 T3) + R3 2 IV from (2°). Then IV is a L—submodule.
Since 12,, s W, 12L = (12,3) 3 WL = W. [3

So L normalizes IV giving A = A1, B 2 B1, and L = L1. Then (h) is proven.
Put E = L/CL(W) and Z = A/CA(W). '

4° AflE=CA(TAflT3) and I71]=IA/AflEI=ITA 0T3].
Proof of (4°). (***) and (2°) give CTA(AflE) 2 [TA, BOEI+RA = (TAflT3)+RA.
Then CA(TA (1 TB) = CA(TA 0 TB + RA) = CA(CTA(A I) E)) = A f) E by 2.8(b).

Now the proof of (b) is complete.
Also, 2.8(b) applied with Y = TA (1 T3 gives IA/A (1 E

 

= IA/CAfTA OTBll =
I((TAOTB)+RA)/RAI = ITAflT3I since RA S TAflT3. Then ((1) is proven. We also
have IAI = IA/CA(RA+(TADT3)+R3)I = IA/CA(TAflT3)I since ADE S CA(R3).
El

50 dimly TA 0 TB 2 2.

Proof of (5°). Observe that N3(A)A is a p—group and N3(A) S A since 1 S N3(A)
and N 3(A) (1 A = 1 as A is a TI-set. Then A is not a Sylow p—subgroup. Choose
1 S s E N3(A). Then 3 ¢ A and A(s) is ap—group. dim TAflTB = dimIA, N3(A)] Z
din1IA,s] Z 2 by 3.2 and (f) is proven. [3

If dimA < 4, then 3.2 gives A E Sylp(C) yielding a contradiction so dimA 2 4

and (g) is proven. CI

Lemma 4.6 Suppose A is a TI-set. Then NA(B) : A n E = CA(W) and A O E is

an IF-subspace of A.

Proof. If NA(B) S 1, this follows from 4.5. So suppose NA(B) = 1. By 2.21,
A n E = A n CL(A(L)) s NA(B) = 1.

41

From 4.3 we have CA(IV) S CA(R3) S NA(R3) S NA(B) = 1. So NA(B) =
A n E = CA(IV) = 1. In particular, A F) E = 1 is IF—subspace of A. D

Lemma 4.7 Suppose A is a TI-set.

(a) If A/(A F] E) is even dimensional, then A F) E = rads = A0.

(b) If A/(A F) E) is odd dimensional, then AO/A I) E is I-dimensional.

Proof. By 4.6, we have A F) E = CA(IV) = CA(R3) = CA(RA + R3). Clearly,
CA0(RA + RB) S CA(RA + RB) as A0 S A. A180. CA(RA + RB) S CA0(RA + RB)
since CA(RA + RB) S NA(RA + R3) = A0. Hence, CA0(RA + R3) = CA(RA + R3)
and AflE = CAO(RA +R3). Also, AO/CA0(RA + R3) is isomorphic to a 2-subgroup
of SL2(IF) making IAO/A n E] S IIFI. Notice that both A0 4S4 rads and A (1 E are
IF -spaces from 2.10 and 4.6. Thus, AO/A [’1 E is an IF—space and either A0 = A (1 E or

AO/A D E is 1-di1nensional. Since A0 = rads is a non—degenerate symplectic space,

A/AO is even dimensional. So if A0 = A F] E, then (a) holds and if AO/A (1 E is

1-dimensional, (b) holds. Cl

Lemma 4.8 Suppose A is a TI -set and A /AflE is odd dimensional. Then L0 induces

S L2(IF) on R A + R 3. In particular, L0 acts transitively on the l-spaces in R A + R 3.

Proof. \Ve have IAO/A (1 E] = IIFI from 4.7(b). Also,
A n E S CAOIRA + RB) = CAOIRB) = NAOIRB) S NA0(B) S A 0 E

as IRA, A0] = 0. So IAO/CAofR/l + R3)] = IIFI. Thus, A0 induces the full centralizer
of R A in SLIF(RA + R3) 011 RA + R3. A similar statement holds for B0 and we
conclude that L0 = (A0, B0) induces SL2(IF) on RA + R3. CI

Lemma 4.9 Suppose A is a TI -set. Then
(a) A1 is an IF-subspace of V.

42

(b) Let a E A1. Then r3 — q(a) E T3a. Moreover, q(a) = 0 if and only ifa E AflE.

{c} Suppose A1 S A (1 E and let D be IF—subspace of A1 maximal with CD(IV) = 1

and 81 ID: 0. Then qID is onto and dimD = l.
{(1) One of the following holds:

1. 81 = 0, A1 = radsl and dim Al/Afl E S 1.

2. 8151'é O,rad51 = AflE and dimAl/AflE = 2.

Proof. (a) By 4.4, A1 = {a E A I [r3,a] E W flTA}. Since W (1 TA is an IF-subspace
of TA, we put U = W 0 TA in 2.10(a) and conclude that A1 is an IF-subspace of A.

(b) Let a E A1. By definition of q and X in 2.11, [r3,a] — q(a) E X. Also,
[r3,a] E IV flTA 2 (TA 0T3) + RA from 4.4. Since q(a) E RA, [r3,a] — q(a) E
((TA 0T3) + RA) (1 X. Since TA 0T3 S X, this means [r3, (1] — q(a) E (TA 0T3) +
(RAflX) 2 TA flT3 +0 2 TAflTB. So r213 -r3 —q(a) E TAflTB and since
r3 E RIB Q T3, r213 — q(a) E T3. Conjugation by a gives, r3 — q(a) E T3a since
q(a) E RA S CV(a).

If a E NA(B), then r3“ = r3. In this case q(a) E TA 0 T3, and since RA 0
(TA (1 T3) = 0, we must have q(a) = 0. If q(a) = O, we conclude that r3 E T3a. If
B“ S B, then R3 S T3a gives a contradiction to 4.1 so we must have B = B“ and
a E NA(B). By 4.6, NA(B) = A n E and so (b) holds.

(0) Notice that D S 0 since 31 vanishes on any 1-dimensiona1 subspace, so ID] 2
IIFI. By 2.11, qID is Z—linear. By (b) and 4.6,. q(d) S 0 for all a gé CA(W) and so for
all at E DII since CD(W) = 1. Thus, qID is one to one. Since Iq(D)I S IRAI S IIFI, we
conclude that IDI S IIFI and so dim D = 1 and q ID is onto.

(d) Notice that CD(IV) = 1 is equivalent to D F) (A (1 E) = 1 and observe that
(D(A fl E))/(A n E) is a maximal isotropic subspace of Al/A n E. Thus, A (1 E S
rad 31 S D(A n E). Then rad 31 = (D(A (1 E)) n rad 31 = (D D rad sl)(A n E).

Moreover, Drad 31/ rad 31 is a maximal isotropic subspace of A1/ rad 31 and since

43

the maximal isotropic subspaces of non-degenerate 2n-dimensional symplectic spaces
have dimension n, we conclude that dim Al/ rad 31 = 2 . dim(D rad sl/ rad sl). This
comes from [Asch, 19.15, 19.16, 20.8].

If D S rad 31, we get dim A1/radsl = 2. Also in this case, if D H rad 31 S 0,
then D = D H rad 31 since dimD S 1. Hence, D S rad 31 which is a contradiction
to our assumption. So D H rad 31 = O and rad 31 = (D D rad 31)(A n E) = A F) E in
this case. If D S rad 81, we get di1nA1/rad 31 = 0, A1 = rad 31 = D(A (1 E) and
dim A1/A F) E = dim D(A fl E)/(A I) E) S 1 since dim D =1. CI

Lemma 4.10 Suppose A is a TI-set, IV is 4-di7nensional, and [r3, A1]+RA = IVflTA
for some 0 7e r3 6 R3. Then A(W) z: {D e A | RD s W} = AL1 = {A} U B41 =
{B} u A131, |A1/A r1 EI = IIFIQ, |A(W)I = |IF|2 + 1, and L1 acts doubly transitive
on A(IV). Also A and B are conjugate and IV = (Rfil). Furthermore, TA (1 IV
is the perp of R A with respect to 5W: where 3W is the symmetric form associate to
qw, and there exists qw : W ——> RA 9’ IF, an L1 -invariant quadratic form of —-
type, ( the maximal singular subspaces of IV with respect to qw are 1-dimensional).
{RD I D E A, RD S W} = {R3 I D E A(IV)} is the set of singular 1-spaces and L1
induces (Z(VV, qw) on W.

