Groups of local characteristic p, The H – structure theorem

U. Meierfrankenfeld, B. Stellmacher, G. Stroth

Preliminary version of May 5, 2004

1 Introduction

In this paper we will consider groups of local characteristic p such that E– uniqueness holds and for some p–local M we have Y_{M} ≰ Q. To be more precise we have to recall some of the definitions of [MeStStr].

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 X containing S, with F^{∗}(X) = O_{p}(X). We will assume that

1) O_{p}(⟨X|X ∈ L(S)⟩) = 1

2) G is a so called K_{p}–group, i.e. any simple group involved in any p-local is abelian, an alternating group, a group of Lie type or one of the 26 sporadic groups.

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 [MeStStr].

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)).

E–uniqueness If X is some p–local of G with E ≤ X, then X ≤ C̃.

In particular C̃ above was uniquely determined. In [MeStStr] we determined the p–locals M which are not contained in C̃.

Let Q = O_{p}(C̃). Then for M ∈ L(S) we set M^{0} = ⟨Q^{M}⟩. As we have seen in [MeStStr] M = M^{0}(M ∩ C̃), we have that M^{0} dominates the structure of M. Set M_{0} = M^{0}S.

Further we will assume that there is at least one p–local M such that Y_{M} ≰ Q. Such p–locals in L(S) which are not in C̃ are described in [MeStStr]. We will use this result freely in what follows. We will determine some subgroup H of G, where H has the following properties.

1) M_{0} ≤ H.

2) O_{p}(H) = 1.

3) F^{∗}(H) is weak parabolic isomorphic to a finite simple group.

A subgroup P ∈ L(S) is called a minimal parabolic if S is contained in a unique maximal subgroup of P and S ⋬P.

We say two groups U, W are weak parabolic isomorphic, if there are Sylow p–subgroups S, T of U, W respectively, minimal parabolic subgroups P_{1}, ⋯, P_{n} ∈ L(S) of U and minimal parabolic subgroups R_{1}, ⋯, R_{n} ∈ L(T) of W such that if we set ⟨P_{1}, ⋯, P_{n−1}⟩ = U_{1}, ⟨R_{1}, ⋯, R_{n−1}⟩ = W_{1} and ⟨P_{2}, ⋯ P_{n}⟩ = U_{n}, ⟨R_{2}, ⋯, R_{n}⟩ = W_{n}, then

• U_{1}/O_{p}(U_{1}) ≅ W_{1}/O_{p}(W_{1}), U_{n}/O_{p}(U_{n}) ≅ W_{n}/O_{p}(W_{n}), where these isomorphisms have the property that they are the same on U_{1} ∩ U_{n}.

• The action on the chieffactors of U_{i}, i = 1, n in O_{p}(U_{i}) is the same as the action of W_{i}, i = 1, n on the chieffactors in O_{p}(W_{i}).

• ⟨P_{1}, P_{n}⟩/O_{p}(⟨P_{1}, P_{n}⟩) ≅ ⟨R_{1}, R_{n}⟩/O_{p}(⟨R_{1}, R_{n}⟩)

This then enables us to introduce some diagram for the parabolic system {P_{1}, ⋯, P_{n}} and {R_{1}, ⋯, R_{n}} as well. If U and W are weak parabolic isomorphic, then it follows that the diagrams are the same. In fact for the groups ocurring in our theorem there are results available which classify them by the diagram. But of course we have much more information, so an easy classification should be available.

To state the theorem in an easy way we just give for the groups occurring, what is meant by weak parabolic isomorphic.

Definition 1.1 Let G be a simple group with parabolic system {P_{1}, ⋯ P_{n}}. Then the groups U_{1}, U_{n} above are the following (1) G ≅ L_{m+r}(q), the parabolics correspond to the A_{n+r−1}-diagram, U_{1} is

the normalizer of a root group and U_{n} is the stabilizer of an n-space in the natural representation.

(2) G ≅ Ω^{±}(n, q), the parabolics correspond to the D_{m}-diagram if G ≅ Ω^{+}(2m, q) or to the C_{i} diagram else. Now U_{1} is the point stabilizer in the natural representation and U_{n} is the normalizer of a root group.

(3) G ≅ Ω^{±}(n, q), the parabolics correspond to the D_{m}-diagram if G ≅ Ω^{+}(2m, q) or to the C_{i} diagram else. Now U_{1} is the point stabilizer and U_{n} is the stabilizer of a maximal isotropic subspace in the natural representation.

(4) G ≅ U_{n}(q), the parabolics correspond to the unitary diagram and U_{1} is point stabilizer U_{n} is the stabilizer of a maximal singular space in the natural representation.

(5) G ≅ E_{6}(q), the parabolics correspond to the E_{6}–diagram and U_{1} is the normalizer of a root group and U_{n} corresponds to one of the maximal parabolics which normalize an elementary abelian group of order q^{16}.

(6) G ≅ E_{7}(q), the parabolics correspond to the E_{7}–diagram and U_{1} is the normalizer of a root group and E(U_{n}/O_{p}(U_{n})) ≅ E_{6}(q).

(7) G ≅ E_{8}(q), the parabolics correspond to the E_{8}–diagram and U_{1} is the normalizer of a root group and E(U_{n}/O_{p}(U_{n})) ≅ Ω^{+}(14, q).

(8) G ≅ Sp(2n, q), q odd, the parabolics correspond to the C_{n} diagram. Now U_{1} is the point stabilizer and U_{n} is the stabilizer of a maximal isotropic subspace in the natural representation.

(9) G ≅ F_{4}(q), q odd. The parabolics correspond to the F_{4}–diagram. Now U_{1} is the normalizer of a root group and E(U_{n}/O_{p}(U_{n})) ≅ Ω(7, q).

(10) G ≅ ^{2}E_{6}(q), the parabolics correspond to the F_{4}–diagram, U_{1} is the normalizer of a root group and E(U_{n}/O_{p}(U_{n})) ≅ Ω^{−}(8, q).

(11) G ≅ U_{4}(3), p = 2. The parabolics correspond to the diagram

1∘  ^{2}∘  ^{3}∘

and U_{1}, U_{n} are extensions of an elementary abelian group of order 16 by A_{6}.

(12) G ≅ Suz, p = 2. The parabolics correspond to the diagram

1∘  ^{2}∘  ^{3}∘

and U_{1} is the centralizer of a 2–central involution and U_{n} is an entension of a 2-group of order 2^{10} by 3A_{6}.

(13) G ≅ J_{4}, p = 2. The parabolics correspond to the diagram

1∘ ^{2}∘ M̃_{22} ^{3}∘

and U_{1} is the centralizer of a 2-central involution and U_{n} = ⟨P_{1}, P_{2}, P_{31}⟩ is an extension of an elementary abelian group of order 2^{11} by M_{24}, where P_{31} is the proper minimal parabolic in P_{3}.

(14) G ≅ Co_{2}, p = 2. The parabolics correspond to the diagram

1∘ ^{2}∘ M_{22} ^{3}∘

and U_{1} is an extension of an elementary abelian group of order 2^{10} by Aut(M_{22}) and U_{n} = ⟨P_{1}, P_{2}, P_{31}⟩, where P_{31} is the proper minimal parabolic in P_{3}, is the centralizer of a 2-central involution.

(15) G ≅ Co_{1}, p = 2. The parabolics correspond to the diagram

1∘  ^{2}∘  ^{3}∘ ∼ ^{4}∘

and U_{1} is an extension of an elementary abelian group of order 2^{11} by M_{24} and U_{n} = ⟨P_{1}, P_{2}, P_{3}, P_{14}⟩, where P_{2}, P_{3}, P_{4}, P_{14} are the four minimal parabolics in U_{1} containing a Sylow 2-subgroup is the centralizer of a 2-central involution.

(16) G ≅ Co_{1}, p = 3. The parabolics correspond to the diagram

1∘ ^{2}∘ M_{12} ^{3}∘

and U_{1} is an extension of an elementary abelian 3-group of order 3^{6} by 2M_{12} and U_{n} is the normalizer of a group of order three in the center of a Sylow 3–subgroup.

(17) G ≅ M(24)′, p = 2. The parabolics correspond to the diagram

1∘  ^{2}∘ ∼ ^{3}∘  ^{4}∘

and U_{1} is the centralizer of a 2-central involution and U_{n} is an extension of an elementary abelian group of order 2^{11} by M_{24}.

(18) G ≅ M(24)′, p = 3. The parabolics correspond to the diagram

1∘ c ^{2}∘  ^{3}∘  ^{4}∘

and U_{1} is an extension of 3^{1+10} by U_{5}(2) : 2 and U_{n} is 3^{7}O(7, 3).

(19) G ≅ He, p = 2. The parabolics correspond to the diagram

1∘ ^{2}∘ ∼ ^{3}∘

and U_{n} is the centralizer of a 2–central involution and U_{1} is an extension of an elementary abelian group of order 2^{6} by 3Σ_{6}.

(20) G ≅ F_{2}, p = 2. The parabolics correspond to the diagram

1∘  ^{2}∘  ^{3}∘ M_{22} ^{4}∘

and U_{1} = ⟨P_{2}, P_{3}, P_{4}⟩ is the centralizer of a 2-central involution and U_{n} = ⟨P_{1}, P_{2}, P_{3}, P_{41}⟩ is an extension of a 2-group of order 2^{25} by Sp(8, 2), where P_{41} is the proper minimal parabolic in P_{4}.

(21) G ≅ F_{1}, p = 2. The parabolics correspond to the diagram

1∘  ^{2}∘  ^{3}∘  ^{4}∘ ∼ ^{5}∘

and U_{1} is the centralizer of a 2–central involution, U_{n}/O_{2}(U_{n}) ≅ Ω^{+}(10, 2) and O_{2}(U_{n}) is an extension of the natural module by the half spin module.

(22) G ≅ F_{1}, p = 3. The parabolics correspond to the diagram

1∘ c ^{2}∘  ^{3}∘  ^{4}∘

and U_{1} ≅ 3^{1+12}2Suz : 2 and U_{n} ≅ 3^{8}Ω^{−}(8, 3).

As one can see from the list of groups above, we are not just aiming for finite simple groups all of whose p-locals are of characteristic p-type. All we assume in this paper is that all p–locals P containing a Sylow p–subgroup S are of characteristic p-type, i.e. P ∈ L(S).

Theorem 1.2 Let p be a prime and G be a finite K_{p}–group and S be a Sylow p–subgroup. Assume that all p-locals P containing S are in L(S). Assume O_{p}(⟨X|X ∈ L(S)⟩) = 1 and further E – uniqueness. Let M with M^{0} be maximal and Y_{M} ≰ Q. Then one of the following holds

(1) There is H with M_{0} ≤ H and O_{p}(H) = 1, such that F^{∗}(H) is weak parabolic isomorphic to one of the following groups

i) a group of Lie type in characteristic p

ii) p = 2, and we have He, Co_{2}, Co_{1}, M(24)′, J_{4}, Suz, F_{2}, F_{1} or U_{4}(3).

iii) p = 3 and we have M(24)′, Co_{1} or F_{1}.

(2) p = 2, M is an extension of an elementary abelian group of order 16 by L_{3}(2), C̃ is an extension of an extraspecial group of order 32 by Σ_{3}×Σ_{3}. Further there are minimal parabolics P_{1}, P_{2} with P_{1}/O_{2}(P_{1}) ≅ P_{2}/O_{2}(P_{2}) ≅ Σ_{3} and O_{2}(⟨P_{1}, P_{2}⟩) = 1.

(3) p = 3, M and C̃ are as in Co_{3}. There are two minimal parabolics P_{1}, P_{2} such that P_{1}/O_{3}(P_{1}) ≅ L_{2}(9), P_{2}/O_{3}(P_{2}) ≅ SL(2, 9) and O_{3}(⟨P_{1}, P_{2}⟩) = 1.

(4) M_{0} is a minimal parabolic.

Recall that in the exceptional case (4) of the theorem, we have that G is a rank two group, i.e. there is a minimal parabolic P ∈ L(S) such that O_{p}(⟨M_{0}, P ⟩) = 1. These groups and also the cases (2) and (3) will be treated somewhere else.

Suppose that our group G is simple, then in most of the cases H = G. Unfortunately there are a few exceptions like Ω^{+}(8, 2) in Ω^{+}(8, 3). In this case p = 2. There are also examples for p = 3 like Ω(7, 3) in M(22). To classify all these exceptions will be the content of a forthcoming paper due to A. Hirn.

As this paper might be part of a revision of the classification of the finite simple groups some words about quotations are in order. To pin down the structure of the p–locals M and C̃, which is the aim of the proof, one needs quite a many results about representations of the finite simple groups, like quadratic representations, F –modules, extensions of modules by trivial modules. These results will all be in [GBSM]. For some of them we have quoted this source but stated the results in this paper, for others we used the original papers. Besides this there is just one further quotation, which is to the Tits classification of buildings. In fact we could have added this to this paper too, as it is an exercise for the reader. We just used from [Tits], that a geometry which has a Dynkin diagram of type A_{n}, n ≥ 3, is a projective space. Then a flag transitive group is a subgroup of P ΓL(n + 1, q). As we have enough information about subgroups of our target group, we then get that this is an automorphism group of P SL(n + 1, q).

2 Some modules

dihed

Lemma 2.1 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: [GBSM]

F F 1

Lemma 2.2 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), q even, 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).

pointstab

Lemma 2.3 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: [GBSM]

2F odd

Lemma 2.4 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: [GM]

2F spor

Lemma 2.5 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: [GM]

quad

Lemma 2.6 Let K be a component of G and V be a GF (p)–module for G with [V, K] ≠ 1. Let A be a quadratic group on V, then one of the following holds

(i) [K, A] ≤ K

(ii) p = 2 A ≠ N_{A}(K), |A/C_{A}(K)| = 2

(iii) p = 2 K ≅ SL_{2}(2^{k}) and |A/N_{A}(K)| = 2

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

Proof: [Cher2]

The next two lemmas are also in [GBSM]. For convenience of the reader we give full proofs here.

split1

Lemma 2.7 Let X ≅ A_{n}, n ≥ 5, V be a GF (2)X-module with [V, X] the natural irreducible permutation module. Assume C_{V} (X) = 1. Then |V : [V, X]| ≤ 2, and V = [V, X] if n is odd. Furthermore V is a factor of the permutation module

Proof: 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 [V, X] = [V, K] is the permutation module for K. By induction V = [V, K] ⨁ T̃. 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 permutation module for K 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.

split

Lemma 2.8 (a) Let L be SL_{n}(q), q = p^{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) Let L ≅ Ω(n, q), q = p^{f}, n ≥ 3 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 ≅ Ω_{6}^{+}(2), and |V | = 2^{7}.

(ii) L ≅ Ω(3, 5) and |V | = 5^{4}

(iii) L ≅ Ω^{−}(4, 3) and |V | = 3^{5}

(iv) L ≅ Ω(5, 3) and |V | = 3^{6}.

(c) Let L be SL_{n}(q), n ≥ 5, q = p^{f}, and V be a module over GF (p) with [V, L] the symmetric square of the natural module and C_{V} (L) = 1. Then V = [V, L].

(d) Let L be SL_{n}(q), n ≥ 3, q = p^{f} be odd, and V be a module over GF (p) with [V, L] the alternating square of the natural module and C_{V} (L) = 1. Then V = [V, L].

(e) Let L ≅ Ω^{±}(2n, q), q = p^{f}, and V be a GF (p)–module with [V, L] a half spin module and C_{V} (L) = 1. Then V = [V, L].

(f) Let L ≅ Sp(2n, q), n ≥ 2, q even, and V be a GF (2)–module with [V, L] the natural module and C_{V} (L) = 1. Then |V : [V, L]| ≤ q.

(g) Let L ≅ E_{6}(q), q = p^{f}, and V be a module over GF (p) with [V, L] be a 27–dimensional module. If C_{V} (L) = 1, then V = [V, L].

Proof: (a) 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 [Hu, (I.17.4)] shows C_{V} (P′) = 1. Hence as |V : C_{V} (O_{p}(P))| = q, we get V = [V, L].

(b) Let first L ≅ Ω(3, q) ≅ L_{2}(q). Let W be a foursgroup in L. Then C_{[V,L]}(W) = 1, hence we get V = [V, L]C_{V} (W). As N_{L}(W) acts on C_{V} (W), we see that V = [V, L]C_{V} (N_{L}(W)). Now let w ∈ W^{♯}, then C_{L}(w) acts on C_{[V,L]}(w). Hence we see that O^{2}(C_{L}(w)) centralizes C_{[V,L]}(w). This now shows that V = [V, L]C_{V} (X), where X = ⟨C_{L}(w), N_{L}(W)⟩. If this is different from L, we must have that C_{L}(w) is a Sylow 2–subgroup. If C_{L}(w) is not of order 4, then there is a fours group W_{1} ≤ C_{L}(w) such that C_{V} (W) = C_{V} (W_{1}). Now V = [V, L]C_{V} (X), where X = ⟨N_{L}(W), N_{L}(W_{1})⟩. But X = L. So we are left with |C_{L}(w)| = 4, and so q = 5, 3. If q = 3, then L = N_{L}(W), a contradiction. So q = 5. In fact in that case the central factor of the permutation module is an example of such a non split extension. Let ρ be an element of order three in N_{L}(W). Then C_{[V,L]}(ρ) is of order five. Let τ be some involution in N_{L}(⟨ρ⟩), then |C_{V} (ρ) : C_{C_{V} (ρ)}(τ)| = 5. As ⟨N_{L}(W), τ⟩ = L, we see that C_{V} (W)∩C_{V} (τ) = 1, and so we get |C_{V} (W)| = 5, i.e. |V | = 5^{4}.

Let L ≅ Ω_{4}^{−}(q). Let ω_{1} ∈ L, o(ω_{1}) = q^{2} − 1/(p, 2). Then we see that there is some power ω of ω_{1} of order q + 1/(p, 2) such that C_{[V,L]}(ω) is of order q^{2}. Now V = [V, L]C_{V} (ω). If there is g ∈ L with ⟨ω, ω^{g}⟩ = L, we get V = [V, L]C_{V} (L), the assertion. So there is no such g. This first shows q is odd. Further we see that there is no Zsygmondi prime which divides q + 1. Hence q + 1 is a power of two. Further we easily see that q + 1/2 has to be of order two, as otherwise some involution in ⟨ω⟩ inverts some element of order q^{2} +1/2 and so two conjugates of ω under this element would generate L. So we have q = 3, and then Ω_{4}^{−}(q) ≅ A_{6} and in fact the permutation module is an example. Here we just have to show that |V : [V, L]| ≤ 3. Let |V : [V, L]| = 9. Let i be some involution, then C_{V} (i) is of order 3^{4}. Let j be some involution centralizing i, then C_{V} (⟨i, j⟩) is of order 27. Now we know that C_{[V,L]}(⟨i, j⟩) is centralized by N_{L}(⟨i, j⟩). Hence there is some element ρ of order three such that C_{V} (ρ) is of order 81. This shows that C_{V} (S) is of order at least 9 for S a sylow 3–subgroup of L. But as C_{[V,L]}(S) is of order 3, so there is some element ν not in [V, L], which is centralized by a Sylow 3–subgroup. This contradicts [Hu, (I.17.4)].

