Non E-uniqueness, the trychonomy theorem

U. Meierfrankenfeld, B. Stellmacher, G. Stroth

1 Introduction

This paper is part of the classification groups of local characteristic p. An overview of this classification can be found in [MSS]. Let G be a group of local characteristic p, i. e. all p-locals are of characteristic p, i. e. for a p-local H we have C I_{H}(O_{p}(H)) ≤ O_{p}(H). Then we like to pin down the structure of G or in a first step the structure of the p-locals containing a given Sylow psubgroup S. We denote by C the normalizer of Ω_{1}(Z(S)). Let now C ≤ C̃, C̃ a maximal p-local subgroup. Set E = O^{p}(F_{p}^{∗}(C IC̃(YC̃))). In [MSS1] we have investigated the structure of all p-locals containing S, which are not in C̃ under the assumption that E is contained in exactly one maximal p-local, which then has to be C̃. In this paper we are going to investigate the case that E is contained in more than one maximal p-local. To state the theorem of this paper we need some further definitions and notations. We introduce the following notation:

Let N ◁ ◁C̃, T ∈ Syl_{p}(N), set
P_{N}(T) = O^{p′}(C I_{N}(Ω_{1}(Z(T)))).

This is also called ”point stabilizer”.

With M(X) we denote the set of maximal p-locals containing X. We consider the following set

Ω = {N ◁ ◁C̃, E ≤ N, M(EP_{N}(T)) ≠ {C̃} for some T ∈ Syl_{p}(N)}.

We call M ∈ Ω ”E-maximal” if

(1) P_{M}(T) is maximal for T ∈ Syl_{p}(M)

(2) M is maximal with respect to (1)

For M ∈ Ω set M_{T} = EP_{N}(T). We have P_{N}(T) = O^{p′}(M_{T}).

Let X be a group. Then define L_{X} = {L ≤ X|C I_{X}(O_{p}(L)) ≤ L}.

If Y ≤ X set L_{X}(Y) = {L|L ∈ L_{X}, Y ≤ L}.

For M ∈ Ω, M E-maximal, set

H(M) = ⋃_{T ∈Syl_{p}(M)}(L_{G}(M_{T}) \ LC̃(M_{T}))

If A, B are groups, then denote by S_{B}(A) the greatest subnormal subgroup of A contained in B. Now we introduce for M ∈ Ω, M E-maximal, H^{∗}(M) = {L ∈ H(M)| |L ∩ C̃|_{p} maximal and then S_{L}(C̃) maximal and

L ∩ C̃ maximal }.

So we see NC̃(O_{p}(L ∩ C̃)) ≤ L. Further we define H_{o}(M) = {L ∈ H(M)| Q ≤ L, L ≰ C̃, S_{L}(C̃) = S_{N_{G}(O_{p}(L∩C̃))}(C̃) and then

L ∩ C̃ maximal }

For L ∈ H^{∗}(M) set N (L) = {N ◁ ◁C̃ | L ∩ C̃ ≤ N_{G}(N) and N is minimal with respect to

N ≰ S_{L}(C̃)}

Now we say

(1) L ∈ H(M) satisfies (P U) if there is no nontrivial characteristic subgroup of O_{p}(L ∩ C̃) which is normal in L.

(2) N∈N (L) satisfies (A), if for P = N(L∩C̃) there is T ∈Syl_{p}(P)∩Syl_{p}(L) with P_{P} (T) = P_{L}(T).

The aim of this paper is to prove Theorem One of the following holds

(1) There is some M and L ∈ H(M) which satisfies (P U)

(2) There is some L ∈ H^{∗}(M) and N ∈ N (L) which satisfies (A)

(3) There is some amalgam P, L, L ∈ H^{∗}(M), P = N(L ∩ C̃), where N ∈ N (L) and b = 1.

In a fortcomming paper [MS] we will determine the simple groups showing up in (3). Hence we are left with a pushing-up situation or a nice amalgam situation.

2 Preliminaries

Lemma 2.1 Let F be of characteristic p-type and K ◁ ◁F. If K/O_{p}(K) possesses no p-chief-factors, then [Y_{K}, K] ≠ 1 iff [Y_{F} , K] ≠ 1.

Proof: Set V = Ω_{1}(Z(O_{p}(F))) and R = [V, K]. We have

[O_{p}(F), O^{p}(K)] ≤ O^{p}(K) ∩ O_{p}(F).

So we may assume K = O^{p}(K)(O_{p}(F) ∩ K). As O_{p}(K) ≤ O_{p}(F) we see R ≤ Z(O_{p}(K)). As K/O_{p}(K) possesses no p-chief-factors we get R ≤ Y_{K}. Further we have [Y_{F} , K] ≤ R. Let first R ≠ 1. As [R, K] = R, we get [Y_{K}, K] ≠ 1. So we have that [Y_{P} , K] ≠ 1 implies [Y_{K}, K] ≠ 1. Let now R = 1. Then [C I_{Y_{K}}(O_{p}(F)), K] = 1. By the P × Q-Lemma we get [Y_{K}, K] = 1. So assume now [Y_{K}, K] ≠ 1. Then R ≠ 1. Now choose N ◁ F, N ≤ K, [N, K] ≠ 1 and N minimal. Set \overline{F} = F/C I_{F} (N). By minimality we see O_{p}(K) ≤ C I_{F} (N). As \overline{K} has no p-chief-factors we get [\overline{K}, O_{p}(\overline{F})] = 1. The P × Q - Lemma now implies [C I_{N}(O_{p}(\overline{F})), \overline{K}] ≠ 1. The minimality now implies O_{p}(\overline{F}) = 1. Hence N ≤ Y_{F} and so [Y_{F} , K] ≠ 1.

Lemma 2.2 Let F be a finite group, p a prime, D ≤ F with p |̸|F : D|. Let further A, B ◁ ◁F and S ∈ Syl_{p}(A) and T ∈ Syl_{p}(B).

(a) We have O^{p′}(⟨A, B⟩ ∩ D) = ⟨O^{p′}(A ∩ D), O^{p′}(B ∩ D)⟩.

(b) We have Syl_{p}(⟨T, S⟩) ≤ Syl_{p}(⟨A, B⟩)

(c) Let F be of characteristic p-type and [A, B] ≤ O_{p}(F). Set P = O^{p′}(C I_{A}(Ω_{1}(Z(S))) and Q = O^{p′}(C I_{B}(Ω_{1}(Z(T))). Let U ∈ Syl_{p}(⟨P, Q⟩) and R = O^{p′}(C I_{⟨P,Q⟩}(Ω_{1}(Z(U))). Then there is some W ∈ Syl_{p}(⟨A, B⟩) such that R = O^{p′}(C I_{⟨A,B⟩}(Ω_{1}(Z(W))).

Proof: We first treat the special case D = S ∈ Syl_{p}(F). We are going to prove (a) and (b) by induction to |F | and then to |F : A| + |F : B|. By induction we may assume F = ⟨A, B⟩. Let now A ≤ M ◁ F, B ≤ N ◁ F such that F/M and F/N are simple. In fact if either A = F or B = F we have nothing to prove.

(a) We have ⟨A ∩ N ∩ S, B ∩ S⟩ ≤ N. Set H = ⟨A ∩ N, B⟩. By induction ⟨A ∩ N ∩ S, B ∩ S⟩ = ⟨A ∩ N, B⟩ ∩ S. Suppose A ∩ N ≰ B. Then by induction on |F : B| + |F : A| we see ⟨A ∩ S, B ∩ S⟩ = ⟨A ∩ S, H ∩ S⟩ = ⟨A, H⟩ ∩ S = ⟨A, B⟩ ∩ S. So we may assume A ∩ N = A ∩ B = B ∩ M. Then A ∩ B ◁ ⟨A, B⟩ = F. By induction we have A∩B = 1, so also A∩N = B∩M = 1. In particular both A and B are simple groups. If A is nonabelian, then either A ≤ B or [A, B] = 1. But A ≤ B is impossible, so [A, B] = 1. Then ⟨A ∩ S, B ∩ S⟩ = AB ∩ S. Let now |A| = r, |B| = s, r, s primes. Then A, B ≤ F (F). This shows F = F (F). If r ≠ s, then again [A, B] = 1 and we are done. If r = s ≠ p, then F is a r-group and so S = 1, again we are done. So we are left with r = s = p. Then A = A ∩ S, B = B ∩ S and S = ⟨A, B⟩.

(b) We have ⟨S ∩ N, T ⟩ ≤ N. By induction we see that ⟨S ∩ N, T ⟩ is a Sylow p-subgroup of H = ⟨A ∩ N, B⟩. Suppose again first A ∩ N ≰ B. Choose T_{1} ∈ Syl_{p}(H). By induction on |F : B| we see that ⟨S, T_{1}⟩ contains a Sylow p-subgroup of ⟨A, H⟩. As ⟨S ∩ N, T ⟩ contains a Sylow p-subgroup of ⟨S, T_{1}⟩, so ⟨S, T ⟩ contains a Sylow p-subgroup of ⟨S, T_{1}⟩ and ⟨S, T_{1}⟩ contains a Sylow p-subgroup of ⟨A, H⟩ and finally ⟨A, H⟩ contains one of ⟨A, B⟩. So again we have A ∩ N = A ∩ B = B ∩ M and A ∩ B ◁ ⟨A, B⟩ = F. By induction again A ∩ B = A ∩ N = B ∩ M = 1. If [A, B] = 1 then ST ∈ Syl_{p}(AB) and we are done. Hence A and B are r-groups. If r ≠ p, so we are done too. If r = p, then S = A, T = B and so ⟨S, T ⟩ = ⟨A, B⟩.

Now we treat the general case. We just have to treat (a). Let S ∈ Syl_{p}(D ∩ ⟨A, B⟩). Now application of the special result to ⟨A, B⟩ shows:

S = ⟨S ∩ A, S ∩ B⟩ ≤ ⟨O^{p′}(A ∩ D), O^{p′}(B ∩ D)⟩.

But ⟨S∩A, S∩B⟩ = O^{p′}(D ∩ ⟨A, B⟩).

c) Let first A = O^{p}(A) and B = O^{p}(B). Then A = O^{p}(AO_{p}(F)) ⊴ ⟨A, B⟩ and also B ⊴⟨A, B⟩. In particular we may always assume A, B ◁F = ⟨A, B⟩. Then O_{p}(A)O_{p}(B) = O_{p}(F). We [P, Q] ≤ O_{p}(A) ≤ P, so [P, Q] ≤ P ∩ Q. Now ST ∈ Syl_{p}(P Q) ≤ Syl_{p}(AB). So

O^{p′}(C I_{A}(Ω_{1}(Z(S))) = P = O^{p′}(C I_{P} (Ω_{1}(Z(ST))) = O^{p′}(C I_{A}(Ω_{1}(Z(ST))))).

In the same way we get Q = O^{p′}(C I_{B}(Ω_{1}(Z(ST)))). Hence

⟨P, Q⟩ = ⟨O^{p′}(C I_{A}(Ω_{1}(Z(ST))), O^{p′}(C I_{B}(Ω_{1}(Z(ST)))⟩ =
O^{p′}(C I_{AB}(Ω_{1}(Z(ST)))).

Lemma 2.3 Let X ◁ H, with H/X be perfect. Then there is a unique minimal subnormal supplement U of X in H.

Proof: Suppose U_{1}X = H = U_{2}X, with U_{1}, U_{2} be subnormal. Let U_{1} ◁ ◁Ũ ◁H. By induction U_{1} is uniquely determined in Ũ with respect to X ∩Ũ. Hence U_{1} ◁ H. The same argument shows U_{2} ◁ H. Then [U_{1}, U_{2}]X = H. Hence U_{1} = [U_{1}, U_{2}] = U_{2}.

Lemma 2.4 Let G be a group, A, B ◁ ◁G. Suppose A to be perfect with an unique maximal normal subgroup M_{A}. Then either A ≤ B or [A, B] ≤ M_{A}.

Proof: Induction to |G|. If ⟨A^{G}⟩ = G, we have A = G. If B ≠ G, then B ≤ M_{A} and so [A, B] ≤ M_{A}. So we have ⟨A^{G}⟩ ≠ G. Let g ∈ G with A^{g} ≠ A. By induction [A, A^{g}] ≤ M_{A}. Hence A ◁ ⟨A^{g}⟩. Set M = ⟨M_{A}^{G}⟩. Then we get [A, M] ≤ A∩M = M_{A}. In particular B ≰ M. We have [A, B] ≤ ⟨A^{G}⟩. Then A ≤ [A, B] ≤ ⟨B^{A}⟩ or [A, [A, B]] ≤ M_{A} ≤ M by induction. In the first case ⟨B^{A}⟩ = G = B and we are done. In the second case as [A, B, A] ≤ M we get with the 3-subgroup lemma [A, A, B] ≤ M. As A is perfect this shows [A, B] ≤ M. By (2.3) we have B ≤ N(AM) and A is the unique minimal subnormal supplement. Hence B ≤ N(A). This gives [B, A] ≤ A∩M = M_{A}.

Let V be a module for a group H, A a p-subgroup of H. The group A is said to act quadratically on V if [V, A, A] = 1. We call an offender on V if

|A/C I_{A}(V)||C I_{V} (A)| ≥ |A_{0}/C I_{A_{0}}(V)||C I_{V} (A_{0})| for all C I_{V} (A) ≤ A_{0} ≤ A.

Lemma 2.5 Let A be a quadratic offender for H on V and U a submodule with [U, A] ≠ 1. Then U is a quadratic offender on U.

Proof: Let C I_{A}(U) ≤ A_{0} ≤ A. Then

|A_{0}||C I_{U}(A_{0})C I_{V} (A)| ≤ |A_{0}||C I_{V} (A_{0})| ≤ |A||C I_{V} (A)|

Hence

|A| ≥ |A_{0}||C I_{U}(A_{0})C I_{V} (A)/C I_{V} (A)| = |A_{0}||C I_{U}(A_{0})/C I_{U}(A)|

and so A is a quadratic offender on U.

Lemma 2.6 Let L ◁ H, W be a quasi irreducible L-submodule of V and A quadratic on V. Set K = [O^{p}(L), A] and A_{0} = C I_{A}(K/C I_{K}(W)). Let O^{p}(L) = KC I_{O^{p}(L)}(W) and O_{p}(K/C I_{K}(W)) = 1. Then W^{A} = W or

|A/C I_{A}(W)| < |W/C I_{W} (K)| and |A||C I_{V} (A)| < |A_{0}||C I_{V} (A_{0})|

Proof: Set \overline{V} = V L/C I_{V} (O^{p}(L)). Then \overline{W} is an irreducible L-module. We have

[C I_{K}(\overline{W}), W] ≤ C I_{V} (O^{p}(L))

So C I_{K}(\overline{W}) acts quadratically on W and then C I_{K}(\overline{W}) ≤ O_{p}(K/C I_{K}(W)). By assumption (1) C I_{K}(\overline{W}) = C I_{K}(W)

This now shows [K, C I_{A}(\overline{W})] ≤ C I_{K}(W) and so C I_{A}(\overline{W}) ≤ A_{0}. By (1) we see [K, A_{0}C I_{K}(\overline{W})] ≤ A_{0}C I_{K}(\overline{W}). We have [\overline{W} , A_{0}C I_{K}(\overline{W})] = [\overline{W} , A_{0}], i. e. [\overline{W} , A_{0}] is invariant under KA. By quadratic action it is centralized by A and so [\overline{W} , A_{0}, K] = 1. As [K, A_{0}, \overline{W}] = 1, the 3-subgroup lemma implies 1 = [\overline{W} , K, A_{0}] = [\overline{W} , A_{0}]. Hence

(2) A_{0} = C I_{A}(\overline{W}).

Let a ∈ A with W^{a} ≠ W. Then also \overline{W}^{a} ≠ \overline{W} as [W C I_{V} (O^{p}(L)), K] = W. Let U = W W^{a}. Then we have

(∗) \overline{U} = \overline{W} ⨁[\overline{W} , a] = \overline{W}^{a} ⨁[\overline{W} , a] = \overline{W} ⨁ \overline{W}^{a}.

In particular N_{A}(\overline{W}) = C I_{A}(\overline{W}) as [\overline{W} , N_{A}(\overline{W})] ≤ \overline{W} ∩ \overline{W}^{a} Now (2) implies

(3) N_{A}(\overline{W}) = C I_{A}(\overline{W}) = C I_{A}(\overline{U}) = A_{0}

With (∗) we get

(4) C I\overline{U}(A) = [\overline{W} , a] and |\overline{U}/C I\overline{U}(A)| = |\overline{W} |.

