Groups of local characteristic p, The structure theorem

U. Meierfrankenfeld, B. Stellmacher, G. Stroth

Preliminary version of November 6, 2002

1 Introduction

This paper is part of the classification of the finite simple groups of local characteristic p. For more information about this program see [MeStSt1]. To state the main result we just need some definitions. In what follows we fix a prime p, a group G and a Sylow p–subgroup S. With L(S) we will denote the set of all subgroups H containing S, with F^{∗}(H) = O_{p}(H). We will assume that

1) Let U be any p–local of G then F^{∗}(U) = O_{p}(U)

2) O_{p}(⟨H|H ∈ L(S)⟩) = 1

3) G is a so called K_{√}–group, i.e. any simple group involved a any p-local

is abelian, an alternating group, a group of Lie type or one of the 26 sporadic groups.

Because of 2) we may assume without loss that G = ⟨H|H ∈ L(S)⟩. If 2) is not satisfied, then we are in the situation of the local C(G, T)–theorem. The situation that G ≠ ⟨H|H ∈ L(S)⟩ will be the purpose of a forthcomming paper.

Let L ∈ L(S) then we denote by Y_{L} the maximal elementary abelian normal p–subgroup of L with O_{p}(L/C_{L}(Y_{L})) = 1. Such a group exists, see 2.2.

Now we denote by C = N_{G}(Ω_{1}(Z(S))). Let C̃ ∈ L(S) with C ≤ C̃ and C̃ be maximal. Set E = O^{p}(F_{p}^{∗}(CC̃(YC̃))), where F_{p}^{∗}(X) is the inverse image of F^{∗}(X/O_{p}(X)).

In a separate paper the case that there is some p–local X containing E with X ≰ C̃ will be treated. So we assume E–uniqueness If X is some p–local of G with E ≤ X, then X ≤ C̃.

In particular C̃ above was uniquely determined. In fact we do not need the full strength of E–uniqueness in this paper, we just need a consequence, which is known as Q–uniqueness, see 2.4.

Now the purpose of this paper is to get hand on p–locals M which are not contained in C̃.

For this we make the following definitions. Let Q = O_{p}(C̃). Then for M ∈ L(S) we set M^{0} = ⟨Q^{M}⟩. As we will see in 2.5 M = M^{0}(M ∩ C̃), we have that M^{0} dominates the structure of M. Hence knowing M^{0} means knowing a great part of M. In a forthcomming paper we will show that using this information is enough to build up a fairly large subgroup of G, which mostly will be G itself.

In this paper we will make one further assumption. A subgroup P ∈ L(S) is called a minimal parabolic if S is contained in a unique maximal subgroup of P and S ≰P. So we will assume that if P_{1}, P_{2} are minimal parabolics in L(S) then O_{p}(⟨P_{1}, P_{2}⟩) ≠ 1. What we really use from this is that M_{0} = M^{0}S is not a minimal parabolic for the M in question (see 2.7).

Before we can state the main theorem of this paper we need a few definitions concerning modules. Let G be a group and V be some nontrivial GF (p)G–module for G. Then

• We call V an F –module, if there is some nontrivial elementary abelian p–subgroup of G with |V : C_{V} (A)| ≤ |A|. The group A then will be called an offender.

• We call V an 2F –module, if there is some nontrivial elementary abelian p–subgroup of G with |V : C_{V} (A)| ≤ |A|^{2}. The group A then will be called an offender.

• An elementary abelian subgroup A of G is said to act quadratically if [V, A, A] = 1 and cubic if [V, A, A, A] = 1.

The purpose of this paper is to prove the following theorem

Theorem 1.1 Let p be a prime and M ∈ L(S), M ≰ C̃, with M^{0} maximal. Assume that Y_{M} ≰ Q. Set M_{0} = M^{0}S and K = F^{∗}(M_{0}/C_{M_{0}}(Y_{M})). Then one of the following holds i) K is quasisimple and isomorph to SL(n, q), Sp(2n, q)′, Ω^{±}(n, q), or

E_{6}(q), q a power of p. In case of K ≅ SL_{n}(q) or E_{6}(q) no element in M_{0}S induces diagram automorphisms.

ii) K ≅ SL_{n}(q)′ ∗ SL_{m}(q)′, q a power of p. Further Y_{M} is the tensor product module.

iii) p = 2 and K ≅ 3A_{6}, M_{22} or M_{24}.

iv) p = 3 and K ≅ M_{11} or 2M_{12}.

v) M_{0} is a minimal parabolic.

Further Y_{M} is an F - module or 2F –module with cubic offender. Finally Y_{M} contains a M_{0} – module V as described in the table below.

In cases i), iii) and iv) C_{M_{0}}(Y_{M})/O_{p}(M_{0}) is in the Schur multiplier of K. In case of ii) the preimage of K in M_{0}/O_{p}(M_{0}) is contained in SL_{n}(q)′ × SL_{m}(q)′ or n = m = q = 2 and the preimage is extraspecial of order 27 and exponent 3.

group prime  module  restriction
SL(n, q)  p  natural
SL(n, q)  p  alternating square
SL(n, q)  p  symmetric square  p odd
SL(n, q^{2})  p  V (λ_{1}) ⊗ V (λ_{1}^{σ})
Sp(2n, q)  p  natural
3A_{6}  2  6-dim
Ω^{±}(n, q)  p  natural
Ω^{+}(10, q)  2  half spin
E_{6}(q)  p  V (λ_{1})
M_{11}  3  5-dimensional
2M_{12}  3  6-dimensional
M_{22}  2  10-dimensional
M_{24}  2  11-dimensional

The proof of the theorem will follow from 4.1 and 5.1.

Together with the result from [MeStStr] we also know the structure of M_{0} if Y_{M} ≤ Q. Hence we have Theorem 1.2 Let p be a prime and M ∈ L(S), M ≰ C̃, with M^{0} maximal. Set K = F^{∗}(M_{0}/C_{M_{0}}(Y_{M})). Then one of the following holds

i) M_{0} is a minimal parabolic.

ii) Y_{M} ≰ Q and the structure of M_{0} is given by Theorem (1.1).

iii) Y_{M} ≤ Q and K ≅ SL_{n}(q) or Sp(2n, q)′ , q a power of p and [Y_{M}, K] is the natural module for M_{0}.

2 Preliminaries

pred

Definition 2.1 Let M ∈ L(S). Then a p–reduced normal subgroup of M is an elementary abelian normal p–subgroup Y of M such that

O_{p}(M/C_{M}(Y)) = 1.
CL

Lemma 2.2 a) Let M ∈ L(S) then there exists a unique maximal p–reduced normal subgroup Y_{M} of M

b) Let S ≤ L ≤ M, M ∈ L(S) and X a p–reduced normal subgroup of L, then ⟨X^{M}⟩ is a p–reduced normal subgroup of M.

c) Let M, L ∈ L(S), L ≤ M, then Y_{L} ≤ Y_{M}.

d) Let M ∈ L(S), C_{M} = C_{M}(Y_{M}) and M^{∗} = N_{M}(S ∩C_{M}). Then S ∩C_{M} = O_{p}(M^{∗}) and Y_{M} = Ω_{1}(Z(S ∩ C_{M})).

e) Let M^{∗} be as in d). Then M = M^{∗}C_{M} and Y_{M} = Y_{M^{∗}}.

Proof: a) Let Y_{M} be the subgroup generated by all p–reduced normal subgroups. If O_{p}(M/C_{M}(Y_{M})) is nontrivial, this also holds for all the generators of Y_{M}, a contradiction.

b) Let Y = ⟨X^{M}⟩ and D = C_{M}(Y). Set N/D = O_{p}(M/D). Then N = (N ∩S)D = (N ∩L)D. As X is p–reduced for R, we have [X, N ∩R] = 1. Further [D, X] = 1, so [N, X] = 1. As N is normal in M, we have [N, Y] = 1, hence Y is p–reduced.

c) Follows from b) with X = Y_{L}.

d) As O_{p}(M/C_{M}) = 1, we have O_{p}(M^{∗}) ≤ C_{M}. So we get O_{p}(M^{∗}) ≤ C_{M} ∩S and so O_{p}(M^{∗}) = C_{M} ∩ S. Set X = Ω_{1}(Z(S ∩ C_{M})). Then Y_{M} ≤ X. Set Y = ⟨X^{M}⟩ = ⟨X^{C_{M}} ⟩ as M = M^{∗}C_{M}. Now X is p–reduced for S and so by b) Y is p–reduced for C_{M}. Set D = C_{M}(Y) and N/D = O_{p}(M/D). Since Y_{M} ≤ X ≤ Y and Y_{M} is p–reduced for M, we get N ≤ C_{M}. As Y is p–reduced for C_{M}, we get [N, Y] = 1. Hence Y is p–reduced for M and so Y ≤ Y_{M} ≤ X. This shows X = Y_{M}, the assertion.

e) The first assertion is just the Frattini argument. Hence now Y_{M} ≤ Y_{M^{∗}}. By c) we have Y_{M^{∗}} ≤ Y_{M}.

unique

Lemma 2.3 We have [E, Ω_{1}(Z(Q))] = 1.

Proof: Let L be a p–component of E with [Ω_{1}(Z(O_{p}(L))), L] = 1. We have O_{p}(L) ≤ Q, so [Ω_{1}(Z(Q)), L] ≤ Ω_{1}(Z(O_{p}(L))) and then

[Ω_{1}(Z(Q)), L, L] = 1.

The 3-subgroup lemma now implies [Ω_{1}(Z(Q)), L] = 1.

Let now [Ω_{1}(Z(O_{p}(L))), L] ≠ 1. Set K = ⟨L, S⟩. We have Ω_{1}(Z(O_{p}(K))) ≤ Y_{K} ≤ YC̃ by 2.2. Now [Ω_{1}(Z(O_{p}(L))), L] ≤ Ω_{1}(Z(O_{p}(K))). But then [Ω_{1}(Z(O_{p}(L))), L, L] = 1, which gives a contradiction with the 3–subgroup lemma.

So we just have to handle F_{p}(E). We have F_{p}(E) ≤ C(Z(S)) and so [Ω_{1}(Z(Q)), F_{p}(E)] ∩ C(S) = 1. Hence [Ω_{1}(Z(Q)), F_{p}(E)] = 1.

Q

Lemma 2.4 Let 1 ≠ U ≤ Z(Q), then N_{G}(U) ≤ C̃.

Proof: As [E, Ω_{1}(Z(Q))] = 1 by 2.3 we get E ≤ N_{G}(U) and so by E–uniqueness we have the assertion.

fac

Lemma 2.5 Let M ∈ L(S), then

M = M^{0}(M ∩ C̃).

Proof: Let S_{1} ∈ Syl_{p}(M^{0}). Then we have M = N_{M}(S_{1})M^{0}. As Q ≤ S_{1} and Z(S_{1}) ≤ Z(Q), we see mit 2.4 that N_{M}(S_{1}) ≤ C̃, the assertion.

Mast

Lemma 2.6 Let M ≰ C̃, then M^{0} ≤ M^{∗}, where M^{∗} is as in 2.2(d).

Proof: We have C_{M} ∩ M^{0} = O_{p}(M^{0}), as this group has to normalize Q. Now [M^{0}, C_{S}(Y_{M})] ≤ O_{p}(M^{0}) ≤ C_{S}(Y_{M}), so M^{0} ≤ M^{∗}.

m

Lemma 2.7 Suppose that there are no two minimal parabolics P_{1}, P_{2} ∈ L(S) with O_{p}(⟨P_{1}, P_{2}⟩) = 1. Let M ∈ L(S) with M^{0} maximal. Then M_{0} is not a minimal parabolic.

Proof: Set R = N_{G}(M^{0}). Then by choice of M we have R^{0} = M^{0}. Let P_{1} ∈ P(S) be a minimal parabolic in ES, which is not in R. Then for H = ⟨M_{0}, P_{1}⟩ we have O_{p}(H) ≠ 1. By maximality of M^{0} we have M^{0} = H^{0}. But then we would have H ≤ R, a contradiction.

quadodd

Lemma 2.8 Let p > 2, |A| = p, A ≤ G acting quadratically on some GF (p)G-vectorspace V. Let X = ⟨A, A^{g}⟩ for some g ∈ G be not a p–group. Then X/O_{p}(X) ≅ SL(2, p) or p = 3 and X/O_{3}(X) ≅ SL(2, 5). In the latter there is some g with X/O_{3}(X) ≅ SL(2, 3).

Proof: [Cher1]

quaderz

Lemma 2.9 Let p be an odd prime, U = ⟨a, b⟩, |U| = p^{2} and assume that a, b and ab acts quadratically on a vectorspace V over GF (p) then U acts quadratically.

Proof: Let x ∈ [V, a] with x^{b} = xy, y ≠ 1. Let r, s ∈ V with r^{a} = rx and s^{a} = sy. Then (r^{b})^{a} = r^{ba} = r^{ab} = (rx)^{b} = r^{b}xy. Hence r^{b} = rst, with t ∈ C_{V} (a). So r^{ab} = rstxy. As ab acts quadratically, we have that (stxy)^{ab} = stxy. But (stxy)^{ab} = (stxy^{2})^{b} = stxy^{3}, in particular y^{2} = 1, which shows y = 1, a contradiction.

SL2p

Lemma 2.10 Let G = SL(2, q), q = p^{f} and V = [V, Z(G)] be a nontrivial module for G, and A ≤ G with [V, A, A] = 1, |A| ≥ p or |A| ≥ 9 for p = 3. Then V is a direct sum of natural modules.

Proof: Let A ≤ S, S a Sylow p–subgroup of G. Set B = N_{G}(S) = SH, with H ∩ S = 1. Let w ∈ N_{G}(H) \ H. Let X be an irreducible B– submodule in V. We have that X is a 1-dimensional module for a splitting field of V, in particular |X| ≤ q. Suppose first that S acts quadratically. We have G = B ∪ BwS and so

Y := ⟨X^{G}⟩ = X + ⟨X^{wS}⟩ = X + X^{w} + [X^{w}, S]

Now [S, X] = 1 = [X^{w}, S, S]. Hence also Y/C_{Y} (S) is irreducible and so |Y/C_{Y} (S)| ≤ q. This shows that |Y | ≤ q^{2} and is 2-dimensional over the splitting field. This now shows that Y is the natural module, the assertion.

Let U = ⟨a, b⟩ ≤ S, |U| = p^{2} and assume that a, b and ab act quadratically on V, then by 2.9 U acts quadratically.