Let L ≅ Ω^{+}(4, q). We write L as L_{1}L_{2}, L_{i} ≅ SL_{2}(q). We may assume q > 2, as the assertion is obvious for q = 2. If q is odd, there is some involution in the center of Ω^{+}(4, q), which acts fixed point freely on [V, L], so V = [V, L]C_{V} (L), the assertion. So we may assume q to be odd. 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.

Let now L ≅ Ω(5, 3). There is a subgroup X ≤ L, X elementary abelian of order 16. Now V = [V, L]C_{V} (X), where C_{[V,L]}(X) = 1. So C_{V} (X) = C_{V} (N_{L}(X)). Let i be some involution and Y = O_{2}(C_{L}(i)). We have |C_{V} (X∩ Y) : C_{V} (X)| = 3|. Hence |C_{V} (Y) : C_{V} (X)| = 3. As |C_{L}(i) : C_{N_{L}(X)}(i)| = 3, we see that |C_{V} (Y) : C_{V} (N_{L}(Y))| ≤ 3 and so |C_{V} (X) : C_{C_{V} (X)}(N_{L}(Y))| ≤ 3, which shows |C_{V} (X)| ≤ 3, so |V | ≤ 3^{6}.

Let next L ≅ Ω_{6}^{+}(q), q > 2. If p > 2, then there is some involution in the center of L, thus we are done. So p = 2. Let P be the parabolic with P′/O_{p}(P) ≅ SL_{2}(q) ∗ SL_{2}(q). Then let V_{1} = C_{[V,L]}(O_{p}(P)). We have |V_{1}| = q. Further as [O_{p}(P), P] = O_{p}(P), we get [V, O_{p}(P)] = [V, L, O_{p}(P)]. Let S be a Sylow p–subgroup of P. Then |C_{[V,L,O_{p}(P)]/V_{1}}(S)| = q. As [P′, V] = [V, O_{p}(P)] we now see that V/V_{1} = [V, L]/V_{1}C_{V/V_{1}}(O_{p}(P)). But then V = [V, L]C_{V} (O_{p}(P)). As [P′, C_{[V,L]}(O_{p}(P))] = 1, we get that V = [V, L]C_{V} (P′), but then by [Hu, (I.17.4)] we get V = [V, L].

Let L ≅ Ω_{6}^{+}(2). Now A_{8} ≅ Ω_{6}^{+}(2) and [V, L] is the permutation module. Hence the assertion follows with 2.7.

Let now finally L ≅ Ω_{2n}^{−}(q), n ≥ 3, L ≅ Ω_{2n}^{+}(q), n ≥ 4, or L ≅ Ω(2n + 1, q), q odd, with n ≥ 2 and q > 3 for n = 2. Set m = 2n, 2n + 1, respectively. Let P be the parabolic with P′/O_{p}(P) ≅ Ω_{m−2}^{±}(q). Set V_{1} = C_{[V,L]}(O_{p}(P)), |V_{1}| = q. We have [P′, V] = [V, O_{p}(P)], as P′ = P′′. Hence we see |V/V_{1} : C_{V/V_{1}}(O_{p}(P))| = q. Furthermore C_{V/V_{1}}(O_{p}(P)) = [C_{V/V_{1}}(O_{p}(P)), P′]C_{V} (O_{p}(P))/V_{1}. Now P acts on C_{V} (O_{p}(P)) and so we get [C_{V} (O_{p}(P)), P′] = 1. By [Hu, (I.17.4)] we have C_{V} (O_{p}(P)) ≤ [V, L] and so [V, L] = V.

(c) Let P be the parabolic in L with O_{p}(L) be the natural module for L/O_{p}(L) ≅ GL(n − 1, q). And assume further that we have choosen P such that C_{[V,L]}(O_{p}(L)) is the natural module for L. In particular C_{[V,L]}(O_{p}(L)) is dual to O_{p}(L) as L/O_{p}(L)–module.

Let P_{1} ≤ P with O_{2}(P) ≤ P_{1} and P_{1}/O_{2}(P) ≅ A_{7} if n = 5 and q = 2 and let P_{1} = P′ else. We have [V, O_{p}(P)] = C_{[V,L]}(O_{p}(P)) as n ≥ 5. If n = 5, q = 2, we see with 2.7 that P_{1} acts on some submodule V_{1} with V = V_{1}[V, L] and V_{1} ∩ [V, L] = C_{[V,L]}(O_{p}(P_{1})). If n = 5 and q > 2, we get the same result with (a). If n > 5, we get the result by induction as [V, L]/C_{[V,L]}(O_{p}(P)) is the symmetric square of the natural module for P/O_{p}(P).

So in any case P_{1} acts on V_{1}, which is a central extension of the natural module. As O_{p}(P) is not isomorphic to C_{[V,L]}(O_{p}(P)), we see that [V_{1}, O_{p}(P)] = 1. Hence we have that V_{1} = C_{V} (O_{p}(L)). This shows that P acts on V_{1} in any case. But now by (a) we have V_{1} = (V_{1} ∩ [V, L])C_{V_{1}}(P′). Application of [Hu, (I.17.4)] shows C_{V} (P′) = 1 and so V = [V, L].

(d) Let P be a parabolic in L corresponding to an end node of the Dynkin diagram such that C_{[V,L]}(O_{p}(P)) is the alternating square for P. Further [V, L]/C_{[V,L]}(O_{p}(P)) is an extension of the trivial module by the natural module. Now we see with (a) that there is some P -module V_{1} with V = [V, L]V_{1} and V_{1} ∩ [V, L] = C_{[V,L]}(O_{p}(P)) and [V_{1}, O_{p}(L)] = 1. Suppose that L ≅ L_{3}(5). Then we have in P some element x of order 4, xO_{p}(L) ≤ Z(P/O_{p}(L)) such that C_{C_{[V,L]}(O_{p}(L))}(x) = 1. As [V_{1}, x] ≤ [V, L], we now see that V_{1} = C_{[V,L]}(O_{p}(P))C_{V_{1}}(P′). Application of [Hu, (I.17.4)] shows C_{V_{1}}(P′) = 1 and so V = [V, L].

Hence assume now that (n, q) ≠ (3, 5). Then for n = 3 we get with (a) and for n > 3 by induction that V_{1} = C_{[V,L]}(O_{p}(P))C_{V_{1}}(P′) . Hence we have the same contradiction as before.

(e) If L ≅ Ω^{+}(6, q) this is (a). Let now L ≅ Ω^{−}(6, q). Then [V, L] is the four dimensional unitary module. We will consider it this way. We have q is even, otherwise there is a fixed point free element in the center of L. Let first q > 2. There are x_{1}, x_{2}, x_{3} ∈ L, (o(ω) = q + 1)

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

x_{3} = ( \begin{matrix}ω \\ & ω \\ & & ω \\ & & & ω^{−3} \end{matrix} ) .

As q > 2, we have [V, L] = [V, x_{i}], i = 1, 2, 3. Now C_{V} (x_{1}) = C_{V} (x_{2}) = C_{V} (x_{3}). As C_{L}(x_{i}) = ⟨x_{i}⟩SU_{3}(q), we get L = ⟨C_{L}(x_{1}), C_{L}(x_{2}), C_{L}(x_{3})⟩. Hence we have the assertion.

Let now q = 2. Let P be the parabolic with P/O_{2}(P) ≅ L_{2}(4). Now |C_{[V,L]}(O_{2}(P))| = 16. As C_{[V,L]}(O_{2}(P)) ≅ O̸_{2}(P) as P/O_{2}(P)-modules, we get V = [V, L] ⨁ C_{V} (O_{2}(P)). We have |C_{V} (O_{2}(P)) : [C_{V} (O_{2}(P)), P]| ≤ 4. Hence |V : [V, L]| ≤ 4 by [Hu, (I.17.4)].

Let first next L ≅ Ω_{8}^{−}(2). Then there are x_{1}, x_{2} ∈ L, o(x_{1}) = o(x_{2}) = 3, x_{1} ∉ ⟨x_{2}⟩, [x_{1}, x_{2}] = 1, [V, L] = [V, x_{i}], i = 1, 2, C_{L}(x_{i}) = ⟨x_{i}⟩ × Ω_{6}^{+}(2). This shows C_{V} (x_{1}) = C_{V} (x_{2}) is invariant under L = ⟨C_{L}(x_{1}), C_{L}(x_{2})⟩. Hence V = [V, L].

Let now L ≅ Ω^{±}(2n, q), n > 3 and q > 2 for Ω^{−}(8, q). Let P be the parabolic with P′/O_{2}(P) ≅ Ω^{±}(2n − 2, q). Then C_{[V,L]}(O_{2}(P)) is the half spin module for Ω^{±}(2n − 2, q) for P′ while O_{2}(P) is the natural Ω^{±}(2n − 2, q)-module. This shows V = [V, L]C_{V} (O_{2}(P)) and [C_{V} (O_{2}(P)), P′] = C_{[V,L]}(O_{2}(P)).

Now the restrictions are made such that the half spin module splits by induction for P′. Hence we have C_{V} (O_{2}(P)) = C_{[V,L]}(O_{2}(P))C_{V} (P′). Application of [Hu, (I.17.4)] shows C_{V} (P′) = 1 and then V = [V, L].

(f) Let now L ≅ Sp_{2n}(q). If L ≅ Sp_{4}(2)′, the assertion follows with 2.7. Let now L ≅ Sp_{2n}(q), q > 2, for n = 2. Let P be the parabolic with P′/O_{2}(P) ≅ Sp_{2n−2}(q), A_{6} for L ≅ Sp_{6}(2). Now V = [V, L]C_{V} (Z(O_{2}(P))).

Set V_{1} = C_{V} (Z(O_{2}(P))). Now [V_{1}, P′] = [[V, L], P′]. Set V_{2} = [V, Z(O_{2}(P))]. Then |V_{2}| = q. We see that |V_{1}/V_{2} : C_{V_{1}/V_{2}}(O_{2}(P))| = q. We have V_{1}/V_{2} = ([[V, L], P′]/V_{2})C_{V_{1}}(O_{2}(P))/V_{2}. Hence we have [C_{V_{1}}(O_{2}(P)), P′] = 1 as P′ = P′′. This shows C_{V_{1}}(O_{2}(P)) ≤ [V, L] and so |V : [V, L]| ≤ q by [Hu, (I.17.4)], or L ≅ Sp_{6}(2) and [P, C_{V_{1}}(O_{2}(P))] ≠ 1. Hence there is some t ∈ P \ P′, o(t) = 2, with [C_{V_{1}}(O_{2}(P)), t] = V_{2}. We may assume that t induces a transvection on the natural module. As [t, [V, L]/V_{2}] ≠ 1 we now see |[V, t]| ≥ 4. But ⟨t⟩ is conjugate to Z(O_{2}(P)) and [V, Z(O_{2}(P))] = V_{2} is of order 2.

(g) Let P be the parabolic in L, which belongs to the Ω^{+}(10, q) – subdiagram. We choose P in such a way that C_{[V,L]}(O_{p}(P)) is the natural module. Then [V, L]/C_{[V,L]}(O_{p}(P)) is an extension of the half spin module by a trivial module. As [[V, L], O_{p}(P)] is of index q in [V, L], we see that there is V_{1} ≤ V such that V = [V, L]V_{1}, where [V_{1}, O_{p}(P)] ≤ V_{1} ∩ [V, L] = C_{[V,L]}(O_{p}(L)). By (e), we get that we may choose V_{1} to be P –invariant. Now V_{1} = C_{V} (O_{p}(P)) and by (b) we get V_{1} = C_{[V,L]}(O_{p}(P))C_{V_{1}}(P′). Application of [Hu, (I.17.4)] shows C_{V} (P′) = 1 and then V = [V, L].

split2

Lemma 2.9 (a) Let L ≅ M_{22} and V be a GF (2)–module with [V, L] an irreducible 10–dimensional module. Assume that C_{V} (L) = 1. Then |V : [V, L]| ≤ 2 and V = [V, L] if the parabolic 2^{4}Σ_{5} has a fixed point in [V, L].

(b) Let L ≅ M_{24} and V be a GF (2)-module with [V, L] an irreducible 11–dimensional module. Assume C_{V} (L) = 1. Then |V : [V, L]| ≤ 2, and V = [V, L] if the parabolic 2^{4}A_{8} in M_{24} has a fixed point in [V, L].

(c) Let L ≅ M_{11} and V be a GF (3)-module with [V, L] an irreducible 5–dimensional module on which L induces orbits of length 55 and 66 on the 1–dimensional subspaces. Assume C_{V} (L) = 1. Then V = [V, L].

Proof: (a) Let P be the parabolic with P/O_{2}(P) ≅ Σ_{5}. Let first C_{[V,L]}(P) ≠ 1. Then [V, L]/[[V, L], O_{2}(P)] is an orthogonal Σ_{5} – module. This now shows with 2.8 that V = [V, L]W, where [W, P] ≤ [[V, L], O_{2}(P)]. Now [[V, L], O_{2}(P)]/C_{[V,L]}(P) is an indecomposable Σ_{5} – module, which is an extension of the trivial module by a 4–dimensional one. As this is indecomposable, we see that [W, O_{2}(P), O_{2}(P)] ≤ C_{[V,L]}(P′). Hence we have that V = [V, L]W_{1}, with [W_{1}, P′] = 1. As C_{[V,L]}(P′) is of order 4, we get that V = [V, L]C_{V} (P). Application of [Hu, (I.17.4)] shows V = [V, L].

So assume now C_{[V,L]}(P) = 1. Let P_{1} be the parabolic with P_{1}/O_{2}(P_{1}) ≅ A_{6}. Then C_{[V,L]}(P_{1}) ≠ 1. Further [V, L]/[V, L, O_{2}(P_{1})] is a nonsplit extension of the trivial module by a 4-dimensional A_{6}–module. Hence by 2.8 there is V_{1} ≥ [V, L] with |V_{1} : [V, L]| ≤ 2 and V = V_{1}W, where [W, P_{1}] ≤ [V, L, O_{2}(P_{1})]. As [V, L, O_{2}(P)]/C_{[V,L]}(P) is not isomorphic to O_{2}(P) as A_{6}–module, we see that W = [V, L, O_{2}(P)]C_{W} (O_{2}(P)) and so we have that V = V_{1}C_{V} (P). By [Hu, (I.17.4)] we get now V = V_{1} and so |V : [V, L]| ≤ 2.

(b) Let P be the parabolic with P/O_{2}(P) ≅ A_{8} Let first C_{[V,L]}(P) ≠ 1. Set U = [[V, L], O_{2}(P)]. Then |U| = 2^{5} and so we see that |[[V, L], x]| = 16 for any involution x ∈ O_{2}(P). Let |V : [V, L]| = 2. Then as [V, L]/U ≅ O_{2}(P), we see that [V, O_{2}(P)] = U. This shows that |[V, x]| = 16 too, recall that [x, U] ≠ 1. Now choose y ∈ P such that y ∼ x and yO_{2}(P) corresponds to (12)(34)(56)(78). Then also [[V, L], y] = [V, y]. This by 2.7 shows that V/U splits as an A_{8}–module. Hence V = [V, L]W, where W is P -invariant and |W : U| = 2. As [U, O_{2}(P)] = C_{[V,L]}(P), we see that U/C_{[V,L]}(P) is dual to O_{2}(P). This shows that [W, O_{2}(P)] = C_{[V,L]}(P) and so W = UC_{W} (O_{2}(P)), where |C_{W} (O_{2}(P))| = 4. But then V = [V, L]C_{V} (P). Application of [Hu, (I.17.4)] gives a contradiction.

Assume now that C_{[V,L]}(P) = 1. Then C_{[V,L]}(O_{2}(P)) is the orthogonal A_{8}–module and [V, L]/C_{[V,L]}(O_{2}(P)) is an extension of the 4-dimensional module by a trivial module. Now with 2.8 we see that V/C_{[V,L]}(O_{2}(P)) = [V, L]W/C_{[V,L]}(O_{2}(P)), where [W, P] ≤ C_{[V,L]}(O_{2}(P)). In particular we have [W, O_{2}(P)] = 1. Now by 2.7 we have W = W_{1}C_{V} (P) where |W_{1}| ≤ 2^{7}. By [Hu, (I.17.4)] we have C_{V} (P) = 1 which gives |V : [V, L]| ≤ 2.

(c) Let U be the subgroup L_{2}(9) in L. Then [V, L] is an extension of the 4-dimensional Ω^{−}(4, 3)–module by the trivial module. With 2.8 we see that V = [V, L]C_{V} (U). Now application of [Hu, (I.17.4)] gives V = [V, L].

schur

Lemma 2.10 Let |Z(L)| = 2 and L/Z(L) ≅ M_{24}, then L ≅ Z_{2} × M_{24}.

Proof: Let P be a parabolic in L such that P/Z(L) is an extension of an elementary abelian group of order 2^{6} by 3Σ_{6}. As 3Σ_{6} is not a subgroup of O(6, 2) we get that the O_{2}(P) is elementary abelian. This shows that for x ∈ L we have that x^{2} = 1 if x^{2} ∈ Z(L). Let now U be a subgroup of order three in O_{2,3}(P). Then N_{P} (U)/Z(L)U ≅ Σ_{6}. As no involutions in Σ_{6} become elements of order 4 in N_{P} (U), we see that N_{P} (U) ≅ Z_{2} × 3Σ_{6}. Then P ≅ Z_{2} × P_{1}, where P_{1} ≅ P/Z(L). Application of [Hu, (I.17.4)] gives the assertion.

3 The nongeometric case

Most of the groups we are aiming at have a geometry. This also means that the 2–locals containing S have a geometry. In this section we deal with those cases, where M_{0} does not necessarily has a geometry. For what follows we assume without saying, that M_{0} is not a minimal parabolic.

lnlm

Lemma 3.1 Let F^{∗}(M_{0}/O_{p}(M_{0})) ≅ SL_{m}(q)′ ∗ SL_{n}(q)′ and Y_{M} be the tensor product module, then O_{p}(M_{0}) = Y_{M} and there is H with M_{0} ≤ H and F^{∗}(H) is weak parabolic isomorphic to L_{n+m}(q).

Proof: Suppose false. Let first m ≥ 3 ≤ n. Set L = C̃ ∩ M_{0}. Then we have QO_{p}(M_{0}) = O_{p}(L). Further L/O_{p}(L) has a normal subgroup isomorphic to L̂ = SL_{m−1}(q) ∗ SL_{n−1}(q). This group induces on Y_{M} the following series

1 < V_{1} < V_{2} < Y_{M},

where V_{1} is the trivial module, V_{2}/V_{1} is a direct sum of the natural m − 1– dimensional module with the natural n − 1–dimensional module and Y_{M}/V_{2} is the thesor product of the two natural modules. Further we see that V_{2} = [Y_{M}, Q]. As (n − 1)(m − 1) ≥ (n − 1) + (m − 1), we see that Q involves an F –module with offender Y_{M}/Y_{M} ∩ Q. Let [Y_{M}, F (C̃/Q)] ≠ 1. Then by [Cher1] p = 2 or 3 and there is a direct product of dihedral groups or central product of SL_{2}(3) which have Y_{M}/Y_{M} ∩ Q as Sylow p–subgroup. But then we see that Y_{M}/Y_{M} ∩ Q is an exact offender, a contradiction. So we have that [Y_{M}, F (C̃/Q)] = 1.

