F -modules for finite simple groups

U. Meierfrankenfeld

G. Stroth

preliminary version of September 29, 2003

In this paper we are going to determine the irreducible faithful F -modules for the finite quasisimple groups. Here we just concern about F -modules over GF (2). A module V is called F -module for G, if there is a nontrivial elementary abelian subgroup A in G such that |V/C_{V} (A)| ≤ |A|. The group A then will be called an offending subgroup or an offender. More precisely we prove

Theorem A Let E(G) = F^{∗}(G) be a quasisimple group and V be an irreducible faithful F -module in characteristic two for G = ⟨A^{G}⟩, A a quadratic offender. Then one of the following holds

(i) E(G)/Z(E(G)) ≅ A_{n} and one of the following is true

(α) V is the natural module (β) n ≤ 8 and |V | = 16 (γ) n = 6, |Z(E(G))| = 3 and |V | = 64

(ii) E(G)/Z(E(G)) = G(q) is a group of Lie type , q = 2^{t}, and one of the following is true

(α) G(q) ≅ L_{n}(q), Sp(2n, q), Ω^{±}(2n, q) or U_{n}(q) and V is the natural or dual module. (β) G(q) ≅ L_{n}(q) and V is the exterior square of the natural or dual module (γ) G(q) ≅ Sp(6, q) and V is the spin module (δ) G(q) ≅ Ω^{+}(8, q) or Ω^{+}(10, q) and V is the half spin module (ϵ) G(q) ≅ G_{2}(q) and V is the natural module.

The theorem will follow from 2.11, 3.1, 4.1, 4.2, and 4.3.

The F -modules for the sporadic groups, alternating groups and groups of Lie type in odd characteristic have been classified by M. Aschbacher [Asch]. The F -modules for the groups of Lie type in characteristic two have been classified by B. Cooperstein [Coop] and in an unpublished paper by B. Cooperstein and G. Mason [CMFF]. Unfortunately in the last two papers never the case that E(G) ≠ G and A ≰ E(G) was handled. The basic idea in all these papers was to get lower bounds for |V : C_{V} (t)| for involutions t and irreducible modules V. This then gives lower bounds for the size of |A| and so one either gets contradictions or is close to the F -module. In contrast to this we use basically the fact that there is always a quadratic offender. Then we apply the classification of quadratic modules in [MeiStr], [MeiStr1] and [Str] to end up with a very short list of modules and so it is easy to detect the F -modules.

If we drop the assumption that V is irreducible for G = ⟨A^{G}⟩, we get direct sums of the modules in (ii)(α), (β), provided n is large enough.

1 Preliminaries

Lemma 1.1 Let G be a finite group and V be a faithful F –module for G. Then there is some offending subgroup A with [V, A, A] = 1.

Proof: This is just Thompson replacement.

Lemma 1.2 Let G = L_{2}(q), q = 2^{n}, and V be a faithful module in characteristic two. Then for any involution a ∈ G we have |V : C_{V} (a)| ≥ q. If G = Sz(q) or U_{3}(q), then for any involution a we get |V : C_{V} (a)| ≥ q^{2}.

Proof: Let p be a Zsigmondy prime which divides q + 1 if G = L_{2}(q) and q^{2} + 1 else. Then we may assume that a inverts some element ω of order p in G. As |[V, ω]| ≥ q^{2}, we get the assertion. Suppose there is no such prime, then q = 8 and q + 1 = 9. Then a inverts some element ω of order nine, whose commutator has to be of order 64 at least.

Lemma 1.3 Let V be an F –module for G = L_{2}(q), Sz(q) or U_{3}(q). Then G = L_{2}(q) and V is the natural module.

Proof: By 1.2 we get G = L_{2}(q) and an offending subgroup A is a Sylow 2-subgroup of G. As there is some g ∈ G with G = ⟨A, A^{g}⟩, we get |V | = q^{2}, and V is the natural module by [Glaub].

Proposition 1.4 Let G = G(q) be a group of Lie type and V be an irreducible module over GF (q). Then

V = V_{1}^{σ_{1}} ⊗ ⋯ ⊗ V_{l}^{σ_{l}}

where the V_{i} are basic irreducible GF (q)G–modules. Further distinct l–tuples (M_{1}, … M_{l}) and (M′_{1}, …, M′_{l}) give nonisomorphic GF (q)G–modules.

Proof: [Stei, Th 41 and 43]

Lemma 1.5 Let G = G(q) be a Chevalley group then GF (q) is a splitting field for any irreducible module.

Proof: [Stei, (7.5)] Lemma 1.6 Let G = G(q) and V an absolutely irreducible KG–module for some K ⊆ G̃F (q). Let {σ_{1}, … σ_{r}} = Gal_{GF (p)}(K). Then K = GF (p)(χ) iff V^{σ_{1}}, …, V^{σ_{r}} are pairwise nonisomorphic KG–modules.

Proof: Choose notation such that σ_{1} = id. Let {f_{1}, … f_{r}} be a GF (p)–basis for K and {v_{1}, …, v_{n}} be a K–basis for V. Then {f_{i}v_{j} | i = 1, …, r, j = 1, … n} is a GF (p)–basis for V. Define

ϕ_{k} : K ⊗ V ⟶ V^{σ_{k}} by f ⊗ f_{i}v_{j} ⟶ f_{i}^{σ_{k}}fv_{j}.

Then ϕ_{k} are epimorphisms. Let n ∈ Kerϕ_{k} for all k. Set

n = ∑_{i,j} f_{ij} ⊗ f_{i}v_{j}

Then

∑_{i,j} f_{ij}f_{i}^{σ_{k}}v_{j} = 0 for all k

This gives

∑_{i} f_{i}^{σ_{k}}f_{ij} = 0 for all k, j

By linear independence of automorphisms we have f_{ij} = 0 for all i, j and so ⋂_{k} Kerϕ_{k} = 0. Set

D_{i} = ⋂_{k≠i} Kerϕ_{k}

Then

K ⊗ V = ⨁_{i=1}^{r} D_{i}

Now as K ⊗ V/Kerϕ_{i} ≅ D_{i}, we have D_{i} ≅ V^{σ_{i}}.

Let now K = GF (p)(χ). Then (Kerϕ_{1})^{σ_{k}} = Kerϕ_{k}, so Gal_{GF (p)}(K) permutes the D_{i} regularly. Hence by [CurRei, 70.15] the D_{i} are pairwise nonisomorphic.

Assume K > GF (p)(χ). Then each V^{σ_{k}} appears |K : GF (p)(χ)| times among the D_{i}. Hence the V^{σ_{i}} are not pairwise nonisomorphic.

We like to consider representations of twisted Chevalley groups as well. Here we fix notation as follows. We have ^{σ}G(q) ≤ G(q^{σ}), where G(q^{σ}) is the corresponding untwisted group, G ≠ F_{4} or B_{2}. Now following [Stei, chapter 9] we see that any basic module for G(q^{σ}) reduced to ^{σ}G(q) remains irreducible. Moreover by [Stei, 9.3] all irreducible modules are given by the tensor product theorem. Further by [Stei, 7.5] GF (q^{σ}) is a splitting field.

It remains the cases G = F_{4}(q) or B_{2}(q). Then there is a duality between the long and short roots. Just take all weights which vanish on all long roots. We call these modules restricted to ^{σ}G(q) the basic modules. Then again the tensor product theorem holds. If the rank is ℓ we now get q^{ℓ/2} modules. By [Stei, 12.2] GF (q) is the field of definition for all these modules.

Proposition 1.7 Let G = A_{ℓ}, D_{ℓ}, E_{6} or D_{4} and V be a basic module with high weight λ. Then the following holds for γ the diagram automorphism

(1) If λ ≠ γ(λ). Then GF (q^{σ}) is the field of definition for V.

(2) If λ = γ(λ) then GF (q) is the field of definition for V.

Proof: As [σγ,^{σ} G] = 1, we have that V and V^{σγ} are isomorphic as GF (q^{σ})-modules. So V ≅ V^{σ} iff V ≅ V^{γ}. Now from 1.4, the fact that GF (q^{σ}) is a splitting field and 1.6 we get that GF (q^{σ}) is the field of definition if V ≅ V̸^{σ} and GF (q) is the field of definition otherwise. So we just have to determine, when V ≅ V^{γ}.

Let C be the Cartan subgroup of G normalized by γ. Let C induce the character λ and λ̃ on the weight space of V and V^{γ}, respectively. Then we see that for c_{α_{i}} ∈ C we have λ̃(c_{α_{i}}) = λ(c_{γ(α_{i})}) = λ^{γ}(c_{α_{i}}). Hence V^{γ} has high weight λ^{γ}, and so V ≅ V^{γ} precisely if λ ≅ λ^{γ}.

Lemma 1.8 Let G = G(q), q = p^{f}, be a Chevalley group and V be an irreducible module in characteristic p. Let P be a parabolic and V_{P} = C_{V} (O_{p}(P)). Then V_{P} is an irreducible P − module.

Proof: This is [Sm].

Lemma 1.9 Let V be an F -module for G and V_{1} be an invariant subspace. Then V_{1} is a trivial subspace or V_{1} is an F -module too.

Proof: Choose A ∈ P(G, V). Set B = C_{A}(V_{1}). Then |V_{1}C_{V} (A)||B| ≤ |C_{V} (A)||A|. Hence |V_{1} : C_{V_{1}}(A)| ≤ |A|/|B| and so A/B offends on V_{1}.

Definition 1.10 Let V be a faithful GF (p)-module for G. Define P(G, V) to be the set of all non-identity p-subgroups X of G such that |X||C_{V} (X)| ≥ |Y ||C_{V} (Y)| for all Y ≤ X (including Y = 1).

If V is an F - module for G then there is always some elementary abelian A in P(G, V).

Lemma 1.11 Let V be an F -module for G.

(a) Let A, B ∈ P(G, V) be with |C_{V} (A)||A| = |C_{V} (B)||B| be maximal, then AB ∈ P(G, V) and |C_{V} (AB)||AB| = |C_{V} (A)||A|.

(b) Let A ∈ P(G, V), A ≤ O_{p}(G). If |C_{V} (A)||A| is maximal among all such A, then ⟨A^{G}⟩ is a (maybe nonabelian) offender on V.