Let now f be odd, then all subgroups of order p in S are conjuagte and so S acts quadratically. Let r be an odd prime divisor of f. Let A_{1} ≤ A, with |A_{1}| = p. Then we may embedd A_{1} into some L ≅ SL(2, p^{r}). Hence a Sylow p-subgroup of L acts quadratically. Further we may choose L in such a way that w ∈ L. Then as seen above X + X^{w} is invariant under L. As B = ⟨A_{1}, H⟩ and G = ⟨H, L⟩, we get that X + X^{w} is invariant under G, the assertion. So we are left with f = 2^{t}. Then we may choose L as SL(2, p^{2}). If t ≠ 2, then we may embedd L/Z(L) into some P GL(2, r) and so all subgroups of order p are conjugate and then a Sylow p – subgroup acts quadratically and we are done again. So we are left with q = p^{2}. If p > 3, then we have p + 1/2 > 2 conjugates of A_{1}. Hence there are a, b with S = ⟨a, b⟩ and a, b, ab act quadratically, so S acts quadratically and we are done. It remains G = SL(2, 9) in which case we assumed that A = S.

quadlie

Lemma 2.11 Let F^{∗}(G) be quasisimple G = ⟨A^{G}⟩, where |A| ≥ p^{2}, podd acts quadratically on some faithful GF (p)G vectorspace V. Then F^{∗}(G) is a group of Lie type in characteristic p.

Proof: [Cher1]

cubicodd

Lemma 2.12 Let K be a component of a group G, acting faithfully on a module V. Assume that A is a cubic acting p–subgroup of G, p odd. If |A : C_{A}(K)| = p^{2}, then A ≤ N_{G}(K).

Proof: We may assume |A| = p^{2}. Suppose first hat A acts regularly on L = {K^{A}}. Let R be a Sylow r–subgroup of L, r ≠ p, 2, A ≤ N(R). Set U_{a} = C_{R}(a), a ∈ A^{♯}. Then U_{a}A acts on [V, a]. As A acts quadratically on [V, a], we see with 2.8 [[U_{a}, A], [V, a]] = 1. Choose 1 ≠ u ∈ U ∩ ⟨K^{⟨a⟩}⟩. By the 3 – subgroup lemma we have [V, u, a] = 1. This shows that any component K_{1} of L not in ⟨K^{⟨a⟩}⟩ centralizes [V, u]. As this can be done for any a ∈ A, we see that V = V_{1} ⊕ V_{2} ⊕ ⋯ V_{p^{2}}, where exactly one component K of L acts nontrivially on V_{i}. But then in V L we have the wreath product Z_{p} ≀ E_{p^{2}}, in which A does not act cubic. So we may assume that N_{A}(K) = ⟨b⟩ is of order p and L = ⟨K^{⟨a⟩}⟩, A = ⟨a, b⟩. If we would have a direct sum of modules as before, we would get the same contradicting, since [V, b]⟨a⟩ contains a wreath product. So we may assume that V = V_{1} ⊗ V_{2} ⊗ ⋯ ⊗ V_{p}, where V_{i} = V_{1}^{a^{i}−1}. Now choose w ∈ V_{1} with w^{b} = w + v, v ≠ 1 and v^{b} = v. Then set x = w ⊗ w^{a} ⊗ v^{a^{2}} ⊗ v^{a^{3}} ⋯ ⊗ v^{a^{p−1}}. Then x^{b} = (w+v)⊗(w+v)^{a}⊗v^{a^{2}}⊗v^{a^{3}} ⋯⊗v^{a^{p−1}}. Hence [x, b] = v⊗w^{a}⊗v^{a^{2}}⊗ v^{a^{3}} ⋯⊗v^{a^{p−1}} +w⊗v^{a}⊗v^{a^{2}} ⊗v^{a^{3}} ⋯⊗v^{a^{p−1}} +v⊗v^{a}⊗v^{a^{2}} ⊗v^{a^{3}} ⋯⊗v^{a^{p−1}}. Now we get that [x, b, a] = w⊗v^{a} ⊗v^{a^{2}} ⊗v^{a^{3}} ⋯⊗v^{a^{p−1}} −v⊗v^{a} ⊗w^{a^{2}} ⊗⋯⊗v^{a^{p−1}}, which is not centralized by a, a contradiction.

module

Lemma 2.13 Let G be a group with F^{∗}(G) = O_{p}(G) ≠ 1 and A ≤ G be elementary abelian with A ≰ O_{p}(G) but A ≤ S for some Sylow p–subgroup S of G. Then there is some g ∈ G such that for X = ⟨A, A^{g}⟩ the following hold

(1) X/O_{p}(X) ≅ SL_{2}(q), or p = 2 and X/O_{2}(X) ≅ Sz(q) or X/O_{2}(X) is a dihedral group of order 2u, u odd.

(2) S ∩ X is a Sylow p–subgroup of X

(3) Y = (A ∩ O_{p}(X))(A^{g} ∩ O_{p}(X)) ≤ X

(4) Y ≠ A ∩ O_{p}(X)

(5) |A : C_{A}(Y)| ≤ |Y : C_{Y} (A)|q ≤ |Y : C_{Y} (A)|^{2}, where q = 2 if X/O_{2}(X) is dihedral.

(6) If X/O_{p}(X) is not dihedral or SL(2, 3), then Y/(A ∩ A^{g}) is a direct sum of natural modules for X/O_{p}(X).

(7) A_{1} = ⟨a | [Y, a]C_{Y} (A) ≠ [Y, A]C_{Y} (A)⟩ ≤ Y and [A_{1}, Y] = 1

or there is some g ∈ G such that B = A^{g} ≤ S, A^{g^{2}} = A, [B, A] ≠ 1 and |A : C_{A}(B)| = |B : C_{B}(A)|.

Proof: We start the proof with some general remarks. Let X be as in (1). Then obviously (3) follows. If (4) would be false, then as [O_{p}(G), A] ≤ O_{p}(G) ∩ A ≤ O_{p}(X) ∩ A, we get that [O_{p}(G), X, X] = 1, which contradicts C_{G}(O_{p}(G)) ≤ O_{p}(G). Hence also (4) holds. Next we see that C_{Y} (A) = A∩Y and so we see that C_{Y/(A∩A^{g})}(A) = (A ∩ Y)/(A ∩ A^{g}) and Y/(A ∩ A^{g}) = (Y ∩A)/(A∩A^{g})⊕(Y ∩A^{g})/(A∩A^{g}). So (5) follows. Let p = 2. Then we see that elements of odd order in X act fixed point freely on Y/(A ∩ A^{g}). Hence [Hig] and [Mar] yield (6). Let now p > 2. We have that A acts quadratically on Y. Then we have that X/O_{p}(X) ≅ SL(2, q), q ≠ 3. Now (6) follows with 2.10, besides |A| = 3 and q = 9. But in that case, we may choose X/O_{3}(X) ≅ SL(2, 3). This now also implies (7). So we see that in case (1) we just have to prove (2) which will be clear by the specific construction.

Set Ḡ = G/O_{p}(G). Let r be some odd prime with [O_{r}(Ḡ), Ā] ≠ 1. As Ā is elementary abelian, we get some g ∈ O_{r}(Ḡ) with |Ā : Ā ∩ O_{p}(X)| = p, where X = ⟨A, A^{g}⟩. If there is some component L with 1 ≠ [L, Ā] and |Ā : CĀ(L)| = p, then again we get X with |Ā : Ā ∩ O_{p}(X)| = p. Now we have (1) with 2.8. In both cases of course S ∩ X is a Sylow p-subgroup of X.

So we may assume that F^{∗}(Ḡ) = E(Ḡ). Further for any component L we may assume that |Ā : CĀ(L)| ≥ p^{2}. Recall that A acts quadratically on O_{p}(G). Let first [Ā, L] ≰ L. Then application of [Cher] shows p = 2 and L ≅ L_{2}(r), r = 2^{t}. Let L_{1} = N_{⟨L^{A}⟩}(S̄) then L_{1}/O_{2}(L_{1}) ≅ Z_{r−1}×Z_{r−1} and Ā acts nontrivially on this group and so we get X with X/O_{2}(X) dihedral and S ∩ X a Sylow 2–subgroup of X. Hence we may assume that Ā normalizes L.

Let B be the projection of Ā onto AutḠ(L). Then |B| ≥ p^{2}. Application of 2.11 shows that L is of Lie type in characteristic p or p = 2.

Assume first that L is of Lie type in p′ - characteristic, which is not also of Lie type in characteristic p. Then p = 2 and by [MeiStr1] we have that L/Z(L) ≅ U_{4}(3). As A≤ S, there is some 2–central involution s in Ā∩L. Let B be the projection of Ā onto L. If B ≰ O_{2}(C_{L}(s)), then there is a conjugate B^{g} such that W = ⟨B, B^{g}⟩ ≅ D_{6} and S̄ ∩ W is a Sylow 2-subgroup of W. Hence we may set X = ⟨A, A^{g}⟩. So we may assume that B ≤ O_{2}(C_{L}(s)). As we may generate C_{L}(s) by elements g with g^{2} ∈ S, the action of C_{L}(s) on O_{2}(C_{L}(s)) gives us some A^{g} ≤ O_{2}(C_{L}(s)) with [Ā, Ā^{g}] ≠ 1, and so we have the second alternative.

Let next L ≅ G(r) be a group of Lie type in characteristic p. Let R be a root subgroup in Z(S̄ ∩ L). Suppose B ≰ O_{2}(N_{L}(R)). Then we have an induction and the lemma holds. So we may assume that B ≤ O_{2}(N_{L}(R)). If B ≤ R, then B ≤ L̃ ≤ L, with L̃ ≅ SL_{2}(r) or Sz(r) and S ∩ L̃ is a Sylow 2–subgroup of L̃. So we have (1) again. We may assume that B ≰ R.

We have that A ≤ O_{p}(N_{L}(R)). If we may generate C_{L}(R) by elements g with g^{2} ∈ N_{L}(A), then we get the second alternative, or ⟨B^{N_{L}(R)}⟩ is abelian. Hence we just have to handle L ≅ L_{n}(r), Sp(2n, r), F_{4}(r), ^{2}F_{4}(r).

Assume first L/Z(L) ≅ L_{n}(r), n ≥ 3. Assume that B acts trivially on the Dynkin diagram. Let P_{1}, P_{n−1} be the two parabolics containig S ∩ L which involve L_{n−1}(r). If B ≰ O_{2}(P_{i}) for one i, then we have induction. So we have B ≤ O_{p}(P_{1}) ∩ O_{p}(P_{n−1}) = R, a contradiction. So let b ∈ B acting nontrivially on the Dynkin diagram, in particular p = 2.

Let first n = 3. Then we see that [b, S̄] is not abelian, which contradicts B ≤ S̄. In any other case we get a parabolic P_{3} such that P_{3}/O_{2}(P_{3}) ≅ L_{2}(r) × L_{2}(r) such that B acts nontrivially on P_{3}/O_{2}(P_{3}). If r > 2, we have induction. If r = 2 this is solvable and we get a dihedral group X/O_{2}(X).

Let next L/Z(L) ≅ Sp(2n, r), n ≥ 2. As ⟨B^{N_{L}(R)}⟩ is abelian, we have again p = 2. We may assume that B ≤ Z(O_{2}(N_{L}(R))). So we may embed B into some L̃ ≅ Sp(4, r) with S ∩ L̃ a Sylow 2–subgroup of L̃. So we may assume L ≅ Sp(4, r). Now we have two parabolics P_{1}, P_{2}, containing S ∩ L. By induction we may assume that B ≤ O_{2}(P_{1}) ∩ O_{2}(P_{2}). As B is not contained in a root subgroup we see that ⟨B^{P_{i}}⟩ = O_{2}(P_{i}) for i = 1, 2. Let H_{i} be the preimage of P_{i}, i.e. H_{i}/O_{2}(G) = P_{i}. Now suppose that ⟨A^{H_{i}}⟩ is abelian. Then we see that O_{2}(H_{i}) ≤ C_{S∩L}(A)O_{2}(G). If this is true for both i, we get S ∩L = C_{S∩L}(A)O_{2}(G). As B acts quadratically on O_{2}(G) there is a chief factor V in O_{2}(G) which is the natural module [Str], or L ≅ 3A_{6} and some 6-dimensional module is involved. In the former we have |[V, B]| = r^{2}, while |C_{V} (S ∩ L)| = r. As [V, B] is covered by A this is a contradiction. So we have the latter. Now by quadratic action B ≤ A_{6} and then |B| = 2, which gives that we have a dihedral group X/O_{2}(X).

Let next L ≅ F_{4}(r). Again p = 2. By induction we may assume that B acts trivially on the Dynkin diagram. We have two root groups R_{1} and R_{2} and we may assume that B ≤ Z(O_{2}(N_{L}(R_{1}))) ∩ Z(O_{2}(N_{L}(R_{2}))). But this group is contained in some Sp(4, r) and we get the assertion by induction.

Let finally L ≅ ^{2}F_{4}(r). As B acts quadratically we get with [Str] B ≤ R, a contradiction.

Let now L/Z(L) ≅ A_{n}, n ≥ 5. So we may assume n = 7, or n ≥ 9. Further we have by 2.11 p = 2. If n = 7 there is some L̃ ≤ L such that L̃/Z(L̃) ≅ A_{6}. Hence we have induction. So we have n ≥ 9. Furthermore by induction we may assume n to be even. Let n = 2^{m}. Then there is a subgroup L̃ ≤ L with S ∩ L ≤ L̃ ≤ Σ_{\frac{n}{2}} . As n ≥ 16 we may apply induction. Let m_{1}, …, m_{r} be the dyadic decomposition of n. Let L̃ be the subgroup of L with S ∩ L ≤ L̃ ≤ Σ_{m_{1}} × ⋯ × Σ_{m_{r}}. Further we may assume that A acts nontrivially on all Σ_{m_{i}}. Hence we have all m_{i} < 8, which has been handled before.

Let finally L be sporadic. By 2.11 p = 2. By [MeiStr2] we get that L/Z(L) ≅ M_{12}, M_{22}, M_{24}, J_{2}, Co_{1}, Co_{2}, or Suz. Now we choose s ∈ Z(S ∩ L ∩ B). If B ≰ O_{2}(C_{L}(s)), then by induction we get the assertion again. If there is some involution g in C_{L}(s) with [B, B^{g}] ≠ 1, we have the second alternative. So we may assume that ⟨B^{C_{L}(s)}⟩ is abelian. This gives L/Z(L) ≅ M_{i}. If L ≅ M_{24} there is a subgroup L̃ ≤ L with S ∩ L ≤ L̃ and L ≅ E_{16}A_{8}. Now by induction we may assume B ≤ O_{2}(L̃). But there is no quadratic foursgroup in O_{2}(L̃).

