2F – modules with quadratic offender for the finite simple groups

G. Stroth

In this paper we are going to determine the irreducible faithful 2F -modules for the finite quasisimple groups. Here we just concern about 2F -modules over GF (2). A module V is called 2F -module for G, if there is a nontrivial elementary abelian subgroup A in G such that |V/C_{V} (A)| ≤ |A|^{2}. The group A then will be called an offending subgroup or an offender. The offender is called quadratic if [V, A, A] = 1. More precisely we prove

Theorem Let G be a quasisimple group and V be an irreducible faithful 2F -module in characteristic two for G with a quadratic offender. Then one of the following holds

(i) G/Z(G) ≅ A_{n} and one of the following is true (a) V is the natural module

(b) n ≤ 8 and |V | = 16

(c) n = 6, |Z(G)| = 3 and |V | = 64

(d) n = 9 and |V | = 2^{8} is the spin module. (ii) G ≅ 3M_{22} or 3U_{4}(3) and V is the 12-dimensional SU_{6}(q)–module. (iii) G/Z(G) = G(q) is a group of Lie type , q = 2^{t}, and one of the following is true

(a) G(q) ≅ L_{n}(q), Sp(2n, q)′, Ω^{±}(2n, q) or U_{n}(q) and V is the natural or dual module.

(b) G(q) ≅ L_{n}(q) and V is the exterior square of the natural or dual module

(c) G(q) ≅ L_{6}(q), or U_{6}(q) and V is the exterior cube of the natural module.

(d) G(q) ≅ Sp(6, q), Sp(8, q) or Sp(10, q) and V is the spin module

(e) G(q) ≅ Ω^{±}(8, q), Ω^{±}(10, q) or Ω^{+}(12, q) and V is the half spin module

(f) G(q) ≅ E_{6}(q) and V = V (λ_{1}) or V (λ_{6}).

(g) G(q) ≅ E_{7}(q) and V = V (λ_{7}).

(h) G(q) ≅ F_{4}(q) and V = V (λ_{1}) or V (λ_{4}).

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

(j) G(q) ≅ Sz(q) and V is the natural module. The proof of the theorem will depend on two main results. First of all we will use

the classification of F –modules. This can be found in an unpublished paper [MeiStr3] due to U. Meierfrankenfeld and the author. A preprint can be found on the homepage (http://coxeter.mathematik.uni-halle.de:8080/∼stroth/rep html). But there is also a classification in the literature. 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The F -modules for the groups of Lie type in characteristic two have been classified by B. Cooperstein [Coop] and in an unpublished paper by B. Cooperstein and G. Mason [CM].

We further will use the classification of quadratic modules in [MeiStr1], [MeiStr2] and [Str] to end up with a very short list of modules and so it is easy to detect the 2F -modules.

In fact there are still some open question. The module V (λ_{1}) for E_{6}(q) is a 2F –module. But we do not know whether it allows a quadratic offender. If not then the cases (iii)(f) - (h) will not show up. Further the paper just considers irreducible modules. So for general module one has to have an overview over the possible offenders, one has to study extensions of the modules above and/or of irreducible 2F –modules by trivial modules. This all has not been done so far but would be very useful for applications.

The main reason for the classification of these modules comes from the application in the so called amalgam method. This method basically provides us with 2F –modules with quadratic offender for the groups involved, provided the parameter b is not 1. So we will use the results in this paper in the revision of the classification of the finite simple groups of characteristic two type.

The notations will be standard. Concerning the representations of the groups of Lie type we follow [Stei1] and will use these results freely. In what follows G will always be a quasisimple group. A Chevalley group or a group of Lie type will always mean a central factor of the corresponding universal Chevalley group.

1 Preliminaries

For convenience of the reader we first state the main resulut of [MeiStr3].

Theorem 1.1 Let E(G) = F^{∗}(G) be a quasisimple group and V be an irreducible faithful F -module in characteristic two for G. 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.