(c) Assume (b). If ⟨A^{G}⟩ is abelian then there is an abelian quadratic offender, which is normal in G.

(d) There is always an A ∈ P(G, V), which is quadratic.

Proof: (a) We have |C_{V} (AB)||AB| = |C_{V} (AB)||A||B|/|A ∩ B|. Furthermore

C_{V} (AB) = C_{V} (A) ∩ C_{V} (B).

So we have

|C_{V} (AB)||AB| = |C_{V} (A) ∩ C_{V} (B)||A||B|/|A ∩ B|
= |C_{V} (A)||C_{V} (B)||A||B|/(|C_{V} (A)C_{V} (B)||A ∩ B|).

As C_{V} (A)C_{V} (B) ≤ C_{V} (A ∩ B), we get

|C_{V} (⟨A, B⟩)||⟨A, B⟩| ≥ |C_{V} (AB)||AB| ≥ |C_{V} (A)||A||C_{V} (B)||B|/(|C_{V} (A ∩ B)||A ∩ B|).

So ⟨A, B⟩ ∈ P(G, V). By maximality we get equality everywhere, the assertion. (b) This follows by repeated application of (a).

(c) This is Timmesfeld replacement.

(d) This is 1.1.

Let G be a Chevalley group with root system Φ and fundamental roots Π. Let l = |Π|, I = {1, 2, … l} and Π = {α_{1}, α_{2}, … α_{l}} where we choose the labeling of the Dynkin diagram as follows:

1∘  ^{2}∘  ^{3}∘ ⋯^{l−2} ∘ ^{l−1} ∘  ^{l}∘
1∘  ^{2}∘  ^{3}∘ ⋯^{l−2} ∘ ^{l−1} ∘  ^{l}∘
1∘  ^{2}∘  ^{3}∘ ⋯^{l−3} ∘ ∘^{l}^{l−2} ∘ ^{l−1} ∘

1∘  ^{2}∘  ∘^{4}^{3} ∘  ^{5}∘  ^{6}∘

1∘  ^{2}∘  ∘^{4}^{3} ∘  ^{5}∘  ^{6}∘  ^{7}∘

1∘  ^{2}∘  ∘^{4}^{3} ∘  ^{5}∘  ^{6}∘  ^{7}∘  ^{8}∘

1∘  ^{2}∘  ^{3}∘  ^{4}∘
1∘ ^{2}∘

For the remainder of this paper we fix the following notation. Let V be some irreducible GF (2) - module for some group of Lie type over GF (q). Then V ⊗ GF (q) is a direct sum of algebraic conjugates of some irreducible GF (q) - module V̂. If this is some algebraic conjugate of some fundamental module V (λ) we write V = V (λ). If this is a tensor product we call V the corresponding tensor product.

Lemma 1.12 Let G = L_{n}(q), n ≥ 5 and V = V (λ_{2}). Then V has an unique offender, which is a transvection group of order q^{n−1}. If n ≤ 4, then an offender intersects a transvection group nontrivially.

Proof: We prove the assertion by induction on n. Let first n = 5. Then V is a 10– dimensional module for L_{5}(q). Let G_{4} be the parabolic which has the following chief series in V

1 < V_{1} < V,

where V_{1} is the 6–dimensional L_{4}(q)–module and V/V_{1} the 4–dimensional one. It means that O_{2}(G_{4}) offends. Let x ∈ O_{2}(G_{4}). We may assume that C_{G}(x) = G_{14}, so |[V, x]| = q^{3}, as all involutions of O_{2}(G_{4}) are conjugate. Let now A_{1} ≤ O_{2}(G_{4}), |A_{1}| ≥ q^{2}. If C_{V} (a) = C_{V} (A_{1}) for all a ∈ A_{1}^{♯}, then C_{V} (A_{1}) is invariant under ⟨C_{G}(a) | a ∈ A_{1}^{♯}⟩ = O′(G_{4}). But as C_{V} (a) ≠ V_{1}, this is impossible. So we have shown that in fact V_{1} = C_{V} (A_{1}). Let now A be any offender. As one can check easily |V_{1} : C_{V_{1}}(A)| ≥ |A/A ∩ O_{2}(G_{4})|. Hence we see A ∩ O_{2}(G_{4}) ≠ 1. So we get |C_{V} (A ∩ O_{2}(G_{4})) : V_{1}| ≤ q. But then we see that |A∩O_{2}(G_{4})| ≥ q^{3}. Now by the above we have V_{1} = C_{V} (A∩O_{2}(G_{4})). This shows A ≥ O_{2}(G_{4}) and so A = O_{2}(G_{4}).

It remains n ≤ 4. For n ≤ 3 there is nothing to prove. So assume n = 4. Then V is the natural Ω^{+}(6, q)–module. Hence |V : C_{V} (A)| ≥ q^{2}. If |A| > q^{2}, we are done. So assume |A| = q^{2}. So we may assume A ∩ O_{2}(G_{1}) = 1. Now G_{1} induces on C_{V} (O_{2}(G_{1})) the natural module and on V/C_{V} (O_{2}(G_{1})) the dual. As A corresponds to a transvection group in G_{1}/O_{2}(G_{1}), we get |V : C_{V} (A)| ≥ q^{3}, a contradiction.

Lemma 1.13 Let G = U_{4}(q) ≅ Ω^{−}(6, q), Sp_{6}(q) or Ω^{+}(10, q) and V = V (λ_{n}). Let A be a quadratic offender, then A is O_{2}(G_{1}) if G ≅ S̸p_{6}(q), while if G ≅ Sp_{6}(q), then |A| ≥ q^{4} and |C_{W} (A)| ≤ q^{2}, where W is the narural Sp_{6}(q)–module. In all cases A contains some root element.

Proof: Obviously O_{2}(G_{1}) is a quadratic offender. So let first A be a quadratic offender which lies in G_{1} but not in O_{2}(G_{1}). Then by 1.9 we may assume that A/A∩O_{2}(G_{1}) offends on C_{V} (O_{2}(G_{1})). Let G ≅ U_{4}(q). Then |A/A ∩ O_{2}(G_{1})| = q^{2} by 1.2 and so |A ∩ O_{2}(G_{1})| ≤ q.

Hence |A| ≤ q^{3}. But as V involves two natural L_{2}(q^{2}) ≅ (G_{1}/O_{2}(G_{1})′–modules we see |V : C_{V} (A)| ≥ q^{4}, a contradiction.

Let now G ≅ Sp_{6}(q). Let A/A ∩ O_{2}(G_{1}) be a transvection group. If C_{V} (A ∩ O_{2}(G_{1})) = C_{V} (O_{2}(G_{1})), then we get |A| = q^{5}. Hence A induces a transvection group on O_{2}(G_{1}) and C_{V} (O_{2}(G_{1})) as well. But this contradicts the action of Sp_{6}(q) on the spin module. So let now |A/A∩O_{2}(G_{1}))| > q. As there is exactly one elementary abelian subgroup of order q^{6} in a Sylow 2-subgroup, which does not offend on the spin module, we get |C_{C_{V} (O_{2}(G_{1}))}(A)| = q^{2}. Let R be the root group normal in G_{1}. Then O_{2}(G_{1})/R ≅ C̸_{V} (R) as an Sp_{4}(q)–module. So we get |A/A ∩ O_{2}(G_{1})| = q^{3} and |A ∩ O_{2}(G_{1})| = q, where A ∩ O_{2}(G_{1}) is a root subgroup. Now we see that in the natural module W we have |C_{W} (A)| = q^{2}.

Let finally G ≅ Ω^{+}(10, q). Let 1 ≠ x ∈ A ∩ O_{2}(G_{1}). Then |[V, x]| = q^{4} or q^{8}. If |[V, x]| = q^{8}, then [V, x] = C_{V} (O_{2}(G_{1})) and as C_{G}(C_{V} (O_{2}(G_{1}))) = O_{2}(G_{1}), we have a contradiction. So we get |[V, x]| = q^{4} and A is in the centralizer of this group. This shows |A/A∩O_{2}(G_{1})| ≤ q^{6} and |V : C_{V} (A)| ≥ q^{8}. But |O_{2}(G_{1}) : C_{O_{2}(G_{1})}(A)| ≥ q|A/A∩O_{2}(G_{1})|, a contradiction. Hence A ∩ O_{2}(G_{1}) = 1 and so |A| ≤ q^{6} and |C_{V} (O_{2}(G_{1})) : C_{C_{V} (O_{2}(G_{1}))}(A)| ≤ q^{3}. This shows that we must have equality everywhere. Now C_{V} (O_{2}(G_{1}))A contains an elementary abelian subgroup of order q^{11}. But C_{V} (O_{2}(G_{1}))Ω^{+}(8, q) is isomorphic to a subgroup of Ω^{+}(10, q) and the latter group contains no elementary abelian 2-subgroups of order greater than q^{10}, a contradiction.

So we are left with A ≤ O_{2}(G_{1}). If A contains some element of type c_{2}, we get the assertion. So assume not. Then |A| ≤ q, q^{4}, q^{2} according to G ≅ U_{4}(q), Ω^{+}(10, q), or Sp_{6}(q). The first case is not possible. In the other two cases we see that C_{V} (A) = C_{V} (a) for all a ∈ A^{♯} and ⟨C_{G}(a) | a ∈ A^{♯}⟩ = G, a contradiction.

Lemma 1.14 Let G = L_{n}(q) or U_{n}(q) and V = V (λ_{1} + λ_{n}). Assume A ≤ G with [V, A, A] = 1. Then |A| ≤ q.

Proof: We have that V is the adjoint module. Let first n = 3 and G = L_{3}(q). then any elementary abelian subgroup intersects a root group. As V is not strong quadratic [Str], we get the assertion. Let now G = L_{n}(q). By induction as V involves the adjoint module for G_{1}/O_{2}(G_{1}) we have |A/A ∩ O_{2}(G_{1})| ≤ q. So either we have the assertion or A ∩ O_{2}(G_{1}) ≠ 1. But then V would be strong quadratic, a contradiction.

