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dissertation entitled

ON CERTAIN PUSHING-UP PROBLEMS RELATED TO
VERTEX TRAN SITIVE GRAPHS

presented by

Matthias Rassy

has been accepted towards fulfillment
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1/” COM“

ON CERTAIN PUSHING-UP PROBLEMS RELATED TO VERTEX
TRANSITIVE GRAPHS

By

Matthias Rassy

A DISSERTATION

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

DOCTOR OF PHILOSOPHY
Department of Mathematics

1998

ABSTRACT

ON CERTAIN PUSHING-UP PROBLEMS RELATED TO
VERTEX TRANSIT IVE GRAPHS

By

Matthias Rassy

Let M be a maximal subgroup of a finite group G such that no nontrivial characteristic
subgroup of M is normal in G. We will consider the problem of determining the
structure of G, in particular its action on the largest normal subgroup R of G that
is contained in M. This is related to a problem about graphs with vertex transitive
automorphism groups that has been considered by V.I. Trovimov and RM. Weiss.

The problem under consideration is similar to the ‘classical’ pushing-up problem,
where the assumption that M is maximal in G is replaced by the assumption that
M is a p-subgroup for some prime p. Using the amalgam method, which has also
been used to solve classical pushing-up problems, we will solve this problem under
some additional assumptions, mainly that the components of G / R are perfect but

not isomorphic to PSLn(q) (n E IN, q a power of a prime).

Acknowledgments

I wish to express my appreciation to my advisor, Prof. Dr. Meierfrankenfeld.

iii

Contents

1 Basic observations, Part 1

1.1 .......................................

2.1 .......................................

3.1 .......................................

4.1 .......................................

5 Determining the action of L on R, Part 3

5.1 .......................................

iv

5 3 ....................................... 113
5 4 ....................................... 117
Proof of Theorem 1 119
General Lemmas 122
A.1 Various Results .............................. 122
A2 FF-Modules ................................ 128
A.3 Modules for central products ....................... 142
A.4 Automorphisms of finite simple groups of Lie Type .......... 148
FF-modules for groups of Lie type 151
B1 Construction of the groups and modules ................ 152
B2 An ..................................... 152

32.1 ................................... 153
B3 Bn ..................................... 155

8.3.1 ................................... 156

B.3.2 ................................... 163
B4 C7, ..................................... 167

8.4.1 ................................... 168
B5 Dn ..................................... 174

B.5.1 ................................... 176

B.5.2 ................................... 183
FF-modules for alternating groups 190
CI ....................................... 190
C 2 ....................................... 192

D Examples
D. 1 .......................................

vi

Introduction

By a pushing-up problem we mean the following: Given a finite group G and a
subgroup M of G such that no nontrivial characteristic subgroup of M is normal in
G, determine the action of G on R 2: 0960 M 9, the largest normal subgroup of G
that is contained in M. Let us explain why this is called a ‘pushing-up’ problem.
The classical pushing-up situation is as follows. Given a p—local subgroup G of a
finite group X and a p—subgroup M of G with 0,,(G) g M where p is a prime,
does N X(M ) Q G imply that G is not a maximal p—local subgroup of X? (I.e.
can G be ‘pushed up’ to a larger p—local subgroup?) Note that if G is a nontrivial
characteristic subgroup of M that is normal in G, then G is properly contained in
the p—local subgroup N X(C) of X. The usual way to treat this pushing-up problem is
by determining as much of the structure of the counterexamples (i.e. G is a maximal
p—local, but N X(M ) Z G) as possible, in particular the action of G on 0,,(G), by

using the property that
(*) no nontrivial characteristic subgroup of M is normal in G.

Note that (*) depends only on G and M, but not on X. Also, (*) still makes sense
if M is not necessarily a p—subgroup of G.

Here we will consider pushing-up problems in which M is a maximal subgroup of G,
but not necessarily a p—group. This is then related to the following graph-theoretic
problem, treated by V.I. Trofimov and RM. Weiss (cf. [17]): Assume that H is a

group of automorphisms of a connected graph F with the following properties:

1. H acts transitively on the vertices of I‘.

1

2. The vertex-stabilizers H7 in H are finite and act primitively on the set of neigh-

bors of the vertex 7.

Then determine the structure of the vertex-stabilizers. Under these assumptions, if
(7, 6) is an edge of I‘, then H, (1 H5 is a maximal subgroup of H7. Moreover, there
exists an element h in the edge-stabilizer HIM} that switches 'y and 6. Since H acts
faithfully on I‘, it follows that no nontrivial subgroup of H7 DH; is normal in both H7
and H {,5}. In particular, no nontrivial characteristic subgroup of H, 0 H5 is normal

in H7. In group-theoretic terms this is equivalent to the following.
(A) G (corresponding to H7) is a finite group,
(B) M (corresponding to H, 0 H5) is a maximal subgroup of G, and

(C) there exists an outer automorphism t of M such that t2 E Inn(M) and no

nontrivial subgroup of M is invariant under both t and G.

Again consider the situation of the first paragraph with M a maximal of G. Let L
be a normal subgroup of G that is minimal subject to LR/R = Soc(G/R) where
Soc(G / R) is the socle of G / R) (i.e. the product of all minimal normal subgroups of
G/R). Then G = ML. Notice that if A is any finite group, then G 2 A and M 2 A
satisfy the same assumptions as G and M, respectively. In particular, there is no hope
to get the action of M on R under control. Hence we will consider the problem of
determining the action of L on R. In the following, assume that [R, L] ;£ 1. Note that
in contrast to the classical pushing-up problem, there is a priory no prime 19 involved
in our pushing-up situation. But using [R, L] aé 1 it can be shown that F ‘(G), the
generalized Fitting subgroup of G, is a p—group for some prime p. We will call p the
characteristic of the pushing-up problem (G, M).

By the O’Nan-Scott-Theorem (cf. [12], e.g.) we have some information about G / R.

Indeed we will Show that under our assumptions not all of the cases listed in this

theorem can occur. In particular, Soc(G / R) is a minimal normal subgroup of G / R.
If Soc(G / R) is solvable, then Soc(G / R) is an elementary abelian q-group and {p, q} 2:
{2, 3}. We will restrict to the more interesting case that Soc(G / R) is perfect.

To treat the pushing-up problem under consideration, we will use the following strat-

egy, which has been successfully applied to classical pushing-up problems (cf. [14],

e.g.):
(1) Reduction to the case that Soc(G/R) is simple.

(2) Solve the problem under the assumption that Soc(G / R) is simple, using the so

called amalgam method.

Note that step (1) is a partial converse of the extension of G to G 2A. In a pushing-up
problem with M E Sylp(G) step ( 1) is staightforward. In the graph-theoretic problem
described above this step seems to be much harder. Our situation (M maximal in G,
satisfying (*)) is somewhere in between. We will reduce this problem to a pushing
up problem in which Soc(G/R) is simple and M is ‘almost’ maximal in G, which
then can be treated by the amalgam method. (Indeed, almost always M will still be
maximal in G.)

In the amalgam method (as introduced in [7]) there is a distinguished prime involved,
which will be the characteristic p of the pushing-up problem in our case. Moreover,

one needs some information about
0 the structure of G / R, and
o FF-modules in characteristic p for G / R.

The O’Nan-Scott-Theorem gives the necessary information about G / R. Some infor-
mation about F F-modules in characteristic p for the minimal normal subgroups of
Soc(G / R) is already needed for step (1). In order to be able to make use of the clas-

sifications of FF-modules in the literature (cf. [14]( 1.2), [6], and [16]) we shall restrict

3

to the case that the minimal normal subgroups of Soc(G/R) belong to the class LP
defined as follows: £2 consists of all finite simple groups of Lie type in characteristic
2, and of all finite alternating groups of degree at least 5. If p is odd, then 5,, consists
of all finite simple groups of Lie type in characteristic p.

In step (2) we will embed G and the semidirect product M Aut(M ) into the free
amalgamated product G * M(M Aut(M )) (therefore the name amalgam method). Then
we will often meet the following situation: X is a normal p—subgroup of G and h E
G *M (M Aut(M )) such that X and X h act on each other. To draw conclusions from
situations like this, eventually leading to a description of the action of L on R, a close
relation between the commutator [X , X h] and the centralizer CX(X") is very useful.
Therefore, we shall further restrict to the case that the minimal normal subgroups of
G / R belong to the class 2,, consisting of all members of L that are not isomorphic
to PSLn(q) (n E IN, q a power of a prime).

Before we can state the result to be proven, let us explain some terminology. By a
p-component of a finite group X we mean a perfect subnormal subgroup Y of X such
that YOP(X)/OP(X) is a component of X / 0,,(X ) The numbering of the nodes in a

Dynkin diagram is always chosen as in [9]. We will prove the following result:

Theorem 1 Let G be a finite group, M a maximal subgroup of G, R the largest
normal subgroup of G that is contained in G, and L a normal subgroup of G that
is minimal subject to LR/R = Soc(G/R). Assume that no nontrivial characteris-
tic subgroup of M is normal in G, [R, L] # 1, and each minimal normal subgroup
of Soc(G/R) belongs to 2,, where p is the characteristic of the pushing-up problem
(G, M). Then L is the central product of the p-components L1, . . . , Lm ofG that are
not contained in R, and M permutes these p-components transitively. Moreover, if

i E {1, . . . , m} then one of the following holds:

1. LiR/R ’=‘=’ PSp’2n(q) (q E {2, 4}), (MflL,)R/R is a parabolic subgroup of cotype
1 in LiR/R, and [R, L,]/C[R,L,](L,-) is a natural Sp2n(q)-m0dule for Li.

4

2. LiR/R ’-—‘-’ PSp'2n(q) {q E {2, 4}), (MflL,)R/R is a parabolic subgroup of cotype
2 in LiR/R, and [R, L,] is a natural Sp2,,(q)-module for L,.

3. LiR/R E“ PSp2n(2), (MflL,)R/R is a parabolic subgroup of cotype 3 in LiR/R,

and [R, L,] is a natural Sp2n(q)-module for L,.

4. LiR/R E Sp6(q) (q = 2“), (M D L,)R/R is a parabolic subgroup of cotype 1
in LiR/R, Z([R, L,]) is a natural 07(q)-module for Li, and [R, L,]/Z([R, L,]) is

an 07(q)-spin module for Li.

5. LiR/R E’ Q§(2), (MflL,)R/R is a parabolic subgroup of cotype {3, 4} in LiR/R,
and [R, L,] is a natural 0;(2)-module for Li.

6. LiR/R ’5 Q§n(2) (n 2 5), (M 0 L,)R/R is a parabolic subgroup of cotype 3 in
L,R/R, and [R, L,] is a natural O§n(2)-module for L,.

7. LiR/R 9-1 Q§n(2) (n 2 3), (M n L,)R/R is a parabolic subgroup of cotype 3 in
LiR/R, and [R, L,] is a natural 02.n(2)-module for Li.

8. LiR/R 3 (23,,(2) (n 2 4), (M0 L,)R/R is the centralizer in LiR/R of a graph
automorphism that switches the nodes n — 1 and n of the Dynkin diagram, and

[R, L,] is a natural O§n(2)-module for L,.

.9. LiR/R E“ Qfo(q) {q = pk), (M D L,)R/R is a parabolic subgroup of cotype I in
LiR/R, and [R, L,] is an 0:0(q)—half spin module for L,.

10. LiR/R 43’ (32(2)’ and [R, L,]/G[R,L,](L,-) is an irreducible FF-module for L, with
HR, L,]] = 26.

Note that (D23) and (D24) show that in the cases 4 and 9 of Theorem 1 no further

restrictions on the field of definition of LiR/ R are possible.

In chapters 1 and 2 we derive some basic results that are needed to get the amalgam
mathod started. In particular, chapter 1 contains the proof that there exists a prime
p such that F *(G) = 0,,(G), and the reduction to the case that Soc(G/R) is simple
is done in chapter 2. One of the fundamental observations in the amalgam method
is that Z := (91(Z(0,,(M)))G) is an FF-module for G. Still in chapter 2 we show
that under the assumption that Soc(G/R) belongs to 2,, Z contains exactly one
noncentral L-chief factor.

In chapters 3—5 we then determine the action of L on R, distinguishing cases by the
isomorphism types of LR/ R, of (M D L)R/ R, and of the noncentral L—chief factor
in Z. Chapter 3 handles the majority of the cases where LR/ R is an orthogonal,
symplectic, or unitary group, (M (I L)R/ R is a parabolic subgroup of LR/ R, and Z
contains a natural module, except the case that LR/ R is symplectic and (M flL)R/ R
is of cotype 1, which is done in chapter 4. Chapter 5 treats the remaining cases, i.e.
‘exceptional’ F F-modules and the case that (M n L)R/ R is not a parabolic subgroup
of LR/ R.

In chapter 6 we summarize the results from chapters 1—5 to prove Theorem 1.
Appendix A contains some general lemmas. In Appendix B we make some concrete
calculations in FF -modules for groups of Lie type, needed in chapters 3—5. Appendix
C lists some properties of FF-modules for alternating groups. Finally, Appendix D

contains some examples for the pushing-up problem under consideration.

Chapter 1

Basic observations, Part 1

Let G be a finite group, M a subgroup of G, R the largest normal subgroup of G
which is contained in M, and L a normal subgroup of G which is minimal with respect

to LR/R = Soc(G/R). Let H be a subgroup of Aut(G), and consider G and H to be

embedded in the semidirect product of G and H. Assume that the following hold:
(I) M and every minimal normal subgroup of G / R are H -invariant.
(II) [UH is a maximal subgroup of GH.

(III) No nontrivial characteristic subgroup of M is normal in G.

(IV) [R, L] 7f 1.

Note that R is H -invariant, since both G and M are H-invariant.

1.1

(1.1.1) One of the following holds:

(a) i. LR/ R is an elementary abelian q-group for some prime q.
ii. M / R acts faithfully and irreducibly on LR/ R.
iii. Oq(G) = Oq(M) =1.

(b) i. L is perfect.
ii. L = 0 {N | N 51 G, NR/R = Soc(G/R)}.

7

Proof. If X is any M H -invariant subgroup of LR, then by (II) either X M = G or
X g M. In particular, L’ M = G or L’ g M, since L’ R is M H -invariant by (I).
First assume that L’ M = G. This implies that Soc(G/R) is perfect, and then by
minimality of L also L is perfect.

Let N be a normal subgroup of G with NR/R = Soc(G/R). Then [L,N]R/R =
Soc(G/R) and [L, N] _<_ L H N. By minimality of L, we get L = [L, N] and, hence,
L S N.

Now assume that L’ g M. Then L’ S R. Let q be a prime that divides |Soc(G/R)|.
Let X be the subgroup of LR such that R S X and X /R = Oq(LR/ R). Note that
X Z M, since X 3 G and X /R 75 1. Since X is MH-invariant, it follows that
X M = G, i.e., X = LR. Hence Soc(G / R) is an abelian q-group. A similar argument
(with X / R 2 @(LR/ R) instead of 0q(LR/R)) shows that Soc(G/ R) is elementary
abelian.

The irreducible M -submodules of LR/ R are precisely the minimal normal subgroups
of G / R, and by (1) these are also H -invariant. Hence again the same argument as
above shows that M acts irreducibly on LR/ R. By definition of R, M / R acts also
faithfully on LR/ R.

Since CLR/R(Oq(M/R)) is M H -invariant, we get similarly to the above Oq(M / R) = 1
and thus 0,,(M) = Oq(R). But then (III) implies

0,,(M) = 0.,(R) :1.
If Oq(G) 76 1, then L = Oq(G) and [L,R] S Oq(R) = 1, a contradiction to (IV). I
(1.1.2) (a) If N is a normal subgroup of G with N D M g R, then L g N.
(b) LR/ R is the unique minimal normal subgroup of G / R.

(c) F*(G) = F*(R) = 0,,(G) and 0,,(M) Z R for some prime p.

(d) [R,L] S 0p(G)-

(e) 0,,(M) n RL = 0,,(M) n 0,,(G)L.

Proof. (a) This can be proven as Lemma 1.2 in [15]. For the convenience of the
reader we repeat the argument here. Pick a subgroup T of G containing R such that
T/R is a minimal normal subgroup of G/R. Then T = (TflL)R, since R S T 5 LR.
IfT S MH, then T S MHflG = M, contrary to T g G and T Z R. Hence, by (11),
(T 0 L)MH = (T H L)RMH : TMH = GH. Therefore,

(*) G=GHnG= (TnL)MHnG= (TflL)M(HflG) =(TnL)M.

If [T H L,N n M] g R, then (*) implies that (N m M)R 31 G, a contradiction to
N n M g R. Thus

(as) [TnL,Nn M] g R.

Since [T D L, N D M] is contained in T and normalized by T n L and M, it follows

from (*) and (an) that
T: [TflL,NflM]R.

Since T/R is an arbitrary minimal normal subgroup or G/R, we get
Soc(G/R) = [L, N]R/R.

By minimality of L, this implies L = [L, N] g N.
(b),(c) If M has a component K which is not contained in R, then (K G) S CG(R)
and, by (a), L g (KG), a contradiction to (IV). Hence E(M) = E(R). Now (111)

implies

Suppose G / R has more than one minimal normal subgroup. Then, by (A.1.2)(b,c),
also GH / R has more than one minimal normal subgroup, where R :2 flgec ”(M H )9 .
Then it follows from [12] that GH / R has exactly two minimal normal subgroups X / R

and Y/ R, so in particular
Soc(GH/R) -—- LR/R.

Moreover, again by [12], M H / R 0 LR/ R is a diagonal between X / R and Y/ R, and
MH/RflLR/R (2 X/R) is perfect. Hence F(MH)R/R centralizes MH/RflLR/R.
But then F (M H )R/ R centralizes LR/ R and, hence, is normal in CH / R Since the
only minimal normal subgroups of GH / R are perfect, it follows that F (M H ) S R

and therefore
F(M) g F(MH)flG _<_ RflG = R,

a contradiction to F (M) # F (R) Hence (b) holds.

Suppose G has a component K which is not contained in R. Then, by (b),
(*) L S (KG)R S 00(3)}?-

If L is perfect, then (1.1.1)(b)(ii) and (*) imply L S CG(R), a contradiction to (IV).
If L is not perfect, then (1.1.1)(a)(i) and (*) imply that LR fl Gg(R) is a nilpotent
normal subgroup of G. Since q (as in (1.1.1)(a)) divides the order of LRfl GG(R), we

get a contradiction to (1.1.1)(a)(iii). Hence

Choose a prime p such that 0,,(R) 75 1. Then 0,,(M) # 0,,(R) by (III). Hence, by
(a), L S (Op(M)G) S CG(O,,I(R)). Suppose 0,(R) 7e 1 for some prime r ¢ p. Then
L S CG(OPI(R)O,:(R)) = CG(F"(R)) and thus

(..) [RM] 3 [RM L] s [CR(F*(R>),L1 s [F*<R),L1 =1.

10

If L is perfect, the Three-Subgroup Lemma and (an) imply [R, L] = 1, a contradiction
to (IV). If L is not perfect, then (1.1.1)(a) and (**) imply [L,L,L,L] = 1, i.e., L is
nilpotent. Since q (as in (1.1.1)(a)) divides the order of L, we get a contradiction to
(1.1.1)(a)(iii).

Hence F‘(R) = 0,,(R) and F‘(G) = F(G). Suppose F(G) Z R. Then (1.1.1)
implies that L is not perfect and F(G)R/R is a q-group (q as in (1.1.1)(a)). But then
Oq(G) 7S 1, a contradiction to (1.1.1)(a)(iii).

(d) Since 0,,(G) is not a characteristic subgroup of M there exists t E Aut(M)
such that 0,,(G)’ Z R. Then [R,Op(G)tl S Rn 0,,(M) = 0,,(R) = 0,,(G) and
L s ((0p(G)‘)G)-

(e) Put X := R 0 0,,(M)L. Let S E Sylp(X). Note that X/(Rfl 0p(G)L) is a

p-group, since
X/(R fl 0,,(G)L) = X/(X fl 0,,(G)L) E XOp(G)L/OP(G)L S
Op(M)L/0p(G)L.
Together with R F] Op(G)L :2 Op(G)(R (1 L) this implies
X = S(R fl 0,,(G)L) = SOP(G)(R n L) = S(R n L).
From (d) and (c) it follows that
[X, R n L] _<_ [R, R n L] n [0,,(M)L, R n L] g 0,,(G) n [0,,(M), R] g 0,,(0).
Therefore S S X, whence S S R. Thus S :2 0,,(G) by (c), and therefore
X = S(R n L) = 0,,(G)(R n L).

Let x E R and y E L with 333/ E 0,,(M). Then at E RflOp(M)L 2: OP(G)(Rfl L) and
hence my 6 0,,(G)(Rfl L)y S 0,,(G)L.

In the following, p denotes the prime defined in (1.1.2)(c).

11

1.2

Let G *M (M Aut(M )) be the free amalgamated product of G and the semidirect
product M Aut(M ) over M. Let F be the coset graph of G *M (M Aut(M )) with
respect to G and M Aut(M ), where we identify G and M Aut(M ) with their images
in G *M (MAut(M)). For vertices 7,)\ E I‘ let d(7,)\) be the distance between 7
and /\ in F, and write 7 ~ A if 7 and /\ are in the same orbit under the action of

G *M (MAut(M)) on F by right multiplication. Put

NW7) :2 {A e r | an, A) = k} for all 7 e r and k E N u {0}, and
A(7) :: A“)(7) for all 7 e r.

If 7 ~ MAut(M), i.e., 7 = MAut(M):r for some a: E G *M (MAut(M)), put

M7 2: MI,

Z7 1: 91(Z(Q7))-

If 7 ~ G, i.e., 7 2 Ga: for some a: E G *M (MAut(M)), put

G, := or,
R. == RI,
L, ;= L“,
Q7 = 0,407),

12

T, := 027(L7) and
CC, 2: G07(Z,/T,).
For each 7 E F put
V7 == (ZA | A E {7} U A‘Q’M), and
W1 == (ZA | /\ E {7} U A‘Q’W) U A”(7))-

Let (a, a’) be a critical pair (i.e., a and oz’ are two vertices of 1" whose distance b is
minimal with respect to Z0 Q (20/). For each i E {0, . . . , b}, we denote the element

of Al’)(a) fl A("_’)(a’) by a +i and the element of A(’)(a') fl A(”“)(a) by oz’ — i.

(MM(vmsasm.
(b) a and a’ are cosets of G.
(c) [Zm Zai] ¢ 1. In particular, (a’, a) is a critical pair.
(d) ZaZa+2 flGa+2 and UaZa+2 5310a”-

(e) Assume that for each critical pair ()1, u’) there exists a critical pair (141/)
with V E A‘2)(u) and V’ E A(”‘2)(u) fl A‘2)(u’). Then Z0Z0+2 SGO, and
ZaUa+2 flGa-

Proof. (b) If a is not a coset of G, then Z, S Z,“ S Q01, a contradiction. If
a’ is not a coset of G, then 2,, S Qar_1 S Q0), again a contradiction. (a) Since
(1.1.2)(c) implies Q, S 62;, for each A E A(a), we get QC, S GO. If Ca Q Ra, then
G0, = COMO,“ and hence Za+1 S Ga, contrary to (III).

(c) Note that (1.1.2)(c) implies that R, 0 Q; = Q,, for all 7 ~ 0: and A E A(7).
Hence the statements Za Q Q0, and [Zm Zai] are equivalent.

(d) Suppose that ZaZa+2 S Ga”. In particular, b > 2. Pick 9 E Ga+2 such that

(0+1)9 2 0+3. Then ZQZQH = 2,9,2,” S Q01, since d(a9, a’) < b, a contradiction.

13

Note that U0 Q Q01, since Z0 2 UOZOH and Zn“ S Qar. Hence similarly to the
above we get UaZa+2 SGO+2.

(e) By the assumption, we can extend the path (a, a + 1, . . . , a’) to a path (a — b+
2, a -— b + 2, . . . , a — 1, a, . . . , a’) such that (a — 2i, (1’ — 22') is a critical pair, for each
i E {0, . . . , b—g—Z}. Then (e) follows from ((1), applied to the critical pair (0+2, oz—b+2)

in place of (0,0’). I

By (1.2.1)(c) we can fix notation such that

|Za/Za n ROI] 3 |ZOI/Za, n Ra].

(1.2.2) Let V be a Ga-submodule of Z, with [Zm LO] Q V. Then Za/V is an FF-module

for G0, and Z0, acts as an offending subgroup on Za/V.

Proof. From (1.1.2)(c) and (1.2.1)(a) it follows that 20 0 R0,, = 20 fl Qar. Since
(Z0 (1 QaI)V/V S CZa/V(Za'), we get

(*l l(Za/V)/CZa/V(Za’)l S lZa/(Za fl Qa’)Vl S lZa/Za rlQo/l :

IZO/Za (1 R0,:

 

Note that (1.1.2)(b) implies
(**) CGo (Za/V) S Ra-

Now the claim follows from (*), (**), and the choice of a and a’. .

(1.2.3) Let A E A(a). Let X be an elementary abelian normal p—subgroup of M,\. Then

the following are equivalent:

(a) lXaLal=1'

(b) Each MA-submodule of X is Ga-invariant.

14

Proof. Clearly (a) implies (b). Assume that (b) holds. In particular, X is a Go,-
module. Let A/B be any Ga-composition factor of X. Since GA/B(Q,\) is nontriv-
ial and M A-invariant, (b) and the irreducibility of A/B as Ga-module imply that
[A, Q,\] S B. Since L0, S ( f“), we get [A, La] S B. Since L0 = 0"(La), it follows

that [X,La] =1. -

For all ,a ~ 0 put
5,, := {,u’ E F | (,u,u’) is critical},
SW, := {,u’ E E“ | d(u, V) + d(u,u’) = b} for all u E F, and

AW, :2 fl [2”, Zui] for all V E F with SW, ¢ (ll.

Meat...
(1.2.4) Let (u,u’) be a critical pair, A E A(u) and 7 E A‘b‘zlm) fl A‘2)(u’). Assume
that the following hold:
(i) b > 2.
(ii) G. = (2,, Mi).
Then [flgEMA Z§,L,,] = 1.

Proof. Put D :2 096MA Z3. Since b > 2 and D S 2,, we get [D, Zul] = 1. Hence the
claim follows from D S MA, (ii) and (1.2.3). I

(1.2.5) Assume that the following hold:
(i) b > 4.
(ii) For any critical pair (a, u’) there exists A E A01) with G“ = (Zflr, M A).
(iii) If (u, u’) is any critical pair and A E A(u) with G“ = (Zfll, My), then
53,, n A9) = {1/ e no) | 2.2,. 210,},
where V’ E A‘b‘2)(u) fl A(2)(,a’).

15

Then [flgeMfil Z3, L0] = 1, for some V E Ema”. In particular, AW,” S Ta.

Proof. Note that {V E A(A) | Z,,Z,1 $167.} 75 0 for all ,u N a and A E A(u), since VA

is a nontrivial characteristic subgroup of M ,\. Hence (iii) and (1.2.1)(e) imply that
(*) ZHZV $0“ for all u ~ a and V E A‘2)(u) with 5”,” 75 0.

Pick a: E L, such that G0 = (Zai, M(a+1)1> and put a—2 :2 (0+2)? By (*) and (iii),
(0: — 2,0/ — 2) is a critical pair. Pick y E La_2 such that Ga_2 = (Zn/4, M(a+1).-.y)
and put a — 4 z: a” and V := (a’ — 4)y_1“’_1. Again by (*) and (iii), (a — 4,a’ —- 4) is
a critical pair. Therefore,

:y—lx—l ,_

V E Ha—4,a—2 : :a,0+‘2'
Define
D 2: m 23.
9€Mo+1

Since ny S 25” = Za,_4, (i) implies that
(**l [nyiza’l : 1'

From (1.2.4) (with a — 2, oz’ — 2, and (a + 1)“! in place of u, u’, and A, respectively)

it follows that L04 centralizes DI”. Hence
(* =I= *) D” = D‘c S1 M(a+1)x.
Now (n), (It * *) and (1.2.3) imply that L0 centralizes Dx. In particular, D = D“c

and hence [D, L0] = 1.

(1.2.6) Let (u,u’) be a critical pair, A E A01), V E A(A) and 7 E A‘b‘2)(u) fl A(2)(u’).
Assume that the following hold:
(0 GIL : (Zu’i MAl'
(ii) Z,,Zfl SG,,.

16

(iii) [Zita Zu’l : [CQ7(Z#)tZl1'l°
Then (V, 7) is a critical pair.

Proof. Suppose that (V, 7) is not a critical pair. Then 2,, S Q, S GM. If b > 2, then
ZV E CQ7(Z,,), since d(V,u) = 2 < b. If b = 2, then 7 = u and again Z” E Q, =

CQ,(ZV), Since Q, = Q“. Now (iii) implies
[Zw 211’] S [0627(211)’ Zu’l : [21“ 211’] S Z!"

Hence Z”, normalizes ZUZ,“ contrary to (i) and (ii). I

17

Chapter 2

Basic observations, Part 2

In addition to (I)—(IV), we assume

(V) The minimal normal subgroups of Soc(G / R) belong to the class LP, as defined

in the introduction. In particular, L is perfect.

2.1

For 7 ~ (1 define
U; := [Z,, Li],

where U”. . . , LI," are the p—components of G, which are not contained in R,. If V is
a finite-dimensional GF(p)-modu1e for a finite group X, let ’P*(X, V) be defined as
in A2.

(2.1.1) (a) If N S G, and Q, S N S Ra, then LAN/N, . . .,L;"N/N are the compo-
nents of Ga/N which are not contained in Ra/N. Moreover, LgN/N #
LiN/N when i 7L j.
(b) Lang....-Lga

(c) Ma+1 acts transitively on {L},, . . . , L31}.

(d) [L3,A] S Li, for all i E {1, . . . ,m} and A E ’P*(Ga, Za/Ta).

18

(e) Let V be a Ga-submodule of 2,, with [ZmLa] Q V. Then for each
i E {1,...,m} there exists A E ’P"(GmZa/V) such that A S Q0“ and
[£3,141 Z R...

(f) CZo(LmTa) = Ta-

(s) [Za/Tm La] = 63211 UéTa/T...

(h) Lmo. = (LtflQa)-~~(L;"HQO)-

(i) LanRaz(LgnRa)-...-(L;"0Ra).

Proof. (a) Clearly LAN/N, . . . ,LZ’N/N are components of Ga/N, since Q0 S N.
Let K be a subgroup of Ga containing N such that K /N is a component of Ga/N
which is not contained in Ra/N. Then K Ra, / RC, is a component of Ga/Ra, i.e., a
minimal normal subgroup of LaRa/Ra. Choose a subnormal subgroup X of La which
is minimal with respect to X Ra /R0 = K Ra / Ra. Then (1.1.2)(d) implies that X is
a p-component of Ga. In particular, X N /N is a component of Ga/N. If X N # K
then [X, K] S N, a contradiction to XRa/Ra = KRa/Ra.

If i,j E {1,...,m} and i # j, then [LL,L{,] S Q, and, hence, [LLN/N,L{,N/N] =1,
which implies LgN/N 3:5 LgN/N.

(b) Let i E {1, . . . , m}. As above, (1.1.2)(d) implies that L0 contains a p-component
X such that X R, / R0 = LLRO/Ra. Since distinct p-components centralize each other
modulo Qa, we get X : Lg. Hence L], ~ . . . - LZ’ S La. Now the claim follows from
the minimality of La.

(c) Let {L2, . . .,L::} be an orbit of Mo,“ on {L3, . . .,L;”}. Then, by (a), (I), and
(II), N :2 L2 . . . . - LfifRa is a normal subgroup of Go, with R0, S N S LaRa and, if
{L2,...,Lff} # {Liv . . .,LZ‘}, then N 75 LaRa. Hence (c) follows from (1.1.2)(b).
(d) This follows from (a) and (A.2.1)(d).

(e) By (1.2.2), there exists A E ’P*(Ga, Za/V) with A S 20’-

19

Suppose that [L;, A] S R, for each i E {1, . . . , m}. Then (b) implies that ARa/Ra E
CGa/Ra (LaRa/Ra). Together with LaRa/Ra = F*(Ga/Ra) it follows that ARa/Ra S
Z(LaRa/Ra) = 1. Then, since A S Z0, S Q0“, we get A S R, H Qa+1 : Q, by
(1.1.2)(c). But then [Za/V, A] : 1 , a contradiction to A E ’P*(Ga, Za/V). Hence

[LENA] Q R0 for some i E {1, . ..,m}.

Now (e) follows from (c).
(f) This follows from [L0, L0,] : La, the definition of Ta, and the Three-Subgroup

Lemma.
(g) From (a) it follows that LiCa/Ga, . . .,LZ’Ca/Ga are components of Ga/Ca.

Note that (e) implies that
LLCa/Ca S (77*(Ga/Ca, Za/Ta)), for each i E {1, . . .,m}.

Now (g) follows from (A.3.3) (with (Ga/Ca,Za/TQ,L(1,Go/Ca,...,LZ’Ga/Ca) in place
of (G, V,L1,...,L,,)), (b), and (f).
(h) Put

7?: = Lax/“Lb; n Qa) ‘ - - - ' (Li? D Qa))°
From (b) and [L3,Lfl] S Q0, 0 Li, (1 L3,, for all i,j E {1, . . . ,m} with i # j, it follows
that I: is the central product of LE, . . . ,Lam. Hence 0,,(L:) is the central product of
0,,(LZ), . . . , Opal?) For each i E {1, . . . , m}, let A,- be the subgroup of La such that
(L3, n Q0) . . . . . (LZ‘ n Q0) 3 A,- and E; = 0,,(L—g). Since

(LgnQQ)-...-(L;"nQa) g A,- g (LgnQO)-...-(LQDQQ)L;,
for each i E {1, . . .,m}, and

LanQa=A1-...-Am,
we get

A: = (LinQalw-v(LZ’DQaXLLO/L‘) = (LloQa)-~--(L;"0Qa),

20

for eachiE {1,...,m}, and thus LaflQa = (LiflQa)-...-(LZ’HQO).

(i) similar to (h). .

(2.1.2) Let i E {1, . . .,m}. Then there exists A S NQ0+,(L:,) such that [Lva] Q R0,
and A E ’P*(Ga, Za/Ta). For any such A and for any irreducible GF(p)(AL;)-
submodule V of UéTa/Ta the following hold:

(a) ACA Lf,(V)/CA L;(V) E 73*(A Lin V).

(b) Li. CA 1.3, (Vl/CA Lg, (V) = F*(A Lb/CAL}, (Vll-
Proof. Choose A as in (2.1.1)(e) (with T0 in place of V). Then A S NQO+,(L:,) by
(2.1.1)(d). From (2.1.1)(a) it follows that Lg/Lf, (1 CO is quasisimple and Z(L’ /L:, 0
Ca) 2 Li, H Ra/Li, (1 Ca. Since Li, 0 Ga S CL;(V) and, by (2.1.1)(f)g)( ,[,V L’] #1,
we get

(*) CL; (V) S Re,

and LL/GLL (V) is quasisimple. Hence L; CA L; (V)/C'AL§1 (V) is a component of
ALL/CAL3(V). Assuming that L; CA Lg(V)/CAL3(V) 7e F*(A Lg/CAL3(V)), we get
0,,(A L;/CAL§'(V)) at 1, since ALL/Li, is a p—group. But 0,,(A LL/CAL3(V)) = 1,
since V is irreducible. Thus (b) holds.
From (*) and [L3, A] Q R, it follows that [V, [L;, A]] 75 1. But then [V, A] # 1, since
Lf, normalizes CA L3(V). Now (a) follows from (A.2.1)(b). I

2.2

In addition to (1)—(V), we assume
(VI) M is a maximal subgroup of G.
(2.2.1) There exists a subgroup E of Aut(LiRa/Ra) and a monomorphism

(:52 Gar/Ra—iElzm

21

such that the following hold:

(a) Inn(LgRa/Ra) S E.

(b) The following diagram commutes:

LLRa/Ra ——+ E

1

00/12,, —"+ E 2 )3",
(c) (Ga/ROW acts transitively on {E1 | x E E 2 Elm}.

(d) ((Ma+1 fl LARa)/Ra)¢’ = X f) (LLRa/ROV’ for some maximal subgroup X
of E.

Proof. Using the notation of [12], Ga/Ra viewed as a permutation group on the right
cosets of Ma+1/Ra is either of type II or of type III(a—c). If Ga/Ra is of type II, i.e.,
m = 1, then the monomorphism <25 : G / R0 —+ Aut(LaRa/Ra) given by the action of
Ga/Ra on LaRa/Ra and E :2 (Ga/Rafi satisfy (a)—(d). Hence we may assume that
Ga/Ra is of type III.

From Z0, S Q0“, (1.2.2), and (2.1.1)(a,d) it follows that Maid/Ra contains a non-
trivial normal p—subgroup Y/Ra which normalizes all components of Ga/Ra. In par-
ticular, Ma+1/ Ra does not act faithfully on the set of components of Ga/Ra, whence
Ga/Ra is not of type III(c).

Suppose Ga/Ra is of type III(a). Then (MGM/Ra) fl (LaRa/Ra) is a diagonal
in the direct product LaRa/Ra = (LiRa/Ra) x x (LERa/Ra). In particular,
(Ma+1/ R0,) 0 (LaRa/Ra) is a nonabelian simple normal subgroup of MC,“ /Ra. But

then

[(Ma+1/Ra) n (LaRa/Rala Y/Ral S

[(Ma+1/Ra) fl (LaRa/Ra)a Qa+1Ra/Ral = 1-

22

Now Y/Ra centralizes a diagonal and normalizes each factor of the direct product
LaRa/Ra = LaRa/Ra x x LZ‘Ra/Ra. Therefore,

[LaRa/Ra,Y/Ra] = 1.
Thus Y/Ra S Ga/Ra, a contradiction to (1.1.2)(b).
Hence Ga/Ra is of type III(b). Thus Ga/Ra is obtained from a primitive permutation
group of type II or III(a) by the construction described in [12]. If Ga/Ra is obtained
from a group of type (II), then (a)-(d) are satisfied. Otherwise Ma+1/ R0, 0 LaRa/Ra

is a direct product of nonabelian simple groups, and we get a similar contradiction

as above. I

(2.2.2) Let i e {1, . . .,m}.
(a) (Ma-+1 fl LgRa)/Ra is maximal (with respect to inclusion) amongst the
proper N M... +1(Lf,,)—invariant subgroups of LLR,1 / Ra.
(b) (Ma-+1 fl LgRa)/Ra is not a p—group.
(c) Qa+1 normalizes L2,.

(d) 0p((M..+1 n Lisa/R.) = (a... n Lima/R..-

Proof. Put X :2 Ma+1 fl LLRO.
(a) Let Y be a proper subgroup of LQRO containing X properly. For each j E

{1,...,m}, put
31,- := {W | a: 6 Ma“, Yr 5 LgRa}.
Then
Ga/Ra =(<3’1>/Ra >< >< <ym>/R.) (Maw/Ra)-

Since by (1.1.2)(b) LaRa/Ra is the only minimal normal subgroup of Ga/Ra, it

follows that

LaRa/Ra = (y1)/Ra x x (ym)/Ra.

23

Hence [3),] > 1 for allj E {1,...,m}. In particular, Y ¢ YJr S L2,Ra, for some
:16 E Illa“. Since L2,,Ra/Ra fl LgRa/Ra =1 ifiaé j E {1, . ..,m}, it follows that :1:
normalizes L2,. Hence Y is not N M0 +1(L2,)-invariant.