Proof. Observe that Al S An E since [73, A1] S 1. Let D E A(I’V) with D S A. By
4.1, RD S IV 0 TA. By assumption we have (W H TA)/RA = ([r3,A1] + RA)/RA.
Let H be a 1-space of IV/RA with H S (W H TA)/RA. As (IV (1 TA)/RA [USED
(R4 + (TA 0 T3))/RA is a hyperplane of IV/RA, H + (IV D TA)/RA = IV/RA. So
there exists a: E (IV flTA)/RA and e E H with e + a: = 7‘3 where 7‘3 = r3 + RA.
Then e — 7‘3 2 —:r E (W flTA)/RA. Now —a: = [7‘3,a] for some a E A1 since
—:1: E (IV flTA)/RA = (Ir3,A1I + RA)/RA. Then 7‘33 = 7‘3 + [7‘3,a] = 7‘3 — a: = e.
Hence, H = ( ‘113 + R A) / R A for some a E A1. So we see that A1 acts transitively on
the 1-spaces of IV/RA that are not in (IV (1 TA)/RA. We know (RA + R3)/RA is a

-1
1-space so there exists a E A1 with RD S R A + RI?‘ Replacing D by D“ we may

44

assume that RD S RA + R3.

Choose 3 E RA with s + r3 E R3. From 4.9(c) we have q [41 is onto. So
there exists d E A1 with q(d) = s so 8 + r3 = q(d) + 73 E Tg from 4.9(b). Thus,
RD S Tg and by 4.1, D = Bd. Hence, A(W) = {A} U BAl. By symmetry,
A(W) = {B} UABI. Then [BA1|= [AlE/EI so |A(W)| = IAlE/EI + 1. Therefore,
L1 acts doubly transitive on A(IV). So A [and B are conjugate and there exists
9 E L1 with A9 = B. By hypothesis, [73, A1] +RA = WflTA 4.321) TAflT3-I-RA so
W S R3 + [r3, A1] + RA S (R?) +RA S (R31) S IV since A and B are conjugate.

So W = (R31).

It remains to show that such a qW exists.

Now let a E A1 \ E. Then by 4.9(d), a E rad 31 and so [IV,a,A1] S 0. Since
A1 normalizes but does not centralize [IV,a], we conclude that [IV, a] is at least 2-
dimensional. Note that IV = R3 EB REL; EB (T3 0 T5). Since [IV, a] is at least 2-
dimensional, we get that [T3 {’1 The] S 0.

Since a E A1 \ E and A (1 E = NA(B), B“ S B. We know B = A9 for some
9 E L1. VVe’ve shown that A and B 2 A9 are in A(W) and since L1 is doubly
transitive on A(W), (Ba)! = A and B1 = B for some l E L1. So we have B” = A.
Conjugating by al 2 l-lal we get Ball—lat = Arlal. Then Bl = Aal. As B1 = B,
we can let to = al and see that there exists an. (17 E L1 such that A” = B and 1.22 = 1.
Since 0 S [T3 (1 Twas], conjugating by l gives 0 S [T31 0 Tgl,WI 2 [T3 0 T‘l‘g’fib] 2
[T3 (1 TA, to].

Choose v0 E R34 and v1 E TA (1 T3 with [v1,w] S 0. Put v2 = v‘f and v3 2 v6”.
Observe that v3 E R3. Then (v1,v2) is an IF—basis for T A 0 T3. Since W = RA +
(TA 0 T3) + R3, (v0,v1,v2,v3) is an IF-basis for IV. Let L1 be the image of L1 in
CL4(IF) and A1 = Al/CA1(W) 4S6 Al/(A (1 E). Since E acts trivially on both W
and A(L), L1 acts on both IV and A(L).

Let C 6 A1. NOW [03,0] E [IV,A1] S I’VOTA = RA+(TAOTB) = IFv0+IFv1 +IF'02

45

so [123, c] = 530(c)v0 +a:1(c)v1 + :rg(c)v2 for some (13,-(0) E IF. Put :1:(c) = ($1(c),x2(c)).
Since RB is l-dimensional, v3 = An; for some 0 S A E IF. Thus, [223,A1] + RA =
[ArB,A1] +RA = /\([7'B, A1] +RA]) = A(IVDTA) by assumption. And A(WOTA) =
WnTA so [223,A1] +RA = WnTA = val +1Fv2+RA. Then :1: : .211 ——> r2 is
onto. Horn 4.9(d), dim A1 /A n E S 2 so we conclude that a: is a bijection and
[Al/A F) El = [F2]. Since A1 acts quadratically on IV/RA, a:(cd) = 513(0) + $(d) for
all c, d E A1. So :1; is an isomorphism from A1 to (F2, +). Let a : 1172 —-+ A1,t -* a(t)
be inverse of :1: : A1 —+ F2. Define q(t) = $0((z.(t)). Since a:(a(t)) = t, we have
[v3,a(t)] = q(t)v0 + tlvl + t2’02. Hence, v3“) = q(t)v0 + tlvl + tgvg + 713. Then
q(t) = 0 if and only if 2);“) E val + vag + va3. Since W 0 TB = IFvl + vag + F113
and va3 = R B: this holds if and only if #113“) S TB which, by 4.1, holds if and only
if a(t) E N/il (B). N31 (B) = 1 since NA(B) = GAO/V) from 4.6 and CA1(W) = 1.
So q(t) = 0 if and only if a(t) = 1. Since a : F2 —> A1 is 1-1, this holds if and only if
t = 0. Then q(t) = 0 if and only ift = O.

O O 1
0 1
We have L?) = 0 X 0 where X = . Let t = (t1,t2) E F2. We
1 0
l O 0

have shown that there exists a(t) E A1, and similarly b(t) E B1, such that

K '1 0 0 0 )
n1(t1,t2) 1 0 0 ..
”Z(tlat2) 0 1 0

 

 

 

 

q(t) t 1
[q(t1,t2) t1 t2 1 j
and
~ (1 t1 t2 6(t1,t2) \
1 t (1(t)
0 1 0 fi1(t1,t2) ~
W): 0 1 fi(t) = 6131.
0 0 1 fi2(t1,t2)
0 0 1
\0 0 0 1 j

46

Observe that a:(a(t)a.(t’)) = t+ t’ so applying a to both sides gives a(t)a(t’) =

a(t + t'). Then we get

(1) 71(t) + n(t') = n(t + t’).
and
(2) W) + WW) + (1(t') = W + t')-

Define a tilde function from F2 \ {O} —> F2 \ {O} by t —-+ {with Bag) = AME) E
A(IV) \ {A, B}. Then F(q(t) t 1) = lF(1 { (76)) and we see that

~ t
3 t: __
( ) (10)
and
.. 1
4 ~ t = —
( ) (1() W)
Now
0 0 1 1 0 0 0 O 1

1 1 0 0
1 t* q(t)
= 0 I n*(t)
0 0 1

47

712051, t2)

where 71*(t) = and t* = (t2,t1). Notice that a(t)“’ E B1 and A1 and

71101, t2)

B1 are conjugate by on so a(t)“’ = b(t*).

Then, since we have a(t)“’ = b(t*), we get

(5) (1115*) = (10‘)
and
(6) fi(t*) = 71*(t).

Thus, (6) together with (3) gives

 

Since Ba“) 2 Ab“), we have AW“) 2 Ab“) so Awa(t)b(£) = A. Therefore, wa(t)b(£) E

NL1(RA) and NL1 (A) so it normalizes [IV, A1]. Hence, wa(t)b(f) must be of the shape

*00
**0

***

We already have

0 0 1 1 0 0

wa(t)b(5)= 0 X 0 n(t) I 0
1 0 0 q(t) t 1

_t_

q(t) t 1 1 q(t)

t l
1 so 7‘)
* t*
0 I n(q(t))
0 0 1
1
q—(tl
t*
((175)
1

Then we get

(9)

 

 

From (7) and (8) we obtain

(10)

 

 

 

 

q(t) t+t1 1+tn*(fi)+1
___ n*(t) ”wag—(ta +X - 12(8) +7t(2f(:t7)
1 -—t— L
(1(1) ‘1“)
(1(t) 0 tll*(at(*T))
= :1: * _t_ Y ”*(t) i
n (t) 7‘ (0“,) +’ (10) +71%”)
1 J— ‘1‘
gm W)
* t* _
m (0(1)) 0
n.*(t) _ 15*
q(t) “ ”(q(t))’
1 t*
tn(t) = 0-

This gives (t + t’)n(t + t') = 0 so (1) and (10) yield

(11)

t'n(t) = tn(t').