Let next L/Z(L) ≅ M_{22}. Then we may embed B into a subgroup SL(3, 4) and again we get the assertion by induction.

So we are left with L ≅ M_{12}. If B ≰ L, then with [MeiStr2] we see that B ≰ S ∩ L, so we have B ≤ L. Now in L there are two parabolics P_{1}, P_{2} such that P_{i}/O_{2}(P_{i}) ≅ Σ_{3}. So if B ≰ O_{2}(P_{i}) for some i we have induction again. Hence we may assume that B is contained in O_{2}(P_{1}) ∩ O_{2}(P_{2}) and ⟨B^{C_{L}(s)}⟩ is elementary abelian of order 8. Then B is centralized by some involution i which is fixed point freely on the 12 points moved by L. So C_{L}(i) ≅ Z_{2} × Σ_{5}. Further S contains a Sylow 2–subgroup of C_{L}(i). As B ≤ C_{L}(i), we get the assertion by induction.

3 Modules dihed

Lemma 3.1 Let V be a module over GF (p) for a solvable group G. Let A ≤ G be an elementary abelian subgroup such that [V, A, A] = 1 and G = O_{p′}(G)A, O_{p′}(G) = [O_{p′}(G), A], V = [V, O_{p′}(G)]. Then V = ⟨v | |A : C_{A}(v)| ≤ p⟩.

Proof: [BaMeiSt]

trans

Lemma 3.2 Let G = O_{p′}(G)A, A an elementary abelian p - group with [O_{p′}(G), A] = O_{p′}(G). Let V be a vectorspace over GF (p) and assume that A induces a group of transvections on V, then |A| = p.

Proof: By duality we may assumme that A induces a group of transvections to a hyperplane. Now the assertion follows with 3.1 as A acts quadratically and for any y in V not in the hyperplane we have C_{A}(y) = 1.

F F 1

Lemma 3.3 Let F^{∗}(H) be quasisimple and V be an irreducible F^{∗}(H)– module over GF (p) which is an F –module. Then F^{∗}(H) is classical, G_{2}(q), A_{n}, or 3A_{6}. and one of the following holds

1) F^{∗}(H) is classical or A_{n} and V is the natural module (in case of A_{n} we have p = 2)

2) F^{∗}(H) ≅ SL_{n}(q) and V is the exterior square of the natural module or its dual 3) F^{∗}(H) ≅ Sp(6, q), Ω(7, q) or Ω^{+}(10, q) and V is the spin module or

half spin module, respectively. If F^{∗}(H) ≅ Ω^{+}(10, q), then this is sharp, i.e. there is no offender A with |V : C_{V} (A)| < |A|.

4) F^{∗}(H) ≅ G_{2}(q) or 3A_{6} , p = 2, and V is the 6–dimensional module

5) H ≅ A_{7} and V is the 4–dimensional module over GF (2).

Proof: [BaMeiSt]

F F sol

Lemma 3.4 Let H = O_{p′}(H)A, A an elementary abelian p-group with [O_{p′}(H), A] = O_{p′}(H). Let V be a faithful GF(p)-module for H such that A is generated by transvections on V. Then p = 2 or 3 and there is characteristic subgroup F = F_{1}×⋯×F_{r}, F_{i} ≅ SL_{2}(p) and V = C_{V} (F)⊕V_{1}⊕⋯⊕V_{r}, V_{i} = [V_{i}, F_{i}] the natural F_{i}–module.

Proof: [BaMeiSt]

• We call V a strong F –module, if there is some nontrivial elementary abelian p–subgroup of G with |V : C_{V} (A)| ≤ |A|, [V, A, A] = 1 and C_{V} (A) = C_{V} (a) for all a ∈ A^{♯}.

• We call V a strong dual F –module, if there is some nontrivial elementary abelian p–subgroup of G with |V : C_{V} (A)| ≤ |A|, [V, A, A] = 1 and [A, v] = [V, A] for all v ∈ V, with [v, A] ≠ 1.

F F

Lemma 3.5 Let F^{∗}(H) be quasissimple and V be an irreducible F^{∗}(H)– module over GF (p) which is a strong (dual) F –module. Then p = 2 and |A| = 2 for an offender, or F^{∗}(H) is isomorphic to SL_{n}(q) or Sp(2n, q) and V is the natural module, or F^{∗}(H) ≅ 3A_{6} and V is one of the two 6-dimensional modules over GF (2) or H ≅ A_{7} and V the 4–dimensional module over GF (2).

Proof: [BaMeiSt]

pointstab

Lemma 3.6 Let F^{∗}(H) quasisimple, V a faithful GF (p)–module for H. Let L = O^{p′}(C_{H}(C_{V} (S))) for a Sylow p–subgroup S of H and A ≤ O_{p}(L). Then |V : C_{V} (A)| ≥ |A|. Assume additionally that V is an irreducible F –module for H with offender A and H = ⟨A^{H}⟩. Then H ≅ SL_{n}(q), Sp(2n, q), G_{2}(q) or Σ_{n}, where p = 2 in the last two cases, and V is the natural module.

Proof: [BaMeiSt]

2F lie

Lemma 3.7 Let F^{∗}(H) be a group of Lie type in characteristic p and V be an irreducible 2F –module in characteristic p. Then V is an F –module, or one of the following holds

(1) p is odd F^{(}∗) = L_{m}(r) and V is the symmetric square of the natural module.

(2) F^{∗}(H) = L_{m}(r^{2}) and V = V (λ_{i}) ⊗ V (λ_{i})^{σ}, where i = 1 or n − 1 and σ is a field automorphism of order two.

(3) F^{∗}(H) = Sp(2n, r) and V = V (λ_{2})

(4) F^{∗}(H) = L_{6}(r) and V = V (λ_{3})

(5) F^{∗}(H) = Sp(2n, r), n = 4, 5, and V = V (λ_{n}).

(6) F^{∗}(H) = Sp(4, r^{2}) and V = V (λ_{1})⊗V (λ_{1})^{σ} or V (λ_{2})⊗V (λ_{2})^{σ}, where σ is a field automorphism of order two

(7) F^{∗}(H) = Ω(7, q), Ω^{−}(8, r), Ω^{−}(10, r), Ω^{+}(12, r) and V is the half spin module

(8) F^{∗}(H) = U_{6}(r) and V = V (λ_{3})

(9) F^{∗}(H) = U_{3}(r) or Sz(r) and V = V (λ_{1})

(10) F^{∗}(H) = G_{2}(q) and V = V (λ_{2})

(11) F^{∗}(H) = F_{4}(q), q = 2^{n}, and V = V (λ_{1})

(12) F^{∗}(H) = E_{6}(r) and V = V (λ_{1}) or V (λ_{6})

(13) F^{∗}(H) = E_{7}(r) and V = V (λ_{7})

Proof: [BaMeiSt]

2F An

Lemma 3.8 Let F^{∗}(H)/Z(F^{∗}(H)) ≅ A_{n} and V be an irreducible faithful 2F –module in characteristic p. Suppose there is an offender A such that [V, A, A, A] = 1. Then V is an F –module, or one of the following holds

(i) p = 2, ⟨A^{H}⟩ = Σ_{7} and V is an eight dimensional module. Further we have |V : C_{V} (A)| = |A|^{2}.

(ii) p = 2, ⟨A^{H}⟩ = A_{9} and V is an eight dimensional module. (iii) p = 3 and V is the permutation module. (iv) p = 3, ⟨A^{H}⟩ = 2A_{5} and V is a four dimensional module.

(v) p = 3, ⟨A^{H}⟩ = 2A_{9} and V is an eight dimensional module.

Proof: [GuMa]

2F odd

Lemma 3.9 Let F^{∗}(H)/Z(F^{∗}(H)), be a group of Lie type in characteristic r and V be an irreducible faithful 2F –module in characteristic p, p ≠ r. Suppose there is an offender A such that [V, A, A, A] = 1. If F^{∗}(H)/Z(F^{∗}(H)) is not a group in characteristic p, too, then one of the following holds

(i) p = 2, H ≅ 3U_{4}(3) and V is a 12 – dimensional module. (ii) p = 3, H ≅ 2L_{3}(4) and V is a 6 – dimensional module. (iii) p = 3, H ≅ Sp(6, 2) and V is a 7 – dimensional module. (iv) p = 3, H ≅ 2Sp(6, 2) or 2Ω^{+}(8, 2) and V is an eight dimensional

module. Proof: [GuMa]

2F spor

Lemma 3.10 Let F^{∗}(H)/Z(F^{∗}(H)), be a sporadic simple group and V be an irreducible faithful 2F –module in characteristic p. Suppose there is an offender A such that [V, A, A, A] = 1. Then one of the following holds

(i) p = 2, H ≅ M_{12}, or M_{22} and V is a 10–dimensional module. (ii) p = 2, H ≅ M_{23} or M_{24} and V is an 11 – dimensional module. (iii) p = 2, H ≅ 3M_{22} or J_{2} and V is a 12 – dimensional module. (iv) p = 2, H ≅ Co_{1}, or Co_{2} and V is a factor of the Leech lattice module.

(v) p = 3, H ≅ M_{11} and V is a five dimensional module. (vi) p = 3, H ≅ M_{12} and V is a six dimensional module.

Proof: [GuMa]

2F odd1

Lemma 3.11 Let F^{∗}(H)/Z(F^{∗}(H)), be a group of Lie type in characteristic r and V be an irreducible faithful 2F –module in characteristic p. Suppose there is an offender A such that [V, A, A, A] = 1. Suppose |A| = q^{s}, q = p^{t}, |V : C_{V} (A)| ≤ |A|q and |V : [V, A]C_{V} (A)| ≤ q. If p = 2 there is an subgroup B of A, |B| = q, which acts quadratically on V. Then p = 3 or F^{∗}(H)/Z(F^{∗}(H)) is a group of Lietype in characteristic p.

Proof: By 3.9 we may assume that p = 2, H ≅ 3U_{4}(3) and V is a 12 – dimensional module. Suppose first q = 2. Then A is an offender as F – module. As there are no transvections we have |A| > 2. Hence we see that |A| = 2^{5}. But then A is uniquely determined and so A acts quadratically on V. Then A would induce transvections, a contradiction.

So we have q > 2 and then we have a quadratic fours group. Now we see that |A| ≥ 16. As |A| ≤ 2^{6}, we get q ≤ 8. Now for q = 8, we get |A| = 64. Again A is uniquely determine and |C_{V} (A)| < 8. So we have q = 4. Hence again |A| ≠ 64. So |A| = 16. In particular |C_{V} (A)| = 64. This shows again that A is uniquely determined and then that A acts quadratically. But then |V : C_{V} (A)| = 4, a contradiction.

2F l3

Lemma 3.12 Let G/Z(G) a group of Lie type in characteristic two, V a faithful GF(3)-module for G. Assume that V is a cubic irreducible 2F – module with offender A such that |[V, a]| ≤ |A| for quadratic acting elements a ∈ A. Then there is no 3–subgroup Q in G with Q∩A ≠ 1, which is normal in C_{G}(x) for all x ∈ V^{♯} with [Q, x] = 1.

Proof: The groups and modules are described in 3.9. If G ≅ 2L_{3}(4), then there is some x ∈ V such that C_{G}(x) ≅ A_{6}, and we are done. Let G ≅ Sp_{6}(2). Then |V | = 3^{7}. If H ≤ G, H contains a Sylow 3-subgroup and O_{3}(H) ≠ 1, then the index of H is divisble by 1120, which is more than the number of subgroups of order three in V. Let next G ≅ 2Ω^{+}(8, 2) and |V | = 3^{8}, then the only subgroup of G whose index is not divisble by three and is smaller than the number of subgroups of order three in V is 3 × U_{4}(2) : 2. So we have |Q| = 3. Now C_{G}(x) ≤ N_{G}(Q) for all x ∈ C_{V} (Q)^{♯}. As |C_{V} (Q)| ≥ 27 by Jordan canonical form, this is impossible. So we are left with G ≅ 2Sp_{6}(2) and |V | = 3^{8}. Now A is uniquely determined of order 27. So there is a ∈ A^{♯} with C_{G}(a) involves ⟨a⟩ × A_{6}. Hence we see that a acts quadratic and |[V, a]| = 3^{4}, a contradiction.

2F spor1

Lemma 3.13 Let F^{∗}(H)/Z(F^{∗}(H)) ≅ M_{12}, 3M_{22}, Co_{1}, Co_{2} or J_{2} and V be an irreducible faithful 2F –module in characteristic 2. Let A be an offender with [V, A, A, A] = 1. If |A| = q^{s}, q = 2^{t}, |V : C_{V} (A)| ≤ |A|q and |V : [V, A]C_{V} (A)| ≤ q, then there is no quadratic subgroup B of A, with |B| = q.

Proof: Let first F^{∗}(H)/Z(F^{∗}(H)) not be a Conway group. By 3.10, we have that V is 10 – dimensional or 12 – dimensional. In any case V is not a F + 1–module, so q > 2. Then we have a quadratic fours group and the groups B are given by [MeiStr2]. If there is no quadratically acting group of order 8, then we get q = 4. Hence |A| ≥ 16.

Suppose F^{∗}(H) = M_{12}. Then |V | = 2^{10}. Further A ≰ F^{∗}(H). Now there is a ∈ A with C(a) involves A_{5} and |[V, a]| = 2^{5}. But then A_{5} would induce transvections on C_{V} (a), a contradiction.

Suppose next F^{∗}(H) = J_{2}. Now A contains a non 2-central involution a. Then |[V, a]| = 2^{6}, which contradicts |[V, x]| ≤ 2^{5} for any x ∈ A.

So we are left with F^{∗}(H) = 3M_{22} and |V | = 2^{12}. As |A| ≤ 2^{5}, we get q = 4 and |A| = 16. Now |V : C_{V} (a)| ≤ 2^{5} for all a ∈ A. Hence we get A ≤ F^{∗}(H) and there are exactly two possiblities. If N(A) involves A_{6}, we get |C_{V} (A)| = 64 and A acts quadratically, a contradiction. So we have that N_{A} involves Σ_{5} and then |C_{V} (A)| = 4. But we know |V : C_{V} (A)| ≤ 2^{6}, a contradiction again.