Lemma 1.2 Let G be quasisimple with G/Z(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/Z(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/Z(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 G be quasisimple with G/Z(G) = Sz(q) or U_{3}(q) and V be an irreducible faithful 2F –module for G with quadratic offender A. Then V is the natural module.

Proof: By 1.2 we get |A| = q. Hence application of [Str, Corollary 1] gives the assertion or G ≅ U_{3}(q) and V is the eight – dimensional module. But in that case there are g, h ∈ G with G = ⟨A, A^{g}, A^{h}⟩. So we get |V | ≤ q^{6}, and so V has to be the natural module.

The next five lemmas are more or less well known results on the representations of groups of Lie type. We will use them freely in the sequel.

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 (V_{1}, … V_{l}), (V′_{1}, …, V′_{l}) give nonisomorphic GF (q)G–modules.

Proof: [Stei1, 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: [?, (7.5)]

Lemma 1.6 Let G/Z(G) = G(q) be a Chevalley group 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}(q) or B_{2}(q). Now following [?, chapter 9] we see that any basic module for G(q^{σ}) reduced to ^{σ}G(q) remains irreducible. Moreover by [?, 9.3] all irreducible modules are given by the tensor product theorem. Further by [?, 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 [?, 12.2] GF (q) is the field of definition for all these modules.

Proposition 1.7 Let G = A_{ℓ}(q), D_{ℓ}(q), E_{6}(q) or D_{4}(q) and V be a basic module with high weight λ for ^{σ}G. 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}.

Lemma 1.8 Let G = G(q), q = p^{f}, be a Chevalley group and V be an irreducible module in characteristic p over the splitting field. 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].

Definition 1.9 Let V be a faithful GF (p)-module for G. For ϵ ∈ R I define P_{ϵ}(G, V) to be the set of all non-identity p-subgroups X of G such that |X|^{ϵ}|C_{V} (X)| ≥ |Y |^{ϵ}|C_{V} (Y)| for all Y ≤ X (including Y = 1).

If V is a 2F –module with quadratic offender there is always some quadratic offender in P_{2}(G, V).

Lemma 1.10 Let V be a 2F -module or F –module for G with quadratic offender and V_{1} be an invariant subspace. Then V_{1} is a trivial subspace or V_{1} is a 2F -module, F –module respectively, with quadratic offender too.

Proof: Set μ = 2 or 1, according to V be a 2F –module or an F –module. Choose A ∈ P_{μ}(G, V), A quadratic on V. Set B = C_{A}(V_{1}). Then |V_{1}C_{V} (A)||B|^{μ} ≤ |C_{V} (A)||A|^{μ}. Hence |V_{1} : C_{V_{1}}(A)| ≤ |A|^{μ}/|B|^{μ} and so A/B offends on V_{1}.

Lemma 1.11 Let V be a GF (p)–module for G.

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

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

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

C_{V} (AB) = C_{V} (A) ∩ C_{V} (B).

So we have

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

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

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

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

Lemma 1.12 Let V be a GF (p)–module for G with quadratically acting elementary abelian p–subgroup A. Let g ∈ G and a ∈ A with a^{g} = az. Then also ⟨z, C_{A}(g)⟩ acts quadratically.

Proof: We have [V, C_{A}(g), a] = 1. So also [V, C_{A}(g), z] = 1. As [C_{A}(g), z] = 1 the three subgroup lemma shows [V, z, C_{A}(g)] = 1, the assertion.

Let W be a Weylgroup with root system Φ and fundamental roots Π. We assume throughout this section that the Dynkin diagram on Π is connected. Let l = |Π|, I = {1, 2, … l} and Π = {α_{1}, α_{2}, … α_{l}} where we choose the labeling as follows:

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

Lemma 1.13 Let G be quasisimple with G/Z(G) = L_{n}(q) or U_{n}(q) and V = V (λ_{1} +λ_{n−1}). 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/Z(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/Z(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/Z(G) ≅ U_{n}(q). If n = 3 there is nothing to prove. Let n = 4. If A ≤ O_{2}(G_{1}), we get with 1.12 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 with 1.12 that |A ∩ O_{2}(G_{1})| ≤ q and A induces transvections on O_{2}(G_{1})/Z(O_{2}(G_{1})) with A ∩ O_{2}(G_{1}) ≤ [A, O_{2}(G_{1})]. But ⟨A, [A, O_{2}(G_{1})]⟩ ≤ U_{4}(q), and so as before we get strong quadratic action, a contradiction.