Since we showed [IV, a.(t')] is 2-din‘1ensional earlier, 1:2 S 0. Consider t’ = (1,0).

49

0

Then 71(t’) = for some IE2 from (10). And t'n(t) + t'n(t') = 0 from (11) so
3:2
7n(t) 0 . .
(1 0 ) + (t1 t2 ) = 0. Hence, n1(t) + t2$2 = 0. Similarly,
712(1) 132
5[:1
consider t’ = (0,1). Then 71(t’) = for some 3:1 from (10). This yields
0
t2$2
n2(t) +t11171 = 0. Thus, 71(t) =
t11131
12$2
Again from (10), tn(t) = 0 so we have (t1 t2 ) = t1t2$2+t2t1x1 = 0.
t1$1

Then t1t2(.7;2 + 1:1) 2 0 and .132 2 11:1. Replacing v1 by 132111 and so also '02 by $2v2

we can let $2 = 11:1 2 1 and discover
(12) 71(t) =

We calculate

H17 71%)) 1 0 0

w(t)== a(t)b(t)a(t)= n(t) I 0 0 I n(t) I 0
q(t) t 1 0 o q(t) t 1

0

[ .

p-a
O
O
F—i

q(t) q(T) 1 0
= 71(t) mega—7+1 ggw: ()5le) n(t) 1
q(t) t+tI 1+tn*( q(t) t 1
t 1
1 RT) EX?) 0 0
= 71(t) n(t);fij+1 O 71(t) I 0
(1(1) 0 0 W) t

50

t t t 1
1+-q—(—5n(t)+1 211171th W

= 71(t) + n(t)a%n(t) + n(t) ”(GI—(£51 + I 0
0

(1(1) 0
1
0 0 W
= 0 mad?) + I 0
(1(1) 0 0

Now we can find

_1_
0 0 W) 0 0 1
12(1) =w(t)w= 0 71(t)q—(%+I 0 0 X 0
q(t) O 0 1 O 0
_1_
W) 0 0

so 11(1) e Nam/1) = N(;(A).

Which allows us to calculate

a(t)h(k) = (h(k))_1a(t)h(lc)

 

q(k) O 0 1 0 O
_ , . 19* —1
— 0 (n.(lo)q(k)+X) 0 n(t) I 0
l
_1_
q ‘l 0

 

 

 

 

 

 

 

fl _1_ _1_
(M) (106) (1(1)
1
EXIT) 0 0
0 n(k)q[;) + X 0
0 0 (1(13)
1 0 0
_ 15* _ 71(5)
_. ("Wm + X) 1M) I 0
((t [5*
3:17))? q—(i—)(71(k)q(k)+X) 1

Now h(t) normalizes a so we have a(t)h(k) = a(r) for some 7‘ E IF X IF which

depends on t and k. Then

(12) 1 ‘32 1
= @011 t2 ) k ((32 k1 )+WQ(k) (t2 t1 )'

1
So 7" = —i—(t1k2 +t2k1) (k. I; ) + —-1—-—q(k) (t t )
q(k‘)2 2 1 (1002 2 1
= (tlkg+t2k1k2+q(k)t2 t1k2k1+t21ef+q(k)tl ) .
(NC)2 We)2

 

 

This gives

151117122 +t2k1k‘2 + t2q(k) tlkgkl + 15216? + t1q(k)
(101")2 ’ (106)2

52

 

 

).

Let q1(k) = q(k,0) and (120;) = q(O, k). If k2 = 0, then

(1(1) :q 15-2 tzkj+t101(k1)
£11051)2 01((31), (11(k‘1)2

 

 

If t2 =2 0, then
(1101) _ t1
(210181)2 QZ(91(k1))'

t
Let or = (12(1) and (11061) = 111- Then (11(11) = (11(ki)2612(m1k1—)) = 1t¥(I2(1) = tia-

 

Similarly, q2(t-2) = tga. (2) and (12) give
(13) (1(t1, t2) = at? +t1t2 + org.

Then q is a quadratic form with associated symplectic form s((t1,t2), (31,32)) =

t152 + t281. Define qw(sov0 + 31121 + 32122 + 33123) = 30.93 + as? + 8182 + asg. qW is

a quadratic form with associated symplectic form sW((t1, t2, t3, t4), (31, 32,33,514» =
O O 1

15134 + t283 + t332 + 16431. Recall w = 0 X 0 so it switches so and 33 and

1 0 0
switches 31 and 32 clearly making (1”; (1)-invariant.

[1000)[1000)

n t,t_ 1 O O t l O 0
Now recall a(t)= 1(1 2) , = 2 .Then

n2(t1,t2) 0 1 0 t1 0 1 0
\WLt‘z) t1 t2 1 j \W) 11 12 1 /

 

 

 

 

(In/((80110 + 81111 + 32212 + 83v3)a(t1’t2))

= (In/(30m + (8112110 + 81111) + (8213100 + 82712) + (539(t)v0 + 831101 + 8312112 + 8303))
= qw((50 + 3112 + 8211 + 83€1(t))”00 + (81 + 8311% + (82 + 8312)’U2 + 83113)

= (so + 31152 + 32t1+ 33q(t))83 + a(51 + .93t1)2

+(81+ 83t1)(82 + .3th) + (1(82 + S3t2)2

53

= 3033 + 31t233 + 32t133 + s§q(t) + 05% + (1312,25? + 5132 + 3183t2 + 3315132
9 o r;
+ Sgtltz + 0.95 + 013.315
0 0 9 9 9
z 5033 + 81t283 + 32t133 + sgat‘l‘ + sgtltg + sgo'tg + (131+ (mat? + 3132
2 ,..2 , 2 2
+ 8133t2 + 53t182 + s3t1t2 + 0.52 + (183t2

= 8083 + as? + 3132 + (13%

= (In/(80m + 81111 + 82112 + 83113)-

So (1W is A1-i11variant as well as w-invariant. L1 = (A1,A‘f) = (A,w) so qW is also
Ll-irwariant.

Consider Rig 2 110i = vao + val + vag. Observe that the definition of SW shows
that TA 0 IV is the perp of R A with respect to 5W. To be singular we must have
0 = qw(t0'vo +t1v1 + tgvg) = 011112 + t1t2 + Orig = q(t1,t2) which implies t1 = t2 = 0
from earlier. Hence, R A is a maximal singular subspace and we see that the maximal
singular subspaces are l-dimensional and (1W is of —-type.

Due to the double transitivity we have R511 2 {RD | D E A(W)}. IV has
[IF I2 + 1 singular l—subspaces and [Bill = IIFI2 + 1 where each RD is singular so
{RD | D E A(W)} is exactly the set of singular l-subspaces of IV.

[Asch, Chap 7] gives us that Q(W, qW) C‘—_.’ SL2(K) where K is a quadratic extension
of IF. A1 is the image of A1 in O(W,qw). Since A1 centralizes [Iii/RA: A1 S
Q(W, qW) and [All = q2, which is the order of the Sylow subgroup in S L2 (K). So A1
and B1 are sent to different Sylow subgroups and S L2(lK) is generated by two Sylow

subgroups so L1 induces {Z(IV, (1w). [3

Lemma 4.11 Suppose that A is a TI-set and NA(B) S 1. Then L = L1, dim A/AflE

is even, and IV flTA = [rB,A[ + RA.

Proof. By 4.5(11), L 2 L1 and therefore 3 = 31. By 4.5(d) and (f), dimIF A/A (1 E =

dim}: TA 0 TB 2 2. Suppose for a contradiction that dim A/A F) E is odd. Then 4.7

54

gives rads S A H E. Since 3 = 31, 4.9(d) implies A = rads and dim A/A n E = 1, a
contradiction to dim A/A (1 E 2 2. Thus, dim A/A n E is even.