Now let b ∈ A, then U ∩ U^{b} is a K-module. Hence either \overline{U} = \overline{U}^{b} or [\overline{W} , a] = \overline{U} ∩ \overline{U}^{b} In the second case [\overline{W} , a] is a K-module. The quadratic action of A now implies [\overline{W} , a, A] = 1 and then [\overline{W} , a, K] = 1. With (∗) we now get the contradiction [\overline{W} , K] = \overline{W} = \overline{W}^{a}. So we have \overline{U}^{A} = \overline{U}. Because of [UC I_{V} (O^{p}(L)), K] = U we get

(5) U^{A} = U.

As [K, A_{0}] is invariant under A,we see

[K, A_{0}] = [K, C I_{A}(\overline{U})] ≤ C I_{K}(U). Hence [U, A_{0}, K] = 1 = [A_{0}, K, U]. By the 3-subgroup lemma we see

(6) C I_{A}(U) = A_{0}.

By (5) \overline{U}^{A} = \overline{U} and according to (∗) there are at most |\overline{W} | + 1 many Aconjugates of \overline{W} in \overline{U}. So with (3), (4) and (6) we see

|A/A_{0}| = |A/N_{A}(\overline{W})| ≤ |\overline{W} | = |\overline{U}/C I\overline{U}(A)| ≤ |U/C I_{U}(A)| ≤ |C I_{V} (A_{0})/C I_{V} (A)| Now either |A||C I_{V} (A)| < |A_{0}||C I_{V} (A_{0})|or |A/N_{A}(\overline{W})| = |\overline{W} |. In the second case there are exactly |\overline{W} | many A-conjugates of \overline{W} in \overline{U}. Together with [\overline{W} , a] they form a partition of \overline{U}. As the set of A-conjugates of \overline{W} is invariant under K, also [\overline{W} , a] is invariant under K. But then [\overline{W} , a, K] = 1 and C I\overline{U}(K) ≠ 1, a contradiction to (∗).

Lemma 2.7 [Cher]. Let A be a quadratic subgroup of H and K be a component of H. Let [V, K] ≠ O. Then one of the following holds

(1) [K, A] ≤ K

(2) p = 2, A ≠ N_{A}(K) and |A/C I_{A}(K)| = 2

(3) p = 2, K = SL_{2}(2^{n}) and |A/N_{A}(K)| = 2

In (2) and (3) A is not an offender as an F -module.

Lemma 2.8 Let K be a component of H and A be a quadratic offender on V with [K, A] ≠ 1. Let W be a quasi irreducible K-submodule of V. Then W^{A} = W.

Proof: By (2.7) K^{A} = K. As K is quasisimple and irreducible on \overline{W}, we see O_{p}(K/C I_{K}(W)) = 1. Set in (2.6) L = K, H = KA and A_{0} = C I_{A}(K/C I_{K}(W)). Then we get either W^{A} = W or

|A||C I_{V} (A)| < |A_{0}||C I_{V} (A_{0})|

But the latter contradicts the fact that A is an offender.

Lemma 2.9 Let N with N/O_{p}(N) ≅ L_{n}(q), n > 2, q = 2^{f}. Let W ◁ N with [W′, O^{p}(N)] = 1. In W/Z(W) there are just natural modules involved. Then [O^{p}(N), W] ≤ Z(W).

Proof: The assertion follows from the fact that in an extraspecial group with V always also V^{∗} is involved, but W involves just natural modules no duals.

3 Reductions

Lemma 3.1 Let M be E-maximal, T ∈ Syl_{p}(M), L ∈ H^{∗}(M), N ∈ N (L).

(a) C I_{M}(Ω_{1}(Z(T))) = M ∩ C^{x} for some x ∈ C̃ with T ≤ S^{x}.

(b) E ≤ C I_{M}(Ω_{1}(Z(T)))

(c) O_{p}(⟨N, L⟩) = 1

(d) Set D = S_{L}(C̃). Then [D, N] ≤ N. If N is perfect, then [D, N] ≤ D.

(e) Let L_{0} ≤ L̃ ∈ H_{0}(M), L_{0} ≰ C̃ but L ∩ C̃ maximal in L_{0}. Then L_{0} ∈ H^{∗}(M) and N ∈ N (L_{0})

Proof: (a) This is just the Kieler Lemma.

(b) follows directly from (1)

(c) Suppose O_{p}(⟨N, L⟩) ≠ 1. Then ⟨N, L⟩ ≤ Y, Y a maximal p-local subgroup. So C I_{G}(O_{p}(Y)) ≤ O_{p}(Y) and then Y ∈ L_{G}.

We have |Y ∩ C̃|_{p} = |L ∩ C̃|_{p}, as L ∈ H^{∗}(M). Now N ≤ S_{Y} (C̃) but N ≰ S_{L}(C̃), contradicting the maximality of L.

(d) As D ≤ L ∩ C̃, the first assertion is clear. So let N be perfect, then the assertion follows with (2.4).

(e) Let T_{1} ∈ Syl_{p}(L ∩ C̃) and T_{1} ≤ T_{0} ∈ Syl_{p}(L_{0} ∩ C̃). Further let T_{0} ≤ T_{2} ∈ Syl_{p}(L̃ ∩ C̃). As L ∈ H^{∗}(M) and D = S_{L}(C̃) ≤ L̃, we see T_{2} = T_{1} and then T_{0} = T_{1} = T_{2}. As L̃ ∈ H_{0}(M) we get [E ∩ T_{2}, Z(T_{2})] = 1 and then [E, Z(T_{2})] = 1 which gives T_{2} ∈ Syl_{p}(L̃). This gives T_{0} ∈ Syl_{p}(L_{0}). Now O_{p}(L̃) ≤ O_{p}(L_{0}) and so L_{0} ∈ L_{G}. This gives L_{0} ∈ H^{∗}(M). As L_{0} ≰ C̃, we see L_{0} ∩ C̃ = L ∩ C̃ and so N ∈ N (L_{0}).

Lemma 3.2 Let M be E-maximal, L ∈ H^{∗}(M). Suppose (P U) does not hold and |M(E)| ≠ 1. Then L ∈ H_{0}(M) ≠ ø.

Proof: Let T_{1} ∈ Syl_{p}(L ∩ C̃). Suppose T_{1} ∉ Syl_{p}(L). By the choice of L and M_{T} ≤ N_{G}(Ω_{1}(Z(T_{1}))) we get N_{G}(Ω_{1}(Z(T_{1}))) ≤ C̃. Hence T_{1} ∈ Syl_{p}(L). As L does not satisfies (P U) there is some 1 ≠ X char O_{p}(L ∩ C̃) with X ◁ L. Set Y = N_{G}(X). Then M_{T} ≤ Y and L ≤ Y. So Y ∈ H(M). We have |Y |_{p} = |L ∩ C̃|_{p}, as L ∈ H^{∗}(M). This shows Q ≤ Y and so Q ≤ L ∩ C̃. Further as L ∈ H^{∗}(M) we have S_{Y} (C̃) = S_{L}(C̃) and so L ∈ H_{0}(M).

Lemma 3.3 Let L_{0} ∈ H^{∗}(M) and (P U) does not hold for L_{0}. If T_{1} ∈ Syl_{p}(L_{0} ∩ C̃), then T_{1} ∈ Syl_{p}(C̃). Further P_{L_{0}}(T_{1}) ≤ C ≤ C̃. If now N ≤ M for some N ∈ N (L_{0}), then (A) holds.

Proof: The first two assertions are clear. We have P = N(L_{0} ∩ C̃). Further P_{N(L_{0}∩C̃)}(T_{1}) = P_{L_{0}}(T_{1}) and T_{1} ∈ Syl_{p}(P) ∩ Syl_{p}(L_{0}), the assertion.

Hypothesis 3.4 L_{0} ∈ H^{∗}(M) and (P U) does not hold for L_{0}. Further N ∈ N (L_{0}) and (A) does not hold for N.

In particular by (3.3) we have N ≰ M for all N ∈ N (L_{0}), if (3.4) holds.

\underline{Notation:} Assume (3.4). Set B_{0} = L_{0} ∩C̃, T = S ∩M ≤ T_{0} ∈ Syl_{p}(B_{0}), Z_{0} = Ω_{1}(Z(T_{0})), N_{0} ∈ N (L_{0}), P_{0} = N_{0}B_{0}, D_{0} = S_{L_{0}}(C̃). Set

H_{0}(M, D_{0}) = {L ∈ H_{0}(M)|D_{0} = S_{L}(C̃)}.

We have shown L_{0} ∈ H_{0}(M).

Lemma 3.5 Assume (3.4). Then

(a) N_{G}(Z_{0}) ≤ C̃

(b) O_{p,p′}(C̃) ≤ M, Q = O_{p}(M)

(c) L_{0} ∈ H_{0}(M, D)

(d) If L̃ ∈ H_{0}(M) and Ñ ∈ N (L̃) with Ñ ≰ M. Then Ñ = [Ñ, O_{p}(L̃ ∩ C̃)] and Ñ = E_{p}(Ñ).

Proof: (a) As M ◁ ◁C̃, this is just the Kieler Lemma

(b) O^{p′}(C I_{O_{p,p′}(C̃)M})(Ω_{1}(Z(T Q))) = P_{O_{p,p′}(C̃)M}(T Q) = QP_{M}(T) ≤ L_{0}. The maximal choice of M implies O_{p,p′}(C̃) ≤ M. In particular Q ≤ M and so Q ≤ O_{p}(M). As M ◁ ◁C̃, we have O_{p}(M) ≤ O_{p}(C̃) = Q

(c) this is obvious

(d) Set B̃ = L̃ ∩ C̃, D̃ = SL̃(C̃). Then Ñ ≰ N_{G}(O_{p}(B̃)) by definition of H_{0}. Otherwise as Ñ ◁ ◁C̃ we have Ñ ≤ D̃ contradicting Ñ ∈ N (L̃). As D̃ ≤ B̃ we have O_{p}(O_{p}(B̃)D̃) = O_{p}(B̃). Hence Ñ ≰ N_{G}(O_{p}(B̃)D̃). This shows [Ñ, O_{p}(B̃)] ≰ D̃. We have [Ñ, O_{p}(B̃)] ⊴ Ñ. Hence the minimal choice of Ñ gives Ñ = [Ñ, O_{p}(B̃)]. Let Ñ_{1} ◁ Ñ with Ñ_{1} is invariant under B̃. Then [Ñ_{1}, O_{p}(B̃)] ≤ Ñ_{1} ∩ O_{p}(B̃) ≤ O_{p}(Ñ_{1}) ≤ Q, as Ñ_{1} ≤ D̃ ≤ B̃. Now we see [Ñ_{1}, Ñ] ≤ Q, as [Ñ, O_{p}(B̃)] = Ñ. This shows Ñ ≤ F_{p}^{∗}(C̃). In particular Ñ = O_{p_{′}p′}(Ñ)E_{p}(Ñ). We have O_{p_{′}p′}(C̃) ≤ M and Ñ ≰ M. Hence E_{p}(Ñ) ≰ M. By the choice of M we have D̃ ≤ M and so E_{p}(Ñ) = Ñ by the choice of Ñ.

Lemma 3.6 Assume (3.4). Let K be a p-component of C̃. Then K ≤ M iff any point stabilizer in K ∩ L_{0} is a point stabilizer in K.

Proof: (a) Let K ≤ M. By the Kieler Lemma we get O^{p′}(C I_{K}(Ω_{1}(Z(S))) = O^{p′}(C I_{K}(Ω_{1}(Z(T)))) ≤ O^{p′}(C I_{M}(Ω_{1}(Z(T)))) ≤ L_{0} as T ≤ M_{T} ≤ L_{0}. Hence O^{p′}(C I_{K∩L_{0}}(Ω_{1}(Z(T))) = O^{p}(C I_{K}(Ω_{1}(Z(T))). As all point stabilizers are conjugate we have that any point stabilizer of K ∩ L_{0} is one of K.

(b) Let P be a point stabilizer of K ∩ L_{0} which is also one of K. Set R = ⟨O^{p′}(C I_{M}(Ω_{1}(Z(T))), P ⟩ ≤ L_{0}. Assume K ≰ M. Then [K, M] ≤ Q. By (2.2) there is some point stabilizer U in KM with U ≤ R ≤ L_{0}. As EU ≤ L_{0}, we have KM ∈ Ω. But this contradicts the maximal choice of M.

Lemma 3.7 Assume (3.4), then M = D.

Proof: Let X be the group generated by the p-components K of C̃ with K ≰ M. Let C_{0} = C IC̃(XQ/Q). Then M ≤ C_{0}. By (3.6) we get that X is invariant under B_{0} = L_{0}∩C̃ as this acts on the point stabilizers. In particular C_{0} is invariant under B_{0}. Set U = ⟨M^{B_{0}C_{0}}⟩. Then U is invariant under B_{0}. Further U ◁ ◁C̃ and U ≤ C_{0}. Suppose U ≰ D. Choose Ñ ∈ N (L_{0}) with Ñ ≤ U. By (3.5)(d) we have Ñ = E_{p}(Ñ). But Ñ ≤ C_{0} and so Ñ ≤ M, a contradiction. Hence U ≤ D. Then M ≤ D ≤ L_{0}. The maximal choice of M now implies M = D.

Lemma 3.8 Assume (3.4). Let K be a p-component of C̃ with K ≰ D. Then for all 1 ≠ x ∈ C I_{S}((S ∩ K)T) we have M(C I_{G}(x)) = {C̃}.

Proof: Set U = KD. Then C I_{S}((S ∩ K)T) = Z(S ∩ U), as Q ≤ T. We have E ≤ D ∩C ≤ U ∩C = C I_{U}(Ω_{1}(Z(S ∩U))) ≤ C I_{G}(x) by the Kieler lemma. Now assume C I_{G}(x) ≤ L̃ ≰ C̃, L̃ ∈ M(C I_{G}(x)). As E ≤ U, M(EO^{p′}(C I_{U}((S ∩ U)T)) ∋ L̃, we get by the maximality of M that D = M = U. Hence K ≤ D, a contradiction.

\underline{Notation.} From now on we fix the following: L ∈ H_{0}(M, D), N ∈ N (L). Now we have (3.6)(d). As M = D we also have N ≰ M. Set P = NB, B = L ∩ C̃. Denote by K(X) the set of p-components of X.

I (C̃) = {x ∈ C̃|M(C I_{G}(x)) = {C̃}}.

For K ∈ K(C̃), K ≰ D let V(K) be the set of quasi irreducible modules V ≤ Y_{K} with [V, T_{K}] = 1 for some T_{K} ∈ Syl_{p}(C IC̃(KQ/Q)) with T_{K} ∩ C I_{P} (KQ/Q) ∈ Syl_{p}(C I_{P} (KQ/Q)). As N = E_{p}(N) we have that K(C̃) ≠ ø.

Lemma 3.9 (a) V(K) ≠ ø

(b) One of the following holds

(1) {K} = K(C̃) \ K(D), K = N = N_{0}, N (L) = {N}

(2) V^{#} ≤ I(C̃) for all V ∈ V(K). If K ∈ K(N), then V ≤ Y_{P} for all V ∈ V(K).