Lemma 1.14 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 1.15 Let X = V ⊗ W, be a faithful GF (q)–modules. Suppose X to be irreducible. If X is a 2F –module with quadratic offender, then G ≅ L_{2}(q) and X = V ⊗ V^{σ}, V the natural module and σ some field automorphism of GF (q).

Proof: Suppose by way of contradiction that X is a 2F –module with quadratic offender A. By 1.14 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}, and so we have equality everywhere. Further we get that |V : C_{V} (a)| = q. This gives |V | = q^{2} and so G ≅ L_{2}(q) and the assertion follows with 1.4.

Lemma 1.16 Let G be quasisimple with G/Z(G) = G(q) be of Lie type of rank at least two.

a) Let A ≤ G, |A| ≤ q and V be an irreducible module with |V : C_{V} (A)| ≤ |A|^{2} and A quadratic, then V is strong quadratic.

b) Let R be a root group in G with |[V, R]| ≤ q^{2} for some irreducible module V, then G/Z(G) ≅ L_{n}(q), U_{n}(q), Ω^{±}(2n, q), Sp(2n, q) and V is the natural module, or G ≅ G_{2}(q) and V is the 6–dimensional module.

Proof: a) Let R be a root group with [R, A] = 1. Suppose first R ≥ A, then obviously [O_{p}(C_{G}(R)), [V, A]] = 1 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^{4}. As C_{G}(r) acts on [V, r] we see again that V is strong quadratic or C_{G}(R) possesses a factor embeddable into GL(4, q). Further all involutions of C_{C_{G}(R)}([V, r]) are in R. So we see that G/Z(G) ≅ L_{4}(q), U_{4}(q), G_{2}(q). Now we have that a ∈ A^{♯} is not conjugate to r. This shows that [[V, a], O_{2}(C_{G}(a))] = 1. Hence V is strong quadratic.

So let now A ≤ Z(O_{2}(C_{G}(R))). Then G ≅ F_{4}(q) or Sp(2n, q). Recall that for L_{3}(q), A intersects always a root group nontrivially. But then there is some g ∈ G such that A^{g} ≤ C_{G}(R) but A^{g} ≰ O_{2}(C_{G}(R)). So we may assume that A ≰ O_{2}(C_{G}(R)) right from the beginning. 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). 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 two nontrivial modules in V. So we may assume that exactly one of those is in [V, R]. This now shows that |O_{2}(P)/R| ≤ q^{5}. In particular G/Z(G) ≅ L_{4}(q), Sp(4, q), U_{4}(q), G_{2}(q). But now we may argue as above. 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^{2}, we see that [[V, R], C_{G}(R)] = 1, or C_{G}(R) has a factor L_{2}(q), which acts nontrivially on [V, R]. Assume the latter. Then G ≅ Sp(2m, q), O^{±}(2m, q), or G_{2}(q). In the first two cases we have the natural module, while for G_{2}(q) we have the 6–dimensional one.

So assume now that λ is the weight corresponding to C_{G}(R), or G/Z(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. Let G be classical. Then we may assume that G ≅ Sp(2m, q), or O^{±}(2m, q) and V = V (λ_{2}). Now in all cases C_{V} (O_{2}(G_{1})) is the natural module for G_{1}/O_{2}(G_{1}). As there is a conjugate of R in G_{1} intersecting O_{2}(G_{1}) trivially, and [G_{1}, V] ≰ C_{V} (O_{2}(G_{1})), we see that G ≅ Sp(2m, q). But now we see that even [O_{2}(C_{G}(R)), [V, R]] = 1, and so V cannot be V (λ_{2}), by [Str]. Let now G/Z(G) ≅ E_{6}(q), E_{7}(q), or F_{4}(q). Then we look at the endnode of the diagram, which does not correspond to λ. Then induction gives us, that we must have G ≅ F_{4}(q) and V = V (λ_{1}). But here the parabolic G_{4} induces more than two nontrivial modules, a contradiction. So we are left with G ≅ G_{2}(q) or ^{3}D_{4}(q). If G ≅ G_{2}(q), then V is the natural module. If G ≅ ^{3}D_{4}(q), then V is the natural Ω^{+}(8, q^{3})-module and so |[V, R]| ≥ q^{3}.