Since A is quadratic on V/RA, the map 6 : A —> TA/RA,a —> [r3,a[ + RA/RA is
a homomorphism. From 4.4(e), A0 = {a E A | [73, a] E RA} so A0 is the kernel of 6.
Thus, [A/AOI = [[r3, A] +RA/RAI. This, along with A0 4S7 ADE and A1 = A, gives

|A/An El = IA/Aol = “713,41 + RA/RAI < lIVflTA/RAI

4.4__(a) 4.5_(d)

lRA+(TAflTB)/RA|= [TAnTBI lA/AHEI-

Hence, [rB,A] + RA 2 IV flTA. 1:]

Lemma 4.12 Suppose A is a TI set, NA(B) = 1, and dim A/A (1 E is even. Then
TA = [Tea/11+ RA, WOT/1 = [T3,/111+ RA, mull/1d: ITA r1TBI-

Proof. We have A04: 7A D E— 4 —6 NA(B) = 1. The map (5 as given in 4.11 is still a

homomorphism with kernel A0. Then
IAI = lA/Aol = lFa./41+ RA/RAI S ITA/RAI = IAI-

Therelore, TA 2 [r3, A] + RA and 6 is an isomorphism. From 4.4(d) we have A1 =
{a E A [ [r3,a] E IV flTA}, so A1 = 6‘1(IVflTA/RA). We conclude that [All 2
[IVHTA/RAIZITAHTBI and IVflTA=[TB,A1[+RA. [:l

Lemma 4.13 Suppose that A is a TI set, dim A/A F) E is even, and TA (1 TB S 0.
Then dim IV: 4, dimTA (1 TB— — 2, and [73, A1] + RA: W (1 TA.

Proof. By 4.11 and 4.12, [T3, A1]+RA = I'VflTA. By 4.5(d) and 4.12, dim Al/AflE =
dimTA (1 TB > 0. By 4.9(d), dimAl/A n E S 2. If dimAl/A F) E = 2, then
dim TA 0 TB = 2, dimW = 4 and we are done in this case.

So suppose for a contradiction, that dim A1/A F) E = dim TA (1 T3 = 1. Then

dim W = 3. If NA(B) S 1, then 4.5(1) gives dimTA (1 TB > 1, a contradiction.

55

Hence, NA(B) = ADE = 1 and dim A1 = 1. Since dim A/AflE is even, 4.7(a) gives
A0 = A fl E = 1.

42(9) {0 e A | [me] e RA} and so [R3,a] 7! RA-

Let 1S a E A1. Then a. E A0 4
From 4.4(a) we see (IV D TA)/RA ”:3 T A (1 TB which is 1-dimensional in this case
and [IV, A1] S W (1 TA, this implies IV (1 T4 = [IV,a] + RA S CW(a). Thus, A1
centralizes IV 0 TA 2 R A + (TA 0 TB) and so also TA 0 TB. By symmetry, B1
centralizes TA 0 TB and so TA (1 TB S CW((A1,Bl)) = (I'M/(Ll). Also, A1 acts
faithfully on W :2 IV / (TA 0T B) and [All = [IF [ Thus, A1 induces the full centralizer
of B A = (R A+(TA 0TB» / (TA 0TB) in S LH:(W) on W. A similar statement holds for
B1 and we conclude that L1 induces S LF(W) on W. In particular, L1 acts transitively
on the 1-spaces of W, and 81 acts transitively on the 1-spaces in W distinct from
B3. Since a E NA(B), RB S RE; and so REL} S WflTB and BC; S 'R—B S E- Thus,

R3“ = RAb for some b E B1. It follows that
Rs 3 1294+ (TA 0 TB) = (RA + (TA n TB))b 3 TE.
and by 4.1, Ba = Ab and 11a = 12),.

Hence, (Ba, B) = (Ab, Bb) = (A, B)b 2 Lb = L and

TA + T3 222 [V, L] = [V, Lb] = T3 + T3 = [T3, o] + T3.

Thus,

|(lTB, 0] + T3)/T131: [(TA + T3)/TBI = [TA/(TA fl TB)| = [TAl/IIFI-

By 2.12 and 2.3, [V, a] is a hyperplane of TA containing RA and so [[V, a]| = [TA[/|lF[.
Then llTesall S |[V,all = [TAl/“Fl = |(lTB,al +TB)/TB| = llTB,al/(1TB,al ”TEN

and so [T3,a] (1 T3 = 0. It follows that [V,a] = [T B,a] and [V, a] (1 T3 = O. In

56

particular, ([V, a] 0 IV) 0 (TB (1 IV) = 0. Thus, dim([V, a] 0 IV) = dim([V, a] 0 IV) +
(W n T3)/(W 0 TB) S dim W/(IV (1 T3) = 1 since IV 0 TB is a hyperplane of W.
Therefore, [V, a] H IV is at most l-dimensional. Since R A S [V, a] D W, this gives
[V,a] (1 IV 2 RA. RB S IV so [R3,a] S [IV,a] S IVfl [V,a] = RA since a E A1.

Hence, a E A0 = 1, a contradiction. [:1

Lemma 4.14 Suppose A is a TI-set, IV is 4-dimensional, and IVflTA = [r3, A1] +
RA. Then there exists an Ll-invariant quadratic lF-for'm, qw, associated with the

symplectic form sW such that
(a) L1 induces (Z(IV, qw) on IV.

(5) (1W 1'3 0f --t:l/pe-
C IV 0 T 2 RJ- with respect to s .
A A W

{(1) Let R be a 1-dimensional subspace of IV. Then R E R if and only if (III/(R) = 0

and if and only if]? E R21.
(e) CL1(IV) = E.

Proof. (a), (b), and (c) are proven in 4.10.

(d) This is shown at the end of the proof of 4.10.

(e) We first show that A1 is a weakly closed subgroup of L1. Recall that CW (A1) =
R A so it’s l-dimensional. Let g E L1 S 1VG(IV) with [A1,Ag] S A1. Then A? S
NG(CW(A1)) = NG(RA). So RA 3 CW(A-‘,’) = 123, and thus 12A = R3,. Then
A = A9 and A1 = A? so A1 is a weakly closed subgroup of L1. From 4.10 we have

A:

Ll/CL1(RL1) ”S 9(W, (1W) g SL2(IF). So we can apply 2.21(g). Hence, E = Cl :=
L 4 10 L 2.24(3.)
CL1(A11). Then CL1(I'V) '2 CL1(RA1) S C1: E. 4.6 gives E S CL1(W) so

E = CL1(W). [:1

Theorem 4.15 Suppose A is a TI-set. Then one of the following holds:

57

1. dimA/Ar) E is even, L 2 L1, dimW = 4, and [rB,A[ + RA = IV flTA.
2. [AI = 24, E = 1, and [lFI = 2.

Proof. If NA(B) S 1, then 4.5, 4.11, and 4.13 Show that (1) holds.

So suppose NA(B) == 1. Then 4.5 gives N3(A) = 1 and A F) E = 1 = B n E
making E = 1. By 2.5, dimA Z 2 and by 4.7, dim A0 S 1. Thus, A S A0.

If A1 S A, pick a E A\A1. If A = A1, pick a E A\A0. Since A0 S A1,
we have a E A0 in either case. Put B = B, A = B“, L = (8,80) = (A, B),
E = (A n Op(L))(B fl Op(L)), and W = R3 + (TB (1 TE.) + RaB = IV(A, B). Since
[TB/CTB(a)l = [TB/(TBflCi/(GDI = |(TB+Cv(a))/Cv(a)| _<_ IV/Cv(a)| = |[V, all =
[TAl/IIFI < [TB[, we have CTB(a) S 0. Since CTB(a) S TB 0 Tg, this implies
T B (1 TE; S 0.

Suppose for a contradiction that dim A / ADE is odd. Then by 4.8, L0 := N L(R 21+
R [3) acts transitively on the l-spaces of R A + RB = R3 + 113%. Hence, there exists
9 e NL(RA + RB) with 12%, = [R3,a]. It follows that 12% 3 TA and so R9 = RA by
4.1. But then [R B, a] = R A and a E A0, a contradiction to the choice of a.