(b) Suppose that X / R, is a p—group. Then, by (a),
X/Ra 6 Sy1p(LiRa/Ra)-
By (V) one of the following holds:
(i) L2Ra/Ra ”:3 An, n 2 7 and p = 2.

(ii) L2Ra/Ra is a group of Lie type in characteristic p (including A5 (’5 PSL2(4))
and A6 (g SP4(2)'))-

In case (i) we get a contradiction from (a) and (A.1.4)(b). Hence (ii) holds. By (a),

the Cartan subgroup of LgRa/Ra is trivial and N M L2,) acts transitively on the

a+1(

nodes of the Dynkin diagram of LLRQ/Ra. Therefore,
L2Ra/Ra 9-: PSL3(2) or Lgna/Ra s: Sp4(2)',

and NMa+,(L2,) contains an element 9 which does not normalize the two minimal
parabolic subgroups of L2,Ra/Ra containing the Sylow 2-subgroup X / Ra.

Put Y := L2,NM0H(L2,). Let V be an irreducible Y-submodule of UgTa/Ta. Then
L2, Q Cy(V) by (2.1.1)(f)(g). From (2.1.1)(a) it follows that LLCa/Ca is a compo-
nent of YCa/Ca. Choose A as in (2.1.2). Then [A,L2,] Q Gy(V). Depending on
whether LgRa/Ra § Sp4(2)’ or L2,Ra/Ra E PSL3(2), either (A.2.7) or (A.2.8) (with
Y/C'y(V), RaCy(V)/Cy(V), L2,C'y(V)/Gy(V), and ACY(V)/CY(V), V) in place of
G, R, L, A, and V, respectively) shows that g normalizes the two minimal parabolic
subgroups of L2,,Ra/Ra containing X / Ra, a contradiction.

(c) follows from (b).

(d) Since R, S M0,“, the claim is equivalent to
011((1’4014-1n [Jill/(Ra fl Lb)):(Qa+1n LLXRa fl [Jill/(Ra fl LII!)

24

Clearly
(Qa+1 fl LLHRa fl [Jill/(Ra 0 Li.) S 012((Ma+10 Lil/(Ra 0 Lil)-

Let X and Y be the subgroups of L2, satisfying Q0, 0 L2, S X, R0 0 L2Jr S Y,
o,,((. 0+1 0 Lin/(Q. o L2.» = XML 0 Li.) and
0p((M.+1 o Lia/(R. 0 L2.» = Y/(R. 0 Li.)-

Note that any two distinct conjugates of X under Mo,“ centralize each other modulo
Q0. Hence X S Q0“, and it suffices to show that Y = X (R0, (1 L2,). Since L2, is a

p—component of G0, which is not contained in Ra,
(R. o ltd/(Q. o Li.) s Z(LL/(Q. o Li»).

Since Op(L2,Qa/Qa) S 0,,(Ga/Qa) = 1, it follows that (R0, 0 L2,)/(Qa 0 L2,) is a
p’—group. Hence Y/(Qa 0 L2,) is the product of (R0, (7 L2,)/(Qa 0 L2,) and a normal

Sylow p—subgroup, which must be X / (Q0, 0 L2,). I

(2.2.3) (a) (ii/10,10 LaRa)/Ra = (M,+1 n L2Ra)/Ra x >< (Ma+1 n LQROVRO.
(b) 0,,((Ma+mLaRa)/Ra) = (Qa+lflLa)Ra/Ra = (Qa+1flL2)Ra/Ra x . . . x

(QOH fl LzllRa/Rcr

Proof. (a) For each i E {1, . . .,m}, let 7r,- be the projection from LaRa/Ra onto
LLRO/Ra. Then

((Ma+1 fl LaRa)/Ra)"‘ = (Ma+1 fl L2R0)/Ra, for each i E {1, . . .,m},

by (2.2.2)(a).
(b) follows from (a) and (2.2.2)(d).

(2.2.4) Let i E {1, . . . , m}. Assume that L2,,Ra/Ra is a group of Lie type. Then one of

the following holds:

25

(a) Qa+1 0 L2, Q R0, and (ll/[0+1 fl L2,Ra)/Ra is a parabolic subgroup of
LLRa/Ra. Moreover, the following hold:

(al) N M. +1 (L2,) is transitive on the set of parabolic subgroups of L2,,Ra/Ra
which contain (MOH fl L2,Ra)/Ra as a maximal subgroup.

(a2) If LgRa/Ra is of type An and (Mm-1 0L2R0)/Ra is of type An._2, then
n = 3 and D'éTa/Ta is the exterior square of a natural SL4(q)-module
for L2,.

(a3) If L2,,Ra/Ra is a rank 2 group, then (Ma+1 fl L2,,Ra)/Ra is a maximal
subgroup of L2, Ra/Ra.

(a4) If LLRa/Ra is of type 2A", then (Ma+1 fl L2,,Ra)/Ra is not of type
2A,,_1.

(b) Qa+1 (1 L2, S Ra. In this case, L2,,Ra/Ra ”-1-“ 95,,(2) (e E {+,—}), and
UéTa/Ta is a natural Q§n(2)-module for L2,.

Proof. Assume that Qa+1 0 L2, Q Ra. Then, by (2.2.2)(d),
X := 0,,((Ma+1 n L2RQ)/RO)

is nontrivial, whence NLgno/RO(X) is a proper NMo +1(L2,)-invariant subgroup of

L2,,Ra/Ra. Hence, by (2.2.2)(a),

(Ma+1 O LbRal/Ra = NL;R../R..(X)-
In particular,

X = 0p(NLi.Ro/R0(X))-

Now [1](47.8) shows that (ll/[0+1 fl L2,,Ra)/Ra is a parabolic subgroup of L2,,Ra/Ra.
(a1) follows from (2.2.2)(a). (a2) follows from (A.2.8). (a3) follows from (a1), (a2)

and (A.2.7). Note that in SUn(p’°) the largest normal p—subgroup of the stabilizer

26

of a 1-dimensional isotropic subspace of the natural module does not contain any
offending subgroups for the natural module. This implies (a4).

Now assume that Q0,“ 0 L2, S Ra. Choose A as in (2.1.2), and let V be an irre-
ducible ALL-submodule of UgTa/Ta. Then ALL/GAL; (V) is one of the groups listed
in (A.2.2), and the intersection of the offending subgroup ACAL3(V)/CAL;(V) with
F‘(AL2,/CAL;(V)) is trivial. This implies (b).

(2.2.5) Let i E {1, . . .,m}. Assume that L2,Ra/Ra E An, for some n E ]N with n = 7

or n > 8. Then one of the following holds:

(a) UgTa/Ta is a natural Zn- or An-module and one of the following holds:
(all (Ma+1 fl Lima/Ra ’i’ 531.42,
(a2) n is even and (MQH fl L2,)Ra/Ra is isomorphic to the stabilizer in An
of a partition of {1, . . . , n} into 2-sets.

(b) n = 7, U2, is the module listed in (A.2.2)(m), and (Ma+1 fl L2,)Ra/Ra is

isomorphic to the stabilizer of {1, 2, 3, 4} in A7.

Proof. From (A.2.2) it follows that UéTa/Ta is an irreducible L2,-module. More
precisely, it is either a natural En- or An-module or the module listed in (A.2.2)(m).
Note that 2,, does not act on the latter.

Assume that not all of the automorphisms of L2,Ra/Ra induced by NMO +1(L2,) are

inner, i.e.,
NM...(L;)L;/CN,.,,,,(L,)L,(Um/T.) '2 2...

Then UéTa/Ta is a natural module by the preceding paragraph. Moreover, by
(2-2-2)(a) NMa-H(LII)(AJO+10LI1)CNMO+I(LL)L21(UcirTQ/Ta)/CNMO+1(L;)LL(UéTa/Ta) is a

maximal subgroup of NMa+1(Li,)L;/CNMO+,(L; )L2, (UéTa/Ta). Hence (a) follows from
(C.1.2).

27

Now assume that all of the automorphisms of L2,,Ra/Ra induced by N Ma +1 (L2,) are
inner. Then (2.2.2)(a) implies that (Ma+1 fl L2,)GL3(UéTa/Ta)/GL3(U2,Ta/Ta) is a
maximal subgroup of L2,/GL,-x (UgTa/Ta) ’5 An. Hence, if U2,Ta/Ta is a natural mod-
ule, then (a) follows from (C.1.2). If UéTa/Ta is the module listed in (A.2.2)(m),
then (b) follows from (C21) and [14](1.5). .

For A ~ 0 + 1 define
NT), 2: n n NM,(L:,).
pEA(A) i=1

(2.2.6) (a) Every p-component of MO“ is contained in R0, or La.

(b) Assume that lilo,“ has p—components. Then Ma“ (1 La S Ma“.

Proof. (a) Let K be a p—component of MO“ with K Q La. Then [K , Ma+1 H La] S
Q0“. Hence by (2.2.3) and (2.2.2)(b) K normalizes L2, for each i E {1,...,m}.
Since K is perfect, it follows that K induces inner automorphisms on LaRa/Ra, i.e.,

K S LaRa. Therefore
K = [K‘IK] S [K, (A[a+l n L0)Ral S Qa+1lKa Ra]-

Since K Q Q0“, we get [K, Ra] Q Q0,“ and hence K S Ra.

(b) Let u E A(a+1) andj E {1, . . . , m}. Since the p—components of Ma+1 generate a
characteristic subgroup of M0,“, there exists a p—component K of Mo“ with K Q R,“
Then K S L” by (a). Moreover, by (2.1.1)(e), (2.2.3) and (1.1.2)(d) we can choose
K such that K S L2,.

Suppose that Ma“ H La does not normalize L2,. Then (2.2.3)(a) implies that
[Ma-+1 0 La: K] g Qa-l-lK-

Hence K is a non-normal p—component of Ma“ (1 La. But by (2.2.3)(a), (2.2.4) and

(2.2.5) Ma“ H La normalizes all of its p—components. I

28

(2.2.7) Assume that p = 3 and there exist integers k1, k2, k3, In; such that the following
hold:

(i) (2110,“ F) L2,)ROQO+1/R0Qa+1) is a central product of k1 copies of GL2(3),

[:2 copies of E4, and k3 copies of SL2(3).
(ii) lZ((1“[a+1 n L;)ROQO+1/ROQO+1))l : 2k4'
Then M,+1 n L, g Lia“.

Proof. Let u E A(a +1). Put

Ala+1 i: A‘Ia+l/RuQa—Hc

Then R0 (1 L0, is an abelian normal subgroup of Mo“. Therefore

 

(4.) Ra n La 0 Ma+1n L, s Z(Ma+1 H Li)-

Assume that k1 > 0 or k2 > 0. Let X be a normal subgroup of (Ma-H flL2)(RaQa+1)
containing ROQOH such that X / ROQOH is one of the factors listed in (i). Suppose
that X has an orbit of size s > 1 on {L2, . . . , L22}. Without loss we may assume that

{L2, . . . , L2,} is such an orbit. Then by (A.1.1)(a)

 

Ma+1fl L2,’ x ... x —_Ma+1r) L";" g [Ma+1 r) L,“ X] and

 

(H) W” x x W” s [M0,+1 nL,,X,X].
Now (*) and (**) imply that

2‘22”“): I HM.“ o L...X,X]R.R.Q.+1/R.R.Q.+1|.
On the other hand, by (2.2.3)(a),

[Ala-H n LuaXaXl S [Ala-H fl LaaXl S X’RaQa+l-

29

Since k1 > 0 or 1:2 > 0, it follows that the order of the Sylow 2-subgroups of
X ’RaQaH /RaQa+1 is at least 22", a contradiction to s > 1 and the choice of X.

Hence

Since this holds for any such X and any u E A(a + 1), we get Ma“ 0 L}, S Ma“.
Assume now that k, 2 k2 = 0. Note that this implies that L2,,Ra/Ra 9.’ PSp4(3) and

(MM) (1 L2,)RO/Ra is a parabolic subgroup of cotype 1. Then
Ma+1n L; S Qa+l(Ra n La)03(CMO+1flL2, (U0 0 Za+1))'

Pick i E {1, . . .,m}. By (2.2.2)(c) Q0,“ normalizes L2,. Since R0, 0 L0, is an abelian

normal subgroup of m, also R0 0 La normalizes L2,. From
[Za+1103(CMO+10L},(Ua O Zo+1))i =
[2041, 03(CM0+10L;(U0 fl Za+1)), 03(CL10+1nL},(Ua O Za+1))l S
an O Za-I-la 03(CM0+10L2,(U01 O Za+1lll = 1

and (2.1.1)(g) it follows that also 03(CMM10LC1, (U0 0 Za+1)) normalizes L2,. Hence

Mai-H n L; S Ma+1~ I

(2.2.8) Assume that p = 2 and (Man 1’) La)RaQa+1/RQQO+1 is a direct product of
23’s. Then M,+1 n L, 3 R0,,

Proof. Let k E IN such that (MM) 0 L2,)RaQa+1/RaQa+1 is a direct product of k

copies of 23. Let ,a E A(a + 1) and put

Ala-+1 3: Ma+l/Ran+l-

Let D E Sy13(Ma+lflL,,). Let A be a 2-subgroup of Ma+1riL2, that is modulo RaQa+1

contained in an 23. Suppose that A has an orbit of size 3 > 1 on {L2, . ..,L,’,"}.

30

Without loss we may assume that {L,1,, . . .,L2,} is such an orbit. Since D is the
only Sylow 3-subgroup of MO,“ 0 L”, A normalizes D. From (A.1.1) it follows that

[[D, A]] _>_ B‘s—”k. Moreover, A acts on [D, A] by inversion. Hence
HE, A,A]l 2 3‘8“.

Together with
[R.. o L}... A] 3 [R.., L..] s Q.

we get that [D, A, A]RaQa+1/ROQO+1 contains a factor group that is a direct product

of at least (3 — 1)k copies of A3 and inverted by A. Now
[D,A,A] 3 WM), M0+1 0 La, Ma,1 0 L2,] 3 Mop,1 r) L},
implies that s = 2 and
[D,A,AlRaQa+1/RQQQ+1 E Sy13((Ma+1 fl LDRQQQH/RaQaH)

Hence A inverts the entire Sylow 3-subgroup of (MQH flL,’,)RaQa+1 /RaQa+1. There-

fore
k = 1,
and then by (2.2.4) and (2.2.5) one of the following holds:
(1) LbRa/Ra °—-’ PSL3(2),
(2) LiRa/Ra 2’ SP4(2)’,
(3) LiRa/Ra ’5 62(2l',

(4) L2,Ra/Ra E“ PSL4(2), (MQH fl L2,)Ra/Ra is a rank 1 parabolic subgroup
of LLRO/Ra corresponding to the middle node of the Dynkin diagram, and

UéTa/Ta is the exterior square of a natural SL4(2)-module.

31

Let X E Syl3([D,A,A]). From [Za+1,X, X] = [Za+1,X] and X S La, it follows that

[Za+la X] : [Zn-H H Um Ar] and
[Za+1,X] “Ta :1.

Since X is a diagonal between the Sylow 3-subgroups of Ma“ (1 L2, and Ma+1 0 L2,
and XRaQaH/RQQOH is a Sylow 3-subgroup of (.MQH r) L2,)RaQaH/ROQOH, we

get

|[Z.+1,Xl| 2 H2... 0 U... XJTt/Ttl = HZ... n UleTa/Tal’ = I[Za+1.X]l’.
Consequently,

(*) [ZO-i-ltXl = L

In particular, LLRa/Ra is not isomorphic to (32(2)’ or PSL4(2).

Suppose that L2Ra/Ra 2’ Sp4(2)’. By (...), (Ma+1flL2,)Ra/Ra is the centralizer ofa 1-
dimensional subspace in the natural Sp4(2)-module. By looking at the 5-dimensional
indecomposable Sp4(2)-module W with a 1-dimensional trivial submodule and a nat-
ural module as factor module, we get that this centralizer even centralizes the cen-

tralizer of its largest normal 2-subgroup in W. Hence
[Za+1 n UaaMa+1 Fl L31] : 1-

Note that if W" is the 5-dimensional indecomposable Sp4(2)—module with a natural
module N as submodule and a trivial module as factor module, then the centralizer
in W‘ of the largest normal 2-subgroup of the centralizer in Sp4(2)’ of a 1-dimensional

subspace in N is contained in N. Therefore
20 = UaTa.
But then

[Za+1,Ma+1fl L211] 2 [(204.10 Ua)Taa A1044 (1 Li] : 1.

32

On the other hand, Ma+1 0 L2, switches Z0,“ 0 U2 and Za+1 0 U3, a contradiction.
Hence L2Ra/Ra 1‘—_’ PSL3(2). Then U2Ta/Ta is a natural SL3(2)-module, and by (...)
Ma“ 0 L2, centralizes a 1-dimensional subspace in this module. Similarly to the

above, we get
2,, = UQTQ.

Then the case U0, 0 To = 1 leads to a similar contradiction as above. Hence U2, is a
4-dimensional indecomposable module for L2,, with a 1-dimensional trivial submodule

and a natural PSL3(2)-module as factor module. Now
(**) [Za-i-la Ma+1 O L3] = U; 0 T0,.

In particular,
llZa+1, Ma+1 H Li“ Z 2'

But also
(* * *) llZa+1 fl l])ia*’l’1cz+ln LilTp/Tpl = 2.

It follows that Ma.“ (1 L2, centralizes U2 (1 T p. If Ma+1 fl L2, normalizes L2,, for some
j E {1, . . . ,m}, then Ma“ 0 L2, also normalizes U2, and thus centralizes U22 0 T u'

Hence

T. s 02.... (M... o L3.) = T.(Z.+1o U2) - - . - - (Z... 0 U3”).
Since T, S Ma“, this implies T,‘ S Ta, i.e.,

T. = T.,
contrary to (**) and (a: a: 2:). I

(2.2.9) Assume that L2Ra/Ra g A7 and U2, is the module listed in (A.2.2)(m). Then

MO+1 n L; S Ma+1-

33

Proof. Let u E A(a + 1). Suppose that Ma“ 0 L2, has an orbit of size 3 > 1 on
{L2, . . . , L,',"}. Without less we may assume that {L2, . . . , L2,} is such an orbit. Then
AL,“ 0 L2, also permutes U2 0 20+], . . . , U; n Za+1 transitively. Then (A.1.1)(b) and

[Za+1,llda+1 O L2,] S U; 0 Z,“ imply that s = 2 and
(*) [(0120 Zed-IND)? fl Zo-+-l)a]lIa-I-1m L2,] = U; n Za+1-

Pick :1: E Ma“ fl L2, with (L2,)I 2 L2,. Note that s = 2 and (A.1.1)(c) imply that

[(U,2 fl Za+1)(U,2 fl Za+1),.llia+1fl L2,] is a diagonal between U2 D Z,“ and U3 0 Z0“.

Together with (L2,)1‘2 = L2,, (Lff’ = L2,, and (*) it follows that
[(U; o Z.+1)(Uff o Z.+1).w"]=1-
Hence 2 acts on (U,2 fl ZO+1)(U,2 fl 20,1) as an involution. Now (*) implies that
[U2 (1 Za+1,:1:]=1,
a contradiction to (C21). I
(2.2.10) M,+1 0 L2, 3 R0,.
Proof. This follows from (2.2.4), (2.2.5), (2.2.6), (2.2.7), (2.2.8), and (2.2.9). I

(2.2.11) Put
G:=Ma+1L2, and

R: 0 193,1.

gEG

Then the following hold:

(a) No nontrivial characteristic subgroup of M0,“ is normal in G.
(b) RnL2, = RanL2.
(c) {L}. R] ¢ 1.

34

(d) L2, is the unique minimal member of {N 31 G | NP/IAZ = Soc(G/R)}. In

particular, Soc(G/P) is a perfect simple group.

Proof. (a) Let C be a characteristic subgroup of 1W0“ that is normal in G. Since

fin“ is a characteristic subgroup of MO“, so is C. From
0.. = <Ma+t L3.) = <Ma+té>
it follows that C is normal in G0,. Hence C = 1.
(b) By (2.2.10), Ra fl L2, 3 Ma+1 (7 L2, 3 Ma“. Since R0 (1 L231 G, we get
R0 0 L2, 3 it n L2,.
Suppose that Ra fl L2, 73 P D L2,. Then L2, 3 it, since L2,/Ra (1 L2, is simple. But
then G0 = (Ma+1,L2,) = M0,“, a contradiction.
(c) Note that (2.2.2)(c) implies 20 S Li. Hence [L2,P] 2 [L2,, Z0] 2 U2, sé 1.

(d) Let L be a normal subgroup of G that is minimal subject to LIT/H 2: Soc(G/IAZ).
Then

[L2,, ore/rt = L2R/R.

In particular, [L2, L] Q Op(G). Since L2, is a p—component of G, this implies
L2, 3 L.

Hence, by (c),
[R, L] ¢ 1.

Let H be he group of automorphisms of G that are induced by N M... +1 (L2,). Note that
(b) and (A.1.2)(b)(d) imply that each minimal normal subgroup of G / P is isomorphic
to L2,Ra/Ra and normalized by H. Now from (2.1.1)(a—c) applied to (G, fia+1,H,L)
instead of (Ga, Ma+1,Ra,La) it follows that

o L is the product of the p-components of G that are not contained in R,

35

o 1W0“ permutes these p—components transitively.
Since L2, is a p—component of G with R 2 L2, 31 G, we get L 2 L2,. .

From (2.2.2)(a) and (2.2.11) it follows that (I)—(V) are satisfied for LTQHL; M0421,
L2,, and the group of automorphisms of fiaHL; that are induced by N Mo. +, (L2,) in

place of G, M, L, and H, respectively.

2.3

In this section we assume (1)—(V) and
(VI) m = 1,

(VII) LaRa/Ra is not isomorphic to PSL,,(q).

(2.3.1) Assume that U0 0 To 5£ 1. Then p = 2 and one of the following holds:

(a) (a1) LaRo/Ra ”E PSp2n(2k)’ for some n, k E IN with n 2 2,
(a2) (MQH fl LO)Ra/Ra is a parabolic subgroup of type Cn_1 of LaRa/Ra,
(a3) UaTa/Ta is a natural Sp2n(2k)-module for La,
(a4) an flTal 3 2k.
(b)(b1) LaRa/Ra E“ G2(2")’, for some k E IN,
(b2) (Man 0 La)Ra/Ra is a rank 1 parabolic subgroup of LaRa/Ra corre-
sponding to the node of the long simple root in the Dynkin diagram,

(b3) UaTa/Ta is the irreducible (52(2")-module described in (A.2.2)(i) for
Lon

(b4) IUa o Tol s 2".

(c) (c1) LORa/Ra E“ A2,, for some n E N with n > 2,

(c2) UaTo/Ta is a natural Agn-module over GF(2) for La,

36

(c3) |Ua flTal = 2.

Proof. First assume that LaRa/Ra is an alternating group. Then p = 2 by (VII). By
(A.2.6) UaTa/Ta is irreducible, and from [14](1.5) it follows that (c) holds.

Now assume that LaRa/Ra is not an alternating group. Let s is the number of
nontrivial composition factors of U0, regarded as a module for La. By [10], [14](1.5),

and (A26) one of the following holds:
(i) LaRa/Ra E“ PSp2,,(2k)’ and |Ua fl Tal g 2"", for some n, k 6 IN with n 2 2.
(ii) LaRa/Ra E” 62(2k)’ and an fl Tal S 2", for some k E IN.
(iii) LaRa/Ra 9—: U4(2) and Wu n Tal _<_ 4.
(iv) LaRa/Ra E“ Un(3), for some n E IN with n 2 4.
(v) LaRa/Ra E“ 92“,,(3), for some n E N with n 2 3.

Put k z: 1 if (iv) or (v) holds, and k := 2 if (iii) holds. By (2.2.4) (MQH flLa)Ra/Ra
is a maximal parabolic subgroup of LaRa/Ra. Let 2' E {1, . . . , n} be such that the
type of this parabolic subgroup is obtained by omitting the i-th node from the Dynkin
diagram. Note that in the cases (i), (iii), (iv), and (v) Q0“ contains an offending
subgroup for at least 2' natural modules only if (i) holds and 2' = 1. Moreover, even
in this case QC.“ contains no offending subgroup for more than one natural module.
(This can be seen by considering matrices, obtained e.g. from the description of the
natural modules given in Appendix B.) In particular, if 2' = 1 then s = 1 and (i) or

(ii) holds, and therefore (a) or (b) holds. Hence suppose that
i > 1.

Since (Uu H T” | ,u E A(a + 1)) is a nontrivial characteristic subgroup of M0,“, there

exists ,u E A(a + 1) such that
U0 0 Ta Q T“.

37

Put
Y := (U0 0 Ta)T#/T#.

Suppose that 2' = 2 and k = 1. By the remark above, 3 = 1. Hence Uqu/T“ contains
a unique irreducible (Ma+1 (mg-submodule, and that one is a natural SL2(p)-modu1e.
Suppose that Y does not contain this module. Then Y is an (Ma+1flL#)-complement
to UMTfl/T,1 in the L,,-submodule (Upr/Tu)Y of Zu/Tp. Since Ma“ 0 Lu con-
tains a Sylow p—subgroup of L“, it follows that Upr/Tp has an Ly-complement in
(Uqu/TH)Y, contrary to CZ,/T,(Lp) = 1.

Hence Y contains a natural SL2(p)-submodule for Ma“ 0 L“. Hence (i) and (ii) are

impossible and p = 3. By the action of SL2(3) on its natural module it follows that
(*) [U0 0 Ta,(1lla+1fl Lu)"] 5f 1.

On the other hand, (Ala-+1 fl Lfl)’ S Ra by the structure of Gal/Ra, and
(Man o L1)” 3 Claw... To),

by (1.1.2)(d). But
CR0(U01T01): CRO(Ua)a

again by (1.1.2)(d) and the Three-Subgroup Lemma, a contradiction to (2)

Hence 2' > 2 or k > 1. Let K be the p—component of Ma“ n L“ that corresponds to
the nodes 1,. . . , 2' —— 1 of the Dynkin diagram of LuRu/Rfl. Then K 3 L0, or K 3 R0
by (2.2.6)(a). In the latter case it follows from K = K’ and arguments similar to the

above that K centralizes U0. Hence in any case
(M) [U0 flTmK] = 1.

If s = 1, then it follows as above that Y contains a nontrivial K -submodule, contrary

to (**). Hence 3 > 1. In particular, (ii) does not hold. Pick an Lu-submodule X of
E ;= ((Uo n Ta)L")T,,/T,,

38

which is maximal subject to
X 0 Y = 1.

Let X1 denote the Lu-submodule of E with X 3 X1 and Xl/X = CE/X(L#). Then
X has an (Maid fl Lp)-complement in the Lu-module X(Y 0 X1), namely Y (1 X1.
Since Ma“ 0 L“ contains a Sylow p—subgroup of L,“ it follows that X has an L”-
complement in X(Y fl X1). Now (2.1.1)(f) and the choice of X imply that X1 = X,
i.e.,

CE/X(Lp)=1.

Pick an Lfl-submodule X2 of E such that X 3 X2 and Xg/X is irreducible. Then
Xg/X is a natural module for Lu- Moreover, YX/X fl Xg/X is a nontrivial K-

submodule of Xg/X, contrary to (an). I

(2.3.2) UaTa/Ta is irreducible as La-module.
Proof. Suppose that UaTa/Ta is not irreducible as La-module. Then by (2.3.1)
U0, 0 To, = 1.

Let X be an irreducible La-submodule of Ua and Y an irreducible Lat-submodule of

U01. From [X, La] 5£ 1 and La 3 (U2?) it follows that X Z Q01. Hence
(*) La: 5 (X Lo").

Note that [Y, L01] # 1 implies that CX(Y) _<_ ROI, i.e.,
(**) Cx(Y) = X “Qa' = CX(Ua’)'

By (A.2.5) X is Ual-invariant. From (VII) and (A26) it follows that there exists a
nondegenerate GF(p)-bilinear form on X that is invariant under U01. Now (A.1.5)(a)

and (an) imply that
(* =1: *) [X, Y] = [X, U01].

39

Again by (A.2.5) Y is X-invariant. But then by (>2) and (>1: :1: *)
[Ua’a La’] S [UOI7 (XLOIH S Ya

i.e., Uar = Y. Hence U0: is irreducible. But then also U0, is irreducible, a contradiction.

40

Chapter 3

Determining the action of L on R,
Part 1

In this chapter we assume (I)~(IV) and
(V) L0 is perfect.
(VI) One of the following holds:

(1) LaRa/Ra 2’ (2;,(q) (q 2 pk) for some n, k E IN with n 2 3, and UaTa/Ta
is a natural O§n(q)-module for GO.

(2) LaRa/Ra 9:“ 017,,(q) (q 2 pk) for some n, k E IN with n 2 3, and UaTa/Ta
is a natural O§n(q)-module for G0.

(3) LaRa/Ra E 92,,+1(q) (q 2 pk odd) for some n, k E IN, and UaTa/Ta is a
natural Ogn+1(q)-module for Go.

(4) LaRa/Ra ’-_‘-’ Un(q) (q 2 19’“) for some 71,]: E IN with r2 2 4, and UaTa/Ta
is a natural SUn(q)-modu1e for La.

(5) LORa/Ra E“ PSp2n(q) (q 2 pk) for some n, k E IN with n _>_ 2, and UaTa/Ta

is a natural Sp2n(q)-module for La.

(VII) (Za+1flUa)To/Ta is a singular subspace of UaTa/Ta with |(Za+1flUa)Ta/Ta| >

q.

41

Let r be the GF(q)—dimension of the singular subspace (Za+1nUa)Ta/Ta of UaTa/Ta.
Since Ma+1 H L0 is the stabilizer in La of this singular subspace, either (Ma+1 fl
La)Ra/Ra is a maximal parabolic subgroup of LaRa/Ra or case (1) in (VI) holds
and (Ma+1 fl La)Ra/Ra is a parabolic subgroup of cotype {r2 — 1, n}. When (Ma+1 fl
L0)Ra/Ra is a maximal parabolic subgroup of LaRa/Ra, let t be its cotype. Note
that (VII) implies r > 1 and hence t # I.

Let s be the number of connected components of the Dynkin diagram of (Ma+1 fl
La)Ra/Ra. Define r1, . . . , r3, corresponding to these connected components, as fol-
lows: If (Ma+l fl L0,)Ra/Ra is a maximal parabolic subgroup of LaRa/Ra, put

{1,...,t—1} if1<t<n—1
{1,...,n—2} iftzn— 1 and LaRa/Ra is not oftype D"

71 “ {1,...,n—2,n} ift=n~1andLaRa/RaisoftypeDn
{1,...,r1—1} ift=n
{t+1,...,n} ift<n—2
._ {n—1,n} ift=n—2andLaRa/RaisnotoftypeDn
72’ {n—l} ift=n—2andLaRa/RaisoftypeDn
{n} ift=n—1and LaRa/Ra is not oftype 0,,

r3 := {n} ift = n — 2 and LORD/Ra is of type D".
If (Ma-H fl La)Ra/Ra is not a maximal parabolic subgroup of LORa/Ra, put
7'12={1,...,n—2}.

For each ,u ~ 02 and A E A01) let LAM, . . . , LAM be subgroups of MAflLfl containing
L,, n R" such that, for each 2' E {1, . . .,s}, Lp,A,,-R,,/R,, is a Levi complement of a
parabolic subgroup of type r, in LflRp/Ru.

For each '7 ~ 01 define
M7) == {u E Amh) | U227 5407},
X.” :2 ([UwQA] I ,u E A(/\) n A(y))Z7 for each /\ E Ah),

X, := (X,,,\ | /\ E 13(7)),

42

Y“ :2 (U,1 | u E A(/\) (1 A(7))Z., for each /\ E 13(7) and
Y7 :: (U21 I H E A(7))27 : (Y7). I A 6 AU»-
3.1

(3.1.1) (a) U0, flTa =1.

(b) [Um U0] is an isotropic subspace of U0.

q3 if LaRa/Ra E“ 92+"(q) and r = n

2 ' N 'l'

, q if LaRa/Ra : 92,,(q) and U0: S LORO

(C) [[10100]] 0 20+] 2 ([2 if LaRa/Ra g 0311((1) and U0! S LaRa '
q else

(d) Let X be a subspace of U0, with X 0 [Ua,Qa+1] = 1. If :1: E X, then

{9 E QO-l-l l :13 E X9} : CQ0+1($)’

Moreover, if a: 72$ 1 then

q2"""l 1f (1) holds in (VI)
q2"”‘1 1f (2) holds in (VI)

|Qa+1 0 La : CQO+,m,O(.r)| 2: q2"" if (3) holds in (VI)
q2""'1 1f (4) holds in (VI)
q2n—r If (5

) holds in (VI)

(e) If A is a normal p—subgroup of MM 1, then one of the following holds:
(81) [Uana+l] : [UaaA]a
(e2) [Um A] is an isotropic subspace of U0.

(f) CQO+1([U01 Qa-HJ) = CQ0+1(Uaa U01 0 Za+l)-

(g) an fl Qa’l : an’ fl Qal-

(h) IfX is a proper MaH-submodule of U0, then UaflZa+1 S X S [Ua,Qa+1].
In particular,
(hI) pr E A(a +1), then U0, 0 Tu = 1 or U0 (1 Za+1 S T“.

(h2) If LaRa/Ra E“ 92+"(q) and r = n, then U0 0 Z0“ is the unique proper
Ma+1-submodule of U0.

43

(i) Assume that X is a proper MOM-submodule of U0, with U0 0 Za+1 75 X #
[Ua,Qa+1]. Then

(i1) LaRa/Ra E“ Q§n(q) (5 E {+, —}) and r : n —— 1,

(i2) X is a maximal isotropic subspace of U0.

Proof. (a) This follows from (VI) and (2.3.1).
(b) This follows from (A.1.5)(a).
(c) Suppose that [UmUaz] fl Za+1 = 1. Since [Ua,Qa+1,Qa+1] g Z0“, it follows

that U0: centralizes [Ua,Qa+1]. Now (a) and (A.1.5)(a) imply that
[UmUa'I S CU..(Qa+1) S Za+1,

contrary to [Um Uar] ¢ 1. Hence
lan, Ua’I fl Za+1l Z (1-

The rest of (c) follows from (B.5.1.6)(a)(c) and (B.5.2.5)(a).
(d) If :L,y E X, g E Q0“ and :1: = y9, then [y,g] E X H [Ua,Qa+1] = 1 and hence

[22, g] = 1-

Since Ma“ acts transitively on {:rC'U, (Qa+1 0 La) | :12 E U0 \ [Um Qa+1 0 La]}, it suf-
fices to check the rest of (d) for only one element a: E U0, \ [Ua,Qa+1 (1 L0]. This can
be done by a simple matrix calculation.

(g) From (A.1.5)(a) it follows that
IU0 0620’] : lCUa(Ua’)l : an 5 lUmUa’Il :
an’ : [UmUa’Il : ICUO,(Ua)| : an’ fl Qal°

(e), (f), (h), and (i) follow by matrix calculations, using e.g. the description of the

natural modules given in Appendix B. .

(3.1.2) If u E A(a +1) and j E {1, . . .,s}, then 0P(L#,a+1,j) S RflRa(Ma+1 0 La).

44

Proof. First assume that LMOHd-Ru/R,‘ is perfect. Then LMQHJ S RpK for some

p—component K of MO“. Hence (2.2.6)(a) implies that
Lu,a+l,j S RuRaLa n Ala+l = RuRa(La n Ma+1)-

Now assume that Lu,a+1,,-R,,/R,, is not perfect, and suppose that OP(L,,,O,+1,,) g
RflRaLa. Then one of the following holds:

(1) q = 2 and Lu,a+1,,R,,/R,, ’-_‘-’ 23,

(2) q = 3 and LMQHJRu/Rfl ”i“ SL2(3),

(3) q = 3 and LMOHJRfl/Ru ”E PSL2(3),

(4) q = 2 and Lu,a+1,,-R,,/R,, § U3(2) or SU3(2).

Suppose that (1) holds. Then LAG“,- induces a 23 of graph automorphisms on
LaRa/Ra. This is only possible if LaRa/Ra is of type D4, (MOM fl La)Ra/Ra is of
type A1 X Al x A1, and LAG“, permutes these three Al’s transitively. On the other
hand, the centralizer of U0 (1 Z0,“ in Ma“ 0 L0, is (modulo Ra) of type A1 X Al, and
Lp,a+1,j normalizes it. This contradiction shows that (1) does not hold.

If (2) or (3) holds, then LMQHJ induces at least an A4 of graph automorphisms on
LaRa/Ra, a contradiction similar to the above.

If (4) holds, then LAO“, induces at least U3(2) or SU3(q) of outer automorphisms

on LaRa/Ra. But Out(LaRa/Ra) is isomorphic to 23 or 02, a contradiction. I

(3.1.3) Let /\ E A(a). Assume that UQIRO fl LQRO = R0 and U0: 9; M,\. Then the

following hold:

(a) Ga : (MAaUa’>'

(b) Assume that b > 2 and p E A(/\) flA(a). Then (,u, 01’ — 2) is a critical pair.

45

Proof. (a) Note that (MM, 0 La)Ro/Ra is a maximal subgroup of LaRa/Ra,
unless LaRa/Ra § (2.2,,(q) and r = n — 1. If LaRa/Ra 2’ $23,,(q) and r = n — 1,
then U0: Z LORO implies that U0: switches the two maximal subgroups of LaRa/Ra
containing (Ma+1 fl La)Ra/Ra. Hence in any case MO“ is a maximal subgroup of
Ga. Then also M ,\ is maximal in GO, and (a) follows from U0: Z M ,\.

(b) Suppose that ([2, 01’ — 2) is not a critical pair.

Assume that [Ua,UO/] Z [U,,,Uar 0 Q0]. Then [UmUal] fl [UWUOI 0 Q0] 2 1, since
|[Ua, U01“ 2 2. Now (A.1.5)(b) implies that U01 normalizes UuUa, a contradiction to

(a). Hence
[U01 Ua’] S [Una Ua’ 0 Q0]-

In particular, [Um U01] 3 UO (1 U,“ whence U0, 0 Up is normalized by U01. But then

(a) implies that Ua = U,“ a contradiction to u E A(a). I

(3.1.4) Let /\ E A(a). Assume that UatRa fl LQRQ 75 RC, and

M W Q+linz.|<{q ifLaRa/Raa-“nmoandrz-n 2

1 else

Then the following hold:

(a) Go, = (AIMUOI).
(b) Assume that b > 2 and |[UO,QO+1] fl ZAI = 1. If u E A()\) (1 A(a) and
[U,,, Q0] = [U,,, (2,], then (p,a’ — 2) is a critical pair.

(c) Assume that b > 2. If 21 E A()\) 0 A(a) and [U,,,Qa] H [Um U01] = 1, then

(u, a’ — 2) is a critical pair.
((1) Assume that the following hold:
(i) b > 2.

(ii) If (7,7’) is any critical pair, then U7U7+2 filG, and U,” 0 Z,“ g T,,

where 7 + 1 E 13(7) 0 [Vb—”(7’) and 7 + 2 E A(2)(y) fl A(b'2)(7’).

46

Then b = 4.
(e) Assume that the following hold:
(i) b = 4.
(ii) There exists )1 E A(/\) such that (u, a’ — 2) is a critical pair.

Then U,, 0 Z), 3 Ta.
Proof. (a) Suppose that
E :2 (AIMUQI) 75 Ga.