This now shows that L̂ is in exactly one component K of C̃/Q, or m = n = 3, q = 2 and there are two components interchanged by L̂, which belongs to the fact that Σ_{3} ≀ Z_{2} contains two Ω^{+}(4, 2), one which induces a tensor product and one which induces a direct sum.

Let F^{∗}(M_{0}/O_{p}(M_{0})) = K_{1}K_{2} and K_{0i}, i = 1, 2, a minimal parabolic in K_{i} which is not in C̃. Set P_{0i} the preimage of K_{0i}N_{S}(K_{i}).

Assume first that L̂ acts on exactly one component. Now 2.2 shows K ≅ L_{t}(q), Sp(2t, q), Ω^{±}(2t, q), U_{t}(q), or A_{t}. Then we see because of the structure of our parabolic L̂, that we have K ≅ L_{t}(q) or n = m = 3, q = 2 and K ≅ Ω^{+}(2t, 2), or A_{t} , further M_{0}/O_{2}(M_{0}) ≅ L_{3}(2) ≀ Z_{2} (recall that Σ_{3} ≀ Z_{2} contains two Ω^{+}(4, 2), one which induces a tensor product and the other a direct sum).

We have [Q′, Y_{M}] = 1, so [K, Q′] = 1. As Ω_{1}(Z(S)) ≤ Q′, we also have [Ω_{1}(Z(S)), K] = 1. Assume first K ≅ L_{t}(q). Now in Q/Q′ centralizes S ∩ K a group of order at least q^{2}, which is in V_{2}/V_{1}. By [Sm] we see that there are two nontrivial irreducible modules involved, which then by 2.2 have to be the natural module and its dual. As [Y_{M}, Q] ≤ V_{2}, we see that Q involves exactly those two nontrivial irreducible modules. Then we see that C̃ ∩ M_{0} has exactly 5 nontrivial modules in QO_{p}(M_{0}), which shows that we have O_{p}(M_{0})/Y_{M} is a trivial M^{0}–module, which shows Y_{M} = O_{p}(M_{0}). Then just by order of S, we get Q is special of order q^{1+2(n−1+m−1)}. Hence t = n + m − 2.

Set T = N_{S}(K_{1}). Assume first that T is a Sylow p–subgroup of T K. Choose P a minimal parabolic of K, which is not in M_{0}. Then we see that it acts trivially on C_{Q/Ω_{1}(Z(S))}(S). As P_{01} and P_{02} both act on this group nontrivially we see that ⟨P_{0i}, P ⟩ = P_{0i}P for i = 1, 2. Let T = N_{S}(P_{01}). Then we see that F^{∗}(⟨N_{M_{0}}(K_{1}), P ⟩) is weak parabolic isomorphic to L_{n+m}(q). As S normalizes this group, we have the assertion.

So we have that T is not a Sylow p–subgroup of KT, in particular S ≠ T and p = 2. Further this gives E(M_{0}/O_{2}(M_{0})) = L_{3}(2) × L_{3}(2). But then some element in M_{0} which does not normalize K_{1} must be inside of K. But then we do not have a tensor product involved in K, which contradicts the fact that K is a linear group.

So we are left with q = 2, n = m = 3 and K ≅ Ω^{+}(2t, 2), or A_{t}. As we must have a nontrivial irreducible L̂–module, we see that also Ω^{+}(2t, 2) is not possible. So we are left with A_{t}. Then again because of F –module we see that exactly one irreducible nontrivial module, the natural one, is involved in Q. The action of Y_{M} on Q shows that |Q : C_{Q/Z(S)}(t)| = 4 for all t ∈ Y_{M} \ Q. But there is no group of order 16 with these property on the natural module.

So we have to handle the case that n = m = 3, q = 2 and L̂ is in two components K_{1} × K_{2}. Now L̂ ∩ K_{1} × K_{2} induces a tensor product in Q, which shows that the same is true for K_{1} × K_{2}. As Y_{M} ∩ K_{1} centralizes a subgroup of index 4 in Q, we see that K_{1} induces at most two nontrivial modules, on which K_{2} acts nontrivially, a contradiction.

So let now n = 2, m ≥ 3. Then we see that M_{0} ∩ C̃/Q is an extension of the natural module with SL_{m−1}(q). Assume first that [F (C̃/Q), Y_{M}] ≠ 1. As before we see p = 2. By 2.1 we see that Q/Q′ is generated by elements inducing a transvection on Y_{M}/Y_{M} ∩ Q′. But for x ∈ Q ∩ P_{01}, we have that C_{Y_{M} /Y_{M} ∩Q′}(x) ≤ Q ∩ Y_{M}, a contradiction.

So we have again a component K on which C̃∩M_{0} acts nontrivially. This now possesses a quadratic group of order p^{2}. Hence by [Cher], [MeiStr1] and [MeiStr2] we see that K is 3U_{4}(3), A_{t}, some sporadic group or a group of Lie type in characteristic p. Suppose we do not have A_{t} or a Lie type group in characteristic p. As for A = Y_{M}/Y_{M} ∩ Q we have |Q : C_{Q}(A)| ≤ |A|q and q ≤ 4 in that cases, we see with 2.4, 2.5 and the fact that A acts quadratically that K ≅ M_{22}, or M_{24} and we have the natural module. By [MeiStr2] we have that |A| = 4, so |Q : C_{Q}(a)| ≤ 4 for all a ∈ A, a contradiction on the natural module.

Let next K = A_{t}. Then by [MeiStr2] we have that Q involves just natural or spin modules. In the latter |A| = 4 and so q = 2. But then |Q : C_{Q}(a)| ≤ 4 for a ∈ Q, and then t = 7. But we have that [[Q, M_{0} ∩ C̃], A] is centralized by O^{2}(M_{0} ∩ C̃), which is not the case for the 4–dimensional module for A_{7}. So we have that just permutation modules are involved. Now q ≤ 4. If t > 12, then we see that a ∈ A has at most 4 cycles. But there is no such group of order 16 on which L_{2}(4) acts. So we have t ≤ 12 for q = 4. Now as A is normal in a Sylow 2–subgroup of K, we see that there is no factor L_{2}(4) in K ∩ M_{0}. This shows q = 2. If t ≠ 8, we see that a ∈ A has at most two cycles. Then A corresponds to ⟨(12)(34), (12)(56)⟩. But now on the permutation module V we see that the element in [V, A] which is centralized by N_{K}(A) is not a commutator, which contradicts the action of Q on Y_{M}. So we get t = 8, but then we have a group of Lie type.

So assume finally that K is a group of Lie type in characteristic p. As Z(S) ≤ Q′, and Y_{M} centralizes Q′, we see that [K, Ω_{1}(Z(S))] = 1. Then as before we see that two nontrivial modules are involved in Q. Hence we get that Y_{M} induces an F -module and so it induces transvections on this modules. Hence we get K = SL_{t}(q). Then we see that P_{01} commutes with the point stabilizer in K. By maximality we have that this point stabilizer is in M_{0}. Hence we have SL_{2}(q) ∗ SL_{t}(q) normal in M_{0}/O_{p}(M_{0}). Further we see that the two modules are in ⟨(Y_{M} ∩ Q)C̃⟩ and so K_{2} ∩ K induces exactly two nontrivial irreducible module in QO_{p}(M_{0}). This shows that [K_{2}, O_{p}(M_{0})] ≤ Y_{M}. As C_{Y_{M}} (K_{2}) = 1, this gives O_{p}(M_{0}) = Y_{M}. Then Q ≅ q^{1+2m} and K = E(C̃/Q). As we also have the dual natural module in Q, we see with the same argument as above, that P_{01} commutes with the point stabilizer in that module. Let U be the group generated by the preimages of K and K_{2} extended by S. Then the maximality of C̃ shows O_{p}(U) = 1. Set R = M_{0} ∩ U. Then we easily see that Y_{R} ⊴ C̃, which gives Y_{R} ⊴ U, a contradiction.

So we are left with n = m = 2. Further as M_{0} is not a minimal parabolic we have that S normalizes both components. Let first V act nontrivially on O_{p′}(C̃/Q). As C_{Q}(x) = C_{Q}(y) for all x, y ∈ V \ Q, we see that q = 2 or 3. Now we get with [Cher1] that [V, O_{p′}(C̃/Q)] ≅ Z_{3} or Q_{8}. Recall that this group has to act on an extraspecial group of order p^{5}. Now we see that V ∩ Φ(O_{p}(M_{0})) = 1. But as V = Y_{M}, this gives Y_{M} = O_{p}(M_{0}) and so Q is extraspecial of order p^{5} and |S| = p^{6}. As there are elements x in Q such that x acts quadratically on V, we see that there are elementary abelian subgoups X of order p^{3}, which are normal in ⟨V C̃⟩. But then we see that N_{M_{0}}(X) contains one of the parabolics, which contradicts E–uniqueness.

So we have that V acts trivially on O_{p′}(C̃/Q). Let L be some component of C̃/Q which is not centralized by V. As V is normal in S, we see that V normalizes L. Set W = ⟨(V ∩ Q)^{L}⟩. As Z(Q) ≤ O_{p}(M_{0}), we see that [L, Z(Q) ∩ V] = 1 and W′ = V ∩ Z(Q). Set M_{0} = P_{01}P_{02} as before. Let first L be of Lie type in characteristic p. Then by quadratic action we have with [Str], [PM] that W/V ∩Z(Q) just has submodules V (λ) for some fundamental weight λ. Let L_{1} be the point stabilizer in L. Then without loss we may assume that L_{1} commutes with P_{01}. By maximality of M_{0} it follows that L_{1} also has to commute with P_{01}, so L_{1} commutes with M_{0}. In particular L_{1} = C_{L}(V Q/Q). Hence we get that L ≅ Sp(2n, q), or U_{n}(q) and V (λ) is the natural module or the exterior square. As |[V, W]Z(Q)/Z(Q)| = q^{2}, we see that just natural modules are involved. Suppose L ≅ U_{n}(q). Then we have some element ω of order q + 1 which acts on V ∩ Q. Let L̂_{1} be the preimage of L. Then ω also acts on C_{O_{p}(L̂_{1})}(V ∩ Q). This shows that it also acts on C_{W} (V ∩ Q)V. Finally it acts on C_{C_{W} (V ∩Q)V} (L) = V. But this is a contradiction. So we have L ≅ Sp(2n, q). Then we must have two natural modules be involved. Set M_{1} = ⟨M_{0}, L_{1}⟩. Then O_{p}(M_{1}) involves just three Sp(2n − 2, q)–modules. This shows that one of the parabolics P_{01}, say, has to centralize all these modules. So [[L_{1}, O_{p}(M_{1})], O^{p}(P_{01})] = 1. Hence V ∩ [L_{1}, O_{p}(M_{1})] = 1, a contradiction as Z(S) ≤ V.

So assume now that L is not of Lie type in characteristic p. If p is odd, this shows that p = 3. If L ≅ U̸_{4}(2), we have that V = O_{3}(N_{L}(V). Further N_{L}(V) commutes with M_{0}. As [Q, W] ≤ V, we now see that for M_{1} = ⟨M_{0}, N_{L}(V)⟩, we have [Q, O_{3}(M_{1})] ≤ V. As Z(S) ≤ V, we see that V = O_{3}(M_{1}), but then V = O_{3}(M) and so C_{M}(O_{3}(M)) ≰ O_{3}(M). So we are left with L ≅ U_{4}(2). But U_{4}(2) ≅ P Sp(4, 3), a case we have treated before.

So we are left with p = 2. Let first L be some alternating group. Then we see that we must have the permutation module or L ≅ A_{7} and we have the 4-dimensional module. Again the point stabilizer L_{1} commutes with both parabolics. Hence V Q/Q ≤ O_{2}(L_{1}). By quadratic action we get that V corresponds to a transvection and so V ≰ O_{2}(M_{0})′ and then V = O_{2}(M_{0}). This shows |S| = 2^{6}, a contradiction.

So let next L be a group of Lie type in odd characteristic. By [MeiStr1] ge get L ≅ 3U_{4}(3). As C_{L}(v) is maximal for v ∈ V \ Q, we see that q = 2. But for any involution i in L, we have |[W, i]| ≥ 2^{4}, a contradiction.

Let now L be sporadic. Then by [MeiStr2] L ≅ M_{12}, M_{22}, 3M_{22}, M_{24}, J_{2}, Co_{1}, Co_{2} or 3Suz. If C_{L}(v) for v ∈ V \ Q is maximal in L, we get a contradiction as before. So we have L ≅ M_{22}, 3M_{22} or M_{24}. As the centralizer of a Sylow 2-subgroup of L in a quadratic module is one dimensional, we get that W involves two such modules W_{1} and W_{2}. But then |W_{1} : C_{W_{1}}(V)| ≤ |V/V ∩ Q| and so W_{1} is an F –module, a contradiction to 2.2.

On

Lemma 3.2 Let F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω^{±}(n, q), n ≥ 5. Assume that Y_{M} contains the natural module V. Then there is some subgroup H of G with M_{0} ≤ H such that one of the following holds

(i) Y_{M} = O_{p}(M_{0}) and F^{∗}(H) is weak parabolic isomorphic to Ω^{±}(n+2, q)

(ii) F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω^{+}(14, q), O_{p}(M_{0})/V is the half spin module and F^{∗}(H) is weak parabolic isomorphic to E_{8}(q)

(iii) q is even, F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω^{+}(8, q) and M_{0} induces O^{+}(8, q), O_{p}(M_{0})/V is a sum of two different half spin modules and F^{∗}(H) is weak parabolic isomorphic to E_{6}(q).

(iv) F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω^{−}(8, q), O_{p}(M_{0})/V is the spin module and F^{∗}(H) is weak parabolic isomorphic to ^{2}E_{6}(q).

(v) q = 2, F^{∗}(M_{0}/O_{2}(M_{0})) ≅ Ω^{+}(10, 2), O_{2}(M_{0})/V is the half spin module and F^{∗}(H) is weak parabolic isomorphic to F_{1}.

(vi) q is odd F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω(7, q), O_{p}(M_{0})/V is the spin module and F^{∗}(H) is weak parabolic isomorphic to F_{4}(q).

(vii) q = 3, F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω(7, 3), V = O_{3}(M_{0}) and F^{∗}(H) is weak parabolic isomorphic to M(24)′.

(viii) q = 3, F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω^{−}(8, 3), V = O_{3}(M_{0}) and F ∗ (H) is weak parabolic isomorphic to F_{1}.

Proof: Set K = F^{∗}(M_{0}/O_{p}(M_{0})). Then K ∩ C̃ ≅ q^{n−2}Ω^{±}(n − 2, q). As there is no offender on V as an F –module in O_{p}(K ∩ C̃)/O_{p}(M_{0}), we see V ≰ Q. Further we have QO_{p}(M_{0}) = O_{p}(K ∩ C̃) and so |V : V ∩ Q| = q.

Set

U = ⟨V C̃⟩Q.

As [V, Q, Q] ≠ 1, we have that Z(Q) ≤ O_{p}(M_{0}). This shows

[Z(Q), U] = 1.

If x, y ∈ V \ Q, then [Q, x]Z(Q) = [Q, y]Z(Q) and C_{Q}(x) = C_{Q}(y). Hence by quadratic action and 2.1 if q ≥ 4, then [V, O_{p′}(C̃/Q)] = 1.

So we first assume that [V, O_{p′}(C̃/Q)] ≠ 1. Then q = p and p = 2 or 3. Set R = [V, O_{p′}(C̃/Q)]. Then R is invariant under K ∩ C̃. Set W = ⟨(V ∩ Q)^{R}⟩. As [Q, V ∩ Q] = Z(Q) ∩ V, we see that W′ = Z(Q) ∩ V. Further as W is invariant under K ∩C̃ we have that W O_{p}(M_{0}) = QO_{p}(M_{0}). As [W, R] = W, we now see that W is extraspecial of width n − 2. This now shows that [W, O_{p}(M_{0})] ≤ V and so O_{p}(M_{0})/V is a trivial K–module. By 2.8 we get O_{p}(M_{0}) = Y_{M} and |Y_{M} : V | ≤ p.

So we see that on R/Φ(R) there acts a group which is a central extension of Ω^{±}(n−2, p). We have that |R/Φ(R)| ≤ r^{n−2}, where r = 4 in case of p = 3. Assume |[R/Φ(R), O^{p}(K ∩C̃)]| ≠ r. Now we easily see that we get n−2 ≤ 6. If n = 6, then we have Ω^{−}(6, p). If p = 3, then R must be extraspecial of width 6, but such a group cannot act on an elementary abelian group of order 3^{12}. So p = 2. Now we have that we have to embedd R into Ω^{+}(12, 2). This shows |R| ≤ 3^{4}. But there is no 3^{4}Sp(4, 3) in Ω^{+}(12, 2). Let n − 2 = 5, then p = 3. Then R has to be extraspecial of width 3. But no Z_{3} × Ω^{−}(6, 2) acts on such a group. So let now n − 2 = 4. As there is no 3^{4}SL_{2}(5) in Ω^{+}(8, 2) and also no 2^{1+6}(Z_{3} × L_{2}(9)), we see that we have Ω^{+}(4, p). Let p = 2, then just by order, we see that |R| ≤ 3^{3}. In any case we get some invariant Σ_{3}. Hence we get H_{0} = 2^{6}A_{8} in M_{0} and H_{1} = 2^{1+8}(Σ_{3} × Σ_{3} × Σ_{3}) in C̃, having a common Sylow 2-subgroup. Let P_{1} be a mininal parabolic in this group with V ≰ O_{2}(P_{1}) and let P_{0} be in 2^{6}A_{8} be a minimal parabolic with P_{0} ≰ C̃. Set Z_{1} = ⟨Ω_{1}(Z(S))^{P_{0}}⟩ and Z_{2} = ⟨Z_{1}^{P_{1}}⟩. Then Z_{2} is elementary abelian of order 8. Further [Z_{2}, O_{2}(P_{1})] = Z(Q). So [Z_{2}, O_{2}(P_{0})] = Z_{1}, in particular Z_{2}O_{2}(M_{0})/O_{2}(M_{0}) = Z(S/O_{2}(M_{0})) and then as Z_{2} is normal in O_{2}(P_{0}), we see that Z_{2} is normal in P_{0}. So we have that Z_{2} is normal in ⟨P_{1}, P_{0}⟩ and so we have ⟨P_{1}, P_{0}⟩/O_{2}(⟨P_{1}, P_{0}⟩) ≅ L_{3}(2). This shows that ⟨H_{0}, H_{1}⟩ has an Ω^{+}(8, 2) diagram, and we are done.

Let now p = 3. Then again we see that there is some normal SL_{2}(3). Now the same argument as before shows that we are weak parabolic isomorphic with Ω^{+}(8, 3).

Let finally n − 2 = 3, then p = 3. As we are inside of Sp(6, 3), we see that we are in SL_{2}(3) ≀ Σ_{3}. This now shows that we must have a subgroup H_{1} inside of C̃ which modulo O_{p}(H_{1}) is SL_{2}(3) × L_{2}(3) and a subgroup H_{0} inside of M_{0} which is 3^{5}Ω(5, 3). As before we get that ⟨H_{0}, H_{1}⟩ has an Ω_{7}(3)–diagram.