Thus, dim A/A n E is even. Since T A 0 T3 S O, we can apply 4.13 to see that
dim IV = 4 and [r3,/11] + RA = W (1 TA' Therefore, we can apply 4.14 to L. So L1
induces 9(W, qW) on IV, where qW is a non-degenerate quadratic form of ——type.
Notice that a normalizes IV and L so qW is a-invariant.

Assume that W 0 TA contains a singular l-space R. Then R 2 Bill for some
9 E L1 from 4.14(d). So R31 S TA and hence, Rfi 4S0 R A yielding A9 = A. Since
A S B and L1 is doubly transitive on the singular l-spaces in W, we may choose 9
such that R5}, = 123. Then 39 = 39 = B and A9 = A. Thus, W = W9 = W since
L1 normalizes W and then a E N A(IV ) = A1. By our choice of a this implies A = A1
and so L = L1 and (1) holds.

Next, assume that W 0 TA contains no singular l-space. Let Y be a isotropic

subspace of IV (1 TA. Let q be the quadratic form on W. Then q [Y is Z—linear. Since

58

Y contains no singular l-spaces, q [y is one—to—one. Thus, [Y] = |q(Y)[ S [IF]. This

implies that W n T A contains no isotropic space of dimension greater than 1.

Since a acts quadratically on IV, [W ,a] is an isotropic subspace of IV. Thus,
[W,a] is a non-singular l-space. Let X1 and X2 be 1-subspaces of [W,a]‘L with
[IV, a]i = X1+X2+[IV, a]. So X1+[IV, a] S X2+[IV, a]. For the following argument
let i E {1,2}. Since X,- and [W,a] are 1-dimensional, they are both isotropic. As
X,- S [II/,a]i we also have [W,a] S Xil. Thus, both X,- and [IV, a] are contained
in X27]- and in [W,a]—L. So X,- + [IV, a] S Xi“L fl [IV,a]i = (X,- + [W,a])i. Then
X,- + [IV, a] is an isotropic 2-space in W and therefore contains a singular 1-space Yi.

Since Y,- is singular, Y,- S [IV, a]. Then Y, + [IV, a] = X, + [IV, a] and so Y1 S Y2.
Also, Y1 = R0 and Y2 2 RD for some C,D E Bil with C S D by 4.14(d). So
BC + RD S [IV, a]i = CW(a) by [Asch, 22.1]. The doubly transitive action of L1 011
BE1 implies IV = W(C’, D) and L = (C, D). Since R0 + RD S [IV,a]‘L = W(a),
a centralizes RC and RD. So a normalizes C and D and we have a E N A(C) n
N A(D). 4.6 states that N A(C) and N A(D) are IF-subspaces of A, so we conclude
that Aa = IFa S NA(C)F1 NA(D) S Ng(W). Since NA(RB) 4[S3 NA(B) = 1, we
have NAa(RB) = 1 and so 4.6 gives CAa(II/) = 1 as well. Also, [Ama] = 1. Then,
since Aa normalizes W and centralizes a, it normalizes the 1-space [W,a] and so
also centralizes it giving [W ,a] S CW(Aa). This gives C'W(Aa)-L S [W,a]‘L. Thus,
[W,Aa] S [W,a]J-. Suppose that [W ,Aa] S [W ,a] and let T be a l-subspace of
[W,Aa] with [W, a] S T. Then [IV,a] + T is an isotropic 2-space in W 0 TA, a
contradiction. Therefore, [IV ,Aa] = [IV, a] is l-dimensional. Notice that this, along
with A5 implies [Aal = 2 and so [IF] = 2.

Since (B, B“) = L = (C, D), we have
TB +Tg = [V,L] = TC+TD.

Let L* = (A, C), E* = (Anop(L*))(Cno,,(L*)), and W* = RA + (TA flTC) +120.

59

Since a E NA(C), NA(C) S 1. So by 4.5(f), TA flTC S 1. By 4.11, dimA/A H E* is
even. Then 4.13 gives dim IV * = 4 and dim TAflTC = 2. Thus, dim[TC, a] S dim T on
TA S 2. Since [IV,a] is non-singular, RC S [IV,a]. Therefore, RC (1 [W,a] = O and
then IV = OJ- 2 (RC (1 [IV, a])J- = RS. + [IV, a]i = (IV 0 Tc) + CW(a). Hence,

[IV,a] 2 [(IV OTC),a] + [C,/9(a),a] S [T0,a].

By symmetry, dim[TD, a] S 2 and [IV, a] S [T0, a] H [T0, a]. Thus, dim[TB +Ta,a] =

dim[TC, a] + [TD, a] S 3. Also, dim T A = dim TE, and dim TB 0 Ta = 2 so we have

dimA —- 1 = dimTA —— 2 = dimTf’; — dim(T3 flTE»)
= dim(TE/(TB f) T1G3» = (1111431? + T3)/TB)
= dim((lTBt a] + T3)/TB) S dimITB,(II/(ITEMr1 TB)

S dim[TB, a] S dim[TB + T5, a] S 3.

Thus, dimA S 4. Observe that a E C as A is a TI-set. And since a normalizes C,
C (a) is a 2-group. It follows that C, and so also A, is not a Sylow 2-subgroup of C.
3.2 now shows that dimA S 3. Thus, dimA = 4 and so [A] 2 [IF]4 = 24. Then (2)
holds. El

60

Chapter 5
Identifying Ln(q2)

Hypothesis 5.1 In this chapter we assume that A is a TI-set, dim A/A D E = 2,
L = L1, dimIV = 4, and [rB,A] + RA = IVflTA for all A S B E A.

Lemma 5.2 IV 2 RL.

Proof. Since L = L1, 4.10 gives IV = (12511) 2 RL. El

Definition 5.3 A point is an element of A. If A and B are distinct points, then
l(A, B) = A((A, B)). Any set of points of the form l(A, B) is called a line. L: is the
set of all lines. A point A is said to be incident to a line l (or lies on a line I) if
A E l. [fl is a line, then L1 = (A [ A El).

A subset B ofA is called a subspace ofA if l(A, B) (_i B for all A S B E B.
The subspace generated by B is the smallest subspace of A containing 3; that is, the

intersection of all the subspaces containing 8. We denote this subspace by [[3].

Lemma 5.4

(a) Let 8 be a subspace ofA and A E S. Then A normalizes S.
(b) Let 0 S B Q A. Then (A((B))) = (B).

(c) Let B Q A. Then [8] = A((B)) and (B) = ([3]).

61

Proof. (a) Let B e 5. If A = B, then A fixes B and so BA g s. If B as A, then A
normalizes l(A, B). Since Sis a subspace, l(A, B) c s and so again BA g s.

(b) (B) S («4MB») S (3)-

(c) IfA, B E A((B)), then (A, B) S (B) and so l(A, B) Q A((B)). Thus, A((B)) is
a subspace of A. If D E B, then D S (B) and so D E A((B)). Therefore, 8 (_3 A((B))
and [B] Q A((B)). By (a), every element in ([8]) normalizes [B] so ([3]) normalizes

[8]. Thus,

MB» 220 M c A<131> c 131 g MB».
Then

A<<B>>=181= AW — A(asm
and

(B) = (A(W)) = (A((IBI))) = ([31). C1

Lemma 5.5 Let t = l(A, B) be a line.

(a) Ll/CLlfl) % agar) 2’ SLsa“).

(b) Ll acts doubly transitively on 1.

(c) IfC El, then C/CC(l) acts regularly on l \ {C}.
(d) L; = (C, D) and l = l(C, D) for all C S D El.

Proof. We may assume I = l(A, B) and so L) = L. By Hypothesis 5.1, 415(1) holds.
So we can apply 4.14. Thus, the map (I) : l ——> {the set of singular 1-spaces of W}
which takes C —> RC is a L—equivariant bijection. Also, the action of 94— (IF) on the
singular 1-spaces is isomorphic to the action of SL2 (F) on the l-spaces of F2. So we

have (a). 4.10 and 4.5 give us A(L) 2 BA U {A} and the doubly transitive action.

62

We also have N A(B) 4:36 A D E = C A(l) so we have (b) and (c). In fact, a line
contains exactly [BA U {A}[ 2 [IF] +1 points. By (b), (A,B) = (C, D) and so also
I = l(A, B) = l(C, D), which gives us ((1). El

Lemma 5.6 Any two distinct points lie on a unique common line.