Proof: Set Y_{0} = C I_{YC̃}(T_{K}), C_{0} = C IC̃(KQ/Q). Without loss we may assume T ≤ T_{K}. As K ≰ E and so [K, YC̃] ≠ 1, we get with P × Q-Lemma that [C I_{YC̃}(T_{K}), K] ≠ 1. Now choose V quasi irreducible with V ≤ [C I_{YC̃}(T_{K}), K] and so we have (a). If {K} = K(C̃) \ K(D) we have (b)(1). So let K̃ ∈ K(C̃) \ K(D), K̃ ≠ K. Then K̃ ≤ C_{0} and T_{K} ∩ K̃T ∈ Syl_{p}(K̃T). By (3.8) we have V^{#} ⊆ Y_{0}^{#} ⊆ C I_{YC̃}(T_{K} ∩ K̃T)^{#} ⊆ I(C̃). Let now K ≤ N. Set P_{0} = ⟨K, S_{0}⟩ where T_{K} ∩P ≤ S_{0} ∈ Syl_{p}(P). Set further Y_{1} = ⟨Y_{0}^{P_{0}}⟩. Let finally Q_{0}/C I_{P_{0}}(Y_{1}) = O_{p}(P_{0}/C I_{P_{0}}(Y_{1})). Then [Y_{1}, K] ≠ 1. As K is a p-component we see [K, Q_{0}] ≤ Q. Hence Q_{0} ≤ C_{0}. Then Q_{0} ∩ S_{0} ≤ C_{0} ∩ S_{0} = P ∩ T_{K} as T_{K} ∈ Syl_{p}(C_{0}). This implies Q_{0} = C I_{P_{0}}(Y_{1})(Q_{0} ∩ S_{0}) = C I_{P_{0}}(Y_{1})(Q_{0} ∩ T_{K}) ≤ C I_{P_{0}}(Y_{0}). Hence Q_{0} = C I_{P_{0}}(Y_{1}) and so Y_{1} ≤ Y_{P_{0}} ≤ Y_{P}.

Lemma 3.10 If N_{1} ⊴ P then either N ≤ N_{1} or N_{1} ≤ B.

Proof: Let N_{1} be a minimal counter example. Then N ≰ N_{1}. Let D_{0} = Core_{P} (B). We have that N_{1}/N_{1} ∩ D_{0} is a minimal normal subgroup in P/N_{1} ∩ D_{0}. In particular N_{1} cannot permute the components of N. The structure of C̃ now implies [N, N_{1}] ≤ Q. Hence [N_{1}, N] ≤ N_{1} ∩ D_{0}. So [N_{1}, O_{p}(B)](N_{1} ∩ D_{0}) ⊴ ⟨N, B⟩ = P. If [N_{1}, O_{p}(B)] ≤ D_{0} so [N_{1}, O_{p}(B)] ≤ O_{p}(B) as [N_{1}, D_{0}O_{p}(B)] ≤ D_{0}O_{p}(B). Hence N_{1} ≤ N_{P} (O_{p}(B)). As L ∈ H_{0} we get N_{P} (O_{p}(B)) ≤ L. In particular N_{1} ≤ B. Let [N_{1}, O_{p}(B)](N_{1} ∩ D_{0}) = N_{1}. As NO_{p}(B) ◁ P we get [N_{1}, O_{p}(B)] ≤ NO_{p}(B) and then [[N_{1}, O_{p}(B)]N/N, O_{p}(B)N/N] = 1. So [N_{1}, O_{p}(B)] ≤ N and then [N_{1}, O_{p}(B)] ≤ Z(N/N ∩Q). In particular [N_{1}, O_{p}(B)](N ∩Q)◁◁C̃. Now B ≤ N([N_{1}, O_{p}(B)]) and so by choice of N we get [N_{1}, O_{p}(B)] ≤ B.

Proposition 3.11 (P, L) is an amalgam will C I_{P} (Y_{P})C I_{L}(Y_{L}) ≤ B.

Proof: As N ≰ E we have [N, Y_{P}] ≠ 1. By (3.10) now C I_{P} (Y_{P}) ≤ B. As L ∩ C̃ contains a Sylow p-subgroup of L and Q ≤ L, there is some x ∈ Y_{L} with C I_{G}(x) ≤ C̃. So C I_{L}(Y_{L}) ≤ L ∩ C̃. We have ⟨N, M⟩ = ⟨N, D⟩ ≤ ⟨P, L⟩. If ⟨P, L⟩ ≤ R ∈ M, then R ≰ C̃. But ⟨N, M⟩ > M contradicts the Emaximality of M. Hence O_{p}(⟨P, L⟩) = 1.

Lemma 3.12 Let K ∈ K(N), V ∈ V(K) and A ≤ N_{P} (K) be a p-group with A ≤ N_{P} ((S ∩ K)T_{K}). If C I_{V} (S ∩ K) ≠ C I_{V} (K), then either A ≤ N_{P} (V) or [V, A] ∩ I(C̃) ≠ ø.

Proof: We have [C I_{V} ((S ∩ K)T_{K}), A] ≤ C I_{Y_{P}} ((S ∩ K)T_{K}). If [C I_{V} ((S ∩ K)T_{K}), A] ≠ 1, the assertion follows with (3.8). If [C I_{V} ((S ∩ K)T_{K}), A] = 1, then as C I_{V} ((S ∩ K)T_{K}) ≰ C I_{V} (K), we see [V, A] ≤ V.

4 The amalgam

According to (3.11) (P, L) is an amalgam. This we now are going to investigate. We introduce the notation as G_{α}, b and so on. If α is of type P, then we denote by N_{α} the group corresponding to N. Further we have C̃_{α}, K(N_{α}) and I(C̃_{α}). If (α, α′) is a critical pair then [Y_{α}, Y_{α′}] ≠ 1, by (3.11). We choose notation such that Y_{α′} is an offender on Y_{α}. Finally we set V_{δ} = ⟨Y_{λ}|λ ∈ Δ(δ)⟩Y_{δ}.

Lemma 4.1 Let α be of type P. Then there is some 1 ≠ A ≤ Y_{α′} and K ∈ K(N_{α}) such that for all V ∈ V(K) we have Y_{α′} ≤ N(K), Y_{α′} ≤ N(V) and further |V/C I_{V} (A)| ≤ |A/C I_{A}(V)| ≠ 1.

Proof: Let A be a minimal offender in Y_{α′}. By (2.7) we have [A, K] ≤ K. Now choose K such that [A, K] ≠ 1 for some minimal offender A. By (2.8) we have that A ≤ N(V) for all V ∈ V(K). As Y_{α′} acts quadratically, then also [Y_{α′}, V] ≤ V and Y_{α′} ≤ N(K). By (2.5) A is an offender on V.

Lemma 4.2 Let b > 1. Then {N} = K(C̃) \ K(D). Further for all K ∈ K(N) and V ∈ V(K) we have V^{#} ⊈ I(C̃).

Proof: By (3.9) it suffices to prove the second assertion. For this we may without loss assume that P ∩ L is maximal in L. Let K ∈ K(N). Suppose first G_{α} = P. Let A, K, V be as in (4.1) and V^{#} ⊆ I(C̃). Assume further b to be even. Then α′ is also of type P. Now choose K̃ ∈ K(N_{α′}) with [Ṽ , K̃] ≤ Q_{α′} for Ṽ ∈ V(K̃). Notice [Y_{α′}, V] ≠ 1. As we may assume that Ṽ is conjugate to V we have Ṽ^{#} ⊆ I(C̃_{α′}). Let V ≤ N_{G}(Ṽ). Then [V, Ṽ] = R ≠ 1. This shows C̃_{α} = C̃_{α′}, but Y_{α} ≤ O_{p}(C̃_{α′}) ≤ O_{p}(G_{α′}) a contradiction. Hence V ≰ N_{G}(Ṽ). As Ṽ ≤ N_{G}(V), we get |[Ṽ , V]| < |V |. For x ∈ V \ Q_{α′} we have |[Ṽ , x]| ≥ |Ṽ /C IṼ (K̃)|, i. e. Ṽ induces on V/C I_{V} (K) transvections. But |Ṽ : Ṽ ∩ Q_{α}| = |V/C I_{V} (K)|, a contradiction. So b is odd and so α′ is of type L. Choose α′ + 1 ∈ Δ(α′) with V ≰ C̃_{α′+1}. We have [[V, Y_{α′}], Y_{α′+1}] = 1 and so Y_{α′+1} ≤ G_{α+1} ∩ C̃_{α} ≤ G_{α}. Let V_{α′+1} ≤ G_{α′+1} be a conjugate of V. If v ∈ V, v ∉ C̃_{α′+1} them C I_{V_{α′+1}}(v) = 1. Hence |[v, V_{α′+1}]| = |V_{α′+1}| = |V |. In particular V_{α′+1} cannot act on V. But [V_{α′+1}, Y_{α′}] = 1 and V_{α′+1}Y_{α′} act quadratically on Y_{α}. If V_{α′+1}Y_{α′} does not normalizes K, we have by (2.7) that K ≅ L_{2}(q) and V is an orthogonal module. But A was an offender, a contradiction. Hence [V_{α′+1}, K] ≤ K. Now [V_{α′+1}, [Y_{α′}, V]] = 1 and so V_{α′+1} ≤ N_{G}(V), a contradiction again. So we have G_{α} = L. Let first α′ be of type P. We have that Y_{α′} is an offender on Y_{α} and Y_{α′} ≤ G_{α} ∩ G_{α+1} which is conjugate to B. As Y_{α′} ≤ O_{p}(G_{α+1}) we may choose notation such that Y_{α′} ≤ O_{p}(B). This gives Y_{α′} ≤ O_{p}(O^{p′}(C I_{N}(Ω_{1}(Z(S))))). By [Stell] we get |Y_{α′}/Y_{α′} ∩ Q_{α}| = |Y_{α}/Y_{α} ∩ Q_{α′}|. But then we are in the case just handled. So we have α′ is of type L too. Let W be a quasi irreducible module for a component of G_{α′} on which Y_{α} acts nontrivialy, i. e. induces an offender. Let first

[Y_{α−1}, W] ≰ [Y_{α}, W][Y_{α−1}, W ∩ Q_{α}].

For the component and the module W we now have the following possibilities SL_{n}(q), Sp(2n, q), G_{2}(2^{m}), Σ_{n} and W the natural module (here we also include SL_{2}(3) and Σ_{3} in the solvable case). But as Y_{α−1}Y_{α} is in the O_{p} of the point stabilizer this is just possible for Sp(2n, q) on the natural module. Assume first [W ∩ Q_{α}, Y_{α−1}] ≠ 1. Then W ∩ Q_{α} is a strong dual offender on Y_{α−1}. By (2.7) we have that W ∩ Q_{α} normalizes any K. As W ∩ Q_{α} acts quadratically (2.8) implies [W ∩ Q_{α}, V] ≤ V. But then Y_{α′} ≤ C̃_{α−1}, a contradiction. Hence we have that [W ∩ Q_{α}, Y_{α−1}] = 1. Set R_{α} = O^{p}(⟨Q_{α}_{−1}^{Gx}⟩). Assume C I_{W} (R_{α′}) ≤ [Y_{α}, W]. Then [W, Y_{α−1}] ≤ [W, Y_{α}]. Hence we have C I_{W} (R_{α′}) ≰ [Y_{α}, W]. Again the component is Sp(2n, q) on the natural module and further q = 2. In particular |Y_{α}/C I_{Y_{α}}(W)| = 2. As [C I_{W} (Y_{α}), Y_{α−1}] = 1 we see Y_{α−1}C I(W) = Y_{α}C I(W). In particular [Y_{α−1}, W] ≤ [Y_{α}, W], a contradiction. So in any case we have W ≤ N(Y_{α−1}Y_{α}). Now choose α − 1 such that ⟨G_{α−1,α}, W ⟩ = G_{α}, which is possible by the minimal choice of L. Then Y_{α−1}Y_{α} ◁ G_{α}. Now also U = [Y_{α−1}, Q_{α}] ◁ G_{α} and [U, Q_{α−1}] = 1. Hence [U, R_{α}] = 1. From [Y_{α−1}Y_{α}, W] ≤ [Y_{α}, W][Y_{α−1}, W ∩ Q_{α}] we get [Y_{α−1}, R_{α}] ≤ U[Y_{α}, R_{α}]. As Y_{α−1}Y_{α} ◁ G_{α}, we have Y_{α−1}Y_{α} = Y_{α}Y_{α+1} and so C I_{W} (Y_{α}) = C I_{W} (Y_{α−1}Y_{α}). This gives [W ∩ Q_{α}, Y_{α−1}] = 1. Further by R_{α} = O^{p}(R_{α}) we see

[Y_{α−1}, Y_{α}, R_{α}] = [Y_{α−1}, Y_{α}, R_{α}, R_{α}] ≤ [U[Y_{α}, R_{α}], R_{α}] = [Y_{α}, R_{α}, R_{α}] = [Y_{α}, R_{α}].

Hence [Y_{α−1}, R_{α}] ≤ [Y_{α}, R_{α}]. Now choose V as before. We have |V | > 2, V ≤ Y_{α−1} but V ∩ Q_{α′} = 1. As Y_{α}C I(W) ≥ Y_{α−1}C I(W), we have |Y_{α} : C I_{Y_{α}}(W)| > 2. If x ∈ W with [x, Y_{α}] ≠ 1, then x cannot centralize any element in V^{#}. Hence |V | ≤ q and as component we have SL_{n}(q), G_{2}(q) or Sp(2n, q), q > 2. Set A = C I_{Y_{α−1}Y_{α}}(R_{α}). Suppose that Y_{α−1}[Y_{α}, R_{α}] would be a nonsplit extension. As q > 2, we have the component Sp(2n, q). But then for a transvection group in Y_{α′} the centralizer is contained in [Y_{α}, R_{α}]C I(R_{α}), a contradiction. Hence the extension is split and Y_{α−1}Y_{α} = A[Y_{α}, R_{α}]. Let δ ∈ Δ(α − 1) and W_{i} the quasi irreducible modules for components K_{i} of G_{δ} which are conjugate to W. We define a set A(Y_{δ}) as follows. Suppose C I_{W_{i}}(K_{i}) = 1, then X_{i} = [C I_{W_{i}}(Q_{α−1,δ}), R_{δ} ∩ G_{α−1}] are the elements of A. Suppose |C I_{W_{i}}(K_{i})| = q so additionally C I_{W_{i}}(K_{i}) is in A. If the component is G_{2}(q), then we just have C I_{W_{i}}(K_{i}) in A. If 1 ≠ |C I_{W_{i}}(K_{i})| < q, then A consists of the centers. To make it clear. If C I_{W_{i}}(K_{i}) ≠ 1, then as q > 2, we have K_{i} ≅ Sp(2n, q) or G_{2}(q). In this case the elements in the subgroups contained in A are exactly the non commutators in C I_{W_{i}}(Q_{α−1,δ}). We next are going to prove b = 2 and [Q′_{α}, R_{α}] = 1. We have [O_{p}(R_{α}), Y_{α}] = 1 and also [O_{p}(R_{α}), Y_{α−1}] = 1. In particular O_{p}(R_{α}) ≤ Q_{α−1} ∩ Q_{α}. As Q_{α−1} ∩Q_{α} = C I_{Q_{α}}(Y_{α−1}Y_{α}) we see Q_{α−1} ∩Q_{α} ⊴⟨Q_{α}_{−1}^{G_{α}}⟩. So modulo Q_{α−1} ∩Q_{α} we have a direct product. Further no nontrivial characteristic subgroup of Q_{α−1} is normal in ⟨Q_{α}_{−1}^{G_{α}}⟩, and Q_{α−1} is normal in the point stabilizer of ⟨Q_{α}_{−1}^{G_{α}}⟩. Hence by [MF] we get b = 2 and [Q′_{α}, R_{α}] = 1. Let now X ∈ A. Then there are δ_{1}, δ_{2} ∈ Δ(α − 1) with X ≤ [Y_{δ_{1}}, Y_{δ_{2}}]. In particular there are quasi irreducible modules I_{1} ≤ Y_{δ_{1}}, I_{2} ≤ Y_{δ_{2}}, with X ≤ [I_{1}, I_{2}]. So A(Y_{δ_{1}}) = A(Y_{δ_{2}}). Let now [I_{1}, Y_{α}] ≠ 1. Then [I_{1}, I] ≠ 1 for some I conjugate to I_{1}. In particular X ≤ [I_{1}, I]. If we have a central extension then the elements in the subgroup of A are just the non commutators. As the Cartan group acts on X, we see X ∈ A(Y_{α}). In particular [R_{α} ∩ G_{α−1}, X] ≤ X. If the center is trivial, so X is uniquely determined and then again [R_{α} ∩ G_{α−1}, X] ≤ X. If [I_{1}, Y_{α}] = 1 = [I_{2}, Y_{α}]. Then [I_{1}, I_{2}] ≤ Q′_{α} and so [X, R_{α}] = 1. So again [R_{α} ∩ G_{α−1}, X] ≤ X. Hence we have that any X ∈ A is invariant under R_{α} ∩ G_{α−1}. As G_{α−1} acts on A, we see that any X ∈ A is invariant under [G_{α−1}, R_{α} ∩ G_{α−1}]. But as N_{α−1} ≤ [G_{α−1}, R_{α} ∩ G_{α−1}], we see that any X ∈ A is invariant under N_{α−1}. If there is a center, i. e. X is in the center, then [Q_{α,α−1}, X] = 1. As [N_{α−1}, Q_{α,α−1}] = N_{α−1}, we see [N_{α−1}, X] = 1. But then X ≤ O_{p}(⟨P, L⟩), a contradiction. So there is no center. As Y_{α−1}Y_{α} = A[Y_{α}, R_{α}] we see that V/X is centralized by the Cartan subgroup Z_{c}. Hence [K, Z_{c}] = 1. Now V ≤ C I_{Y_{α}Y_{α−1}}(Z_{c}) ≤ C I_{Y_{α}Y_{α−1}}(R_{α}). This shows V ◁ ⟨N_{α−1}, R_{α}⟩ = ⟨P, L⟩, a contradiction.