So we may assume that |[O^{p}(K ∩ C̃), R/Φ(R)]| = r. This again shows that we have a subgroup H_{1} of C̃, with H_{1}/O_{p}(H_{1}) ≅ SL_{2}(p)×Ω^{±}(n−2, p). If n − 2 ≤ 4, we may argue as before. So let n − 2 ≥ 5. Then P_{0} commutes with the point stabilizer on W. This shows that ⟨P_{0}, P_{1}⟩ commutes with some minimal parabolic, where P_{0}, P_{1} are as before. Hence we have that O_{p}(⟨P_{0}, P_{1}⟩) ≠ 1. As above we see again, that this group acts on p^{3}, and so we have the diagram of Ω^{±}(n + 2, p).

From now on we will assume that [O_{p′}(C̃/Q), V] = 1.

Let L be some component in C̃/Q. Then as V is normal in S, we see that [V, L] ≤ L. Let now L be some component with [L, V] ≠ 1. Suppose first that O^{p}(K ∩ C̃) does not normalize L. Then we see that L^{(K∩C̃)} has at most n − 2 components. This shows that Ω^{±}(n − 2, q) possesses a proper subgroup of index at most n − 2. This implies q = p = 2 or 3 and n − 2 ≤ 4. Now as [Q, V, O_{p}(M_{0})] = 1, we see that W is a sum of modules for the components, i.e. [L, W] is centralized by the other components in L^{(K∩C̃)}. But [V, W]/Z(W) is an irreducible module for K ∩ C̃, a contradiction. So we may assume that K ≅ Ω^{+}(6, 2) and M induces O^{+}(6, 2). Further we have that L^{S} = L × L^{s} for some s ∈ S. Then by quadratic action we get that W/V ∩ Z(S) is a direct sum of two modules W_{1} and W_{1}^{s}. We have that |[W_{1}, V]| = 4. With [MeiStr1], [MeiStr2] we see that L is a group of Lie type in characteristic two or A_{7}. We have that W acts quadratically on O_{2}(M_{0})/V and W O_{2}(M_{0}) = QO_{2}(M_{0}). Now in O_{2}(M)/V the only nontrivial modules involved are the natural L_{4}(2)-modules and so they all occure in pairs. Now we have O^{2}(K ∩ C̃/O_{2}(K ∩ C̃)) = ⟨ρ_{1}⟩ × ⟨ρ_{2}⟩, where [W_{1}, ρ_{2}] = 1 and ρ_{1} is conjugated to ρ_{2} inside of L_{4}(2) = K. This shows that both elements are fixed point freely on the natural module. Hence [O_{2}(M_{0})/V, ρ_{1}] = [O_{2}(M_{0})/V, ρ_{2}]. But as [W_{1}, ρ_{2}] = 1, we see that [O_{2}(M_{0})/V, ρ_{1}] = 1 and so O_{2}(M_{0})/V is the trivial module for K. Hence [ρ_{1}, O_{2}(M_{0})Q/Q] = 1. As N_{L}(K ∩ C̃) acts irreducibly on [V, W_{1}], we see with [Str] that L ≅ Ω^{±}(2n, 2), Sp(2n, 2), U_{n}(2), G_{2}(2)′ or A_{7}. But as [O_{2}(M_{0})Q/Q, ρ_{1}] = 1 we see L ≅ A_{7} and in W we have two 4-dimensional modules. In this case we see that [W, O_{2}(M_{0})] ≤ V. By 2.8 we get |O_{2}(M_{0})| ≤ 2^{7}. So |S| ≤ 2^{14}. As |S/Q| ≥ 2^{7}, we see that |Q| ≤ 2^{7}, but then A_{7} × A_{7} cannot act faithfully on Q.

So we may assume that L is normalized by K ∩ C̃. Assume first that [O^{p}(K ∩ C̃), L] = 1. Then we now see that in W/Z(W) we do have a tensor product module. As W O_{p}(M_{0}) = QO_{p}(M_{0}) and [O_{p}(M_{0}), W, W] ≤ V, we see that O_{p}(M_{0})/V just involves trivial and half spin modules. Now [Q, L] ≤ W. So if there is a nontrivial module in O_{p}(M_{0})/V, then L acts on the intersection of this module with W. As W is a tensor product module, we get that this is not possible. So we have that O_{p}(M_{0})/V is a trivial module and W/Z(W) is a direct sum of two orthogonal modules. Hence L ≅ SL_{2}(q). As L is nonsolvable q > 3. But now we may argue as in the solvable case to see that we have F^{∗}(H) is weak parabolic isomorphic to Ω^{±}(n + 2, q).

So we now may assume that O^{p}(K ∩ C̃) induces inner automorphisms on L. Let first L be of Lie type in characteristic p. As K ∩ C̃ acts irreducibly on [V, W] we see that C_{L}(V)/C_{C_{L}(V)}([V, W]) has a normal subgroup K ∩ C̃. Suppose first that K ∩ C̃ corresponds to a normal subgroup of C_{L}(V)/O_{p}(C_{L}(V)). By [Str], [PM] W has a submodule V (λ) for some fundamental weight λ. Hence K ∩ C̃ ≅ Ω^{+}(4, q) is not possible. Let K ∩ C̃ be nonsolvable, then we get that L ≅ Ω^{±}(n + 2, q), E_{7}(q), L_{6}(q), U_{6}(q), Sp(6, q), q odd, G_{2}(q^{2}), Sp(4, q^{2}), G_{2}(q), q odd, Sp(4, q), q odd.

Let L ≅ Ω^{±}(n + 2, q). Again we know that W acts quadratically on O_{p}(M_{0})/V and so there are just half spin modules involved. As K∩C̃ induces natural modules in O_{p}(M_{0})Q/Q, we see that natural modules have to be spin modules too. The only case where this happens is n − 2 = 8. Hence we have L ≅ Ω^{+}(12, q). As W is a strong quadratic module for L, we get that W is a half spin module for L. This now shows that in O_{p}(M_{0})/V we just have two half spin modules involved. Further as O_{p}(M_{0})Q/QV involves a trivial K∩C̃ – module, we see that also O_{p}(M_{0}) involves a trivial K – module. Hence with 2.8 we see that there is a normal subgroup X in O_{p}(M_{0}) such that X/V is the extension of the half spin module [O_{p}(M_{0}), X]/V by a trivial module. In particular X is elementary abelian of order q^{27} and |O_{p}(M_{0})| = q^{43}, which shows that |Q| = q^{33}. Now we see that XQ/Q corresponds to an elementary abelian group of order q^{10} in L which is normalized by some Ω^{+}(10, q). But in QX there is a unique elementary abelian subgroup of oder q^{27}, which shows that N_{G}(X)_{0} > M_{0}, a contradiction.

Let next L ≅ E_{7}(q). Then W is V (λ_{1}) the 56–dimensional module. Now K ∩ C̃ involves Ω^{+}(12, q), which in W ∩ O_{p}(M_{0})/V ∩ W induces the half spin module. Hence in O_{p}(M_{0})/V there is just one further nontrivial module involved, the half spin module. Application of 2.8 shows that O_{p}(M_{0})/V is the half spin module and so |O_{p}(M_{0})| = q^{78}. This gives |Q| = q^{57}. Let U be the point stablizer in L and P_{0} be as before, then again P_{0} commutes with U. Hence we see that H = ⟨M_{0}, L⟩ has an E_{8} - diagram. So F^{∗}(H) is weak parabolic isomorphic to E_{8}(q).

Let next L ≅ L_{6}(q). Then K ∩ C̃ induces on [V, W] the orthogonal module and so W is V (λ_{3}). Now as before, we get that O_{p}(M_{0})/V involves two half spin modules. So we see that there is an elemtary abelian subgroup X of order q^{16} which is normalized by K. In L we see that this group is normalized by some L_{5}(q). If this parabolic contains a Sylow p – subgroup of G, we get again a contradiction to the maximality of M_{0}. So we get that p = 2, M_{0} induces O^{+}(8, q) and in C̃ we have some diagram automorphism involved. As this automorphism also induces an automorphism on K ∩ C̃, we can produce subgroups M_{1} and C̃_{1}, such that M_{0} ≤ M_{1}, |S : S ∩M_{1}| = 2 and S ∩ M_{1} is also a Sylow 2-subgroup of C̃_{1}. Now we investigate N_{G}(X). As X is a quadratic module for N_{G}(X), we see with [MeiStr1], [MeiStr2] that N_{G}(X)^{0}/X is a group of Lie type in characteristic two, recall that O_{2}(M_{0})/X acts quadratically on X. Now we have N_{G}(X)^{0}/X ≅ Ω^{+}(10, q), as this group has a parabolic q^{8}Ω^{+}(8, q). Now we see that H_{1} = ⟨N_{G}(X)_{0}, L⟩ has an E_{6}(q) diagram. Further we see that S acts on this group, so we have that for H = H_{1}S F^{∗}(H) is weak parabolic isomorphic to E_{6}(q).

Let next L ≅ U_{6}(q). Again W is of type V (λ_{3}). Then we see that O_{p}(M_{0})/V involves the spin module just once. By 2.8 we get that O_{p}(M_{0})/V is the spin module and so |O_{p}(M_{0})| = q^{24} and then |Q| = q^{21}. Now again P_{0} commutes with the point stabilizer and so we get a diagram of type ^{2}E_{6}(q), i.e. for H = ⟨M_{0}, L⟩ we have that F^{∗}(H) is weak parabolic isomorphic to 2E_{6}(q).

Let L ≅ Sp(6, q), q odd. Then W is the spin module, i.e W/V ∩ Z(Q) is of order q^{14}. Now in O_{p}(M_{0})/V we get the spin module for Ω(7, q) just once. By 2.8 we see that |O_{p}(M_{0})| = q^{15} and so O_{p}(M_{0}) is an extension of the natural module by the spin module. Further we have Q = W. Again the point stabilizer in L commutes with P_{0} and so we get a diagram of type F_{4}(q), and so for H = ⟨M_{0}, L⟩ we have that F^{∗}(H) is weak parabolic isomorphic F_{4}(q).

If we have G_{2}(r) or Sp(4, r), r = q, q^{2}, then K ∩C̃ has to induce the natural module on [V, W] by as W is some V (λ). But it induces the orthogonal module, which is different.

So it remains the case that K ∩ C̃ is solvable, so K ≅ Ω^{+}(6, 2), Ω^{+}(6, 3) or Ω(5, 3). The first two cases again are out as W is some V (λ). So we have K ≅ Ω(5, 3). Then we may assume that 3-elements in K ∩ C̃ induce outer automorphisms on L. Hence they either induce field automorphisms or L ≅ Ω^{+}(8, 3) and they induce a triallity. In the latter case we must have that there is a submodule of W on which V induces a GF (3)-transvection, which is impossible. In the former case L is defined over GF (3^{3}) and V induces a GF (3^{3})-transvection. So we have that L ≅ SL(m, 3^{3}) or Sp(2m, 3^{3}). But then C(V) cannot act irreducibly over GF (3) on [V, W].

Let next L be a group of Lie type in cross characteristic. Suppose first p = 2. If there is a quadratic fours group in L, we see with [MeiStr1] that L ≅ 3U_{4}(3) and W is the 12-dimensional module. Again we see that in O_{2}(M_{0})/V we just have two natural modules. As [V, W] is an irreducible K∩C̃ - module, we see that Z(L) ≤ K. Hence we see that |O_{2}(M_{0})Q/Q ∩ L| = 2^{6}. This shows that O_{2}(M_{0}) also has to involve a trivial K–module. Now we see that there is a subgroup X in O_{2}(M_{0}), which is normal in M_{0} such that X/V is a 5-dimensional module and [O_{2}(M_{0}), X]/V is a four dimensional module, or a one dimensional module. Hence X is elementary abelian of order 2^{11}. Further X is in the preimage of L. Then we see that N_{L}(XQ/Q) is 2^{4}3A_{6}. But then NC̃(X) covers this group and contains M_{0}, which contradicts the fact that M_{0} is maximal. Recall that O^{+}(6, 2) cannot be involved in M as Z(L) is S - invariant.

So we may assume that there is no quadratic fours group in L. In particular we have that V Q = O_{2}(M_{0})Q. Further K ∩ C̃ must be L_{2}(4) and so we have C_{L}(V) ≅ SL_{2}(5) : 2. We now see that O_{2}(M_{0})/V is the trivial module and so by 2.8 V = O_{2}(M_{0}) and then Q is extraspecial of order 2^{9}, i.e. L is a subgroup of Ω^{+}(8, 2), which is not possible.

So let now p be odd. Then by quadratic action of an element in the center of a Sylow p – subgroup we get with [Cher, Theorem B] that L ≅ U_{5}(2) and K ≅ Ω(7, 3). Further we have that W is the 10 - dimensional module. But then again O_{3}(M_{0})/V is a trivial module. By 2.8 we get that V = O_{3}(M_{0}) and Q is extraspecial of order 3^{11}. As the point stabilizer commutes with P_{0}, we get a diagram of type ^{1}∘ c ^{2}∘ ^{3}∘ ^{4}∘. By definition F^{∗}(⟨M_{0}, L⟩) is weak parabolic isomorphic to M(24)′.

So let next L be alternating. If p is odd then by [Cher, Theorem B] V corresponds to (1, 2, 3). This has to be in the center of a Sylow 3-subgroup, so we get L ≅ Â_{7} or Â_{8}. Then we have K ≅ Ω(5, 3). As C_{L}(V) involves A_{5}, but C_{L}(V) has to act nontrivially on [V, W], we get L ≅ Â_{7}. Further as there is no nontrivial K ∩C̃ –module in O_{3}(M_{0})Q/Q, we see that O_{3}(M_{0})/V is the trivial module. Then by 2.8 we get |O_{3}(M_{0})| ≤ 3^{6}. So we see that M_{0}/O_{3}(M_{0}) ≅ Ω(5, 3), so M_{0} ∩ L ≅ Z_{3}Σ_{3}, which contradicts the structure of Â_{7}, where we have Z_{3}GL(2, 3).

Let p = 2. Now we have that some alternating group in C_{L}(V) has to be an orthogonal group at the same time. This is just possible for A_{8} and A_{5}. Suppose W is the natural module. Then we see that K ∩ C̃ acts trivially on W ∩ O_{2}(M_{0})/W ∩ V. Hence we have that O_{2}(M_{0})/V is a trivial module. So we get that |S| ≤ 2^{14}, 2^{22}, respectively. But we have that L ≅ A_{10}, A_{16}, at least. This gives |S| ≥ 2^{16}, 2^{27}, respectively, a contradiction. So we do not have the natural module. If there is a quadratic fours group, then by [MeiStr2] we get that V corresponds to (1, 2)(3, 4) and as V is in the center, we see that L ≅ A_{12}. Now K ∩ C̃ acts trivially on O_{2}(M_{0})Q/Q, so we see that O_{2}(M_{0})/V is a trivial module, and so again |S| ≤ 2^{22}. This now shows equality and we have that Q is extraspecial of order 2^{13}. But now we get that there is a fours group V_{1} containing V such that C_{W} (V_{1}) = C_{W} (V). Then there is some element ρ of order three in L normalizing V_{1} and centralizing K ∩ C̃. As C_{W} (V) is the orthogonal module, we get that [C_{W} (V), ρ] = 1 and so [W, ρ] = 1, a contradiction.

Hence we have no quadratic fours group and so V Q/Q = O_{2}(M_{0})Q/Q. This shows L ≅ A_{7}, or A_{10}. In both cases we see that O_{2}(M_{0})/V is a trivial module. Hence we get |W | = 2^{8}, 2^{12}, respectively. So in the first case W is a direct sum of two 4-dimensional A_{7}–modules. But as C_{L}(V) cotains an element of type (1, 2, 3), we see that this element acts fixed point freely, which contradicts that C_{L}(V) has to induce Ω^{−}(4, 2). If we have A_{10}, then we have a subgroup Σ_{3} × Σ_{7}, where the Σ_{7} is contained in C_{L}(V). Hence we have that also W contains two 6–dimensional A_{7}–modules, and so the element of order three centralizing the A_{7} has to be fixed point freely. But it is conjugated to an element inside of C_{L}(V), which has fixed points on the natural orthogonal module.

Let now L be sporadic. Let first p = 2 and assume there is a quadratic fours group in L. Then by [MeiStr2] we have L ≅ M_{12}, M_{22}, M_{23}, M_{24}, J_{2}, Suz, Co_{2}, or Co_{1}. As C_{L}(V) has to involve an orthogonal group we get L ≅ J_{2}, Suz or Co_{1}.

As W has to act quadratically on O_{2}(M_{0})/V, we again see that there are just spin modules involved. Hence in O_{2}(M_{0})Q/Q there are just half spin modules for C_{L}(V) involved, which shows that we have L ≅ Co_{1} and K ≅ Ω^{+}(10, 2). Now we have that there is exactly one nontrivial module, the half spin module in O_{2}(M_{0})/V. By 2.8 we see that |O_{2}(M)| = 2^{10+16} and so |Q| = 2^{1+24}. Again P_{0} commutes with the point stabilizer in L, and so we get a diagram for ⟨L, P_{0}⟩ of type C̃_{5} and so of type F_{1}. Further M_{0} ≤ ⟨L, P_{0}⟩ and F^{∗}(⟨L, P_{0}⟩) is weak parabolic to F_{1}.

Assume now that we have no quadratic fours group. Then V Q = O_{2}(M_{0})Q and so we just have L = J_{1} or McL. In the first case we get with 2.8 that |S| ≤ 2^{13}, which gives |W | ≤ 2^{9}. But L contains elements of order 11, a contradiction. In the second case we get |S| ≤ 2^{21} and so |W | ≤ 2^{13}. Now there is an extraspecial group of order 5^{3} in L, which is impossible.

Let now p > 2. Then with [Cher] we get that p = 3 and L ≅ 2J_{2}, 2Suz or 2Co_{1}. As C_{L}(V) acts nontrivially on [V, W], we see that 2Co_{1} is not possible. Assume first L ≅ 2Suz. This then shows that K ≅ Ω^{−}(8, 3). Further we see that W is 12–dimensional. As V Q = O_{3}(M_{0})Q, we also see with 2.8 that O_{3}(M) = V and so Q is extraspecial of order 3^{13}. Again P_{0} commutes with the point stabilizer in L which is 3^{5}M_{11}, and so ⟨L, M_{0}⟩ has a diagram of type ^{1}∘ c ^{2}∘ ^{3}∘ ^{4}∘, so F^{∗}(⟨M_{0}, L⟩) is weak parabolic isomorphic to F_{1}.

Assume now finally L ≅ 2J_{2}, then we get that K ≅ Ω^{−}(6, 3). Again V Q = O_{3}(M_{0})Q and so again by 2.8 we get O_{3}(M_{0}) = V. As the smallest representation of 2J_{2} is 12–dimensional, this just follows from the fact that an element of order three acts on a Sylow 5–subgroup, we get |Q| ≥ 3^{13}, a contradiction.