First assume that |[Ua,Qa+1] fl Z,\| 75 1. Then LORD/Ra E“ 0;,(q) and r = 71. By
(3.1.1)(h), this implies that

[UaaQa+1] 2 U0 n Za+1

Since M A is a maximal subgroup of Ga, it follows that Ua: is contained in MA and

hence normalizes Z A. Thus
[(Ua n Za+1)(Ua n ZA): Ua’] : [U0 n Z)” Ua’] S [U02 Ua’] n ZA S U0 0 Za+l n ZA-

This means that U01 Ra / R, is contained in the largest normal p—subgroup of N La (Uafl
Za+1 0 Z ”Ra/Ra, which is a parabolic subgroup of cotype 1 of LaRa/Ra. But in
O§n(q) no offending subgroup for the natural module is contained in the largest normal
p—subgroup of a parabolic subgroup of cotype 1, by (B.5.1.4). This contradiction shows

that
(*I) U01 : [U07Q0+1] X (U0 0 ZA)

Assume that U0, is contained in M A and hence normalizes Z A. Since U0: 3 Q0“, it

follows from (*’) that

(H) [UmUa’IlemQa+1,Ua'l-

47

Note that [Ua,Qa+1,Qa+1 fl LaRa] g ZQH. Then (**) and (3.1.1)(f) imply that
[Um U0: 0 LaRa] = 1, a contradiction to UarRa fl LaRa 7L Ra.

Now assume that U0; is not contained in M A. Then M ,\ is not a maximal subgroup
of Ga. It follows that LaRa/Ra E” Q§n(q), r = n — 1, and E is the stabilizer in 0,, of
a maximal singular subspace X of U0 with U0 fl Z ,\ 5 X. Moreover, as in the proof
of (3.1.3) it follows that U0: 3 LORO, since MO“ is not a maximal subgroup of Ga.

Note that (B.5.1.5) implies that
Ua’Ra/Ra O Z(Qa+1Ra/Ra)7£1,

since U0: acts as an offending subgroup on U0. Pick 21 E U01 \Ra such that [12, Qa+1] g
Ra. Then [Um u] is a subspace of UaflZa+1 with |[Ua, u]| 2 q2. Since [Ua,Qa+1,u] = 1,

it follows from (*’) that [Um u] g X. But then
(Ua O Za+1) O (Ua O ZA) ¢ 1,

since |X : Ua fl Z,\| = q. Now (*') implies that
Ua O Za+1 Z [Um Qa+1I,

a contradiction.
(b),(c) Suppose this is false. First assume that [U,,, Q0] 2 [U,,,Q,\] and [Ua,Qa+1] D
Z,\ = 1. Then (A.1.5)(a) implies that CU,(QO) = U” n ZA. Hence

Uana =UpflUaflZA.
Together with [Um Qa+1l D Z A = 1 we get
(**) [CUa'(U0)’ Ufll fl [U05 U0] S [Q09 Up] 0 [Ua’a UaI S UH H [Um Qa+1I :1:

Since ([2, a’ — 2) is not a critical pair, U” S Qa:_2 _<_ Gal. Then it follows from (H)

and (A.1.5)(b) that

[Uar, UflUa] S UflUa.

48

Now p E A(a) implies that LO, Q (MA, U01), contrary to (a).

Now assume that [U,,, Q0] H [Um U01] 2 1. It follows that
[CUOI (U0), Up] 0 [Ua’a U0] : [Ua’ fl Qua Up] 0 [Ua’a U0] : 11

and we get the same contradiction as above.

(d) Suppose that b > 4. Pick ha E L, such that A = (a + 1)"°. Put a - 1 := A and
a — 2 := (a + 2)"°. If U0+2 (1 Ua # 1, then Ua+2 0 Z,“ 3 U, by (3.1.1)(h), contrary
to (ii) and (3.1.1)(a). Thus

Ua+2 fl U0 2 1.

Note that (*) is satisfied for A9 in place of A for each 9 E Ma“. Together with
Ua_2 (1 U0, = (Ua+2 (1 U0)"(1 : 1 and (c) it follows that ((a — 2)9, a’ — 2) is a critical
pair for each 9 E Ma“.

Since (a — 2, a — 1) is conjugate to (a, a +1), there exists a — 3 E A(a — 2) such that

if LaRa/Ra ’5 (23",,(q) and r = 71
else '

(=1!) HUG—2.620-110 z._.| s { 3
Pick ha_2 E La_2 such that a — 3 = (a — 1)"°-2. Put a — 4 := aha-2. As above we get
Ua_4 flUa_2 = 1. Note that (*’) is satisfied for ((a — 3)9, (a — 2)9, (a — 1)9) in place of
(a - 3,a — 2, a — 1). Hence, if g E Ma“, then (c) implies that ((a — 4)”, a’ —— 4) is a
critical pair, provided Uar_2Rf,_2flLf,_2Rf,_2 76 R2514. Note that a—4 E A(a—2), since
a E A(a + 2) and (a — 4, a — 2) = (Ci/“”0“”, (a + 2)"°"°-2). Hence, ifg E Ma“, then
(3.1.3) implies that ((a—4)9, a’—4) is a critical pair, provided Uat_2Rfl_2flLf,_2Rf,_2 2

R24. For each 9 E MO,“ define

Y ,___ U2-.. if[U£_4,Ual_4]SZZ_3 _
9 [Ug-.,c2.-.1 ir1Ug_.,U.1-.1zz:_.

Let A be a set of representatives for the cosets of QC, (1 L0 in Qa+1 H La. Let E be a

complement to U0 fl 20,11 (1 Za_1 in U0, fl Za_1. Hence, by (alt),
We 0 Za_1:E|§ q and

49

E = U0 0 Za_1 unless LaRa/Ra E’ (23,,(q) and r : r1.

Choose 6 E IN such that
qe = min {|[Ya, Ua:_4] fl Ea] | a E A}.
Define

X, :2 {(1:,a) |19é :1: E [Y,, U014] 0 E“} for each a E A and

X :2 U[Y,,U,,,_,]nEa\{1}.

aEA

Then

I U Xal = Z IXal Z |A|(qe -1)-

aEA aEA

Note that, by (3.1.1)(h), [Ua,Qa+1] 2 U0 0 Z0“, if [UmQaH] (1 20-1 76 1. Hence
E“ F) [Ua,Qa+1] =1 for each a E A.

Therefore by (3.1.1)(d) each :1: E X is contained in at most q’2"+"+1|A| of the sub-
groups [Ya,Ua1_4] H E“ with a E A, if (I), (2) or (4) holds in (VI), and in at most
q‘2"+"]A| of these subgroups, if (3) or (5) holds in (VI). Hence

IUXa|= 2|{GEA|(I,G)€Xa}|S

aEA xEX

{ q‘2"+"+1|A| |X| , in the cases (1),(2) and (4)

q‘2"+"|A| |X| , in the cases (3) and (5)
Therefore,
(qe — 1)q2""“1 , in the cases (1),(2) and (4)
(**l le Z { (qe _ 1)q2"—' , in the cases (3) and (5)

Suppose that b > 6. Then

W:=(Ya|aEA)

50

acts quadratically or trivially on Uar_4, even if b = 8. By (A.1.5)(a), it follows that

[W, U014] is an isotropic subspace of Uar_4. Hence
(:1: =1: *) IX] S |[W,Ua/_4]| -1 S q" — 1.

From (an) and (a: * *) it follows that r = n, e = 1 and in (VI) neither (3) nor (5)
holds. Since r = n is impossible in case (2), either (1) or (4) holds. Also r = 71 implies

that
[U3_4, U014] g Zf,__3, for each a. E A.

In particular, Y, = Ug_4 for each a E A. From (ii) we get
Ua—4 O 20—3 = (Ua O Za——1)h°_2 = Ua O 20—1

and hence
[Ym U014] 3 U0, 0 22,1, for each a E A.

Therefore (3.1.1)(c) gives a contradiction to e = 1 in case (1), even if E 75 U0, 0 20-1.
Hence (4) holds and

an’—4/Uo’—4 fl Qa—4l S q < (12 : HUG—4’ Ua’-4H : [Ua—4/CUO_4(Ua’—4)|a

a contradiction.

Hence b = 6. In particular, Yag : Yag, for all (2,9 E Ma“. Thus
Y 3: ([YhaUa+2] l h E Ma+1>

is an Meal-submodule of U0”. On the other hand, since Ua_4 fl Za_3 2 (U0 0
Zo,_1)""—2 g T04 and a is conjugate to a — 4 under La_2, Y is an Mom-submodule
of U0. Moreover, (=2) implies that this submodule is not contained in [Um Qa+1], since

U0 0 20-1 2 Ua_.4 (1 20-3 and

3 ' 5:! + _
I[Y1,U..+2] r1 z,_,| 2 { <1 If LaRa/R. _ 92,,(q) and .~ _ n _
q else

51

But [Um Qa+1l is the unique maximal MGM-submodule of U0, by (3.1.1)(h). Hence
U01 = Y S Ua+2.

a contradiction.

(e) Suppose that U” (W Zr 2 Ta. Then

by (3.1.1)(h). Put
E 3: (([U,,, Ua+2l O leMafllTa-

Note that E 5 20,12 and so Ua Q E. Hence (3.1.1)(h) implies that
E _<_ [UaaQa-HITQ-

Together with (*) it follows that

l(lU/11Ua+2] O Z),)Ta/Ta| S

q if LaRa/Ra ’5’ (23,,(q) and r = n
|[UaaQa+1]Ta/Ta fl ZA/Ta] S { 1 8188 -

Therefore,

' 2 + _
(* * *) llU...U..+2] n Zl| s { ‘2 glider/Ra - 92.01) and r — n

by (401:). Since (3.1.1)(g) implies that (a, 01’) is an arbitrary critical pair, we can apply

(3.1.1)(c) to (21,0 + 2) in place of (Oz, 0’) and get a contradiction to (=1: :1: *). I

(3.1.5) Let A E A(a) and p E A(A) such that U“ 0 Z,\ 5 To.

(a) 02(L,,.,1) g 12.12,.

(1)) Up S ”(Ba)-

52

Proof. Put A 2: OP(L,,,,\,1)R,,RQ. Then (3.1.2) implies that
A : RuRa(A (1 La).

Note that B :2 LMMQ), is a normal subgroup of MA with CB(U,, fl ZA) 2 Q).
Therefore, by U, 0 Z A 3 To,

A = [A, L,,,,1]R,Ra = [A n La, L,,.,1]R,Ra g CB(U,, r1 2912,12,, = 62,12,120.

Since A/RflRa has no nontrivial p—factor group, we get A = RpRa. Hence (a) holds.
Note that r > 1 implies that [U,,,OP(L,,,A,1] Z [UwQA]. Therefore, by (3.1.1)(h),

Uu : [Um <0p(Lu./\,1)MA>I'

Hence (b) follows from (a). .

3.2

In addition to (1)—(VII) we now assume
(VIII) If LaRa/Ra E Q§n(q) (E E {+, —}), then r < r1 — I.

In particular, Ma+1 is a maximal subgroup of Ga.

(3.21) (a) Q0.“ S QaLa-

(b) If LaRa/Ra ”_- Q§n(q) (E E {+, —}), then 72 Z 4.

Proof. (a) Suppose that Q0“ Z QOLO. Then QC,“ g RaLa by (1.1.2)(e). Pick a E
Qa+1 \RaLa. Then a induces an outer automorphism on LaRa/Ra witch centralizes
the parabolic subgroup (MGM fl La)Ra/Ra modulo its largest normal p—subgroup.
Hence (A.4.1) applies. Since a is a p-element, case (a) of (A41) does not hold. Thus
a induces a graph automorphism that fixes each node in the Dynkin diagram that
belongs to (Ma-+1 flLa)Ra/Ra. But then (VIII) implies that this graph automorphism
fixes each node in the Dynkin diagram and hence is trivial, a contradiction.

(b) Since r > 1, this follows from (VIII). .

53

(3.2.2) Let ('y, 7 + 1,. . .,y + b) be a path such that (y, 7 + b) is a critical pair. Then

there exists 9 E L, such that the following hold:

(a) The intersection of the parabolic subgroups (M,+1flL,)R,/R, and (M,+lfl
L,)9R,/R,, of L,R,/R, is the product of a Cartan subgroup and a Levi

complement in both of them.

(b) If b > 2, then (12,7 + b — 2) is a critical pair, for all :1: E Q,+1 and V E

A((7 +1)9:t)n AW)-

Proof. Note that (3.1.1)(g) implies that (a, 01’) is an arbitrary critical pair. Hence it
suffices to prove this for (7, 7’) = (01,01’)

Since (M,+1 fl L,)R,/R, is a parabolic subgroup of L,R,/R,, there exists a root
system <I> for L,R,/R, such that

(Mm r1 Lam/R. = QLH,

where Q is the product of the root subgroups corresponding to the positive roots
which are not contained in the root system \II spanned by the simple roots that
belong to the nodes 1,. .. ,t —— 1,t + 1,. . .,n of the Dynkin diagram, L is generated
by the root subgroups corresponding to the roots in ‘11, and H is a Cartan subgroup
in the normalizer of the product of the root subgroups corresponding to the positive
roots.

Let Q‘ be the product of the root subgroups corresponding to the negative roots
which are not contained in \II. Then Q‘LH is a parabolic subgroup of L,R,/R, that

has the same type as QLH by (VIII). Hence there exists 9 E L, such that

(M,+1r1 L,)9R,/R, = Q‘LH.

Then (a) holds.

Assume that b > 2. Pick :1: E Q0“. Note that by (a)

(*) [U72 Q7+1] fl Z’gil = 1.

54

Let 1/ E A((7 +1)9)fl A(7). From (3.1.4)(b) and (*) it follows that (V, 7 + b -— 2) is a

critical pair, provided [Um ,+1 — -.[U,,,Q,] Assume that
[Um 3+1] 75 [Uval
Then (VIII) and (3.1.1)(i) imply that
[Ur/1627] 2 UV O Z3“
Together with (*) it follows that
[UV1Q7IO [U71U7+bl _<— lUmQ7I O [U71Q7+1I <Z 9+1 O [U71Q7+1I_ *1
Now (3.1.4)(c) implies that (V, 7 + b — 2) is a critical pair. I

Choose go, E La such that (a) and (b) of (3.2.2) are satisfied for (a, a’, go) in place of
(7,7 ,9.) Define

a —1:=(a + I)“ and
a—2 :2 (a+2)9°
(3.2.3) b _<_ 4.

Proof. Suppose that b > 4. Note that (3.2.2) and (1.2.1)(e) imply that if (7,7’) is
any critical pair and p E A(2)(7 )0 WA“ 2)(7 ’,) then U, U ilG, Hence it follows from
b > 4 and (3.1.4)(d) that there exists a critical pair (7, 7’) such that U,, D Z; Z T,,
where A E A(7) 0 A“ 1)(7 ’) and 12 E A (”(7 )0 A“ ”2 (7 ’.) Since (3.1.1)(g) implies

that (a,a’) is an arbitrary critical pair, we may assume that this is the case for

(a,a’) = (7, 7’). Then, by (3.1.1)(h),
(*) (10.2 H To =1.

Let A be a set of representatives for the cosets of QC, in Q0“. Choose 6 E IN such

that

qezminUan—U2a 01’ -Za:2In a_.1|ICl€A}

55

Define

X0 =,U2,‘{(zo)|19é;1:E[_2, U0 _3‘2]flZ_ }, foreachaEA,and

:UUla U’a —I2OZa—1\{1}

aEA

Then

I U Xal = Z lXal 2 WM" -1)-

aEA aEA

On the other hand, (3.2.1)(a), (3.1.1)(d), and (*) imply that each :1: E X is contained
in at most q‘2"+"+1|A| of the subgroups [U _2, U0: -2] with a E A, if (1 ), (2 ) or (4)
holds in (VI), and in at most q’2"+"|A| of these subgroups, if (3) or (5) holds in (VI).

Hence

IUXa|=

aEA

Z Mae A I (w) EX.}| s

zEX

q"2"+’+1|A| |X| ,in the cases (1),(2) and (4)
q‘2"+"]A| |X| ,in the cases (3) and (5) '

Therefore,

(qe _ 1)q2"—'“1 , in the cases (1),(2) and (4)
(**) [XI 2 [ (qe ._ 1)q?"—r , in the cases (3) and (5) '

Since b > 4, Va acts quadratically on U014. By (A.1.5)(a), it follows that [V0, U014]

is an isotropic subspace of U014. Hence
(***) |X|<|lVa Ua —2l|-1<q” -1

From (21*) and (>1: >1: *) it follows that r = n, e = 1 and in (VI) neither (3) nor (5)

holds. By (VIII) and r = n we get that (4) holds. Then 6 = 1 implies that
[Uo’—2/Ua’—2 fl Qa—2l S q < Q2
= [[Ua—2a Ua’—2][ : [Ua—2/CUO_2(Ua’—2)| : [Ua—2/Ua-2 n Qa’—2lv

a contradiction. .

56

Choose ga_2 E La_2 such that (a) and (b) of (3.2.2) are satisfied for (a—2, a’—2, 90-2)

in place of (7,7’,g). Define

a — 3 z: (a —1)9‘“2 and
02 — 4 2: 019°”.
(3.2.4) Assume that b = 4. Let 2' 6 {1,2}, j E {1, . ..,s} and p E A(a — 21+1).

(a) O”(L#,O_2.-+1J-) S RyQa_2,+1(Ma_2,-+1flMa_2,-+3). Hence all p’-elements of
some Levi complement of (1\Ia_2,+1flLp)Ru/R# in LuRy/R,‘ are contained

in (1 a—2i+1 fl A/IO_21‘+3)Ru/R#.

(b) 0P(La—2i+4,a—2i+3.j) S Ra—2i+4Qa-2i+3(Ma—2i+1 fl Ma—2i+3)- Hence all P"
elements of some Levi complement of (Ma_2,-+3 fl La_2,-+4)Ra_2,+4/Ra_2i+4
in Lo—2i+4Ra-2i+4/Ra—2i+4 are contained in (Ma—2H1 fl Mo—2z‘+3)Ra—2i+4/

Ra—2H-4-

Proof. Since the choice of 90-2,” implies that Ma_2,+1 fl La_2,-+2 S (Ma_2,-+1 fl

Ma_2,-+3)Qa_2,-+1, it follows from (3.1.2) that
0P(Lp,a-2i+l,j) _<_ Ran—2i+1(Ma—2i+1 fl Ma—2i+3)'
Hence (a) holds. The proof of (b) is similar. I
(3.2.5) Assume that b = 4. Let 2' 6 {1,2} and ,u E A(a — 22' + 1) fl A(a — 21' + 2).

(a) Up 0 Za—2i+l : [Um Ua—21+4]-
(b) U); n Za-QH-l S Ta—2i+2-
(C) Op(Lu,a—2i+l,1) S RpRa—QH-Q'

(d) Qa—2i+1 = QuQa—2H—2o

57

Proof. (a) By (3.2.2), (p,a — 22' + 4) is a critical pair. Hence Ua_2,-+4Ru/R,, is a
nontrivial elementary abelian p—subgroup of (12104,“ 0 L#)R#/R,,. Moreover, (3.2.4)
implies that Ua_2,~+4R,,/R,, is normalized by all p’—elements of some Levi complement
of (Ma_2,-+1 fl L,,)R,,/Rp. Since 2' > 1, it follows that (a) holds.

(b) Note that we can apply (3.1.4) with 02,0 — 22' + 1, a — 22+ 2,0’ — 22' + 2) in
place of (u,a,a’). Hence (b) follows from (3.1.4)(c).

(c) This follows from (b) and (3.1.5).

(d) Note that r > 1 implies that
Qo—22+1Rp = [Qa—QH—la OP(L;1,a—2i+l,1)]Ru-

Also, by (c),
[Qa—zz‘H,0P(Lp,a—2i+1,1)l S [Qa—22+11Ra~22+2Rp] S [Qa—2i+1aRa—2i+2lRu S
Qa—22+2Rp°

Therefore,

Qa—22'+1 = Qa—2i+2(Rp fl Qa—Zi-H = Qa-ZH—2Qp'

(3.2.6) Assume that b = 4. Then [Xm Q0] S T,,.

Proof. From (3.2.5)(b) we get [X0,0_1,Qa] S To. Now the claim follows from the

definition of X0. I

(3.2.7) Assume that b = 4. Let ,u and u be vertices with )2 ~ Q ~ V and d(u, V) = 2.
Then X” S 62,, or XV S Q“.

Proof. Suppose that X p Z 62,, and X, 2 Q”. Let A be the common neighbor of p

and V. Without loss we may assume that
IXu n 62.43122 n 62.11.

58

Note that (3.2.6) implies that Xfl/Zu is a module for Gu/Qp.
Suppose that [XwLu] S Z“. Then Xu 2 X”. Now b > 2 implies X“ S Q”, a

contradiction. Hence
[Xw Lu] 2 Zu-

From this we get that CG#(X,,, Z“) S R”, and so
Cxu(Xw Zu) = XV fl Qu-

By (3.2.6),
X22 0 Qu S CX,.(XV, Z”).

Hence Xfl/Zp is an FF—module for (ifi/Q“, and X, acts as an offending subgroup on
X ,, /Z,,.

Let E be a subgroup of [X,“ L#]Z# containing 2,, such that E /Z,, is a maximal LuRu'
submodule of [X#,L,,]Z,,/Z,,. Then [X,,,L,,]Z,,/E is an irreducible FF—module for
LfiR“, and XV acts as an offending subgroup on [X,“ LylZfl/E.

Suppose that V ¢ A02). Then (V, p) is not a critical pair, for any p E A‘le), whence
V” S Q”. In particular, X,, S Q”, a contradiction. Hence V 6 A02). Now (3.2.5)(d)

implies that
0”(Lu,1,1) S RuRp'

Suppose that lle Ln],OP(LW\,1)] 2 ER”. Then it follows from OP(L,,,,\,1) S RVRM
(1.1.2)(d), and (A31) that [X,“ L,,]Z,,/E, regarded as Lp-module, is the direct sum of
at least 2' irreducible submodules. But then Q ,\, being the largest normal p—subgroup
of the stabilizer of an r-dimensional singular subspace of the natural module, contains
no offending subgroup for [X m L,,]Z,1 / E , a contradiction.

Hence O”(L,,,,\,1) centralizes [X,“ L#]Z,,/ E modulo RV. Note that

X22 S CQA (QM Q11),

59

since X” S M,\ and, by (3.2.6), X; S Q”. Now the structure of CQA(QA,Q,,)/Qu as
OP(LW\,1)-module implies that EQu/QV is a hyperplane in [X,“ Lu]Q,,/Qu. Therefore
(E62,, 0 [X#,L,,]Z,,)/E is a hyperplane in [X“,Lu]Zp/E, and XV centralizes it by

(3.2.6). But this is a contradiction to X, S MA and (VII). I

(3.2.8) Assume that b = 4.

(a) X0 S Q“ for each ,u E A‘2)(a). In particular, X0, is elementary abelian.

(b) [YaaQa] S Xa-

Proof. (a) Suppose that X0 Q Q,“ for some [2 E A(2)(a). Then (1.1.2)(a) implies

that

From (3.2.7) and (3.2.6) we get [X,“ X0] S [Qm X0] S Ta S Z“, and therefore
[X2 L21 3 [X2,<X.52>1 s 22-

Since 12 is conjugate to a, this implies
leLOl S Za-

Hence, if /\ is the common neighbor of a and ,u, then X0 = X0), a contradiction to

Xa Z Qu-
(b) Since Q0 S QA, for each A E A(02), (b) follows from the definition of Y0, and
X

0' I
(3.2.9) Assume that b = 4.
(a) Y; S T,,.

(b) (Ya n Qa—2)Ro-4/Ra—4 _<_ Z(Qa—3Ra—4/Ra—4)-

60

Proof. (a) From (3.2.8)(b) it follows that
Yo'l S X0.

Pick 22 E A(a - 1) fl A(a). Then Y; S X0 and (3.2.8)(a) imply that YaRu/R“ is an

elementary abelian normal p—subgroup of AIQ_1/Ru. Hence by r > 1
Yam/R. s Z(Q._.R../R.>.

Now it follows from (3.2.5) that
[Um Ya] S U“ H Z04 S T,,.

Since La acts transitively on [3(0) and normalizes Ya, (a) follows.

(b) Note that by (a) (Ya fl Qa_2)RO_4/RO_4 is an elementary abelian subgroup
of QO-3RO_4/Ra_4, and (3.2.4) implies that (Ya fl Qa_2)Ra_4/Ra_4 is normalized by
each p’—element of some Levi complement of (Ala_3flLa_4)Ra_4 /RO_4. Now the claim

follows from r > 1. I
(3.2.10) b : 2.

Proof. Suppose that b > 2. From (3.2.9)(b) and (3.2.5)(b) it follows that
[Ya n Qa—2a Ua—4l S Tar—2 S Xa-

Note that (3.2.9)(a) implies that YOR0_2/ 120-2 is an elementary abelian normal p-

subgroup of Ma_1/Ra_2. By r > 1, we get that
lYa 1 Ya fl Qa—2l = lYaRa—2/Ra—2l S firm”)-

Therefore,
e) In : CY.(U._.,X..)I s 212*“).

If [Ym La] S X0, then Ua_2 S Y0, = Ya,a+1 S Q0”, a contradiction. Hence
[Ym La] Z X0.

61

Let W be a GF(p)LaRa-composition factor of Ya/Xa that is not centralized by La.
Let D be an irreducible GF(p)Ra—submodule of W. From r > 1 and (3.2.5)(c) it
follows that Ya / X a is the direct sum of r-dimensional absolutely irreducible GF(q)Ra-

modules, each of which is irreducible as GF(p)Ra-module. Hence |D| = q' and
EndaF<p>Ra(D) “=" GFUI).

From (A.3.1) (with RaLa/Qa, Ra/Qa, LaQa/Qa, GF(-p), G'F(q), W and D in
place of G, A, B, F, K, V and X, respectively) we get that

W 3’ D ®Gp(q) E

for some irreducible GF(q)La-module E. Thus, regarded as GF(p)La-module, W is

the direct sum of r copies of E. Together with (*) this implies that
(...) IE: 05w.-.» 3 (1%“).

On the other hand,
an_4 = Cu.._4(E)l = |Ua_4Ra/Ral Z (1'.

since (3.2.4) implies that Ua_4Ra/Ra is normalized by each p’—element of some Levi
complement of Ma_1/Ra. In particular, E is an FF-module and Ua_4 acts as an
offending subgroup. Note that the ‘exceptional’ irreducible FF-modules for the or-
thogonal groups ( cases (g) and (n) in (A.2.2) ) have no quadratically acting of—
fender in common with the natural module. Moreover, if E is as in (A.2.2)(d) then
lE : CE(UO_4)| = q4 and r S 3 = %, contrary to (202). Hence E is a natural module.

But then
(* * *) IE = CE(U0_4)I = 9',

again since (3.2.4) implies that Ua_4Ra/Ra is normalized by each p’-element of some
Levi complement of Ma_1/Ra. From (**) and (=2 * >2) it follows that r = 1, a contra-

dlctlon. I

62

3.3

In addition to (I)—(VII) we now assume
(VIII) LaRa/Ra E Q§n(q) (e E {+, —}) and r = n — 1.
(3.3.1) Assume that UatRo, fl LORQ = Ra.
(a) Ian,Ua'l| = 2-
(b) b > 2.

(C) U0 2 (U0 n Qa’)[Uay Qa+1]'

Proof. Note that UalRa 0 L012.) = R0 and U02 S Q0“ imply that U0: induces
outer automorphisms on LaRa/Ra that centralize the parabolic subgroup (MGM fl

L0)Ra/Ra modulo its largest normal p-subgroup. Hence, by (A.4.1),
an’Ra/Ral = 2-

Therefore
(*) |[U0,Ua'l| = My :CUOI(Ua)l = anI : Ua’ 0620! = an’ = Ua’ fl Ral = 2-

Thus (a) holds.

Let P1 and P2 be the two maximal subgroups of LaRa that contain Mo,“ 0 LaRa.
Since Ua: induces on LaRa/Ra a graph automorphism that switches P1 /Ra and
Pg/Ra, it follows that U02 switches 0,,(P1/Ra) and 0,,(P2/Ra), and hence acts non-
trivially on Qa+1Ra/Ra Thus UalRa/Ra is not normalized by Mom/Ra, whence
b > 2. Also the two maximal isotropic subspaces [Um 0,,(P1)] and [Um 0,,(P2)] of U0,
are switched by U02, whence [Ua,Qa+1] Z Q02. Since (*) and (3.1.1)(g) implies that

an : U0 flQall : 2,
it follows that (c) holds. I

63

(3.3.2) Let ('y, 7’) be a critical pair, A E A(y)flA(b‘1)(7’), )2 E A(2)(7) flA(b‘2)(7’) and

6 E A(2)(7’) fl A(b‘2)(y). Then there exists 9 E L, such that the following hold:

(a) The intersection of the parabolic subgroups (M ,\ fl L7)R,/R,, and (M ,\ fl
L7)9R2,/R2, of L7R,/R,, is the product of a Cartan subgroup and a Levi

complement in both of them.

(b) Assume that M)R/7. 0 L711, aé R2,. Then one of the following holds:

(b1) ()29‘”, (5) is a critical pair for each a: E M,\.

(b2) [Uw Qu] is a maximal isotropic subspace of U7, and ()293, 6) is a critical

pair for each a: E CMA([U7, Q#]).

Proof. Note that (3.1.1)(g) implies that (02,02') is an arbitrary critical pair. Hence
it suffices to prove this for (7, 7’) = (a, a’). Since (Ma+1 fl La)Ra/Ra is a parabolic

subgroup of LaRa/Ra, there exists a root system <I> for LaRa/Ra such that
(fl/[0+1 fl La)Ra/Ra = QLH,

where Q is the product of the root subgroups corresponding to the positive roots which
are not contained in the root system \II spanned by the simple roots that belong to
the nodes 1,. . . , n - 2 of the Dynkin diagram, L is generated by the root subgroups
corresponding to the roots in \II, and H is a Cartan subgroup in the normalizer of the
product of the root subgroups corresponding to the positive roots.

Let Q“ be the product of the root subgroups corresponding to the negative roots
which are not contained in \II. Then Q‘LH is a parabolic subgroup of LaRa/Ra that

has the same type as QLH. Hence there exists h E L, such that
(Ma+1 n La)hRa/Ra = Q_LH

Then (a) is satisfied for each 9 E h(Ma+1 0 La).

64

Assume that UyR, 0 L751, # R4. Clearly (b1) holds if b = 2. Hence assume that

b > 2. Note that
(*) [UmQaH] o 23,, =1, and
(*') [UinHl fl Za+1 = 1, for each 9 E hMOH.

From (3.1.4)(b) and (*) it follows that (b1) holds, provided [Ua+2,Qa+1] = [Ua+2, Q0].

Assume that

(**) [UcH-Za Qa+1l 7E [Ua+2a Q0]-

Then (3.1.1)(c) implies that [UO,QO+2] is an isotropic subspace of U0. Note that
[Ugfiz,Qa] 0 U0, is a proper Mgfil-submodule of U0, for each as E Ga. Hence, if

[UmUaI] S Z0“, then by (3.1.1)(h) and (*’)
[US—:29 Q0] 0 [Uaa Ua’] S [US—12:2: Q0] 0 U0 0 Za+1 S [Uaa :11] n Za+1 : 1:

for each :1: E Ma“. Therefore, (b1) follows from (3.1.4)(c) in this case. Hence we

may assume that

[U02 Ua’] g Za+l~

In particular, [Um 620,22] 75 UaflZa+1. Then [Um Qa+2] is a maximal isotropic subspace
by (3.1.1)(h)(i).
If [Ug+2, Q0] H [Um U02] : 1 for some 9 E h(Ma+1 H La), then also

[UL-€12,620] H [Um U02] 2 1 for each a: E CMO+,([UQ,QQ+2]),

since [Um U01] S [Um Qa+2]. Therefore, (b2) follows from (3.1.4)(c) in this case.

Assume that
(* * *) [U§+2,Qa] H [Um U02] ¢ 1 for each 9 E h(Ma.+1 (7 La),
Note that |[Ua+2,Qa]| S q" by (**) and (3.1.1)(e). In particular,

[Ua+21 Q0] 0 U0 75 [U01 Qa-l—ll-

65

Note that [Ua+2, Qa]flUa Q Za+1 by (=2) and (***). Then by (3.1.1)(h)(i) [Ua+2,QO]fl
U0 is a maximal isotropic subspace of U0, that contains U0 (1 Z0“. Together with (*)

we get
llU§+2=Qal fl lUOaQo+lll S (I, for each 9 E h(Ma+1 0 La),
and hence, by (=2 =2 =2),
(= = = =) «(12.26201 0 [UmQaHDI s [UmUd], for each a: 6 M0,, n L...
Note that
U. n 2....) s (HI/£20210[U..,Q..+11)M0+*”L°>.
Since [Um U02] Z Z0“, it follows from (=2 =2: =2 =2) that
lan, Ua’ll = q".
Hence [Um U02] 2 [Um Qa+2]- In particular, anl H Qal = q" by (3.1.1)(g). Therefore,
(=1: =2 =2: =1: =2) U0: 0 Q0, 2 [Um Uw].
Since (UV | V E A(a — 1)) is a nontrivial characteristic subgroup of Ma_1,
UVUO flGa, for some V E A(oz +1).

Suppose that (V’n,a’ — 2) is not a critical pair for some a: E Ma“. Then U3,” S

Q0242 S Gar. Together with (* =1: * =2: =2) we get
[ngan’l S Q0: n Ua’ S Um

contrary to (*), (3.1.4)(a) and the choice of V. Hence (Mn, 02’ — 2) is a critical pair

for each :2: E Ma“. Now (1.2.1)(e) implies that Ua+2Ua $00. Hence we can choose
V = a + 2. Then (b1) holds. I

(3.3.3) Assume that in (3.3.2) (b2) is satisfied. Then |Q : CQ(:1:)| = q"’1 for each
a: E U7 0 Zi’, where Q :2 CQmL7([U7,Q#]).

66

Proof. Similarly to (3.1.1)(d), this follows by a simple matrix calculation. .

(3.3.4) Let (7, 7’) be a critical pair, A E Ah) flA(b‘1)('y'), )2 E A(2)(7) HAW-”(7’), and
6 E Amh’) fl A(”‘2)(y). Choose 9 E L, as in (3.3.2), if U72R7 fl L712, ¢ [L,.
Choose g E L, with U7: Q Mi, if U7=Rq fl [4de = R7.

(81) G») = (MiUvA
(b) Assume that b > 2, U, H U), S ZA, and Q), 75 Q7Qu. Then U6 S LfiRfi.

(c) Assume that b > 2 and U, D U), S ZA. UgRfi n Lng, 7E Ril'

Proof. (a) This follows from (3.1.4)(a), (3.1.3)(a), and the choice of g.
(b) Suppose that U5 9; LfiRZ. In particular, )2 = 2.
Suppose that U), (1 Z A Z U7. Since U), (W Z ,\ is irreducible as M A-module, it follows

that U), D Z A is isomorphic to an M A-submodule of Z7/U7. Hence L7,,” centralizes

U“ fl 2;. But then OP(L,,A,1) S RqRu, contrary to Q.,Q,u aé Q). Hence
(*) Ul‘nZAZU’YflZAzU'yr-AUIP

Since [U3, 3", Q3] = (U3 (1 Z§)[U3, U5], it follows from (=2) and b > 2 that U7: normal-

izes the M g-module
Y==1U2 3.622112.

Hence Y is Cry-invariant by (a). Note that Q, centralizes Y, since Q.,Q,1 32$ Q A. Hence

Y is a module for L7Q7/Q7. Now
[Y2 U’r’l : llUwQAwQAlUw U'r’l : lUviUv’l

and |[U.,,U,=]| = U7=Q.,/Q., imply that the dual of Y is an F F-module for L7,
and UV acts as an offending subgroup. Thus [14](1.5) shows that Y 2 U7. Since
[Um QA, QA] Z Z,\, this is a contradiction to (*).

(c) Suppose that U5Rf, fl LfiRfi = R3,. Then Q.,Qp = Q) by (3.3.1)(b) and (b). If
U721"):7 0 L711, yé R4, then [U,,Q,\] (1 ZS’ == 1 by the choice of g. If U,=R, 0 L711, :-

67

, then U ,U , is not sin ular, but U fl 29 is a sin ular subspace of U , whence
7 7 S 7 A g 7

U ,U= 0 Z9 : 1. Therefore, in any case U D U S Z) implies that
'7 '7 A '7 l1
[U,,U,=] n U; = 1.

From (A.1.5)(b) (with U,, (U3 fl Q5)U,, and U, in place of A, B, and V) it follows

that
(*) [(113 o Q6)U7= U2] 3 U321,

Put
Y ;= (va> U,.

From Q; = Q,Q,,, (2) (a), and (3.3.1)(c) we get
Y = [Y,Q,]U,Uf,’.

But then Y/(U,Ug) = [Y/(U,Ug), Q,] and hence
U,Ug = Y S 0,,

a contradiction. I

(3.3.5) Assume that if (7, 7 + 1,. . . ,7 + b) is any path such that (7, 7 + b) is a critical

pair, then U, D U,+2 = 1 or U, (1 U,+2 Q Z,+1. Then b S 4.

Proof. Suppose that b > 4. Note that (3.3.2) and (3.1.3) imply that if (7,7’) is any
critical pair and )2 E N2) (7) fl A(b’2)(7’), then U,U,, S0,. Hence it follows from
b > 4 and (3.1.4)(d) that there exists a path (7, 7 + 1,. . . ,7 + b) such that (7, 7 + b)

is a critical pair and
U, n U,+2 # 1.

Hence U, D U,+2 Q Z,+1. Note that this means that

U7 n U7+2 : [U72 Q7+12Q7+1l = [U7+2a Q7+12Q7+1l

68

Now extend (7, . . . ,7+b) to a path (7—b—2, 7—b+1, . . . ,7+b) such that (7—22', 7+b—
22') is a critical pair, for each 2' E {0, . . . , 975—4}, as follows. Assume that (7—22', . . . , 7+b)
has already been defined, for some 2' E {0, . . . , g}. If U,+b_2,~R,_2,- fl L,_2,-R,_2, 75
R,_2,-, then choose g,_2,- E L,_2, such that (3.3.2) is satisfied for (7—22', 7+b—22) and
g,_2,- in place of (7,7’) and g. If U,+b_2,R,,_2,- fl L,_2,-R,,_2,- = R,+b_2,-, then choose
g,_2.- E L,_2, such that U,_2,- = (U,_2,- fl U,_2,-+2) x (U,_2,- fl U,_2,-+2)97-"). (A simple

calculation shows that this is possible.) Then put
7 — 22' — 1 := (7 — 22' + 1)9"’2‘ and
7 — 22' — 2 z: (7 — 22' + 2)97"".

Suppose that (7 — 22' — 2,7 + b —- 22' — 2) is not a critical pair. Then (3.3.2) and
(3.1.3) Imply that U,+b_2,-R,,_2,-flL,_2,-R,_2,- = R7+b—2i and U7+b—2i < M,_2i_1. Since

U,_2,- fl Z,_2,-_1 is the unique proper M,_2,-_1-submodule of U,__2,- fl U,_2,-_2, it follows

that
[U7—22'2 U7+b—22'] = [(U7—22' fl U7—22'+2)(U7—22' 0 (1742—2), U7+b—22'l S
U7—22 fl 27—22—1-

Since U,_2,- fl Z,_2,-_1 is a singular subspace of U,-2,s, but U,+b_2,-R,_2,- flL,_2,-R,,_2,- =
R,+b_2,- implies that [U,_2,-, U,+b_2,] is not singular, this is impossible. Hence (7 —
22 — 2, 7 + b — 22' — 2) is a critical pair.