From now on we choose L minimal.

Lemma 4.3 If b > 1 be odd, then α is of type P and N_{α} ≅ A_{n} or Ω^{±}(2n, 2) and Y_{α}/C I_{Y_{α}}(N_{α}) is the natural module.

Proof: As there are no over-offender in L we see that α has to be of type P. Let V ∈ V(N_{α}) set F_{α} = N_{α}Y_{α′}Q_{α}/Q_{α} and R = [V, Y_{α′}] ≠ 1. If R ∩ I(C̃_{α}) = ø, then by (3.8) R contains no p-central elements. The list of F -modules [GBSM] now yields the assertion. Hence we may assume R ∩ I(C̃_{α}) ≠ ø. If all elements in (V/C I_{V} (N_{α}))^{#} are conjugate, we see C I_{V} (S ∩ N_{α}) = C I_{V} (N_{α}). Application of [GBSM] yields K ≅ A_{n} and we have the assertion or K ≅ L_{3}(2) and |V | = 16. So we have to consider this case. As b > 1 and R contains uniqueness elements, we see [Y_{α′+1}, R] = 1, i. e. Y_{α′+1} ≤ G_{α}. Now [Y_{α′+1}, V] = [Y_{α′}, V] and so Y_{α′}Y_{α′+1}⊴⟨Y_{α}^{G_{α′}}⟩ = H. So Y_{α′}Y_{α′+1} = Y_{α′}Y_{α′−1}. This gives [Y_{α′−1}, Q_{α′}]⊴H. As [Y_{α}, Y_{α′−1}] = 1, we see [[Y_{α′−1}, Q_{α′}], O^{2}(H)] = 1. But then O^{2}(H) ≤ C̃_{α′−1}, a contradiction. So we have C I_{V} (S ∩ N_{α}) ≰ C I_{V} (N_{α}) and in particular N_{α} does not act transitively on (V/C I_{V} (N_{α}))^{#}. Suppose that V/C I_{V} (N_{α}) carries a nontrivial form. Then the singular vectors are the uniqueness elements. Let V_{α′+1} conjugate to V. Let further x ∈ V \ C̃_{α′+1}. Then as V_{α′+1} ≤ P we get |C I_{V_{α′+1}}(x)| ≤ q^{2}. Hence if |V_{α′+1}| = q^{n}, then |[x, V_{α′+1}]| ≥ q^{n−2}. But this is impossible for spin modules or half spin modules. For orthogonal or unitary modules we get n − 2 ≤ ^{n}/_{2} and so n ≤ 4. This shows K ≅ Ω^{−}(4, q) ≅ L_{2}(q^{2}). As we have an F -module, we get q = 2 and so |Y_{α′}C I(V)/C I(V)| = 2 = |V/C I_{V} (Y_{α′})|, in particular in Y_{α} there is just one nontrivial module. Let next N_{α} = SL_{n}(q) and V/C I_{V} (N_{α}) be the exterior square. Then there is a subgroup W in V_{α′+1} with |W | = q^{n−1} which just consists of uniqueness elements. Hence W ∩ Q_{α} = 1. In particular W Q_{α} = Y_{α′}Q_{α}. Now |[V, t]| = q^{n−2} for all t ∈ W^{♯}. Hence x ∈ V \ Q_{α′} centralizes some t ∈ W, a contradiction. So we are left with N_{α} ≅ 3A_{6}, and V be the 6-dimensional module. As before we see Y_{α′+1}Y_{α′} ◁ ⟨Y_{α}^{G_{α′}}⟩ = H. Then also [Y_{α′−1}, Q_{α′}] ◁ H and [O^{2}(H), [Y_{α′−1}, Q_{α′}]] = 1. This gives the contradiction O^{2}(H) ≤ C̃_{α′−1}.

Lemma 4.4 Let b > 1. Then there is some critical pair (α, α′) such that one of the following holds

(1) The structure of P is as in (4.3) (b might be even)

(2) K ≅ L_{3}(2), α is of type P, and K acts on V, |V | = 2^{4}. Further b is even

(3) α is of type L and Y_{α−1}Y_{α} ◁ G_{α}

Proof: Assume that for all critical pairs (α, α′) we always have that α is of type L and b is even. Then Y_{α−1} ≤ G_{α′}. According to [Stell] if K ≅ S̸p(2n, q), we get [Y_{α−1}, Y_{α′}] = [Y_{α}, Y_{α′}] and so Y_{α−1}Y_{α} ◁ G_{α} which is (3). So let K ≅ Sp(2n, q) on the natural module. If [Y_{α−1}, Y_{α′}] ≰ [Y_{α}, Y_{α′}] then [Y_{α′} ∩ Q_{α}, Y_{α}] ≠ 1. Now Y_{α′} ∩ Q_{α} is a strong dual offender on Y_{α−1}. Further [Y_{α′} ∩ Q_{α}, Y_{α−1}] contains no uniqueness elements so as in (4.3) we get (1), (2) or (3). Hence we may assume that α is of type P. If b is odd (1) follows with (4.3). So assume b to be even. Set V = V_{α} and V_{α′} be the corresponding module with [V_{α}, V_{α′}] ≠ 1. As C̃_{α} ≠ C̃_{α′}, there is no x ∈ [V_{α}, V_{α′}] which is a uniqueness element on both sides. In particular N_{α} cannot act transitively on (V/C I_{V} (N_{α}))^{#}. If there is no uniqueness element in V \ C I_{V} (N_{α}), then we get (1) or (2). If K_{α} = A_{n}, then there is just one nontrivial module, the natural one, and again we have (1). Let K_{α} ≅ Ω^{±}(2n, q) on the natural module. As V_{α′} acts quadratically we get that [V_{α}, V_{α′}] is a singular space. So |V_{α′} : C I_{V_{α′}}(V_{α})| = 2 and again we just have one module, which is (1). If K ≅ U_{n}(q) on the natural module. Then [V_{α}, V_{α′}] does not contain vectors which are isotrop in G_{α} and G_{α′} as well. Hence |[V_{α}, V_{α′}]| ≤ q^{2}. But we have an F -module offender V_{α′}, a contradiction. Let K ≅ Sp(6, q) and V be the spin module. Then [V_{α}, V_{α′}] is totally isotrop on both sides, a contradiction. Let K ≅ Ω^{+}(10, q) on the half spin module. Now choose i ∈ V_{α′}, j ∈ V_{α} such that i, j are 2-central in N_{α′} and N_{α}, respectively. Then [i, j] is isotrop in both modules, a contradiction. Let next K ≅ SL_{n}(q) and V/C I_{V} (N_{α}) be the exterior square. But then for t ∈ V_{α′} all elements in [V_{α}, t] are uniqueness elements. Let finally K ≅ 3A_{6} and |V | = 2^{6}. By [GBSM] exactly the commutators are the uniqueness elements. This proves the lemma.

Lemma 4.5 Let b > 1 and Y_{α}Y_{α−1} ⋬ G_{α} if α is of type L. Then K ≅ L̸_{3}(2).

Proof: By (4.4) then b is even. Further α is of type P. Then [V_{α}, V_{α′}] contains a uniqueness element in V_{α}(|V_{α}| = 16). So Y_{α′+1} ≤ G_{α+1} ∩C̃_{α} ≤ G_{α}. Then [Y_{α′+1}, V_{α}] = [V_{α}, V_{α′}] and so Y_{α′}Y_{α′+1} ◁ G_{α′} and then also Q_{α} ∩ Q_{α−1} ◁ G_{α}. Now choose α − 2 ∈ Δ^{2}(α) with N_{α} ≰ N(Y_{α}Y_{α−2}). Set U = {Y_{α′}^{g}|g ∈ G_{α}, Y_{α′}^{g} ≰ N(Y_{α}Y_{α−2})}. If x ∈ N(Y_{α}Y_{α−2}) then x acts on ⟨U⟩. Hence ⟨U⟩ ⊴ ⟨U⟩N(Y_{α}Y_{α−2}). As ⟨U⟩N(Y_{α}Y_{α−2}) contains ⟨Y_{α′}^{g}|g ∈ G_{α}⟩ ≥ N_{α}, we see that N_{α} ≤ ⟨(Y_{α′}^{g})^{N_{α}}⟩ ≤ ⟨U⟩. Now let Y_{α′}^{g} ∈ U. Then [Y_{α−2}, Y_{α′−2}^{g}] ≠ 1, as otherwise Y_{α−2} ≤ Q_{α′−2}^{g} ∩ Q_{α′−3}^{g} = Q_{α′−1}^{g} ∩ Q_{α′−2}^{g} ≤ Q_{α′−1}^{g} and then [Y_{α−2}, Y_{α′}^{g}] ≤ [Y_{α′}, Y_{α′}^{g}] ≤ Y_{α}, a contradiction to Y_{α′}^{g} ≰ N(Y_{α−2}Y_{α}). Hence [Y_{α−2}, Y_{α′−2}^{g}] ≠ 1. Set A_{α,α+1} = [Y_{α}, A_{α+1}], where A_{α+1} is an offender in Q_{α+1}. We have A_{α−2,α−1} = [Y_{α−2}, Y_{α′−2}^{g}]. Assume first b > 2. Then [A_{α−2,α−1}, Y_{α′}^{g}] = 1 and so [A_{α−2,α−1}, N_{α}] = 1. Choose α − 4 ∈ α^{N_{α−2}} with A_{α−4,a−3} = [Y_{α−4}, Y_{α′−4}]. If b > 4, then [A_{α−4,α−3}, Y_{α′}] = 1 and as [N_{α−2}, A_{α−4,α−3}] = 1 we get A_{α−4,α−3} = A_{α,α−1}. Hence [A_{α,α−1}, Y_{α′}] = 1. As A_{α,α−1} ≤ [Y_{α}, Q_{α−1}] we may choose α − 1 such that [Y_{α′}, A_{α,α−1}] ≠ 1, a contradiction. So b = 4. Set W_{α} = ⟨Y_{α−2}^{g}|g ∈ G_{α}⟩. Then W_{α} ≤ G_{α+2} and |W_{α} : W_{α} ∩ Q_{α+2}| ≤ 4. As Q_{α+1} ∩ Q_{α+2} ≤ Q_{α+3} we have W_{α} ∩ Q_{α+2} ≤ G_{α′}. So |C I_{W_{α}/Y_{α}}(Y_{α′})| ≤ 4. Hence just natural modules are in W_{α}. As [Y_{α+2}, Y_{α−2}] = A_{α−2,α−1} and [N_{α}, A_{α−2,α−1}] = 1, we see [Wα^{′}, N_{α}] = 1. By (2.9) we get [W_{α}, O^{p}(N_{α})] ≤ Z(W_{α}). This implies W_{α} = V_{α−1}Z(W_{α}) and so W_{α} is abelian, a contradiction.

So we are left with b = 2. Then [Q_{α} ∩ Q_{α+1}, Y_{α′}] ≤ [Y_{α}, Y_{α′}] ≤ Y_{α}, as Y_{α} ≤ Z(Q_{α} ∩ Q_{α+1}). From Q_{α} ∩ Q_{α+1} ⊴ G_{α} we get [Q_{α} ∩ Q_{α+1}, N_{α}] ≤ Y_{α} and so [O^{2}(N_{α}), Q_{α}] ≤ Y_{α}.

Set D_{α} = C I_{Q_{α}∩Q_{α+1}}(O^{2}(N_{α})). Then Q_{α} ∩ Q_{α+1} = D_{α}Y_{α}. It D_{α} ≰ Q_{α′} then D_{α} ≥ [D_{α}, Y_{α′}] = [Y_{α}, Y_{α′}]. But this contradicts [D_{α}, O^{2}(N_{α})] = 1. Hence D_{α} ≤ Q_{α′}. Then Q_{α} ∩ Q_{α+1} = Y_{α}(Q_{α} ∩ Q_{α+1} ∩ Q_{α′}). Now ϕ(Q_{α} ∩ Q_{α+1}) = ϕ(Q_{α}∩Q_{α+1}∩Q_{α′}) = ϕ(Q_{α+1}∩Q_{α′}). Then ϕ(Q_{α}∩Q_{α+1})◁⟨G_{α}, G_{α′}⟩ ≥ G_{α+1}, which implies ϕ(Q_{α} ∩ Q_{α+1}) = 1. Set W_{α+1} = ⟨Y_{α}^{g}|g ∈ G_{α+1}⟩. Then [W_{α+1}∩Q_{α}, N_{α}] ≤ Y_{α} ≤ W_{α+1}∩Q_{α}. Hence W_{α+1}∩Q_{α}◁N_{α}G_{α,α+1} = G_{α}. We know N_{G_{α+1}}(W_{α+1}∩Q_{α}) = G_{α,α+1}. We have W_{α+1} = (W_{α+1} ∩ Q_{α})Y_{α′}. Hence [W_{α+1}, Y_{α′}] = [Y_{α}, Y_{α′}] and so ϕ(W_{α+1}) = [Y_{α}, Y_{α′}]. Now W_{α+1} ∩ Q_{α} and W_{α+1} ∩ Q_{α′} are the only maximal elementary abelian subgroups in W_{α+1}. This gives |G_{α+1} : N_{G_{α+1}}(W_{α+1} ∩ Q_{α})| ≤ 2. But G_{α,α+1} contains a Sylow 2-subgroup of G_{α+1}, which shows W_{α+1} ∩Q_{α} ◁G_{α+1}, a contradiction.

Lemma 4.6 Let b > 1 be even. If K ≅ Ω^{±}(2n, 2) not Ω^{−}(4, 2) or Ω^{+}(6, 2) and Y_{P} Y_{L} ⋬ L, then there is a critical pair (α, α′) with α of type P.

Proof: Suppose α of type L for all critical pairs. Let [Y_{α−1}, Y_{α′} ∩ Q_{α}] ≠ 1. Then Y_{α′} ∩ Q_{α} is a strong dual offender and so |[Y_{α′} ∩ Q_{α}, Y_{α−1}]| = 2. In particular N_{α′} ≅ Sp(2n, 2). Choose α − 1 ∈ △(α) with Y_{α′} ≰ G_{α−1}. Let V ≤ Y_{α−1} be the natural module. Then the uniqueness elements are the singular ones. Hence |V ∩ Q_{α′}| ≤ 4. In particular |V : V ∩ Q_{α′}| ≥ 2^{2n−2}. On the other hand the hyperplane in Y_{α′} induces transvections on Y_{α}. This shows |Y_{α′} : Y_{α′} ∩ Q_{α}| ≤ 4 and so |Y_{α} : Y_{α} ∩ C_{α′}| ≤ 4. Now |Y_{α−1}Y_{α} : Y_{α}| ≤ 2 so |Y_{α−1} : Y_{α−1} ∩ Q_{α′}| ≤ 8. This shows |V | ≤ 2^{5} and so |V | = 16 and K ≅ Ω^{−}(4, 2) a contradiction. Hence [Y_{α′} ∩ Q_{α}, Y_{α−1}] = 1 and so Y_{α−1}Y_{α} ⊴ G_{α}, a contradiction.

Lemma 4.7 Let b > 1, K ≅ Ω^{±}(2n, 2) but K ≅ Ω̸^{−}(4, 2) or Ω^{+}(6, 2). If Y_{L}Y_{P} ⋬ L, then b is odd.

Proof: Let b be even. By (4.6) there is a critical pair (α, α′) with α of type P. Then Y_{α} induces transvections on Y_{α′} otherwise there are common uniqueness elements.