Suppose next that F^{∗}(H) ≅ Co_{1}. Suppose there is a 2–central involution z in A. Then as V is the Leech lattice module, we see |[V, a]| = 2^{8}. Hence 2^{8} ≤ |A| ≤ 2^{10}, remember |A| is a power of 4. We have |[V, a] : C_{[V,a]}(A)| ≤ 4. As C_{F^{∗}(H)}(a)/O_{2}(C_{F^{∗}(H)}(a)) = Ω^{+}(8, 2), we get |A : A ∩ O_{2}(C_{F^{∗}(H)}(a))| ≤ 2. But as O_{2}(C_{F^{∗}(H)}(a)) is extraspecial of order 2^{9} this is impossible. Hence we have that there is no 2–central involution in A. Let a be as before, we get A ∩ O_{2}(C_{F^{∗}(H)}(a)) = 1. In particular |A| ≤ 2^{6}, and so |[V, z]| ≤ 2^{7} for z ∈ A. But |[V, x]| ≥ 2^{8} for all involutions in H.

Let next F^{∗}(H) ≅ Co_{2}. Now any quadratic fours group contains a 2– central involution, in particular A contains a 2–central involution a. Now we see that |A| ≥ 2^{8}, while |A : A ∩ O_{2}(C_{F^{∗}(H)}(a))| ≤ 2, and so we have a contradiction as before.

An

Lemma 3.14 Let F^{∗}(H) ≅ A_{n}, n ≠ 5, 6, 8, and V be the natural module. Then there is no 1 ≠ Q in H, Q a 2–group such that Q ≤ O_{2}(C_{H}(x)) for all 1 ≠ x ∈ C_{V} (Q).

Proof: Let S be a Sylow 2–subgroup with Q ≤ S. Then Q ≤ O_{2}(C_{H}(x)) for all 1 ≠ x ∈ C_{V} (S). In particular O_{2}(C_{H}(x)) ≠ 1. Hence we have that Q is conjugate to a subgroup of Σ_{4} acting on {1, 2, 3, 4}, say. But then we see that H = ⟨C_{H}(x) | 1 ≠ x ∈ C_{V} (Q)⟩.

2F An3

Lemma 3.15 Let G/Z(G) ≅ A_{n}, n ≥ 5, n ≠ 6, 9 and V an irreducible faithful cubic 2F –module over GF(3) with offender A, such that for a ∈ A^{♯} which acts quadratically we have |[V, a]| ≤ |A|. Then there is no subgroup Q of G with A ∩ Q ≠ 1 which is normal in C_{G}(x) for all x ∈ C_{V} (Q)^{∗}.

Proof: The structure of G and V is given in 3.8. Let first V be the natural module. If Q does not act transitively on {1, …, n}, then Q is normal in some A_{m} × A_{n−m}. Hence without loss we may assume m = 3 and so |Q| = 3 Let n ≠ 5. Then we see that G is generated by C_{G}(x) with x ∈ C_{V} (Q)^{♯}. So let n = 5. In the 4 - dimensional permutation module for A_{5} there is a vector centralized by A_{4} and so Q would be not normal.

So we may assume that Q acts transitively. Then some element which is centralized by a Sylow 3–subgroup is centralized by A_{m} ≀ 3. This shows m = 3 and so n = 9, a contradiction.

So we are left with G ≅ 2A_{5} and V the 2–dimensional GF (9)–module. Then A acts quadratic, but V is not an F –module.

H11

Lemma 3.16 Let X ≅ A_{n}, n ≥ 5, V be a GF (2)X-module with V/C_{V} (X) the natural irreducible permutation module. Assume [V, X] = V. Then |C_{V} (X)| ≤ 2, and |C_{V} (X)| = 1 if n is odd. Furthermore V is a submodule of the permutation module.

Proof: For simplicity we prove the dual statement. Let C_{V} (X) = 1 and [V, X] be the natural module. Then |V : [V, X]| ≤ 2, |V : [V, X]| = 1 for n odd, and V is a factor of the permutation module.

This will be proved by induction on n. For n = 5 this is well known. So let n > 5, K ≅ A_{n−1}, K ≤ X. If n − 1 is odd, then we have that [V, X] = T ⨁ T_{1}, T_{1} the irreducible permutation module for K, [T, K] = 1. By induction V = [V, K] ⨁ T̃, [V, K] the irreducible permutation module. Hence there is v ∈ V \ [V, X], [v, K] = 1, i.e. ⟨v^{X}⟩ = V is a factor of the permutation module.

Let n − 1 be even. Then we have a K-chain. 1 < T < T_{1} < [V, X] < V, with |T | = 2, T_{1}/T the irreducible K-module and |[V, X]/T_{1}| = 2. Now by induction C_{V/T} (K) ≠ 1. As C_{V/T} (K) ≰ [V, X]/T, we again get some v ∈ V \ [V, X], [v, K] = 1, and so V is a factor of the permutation module.

s

Lemma 3.17 (a) Let L be SL_{n}(q), q = p^{f} or Ω^{+}(4, q), q = 2^{f}, and V be a module over GF (p) with [V, L] the natural module and C_{V} (L) = 1. Then V = [V, L], or one of the following holds:

(i) L ≅ L_{2}(q), q even, and [V : [V, L]] ≤ q

(ii) L ≅ L_{3}(2), and |V | = 16

(b) If in (a) we have V = [V, L] and V/C_{V} (L) is the natural module, then we either get C_{V} (L) = 1 or we have the exceptions as in (a) and the bound for |C_{V} (L)| is the same as the bound for |V : [V, L]| in (a).

Proof: (b) follows by duality. Hence we just have to prove (a).

(1) Let first L ≅ SL_{n}(q):

Obviously we may assume Z(L) = 1 as otherwise [V, L] = [V, Z(L)] and V = [V, Z(L)] ⊕ C_{V} (Z(L)).

If L ≅ L_{2}(q), then q is even and for x ∈ L, o(x) = 2, we have |V : C_{V} (x)| = q. We have that L is generated by three conjugates of x. Hence |V | ≤ q^{3}.

Let now L ≅ SL_{3}(q). If q ≠ 2, then there are three elements x_{1}, x_{2}, x_{3} in L acting fixed point freely

x_{1} = ( \begin{matrix}ω^{−2} \\ & ω \\ & & ω \end{matrix} ), x_{2} = ( \begin{matrix}ω \\ & ω^{−2} \\ & & ω \end{matrix} ), x_{3} = ( \begin{matrix}ω \\ & ω \\ & & ω^{−2} \end{matrix} ),

o(ω) = q − 1.

We have [x_{i}, V] = [V, L], i = 1, 2, 3, and so as [x_{i}, x_{j}] = 1 for all i, j, we get C_{V} (x_{1}) = C_{V} (x_{2}) = C_{V} (x_{3}). But L = ⟨C_{L}(x_{i})|i = 1, 2, 3⟩, the assertion.

So let q = 2. Then there is x ∈ V \ [V, L], |x^{L}| = 8. Hence V is a factor module of the permutation module, which shows |V | ≤ 16.

Let next L ≅ L_{4}(2). There is ⟨ρ⟩ × A_{5} ≤ L, o(ρ) = 3, [V, L] = [V, ρ] and [V, L] = [V, γ], γ ∈ A_{5}, o(γ) = 3. Hence C_{V} (ρ) = C_{V} (γ). Now as ⟨C_{L}(γ), C_{L}(ρ)⟩ = L, we get V = [V, L].

Let now n ≥ 4, q > 2 for n = 4. Let P be the parabolic in L with |O_{p}(P)| = q^{n−1} and P′/O_{p}(P) ≅ SL_{n−1}(q). Assume |C_{[V,L]}(O_{p}(P))| = q^{n−1}. Then [P′, V] = C_{[V,L]}(O_{p}(P)), as P′ = P′′. This now shows that [C_{V} (O_{p}(P)), P′] is the natural module. By induction on n we have that C_{V} (O_{p}(P)) = C_{V} (P′) ⨁ C_{[V,L]}(O_{p}(P)). Now application of Gaschütz [Hu] shows C_{V} (P′) = 1. Hence as |V : C_{V} (O_{p}(P))| = q, we get V = [V, L].

(2) Let L ≅ Ω^{+}(4, q):

We write L as L_{1} × L_{2}, L_{i} ≅ L_{2}(q). We may assume q > 2, as the assertion is obvious for q = 2. Now there is ω_{i} ∈ L_{i}, with o(ω_{i}) = q + 1 and ω_{i} acting fixed point freely on [V, L], i = 1, 2. Now choose v ∈ V \ [V, L] with [v_{1}, ω_{1}] = 1. Then v_{1} is uniquely determined in [V, L]v_{1}. So K_{2} centralizes v_{1} too. Hence [ω_{2}, v_{1}] = 1. As there is a unique fixed point of ω_{2} in [V, L]v_{1}, we see [K_{1}, v_{1}] = 1, so [v_{1}, L] = 1, a contradiction. This proves (a).

l32

Lemma 3.18 Let G = L_{3}(2) and V some GF (2)–module such that [V, G] is the natural module. If |V/[V, G]| = 2, and some element in G induces a transvection on V, then C_{V} (G) ≠ 1. By duality if V/C_{V} (G) is the natural module and |C_{V} (G)| = 2 and some element induces a transvection, then [V, G] is the natural module.

Proof: Let x ∈ V \ [V, G] with |C_{G}(x)| = 21, then |x^{G}| = 8 and so x[V, G] = x^{G}. But if there are transvections in G they have to centralize some elements in V \ [V, G], a contradiction.

weak

Lemma 3.19 Let M ∈ L(S), M ≰ C̃, and M^{0} be maximal. Assume that M_{0} is not a minimal parabolic. Let Y_{M} be a strong dual F –module with offender A, then E_{p}(M_{0}) has just one component. Further the offender acts nontrivially on E_{p}(M_{0}) and F_{p}^{∗}(M_{0}) = E_{p}(M_{0}).

Proof: We may assume M^{∗} = M. Let now K = [F_{p}^{∗}(M/C_{M}(Y_{M})), A]. The assumption yields that there is exactly one nontrivial K–modules in Y_{M}. Assume next [K, M_{0}] ≤ C_{M_{0}}(Y_{M}). So F_{p}^{∗}(M_{0}) has to act nontrivially on this module. But [Q, [Y_{M}, K]] = 1, a contradiction. So we get that K^{Q} ≤ M_{0}.

Let K_{1} × … × K_{r} = ⟨K^{Q}⟩. Let r > 1. Then by [BaMeiSt] we have that [Y_{M}, K_{1}K_{2} … K_{r}] = V_{1} ⊕ … ⊕ V_{r}, where [V_{i}, K_{i}] = V_{i} is a strong F –module for K_{i} and [V_{i}, K_{j}] = 1 for i ≠ j. Now let 1 ≠ x_{i} ∈ C_{V_{i}}(S ∩ K_{i}). Hence we may assume that y = x_{1}x_{2} ⋯ x_{r} ∈ C_{Y_{M}} (Q). Now we have with 2.4 that [C_{K_{1}}(y), Q] ≤ Q. Hence we have C_{K_{1}}(y) is a p–group. So application of 3.5 shows that K/O_{p}(K) = SL(2, q) and [Y_{M}, K] is the natural module. Now we see that C_{M_{0}/C_{M_{0}}(Y_{M})}(K^{S}) = F (M_{0}/C_{M_{0}}(Y_{M})). Set U = [F_{p}(M_{0}), Q]. Then we see that U⟨K^{Q}⟩ = C_{U⟨K^{Q}⟩}(C_{[Y_{M} ,K^{Q}]}(Q))⟨K^{Q}⟩. Hence we have U⟨K^{Q}⟩ = ⟨K^{Q}⟩(C̃ ∩ (U⟨K^{Q}⟩)). Hence M_{0} is a minimal parabolic, a contradiction.

So ⟨K^{Q}⟩ = K. Now by 2.4 K is the only component of M_{0}/C_{M_{0}}(Y_{M}) which is not centralized by A. Suppose F_{p}(M_{0}) ≠ 1, then there is some t ∈ F^{∗}(M_{0}/C_{M_{0}}(Y_{M})) \ E(M_{0}/C_{M_{0}}(Y_{M})) such that C_{Y} (t) ∩ C_{Y} (Q) ≠ 1, which shows [Q, t] = 1. Hence we get that F_{p}^{∗}(M_{0}) = E_{p}(M_{0}). So we have that K is solvable. Because of quadratic action we get that |A| = p and so A induces a GF (p) transvection. Hence we see that p = 2, or 3. In particular we have that K is a 3–group or 2–group respectively. Set M̄_{0} = M_{0}/C_{M_{0}}(Y_{M}). Then we see that X̄ = ⟨ĀK̄⟩ = Σ_{3} or SL_{2}(3). Now ⟨X̄^{Q}⟩ ≅ K_{1} × ⋯ × K_{r} with K_{i} ≅ Σ_{3} or SL_{2}(3). With 2.4 we get that X = M_{0}. As [C_{M_{0}}(Y_{M}), Q] = 1, we see that M_{0} is a minimal parabolic, a contradiction.

4 The structure of M_{0}

The aim of this paragraph is to prove the following proposition.

strub1

Proposition 4.1 Let M ∈ L(S), M ≰ C̃, with M^{0} maximal. Set M_{0} = M^{0}S. If Y_{M} ≰ Q then one of the following holds

a) F^{∗}(M_{0}/C_{M_{0}}(Y_{M})) is quasisimple and Y_{M} is a 2F –module with quadratic or cubic offender. Further

C_{M^{0}}(O^{p}(E(M^{0}/C_{M^{0}}(Y_{M}))) = Z(O^{p}(M^{0}/C_{M^{0}}(Y_{M}))∩(O^{p}(M^{0}/C_{M^{0}}(Y_{M}))′.

b) F^{∗}(M_{0}/C_{M_{0}}(Y_{M})) ≅ SL_{n}(r)′ ∗ SL_{m}(r)′ and Y_{M} is the tensor product module. Further the preimage of F^{∗}(M_{0}/C_{M_{0}}(Y_{M})) in M_{0}/O_{p}(M_{0}) is contained in SL_{n}(r)′ × SL_{m}(r)′ or n = m = r = 2 and the preimage is extraspecial of order 27 and exponent 3.

We will prove 4.1 in a series of lemmas.

2F module

Lemma 4.2 Let M ∈ L(S), M ≰ C̃, with M_{0} maximal. Suppose Y_{M} ≰ Q. Then one of the following holds.

(1) There is some g ∈ C̃ with Y_{M}^{g} ≤ S ≤ M, Y_{M} ≤ M^{g} and Y_{M}^{g^{2}} = Y_{M}. Further 1 ≠ |Y_{M} : C_{Y_{M}} (Y_{M}^{g})| = |Y_{M}^{g} : C_{Y_{M}^{g}} (Y_{M})|. In particular Y_{M} is an F –module.