Define

V := (U), l )2 E A(2)(7 — b + 2), (7, )2) is a critical pair).

By a counting argument as in the proof of (3.1.4)(d) (and again shown in detail in

the proof of (33.7)) we get
(*) llva‘rll 2 <1"-

69

Note that U,_2 O U, Q Z,+1 implies that [U,, Q,_2] S U,_2 and hence
[V, U7l S [(27—22 U7l S U7 0 U7-2-

Together with (*) it follows that [V, U,] = U, 0 U,_2. Since [V, Q,_b+2] S U,_b+2 and

b > 4, we get
U7 n U7-2 S U7—b+2 S Q7-b—2-

Since (7 — b — 2, 7 —— 2) is a critical pair, it now follows the construction of the path

(7—b—2,...,7+b) that

U7+b—2R/7—2 fl L7+b—2R7+b—2 ¢ R7+b—2-

Hence, if ()2 — 2, )2 — 1, )2, . . . , )2 + b) is any path Such that ()2, )2 + b) is a critical pair
with U,,flUwg 5:5 1 and ()2— 2,)2 — 1,. . .,)2+b) is constructed from ()2, . . .,)2+b) by

the method above, then

(**) Uu+b-2Ru-2 fl Lu+b—2R)2+b—2 3'6 Row—2-

Note that in the construction of (7 — b — 2,. . . ,7 + b), we can replace g,_b by g,_b:z:,

for each a: E C'Q7_b+,nL7_b([U,_b, Q,_b+2]). Define

Y :2 (U.f_,,_2 I a: E CQ,_,+,nL,_,([U—y—baQy—b+2l))-
Then again a counting argument as in the proof of (3.1.4)(d) shows that
llY2U7ll > (Q2 _1)qn—1,

using (**) and (3.1.1)(c). But since Y acts quadratically on U, and hence [Y, U,] is

an isotropic subspace of U,, this is impossible. Hence b S 4. I

(3.3.6) There exists a path (7 — 4, 7 — 3, . . . ,7 + b) with the following properties:
(a) 7 — 2 = (7 + 2)97 for some 9, E L,.
(b) 7 — 4 = 797‘2 for some g,_2 E L,_2.

70

(c) (7, 7 + b) is a critical pair and U,+bR,, fl L,R,, 7é R,

(d) If Q,+1 = Q,Q,+2, then for each a: E M,“ the following hold:
(d1) ((7 — 2)$, 7 + b — 2) is a critical pair.
(d2) If U, H U,+2 S Z,+1, then U,+b_2R‘f,_2 fl L:_2R?,_2 75 R312.

(e) If Q,+1 # Q,Q,+2, then for each a: E C217+,([U,, Q,+1, Q,+1]) the following
hold:

(e1) ((7 — 2):”, 7 + b — 2) is a critical pair.
(02) If U, H U,+2 S 2,“, then U,+b_2 S Liz :4.

(f) If there exists a critical pair ()2, )2’) such that 1 ¢ U,, H U,, S Z ,\ where V E

A(2)()2)flA(b’2)()2') and A E A()2)flA(b‘1)()2'), then 1 yé U,flU,+2 S 2,“.

Proof. This follows from (3.3.2), (3.1.3), and (3.3.4). I

Assume that the critical pair (a, a’) is chosen such that the path (a, . . .,a’) can be
extended to a path (a — 4,. ..,a’) as in (3.3.6). Choose 90 E La and ga_2 E La_2

such that
a—2= (a+2)9°‘ and
a — 4 = 09"".
(3.3.7) b S 4.

Proof. Suppose that b > 4. Then it follows from (3.3.5), (3.3.6)(f), and the choice of

the path (a — 4,. . . , 02’) that
1 79 Ua+2 (1 U0 S Za+l-

In particular, by (3.1.1)(a)(h)
Ua+2 H To, = 1.

71

Put

Q .2 { 620+) 0 La if 620.1 = 62.62....
' CQaHflLa<ana Qo+2l) if Qa+1 75 QaQa+2-

Let A be a set of representatives for the cosets of Q P! Q, in Q. Define

X0 :2 {(x,a) | 1% :1: E [Ug_2,Ua=_-2] fl Z3_1}, for each a E A, and

X == U 111:-.. UN] n Z:-1\{1}-

aEA

Then by (3.1.1)(c) and Ua+2 H U,, S 20“

_ lAl(q _ 1) if QcH-l : QaQa+2
laLEJA XGI — aEZA “Yul 2{ lAl(q2 _' 1) if Qa-H 3'5 QaQa+2~

 

Note that Ua+2 F) To, = 1 implies that
UC‘,’_2 fl Ta = 1 for each a E A.

Therefore by (3.1.1)(d) and (3.3.3), each .2: E X is contained in at most q‘"|A| of the
subgroups [Ug_2,Ua=_2] with a E A, if Qa+1 : QOQMQ, and in at most ql‘”|A| of
these subgroups, if Q0“ 75 QaQa+2~ Hence

{ q—nIAI l‘Yl if Qa+1 : QaQa+2

X0 = E A ' , E Xa S —n '
l U l 2 Ha I (J: a) }l q1 IAI I)“ 1ch1-1-17'é QaQa+2-

aEA IE X

Therefore,

((12 _ 1)qn—1 ich1+l 7E QaQa+2o

Since (2 > 4, Va acts quadratically on U024. By (A.1.5)(a), it follows that [V0, U024]

(*) |X| Z { (q _1)qn if Qa+1 : QaQo+2

iS an isotropic subspace of Ua=_2. Hence
(**) |X| g |[Va,Ua=_2]| — 1 s q" — 1.

From (*) and (202) it follows that (1"+1 — (In S q” — 1a a contradiction. I

(3.3.8) Assume that b = 4.

72

(a) IN E {1, 2}, then U04),- 0 Z04,“ S Ta_2,+2.

(b) In e {1. 2}, then (0.522% s n.6,-..” 22-2.1222.

(c) Ifz' E {1, 2} and :1: E L04,” with U012,- Q QC,_2,-, then
Z(Qa—2i+1Ra—2i/Ra—22) S U:_2iRa—2i/Ra—2i-

(d) n S 4.

Proof. (a) By the choice of the path (a —— 4, . . .,a’) we can apply (3.1.4)(c) with
(a — 22', a — 22+ 2,a’ — 22' + 2) in place of ()2,a,a').
(b) Suppose this is false. Then U04,- 2 Lx R‘r for some :1: E La_2,-+2. In

a—22 02—22

particular,
p = 2.

Suppose that 1110-2,“ 0 L04, has a p—component K. Then Ua_2,- S K and, by (a)
and (3.1.5)(a), K S R0_2,+2. Since (1.1.2)(d) implies that L04,” normalizes each

p—C0mponent of 1204,22, it follows that
Ua—zz' S KI S L2,—222

a contradiction.

Hence n = 3 and q = 2. Then 02(G“

a—22'

) S L35 R35 But (a) and (3.1.5)(b) imply

a—22 a—22"
that Ua—22 S 02(RO_2,‘+2). Since R0_2,‘+2 S G:_2i, we get
(Ia—m SELF x

02—22 02—22"

a Contradiction.

(C),(d) Note that (B.5.1.5) implies that

Ug~2iRa—2i/Ra—2i fl Z(Qa—2H—1Ro—22/Ra—2i) $5 1,

73

since U34,- acts as an offending subgroup on Ua_2,- and is contained in La_2,Ra_2,
by (b). Since Z(QO_2,~+1RO_2,/Ra_2,~) is an irreducible OP(L0_2,,O_22+1,1)-module and

’32 2,. is Rm 2,+2—invariant, (a) and (3.1.5)(a) imply that
Z(Qa—22+1Ro—2i/Ra—2z) S U2- 2,- Ra—22/Ra—2i~
Hence (c) holds. I\=’Ioreover,
(*) IUO_ 2.3 o—22flQO-21lzlljgf2ifg-O zz/Ra— 21 >
lZ(Qa—22+1Ra—2i/Ra—22)l = (1%(7' ”(n—2).

Since [U§_2,-, Ua_2,-] is an isotropic subspace of U322“ we also have

(**) WET-22: a— 22' nQ0421: llfla—221Ua-2ill S 4"-
From (*) and (22*) it follows that n S 4. I

Define
Y': (Ua+2>'
(3.3.9) Assume that b = 4 and n = 4.

(a) Y Ra—2/Ra—2 : Uo+2Ra—2/Ra—2 Z Z(Qa—lRa—2/Ra—2)-

(b) Y, = [Ua+21Ua—2] S Ta-

Proof. Pick a: E La with U§+2 Z Q0- 2. Suppose that |[U"” 0+2, Ua_2]| = q". Then, since

OP(L0_2,O_1,1) normalizes U§+2Ra_2/Ra_2 by (3.3.8)(a) and (3.1.5)(a), it follows from
(3.3.8)(c) that |U§+2RO_2/RO_2| = q6. But then

Q4 — _lel 0+2, U-0 21': lUcf+2 : Clx’§+2(Ua—2)l =

lUg+2 3 U5+2 fl Qa—Zl = lUcf+2Ra-2/Ra-2l : (162

74

a contradiction. Hence, by (3.3.8)(c)
U5+2Ra-2/Ra—2 = Z(Qa_1Ra_2/Ra_2).

Since this holds for each :1: E La with Ugf+2 Z Qa_2, (a) follows. Moreover,
[Ua—21 Y] = an—2, Ua+2l S Ta

by (3.1.4)(e). Hence (b) holds. .

Define
X1: CQa(QmTa)'

(3.3.10) Assume that b = 4 and 72 = 3. Let 2 E {1, 2}.

(a) [Ua_2,-, Ua_2,~+4] is a maximal isotropic subspace of Ua_2,-.
(b) Ua—2i+4Ra—22/Ra—2i fl Op(0pl((Ma—22+l n La—22)Ra—2i/Ra—2i))-

(C) X Ra—2/Ra—2 : [Ua+2a QalRa—2/Ra—2 : Z(Qa—lRa—2/Ra—2)-

Proof. (a) By (A.1.5)(a) [Ua_2,-, Ua_2,-+2] is an isotropic subspace of Ua_2,-. If it is not
maximal, then (3.3.8)(C) implies that Ua—2i+4Ra—2i/Ra—2i = Z(Qa—2i+1Ra—2i/Ra-2i)

and hence
an—2i+4Ra—2i/Ra—2il = q < Q2 = lan—21,Ua—2i+4ll,

contrary to
lan—2i+4a Ua—zz'll = an—2i+4 3 CUO_2.-+4(Ua—22)l = an-2i+4 3 Ua—22'+4 fl Qa—22‘l =
an—22+4Ra-22/Ra—2il-

Hence (a) holds.
(b) follows from (3.3.8)(a) and (3.1.5)(a).

75

(c) Note that [U,,-2, X] S Ta S Zm, since b > 2. Therefore,

XRa-2/Ra—2 S CQO_1(Ua—2a Ua—2 fl Za—1)Ra—2/Ra—2 = Z(Qa—lRa—2/Ra—2)-
Moreover, by (a) and (b),

[Ua+2a QalRa—2/Ra-2 = Z(Qa—lRa—2/Ra—2)-

Since (3.1.4)(e) implies that [Ua+2,Qa] S X, (c) holds. I
(aan)b=2

Proof. Suppose that b > 2 and n = 4. From (3.3.9)(a) it follows that
(*) [Y : Y F) Qa_2| = q3.

Let A be the subgroup of 1110-3 with 120...; S A and A/Ra_4 = Z(Qa_3Ra_4/Ra_4).
By (3.3.9)(b), Y n Qa_2 acts quadratically on U04, whence

(**) lY fl Qa—2 3 Y n Qa—2 H Al S (13-
Note that, by (3.3.8)(a) and the definition of A,
[Ua_4, Y r7 Qa_2 r1 A] g Tc,_2.
Thus (=2) and (**) imply that
(* * *) IY : Cy(Ua_4, ZalY1Qal)| S (16-

Suppose that [Y,La] S ZO[Y,QO]. Then Y S [Y,Qa]ZaUa+2. Since [Y,Qa] S Z(Y)
by (3.3.9)(a), it follows that Y is abelian, contrary to (3.3.9)(b). Hence

[Y, La] 9; 2.1K Q0].
Let W be a GF(p)LaRa-composition factor of YZa/[Y,Qa]Za that is not central-

ized by La. Let D be an irreducible GF(p)Ra-submodule of W. From (3.3.8)(a)

76

and (3.1.5)(a) it follows that Y Za/[Y, Q0]Za is the direct sum of 3-dimensional abso-
lutely irreducible G F (q)Ra-modules, each of which is irreducible as GF(p)Ra-module.
Hence |D| = q3 and

EndGF1p112..(D) ’5 GF(<1)-

From (A.3.1) (with RaLa/Qa, Ra/Qa, LaQa/Qa, GF(p), GF(q), W and D in
place of G, A, B, F, K, V and X, respectively) we get that

W '73 D ®GF<<1> E

for some irreducible GF(q)La-module E. Thus, regarded as GF(p)La-module, W is

the direct sum of three copies of E. Together with (=1: =1: =1:) this implies that
(=1: =1: =1: =1:) |E : CE(UO_4)| S q2.

On the other hand, it follows from (3.3.8)(c) that
an—4Ra/Ral = an—4 3 Ua-4 r7622' 2 lan—41Uall =
We 1 Ua O Qa—4l = anRa—4/Ra—4l Z (13-

Since Q§(q) has no nontrivial irreducible module over GF (q) in which the index of
the centralizer of a subgroup of order q3 is at most q2, this contradicts (=1: =1: =1: =1:).
Suppose that b > 2 and n == 3. If [X,La] S Z0, then [Ua+2,Qa] S [Ua_2,Qa]Za S
120-2, contrary to (3.3.10)(c). Hence

[XaLal Z Za-
From (3.3.10)(a),(b) it follows that

UaRa—4/Ra—4 = CQO_3RO_4/Ra_4(UaRa—4/Ra—4)-
Therefore,

[Ua—41X O Qa—2l = [Ua—41Ua] S Ua-

77

Together with (3.3.10)(c) we get
lX I CX(Ua—4aZa)l S IX 1 X r1Q11—2l S (1-

From [Ua_4RO/Ra| = IUORO_4/RO_4 and (3.3.10)(a),(b) it follows that
an—4Ra/Ral = (13-

Hence X /Za contains exactly one non-central La-chief factor E, and E is a natural
SL4(q)-module for La.

Let a be an automorphism of Ga which normalizes Ma_1. Then X, 20 and La are a.-
invariant. Hence also E is a-invariant. Thus a normalizes the two maximal parabolic
subgroups of LaRa/Ra that contain (ll/[0,4 fl La)Ra/Ra. Since this holds for each
a E Aut(Ga) with Mg_1 2 MO-“ we get a contradiction to (11).

Now the claim follows from (3.3.8)(c). I

3.4

In addition to (1)—(VII) we now assume

(VIII) LaRa/Ra 9: 0;,(q) and r = 72.

(3.4.1) Let (7,7’) be a critical pair, A E A(7) flA(b‘1)(7’), )2 E A(2)(7) flA(b’2)(7’) and
6 E A(2)(7’) fl A(b‘2)(7). Then there exists 9 E L, such that the following hold:
enwnanfimq

(b) ()2”, 6) is a critical pair for each a: E Q1.

Proof. If n is even, then the proof for this is similar to the proof of (3.3.2). Hence
assume that n is odd. Let P be a subgroup of M1 containing R, such that P/R, is
a parabolic subgroup of L,R,/R, of type {1, . . . , n — 2}. As in the proof of (3.3.2)

78

we can pick 9 E L, such that the intersection of the parabolic subgroups P/ R, and

P9 / R, is a Levi complement in both of them. Then
(*) [U7 O ZA O Zi’xl = q, for each :1: E M).

Suppose that G, 79 (fl/If, U,:) for some :1; E M). Since Mint is a maximal subgroup

of G,, it follows that U,: is contained in M in and hence normalizes Z 3”. Thus
[(U7 O ZA)(U7 O 21$)» U7’l 2 [U7 O 2A”, U7’l S [U72 U7’l O 211 S U7 O ZA O 2235'

This means that U,=R, / R, is contained in the largest normal p—subgroup of N12,,(U, fl
Z10Z§I)R, / R,, which is a parabolic subgroup of cotype 1 of L,R,/R,. But in 03,,(q)
no offending subgroup for the natural module is contained in the largest normal 19-
subgroup of a parabolic subgroup of cotype 1, by (B.5.1.4). This contradiction shows

that
(=1:=1:) G, = (.lIfI,U,=), for each :1: E M,\.

Since (U,, I V E A(A)) is a nontrivial characteristic subgroup of MA, we can choose

V E A(A) such that
U,,U, SG,

Assume that |[U,, U,“ = q". Then
[U7=U7'l = [Q61U7’l'

Hence, if (V91, 6) is not a critical pair for some :1: E Q,\, then
[U311U7'l S [Q61U7’l S U7,

contrary to (=1:=1:) and the choice of V. Now (1.2.1)(c) implies that we can choose V = )2.
Now assume that |[U,,U,1]| 75 q". Then [U,,U,=] is a proper GF(q)-subspace of
U, 0Z1. Since M ,\ 0 L, acts transitively on the set of 1-dimensional GF(q)-subspaces

Of U, H Z2, (=1:) implies that there exists y E A11 H L, such that
[U,, U,=] r1 Z) r1 Ziy =1.

79

Since [U,, U,:] S 21 and [Z1,Q,\] z: 1, it follows that
[U,, U,=] fl Zi’yx =1, for each :1: E Q,\.

Therefore,
(* * *) lU7=U7'l O [L73y11U2' O Q] S [U,,U,=] O [Ugyx1Q7l S
[U,, U,=] fl Zim =1, for each :1: E Q).

From (=1<=1:), (=1: =1: =1:), (A.1.5)(b), and the choice of V it follows that (ngz, 6) is a critical
pair for each :1: E Q,\. As above, (1.2.1)(e) implies that we can choose V = )2. Now

the claim holds with gy in place of g. I

Choose ga E La such that (a) and (b) of (3.4.1) are satisfied for (a, a’,ga) in place of

(7, 7’, g)- Define

a —1:=(a+1)9° and
a— 2 :2 (a+2)g°.
(3.4.2) b S 4.

Proof. Suppose that b > 4. Note that (3.4.1) implies that if (7, 7’) is any critical pair
and )2 E A(2)(7) fl A(b‘2)(7’), then U,U,, SIG, As in the proof of (3.2.2) it follows

that we may assume that
Ua+2 O Ta = 1.

Let A be a set of representatives for the cosets of Q, in Q0“. Let E be a complement

to U0, (1 20,21 0 ZO_1 in U0 0 Za_1. Define

Xa:={(:1:,a) [176 :1: E [U“_2,Ua=_2] fl TOE}, for each a E A, and

O

X 2: U [Ug_,, U014] r7 TaE“ \ {1}.

aEA

80

Then by (3.1.1)(c)

| U Xal = Z lXal Z |A|(q2 -1)-

aEA aEA

Note that Ua+2 0 Ta = 1 implies that
U:_2 D To =1, for each a E A.

Therefore by (3.1.1)(d) each :1: E X is contained in at most ql‘"|A| of the subgroups

l 3-21UQI—2] O TQE“ with a E A. Hence

I U Xal = 2 H0 E A | (1710) E Xall S q1’"|A| |X|-

aEA xEX

Therefore,
(*) |X| Z ((12 -1)(1”‘1-

Since b > 4, 1’}, acts quadratically on Ua=_2. By (A.1.5)(a), it follows that [V0, Uar_2]
is an isotrOpic subspace of U01_2. Hence

(**) |X| S llVaan’—2ll — 1 S q" — 1.

From (=1:) and (=1:=1:) it follows that 11"“ — q"‘1 S q” — 1, a contradiction. I

(3.4.3) Assume that b > 2.
(a) Ua—2 O 20—] S Ta-
(b) La-2,a—1,1 S Ra—2Ra-
(C) Ua+2Ra-2/Ra—2 = Qa—lRa—Z/Ra—2-

(d) 7223.

Proof. (a) and (1)) follow from (3.4.2) and (3.1.4)(e). Since Qa_1Ra_2/Ra-2 is an
irreducible La_2,a_1,1-modu1e, (c) follows from (b) and (3.4.2). Note that (c) implies

that

n in n—
an—2 : Ua—‘Z fl Qa+2l : (I and an+2 3 Ua+2 n Qa—2l : (12 ( l)-

81

Hence ((1) follows from (3.1.1)(g).

Choose 90-2 E L042 such that (a) and (b) of (3.4.1) are satisfied for (oz—2, a’—2, 90-2)

in place of (7,7’,g). Define
a — 3 z: (02 —1)9"‘2 and
a — 4 := 09"‘2.

(3.4.4) b = 2.

Proof. Suppose that b > 2. Define
Y 2: (U532)Ua.

Then, by (3.4.3)(c)(d)
lY 3 Y O Qa—2l Z (13-

Since a — 4,. . . , a satisfy the same assumptions as a — 2,. . . , a + 2, (3.4.3)(c) implies

that
[Ua—41 Y O Qa—2l S [Ha—41Qa—Sl = [Ua—41Ua] S Ua-

Therefore,
(*) IYICY(Ua—41[Y1Qaan)| S (13-

If [Y, La] S [Y,QO]UQ, then Ua+2 S [Y, Qa]UaUa_2 S Qa_2, a contradiction. Hence
71: Y/CY(L011Y1Qaan) # 1-

Let W be a subgroup of Y containing (3)/(La, [Y, Q0]Ua) such that

W :2 W/Cy(La, [Y, Q0]Ua)

82

is an irreducible LaRa-submodule of 7. Note that (3.4.3)(b) implies that 7, regarded
as Ra-module, is a direct sum of natural SL3(q)-modules. As in the proof of (3.3.11),

it follows from (A.3.1) that
W 3 E ®GF(<1) D,

where E is an irreducible La-submodule of W and D is a natural SL3(q)-module for

Ra. Now (*) implies that

 

(**) [7, Lal = W
and E is a natural SL4(q)—module for La with

ICE(Qa-1)l = ICE(Ua—4)| = (13-

In particular, CW(QO_1) is the unique irreducible Ma_1 fl La-submodule of W and

 

has order q9. Since Ua_2 :2 Ua_2Cy(La, [Y,Qa]Ua)/Cy(La, [Y,Qa]Ua) is an Ma_1-

submodule of 7 with |U0_2| = (13, it follows that

 

(* * *) W0 Ua_2 1.

From (**) we get that l—VUO_2 is an La-submodule of 7. Then (* a: *) implies that,
regarded as La-module, W has a complement in W U04, since Ma_1 H La contains a
Sylow p—subgroup of La. By (**), this complement is a trivial La-module of order q3.

But the definition of 7 implies that 7 has no trivial La-submodule, a contradiction.

3.5

In this section we assume that (1)—(VII) hold. Note that by the results of the previous

sections we have b : 2.

(35.1) [Ua’,Qa+1,La] : U0.

83

Proof. Suppose that [Ua:, Qa+1, La] 79 U0. In particular, we are not in the situation

of section (3.4). Then by (3.2.2), (3.3.2), and (3.3.4) we can choose 9 G La such that
G0 = (Mg+1,Ua:) and
Z3,+1 nZ.1+1 0 U0 =1.

Note that with respect to the nondegenerate Lar-invariant bilinear form on U0: we

have
[Uau Qa+1lL = U0’ 0 Za+1 = an', Ual,
by (A.1.5)(a). Again by (A.1.5)(a) and
“Us" 31H], an',Qa+1ll S [Ugo 3+11Qal fl [Ua',Qa+1,Qal _<_
Z5},+1 0th1 nUa =1
it follows that
[Uau [U2], gull S [Uo’aUal

Now the choice of 9 implies that [U3 i +1]Ua and hence also [Uar,Qa+1]Ua is nor-
malized by Ca. But then [[Ua:,Qa+1]Ua,UaI] g U0 and La 3 (Ugo) imply that

[Ua:,Qa+1, La] S U0, a contradiction. I

(3.5.2) Assume that LaRa/Ra 36 Q; (q). Then one of the following holds:
(a) Case (1),(2), or (5) holds in (VI), r = 3, q = 2, and [Qm L0] 2 U0.

(b) Case (5) holds in (VI), r = 2, q 6 {2,4}, and [QmLa] 2 U0.

Proof. Since b = 2, it follows as in [14](4.1) that case (4) in (VI) is impossible.
From (3.5.1) and L; 2 La S (U?) it follows that

[QaaLa] : [QaaLaa La] S ([Qaa Ua’aLalLO> : Ua-

84

If Qa+1 = QaQa’a then lQa+hUal : lQa',Ual S Qa'- If Qa+1 75 QaQa’a then
[Qa+1,Qa] S Q01, since QOQOI S Ma“. Hence in any case UalRa/Ra is an M0“-
submodule of Z(Qa+1Ra/RO).

Note that UarRa/Ra, regarded as GF(p)(Mo+1 fl Lad-module, is isomorphic to the
dual of U0, OZOH. By (B.3.1.5), (8.4.1.5), (B.5.1.7), and (B.5.2.6) we get the follow-

ing:

r S 3 and
r = 3, unless case (5) holds in (IV) and p = 2.

Suppose that neither (a) nor (b) holds. In particular, (r, q) 76 (2,2). Let K be the
p—component of Ma“ 0 La that acts nontrivially on U0 0 ZQH. Then K S La],
for otherwise K S Rd and hence Ua S [UmK] S Rd, a contradiction. Since
LaRa/Ra ’3?! Q; (q), each automorphism of K /Op(K ) is induced by some element of
La. Together with (A.4.2) this implies that Ma“ induces an inner automorphism on

L0, Ra /Ra. Hence

Since Ma+1 induces GL3(q) on Uaz/[Ua:,Qa+1], it follows from (*) and U0: 0 R0 =
[Uar,Qa+1] that Ma“ 0L0, induces GL3(q) on Ua/Ra/Ra. By the structure of (Ma+1fl

L0,)Ra/Ra this implies that
p = 2.

suppose that UaflZcH-l 79 [U03 Qa+ll- Let g E A :: CMO+1flLaI([Ua’a Qa+1lv Ua’flZa+1)-
If [Ua’aQa-H] = [UaaQa+1la then [UaanH—lagl S Ua n Za+1 by the ChOiCB 0f 9° If

[Ua’an-l-Il :I'é [UaaQa+lla then [UaaQa-Hl fl Ua’ S Za-l-ly and since [Ua’Qa+1l S Qa'a
we get [Uaa Qa+lvgl S [Uana+ll fl [Qa’a Lo’] S U0 fl Za+1' Hence III any case

(**) [U03 Qo+lagl S U0 0 Za+l-

85

From (*) it follows that there exist 91 6 AL,“ 0 La and 92 6 R0 such that g = 9192.
By (1.1.2)(d), 92 induces an GF(p)La—endomorphism on U0. Together with p = 2
and (u) it follows that [Um Qa+1,gll S U0, D 20“. Now it follows from (**) that 92
centralizes [Um QC,+1]/(UO (1 20“). Since Ua fl Za+1 # [Um Qa+1l, 92 centralizes also

U0. Therefore,
A S (11104.1 (7 LO)C(;O(UQ).

In particular, Ua/[UO,QQ+1] is dual to UoflZaH as a module for A. But now (B.4.1.5),
(B.5.1.7), and (B.5.2.6) imply that (a) or (b) holds, a contradiction.

Hence U0 0 20H 2 [Ua,Qa+1]. This implies that r = 3 and (5) holds in (IV) Note
that Ra centralizes Uaz/[UOI,QO+1], since [UMRO] S Uar D R, 2 U0: (1 Q0. Then
in Gal/Rat we see that this implies that Ra does not centralize QOH/QOQGI unless

Ra S Qa+lRo’- Since [QO+19 Ra] S Qa, we get
Ra S Qa+1Ra’-

In particular, Ra centralizes UaRai/ROI, i.e.,
[Um R0] 3 U. o R0, = U... o Qai.

Since U0 is an irreducible La-module, it follows that [Um R0] = 1. Together with (*)
this implies that U0 (1 Z0,“ is dual to U,,/[Um Qa+1l, regarded as a module for Ma“.

Now (B.4.1.5) implies that (a) holds, a contradiction. I

Note that (D22) shows that the assumption LORa/RQ $3 Qflq) in (3.5.2) is neces-

sary.

86

Chapter 4

Determining the action of L on R,
Part 2

In this chapter we assume (1)—(IV) and
(V) LORa/Ra E“ PSp2n(q)' (q 2 pk) for some n, k E IN with n 2 2.
(VI) UaTa/Ta is a natural Sp2n(q)-module for La.
(VII) (MQH fl La)Ra/Ra is a parabolic subgroup of cotype 1 in LaRa/Ra.

For each u ~ (1 and /\ 6 AM), let L,“ be a subgroup of [VIA H L“ containing L“ (1 R”
such that LMARfl/Rp is a Levi complement of the parabolic subgroup (M AflLfl)R# / R”
in LuRfl/Ry.

For each 7 ~ a define
A(7) I: {M E NW7) I Uqu 2107},
X7) := ([Uinl l u E A(A) fl A(7))Zva for each /\ 6 Ah),
X71: (X7). |)\ 6 AW),
Y“ := (U,, I n E A()\) 0 11(7))27, for each A 6 AW), and

Y. := <U. | u 6 mm

87

4.1

(4.1.1) (a) w, n Tal e {1,q}.

(b) If A is an offending subgroup for U0 with A S Q0“, then

CQ0+1(A[0+19Q0) S AQo-

Moreover, U0 0 To, Z [Um/l] if and only if q = 2, U0 (1 Ta aé 1, and
CQO+1(M0+11Q0) = Ach

(c) an : U0 0 Qatl = IUOI : U01 flQal.

Proof. (a) Assume that U0 (1 To 75 1. Then (Up 0 T,, | u E A(oz + 1)) is a nontrivial
characteristic subgroup of Mo“. Hence there exists ,u E A(a+ 1) such that UyflTfl Z
Ta. Then (U,, D T,,)Ta/Ta is an Ma+1-submodule of Za/Ta.

Suppose that (U,, (1 T,,) n UaTa S Ta, i.e.,
(*) Ua(U.. fl Tl.)Ta/Ta = (UaTa/TO) x (U,, n T,,)Ta/Ta.

Note that UO(Up fl T,,)Ta/Ta is an La-submodule of Za/Ta. By (*), UaTa/Ta has
an (MOM fl La)-complement in this module. Since Ma+1 H La contains a Sylow 2-
subgroup of La, it follows that UaTa/Ta has also an La—complement in this module.
Now [Zm La] :2 U0 and CZO(LQ,TO) = To, imply that U,, n T,, S Ta, contrary to the
choice of #-

Hence (U# H T,,) (1 UOTa g To. Since (U0 0 Za+1)Ta/Ta is the only minimal Ma+1'

submodule of UaTa/Ta and has order q, it follows that
lUIJ fl Tul Z “(Uri 0 Tu) (W UaTalTa/Tal = (I-

By [10] we also have IU,1 n T,,] S q. Hence We 0 To] 2 Wu 0 Tul = q.
(b) This follows from (B.3.1.4)(b) and (B.4.1.3)(b).

(C) The proof for this is tha same as the proof for (3.1.1)(g). .

88

(4.1.2) If p E A(a +1), then OP(L,,,Q+1) S RuRa(Ma+1fl La).
Proof. This follows as in the proof of (3.1.2). I

(4.1.3) Assume that b > 2. Let (7,7’) be a critical pair and ’7 +2' 2 7’ — b +2' 6

A(‘)(7) fl A(b“)(’y’) for each 2' E {1, . . . , b}. Then there exists 9 E L, such that

(a) The intersection of the parabolic subgroups (M,+ 10L,)R,, / R, and (M 9+lfl

’7

L,)R,/R,, of L,R,,/R, is the product of a Levi complement and a Cartan

subgroup in both of them.
Moreover, for any such 9 E L, the following hold:

(b) G7 : (Maul/7')-

(0) Assume that (my + b — 2) is not a critical pair for some ,u E A(('y +
1)”) fl A(7). Then U, m T, 75 1, q = 2, 1 75 [U,,,U, n 62,] S T,, and
[U,,,Q§+1]U, S G,

(d) (p, ’7 + b — 2) is a critical pair for some [1 E A((7 + 1)”).

Proof. By (4.1.1)(c) it suffices to prove this for (7,7’) = (a, oz’). As in the proof of

(3.2.2) it follows that we can choose 9 E La such that (a) holds. Then
(*) UOTa/Ta : [Um Qa-l—llTa/Ta X (U0 0 ZZ+1)Ta/Ta.

Since [Um Uar] Z T a, it follows from (*) that U,,: does not normalize 23 +1 and hence
U0: Z Mg“. Thus (b) holds.
From (4.1.1)(c) it follows that we can apply (4.1.1)(b) with (a’, a) in place of (a, a’)

and get

CQa;_1(NIa’—1aQo/) S UaQa’a

Therefore,

(**) UO’ 0 Q0 : CUOI(UaQa’) S CUaI(CQar_1(Ma’—la Qa’)) : [Ua’aQa’-ll-

89

Assume that U,1 S Qal_2 for some n E A((a +1)9)fl A(o). From U,, S 6201-2 and

(**) we get

|[Ua' 0 Q0101!“ S lan’aQa’-11Qa’—Ill : q

From this it follows that (U0, 0 Q0)Q,,/Q,, is a subgroup of order at most 2 in
CQZH(OP'(MC€+1),Q,,)/Qu, centralizing [U,,,Qf,+1]. Therefore (A.1.5)(b) (applied to
[U,,, 3,“an, [U,,, 3+1], and UazTal/Tar in place of A, B, and V, respectively) shows

that U01 normalizes [U,,, 2&ij Then (b) implies that
[Uflv Qg+lan fl Ga-

If [U,,, Uar (1 Q0] 2 1, then again (A.1.5)(b) (with U,,Ua, U,,, and UalTaI/Ta: in place
of A, B, and V, respectively) shows that U01 normalizes U,,Ua, contrary to (b) and

the choice of p. Hence

[U,,,Ua, nQa] ¢ 1.

Together with |(Uar fl QO)Qp/Q#| S 2 it follows that
IUOI flQa : Uar 0 Q0 (1le = 2.

Note that To, 3 QamQu, (U0, nQQ/(Ual on) = CUa,/(Ua,nTa,)(Ua), and (U0, flQafl
Q#)/(Uar 0 Ta!) = CUaI/(LlalnTa!)(UaU[J)‘ Since EndGF(2)Ga, (Uni/(U0! flTaI» g GF(Q),

it follows that both IUQI fl Qal and Wu: H Q0 0 QMI are powers of q. Hence
q = 2.

Now (U0; flQa)Q,, = CQ§+1(MO+1,Q,,) S1 Mg“, and hence [U,,, U01 flQa] is a subgroup
of 23“ that is normalized by (Mg+1, U01) 2: Ga. Thus [U,,, U0: (1 Q0] S Ta.
Suppose that UaflTa = 1. Then [Um U01] 2 [U,,, U0: (1620], and once again (A.1.5)(b)

shows that U, normalizes U0, U,,, contrary to (b) and the choice of ,u. Hence (c) holds.

90

Suppose that (d) is false. Then (c) implies that [U“,Qf,+1]Ua S Go, for each u E

A((oz + 1)9) fl A(a). Hence, by conjugation we get for each I! ~ 0 that
(:1: =1: =1:) [U,,, QflUu S G”, for all /\ E A(1/) and 77 E A(/\) O A(1/).

Pick p E A((a + 1)9) fl A(o). Choose :1: E L014 such that a’ — 3 = (a’ — 1)? Since
[Ua’a Qa’—1]Ura’—2 g Ga’—2 by (* * *)a we get

[Ua’a Qa’—Il S [Ugh Qa’—3an’—2-

Now (c) shows that U,, does not centralize {US}, Qai_3]. But (=1: =1: =1:) (with a’ — 4, a’ —3,
and (01'): in place of u, A, and 17, respectively) implies that [U§,,Qar_3] S Ua:_4U5,

for some (5 E A(2)(a’ — 4) fl A(b"2)(/1), a contradiction. .

(4.1.4) Assume that b > 2. There exists a path (7—4, 7—3, . . . , 7+b) with the following

properties:
(a) (7, 7 + b), (7 — 2, 7 + b — 2) and (7 — 4, *7 + b — 4) are critical pairs.
(b) '7 — 4 : 79 for some 9 E L,_2.
(c) For each 2' E {0, . . . , g}, the intersection of the parabolic subgroups

(‘szi—l fl L7+2i—2)R/7+2i—2/R’r+2i-2 and
(M7+21—3 fl L7+2i—2)Ry+2i—2/R’v+2i—2

of L7+2i—2Ry+2i—2 / 314.2142 is the product of a Levi complement and a Car-

tan subgroup in both of them.

Proof. By (4.1.3), we can inductively choose 90.2,— E La_2.- and a — 22' — 2 E A(a —
2i+1)9°-2", for each 2' E {0, . . . , b}, such that (a — 21' — 2, a + b — 22’ — 2) is a critical
pair, and (a)—(c) of (4.1.3) are satisfied for ga_2i,a — 2i, . . .,a — 22' + b in place of
g,7,...,7+b, and then put

oz — 21-1 :2 (a — 2'z'+1)g°‘2".

91

If there exists 2' E {3, . . . , b — 1} such that ((o — 21+ 2)9°-’~"', Oz + b — 22' — 2) is a critical
pair, then a — 22' — 2 can be chosen as (a —- 2i + 2)9°—2‘, and then (a — 22' — 2, a — 22' —
1, . . .a + b — 22' -+- 2) is a path with the desired properties.

Suppose that ((a — 21' + 2)9°’2",o + b — 2i — 2) is not a critical pair for any 2' E

{3, . . .,b — 1}. In particular, by (4.1.3)(c),
(*) [Ua—b+23 Qa—b+1an—b Sl Go—b and
(**) [073:322421Qa—2b+1an—2b+2 : [Ua-2b-l-4a Qa—2b+3]Ua—2b+2 g Ga—2b+2-

Put p z: (a — 2b + 4)90-2b+2 and V := (a - b + 2)9°-b. Since U,, does not centralize
[Ua—b+2,Qa—b+1] by (4.1.3)(c), it follows from (at) that (11,11) is a critical pair with
|U,, : U,, (1 le > q. But (**) implies that |Uu : U,, (1 Qul = q, a contradiction to

(4.1.1)(c). .

In the following, assume that b > 2 and (a, 01’) is chosen such that the path (a, . . . , a')
between a and o/ can be extented to a path (a — 4,a — 3,01 — 2,0 —1,a,...,a’) as

in (4.1.4). Then
a — 4 = 0190-2
for some 90-2 E La_.2.

(4.1.5) b g 4.

Proof. Suppose that b > 4. Then the same argument as in the proof of (1.2.5) shows

that
Aa,a+2 S Ta-
On the other hand, (4.1.1)(b) implies that

[U01 CQ0+1(MO+1? Q01” S Aa,a+21

a contradiction. I

92

4.2
(4.2.1) Assume that b = 4.
(a) [XaaQa] E To-
(b) [Xm L0] 2 U0.