So let now G ≅ U_{n}(q). If n = 3 there is nothing to prove. Let n = 4. If A ≤ O_{2}(G_{1}), we get that V is strong quadratic or |A| ≤ q. So let A ≰ O_{2}(G_{1}). Then we have |A/A ∩ O_{2}(G_{1})| ≤ q. Let A ∩ O_{2}(G_{1}) ≠ 1. As V is not strong quadratic we get |A| ≤ q^{2} and A ≤ O_{2}(G_{2}). So let x ∈ A^{♯}. Then [⟨A^{C_{G}(x)}⟩, x] = 1, and we get a quadratic fours group which intersects a root group and contains x, a contradiction. Let now n ≥ 5. Then by induction |A : A ∩ O_{2}(G_{1})| ≤ q. As V is not strong quadratic, we see that |A ∩ O_{2}(G_{1})| ≤ q and A induces GF (q^{2})–transvections on O_{2}(G_{1})/Z(O_{2}(G_{1})) with A∩O_{2}(G_{1}) ≤ [A, O_{2}(G_{1})]. But now ⟨A, [A, O_{2}(G_{1})]⟩ ≤ U_{4}(q), and so as before we get strong quadratic action, a contradiction.

Lemma 1.15 Let X = V ⊗ W be a faithful GF (q)–modules and A be quadratic on X . Then |A| ≤ q.

Proof: Let v⊗w ∈ V ⊗W, a ∈ A. Then [v⊗w, a] = [v, a]⊗w+v⊗[w, a]+[v, a]⊗[w, a]. Let b ∈ A then 0 = [v ⊗ w, a, b] = [v, a] ⊗ [w, b] + [v, b] ⊗ [w, a], hence [v, a] = λ[v, b] and [w, b] = λ^{−1}[w, a], for some λ ∈ GF (q)^{♯}. In particular |A| ≤ q.

Lemma 1.16 Let X = V ⊗W, be a faithful GF (q)–modules. Then X is not an F –module.

Proof: Suppose by way of contradiction that X is an F –module. By 1.1 we have a quadratic offender A. By 1.15 we have |A| ≤ q. Now we get that for a ∈ A^{♯} the codimension of C_{X}(a) is at least twice the codimension of the centralizers in the modules V and W. This shows that the codimension is at least two. Hence we get |X : C_{X}(a)| ≥ q^{2}, a contradiction.

Lemma 1.17 (a) Let G = G(q) be of Lie rank at least two and A ≤ G, |A| ≤ q. Let V be an irreducible module with |V : C_{V} (A)| ≤ |A|, then V is strong quadratic.

(b) Let R be a root group in G with |[V, R]| ≤ q for some irreducible module V, then G ≅ L_{n}(q) or Sp(2n, q) and V is the natural module.

Proof: (a) Let R be a root group with [R, A] = 1. Suppose first R ≥ A, then obviously [C_{G}(a)′, [a, V]] = 1 for a ∈ A^{♯} and so V is strong quadratic. So let R ≱ A. If A ≤ O_{2}(C_{G}(R)) but A ≰ Z(O_{2}(C_{G}(R))), then there is some a ∈ A and r ∈ R such that a ∼ ar. So |[V, r]| ≤ q^{2}. As C_{G}(r) acts on [V, r] we see again that V is strong quadratic. So let now A ≤ Z(O_{2}(C_{G}(R))). Then G ≅ F_{4}(q) or Sp(2n, q). But then there is some g ∈ G and a ∈ A such that a^{g} ∈ C_{G}(R) \ O_{2}(C_{G}(R)). So we may assume that A ≰ O_{2}(C_{G}(R)). Then there is even some h ∈ C_{G}(R) and some a ∈ A such that a^{h} ∈ P \ O_{2}(P), where P is some minimal parabolic of C_{G}(R). (Recall that for G = L_{3}(q) of course A intersects some root subgroup nontrivially and so V is strong quadratic. ) Now P acts on [V, R] and V/C_{V} (R). But by 1.2 we have that P/O_{2}(P) can just induce one nontrivial module in V, so we see that [[V, R], O^{2}(P/O_{2}(P))] = 1. Hence V is strong quadratic.

(b) We may argue by induction. By (a) we have that V is strong quadratic. So by [Str] V = V (λ) for some weight λ. Further as |[V, R]| = q, we see that [[V, R], C_{G}(R)′] = 1, so λ is the weight corresponding to C_{G}(R), or G ≅ L_{n}(q) and λ = λ_{1} or λ_{n−1}. But the latter are the natural modules and we are done. So assume the former. By induction we have that any maximal parabolic with connected diagram, which does not belong to λ is of type L_{m} or Sp_{2m}. Hence we get that G ≅ Sp(2n, q) and λ = λ_{1} or λ_{2}, G ≅ F_{4}(q), λ = λ_{1} or λ_{4}, or G ≅ G_{2}(q) or ^{3}D_{4}(q). If G ≅ G_{2}(q), then V is the natural Sp(6, q)–module and R corresponds to an element of type a_{2}, i.e. |[V, R]| = q^{2}. If G ≅ ^{3}D_{4}(q), then V is the natural Ω^{+}(8, q^{3})-module and so |[V, R]| ≥ q^{3}. For G ≅ F_{4}(q) we may assume by duality that V = V (λ_{4}). If G ≅ Sp(2n, q) we may assume V = V (λ_{2}), as V (λ_{1}) is the natural module. Now we have G_{1}/O_{2}(G_{1}) contains Sp(2m, q) for some m. By induction C_{V} (O_{2}(G_{1})) is the natural module. Let R_{1} be the root group normal in G_{1}. As there is some conjugate of R in G_{1} \ O_{2}(G_{1}), we see that [V, R_{1}] ≤ C_{V} (O_{2}(G_{1})) and [O^{2}(C_{G}(R_{1})), V] ≤ C_{V} (O_{2}(G_{1})), a contradiction.

2 F –modules in even characteristic for Lie type groups

Throughout this chapter we will assume that G is a group of Lie type over a field with q = 2^{n} elements. Further V is an F -module and A ∈ P(G, V) is an offending subgroup. By 1.11 we may assume that A acts quadratically. If G_{i} is a maximal parabolic in G we set Q_{i} = O_{p}(G_{i}).

Lemma 2.1 Let G = G(q) be classical or F_{4}(q) and V be an F -module. If V = V (λ_{2}), then G ≅ L_{n}(q), Ω^{+}(6, q) or Sp_{4}(q).

Proof: If G ≅ Sp_{2n}(q), n > 2 or Ω^{±}(2n, q), n > 3, then V is the adjoint module. So for G_{n}/Q_{n}, we have some adjoint module involved too. Now C_{V} (Q_{n}) ≅ V (λ_{2}) for G_{n}/Q_{n}. Suppose A ≰ Q_{n}. Then by 1.9 and 1.12 we have that AQ_{n}/Q_{n} is a transvection group in G_{n}/Q_{n} or G_{n}/Q_{n} induces L_{4}(q) or L_{3}(q). Further AQ_{n}/Q_{n} intersects a root group. Assume that |AQ_{n}/Q_{n}| > q. Now A acts quadratically on any G_{n}/Q_{n}–chieffactor in V. This contradicts 1.14. So we are left with G = Sp_{6}(q) or Ω^{−}(8, q) and |AQ_{n}/Q_{n}| = q. If G = Sp_{6}(q) we so get |A| ≤ q^{5}. But the codimension of the centralizer of A in the adjoint module for L_{3}(q) is four and so |V : C_{V} (A)| ≥ q^{6}, a contradiction. Let now G = Ω^{−}(8, q). Again |V : C_{V} (A)| ≥ q^{6}. As V is not strong quadratic, we get A ≰ Q_{1}. So we see with 1.14 that |AQ_{1}/Q_{1}| ≤ q and so again |A| ≤ q^{5}, a contradiction.

So A ≤ Q_{n}. Then by 1.11 we may assume that A ≤ G_{n}. Then there is a root group R in A, where R ≤ G_{2}. From this by quadratic action we see that [[V, R], ⟨A^{G_{2}}⟩] = 1, i.e. [V, R] ≤ C_{V} (Q_{2}). But |C_{V} (Q_{2})| = q, contradicting 1.17.

Let now G ≅ U_{n}(q). Then G is embedded in L_{n}(q^{2}) and the modules V (λ_{2}) are the same for both groups. Let n ≥ 5. By 1.12 L_{n}(q^{2}) has an unique offender, the transvection group on V (λ_{2}). But this group is not in U_{n}(q).

So let finally G ≅ F_{4}(q). Then C_{V} (Q_{4}) ≅ V (λ_{2}) for Sp_{6}(q). So we have |C_{V} (Q_{4})| = q^{14} and A ≤ Q_{4}, as V (λ_{2}) is not an F –module for Sp_{6}(q). By 1.17 we have |A| > q. Let |A| = q^{7}, A≤ G_{4}. Then |AQ_{1}/Q_{1}| = q. But C_{V} (Q_{1}) is the spin module for G_{1}/Q_{1}. Now we get a contradiction with 1.9 and 1.17. So we have ⟨A_{4}^{G}⟩ = Q_{4} and then |V : C_{V} (Q_{4})| ≤ |Q_{4}| by 1.11. This shows |V | ≤ q^{29}. As |V : C_{V} (Q′_{4})| = q^{14}, we get |V | = q^{28} or q^{29}. As Q_{4}/Q′_{4} is not irreducible and V/C_{V} (Q′_{4}) ≅ C_{V} (Q_{4}), we see [V, Q_{4}, Q_{4}] = 1, a contradiction as Q_{4} contains elements of order four.

Lemma 2.2 Let G = G(q) and V be an F –module. If C_{V} (Q_{i}) = C_{V} (Z(S)) for some i, then V = V (λ_{i}).

Proof: By 1.4 we have V ≅ V (λ_{i})^{σ_{1}} ⊗ V (λ_{i})^{σ_{2}} ⊗ ⋯ ⊗ V (λ_{i})^{σ_{k}}, for some field automorphisms σ_{1}, …, σ_{k}. By 1.16 we get k = 1.