(2) There is some g ∈ C̃ such that for L = ⟨Y_{M}, Y_{M}^{g}⟩ we have L/O_{p}(L) ≅ SL_{2}(q), Sz(q), q even or D_{2u}, a dihedral group of order 2u, u odd. Set q = 2 in the latter. Further we have that A = Y_{M}^{g} ∩ O_{p}(L) ≤ S ≤ M. For the action of A on Y_{M} we have [Y_{M}, A, A, A] = 1, If x ∈ Y_{M} \ O_{p}(L), then C_{A}(x) = A ∩ Y_{M}, and |Y_{M} : C_{Y_{M}} (A)| ≤ q|A/(A ∩ Y_{M})|. In particular Y_{M} is a 2F –module with offender A/(A ∩ Y_{M}) and an F + 1–module in case of q = p. Further we have U = ⟨x | x ∈ Y_{M} : [x, A]C_{Y_{M}} (A) ≠ [Y_{M}, A]C_{Y_{M}} (A)⟩ = Y_{M} ∩ O_{p}(L) and if [Y_{M}, A, A] = 1, then Y_{M} is an F –module. If |Y_{M} : C_{Y_{M}} (A)| = |A|^{2}, then p > 2.

Proof: We find most in 2.13 where G = C̃ and A = Y_{M}. Suppose [Y_{M}, A, A] = 1. Then C_{Y_{M}} (A) = Y_{M} ∩ O_{p}(L). By 2.13 we have |A/A ∩ O_{p}(L)| ≤ q, while |A| ≥ q, so Y_{M} is an F –module. Let |Y_{M} : C_{Y_{M}} (A)| = |A|^{2} and p = 2. Then as |A| is a power of q, and |Y_{M}/Y_{M} ∩ O_{2}(L)| ≤ q, we see |A| = q and so ⟨Y_{M} ∩ O_{2}(L), Y_{M}^{g} ∩ O_{2}(L)⟩/Y_{M} ∩ Y_{M}^{g} is the natural module for L. Hence L acts transitively and so as Y_{M} is elementary abelian, we see that ⟨Y_{M} ∩ O_{2}(L), Y_{M}^{g} ∩ O_{2}(L)⟩ is abelian. Then |Y_{M} : C_{Y_{M}} (A)| ≤ |A|, a contradiction.

For the remainder of this section we will fix the following notation from 4.2. If we are in 4.2(1) then we set A = Y_{M}^{g}. The structure of L tells us that we may assume that for p = 2 g^{2} ∈ S. By 2.6 we have M_{0} ≤ M^{∗}. Further A ≤ S. Hence again we may assume M = M^{∗}.

M0

Lemma 4.3 Let [A, O_{p′}(M_{0}/C_{M_{0}}(Y_{M}))] = 1, then there is some component K of M_{0} such that [A, K] ≠ 1.

Proof: Suppose that [F_{p}^{∗}(M_{0}), A] ≤ C_{M}(Y_{M}). By the P × Q–lemma, A acts nontrivially on C_{Y_{M}} (Q). Suppose first that we have 4.2(1). Choose x ∈ C_{Y_{M}} (Q). Then x^{g} ∈ C_{A}(Q) ≤ A ∩ Q ≤ A ∩ O_{p}(M) by assumption. So we have [x^{g}, Y_{M}] = 1, but then also [x, A] = 1, a contradiction.

So we have 4.2(2). Then C_{Y_{M}} (Q) ≤ O_{p}(L) and so again C_{Y_{M}} (Q)^{g} ≤ C_{A}(Q) ≤ A ∩ O_{p}(M). Now we get the contradiction as before.

Knormal

Lemma 4.4 Let K be a p-component of M_{0} with [K, A] ≰ C_{M_{0}}(Y_{M}), then [K, A] ≤ K.

Proof: Set K_{1} = KC_{M_{0}}(Y_{M})/C_{M_{0}}(Y_{M}). We may assume that K_{1} is nonsolvable. Let K_{1}^{A} ≠ K_{1}. Suppose first that we are in 4.2(1). Then by [Cher] we have p = 2 and either |A : C_{A}(K_{1})| = 2 or K_{1} ≅ L_{2}(2^{n}). In the first case by quadratic action there is some x ∈ Y_{M} which induces a transvection on A. Hence there is also some a ∈ A, which induces a transvection on Y_{M}. But this contradicts 3.19. So we have the latter. Then A does not offend on [Y_{M}, K^{A}]. Hence there are other components which are normalized by A. Now as Q normalizes A, we get that Q acts on the components normalized by A and those which are not. In particular by quadratic action we get some V ≤ Y_{M} with [V, Q] ≤ V and [K, V] = 1. Then by 2.4 K ≤ C̃, a contradiction.

So we may assume that we are in 4.2(2). Suppose first q > p. Assume first p to be odd. By 2.12 we have |A : C_{A}(K)| = p. Set A_{1} = C_{A}(K). As A acts quadratically on [Y_{M}, A_{1}], we see with [Cher] [K, [Y_{M}, A_{1}]] = 1. By the 3–subgroup Lemma we get [Y_{M}, K] ≤ O_{p}(L). By 4.2 we have |[Y_{M}, A]C_{Y_{M}} (A) : [Y_{M}, A_{1}]C_{Y_{M}} (A)| ≤ p. Hence A induces a transvection on [Y_{M}, ⟨K^{A}⟩]/C_{Y_{M}} (⟨K^{A}⟩). But this contradicts [Cher]. So we have p = 2

By 2.13 we know that Y := (Y_{M} ∩ O_{2}(L))(Y_{M}^{g} ∩ O_{2}(L))/(Y_{M} ∩ Y_{M}^{g}) is a direct sum of natural modules. So let A_{1} be contained in the intersection of A with one of this modules, with |A_{1} : A_{1} ∩ Y_{M}| = q. Then we see that A_{1} acts quadratically on Y_{M}. Hence by [Cher] we have three posibilities

(1) [K, A_{1}] ≤ K

(2) |A_{1} : C_{A_{1}}(K_{1})| > 2 and [K_{1}, A_{1}] ≰ K_{1} and K_{1} ≅ L_{2}(2^{n})

(3) |A_{1} : C_{A_{1}}(K_{1})| = 2 and [K_{1}, A_{1}] ≰ K_{1}.

Assume (3). Let a ∈ C_{A_{1}}(K_{1}). Then K_{1}^{A_{1}} acts on [Y_{M}, a]. By quadratic action we have [Y_{M}, a, K_{1}] = 1. In particular we get that [Y_{M}, K_{1}] is centralized by a and so by 4.2 [Y_{M}, K_{1}] ≤ O_{2}(L). Hence A acts quadratically on [Y_{M}, K_{1}^{A}]. Assume now |A : C_{A}(K_{1})| = 2. Then [Y_{M}, K_{1}] is centralized by a subgroup of index two in A. Let B be a complement in [Y_{M}, K_{1}] of C_{[Y_{M} ,K_{1}]}(A). Then all b ∈ B have the same centralizer in Y_{M}^{g}. As we have that |B| > 2, we see that B has to normalize any 2–component of M_{0}^{g}.

Hence there are also 2–components in M_{0} normalized by A, which gives a contradiction as before.

So we may assume that K_{1}^{A} = Ω^{+}(4, 2^{n}). In particular we may assume that we are in (2). Suppose first [Y_{M}, K_{1}] ≤ O_{2}(L). By ?? there is some y ∈ C_{Y_{M}} (K_{1}) \ O_{2}(L). Hence we see [y, A](Y_{M} ∩ A) = Y_{M} ∩ O_{2}(L). Now [Y_{M}, K_{1}, A] ≤ [Y_{M} ∩ O_{2}(L), A] = [y, A, A]. But [y, A] ≤ C_{Y_{M}} (K_{1}), a contradiction. So we have [Y_{M}, K_{1}] ≰ O_{2}(L). In particular there is some minimal module V such that V ≰ O_{2}(L) and V/C_{V} (K_{1}) is the natural Ω^{+}(4, 2^{n}) – module. As above we see that C_{A_{1}}(K_{1}) = 1. Choose y ∈ V \ O_{2}(L), then we get [y, A_{1}](Y_{M} ∩ A) = [Y_{M}, A_{1}]. As |[V/C_{V} (K_{1}), A_{1}]| > |A_{1}|, we see that V ∩ A ≰ C_{V} (K_{1}). Let a ∈ C_{A}(K_{1}). Then V^{a} = V, and so [V, a] = 1, which gives a = 1, as V ≰ O_{2}(L). So we have C_{A}(K_{1}) = 1. But then A acts quadratically on V, which then gives that V is an F –module, a contradiction.

So we have shown that K_{1}^{A_{1}} = K_{1}. As A is generated by such subgroups A_{1}, we have the assertion [K_{1}, A] ≤ K_{1}.

So now we have q = p and then Y_{M} is an F +1–module. Let first p be odd. Then as above we have |A| = p. Hence we have that |Y_{M} : C_{Y_{M}} (A)| = p^{2}. Let r be a prime, r ≠ p, 2, which divides K_{1}. Then we have a subgroup of order r^{p−1}, which is generated by two conjuagtes of A. Hence this group acts faitfully on a four dimensional GF (p) space. This in the first place shows p ≤ 5. But then r^{4} does not divide the order of GL(4, p). So we have p = 2.

If [Y_{M}, K_{1}] ≤ O_{2}(L), then again A acts quadratically on [Y_{M}, K_{1}]. If we have |A : C_{A}(K_{1})| > 2, we may argue as before. So let |A : C_{A}(K_{1})| = 2. Then as before [Y_{M}, K_{1}] centralizes a subgroup of index 2 in A and a subgroup of index 4 in Y_{M}^{g}. As before we may assume that [Y_{M}, K_{1}] does not normalizes any component in M_{0}^{g}. Hence as we now get that A induces transvections on [Y_{M}, K_{1}], a contradiction.

So we have [Y_{M}, K_{1}] ≰ O_{2}(L). Let a ∈ A with K_{1}^{a} ≠ K_{1}. Assume first [Y_{M}, a, A] = 1, then by [Cher] either |A : C_{A}(K_{1})| = 2, or K_{1} ≅ L_{2}(r) and orthogonal Ω^{+}(4, r)–modules are involved. Suppose the latter. Then as before we get a contradiction. So we have |A : C_{A}(K_{1})| = 2. Now we have that [K_{1}, Y_{M}, C_{A}(K_{1})] = 1, but as [Y_{M}, K_{1}] ≰ O_{2}(L) this shows |A : A ∩ Y_{M}| = 2. Now A induces transvections on Y_{M} and the assertion follows with 3.19.

So we have that [Y_{M}, a, A] ≠ 1. Now A induces transvections on [Y_{M}, a] to a hyperplane. Let b ∈ C_{A}(K_{1}) and assume first that [Y_{M}, b, K_{1}] = 1. Then also [Y_{M}, K_{1}, b] = 1 and so [Y_{M}, K_{1}] ≤ O_{2}(L), a contradiction. Hence K_{1} acts nontrivially on [Y_{M}, b]. But b induces a transvection on [Y_{M}, a] , a contradiction. So we have that C_{A}(K_{1}) = 1. Further K_{1}^{A} = K_{1} × K_{1}^{a}. As [Y_{M}, a, A] ≠ 1, we see that K = C_{K_{1}×K_{1}^{a}}(a) ≅ K_{1} acts faithfully on [Y_{M}, a], and so as A induces transvections to a hyperplane, we get with [BaMeiSt] K_{1} ≅ L_{n}(2), Sp(2n, 2), Ω^{±}(2n, 2) or A_{n}. We have |A| > 2. So assume first |A| = 4. Then |[Y_{M}, a]| ≤ 4, but K has to act nontrivially on [Y_{M}, a].

So we have |A| > 4 and then K_{1} ≅ L_{n}(2). Further |[Y_{M}, a]| ≥ 2^{n}. As |A| ≤ 2^{n}, we get equality everywhere and N_{A}(K_{1}) induces the full transvection group on [Y_{M}, a]. This shows that [Y_{M}, K_{1}] just involves the natural module. By ??, we get [Y_{M}, K_{1}K_{1}^{a}] = [Y_{M}, K_{1}] ⊕ [Y_{M}, K_{1}^{a}]. We see that Q normalizes K_{1}K_{1}^{a} and so, as Q has to normalize K_{1} and K_{1}^{a} , we get a contradiction by 2.4.

Qfix1

Lemma 4.5 Let K be a p – component of M/C_{M}(Y_{M}) with [K, A] ≠ 1, then [K, Q] ≤ K, or KS is a minimal parabolic.

Proof: First of all we may assume that K ≤ M_{0}C_{M}(Y_{M})/C_{M}(Y_{M}). Suppose that KS is not a minimal parabolic. Set U = ⟨K^{Q}⟩. Then by 2.4 we get that C_{Y_{M}} (U) = 1. Let T be a Sylow p–subgroup of UQ. Let V_{1} be some irreducible KA-submodule in Y_{M}. Suppose first that V_{1} ≰ O_{p}(L). Then there is some x ∈ V_{1} with [x, A]C_{Y_{M}} (A) = [Y_{M}, A]C_{Y_{M}} (A). Let K_{2} ≤ U be a component different from K. Then we now see that [V_{1}, K_{2}A] = V_{1}. But this now shows that V_{1} is a tensor product of irreducible modules W for K and K_{2}, which are 2F–modules. The structure of W is given by 3.7 - 3.13. Now choose 1 ≠ x ∈ C_{V} (T). Then we see that N_{K}(⟨x⟩)/Z(K) is not a p-group. As N_{UQ}(⟨x⟩) contains N_{K}(⟨x⟩)^{Q}, which is not a p–group, this contradicts 2.4.

So we may assume that V_{1} ≤ O_{p}(L). Then A acts quadratically. By 2.4 again we have that A acts faithfully on U. So V is an F -module for UA by 4.2. Hence V is a direct sum of irreducible modules, which are permuted by Q. But now we get a contradiction as before.

Qfix2

Lemma 4.6 Let the assumption be as in 4.5. Then [K, Q] ≤ K.