Proof. (a) Suppose that [XmQa] Q T0,. Pick [1 E A(a + 1) fl A(a) such that

[U111 Qa+11 Q01] Z Ta-
If [U,,, U04} 2 U“ (1 Zn“, then (1.2.3) implies that U,, H Za+1 S Ta, contrary to the

choice of )1. Hence
[Um Ua—2l .71" Up 0 Za+1~

i.e., either (a — 2,,11) is not a critical pair, or q = 2, U0 0 Ta 91$ 1, and U,, and Ua_2

act as transvections on each other. In any case we have

(*) [UpaQa+1l S Qa-2-

From [Ua,Ua_4,UO_4] = 1, (4.1.4)(c) and the choice of the path (a - 4,...,a’) it

follows that
[Ua—41Qa—3] S Qa-
Suppose that [[Um Qa+1la [Ua—41Qa—3ll 7E 1' Then

[[Upa Qa+1]a[Ua~4a Qa—3llTa/Ta : [Um Qa+l1QalTa/Ta : C(Qa+1)~

Since CUaTa/TO(QO+1) is not contained in [Um Qa-1]Ta/Ta, which is the unique max-

imal Ma_1-submodule of UaTa/Ta, it follows that
UaTa S ([[UpaQa+1]1[Ua—41Qa—3]]MO_I>T0 S [XaaXa—2lTa S Xa—Z-
Since Q04 acts quadratically on X04, but not on Ua, this is impossible. Hence

[[quQa+1laan-41Qa—3ll = 1.

93

Together with (*) we get

[Um-4, [Uana+lll S an—41Ual S Ua-
Since Ga : (MaH, Ua_4), it follows that

[Um Qa+1an S Ga.
But then

[UwQa+11Qal =[lUana+1lUmQalS Go-

Since [U,,,QMhQOH] S Z0“, it follows that [U,,,Qa+1,Qa] S Ta, contrary to the
choice of u.
(b) Pick ga_4 E LO_4 such that (a)—(c) of (4.1.3) are satisfied for ga_4,a — 4,. . . ,a

in place of 9,7, . . .,7’. Put
a — 5 := (a — 3)g°“‘.

Suppose that [U,,, Qa_5]Ua_4 $00-4 for some n E A(a—5)flA(a—4). Then (p, a—2)

is a critical pair by (4.1.3)(c). Note that (4.1.3)(a) implies that
Ma—S S Ala—3ch—5-

Together with Ua_2 S Ma_.3 and [U,,, U04, U042] = 1 it follows that Ua_2 is contained

in CQO_5(OPI(MO_5), Q“) and hence

[UwQa—S] S Qa—2-

Note that
U0 (1 Qa—4 S [UmQa—ll S Xa—2-
Now (a) and o — 4 = 0190‘? imply that
(*) llUu, Qa—sl, U0 0 Qa—4l S [Qa—21Xa—2l fl [Xa-41Qa—4] S Ta~2 fl Ta—4 S To.

94

From (*) and (A.1.5)(b) (with [U,,, Qa_5]Ua_4, Ua_4, and U,,/(U0, 0T0) in place of A,
B, and V, respectively) it follows that

[[Upa Qa—San—‘h U0] S [[Uua Qa—San—41 U0 0 Qa—4lan—41UQKUQ 0 Ta)-

Since [U,,, Qa_5]Ua_4 SGa_4 = (Ma_5, U0), this implies that UaflTa g [U#,Q0_5]Ua_4.
Together with (4.1.1)(b) we get [Up,Qa_5,Ua] S [Ua_4,Ua]. Thus U0, normalizes

[U,,, Qa+5]Ua_4, a contradiction. Hence

[Um Qa—San—4 g 00-4)

for each u E A(oz - 5) fl A(a — 4). Since La_4 is transitive on A(a —— 4), we get

Xa_4 : Xa_4,a_3 and hence

[Xa—41Ual : [Xa—4,a—31 U0] _<_ [Za—4lva—31Qa—3la U01] S

[Za—41 Uallva—31Q0—31Qa—3] S 20—4-

Now the claim follows from a — 4 ~ 0 and L2H, 2: La_4 S (Ufa—4). -

(4.2.2) Assume that b = 4. Let (7, 7+1, 7+2, 7+3, 7+4) be a path such that (7, 7+4)
is a critical pair.
(a) U7+4Q7 : CQ7+1(OPI(]W7+1)1Q7)
(b) [U71U7+4l S 27-H-

(c) Assume that g E L, satisfies (4.1.3)(a). Then (,u.,7 + 2) is a critical pair,

for each u E A((7 +1)9)fl M7)

Proof. (a) follows from (4.1.1)(b) and (4.2.1)(b). (b) follows from (a).
(c) Suppose that 01,7 + 2) is not a critical pair. Then U,, does not centralize
[U,+4, Q,+3] by (4.1.3)(c). Pick :1: E L,+2 such that (7+ 3)$ = 7 +1. Since (b) ( with

95

7 + 2 in place of a ) implies that [U,+4, Q,+3]U,+2 S G,+2, it follows that U,, does not

centralize [Ii/3+4, Q,+1]. Since (b) and b > 2 imply that

X7 : U7Xg _<- Vyg+l S Q!“

7.7+1

we get [Uj,‘+4, Q,+1] Z X,, a contradiction. I
(4.2.3) Assume that b = 4.
(a) Y; s :r...
(b) Ya S CQa—1(Op’(A/Ia—1)1Qa—2)°
(C) Ya fl Qa—2 S CQO_3(Op,(A/Ia-3)1Qa—4)1

Proof. (a) Pick A E A(a), u E A()\), and 7 E A(a — 1) flA(a). By the choice of the

path (a — 4, . . . , a’), it follows from (4.2.2)(c) that (7, a + 2) is a critical pair. Hence

[U71 U11] S [U,,CQO_,(OPI(./Wa_l),Q,)] : n [U71Ug+‘2l S Tm

QEMa—l
by (4.2.2)(a)(b) and (1.2.4). Therefore, [Ya,0_1,Ya] S Ta. Since L0 is transitive on

A(a), (a) holds.

(b) From (a) we get [Ua_2,Ya,Ya] S [YmYmYa] = 1. Since Ya S Ma_1, it follows
that (b) holds.

(c) Since a - 4 E A(o — 2), (4.2.1)(a) implies that

[Ila—4, Ya fl Qa—21Ya fl Qa~2l S [Y —21Qa—21Qa—2l S [Xe—21Qa—2] S Tel—2-
From (a) we get

[Ua—4, Ya fl Qa—21Ya fl Qa-2l S Y; S Ta
and therefore

[Ua—4, Ya fl Qa—Qa Ya fl Qa—2] S To C Tar—2 = (Ta—4 fl Tel—fly"-2 = Ta—4 fl Tar—2°
Hence Ya 0620-2 acts quadratically or trivially on Ua_4Ta-4/Ta_4. Since Ya (1 (20.2 S

Ma_1, (c) now follows from (4.1.2), (4.1.4)(c) and the choice of (a — 4,. .. ,o’). -

96

(4.2.4) Assume that b = 4.

(a) Ya/Xa is an FF-module for LaQa/Qa, and [Ym La]Xa/C[ymLa]Xa(La,Xa)

is a natural Sp2n(q)-module for La.

(b) If p E A(o -— 1) fl A(a), then [U,,,Qa, L0,] = 1.

Proof. (a) Note that Ya/Xa is a module for LaQa/Qa. From (4.1.1)b) and (4.2.3)(c)
it follows that [Ua_4,Ya 0 620-2] S [Ua_4,Ua] S Ua. Thus Ua_4 centralizes (Ya fl

Qa_2)Xa/Xa. Hence
lYa 3 CYQ (Ua—41Xa)l _<_ q,

by (4.2.3)(b), and then (4.1.1)(b) implies that Ya/Xa is an FF-module for LaQa and
Ua_4 acts as an offending subgroup. Moreover, (4.2.2)(a) implies that if Ua_4 acts as
an offending subgroup on an FF-module W for LaQa/Qa, then W contains exactly
one nontrivial La-composition factor and that one is a natural Sp2,,(q)-module for
La.

(b) Pick u E A(a —— 1) fl A(a). By the choice of the path (oz -— 4,. . .,a’), it follows

from (4.2.3)(c) that (p, a + 2) is a critical pair. Hence

(*) [UmCQa—1(OPI(Ma—l)aQu)l S Tm

by (4.2.2)(a) and (4.2.3)(a). If Q06)“ 75 Qa_1, then (*) implies that [U,,,Qa, L0] = 1.

Hence we may assume that
(**) Qan : Qa—l-

By (4.2.1)(a),
A 2: Xa/Ta

is a module for Ga/Qa. From (4.2.1)(b) and [10] it follows that
(* * *) IA : CAI/(1)114, Loll S q-

97

Note that, if U,, H T,, = 1 or q # 2, then (*) implies that [U,,, Qa_1]Ta/Ta is an irre-
ducible MGM-module of order q2”“2. If U,, 0T” aé 1 and q = 2, then [U,,, Qa_1]Ta/Ta
is an indecomposable Ma_1-module of order 22"”, (U,, D Za_1)Ta/Ta is the unique

proper Ma_1—submod111e and has order 2. Hence in any case (:1: :1: =1:) implies that
[UmQa—llTa/Ta S CA(La)lA1Lal‘

Suppose that [U,,, Qa_1]Ta/Ta Z CA(LQ). Since
CZQ_1(LaaTo) S To,

the stucture of [U,,, Qa_1]Ta/Ta as Ma_1-module implies that
[Uu1Qa-llTa/Ta fl CA(La) = 1-

Hence [Uu,Qa_1]Ta/Ta is isomorphic to an Ma_1-submodule of [A, La]/C[A,La](La).
Note that (4.2.1)(b) implies that

[A, L0] = UaTa/Ta T—i Ua/(Ua 0 To).
In particular, q = 2 and
(* * * *) |[U”, Qa—1lTa/Tal = 612"“,

since Ua/(Ua 0 Ta) contains no irreducible MGM-submodule of order q2n‘2. Further-
more, the Ma_1-composition factor of order q""’2 in [U,,, Qa_1]Ta/Ta also appears in
U,,, whence M04 0 La acts nontrivially on this composition factor. Now (4.2.1)(b)

implies that this composition factor does not appear in A / [A, La]. Hence
[Uana—liTa/Ta S [A,La].

Since Q0 centralizes [A, La], this is a contradiction to (**) and (* * :1: =1:). Thus
[UpaQa—llTa/Ta S CA(La)a

i.e., [Up,Qa_1,La] S T,,. Now the claim follows from La : L’a and the Three-

Subgroup Lemma. I

98

(4.2.5) Assume that b = 4. Then [Ua+2, Ua_4, Ua_4] = 1.

Proof. Suppose that [Ua+2, Ua_4, Ua_4] 75 1. Define
Y 1: (ULi'zl-

0

Since [Y, Ua_4] S YO_2, (4.2.3)(b) applied to (Oz—2, o—4) in place of (01,0—2) implies

that
(*) [YaUa-41Ua—4] S Ta—2-

Suppose that [Y, Ua_4, Ua_4] S T,,. Then (*) implies that
[YaUa—41Ua—4l S T304 = Tar—4°

But then [Y, Ua_4] S 620-4, contrary to [Y, Ua_4, U04] # 1. Hence
(**) [Yan-41Ua—4] Z Ta-

Since (*) implies that
[Yan—41Ua—4] S U312? = Um

it follows from (an) that
U... s [Y,Lal-

Put

A I: [Y, La]/C[Y,LOJ(LQ) and
B I: UaC[Y,La](La)/C[Y,La](La)-

Suppose that A = B. Pick 9 E La such that (a+1)9 : a—l. Then Y = UZ+2[Y, L0] =
Ug+2UaC[y,La](La). Since d(a — 4, (a + 2)9) S 4 and Ua_4 S LOQO, it follows that

[Y, Ua-4a Ua—4] S [[Ug+21Ua—4][Uaa Ua—4]C[Y,L01(La)aUa—4] : 1)

99

a contradiction. Therefore
A ¢ B.

Now (4.2.4)(a) and (4.2.1)(b) imply that A/CA(LQ,B) is a natural Sp2n(q)-module
for LaQa/Qa. Put

C I: CA(UQ_4).

Note that C is an Ma_1-submodule of A by (4.2.2)(a). From (4.2.3)(b)(c) it follows

that IYa : Ya 0 620-2 (1 Qa_4| S q.2 and hence
(***) |A:C| S (12.

Suppose that LaRa/Ra 9—“ Sp4(2)’. Pick 513,31 E La such that KRa/Ra is a Levi
complement of Qa_1RO/Ra in (Ma_1 fl La)Ra/Ra. where K :2 (U§_4,U§,’_4). Note
that A, regarded as a module for K /CA(K ) (2 23), consists of two natural 23-
modules and trivial modules. Hence (* :1: =1:) and the projectivity of the natural 23-
module imply that A is completely reducible as K -module. But then U54 and hence
also Ua_4 acts quadratically on A. Since Ta_2 S Z0, this is a contradiction to (*)

and (**). Hence

(1k :1: *) LaRa/Ra '57—" Sp4(2)'.

Since U04 does not centralize the natural Sp2n(q)-modules B and A/CA(LO,,B),
(>1: =1: =1:) implies that A / C picks up the two Ma_1-composition factors on top of B and
A/CA(LO,B). Hence

[A,O”'(Ma_1),0”'(Ma_1)] g C.

Since (4: =1: *) implies that Ua_4 S 620-1 S QOOP'(MO_1)’, it follows that Ua_4 central-

izes A/C, a contradiction to (*) and (**). .

(4.2.6) b = 2.

100

Proof. Suppose that b > 2. Then b = 4 by (4.1.5). By (4.2.3)(c) there exist a — 5 E
A(a — 4) and a - 6 E A(a — 5) such that (o — 6, a — 2) is a critical pair. Note that it
follows from (4.1.1)(b) and the choice of the path (a — 4,. . . , 01’) that we can choose
:1: E Ua+2 such that (4.1.3)(a) is satisfied for (a — 2, a — 6) in place of (7, 7’) and some

9 E :cRa_2 fl La_2. Then (4.2.3)(c) implies that (a — 4, (a — 4)$) is a critical pair, i.e.,

Uri—4 aq— Qa—4-

But from (4.2.5) we get
(US/3:2) = Ua-4[Ua—41Ua+2] S Qo—47
a contradiction. I

4.3

(4.3.1) (.) X. s Q-

(mazmn.
(C) p = 2-
((1) G. = LQRO.

(e) l ana] 21'
(f) [XmQal S Ta-

(g) [Q.,L.] S X.-
Proof. (a),(b) Note that
(4:) [ZmOp’Ul/Ifl] S Q0 for all A E A(oz) and u E A(A),

since Z“ S M), and [UmeZu] = 1. In particular, (a) holds. If Z, 79 UaTa, then
U01 S [Zai,0p'(.M,\)], contrary to (*) and U,,: Q Q0.

101

(c) Note that each eigenvalue of an element of Ma+1flLa on C'Qa +1 (09' (Ma-+1): Qa) / Qa
is a square in CF (q), but each element of GP (q) is an eigenvalue of some element of
Ma+1 (1 La’ 011 Ud/[Ué’ Qa+1l-

Suppose that p is odd. Then there exists A E GF (q) such that A is not a square in
CF (q) Pick g E Ma“ 0 La, such that 9 acts on U01 /[U01, Qa+1l by multiplication by
A.

Suppose that g induces an outer automorphism on LaRa/Ra. Put
A :2 01”((M,,+1 r) Lama/Ra).

Assume that n > 2 or q > 3, and hence A = K R, / Ra for some p—component K of
MO“. Then K S R0, or K S La, by (2.2.6)(a). In both cases there exists h E La
that induces tha same automorphism on A/Op(A) as 9. But then gh'l induces an
outer automorphism on LaRa/Ra that centralizes A/0p(A), contrary to (A.4.2).
Assume now that n = 2 and q = 3. In this case A = —1 and we may choose 9
such that g E Ra: (1 L0,. If g induces an outer automorphism on A/Op(A), then
A :2 [A,g]OP(A) S (RarRa/Ra) 0,,(A) and hence

A : [Aagl0p(A) S ([Ra’a La’lRa/Ra) 0,,(14) S Qa’Ra/Raa

a contradiction. Hence 9 induces an inner automorphism on A/OP(A), contrary to

(A.4.2).
Thus 9 induces an inner automorphism on LaRa/Ra. Since LaRa/Ra = F *(G'a / Ra),

it follows that
g = 9192 for some 91 E (Ma+1 0 La) and 92 E R0.

Since Ua/[Um Qa+1] is dual to UaflUa/ as (Ma+lflLa)-module, there exists A1 E GF(q)
such that gl acts on Ua/[Um 6204.1] by multiplication by Al'1 and on U0, 0 U0: by mul-

tiplication by A1. Note that this implies that 91 acts on CQO+,(OP'(MO+1),Qa)/Qa by

102

multiplication by Af. Since [Qa+1,Ra] S Q0, also 9 acts on C'Qa+1 (OP'(MO+1), Qa)/Qa

by multiplication by A? Since (a) implies that

U.1Q. = Co...(0"'(M..1),Q.),

it follows that A is a square in CF (q), a contradiction.

((1) By (c), Q; S Q0: and hence [C22, U01] 2 1. Now ((1) follows from La S (Ucff).
(e) Note that [X,), Q0] S Z A for each A E A(a). Hence (e) follows from (d).

(f) From the definition of X0, and Ya it follows that [YmQa] S X... Since (4.2.6)

implies that La S Ya, (f) holds. I
(4.3.2) Assume that [Qm La] aé U,,.

(a) Qa+1 : Xan' = XOIQQ.
(b) U. o T. 751-
(C) n = 3.

(d) [Qm Lal/C[Q.,L.](Lm U0) is an 07(q)-spin-module for La.

Proof. (a) Suppose that Q0,“ 75 X06201. Then X0, S UaQal, since X0, S Ma“.
Hence [Xm U01] S Ua. Since La S (U53), we get

[X.,L.] S U01

contrary to (4.3.1)(f) and [Qm La] ¢ U0. The proof of Q0“ 2 XalQa is similar.
(b) Suppose that U0, 0 To, = 1. Then

Ua’ n Za+1 :- [Ua’v Ual S U01-
But also, by (a),
Ua’ n Za+1 : [Ua’aQa+11Xa] _<_ X;

Now (4.3.1)(e) implies that U0: (1 Za+1 S U0, 0 To, a contradiction.

103

(C),(d) By (4-3.1)(e),
14::[JYOaLOI/C[XO,L0}(L07U0)

is a module for Ga/Qa. Since a is conjugate to a’, (4.3.1)(f) implies that [X,, Q01] S

Ta, and hence

I4 3 CA(4Y0’)I S

 

[43(0) La] : [dyayLa] n Qa’l _<_ lXa : X01 n Qa’l-
Note that
lXa 3 X0 0 Qa’l Z 4211—1le0, 3X0’ 0 Qal-

Hence A is an FF-module for L, on which X0, acts as an offending subgroup. More-

over, by (4.3.1)(a)(e),
[A,XOI,XOI] = 1.

Since (a) implies that X,,: does not act quadratically on the natural Sp2n(q)-module,

it follows that n = 3 and A is an 07(q)-spin-module for La. I
(4.3.3) Assume that [(20, L0,] 2 U0. Then q E {2,4}.

PTOOf. Let g E A :2 CAIO+10LOI([UO’3QO+1]1Ua’ n Za+l)- Ifan’1Qa-t-l] : [UO)QO+1])
then [UayQa-Hagl S U0 0 Za+1 by the ChOICC 0f .9 If [Ua’1Qa-HI # [U01Qa+lla then
[UmQaH] (‘1 U,,: S Z0“, and since [Ua,Qa+1] S Q0: by (4.3.1)(a), we get

[UmQa+1,9lS[UmQa+1lfllQa',La'l S Ua fl Za+1'

Hence in any case

(*) an1Qa+119l S Ua n Za+1.

In particular, 9 centralizes 01"(MOH fl L0)Qa+1/Qa+1 and hence induces an inner
automorphism on LORa/Ra, by (A.4.2). Thus there exist 91 E A/IaHflLa and 92 E R0
such that g = 9192. By (1.1.2)(d), 92 induces GF(p)La-endomorphisms on Ua/(Ua m

104

Ta). Together with (*) it follows that g1 centralizes OP'(A/Ia+1flLa)Qa+1/Qa+1. Then
[Ua,Qa+1,gl] S U0, (1 Z,“ by (4.3.1)(e). Now it follows from (*) that 92 centralizes
[Um Qa+1]/(Ua fl Za+l) and hence also Ua/(Ua D To). Therefore,

(**) A S (M.+1 o L.)CGO(U., U. 0 T0,).

In particular, Ua/[Um Qa+1l is dual to (U0, 0 Za+1)/(Ua (1 Ta) as a module for A.
Assume that U0 0 Za+l S TaTau Then A centralizes (UaflZa+1)/(UaflTa) and hence
also Ua/[Ua,Qa+1]. Since CQO+1(OPI(A’IQ+1, Q01) is isomorphic to Ua/[Um Qa+1l as A-
module, it follows that q = 2.

Assume that U0 (1 Za+1 Z TOTO,“ Then (U0 (1 Za+1)/(Ua flTa) is isomorphic to (U01 0
Za+1)/(Uar 0T0) as A-module. Hence Uo/[Um QQH] is dual to (Ua: flZa+1)/(Uar 0T0)
and isomorphic to UaI/[Ua’1Qa+1l as A-module. Now (B.4.1.4)(b) implies that q E

{2,4}. _

105

Chapter 5

Determining the action of L on R,
Part 3

5.1

In this section we assume (1)—(IV) and
(V) LaRa/Ra E“ Gg(q)’ (q = 2" for some k E IN), and p = 2.

Note that (2.3.2) implies that U,,,TCJTor is the Gg(q)-module listed in (A.2.2)(i).

(5.1.1) Let (7, y) be a critical pair.
(a) [U71U7’I = Uv 0 Q) = U7 0 Q7“
0)) IU) 1 U7 0 Qvl = lUv’ 3 U7’ 0 Q7l = Q3-
(0) Uv'Rq/Ra = CC,/R,(U,R,/R,).
Proof. [14](1.2). I
(5.1.2) b g 4.

Proof. Suppose that b > 4. By (5.1.1)(c) and (1.2.6) assumption (iii) of (1.2.5) is
satisfied. Since M0,“ is a maximal subgroup of Ga, also assumption (ii) of (1.2.5) is

satisfied. Hence (1.2.5) implies that there exists 1/ E Ema” such that

I H ZB,L.1=1.

9€M0+1

106

Define

D2: fl [UmUS].

96M0+1
Then D S Ta. But either U,,Ra/Ra is normalized by Ma+1/Ra or IUuRa/Ra fl

Z(AIaH/RQM = q, and therefore in any case D Z Ta, a contradiction. I

Choose 90 E L, such that Uar Q A131,, i.e.,
G0 = (McicliaUa'A
Define
a — 1 2: (01+ 1)“ and
a — 2 :2 (a+2)9°.
(5.1.3) Assume that b = 4.
(a) (a — 2, a + 2) is a critical pair.
(1)) Qa—l = Qa—2Qa-

Proof. (a) follows from (1.2.6), (1.2.1)(e) and (5.1.2).

(b) Assume first that Ua+2R0_2 SM0_1. Then Z(Ma_.1/Ra_2) S Ua+2Ra_2/Ra_2.
Since Qa_1Ra-2/CQO_,RO_2(M0_1, 120-2) consists of two irreducible MGM-modules of
order q2, it follows that 620-1 S (Ufi’2")Ra_2. Now QaRa_2 S Ma_1 implies that
Qa-l S QaRa—2 and hence Qa—l = Qa(Qa—1 fl Ra—2) = QaQa—2-

Hence we may assume that

(*) Ua+2Ra—2 g Ma—l-

Let K be a Sylow 3—subgroup of Ma_1 (1 L0 or the 2-component of M04 (1 La,
depending on whether q = 2 or q > 2. From Ga 2 (Ma_1,Ua,), b = 4, and (*) it

follows that
[Ua—21Ua+2l E To-

107

Hence K centralizes [Ua-2, UQH], which implies that
K S R04.

Since the two Ma_1/Ra-chief factors of order q2 in Qa_1Ra/Ra 2: [Qa_1, Ma_1]Ra/Ra
are contained in [Qa_1,K]Ra/Ra, we get Qa_1 S [Qa_1,R0_2]Ra S Qa_2Ra and
hence Qa—l : Qa—2(Qa—l (1 Ba) : Qo-2Qa- I

(5.1.4) b = 2.

Proof. Suppose that b > 2. Define
Y := [<U£:2>. Q.lU..

Note that [Y,Qa] S R04 by (5.1.1)(b). Hence
(*) IY : Yo Q._2l s 42

by (5.1.3)(b). Pick ga_2 E Lo,_2 such that Ua+2 Z M3112 and put
a — 4 2: 09°"?

As in (5.1.3)(a) it follows that (a — 4,01) is a critical pair. Since Y centralizes U0,

(5.1.1)(c) implies that

(**) Y n Qa—2 S CY(Ua—41[Y1Qaan)-

Since (52(q) does not have an FF -module in which the index of the centralizer of an

offending subgroup is smaller than q3, it follows from (*), (*4), and (5.1.1)(b) that
[Y,L.1 g [Y, Q.]U..
But then

[Ua+2a Q01] S [Ua—21Qally1Qaan S Qa—21

contrary to (5.1.1)(b) and (5.1.3)(b).

108

(5.1.5) (a) [Um gm] = U... n Q0, = U... n U0: = U0, n Q... = [UOI,Q;+1].

(b) [620, La] = U...

Proof. (a) This follows from (5.1.1)(a)(b) and (5.1.4).
(b) From (a) and L0 2 (U5?) we get

[QaaLa] S ([QaaUa’lLo) : ((U0 0 Ua’)LO> 2 U0:-

Hence (b) holds. I

(5.1.6) Assume that q ¢ 2. Let K be the 2-component of N10“ (1 La.

(a) K S L0,.

(b) Ra! = 620.4120, 0 Ba).

(C) [Uau 80’] = 1-
Proof. (a) Note that (5.1.5)(a) implies that [UmK] Z U0, 0 Qat. Hence K Z Rat.
Now the claim follows from (2.2.6)(a).
(b) Suppose that Ra, Z Qal(Rar ORG). Since RalRa/Ra S1 Ma+1/Ra, it follows that

QC,“ S RaRaI. Hence [K, Rat] Z Q01, contrary to (a) and (1.1.2)(d).
(c) From (b) and (5.1.4) it follows that [U05 Bar] 76 Uat, i.e.,

[Ua’a Ra’] S Ta’ °
Now the claim follows from (1.1.2)(d) and the Three-Subgroup Lemma. I

(5.1.7) q = 2.

Proof. Suppose that q > 2. Let S be a Sylow 2-subgroup of MO“. Let H be a sub-
group of Ma+1flLa such that H Ra / R0 is the intersection of Can / Ra (Z (S Ra / Ra)) and
a Cartan subgroup in NMa+1/R..(SRa/Ra). Then q > 2 implies that [Um Qg+1]Ta/Ta

contains no trivial H -composition factor.

109

Suppose that H Z LalRal. Define
71173:: Alo+1 ma...

Let K be the 2-component of Mo,“ H La. Then
K E“ SL2(q) and HK ”-3 GL2(q).

Pick h E H and k E K such that h ¢ LOIRQI and 1175 E Z(HK). From (5.1.6)(a)
it follows that hk g! LQIROI, i.e., hk induces an outer automorphism on LazRal/Ral.
But also by (5.1.6)(a), K is the 2-component of Ma+1flLaI and hence [K, hk] S Q0“
implies that his induces an inner automorphism on LaiRal/Rar. This contradiction

shows that
(*) H S (Ma+1 fl L0,)Rar.

Since, regarded as (MQH fl Lad-module, [Uar, Qg+l]Za+1/Za+1 is dual to the central-
izer of Q0“ in UaI/[UOI,Q:1+1], it follows from (*), (5.1.6)(c)(a), and the first para-
graph that the centralizer of Qa+l in U0, / [U05 Qh+1l Contains no trivial H -composition

factor. But
Z(SRa/Ra) S CUa; (Qa-Ha [Ua’i Qg+1])Ra/Raa

and H centralizes Z(SRa/Ra), a contradiction. .

5.2
In this section we assume (I)*(IV) and
(V) LaRa/Ra E“ Qf0(q) (q : pk) for some k E IN.
(VI) U0 is a half-spin module for L0.
(5.2.1) Let (7,7’) be a critical pair.

(a) [U., U71] = U7, n Q. 2 U7 n Q...

110

(b) lU7 : U7 flQ71|=|U7r :Uv’ nQvl = (18-
(c) U716)? = Q; where A 6 Ah) 0 A(b‘1)(7’).

(d) (MA 0 L7)R,,/R,7 is a parabolic subgroup of cotype 1 in LJLJR, where
A 6 N?) n Nb"”(7’)-

Proof. [14](1.2) and [16]. I
(5.2.2) b g 4.

Proof. Suppose that b > 4. By (5.2.1)(c) and (1.2.6) assumption (iii) of (1.2.5) is
satisfied. Since Mo,“ is a maximal subgroup of Ga, also assumption (ii) of (1.2.5) is

satisfied. Hence (1.2.5) and (5.2.1)(c) imply that
W... Hal] 3 T...
contrary to La S (Hi?) I
Choose go 6 LOr such that U0: Q 5131., i.e.,
Ga =<A1£11anl
Define
a —1:=(oz+1)9° and
a— 2 2: (0+2)9°.
(5.2.3) Assume that b > 2.

(a) (a — 2, a + 2) is a critical pair.
(b) [Ua-Q) Ua+2l S To.

(C) OP'(MO,_1) 5 120-2120.

111

Proof. (a) follows from (1.2.6), (1.2.1)(e) and (5.2.2).
(b) From (1.2.4) and (5.2.1)(c) it follows that LG centralizes [Ua_2,Uo+2]. Also
(5.2.1)(c) implies that (20-1 centralizes [Ua_2, Ua+2].
(c) Let K be the (by (5.2.1)(d) unique) p—component of Ma_1 fl LO_2. Then
WWI/0.2.15) ,2 1.
Thus (b) implies that K Z L0,. Hence K S Ra by (2.2.6)(a). .
(5.2.4) b = 2.
Proof. Suppose that b > 2. Define
Y 3: (U512) Za-
From (5.2.1)(c) we get
(*) [Y : Y 0 620-2] S q8.
Pick 90-2 6 La_2 such that Ua+2 Z Mgr," and put

a — 4 :2 029°”.

As in (5.2.3)(a) it follows that (oz — 4, oz) is a critical pair. From (5.2.1)(c) it follows

that
(H) Y D Qa—2 S CY(Ua—4i U0)’

Note that Y/Za is a module for Ga/Qa by (5.2.3)(b). If [Y,La] S Z0, then Ua+2 S

Ua_2Za S Q04, contrary to (5.2.3)(a). Hence
[Y, La] Z Z0.

Let W be a GF(p)(LaRa)-composition factor of Y/Za that is not centralized by La.
By the same argument as in the proof of (3.2.10) it follows that, regarded as GF(p)La—
module, W is the direct sum of 8 copies of some irreducible GF(p)La-submodule E
of W. Then (*) and (an) imply that

(an: :1: *) IE : CE(UQ_4)| S q.

112

In particular, E is an FF -module for LOUQ_4 and Ua_4 acts as an offending subgroup.
By (5.2.1)(c), Ua_4 does not act as an offending subgroup on the natural module.

Hence (A.2.2) implies that E is a half-spin module. But then (4: * *) contradicts

(5.2.1)(5). _

(5.2.5) [em/La] = U...

Proof. From (5.2.1)(c) (with (a',a) in place of (7,7’)), (5.2.4), and La S (U?) it
follows that

[QaaLal S ([QaaUa’]La> S ([Qa-l—laUa’lLa) Z ([UaQa’aUa’]La) S Ua-

5.3
In this section we assume (1)—(IV) and
(V) Qa+1fl La S Ra-

(5.3.1) (a) One of the following holds:

(al) 00/120 2’ 03,,(2) and U0, is a natural Q;,,(2)-module for La for some

€E{+,—} andnEleithn24.

(a2) Ga/Ra E“ 2,, and UaTa/Ta is a natural An-module for La (n E ]N
with n 2 5).

(b) Qa+1 : QaUa’-
(c) Qa+l acts as a transvection on U0. In particular, |Ua : U0, 0 Za+1| = 2.
(d) Ma-H : 000([UaaQa+llTa/Ta)-

(e) MO“ is a maximal subgroup of Ga.

113

Proof. This follows from (2.2.4) and (2.3.1). .

In chapter 1 the critical pair (a,a’) was chosen such that Zar acts as an offending

subgroup on Z0. By (5.3.1)(c) we now have symmetry in a and a’.

(5.3.2) Let (7, 7’) be a critical pair and 7+2 = 7’ — b+i E Alilh) fl Alb")(7’) for each
2'6 {1,...,b}. Let A 6 Ah)

(a) G') 2 (AI), Q'y-H) if and only if lQAaQ7+ll Z Q7-
(b) Assume that G, 2 (MA, Q7“). Then (,u, 7’ — 2) is a critical pair for each
,u E A()\) with ZflZ, 2107.

Proof. Since (5.3.1)(c) implies that (a, a’) is an arbitrary critical pair, it suffices to
prove this for (7,7’) = (a, a’).

(a) Note that [QM QQH] S QC, is equivalent to Q0“ S Nga (QA), since Q0 has index
2 in QA and is normalized by Q0“. Hence (a) follows from (5.3.1)(e).

(b) Since ZflZa is normalized by MA, (5.3.1)(b) implies
[ZanI] Z ZHZO.
Again by (5.3.1)(b) and the remark following (5.3.1),
[Qa’—1aUa’l = anQa’aUa'l = [UaaUa’l S Z0.
Hence Z), Z 6201-1. In particular, 2,, Z Qa:_2. .
(5.3.3) Assume that b > 2.
(a) b = 4.
(b) [UaaQa-Hl S Ta+2-

Proof. Note that (5.3.1)(a) implies A010” 2 [Ua,Qa+1] Z T,,. Moreover, by (5.3.2)
assumptions (ii) and (iii) of (1.2.5) are satisfied. Hence the claim follows from from

(1.2.4) and (1.2.5). _

114

For ’7 ~ a define
X7 3: ZV<U6 l (“745) = 2)-
(5.3.4) Assume that b = 4.

(a) [XmQa] E To-

(b) Q2, +1 is elementary abelian.

Proof. (a) follows from (5.3.3)(b).
(b) From I) = 4 and (5.3.1)(b) it follows that

Qa-H = QaXa+2
and, hence,

Q2,“ = QLXL+2lQm Xa+2l = leQO-l-Ia Xa+2l~
Note that

[Qa+1,Xa+2l = [XaQa+2,Xa+2l S [XmXa + 2]Ta+2 S Xa-
Hence

[Qa-l-laXa-i-QaQLJ S [1¥0,Qa] 5 Ta and
(*) [Qa+1,Xo+2ll S X; S Ta,
by (a). Clearly

A :2 [Qa+1a Xa+2a lelQa+lv Xa+2]’

is a normal subgroup of MO“. Since [A,La] S [TmLa] = 1, we get A S Ga. Now

3+1 = QZA implies that
Q2,“ : 1-

115

In particular,

(**) ¢(Q;+1)= ¢(Q;)¢(lQa+1aXa+2l)-

Since X0, is generated by involutions, (*) implies

(I)([QO+13XO+2]) S Ta-

As above we get that (I>([Qa+1, Xa+2]) S Ga, and then (b) follows from (**). I

(5.3.5) There exists /\ E A(a) such that (Qa+1,QA)Ra/Ra E’ 23.

Proof. It suffices to show that QaHRa/Ra is contained in a subgroup of 670/120, that
is isomorphic to 23. This is evident if (a2) holds in (5.3.1)(a).

Assume first that Ga/Ra is isomorphic to 03,,(2) for some n 2 4. Then Q0“ induces
a nontrivial graph automorphism on a Levi complement X of a parabolic subgroup
of type {71 — 2,72 — 1,71} of LaRa/Ra. Since X 2’ PSL4(2) ’-_‘-’ A3, it follows that
(Q0+1Ra/RO)X E” 238 and then the claim is obvious.

Assume now that Ga/Ra is isomorphic to 02",,(2) for some n 2 4. Then Q0,“ in-

duces a nontrivial graph automorphism on a Levi complement X of a parabolic sub-
group of type {n — 1,72} of LaRa/Ra. Since X ’_—‘1 PSL2(4) ’:‘:’ A5, it follows that
(Qa+1Ra/RO)X E“ 25 and again the claim is obvious. I

Choose go, 6 Lo, such that (Qa+1, 3:1)Ra/Ra E 23 and put
a —1:=(a +1)9° and
a— 2 :2 (a+2)9°.

(5.3.5) b = 2.

Proof. Suppose b ¢ 2. Then b = 4 by (5.3.3)(a). Pick 11. 6 U0), \Qa. Note that the

choice of go, implies that

lQa—la 31—1] g Qa-

116

Therefore
Ga = (U353 ‘Mg—i)

by (5.3.2)(a). From (5.3.2) and (1.2.1)(e) it follows that Ua+2Ua SGO, and hence
L7(a_2)u U0, SGO.

Now from (5.3.2) (with (a, (a’)9“, (a -— 1)“, (a — 2)“) in place of (7,7’,/\,p)) we get
that (a — 2, (a — 2)“) is a critical pair.
Pick 2) E U04 \ Q34. Then U“ ¢ 620-2, i.e., [U,U“] 76 1. But then [22,21] is not an

involution. Since U, u 6 Q0“, this is a contradiction to (5.3.4)(b). I

(5.3.7) [Q..L.] = U...

Proof. This follows from (5.3.1)(b) and (5.3.6) the same way (5.2.5) follows from
(5.2.1)(c) and (5.2.4).

5.4

In this section we assume (1)—(IV) and
(V) LaRa/Ra T—l 97(q) (q 2 pk) for some k E IN.
(VI) U0 is a spin module for La.
(5.4.1) Let (7,7’) be a critical pair.

(a) [U72 U71] = [U., Q] = U7’ 0 Q = U7 0 Q7! where A E Ah)r1 A(”“)(7’)
(b) IU, : U, D Q71] 2 |U7: : U7: 0Q7| = q4.
(c) (MA fl L,)R,,/R,, is a parabolic subgroup of type 82 in L7117 /R,7 where
/\ 6 AW) 0 A‘b‘llh’)
Proof. [14](1.2), [16], and (B.3.2.1). .

117

(5.4.2) UaIRa/Ra SMa+1/Ra. In particular, b yé 2.

Proof. By (5.4.1)(c) QOHRa/Ra contains no MON-submodule of order q“. Hence the

claim follows from (5.4.1)(b). I

(5.4.3) A contradiction.

Proof. Using (5.4.1) in place of (5.2.1), it follows as in the proof of (5.2.2) that b S 4.
Hence, by (5.4.2),

b=4.

Pick g 6 La such that U0: Z Mg“. Then (5.4.1) and (1.2.6) imply that ((a+2)9, 0+2)
is a critical pair.

Suppose that [U§+2,Ua+2] S Ua. Then (5.4.1)(a) and [Ua+2,Uar] 2: 1 imply that
[UmQflH] = [Um QMQ]. Since NI... ([Ua, Qa+1]) = Ma“ 0 La, this is a contradiction

to the choice of 9. Hence
[Ug+21 Ua+2l S Ta-

But this implies that the p—component of Mg“ (‘1 L3,” is contained in R0, and hence

normalizes Ua+2, contrary to (5.4.2) (applied to ((a + 2)9, a + 2) in place of (a, (1’)).

118

Chapter 6
Proof of Theorem 1

Let G, M, R, L, L1,...,L,,,, and p satisfy the assumptions of Theorem 1. By

(2.2.1)(b)(c) L is the product of L1, . . . , Lm, and M permutes L1, . . . , Lm transitively.