We first assume that (α−1, α′−1) is also a critical pair. Set A_{αβ} = [Y_{α}, V_{β}^{(b−1)}] where β ∈ Δ(α) and d(β, α′) = b − 1. Now choose α − 2 ∈ Δ(α − 1) with Y_{α′−1} ≰ G_{α−2,α−1}. Let W_{α′−1} ∈ V(K_{α′−1}) with W_{α′−1} ≰ G_{α−2,α−1}. Let Y_{α−2} ≤ Q_{α′−2}. As Y_{α′−1} is an F -module with offender in the O_{p} of the point stabilizer, we see that for K ∈ K(L) we have K ≅ L_{n}(q), Sp(2n, q), G_{2}(q) or A_{n}. As Y_{α−2}Y_{α−1} ⋬ G_{α−1} we get K ≅ Sp(2n, q) on the symplectic module. Let H_{α′−1} be the hyperplane in W_{α′−1} corresponding to G_{α′−1,α′−2}. If [H_{α′−1}, Y_{α−1}] ≠ 1, then [Y_{α−2}, H_{α′−1}] = [Y_{α−1}, W_{α′−1}] and so W_{α′−1} ≤ N_{G_{α−1}}(Y_{α−1}Y_{α−2}) = G_{α−1,α−2}, a contradiction. Hence we have [H_{α′−1}, Y_{α−1}] = 1. Then H_{α′−1} is a strong offender on Y_{α−2}. This implies |H_{α′−1}/H_{α′−1} ∩ Q_{α−2}| = 2, in particular q = 2 and then |Y_{α−2} : Y_{α−2} ∩ C I(W_{α′−1})| ≤ 4. Because of the uniqueness elements we now get |Y_{α−2}| ≤ 16, a contradiction. So we have Y_{α−2} ≰ Q_{α′−2}. We now consider [Y_{α−2} ∩ Q_{α′−2}, W_{α′−1}]. Assume that [Y_{α−2} ∩ Q_{α′−2}, W_{α′−1}] ≤ [Y_{α−1}, W_{α′−1}]. Then (Y_{α−2} ∩ Q_{α′−2})W_{α−1} is normalized by W_{α′−1}. Hence

W_{α′−1} ≤ N_{G_{α−1}}([Y_{α−2} ∩ Q_{α′−2}, V_{α−1}^{(b−2)}][Y_{α−2}, Y_{α′−2}]) ≤ N_{G_{α−1}}(A_{α−2,α−1}) ≤ G_{α−2,α−1}

a contradiction. So we have [Y_{α−2} ∩ Q_{α′−2}, W_{α′−1}] ≰ [Y_{α−1}, W_{α′−1}]. In particular R = [Y_{α−2} ∩ Q_{α′−2}, W_{α′−1} ∩ Q_{α−1}] ≠ 1. As W_{α′−1} is an F -module with offender in the O_{p} of the point stabilizer this is just possible for symplectic and alternating groups. In the last case we get |Y_{α−2} : Y_{α−2} ∩ C I(W_{α′−1})| ≤ 8. As Y_{α−2} ∩ C I(W_{α′−1}) is totally isotrop this implies |Y_{α−2}| ≤ 32, a contradiction. So we have K_{α′−1} ≅ Sp(2n, q) and |R| = q. Now consider [Y_{α−2}, Y_{α′−2}][Y_{α−2} ∩ Q_{α′−2}, Y_{α′−1} ∩ Q_{α−1}] = W_{α−2}. As Y_{α′−1} ≰ G_{α−1,α−2} we see that W_{α−2} cannot contain uniqueness elements. So [Y_{α−2} ∩ Q_{α′−2}, Y_{α′−1} ∩ Q_{α−1}] ≤ [Y_{α−2}, Y_{α′−2}]. But Y_{α−2} ∩ Q_{α′−2} is the hyperplane which is centralized by Y_{α′−2}. This also must be centralized by Y_{α′−1} ∩ Q_{α−2}, so [Y_{α−2} ∩ Q_{α′−2}, Y_{α′−1} ∩ Q_{α−1}] = 1, a contradiction. Hence we have shown that (α − 1, α′ − 1) is not a critical pair. In particular Y_{α−1} ≤ G_{α′}. Then [Y_{α−1}, Y_{α′}] ≤ [Y_{α}, Y_{α′}][Y_{α−1}, Y_{α′} ∩ Q_{α}] by the action of the symplectic group. So we see Y_{α−1}Y_{α} ◁ G_{α}. Now we set H_{α−2} = C I_{Y_{α−2}}(V_{α}^{(b−2)}), which is a hyperplane in Y_{α−2}. We have [H_{α−2}, Y_{α′−3}Y_{α′−2}] = 1. As Y_{α′−3}Y_{α′−2} = Y_{α′−2}Y_{α′−1} we see [H_{α−2}, Y_{α′−1}] = 1, so H_{α−2} ≤ G_{α′}. In particular H_{α−2} ≤ G_{α}^{g}, for all g ∈ G_{α}. The structure of the symplectic group now implies [H_{α−2}, Y_{α′}^{g}] = [H_{α−2}, Y_{α′}^{g} ∩ Q_{α}][Y_{α}, Y_{α′}^{g}]. Hence N_{α} ≤ N(H_{α−2}Y_{α}). As Q_{α−2} ≰ Q_{α} we have [O^{p}(N_{α}), H_{α−2}] ≤ Y_{α}. This now implies with [O^{p}(N_{α}), O^{p}(N_{α})] = O^{p}(N_{α}) that H_{α−2}Y_{α} = C I_{H_{α−2}Y_{α}}(N_{α})Y_{α}. Let now W_{α−2} be a 2-dimensional singular subspace of H_{α−2}. Suppose W_{α−2} ∩ C I_{H_{α−2}Y_{α}}(N_{α}) ≠ 1. Then N_{α} ≤ C̃_{α−2}. If W_{α−2} ∩ C I_{H_{α−2}Y_{α}}(N_{α}) = 1, then W_{α−2} corresponds to a 2-dimensional subspace W_{α} in Y_{α}. But N_{α} = ⟨C I_{N_{α}}(x)|x ∈ W_{α}^{#}⟩. In particular N_{α} = ⟨C I_{N_{α}}(g)|g ∈ W_{α−2}^{#}⟩ ≤ C̃_{α−2}. Hence we

have N_{α} ≤ ⋂_{α−2∈Δ(α−1)} C̃_{α−2} = R. Obvisously G_{α−1} ≤ N(R). As N_{α}◁◁C̃_{α}, we

have that N_{α}◁◁R. In particular O_{p}(R) ≠ 1 and so ⟨G_{α−1}, N_{α}⟩ ≤ N(O_{p}(R)), a contradiction.

Proposition 4.8 Let b > 1 and K ≅ Ω^{±}(2n, 2), K ≅ Ω̸^{−}(4, 2) or Ω^{+}(6, 2), then Y_{L}Y_{P} ◁ L.

Proof: By (4.7) we may assume b to be odd, in particular b ≥ 3. Let Y_{α′+1} ≤ G_{α}, where α′ + 1 is such that Y_{α} ≰ C̃_{α′+1}. Then |Y_{α′+1} ∩ Q_{α}| ≤ 4, as we cannot have uniqueness elements in Y_{α′+1} ∩ Q_{α}. If x ∈ Y_{α} \ C̃_{α′+1} then by the same reason |C I_{Y_{α′+1}}(x)| ≤ 4. Hence |[Y_{α′+1}, x]| ≥ 2^{2n−2}. As Y_{α′+1} acts quadratically on Y_{α} we get |[Y_{α′+1}, x]| ≤ 2^{n}. Hence n ≥ 2n − 2 and so n ≤ 2, a contradiction. Hence we have Y_{α′+1} ≰ G_{α}. Then [Y_{α}, Y_{α′}] does not contain uniqueness elements in G_{α}. This shows |[Y_{α}, Y_{α′}]| = 2. So we get |Y_{α′} : C I_{Y_{α′}}(Y_{α})| = 2 and then Y_{α} acts on just one component which gives |Y_{α} : Y_{α} ∩ Q_{α′}| = 2 too. We have [Y_{α′+1}, [Y_{α}, Y_{α′}][Y_{α′+1} ∩ Q_{α+1}, Y_{α} ∩ Q_{α′}]] = 1. In particular [Y_{α}, Y_{α′}][Y_{α′+1} ∩ Q_{α+1}, Y_{α} ∩ Q_{α′}] contains no uniqueness element. This shows [Y_{α}, Y_{α′}] ≥ [Y_{α′+1} ∩ Q_{α+1}, Y_{α} ∩ Q_{α′}]. But then the action of the symplectic group shows [Y_{α′+1} ∩ Q_{α+1}, Y_{α} ∩ Q_{α′}] = 1. Hence Y_{α} ∩ Q_{α′} induces a transvection on Y_{α′+1}, which gives |Y_{α} : C I_{Y_{α}}(Y_{α′+1})| ≤ 4. Again we get |Y_{α}| ≤ 16, a contradiction.

5 The alternating groups

From now on we will assume that K ≅ A_{n}.

Lemma 5.1 Let b > 1 and (α, α′) be a critical pair with α of type P and α′ of type L. Then on both sides we have an F -module.

Proof: Let Y_{α} be an offender on Y_{α′} then by [GBSM] also Y_{α′} is an offender on Y_{α}. Hence we may assume that Y_{α} is not an offender on Y_{α′}. Then Y_{α′} is an over-offender and again by [GBSM] Y_{α′} is the full transvection group for K ≅ A_{n}. Now [Y_{α′}, Y_{α}] is normalized by S ∩ N_{α} and so contains uniqueness elements by (3.8). As b > 1, we get V_{α′} ≤ G_{α}. Then V_{α′}Q_{α} = Y_{α′}Q_{α} and then [V_{α′}, Y_{α}] = [Y_{α′}, Y_{α}]. This gives [N_{α′}, V_{α′}] ≤ Y_{α′} and so Y_{α′−1}Y_{α′} ◁ G_{α′}. Hence Y_{α}Y_{α+1} = Y_{α+1}Y_{α+2} ◁ G_{α+1} which contradicts [Y_{α′}, Y_{α}] ≠ 1.

Lemma 5.2 Let (α, α′) be a critical pair and assume that G_{α′} induces Σ_{n} on the natural module. Assume further that Y_{α} induces ⟨(1, 2), (3, 4), ...(k − 1, k)⟩ ≠ ⟨(1, 2), (3, 4), ..., (n − 1, n)⟩ and |Y_{α}/C I_{Y_{α}}(Y_{α′})| ≥ 4. If we have the basis a_{ij} for the permutation module Y_{α′} set X = ⟨a_{13}, a_{15}, ..., a_{k−3k−2}⟩. Then |Y_{α′}/X(Y_{α′} ∩ Q_{α})| ≤ 2 and [C I_{G_{α′}/Q_{α′}}(Y_{α}Q_{α′}/Q_{α′}), X] = [Y_{α}, Y_{α′}].

Proof: The first assertion is clear as XC I_{Y_{α′}}(Y_{α}) is a hyperplane. We have that C I_{G_{α′}/Q_{α′}}(Y_{α}Q_{α′}/Q_{α′}) ≅ Y_{α}Q_{α′}/Q_{α} × Σ_{r} and Σ_{r} acts trivially on X.

Lemma 5.3 Let [Y_{P} , YP̃] ≠ 1, P ∼ P̃ in G, then Y_{P} corresponds to a hyperplane in the transposition group which is generated by transpositions or n is odd and Y_{P} corresponds to the full transposition group.

Proof: Let first n = 8 and Y_{P} corresponds to a group of order 8 in L_{4}(2) (transvection group on a natural module). Then [Y_{P} , YP̃] contains uniqueness elements in P and P̃ as well, as Y_{P} acts quadratically on YP̃ and so C I_{YP̃}(NP̃) = 1, a contradiction. Hence we have that in any case Y_{P} is in a transposition group. Suppose first that Y_{P} corresponds to the full transposition group and n be even. Then there are no transvection in YP̃ on Y_{P}. So YP̃ corresponds to the uniquely detemined hyperplane in the transvection group in P. Hence |Y_{P} : C I_{Y_{P}} (YP̃)| = |Ỹ_{P} Q_{P} /Q_{P} |, a contradiction.

So we have that both are hyperplanes. Assume that Y_{P} corresponds to ⟨(1, 2), (3, 4), ..., (k, k−1)⟩×⟨some even permutations⟩. Then as above we see that there are no transvections in YP̃. Hence again YP̃ is uniquely determined. But then also Y_{P} induces no transvections on YP̃ and so also Y_{P} is the unique offender. Set AP̃ = {u|1 ≠ u ∈ [YP̃, x_{i}] for at least three different x_{i} with x_{i} ∈ Y_{P} and |[YP̃, x_{i}]| = 4}. We define A_{P} in the corresponding way. If u ∈ A_{P}, then u is of type a_{12} (a_{ij} generators of the permutation module). Then ∏_{u∈A_{P}} u is the element of longest length in Y_{P}, i. e. a uniqueness element. We have YP̃ = ⟨a_{13}, a_{15}, ..., a_{1n−1}⟩. The corresponding elements in Y_{P} are of the same type, hence the uniqueness element is the same, a contradiction.

Definition 5.4 Let T_{α} be the largest subgroup of Q_{α} with the following property: If δ ∈ Δ(α), then T_{α}/C I_{T_{α}}(Y_{δ}) is generated by minimal offenders. Denote by T_{αβ} the group generated by the offenders on Y_{β} which are contained in Q_{α}, β ∈ Δ(α). Finally let A_{βα} = [Y_{β}, T_{α}].

Lemma 5.5 Let b > 1 and (α, α′) be a critical pair. Let β ∈ Δ(α). Then A_{βα} ⊴ G_{α} and C I_{Y_{β}}(T_{αβ})Y_{α} ⊴ G_{α}.

Proof: We will assume that d(β, α′) = d(α, α′) + 1. If Y_{β}Y_{α} ◁ G_{α}, then A_{βα} = [Y_{β}Y_{α}, T_{α}]⊴G_{α}. Further C I_{Y_{β}}(T_{αβ})Y_{α} = C I_{Y_{β}Y_{α}}(T_{αβ}) and so C I_{Y_{β}}(T_{β})Y_{α}⊴ G_{α}. So we may assume Y_{β}Y_{α} ⋪ G_{α}. If [Y_{β}, T_{α}] = 1, then A_{βα} = 1. It remains to prove C I_{Y_{β}}(T_{αβ})Y_{α} ◁ G_{α} for T_{αβ} = 1. Then we have to prove that Y_{β}Y_{α} ◁ G_{α}. In particular we now by (5.1) have (β, α′ − 1) is not a critical pair. So Y_{β} ≤ G_{α′}. Let α′ be of type L, then [Y_{β}, Y_{α′}] ≤ [Y_{α′}, Y_{α}] and so Y_{β}Y_{α} ◁ G_{α} and we are done, or we have Σ_{m} or Sp(2n, q) and Y_{α′} ∩ Q_{α} is an offender on Y_{β}, which does not exist. If α′ is of type P, then G_{α′} is of type Σ_{n}. By (5.3) and (5.1) we see that Y_{α} is a transposition group and also Y_{β}Y_{α} is a transposition group. It follows that Y_{β}Y_{α} is the full transposition group. Now choose again X as in (5.2). Then we get [Y_{β}Y_{α}, X] ≤ [Y_{α}, Y_{α′}] = [Y_{α}, X]. Then Y_{β}Y_{α} ◁ G_{α}, the assertion. So we may assume [Y_{β}, T_{α}] ≠ 1. In particular we have the structure of G_{β}. If β ≁ α, then we have SL_{n}(q) or Sp(2n, q) or A_{n} on the natural module by [GBSM]. Let V be a quasi irreducible module for a p-component of G_{β}. Set

A = V ∩ A_{βα}.

By (5.1) Y_{α′} is a F -module as well. Let W be a quasi irreducible submodule of Y_{α′} with [Y_{α}, W] ≠ 1. If α′ is of type P and |[Y_{α}, W]| > 2, so let X be the hyperplane from lemma (5.2) otherwise let X = W. Checking the cases for α′ we see that either [V ∩ Q_{α′−1}, X ∩ Q_{α}] = 1 or X induces a strong dual offender on V ∩ Q_{α′−1}, or |[V ∩ Q_{α′−1}, X ∩ Q_{α}]| = 2. Suppose that the following holds

(∗) A = A ∩ ([V, Y_{α′−1}][V ∩ Q_{α′−1}, X ∩ Q_{α}])[C I_{V_{∩}Q_{α′−1}}(X ∩ Q_{α}), T_{α}].