Proof: By 4.5 we may assume that KS is a minimal parabolic and K is involved in M_{0}. Let K_{1} be some component in M_{0}/C_{M_{0}}(Y_{M}) or Sylow r–subgroup of F (M_{0}/C_{M_{0}}(Y_{M})) not in KS. If [[K, Y_{M}], K_{1}] = 1, then also [K_{1}^{Q}, [K^{Q}, Y_{M}]] = 1 and so by 2.4 we get K ≤ C̃ ∩ M/O_{p}(M) and then [K, Q] = 1, the assertion. So we have [[K, Y_{M}], K_{1}] ≠ 1. By 4.4 we have [K, A] ≤ K. If [K, Y_{M}] is irreducible, then with 3.7 - 3.13 we see that K_{1} has to induce field multiplications on [K, Y_{M}]. But then we find some t ∈ K_{1}KS, which centralizes some element in C_{[Y_{M} ,K^{S}]}(Q), but does not normalize Q, a contradiction to 2.4. Hence we have that [K, Y_{M}] involves at least two nontrivial modules. Further we may assume that K_{1} acts on these modules nontrivially. This shows K ≅ L_{2}(q) and [Y_{M}, K] is a direct sum of two modules on which A acts quadratically. Suppose first that A acts quadratically on Y_{M}, then with 4.2 we see that Y_{M} is an F –module with offender A. But this is not possible. Hence A does not act quadratically and so [Y_{M}, K^{S}] ≤ O_{2}(L). But then A has to induce an F –module, i.e. B = C_{A}(K^{S}) ≠ 1. Now we see that [B, [Y_{M}, K]] = 1, as [B, Y_{M} ∩ Y_{M}^{g}] = 1. Let K_{2} be a p-component of M with [B, K_{2}] ≠ 1 , then [K_{2}, [Y_{M}, K^{S}]] = 1. But this contradicts 2.4. So we have seen that there in no K_{1}, in particular M_{0} is a minimal parabolic contradicting 2.7.

Qfix

Lemma 4.7 Let K_{1}, K_{2} be subgroups of M/C_{M}(Y_{M}) with [K_{i}, Q] ≤ K_{i}, i = 1, 2. Suppose there are V_{i} ≤ Y_{M}, V_{i}^{Q} = V_{i} and [V_{i}, K_{3−i}] = 1, i = 1, 2. Then K_{i} ≤ (M ∩ C̃)/O_{p}(M).

Proof: This follows immediatly from 2.4

2F unique

Lemma 4.8 Assume 4.2(2). Let K be a component of E(M_{0}/C_{M_{0}}(Y_{M})), which is not centralized by A. Suppose K_{1} is a second component, or a Sylow u–subgroup of F (M_{0}/C_{M_{0}}(Y_{M})) with the same property. Then K ≅ SL_{n}(r), K_{1} ≅ SL_{m}(r)′ for some some n, m and r = p^{t}. Further Y_{M} is the tensor product module for K × K_{1} and KK_{1} = F^{∗}(M_{0}/C_{M_{0}}(Y_{M})).

Proof: Let KA = K ×T. By abuse of notation we denote with K also KOut_{M}(K). Let first T = 1. Now we have that there is some a ∈ A which induces an outer automorphism on K, otherwise A ≤ K and so [K_{1}, A] = 1. Let V_{1} be an irreducible E(K)–module. Assume further that V_{1}^{a} = V_{2} ≠ V_{1}. Then we have that |V_{1}| < |A|^{2}. In particular we see that A ∩ E(K) ≠ 1. If V_{1} ≤ O_{p}(L), then we get [A ∩ K, V_{1}, a] = 1, a contradiction. So we have V_{1} ≰ O_{p}(L). Let |V_{1}| = r^{t} and |V_{1} : C_{V_{1}}(A∩E(K))| = r^{t−e}. As V_{1} ∩Y_{M}^{g} = 1, we see that A ∩ E(K) acts quadratically on V_{1}. There is v ∈ V_{1} such that [v, A ∩ E(K)] ≅ A ∩ E(K). Then we have |A ∩ E(K)| ≤ r^{e}. Now we see |V_{1}⊕V_{2} : C_{V_{1}⊕V_{2}}(A)| ≥ r^{2t−e}. On the other hand |V_{1}⊕V_{2} : C_{V_{1}⊕V_{2}}(A)| ≤ |A : A∩E(K)||A∩E(K)|r^{t−e} ≤ |A : A∩E(K)|r^{t}, as |V_{1} ⊕V_{2} : V_{1} ⊕V_{2} ∩O_{p}(L)| = |V_{1} : C_{V_{1}}(A ∩ E(K)|. Hence we see r^{t−e} ≤ |A : A ∩ E(K)| ≤ r. So we get t = e + 1 and then A ∩ E(K) induces the full transvection group to a point, i.e. K ≅ SL_{t}(r). But as |A/A ∩ E(K)| = r, we have that A contains some element which does not act trivially on the Dynkin diagram. But this is not possible, as A∩E(K) has to induce transvections to a point on V_{2} too. So we see that V_{1}^{A} = V_{1} and so V_{1} is invariant under K. Now we see that K_{1} has to induce field multiplications on V_{1} and then we get some 1 ≠ t ∈ K_{1} × K \ K which centralizes some nontrivial element in C_{V_{1}}(Q), in particular t ∈ C̃ by 2.4, a contradiction. So we have |T | > 1.

Let V_{1} be as before. By 2.4 and 4.6 we get that V_{1} is nontrivial. Let W = V_{1} ⊕ V_{2} ⊕ ⋯ ⊕ V_{s} = V_{1}^{T}. If s = 1, then we get [V_{1}, T] = 1 By 4.7 this shows T = 1. So we have s > 1.

Assume first that [V_{1}, T] = 1 for any irreducible K–module V_{1}. Then K_{1} centralizes the socle of K in Y_{M}. Hence C_{Y_{M}} (K) ≠ 1, which contradicts 2.4 and 4.6. So we may assume that [V_{1}, T] ≠ 1.

We first assume C_{A}(K) = 1. Let t ∈ T with [V_{1}, t] = 1. There is some b ∈ K with a = bt ∈ A. Hence the action of a and b are the same on W. So we may replace a by b without changing any action. As [V_{1}, T] ≠ 1, still |T | will be greater then 1. So we may assume that C_{T} (V_{1}) = 1. Finally we may choose notation such that V_{i} = V_{1}^{t_{i}} for certain t_{i} ∈ T, i = 1, …, s.

Assume first V_{1} ≤ O_{p}(L). We have W = V_{1}^{A}, so W ≤ O_{p}(L). In particular A acts quadratically on W. But then we must have N_{A}(V_{1}) = 1. We have |V_{1}| ≤ |A|. Further by quadratic action we have |W : C_{W} (A)| = |V_{1}|. Let |V_{1}| = r^{t}. Then |W | = r^{st} and we get |A| ≤ r^{s−1}. But in W we have a chain of AK – modules of length s

1 < V̂_{1} < V̂_{2} < ⋯ W

with V̂_{i}/V̂_{i−1} ≅ V_{1}. But this implies |W : C_{W} (A)| ≥ r^{s} > |A|, a contradiction.

So we have V_{1} ≰ O_{p}(L). As V_{1}^{a} = V_{2} for some a ∈ A, we have that V_{1} ∩ Y_{M}^{g} = 1. Assume first O_{p}(L) ∩ V_{1} ≠ 1. Then V_{1} ⊕ V_{2} ∩ Y_{M}^{g} ≠ 1. So a normalizes V_{1} ⊕ V_{2}. Now let t_{1} ∈ T \ ⟨t⟩, where t ∈ T corresponds a. We see that V = (V_{1} ⊕ V_{2})^{t_{1}} ∩ (V_{1} ⊕ V_{2}) ≠ 1. As V is a K–module, we have that either V = V_{1} ⊕ V_{2} or V ≅ V_{1}. Now always V ≥ V_{1} ⊕ V_{2} ∩ Y_{M}^{g}. Let b ∈ A then b = ut_{2}, u ∈ K, t_{2} ∈ T. Then V^{t_{2}} ∩ V ≠ 1. Suppose V ≠ V_{1} ⊕ V_{2}, then [V, t_{2}] = 1, i.e. [V, T] = 1. This shows V = [W, T] and T acts quadratically on W. If V = V_{1} ⊕ V_{2}, then [W, t] = [V_{1}, t] and so [W, t, t_{1}] = 1 and again T acts quadratically on W. Now by construction we have W = [V_{1}, T]V_{1} and so T induces some group of transvections over some field GF (r), where V_{1} is a module over GF (r). Let |V_{1}| = r^{x}. As a acts on V_{1} ⊕ V_{2} we see that for |[V_{1}, t] : [V_{1}, t, a]| = r^{y} we have x = 2y and so |[V_{1}, t] : C_{[V_{1},t]}(A)| ≥ r^{y} Hence we now get that |W : C_{W} (A)| ≥ r^{ys}. Further |T | ≤ r^{s−1}.

Suppose first A ∩ K = 1. Then |A| = |T |. Further |[W, T]| = |[W, A]|, [W, A] ≤ O_{p}(L) and r^{y(s−1)} ≤ |[W, A] : C_{[W,A]}(A)| ≤ |A| ≤ r^{s−1}. This shows y = 1. Now x = 2 and so E(K) ≤ SL_{2}(r). This shows |A| ≤ r and so s = 2. Then W = V_{1} ⊕V_{2} and |W | = r^{4}. As Y_{M} ∩O_{p}(L) = (W ∩O_{p}(L))(Y_{M} ∩Y_{M}^{g}), we have [Y_{M}, T] ≤ W by order and so W ≥ [K_{1}, Y_{M}]. This yields K_{1} ≅ SL_{2}(r) and W is the tensor product module. As C_{Y_{M}} (K) = 1, we have W = Y_{M}, the assertion.

So we may assume A ∩ K ≠ 1. Let v ∈ V_{1} \ O_{p}(L). Then as V_{1} ∩ Y_{M}^{g} = 1, we see that [v, A ∩ K] ≅ A ∩ K. This shows that |V_{1} ∩ O_{p}(L)| ≥ |A ∩ K|. Now let a ∈ A \ K with V_{1}^{a} ≠ V_{1}. Then [V_{1}, a] ≤ O_{p}(L) and |[V_{1}, a] ∩ Y_{M}^{g}| = |V_{1} ∩ O_{p}(L)| ≥ |A ∩ K|. We have W = V_{1}(W ∩ O_{p}(L)), so

|W : C_{W} (A)| = |A||V_{1} : V_{1} ∩ O_{p}(L)| ≤ |A||V_{1}|/|A ∩ K| = |T ||V_{1}|

Now again for |V_{1}| = r^{2y}, we have |W : C_{W} (A)| ≥ r^{ys}. So we get

ys ≤ 2y + (s − 1).

Let first s = 2. Then we see that |[W, A]/W ∩ Y_{M}^{g}| = |V_{1}|, hence |A| = |V_{1}|, which shows that |A∩K| = |V_{1}∩O_{p}(L)| and |V_{1}/V_{1}∩O_{p}(L)| = r. Hence A∩K induces the full group of transvections to a hyperplane on V_{1}. In particular C_{K}(A ∩ K) = A ∩ K. Now for a ∈ A \ K, we have a = ut with u ∈ K and so [u, A] = 1, a contradiction. So we have s > 2 and so y = 1 or y = 2. If y = 1, we get again E(K) ≅ SL_{2}(r) and a contradiction as |A| = r. So let y = 2. Now s = 3. We have |A| ≤ r^{4} and |W : C_{W} (A)| ≥ r^{6}. We see that |V_{1} : V_{1} ∩ O_{p}(L)| = r^{2} and |A| = r^{4}. Hence we have |W : C_{W} (A ∩ K)| = r^{6}. As A ≠ A ∩ K, this is not possible.

So we have V_{1} ∩O_{p}(L) = 1. In particular A∩K = 1. Let t ∈ T. Then K acts on [V_{1}, t]. Let a ∈ A such that t = ua with u ∈ K. Hence [[V_{1}, t], a] ≠ 1. In particular [V_{1}, t] ∩ O_{p}(L) ≠ 1. But then as before we get [[V_{1}, t], T] = 1, in particular T acts quadratically. Now |T | = |A| ≤ r^{s−1}. Further again |V_{1}| = r^{2y} and |W : C_{W} (A)| ≥ r^{sy}. So we see that sy ≤ 2(s − 1), which gives y = 1. But then E(K) ≤ SL_{2}(r) and so |A| ≤ r, which shows s = 2. But |A| ≥ |Y_{M} : Y_{M} ∩ O_{p}(L)| ≥ |V_{1}| = r^{2}, a contradiction.

Hence we have C_{A}(K) ≠ 1. In particular we may assume A ∩ T ≠ 1. Let 1 ≠ a ∈ T ∩ A and [V_{1}, a] ≤ V_{1}, then we see that [V_{1}, a] = 1. Hence V_{1} ≤ O_{p}(L) and then W ≤ O_{p}(L). Now A acts quadratically and W = V_{1}(W ∩ Y_{M}^{g}). Let t = bu ∈ T \ A with u ∈ K and b ∈ A. Now A acts on [V_{1}, t] and so [V_{1}, t] ∩ Y_{M}^{g} ≠ 1. This shows that [V_{1}, t, T] = 1 and so also T acts quadratically. Now b acts as u^{−1} on [V_{1}, t], so u acts quadratically and then also on V_{1}. But then V_{1}^{t} = V_{1}, a contradiction. So we again see that T acts regular on V_{1}^{T} = V_{1}^{A}. In particular V_{1} ≰ O_{p}(L). Then as above we get that T acts quadratically on W and T induces transvections. Hence |T | ≤ r^{s−1} again.