Put
M 1: n n NM(L,‘)¢,
aseAutw) i=1
6' I: A7111,
H 2: fl 1W9,

965

Z := (91(Z(0p(M)))G>.
I7 2: [2,L1], and

By (2.2.2)(a)(c) and (2.2.11) assumptions (1)—(IV) of chapter 1 are satisfied for 5',
A7, H, L1, and the group of automorphisms of C? that are induced by NM(L1) in
place of G, M, R, L, and H. Moreover, 2, l7, and T play the role of Za, U0, and

To, respectively. Note that 0,,(M) = 0,,(M) by (2.2.2)(c), and hence 0,,(G) S
Rfl CPU?) 2 0,,(6'). Now L1 2 U1 and (1.1.2)(d) imply that

(*) [R,L1l=lR,L1,L1l S [0,,(G),L1] S [01450.15]-

119

Assume LIE/H is not an orthogonal, symplectic, or unitary group. Then (A.2.2)
implies that LlH/H E“ 62(2k)’ for some k E IN, since U1 is an FF-module for L1 and
we assume that LIE/H is a group of Lie type. Now (5.1.5), (5.1.7), and (*) show
that case 10 of Theorem 1 holds for 2 = 1.

Assume LIE/H is an orthogonal, symplectic, or unitary group. If 0 HT ¢ 1, then by
(2.2.11)(d) and the assumption that LlR/R belongs to 2,, we can apply (2.3.1), and
thus (4.3.2), (4.3.3), and (*) show that one of the cases 1 or 4 of Theorem 1 holds for
2 = 1.

Assume now that (7 0T 2 1. If (M n M)R/H is not a parabolic subgroup of LIE/H,
then (2.2.4), (5.3.6), (5.3.7), and (air) imply that case 8 of Theorem 1 holds for 2 = 1.
Hence assume that (H F) L1)E/ B is a parabolic subgroup of LlH/H. If U1 is not a
natural module for L1, then (A.2.2), (5.2.5), (5.4.3), and (*) show that one of the
case 9 of Theorem 1 holds for 2 = 1.

Hence assume that U1 is a natural module for L1. Suppose that A? is not the sta-
bilizer of a singular subspace of U1. Then by (2.2.4) LIE/H is of type D4, and the
parabolic subgroup (ll/7 flLQH/H belongs to the interior node of the Dynkin diagram.
Moreover, N M(L1) permutes the maximal parabolic subgroups of L1H/ E containing
(M n M)R/H transitively. But this is a contradiction to (A.2.9). Hence A? is the
stabilizer ofa singular subspace of U1. Now (3.5.2), (4.3.2), (4.3.3), and (*) show that
one of the cases 1,2,3,5,6, or 7 of Theorem 1 holds for 2' = 1.

Since M permutes L1, . . . , Lm transitively, all of this holds for any 2' E { 1, . . .,m} in
place of 1.

Suppose that [L,, Lj] ¢ 1, for some 2,j E {1, . . .,m} with 2 75 j. Since L,- and L,- are

p—components,

[Li’ 1’1"le : [L15 Li, Lil : [[42le S 012(0) S R-

120

Note that in the cases 1-3 and 5—10 of Theorem 1 we have [[R, Li] 0 [R, Lj], L] = 1
by (2.1.1)(g), and hence

114,111,“: [[L,-,LJ-,L,-]n[LJ-,L.-,L,-],L] S [[R,lefllR,L.r],L] =1,

contrary to 1 76 [L,-,Lj] = [L,, Lj,LJ-]. Hence we are in case 4 of Theorem 1. Pick
a subgroup X of [L,-,Lj] such that [L,-,Lj]/X is an irreducible LiLj’HlOdUle. Then
(A.3.1) implies that [L,-,Lj]/X contains 8 nontrivial Li-composition factors. But
0,,(6') contains only 2 nontrivial Li-chief factors. This contradiction shows that L is

the central product of L1, . . . , Lm.

121

Appendix A

General Lemmas

A. 1 Various Results

(A.1.1) Let X and Y be finite groups such that Y acts on X. Assume that X is the

direct product
JYZXI X... XX1n

of at least two subgroups X1, . . . , Xm, that are transitively permuted by Y.

(a) X’ g [X, Y].

(b) [X , Y] /X ’ contains a subgroup that is isomorphic to a direct product of

m — 1 copies of Xl/Xi.

(c) X = [X,Y]X,, for each 2 E {1, . . .,m}.

Proof. (a) Let 2 E {1, . ..,m}. Pick y E Y such that Xg" 75 X,. Since [X,-,Xf’] =1,

we get
(a. ylib, yiiab. y)" = a'1ayb‘1by<(ab>y)—‘ab = a—‘b—labaybymyby)” = [.,b],

for all a, b E X,.
(b) For each 2 E {1, . . .,m — 1}, pick y, E Y with X3" = Xm. Then

A2 3: [XiayilX’

122

is a subgroup of [X, Y] with Ai/X’ ’-_‘-’ Xl/Xi and
Aim/ll....-A,-_1A,-+1-...-A,,,_1SAiflX1-...-X,-_1X;X,-+1-...-X,,,SX'

for each 2 E {1, . . . , m — 1}. Hence A1 - ...-Am_1/X’ has the desired properties.
(c) If i,j E {1, . ..,m}, then there exists y E Y such that X,” = X,- and hence

)(J' Z [)(J‘, y]4Yi Z LY, Y]4Yi. I

(A.1.2) Let G be a finite group, A! a subgroup of G, and H a subgroup of Aut(G).
Assume that the following hold:
(i) A! is H-invariant.
(ii) MH is a maximal subgroup of GH.

Put R :2 ngEG Mg and H :: flg€GH(MH)9. Let N be the set of all minimal

normal subgroups of G / R. Put

No := {N0 ENINfl(N0”) = 1 for each N EN\N0H}.
For each N E N define

(MN) 1: (ané/R,

whereRSXSGand X/RzN.

(a) If N E N is perfect, then N E No.

(b) Let N E No. Then ¢(N) is a minimal normal subgroup of GH/H. More-
over, if the H—orbit of N in N has size k, then ¢(N) is isomorphic to a
direct product of at least min{2, k} and at most k copies of N. In partic-
ular, ¢(N) is perfect if and only if N is perfect.

(c) If N1 6 No and N2 6 N, then ¢(N1) = ¢(N2) if and only if N1 and N2 are

conjugate under H.

123

(d) The elements of N are pairwise isomorphic.

(e) If GH/H has a nonabelian simple normal subgroup A, then each elements

of N is isomorphic to A and normalized by H.

Proof. (a) This is clear, since the set of components of a direct product is the union

of the sets of components of the direct factors.
(b) Let X g G such that R g X and X/R = N. Let Y s (X”)B such that R g Y
and Y/H is a minimal normal subgroup of GH/E. Then

Y=(Yn(x”))f2 and YZR

Hence Y 0 (XH) Z R. Choose X1 S Y O (X") such that R S X1 and Xl/R E N.
Then N and Xl/R are in the same H-orbit of N, since N E No. Hence ¢(N) =
¢(X1/R) = Y/H is a minimal normal subgroup of GH/H. The second assertion

follows from
¢(N)=(X”>1~?/R ’5 <X”>/<X"> 0 i? = <X”>/R-

(c) Clearly ¢(N1) = ¢(N2) if N1 and N2 are conjugate under H. Now assume that
(fit/VI) 2: $(N2). LCL X1,X2 _<_ C SUCh that R S X1,X2,X1/R : N1 and Xz/R = N2.
Then (XIH) S (XQH)H. Since Rn G = R, we get

(X1H>.<_ (XzfllR = (X21),
and likewise (X?) S (XIH). Hence
Wt”) = (N251).

Since N1 6 N o, it follows that N2 is conjugate to N1 under H.

((1) Note that if N E N is abelian, then ¢(N) is an abelian normal subgroup of
GH/E and hence GH/H has an abelian minimal normal subgroup. Since, by [12],
the minimal normal subgroups of GH/E (at most two) are pairwise isomorphic, (d)

now follows from (a) and (b). I

124

(A.1.3) Let G be a symmetric group of degree 2“ for some 12 E IN \ {0}. Put
X.- z: ({r | 2"(j — 1) <r g 2"j}|1SjS 2H}
for each 2 E {0, . . . , k}. Then
(a) Nc:(U1-°=o X 2') 6 8.142(0).
0)) Z(NG(Ulc=0Xi)) = ((1 2) (3 4)-~(2'°-1 2"»
Proof. This is obvious when k = 1, so assume that k > 1. Put
A 2: (30({2’c-1 + 1,. . .,2“)),
B :2 CG({1, . . . ,2k—1}) and
:2: := (1 2‘“1 +1) (2 2’“1 + 2) (2"-1 2k).

By induction on k we may assume that

k
NA(U A1226 Sy12(A),

k k

k k
NG(U X.) = (NA(U x.) x NB(U X,))(;t) e NA(U 21’.)ng
i=0 2:0 2:0 2:0
is a Sylow 2-subgroup, since it has the appropriate order. Moreover,

I:
Z(NG(U 9(2)) = C

H) z(N,t(Uf=0x.))xz(1vB( i=0 “)(2) =

((1 2) (3 4)...(2k—1 2k)).

125

(A.1.4) Let G be a finite symmetric group of degree n 2 7. Then for each 5' E Syl2(G)

the following hold:
(a) Z(SflG’) S Z(S).
(b) CGI(Z(S n G')) 75 3 F) G’.

Proof. (a) Assume that n = 2k for some k E IN. Put
H :2 C(;({1, . . . , 2k‘1}) and
x:(1?*+u(2fl*+2)u.m“1m.

Let T be a Sylow 2-subgroup of H. Then
S := (T, :r) = (T1 X T)(2:)

is a Sylow 2-subgroup of G, and
(*) Z(S) = CZ(T’)><Z(T)(I)-

If Z(SflG’) Z T1 x T then TflG’SS, a contradiction to TINT = land lTflG'I Z
2k'2_>_2. Hence Z(SflG’) STIXT. Leta: TxxT—>T‘” andfi: T’XT—>Tbe
defined by

y = yayfl for each y E T“C x T.

Then (TI x T) HG" centralizes ya and y5 for each y E Z(SflG’). Since Z(SflG’) HT
is centralized by ((Tz x T) 0 (1’)" = T, we get

Z(S r1 0’) r1 T g Z(T)
and likewise

msnaynvgzafl.

126

Hence
Z(SnG’) = Z(SnG’) r1 (T1r x T) g
(Z(SflG’) 0 (Tx x T))" x (Z(SflG’) F) (T1 x T))3 S
(Z(SriG’) n TI) >< (Z(Sn G’) n T) g Z(T“) x Z(T).

Since 2"”1 is even, S H G’ contains 1:. Now (a) follows from (*).
Assume that n = 21:12“ for some integers k1,...,km with m > 1 and 1 S k1 <
< km. For each 2 E {1, . . . , m} let T,- be a Sylow 2-subgroup of
H,- :2 00({r | 231:, 2*: < r g 231:, 2’2}).
Then
5:: (T1,...,Tm) :2“, x me

is a Sylow 2-subgroup of G. Define 02,-: S —> T,- (2 E {1, . . .,m}) by

01 02 a

yzy y -...~y'" foreachyES.

Note that m > 1 implies (SflG’)"‘ = T,- for each 2 E {1, . . . ,m}, whence
Z(SnG’) g Z(SnG’)°1 x x Z(SflG’)“’" g
CT,((SflG')‘“) x x CTm((SflG')""‘) = Z(Tl) x x Z(Tm) = Z(S).

(b) If n = 7 and S 6 Syl2(G) then Z(S (1 G’) has 3 fixed points on {1, . ..,7}, so
CGI(S 0 G’) contains an element of order 3 permuting the fixed points of Z (S n G’).
Now assume n 2 8. Let k1, . . .,km,T1, . . . ,Tm and S be defined as in the proof of
(a) (except that now possibly m = 1). From (A.1.3) and n _>_ 8 it follows that Z (T,,)
has order 2 and at least 4 orbits of size 2 on {1, . . . , 72}. Hence CGI(Z(S)) contains
a subgroup isomorphic to A4 permuting the Z (Tm)-orbits of size 2. Now the claim

follows from (a). I

The following lemma from [14] will be particularly useful.

127

(A.1.5) Let G be a group, p a prime, and V a faithful GF(p)G-module. Assume that

there exists a non-degenerate G-invariant symmetric GF(p)-bilinear form on V.

(a) [V, A]i = CV(A), for each A S G.

(b) Assume that B S A S G, [V,B,A] = 1, and [Ct/(B),A]fl[V, B] = 1. Then
[V2 Al : [CV(B): A]]V, Bl'

Proof. If p = 2, then this is [14](1.7). The proof for an arbitrary prime is the same. -

A.2 FF-Modules

Let p be prime and G a finite group. Define
8,,(G) := {A S G I A is a nontrivial elementary abelian p—group}.

If V is a faithful finite-dimensional GF(p)G-module, put
P(G, V) I: {A E 8,,(G) | B S A => IAI le(A)| Z IBI le(B)|} and
P‘(G, V) := {A e 19(G, V) | A 2 B e ’P(G, V) => A : B}.

If V is any finite-dimensional GF(p)G—module, put
P(G,V) == {A S G l ACG(V)/CG(V) E 1’(G/CGW), V)} and
BAG, V) 3: {A S G I ACG(V)/CG(V) E P‘(G/CG(V),V)}.

(A.2.1) Let G be a finite group, p a prime, V a faithful GF(p)G—module, and A E
’P*(G, V).
(a) [V,A, A] = 0.
(b) IfV 2 W 2 [W, A] aé 0, then ACG(W)/CG(W) E ’P“(NG(W)/CG(W), W).

(c) If W is an A-invariant subspace of V such that CA(V/W) = 1, then
V/W is an FF-module for NG(W)/CNG(W)(V/W) and ACNG(w)(V/W)/

CNG(W)(V/ W) is an offending subgroup acting quadratically on V/ W.

128

(d) A normalizes each component of G.

Proof. (a) This follows from [5](4.2).
(b) If B S A, then

IBCMWN ICE/(B CA(W))| S W ION/1)],
since A E 'P(G, V), and therefore

IBCG(W)/CG(W)| lCW(B)l = lBCA(W)CG(W)/CG(W)I lCW(B CA(W))| =
lBCA(W)l
lCBCA(W)(W)l

lBCA(W)l IBCA(W)I le(BCA(W))l|Wl
ICA(W)l ICA(W)] le(B CA(W)) +Wl -

IBCA(W)| ICV(BCA(VV))I lWl < IA] lot/(All “W =
ICA(W)| IONA) + W] _ IC.4(W)I le(A) + W]

__ IBCA(W)I

lCu/(BCA(W))I— ICA(W)| ICW(BCA(W))l=

 

le(B CA(W)) 0 WI =

 

 

 

IACa(W)/CG(W)| ICw(A)|-

(c) This is obvious.

(d) [4] .

(A.2.2) Let G be a finite group, p a prime, and V a faithful and irreducible FF-module
for GF(p)G. Assume that the following hold:

(i) F‘(G) is quasisimple, and F‘(G)/Z(F*(G)) belongs to the class [,p, as

defined in the introduction.

(ii) G = AF”(G) for some A 6 ’P“(G, V).
Then one of the following holds:

(a) G E’ SLn(q) (q 2 pk), and V is a natural SLn(q)-module.

(b) G E“ SLn(q) (q 2 12", 72 Z 4), and V is the second exterior power of a

natural SL,,(q)-module.

129

(c) G ”-3 Sp2,,(q) (q = pk), and V is a natural Sp2n(q)-module.

(d) C ’-‘—-’ Sps(q) (p = 2, q = 2’“), M = (18, lCi/(All = lthAll = q“, and
[V,Al = [V,CcfiVtAlll.

(e) G ”3:“ Q:(q) (q 2 1.)", E E {0, +, —}), and V is a natural Qf,(q)-module.

(f) G ’55 03,,(q) (p = 2, q = 2", 5 E {+,—}), and V is a natural O§n(q)-

module.
(g) G ”—3 Qfo(q) (q 2 pk), and V is a half-spin module.
(h) G ’-_‘-’ SUn(q) (q = pk, 22 Z 4), and V is a natural SUn(q)-module.

(i) G ’5 62((1) (p Z 22 q = 2k): [V] 2 (16a [Cl/(44)] : llVrAll : [A] = (13’ and
00(4) = A.

(j) G ”E X,,, p = 2, and V is a natural X,,-module.
(k) G 91 A2", p = 2, and V is a natural Agn-module.

(1) G g .46, p = 2, |V| = 26, |V/CV(A)| = |A| = 4, |[V,A]| = 15, and
CG(A) = Z(G)A.

(m) G 2 47. p = 2. and IVI = 24. lV/Cv(A)| = IAI : NV. All = 4.

(n) G § 09(q) (p 95 2, q 2 pk), and V is a spin module.
Proof. This follows from [14](1.2), [6], and [16]. I

(A.2.3) Let G be a finite group and V a faithful FF-module for GF(p)G. Let L be a
component of G. Assume that [A,L] Z Z (L) for some A E ’P‘(G, V). Then
for each GF(p)(AL)-composition factor W of V with [W, L] 76 0 the following
hold:

(a) ACAL(IV)/CAL(YV) E PIX/4L, VV),

(b) LCAL(VV)/CAL(VV) = F*(AL/CAL(W)),

130

(c) [L, CAL(W)] g Z(L).

Proof. From [W, L] ¢ 0 it follows that LGAL(W)/GAL(W) is a component of
Anchor). If Lcs,,(w">/CAL(W) 79 F*(AL/CAL(W)). then optAL/CAAW»
¢ 1, since AL/L is a p—group. But Op(AL/CAL(W)) = 1, since W is irreducible.
Thus (b) holds.

(c) follows from [L, CAL(W)] S CL(W) and [W, L] 75 0.

(a) follows from (c), [L, A] Z Z(L) and (A.2.1)(b).

(A.2.4) Let G be a finite group, F a finite field of characteristic p, and V an irreducible
F G—module. Assume that there exist subgroups A and L of G such that G 2

AL, [V, A, A] = 0 and L S G. Then one of the following holds:
(a) V is irreducible as FL-module.
(b) lA/CA(V)| = [G : CG(V)L| = 2 and
V "-3 IV ®F(CG(V)L) FG
for some irreducible FL-submodule W of V. In particular, [A/GA(V)| <

lV/Cv'(A)l-

Proof. Without loss we may assume that G acts faithfully on V. Let W be an
irreducible F L-submodule of V and assume that W 3i V. Then N A(W) is a proper
subgroup of A.

Choose a E A \ N A(W). Then W + W“ is the direct sum of W and W“. Since
A, acting faithfully and quadratically on V, is abelian, NA(W) normalizes W“.
Now [W,a,NA(W)] S [V,A,A] = 0 implies NA(W) = CA(W). Since CG(V) =
1, NA(W) S A and (WA) 2 V, we get

(*) NA(W) =1.

131

In particular,
A n L = 1.
As FL—module,
V: W183...€BW,,,

where W1, . . .,Wn are irreducible F L-submodules conjugate to W. Without loss,

W,- E W precisely when 1 S 2 S m. Put X := W1+...+ Wm. Since

HompL(W, X) E EBHomFL(W, W,) E @EndpL(W)
i=1 2:1
is an m-dimensional vector space over EndpL(W), the number of irreducible F L—
submodules of X is Elf-:51 [EndpL(W)|‘, which is congruent 1 modulo 19. Consequently,

the p—group N A(X ) fixes some irreducible F L-submodule of X. But then, since W is

an arbitrary irreducible FL-submodule of V, (*) implies
N A(X ) = 1.
From this it follows that

V E ® W“ as F L-module.
aEA

Hence dim V 2 [Al dimW = [G : L] dimW = dim(W (EFL FG). Since V is generated
as F G-module by the FL-submodule W, we get

V§W®FLFG.

From A 0 L = 1 it follows that V is a free F A-module. Since [V,A,A] = 0, this
implies [A] :2 2.

Suppose |A| Z [V/GV(A)|. Since V is free as FA-module and [A] = 2, we get [Fl 2 2
and dimV = 2. Then W is a trivial F L-module and [V, L] = 0, whence L = 1 and

G = A. But this is a contradiction to the irreducibility of V as FG-module. I

132

(A.2.5) Let G be a finite group, K a finite field of characteristic p, and V a faithful
KG-module. Assume that there exist subgroups A and L of G such that
(i) L S G and G 2 AL,
(ii) V is completely reducible as KL-module,
(iii) [V, L] = V,
(iv) [V,A, A] = 0,
(V) ll""/Cv(x4)| .<_ lA/CA(V)|,
(vi) L is quasisimple,
(vii) 0,,(G) = 1.

Then V is completely reducible as K G-module. Moreover, each irreducible

K G-submodule of V is also irreducible as K L-module.

Proof. Suppose this is false. Let (G, V) be a counterexample with [G] + [V] minimal.
Then

000/) = 1.

Suppose that there exists an irreducible K G-submodule X of V which is also irre-

ducible as KL-module. Let Y be any KL-submodule of V with X sé Y. Then
V = X 619 Y

and Y is not A-invariant, for otherwise there would be a counterexample for the same
group G and a K G-module W strictly smaller than V. Pick a E A such that Y 75 Y“.
Then Y“ is a diagonal between X and Y. Hence there exist bijective K -linear maps

a:Y—>Xand)8: Y—>Ysuchthat

ya=ya+yfl foreachyEY

133

and B’la is a K L-isomorphism from Y onto X. Then
Cv(a) Z {r + y l :1: E X, y = —[.r,a]a‘1} and
[V,a] = {((an + ya) + (2/3 — y) I :v e X, y 6 Y}.
Since a is bijective, we get
Iowan s |X| = IYI 3 (iv, all.
Now [V,a] S Cy(A) S GV((2) implies
[V, a] = Cv(a) = CV(A).
Then
[Y, NAG/ll S CY(A) = CYW) = 0-

Since A is abelian, NA(Y) = NA(Y“). Hence we also have [Y“,NA(Y)] = 0. Now

V = Y + Y“ implies
NA(Y) S CG(V):1°

Thus A acts fixed point freely on the set y of irreducible K L-submodules of V other

than X. Put E :2 EndKL(X). Then 32 2 [El and therefore
W S IEI-

On the other hand,

_ IV] _ [V] _ _ dimgX
’ (Ct/(an ‘ |X| ‘ 'X' ‘ 'E' '

 

IAI Z IV/Cv(A)|

Hence X is a 1-dimensional EL-module, contrary to (iii).
Thus no irreducible K G-submodule of V is irreducible as K L-module. Let X be any
irreducible KG-submodule of V. Then X is like V in (A.2.4)(b). In particular,

|A : CA(X)| = 2.

134

Put H :2 GA(X)L. Then X is not irreducible as K H-module. Hence H is a
proper subgroup of G. Note that the assumptions of the lemma are satisfied for
H, CA(X),L,V in place of G,A,L,V, respectively. Therefore the minimality of

|G| + V

 

 

implies that
VZW1®...€BI/Vm

for some irreducible K H -submodules W1, . . . , Wm which are also irreducible as K L-
modules. Note that V is indecomposable as KG-module by the minimality of |GI+|V|.

Since H has index 2 in G, it follows that m = 2, i.e.,

V = X,
a contradiction to (A.2.4)(b) and (v). I
(A.2.6) Let G be a finite group, :: F*(G), p a prime, and V a faithful FF-module

for GF(p)G. Assume that the following hold:

(i) L is quasisimple,

(ii) G 2 AL for some A E ’P*(G, V),

(iV) 0p(G)=1,

(v) V contains more than one nontrivial G-composition factor.

Then one of the following holds:

(a) G E SLn(Q) (q = pk)-
(b) G E Sp2n(q) (22 Z 3, q 2 pk), and each nontrivial G-composition factor of

V is a natural Sp2n(q)-module.

(c) G E Qf,(q) (q = pk, E E {0,+, -}), and each nontrivial G-composition

factor of V is a natural Qf,(q)-module.

135

(d) G E 03,,(q) (p = 2, q = 2", E E {+,—}), and each nontrivial G-
composition factor of V is a natural O§n(q)-module.
(e) G E SUn(q) (q 2 pk, n 2 4), and each nontrivial G-composition factor of

V is a natural SU,,(q)-module.

Proof. Let IV be a a nontrivial G-composition factor of V. By (A23) and (iii),
[L,GG(W)] S Z(G). Hence GA(W)Z(G) S G, and then (iv) implies that

(*) CG(W) S Z(L).
Note that (G / GG(W), W) appears in (A.2.2). Since (*) implies that A acts faithfully

on each nontrivial G-composition factor, (v) excludes the cases (d), (g) and (i)-(n) in

(A.2.2). Thus one of the following holds:
(1) G/CG(W) 3 SL110!) (q = 10"),
(2) G/GG(W) E Sp2n(q) (q 2 pk), and W is a natural Sp2n(q)-module,
(3) G/G(;(W) E Qf,(q) (q 2 pk, 5 E {0, +, —}), and W is a natural Qf,(q)-module,

(4) G/CG(W) E 05,,(q) (p = 2, q = 2", 6 E {+, —}), and W is a natural O§n(q)-

module,

(5) G/GG(W) E SUn(q) (q = pk, n 2 4), and W is a natural SUn(q)-module.

Note that the isomorphism type of L/GL(W) does not depend on the choice of W.
Since Z (L) is cyclic, it follows that CG(W) (= CL(W)) does not depend on the
choice of W. Hence CG(W) = GG(V) = 1, since Z (L) is a p’-group. The restriction
n 2 3 in (b) follows from (B.4.1.6). I

(A.2.7) Let G be a finite group, R a normal subgroup of G, L a component of G, and
V a faithful irreducible FF-module for GF(2)G. Let S be a subgroup of LR
containing R such that S/ R is a Sylow 2-subgroup of LR/ R. Assume that the

following hold:

136

(i) LR/R = F"(G/R) E Sp4(2’“)’ for some k E IN.
(ii) There exists A E ’P"(NG(S), V) such that [A, L] Z Z(L).
(iii) 02(G) = 1.
Let P1, P2 be subgroups of LR containing R such that P1 / R and P2/ R are the

minimal parabolic subgroups of LR/R containing S/R. Then NG(S) S NG(P,-)
for each 2 6 {1,2}.

Proof. If K is any component of G with K Z R then, by (i), KR/R = LR/R and,
hence, [K , L] 329 1. It follows that L is the unique component of G which is not

contained in R. In particular,
L S G.

Put H :: AL. Let W be an irreducible H-submodule of V. Note that

(*) fl 00(ng =1,

966‘

since GG(V) = 1 and V 2 (WC) S Cy(ngec CG(W)9). In particular, [W, L] ¢ 0
and, hence, GL(W) S Z (L) Since G acts faithfully and irreducibly on V, 02(L) S
02(G) = 1. Then by tables 6.1.2 and 6.1.3 in [8], Z(L) is a cyclic group of order 1 or
3. Thus GL(W), being a characteristic subgroup of Z (L), is normal in G. Now (at)
implies

(**) CL(W) = 1.
But then [GG(W),L] S CLO/V) = 1. Since (i) implies Cg(L)R/R S GG/R(LR/R) =
1, we get

(at: a: at) GG(W) S CG(L) S R.

Since [W, L] E 0, (A23) implies that (H/GH(W), W) appears in (A.2.2). Note that
the natural EEG-module is isomorphic to a natural Sp4(2)—module. Hence we are left

with the following cases:

137

(1) H/C'H(W') E Sp4(2’“), and W is a natural Sp4(2k)-module.
(2) H/GH(W) E Sp4(2)’, and W is a natural Sp4(2)-module.

(3) H/CH(W) Eras, |W

 

= 26 and (W/CW(A)| = (A( = 4.

Together with (a: * =1:) it follows that H n R/GH(W) is a 2’-group. Since W is an
arbitrary irreducible H -submodule of V and (A.2.5) implies that V is the direct sum
of its irreducible H-submodules, it follows that HflR is a 2’-group. Hence HflR S L

and therefore by (**)
CHU/V) = 1.

Note that by (A.2.6) none of Sp4(2"), Sp4(2)’ or A6 has an FF-module with more
than one nontrivial composition factor. Thus in each of the cases (1)—(3) GH(W) = 1

implies that
V = W.

But if g E NG(S) \ NG(P1), then W9 is not isomorphic to W as L-module. Hence
NG(S) = NG(P1)- .

(A.2.8) Let G be a finite group, R a normal subgroup of G, L a component of G, and

p a prime such that

(i) LR/R = F*(G/R) E PSL,,(pk), for some n, k E IN with n 2 3,

(ii) 0,,(G) = 1.
Let M, Q, P1 and P2 be subgroups of LR containing R such that

(iii) M / R is a parabolic subgroup of LR/ R corresponding to the n — 2 interior

nodes of the Dynkin diagram,

(iv) 62/}? = 0,.(12/3),

138

(v) P1 / R and P2 / R are the maximal parabolic subgroups of LR/ R containing
M / R.

Let V be a faithful irreducible FF-module for GF(p)G. Assume that there
exists A E ’P"(NG(M), V) such that

(vi) [A, L] g Z(L) and

(vii) [A, AI] S QA.
Then one of the following holds:

(a) NG(1\I)S N(;(P,) for each 2 6 {1,2}.

(b) L E SL401" ), and V is the exterior square of a natural module for L.

Proof. Put H :2 AL. Let W be an irreducible H-submodule of V. The same

argument as in the proof of (A.2.7) shows that
(at) GLUV) : 1 and

(4*) CG(W) g R.

Since [W, L] 74$ 0, (A23) implies that (H/CH(W), W) appears in (A.2.2). Hence one
of the following holds:
(1) H/GHU/V) E SLn(pk), and W is a natural SLn(pk)-module,

(2) H/CH(W

) N SLn(p’°), and W is the exterior square of a natural SLn(p")-

module,

(3) p" = 2, n = 4, H/GH(W) E 28, and W is a natural {lg-module.

Together with (**) it follows that H F) R/GH(W) is a p’-group. Since W is an

arbitrary irreducible H -submodule of V and (A.2.5) implies that V is the direct sum

139

of its irreducible H —submodules, it follows that H OR is a p’-group. Hence H flR S L

and therefore by (*)
GH(W) = 1.

Assume that (3) holds. Since V is the direct sum of L-submodules conjugate to W
and also the direct sum of irreducible H-submodules, we get that V is a direct sum
of natural 28-modules for H. Note that no subgroup of 28 acts as an offender on a
direct sum of two natural modules. Hence V = W and (b) holds.

Assume that (1) or (2) holds. Similarly to the above we get that as H-module V is
a direct sum of irreducible H -submodules that are isomorphic to W and irreducible
H -submodules that are dual to W. If no irreducible H -submodule of V is dual to W,
then (a) holds. Hence assume that both types occur. Note that A S CL(M flL, QnL)
implies that A does not act as an offender on a direct sum of a natural module and

its dual. Thus W is isomorphic to its dual and V = W. Hence (b) holds. I

(A.2.9) Let G be a finite group, R a normal subgroup of G, L a component of G, and
p a prime such that
(i) LR/R = F’(G/R) E Qflpk), for some k E IN,

(ii) OP(G) =1.
Let M, Q, P1, P2, and P3 be subgroups of LR containing R such that

(iii) M / R is a rank 1 parabolic subgroup of LR/ R corresponding to the interior

node of the Dynkin diagram,
0") Q/R = CAM/R),
(v) P1 /R, Pz/R and P3/R are the maximal parabolic subgroups of LR/R

containing .M/R.

Let V be a faithful irreducible FF-module for GF(p)G. Assume that there
exists A E ’P*(NG(M), V) such that

140

(vi) [A,L] Z Z(L) and

(vii) [A, M] S QA.
Then NG(M) S NG(P,-) for some 2 6 {1,2,3}.

Proof. Put H :2 AL. Let W be an irreducible H-submodule of V. As in the proof
of (A.2.7) we get

CLO/V) =1 and
CCU/V) S R.

Since [W, L] 75 0, (A23) implies that (H/CH(W),W) appears in (A.2.2). Hence
H / GW(H ) is isomorphic to Qflq) or O§(q), and W is a natural Q§(q)-module. As

in the proof of (A28) it follows that
GH(W) = 1.

Suppose that NG(M) does not normalize any one of P1, P2, or P3. Then V contains
at least three pairwise non-isomorphic natural Q; (q)—modules for H. Without loss,
we may assume that W is chosen such that [W : CW(A)] is minimal. Note that

[AzAflLl Sp. Then

IA 0 LI 2 p"1|A| Z p‘llV/Cv(A)| 2 p‘IIW/Cw(A)|3,
and hence

IA 0 LI 2 IW/C'w(r4)|2 = llwsAllz,

contrary to (B.5.1.8). I

141

A.3 Modules for central products
Let p be a prime.

(A.3.1) Let G be a finite group which is the central product of two subgroups A and
B. Let K be a field, F a subfield of K, X an irreducible K A-module, and V

an irreducible F G-module. Assume that the following hold:

(1) There exists an FA-rnonomorphism (15 : X —> V.
(ii) X is irreducible as FA-module.

(111) EndpA(.¥) (E K.
Then there exists a K B-module Y such that the following hold:

(a) X (81,, Y is a KG-module, where G acts on X ®K Y as follows:
(:1: (SK y)(2b = (2:22) ®K (yb), for all :c E X, y E Y, a E A, and b E B.

(b) X ®K Y and V are isomorphic as F G-modules.

(c) Y is irreducible as FB-module.

Proof. Let H be the direct product of A and B. Clearly X ®K KB is a KH-module,

where the action of H is given by
(:1: ®K y)ab =(:1:a)®K(yb), for all z E X, y 6 KB, a E A, and b E B.

Note that :1: (Ex b 1—+ :1: (811” b (:r E X, b E B) defines a KH-isomorphism a from
X 81K KB onto X ®KA KH.

Since X ®KA K H is generated as F H -module by the F A-submodule X ®KA 1, there
exists an F H -epimorphism B from X ®FA F H onto X ®KA K H which maps :1: (81;: A 1

tox®KA1foreachzr€X. As
dimp(X ®FA PH) 2 [Bldlme = [BldlmpK dimKX =

142

dimpK dimK(X ®KA KH) 1' dimp(X ®KA KH),

)3 is an F H -isomorphism.
Hence 016-1 is an F H -isomorphism from X ®K KB to X ®FA F H . Note that there
is an epimorphism 7 from H onto G. Regarding V via 7 as an F H -module, it follows

that the map
an®KKB—>V,$®Kbr—+(:1:¢)b (:rEX,b€B)
is an F H -epimorphism. Let U be the kernel of 7). Then
W:= {bEKB|X®KbSU}

is a K B-module. As F A-module, X ®K KB is the direct sum of the submodules
X ®K b with b E B. Now it follows from (ii) and (iii) that any irreducible FA-
submodule S of X ®K KB is of the form

S=X®K Zkbb
beB

for some kb 6 K. Since U is the sum of its irreducible F A-submodules, we get
U = X ®K W.
Put Y := KB/W. Then
V E (X ®K KB)/(X ®K W) E X ®K (KB/W) = X ®K Y (as FH-modules).

By definition of the action of H on X ®K KB, (a) and (b) are satisfied with H instead
of G. But since H acts on V via 7, it follows that (a) and (b) hold.

The irreducibility of V implies (c). I

(A.3.2) Let G be a finite group which is the central product of two subgroups A and B.
Let F be a field, V an irreducible FG-module, and K 2: End pG(V). Let X be
a K A-submodule and Y a K B-submodule of V.

143

(a) X ®K Y is a KG-module, where G acts on X (81K Y as follows:
(:1: 81K y)ab 2 (23a) ®K (yb), for all 2: E X, y E Y, a E A, and b E B.
(b) If X and Y are irreducible, then V is isomorphic to X ®KY as K G—module.

Proof. (a) It suffices to show that (re) ®K y = :1: (SK (yc), for all :1: E X, y E Y, and
cE AflB. Since AflB S Z(G), for each cE AflB the map

(be: V—)V,’U1—-)UC

is an F G—endomorphism. Hence

(Inc) ®K y = (we) ®K y = I 6% (3105.) = :1: 69K (216).

for alleX, yE Y, anchAflB.

(b) Let X0 be an irreducible FA—submodule of X. From (A.3.1) (with G, A,B,
EndpAXo, F, X0 in place of G, A, B, K, F,X, respectively) it follows that V has the
structure of an EndpAXo-module. Hence E :2 NK(X0) is a subfield of K with
E E EndpAXo. Since X0 is absolutely irreducible as EA-module, X (E K 6813 X0) is
absolutely irreducible as K A-module. From (A.3.1) (with G, A, B, K, K,X in place
of G, A, B, K, F, X, respectively) it follows that V E X 81K W for some irreducible
KB-module W, and G acts on X 81K W as described in (A.3.1)(a). Hence, by (a),
V is isomorphic to X 81K U for any K B-submodule U of V with U E W. But any
irreducible K B-submodule of X ®K W is isomorphic to W. In particular, Y E W
and (b) holds.

(A.3.3) Let G be a finite group, F a finite field of characteristic p, and V an F F-module

for FG. Let {L1, . . . , Ln} be a G-invariant set of components of G satisfying
L _<_ (FTC, V/Cv(L))>

where L 2: (L1, . . . , Ln). Then

71

[wot/(L114 = EBB/MAL), 2,).

2:1

144

Proof. Let (G,V,L1,...,L,,) be a counterexample with [G] + [V] minimal. Then
GV(L) = 0, CG(V) = 1, and G = (’P“(G, V)). H2 753' implies [V,L,, Lj] = 0, for all
2,j E {1,...,n}, then [V,L,] fl Shh-[V,L]] S GV(L) = 0, for all 2 E {1,...,12}, and

(G, V, L1, . . . , L,,) is not a counterexample. Hence, without loss we may assume that
[V.L1,Ls] 75 0.

From G = (79"(G, V)) and (A.2.1)(d) it follows that L), S G for each k E {1, . . .,n}.
Suppose that V is not irreducible. Let W 75 0 be a prOper submodule of V. Choose
i,j E {1,...,72} such that 2 76 j. If L), S (’P‘(G,W)) for each k E {2,j}, then
[W, L,, Lj] z: 0 by the minimality of |G|+IV|. If Lk Z (’P‘(G, W)) for some 12 E {2,j},
then, by (A.2.1)(b), [IV, A] = 0, for each A E ’P*(G, V) with [Lk,A] Z Z(Lk), and

therefore [WC L1,] = 0. Hence in any case we get
[W, Li, Lj] = 0.

A similar argument (with V/W instead of W, using (A.2.1)(c) instead of (A.2.1)(b))

shows that
[V,L,,le s W.

Note that
[V,L1,L2,L2] = [V,L1,L2] = [V,L2,L1] = [V,L2,L1,L1]

by the Three-Subgroup Lemma. Hence
[V,Ll, L2, L] = [V,L1,L2,L1] + [V,L2,L1,L2] + [V, L1, L2, L3 - . . . - Ln] =
[V, L1, L2, L2, L] + [V, L2, L1, L1, L2] + [V, L1, L2, L2, L3 - . . . - Ln] S
[W,Lg, L] + [W,L1,L2] + [W, L2,L3 - . . . - Ln] = 0,

contrary to [V, L1, L2] 75 0 = Gv(L). Hence V is irreducible.

145

For the rest of the proof let

{i,j} = {1,2}-

Put

A

L, :2 LjL3 ° ' ° L",
and choose A, E ’P‘(G, V) such that
[L,,A,] Z Z(L,).