Then we have [X, A ∩ ([V, Y_{α′−1}][V ∩ Q_{α′−1}, X ∩ Q_{α}])] = 1 and X normalizes [C I_{V_{∩}Q_{α′−1}}(X ∩ Q_{α′})Y_{α}, T_{α}]. Hence X ≤ N(A). Then also X ≤ N(⟨A^{G_{α}}⟩). This shows ⟨X^{G_{α}}⟩ ≤ N(A). As G_{αβ} ≤ N(A), we have G_{α} ≤ N(A), the first assertion. Now if P is of type L, then C I_{V} (T_{αβ}) is one dimensional. If [V, Y_{α′−1}] ≠ 1, then C I_{V} (T_{αβ}) is centralized by X ∩ Q_{α}. If [V, Y_{α′−1}] = 1, then X ∩ Q_{α} is an offender on V and so also [C I_{V} (T_{αβ}), X ∩ Q_{α}] = 1. If β is of type P, then Y_{α′−1}(X ∩ Q_{α}) is in the transvection group. Hence [Y_{α′−1}(X ∩ Q_{α}), C I_{V} (T_{αβ})] = 1. So in any case [(X ∩Q_{α}), C I_{V} (T_{αβ})] = 1. We have C I_{V} (T_{αβ}) ≤ C I_{V ∩Q_{α′}}(X ∩Q_{α}). Hence X ≤ N(C I_{V} (T_{αβ})Y_{α}). As G_{αβ} ≤ N(C I_{V} (T_{αβ})V_{α}), we get C I_{V} (T_{αβ})Y_{α} ◁ G_{α}, the second assertion. So we show what happens if (∗) does not hold. Let first β be of type L. In SL_{n}(q), Sp(2n, q) or Σ_{n} we see that A is contained in [V, Y_{α′−1}] for all Y_{α′−1} with [Y_{α′−1}, V] ≠ 1, so (∗) holds. If we have G_{2}(q) then also [V, Y_{α′−1}] = A. Otherwise Y_{α′−1} is not an offender on V. But then V is an offender on Y_{α′−1}. If α′ − 1 is of type L, then this is not possible as there is a unique offender in G_{2}(q). Hence α′ − 1 is of type P. But then there are transvections on V, a contradiction. So we are left with [V, Y_{α′−1}] = 1. We may assume [V, X ∩Q_{α}] ≠ 1. But then by the above X ∩ Q_{α} is a minimal offender and so (∗) holds. Let now β be of type P. If [V, Y_{α′−1}] = 1 then as above. Hence we may assume [V, Y_{α′−1}] ≠ 1. Then also (β, α′ − 1) is a critical pair. By (5.1) and (5.3) we get that Y_{α′−1} is a transposition group. We have (X ∩Q_{α′})Q_{β}/Q_{β} ≤ C I(Y_{α′−1}Q_{β}/Q_{β}) = Y_{α′−1}Q_{β}/Q_{β} ×Σ_{k}. Then we have (∗) or Y_{α′−1}(X ∩Q_{α′}) is of index two in T_{α} = ⟨(1, 2), (3, 4), ..., (2n−1, 2n)⟩ and P induces Σ_{2n}. Finally we have a central extension of the permutation module and |Y_{α′−1}Q_{β}/Q_{β}| ≥ 4. So we have to treat the latter. We have that G_{β} acts on A_{αβ} = [Y_{α}, T_{β}]. As Q_{α} ≰ Q_{β} we have [N_{β}, A_{αβ}] = 1. Hence N_{β} ∩ G_{α} acts on any p-component K of G_{α}. Let [Y_{α′−1}(X ∩ Q_{α}), Y_{β}] = A_{β} which is a hyperplane. Assume that there is some p-component K in G_{α} with [A_{β}, K] = 1. Then also [⟨A_{β}^{N_{β}∩G_{α}}⟩, K] = 1. If ⟨A_{β}^{Nβ∩G_{α}}⟩ = A_{βα}, so A_{βα} ◁ ⟨K, G_{αβ}⟩ = G_{α}, and we are done. So we have N_{β} ∩ G_{α} ≤ N(A_{β}). Let as in the proof of (5.3) a be the element of longest length in A_{β}. Then [N_{β} ∩ G_{α}, a] = 1. Further G_{αβ} ≤ C I(a), as G_{αβ} = C I_{G_{β}}([V, N_{β}])T_{α}(N_{β} ∩ G_{α}). As K ≤ C I(a) we get C I_{G_{β}}(a) = C I_{G_{αβ}}(a) by the choice of L with L∩C̃ maximal. But now we get |A_{β}| = 2, a contradiction as P is not of type Σ_{6} ≅ Sp(4, 2). So we have that there is no such component. As [A_{β}, Q_{β}] = 1 = [A_{β}, X], we see that ⟨Q_{β}, X⟩ does not contain a component independent of the choice of β. This is just possible for L of type Σ_{m} or L of type Sp(2n, q) and |Q_{β}Q_{α}/Q_{α}| = q. Let first L be of type Σ_{m}. For α we now have C I_{Y_{β}}(T_{αβ})Y_{α} ◁G_{α} as C I_{Y_{β}}(T_{αβ}) ≤ G_{α′} and centralizes in W the hyperplane W ∩ Q_{α}. So we have C I_{Y_{β}}(T_{αβ})Y_{α} ◁ ⟨G_{αβ}, Y_{α′}⟩ = G_{α}. As Q_{β} ≰ Q_{α} we see [N_{β}, C I_{Y_{β}}(T_{αβ})] ≤ Y_{α}, so [A_{βα}, O^{2}(N_{β} ∩ G_{α})] ≤ Y_{α}.

We have [A_{βα}, O^{2}(N_{β} ∩G_{α})] ≠ 1. As Y_{α} possesses just one nontrivial O^{2}(N_{β} ∩ G_{α})-chief factor we get

[Y_{α}, O^{2}(N_{β} ∩ G_{α})] = [A_{βα}, O^{2}(N_{β} ∩ G_{α})]

Let now K be a p-component in G_{α}. Then |(Q_{αβ} ∩ K)Q_{α} : Q_{α}| = 2 besides for K/O_{2}(K) ≅ Σ_{6}, where |(Q_{αβ} ∩K)Q_{α} : Q_{α}| = 8. Let first K/O_{2}(K) ≅ Σ̸_{6}. Then [Q_{αβ}, [A_{βα}, O^{2}(N_{β} ∩ G_{α})]] = 1. As [[A_{β}, O^{2}(N_{β} ∩ G_{α})], X] = 1, we see that [A_{βα}, O^{2}(N_{β} ∩ G_{α})] ≠ [A_{β}, O^{2}(N_{β} ∩ G_{α})]. This gives Q_{αβ} = Q_{β}T_{α}. So G_{αβ} = N_{G_{β}}(Q_{αβ}) induces Z_{2} ≀ Σ_{n}. Next we show that in both cases L is of type Σ_{m} and L is of type Sp(2n, q) we just have one component. Let D_{β} = C I_{G_{β}}(Y_{β}). Then G_{αβ} = D_{β}(N_{β} ∩ G_{α})Q_{α}. As N_{β} ∩ G_{α} ≤ N(K), we see that D_{β} acts transitively on the p-components. Assume O^{2}(N_{α} ∩ G_{β}) ≤ D_{β}. Then [A_{βα}, Q_{β}O^{2}(N_{α} ∩ G_{β})D_{β}] = 1. Set F = A_{βα}Y_{α}. We have A_{βα}Y_{α} ≤ C I_{Y_{β}}(T_{α})Y_{α}. In particular F ◁ G_{α}. Set \overline{F} = F/C I_{F} (N_{α}). Then \overline{A}_{βα} ≤ \overline{A}_{αβ}. As [N_{β}, A_{αβ}] = 1, we see that [A_{βα}, N_{β} ∩ G_{α}] ≤ C I_{F} (N_{α}). This shows A_{β}C I_{F} (N_{α})⊴⟨D_{β}(N_{β} ∩G_{α})Q_{α}, Y_{α′}⟩ = G_{α}. Notice that [F, G_{α}] ≤ C I_{F} (N_{α}) as [A_{βα}, Q_{α}] ◁ N_{α}. So we see [A_{β}, N_{α}] = 1. But then A_{β}[A_{β}, G_{αβ}] contains no element a whose centralizer is not in G_{α}. This implies |A_{β}| = 2, a contradiction. So we have O^{2}(N_{α} ∩ G_{β}) ≰ D_{β}. As D_{β} ◁ G_{β} we get [O^{2}(N_{α} ∩ G_{β}), D_{β}] ≤ D_{β}. Hence [O^{2}(N_{α} ∩ G_{β}), D_{β}] ≤ Q_{αβ} and [D_{β}, A_{βα}] = 1. If q > 2 (in case of Sp(2n, q)) we know [\overline{A}_{αβ}, G_{αβ}] ≠ 1 and so [\overline{A}_{αβ}, D_{β}] ≠ 1. In particular \overline{A}_{αβ} ≰ C I\overline{F} (D_{β}). As \overline{A}_{αβ} is the only minimal normal subgroup in \overline{G}_{αβ}, we see C I\overline{F} (D_{β}) = 1, a contradiction. Hence q = 2 and D_{β} centralizes all components. Hence there is just one component.

Now we come back to the case of Σ_{m}. We then have C I_{Y_{β}}(T_{α}) = Y_{αβ} = C I_{Y_{α}}(Q_{β}). As A_{β} ≠ A_{βα} we know that Y_{β} is a central extension with the natural module. Furthermore there is no extension with the trivial module, as otherwise we would get all commutators. Hence we have |Y_{β}| = 2^{2n−1} and |Y_{αβ}| = 2^{n}. This shows |Y_{α}| = 2^{n+1}, as Y_{αβ} was a hyperplane in Y_{α}. Now C I_{Y_{αβ}}(N_{α}) ≤ C I_{Y_{β}}(N_{α} ∩ G_{β}) ≤ C I_{Y_{β}}(N_{β}) and so C I_{Y_{αβ}}(N_{α}) = 1. Now we see Y_{α′} ∩ Q_{α} ≤ Y_{α′−1} and so |Y_{α′−1}Q_{β}/Q_{β}| = 2^{n} − 1. Let δ be a neighbour of β. We have Y_{δ} ≤ G_{α′−2}. Assume Y_{δ} ≤ Q_{α′−2} for all δ. Now choose X_{α′−1} as in (5.2). Then [Y_{δ}Y_{β}, X_{α′−1}] = [Y_{β}, X_{α′−1}] ≤ Y_{β}. Then X_{α′−1} ≤ N(Y_{δ}Y_{β}). Now choose δ such that ⟨G_{δβ}, X_{α′−1}⟩ = G_{β}. Then we get Y_{δ}Y_{β} ⊴ G_{β}. Then also Y_{α′}Y_{α′−1} = Y_{α′−1}Y_{α′−2}, contradicting [Y_{α}, Y_{α′}] ≠ 1. Hence there is some δ with Y_{δ} ≰ Q_{α′−2}. As for groups of type L the assertion is proved we get [Y_{α−2}, Y_{α′−2}] ≤ Z(G_{β}) and [Y_{α−2}, Y_{α′−2}] = [Y_{α−2}, T_{α−1}]. If b > 2, then [[Y_{α′−2}, Y_{α−2}], ⟨G_{β}, Y_{α′}⟩] = 1. But ⟨G_{β}, Y_{α′}⟩ contains ⟨G_{αβ}, Y_{α′}⟩ = G_{α}, a contradiction. Hence we have b = 2. Now choose notation as follows δ ∈ Δ(α) ∩ Δ(α′) and β ∈ Δ(α) \ Δ(α′). We have [V_{β} ∩ Q_{α}, Y_{δ}] ≤ Y_{αδ} ≤ Y_{α}. By quadratic action we get [Q_{αβ}, Y_{β}] ≤ Y_{αβ} ≤ Y_{α} and so [Q_{α}, V_{α}] ≤ Y_{α}. Then [V_{β} ∩ Q_{α}/Y_{β}, V_{α}/Y_{β}] ≤ Y_{α}Y_{β}/Y_{β}, where |Y_{β}Y_{α}/Y_{α}| = 2. Finally |V_{β}/V_{β} ∩ Q_{α}| ≤ 2. We have |Y_{β}Q_{α′−1}/Q_{α′−1}| ≥ 4. Now choose X_{α′−1} ≤ Y_{α′−1} according to (5.1). Then [C I_{G_{α′−1}}(Y_{β}), X_{α′−1}] ≤ [Y_{β}, Y_{α′−1}] ≤ Y_{β}. Hence [V_{β} ∩ Q_{α′−2}, X] ≤ Y_{β}. This says that X_{α′−1} induces transvections on V_{β} to the hyperplane V_{β} ∩ Q_{α′−2}. But then |X_{α′−1}| = 2, a contradiction. Assume next that L is of type Sp(2n, q) ≠ Σ_{6}. Then |Q_{β}Q_{α}/Q_{α}| = q. Hence Q_{αβ} acts trivially on A_{βα}. If [A_{βα}, Q_{αβ}, O^{2}(N_{β} ∩ G_{α})] = 1, then O^{2}(N_{β} ∩ G_{α}) has orbits of length at most two, a contradiction. Hence [A_{βα}, Q_{αβ}] is a hyperplane in Y_{α} and we have an extension of the trivial module by the natural one. Now we have [A_{βα}, Q_{αβ}, Q_{αβ}, O^{2}(N_{β} ∩ G_{α})] = 1. In this case we have orbits of length at most 4. Thus O^{2}(N_{β} ∩G_{α}) ≅ A_{4}. As N_{β} ∩G_{α}◁◁G_{αβ} we get the contradiction K ≅ Sp(4, 2). So we are left with K ≅ Σ_{6} and |Q_{αβ}/Q_{α}| = 8. As K is normalized by O^{2}(N_{β} ∩ G_{α}) but K is not trivial on A_{βα}, we get O^{2}(N_{β} ∩ G_{α}) ≅ Z_{3}. This shows P/O_{2}(P) ≅ Σ_{8}. Then |Y_{P} | = 2^{7}, |Y_{α}| = 2^{5}, |Y_{αβ}| = 2 and |A_{βα}| = 16. We have A_{βα} = Y_{α} ∩ Y_{β} and as before we get b = 2. Then again |[V_{β} ∩ Q_{α}/Y_{β}, V_{α}/Y_{β}]| = 2, the same contradiction as before.

Lemma 5.6 Let b > 1 and (α, α′) be a critical pair such that α is of type A_{n}. Then either Y_{P} Y_{L} ◁ L or b is even.