Lemma 2.3 Let G = G(q) ≅ L̸_{n}(q) be classical and V be an F –module with A ≤ Q_{1}. Then V ≅ V (λ_{1}) or G ≅ Sp_{4}(q), Sp_{6}(q), Ω^{+}(8, q), Ω^{+}(10, q) or Ω^{−}(6, q) and V ≅ V (λ_{n}).

Proof: Application of 1.17 shows that we may assume that |A| > q. Now application of 1.11 shows that Q_{1} offends on V. So |V : C_{V} (Q_{1})| ≤ |Q_{1}|. Furthermore by 2.2 we may assume that C_{V} (Q_{1}) is a nontrivial G_{1}/Q_{1}–module. If G ≅ Ω^{±}(2n, q), then again by 1.11 Q_{1} acts quadratically on V. So we have that V is strongly quadratic and so by [Str] V ≅ V (λ_{n}). But now also |V/C_{V} (Q_{1})| = |C_{V} (Q_{1})| and |C_{V} (Q_{1}) is the spin module for G_{1}/Q_{1}. So |C_{V} (Q_{1})| ≤ |Q_{1}|, which shows G ≅ Ω^{+}(2n, q), n = 3, 4, 5.

Let now G ≅ Sp_{2n}(q) or U_{n}(q). Let R≤ G_{1} be a root group. Then we see that V/C_{V} (Q_{1}) involves some module isomorphic to C_{V} (Q_{1}). As G acts strongly quadratic and V ≅ V̸ (λ_{2}) by 2.1, we get with [Str] that C_{V} (Q_{1}) is not the natural G_{1}/Q_{1}–module. Hence V/C_{V} (Q_{1}) is not the natural module and so [V, Q_{1}, Q_{1}] = 1, in particular G ≅ U̸_{n}(q). Let now R_{1} be a root group in Q_{1}, R_{1} ≠ R. Then we may assume that C_{G}(R_{1}) ≤ G_{2}. As we have by [Str] n ≥ 4, we see that V/C_{V} (Q_{1}) involves at least one nontrivial Sp_{2n−4}(q)–module. If C_{V} (R_{1}) covers this module we get |V : C_{V} (R_{1})| ≤ q^{3}. Now there is a conjugate R_{2} of R_{1} in G_{1} \ Q_{1}. This shows that |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(R_{2})| ≤ q. But now application of 1.17 gives a contradiction.

Lemma 2.4 Let G = G(q) of rank at most two and V be an F –module. Then G ≅ L_{2}(q), L_{3}(q), U_{4}(q), Ω^{−}(6) or G_{2}(q) and V is the natural module.

Proof: If G is of rank one the assertion follows with 1.3. So let G be of rank two. If G ≅ L_{3}(q), then by 1.17 V is strong quadratic and the assertion follows with [Str].

Let G ≅ Sp_{4}(q), U_{4}(q), U_{5}(q), or Ω^{−}(6, q). Then by 2.3 A ≰ Q_{1}. Furthermore by 2.2 V_{1} = C_{V} (Q_{1}) is a nontrivial G_{1}/Q_{1}–module. So it is the natural module in all cases. In particular |A : A ∩ Q_{1}| = q and G ≅ U̸_{5}(q). If G ≅ Sp_{4}(q), we have A ≤ Q_{2} and we get the assertion with 2.3. Now in all cases we have by [Str] that V is not strongly quadratic and so G ≅ U_{4}(q), q < |A| ≤ q^{2}. Further A ≰ Q_{2}. Now we get by 1.9 that |A| = q^{2} and A ∩ Q_{2} = 1. But then C_{V} (Q_{2}) is the natural G_{2}/Q_{2}–module and [Q_{2}, V] ≤ C_{V} (Q_{2}), i.e. Q_{2} acts quadratically, contradicting the fact that V is not strongly quadratic.

Let now G ≅ G_{2}(q). If V is strongly quadratic then by [Str] V ≅ V (λ_{1}), the assertion. So we may assume that V is not strongly quadratic. Now by 1.17 |A| > q. Hence A ≰ Q_{1}. So we get |A| ≤ q^{2}. If [A, Z(Q_{2})] = 1, we have A ∩ Z(Q_{2}) ≠ 1, and so V would be strong quadratic. So we have that |A : C_{A}(Z(Q_{2}))| = q and A ∩ Z(Q_{2}) = 1. Then there is some x ∈ Z(Q_{2}) and a ∈ A^{♯} with [a, x] ≠ 1. Now for b ∈ C_{A}(Z(Q_{2}))^{♯} we get that ⟨[a, x], b⟩ is quadratic on V and so V is strong quadratic, a contradiction.

Let finally G ≅ ^{3}D_{4}(q) or ^{2}F_{4}(q). Then in both cases |A| ≤ q^{5}. If V is strongly quadratic, we get with [Str] that G ≅ ^{3}D_{4}(q) and V is the natural Ω^{+}(8, q^{3})–module. But in this case |[V, t]| ≥ q^{6} for any involution t ∈ G, contradicting |A| ≤ q^{5}. So we have that V is not strongly quadratic. In particular by 1.17 |A| > q and then A ≰ Q_{1}. Let C_{V} (Q_{1}) be a trivial G_{1}–module. Let R be a root group normal in G_{1}. Then we see that [V, R] involves at least the eight-dimensional L_{2}(q^{3})–module for G ≅ ^{3}D_{4}(q) or twice the natural Sz(q)–module for G ≅ ^{2}F_{4}(q). Now |[V, t]| ≥ q^{8} for involutions t ∈ G_{1} \ Q_{1}, contradicting |A| ≤ q^{5}. So we have with 1.9 that C_{V} (Q_{1}) is an F –module for G_{1}/Q_{1}, and so by 1.3 we get G ≅ ^{3}D_{4}(q) and C_{V} (Q_{1}) is the natural L_{2}(q^{3})–module. But then there is exactly one such module involved, i.e. [V, G_{1}] ≤ C_{V} (Q_{1}) and so V would be strongly quadratic, a contradiction.

Lemma 2.5 Let G = G(q) ≅ L̸_{n}(q) of rank at least three be classical and V = V (λ_{n}) be an F –module. Then G ≅ Sp_{6}(q), Ω^{+}(8, q) or Ω^{+}(10, q).

Proof: By 2.3 we may assume A ≰ Q_{1}. So by 1.9 it is enough to prove that V (λ_{n}) is not an F –module for G ≅ Sp_{8}(q), Ω^{+}(12, q), Ω^{−}(8, q), U_{5}(q) and U_{6}(q). The case G ≅ U_{5}(q) follows from 2.4. As the spin modules for Sp_{8}(q) and Ω^{−}(8, q) are the same, it is enough to handle G ≅ Sp_{8}(q), Ω^{+}(12, q) and U_{6}(q). Now by 1.13 we have AQ_{1}/Q_{1} = O_{2}(G_{12}/Q_{1}) or G ≅ Sp_{8}(q) and |AQ_{1}/Q_{1}| ≥ q^{4}. Let first G ≅ Ω^{+}(12, q). Then |A| ≤ q|A/A ∩ Q_{1}| and |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = |A/A ∩ Q_{1}|. Hence we see that [G_{1}, V] = C_{V} (Q_{1}). But then [Q_{1}, V] = 1 as C_{V} (Q_{1}) ≅ Q̸_{1} as G_{1}–modules.

So let now G ≅ Sp_{8}(q) or U_{6}(q). Then we see that |C_{Q_{1}}(A)| ≤ q^{3} and so |A| < |A/A ∩ Q_{1}|^{2}. Let R be the root group normal in G_{1}. We have C_{V} (Q_{1}) ≤ [V, R], so there are at least two such modules involved. But |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = |A/A∩Q_{1}| or G ≅ Sp_{8}(q) and AQ_{1}/Q_{1} = O_{2}(G_{12})/Q_{1}, in which case we have |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = |A/A∩Q_{1}|q^{−1} and |A| ≤ q^{2}|A/A ∩ Q_{1}|. In both cases we see that |V : C_{V} (A)| > |A|, a contradiction.

Lemma 2.6 Let G ≅ Ω^{−}(2n, q), n > 3, and V be an F –module. Then V ≅ V (λ_{1}).

Proof: By [Str] it is enough to prove that V is strongly quadratic. So assume not. Then by 1.11 A ≰ Q_{1}. If A ∩ Q_{1} ≠ 1, there is some t ∈ A \ Q_{1} such that [t, Q_{1}] contains some root element and so V is strongly quadratic. Hence we may assume A ∩ Q_{1} = 1. If there are s, t ∈ A^{♯} with C_{Q_{1}}(s) ≠ C_{Q_{1}}(t), then there is some g ∈ C_{Q_{1}}(s), say, with t^{g} = t and s^{g} = sa ≠ s. Now ⟨t, a⟩ acts quadratically and if a is not in a root subgroup there is h ∈ Q_{1} with t^{h} = tb ≠ t, b in a root subgroup. Now ⟨a, b⟩ acts quadratically and so V is strongly quadratic. So we may assume that C_{Q_{1}}(A) = C_{Q_{1}}(t) for all t ∈ A^{♯}. Hence |A| ≤ q^{2n−4}. So |V : C_{V} (t)| ≤ q^{2n−4}. For a root element a we then get |V : C_{V} (a)| ≤ q^{4n−8}. But as V is not strongly quadratic, we get |[V, a]| > |Q_{2}/Q′_{2}| = q^{4n−8}, a contradiction.

Lemma 2.7 Let G ≅ Sp_{2n}(q), n = 3, 4, and V ≅ V (λ_{1} +λ_{n}). Then V is not an F –module.

Proof: By [Str] we know that V is not strong qudartic. Let first A ∩ Q_{n} ≠ 1. There is U ≤ G_{n}, U is the natural L_{3}(q)–module or the natural Ω^{+}(6, q)–module. So by 1.11 we have A ≰ Q_{n}, as otherwise U ≤ A, and U contains root elements. But now there is some t ∈ A \ Q_{n}, with [t, U] contains a root element. As A does not contain root elements, we see that V is strongly quadratic. So we have A ∩ Q_{n} = 1. Now we have C_{Q_{n}}(A) = C_{Q_{n}}(a) for all a ∈ A^{♯}. This shows |A| ≤ q^{2} for n = 3 and |A| ≤ q for n = 4. But as V is not strongly quadratic we get with 1.17 that n = 3. Now for a root element r we have |[V, r]| ≤ q^{4}. As C_{V} (O_{2}(C_{G}(r))) is a non trivial module for G_{2}, we see that C_{O_{2}(C_{G}(r))}([V, r]) > R, R the root group containing r, which shows that V is strongly quadratic, a contradiction.