Proof. By 5.5(d), l(A, B) is the unique line incident with A and B. C]
Lemma 5.7

(a) Let m be a line. Then m is A—invariant if and only if A E m.

( b ) Let a E A and m an a-invariant line. Then one of the following holds:

1. A E m and A is the unique fired-point ofa on m.

2. a fires all points on m.

(e) Let J Q A. Then CA(J) is a subspace of A.

Proof. (a) Suppose first that A E m and let D be a point that lies on m with A S D.
Then 5.6 gives m = l(A, D) = A((A, D)) and so m is invariant under A.

Suppose next that m is A-invariant. Let C be a point incident to m. Then both
A and C are contained in N(;(m). A = C9 E m by 2.20.

(b) We may assume that there exists D, D“ E m with D S D“ or else a fixes all
points on m. Put 1 = l(A, D). Since I is A-invariant by (a), D“ E I. Then 5.5(d) says
I = l(D, Da). Also, 5.5(c) says A/CA(l) acts regularly 011 m\ {A} and so A is the
only fixed point of a on l .

(e) Let a E J. Let C S D E CA(J). Then m = l(C, D) is invariant under a and

a has at least two fixed—points on m. Thus by (b), m g CA(a) so 7n (_3 CA(J). El

Lemma 5.8 Let A, B,C be non-collinear points. Then IV 0 TC is at most two-

dimensional and there exists D E A(L) with TD (1 TC S IV.

63

Proof. Suppose dimW 0 TC 2 3. Then IV (1 TC contains an isotropic 2—space and
then also a singular 1-space. So R3, S T0 for some g E L. But then C = A9 E A(L),
contrary to the assumption that A, B, C are non-collinear. So dim W 0 TC S 2.
Suppose next TD 0 T C S IV for all D E A(L). Since 4.13 states that TD 0 TC
is 2-dimensional, we conclude that IV 0 TC 2 TD (1 T C S Rb. But then IV 0 TC S

(RD [ D E A(L)}‘L = IVJ' = 0 and IV (1 T0 = 0, a contradiction. [3

Lemma 5.9 Suppose that NA(B) = 1. Then CO = L, R0 = IV, and RC is natural
Q;(lF)-module for C0.

Proof. Since NA(B) 4S6

A n E = 1, we have dimA = dim A/(A I) E) = 2. Thus,
dim TA 2 3 and TA = IV 0 TA S IV. Let C E A and suppose that C E A(L).
Then by 5.8, TD 0 T C S IV for some D E A(L). So D E A1 for some I E L making
TD = TAI S IV1 2 IV, a contradiction. Thus, C E A(L). Hence, A = A(L), Co = L,

and IV = RL = 130- We have RC is natural f2;(lF)-module for CO from 4.10. CI

Lemma 5.10 E = CL(IV) and L = L’.

Proof. For all A E A(L), E s NG(A) so E g Cc(RA). Then we have E g CL(W) 5:2

2.24 2.21
CL(RL) S CL(-A(L)) =
L/CL(W) 2’ SI; (IF) '5 SL2(IF) from 4.10. So L/CL(IV) is simple, L = L’CL(W),

E.

and L = L’ E. It remains to show that E S L’. Let a E A \ E. Since CB(a) S

B (1 B“ = 1, we have
013(0) = C(AnB)(BnB)(a) = (A 11 E)CBnB(a) = A I) E-

Tlius, [E/CE(a)| = [E/ADEI 2 [B0 E] = [A 0 E] and so also |[E,a][ = [A0 El.
On the other hand, a acts quadratically on E and so [E, a] S CE(a) :2 A n E. As
they have the same order, A n E = [E, a] S L’. By symmetry, B n E S L' and so

E=(AflE)(BnE)SL’. [I]

64

Lemma 5.11 Let A, B, C be non-collinear points. Then A = NA(B)NA(C).

Proof. Suppose NA(B) = 1. Then 5.9 gives A = A(L), a contradiction to A, B,C
non-collinear. Hence, N A(B) S 1 and similarly N A(C) S 1.
Suppose first that NA(B) 01VA(C) = 1. In 4.5, (g) gives dimA Z 4 while
(b), (d), and (f), together with the fact that dim IV = 4 give dim A/NA(B) =
= dimA/NA(C). So 4 g dimA = dim A/(NA(B) n NA(C)) s dimA/NA(B) +
dim A/NA(C) = 4. It follows that dimA = 4, dim NA(B) = dim NA(C) = 2 and
A = NA(B)NA(C) in this case.

So now suppose that NA(B) fl NA(C) S 1 and A S NA(B)NB(C). Let P =
[A, B, C]. Notice that NA(B)flNA(C) fixes A, B and C. Since CA(NA(B)DNA(C))
is a subspace of A from 5.7(c), we conclude that N A(B) (1 N A(C ) fixes all points in
’P and so

CAU’) = NA(B) F1 NB(C) S 1-

Put H = (A, B, C), Q = CA(’P)CB(P)CC(P), and Y = IV(A, B)IV(A, C)W(B, C).
By (>1: =1: as), (b), and (e) in 4.5 we have
CTA(NA(B)) = TA ‘1 TB + RA-
Since N A(B)N A(C) is a proper IF -subspace of A, this means
2.8
CTA(NA(B)) r1C'rA(NB(C)) = CTA(NA(B)NA(C)) S RA
and so
((TA 0 TB) + RA) r1((TA 0 TC) + RA) S RA-

Thus, there exists a 1-subspace, J, of TA with J S R A and J S ((TA (1 TB) + R A) 0
((TA OTC) + RA). So J S TB + RA and therefore J + RA S TB + RA. Since TB is

a hyperplane of TB + RA, (J + RA) 0 TB is a hyperplane of J + R A and therefore

65

l-dimensional. Since A acts transitively on the 1-subspaces of J + R A different from

RA, there exists :1: E A with J” = (J + RA) 0TB so J‘” S TB. Similarly there exists
' —-l -1

y E A with Jy = (J + RA) OTC S TC. Replacing B by B“: and C by Cy we

may assume that J S TB and J S TC. So J S TA flTB (1 Tc-

Note that

Y:=RA+RB+RC+(TAnTB)+(TAnTC)+(TBnTC)

2 IV + R0 + (TA fl Tc) + (T3 fl To).

Since J S IV, dim((TB 0T0) + IV)/W = dim(TB flTC)/(TB 0T0 H W) S dim(T3 0
TC) / J S 1 as TB (1 TC is 2-dimensional. So both ((TB 0 TC) + W) / W and similarly
((TA 0 TC) + IV) / IV are at most l-dimensional. Thus, dim Y/ IV S 3.

We will now show that By S Y. By 5.4, H = (P). In particular, P is H—
invariant. Observe that CA(P) S N0(B) and so [CA(P),B] S CB(P). It follows
that H normalizes Q. Since C A(P) S 1, we have R A S [TA, C A(P)] from 2.3. Also,
[TB,CA(P)] S TA 0TB S W(A,B) S Y and so

RA S ITATBTCv Q] S V-

2.20 says A(H) 2 AH so R11 2 (R511). Since H normalizes TATBTC and Q, this
implies that
as = <Rfil> s [TATeTa Q] s K

As L S H, we conclude that L acts on RH/IV. Since [L/CL(IV)| = [93013)], we

have that [IF]2 + 1 divides [L/CL(W)| 5S0 [L/E]. On the other hand, [IF]2 +1 does

not divide |CL3(IF)|. So if K := CL(RH/IV) S E, then

|lF|2 +1

66

divides

IL/CL(I'V)| = lL/El

divides
lL/EllE/Kl = IL/Kl
divides
IGLfRH/WH
which divides
[GL3(F)|

as dim RH / IV S dim Y/ IV S 3. This is a contradiction, so it follows that K S E.
Since L/E is simple, this means L = EK. Since E is abelian, we get L/K is abelian,
and since L is perfect, L = K. Thus, [R11,L] S IV. Let D E A(L) with D S C.
Then

TD flTC g W(D, C) nTD = [120,191+ RD 3 [RH,L] + RD 3 W.

But this contradicts 5.8. [:1

Corollary 5.12 Let A,B and C be non-collinear. Then NA(B) fires all points on
l(A, B), fixes all lines through A, and acts transitively on the points of l(A, C) distinct

from A.