Proof: By (5.1) we have that Y_{α} is an offender on Y_{α′} and both Y_{α′} and Y_{α} are known. Now Y_{α} is an exact offender this implies |Y_{α′} : Y_{α′} ∩ Q_{α}| = |Y_{α} : C I_{Y_{α}}(Y_{α′})|. Then either n = 8 or Y_{α′} is a transposition group. In the first case [Y_{α}, Y_{α′}] contains just uniqueness elements. Hence V_{α′} ≤ G_{α} and then V_{α′}Q_{α} = Y_{α′}Q_{α}. So Y_{α+1}Y_{α} ◁ G_{α}, a contradiction. So we have the first case. Let A_{αβ} be as in (5.5). Then A_{αβ} ◁ G_{αβ}N_{β} = G_{β}. As N_{β} ≤ ⟨Q_{α}^{N_{β}}⟩ we see [A_{αβ}, N_{β}] = 1. So we may set A_{β} = A_{αβ}. We have G_{αβ} = N_{G_{α}}(A_{β}), further A_{β} does not contain uniqueness elements. Let T_{β}Q_{α}/Q_{α} = ⟨(1, 2), (3, 4), ..., (k − 1, k)⟩. Then x = a_{12} ⋯ a_{k−1k} ∈ A_{β}. We have G_{β} ≤ C I_{G}(x) but C I_{N_{α}}(x) ≅ S_{k} × S_{e}. The maximality of L ∩ C̃ now implies k = 2, so T_{β}Q_{α}/Q_{α} = ⟨(1, 2)⟩ and N_{α} ∩ G_{β}/Q_{αβ} ≅ S_{n−2}. By (5.5) we have that N_{α} ∩ G_{β} centralizes A_{β} and so it normalizes all components in G_{β}. Let first N_{β} be an alternating group. We have that Y_{α} ∩ Y_{αβ} = C I_{Y_{α}}(Q_{αβ}) is a hyperplane in Y_{α}. Let α − 1 ∈ Δ(α). If Y_{α−1} ≤ Q_{α′−1} for all α − 1, so choose Ṽ ≤ Y_{α′}, Ṽ ∈ V(K_{α′}) with [Ṽ , Y_{α}] ≠ 1. Then [Y_{α−1}, Ṽ] ≤ [Y_{α}, Y_{α′}] ≤ Y_{α}. Hence Y_{α−1}Y_{α} ◁ G_{α}, a contradiction. Hence there is α − 1 ∈ Δ(α) with Y_{α−1} ≰ Q_{α′−1}. We have A_{α−1} ≤ Y_{α−2}. Let [Y_{α−2}, Y_{α′−2}] ≠ 1, then [Y_{α−2}, Y_{α′−2}] = A_{α−1}. As b ≥ 3 we now get [Ṽ , A_{α−1}] = 1 and then G_{α} ≤ N(A_{α−1}). As N_{α−1} ≤ N(A_{α−1}) this is not possible. Hence [Y_{α−2}, Y_{α′−2}] = 1. Then Y_{α−2} ≤ Q_{α′−2,α′−1}. This implies Q_{α′−1}Y_{α−1} = Q_{α′−1}Y_{α−2} and so Y_{α−2}Y_{α−1} ◁ G_{α−1}, a contradiction. Let now N_{β} be a group of Lie type. Then in A_{β}[Y_{αβ} ∩ Y_{α}, O^{p}(N_{α} ∩ G_{β})] there are always uniqueness elements. Hence [Y_{αβ}, O^{p}(N_{α} ∩ G_{β})] ≰ C I_{Y_{β}}(N_{β}). Then also [C I_{Y_{β}}(Q_{αβ}), O^{p}(N_{α} ∩ G_{β})] ≰ C I_{Y_{β}}(N_{β}). As C I_{Y_{β}}(Q_{αβ}) is one dimensional we see [C I_{Y_{β}}(Q_{αβ}), O^{p}(N_{α} ∩ G_{β})] = 1, a contradiction.

Lemma 5.7 Let b > 1 and Y_{P} Y_{L} ≰ L. Then for any critical pair (α, α′) we have α of type P.

Proof: Suppose false. Then there is a critical pair (α, α′) with α and α′ of type L. Let T_{α} and A_{αβ} be defined as before.

(1) V_{α}^{(b−1)} ≤ T_{α}

Let Y_{μ} ≤ V_{α}^{(b−1)} with d(α, μ) = b − 1. Let δ ∈ Δ(α). We assume [Y_{δ}, Y_{μ}] ≠ 1. By (5.3) we then have Y_{μ} ≤ T_{α}. Let [T_{α}, Y_{δ}] = 1 for all δ. Then Y_{α−1} ≤ G_{α′}. As Y_{α−1}Y_{α} ⋬ G_{α} we get [Y_{α−1}, Y_{α}] ≰ [Y_{α}, Y_{α′}]. This is just possible for G_{α′} of type Sp(2n, q) or Σ_{n}. Further [Y_{α′} ∩ Q_{α}, Y_{α−1}] ≠ 1. In the second case the group Y_{α′} ∩ Q_{α} induces transvections, a contradiction. So we have the first case. Now Y_{α′} ∩ Q_{α} is a strong offender on Y_{α−1} and again we have transvections.

(2) There is a quasi irreducible module W_{α′} in Y_{α′} with W_{α′} ∩ Q_{α} ≤ T_{α}. As b > 1 we have W_{α′} ∩ Q_{α} ≤ C I_{G_{α−1}}(Y_{α′−1}Q_{α−1}/Q_{α−1}). Further C I_{G_{α−1}}(Y_{α′−1}Q_{α−1}/Q_{α−1}) ≅ Y_{α′−1}Q_{α−1}/Q_{α−1}×Σ_{t} Finally we have Y_{α}∩G_{α′−1} ≤ G_{α′}. If L is of type G_{2}(q), then [Y_{α}, W_{α′}] = W_{α′} ∩ Q_{α} ≤ Y_{α}, so [Y_{α−1} ∩ Q_{α′−1}, W_{α′} ∩ Q_{α}] = 1. If L is of type Σ_{m}, then |[W_{α′} ∩ Q_{α}, Y_{α−1} ∩ Q_{α′−1}]| ≤ 2. If L is of type SL_{n}(q) or Sp(2n, q), then W_{α′} ∩Q_{α} is a strong dual offender on Y_{α−1} ∩ Q_{α′−1}. So in all cases the projection of W_{α′} ∩ Q_{α} onto Σ_{t} is generated by transpositions. Hence Y_{α′−1}(W_{α′} ∩ Q_{α}) is generated by transpositions. This shows W_{α′} ∩ Q_{α} ≤ T_{α}. As [Q_{αα−1}, C I_{Y_{α−1}}(T_{α})] = 1 we see that C I_{Y_{α−1}}(T_{α})Y_{α} = D_{α}Y_{α} with D_{α} = C I_{C I_{Y_{α−1}}(T_{α})Y_{α}}(N_{α}). We have Q_{αα−1} ≤ N(A_{αα−1}) and by (5.5) [A_{αα−1}, N_{α}] = 1. This shows G_{αα−1} = N_{G_{α−1}}(A_{αα−1}), as L ∩ C̃ is maximal. We have N_{G_{α−1}}(A_{αα−1})/Q_{α−1} ≅ Σ_{m}×Σ_{l} but G_{αα−1}/Q_{α−1} ≅ Z_{2}≀Σ_{r}×Σ_{l}. This shows |T_{α}Q_{α−1}/Q_{α−1}| = 2 and |A_{αα−1}| = 2. Then G_{αα−1}/O_{2}(G_{αα−1}) ≅ Σ_{n−2}. Now let first L be of type Lie(q). In A_{αα−1}[Y_{αα−1}∩Y_{α−1}, O^{p}(N_{α−1}∩G_{α})] there are always uniqueness elements. So we get [Y_{αα−1}, O^{p}(N_{α−1}∩G_{α})] ≰ C I_{Y_{α}}(N_{α}). Then also [C I_{Y_{α}}(Q_{αα−1}), O^{p}(N_{α−1} ∩ G_{α})] ≰ C I_{Y_{α}}(N_{α}), but the commutator is trivial as C I_{Y_{α}}(Q_{αα−1}) is one dimensional. So let now L be of type Σ_{m}. We have that G_{αα−1} is maximal in G_{α}, O_{p}(G_{αα−1}) contains an offender and G_{αα−1} contains a point stabilizer. This shows N_{α} ∩ G_{α−1}/Q_{α} ≅ Z_{2} × Σ_{m−2}. Then we have n = m and so L is of type Σ_{n} too. Further we now have that G_{α} and G_{α−1} have a common Sylow 2subgroup. Now A_{αβ} contains uniqueness elements as [A_{αα−1}, G_{αα−1}] = 1 and (3.8). But we had [N_{α}, A_{αα−1}] = 1 and so N_{α} ≤ G_{α−1}, a contradiction.

6 The conclusion

Lemma 6.1 Let b > 2, then Y_{L}Y_{P} ◁ L Proof: By (4.4), (4.5), (4.8) and (5.7) we have that for a critical pair (α, α′) b is even, α is of type P with P of type Σ_{n} and there are no critical pairs with α of type L. First we prove

(1) If β ∈ Δ(α) then Y_{β}Y_{α} ⊴ G_{α}.

We have Y_{β} ≤ G_{α′}. Now choose X_{α′} as in (5.2). If |Y_{α}Q_{α′}/Q_{α′}| > 2, then (5.2) implies [Y_{β}, X] ≤ [Y_{α}, Y_{α′}], the assertion. Hence we may assume that |Y_{α}Q_{α′}/Q_{α′}| = 2. Then Y_{α′} ∩ Q_{α} is a hyperplane in Y_{α′} and so [Y_{β}, Y_{α′}] ≤ [Y_{α}, Y_{α′}][Y_{β}, Y_{α′} ∩ Q_{α}] ≤ Y_{α}Y_{β}. Hence again Y_{α}Y_{β} ◁ G_{α}. Now for δ ∈ Δ^{2}(α) set A_{δα} = [Y_{δ}, V_{α}^{(b−2)}]. Choose (α, α′) such that |Y_{α}Q_{α′}/Q_{α′}| is maximal. We now are going to show that one of the following cases holds.

\underline{Case 1:} [A_{δα}, N_{α}] = 1 \underline{Case 2:} n = 2m, |Y_{α}Q_{α′}/Q_{α′}| = 2^{m−1} and X_{α′} ≤ N((Y_{δ} ∩ Q_{α′−1})Y_{α}) for X_{α′} as in (5.2) or X_{α′} = Y_{α′}.

Let X_{α′} ≰ N((Y_{δ}∩Q_{α′−1})Y_{α}). Then because of (5.2) we have |Y_{α}Q_{α′}/Q_{α′}| = 2 and so X_{α′} = Y_{α′}. Now [Y_{δ} ∩ Q_{α′−1}, Y_{α′}] ≰ [Y_{α}, Y_{α′}][Y_{δ} ∩ Q_{α′−1}, Y_{α′} ∩ Q_{α}]. The only possibility for Y_{α}(Y_{δ} ∩ Q_{α′−1}) is to be ⟨(1, 2), (3, 4) … (n − 1, n)⟩. Then |Y_{δ} ∩ Q_{α′−1}/C I_{Y_{δ}∩Q_{α′−1}}(Y_{α′} ∩ Q_{α})| = 2. But |[Y_{δ} ∩ Q_{α′−1}, Y_{α′} ∩ Q_{α}]| > 2, a contradiction. Hence in any case we have X_{α′} ≤ N(Y_{α}(Y_{δ} ∩Q_{α′−1})). Set A = [Y_{δ}, Y_{α′−2}][(Y_{δ} ∩ Q_{α′−1})Y_{α}, V_{α}^{(b−2)}]. Then X ≤ N(A). If A = A_{δα}, then because of Q_{β} ≤ Q_{α}, we get case 2. So let A ≠ A_{δα}. Then the only possibility again is Y_{α′−2} corresponds to ⟨(1, 2), ⋯, (m − 3, m − 2)⟩ thus |Y_{α′−2}Q_{δ}/Q_{δ}| = 2^{m−1}. But then this is also true for the pair (α, α′).

(2) In V_{α}^{(2)}/Y_{α} there is exactly one non trivial chief factor which is an F - module.

Let again T_{δα} be the group generated by the offenders in Q_{αδ}. Then as usual we get [C I_{Y_{δ}}(T_{δα}), N_{α}] ≤ Y_{α}, as [C I_{Y_{δ}}(T_{δ_{α}})Y_{α}, X_{α′}] ≤ [Y_{α}, Y_{α′}]. Set W_{α}/Y_{α} = C I_{V_{α}^{(2)}/Y_{α}}(N_{α}). We have that C I_{Y_{δ}}(T_{δα})Y_{α} = C I_{Y_{μ}}(T_{μα})Y_{α} for some μ ∈ Δ(α+1).

As V_{α′−2}^{(2)} induces some offender on Y_{μ} we see [V_{α′−2}^{(2)} ∩ Q_{α}, C I_{Y_{μ}}(T_{μα})Y_{α}] = 1, and then [V_{α′−2}^{(2)} ∩ Q_{α}, C I_{Y_{δ}}(T_{δα})] = 1. The action on the permutation module shows [Y_{δ}, V_{α′−2}^{(2)} ∩ Q_{α}] ≤ C I_{Y_{δ}}(T_{δα}). Hence [V_{α}^{(2)}, V_{α′−2}^{(2)} ∩ Q_{α}] ≤ W_{α}. In the same way we obtain [V_{α−2}^{(2)} ∩ Q_{α′−2}, V_{α′−2}^{(2)}] ≤ W_{α′−2}. Hence V_{α}^{(2)}/W_{α} is an F -module with offender V_{α′−2}^{(2)}. In particular there is just one nontrivial chief factor. Let now first case 1. We are going to show that |[V_{α}^{(2)}, V_{α}^{(b−2)}]| = 2. For this we choose δ ∈ Δ^{2}(α) with [Y_{δ}, V_{α}^{(b−2)}] ≠ 1. Let x ∈ Y_{δ} with [x, V_{α}^{(b−2)}] ≠ 1 and |[x, V_{α}^{(b−2)}]| minimal. Then the action of V_{α}^{(b−2)} on Y_{δ} shows that |[x, V_{α}^{(b−2)}]| ≤ 4. Then [[V_{α}^{(b−2)}, Y_{δ}], N_{α}] = 1. This implies [⟨x^{N_{α}}⟩, V_{α}^{(b−2)}] = [x, V_{α}^{(b−2)}]. Let again W_{α}/Y_{α} = C I_{V_{α}^{(2)}/Y_{α}}(N_{α}). If x ∈ W_{α}, then x ∈ V_{μ} for all

μ ∈ Δ(α). Hence [x, V_{μ}^{(b−3)}] = 1 and then [x, V_{α}^{(b−2)}] = 1, a contradiction. So we have x ∉ W_{α}. Then by (2) we have V_{α}^{(2)} = ⟨x^{N_{α}}⟩V_{β}. This shows [V_{α}^{(2)}, V_{β}^{(b−3)}] = [⟨x^{N_{α}}⟩, V_{β}^{(b−3)}] ≤ [x, V_{α}^{(b−2)}]. Conjugation with N_{α} shows [V_{α}^{(2)}, V_{α}^{(b−2)}] ≤ [x, V_{α}^{(b−2)}] and so |[Y_{δ}, V_{α}^{(b−2)}]| ≤ 4. But in that case we may choose x such that |[x, V_{α}^{(b−2)}]| = 2. This now implies |[V_{α}^{(2)}, V_{α}^{(b−2)}]| = 2. Next we are going to prove that b = 4. For this assume b > 4. Set A_{α} = [V_{α}^{(2)}, V_{α}^{(b−2)}]. As just shown A_{α} ≤ Z(G_{α}). As X_{α′} ≤ N((Y_{α−2} ∩ Q_{α′−1})Y_{α}) we have Y_{α−2} ≰ Q_{α′−2}, otherwise [Y_{α−2}, Y_{α′−2}Y_{α′−3}] = 1. By (1) Y_{α′−2}Y_{α′−3} = Y_{α′−2}Y_{α′−1}. Hence Y_{α−2} ≤ Q_{α′−1} and then X_{α′} ≤ N(Y_{α−2}Y_{α}) contradicting (2). So we have that also (α − 2, α′ − 2) is critical. In that way we get a critical pair (α − b, α). Then A_{α−2} = [V_{α−2}^{(b−2)}, Y_{α}]. Hence [A_{α−2}, N_{α}] ≠ 1. Now choose α − 2 such that [A_{α−2}, Y_{α′}] ≠ 1. Let (α − 4, α′ − 4) be critical. Then [Y_{α−4}, Y_{α′−4}] = A_{α−2}, which is centralized by Y_{α′}, a contradiction.