4 The generic cases

In this chapter we investigate those groups M_{0} which have a geometry and lead to the series of groups of Lie type.

halfspin

Lemma 4.1 Let F^{∗}(M_{0}/O_{p}(M_{0})) ≅ Ω^{+}(10, q) and assume that Y_{M} contains the half spin module V, then there is a subgroup H of G with M_{0} ≤ H such that F^{∗}(H) is weak parabolic isomorphic to E_{6}(q).

Proof: As V is not an F -module with an offender in QO_{p}(M_{0})/O_{p}(M_{0}) by 2.3, we see that V ≰ Q. Now for K ∩ C̃ we have th following series

1 < V_{1} < V_{2} < V,

where V_{1} is trivial, V_{2}/V_{1} is 10–dimensional and V/V_{2} is 5-dimensional. Further V_{2} = V ∩Q. We have that K ∩C̃ acts transitively on the nontrivial elements in V/V ∩Q. As |Q : C_{Q}(V)| ≤ q^{10}, we see that V cannot act faithfully on an elementary abelian p′–group. So by 2.1 we see that [V, O_{p′}(C̃/Q)] = 1, further by 2.6 V normalizes any component of C̃/Q. Let L be some component in C̃/Q, which is not centralized by V.

Set L_{1} = L^{(C̃∩K)} and assume L ≠ L_{1}. If |V : C_{V} (L)| = p, then V acts faithfully on some elemenary abelian p′– group, a contradiction as before. So we have that |V : C_{V} (L)| > p. Now as V is in the center of S ∩ L_{1}, we get with [Cher], [MeiStr1], [MeiStr2] that L is a group of Lie type in characteristic over a field with r = p^{f} elements. Further we get a direct sum U_{1} ⊕ … ⊕ U_{t} of modules for the components of L_{1}, t the number of components, where any component acts nontrivially on exactly one module. For x ∈ Q \ O_{p}(M_{0}), we have |V : C_{V} (x)| ≥ q^{4}. This now shows that we have at most three modules, i.e. t ≤ 3. As L_{5}(q) does not have a subgroup of index two or three, we get L = L_{1}.

As |V : V ∩ Q| = q^{5} and V acts quadratically on Q, we get with [Cher], [MeiStr1], [MeiStr2] that L is a group of Lie type in characteristic p. As Q/Q ∩ O_{p}(M_{0}) is an irreducible K ∩ C̃–module and [V_{2}, Q] ≠ 1, we see that [Z(Q), L] = 1. So V_{1} is normalized by L. Set W = ⟨V_{2}^{L}⟩/V_{1}. The W O_{p}(M_{0}) = QO_{p}(M_{0}) acts quadratically on O_{p}(M_{0})/V. This with [Str],

[PM] shows that we just have natural modules in O_{p}(M_{0})/V, and so in O_{p}(M_{0})Q/Q, we just have natural SL_{5}(q)–modules. Now by [Str] and [PM] we have that W = V (λ) for some fundamental weight λ. As [V, W] is V (λ_{2}) for the SL_{5}(q) in C_{L}(V), we see that the weight λ does not correspond to an end node of the Dynkin diagram. Hence we have that L ≅ SL_{n}(q), Sp(2n, q) or U_{n}(q). But then, because of the involved L_{5}(q)–modules, we see that we must have SL_{n}(q). As N_{L}(V) acts on [W, V], which is irreducible under the SL_{5}(q), we see that E(N_{L}(V)/C_{L}(V)) ≅ SL_{5}(q). Hence we get that W = V (λ_{2}) or V (λ_{3}). Now for v ∈ V^{♯} we see that |[W, v]| ≥ q^{n−2}, which gives that n − 2 ≤ 6 and C_{K∩L}(v) induces the orthogonal module on [W, v], so we have L ≅ SL_{6}(q), and W = V (λ_{3}). As V Q = O_{p}(M_{0})Q, we see that O_{p}(M_{0})/Q is the trivial module and with 2.8 we get that V = O_{p}(M_{0}), Q = W, recall that for p odd we have an involution in Z(K), which is fixed point freely on V. Again P_{0} commutes with the point stabilizer in L. No diagram automorphisms could be involved and field automorphisms can be seen on both sides at the same time, i.e. there is M_{1} ≤ M and C_{1} ≤ C̃ with H_{1} = ⟨M_{1}, C_{1}⟩ has an E_{6}(q) diagram. As both groups share a Sylow p–subgroup and are normalized by S we get for H = H_{1}S that F^{∗}(H) is weak parabolic isomorphic to E_{6}(q).

symmetric

Lemma 4.2 Let K = F^{∗}(M_{0}/O_{p}(M_{0})) ≅ SL(n, q), n > 4, and assume that Y_{M} contains the symmetric square V, then there is a subgroup H with M_{0} ≤ H and F^{∗}(H) is weak parabolic isomorphic to Ω^{±}(m, q). If n ≥ 3, q is odd and V is the alternating square, then there is H with M_{0} ≤ H and F^{∗}(H) is weak parabolic isomorphic to Sp(2n, q).

Proof: As V is not an F – module, we see V ≰ Q. Further we see that V/V ∩ Q is the symmetric square or alternating square for the SL(n − 2, q), SL(n − 1, q) respectively, in C̃ ∩ K. If we have the symmetric square there is a group SL_{2}(q) in K ∩C̃ which is trivial on V/V ∩Q and V ∩Q/Z(Q)∩V is a tensor product of the natural n − 2–dimensional module with the 2– dimensional module. If we have the alternating square, then V ∩Q/V ∩Z(Q) is the natural module for K ∩ C̃, which induces SL_{n−1}(q).

We have that K ∩ C̃ acts irreducibly on V/V ∩ Q and so we see that [V, O_{p′}(C̃/Q)] = 1. Further again by 2.6 we see that V normalizes any component. Let L be some component with [V, L] ≠ 1 and set L_{1} = L^{(C̃∩K)}. If L ≠ L_{1}, we see as in the previous lemma that L is a group of Lie type in characteristic p. Furthermore as QO_{p}(M_{0})/O_{p}(M_{0}) is an irreducible K ∩C̃– module and [V, Q, Q] ≠ 1, we see that [L, Z(Q)] = 1. Now let W = ⟨(V ∩ Q)^{L_{1}}⟩/V ∩ Z(Q). Then we see again that W is a direct sum of modules U_{1} up to U_{t}, t the number of components in L_{1}. But as Q/Q ∩ C(V) is an irreducible K ∩ C̃–module, we see L = L_{1}.

Suppose first that L is alternating. As V acts quadratically we easily see that we cannot have the permutation module. Then we get n = 5 and L ≅ A_{8} or A_{9}. The former we will handle as a group of Lie type Ω^{+}(6, 2) lateron. So let L ≅ A_{9}. With 2.8 we see that V = O_{2}(M_{0}) and so Q is of order 2^{13}. Now we see some Σ_{3} centralizing L and so W is a direct sum of two 6-dimensional modules for A_{9}, which do not exist.

Now application of [Cher], [MeiStr1] [MeiStr2] shows that we have that L is a group of Lie type in characteristic p. By [Str] and [PM] we also have that W is V (λ) for some fundamental weight λ or a direct sum of two copies of them. The latter just occurs if we have the symmetric square and the SL_{2}(q) in K ∩ C̃ centralizes L. As a parabolic can never induce a tensor product on the centralizer of its O_{p} in some V (λ), we see that the SL_{2}(q) centralizes L. Now L has a parabolic which induces a symmetric or alternating square. This shows L ≅ Ω^{±}(m, q) or Sp(2m, q), where q is even for the symmetric square and odd for the alternating square, or n = 5. In the latter let t ∈ V \ Q. Then |W : C_{W} (t)| ≤ q^{6}. As W is a direct sum of two modules we get that C_{L}(t)/O_{p}(C_{L}(t)) has a component in SL_{3}(q). This again shows that L is orthogonal or symplectic, or SL_{5}(q). In the latter elements in V \ Q have the same centralizer in W. But then this centralizer would be normalized by L_{1} some extension of q^{4} by SL_{4}(q). This means that W ∩ O_{p}(M_{0}) is normal in this group. But then also the factor would be, which is a direct sum of two nontrivial 3-dimensional modules for K ∩L. This now has to be a 6–dimensional L_{1}–module. But we also know that some SL_{2}(q) centralizing L has to act nontrivially on this module, a contradiction.

Let P_{0} in M_{0} as usual. And let L_{1} be the point stabilizer in L, then ⟨P_{0}, L_{1}⟩ = P_{0}L_{1}. Let first L be an orthogonal group or a symplectic group and let W contain the natural module. Then we have in L_{1} a maximal linear group L_{2} and so ⟨M_{0}, L_{2}⟩ has a string diagram of type A_{t}. If M_{0} is a proper subgroup we see by maximality that O_{p}(⟨M_{0}, L_{2}⟩) = 1. Hence by [Tits] we get E(⟨M_{0}, L_{2}⟩) = L_{t+1}(q). But then M_{0} cannot act on the symmetric or alternating square. So we have that L_{2} ≤ M_{0}.

Let L = Ω^{+}(2(n − 2), q). Then we have with 2.8 that O_{p}(M_{0}) = V and so Q = W is of order q^{1+4(n−2)}. As P_{0} commutes with L_{1}, we see that for H as before F^{∗}(H) is weak parabolic isomorphic to Ω^{+}(2n, q).

Let now L = Ω^{−}(2(n − 1), q). Then we see that in O_{p}(M_{0})/V there are exactly two natural SL_{n}(q) modules involved, as the same is true for L ∩ K. By 2.8 we see that there is X ≤ O_{p}(M_{0}) with X/V is a direct sum of two natural SL_{n}(q)–modules and [M_{0}, O_{p}(M_{0})] = X. Let O_{p}(M_{0}) > X, then we see as n > 3, that [O_{p}(M_{0}), X]V/V is a natural module. In particular O_{p}(M_{0}) ∩ Q = O_{p}(M_{0}) ∩ X. But then O_{p}(M_{0}) acts accordingly on L, which gives p = 2 and |O_{2}(M_{0}) : X| = 2. Now in both cases we first see that W = Q and Q/Z(Q) is a tensorproduct of the orthogonal module with a 2-dimensional SL_{2}(q)–module. If O_{2}(M_{0}) ≠ X, we also see that we have subgroups M_{1} < M_{0} and L̂ < LS, sharing a Sylow 2-subgoup T which is of index two in S. Now in in any case we get that P_{0} commutes with the point stabilizer in L_{1} and so for H = ⟨M_{0}, L⟩ or H = ⟨M_{1}, L̂⟩ we have that F^{∗}(H) is weak parabolic isomorphic to Ω^{−}(2(n + 1), q).

Let next L ≅ Ω(2(n−2)+1, q), q odd, or Sp(2(n−2), q), q even. Then as in L there is exactly one nontrivial SL(n − 2, q)–module in O_{p}(M_{0})Q/QV, we see that there is X ≤ O_{p}(M_{0}) with X/V is the natural module and [M_{0}, O_{p}(M_{0})] ≤ X. As n > 3 for q even, we now see with 2.8 X = O_{p}(M_{0}). If we are in the orthogonal case we see that Q/Z(Q) is a tensor product of the natural module with the two dimensional module. Then as above we get that F^{∗}(H) is weak parabolic isomorphic to Ω(2n + 1, q).

So assume we have the symplectic type. First of all we see as SL(n − 1, q) has no fixed points in V and q is even, we must have that O_{2}(M_{0}) is elementary abelian. Further as L-modules in Q/V ∩ Z(Q) there are two symplectic modules and two trivial modules involved. Now we have that (Q ∩ O_{2}(M_{0}))/V ∩ Q is the 2-dimensional SL_{2}(q)–module. As n > 4, we see with 2.8 that this module splits or n = 5, q = 2. Suppose that the module splits. Then we have a direct sum in O_{2}(M_{0})∩Q/Z(Q)∩V of a 2-dimensional module V_{1} with the tensor product of the natural (n−2)-dimensional natural SL_{n−2}(q)-module with the 2-dimensional SL_{2}(q)–module. Now as O_{2}(M_{0}) is elementary abelian, the preimage W_{1} of V_{1} is elementary abelian and so W_{1} = Z(Q).

Assume now that we have n = 5, q = 2. Then O_{2}(M_{0}) ∩ Q = C_{Q}(V ∩ Q) and as Q′ = V ∩ Z(Q) and O_{2}(M_{0}) is elementary abelian, we again gat |Z(Q)| = 8.

Now we see that Z(S) ≰ V. Let s ∈ Z(S) \ V. Then C_{M_{0}}(sV) is a SL_{n−1}(q)–parabolic. For this parabolic V is an extension of the natural module by the symmetric square. Hence by 2.8 we see that this parabolic centralizes some nontrivial element in Z(S). By E–uniqueness it should be in C̃, which implies M_{0} ≤ C̃, a contradiction.

Let now L ≅ Sp(2(n − 1), q), q odd, and so V be the alternating square. Now we have V Q = O_{p}(M_{0})Q, so by 2.8 V = O_{p}(M_{0}). Then we see that Q/Z(Q) is the natural module. As P_{0} again commutes with the point stabilizer, we see that we get a Sp(2n, q)–diagram, and so F^{∗}(H), H as before, is weak parabolic isomorphic to Sp(2n, q).

So assume now that we do not have the natural module. Then we have Ω^{+}(2m, q) and the half spin module, or Sp(2m, q) and the V (λ_{m−1}). In the former W also involves the symmetric square for K ∩ L. But as W acts quadratically on O_{p}(M_{0})/V just natural modules are involved. This shows that n − 2 = 3. As before L ∩ K is a maximal parabolic, so we get L ≅ Ω^{+}(6, q). Then we would get that the natural SL_{4}(q) module is involved, but we must have two natural SL_{3}(q)–modules inside, which is not possible. So let now L ≅ Sp(2m, q) and V (λ_{m−1}) be involved. We also have that the order of a symmetric square is at least the order of the natural module. We see that we just have F -modules involved and so by 2.2 we cannot have V (λ_{m−1}) to be involved.

tensor

Lemma 4.3 Let F^{∗}(M_{0}/O_{p}(M_{0})) ≅ SL_{n}(q^{2}) and assume that Y_{M} contains V = V (λ_{1}) ⊗ V (λ_{1})^{σ}, σ the field automorphism of order two, then there is a subgroup H of G with M_{0} ≤ H and F^{∗}(H) is weak parabolic isomorphic to U_{m}(q), m = 2n, or 2n + 1.

Proof: As V is not an F –module, we have V ≰ Q. We get V/V ∩ Q is the tensor product module for SL_{n−1}(q^{2}). No we have that V is an F – module offender on Q, as |Q/Q∩O_{p}(M)| = q^{2(n−1)}. As K∩C̃ acts irreducibly on V/V ∩ Q, we get again that [V, O_{p′}(C̃/Q)] = 1. Further we get that V normalizes all components of C̃/Q and then even K ∩ C̃ does it. If n > 3, then we see that L ≅ U_{m}(q) for some m, where L is some component of C̃/Q with [V, L] ≠ 1. Let n = 3 then we also have the possibilty of Ω^{−}(m, q) and A_{m}.

Assume L ≅ A_{m}. But as W = ⟨(V ∩Q)^{L}⟩/(V ∩Z(Q)) is an F –module, we have just permutation modules involved by 2.2. But then on C_{W} (O_{p}(M_{0})) the group L_{2}(4) induces an permutation module while in V ∩ Q it is the natural L_{2}(4)–module.

Hence we have that L is of Lie type in characteristic p. Let first L ≅ Ω^{−}(m, q). If m ≠ 6, then K ∩ C̃ centralizes Z(S/Q) ∩ L, but it has to induce the natural module, a contradiction. So we have m = 6 and then Ω^{−}(6, q) = SU_{4}(q). Hence in any case we have an unitary group. Let V_{1} = Z(S) ∩ V and V_{2} = ⟨V_{1}^{P_{0}}⟩. Then |V_{2}| = q^{4} and |V_{2} ∩ Q| = q^{3}. Let X = C_{O_{p}(M_{0})Q}(V_{2} ∩ Q). Then |O_{p}(M_{0})Q : X| = q^{2}. Further V_{2} ≤ Z(X). Now let L_{1} be the point stabilizer in W to the point V_{2}. Then we see that L_{1} acts on QV_{2}. Now it acts on C_{QV_{2}}(V_{2} ∩ Q) which is (X ∩ Q)V_{2}. As [O^{p}(L_{1}), Q] ≤ W, we see that O^{p}(L_{1}) is trivial on Z((X ∩ Q)V_{2}). But V_{2} ≤ Z((X ∩ Q)V_{2}) and so [O^{p}(L_{1}), V_{2}] = 1, which shows that L_{1} commutes with P_{0}. If now K ∩ C̃ does not correspond to a maximal SL_{t}(q^{2})–parabolic in L, we would get a subgroup U which contains M_{0} properly and has some A_{r}–diagram. By maximality of M_{0}, we get O_{p}(U) = 1 and so by [Tits] we get that E(U) ≅ SL_{r+1}(q^{2}). But then in U there is no parabolic which could induce the tensor product module in its p-radical. So we have that L ≅ U_{2(n−1)}(q) or U_{2n−1}(q).

Let first L ≅ U_{2(n−1)}(q). Then V Q = O_{p}(M_{0})Q. This shows that O_{p}(M_{0})/V is a trivial module. Hence Y_{M} = O_{p}(M_{0}) and Y_{M}/V is the trivial module. Further we have that Y_{M} = V (Y_{M} ∩ Q). Let W = ⟨(V ∩ Q)^{L}⟩. Then [W, Y_{M}] ≤ V ∩ Q. Let L_{1} ≤ K ∩ C̃ such that O_{p}(L_{1}) = O_{p}(K ∩ C̃) and L_{1}/O_{p}(L_{1}) ≅ SL(n, q). Then we have hat Y_{M}/V ∩ Q = (V/V ∩ Q)C_{Y_{M} /V ∩Q}(L_{1}). Further we see that the preimage W_{1} of C_{Y_{M} /V ∩Q}(L_{1}) is in Q since L_{1}/Q has no center. Now we see that [W, W_{1}] ≤ V ∩ Z(Q). Hence W_{1} = (V ∩ Q)C_{W_{1}}(Q). In particular Y_{M} = V C_{Y_{M}} (Q). But then K ∩ C̃ centralizes C_{Y_{M}} (Q), which with [Hu, (I.17.4)] yields C_{Y_{M}} (Q) ≤ V and so Y_{M} = V. So Q/Z(Q) is the natural module. Hence the parabolics are as in U_{2n}(q) and as P_{0} commutes with the point stabilizer, we also get that F^{∗}(⟨M_{0}, L⟩) is weak parabolic isomorphic to U_{2n}(q).

Let now L ≅ U_{2n−1}(q). Then O_{p}(M_{0})Q/V Q is the natural module for K ∩ C̃. As W acts quadratically on O_{p}(M_{0}) and K ∩ C̃ induces in W ∩ O_{p}(M_{0})/V ∩ W just trivial modules, we see that in O_{p}(M_{0})/V there is exactly one nontrivial module, the natural module, involved. As above we see that Y_{M}/V involves no trivial submodule. Hence O_{p}(M_{0})/V has a natural module as submodule. Now as n ≠ 2 and q^{2} > 2, we see with 2.8 that O_{p}(M_{0})/V is the natural module. This shows just by order that Q/Z(Q) is the natural unitary module and so we have again as before that F^{∗}(⟨M_{0}, L⟩) is weak parabolic isomorphic to U_{2n+1}(q).