Lemma 1.17 Let be quasisimple with G/Z(G) ≅ L_{n}(q), n ≥ 5, let V = V (λ_{2}) and A be an offender as F -module. Then |A| = q^{n−1} = |V : C_{V} (A)|.

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})^{♯} 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}).

Let now n > 5. We first show A ≤ O_{2}(G_{n−1}). Suppose false then we have by induction and 1.10 that A/A∩O_{2}(G_{n−1}) is uniqueley determined. So O_{2}(G_{n−1})A = O_{2}(G_{n−1}∩G_{n−2}). This shows that also A∩O_{2}(G_{n−1}) offends. So we may just prove that O_{2}(G_{n−1}) is a minimal offender. Let x ∈ O_{2}(G_{n−1})^{♯}, then we get that |[V, x]| = q^{n−2}. Hence for A ≤ O_{2}(G_{n−1}) we have |A| ≥ q^{n−2}. If |A| = q^{n−2} then C_{V} (A) = C_{V} (a) for all a ∈ A^{♯}. But then G_{n−1} = ⟨C_{G_{1}}(a) | a ∈ A^{♯}⟩ would act on C_{V} (A), a contradiction. This shows A = O_{2}(G_{n−1}) and we are done.

2 2F –modules for Lie type groups in even characteristic

Throughout this chapter we will assume that G is a quasisimple group with G/Z(G) a group of Lie type over a field with q = 2^{n} elements, including G(2)′ and A_{6}. We additionally assume that G′ ≅ 3̸ · A_{6}, as this will be handled in the last chapter together with the alternating groups. As G will act faithfully on a GF (2)–module, G will always be a factor of the universal group.

Further V is an 2F -module and A ∈ P_{2}(G, V) is a quadratic offending subgroup. If G_{i} is a maximal parabolic in G we set Q_{i} = O_{2}(G_{i}).

Lemma 2.1 Let G/Z(G) = G(q) be classical or F_{4}(q) and V = V (λ_{2}) be a 2F -module with quadratic offender A. Then V is an F –module too.