Proposition 2.8 Let G = G(q) be classical and V be an F –module. Then one of the following holds (i) G ≅ L_{n}(q), V ≅ V (λ_{1}), V (λ_{2}), V (λ_{n−2}), V (λ_{n−1})

(ii) G ≅ Sp_{2n}(q), Ω^{±}(2n, q), or U_{n}(q) and V ≅ V (λ_{1})

(iii) G ≅ Sp_{6}(q), V ≅ V (λ_{3})

(iv) G ≅ Ω^{+}(2n, q), n = 4, 5 and V ≅ V (λ_{n}).

Proof: By 2.4 we may assume that the rank of G is at least three. By 2.6 we have G ≅ Ω̸^{−}(2n, q). By 2.3 and symmetry for G ≅ L_{n}(q) we may assume that A ≰ Q_{1}. By 2.2 we have that C_{V} (Q_{1}) is an F –module. So by induction we first assume that C_{V} (Q_{1}) is the natural module. Let A ≤ Q_{n}, then by 1.17 and 1.11 we have that V is strongly quadratic and so V ≅ V (λ_{2}). If A ≰ Q_{n}, we have that C_{V} (Q_{n}) is an F –module. As G_{n}/Q_{n} is linear we get V ≅ V (λ_{2}) again. Now the assertion follows with 2.1.

So we may assume that C_{V} (Q_{1}) is not natural. Let first this module correspond to λ_{n}. If A ≤ Q_{n}, we get V ≅ V (λ_{n}). If A ≰ Q_{n} and C_{V} (Q_{n}) is the trivial module, we get V ≅ V (λ_{n}) with 2.2 again. So by 2.5 we have the assertion.

So let C_{V} (Q_{n}) be an F –module too. Then we get V ≅ V (λ_{1} + λ_{n}). Now 1.14 and 1.17 yield G ≅ Sp_{2n}(q) or Ω^{+}(2n, q). If G ≅ Sp_{2n}(q) then by induction we have n = 3 or 4 and we get a contradiction with 2.7. So let G ≅ Ω^{+}(2n, q). Now for G_{n−1} we get that C_{V} (Q_{n−1}) is V (λ_{1} + λ_{n}) for G_{n−1}/Q_{n−1}. As V is not strongly quadratic by [Str] we get with 1.11 that A ≰ Q_{n−1}. But now we have a contradiction with 1.9 and induction.

So by induction the only case left is G ≅ L_{n}(q) and C_{V} (Q_{1}) corresponds to λ_{n−2} or λ_{3}. assume the former. If V is strongly quadratic we get with [Str] that V ≅ V (λ_{n−2}). So let V not be strongly quadratic. Then C_{V} (Q_{n−1}) is not the natural module. In particular it is not an F –module. Hence by 1.9 A ≤ Q_{n−1} which contradicts 1.11 and the fact that V is not strongly quadratic.

So let C_{V} (Q_{1}) correspond to λ_{3}. Then either C_{V} (Q_{n−1}) is an F –module or V is strong quadratic. In both cases we get V ≅ V (λ_{3}). Further we may assume that n > 5, otherwise λ_{3} = λ_{n−2} or λ_{n−1}. Now by 1.12 we have |A : A ∩ Q_{1}| = q^{n−2}. Further |A ∩ Q_{1}| ≤ q. As |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = |A/A ∩ Q_{1}|, we get that either |A ∩ Q_{1}| = q or C_{V} (Q_{1}) = [V, G_{1}]. As C_{V} (Q_{1}) ≅ Q̸_{1} as G_{1}/Q_{1}–modules, the latter is not possible. Now for r ∈ A ∩ Q_{1} we see that C_{V} (Q_{1}) ≤ [V, r]. Hence \binom{n−2}{2} ≤ n − 1, a contradiction.

Proposition 2.9 There is no F –module for G = E_{n}(q), n = 6, 7, 8.

Proof: Suppose false. Let first G = E_{6}(q). If A ⊆ Q_{1}, then by 1.11 we get A = Q_{1} and V is strongly quadratic. Now by [Str] we have V ≅ V (λ_{1}) or V (λ_{6}). Let A ≰ Q_{1} and V ≅ V̸ (λ_{1}). Then C_{V} (Q_{1}) is an F –module by 1.9 for Ω^{+}(10, q). By 2.8 this is the natural or half spin module. Suppose it is not the natural one. Then we get that V is not strong quadratic and so A ≰ Q_{6}. Hence C_{V} (Q_{6}) is an F –module, which now has to be a half spin module and we have V ≅ V (λ_{4}).

So let now C_{V} (Q_{1}) be the natural module. If V ≅ V̸ (λ_{6}), we may assume by symmetry that also C_{V} (Q_{6}) is the natural module and so V ≅ V (λ_{1} + λ_{6}). But now C_{V} (Q_{4}) is not an F –module by 2.8, and so A ≤ Q_{4}, which shows that V is strongly quadratic, contradicting [Str].

So in any case we have that either V ≅ V (λ_{1}) or V (λ_{4}). Let first V ≅ V (λ_{4}). Then V is the adjoint module, V is not strongly quadratic and so A ≰ Q_{1}. For G_{1} we have that C_{V} (Q_{1}) is the half spin module. By 2.5 we know A/A ∩ Q_{1}. So A/A ∩ Q_{1} acts strongly quadratic on any G_{1}/Q_{1}–module involved in V. But as an adjoint module it involves V (λ_{2}) for G_{1}/Q_{1}, which is not strongly quadratic by [Str].

So let V ≅ V (λ_{1}). We may assume that we have the following G_{1}–series on V

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

where V_{1} is a trivial module, V_{2}/V_{1} ≅ Q_{1} and V/V_{2} is the natural module. As Q_{1} does not offend, we have A ≰ Q_{1} by 1.11. Now in V/V_{2} there are two G_{1}–classes of vectors v_{1}V_{2}, v_{2}V_{2}. We have that v_{1}V_{2} is centralized by H_{1} ≅ Sp_{8}(q) in G_{1}/Q_{1} and H_{1} induces the spin module on Q_{1}, so C_{⟨v_{1},V_{2}⟩/V_{1}}(Q_{1}) = V_{2}/V_{1}. Now for x ∈ Q_{1} we have that C_{V} (x)V_{2}/V_{2} just contains vectors of type v_{2}V_{2}. In particular |C_{V} (x)V_{2}/V_{2}| ≤ q^{5}. By 2.5 we have |V_{2}/V_{1} : C_{V_{2}/V_{1}}(A/A∩Q_{1})| ≥ |A/A∩Q_{1}|, we see that |A∩Q_{1}| ≥ q^{5}. Now v_{2}V_{2} is centralized by some H_{2} ≅ Ω^{+}(8, q) and Q_{1} is an extension of an 8–dimensional H_{2}–module by an 8– dimensional one, i.e. |C_{Q_{1}}(⟨v_{2}, V_{2}⟩/V_{1})| ≤ q^{8}. Let W be a fours group of vectors of type v_{2}V_{2} in V/V_{2}. Then W is centralized by Ω^{+}(6, q), which shows that the elements in W cannot have the same centralizer. Let W_{1} be the preimage of W in V/V_{1}. then we see that |C_{Q_{1}}(W_{1})| = q^{4}. Hence |C_{V} (A ∩ Q_{1})V_{2}/V_{2}| ≤ q. This shows |A ∩ Q_{1}| ≥ q^{9}. But now also |V_{2} : C_{V_{2}}(A ∩ Q_{1})| ≥ q^{9}, and so |A| ≥ q^{18}, a contradiction to the 2-rank of E_{6}(q).

Let G ≅ E_{7}(q). Let first A ≤ Q_{7}. If |A| > q, then V is strongly quadratic and the same holds for |A| ≤ q by 1.17. By [Str] V ≅ V (λ_{7}). Assume A ≰ Q_{7}. As E_{6}(q) does not posses F –modules, we get with 1.9 that C_{V} (Q_{7}) is a trivial G_{7}–module. Now by 2.2 V ≅ V (λ_{7}) again. As a G_{7}/Q_{7}–module we have the following series

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

where V_{1} and V/V_{3} are trivial modules and V_{2}/V_{1}, V_{3}/V_{2} are 27–dimensional E_{6}(q)–modules. As [Q_{7}, V_{2}] = V_{1}, we see that A ≰ Q_{7}. But on V_{3}/V_{2} we have |V_{3}/V_{2} : C_{V_{3}/V_{2}}(A/A ∩ Q_{7})| > |A/A ∩ Q_{7}|. Hence as |V_{2} : C_{V_{2}}(A ∩ Q_{7})| = |A ∩ Q_{7}|, we see that |V : C_{V} (A)| > |A|, a contradiction.

Let finally G ≅ E_{8}(q). If A ≤ Q_{8}, then V is strongly quadratic, contradicting [Str]. So A ≰ Q_{8}. As E_{7}(q) does not posses an F –module we get with 1.9 and 2.2 that V ≅ V (λ_{8}). As V is not strong quadratic, we have |A ∩ Q_{8}| ≤ q. Now for the root group R normal in G_{8}, we have that [V, R] and V/C_{V} (R) involve nontrivial E_{7}(q)–modules. Hence |V : C_{V} (A/A ∩ Q_{8})| ≥ q^{2}|A/A ∩ Q_{8}|^{2}. But then q ≥ q|A/A ∩ Q_{8}|, a contradiction.

Proposition 2.10 The groups ^{2}E_{6}(q) and F_{4}(q) do not posses F –modules.