E6

Lemma 4.4 Let F^{∗}(M_{0}/O_{p}(M_{0})) ≅ E_{6}(q) and assume that Y_{M} contains V, the 27–dimensional module. Then there is a subgroup H of G such that F^{∗}(H) is weak parabolic isomorphic to E_{7}(q).

Proof: As V is not an F –module, we see V ≰ Q. We have K ∩ C̃/O_{p}(M_{0}) is an extension of an elementary abelian group of order q^{16} by Ω^{+}(10, q), where the abelian group is the half spin module. Hence O_{p}(M_{0})Q = O_{p}(C̃ ∩ M_{0}). For K ∩ C̃ we have the following series

1 < V_{1} < V_{2} < V,

where V_{1} = V ∩Z(Q), V_{2}/V_{1} is the half spin module and V/V_{2} is the natural module. Further V_{2} = [V, Q] = V ∩Q. Let W = ⟨(V ∩Q)^{L}⟩, where L = ⟨V C̃⟩. As [V, Z(Q)] = 1, we have [L, Z(Q)] = 1. So we see that [W, Q] = V_{1}. Hence W O_{p}(M_{0}) = QO_{p}(M_{0}) acts quadratically on O_{p}(M_{0})/V. But then by [Str] and [PM] O_{p}(M_{0})/V is the trivial module and then by 2.8 V = O_{p}(M_{0}) and then V_{1} = Z(Q) and |Q/Z(Q)| = q^{32}. By the action of V on Q, we see that [V, O_{p′}(C̃/Q)] = 1. As V acts quadratically, we have that V normalizes any component of C̃/Q. Let L be some component of C̃/Q with [L, V] ≠ 1, then as L^{K∩C̃} has to act on a 32–dimensional module, we see that L = L^{K∩C̃}.

As L possesses a quadratic group of order q^{10}, we see with [Cher], [MeiStr1], [MeiStr2] that L is alternating or of Lie type in characteristic p. As L possesses a parabolic which involves Ω^{+}(10, q), we see that L is of Lie type. Now we get that L is of orthogonal type Ω^{+}(2n, q) or E_{n}(q). By quadratic action and dimension of W equal to 32, we see that just Ω^{+}(12, q) and the half spin module is possible. Hence the parabolics are as in E_{7}(q). Now let L_{1} be the point stabilizer in L. Then L_{1} involves SL_{6}(q). Further P_{0} commutes with L_{1}. This shows that F^{∗}(⟨P_{0}, L⟩) is weak parabolic isomorphic to E_{7}(q). Notice that field automorphisms are also field automorphisms on K ∩ L and can be factored out.

5 The sporadic cases

In this chapter we treat those groups M_{0} which lead to sporadic simple groups.

Spn

Lemma 5.1 Let F^{∗}(M_{0}/O_{p}(M_{0}))/Z(F^{∗}(M_{0}/O_{p}(M_{0}))) ≅ Sp(2n, q)′, n ≥ 2, and assume that Y_{M} contains V, the natural module. Then q = 2 and there is a subgroup H of G with M_{0} ≤ H such that one of the following holds

(i) n = 2 and F^{∗}(H) is weak parabolic isomorphic to U_{4}(3)

(ii) n = 2 and F^{∗}(H) is weak parabolic isomorphic to Suz

(iii) n = 4 and F^{∗}(H) is weak parabolic isomorphic to F_{2}

Proof: Let first V ≤ Q. Set W = ⟨V C̃⟩. Then W′ ≠ 1. Suppose [Y_{M}, W′] = 1. In particular [W, W′] = 1, so W O_{p}(M_{0}) induces a transvection group on V. Now [⟨Y C̃ ; M⟩, Z(W)] = 1. Let first Y_{M}^{g} be some F –module offender on Y_{M}. By quadratic action we get |Y_{M}^{g} : Y_{M}^{g} ∩ O_{p}(M_{0})| = q, so we have that Y_{M}/V is the trivial module. But then [W, Y_{M}] ≤ Z(W), a contradiction. So we have with [MeStStr, (4.2)] that A is a 2F–module offender, and also |A : A ∩ O_{p}(M_{0})| = q. So we get that Y_{M}/V involves the natural module just once. Then Y_{M} ∩ O_{p}(⟨Y_{M}, Y_{M}^{g}⟩) = V C_{Y_{M}} (A). Now we see that |Y_{M} : O_{p}(⟨Y_{M}, Y_{M}^{g}⟩)| = q and ⟨Y_{M}, Y_{M}^{g}⟩ induces SL_{2}(q) on (Y_{M} ∩ O_{p}(⟨Y_{M}, Y_{M}^{g}⟩))(Y_{M}^{g} ∩ O_{p}(⟨Y_{M}, Y_{M}^{g}⟩))/(Y_{M} ∩ Y_{M}^{g}). As A is not an F –module offender, we get p is odd and so Y_{M}/V is the natural module. But now A ≤ C_{K}(A)′. As C_{K}(A)′ centralizes Y_{M}/C_{Y_{M}} (A), we see that [A, Y_{M}] ≤ C_{Y_{M}} (A) and so [W, Y_{M}] ≤ Z(W), a contradiction.

So we have that [W, W′] ≠ 1. In particular p is odd and |W : W ∩ O_{p}(M_{0})| = q^{2n−1}. Now we get again that Y_{M}/V is the natural module. Then |[W′, Y_{M}]V : V | = q. As [V, W′] ≤ Z(Q) ∩ V, we see that [W, W′] ≤ Z(Q). Hence we get that [W′, Y_{M}]V ≤ V Z(Q). This shows that we have [Z(Q), Y_{M}] = 1 and [W′, Y_{M}] ≤ Z(Q). But then we see for X = O^{p}(⟨Y C̃ ; M⟩), that [X, W′] = 1. Now we have V ∩ W′ = V^{g} ∩ W′ for all g ∈ X. But |V : V ∩ W′| = q, so |V^{g} : V^{g} ∩ O_{p}(M_{0})| = q, which shows that |W : W ∩ O_{p}(M)| = q, as V^{g} is normal in W. But this is a contradiction.

So we have shown V ≰ Q. Set W = ⟨(V ∩ Q)^{L}⟩, where L = ⟨V C̃⟩. Next we are going to show that p = 2. So let p be odd. Assume first that |W : W ∩ O_{p}(M_{0})| = q. Then W is abelian. We have [V, W] ≤ Z(Q) and so [W, L] ≤ Z(Q) ∩ W. So [Q, W] = [V ∩ Q, Q] = Z(Q) ∩ V. Hence Z(Q) ∩ V is normalized by L and so we see that [O^{p}(L), W] = 1, a contradiction. So we have that W O_{p}(M_{0}) = QO_{p}(M_{0}). We have |[V ∩ Q, W]| = q. Let (V ∩ Q)^{g} ≤ W with O_{p}(M_{0})(V ∩ Q)^{g}/O_{p}(M_{0}) not in Z(W O_{p}(M_{0})/O_{p}(M_{0}). Then [V ∩ Q, (V ∩ Q)^{g}] ≠ 1 and [(V ∩ Q)^{g}, W] ≰ O_{p}(M_{0}), so we have |[(V ∩ Q)^{g}, W]| > q, a contradiction.

Hence we have p = 2. We first assume that [F (C̃/Q), V] ≠ 1. As C_{Q}(x) = C_{Q}(y) for all x, y ∈ V \ Q, we see by quadratic action and 2.1 that q = 2. As O^{2}(K ∩ C̃) does not possess a proper subgroup of index at most 2n − 2, and [V, W]Z(W)/Z(W) is of order 2^{2n−2} we see that [[F (C̃/Q), V], O^{2}(K ∩ C̃)] = 1. Then for x ∈ [F (C̃/Q), V]^{♯} we get that ([V, W]Z(W)/Z(W))^{x} is isomorphic to [V, W]Z(W)/Z(W) as an O^{2}(K ∩ L)

– module. If n > 2, then W O_{2}(M_{0})/O_{2}(M_{0}) is a nonsplit extension, while [V, W]/Z(W) ∩ V is the natural module. This would imply that Z(W) ≰ O_{2}(M_{0}). But then [Z(W), V] ≤ Z(W) ∩ V ≤ W′. As [V, W′] = 1, we see that [x, Z(W)] = 1. But then V ∩ W′ ≤ V^{x} and so [W, V]/(V ∩ W′) is the same module as [W, V]^{x}/(V ∩ W′), a contradiction.

So we have n = 2. Further |W O_{2}(M_{0})/O_{2}(M_{0})| = 4. Now we see that [F (C̃/Q), V] is of order three and W is extraspecial of order 32. As W acts quadratically on O_{2}(M_{0})/V, we see that in that group there are only natural Sp_{4}(2)– modules or 6–dimensional 3A_{6}–modules involved. In both cases we have some nontrivial K ∩C̃–module in O_{2}(M_{0})Q/Q, which is not possible as this module has to act nontrivally on W. So we have that O_{2}(M_{0})/V is the trivial module, and so by 2.8 |O_{2}(M_{0})| ≤ 32. This shows that in any case we have a subgroup M_{1} in M_{0} of index at most 4 such that M_{1} is an extension of an elementary abelian subgroup of order 16 by A_{6}. Further W ≤ M_{1} now we also see that in C̃ we have a subgroup C_{1} such that O_{2}(C_{1}) = W and C_{1}/W ≅ Σ_{3} × Σ_{3}. Further C_{1} and M_{1} share a parabolic in M_{1} and both are normalized by S. Let U be the elementary abelian subgroup of order 16 in S ∩ M_{1}, U ≠ V, which is normal in P_{0}. Let P be the parabolic in C_{1} which is not in M_{1}. We have that U ∩ W is elementary abelian of order 8. In W there are exactly two elementary abelian subgroups of order 8 containing ⟨Z(S)^{P_{0}}⟩. As an element of order 3 acts fixed point freely on W/Z(W), we see that ⟨(⟨Z(S)^{P_{0}}⟩)^{P} ⟩ = U ∩ W. So C_{C_{1}}(U ∩ W) = U is normal in P. Hence U is normalized by ⟨P, P_{0}⟩ = M_{2}. As U is an irreducible M_{2}–module, the structure of GL_{4}(2) shows that M_{2}/U ≅ A_{6}. Hence H = ⟨M_{0}, C_{1}⟩ has an U_{4}(3)–diagram of type ∘ ∘ ∘ and so F^{∗}(H) is weak parabolic isomorphic to U_{4}(3).

So we now have that [V, F (C̃/Q)] = 1. As usual we see that V normalizes any component of C̃/Q. Let L be some component with [V, L] ≠ 1. Let L_{1} = L^{O^{2}(K∩C̃)}. By quadratic action we have in W/Z(W) a direct sum of modules for the components of L_{1}. Then as before we get L_{1} = L.

Assume first that [L, O^{2}(K ∩ C̃)] = 1. Then for x ∈ L, we have that [V, W]^{x}Z(Q)/Z(Q) is isomorphic to [V, W]Z(Q)/Z(Q). As before we get q = 2, n = 2. But then Aut(W) is solvable, a contradiction.

So we have [L, O^{2}(K ∩C̃)] ≠ 1. Let first L be of Lie type in characteristic two. Then N_{L}(V)/O_{2}(N_{L}(V)) has a normal subgroup Sp(2n−2, q)′. Further we have that W/Z(W) involves a strong quadratic module. Suppose first 2n−2 > 2. Then L ≅ Sp(2r, q) or F_{4}(q). We know that W acts quadratically on O_{2}(M_{0})/V, which gives with [Str] that O_{2}(M_{0})/V just involves spin modules. This shows K ∩ C̃ has a factor Sp(4, q). So L ≅ Sp(6, q), Sp(8, q) or F_{4}(q). If we have F_{4}(q), then W involves V (λ_{1}) and so C_{W} (O_{2}(M_{0}) is a trivial module for K ∩ C̃, a contradiction. Let W involve the spin module for L. The spin module is a sum of 4-dimensional Sp(4, q)–modules. But in QO_{2}(M_{0})(/O_{2}(M_{0}) we have a nonsplit extension of the trivial module with the 4-dimensional module, a contradiction as before. So we are left with L ≅ Sp(2n, q) and W involves V (λ_{n−1}). Now in O_{2}(M_{0})Q/Q we have at most 2 nontrivial Sp(4, q)–modules, so in W we have at most 4. This shows L ≅ Sp(6, q) and W involves V (λ_{2}). Now O_{2}(M_{0})/V involves the spin module just once. But then in C̃ we see that we both types of 4– dimensional modules, three of them are of the same type, while in M_{0} these modules come in pairs, a contradiction.

Let now n = 2. Then in N_{L}(V)/O_{2}(N_{L}(V)) we have a normal subgroup L_{2}(q). Assume that there is no module in O_{2}(M_{0})/V on which M_{0} induces the 6–dimensional 3A_{6} – module. Let first L ≅ Sp(2m, q) or Ω^{±}(2m, q). Then we have in W the natural module. But there we have just two nontrivial L_{2}(q)–modules, which shows that O_{2}(M_{0})/V is a trivial module, contradicting O_{2}(M_{0})Q/Q contains nontrivial modules.

So we have that L is a rank two group. Hence we have G_{2}(q)′ or ^{3}D_{4}(r), r^{3} = q. As there are just natural L_{2}(q)–modules involved, the latter is not possible. So we have G_{2}(q)′. Then W involves the natural module. If q ≠ 2, we have two L_{2}(q)–modules in O_{2}(M_{0})Q/Q, which shows that we must have two 4-dimensional modules in O_{2}(M_{0})/V, but we just have three of them in W, a contradiction. So we have q = 2. But then L acts transitively on this module, so W/W′ cannot be just the natural module. Further we see that O_{2}(M_{0}) just involves two 4-dimensional modules, and so by 2.8 we get |O_{2}(M)| ≤ 2^{10}. This now shows that there is a cyclic group U of order 4, centralizing W and normalized by L. Then we have U ≤ O_{2}(M) and [M_{0}, U] ≤ V. But now U ∩ V would be normal in M_{0}, a contradiction.

So we have shown that M_{0}/O_{2}(M_{0}) contains a normal subgroup 3A_{6} and O_{2}(M_{0})/V involves the 6-dimensional module. In particular p = q = 2. Let ρ be an element of order three in the center of the 3A_{6}. The existence of the 6-dimensional module and the fact that [W, O_{2}(M_{0})]V/V contains elements from this module shows, that ρ acts nontrivially on W. As [V, ρ] = 1, we see that [L, ρ] ≠ 1, and so ρ induces an automorphism on L. This first shows that we have L ≅ Sp(2m, 2) or Ω^{±}(2m, 2). Further now QO_{2}(M_{0})/O_{2}(M_{0}) is of order 4 and so C_{O_{2}(M_{0})Q}(V ∩Q) = O_{2}(M_{0}). As C_{L}(V) acts on V ∩W, we see that it normalizes O_{2}(M_{0} ∩Q)V. We further see that C_{L}(V ∩W) induces just one nontrivial irreducible module in this group, which is in W and so O^{2}(C_{L}(V ∩W)) centralizes Z((O_{2}(M_{0})∩Q)V), which contains V, hence is in M. Now the structure of M_{0} shows that O^{2}(C_{L}(V ∩Q))/O_{2}(O^{2}(C_{L}(V ∩Q))) has a normal subgroup of order three. This shows L ≅ Sp(6, 2), Ω^{+}(8, 2), or Ω^{−}(6, 2). We have that in O_{2}(M_{0})Q the number of nontrivial irreducible ρ-modules is a multiple of three. So Sp(6, 2) is not possible.

Let L = Ω^{+}(8, 2). Then first M_{0}/O_{2}(M_{0}) ≅ 3Σ_{6}, and in O_{2}(M_{0})/V we have just two 6-dimensional modules. The structure of O_{2}(M_{0})Q/Q shows that these two modules are interchanged in O_{2}(M_{0}). By 2.8 we now get that |O_{2}(M_{0})| ≤ 2^{18}. If we have equality, we see that Z(Q) is of order 4 and so V Z(Q) is normal in M_{0}. Now [L, Z(Q)] = 1, and as O^{+}(8, 2) cannot be involved in C̃, we see that [S, Z(Q)] = 1, a contradiction. So we have that |O_{2}(M_{0})| = 2^{17} and Q = W is extraspecial of order 2^{9}. Let X ≤ O_{2}(M_{0}) with X/V is an extension of the trivial module by an 6-dimensional module and X normal in M_{0}. Then as |[O_{2}(M), X]V | = 2^{10}, we see that Z(X) > V. But then X is abelian. We have X ∩ Q is of order 2^{5} and so |XQ/Q| = 2^{6} This shows that N_{L}(XQ/Q) contains A_{8}. Now we see, as C_{Q}(X) = X ∩ Q, that X is the unique elementary abelian subgroup of order 2^{11} in XQ, which shows that NC̃(X) covers N_{L}(XQ/Q). As X was normal in M_{0}, we see that N_{G}(X)_{0} > M_{0}, a contradiction.

Suppose that we have L ≅ Ω^{−}(6, 2). Then we have just one 6-dimensional module involved. Hence by 2.8 we see that |O_{2}(M_{0})| ≤ 2^{11}, and that |Q| ≤ 2^{8}. If |Q| = 2^{8}, then |Z(Q)| = 4. As [O^{2}(C̃ ∩ M_{0}), Z(Q)] = 1, see that V Z(Q) is normal in M_{0} and further as S ∩ L centralizes Z(Q), we have that C_{M_{0}}(ρ) centralizes Z(Q). But then we have that V Z(Q) splits and so Z(M_{0}) ≠ 1, a contradiction. So we have that |O_{2}(M_{0})| = 2^{10} and W = Q is extraspecial of order 2^{7}. This now also shows that there are subgroups M_{1} of M_{0} and C_{1} of C̃ such that M_{1}/O_{2}(M_{1}) ≅ 3A_{6}, C_{1}/W ≅ Ω^{−}(6, 2) and both are normalized by S with S ∩ M_{1} = S ∩ C_{1}. Hence these groups are of type Suz. We also see that P_{0} commutes with the point stabilizer in C_{1} and so H = ⟨M_{1}, C_{1}⟩ has a diagram of type ∘ ∘ ∘, i.e. F^{∗}(H) is weak parabolic isomorphic to Suz.

Let next L ≅ A_{m}. Then we have q = 2 and n = 2 or 3. Suppose that W involves the permutation module. As [V, W]Z(W)/Z(W) has no trivial module, we see that m = 12 or m = 6. In L = A_{12} there are excactly two A_{6}–modules in W. This shows that O_{2}(M_{0})/V is the trivial module, which contradicts the fact that O_{2}(M_{0})Q/Q involves a nontrivial module. If L = A_{6} the same applies. So we have that W involves the spin module. Then V corresponds to (1, 2)(3, 4). Then we get m = 7 and V involves the 4–dimensional module. Now we see that O_{2}(M_{0})/V is the trivial module and so by 2.8 |O_{2}(M_{0})| ≤ 2^{5}, and |S| ≤ 2^{9}. This gives |Q| ≤ 2^{6}. But Q is nonabelian, and so A_{7} cannot act.