Proof: If G ≅ Sp(2n, q), n > 2, or Ω^{±}(2n, q), n > 3, then V is the adjoint module. Let first A ≤ Q_{2}. As V is not strong quadratic by [Str], we see with 1.12 that |A| ≤ q or A ≤ Z(Q_{2}). By 1.16 we get A ≤ Z(Q_{2}) and G ≅ Sp(2n, q). Further q^{2} ≥ |A| > q. As [[V, A], Q_{2}′] ≠ 1, we see G ≅ Sp(6, q), |A| = q^{2}, and |[V, A]| = q^{4}. Let g ∈ G_{2} with g ∉ N_{G_{2}}(A), which must exist. Then |A ∩ A^{g}| ≥ q and [V, A ∩ A^{g}, ⟨A, A^{g}⟩] = 1. As ⟨A, A^{g}⟩ contains root elements, we get that V is strong quadratic, a contradiction. So we have A ≰ Q_{2}. By 1.12 we see that |A ∩ Q_{2} : A ∩ Z(Q_{2})| ≤ q. If A ∩ Q_{2} ≰ Z(Q_{2}), we even get A ∩ Z(Q_{2}) = 1. Now for a ∈ A and b ∈ A ∩ Q_{2}, we get with 1.12 [C_{Q_{2}}(b), a] ≤ A ∩ Q_{2}. This shows |A : A ∩ Q_{2}| ≤ q and then |A| ≤ q^{2} and G ≅ Sp(2n, q). Now for the root group R normal in G_{2} we see that [V, R] involves a module isomorphic to Q_{2}/Z(Q_{2}). Further C_{V} (R) involves some module V (λ_{2}) for Sp(2n−4, q). This shows |V : C_{V} (A)| ≥ q^{5}, a contradiction. So we have A ∩ Q_{2} ≤ Z(Q_{2}). Now with 1.12 we get A ∩ Q_{2} = 1. Let a, b ∈ A^{♯}, both not transvections on the natural module. Let x ∈ C_{Q_{2}}(b), x ∉ C_{Q_{2}}(a). Then by 1.12 ⟨b, [x, a]⟩ acts quadratically on V. Now [C_{Q_{2}}([x, a]), b] ≰ Z(C_{Q_{2}}([x, a])), as |[Q_{2}, b]| ≥ q^{4}. Hence there is ⟨c, d⟩ ≤ Q_{2}, with C_{Q_{2}}(c) ≠ C_{Q_{2}}(d), which acts quadratically on V. But then by 1.12 V is strong quadratic, a contradiction. So we see that A contains a subgroup B of index at most q, which does not contain any transvection such that C_{Q_{2}}(b) = C_{Q_{2}}(B) for any b ∈ B^{♯}. Hence |Q_{2}/Z(Q_{2}) : C_{Q_{2}/Z(Q_{2})}(B)| ≥ |B|^{2}, as Q_{2}/Z(Q_{2}) is a direct sum of two natural modules for Sp(2n − 4, q) or Ω^{±}(2n − 4, q). Again as [V, R] involves a module isomorphic to Q_{2}/Z(Q_{2}), we see that |V : C_{V} (B)| ≥ |B|^{4}. This now shows |B| ≤ q, as |A| ≤ |B|q. Now by 1.16 we have that A contains elements which are not transvections on the natural module. Then |V : C_{V} (A)| ≥ q^{8}, which contradicts |A| ≤ q^{2}.

Let now G/Z(G) ≅ U_{n}(q). Then G is embedded in SL_{n}(q^{2}) and the modules V (λ_{2}) are the same for both groups. Let n = 5. If A ≤ Q_{1}, then we get |A| = q^{3} and so C_{V} (A) = C_{V} (R), R a root group. But then we get C_{V} (R) = C_{V} (Q_{1}), a contradiction. So we have A ≰ Q_{1}. Then |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| ≥ |A/A ∩ Q_{1}|^{2}. This shows that also A ∩ Q_{1} is an offender, a contradiction.

Let next n = 6. If A ≤ Q_{1}, then by 1.11 and 1.16 we get |V : C_{V} (Q_{1})| ≤ |Q_{1}|^{2}, a contradiction. So we have A ≰ Q_{1}. The same argument shows A ≰ Q_{3} and so |A| ≤ q^{8}. If x ∈ A ∩ Q_{1} \ R, we have that x is conjugated to an element of type c_{2} in G_{1}/Q_{1}. This shows |V : C_{V} (x)| ≥ q^{12}. If now |A : A ∩ Q_{1}| > q, we see that |V : C_{V} (A)| ≥ |V : C_{V} (x)||C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| ≥ q^{12}q^{4}. Hence |A| = q^{8} and we have equality everywhere. As now |A ∩ Q_{1}| ≥ q^{4}, we get x, y ∈ A ∩ Q_{1} \ R with G_{1} = ⟨C_{G_{1}}(x), C_{G_{1}}(y)⟩. This now shows C_{V} (x) ≠ C_{V} (y). But then |V : C_{V} (A)| > q^{16}. So we either have A ∩ Q_{1} ≤ R or |A : A ∩ Q_{1}| ≤ q. Suppose A ∩ Q_{1} ≰ R. Then we have |V : C_{V} (x)| = q^{12} = |V : C_{V} (A)|, as |A| ≤ q^{6}. But now we argue as before. So we have A ∩ Q_{1} ≤ R and so |A| ≤ q^{5}. Now A/A ∩ Q_{1} cannot contain any element of type c_{2} in G_{1}/Q_{1}. This shows |A| ≤ q^{2}. But this contradicts |V : C_{V} (a)| ≥ q^{6} for any a ∈ A^{♯}.