Let W be an irreducible F (A,L)-submodule of V. Put K :2 Endp(A,L)W, and let X
be an irreducible K (A,L,)-submodule of W. Note that (A.2.1)(b) and (A24) imply
that W is irreducible as FL-module and X is irreducible as K L,-module. Thus, by
(A.3.2),

W E X (8),, Y, for some irreducible K L,-submodule Y of W.

Since L SG and V is an irreducible F G-module, we get that, regarded as F L-module,

V is a direct sum of conjugates of W. In particular, X is not a trivial F L,-module,

since [V,L,] 750. Let
O=W0<W1<...<Wm=W

be a K (A,L,)-composition series of W.

We assert that GA,(Wk/Wk_1) S CA,(W) for each k E {1, . . . ,m}. As KL,-modules
both Wk/Wk-1 and W are direct sums of submodules isomorphic to X, whence
CL,(Wk/Wk_1) centralizes W. Then GL,(Wk/Wk_1) centralizes V, since as FL-
module V is a direct sum of conjugates of W. Since G acts faithfully on V, we
get GL,(Wk/Wk_1) = 1. In particular, L, centralizes GA,(Wk/Wk_1). Since A,, acting
faithfully and quadratically on V, is elementary abelian, we get GA,(Wk/Wk_1) S
Z(A,L,). Hence Gw(GA,(Wk/Wk_1)) is a K(A,L,)-submodule of W. From the

quadratic action of A, on W it follows that A, centralizes the K (A,L,)-module

146

W/CW(CA,(Wk/Wk_1)). But L(A,~L") contains L,, since [A,,L,] g; Z(L,). There-
fore, W/C'u/(G,4,(Wk/Wk_1)) is centralized by L,. Since all K L,-composition fac-
tors of W are isomorphic to X, which is a nontrivial K L,-module, this implies
W : CW(GA,(Wk/1Vk_1)).

Now it follows from (A.2.1)(c) and (A24) that VVk/Wk_1 E X as KL,-modules for

each k E {1, . . . , m}. In particular,
:1: :2 dimKX = dimK Wk/Wk_1, for each k E {1, . . .,m}.

Since, for each k E {1,...,m}, A,/CA,(W) acts faithfully and quadratically on

Wk/Wk_1, we get

(*) IAi/CA.(W)| S Win—6")“,
where

ck 2: dimK ka/wk_,(A,-), for each k E {1, . . .,m}.
From (A.2.1)(b) it follows that

Ai/CA.(W) 3" AtCA,L.-(W)/CA,L.(W) E p'(AtLi/CA,L,(W)1W)-
Since

le(A.-)| 3 fi le./w._.(At)| = IKIC‘+'"+C'",
this implies

(...) lAt/CA.(W)I 2 lW/Cw(At)l 2 IKI'M‘CM 2 (KW-c),
where

c :2 max{c1, . . . ,cm}.

Thus m S c by (at) and (22*). Since Wk/Ii/'k_1 E X for each k E {1,...,m} and
WEX®K Y, we get

dimK Y < dimK X.

147

Since V, regarded as KL-module, is a direct sum of conjugates of W, this implies that
the dimension of any K L,-composition factor of V is smaller than the dimension of any
K L,-composition factor of V. In particular, the dimension of any K Lj-composition
factor of V is smaller than the dimension of any K L,-composition factor of V. Since

all assumptions are symmetric in L1 and L2, this is a contradiction. I

A.4 Automorphisms of finite simple groups of Lie
Type

In this section we adopt the notation of chapter 2 in [8]. In particular, let r be a
prime and K a finite simple group of Lie type in characteristic r. Let q be the power
of r such that GF (q) is the field of definition for K. Let a be an automorphism of
K. As in Theorem 2.5.1 of [8], write a = 2dfg where 2 is an inner automorphism, d is
a diagonal automorphism, f is a field automorphism and g is a graph automorphism.

Moreover, if K is a twisted group then 9 is the identity.

(A.4.1) Let P be a parabolic subgroup of K containing the Borel subgroup B. Assume
that [P, a] S 0,.(P). Then one of the following holds:

(a) a = 2d,

(b) q = 2, a 2 2g, and g fixes each node of the Dynkin diagram that belongs
to P.

Proof. Note that d, f and g all normalize B. Since [B,a] S [P,a] S O,(P) S B, it

follows that also 2 normalizes B. Hence 2 is induced by an element of B. In particular,
(*) [h3(t),2] S U, for all 51 E B and 0 315 t E GF(q).
Note that we also have

(**) [hA(t),d] = 1, for all 51 E 5: and 0 ¢ t E GF(q), and

O

148

f e NK((h;,(t) (0 at t e GF(q))), for all a e 2.
Since

[023(2) ( t e GF(q))U,a] g [P,a] g 0,(P) s U for all a e )5,
it follows that

(:1: =1: 21:) ((12,3(2) ( t e GF(q))U,g] s U, for all a e 2.

Assume that g is not the identity. Then (=1: * =1:) implies that q = 2. Hence both (1 and
f are the identity. If a E Z is a fundamental root corresponding to a node in the

Dynkin diagram of P, then

[20(2),2] S [U, B] = [U, U] and

[230(2),a] S OT(P), for each 2 E GF(q).
Hence

[:ra(t),g] S O,(P)[U, U], for each 2 E GF(q).

This implies that g fixes the node corresponding to 02 in the Dynkin diagram. Hence
(b) holds.
Now assume that g is the identity. As in Theorem 2.5.1 of [8], let (I) be the automor-

phism of GF(q) that induces f on K. Then

(123(2), f] = 113(t)-1h3(t¢) e H, for all a e i and 0 ,2 t e GF(q).
On the other hand, by (*) and (22*),

[h3(t), f] 3 U, for all a e f: and t e GF(q).

Since U n H = 1, it follows that (b is the identity. Hence (a) holds. I

149

(A.4.2) Assume that K is of type Cn with n 2 2. Let P be the parabolic subgroup of
cotype 1 in K containing the Borel subgroup B. Assume that [P, a] S P. Then

a E Inn(K) if and only if a. induces an outer automorphism on 0"(P)/0,(P).

Proof. Let A be the normalizer of P in Aut(K). Let A1 be the intersection of
A and Inn(K). Let A2 be the subgroup of A that induces inner automorphisms on
0"(P)/O,(P). Since P = N [((P), each element of A1 induces an inner automorphism
on P/O,(P). Together with

P/O,(P) = 0"(P)/O.(P)Cp/o,(p)(O"(P)/O,(P))

we get that
(*) A2 S A1—

If (15 E Aut(GF(q)), then the field automorphism of K defined by
2:0(2) +—> 5170(2‘2’) for each 02 E E and t E GF(q)

induces on O”(P)/O,.(P) a field automorphism of the same type. If q is odd, then
it follows from section 7.1 in [3] that K has an outer diagonal automorphism which
acts as follows:
1700) +——> { 230(2) if the long simple root is not involved in a
za(—t) 1f the long Simple root 1S involved in a
for each a E 2 and t E GF (q) Note that this automorphism induces on O" (P) / Or(P),

which is of type Cn_1, a diagonal automorphism of the same type. Thus

_ 2|Aut(GF(q))| if q is odd
(H) IA ' A2l Z { [Aut(GF(q))l ifq is even

Since A fixes P, no element of A involves a graph automorphism of K. Hence by
Table 2.1.C in [11],

(A - A (_ 2l/‘tut(GF(q))| ifq is odd
I l _ IAUt(GF(Q))l ifqis even '

Together with (*) and (**) it follows that A, = A2. I

150

Appendix B

FF-modules for groups of Lie type

Let .C be a complex simple Lie algebra of type T, where T is one of An, Ban or
D". Let r) : f, ——> U be the embedding of L in its universal enveloping algabra Ll.
Let H be a Cartan subalgebra. Let (D = <I>(T) be the corresponding root system
and I1 = (a,)15,sn a system of fundamental roots, where a, corresponds to the node
2 in the Dynkin diagram, and the numbering of the nodes is as in [9]. Also, the
fundamental weights are labeled as in Table 1 of [9]. We denote the height of a root
S by ht(fi). Let < be the total ordering on the euclidean space spanned by (I) defined
by

222,01, '< sz‘a, 42>
2:1 221
there exists j E {1,...,m} with a, = b, for all 2 E {1,...,j — 1} and a, < b,.

Let ha (a E H), 85 (S E (1)) be a Chevalley basis of L. Since we will write modules
as right modules, we choose this Chevalley basis such that, for each a E II, we have
ha 2 [e-mea] rather than ha 2 [ea,e_a] as in [9]. (Pick a Chevalley basis as in [9]
and replace ha by —ha).

Let le be the Z-form of u corresponding to this Chevalley basis, i.e., Hz is the
Z-subalgebra of U generated by the elements fibeaq)’c (k E N, a E (1)).

Let K be a finite field of characteristic p. Put q := [K].

151

B.1 Construction of the groups and modules

Let A be a dominant integral weight, W = W()\) the irreducible £—module of highest
weight A, and A the set of weights of W with respect to 72. Let (b be the representation
OH! on W such that 21(211721) : 21a, for all 21 E W and a E L.

W has a C-basis B which is also a Z-basis for a 112-submodule Wz of W.

Put V 2: V(T, K, A) :2 K®z Wz. For all t E K and a E (I) put

(l,(ean) )4) (acme/)1.

20(2):: 1:0,;(2 T, K, A) :2: (tkidK @z( k

kzl
Define

Xasza,(TK,)\):——— (:1: C,(2‘,)|tEK) foreachaECI),and
G :2 C(T, K, A) := (X,, ( a 6 <2).

Let <I>+ = (I>(T)+ be the set of positive roots. For each J Z {1, . . . ,n}, let in = <I>J(T)

be the root system spanned by (a, I j E J}, and put
Q12: QJ(T) :2 (Xa [ a E (I)+ \(PJ),
LJ 2: LJ(T) 2:- (X0 I a E (DJ) and
PJ:: PJ(T) :2 NG(QJ).
3.2 An
Assume that L is of type An. Then (I)+ = {Sm- | 2,j E {1, . . .,n}, 2 S j}, where
j
S] 220),, forall2,j E {1,...,72} with2Sj.

Note that with respect to < the pairs ((1,, 3,110,) are extraspecial in the sense of [3]

for all 2, j E {1, . . . ,22} with 2 < j. Hence without loss we may assume that
65,0. 2 -[ea,,e_5,+,d.], for all i,j E {1, . . . ,n} with 2 < j.

152

B.2.1

In the construction of section B.1, let

 

I n .
A = A1(A,,) = ”+1201.— 1+ 1)a,.
i=1

Then A = {221, . . . , Illn+1}, where

1 1—1 71
<—Zkak+2(n—k+1)ak), for all2E {1,...,n+1}.
n+1 k=l [:22

be a nonzero weight vector of weight 111. Then the basis B can be chosen as

 

#1 I:

Let U,,,

follows: Put

U,,,+1 :2 U,,,em” for each 2 E {1, . . .,n}.

Then {U,m . . . , U,,n} is a basis with the desired properties. We obtain a group G with

generators x,,(t) ([3 E (I), t E K) acting on a module V.

(B.2.1.1) Let i,j e {1,...,12} with i g j.

(a) 51¢ = #2 — #141.

(b) Let 12 E A. Then
_ Um if ,1 = .uj+1
12,165,, _ { 0 else.
and

_ Us... “#:111-
U,,(ng. _ { 0 else.

Proof. (a) This is clear.

(b) From (a) it follows that

U,,egw. = 0 unless ,u : 121-+1, and

12,115,”. = 0 unless u = ,11,.

153

First assume that 2 = j. Then U,,, = (1,, and from the definition of U,,“,1 we get
vine-51d : Um“ and
vut+tex3aj : ”me—0.80:. : ”what + "meme—a. : (Hit ai>vut + 0 : ”#1-

Hence (1)) holds in this case.

If2 < j, then by induction on ht(,13,,J-) we get

vflj+1e131,j : _v#,+1lean 65.2213] : —v#j+1eateflt+1.j + vl‘j-HefiH-Ljeai = vut+ieat : U11;
and
Dike—5:.) : v”1[e—ai’e—BI+I,J] : vute—Ote—fiz-HJ — vflie-fiH-Lje—Oi : vut+ie-Bt+1,j =
vflj+1'
I
Put

U, :2 1®z U,,], for each 2 E {1,...,n+1}.

(B.2.1.2) LettE K, i,j E {1,...,n} withiSj, and k E {1,...,n+1}. Then

11 +211, ifk= '+1
vkxfit,1(t):{ k J

21,c else.

and

Uk+tv-1ifk=2
21127-5“).(2) = { 211 H else.

Proof. This follows from (8.2.1.1). I

154

3.3 8,,

Assume that .C is of type B,,. Then
<I>+={)3,,,|2,jE{1,...,n},2Sj}U{)3f,j|2,jE{1,...,n—1},2Sj} where

j
13,]- := 20,, for all 2,j E {1, . ..,n} with 2 S j, and
k=2

.7 72
BL]- :=Zak+2 2 02k, for all 2,jE {1,...,r1—1} with2Sj.
kZi k:j+1

Note that with respect to -< the following pairs are extraspecial in the sense of [3]:
((,1,,)3,~+1,j), for all i,j E {1,...,12} with 2 < j,
(a,,f3£+1,j), for all 2,j E {1, . . .,12 — 1} with 2 < j,
(a,+1,fif’,+1), for all2E {1,...,12—2}, and

(am/3114,12).
Hence without loss we may assume that
(i) 63”. = —[ca,,e/3,+,J], for all 2,j E {1, . ..,n} with 2 < j,
(ii) 65:.) = —[8m,€g:+w], for all i,j E {1, . ..,n — 1} with 2 < j,
(iii) 85:“, : —[602+1’851..+1]’ for all 2 E {1, . . .,n — 2},

(iv) 283 = —[ean,egn_,m].

n—I,n-1

155

B.3.1

This subsection is about the natural Ogn+1(q)-module. In the construction of section

B.1, let
/\ I A1(Bn) I: Z Gk.
11:1
Then A = {11.0,121,...,pn,—121,...,—/2,,}, where

[20 2: 0, and

p, :2 Zak, for all 2 E {1, . . .,n}.
k=2

Let U,,, be a nonzero weight vector of weight ,111. Then the basis B can be chosen as

follows: Put

"U,,,+1 :2 U,,,e_a,, for each 2 E {1, . . .,n — 1},

2A,,“ z: U,,ne_an,
1

2A,," .: —§U,,Oe_an, and

v_,,n_, 2: —U_,,n_,+,e_on_,, for each 2 E {1,...,12 — 1}.

Then {21%, Um, . . . ,11,,n,21_,,,, . . . , own} is a basis with the desired properties.

(8.3.1.1) (a) S,”- = 12, — 121-+1, for all 2,j E {1,...,72 — 1} with 2 S j.
(b) flan = [21' — [20, for all 71E {1,. . . ,Tl}.
(c) [J- = 22, + HJ+11 for all i,j E {1, . . .,72 — 1} with 2 S j.

(d) LetuEA,€E {1,—1},and2,jE {1,...,12—1} with2Sj. Then

5’05“, If ft = 5flj+1
0,1853”. 2 —€U_E,,J+, 1f )2 = —€12,
0 else.

156

(e) Let )2 E A, E E {1, +1}, and 2 E {1,...,22}. Then

25215“, if 12 = 5,110
24,655“ = -—€11,,0 if u = —5/2,
0 else.

(f) Let 21 E A, 5E {1,—1} and 2,j E {1,...,72— 1} with2Sj. Then

eve... if u = —€uj+t
U#659:J = -5U,,,J.+, if p = —-511,
0 else.

Proof. (a),(b),(c) This is clear.

((1) From (a) it follows that
U,,eww = 0 unless u E {—5,11,,512j+1}.

First assume that 2 = 3'. Then fit, 2 C1,. From the definition of '11,),+1 and 21-,“ we get

”11.421601. : ”use—0.80:.- = ”111-ha.- + ”meme—a : (Hit 022%. + 0 : ”#11 and

”-11.80. v—u.+1e—a.ea. v-us+1ha. v-u.+1eate—at :
_<_/1H1, Gav-#1“ _ 0 = —v—#t‘+1°
Hence (d) holds if2 = j. ‘

Now assume that 2 < j. Then by (i) and induction on ht(fl,,,) we get
v_uiegi.j = —v-#teatefit+1,j + v-ute[31+1.j 801' = v-ut+1efit+1.j + 0 : —v-#j+11
vflj+1eflt,j : —vflj+1eot'efit+1,j + vflj+1eflt+i.jeat : 0 + vflt+ieai : U111?
UMP—514 = vflie-aie—BH-Lj _ vflte—31+1,je-at : vm+1e—3t+1,j - 0 : vflj+11

and

v‘WHe—fiid Z v—fljHe—Oie-Bms — v—#1+le—51+1.je-at = 0 + v—Mi+le-Oi = ‘U-m-

157

(e) From (b) it follows that

First assume that 2 : n. Then [3,," = an. From the definition of 21m, and 1A,,” we get
U,,ne_an = 21,,0,
Woman 2 —2U_,,n,

omega 2 U,,ne_anean = U,,nhan + ouneanefln = (,12man)v,,n + 0 = 2’11“", and

1 1 1
apnea, = —§'U,,0e_anean = —-2-U,,Ohan — §vpoeane—an =
1
-§(;20,01,)U_,,0 — vflne“an — 0 _ vuo — ’vuo-

Hence (e) holds if2 = 72.

Now assume that 2 < 12. Then by (i), (d), and induction on ht(fl,,n) we get
v-utefit,n : _‘U—uteatefith + v—mefit+i.neat : v-#i+ieflt+1.n + 0 : —U_,,0,
omega,” 2 -U“080‘65,+1'n + U,,Oegwmea, = 0 + QUMH—leai = 221,“,
vine-31.7. : Uflie_aie—Bi+l.n _ vute-Bt+i.ne-at : vut+ie-5t+i.n — O = UHO’

and
vitae-[31,71 : Ufloe—Ote-Bth _ vfioe—Bt+i,ne-Oi : O + 2v—#t+ie-Ot : —2v—#i'

(f) From (c) it follows that
U,,eggz‘j = 0 unless 12 E {-€12,, —5,12j+1}.

From (iv), ((1), and (e) it follows that

v_l‘nefl;,_1'n_1 : —§v‘#neanefin—1,n + iv—flnefin—lmean : avfloefin-lm + O = vI-‘n-l’

158

1
v_#n-lefi:1_1‘n_1 Z -—v"#n—leanefin-l,n + —v“#n—lefin—l,nean : 0 — 511/1060”; :

2 2

—’U“n ,

1 1 I
vflne—fl;,_lyn_1 — ivflne-ane-fin—lm — ivflne‘fin—l,ne‘an — §v#06“3n—1,n - O —

__v‘l‘n—l’

and

1 1
vfln—le-fi;-1'n_.1 : ivfln-le“ane-fin-l.n — gvfln~le—Bn-l.ne’an : 0 — EUHOe-‘an

v—un 0

Hence (f) holds if2 23' = 72 — 1.

H2 < j, then by (ii), ((1), and induction on ht( :1) we get

the,“ = —U—u,€o,813;+,,j + 21_,,,e,3£+l‘jeo,i = v-,,,+,efl;+1’j + 0 = —v,,j+,,

U—MH-lefiz'j = —v—#j+leaiefi:+1'j + v-Iljstiefififld-eat : 0 + vflt+ieai = Um,

vine-BL, : vl‘te’aie“fi:+l,3~ — vute—fizfl‘je-Ot : vflt+ie-5:+1J — 0 = ”—111“,
and

”mug-[3L]- = vuj+ie—Ote—5£+l_, — vflj+le—5;+1,je“ai : O + v-ut+ie—at = —v-ut'

If2 < n -— 1, then by (iii), ((1), and induction on ht(flfJ) we get

U‘Hfififi, : _v-#teat+iefig,,+1 + U‘fliefiz,,+leai+l = 0 — 711114—2601“ : _v#t+ii
’U—I‘i-Hefii‘, : —U'Hi+leai+lefl:'i+1 + v—I‘i+lefi:,,+leai+l : U—ut+2efifi,,+1 + 0 = 7112,,
vflte‘5£,, = Uflte—OHIe-fifiyfl — vflte-flfi'wle-Oifl = 0 _ v—uiHe-Ot’H : v—fliH’

159

and

UMHBSL, : v211+18"ai+le—J3:',+1 — UH2+Ie-B:I,+le—ai+l : ”Mme-fig,“ _ 0 _—_ —v—l‘i'

Put
21, := 1 ®z U,,, and
11-, z: 1 (812 “—11.

for each 2 E {0, ....,72}

(B.3.1.2) (a) Let k E {—72,...,72}, 5 E {1,—1},2,j E {1,...,72 — 1} with 2 S j, and

t E K. Then
21,, + 52115, if k = 5(j + 1)
1151355,.) (2) — ’Uk — €t’U_E(j+1) 1f 16 = —82
21,, else.
(b) Let k E {—7i,...,72}, E E {1,—1}, 2 E {1,...,72}, and t E K. Then
’01, + 2t8’UE,‘ If k = 0
magma) = U,c — 25210 — 22215, if k = —52
11;, else.
(0) Let k E {—72,...,72}, 5 E {1,—1},2,j E {1,...,72— 1} with 2 S j, and
t E K. Then
21,, + tevs, if k = —€(j + 1)
kagg:j(t) — 21,c — 252150“) if k = —52
21,, else.

Proof. This follows from (B.3.1.1). I

(B.3.1.3) (a) Let r E {1,...,72} and put J :2 {1,...,72} \ {r}.

(31) [V1 QJI = $220 K7127 EB $234.1 KU—z‘a
(32) [‘fi QJi QJ] Z CV(QJ) = $5=1Kv,.
(a3) [ll/10“] O CV(QJ)[ 7é 1, for each a E QJ \ {1}

160

(b) If q is odd, then the map
{I V X V -) K, (2 (2,21,, 2: bi’Ui) H 2041)-, + 01-,‘(21'
2=—n i=—n '=0

is a nondegenarete G-invariant symmetric bilinear form.
Proof. This follows from (8.3.1.2). I

(8.3.1.4) Let J :2 {2,...,72}. Assume that A is a subgroup of QJ with 1 76 |A| 2 W:
C,.(A)|.

(a) q is even.

(b) K'UO Z [V, A] if and only if q = 2 and A = Xgm.

Proof. Note that (8.3.1.3) implies that

 

(*) m = [U_1,a][11_1,b] for all a,b E Q; where
V 2: V/K’Ul.
From (B.3.1.2) we get that
we — 13201 = [U—11$61.n(-tll E [U—liQJIa
211-, = [U_1,:1:5,',_,(—t)] E [U_1,QJ], and

211,: [11-1,ng , (—t)] E [U-1,QJ] for all t E K and 2 E {2, . ...,72}

1.2-l

Then (at) implies that the map

832 QJ —> [V,QJ], (21—1 [U_1,a]

is surjective. Since [QJI = [[V, QJH by (8.3.1.3)(a1), it follows that K. is bijective. In

particular,

|[V,Al| 2 |K(A)| = IAI-

161

Assume that [[V,A]| S [A]. Then [[V,A]] = |h(A)| and hence [V, A] (1 K21, = 0. Now
(8.3.1.2) implies that q is even, A S X51,” and [A] = 2. Since [V : GV(A)| is a power
of q, it follows that q = 2.

Assume that [[V, A]| > [A]. Then (8.3.1.3)(b) and the assumption |V : Gv(A)| S [A]
imply that q is even. Since [[V/Kvo, A]] = |(V/KU0) : CV/Kv0(A)[ S [A] < [[V,A]], it
follows that K110 S [V, A].

From (B.3.1.2) it follows that Xgm is a subgroup of QJ with IXgml 2 W : CV(Xfil.n)[
and K110 Z [V, X,,,JJ, provided q = 2. I

(8.3.1.5) Assume that q is odd. Let r E {2, . . .,72}. Put J := {1 . . .,72} \ {r}.
(a) Z(QJ) is a vector space over K with basis {$g:}(1) l 1 S 2 Sj S r — 1}
where
2:23;],(1) = 27,3;j(t), for all t E K and 1 S 2 S j S r— 1.

(b) Z (Q J) and CV(QJ) /\ CV(QJ) are isomorphic as K PJ-modules.

(c) V/[V, Q1] and Z(QJ) are not isomorphic as GF(p)PJ-modules.
Proof. (a) follows from (B.3.1.2). Again by (B.3.1.2), the K—linear map which sends
$g:}(1) to v, /\ U,, for all 1 S 2 S j S r, is a KPJ-isomorphism from Z(QJ) to
CV(QJ) /\ CV(QJ). Hence (b) hOldS.
Suppose that Z (Q J) and V/[V, Q1] are isomorphic as GF(p)PJ-modu1es. Then r = 3,
since dimK Z(QJ) : 5527—12 and dimK V/[V,QJ] = r. For each t E K \ {0}, put

W) I: 11711;,(0113—27,,(-t_1)$22,,(t)$2;,,(1)$—2;,,(-1)$2'2,,(1).

Then h(t) has on Z(QJ) the eigenvalues t‘1 and 2'2, and on V/[V, QJ] the eigenvalues

1 and t, for each 2 E K \ {0}, contrary to the assumption that q is odd. I

162

B.3.2

This subsection is about the spin module for O7(q). Assume that 72 = 3. In the

construction of section 8.1, let

1
A = A3(Bg) 2’ '2-(01‘1' 202 + 303).

Then A =(121,...,,12.,,—,121,...,—f24}, where
#1 '_ A:
1
p2 := 5(01-1- 202 + (13),
I
123:: -2-(021+ (13), and
1( 1
:= — - -a' .
#4 2 011 3

Let U,,, be a nonzero weight vector of weight [21. Then the basis B can be chosen as

follows: Put
U,,2 :2 U,,,e_a3,
21,,3 :2 11,128-02,
”/24 ;: 722138-031
”-114 :2 vitae-Oi?
”—123 ;: ’U..MC_03,
’U_#2 := 11_,,3e_a,, and
”-221 3: v-uze—aa-

Then (Um, . . . , 11m, 21_,,,, . . . , U.,,,,} is a basis with the desired properties. We obtain

a group G with generators 23(2) (6 E (I), t E K) acting on a module V.

163

Using (i)—(iv) and the definition of 21,“, . . . , 21,“, 21_,,,, . . . , ”—124 we get
v—Meai Z vitae-01601 Z vuahai Z (21310071213 Z ”223:
v-meai Z v‘fl4e'03801 Z U—meaie-aa Z vitae—as Z ”#41
vfl4e‘al Z Ulise-Ose-Oi Z vflse—Oie-Oa Z ’U__,,4€_03 Z ”-2231
”#3602 Z ”2126-02602 Z Umhaz Z (2‘21a2)v#2 Z U212,
v-fl'zeaz Z v-Hse-azea-z Z U-mhm Z (_H3202)U-#3 Z 73—223,
71222603 Z ’Um€_0,3803 Z vuihaa Z (#1103>vui Z vfll’
”#11803 Z v-Me-anaa Z U-uzhaz Z (ZH21 023)U._,,2 Z v—uzi
”#4803 Z vitae-03803 Z 'Uuahoa Z (2‘31a3>U#3 Z 71123,

v-flaeas Z ”flue—03603 Z v-mhas Z (—,224,023)'U_,14 Z ”-2141

”#146512 Z ”-2” [8021 601] Z v-HAeazeai _ U-fl4801602 Z 0 — ”#3602 Z Zvuw
v-flzefim Z v-u2lea2i 601] Z U-Meazeai Z v-flaeai Z UM“
711136523 Z U113 [8031 802] Z Zvltseazeas Z vaeaa Z Zvflia

v-u1862,3 Z ”—111 [8031 802] Z v-measeaz Z v-meaz Z ”-1231
vu4efi2,3 Z ”#4 [60131 802] Z ”#4603602 Z 11113602 Z vflw

v-Mefiaa Z ”—112 [8031802] Z Zv-Meazeaa Z Zv—uaeaa Z Zv-Mv
v-meBm Z ”U-“ [8132.31 601] Z _v-meaiefizs Z Zvuaefi'zs Z villi,

v—Miefiin Z v-ui [€132.31 earl Z v-#1652,3eai Z ”-21an1 Z ”#41

164

v—#3651.3 Z ”-213 [652.31 801] Z Zvfltseaiefim Z Zvu4652.3 Z Zvuzi

v-medis Z ”-212 [6132.31 801] Z v-mefimeai Z Zv—Meai Z Zvuai
I 1 1 I 1

2221481333 Z gum [8132.31 803] Z ivlttiefizseaa Z Evmeasefias Z ivflzeaa Z ivflaefizs Z

vl‘l’

1
”(j-#162723 Z iv-fllleBZS’ 603] Z Ev—fllefiZSGOS Z 511—2116036323 Z

1 1

2v—useas Z év-Itzefiss = ”-2141

v-mefi'm Z Zv—Mi [801’ 8233.2] Z Uruiefigeeai Z U—meai Z ”#31

v-fisefi'm Z Zv—us [8011823223] Z —vZ’—‘3eale)3,2'2 Z vaefifm Z Zvuii

v—l‘lefij‘l Z Zv—ui [802181333] Z le‘lefi’l'geOQ Z vitaeaz Z ”#21 and

v—Mefi'm Z Z/U-M [602181313] Z —U_,,,ea,e)311.2 Z Zv-213613[_2 Z “#1'
Put

21,- := 1 ®z U,,, and
11-, :2 1®z 21-“,
for each 2 E {1,...,4}.

(8.3.2.1) Let J := {2,3}. Assume that A is asubgroup of Q1 with [V : GV(A)[ S [A[ aé 1.
Then [V, A] = [V, Q1].

165

Proof. From the calculations above it follows that with respect to the basis ’01,. . . , ’04,
21-4,...,v_1 of V the action of the generators 2:0,,(2), 2:51,,(2), 52,313“), 9351 1(2), and

2312(2) (t E K) of Q J is given by the following matrices:

 

 

 

 

 

 

 

 

 

 

/ 1 ) f 1 ' )
l 1
1 1
some) , 1 , 5051,20): _, 1 ,
t 1 1
1 t 1
K 12 K 1}
1 1
1 1 i ( 1 )
1 1
3331.3“): t 1 1 $511“): 1 1
—t 1 1
—t 1 t 1
K t 1 j K t 1)
and
I
r 1 )
I
2351,20): 1 1
-t 1
1
K t 1)

Suppose that [V, A] yé [V, Q1]. Since L J induces Sp4(q) on [V, Q1] and therefore acts

tranisively on the 3-dimensional subspaces of [V, Q J], we may assume that [V, A] S

K111 + K212 + K213. By the matrices above it follows that
A S {$311.1(8) $131.2“) [ 8,2 6 K}.
In particular, [A[ S [K]2 and hence

(*) Cv(A) 24 [V, Q1]-

166

Pick a E A\ {1}. Then a = 123311 l(3)1231l 2(t), for some 3,2 E K. By the matrices above

we get
(**) Cv(a) = [V, QJ] + KU_4 + K(t’U_2 — 311-3).

IfA S {2:5[II(512) 253,2(222) [ 22 E K}, then [A[ S [K] and, by (**), [V : GV(A)[ = [K[2,
a contradiction. If A Z {2511(su)x,311‘2(212) [u E K}, then [A[ S [KI2 and, by (>102),

[V : CV(A)[ = |K[3, again a contradiction. I
B.4 Cn

Assume that L is of type Cu, and

n l 1

A: :a,+—an.
2:1 2
Then
(13+: {)3,,j[i,j€ {1,...,772—1},ZSj}U{fl;J-[Z,j€ {l,...,7l},’l£j} and
A: {u1"°',lJ’n1—/’l’1)"'2—Mn}i

where

1'
Hm- := Zak, for all i,j E {1,...,72— 1} with 2 S j,
k=2
jZl n—l
2;, 2:2ak+22ak+an, for all i,j e {1,...,72} with2Sj, and
2:22 k=j

71—1
1
22, :22 a), + 5am for all 2 E {1,...,72}.
k=i

Let U,,, be a nonzero weight vector of weight 121. Then the basis B can be chosen as

follows: Put
Um+122 owed,“ for each 2 E {1, . . . ,72 — 1},

”-221. 2: U,,ne_an, and

167

21_,,n_i :2 —v_,,n_,+,e_an_,, for each 2 E {1, . . .,72 — 1}.

Then (21,“, . . . ,vfln,11_,,,, . . . , U_,,n} is a basis with the desired properties. Note that

with respect to < the following pairs are extraspecial in the sense of [3]:
(a,,fl,+1,j), for all 2,j E {1, . ..,72 — 1} with 2 < j,
((1,,fizfld), for all 2,j E {1, . . .,72} with 2 < j, and
(abflfifi-H), for all 2 E {1, . . .,72 — 1}.
Hence without loss we may assume that
(i) egw. = —[ea,,eg,+,‘j], for all i,j E {1,...,r2 — 1} with 2 < j,
(ii) 65:.) = —[ea,,eg:+l‘1], for all 2,j E {1, . . .,72} with 2 < j,
(iii) 263;] = —[ea,,eg:‘i+l], for all 2 E {1,...,12 — 1}.
13.4.1

This subsection is about the natural Sp2,,(q)-module.

(8.4.1.1) (a) [3,, = 12, —— 22,-“, for all 2,j E {1, . . .,72 — 1} with 2 S j.
(b) figd- = M, +ftj, for all i,j E {1,. . .,72} With 2 S j.

(c) LetpEA,5E {1,—1},and2,jE {1,...,72—1}with2Sj. Then

Even, if u = 5,2,1,
byes/31‘]. = —€U_E,,j+, if )2 = —5)i2,
0 else.

(d) Let 22 E A, E E {1,—1} and 2,j E {1,...,72} with 2 Sj. Then
Us,“ if12=—€;2,

U,,eggzd 2 Us“). ifp = —€/2,
0 else.

168

Proof. (a),(b) This is clear.

(c) From (a) it follows that
11,335,,” = 0 unless ,u E {—€/2,,5/2,+1}.

First assume that 2 = j. Then 5,), = 02,. From the definition of 21,,i+l and 21-,“
U,,,ew, = Um“,
v—#t+ie-at Z Zv-lm

U,,,Hea, : U,,,e_a,ea, = U,,,ha, + U,,,eaiew, = (,12,,c2,)21,,i + 0 = 21,“, and

v-fls’ea.‘ Z Zv—#i+le-Oieai Z Zv-#t+ihat Z IUZ#i+leaie-ai Z

Z<Z#i+la GNU-#2“ Z 0 Z Zv-#t+i'

Hence (c) holds if 2 = j.

Now assume that 2 < j. Then by (i) and induction on ht(/8,0) we get
[DZ-#2662.) Z Zv—uieatefit+i.j + v—flte132+i,jeat Z v-#t+ieflt+i,j + 0 Z Zv-uj+ii
vuj-f-lefilJ Z Zvflj+ieatefii+i,j + vflj+ieflt+i,jeai Z 0 + vfli+ieai Z UH."
vine-[32,1 Z vflie—Oie"52+l,j Z vflte-flt+i,je-at Z UHi-t—le-fli-i-IJ Z 0 Z vuj+ii

and

v-u1+ie—fit.1 Z v—#1+ie—ate-Bt+i.j Z v-#1+ie-fit+i.je-at Z 0 + v—#i+le—ai Z Z

(d) From (b) it follows that

Duet-5;], = 0 unless 22 E {—€/2,, —5;2j}.

we get

v_,,,.

First assume that 2 = j = 72. Then 8;, 2 an, and from the definition of 11-,” we get

Wham,“ : ”—21,. and

169

’U—unefig," = Unnewnean = ”U,,“ ha" + U,," canewn : ()2,” 0,021,," — 0 = U,,".

Hence ((1) holds in this case.

Now assume that 2 < j. Then by (ii), (c), and induction on ht( :3.) we get

v_“1€[3:'1 : ZvZuieaie + UZl‘iefi

(32+IJ 2+1,jeai Z v—fli-HefiLH'j + O _ vuji

v-m 8131‘, Z Zv—w 802851,,” + ”-241 efififlJeat Z 0 + vm+ieat Z ”(In

fume-£31.,- Z ”use-aie-AIHJ Z ”fire—Bfiflde—Oi Z ”Hue-[31“,], Z 0 Z v“#j’
and

ville-51',- Z vuje—Oie-fifflJ Z L,#]eZfi:+l‘jeZai Z 0 Z U-#2+ie-as Z ”-212:
Now assume that 2 23' < 72. Then by (iii), (c) and induction on ht(fifJ) we get
1
v—l‘26315' Z E(ZvZfl‘ea‘efilM-i + v‘“iefii.t+iea‘) Z 5(v_“i+lefii,t+i + viii-Heat) ‘—

1
E(Ufli + ”24.) Z ”22.

and

1
vflteZfi1'j : E(vuieZOICZHI'Jqd — Uflie_fl:‘i+le_ai) :

E(v—Ili + ”#2) Z +v-ut'

Put
21, := 1 ®z U,,, and
21_, := 1 ®z 21-“,

for each 2 E {1, . . .,72}.

170

E(vflwie-MJH Z v-#t+ie-Ot) —

(8.4.1.2) LetkE {1,...,72}U{—1,...,—72},2,jE {1,...,72—1}with2Sj,€E {1,—1},
anth K. Then

”Uk-l-Etvs,‘ lfk=€(j+1)
kaffii.) (t) Z ”0],; '— 8211-504.” If k = —571
21,, else.
and
’0}: ‘2' give, If k = —€j
kaefi:j(t) = ”U,, + Et’UEj If k = —52
11)c else.

Proof. This follows from (8.4.1.1). I
(8.4.1.3) Let J 2: {2, . . .,72}.

(a) CV(Q_]) S [V, a], for each a E Q) \ {1}.

(b) Assume that 1 E A S QJ and [A[ Z [[V,A]]. Then CQJ([V,QJ]) S A.
Proof. Note that (8.4.1.2) this implies that
[V, QJ] = é} KU, EB 6719 KU_,, and
2:1 2:2
[VtQJtQJl = CVKQJ) = K”Ui-

In particular,

 

(*) [U_1,ab] = [21_1,a] [’U_1,b] for all a,b E QJ where
V 2: V/Kvl.
(a) Pick a E Q1 \ {1}. Then there exists 21,. . .,tn_1,t’1, . . .,tfi, E K such that

a Z $52,101) ' ' - ° ° xfit’m—i (tn—1) ' ”IE/31,101) ' ' ' ' ' 335'- (th)

1.12—1
Ift,7€0for some2E {1,...,72— 1}, then
2211 = [ti-12m“, a] E [V,a] for each 2 E K.

171

If t: # 0 for some 2 E {2, . . .,72}, then
2211 = [(t;)_lt’v-,,a] E [V, a] for each 2 E K.
If t, = 2;“ = 0 for each 2 E {1,...,72 — 1}, then 21 E 0 and
to, = [(t’l)‘ltv_1,a] E [V,a] for each 2 E K.
(b) From (8.4.1.2) we get that
tv_, 2 [2.1-1,xgl‘,_,(—t)] E [U_1,QJ] and
to, = [’U_1,l'3'm(—t)] E ['U_1,QJ] for all t E K and 2 E {2, . . .,72}.
Then (*) implies that the map
It: QJ —> W, a 1—1 m

is a surjective homomorphism. Note that GQ J([V, Q J]) is the kernel of K. and has

order q. Together with (a) and [[V, A][ S [A[ we get
[A[ = [A0 K€f('€)l ht(/1)] S (IMAM S Ill/#1“ S [A[-

Hence [A F) Ker(n')[ = q, i.e., CQJ([V,QJ]) S A.

(3.4.1.4) Put J :2 (2...,12}.