So we have that L is sporadic or a group of Lie type in odd characteristic. Suppose that there is a quadratic fours group. Then by [MeiStr1],

[MeiStr2] L ≅ 3U_{4}(3), M_{12}, M_{22}, or Co_{2}. In the first three cases we have that [W, V]Z(W)/Z(W) is of order 4. But from the modules we see that this is not the case. So we have L ≅ Co_{2}, |[W, V]Z(W)/Z(W)| = 2^{6} and W involves the 22–dimensional Leech lattice module. Now in QO_{2}(M_{0}) there are exactly two spin modules for Sp(6, 2). Hence in O_{2}(M_{0})/V there is exactly one nontrivial module involved, the spin module. Now by 2.8 we see that |O_{2}(M_{0})| ≤ 2^{25}, and so |S| ≤ 2^{41}. As |Q| ≥ 2^{23}, we get equality everywhere. So both parabolics are as in F_{2}. It remains to show that we have a diagram like in F_{2}. The point stabilizer in Co_{2} is the M_{22}–parabolic. This commutes with P_{0}. Hence we see that we have a diagram of type ∘ ∘ ∘ M^{22} ∘ for H = ⟨M_{0}, L⟩, and so F^{∗}(H) is weak parabolic isomorphic to F_{2}.

So we have that there is no quadratic fours group. If there is no nontrivial module for O^{2}(K ∩ C̃) in O_{2}(M_{0})Q/Q, we get that O_{2}(M_{0})/V is the trivial module. In particular O_{2}(M_{0}) = Y_{M}. So |Y_{M} : Y_{M} ∩ Q| = 2. Hence L ≅ L_{3}(r), L_{2}(r) or U_{3}(r), r odd, or Co_{3}. In the first three case M_{0}/C_{M_{0}}(V) ≅ Sp(4, 2) or Sp_{6}(2). In the first case |S| ≤ 2^{9}, so |Q| ≤ 2^{7}. This shows L ≅ L_{2}(r). Now we see that |Q/Z(Q)| = 2^{4}, so L ≤ A_{8}, but then there is no element of order three in C_{L}(V Q/Q). So assume that we have Sp(6, 2). Then we have L ≅ L_{3}(9) or U_{3}(9). But |Q/Z(Q)| ≤ 2^{9}, a contradiction. So we are left with L ≅ Co_{3}. Then M_{0}/O_{2}(M_{0}) ≅ Sp(8, 2). Then by 2.8 |O_{2}(M_{0})| ≤ 2^{9} and so also |Q| ≤ 2^{15}. But Co_{3} has no 14–dimensional representation as it contains an extraspecial group of order 5^{3}.

So we are left with K ∩ C̃ acting nontrivially on O_{2}(M_{0})Q/Q. which shows that O_{2}(M_{0})Q/Q ≅ Q_{8}, and so L has a semidihedral Sylow 2subgroup. Again L ≅ L_{3}(r) or U_{3}(r), or L ≅ M_{11}. Now in any case we have that E(M_{0}/C_{M_{0}}(V)) ≅ A_{6}. Further there cannot be some 6-dimensional module for 3A_{6} in O_{2}(M_{0}) as otherwise an element of order three in C(V) has to centralize L and W, a contradiction. So we have that |O_{2}(M_{0})| ≤ 2^{10} and then |Q/Z(Q)| ≤ 2^{9}. In particular L ≅ U_{3}(3) ≅ G_{2}(2)′, a contradiction.

Sln

Lemma 5.2 Let F^{∗}(M_{0}/O_{p}(M_{0}))/Z(F^{∗}(M_{0}/O_{p}(M_{0}))) ≅ SL_{n}(q), n ≥ 3 and assume that Y_{M} contains V, the natural module. Then q = 2 and n = 3. Further |O_{2}(M_{0})| = 2^{4}, Q is extraspecial of order 32 and C̃/Q ≅ Σ_{3} × Σ_{3}. Further there are two minimal parabolics P_{1}, P_{2} with P_{i}/O_{2}(P_{i} ≅ Σ_{3}, i = 1, 2, and O_{2}(⟨P_{1}, P_{2}⟩) = 1.

Proof: Suppose V ≰ Q. Then we have that V acts nontrivially on Z(Q). Now as [Q, x] = [Q, y] for all x, y ∈ V \Q, we see that [O_{p′}(C̃/Q), V] = 1 and by quadratic action we then see that there is a unique component L of C̃/Q, not centralized by V. As |V : V ∩ Q| ≥ p^{2}, we see with [Cher],

[MeiStr1] and [MeiStr2] that L is of Lietype in characteristic p, or p = 2 = q and it is alternating, 3U_{4}(3) or some sporadic group. If p = q = 2, then V induces on Z(Q) a group of transvections to Z(S) ∩ V, which shows that L = L_{m}(2) for suitable m. Hence in any case L is of Lietype in characteristic p. Let x ∈ V \ Q, then C_{L}(x) cannot be a maximal subgroup as |V : V ∩ Q| > q. Hence we get L ≅ SL_{m}(q). Further V induces a group of transvections to a point. We see that [O^{p}(C̃ ∩ M_{0}), V Q] is extraspecial. As [V, E] ≤ Q, this group is normalized by E and as [E, Z(Q)] = 1, we have [E, V] = 1. This shows E ≤ M, a contradiction to E–uniqueness.

So we have shown that V ≤ Q. In particular [L, Z(Q)] = 1, where L = ⟨Y C̃ ; M⟩. Set W = ⟨V^{L}⟩, then [W, Q] ≤ V ∩ Z(Q). As [W, Y_{M}] ≠ 1, we see that [W, V] ≠ 1, so W O_{p}(M_{0}) = QO_{p}(M_{0}). We have that W acts quadratically on O_{p}(M_{0})/V and so also on Y_{M}/V. As Y_{M} is a 2F–module we see that Y_{M}/V just involves trivial modules, natural modules, alternating or symmetric squares of natural modules. Set X_{0} = [Y_{M}, W]V. As C_{W} (V) = C_{W} (Y_{M}), we see that X_{0} = XV, where X ≤ Z(W). As [L, Z(W) ∩ V] = 1, we see that we may choose X such that [X, L] = 1. Now let Ṽ ≤ Y_{M}, such that Ṽ /V is an irreducible M_{0}–module. As [W, Ṽ V/V] is centralized by M_{0} ∩ C̃, we see that Ṽ /V is the trivial module or the natural module isomorphic to V. Suppose that we have the natural module. Then we have [W, Ṽ]V = V U, where [U, C̃ ∩ M_{0}] = 1. So we have that C_{Ṽ /U}(W) is an extension of the trivial module by a natural module. If this splits, then we get with [Hu, (I.17.4)], that we have that Ṽ splits over V. So assume not, then either n = 3, q is even, or n = 4, q = 2. Let first n = 4. As there is some x ∈ U^{♯} such that x is the only conjugate of x in xV, this holds for any coset of V, but if C_{Ṽ /U}(W) is nonsplit this is not true here, as we have at least two conjugates for any representative. So also in that case Ṽ splits. It remains n = 3, q even. But for a root group R in W we have [Ṽ , R] is of order q^{2} and so this is also true for a root group in C̃ ∩M_{0} which is not in W. But this implies that C_{Ṽ /U}(W) has to split. So we have a split extension in any case. Now as before we get that Ṽ ≤ Q. But then [W, Ṽ] ≤ [W, Q] ≤ V, a contradiction.

So we have that Ṽ /V is the trivial module. Then by 2.8 we get that p = 2 and M_{0}/O_{2}(M_{0}) ≅ L_{3}(2). Let Y_{M} ≥ V_{1} > Ṽ, V_{1}/Ṽ a nontrivial irreducible M_{0}–module. Then this module has to be a natural one. But as CṼ (w) ≤ V for any w ∈ W \ O_{2}(M_{0}), we see that V_{1}/V has to split, a contradiction. So we have that Y_{M} is of order 16 and |Y_{M} : Y_{M} ∩ Q| = 2.

Let y ∈ Y_{M} \ Q, then [Q, y] is of order 4. Suppose R = [y, F (C̃/Q)] ≠ 1. Then R is a 3–group or a 5–group. Set W̃ = ⟨V^{R}⟩. If R is a 5-group, we see that W̃ is extraspecial of order 32. But then [y, W̃] cannot be elementary abelian. So we have that R is a 3–group. Let R_{1} be a critical subgroup in R and R_{2} = [R_{1}, y]. Suppose R_{2} is nonabelian. Then we see that R_{2} is extraspecial of order 27. This now shows that ⟨V^{R_{2}}⟩ is extraspecial of order 2^{7}. Now R_{2} contains some element of order three in the center, which acts on V nontrivially, so it has to be in M_{0}. This shows that there is one further natural module in O_{2}(M_{0}), which is dual to V. Hence we see that |O_{2}(M_{0})| = 2^{7} or 2^{8}. In any case O_{2}(M_{0}) is abelian, a contradiction. So R_{2} is elementary abelian. In particular |R_{2}| ≤ 9. Suppose |R_{2}| = 9. Then ⟨V^{R_{2}}⟩ is extraspecial of order 32. But also C̃ ∩M_{0} act on this group. As there is no 3-group of order 27 in the automorphism group, we get some element of order three centralizing V. But this element is neither in R_{2} nor in M_{0}. As any y-invariant group of order three in R_{2}(M_{0}∩C̃) is either in M_{0} or R_{2}, we have a contradiction. So we are left with |R_{2}| = 3. Again ⟨V^{R_{2}}⟩ is extraspecial of order 32. Hence for x ∈ W, we have that |O_{2}(M_{0}) : C_{O_{2}(M_{0})}(x)| ≤ 4. This shows that [O_{2}(M_{0}), W] ≤ Y_{M} and so O_{2}(M_{0})/Y_{M} is the trivial module. Hnece we get O_{2}(M_{0}) = Y_{M} and |S| = 2^{7}. This also shows that |Q| = 2^{5} and then R = R_{1} and C̃/Q ≅ Σ_{3} × Σ_{3}. Let P_{0} be the minimal parabolic in M_{0} not in C̃ and P_{1} the parabolic in C̃ not in M_{0}. Set U = ⟨P_{0}, P_{1}⟩. We have that W_{1} = ⟨Z(S)^{P_{0}}⟩ is of order 4. Now W_{2} = ⟨W_{1}^{P_{1}}⟩ ≤ Q, is of order 8. But W_{2} ≠ V, so W_{2} ≰ Y_{M}. Hence W_{3} = ⟨W_{2}^{P_{0}}⟩ is of order at least 16. As W_{3} ≠ Y_{M}, we see that W_{3} ≰ O_{2}(P_{1}) and so O_{2}(U) = 1.

So assume now that [y, F (C̃/Q)] = 1. Then we have some component L of C̃/Q with [L, y] ≠ 1. As |[Q, y]| = 4, we see that y normalizes L and the same is true for M_{0} ∩ C̃. By 2.5 and 2.4 we see that L is of Lie type in characteristic two, 3A_{6} or alternating.

Suppose first that L is of Lie type in characteristic two. As C_{L}(y) acts on V, we have that C_{L}(v) has a factor group Σ_{3}, or L ≅ SL(m, 4) and M_{0} ∩ C̃ induces field automorphisms. Assume the latter. Then W/C_{W} (L) is the natural module. But we now have some element ρ of order three in Z(L) and then W = [W, ρ], which implies W to be abelian, a contradiction. So we have that L ≅ Ω^{±}(2n, 2), Sp(2n, 2) or G_{2}(2)′. In any case W involves just the natural module, or we have Sp(6, 2) or Ω^{−}(6, 2) and the spin module is involved. Let L_{1} be the point stabilizer. Then we see that L_{1} commutes with P_{0}. If we have the orthogonal or symplectic module, then |⟨V^{O^{2}(L_{1})}⟩O_{2}(M_{0}) : O_{2}(M_{0})| = 2. As this group is P_{0}–invariant, we have a contradiction. So let L ≅ Ω^{−}(6, 2) and the half spin module is involved. Then an element ρ of order three in C̃ ∩ M_{0} acts fixed point freely on this module. This shows that in O_{2}(M_{0})/Y_{M} there are two natural modules involved which are dual to V. But this implies that C_{O_{2}(C_{L}(y))}(ρ) is of order at least 8, a contradiction. So let L ≅ Sp(6, 2) and the spin module be involved. Then we have some element τ of order three in C_{L}(y) such that τ commutes with M_{0}/O_{2}(M_{0}). Now we see again that in O_{2}(M_{0})/O_{2}(M_{0})′ we have two modules which are dual to V. Further as [ρ, O_{2}(M_{0})Q/Q] is extraspecial of order 32, ρ as before, we see that there are also two natural modules which are isomorphic to V. Now as [O_{2}(M_{0})Q/Q, τ] = O_{2}(M_{0})Q/Q, we see that all these modules are in [O_{2}(M_{0})/O_{2}(M_{0})′, τ]. But |[Q, τ]/[Q, τ]′| = 16, so |[O_{2}(M_{0})/O_{2}(M_{0})′, τ]| = 2^{10}, a contradiction, as this group has to contain a direct sum of 4 natural L_{3}(2)–modules.

So we are left with L ≅ G_{2}(2)′. Then W involves the 6-dimensional module. As G_{2}(2)′ acts transitively on the nontrivial vectors of the natural module, we see that W ≅ Z_{4} ∗ Q_{8} ∗ Q_{8} ∗ Q_{8}. Now let τ be some element of order three in the point stabilizer and V_{1} = C_{V} (τ), then ⟨τ, P_{0}⟩ acts on V_{1}. Let g ∈ P_{0}, g^{2} ∈ C(V_{1}), g ∉ C(V_{1}). We may assume that g commutes with τ. So we see that V ≤ W ∩ W^{g} and so as Φ(W ∩ W^{g}) = 1 and τ acts on W ∩ W^{g}, we see that W ∩ W^{g} is elementary abelian of order 16. As [W, Q] = W′, we see that W^{g} ∩ Q = W^{g} ∩ W. As |W^{g} : W^{g} ∩ C̃| = 2, we see that |W^{g} ∩ C̃ : W^{g} ∩ Q| = 8. In particular Z(W^{g}) ≰ Q and [τ, Z(W^{g})] ≤ Q. This shows that [C_{W} (V_{1}), Z(W^{g})] ≰ V_{1}, but [Z(W), C_{W^{g}}(V_{1})] ≤ Z(W), a contradiction.

Let now L ≅ 3A_{6}. Then W involves the 6–dimensional module. Because of the central element of order three, we see that W is extraspecial of order 2^{7}. But 3A_{6} is not a subgroup of O^{−}(6, 2).

So let finally L be alternating. As [W, y] contains no trivial C_{L}(y)– module, we see that L ≅ A_{7} and we have the 4–dimensional module. Again P_{0} commutes with the point stabilizer, whose maximal normal 2-subgroup is Q, which shows that P_{0} normalizes Q, a contradiction.

3A6

Lemma 5.3 Let F^{∗}(M_{0}/O_{2}(M_{0})) ≅ 3A_{6} and assume that Y_{M} contains V, the 6–dimensional module. Then there is a subgroup H of G such that F^{∗}(H) is weak parabolic isomorphic to He.

Proof: As there is no offender in QO_{2}(M_{0}) on V, we have V ≰ Q. Suppose QO_{2}(M_{0}) ≤ M′_{0}O_{2}(M_{0}). Then Z(M_{0}/O_{2}(M_{0})) acts on a subgroup of Z(Q) and so we have Z(M_{0}/O_{2}(M_{0})) ≤ C̃ ∩ M_{0}/O_{2}(M_{0}). Hence we have |V/V ∩ Q| = 16 and so V ∩ Q ≤ Z(Q). Now we have fours groups of transvections on Z(Q). As Σ_{3} × Σ_{3} acts on V/V ∩ Q we see that V has to centralize F (C̃/Q). Let now L be a component with [V, L] ≠ 1. By quadratic action we see that [L, V] ≤ L. As we have fours groups of transvections, we see L ≅ L_{n}(2) for suitable n. Further W = ⟨(V ∩ Q)^{L}⟩ ≤ Z(Q), is the natural module. As all submodules of Y_{M} are 6-dimensional modules by 5.1, and Y_{M} is better than 2F by [MeStStr, (4.2)], we have that Y_{M} = V, and so Ω_{1}(Z(S)) ≤ W. As [O^{2}(L), Q] ≤ W, we see W = Q. But then E = 1, a contradiction to E–uniqueness.

So we have that Q ≰ M′_{0}O_{2}(M_{0}). Then we also see that |V/V ∩ Q| = 4 and |V ∩ Q| = 16. Further [Q, V ∩ Q] = Ω_{1}(Z(S)). Set C_{1} = ⟨V C̃⟩. Assume that [C_{1}, Ω_{1}(Z(Q))] ≠ 1. Then Ω_{1}(Z(Q)) ≰ O_{2}(M_{0}). Then as [V, Q] = V ∩Q, we get that |Z(Q) : Z(Q)∩O_{2}(M_{0})| = 4. But then [V, Z(Q)] is invariant under Z(M_{0}/O_{2}(M_{0})) and so Q ≤ M′_{0}O_{2}(M_{0}), a contradiction. So [C_{1}, Ω_{1}(Z(Q))] = 1. Set W = ⟨(V ∩ Q)^{C_{1}}⟩. Then W′ = Ω_{1}(Z(S)).

Assume first that W O_{2}(M_{0}) ≤ M′_{0}O_{2}(M_{0}). Then |Z(W) ∩ V | = 4. Let τ be an element of order three with τO_{2}(M_{0}) ∈ Z((M_{0}/O_{2}(M_{0}))′). Then it acts on Z(W) ∩ V. Suppose [C_{1}, Z(W)] = 1, then we see that N_{G}(Z(W) ∩ V) ≥ ⟨C_{1}, τ⟩.

We have that C_{W} (x) = C_{W} (y) for all x, y ∈ V \Q. Hence [V, F (C̃/Q)] = 1. Now let L ≤ C_{1}Q/Q be a component with [L, V] ≰ Q. Then L ≤ C(Z(W)∩ V)/Q and L^{τ} ≤ C_{1}/Q. Set U = O^{2}(⟨L^{⟨τ⟩}⟩). Then U is normalized by E and τ, so τ ∈ C̃ by E–uniqueness, a contradiction.

As there are no transvections on V, we see that |W : W ∩ O_{2}(M_{0})| > 2. This now shows W O_{2}(M_{0}) = QO_{2}(M_{0}) and W′ = Ω_{1}(Z(S) ∩ V). Hence W acts quadratically on O_{2}(M_{0})/V. In particular no 6–dimensional 3A_{6}– modules is involved. Let τ be as before, then we have V = [O_{2}(M_{0}), τ]. As Y_{M} = V, we see that V = O_{2}(M_{0}). Hence |S| = 2^{10}, Q = W is of order 2^{7}. So C̃/Q is a subgroup of Σ_{8} which has a dihedral Sylow 2-subgroup. Let R ≤ M_{0} ∩ C̃, R/Q ≅ A_{4}. Assume R ≤ R_{1} with R_{1}/Q ≅ A_{5}. As R_{1} is not irreducible, we have that R_{1} has a fixed point on W/Z(W). But R has just one fixed point whose preimage is a fours group. Then R_{1} acts on Z_{2} × D_{8} ∗ D_{8}, which is not possible. Hence we have that R/Q is not in some A_{5}. Inspection of the subgroups of A_{8} shows that we have C̃/Q ≅ L_{3}(2). Hence the parabolics are as in He. Let P_{1} be a minimal parabolic in M_{0} such that ⟨Z(S)^{P_{1}}⟩ is centralized by a minimal parabolic P_{3} in C̃. Hence ⟨P_{1}, P_{3}⟩ = P_{1}P_{3}. So H = ⟨P_{1}, P_{3}, C̃ ∩ M_{0}⟩ has diagram ∘ ∘ ∼ ∘, which shows that F^{∗}(H) is weak parabbolic isomorphic to He.