Proof. Since NA(B) 4S6 A 0 E and E fixes A(L), NA(B) fixes all points 011 l(A, B).
By 5.5, A fixes all lines through A and acts transitively on I (A, C)\{A}. Also, N A(C )
fixes all points in l(A, C). As A = NA(B)NA(C) from 5.11, this implies that NA(B)
acts transitively on l(A, C) \ {A}. [:1

Lemma 5.13 Let A, B, C be non-collinear points and P the subspace of A generated
by A, B and C. Then P is a Moufang plane.

67

Proof. Let PA be set of points which lie on a line from A to a point on l(B,C).
Similarly, let PB the set of points which lie on a line from B to a point on l(A, C)

and let PC be the set of points which lie on a line from C to a point on 1 (A, B).
1° P = PA

Proof of { 1° ) We will first show that PA Q PB.

Let D be a point on l(B, C). If D = B, then l(A, D) g PB. So suppose D S B.
Then l(D,B) = l(C, B) S l(A, B) and D,A, B are non-collinear. Let F be a point
on l(D, A). So F is an arbitrary element in PA. If F = A, then F E PB, so we may
assume that F S A. By 5.12 there exists y E NA(B) with D31 = F. Since y E NC;(B)
and D lies on l(C, B), F = Dy lies on l(Cy,B). Since y E A and A normalizes
l(A,C), Cy lies on l(A, C). So F E PB. This completes the proof that PA 9 PB.
By symmetry, PB _C_ PA. Hence, PA = P3 and by symmetry, PA = P3 = PC.

Since PA is the set of points from a union of lines through A and A normalizes
every line through A by 5.7(a), A normalizes PA. Similarly B normalizes PB and C
normalizes 730- It follows that H := (A, B, C) normalizes PA = P3 = P0. Clearly
PA Q P. By 5.4(c),

73 = [A, B, e] = $100220 AH g 79,,

and so P = PA and (1°) holds. El
2° Put n = [IF[2. Then there are n2 + n + 1 points and n2 + n + 1 lines in P.

Proof of (2°). 4.10 gives [A U BA] = [IF]2 + 1 so every line contains n + 1 points.
Therefore, there are n + 1 lines from A to a point on l(B, C). Each of these lines
contains n points other than A and so there are (n + 1)n + 1 = n2 + n + 1 points
in PA = P. There are (n2 + n + 1)(n2 + 71.) pairs of points in P. Each line contains

(n + 1)n pairs of points and each pair of points uniquely determines a line, so there

68

are
(n2 + n -I-1)(n2 + n)

2
= 1
(n+1)n n +n+

 

lines. So (2°) holds. [3

We conclude from (2°) and [Ha, Theorem 20.8.1] that P is a projective plane. Let
P be a point and l a line with P E I. An elation on P with center P and axis 1 is
an automorphism of P which fixes all points on 1 and all lines in P through P. Let
q be a line in P through P distinct from 1. By definition the projective plane P is a
Moufang plane if for all such P,l and q, the group of elations with center P and axis
l acts transitively on q \ {P}.

Let R be a point on t distinct from P. By 5.12, N p(R) acts as a group of elations
with center P and axis 1 on P. Moreover, N p(R) acts transitively on q\ {P}. So P

is indeed a Moufang plane. El
Corollary 5.14 P is isomorphic to the projective plane defined over IF.

Proof. Since P is Moufang plane, [Ha, Theorem 20.5.3] shows that the ternary ring R
associated to P is an alternate division ring. Since P, and therefore R, is finite, [Ha,
Theorem 20.6.2] shows that R is a field. Since [Ha] has also shown us that [R] + 1 is
the number of points on a line, we get R 2 [IF [2 = [F]. Any two finite fields of the

same order are isomorphic and so R 9—4 F. El
Proposition 5.15 A is a projective space defined over IF.

Proof. According to the Veblen-Young axioms a projective space is set of points and

lines such that

0 Any two distinct points lie on a unique common line; and

o If A, B and C are non-collinear points and D and E are distinct points such
that A, C, E and B, C, D are collinear, then the line through D and E intersects

the line though A and B in a point F.

69

The first statement we already have in 5.6. Let P = (A, B, C). For the second,
B, C, D collinear gives us that D lies in P and A, C, E collinear gives us that E lies
in P. So the line through D and E lies in P. The line through A and B also lies in
P so it intersects the line through D and E in a point F.

Then A is a projective space. If A is 2-di1nensional, then A is generated by three
points and 5.14 gives us the result. So assume that A is at least 3-dimensional. It
follows that A is Desarguesian and therefore a projective space defined over a field

R. From 5.14 we see that R E“ IF and the Proposition is proven. CI
Theorem 5.16 Part 5’ of Theorem 2.2 holds.

Proof. For H S G’ let H I be the image of H in the automorphism group, Aut(A),
of the projective space A, and let H I be the image of H in GLIF(R0). Let I be
the subgroup of Aut(A) consisting of the identity element and all transvections with
center A.

Claim: AI = I.

Proof of Claim. Suppose for a contradiction that Al S I. Then there exists an
IF g-hyperplane, [1, of I with AI S I1. Notice that I1 contains an IF-hyperplane, 12,
of I and that 12 = CI(D) for some point A S D E A. Then AIC1(D) S 11 and
SO lA/NA(D)| = IAI/NAHUM < II/CI(D)| = IFI- Hence, IDA] < [Fl = |l(A,D)\
{A}| 5.5:“) [DA], a contradiction. Thus, AI = I and the claim holds.

Since G acts transitively on A, we conclude that C]; is the subgroup of Aut(A)
generated by the transvections. Hence, GO/CGO (A) '5 CI ’=-‘—’ PS Lm(F), where m — 1
is the dimension of the projective space A. Let :1: be a p—element in Cc(A). Then a:
centralizes all D E A so we have m E C0(RG). Thus, 001 (A) = C00 (A) / Cg(R(;) is
a p’ -group. In particular, [A, Cg(A)] S AflCg(A) S C0(RG) and cab (A) S Z (C3)
[Griess] now shows that C1 ’5 SLm(IF)/ZO for some subgroup Z0 S Z (SLm(IF)).

For an lF-subspace, X, of V put J? 2 IF ®IF X. From CV(A) 2S3 RA we get
Cf/(A) = RA. Let 2? be a non-zero FCO-submodule of RC. Then 0 S CX(A) S R4,

70

and since R A is 1-dimensional over IF, RA = C X(A) S 2?. Thus, RC = (Rio) S
2?. Hence, RC is a simple FGo-module. Let A0 < A1 < < Am_1 = A be
a chain of subspaces of A with dimA,- = i, A0 = {A}, and A1 = l(A, B). Let
P, = fl{NGO(A_7-)iL [ 0 Sj < rn—1,i S j} and L,- = OPI(P,-). Fori > 0, P,- S NGI(A)
so L,- centralizes R A- N ow L0 normalizes l(A, B) and so also L and R L- It follows that
L0 = LICL0(RL) and so RL E NO (8) N8 where N0 is a natural SL2(F)-module for
L0 and a is the field automorphism of order 2 of IF. Curtis’ Lemma [MS] now shows
that RC is uniquely determined up to isomorphism as an FSLm(F)-module and that
RG E’ N @135 N0 for some natural 1F SLm(F)-module, N. Let 2 E Z (SLm(F)). Then
2: = A >1: id for some A E IF with Am = 1. Moreover, 2 acts as AA” =1: id on N @135 N ‘7 and
so 2 E Z0 if and only if A” = A‘l. Thus ZO = {A * id | A E IF, Am =1,A"r = A'l}

and all parts of the theorem are proved. [:1

71

Chapter 6

Main Theorem

We are now able to prove our main theorem.

Proof of Theorem 2.2. \Ve have Cg(A)/A a p’-group from 2.16. Also, 2.19 states that
A is a weakly closed subgroup of C. \Ve have R A S R B from 2.23. If TA = TB, then

RA S TB and 3.1 applies making TA S TB. So we have TA S TB-

Case 1: Suppose A is not a TI—set. Then there exists A S B E A such that
[AflB] S 1. Hence, RA 2 [TA,AflB] S TB-

Case 1a: Suppose [IF] > 2. Then 3.1(a) holds. S0 [A] = [IF]2 making [A] _>_ 4, and
[A (1 B] :2 [IF]. We are then able to apply 3.2 to get A E Sylp(G'). We can also apply
3.3 to see that CO/CGO(RC;) E’ SL2(IF) or CO/CGO (RC) E“ 91(IF) and in either case
RC is the corresponding natural module. So in this situation 2.2(1) holds.