(a) CQJ([V,QJ]) is a 1-dimensional vector space over K spanned by 251.1(1)

where
2.2311(1): $g:1(t), for all t E K.

(b) Let A :2 CPJ([V,QJ],CV(QJ)). If GQJ([V,QJ]) and V/[V,QJ] are isomor-
phic as GF(p)A-modules, then q E {2, 4}.

172

Proof. (a) This follows from (8.4.1.2).
(b) Assume that GQ J([V, Q J]) and V/ [V, Q J] are isomorphic as GF(p)PJ-modules.
For each t E K \ {0}, put

Mt) == $23,,(t)1‘—5',,,(—t_1)$2;,,(t)$131,,(‘1)513—61,(1)1721,(—1)-

Then h(t) E A has on GQ, ([V, QJ]) the eigenvalue 2‘2, and on V/[V, Q1] the eigenvalue
2, for each 2 E K \ {0}. Put H := (}2(t) [ t E K). Then K ®cp(p) (V/[V,QJ]) is the
direct sum of 1-dimensional K H-modules W0, . . . , Wk_1 where h(t) has on W, the
eigenvalue 2’", for all t E K and 2 E {0, . . . , k — 1}. Since GQJ([V, QJ]) and V/[V, Q1]
are isomorphic as GF(p)H-modules, it follows that there exists m E {0, . . . , k — 1}

such that
tpm 2 2‘2, for each 2 E K.

Hence pk -— 1 divides pm + 2. Since 772 < k, this implies p = 2 and k S 2. I

(8.4.1.5) Let r E {2,...,n}. Put J I: {1 ...,72} \ {r}.

(a) Z(QJ) is a vector space over K with basis {I52J(1) [ 1 S 2 S j S r} where
2:133:1(1): 25;.)(2), for all t E K and 1 S 2 Sj S r.
(b) Assume that q is odd. Then Z (Q J) and the symmetric square of GV(Q J)
are isomorphic as K PJ-modules.
(c) Assume that q is even. Put W 2: (Xfiis | 1 S 2 < j S r).

(c1) 14’ is a K PJ'SmeOdUIC of Z (Q J) that is isomorphic to the exterior

square of CV(QJ).

(c2) If Z(QJ)/W and V/[V, QJ] are isomorphic as KPJ-modules, then q E
{2, 4}.

Proof. (a) This follows from (8.4.1.2).

173

(b) Let ”y be the natural homomorphism from CV(QJ)®KCV(QJ) onto the symmetric

square of CV(QJ). Then (8.4.1.2) implies that the K -linear map defined by

7('U,®U,) if1S2<er
xfi"’(1)Z{%7KU2®vi) if1S2=er

is a K PJ-isomorphism.
(C) By (8.4.1.2) the K-linear map that sends 25:1(1) to 21, A21], for all 1 S 2 < j S r,
is a KPJ-isomorphism from W to CV(QJ) /\ CV(QJ). Hence (c1) holds. The proof of

(c2) is similar to the proof of (B.4.1.4)(b). I

(8.4.1.6) Assume that A is a nontrivial subgroup ofG with [A[ Z [[V, A]]2 and [V, A, A] =
0. Then 72 2 3.

Proof. Note that (A.1.5)(a) implies that [V, A] is an isotropic subspace of V. Let r
be the K -dimension of [V, A]. Then r S 72, since [V, A] is isotropic. Moreover, [V, A]
is conjugate to the span of U1, . . . , 21,, and hence [A[ S qZ(2le. Thus 10%) Z 27, i.e.,

r23.I

8.5 D,

Assume that 72 2 3, L is of type D", and
n-2 1
A = Z (Y, + 5(On_1+ (in).
2:1

Then <i>+ = {2,,,(2,je{1,...,n—1},231}U{13.2,-li,j€{1,...,n-1},isj}

and A = {221,...,12",—;2(,...,—;2,,}, where

1
fl,”- := Zak, for all 2,j E {1,...,72— 1} with 2 Sj,
k=i

j n—2
3,2]- :=Zak+2 Z ak+an_1+a,,, for all 2,j E {1,...,72—2} with2Sj,
k=i k=j+l
n—2
8,2,“, :2 Zak + an, for all 2 E {1,...,72 — 1},
2:22

174

n—2
I
)2, := Z 01,, + E(an_1+ on), for all 2 E {1, . . .,72 — 1}, and
1:22

1
#72 -— 'éKan Z (172—1)-

Let U,,, be a nonzero weight vector of weight 121. Then the basis B can be chosen as

follows: Put
for each 2 E {1,...,72 —- 1},

viii-+1 :2 v“: 6Zai7

'11-”, :2 U,,n_16_an, and
for each 2 E {1,...,72 — 1}.

vZI-‘n-i :2 ZUI‘n—2+18Zan-i’

Then {U,,1, . . .,U,,n,21_,,,,...,21_,,n} is a basis with the desired properties. Note that

with respect to < the following pairs are extraspecial in the sense of [3]:
(a,,fl,+1,j), for all 2,j E {1, . ..,72 — 1} with 2 < j,
(abfifflJ), for all 2,j E {1, . . .,72 - 2} with 2 < j, and
(fl,,,+1,/3:+1’,+1), for all 2 E {1, . . .,72 — 2}.
Hence without loss we may assume that
(i) 85"}. = —[60i,€5‘+1'j], for all 2,j E {1, . ..,72 — 1} with 2 < j,
(ii) em}, = Zleavefii+t,,]’ for all 2,j E {1, . ..,72 — 2} with 2 < j,

(iii) 83].] = —[€gm+l,€g:+m+l], for all 2 E {1, . . .,72 — 2}.

175

B.5.1

This subsection is about the natural O§n(q)-module. The information given in the

lemmas below will also help to construct the natural O§n(q)-module later.

(8.5.1.1) (a) fiio‘ = 22, — ,12j+1,for all 2,j E {1, . ..,72 — 1} with 2 S j.
(b) {J = ,12, + 22,“, for all i,j E {1,...,72 — 1} with 2 S j.

(c) Let )2 E A and S E (I). Then

U,,“; if 2/2 + S is a sum of pos. roots and 22 + S E A
one), = ——21,,+5 if 212 + 5 is a sum of neg. roots and 22 + B E A
0 else.

Proof. (a),(b) This is clear.
(c) Iffi = —a, for some 2 E {1, . . . , 72— 1}, then (c) holds by definition ofv,,,,...,11,,n
and v_,,,,_,,...,v_,,,. Note that 2,12,,_1 — on = 02"-, is a sum of positive roots and

222,l — 01,, = —02,,_1 is a sum of negative roots. Hence the definition of 21-,” and

vflneZan : vfln—leZan—leZan : vl‘n—I [eZan-l ’ eZan] + vfln-leZaneZan—l —
O + le‘neZan—l : Zle‘n—l’

imply that (c) holds if B = —a,,.

Now assume that S = a,, for some 2 E {1, . . .,72 — 1}. Then
vm+iefi Z ”me—0.80. Z “what + ”meme—a.- Z 0121020741.- + 0 Z ”22.-

and
”-21.85 Z Zv-ut+ie-Oteat Z Zv-uH-ihat Z vu2+ieate-Ot Z Z<Zfli+liai>vm+i Z 0 Z
Zv#a+i '

Hence (c) also holds in this case.

Let 2,j E {1, . . .,72 — 1} with 2 < j. Then, by (i) and induction on htflm-

vflj+ie132.j Z Zvflj+i lean effing] Z vl‘j+lefii+l,jeai Z viii-Heat Z vflt‘i

176

and

v-fliefiw Z Zv—ut [8021 efit+i.,l Z Zv—flteatefit+i,j Z v—#t+iefit+i,j Z Zv‘flj-i—ii

v—fljHe—fiw Z v—#1+ile—0ne—fit+i.jl Z ZUZ/‘j-He'flH-Lje—ai Z v—#t+ie-Ot Z Zv—llt'i

vine-32,; Z vutlc-Oti e-fit+i.)I Z Uflie‘aie—fiH-Lj Z Ullt+ie-Bt+i.j Z vflj+i'

Hence (c) also holds if 13 2 fit,- or B = —)3,,,.

Let 2,j E {1, . . .,72 — 1} with 2 S j. By (ii), (iii), induction on htflfJ, and the parts

of (c

and

) proven so far, if2 < j then

v-fljHe/XJ Z Zv—flfiileanefiiflJl Z v—I‘j+lefi;+1J-eat Z vui+ieai Z Ulla"
vZflleflz'J Z Zval [8021613:+1J] Z ZvZuteaie/B:+l'j Z vai+lefiz+lJ Z Zvflj'l’l’
”0254216431.,- Z vflj+l[6“at’e—‘fi:+1'j] Z Zvflj+le-fi:+1‘je—Oi Z v-#2+ie-at Z Zv-flti

fume-6:“) Z vflxle—Oti e—fi:+i.j] Z vflne-aieZBl-HJ Z v#‘+leZfi:+l.J Z UZMTI;

andif2=j<72—1thcn

and

Hence (c) also holds if 5 2 3:0- or H = —B’

vZutefifiJ Z Zv-I‘: [efii,i+l’efl:+l‘i+1] Z ZUZI‘! e232,i+1816:+1‘i+1 Z v—#t+2efi:+1.,+1 Z vflH-i’
vZfli+1631J Z ZvZfli+llefii,2+l’eflz+l‘,+1] Z UZH2+lefiz+1J+lefiin+l Z va+2efit.2+i Z Zvflti
vute-Bfi‘, Z U212 [e—Bi,i+l’e-5:+l.,+l] Z vflie—Bi.i+le—6:+l'i+l Z vut+26-B:+l‘,+l Z Zv-IItHi
vfli+le—13:'J Z ”Mule-513+“e—Bfifl'FHI Z Zvfll+le_fl:+l‘i+le_fit,i+l Z Zv-#t+2e-52,t+i Z
1A,“.

1.1" I

177

Put
1),- := 1®z v,“ and
214:: 1 ®z v-“

for each i E {1, . ..,72}.

(8.5.1.2) Let/CE {1,...,n}U{—1,...,—n},i,j E {1,...,72—1} withigj,86 {1,—1},
and t E K. Then

vk + stud if k 2 EU + 1)
vkxefiwU) = vk — €tv_€(j+1) if k :2 —5i
vk else.
and
vk + 815in if k = —€(j + 1)
Ukl‘efiz‘j (t) ’0]; — €tU5(j+1) If k = —€?:
vk else.

Proof. This follows from (B.5.1.1)(c). .

(13.5.1.3) (a) Let r e {1, . . .,n — 2} and put J := {1, . . .,n} \ {1”}. Then
(a1) . [V, Q1] = 2;. m,- 69 e?=.+1Kv_.,
(a?) [V, 621,621] = CV(QJ) = {:1Kvi.
(b) IfJ = {1,...,n—2}, then
(b1) [V, Q1] = 69;. Xe. ea Kv_m
(b2) [V, Q1, Q1] = CV(QJ) = egg; Kvi.

(c) IfJ:{1,...,n—1},then

[VaQJl = CV(QJ) = $1011..

178

(d) The map
é : V x V ——) K,
(En: (Ll-v,- + a_iv_,-, i bivi + b_,-v_,-) +—> 2n: aib_,- + (1-1-b,-
2:1 1:1 i=1
is a nondegenarete G—invariant symmetric bilinear form.
Proof. This follows from (8.5.1.2) '

(8.5.1.4) Let J := {2,...,n}. If13aé A S Q1, then IAI < |[V,A]|.

Proof. Note that (8.5.1.3) implies that

 

 

(*) [v_1,ob] = [v_1,a] [v_1, b] for all a,b 6 Q1 where
V := V/Kvl.

From (8.5.1.2) we get that
tv_,v = [’U_1,£L‘gm_l(—t)] E [U..1,QJ] and

to; = [11-1,ng (—t)] E (v-1,QJ] for all t E K and i E {2, . ..,.n}

l,i-1
Then (*) implies that the map

n: QJ —> [V,QJ], (1H [v-1,a]

is surjective. Since IQJI = |[V,QJH by (8.5.1.3)(a1), it follows that It is bijective. In

particular,
|[V, All 2 |K(A)l = IAI-

Suppose that |[V,A]| g |A|. Then [[V,A]I = |K(A)| and hence [V, A] 0 Km 2 0. Now

(8.5.1.2) implies that A = 1, a contradiction. I

179

(8.5.1.5) Let J z: {1, . . .,n —- 2}. Assume that A is a nontrivial subgroup of Q1 with
[V, A, A] = 0 and |A| 3 ”V, A]|. Then A n CQJ([V, Q11) es 1.

Proof. Put Z 2: CQJ([V,QJ]). Suppose that A 0 Z =2 1. Note that, by (8.5.1.2),
[K’Un + K'U_n,LJ] = 0.

Suppose that there exists a E A such that [K11n + K v_n,a] is 2-dimensional. Since
LJ is 2-transitive on CV(QJ) \ {O}, we may assume that [U,,, a] 2 v1 and [mm a] 2 v2,

i.e.,

0' E xfil,n—l(1)‘T/3.I (DZ

But then, by (8.5.1.2),
[—v_1,a,a] = [U—naa] = 222.

contrary to [V, A, A] = 0.
Now let a be any nontrivial element of A. From AflZ = 1 and the previous paragraph
it follows that [K vn+K 1)-“, a] is l-dimensional. Since L J is transitive on CV(QJ)\{O},

we may assume that [K21n + Kv_n, a] = K221, i.e.,

aEx _ 522' tZ, forsomes,t€Kwiths¢00rt#0.
a1,n1()a ()

Then, by (8.5.1.2),
[U_1, a] E —8’U_n — ton + CV(QJ).

In particular, [V, A] Z CV(QJ). Together with |[V,A]| S |A| it follows that the

following homomorphisms are not injective:

Kan: A —> CV(QJ), g r—> [21mg] and

K._n2 A ——> CV(QJ), g +—-> [v-mg].

180

Pick b E Ker(rc,,) \ {1}. Then

b E Slig’l‘n_l (t1)$'3’2'n_1 (t2) ° . . . ' (Egl (tn_1)Z,

n-1,n_1
for some t1, . . .,tn_1 E K. Since Afl Z = 1, there existsj E {1, . ..,n — 1} such that
13,- 74 0. Then

[v_j,b,a] = [—tjv,,,a] = —tjsvl.
Since [V, A, A] = 0, this implies that s = O. A similar argument, using Kerb-Ln) 7:9 1,

shows that t = 0. But then a E Z, a contradiction. I

(8.5.1.6) Let J z: {1 . . .,n} \ {r}, for some r E {2, . . .,n —— 2} U {n}. Assume that A is
a nontrivial subgroup of Q1 with [V, A, A] = 0 and [Al 2 |[V, A]].

(a) |[V,/1W) CV(QJ)| Z (12-
(b) |[V,/1H 2 (13-

(c) If r = n, then |[V, A] F) CV(QJ)] 2 Q3.

Proof. (a) Suppose that |[V, A] 0 CV(QJ)| s q. Since L J acts transitively on

C'V(QJ) \ {0}, we may assume that
(*) [V,A] r‘Cv(QJ) S Kvt.

Since [V,QJ, A] g CV(QJ) by (B.5.1.3)(a)(c), we get
[V,QJ, A] g Kvl.

Note that (B.5.1.3)(d) implies that there exists a G-isomorphism p from [V, Q1] to
HomK(V/CV(QJ), K) defined by

(v + CV(QJ))(wp) := {(12,111) , for all 2) E V and w E [V,QJ].

Together with [V, Q], A] 3 Km it follows that A centralizes v-2 + CV(QJ), . . . , v_n +

CV(QJ) in V/CV(Q_]). Now (*) implies that

[69100.69 Kv-,-,A] 3 Km.

i=2

181

By (8.5.1.2), this means that A g Q{2 ..... n}, contrary to (8.5.1.4).
(b) Suppose that |[V,/1]] 3 Q2. Then [V, A] g CV(QJ) by (a). Since LJ acts
2-transitively on CV(QJ) \ {0}, we may assume that
[V, A] 3 Km + K212.
Now that (8.5.1.3)(d) implies that
{an} K1569 Kv_,-, A] = 0.
i=3
Together with [V, A] 3 Km + K212 and (8.5.1.2) it follows that A 3 X511 and hence
|[V,/4H 5 HI 3 Ixm = q.
contrary to (a).

(c) follows from (b) and (8.5.1.3)(c). I

(8.5.1.7) Let r E {2,...,n}. Put J := {1 . ..,n}\{7‘}, ifr 75 n—1, and J := {1,...,n—2},
ifr = n — 1.

(a) Z(QJ) is a vector space over K with basis {$fi:j(1) | 1 S i gj S r — 1}

where
t$g:1(1) = $5;J(t), for all t E K and 1 S i S j g r— 1.

(b) Z (Q J) and the exterior square of CV(QJ) are isomorphic as K PJ-modules.

(c) If V/[V,QJ] and Z(QJ) are isomorphic as GF(p)PJ-modules, then r = 3

and q: 2.

Proof. (a) follows from (8.5.1.2). Again by (8.5.1.2), the K-linear map which sends
$5:‘1(1) to 1),-A 2),-+1, for all 1 g 2' g j g r — 1, is a KPJ—isomorphism from Z(QJ) to
CV(QJ) /\ CV(QJ). Hence (b) holds.

Assume that Z (Q J) and V/ [V, Q J] are isomorphic as GF(p)PJ-modules. Then T = 3,
since dimK Z(QJ) = 51%) and dimK V/[V,QJ] = T. For each t E K \ {0}, put

W) 1= 13%,,(055-[333(*Fllxfig,(t)$ag.2(-1)$—Ag,,(1)173;,,(—1)-

182

Then h(t) has on Z (Q J) the eigenvalues It“1 and fl, and on V/[V, Q1] the eigenvalues

1 and t, for each t E K \ {0}. Hence q = 2. I

(8.5.1.8) Assume that A is a nontrivial subgroup ofG with [AI 2 |[V, A]]2 and [V, A, A] =
0. Then n 2 5.

Proof. Note that (A.1.5)(a) implies that [V, A] is an isotropic subspace of V. Let r
be the K -dimension of [V, A]. Then r g 71, since [V, A] is isotropic.

Assume first that [V, A] is singular. Then [V, A] is conjugate to the span of m, . . . , v,.,
and hence |A| g qu—ll. Thus 5(ng 2 2r, i.e., r 2 5.

Assume now that [V, A] is not singular. Then [V, A] is conjugate to the span of
v1,...,v,._1,vn + v_,,, and hence |A| g QW”. Thus $592153) + r 2 2r. It

follows that r(r — 2) > (r —-1)(r — 2) 2 2r, i.e., r > 4. I
3.5.2

This subsection is about the natural O§n(q)-module. Assume that K = GF(qz) for

some prime power q. Let r : V —> V be the K -linear map defined by
v,- ifz'E{1,...,n-1}U{—-1,...,—n}
’UiT = .

v-” ifizn
U,, ifz'z—n

Put K := {t E K | t = t"}. Let a: V —> V be the K—linear map defined by
tviathvi, for alliE {1,...,n}U{——1,...,—-n} anth K.

Put 17 :2 Cv(or). Choose an element 3 in K \ K. For each i E {1,...,m} U

{—1,...,—n} define

v, ifiE{1,...,n—1}U{—1,...,—n}
17,-: vn+v_n ifi=n .
svn+sqv_n ifi: —n

For all i,je {1,...,n—1} withigjdefine

W— ,Big' ifjgn—Z
1.] ' %(fii,n—1 + mat—1) ifj = n ‘1

183

and

It;

',. ifjgn—2

1

n7
%(/3i,n—l + fl:,n_1) ifj : n —1

Us)

I .—
i,j ~—

<i>+ := {A.A. l 5,5- 6 {1,...,5 —1}, 2'51},

Put

£13- ;={_B,,j,)3’;;|t',je {1,...,n—1},igj},

A

(D :

(T)+ U (TV, and

A

H 3: {81,1,---t/§n—1,n—1}-

Note that <13 is a root system of type 855-1 and fl is a system of fundamental roots
in such a way that 35,,- corresponds to the node i in the Dynkin diagram. For all

iE {1,...,n— 1} and t E K define
5_ (t) == we.,._.(t> xe;,,_,(t") and
17-51,“, (t) I: 55-53114“) $—ag,n_,(tq)-

For all 3 E <T> let 5% be the subgroup of G consisting of all 2:5(t) where t ranges over
K or K, depending on whether 3 E {35354, ~3,,,,_1} for some i E {1, . . .,n — 1} or

not. Define

A

G:=()T§|§e<i>).

For each J Q {1, . . .,n — 1}, let in be the root system spanned by {3333‘ I j 6 J}, and

put

(8.5.2.1) (a) 5‘1, . . .,17n,17_1, . . . ,27_n is a K—basis of v.

184

(b) V is a Ka-module.

(c) Let k E {1,...,n}U{—1,...,—n}, i,j E {1,...,n—2} withiSj, 5 E
{1,—1}, and t E K. Then

27k+€t2755~ ifk=€(j+1)

fikl‘gb‘w (t) 2 13k — Et17_5(j+1) if k 2 —5i
'flk else.

and

27k + 5t2755- if k = —5(j +1)

fikflieb‘, (f) _ 17k — Et6€(j+1) If k = —Ei

U 1715 else.
((1) Let k E {1,...,n}U{—1,...,—n}, i E {1,...,n—1},£ E {1,—1}, and

t E K. Write t = t1 + tgs with t1,t2 E K. Then

ilkxgb‘mhl (t) =
’ 17n+5(t+tq)275,- isz n
27-” + (st+(st)q)1’3,- if k = —n and 5 = 1
27.” + (3% + stq)z7_,- if k = —n and s z: —1
< 17_i-’tqt’(71'— (t1+t2(8q + 8))1771 + 15217." if k = —i and 5 =1
17,- - tqtzl, —t127,, — t23_n if k = 2' and e = —1
( 27;. else.

 

Proof. This follows from the definitions and (8.5.1.2). I

(8.5.2.2) (a) Let r E {1,...,n — 1} and put J := {1,...,n — 1} \ {r}. Then
(8.1) [V’QJ] : {1212181 69 ®?=r+l 26““
(a2) [V,QJ,Q1]= CQ(QJ) = {:1K175.

(b) The map

A A

6 : 17 x l7 —> K,
n n 71

(Z (1161‘ + a411,, 2 (25-17,- + (9-16-5) H 2 a,b_,- + 0...,‘b,’
i=1 i=1 i=1

is a nondegenarete G-invariant symmetric bilinear form.

Proof. This follows from (8.5.2.1) I

185

(13.5.2.3) Let J z: {2, . . .,n — 1}. If 1 5a A 3 Q5, then |A] < |[V,A]].

Proof. This follows from (8.5.2.1) and (8.5.1.2) the same way as (8.5.1.4) follows
from (8.5.1.2) and (8.5.1.3). I

(8.5.2.4) Let J := {1, . . .,n — 2}. Assume that A is a nontrivial subgroup of Q, with
[17, A, A] = 0 and |A] 2 |[V, A]]. Then A n Céj(]t7,QJ]) 5e 1.

Proof. Put Z 2: CéJ([l7, Qfl). Suppose that A (1 Z = 1. Note that, by (8.5.2.1),
[K65 + i621", L5] = 0.

Suppose that there exists a E A such that [K2755 + K21", a] is 2-dimensional over K.
Since LJ is 2-transitive on 09(Q1) \ {0}, we may assume that [85,, a] = (s + sq)271 and
['8_,,,a] = (s + sq)272, i.e.,

a E $51,n_1(5)$5§.n—1(1)Z'
But then, by (8.5.2.1),

[111, a, a] = [—sq3271—(s + sq)27,, + 27-5,,a] = —(s + sq)62,

contrary to [17, A, A] = 0.
Now let a be any nontrivial element of A. From AflZ : 1 and the previous paragraph
it follows that [K 175+K 27-5,, a] is l-dimensional. Since id is transitive on C")(QJ)\{O},

we may assume that [K2755 + Kiln, a] 2 K17], i.e.,
a E x5,'n_,(t)Z, for some t E K with t 75 0.

Write t = t1 + t28 for some t1,t2 E K. Then, by (8.5.2.1),
[17-1,a] = —tqt61—(t1+t2(sq + 3))275 + 25227.55 9! 09(635).

Together with |[€,A]| g |A| it follows that the following homomorphisms are not

injective:
Kn: A —> 09(Q1),g r—> [17mg] and

186

K._n2 A ——> CV(QJ), g H [17-n,g].
Pick b E Ker(n,,) \ {1}. Then

b E (tam—1 (U1):17§2‘n_1(u2) ° . . . ° 1“ (Un_1)Z,

fin-im—l
for some u], . . . , 21.5-1 E K with

uf+u,-=O, foreachz'E{1,...,n—1}.

Since AflZ = 1, there existsj E {1, . . . , n—l} such that u]- # 0. Write a, = uj,1+uj,2s

with uj,1,uj,2 E K. Then
[17-], b, a] = [—ujujz7j — (217,] + uj,2(sq + 3))175, + 211-3215,, a] =
(‘(W.1 + “j,2(3q + 5))(t + tql + ”1.203154“ (3t)q))v1 =
—(uj,1(t+tq) + uj,2(sqt + tqs))v1 =
—(u3’-t + tquj)v1 = uJ-(t — tq)v1.

Since [17, A, A] = 0 and uj 9E 0, this implies that t = t". A similar argument, using

Ker(r~c_n) 9f 1, shows that st 2 (st)". But then
s = (st)"t'1 = 3%th : sqtt'1 = 3",

a contradiction. I

(8.5.2.5) Let J := {1 . ..,n — 1} \ {r}, for some r E {2,...,n — 1}. Assume that A is a
nontrivial subgroup of 521 with [17, A, A] = 0 and IA] 2 |[V, A“.

(a) |[V, Al 0 CMQAI 2 (12.

(b) |[V,/1H2 (13-

187

Proof. (a) Suppose that [[V,A] 0 09K? 1)] S q. Since LJ acts transitively on

CfiQJ) \ {0}, we may assume that

A

(*) [V,AlflCleJ) K271.

l/\

Since [V,QJ, A] S CfiQJ) by (8.5.2.2)(a), we get

AA A

[V)QJaA] S K61

Note that (8.5.2.2)(b) implies that there is a G-isomorphism ,5 from [17,631] to
HomR(V/CG(QJ), R) defined by

(v + 09(QJ))(wp) :2 E(vyw) , for all 2) E l7 and w E [V,QJ].

Together with [17, CL, A] S K271 it follows that A centralizes 6-2 +C";(QJ), . . . , 27.5, +
CV(QJ) in V/Ci/‘(QJ). Now (at) implies that
[6911527, 6]; 12274, A] g i527].
i=2
8y (8.5.2.1), this means that A S CA2” ,,,,, n_1}, contrary to (8.5.2.3).
(b) Suppose that |[V,A]| S q2. Then [V,A] S 09(631) by (a). Since f4 acts

2-transitively on C";(QJ) \ {0}, we may assume that
[V,A] 5 R275 + K272.
Now that (8.5.2.2)(b) implies that

[$272 $Rfi_i,A] = 0.

i=3

Together with [17, A] S K271 + K272 and (8.5.2.1) it follows that A S 255,} and hence

I

|[V,/1H 3 IA! 3 15554 = q,

contrary to (a). I

(3.5.2.6) Let J2: {1...,n—1}\{r}, for some r E {2,...,n-— 1}.

188

(a) Z(QJ) is a vector space over K with basis {1:35 (1) I 1 S 2' Sj S r — 1}
‘J

where

“7315(1): 1:31.1(t) for all t E K and 1 S 2' Sj S r — 1.
(b) Z (Q1) and the exterior square of CfiQJ) are isomorphic as K PJ-modules.
(c) If 17/ [9,52 J] and 00(QJ) are isomorphic as PJ-modules, then r = 3 and
|X| = 2.

Proof. This follows from (8.5.2.1) the same way (8.5.1.7) follows from (8.5.1.2). I

189

Appendix C

FF-modules for alternating groups

(3.1

Let G := 2,, for some n E IN with n 2 5 and 77. ¢ 8. Let V be the permutation
module for GF(2)G' constructed from the natural action of G on {1, . . .,72}. Put
W := [V, G]CV(G)/CV(G). Define

if n is even

m‘={ —;—1 ifnisodd.

3N”:

(C.1.1) Let A E ’P‘(G, W). Then there exist pairwise commuting transpositions t1, . . .,

tm in G such that one of the following holds:

(a) A: (t1,...,tk), for somekE {1,...,m}.

(b) n is even, and A is a subgroup of index 2 in (t1, . . . , tm).
Proof. [14](1.2) I

(C.1.2) Let H E {G,G’}. Let M be a maximal subgroup of H such that A S 02(M)
for some A E ’P‘(H, W). Then one of the following holds:

(a) H = G and M = CG(t), for some transposition t E G.

(b) n is even, and M is the stabilizer in H of a partition of {1, . . .,n} into

2—sets.

Proof. From [13] and 02(M) ;£ 1 it follows that one of the following holds:

190

(1) M is conjugate to the stabilizer of {1, 2} in H,

(2) AI is conjugate to the stabilizer of {1,2, 3, 4} in H,

(3) n is even and M is conjugate to the stabilizer of {{22' — 1, 22'} | i E {1, . . . , m}}
in H,

(4) n is divisible by 4, and M is conjugate to the stabilizer in H of the partition

{{4i—3,4i-2,4i—1,4i}li€ {1,...,§}}

of{1,...,n},

(5) n is a power of 2, and M is the normalizer in H of an elementary abelian

2—subgroup of G acting regularly on {1, . . . , 17.}.

If A is as in (C.1.1)(a), then H = G and (1) or (3) holds, since these are the only
cases in which 02(M) contains transpositions. Now assume that A is as in (C.1.1)(b).
Note that A has m orbits of size 2 on {1, . . . , n}. In case (2), 02(M) does not have
a subgroup with this property unless n = 8, and then such a subgroup has order 4.
In case (4), the only subgroups of 02(M) with this property have oder 2%. In case
(5), the only subgroups of 02(M) acting quadratically on W are of order 2. Since

|A| = 2””, this excludes cases (2), (4) and (5). I

Let ( , ) be the bilinear form on V which has 121, . . . , U,, as an orthogonal basis, where

'0,- denotes the vector of V that corresponds to the element 2' in {1, . . . , n}.
(C.1.3) ( , ) is a non-degenerate symmetric G-invariant bilinear form.

Proof. This is obvious. I

191

C.2

Let G 2: A7. Since A7 S A8 E SL4(2), the natural SL4(2)-module V is also a

G-module.

(C21) Let M be a maximal subgroup of G. Assume that |A] < |V/CV(A)], for some

nontrivial elementary abelian subgroup A of 02(.M).

(a) M is conjugate to the stabilizer of {1,2, 3,4} in G.

(b) If a: E M \ 02(M), then a: does not centralize [V, A].

Proof. (a) Since 02(M) aé 1, this follows from [13].

(b) By (A.2.2)(m) [V, A] is a 2-dimensional subspace of V. Hence the centralizer of
[V, A] in SL4(2) has order 25 - 3. Since (a) implies that |M] = 23 - 32, it follows that M
does not centralize [V, A]. If a: E M \ 02(M), then also by (a) M = (56“) and hence

a: does not centralize [V, A]. I

192

Appendix D

Examples

D.1

Let n E N with n 2 3. For each j E {0, . . .,n — 1} let UJ- be a 2-dimensional vector

(0) um

space over GF(3) with basis a, , J

and define 13,-, y]- E GL(U,-) whose action on U,-

(0) um.
, .

is given by the following matrices with respect to the basis uj ,

. 01 .10
“’1" -—10 ”J" 0—1'

VZ=U0®...®Un_1 and

Put

H z: GL(V).
For each j E {0, . . . , n — 1} let ¢j be the homomorphism from GL(UJ-) to H satisfying
(a0 (8) . . . <8) an_1)(¢j(a:)) = (a0 <8) . . . <8) aj_1® aja: ® aj+1®...® a5,_1)
for all an E U0 . . .,an_1 E U,,-1 and put
63‘ I: ¢j(yj$j) and fj 3: (9313/1)-
Then
Q== (655fj HE {0,...,n—1})

. .- 1+2n
1S extraspecial of type 2 + .

193

For each m 2 23:6 mJ-2j with 172.0,. . .,mn_1 E {0,1} put

n—l
'Um :: ®U§mj).
i=0
Then U0,...,U2n_1 is a basis of V. For all r E {0,...,n} and k E {0,...,2’ — 1}
define

w,,,. ;= (n, ] 2""k g m < 2""‘(k + 1)).

(D11) (a) Ifj E {O,...,n — 1}, then f, centralizes ®::Bj—l’1Wn-j,2k and inverts

2"-J'—1—1
k=0 Wn—j.2k+1-

(b) Ifr E {0,...,n}, then

2"—1

CH((fj | n - 7‘ S j S n —1))S fl NH(Wr,k)'
k=O

Proof. (a) is an immediate consequence of the definitions. (b) follows from (a). I

Let 2/2 be the homomorphism from 23 to H defined as follows. If 7r E 23 and m =

31;..(1m12j With m0, ' ' ' ) mn—l 6 {0,1}, then

vmw(7r) := 2155,: where

71—41
m’ = Z mj2j + 2"_3((mn_3 + 2mn_2 + 4m5,_1+1)7r — 1).
i=0
Define
G 1= NH(Q),

F I: <fn-3afn—2afn—1>Z(H)a

M :2 N0(F),
7
X m NH(W3,k), and
lc=0

Y==<1/2((2 4)(6 8)),1/J((3 7)(4 8)), 111((3 4)(7 8))71/J((5 7)(6 8D)-

Let 2. be the involution in Z(NH(W3,0) fl 0,7,:1 CH(VV3,;5)). Let c be the involution in

194

(D.1.2) (a) NH(F) S XYQ.
(b) [z,XYQ] S AI.
(c) z E G.

Proof. (a) From (D.1.1)(b) it follows that CH(F) S X. Since IGH(F,Z(H)) :
C;,(F)| S 8 = IQ 3 0007)], we get

Regarding F/Z(H) as a vector space over GF(2) with basis fn_3Z(H), fn_2Z(H),
fn—1Z(H)1the action of 1W2 4)(6 8)), 111((3 7)(4 8))t1/J((3 4)(7 8)), and ¢((5 7)(5 8))

is described by the following matrices:

1 O 0 1 0 0 1 1 O 1 0 0

1 1 O , 0 1 0 , O 1 0 , 0 1 1 .

0 0 1 0 1 1 O 0 1 O 0 1
Hence Y/Gy(F, Z(H)) ”2‘5 PSL3(2) E Aut(F/Z(H)) and therefore

NH(F) S CH(F,Z(H))Y S XQY-

(b),(c) Note that z is centralized by X, Y, f0, . . . , fn_1, and 60, . . . , en_4. Moreover,

[en_3, 2], [625-2, z], and [en_1, 2] act on V as follows:

”mm-3’ Z] _ v if m > 2"-2 ’
m _

_ —vm if m < 2'“3 or 2""2 g m < 211—3 + 271—2
’Umlen—2a 2] _ Um if 211—3 < m < 271—2 01' m > 211—3 + 211—2 ,

—vm if m < 2’“3 or 2’“1 S m < 2"“3 + 2"“1
”Men-1’2] : em if 2'“3 < m < 2’“1 or m > 2"‘3 + 2"“1 ’

for each m E {0, . . . , 2” — 1}. Therefore,

1 ifj S n — 3
[€n_3, 2,83] : €n_2fn_1C ifj : Tl _ 2 a
en—lfn—2C ifj = n - 1

195

en—3fn—lc ifj : n '— 3 a

l iijn—4orjzn—2
[en—QaZ’ej] :
en—lfn—3C ifj 271—1

1 iijn—4orjzn—1
[en_1,z,ej] = en_3fn_2c ifj = n — 3 ,
en—2fn—3C ifj 272—2

for each j E {0, . . .,n — 1}. In particular, [Q,z,Q] g; Z(Q) and hence z E G. Since

f0, . . . , fn_1 are centralized by [en_3, 2], [655-2, 2], and [en_1, 2:], we also get [62, 2] S M.
I
Put

G :2 G/Z(H).

(D-1-3) (a) Fifi/a) '5’ 9371(2)-
(b) M is a maximal subgroup of G.

(c) No nontrivial characteristic subgroup of M is normal in G.

Proof. (a) From (D.1.1)(b) (with r = n) it follows that GH(Q) normalizes (Um) for

each m E {0,...,2" — 1}. Since (e0,...,e,,_1) acts transitively on (v0),...,(v2n_1),
we get
CH(Q) = Z(H)

Hence G/@ is isomorphic to a subgroup of Out(Q), which is isomorphic to 03,,(2) by
[11], Table 4.6.A. Moreover, G induces at least (23”,,(2) on Q by Proposition 4.6.8(II)
of [11].

(b) With respect to the non-degenerate G—invariant quadratic form n on Z2- defined

by

1 ifa2=c

”(5):={0 ifa2=1

196

for each a E Q, F— is a singular subspace. Since the stabilizer of any singular subspace
of a in G is maximal in G, (b) holds.

(C) By (a), (2- is the only nontrivial normal subgroup of G that is contained in M.
From (D.1.2)(a)(b) it follows that z induces an automorphism on M, and (D.1.2)(c)

implies that a is not invariant under this automorphism. I

D.2

(D.2.1) Let X be a finite simple group, G a subgroup of H, M a proper subgroup of G,
and a E Aut(H). Assume that X = (G“, G) and M a = M. Then no nontrivial

characteristic subgroup of M is normal in G.

Proof. Let G be a characteristic subgroup of M that is normal in G. Then G =

C“ S (G, G“) = X. Since H is simple and G S M 74 H, it follows that G = 1. .

(D.2.2) Let X := Q; (pk), for some k E N and some prime p. Let B be a Borel subgroup
of X, G the parabolic subgroup of type {1,2,3} and M the parabolic subgroup
of type {1, 2} of X containing B.

(a) No nontrivial characteristic subgroup of M is normal in G.

(b) 0,,(G) is a natural 93(q)-module for 01"(G).

Proof. (a) This follows from (D.2.1), since X has a graph automorphism that

normalizes B and acts on the Dynkin diagram by switching the nodes 3 and 4.

(b) This follows from [2]. I

(D.2.3) Let X be a simple group of type E5(pk), for some I: E IN and some prime p.
Let B be a Borel subgroup of X, G the parabolic subgroup of type {1, 2, 3, 4, 5}

and M the parabolic subgroup of type {2, 3, 4, 5} of X containing B.

(a) No nontrivial characteristic subgroup of M is normal in G.

197

b) O G) is an Qfifiq -half spin module for Op'(G).
:2

Proof. (a) This follows from (D.2.1), since X has a graph automorphism that
normalizes B and induces on the Dynkin diagram the permutation (1 6)(3 5).

(b) This follows from [2]. I

(D.2.4) Let X be a simple group of type F4(2'°), for some k E IN. Let B be a Borel
subgroup of X, G the parabolic subgroup of type {1,2,3} and M the parabolic

subgroup of type {2,3} of X containing B.

(a) No nontrivial characteristic subgroup of M is normal in G.

(b) 0,,(G) contains exactly two noncentral OPI(G)-chief factors, one natural

Sp6(2“)-module and one 07(2")-spin module.

Proof. (a) This follows from (D.2.1), since X has a graph automorphism that
normalizes B and induces on the Dynkin diagram the permutation (1 4)(2 3).

(b) If k > 2, then this follows from (a) and Theorem 1. From the way F4(2) and
F4(4) are embedded in F4(8), this implies that (b) also holds if k E { 1, 2}. I

198

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200

MIC

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