M24

Lemma 5.4 Let F^{∗}(M_{0}/O_{2}(M_{0})) ≅ M_{24} and assume that Y_{M} contains V, one of the two 11–dimensional modules. Then there is a subgroup H of G such that F^{∗}(H) is weak parabolic isomorphic to M(24)′, Co_{1} or J_{4}.

Proof: There are two 11–dimensional modules Assume first that C̃ ∩ M_{0}/O_{2}(C̃ ∩ M_{0}) ≅ 3Σ_{6}. We have for C̃ ∩ M_{0} the following series in V

1 < V_{1} < V_{2} < V,

where V_{1} = Z(S) ∩ V, V_{2}/V_{1} is the 6–dimensional 3Σ_{6}–module and V/V_{2} is a 4–dimensional module. As V is not an F –module we have V ≰ Q and so |V/V ∩ Q| = 2^{4}. As Z(Q) ≤ O_{2}(M_{0}), we see that for W = ⟨(V ∩ Q)^{C_{1}}⟩ that W′ = Z(S) ∩ V = [Q, W], where C_{1} = ⟨Y C̃ ; M⟩(M_{0} ∩ C̃). Now we see that W acts quadratically on O_{2}(M_{0})/V. As |W : W ∩ O_{2}(M_{0})| = 2^{6}, we see with [MeiStr1] that O_{2}(M_{0})/V is the trivial module. Hence O_{2}(M_{0}) = Y_{M} and by 2.8 |Y_{M} : V | ≤ 2. This shows that |Q : W | ≤ 2 and W is extraspecial of order 2^{13}. Suppose that Q > W. Then also Y_{M} ∩ Q > V ∩ Q and there is some x ∈ Y_{M} \ V, which is centralized by (M_{0} ∩ C̃)′. As there are 2048 elements in Y_{M} \ V, we see that x is centralized by S, which contradicts [Hu, (I.17.4)] and Z(M_{0}) = 1. So we have Q = W is extraspecial. Now C̃/Q is a subgroup of Ω^{+}(12, 2). As 2^{4}3Σ_{6} is a subgroup we first see that F (C̃/Q) is centralized by V and that there is exactly one component which is not centralized by V. This then shows that E(C̃/Q) is a group of Lie type in characteristic two or by [MeiStr1], [MeiStr2] it is 3U_{4}(3) or 3M_{22}, or alternating. The possible groups of Lie type are symplectic groups. For the alternating groups and for the groups of Lie type we get that there is a subgroup of order three in the Schurmultiplier, a contradiction. Suppose that we have 3M_{22} in C̃/Q, then as M_{0} ∩ C̃ involves Σ_{6}, we get C̃/Q ≅ 3Aut(M_{22}) and so Y_{M} = V. Hence we have the parabolics as in J_{4}. Set H = ⟨M_{0}, C̃⟩. In C̃ the point stabilizer on Q/Z(Q) is the Σ_{5} parabolic with a Σ_{3}–subparabolic in M_{0} ∩ C̃, which together with P_{0} is an L_{3}(2)–parabolic in M_{24}. So we get the diagram

∘ ∘ M̃_{22} ∘, hence F^{∗}(H) is weak parabolic isomorphic to J_{4}.

Suppose now that we have a normal subgroup 3U_{4}(3). Suppose Y_{M} ≠ V, then we see that Y_{M} ≰ C̃′. By 2.10 we have that M_{0}/V ≅ Z_{2} × M_{24}. Set M_{1} = M′_{0} and C_{1} = C̃′T, where T is a Sylow 2-subgroup of M_{1} ∩ C̃, then both groups have a common Sylow 2–subgroup. Set H = ⟨M_{1}, C_{1}⟩. Then the M_{1} and C_{1} are as in M(24)′. Let L be the point stabilizer in C̃. Then we see that L commutes with P_{0}. Hence we get a diagram ∘ ∘ ∼ ∘ ∘ for H and so F^{∗}(H) is weak parabolic isomorhic to M(24)′.

So assume now that M_{0}∩C̃/O_{2}(M_{0}∩C̃) ≅ A_{8}. Then we get the following M_{0} ∩ C̃–series in V

1 < V_{1} < V_{2} < V,

where V_{1} = Z(S) ∩ V, V_{2}/V_{1} is the L_{4}(2)–module and V/V_{2} is the Ω^{+}(6, 2)– module. Again V ≰ Q and V_{2} = V ∩ Q. Define W as before, then we see that W acts quadratically on O_{2}(M_{0})/V. Again by [MeiStr2] we get that O_{2}(M_{0})/V is the trivial module. By 2.8 we have V = O_{2}(M_{0}). This now shows that W = Q is extraspecial of order 2^{9} and so C̃/Q is a subgroup of O^{+}(8, 2). As it contains a maximal parabolic in Ω^{+}(8, 2), we get C̃/Q ≅ Ω^{+}(8, 2). In particular we have parabolics as in Co_{1}. We have in C̃ the following

1∘  ∘^{3}^{2} ∘  ^{4}∘

diagram where we choose notation such that ⟨P_{1}, P_{2}, P_{3}⟩ = M_{0} ∩ C̃. Further we may choose notation such that M_{0} has the following diagram ^{2}∘ ^{1}∘ ∼ ^{0}∘. Now P_{4} is in the point stabilizer, which commutes with P_{0}. This shows that

H = ⟨M_{0}, C̃⟩ has the following diagram ^{4}∘  ^{2}∘  ^{1}∘ ∼ ^{0}∘, so F^{∗}(H) is

of weak parabolic isomorphic to Co_{1}.

M22

Lemma 5.5 Let F^{∗}(M_{0}/O_{2}(M_{0})) ≅ M_{22} or 3M_{22}, and assume that Y_{M} contains V, one of the two 10–dimensional modules. Then there is a subgroup H of G such that F^{∗}(H) is weak parabolic isomorphic to Co_{2}.

Proof: As V is not an F –module, we have V ≰ Q. Assume first that M_{0} ∩ C̃/O_{2}(M_{0} ∩ C̃) ≅ A_{6}, Σ_{6}, 3A_{6}, or 3Σ_{6}. As an M_{0} ∩ C̃–module we have

1 < V_{1} < V_{2} < V_{3} < V,

where V_{1} = Z(S) ∩ V, V_{2}/V_{1} and V/V_{3} are 4–dimensional modules and V_{3}/V_{2} is a trivial module. So V ∩ Q = V_{2} or V_{3}. Suppose V_{3} = V ∩ Q. Set W = ⟨(V ∩ Q)^{C_{1}}⟩, where C_{1} = ⟨V C̃⟩(M_{0} ∩ C̃). As [V_{3}, Q] = V_{2}, we see that there are two nontrivial C_{1}–modules in W. Hence we also must have two modules in QO_{2}(M_{0})/O_{2}(M_{0}). But this is the 4-dimensional A_{6}–module, a contradiction.

So we have that V_{2} = V ∩ Q. We see that Z(Q) ≤ O_{2}(M_{0}) and so [C_{1}, Z(Q) ∩ V] = 1. Hence we see that [Q, W] = Z(S) ∩ V. Further W O_{2}(M_{0}) = QO_{2}(M_{0}) and W acts quadratically on O_{2}(M_{0})/V. Suppose that O_{2}(M_{0})/V is the trivial module, so O_{2}(M_{0}) = Y_{M}. By 2.8 we see that |Y_{M} : V | ≤ 2. Hence we get that |Q| ≤ 2^{10}. Further we have that W is extraspecial of order 2^{9} or a symplectic group of order 2^{10}. So C̃/Q is a subgroup of Sp(8, 2). This first shows that [V, F (C̃/Q)] = 1. Let L be some component in C̃/Q, with [V, L] ≰ Q, then we see by quadratic action of V, which is of order 32, and [MeiStr1], [MeiStr2] that L is a group of Lie type in characteristic two or alternating. As 11 does not divide the order of Sp(8, 2) and no A_{n}, n < 11, has a subgroup 2^{5}A_{6}, we see that L is of Lie type in characteristic two with a Sp(4, 2), parabolic. As |S/Q| ≤ 2^{10}, we see L ≅ Sp(6, 2). Assume now that |Q| = 2^{10}, then |Z(S)| = 4. Hence Y_{M} = V Z(S). But then by [Hu, (I.17.4)] the extension splits, a contradiction. Hence we have that V = Y_{M}, M_{0}/V ≅ Aut(M_{22}) and Q is extraspecial of order 2^{9}, and C̃ induces the spin module in Q/Z(Q). So these parabolics are as in Co_{2}. Let P_{0} be the Σ_{5}–parabolic in M_{22} such that |P_{0} : P_{0} ∩ C̃| = 5. Let P_{1} be the minimal parabolic in C̃, P_{1} > S, P_{1} ≰ M_{0}. Then first ⟨P_{1}, P_{0} ∩ C̃⟩ = P_{1}(P_{0} ∩ C̃), W_{1} = ⟨Z(S)^{P_{0}}⟩ is the permutation module and |W_{1} : W_{1} ∩ Q| = 2. So we see that P_{1} acts on W_{1}Q. As we also have that W_{1}/Z(S) ∩ Z_{2}(S)/Z(S) ≠ 1 and P_{1} centralizes Z_{2}(S)/Z(S), we may assume that for some element ρ of order three in P_{1} we have |C_{W_{1}}(ρ)| ≥ 8. But then there is an element τ of order three in P_{0} ∩ C̃ with [ρ, τ] = 1 and ⟨C_{W_{1}}(ρ)^{⟨τ⟩}⟩ = W_{1}. This shows that ⟨P_{0}, P_{1}⟩ ≤ N_{G}(W_{1}) and so they commute. Hence for H = ⟨M_{0}, C̃⟩ we get the Co_{2}–diagram ∘ ∘ M^{22} ∘ and so F^{∗}(H) is weak parabolic isomorphic to Co_{2}.

So assume now that O_{2}(M_{0})/V is not a trivial module. Then by [MeiStr2] we see that we have 3M_{22} involved and the 12–dimensional module is induced. As the central element of order three in 3M_{22} is in C̃, we see that this module is covered by Q. But [W, Q] ≤ V, a contradiction.

Next we assume that the point stabilizer in V is the Σ_{5}–parabolic in M_{22}. If |QO_{2}(M_{0})/O_{2}(M_{0})| = 16, we get |C_{V} (Q)| = 4. But then we get a contradiction to E–uniqueness as ⟨C_{M_{0}}(x) | x ∈ C_{V} (Q)^{♯}⟩ = M_{0}. So we have that |QO_{2}(M_{0})/O_{2}(M_{0})| = 32 and so M_{0}/O_{2}(M_{0}) ≅ Aut(M_{22}) or Aut(3M_{22}). As an M_{0} ∩ C̃–module, we have

1 < V_{1} < V_{2} < V_{3} < V,

where V_{1} and V_{2}/V_{1} are trivial modules and V_{3}/V_{2} and V/V_{3} are 4–dimensional Σ_{5}–modules, V_{3}/V_{2} the L_{2}(4)–module and V/V_{3} the permutation module. Hence V ∩Q = V_{3}. As C_{S}(V_{3}) = O_{2}(M_{0}), see Z(Q) ≤ O_{2}(M_{0}). Now set again W = ⟨(V ∩ Q)^{C_{1}}⟩. Then [Q, W] = Z(Q) ∩ V. This shows that W acts quadratically on O_{2}(M_{0})/V. With [MeiStr2] we get that O_{2}(M_{0})/V is the trivial module. By 2.8 we now see that O_{2}(M_{0}) = V and M_{0}/V ≅ Aut(M_{22}). Hence Q is extraspecial of order 2^{11} and C̃/Q is a subgroup of O^{+}(10, 2). Again we see that [V, F (C̃/Q)] = 1 and there is a component L of C̃/Q, normalized by M_{0} ∩ C̃, with [V, L] ≠ 1. Application of [MeiStr2] and [MeiStr1] shows that L is a group of Lie type in characteristic two or alternating. As |S/Q| = 2^{7}, and we have a parabolic 2^{4}Σ_{5}, the only possible alternating groups are A_{10} or A_{11}. Then Q/Z(Q) involves the permutation module, hence the same is true for C_{Q}(V), which is V_{3}, a contradiction as there the L_{2}(4)–module is involved. So we have a group of Lie type in characteristic two. Now C_{Q/Z(Q)}(V) = V_{3}/V_{1}, which is not an irreducible M_{0} ∩ C̃–module. As O_{2}(M_{0} ∩ C̃)/Q = V Q/Q, we get with [Sm] that Q/Z(Q) is not an irreducible L–module. But then we see that in Q/Z(Q) there is exactly one nontrivial irreducible module involved and as M_{0} ∩ C̃ acts indecomposably on V_{3}/V_{1}, we get that there is some trivial submodule, i.e. C_{Q/Z(Q)}(L) ≠ 1. Let L_{1} be the preimage of L, then [V_{2}, O^{2}(L_{1})] = 1. As O^{2}(L_{1}) = E, we again see that C_{G}(x) ≤ C̃ for any x ∈ V_{2}^{♯}. But ⟨C_{M_{0}}(x) | x ∈ V_{2}^{♯}⟩ = M_{0}, a contradiction.

M11

Lemma 5.6 Let F^{∗}(M_{0}/O_{3}(M_{0})) ≅ M_{11}, and assume that Y_{M} contains V, the 5–dimensional GF (3)–module. Then M/O_{3}(M) ≅ Z_{2} × M_{11}, C̃/Q ≅ 4SL_{2}(9) : 2 (both groups are as in Co_{3}) and there are two minimal parabolics P_{1}, P_{2}, sharing a Sylow 3–subgroup of G such that P_{1}/O_{3}(P_{1}) ≅ A_{6} and P_{2}/O_{3}(P_{2}) ≅ SL(2, 9) and O_{3}(⟨P_{1}, P_{2}⟩) = 1.

Proof: There are two 5-dimensional M_{11}–modules V. But by E– uniqueness we see that we must have the orbits of length 55 and 66 on the 1–dimensional subspaces. Hence we see that M_{0} ∩ C̃/O_{3}(M_{0}) ≅ 3^{2}QD_{16}, the Sylow 3-normalizer in M_{11}. Now we also see that |V : V ∩ Q| = 3^{2} and [V, Z(Q)] = 1. Set C_{1} = ⟨V C̃⟩(M_{0} ∩ C̃), then we have that C_{1} normalizes V ∩ Z(S). Hence [W, Q] = V ∩ Z(S), where W = ⟨(V ∩ Q)^{C_{1}}⟩. This shows that W acts quadratically on O_{3}(M_{0})/V and then by [Cher] O_{3}(M_{0})/V is a trivial module. By 2.8 we see that V = O_{3}(M_{0}). Hence |S| = 3^{7} and Q is extraspecial of order 3^{5}. This shows that CC̃(Z(Q))/Q is a subgroup of Sp(4, 3). Now the only possibility is that E(C̃/Q) ≅ SL_{2}(9). As NC̃(S) involves some quasidihedral group of order 16, we see that SL_{2}(9) : 2 is involved. In M_{0} we see that C_{M_{0}∩C̃}(Z(S))/S ≅ D_{8}, while in C̃ we see that CC̃(Z(S))/Q ∩ E(C̃/Q) it is an extension of a 3-group by Z_{8}. Hence we get that C̃/Q ≅ 4SL_{2}(9) : 2. Then we see that M/V = N_{G}(V)/V ≅ Z_{2} × M_{11}. Hence both groups are as in Co_{3}.

Now choose a subgroup M_{1} in M_{0} with O_{3}(M_{0}) ≤ M_{1} and M_{1}/O_{3}(M_{0}) ≅ L_{2}(9). Then M_{1} is a minimal parabolic. Now choose C_{1} ≤ C̃ such that Q ≤ C_{1} and C_{1}/Q ≅ SL_{2}(9). Then also C_{1} is a minimal parabolic. Set U = ⟨M_{1}, C_{1}⟩. We see that O_{3}(U) ≤ O_{3}(M_{0}) ∩ Q and so O_{3}(U) = 1.

M12

Lemma 5.7 Let F^{∗}(M_{0}/O_{3}(M_{0})) ≅ 2M_{12}, and assume that Y_{M} contains V, the 6–dimensional GF (3)–module. Then there is a subgroup H of G such that F^{∗}(H) is weak parabolic isomorphic to Co_{1}.

Proof: We have that M_{0} ∩ C̃/O_{3}(M_{0} ∩ C̃) ≅ Z_{2}3^{2}GL_{2}(3). So we see that |V ∩ Q| = 3^{3}. Further V/V ∩ Q is the O(3, 3)-module for M_{0} ∩ C̃. Now again we see that Q acts quadratically on O_{3}(M_{0})/V and then with [Cher] we see that M_{0} acts trivally on O_{3}(M_{0})/V. As Z(M_{0}/O_{3}(M_{0})) acts fixed point freely on V we get O_{3}(M_{0}) = V. This shows that Q is extraspecial of order 3^{5}, so CC̃(Z(Q))/Q is a subgroup of Sp(4, 3). But M_{0} ∩C̃ corresponds to a maximal parabolic in Sp(4, 3) and so we see that C̃/Q ≅ Sp(4, 3) : 2. Now we have that both M_{0} and C̃ are as in Co_{1}.

Let P_{1} be the point stabilizer in C̃ on Q/Z(Q) and P_{0} the minimal parabolic in M_{0} not in C̃. Set X = ⟨Z(S)^{P_{0}}⟩. Then |X| = 3^{3} and X is the O(3, 3)–module for P_{0}. This shows |X ∩Q| = 3^{2}, which is normalized by P_{1}. As XQ/Q ⊴ S/Q, we see that XQ/Q ⊴ P_{1}/Q. Let t be an involution with [t, C̃] ≤ Q. Then t ∈ NC̃(S) ≤ P_{0} and so t ∈ N_{G}(X). As [t, Q] = Q, we see that X = C_{XQ}(t)(X ∩Q). As P_{1} normalizes C_{XQ}(t) and X ∩Q , we see that P_{1} normalizes X. But then O^{3}(P_{1}) centralizes X and so ⟨P_{1}, P_{0}⟩ = P_{1}P_{0}. Hence for H = ⟨M_{0}, C̃⟩ we have a diagram of type ∘ ∘ M^{12} ∘, and so F^{∗}(H) is weak parabolic isomorphic to Co_{1}.

Proof of the main theorem : The theorem follows with [MeStStr] and 3.1, 3.2, 4.1 - 4.4, 5.1 - 5.7.

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