Case 1b: Suppose [IF] = 2. Then 3.1(b) holds. So [A n B] S 2, and [A] S 24. In
this situation 2.2(2) holds.

Case 2: Suppose A is a TI -set.

Case 2a: Suppose N A(B) = 1 for some A S B E A. Then we can apply 415(2)
to get [IF] = 2 and [A] = 24. This situation also gives 2.2(2).

Case 2b: Suppose NA(B) S 1 for all A S B E A. Then we can apply 415(1) to
get dim A/AflE = 2, L = L1, dim IV = 4, and [r3, A]+RA = IVflTA. So hypothesis

72

5.1 is fulfilled and we can apply 5.16. Hence, 2.2(3) holds in this case.

73

Appendix A

Background Lemmas

Lemma A.1 Let P be a finite p-group and H a finite group acting on P. If H
stabilizes a subnormal series on P, then H / CH(P) is a p-group and [P, Op(H)] = 1.
In particular, if [P, H, H, . . . H] = 1, then H/CH(P) is ap-group and [P, OP(H)] = 1.

Proof. [Gor, 5.3.3] gives the main result. In particular, if [P, H, H, . ..H] = 1, then
we have a subnormal series 1 = [P,H,H,...H] Q Q [P,H,H] Q [P, H] Q P
stabilized by H/CH(P). [3

Lemma A.2 [V, OP(C), OP(C)] = [V,OP(C),C] :2 [V,Op(C)].

Proof. [V, Op(C), Op(G)] S [V, OP(GII S V is a subnormal series stabilized by 013(0)
so it’s centralized by OP(OP(C)) = OP(C). Then [V,Op(G')] S [V, Op(C),Op(G)] S
[Vi0p(G)iGl S [V, 019(0)]-

El

Lemma A.3 IfN S C, then Ol’(C/N) = Op(C)N/N.

P’I'00f- (G/OI’(G))/(OP(G)N/01’(G)) '5 G/OP(G)N '5 (G/N)/(OP(G)N/N) by the
third isomorphism theorem. Then C/OP(C)N is a p—group since (C/OP(C)) is a
p—group by definition. Also by definition, Op(C/N) is the smallest normal subgroup
of G/N with a p—group as its quotient so Op(C/N) S OP(G)N/N. 0P(C)N/N is
generated by the 19’ elements so OP(G)N/N S OP(C/N). El

74

Lemma A.4 IfV = [V, L], then V = [V, Op(L)].

Proof. Let V = V/[V, OP(L)]. Then L/CL(V) is ap—group so ifV S 1, then CV*(L) S
0. Hence, [V, L] S V. This is a contradiction so V = 1. III

Lemma A.5 Let W be an orthogonal space. Let X S IV be a 1-dimensional non-

singular subspace. Then there exists at most one a E O(IV) with [IV, a] = X.

Proof. Let w E IV \ Xi and :L‘ E X with w“ = w + ha‘ where k S 0. So q(w) =
q(wa) = q(w) + ks(w,:r) + k2q(a:). Then k(s(w,:1:) + kq(:r)) = 0. Since k S 0,

 

k = 19M. So w“ = w — s(w’x)a: and we see that a is unique. D
(10?) 9(56)

75

Appendix B
Definitions

Let C be a finite group and p a prime for all the following definitions.

Definition B.1 A group which is abelian and all nontrivial elements have order p is

called an elementary abelian p-group.

Definition B.2 C has characteristic p if Cc(0p(C)) S Op(C). C has local

characteristic p if all the p-local subgroups have characteristic p.

Definition B.3 Q is a large p—subgroup ofG ifQ is a p-subgroup, Q S N(;(A) for
all 1 S A S Z(Q), and C0(Q) S Q.

Definition B.4 A subgroup P is a parabolic subgroup of C if P contains a Sylow
p-subgroup of C. A subgroup P containing a Sylow p-subgroup of C is a p-parabolic

subgroup of G, and P is a local p-parabolie subgroup if, in addition, 0p(P) S 1.

Definition B.5 A p-subgroup Y ofC is called p—reduced {for C) ifY is elementary
abelian and normal in C, and Op(G/C(;(Y)) = 1. The largest p-reduced subgroup of
C is denoted by Y0.

Definition B.6 Let A be an elementary abelian p-group and V a finite dimensional

CF(p)A-module. Then A is

(a) quadratic on V if [V, A,A] = 0.

76

(b) a 2F—ofi'ender on v if |V/CV(A)| g |A/CA(V)[2.
(c) non-trivial on V if [V, A] S 0.

Definition B.7 A homomorphism, 7r : C -—+ SQ is an action of C on Q defined by
a9 = 0.9”. If the kernel ofrr is I, then G acts faithfully on Q. If Kern = G, then

G acts trivially on 9.

Definition B.8 If the action is transitive and no element other than the identity

fixes any other element, then the action is called regular.

Definition B.9 If G is faithful on an elementary abelian p-group, V, and there exists
and elementary abelian p-group, A with 1 S A S G with [A|[CV(A)| Z [V], then V
is called a failure of factorization module or F F -module for C. The subgroup A is
called an offending subgroup.

Definition B.10 Op(M) is the largest normal p-subgroup of AI.

Definition B.11 Op(M) is the smallest normal subgroup of III such that III/OHM)
is a p—group.

Definition B.12 ForX Q V, Xi = {v E V [ x J. v for allx E X} where x 1 v if
s(x, v) = 0.

Definition B.13 A subspace U is isotropic if the symplectic form vanishes on U;
that is, ifU S Ui (s(u,u) = 0).

Definition B.14 A vector v E V is singular ifv is isotropic (s(v,v) = 0) and
q(v) = 0 where V is an orthogonal space and q is the quadratic form.

Definition B.15 T Q C is a TI-set ifT F) T9 Q {1} for all g E C \ Ng(T).
Definition B.16 Let C be a finite group with C’ = C. If there exists a largest group
H (unique up to isomorphism) such that H /Z (H ) ’5 C with H = H ’ , then Z (H) is
the Schur multiplier. Note that the Schur multiplier is the largest perfect central

extension.

77

Definition B.17 If we have C acting on Q, then Cw is transitive on Q \ w and
C is doubly transitive on {(w1,w2)[w1 S cog}. An action is transitive if there is

only one orbit. The action is doubly transitive if some permutation takes any pair of

elements to any other pair.

78

BIBLIOGRAPHY

[Asch] M. Asehbacher, Finite Group Theory Second Edition. Cambridge University
Press 2000.

[GL1] RM. Guralnick, G. Malle, Classification of 2F-modules, I, J. Algebra 257,
2002, 348-372.

[GL2] R.M. Guralnick, G. Malle, Classification of 2F-modules, II. Finite groups
2003, 117-183, Walter de Gruyter GmbH & Co. KG, Berlin, 2004

[GLM] R.M. Guralnick, R. Lawther, G. Malle, 2F-modules for nearly simple groups,
J. Algebra 307 (2007), no. 2, 643-676.

[Gor] Daniel Gorenstein, Finite Groups. Harper and Row 1968.

[Griess] Robert Griess, Schur Multipliers of the Known Finite Simple Groups. Bul-
letin of the American Mathematical Society, Volume 78, Number 1, January
1972.

[Ha] M. Hall JR., The Theory of groups. AMS Chelsea Publishing, 1976.
[Huppert] B. Huppert, N. Blackburn, Finite Groups 111. Springer-Verlag 1982.

[KS] H. Kurzweil, B. Stellmacher, The Theory of Finite Groups. Springer-Verlag,
New York, 2004.

[MS] U. Meierfrankenfeld, B. Stellmacher, The General FF -module Theorem, In
preparation.

[MSS] U. Meierfrankenfeld, B. Stellmacher, G. Stroth, The Struc-
ture Theorem, In preparation (see: www.math.msu.edu/~

meier / Preprints / CGP / Mb / mb.html)

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