Suppose first A ∩ K ≠ 1. Then V_{1} ∩ O_{p}(L) ≠ 1 and we get |W : C_{W} (A)| = |V_{1}/V_{1} ∩ O_{p}(L)|^{s−1}|V_{1}|. We have |A| ≤ |A ∩ K|r^{s−1}. Further we see that W = V_{1}(W ∩ O_{p}(L)) and so |W : C_{W} (A)| ≤ |V_{1}V_{1}/ ∩ O_{p}(L)||A|. Now |V_{1}/V_{1} ∩ O_{p}(L)|^{s−1}|V_{1}| ≤ |V_{1}/V_{1} ∩ O_{p}(L)||V_{1} ∩ O_{p}(L)|r^{s−1}. This shows |V_{1}/V_{1} ∩ O_{p}(L)| = r and |T | = r^{s−1}. Finally |A ∩ K| = |V_{1} ∩ O_{p}(L)| and so A ∩ K is the full transvection group to V_{1} ∩ O_{p}(L), which shows K ≅ SL_{m}(r). But as C_{K}(A ∩ K) = A ∩ K we get T = T ∩ A. Let K_{1} be a component or a t – group with [T, K_{1}] ≠ 1 and m ∈ K_{1} with V_{1}^{m} ≰ W. If v ∈ V_{1} \ O_{p}(L), then we have [v, A](Y_{M} ∩ Y_{M}^{g}) = Y_{M} ∩ O_{p}(L). Hence by |Y_{M} ∩ O_{p}(L) : C_{Y_{M} ∩O_{p}(L)}(A)| = |A| = |W ∩ O_{p}(L) : C_{W ∩O_{p}(L)}(A)| we get that V_{1}^{m} ≰ O_{p}(L), otherwise V_{1}^{m} ≤ W (Y_{M} ∩ Y_{M}^{g}) and so [V_{1}^{m}, A ∩ K] = 1, a contradiction. Now (V_{1}^{m})^{T} = V_{1}^{m} and so [V_{1}^{m}, T] = 1, which is impossible. So we have that [W, K_{1}] = W and then because T is the full transvection group, we get that K_{1} = SL_{s}(r) or r = 2, or 3 and K_{1}/C_{K_{1}}(W) ≅ SL_{2}(r) and W is the tensor product module. Suppose W ≠ Y_{M}. By 2.4 C_{Y_{M}} (K) = 1. Further the same argument as above shows that W is the socle of K in Y_{M}. So let W_{1}/W be an irreducible module in Y_{M}/W. If K acts trivially then by ?? and 3.18 we get a contradiction, or K ≅ SL_{2}(r). So assume K ≅ S̸L_{2}(r). So W_{1}/W is a nontrivial module. Again we see that W_{1}/W is covered by Y_{M} ∩ Y_{M}^{g}, which shows that A acts trivially, a contradiction, so W = Y_{M}. Hence we are left with K ≅ SL_{2}(r)×SL_{2}(r), and W = [K ×K_{1}, W], if K_{1} is nonsolvable. But then W is the O^{+}(4, r)–module and again W = Y_{M} follows with ?? for r even, and it is obviously true for r odd, as W = [Y_{M}, Z(K)] in that case. Let K_{1} be solvable and X = C_{K_{1}}(W) = C_{K_{1}}(Y_{M}). Then X ≤ C̃ and so [Q, X] = 1. Let R be a Sylow t – subgroup of K_{1}. Then [Q, R] ≠ 1. Now we see that Q acts nontrivially on R/ϕ(R). Hence for p = 2 we see |R| = 3 and for p = 3, we get R/X ≅ Q_{8} and so R ≅ Q_{8}. Hence in both cases we get K_{1} ≅ SL_{2}(r), too, and we may argue as before.

If A ∩ K = 1, then as above we get K ≅ SL_{2}(r) and |T : T ∩ A| ≤ r. We have Y_{M} ∩ O_{p}(L) = (W ∩ O_{p}(L))(Y_{M} ∩ Y_{M}^{g}). This shows that [Y_{M}, T] ≤ W. Let K_{1} be another component of M_{0}/O_{p}(M_{0}) with [K_{1}, A] ≠ 1. Then we see that W ≥ [K_{1}, Y_{M}] and W is the tensor product module. Now if K_{1} is nonsolvable we get as before that K_{1} ≅ SL_{s}(r). But then as A∩C(K_{1}) ≠ 1, we get A ∩ K ≠ 1, a contradiction. So we have K_{1} to be solvable. As T acts quadratic we see with 2.8 that p = 2 or 3. Now as T is the full transvection group, we see that |T | = p, and so we get as above K_{1}A ≅ SL_{2}(p) ≅ K. But then W is the tensorproduct module and so it is not a 2F – module, a contradiction.

2F M0

Lemma 4.9 Assume 4.2(2). Let K be a component of E(M_{0}/C_{M_{0}}(Y_{M})), which is not centralized by A. Then K = F^{∗}(M_{0}C_{M}(Y_{M})/C_{M}(Y_{M})), or F^{∗}(M_{0}C_{M}(Y_{M})/C_{M}(Y_{M})) = K ∗ K_{1} ≅ SL_{n}(r)′ ∗ SL_{m}(r)′ for some n, m. Further Y_{M} is the tensor product module for K × K_{1}.

Proof: Suppose first that we are in the situation of 4.8. Then we have that Y_{M} is the tensorproduct module for SL_{n}(r)′ ∗ SL_{m}(r)′. Now as Y_{M} is irreducible any other component has to centralize Y_{M} and so we have the assertion.

So we may assume that there is a unique component K which is not centralized by A. Let K_{1} be a further component in M_{0}C_{M}(Y_{M})/C_{M}(Y_{M}) or some r–group. Let V_{1} be an irreducible module of KA in Y_{M}. By 4.7 we see that [V_{1}, K_{1}] ≠ 1. Let first [V_{1}, K_{1}] ≰ V_{1}, hence W = V_{1}^{K_{1}} = V_{1}⊕V_{2}⊕⋯⊕V_{s}, s > 1.

Suppose first that V_{1} ≰ O_{p}(L). Choose v ∈ V_{1} with C_{A}(v) = 1. Then Y_{M} ∩ O_{p}(L) = (V_{1} ∩ O_{p}(L))(Y_{M} ∩ Y_{M}^{g}) and so [A, Y_{M} ∩ O_{p}(L)] ≤ V_{1}. In particular A acts quadratically on any irreducible submodule besides V_{1}. As V_{1}^{K_{1}} ≠ V_{1} and [A, V_{1}] = 1, we see that A acts quadratically on V_{1} too. So A acts quadratically on W and Y_{M} is an F–module with offender A. We have [Q, Y_{M}] ≤ O_{p}(L), so [Q, Y_{M}, A] = 1, by quadratic action of A on Y_{M}. Suppose there is some u ∈ Q and V an irreducible module with V^{u} ≠ V. Let t^{u} ∈ V^{u}. Then [t^{−1}t^{u}, A] = 1. But this implies that [t, a] = [t^{u}, a] for all a ∈ A. Then [t, a] ∈ V ∩ V^{u} = 1, a contradiction. Hence Q fixes any irreducible submodule of K. As in 4.8 we see that V_{1} as an K–module is irreducible. As Q ≤ C_{KQ}(C_{V_{1}}(Q)), and so we see that K is a group of Lie type and so [C_{V_{i}}(S ∩ K), Q] = 1. So K_{1} acts on a nontrivial subgroup of C_{W} (Q), which gives K_{1} ≤ C̃, a contradiction.

So we have that W, the sum of all irreducible K–submodules, is contained in O_{p}(L). Now let W̃ be maximal such that [W̃ , K × K_{1}] ≤ W̃ and W̃ ≤ O_{p}(L). Let Ŵ be such that Ŵ /W̃ is an irreducible K × K_{1}–module.

Suppose first [K_{1}, Ŵ] ≤ W̃. Now choose ω ∈ K_{1}, ω a p′–element. Let x ∈ Ŵ \ O_{p}(L), with [ω, x] = 1, this exists. Now as [A, ω] = 1, we see that [x, A] ≤ CŴ (ω). We know that [x, A] covers V_{1}, i.e. V_{1} ≤ [x, A](Y_{M} ∩ Y_{M}^{g}). So let v_{1} ∈ V_{1}, then there is w ∈ W ∩[x, A] and w̃ ∈ Y_{M} ∩Y_{M}^{g} with v_{1} = ww̃. Now [w, A] ≤ CŴ (ω) and [w, A] = [ww̃, A] = [v_{1}, A] ∈ V_{1}. But we may choose ω such that C_{V_{1}}(ω) = 1. This shows [V_{1}, A] = 1, a contradiction.

So we have [Ŵ , K_{1}] ≰ W̃. Suppose further [K, Ŵ] ≤ W̃. As [K_{1}, A] = 1, we have [Ŵ , A] ≤ W̃. As K_{1} acts on CŴ (A) we see that A ∩ Y_{M} ∩ Ŵ ≤ W̃. So Ŵ ∩ O_{p}(L) = W̃. Now [Ŵ , Q] ≤ O_{p}(L) and so [Ŵ , Q] ≤ W̃. Then as [Q, K_{1}] ≠ 1, we get the contradiction [Ŵ , K_{1}] ≤ W̃.

So we have [K, Ŵ] ≰ W̃. Now we write Ŵ /W̃ = V̂_{1} ⊕ … ⊕ V̂_{t}. As before we see that A acts quadratically on Ŵ /W̃. This shows Ŵ /W̃ = V̂_{1}(Y_{M} ∩ A)/W̃. But then |Ŵ : CŴ (A)| = |V̂_{1} : CV̂_{1}(A)|, so s = 1 and [V̂_{1}, K_{1}] ≤ V̂_{1}. If [V̂_{1}, K_{1}] = 1, we would get [Ŵ_{1}, K_{1}] ≤ W̃. So we have [V̂_{1}, K_{1}Q] = V̂_{1}. Then we have that K_{1} induces field multiplications on V̂_{1}, in particular K_{1} is solvable. Suppose V̂_{1}^{Q} ≠ V̂_{1}. Then as before we have that A acts quardtatically on ⟨V̂_{1}^{Q}⟩. Now we get again as before that [V̂_{1}, Q] ≤ V̂_{1}. As [K_{1}, Q] ≠ 1, Q has to induce an outer automorphism on K. We have that W̃ ≤ O_{p}(L) and so W̃ is an F –module with offender A. We have [V̂_{1}, A] ≠ 1. So W is better than F –module. By 3.3 we see that K is a Lie type group. Set W_{1} = V_{1}^{Q}. Then we have that Q ≤ O_{p}(C_{K}(C_{W_{1}}(Q))). As Q cannot induce field automorphisms we see that K ≅ SL_{4}(q) and then W involves both 4–dimensional modules. But then W̃ is not better than F –module.

So we are left with [V_{1}, K_{1}] ≤ V_{1} for any irreducible K–submodule V_{1}. In particular K_{1} has to induce field multiplications on V_{1}. If V_{1} is Q–invariant, we get 1 ≠ tt_{1}, t ∈ K, t_{1} ∈ K_{1} with [C_{V_{1}}(Q), tt_{1}] = 1, a contradiction. So set W_{1} = V_{1}^{Q}. Now V_{1} is an F –module and Q induces outer automorphism on K. So again we see that K ≅ Ω^{+}(n, q), n = 6, 8, 10, and V_{1} is the half spin module. But then in C_{W_{1}}(S ∩ K)^{♯} we have exactly three K–orbits and two KQ–orbits. But they are of different length. So K_{1} contains elements which are conjugate to some element in C(Q), a contradiction.

2F sol

Lemma 4.10 Let 4.2(2) and [A, E(M_{0}C_{M_{0}}(Y_{M})/C_{M_{0}}M(Y_{M}))] ≤ C_{M_{0}}(Y_{M}), Then p = 3 and M_{0}/C_{M_{0}}(Y_{M}) is conjugated to SL_{2}(3) ∗ SL_{2}(3) and Y_{M} is the orthogonal 4–dimensional module.

Proof: By 4.3 we know that [F_{p}(M_{0}), A] ≰ C_{M_{0}}(Y_{M}). Let R_{p} be a Sylow r–subgroup of F (M_{0}/C_{M_{0}}(Y_{M})), which is normalized by A. Let further R = [R_{1}, A]. Hence R = [R, A] and R = [R, Q]. Finally C_{Y_{M}} (R) = 1. Let B ≤ A, |B| = q, with [Q, B] ≤ Y_{M}. We may assume that B ≤ Q. If b ∈ B^{♯} with [b, R] = 1. Then as [Y_{M}, b, b, A] = 1, we see that b acts quadratically on Y_{M}. As we may choose B in such a way that some elment of order q−1 acting on Y_{M} ∩O_{p}(L) acts transitivelly on B^{♯}, we have that B acts quadratically. Now [Y_{M}, b, B] = 1 and so [R, B] = 1. Now B acts on some other Sylow t–subgroup T of F (M_{0}/C_{M_{0}}(Y_{M})) faithfully. But by quadratic action and 3.1 we have [Y_{M}, [T, B]] = ⟨y | |B : C_{B}(y)| ≤ p⟩. If [Y_{M}, [T, B]] ≰ O_{p}(L), then q = p and p = 3, t = 2 or p = 2. Now A induces transvections on [Y_{M}, b] to a hyperplane. Then we have |A : C_{A}([Y_{M}, b])| = p. This now shows that |A : C_{A}(R)| = p. Finally, as C_{R}([Y_{M}, b]) is centralized by Q, we see that [R, A]ϕ(R)/ϕ(R) is of order 3 or 4. In the first case we see that |R| = 3 while in the second R ≅ Q_{8}. As t ≠ r this gives p = 2. So t = 5 and |[b, T]| = 5. As Q acts on [b, T], we see that |Y_{M}| = 16. But there is no Σ_{3} × F_{10} acting faithfully on a 4–dimensional vectorspace over GF (2). So let [Y_{M}, [T, B]] ≤ O_{p}(L). As A acts on [T, B], we get that [Y_{M}, [T, B]] ≤ Y_{M}^{g} and so is centralized by B, a contradiction.

This proves C_{B}(R) = 1. If B acts quadratically on Y_{M}, we get as above q = p. So assume q > p. Then we have p > 2. Now choose b ∈ B^{♯} with C_{R}(b) ≰ ϕ(R). Then as C_{R}(b) ≰ C̃, we get as above that b and then B acts quadratically. So in any case we have p = q and so Y_{M} is an F+1 – module with offender A.

Let |A| = p. Then |Y_{M} : C_{Y_{M}} (A)| ≤ p^{2}. If A acts quadratically then, we get |Y_{M}| = p^{2} and so RQ = M_{0}/C_{M_{0}}(Y_{M}) is isomorphic to SL_{2}(p). But this contradicts 2.7. So we have p > 2 and |Y_{M} : C_{Y_{M}} (A)| = p^{2}. Let U be a critical subgroup in R/C_{R}(Y_{M}). Suppose that U is not abelian. Then [A, Z(U)] = 1 and |Z(U)| = r. Let U_{1} ≤ U, such that U_{1}/Z(U) is an irreducible A–module. Then U_{1} is conjugated to a subgroup of SL_{4}(p) and so we see that p = 3, r = 2 and U_{1} ≅ V_{4} or Q_{8}. Further there is some U_{1} ≅ Q_{8}. Then |[U_{1}, Y_{M}]| = 3^{4} and so as [Z(U), Y_{M}] = Y_{M}, we get |Y_{M}| = 3^{4}. We see that R = U ≅ Q_{8} ∗ Q_{8}, the assertion. If A = Q, then A acts on some elementary abelian subgroup of order 4, which centralize elements in C_{Y_{M}} (A) and so would be in C̃, a contradiction. So RQ is conjugated to SL_{2}(3) ∗ SL_{2}(3). It remains to show that U is abelian. Then with U_{1} as above, we get that U acts on C_{Y_{M}} (U_{1}), so we have that C_{Y_{M}} (U_{1}) = 1 and then |Y_{M}| = p^{3}. But then M_{0} is a minimalparabolic, a contradiction to 2.7.