So let now n > 6. Again A ≰ Q_{1}. As C_{V} (Z(Q_{1})) involves V (λ_{2}) for G_{1}/Q_{1}, we get by induction that |A ∩ Q_{1}| ≠ 1. We have |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| ≥ q^{2}. So we see that |V : C_{V} (A)| > q^{2}|A/A ∩ Q_{1}|^{2}q^{2(n−2)}. This shows that |A ∩ Q_{1}| > q^{n−1}, a contradiction, as Q_{1} does not contain abelian subgroups of that order.

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 a 2F –module with quadratic offender for Sp_{6}(q). By 1.16 we have |A| > q. Let A ≤ Z(Q_{4}). Then with 1.11 we get that also Z(Q_{4}) is an offender in P_{2}(G, V). Now |Z(Q_{4})Q_{1}/Q_{1}| = q is a long root group. But C_{V} (Q_{1}) is the spin module for G_{1}/Q_{1}, and so |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(Z(Q_{4}))| = q^{4}, which contradicts 1.10. So we have A ≰ Z(Q_{4}). As [V, Q_{4}, Q′_{4}] ≠ 1 by [Str], we get |A : A ∩ Z(Q_{4})| ≤ q. Now the same applies for A∩Q_{1}, i.e. |A∩Q_{1} : A∩Z(Q_{1})| ≤ q. Hence we see that |A| ≤ q^{4}. Recall that Q′_{4} ∩A = 1 by [Str] and the same applies for Q′_{1} if A ∩ Q_{1} ≰ Z(Q_{1}). Now as for x ∈ Z(Q_{1}) we have |V : C_{V} (x)| ≥ q^{8}, as a conjugate of x acts on two modules V (λ_{2}) for Sp_{6}(q), we get |A| = q^{4}. But as A ≰ Q_{1}, we get that C_{C_{V} (x)}(A) ≠ C_{V} (x) for x ∈ A ∩ Z(Q_{1}), a contradiction.

Lemma 2.2 Let G/Z(G) = G(q) and V be an irreducible faithful 2F –module with quadratic offender. If C_{V} (Q_{i}) = C_{V} (Z(S)) for some i, then V = V (λ_{i}) or G/Z(G) ≅ L_{2}(q) and V = X ⊗ X^{σ} for some field automorphism σ, X the natural module.

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 1.15 we get the assertion.

Lemma 2.3 Let G/Z(G) = G(q) ≅ L̸_{n}(q) be classical and V be an irreducible faithful 2F –module with quadratic offender A ≤ Q_{1}. Then then pair (G, V) is one of the theorem.

Proof: Suppose false. Application of 1.16 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}|^{2}. Furthermore by 2.2 we may assume that C_{V} (Q_{1}) is a nontrivial G_{1}/Q_{1}–module. Let G ≅ Ω^{±}(2n, q). Choose a ∈ A^{♯}. Set B = ⟨A^{C_{G_{1}}(a)}⟩. If B does not contain root elements, we see |B| ≤ q^{2}. But then |A| ≤ q. So by 1.16 V is strong quadratic. So in any case V is strong quadratic and then 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}|^{2}, which shows G ≅ Ω^{+}(2n, q), n = 3, 4, 5, 6 or Ω^{−}(2n, q), n = 3, 4, 5.