Proof: Suppose false. If V is strongly quadratic we may assume V ≅ V (λ_{1}) or V (λ_{2}) in case of F_{4}(q) by [Str]. So assume first that V is not strong quadratic. If A ≤ Q_{1} then by 1.17, we have |A| > q. Now we see that V is strongly quadratic, a contradiction, where we use 1.11 in case of G ≅ F_{4}(q) and A ≤ Z(Q_{1}). So we have A ≰ Q_{1}. Further |A ∩ Q_{1}| ≤ q for G ≅ ^{2}E_{6}(q) and A ∩ Q_{1} ≤ Z(Q_{1}) for G ≅ F_{4}(q). So for G ≅ F_{4}(q) we have some t ∈ A \ Q_{1} and x ∈ Q_{1} with t^{x} = ta, a ∈ Q_{1} \ Z(Q_{1}). If A ∩ Q_{1} ≠ 1, we get that V is strongly quadratic. So we may assume that A ∩ Q_{1} = 1 in that case.

If G ≅ F_{4}(q), then V involves at least four nontrivial irreducible G_{1}/Q_{1}–modules, at most two of them are natural. So |V : C_{V} (A)| > q^{6}, a contradiction.

If G ≅ ^{2}E_{6}(q), we get that there are at least two nontrivial G_{1}/Q_{1}–modules involved, which are isomorphic to Q_{1}/Z(Q_{1}), or at least four nontrivial irreducible modules are involved. As Q_{1}/Z(Q_{1}) is not an F –module for U_{6}(q), we would get q|A/A ∩ Q_{1}| ≥ |V : C_{V} (A)| ≥ q^{2}|A/A ∩ Q_{1}|^{2} in the former, a contradiction. In the latter we get for the natural module V_{1} that |V_{1} : C_{V_{1}}(A/A ∩ Q_{1})| < q^{3}. This shows |A/A ∩ Q_{1}| ≤ q. But then |A| ≤ q^{2}, a contradiction again.

Let V = V (λ_{2}) and G = F_{4}(q). Then C_{V} (Q_{1}) is the V (λ_{2})–module for Sp(6, q). Now by 2.8 we have A ≤ Q_{1}. Let R be the root group normal in G_{1}. Then [V, R] ≥ C_{V} (R) and so |[V, R]| ≥ q^{14}. Hence R ≰ A. By 1.11 we have ⟨A^{G_{1}}⟩ = Q_{1} and so Q_{1} offends. Hence |V : C_{V} (Q_{1})| ≤ |Q_{1}|, and so |V | ≤ q^{29}, a contradiction.

So we have that V ≅ V (λ_{1}). Now G ≤ E_{6}(q^{2}) or E_{6}(q) and the module V (λ_{1}) of those groups restricted to G is either irreducible, i.e. V, or a direct sum of a trivial module with V. Now by 2.9 V is not an F –module.

So we have shown

Proposition 2.11 Let G = G(q) be of Lie type, V be an F –module with offending subgroup A ≤ G. Then one of the following holds

(i) G ≅ L_{n}(q), V ≅ V (λ_{1}), V (λ_{2}), V (λ_{n−2}), V (λ_{n−1})

(ii) G ≅ Sp_{2n}(q), Ω^{±}(2n, q), or U_{n}(q) and V ≅ V (λ_{1})

(iii) G ≅ Sp_{6}(q), V ≅ V (λ_{3})

(iv) G ≅ Ω^{+}(2n, q), n = 4, 5 and V ≅ V (λ_{n}).

(v) G ≅ G_{2}(q) and V is the natural module.

3 F -modules for Lie type groups in characteristic two with offending subgroup not in E(G)

In this section we will prove the following

Proposition 3.1 Let G = G(q), q = 2^{t}, be of Lie type, V be an irreducible F -module with quadratic offending subgroup A ∈ P(G, V), A ≰ E(G). Then V is the natural module for E(G) = Ω^{±}(2n, q).

We will establish this proposition in a series of lemmas.

Lemma 3.2 We have that V is irreducible for E(G).

Proof: Suppose there is t ∈ A such that V = V_{1} ⊕ V_{1}^{t}. Then by quadratic action we have |A| = 2. Now |V | = 4, a contradiction.

Lemma 3.3 Let G = Ω^{±}(2n, q), V be the natural module and A ≤ G be a 2-group such that C_{V} (a) = C_{V} (A) for all a ∈ A^{♯}, then |A| ≤ |V : C_{V} (A)|.

Proof: Let W = C_{V} (a) = C_{V} (A). Then [V, a] = W^{⊥}. This shows [V, a] = [V, b] for all a, b ∈ A^{♯}, hence |A| ≤ |W^{⊥}| = |V : C_{V} (A)|.

Lemma 3.4 Let G = Ω^{±}(2n, q), G_{1} be the point stabilizer in the natural representation of G and E = O_{2}(G_{1}). Let ⟨a, b⟩ be a fours group in G_{1} with E ∩⟨a, b⟩ = 1. Suppose that ⟨a, b⟩ acts quadratically on some GF (2)–module V. If C_{E}(a) ≠ C_{E}(b) then V is strong quadratic.

Proof: We may assume that there is some e ∈ E^{♯} with [e, a] = 1 ≠ [e, b]. Then ⟨a, [e, b]b⟩ acts quadratically and so by the 3-subgroup lemma also ⟨a, [e, b]⟩ acts quadratically. Now as [E, b] always contains root elements, we get the assertion.

From now on let G be a minimal counterexample to 3.1. By 3.2 we may assume that V is an irreducible E(G)–module.

Lemma 3.5 Let E(G) be of rank at most two, then E(G) ≅ L_{2}(4) and V is the natural Ω^{−}(4, 2)-module, or E(G) ≅ Ω^{−}(6, q) and V is the natural module, or G ≅ Σ_{6} and V is the irreducible permutation module or its dual.

Proof: Let first E(G) be of rank one. Then E(G) ≅ L_{2}(q), Sz(q), or U_{3}(q). If E(G) ≅ L̸_{2}(q), then for each a ∈ A ∩ E(G)^{♯} we have |V : C_{V} (a)| ≥ q^{2} by 1.2. So suppose A∩E(G) ≠ 1, then we get E(G) ≅ L_{2}(q). By 1.2 we have |V : C_{V} (a)| ≥ q for a ≤ A∩E(G)^{♯}. As |A| ≤ 2\sqrt{q}, we see q = 4 and we have the first assertion. So let A ∩ E(G) = 1, then |A| = 2. Then for any involution x ∈ E(G), we get |V : C_{V} (x)| ≤ 4. This with 1.2 shows q = 4 and E(G) ≅ L_{2}(4), the conclusion.

Let next E(G) ≅ ^{3}D_{4}(q), ^{2}F_{4}(q), or G_{2}(q). As ^{2}F_{4}(q) does not have an outer automorphism of order two, the second case does not occur. Now t induces a field automorphisms. So by induction and 1.9 we get q = 4. Furthermore as the L_{2}(4)–module is a tensor product we see that V is a tensor product. Hence by 1.15, we see |A ∩ E(G)| ≤ 4, so |A| ≤ 8. So for a root element r we have |V : C_{V} (r)| ≤ 64. If E(G) ≅ ^{3}D_{4}(q), then this would imply that [[V, r], O_{2}(C_{G}(r))] = 1, and so V is strong quadratic, contradicting [Str]. So we have E(G) ≅ G_{2}(4). Further we get |V : C_{V} (r)| = 64 and so |A| = 8. Finally A ∩ E(G) just contains short root elements. Let s be such an element, then C_{E(G)}(s) ≅ S × E_{16}L_{2}(4), |S| = 4. As |[V, s]| ≤ 8, we get [[V, s], E_{16}L_{2}(4)] = 1, and so V is strong quadratic, a contradiction.

Let E(G) ≅ L_{3}(q). Then V is not strong quadratic and so A ∩ E(G) ≤ R, where R is a root group. Suppose that t induces a graph automorphism, then C_{E(G)}(t) ≅ L_{2}(q). By the A × B-lemma and 1.2, we see that |V : C_{V} (t)| ≤ 2. But then for r ∈ R^{♯} we have |V : C_{V} (r)| ≤ 4 and so by 1.2 q ≤ 4. Suppose t does not induce a graph automorphism, then |A ∩ E(G)| ≤ \sqrt{q} and so we see with 1.2 that q ≤ 4 again. Let q = 2, then t inverts an element of order 7, a contradiction. Let q = 4. Then |[V, r]| ≤ 4 and so an elementary abelian group of order 8 centralizes [V, r], whence V is strong quadratic, a contradiction.

Let E(G) ≅ Sp(4, q). Let first q = 2. Then E(G) ≅ A_{6}. If t induces an outer automorphism on Σ_{6}, then |A| ≤ 4. As t inverts some element of order 5, we get |A| = 4 and C_{V} (A) = C_{V} (t). But some a ∈ A inverts the element of order 5 centralized by t, contradicting the A × B - Lemma. So we have G ≅ Σ_{6}. If A ∩ E(G) = 1, we have the assertion. So let A ∩ E(G) ≠ 1. As the dimensions for Σ_{6}-modules are 1,4,4,16, |V : C_{V} (t)| ≤ 8 and G is generated by 5 conjugates of t, we get |V | = 16. This is the assertion.

So let now q > 2. Suppose first that t normalizes both parabolics. Then by induction and 1.9 we get q = 4. Further V is a tensorproduct and so by 1.15 we see |A ∩ E(G)| ≤ 4. Let c ∈ A ∩ E(G)^{♯}. Then |[V, c]| ≤ 8. So we see that there is some Sylow 2 -subgroup S such that [[V, c], Z(S)] = 1. But then V is strong quadratic, a contradiction. So we have |A| = 2 and so for some root element r we get |[V, r]| ≤ 4, a contradiction as before.

So we may assume that t does not normalize the parabolics. Then |A ∩ E(G)| ≤ q. Let A ∩ E(G) ≠ 1. Then elements in A ∩ E(G)^{♯} invert elements of order q^{2} + 1 in G. So |V : C_{V} (a)| ≥ q^{2}. Hence again we see q = 4 and then |A : A ∩ E(G)| = 4. But this is not possible.

Let next E(G) ≅ U_{5}(q). Then by induction we get q = 2. Further |A| ≤ 16. As V is not strongly quadratic we even get |A| ≤ 8. Let P be the normalizer of a root element r. We have |V : C_{V} (r)| ≤ 64. But then [[V, r], O_{2}(P)] = 1, and so V is strong quadratic, a contradiction.