(3) b = 4

We just have to consider case 2. We saw that with (α, α′) critical also (α−2, α′−2) was critical of the same type. Now choose (α−b, α) be to critical of the same type. Then |(Q_{α−2} ∩ Q_{β})Q_{α}/Q_{α}| ≥ 2^{m−1}. Hence we may choose Y_{α′} such that ⟨Q_{α−2} ∩ Q_{β}, Y_{α′}⟩ ≥ N_{α}. In particular [[Y_{α−2}, Y_{α′−2}], N_{α}] = 1. Let α − 4 be conjugate under N_{α−2} to α such that (α − 4, α′ − 4) is critical. Then we have [[Y_{α−4}, Y_{α′−4}], N_{α−2}] = 1. As α is conjugate to α − 4 under N_{α−2} we see O_{2}(N_{α−2}) ≤ Q_{α−4} ∩ Q_{α−3} and so [Y_{α−4}, Y_{α′−4}] is centralized by ⟨O_{2}N_{α−2}, Y_{α′}⟩ ≥ N_{α}. Now again as α ∼ α − 4 under N_{α−2} we get [[Y_{α−4}, Y_{α′−4}], N_{α−4}] = 1, a contradiction. So we have (3). Now let again be case 1. By (2) we have in V_{α}^{(2)}/Y_{α} just one nontrivial chief factor. Let first [Y_{δ}, Q_{α}]Y_{α} be normalized by N_{α}. Then [Y_{δ}, Q_{α}] ≤ ⋂ ; μ∈Δ(α)

V_{μ} ≤ Ω_{1}(Z(V_{α}^{(2)})). If N_{α} does not normalize [Y_{δ}, Q_{α}]Y_{α}, then [V_{α}^{(2)}, N_{α}] ≤ [V_{α}^{(2)}, Q_{α}] as there is just one chief factor. Then V_{α}^{(2)} = V_{β}[V_{α}^{(2)}, Q_{α}] and so V_{α}^{(2)} = V_{β}, a contradiction. Hence we have [V_{α}^{(2)}, Q_{α}] ≤ Ω_{1}(Z(V_{α}^{(2)})). In the same way one can see C I_{Y_{δ}}(T_{δα}) ≤ Ω_{1}(Z(V_{α}^{(2)})). As we know [V_{α}^{(2)}, V_{α′}^{(2)} ∩ Q_{α}] ≤ C I_{Y_{δ}}(T_{δα}) ≤ Ω_{1}(Z (V_{α}^{(2)})). Hence \overline{V}_{α} = V_{α}^{(2)}/Ω_{1}(Z(V_{α}^{(2)})) is an F -module and [\overline{V}_{α}, N_{α}, V_{α+2}^{(2)}] ≤ [\overline{V}_{α}, V_{α+2}^{(2)}] has order 2. As |V_{α}^{(2)′}| = 2, we see that \overline{V}_{α} carries a form. As the coset [\overline{V}_{α}, V_{α+2}] contains involutions we get that [V_{α}^{(2)}, N_{α}] is abelian. The 3subgroup lemma shows [V_{α}^{(2)}, N_{α}, [V_{α}^{(2)}, N_{α}]] = 1 and [[V_{α}^{(2)}, [V_{α}^{(2)}, N_{α}]], N_{α}] = 1, so also [[V_{α}^{(2)}, N_{α}], N_{α}, V_{α}^{(2)}] = 1. But [V_{α}^{(2)}, N_{α}, N_{α}] = [V_{α}^{(2)}, N_{α}], so [V_{α}^{(2)}, N_{α}, V_{α}^{(2)}] = 1. Hence [V_{α}^{(2)}, N_{α}] ≤ Z(V_{α}^{(2)}). But then V_{α}^{(2)} = Z(V_{α}^{(2)})V_{β}, which gives the contradiction V_{α}^{(2)} = Z(V_{α}^{(2)}). So let finally be case 2. Again as there is just one chief factor we have V_{α}^{(2)} = [V_{α}^{(2)}, N_{α}]V_{α+1}. Further [V_{α+1}, Y_{α+2}] = 1 and |V_{α}^{(2)}/C I_{V_{α}^{(2)}}(Y_{α+2})| =

|[V_{α}^{(2)}, N_{α}]/C I_{[V_{α}^{(2)},N_{α}]}(Y_{α+2})|. As V_{α}^{(2)} involves an F -module we see that |V_{α}^{(2)}/

C I_{V_{α}^{(2)}}(Y_{α+2})| ≤ 2^{m−1}. In particular we have V_{α}^{(2)} = [V_{α}^{(2)}, N_{α}]Ω_{1}(Z(V_{α}^{(2)}))Y_{α−2}

and so V_{α}^{(2)} = Y_{α−2}C I_{V_{α}^{(2)}}(Y_{α+2}). Hence we have, because of |Y_{α−2}Q_{α+2}/Q_{α+2}|

= 2^{m−1}, that |V_{α}^{(2)}Q_{α+2}/Q_{α+2}| = 2^{m}, a contradiction.

Proposition 6.2 If b > 1, then Y_{P} Y_{L} ◁ L

Proof: Suppose false. By (6.1) we are left with b = 2, (α, α′) is critical with α of type P. As in (6.1) we see [C I_{Y_{δ}}(T_{δ_{α}})Y_{α}, N_{α}] ≤ Y_{α}. Further we have as in (6.1) Y_{α}Y_{β} ◁ G_{α} and so Q_{α} ∩ Q_{β} ◁ G_{α}. Set W_{α} = ⟨C I_{Y_{δ}}(T_{δ_{α}})Y_{α}|δ ∈ Δ^{2}(α)⟩. Then [W_{α}, Q_{α} ∩ Q_{β}, N_{α}] = 1. This can be seen as follows. [C I_{Y_{δ}}(T_{δ_{α}}), Q_{α} ∩ Q_{β}] is a normal subgroup different from Y_{α}. As Q_{α} ∩ Q_{β} ◁ G_{α}, this group has to be centralized by N_{α}. We also have [W_{α}, N_{α}, Q_{α} ∩ Q_{β}] ≤ [Y_{α}, Q_{α} ∩ Q_{β}] = 1. By the 3-subgroup lemma we get [Q_{α} ∩ Q_{β}, N_{α}, W_{α}] = 1. As [Q_{α}, N_{α}] ≤ Q_{α} ∩ Q_{β}, we see [Q_{α}, N_{α}] ≤ C I_{Q_{α}∩Q_{β}}(W_{α}). Set F_{α} = C I_{Q_{α}∩Q_{β}}(W_{α}). Then by (5.2) we have [F_{α}, Y_{α′}] ≤ C I_{Y_{α′}}(T_{αα′}) ≤ W_{α}. Hence [F_{α}, N_{α}] ≤ W_{α} and [W_{α}, N_{α}] ≤ Y_{α}. So [Q_{α}, N_{α}] ≤ F_{α}. This shows [Q_{α}, N_{α}] = [Y_{α}, N_{α}] and so Q_{α} = C I_{Q_{α}}(N_{α})Y_{α}. Let now u ∈ Q_{α} ∩ Q_{α′} be an uniqueness element for G_{α}. Then uC I_{Q_{α′}}(N_{α′}) is an element of Y_{α′}C I_{Q_{α′}}(N_{α′}), which has Y_{α}Q_{α′}/Q_{α′} in the largest normal 2-subgroup of its centralizer and also an offender in the O_{2} of the point stabilizer. As u cannot be an uniqueness element in G_{α′} as well, we get for the centralizer Z_{2} × H_{α′} where H_{α′} ≅ Σ_{n}. Further there is no fours group of uniqueness elements in Q_{α} ∩ Q_{α′}, as their centralizers would generate N_{α′}. Then N_{α′} ≤ C̃_{α} and so Y_{α} ≤ O_{2}(G_{α′}), a contradiction. Hence there is no fours group centralized by S according to (3.8). This shows N_{α} ≅ Σ_{m}, m = 2^{t}, n = 2^{t} − 2. Now H_{α′} ≤ C̃_{α}. We have [Q_{α}, N_{α}] = [Y_{α}, N_{α}] and [[Y_{α}, N_{α}], H_{α′}] ≤ [Y_{α′}, H_{α′}]. So we get [Y_{α}, N_{α}][Y_{α′}, H_{α′}] ⊴ ⟨N_{α}, H_{α′}⟩. Assume first [Y_{α′}, H_{α′}] ≤ [Y_{α}, N_{α}]. Let ⟨y⟩ = [Y_{α′}, Y_{α}] and x be the central element in Y_{α}, x′ the one in Y_{α′}. Then x is the uniqueness element in Y_{α}. Thus x′ ∈ ⟨x, y⟩, x′ ∈ C I(N_{α′}). Then as [H_{α′}, Y_{α′}] ≤ [Y_{α}, N_{α}] this yields yx = yx′ and so x = x′, a contradiction. So we have [[Y_{α}, N_{α}], H_{α′}] = 1. As N_{α} = O^{2}(N_{α}) ◁ ⟨N_{α}, H_{α′}⟩, we see [N_{α}, H_{α′}] = 1. Denote by U_{α} = [Y_{α}, N_{α}] and Z_{α} = C I_{Y_{α′}}(N_{α}), R = [U_{α}, U_{α′}]. From [U_{α}, H_{α′}] = 1 we get

(1) U_{α} ∩ U_{α′} ≤ [U_{α}, U_{α′}]Z_{α}

(2) It δ ∈ Δ(β) with R = [U_{δ}, U_{α′}], then [U_{δ}, U_{α}] = 1.

Let (δ, α) be a critical pair. Then R ≤ U_{δ} ∩ U_{α} ≤ [U_{δ}, U_{α}]Z_{α}. This shows U_{δ}Q_{α} = U_{α′}Q_{α} as [U_{δ}, U_{α}] = [U_{α′}, U_{α}] and there is just one transvection to a point. Set F_{α} = C I_{Q_{α}}(N_{α}), \overline{G}_{α} = G_{α}/F_{α}. Then \overline{U_{δ}Q_{α}} possesses exactly two maximal elementary abelian subgroups. Thus U_{δ}, U_{α′} are in one of them. This means R = [U_{δ}, U_{α′}] ≤ F_{α}. But R = [U_{α}, U_{α′}] was not in F_{α}, a contradiction. This shows (2).

(3) Let δ be as in (2) then δ = α.

Let μ, μ′ ∈ Δ(β) with [U_{μ}, U_{μ′}] = R. Then, maybe after changing μ and μ′, we get

(∗) [U_{μ}, U_{μ′}] = R = [U_{α}, U_{α′}] = [U_{μ}, U_{α′}] = [U_{μ′}, U_{α}] and [U_{μ}, U_{α}] = 1 = [U_{μ′}, U_{α′}].

Let [U_{α}, U_{μ}] ≠ 1. Then, as [U_{α}, U_{α′}] = [U_{μ}, U_{μ′}] ≤ U_{α} ∩ U_{μ}, we get R ≤ U_{α} ∩ U_{μ} ≤ [U_{α}, U_{μ}]Z_{α}. Thus R = [U_{α}, U_{μ}] and with (2) we get [U_{μ}, U_{α′}] = 1. Thus without loss we may assume [U_{α}, U_{μ}] = 1. If also [U_{α}, U_{μ′}] = 1, then R = [U_{μ}, U_{μ′}] ≤ Qα^{′} ≤ F_{α}, a contradiction. Thus (α, μ′) is a critical pair. This shows R ≤ U_{α} ∩ U_{μ′} ≤ [U_{α}, U_{μ′}]Z_{α} and so R = [U_{α}, U_{μ′}]. By (2) we get [U_{μ′}, U_{α′}] = 1. By symmetry we get also [U_{μ}, U_{α′}] = R. We define a graph Γ by the following. The vertices of Γ are those δ ∈ Δ(β). We have an edge (δ, δ′) if there is a critical pair (δ, δ′) with R = [U_{δ}, U_{δ′}]. We just have shown that Γ is a complete bipartite graph with partition Γ^{+}, Γ^{−} where [U_{δ}, U_{δ′}] = 1 iff δ, δ′ ∈ Γ^{+} or δ, δ′ ∈ Γ^{−}. Let α ∈ Γ^{+}. Set V = ⟨U_{δ}|δ ∈ Γ^{+}⟩. As [D_{α}, R] = 1 we see that D_{α} acts on Γ. (Recall D_{α} centralizes Y_{P} , P = N(L ∩ C̃), N ◁ ◁C̃, D = S_{L}(C̃), L ≰ C̃). As α^{D_{α}} = α we see that D_{α} acts on Γ^{+}. Thus D_{α} ≤ N_{G_{α}}(V). We have V ≤ C I(U_{α}) ≤ Q_{α} and so [V, N_{α}] ≤ U_{α} ≤ V. Thus N_{α} ≤ N_{G_{α}}(V). As α^{Q_{β}} = α, and [Q_{β}, R] = 1, we have Q_{β} ≤ N(V). Thus V ⊴D_{α}N_{α}Q_{β} = G_{α}. This shows N_{G_{β}}(V) = G_{αβ}. As V is independent of α we see N_{G_{β}}(V) = G_{αβ} = G_{δβ} for all δ ∈ Γ^{+}. As G_{αβ} ≰ G_{β}, we get Γ^{+} = {α}. This shows δ = α and so (3).

(4) Let g ∈ N_{α} ∩ G_{β} then (α′, α^{′g}) is not a critical pair.

Consider again [U_{α′}, H_{α′}]. This is centralized by N_{α}. Thus [U_{α′}, H_{α′}] ≤ U_{α′} ∩ U_{α′}^{g}. As n + 2 is a power of 2, [U_{α′}, H_{α′}] is to large. (Recall if (α′, α^{′g}) is critical, U_{α′} ∩ U_{α′}^{g} ≤ [U_{α′}, U_{α′}^{g}]Z_{α′}and so (|U_{α} ∩ U_{α′}^{g}| ≤ 4). Hence we have (4).

(5) G_{αβ} = N_{G_{α}}([U_{α}, V_{β}]) and G_{αβ} ∩ N_{α}/Q_{α} ≅ Σ_{k} × (Z_{2} ≀ Σ_{m}).

We know that V_{β}Q_{α}/Q_{α} is contained in a transposition group. This im-

plies Vβ^{′}F_{α} = [U_{α}, V_{β}]F_{α}. Now set H = N_{N_{α}}(Vβ^{′}F_{α}/F_{α}). We have Vβ^{′} =

[U_{α}, V_{β}](Vβ^{′}∩F_{α}) = (Vβ^{′}∩U_{α})(Vβ^{′}∩F_{α}). This implies N_{N_{α}}(Vβ^{′}) = N_{N_{α}}(Vβ^{′}∩U_{α}).

Hence G_{αβ} = N_{G_{α}}(Vβ^{′} ∩U_{α}) as G_{αβ} = N_{G_{α}}(Vβ^{′}). This shows Vβ^{′} ∩U_{α} = [U_{α}, V_{β}]

or [U_{α}, V_{β}]Z_{α}. Thus G_{αβ} = N_{G_{α}}([U_{α}, V_{β}]). This shows G_{αβ} ∩ N_{α}/Q_{α} ≅ Σ_{k} × (Z_{2} ≀ Σ_{m}). This is (5). If m = 1, then V_{β}Q_{α} = Y_{α′}Q_{α} and so Δ(β) = {α, α′}. But G_{αβ} ≰ G_{β}. Thus m > 1. Set V = ⟨Y_{α′}^{N_{α}∩G_{β}}⟩. If g ∈ N_{α} ∩ G_{β} then [Y_{α′}^{g}, Y_{α′}] = 1 by (4). Thus V is abelian and V ≤ Q_{α′}. Let g ∈ N_{α} ∩ G_{β}. Then (α, α^{′g}) is a critical pair. By (3) we have [[U_{α}, U_{α′}^{g}], D_{α}] = 1. We have D_{α} ≤ G_{α} ∩ G_{α′}^{g} and so D_{α} ≤ N_{G}(V). Then G_{α} = D_{α}(N_{α} ∩ G_{β})Q_{β} ≤ N_{G}(V). This implies [V, N_{α′}] ≤ [Q_{α′}, N_{α′}] ≤ U_{α′} ≤ Y_{α′} ≤ V. Hence N_{α′} ≤ N_{G}(V). Then also N_{α′} ∩ G_{β} ≤ N_{G_{β}}(V) = G_{αβ}, as G_{β} ≰ N_{G}(V). Hence N_{α′} ∩ G_{β} ≤ G_{α}. But as m > 1 we have that N_{α′} ∩ G_{β} cannot normalize Y_{α}, a contradiction which proves the proposition.

Now in any case we have that Y_{P} Y_{L} ◁ L. We easily see that O_{p}(⟨O_{p}(P)^{L}⟩) ≤ O_{p}(P), as O_{p}(L) ∩ O_{p}(P) ◁ L. But then we have (P U) for L̃ = ⟨O_{p}(P)^{L}⟩, a contradiction. This proves the theorem.

References

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[Cher] A. Chermak, An elementary proof of the R(G, V)-theorem, Journal of Group Theory 2 (1999), 1 - 13.

[MF] U. Meierfrankenfeld, pushing-up

[MSS] U. Meierfrankenfeld, B. Stellmacher, G. Stroth, Finite groups of local characteristic p, An overview, Procecchings LMS, to appear

[MSS1] U. Meierfrankenfeld, B. Stellmacher, G. Stroth, Groups of local characteristic p, The strucutre theorem, preprint [MS] U. Meierfrankenfeld, G. Stroth, The non E-uniqueness case,

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[Stell] B. Stellmacher