So we may assume |A| > p. Let a ∈ A^{♯} with C_{R}(a) ≰ ϕ(R). Then A induces transvections on [Y_{M}, a]. This shows |A : C_{A}([Y_{M}, a])| ≤ p. Further we see that a acts quadratically. Let p > 2. As this is true for any such a we see that any element in C_{A}([Y_{M}, a]) acts quadratically and so by 2.9 C_{A}([Y_{M}, a]) acts quadratically. Now by 3.1 we have that [Y_{M}, [R, C_{A}([Y_{M}, a])]] = ⟨y | |A : C_{A}(y)| ≤ p^{2}⟩. This gives |A| = p^{2} or [Y_{M}, [R, C_{A}([Y_{M}, a])]] ≤ O_{p}(L). In the latter we see that |[Y_{M}, [R, C_{A}([Y_{M}, a])]] : [Y_{M}, [R, C_{A}([Y_{M}, a])]] ∩ Y_{M}^{g}| ≤ p. Now by 3.2 a subgroup of index p in A centralizes [Y_{M}, [R, C_{A}([Y_{M}, a])]].

This now gives that [Y_{M}, [R, C_{A}([Y_{M}, a])]]/C_{[Y_{M} ,[R,C_{A}([Y_{M} ,a])]]}(Y_{M}) ≅ Z_{3} or Q_{8}. So |C_{A}([Y_{M}, a])| = p and then again |A| = p^{2}.

So let p = 2. Then set R_{1} = C_{R}([Y_{M}, a]). There is y ∈ Y_{M} \ O_{2}(L) with [y, R_{1}] = 1. As R_{1} is A–invariant, we see that [R_{1}, Y_{M}] ≤ Y_{M} ∩ A. But C_{R}(Y_{M}) = 1 and so we see that R_{1} = 1, which gives |A| ≤ 4.

Now we have that |Y_{M} : C_{Y_{M}} (A)| = p^{3}, p = 2 or 3 and A is generated by elements a, acting quadratically and |Y_{M} : C_{Y_{M}} (a)| ≤ p^{2}. Let first p = 3. If there are at least three such elements a, then by 2.9 we have that A acts quadratically and so A induces a group of transvections, a contradiction to 3.2. So we have exactly two such elements a_{1}, a_{2} and then R = C_{R}(a_{1})C_{R}(a_{2}). Further [a_{i}, Q] ≤ O_{p}(M_{0}) for i = 1, 2. Then we see that C_{R/C_{R}(Y_{M})}(a_{i}) ≅ Q_{8} and so R/C_{R}(Y_{M}) ≅ Q_{8} ∗ Q_{8} and RA/C_{R}(Y_{M}) ≅ SL_{2}(3) ∗ SL_{2}(3), RA = RQ, i.e. [C_{R}(Y_{M}), A] = 1. But now choose U as before, then we see that U ≅ Q_{8} ∗ Q_{8} and then R = UC_{R}(U). Now [C_{R}(U), A] = 1 and as R = [R, A], we see U = R. Now |Y_{M}| = 3^{4} and then M_{0}/C_{M_{0}}(Y_{M}) = UA, the assertion.

Let next p = 2. Again A cannot act quadratically. So |Y_{M} : C_{Y_{M}} (a)| = 4 for all a ∈ A^{♯}. Further we have that |C_{R/ϕ(R)}(a)| ≤ 3 for a ∈ A^{♯}. This shows |R/ϕ(R)| ≤ 27. Let |R/ϕ(R)| = 27. Let a ∈ A^{♯} with [a, Q] ≤ C_{M_{0}}(Y_{M}). Then C_{R}(a) is Q - invariant. As A induces a transvection on C_{Y_{M}} (a), we see that [C_{R}(a), A] induces Z_{3} on C_{Y_{M}} (a) and so on Y_{M}. As C_{Y_{M}} (C_{R}(a)) = 1, we see |C_{Y_{M}} (a)| = 4 and so |Y_{M}| = 16, which contradicts |R/ϕ(R)| = 27. So we have |R/ϕ(R)| = 9. Now RA is generated by A and some conjugate of a certain a ∈ A, which shows |Y_{M}| ≤ 2^{5}. Hence |R/C_{R}(Y_{M})| = 9 and so ϕ(R) = C_{R}(Y_{M}). As above choose U, so U is elementary abelian of order 9 and then again U = R. Now RA ≅ SL_{2}(2) ∗ SL_{2}(2), |Y_{M}| = 16 and Y_{M} is the tensor product module. Now RQ ≅ M_{0}/C_{M_{0}}(Y_{M}) and so by 2.7 we have that RQ = RA, the assertion.

F unique

Lemma 4.11 Assume 4.2(1). Let K be a component of E(M_{0}/C_{M_{0}}(Y_{M})), which is not centralized by A. Then K = [F^{∗}(M_{0}/C_{M_{0}}(Y_{M})), A].

Proof: Suppose false. As in the previous lemma set KA = K × T. Further set W = V_{1} ⊕ ⋯ ⊕ V_{s}, where V_{i} is a nontrivial irreducible K– submodule of Y_{M}, and W = V_{1}^{T}. Suppose first s > 1. As A is a quadratic offender as an F –module, we have that [V_{1}, A] ∩ V_{1} = 1 and so |V_{1}| ≤ |A|. Further by quadratic action we see that A and T both induce transvection groups on W. Again |V_{1}| = r^{2y} for some r, we see that |W : C_{W} (A)| = r^{sy}, which shows s = 2 and then |T : T ∩ A| ≤ r. Let T ∩ A ≠ 1. Then there is some component K_{1} or r–group K_{1} in F (M_{0}/C_{M_{0}}(Y_{M})) which is normalized but not centralized by A. So we may assume that [K_{1}, T ∩ A] ≠ 1. Now quadratic action and 4.6 and 4.7 yield a contradiction. So we have that A ∩ T = 1. We see that W centralizes a subgroup of index r in A and so W^{g} centralizes a subgroup of index r in Y_{M}. In particular [W, W^{g}] = 1. Hence [W^{g}, Y_{M}] = 1 and then [A, W] = 1, a contradiction.

So we are left with s = 1. Let T ≠ 1. Then [V_{1}, T] ≤ V_{1} for all irreducible K–submodules in Y_{M}. This shows again that [K_{1}, V_{1}] = 1 for all irreducible K-submodules in Y_{M}. But then again 4.7 yields the contradiction.

So we have T = 1. By quadratic action we have that A ∩ E(K) = 1 and so |A| ≤ 4 for p = 2 and |A| ≤ p else. In particular V_{1} is an irreducible E(K)–module. Then we get |A| = p and A induces a transvection on Y_{M}. But then [A, K_{1}] = 1, K_{1} as above, a contradiction.

F M0

Lemma 4.12 Assume 4.2(1). Let K be a component of E(M_{0}/C_{M_{0}}(Y_{M})), which is not centralized by A. Then either K = F^{∗}(M_{0}/C_{M_{0}}(Y_{M})), or F^{∗}(M_{0}/C_{M_{0}}(Y_{M})) = K ∗ K_{1} ≅ SL_{n}(r)′ ∗ SL_{m}(r)′ for some some n, m and r = p^{t}. Further Y_{M} is the tensor product module for K × K_{1}.

Proof: We have [K, A] ≠ 1 and [K_{1}, A] = 1. Let V_{1} be an irreducible K–submodule of Y_{M} and W = V_{1} ⊕ ⋯ ⊕ V_{s} = V_{1}^{K_{1}}. Clearly by 4.7 we may assume that s > 1. Now as Y_{M} is an F –module with offender A we have that V_{1} is better than an F –module which with 3.3 gives that K ≅ SL_{n}(r), Sp(2n, r), SU_{n}(r) or Ω^{±}(n, r) and V_{1} is the natural module. As K ≤ M_{0}, we have [K, Q] ≠ 1. Let v ∈ V_{1} such that v is centralized by S∩K, then we have that Q is normalized by N_{K}(⟨v⟩). As also K_{1} ≤ M_{0}, we have that Q ≰ K. Now choose x ∈ Q \ K. Then [x, N_{K}(Q)] ≤ O_{p}(N_{K}(Q)). But this shows that x induces an inner automorphism on K and so KQ = K × U. Suppose U ∩Q = 1. Then we have K ≅ Sp(2n, p) or SU_{n}(p) and |Q : Q∩O_{p}(M)| = p. Then |Q : C_{Q}(Y_{M})| = p. Hence also |Q : C_{Q}(A)| = p. Now we may assume that [V_{1}, U] ≠ 1. Then we see that [V_{1}, x] ≅ V_{1} as an N_{K}(Q)–module. As [V_{1}, x] ≤ Q this shows that A induces a transvection group on V_{1} and so |A| = p, which contradicts the fact that V_{1} is better than F –module. Hence we have U ∩Q ≠ 1 and we may assume that x ∈ U. Now [W, x] is a nontrivial K–module. Let W_{1} ≤ [W, x] be an irreducible nontrivial K-module, which is centralized by U. In particular A acts nontrivially and so W_{1} ≰ Y_{M}^{g}. We have that W_{1}^{g} ≤ Q. If [W_{1}^{g}, W_{1}] = 1, then we have W_{1}^{g} ≤ O_{p}(M) and then we see that [A, W_{1}] = 1, a contradiction. Hence we have that there is an offender on W_{1} in A∩Q, which means that there is an offender on V_{1}, which is contained in O_{p}(N_{K}(Q)). Now we see that K ≅ SL_{n}(r) or Sp(2n, r). Further by 2.4 we get that |C_{W_{1}}(Q)| = r and so Q ∩ K = O_{p}(N_{K}(Q)). Now [W, x]^{g} ≤ Q ∩ K ∩ A. But Q ∩ K is an exact offender on W_{1}, which shows that W_{1} = [W, x]. The same argument also implies that [W, y] ≤ [W, x] for all y ∈ U. Hence U is a transvection group to a point.

We have |[[W_{1}, A], Q]| ≤ r, hence |[[W_{1}^{g}, Y_{M}], Q]| ≤ r. As [W_{1}^{g}, V_{1}] ≠ 1, we get |[[W_{1}^{g}, Y_{M}], Q]| = r and |[W_{1}^{g}, V_{1}]| = r, hence |[W_{1}^{g}, W]| = r^{s}. and so |[W_{1}, A]| ≥ r^{s}. Let y ∈ C_{W_{1}}(W_{1}^{g}) with [y, A] ≠ 1, then [y^{g}, Y_{M}] ≠ 1. But y^{g} ∈ C_{W_{1}^{g}}(W_{1}) ≤ C_{M}(Y_{M}), a contradiction. So we have [C_{W_{1}}(W_{1}^{g}), A] = 1. If K ≅ Sp(2n, r), then we get A = W_{1}^{g} is of order at most r, which contradicts the fact that A is an offender and s > 1. So we have K ≅ SL_{n}(r) and A is contained in a transvection group to a hyperplane. Now we see that the largest completely reducible K–submodule of Y_{M} is a direct sum of natural modules. So without loss we may assume that this is W. As U is a transvection group to a point, we see with 2.4 that K_{1} ≅ SL_{s}(r) and U is the full transvection group. Finally W is the tensor product of the corresponding natural modules.

What is left is to show that W = Y_{M}. Otherwise by ?? we get a submodule W̃ > W such that W̃ /W ≅ W as K × K_{1}–module. In particular there is x ∈ U such that [W̃ , x] is an extension of the natural K–module by the natural K–module. Now we know that |[W̃ , x]^{g} : [W̃ , x]^{g} ∩ O_{2}(M)]| ≤ r. Hence |[W̃ , x] : C_{[W̃ ,x]}(A)| ≤ r, a contradiction. So we have W = Y_{M}.

F sol

Lemma 4.13 Assume 4.2(1). Then [A, E(M_{0}/C_{M_{0}}(Y_{M}))] ≰ C_{M_{0}}(Y_{M}).

Proof: Suppose false. By 4.3 we have that [A, F_{p}(M_{0})] ≰ C_{M_{0}}(Y_{M}). As A acts quadratically on Y_{M} we see with 2.4 that A acts nontrivially on exactly one Sylow r-subgroup R of F (M_{0}/C_{M_{0}}(Y_{M})). Further p = 2 or 3 by 2.8. As A is an F –module offender, we see that R is a 3–group in case of p = 2. As we have Y_{M} = [Y_{M}, [R, A]] by 2.4 we see with 3.1 we get Y_{M} = ⟨y | |A : C_{A}(y)| ≤ p⟩. So A is generated by elements which induces transvections on Y_{M}. By 3.4 we see [R, A]A/C_{[R,A]}(Y_{M}) ≅ SL_{2}(p) × ⋯ × SL_{2}(p). As Q normalizes [R, A] we get with 2.4 that R = [R, A] = F^{∗}(M_{0}/C_{M_{0}}(Y_{M})). Now Q acts transitively on the SL_{2}(p)′s, by 2.4. This shows that M_{0} is a minimal parabolic, contradicting 2.7.

Now the structure of M_{0}/C_{M_{0}}(Y_{M}) in 4.1 follows from 4.12, 4.13, 4.10 and 4.9.

What is left is to show that C_{M_{0}}(Y_{M})/O_{p}(M_{0}) is in the Schur multiplier of the component K. As [C_{M_{0}}(Y_{M}), Q] ≤ Q ∩ C_{M_{0}}(Y_{M}) ≤ O_{p}(M_{0}), we see that C_{M_{0}}(Y_{M})/O_{p}(M_{0}) ≤ Z(M_{0}/O_{p}(M_{0})). Now if K is quasisimple we have the assertion.

If we have K ≅ SL_{n}(q)′ ∗ SL_{m}(q)′ and q > 3, we see F^{∗}(M_{0}/O_{p}(M_{0})) ≤ Sl_{n}(q) × SL_{m}(q). Let now q ≤ 3. If m and n both are at least 3, we again have the assertion. So assume that m = 2. As both Z_{3} and Q_{8} have a trivial Schur multiplier, we have the assertion if n > 2, so assume n = m = 2. Let D be a critical group in the preimage U of K in M_{0}/O_{p}(M_{0}). Then U = DC_{M_{0}}(Y_{M})/O_{p}(M_{0}). So, we see that D = U and then in case of p = 2, we have D ≅ 3^{1+2}. If p = 3, then |Q : Q ∩ O_{3}(M_{0})| = 9 and so D ≅ Q_{8} × Q_{8} or Q_{8} ∗ Q_{8}. This is the assertion.

5 The modules in 4.1