Let finally E(G) ≅ Ω^{−}(6, q). If V is a tensor product, then by 1.15 we get |A∩E(G)| ≤ q. If there is a root element in A, then V is strongly quadratic, contradicting [Str]. Let some c ∈ A of type c_{2}. Then c inverts some element of order q^{2} + 1, so |[V, c]| ≥ q^{2} and then q = 2. Now |A| ≤ 8 and so |[V, c]| ≤ 8. If A ∩ E(G)| = 1, then for some c ∈ E(G), c of type c_{2}, we have |[V, c]| ≤ 4. Let G_{1} be the stabilizer of an isotropic vector in the natural representation. Then [G_{1}, C_{V} (O_{2}(G_{1}))] ≠ 1. Hence this is an irreducible A_{5}-module. As |[V, c]| ≤ 8 for any involution in A_{5}, we see that [O^{2}(G_{1}), V] = C_{V} (O_{2}(G_{1})). But then O_{2}(G_{1}) acts quadratically , and so V is strong quadratic, contradicting [Str].

So we have that V is not a tensor product. Then by induction and 1.9 we see that C_{V} (O_{2}(G_{1})) (notation as before) has to be an F -module. As V is not a tensor product it has to be the trivial module, so V is the natural module, the conclusion.

First we study the action of A on the Dynkin diagram.

Lemma 3.6 Let the rank of E(G) be at least three and G_{1} be one of the maximal parabolics with connected Dynkin diagram. Assume that [A, G_{1}] ≤ G_{1}. Let Q_{1} = O_{2}(G_{1}). Then either C_{V} (Q_{1}) is a trivial G_{1}/Q_{1}–module or A induces inner automorphisms on G_{1}/Q_{1}.

Proof: Suppose false. Then by induction we may assume that G_{1}/Q_{1} is Ω^{±}(2n−2, q). This shows that E(G) ≅ Ω^{±}(2n, q), E_{n}(q), U_{6}(q) or ^{2}E_{6}(q). Now in any case the action of A on E(G) is known. Suppose E(G) ≅ Ω̸^{±}(2n, q). Then there is a second parabolic on which A induces an outer automorphism. But then by induction we have a contradiction. So we have E(G) ≅ Ω^{±}(2n, q). Further C_{V} (Q_{1}) is the natural G_{1}/Q_{1}–module. Hence by [Str] we have that V is not strongly quadratic. This shows that A ∩ Q_{1} = 1. By 3.3 and 3.4 we have that |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A ∩ E(G))| ≥ |A ∩ E(G)|. Recall that C_{V} (Q_{1}) is isomorphic to Q_{1} as G_{1}/Q_{1}–module. So we see that [V, A] ≤ C_{V} (Q_{1}). Then [V, Q_{1}, Q_{1}] = 1, a contradiction.

Lemma 3.7 Let the notation be as in 3.6. Then A induces inner automorphisms on G_{1}/Q_{1}.

Proof: Suppose false. Then by 3.6 we have that C_{V} (Q_{1}) is the trivial module. Let G_{2} be a second maximal parabolic with connected diagram with [G_{2}, A] ≤ G_{2}. Then by 3.6 we have that A induces inner automorphisms on G_{2}/Q_{2}. In particular we see that A does not induce field automorphisms on E(G). This shows that E(G) ≅ Ω^{±}(2n, q) or E_{6}(q).

Suppose first that V ≅ V̸ (λ) for some fundamental weight λ. Then it is a tensor product and by 1.15 we have |A ∩ E(G)| ≤ q, so |A| ≤ 4q. This shows that for some root element r we have |[V, r]| ≤ q^{8}, As the q - rank of L_{8}(q) is 16 but in E_{6}(q) we have |O_{2}(C_{G}(r))/Z(O_{2}(C_{G}(r))| = q^{20}, we see that V would be strong quadratic, a contradiction to [Str]. Hence E(G) ≅ E̸_{6}(q). In the same way we see that for E(G) ≅ Ω^{±}(2n, q) we have n ≤ 6. But in that case we get |A| ≤ q^{2} and so |[V, r]| ≤ q^{4} and then n ≤ 3. This shows G ≅ Ω^{±}(6, q), q ≤ 4. But then for some w ∈ E(G), w of type c_{2} we see |V : C_{V} (w)| ≤ q^{2}, which now easily shows that V is strongly quadratic a contradiction.

So V = V (λ), and so we have G = E_{6}(q) and V = V (λ_{4}). But now for a root group R we have |[V, R]| = q^{21} and so |C_{V} (R) : C_{C_{V} (R)}(A ∩ Q_{1})| ≥ |A ∩ Q_{1}|, hence V/C_{V} (R) has to be an F -module for G_{1}/Q_{1} ≅ L_{6}(q) : 2, a contradiction.

Lemma 3.8 Let the rank of E(G) be at least three. Then A does not fix any maximal parabolic with connected diagram.

Proof: Suppose false. Then by 3.7 A has to induce inner automorphisms on any such parabolic. This shows that A does not induce field automorphisms. Hence we see that A induces graph or graph-field automorphisms, which have to be trivial on any such parabolic which is left invariant, a contradiction.

So for the remainder of this section we have to consider E(G) ≅ L_{n}(q) or F_{4}(q) and A induces a diagram automorphism.

Lemma 3.9 Let E(G) ≅ L_{n}(q), n ≥ 4, or F_{4}(q) and assume that A contains some t which acts nontrivially on the Dynkin diagram. Then G ≅ L_{4}(q) ≅ Ω^{+}(6, q) and V is the natural orthogonal module.

Proof: Let first E(G) ≅ L_{4}(q). Suppose that V is not strong quadratic. Then |A ∩ E(G)| ≤ q^{2}. Choose s ∈ A ∩ E(G), s ≠ 1. Then |[V, s]| ≤ 2q^{2}. If q ≠ 2 we see that V has to be strong quadratic. If A ∩ E(G) = 1, then |A| ≤ q and so again there is some s ∈ E(G) with |[V, s]| ≤ q^{2}. Hence in any case we may assume that q = 2 and so G ≅ Σ_{8} and |V : C_{V} (x)| = 8 for all x ∈ A∩E(G)^{♯}. But then A∩E(G) just contains short involutions and so A contains a transposition t. Now [C_{V} (t), C_{E(G)}(t)′] = 1, contradicting the A×B-lemma. So we have that V is strongly quadratic and the assertion follows with [Str].

Let next E(G) ≅ L_{5}(q). Then by [Str] V is not strongly quadratic. So for any involution s ∈ E(G) we have |[V, s]| ≥ q^{5}. From this we see that A ∩ E(G) does not contain root elements and so |A ∩ E(G)| ≤ q^{4}. Let s ∈ A ∩ E(G)^{♯}. Then C_{E(G)}(s) contains an extension of a class two group of order q^{8} by L_{2}(q). This shows that |[V, s]| ≥ q^{5}. Now we get q = 2 and |A ∩ E(G)| = 16. Let r be a root element in E(G), which is centralized by A. Then |A ∩ O_{2}(C_{G}(r))| ≥ 8, but then there is a quadratic fours group in E(G) which contains r, a contradiction.

Let next E(G) ≅ L_{6}(q) and V = V (λ_{3}). Then dim V = 20 and for the parabolic G_{15} we get the following series

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

dim V_{1} = dim V/V_{3} = 6, dim V_{2}/V_{1} = dim V_{3}/V_{2} = 4. Further [t, V_{1}] ≤ V_{1}, [t, V/V_{3}] ≤ V/V_{3} and [t, V_{2}/V_{1}] = V_{3}/V_{2}. This shows |[V, t]| ≥ q^{6}. Further we see that A ∩ E(G) ≤ Q_{15}. This shows |A| ≤ 2q^{5}. Hence q = 2 and |A| = 64. Further Z(G_{15}) ≤ A, but [t, [V, Z(G_{15})]] = [t, V_{1}] ≠ 1, a contradiction.

Let now E(G) ≅ L_{n}(q), n ≥ 6. Assume first that V is not strongly quadratic. Then [G_{1,n−1}, C_{V} (Q_{1,n−1})] ≠ 1, and so by induction we see E(G_{1,n−1}/Q_{1,n−1}) ≅ L_{4}(q) and C_{V} (Q_{1,n−1}) is the natural module. So we have n = 6 and V ≅ V (λ_{3}), a contradiction. So we have that V is not strongly quadratic.

Let ⟨c, d⟩ ≤ A ∩ E(G) with c of type a_{2} and d ∉ Z(C_{E(G)}(c)). We have that

O^{2′}(C_{E(G)}(c)/O_{2}(C_{E(G)}(c))) ≅ L_{n−4}(q) × L_{2}(q)

and O_{2}(C_{E(G)}(c)) is an eytension of Z(O_{2}(C_{E(G)}(c))) by a group of order q^{4(n−4)}, which is a sum of tensor products of the two natural modules for L_{2}(q) and L_{n−4}(q). As C_{E(G)}(c) acts indecomposably on Z(O_{2}(C_{E(G)}(c))) for q > 2 and V is not strongly quadratic, we get that q = 2 and d ∈ Z(O_{2}(C_{E(G)}(c))) = E.

Set X = ⟨P_{1}, P_{n−1}⟩. Let E ≤ X, where E is as before. Let first A ∩ O_{2}(X) ≠ 1. As E = Z(O_{2}(X)), we see that [t, E] contains some element 1 ≠ c ∈ E. As V is not strongly quadratic, this is not a transvection. So it is of type a_{2}. Now by the above we get q = 2 and A ∩ O_{2}(X) ≤ E. In particular |A ∩ O_{2}(X)| ≤ 4. Let s ∈ A ∩ X, s ∉ O_{2}(X). Then [s, E] contains transvections, hence A ∩ O_{2}(X) = 1. So we have |A| ≤ 8 or A ∩ O_{2}(X) = 1. Let |A| ≤ 8. Then |[V, c]| ≤ 8 and so we get n ≤ 4, a contradiction.

So we have A ∩ O_{2}(X) = 1. If q > 2, we see that C_{V} (O_{2}(X)) is trivial, as it has to be an F -module. Now by induction applied to G_{1,n−1} we get again n = 6 and V = V (λ_{3}), a contradiction. So we are left with q = 2 and |A| ≤ 4. As n ≥ 6, we see that t inverts an element of order 7 in ⟨P_{1}, P_{2}, P_{n−2}, P_{n−1}⟩ = T. Hence we get that C_{V} (O_{2}(T)) is a trivial module. Now again induction on G_{1,n−1}, yields the contradiction.

So let now E(G) ≅ F_{4}(q). By [Str] we have that V is not strongly quadratic. By 3.5 we have that G_{1,4} acts trivially on C_{V} (O_{2}(G_{1,4})). Again set X = ⟨P_{1}, P_{4}⟩. Let first A ∩ O_{2}(X) = 1. Then |A| ≤ 2q. Hence for some root element r ∈ E(G) we get |V : C_{V} (r)| ≤ 16q^{4} ≤ q^{8}. From this we see that [O_{2}(C(r)), [V, r]] = 1, contradicting V to be not strongly quadratic. So we have A ∩ O_{2}(X) ≠ 1. Hence again there is some quadratic fours group ⟨c, d⟩ ≤ O_{2}(X) with c ≤ Z(O_{2}(X)). Let r be a root element with Z(O_{2}(X)) ≤ O_{2}(C(r)) = Y. If d ∉ Y, we easily get a strongly quadratic module. So d ∈ Y and then in fact d ∈ Z(Y). But in Z(O_{2}(X)) there are root elements r and s of both types. So the same is true for Y_{1} = O_{2}(C(s)). Hence d ∈ Z(Y) ∩ Z(Y_{1}), which is of order q^{2}. As d is not a root element, we get that ⟨c, d⟩ is in Z(Y) ∩ Z(Y_{1}) and so d ∈ Z(C_{E(G)}(c)), a contradiction.

Now the proposition follows from 3.5, 3.8 and 3.9.

4 F -modules for alternating, sporadic and Lie type groups in odd characteristic

Throughout this chapter we will assume that E(G) is a perfect central extension of an alternating group, a sporadic group or a group of Lie type over a field of odd characteristic, which is not a group of Lie type over a field of characteristic 2 too. Further V is an F -module and A ∈ P(G, V) a quadratic offending subgroup.

Lemma 4.1 Let E(G) = A_{n} or E(G) = 3A_{6}, then either V is the permutation module or

(i) G ≅ A_{8} ≅ L_{4}(2) and V is the natural L_{4}(2)-module.

(ii) G ≅ A_{7} and V is as in (i).

(iii) E(G) ≅ 3A_{6} and |V | = 2^{6}.

(iv) E(G) ≅ A_{5} ≅ L_{2}(4) and V is the natural L_{2}(4)-module.

Proof: Suppose first |A| = 2. Then A induces transvections. Hence A just inverts elements of order three. So either A = ⟨(1, 2)⟩ and so V is the permutation module, or A = ⟨(1, 2)(3, 4) ⋯ (n − 1, n)⟩, n is even, and n ≤ 8. As A_{8} ≅ L_{4}(2) and A_{6} ≅ Sp_{4}(2)′, we have handled these cases before. So let E(G) ≅ 3A_{6}. But then A inverts the center of E(G), which has to act fixed point freely on V, a contradiction.

So let |A| ≥ 4. Let furthermore E(G) ≅ 3̸A_{6}. By [MeiStr] we have that V is either the natural module or the spin module. So assume V to be the spin module. Let a ∈ A^{♯}. Let k be the number of fixed points of a. Then there is K ≤ C_{G}(a), K ≅ Σ_{k}. Furthermore C_{C_{G}(a)}(K′) is an extension of a 2–group by Σ_{m}, m = (n − k)/2. Now choose a ∈ A with m > 2 if possible. Suppose first [A, C_{C_{G}(a)}(K′)] ≠ 1. If m ≥ 5, then Σ_{m} is nonsolvable and so C_{C_{G}(a)}([V, a]) contains an elementary abelian subgroup of O_{2}(C_{G}(a)) of order 2^{m−1}. But then this group contains a conjugate of (12)(34) which contradicts [MeiStr, (4.3)].

Let m = 4. Then a ∼ (12)(34)(56)(78). Furthermore as we may assume that there is no x ∼ (12)(34) which is contained in ⟨A^{C_{G}(a)}⟩ we see that A is conjugate to a subgroup of ⟨(12)(34)(56)(78), (13)(24)(57)(68), (15)(26)(37)(48)⟩.

Let m = 3. Then C(K′) ≤ Σ_{6} and a ∼ (12)(34)(56). Then ⟨A^{C_{G}(a)}⟩ contains some x ∼ (12)(34), contradicting [MeiStr, (4.3)].

So let [A, O^{2′}(C_{C_{G}(a)}(K′))] = 1. If [A, K′] ≠ 1, then [K′, [V, a]] = 0. If k ≥ 4, then K′ contains some x ∼ (12)(34). This again contradicts [MeiStr, (4.3)]. Let k ≤ 3. As [A, O^{2′}(C_{C_{G}(a)}(K′))] = 1 and m > 2, there is x ∼ (12) in A, a contradiction. So we are left with [A, K′] = 1 = [A, C_{C_{G}(a)}(K′)]. But this is impossible with m > 2.

So we have m ≤ 2 for all a ∈ A^{♯}. As there is no fours group of transvections we may assume a = (12)(34) ∈ A. Now A ≥ ⟨a, b⟩, b = (13)(24), (12)(56) or (34). Let [b, K′] ≠ 1. Then b = (12)(56) and so K′ contains no involutions by [MeiStr, (4.3)]. This shows k ≤ 3 and so A ≤ Σ_{7}. But this group A = ⟨(12)(34), (12)(56)⟩ does not act quadratically on the four dimensional module.

Assume now b = (34). Then [a, E(C_{G}(b))] ≠ 1. Now E(C_{G}(b)) ≅ Σ_{n−2}, which is nonsolvable. But then ⟨(34), (12)(56)⟩ acts quadratically, a contradiction.

So we have seen that |A| ≤ 8 and so |V : C_{V} (a)| ≤ 8 for all a ∈ A^{♯}. Furthermore for n > 8, we see that there is some a ∈ A^{♯} which inverts (1, 2, 3) = ρ. But ρ is fixed point freely on V. This shows |V | ≤ 2^{6}, a contradiction. So we have n ≤ 8.. The case of n = 5, 6, 8 was done before. So we have n = 7. Then |A| = 4, and |V | = 16, which is (ii).

So we are left with E(G) ≅ 3A_{6} and |A| = 4, A ≤ E(G). Now we see that E(G) is generated by three conjugates of A and so |V | ≤ 2^{6}. As Z(E(G)) has to act fixed point freely, we see |V | = 2^{6}.

Lemma 4.2 There are no F -modules for E(G) sporadic.

Proof: If there is a transvection, then there is also a quadratic fours group. So by [MeiStr] we just have to handle the cases E(G) ≅ M_{12}, M_{22}, M_{24}, 3Suz, 3M_{22}, J_{2}, Co_{1}, Co_{2}. Furthermore V is known. In all cases besides E(G) ≅ 3M_{22} we have |A| ≤ 4.

Let |A| = 2, then any fours group containing A would be quadratic, which contradicts [MeiStr]. So we have |A| = 4 and if E(G) ≅ M̸_{12} or M_{22}, we see that for each a ∈ A^{♯} we have [O^{2}(C_{E(G)}(a)), [V, a]] = 1. But then we get to many quadratic fours groups which contradicts [MeiStr] again. So we are left with E(G) ≅ M_{12} or M_{22} and C_{E(G)}(a) induces Σ_{3} on [V, a]. In both cases there is some a ∈ A^{♯} which inverts some element ω of order 5 in E(G) which acts on some element ν of order eleven. Hence |[V, a]| ≥ 2^{4}, a contradiction.

So we are left with E(G) ≅ 3M_{22} and |V | = 2^{12}. As can be seen from the action described in [MeiStr] |A| ≠ 16. As V is defined over GF (4), we see that the elements in A induce GF (4)-transvections on V. But any a inverts some element ω of order 5 which in turn normalizes some element ν of order 11. Hence |[V, ω]| = 2^{8} and so |[V, a]| = 2^{4}, a contradiction.

Lemma 4.3 If E(G) is some group of Lie type in odd characteristic then it is a group of Lie type in even characteristic too.

Proof: Suppose false. We have a quadratic fours group. So by [MeiStr1] we get E(G) ≅ 3U_{4}(3) and |V | = 2^{12}. As V is defined over GF (4), we see that A ∩ E(G) ≠ 1. For a ∈ A ∩ E(G)^{♯} , we have |V : C_{V} (a)| = 16, so we have |A| ≥ 16. As C_{E(G)}(a) is a maximal subgroup in E(G), we see that |A| = 2^{5} and so A is uniquely determined. But now A ≰ E(G) and so |V : C_{V} (A)| = 2^{6}, a contradiction.

References

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[CurRei] C.W.Curtis, I. Reiner Representation Theory of finite groups and associative algebras, John Wiley and Sons, London.

[Glaub] G. Glauberman, A sufficient condition for p-stability, Proc. London Math. Soc. 25, 253 - 287, 1972

[MeiStr] U. Meierfrankenfeld, G. Stroth, Quadratic GF (2) - modules for sporadic simple groups and alternating groups, Comm. Algebra 18, 2099 - 2139, 1990.

[MeiStr1] U. Meierfrankenfeld, G. Stroth, On quadratic GF (2)-modules for chevalley groups over fields of odd order, Arch. Math. 55, 105 - 110, 1990

[Sm] S. Smith, Irreducible modules and parabolic subgroups, J. Algebra 75, 286 - 289, 1982

[Stei] R. Steinberg, Lectures on Chevalley groups, Yale University 1967.

[Str] G. Stroth, Strong quadratic modules, Israel J. Math. 79, 257 -279, 1992

Department of Mathematics

Michigan State University

East Lansing, Michigan 48824

USA

meier@math.msu.edu

Fachbereich Mathematik und Informatik Institut für Algebra und Geometrie Martin-Luther-Universität Halle Wittenberg 06099 Halle Germany stroth@coxeter.mathematik.uni-halle.de