Let now G/Z(G) ≅ Sp(2n, q) or U_{n}(q). Let first G ≅ S̸p(4, q). Let R be a root group normal in G_{1}. Then we see that G acts strongly quadratic, as |A| > q by 1.16. Further we see that V/C_{V} (Q_{1}) involves some module isomorphic to C_{V} (Q_{1}), as C_{V} (Q_{1}) ≤ [R, V] by 1.8. As 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. Further by 1.11 we see that |V : C_{V} (Q_{1})| ≤ |Q_{1}|^{2}. As C_{V} (Q_{1}) is some V (λ_{i}), i ≠ 1, for G_{1}/Q_{1}, we see that either [V, Q_{1}, Q_{1}] = 1 or G = Sp(2n, q), n ≤ 5, and V is the spin module, or G/Z(G) = U_{n}(q), n ≤ 6, and V = V (λ_{[n/2]}). Suppose first [V, Q_{1}, Q_{1}] = 1. Then Q_{1} is abelian and so G ≅ Sp(2n, q). Now we get that for G_{2}, we also have [V, Q′_{2}, Q_{2}] = 1. So by [Str], again V is the spinmodule and we are done. So we are left with U_{5}(q), V = V (λ_{2}). Now |[V, R]| = q^{6} and for an involution x ∈ Q_{1} \ Q′_{1}, we have |[V, x]| = q^{8}. This shows |A| = q^{4}, but there is no such elementary abelian subgroup in Q_{1}.

So we have to treat G ≅ Sp(4, q). Further we may assume that V is not strong quadratic. So we have that A does not intersect any root group. In particular |A| ≤ q^{2}. Using the closure under G_{1}, we even see that |A| ≤ 2q. If q > 4, we see with 1.2 that V involves at most two nontrivial G_{1}/Q_{1}–modules and also two G_{2}/Q_{2}–modules. This shows that |[V, R]| ≤ q^{3} and there is just once the natural module involved in [V, Q]. Now |C_{V} (Q_{1})| = q, or q^{2}. As |V : C_{V} (Q_{1})| ≤ q^{6}, we get that |V | ≤ q^{8}, a contradiction, as by 1.4 then V has to be the natural module. So we may assume q = 4 and G_{1} induces three nontrivial modules. But then again [V, R] can just involve one such module, and so |[V, R]| ≤ q^{3}. Again as before, we get |V | ≤ q^{8}.

Lemma 2.4 Let G/Z(G) = G(q) be of rank at most two and V be a an irreducible faithful 2F –module with quadratic offender A. Then G/Z(G) ≅ L_{2}(q), L_{3}(q), Sp(4, q), U_{4}(q), U_{5}(q), Ω^{−}(6, q) or G_{2}(q) and V is the natural module, or G ≅ L_{2}(q) and V is a tensor product.

Proof: If G is of rank one the assertion follows with 1.3 and 1.15. So let G be of rank two. If G/Z(G) ≅ L_{3}(q), then by 1.16 V is strong quadratic and the assertion follows with [Str].

Let G/Z(G) ≅ Sp(4, q), U_{4}(q), U_{5}(q), or Ω^{−}(6, q). Then by 2.3 A ≰ Q_{1}. If G ≅ Sp_{4}(q), we have A ≤ Q_{2} and we get the assertion with 2.3. So let G/Z(G) ≅ U_{4}(q), U_{5}(q), or Ω^{−}(6, q). By 2.2 V_{1} = C_{V} (Q_{1}) is a nontrivial G_{1}/Q_{1}–module. So by 1.8 it is the natural module or the tensor product in all cases. In particular |A : A∩Q_{1}| ≤ q^{2} and |A : A∩Q_{1}| = q if G/Z(G) ≅ U_{5}(q). Now in all cases we have by [Str] that V is not strong quadratic. So we see that |A| ≤ q^{2}. Now by 2.2 we have that |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = q^{2}, q, respectively. This shows that in case of U_{5}(q), we have |V : C_{V} (A ∩ Q_{1})| = q^{2}, and so for the root group R, we get |V : C_{V} (R)| ≤ q^{4}, which contradicts the fact that [V, R] contains a nontrivial U_{3}(q)–module. So we have G ≅ U_{4}(q). Further A ≰ Q_{2}. As V is not strong quadratic, we get that A ∩ Q_{2} = 1. As |A| > q, there are x, y ∈ A^{♯} such that [C_{Q_{2}}(x), y] contains a root element. But then V would be strong quadratic by 1.12, a contradiction

Let now G ≅ G_{2}(q). If V is strong quadratic then by [Str] V ≅ V (λ_{1}), the assertion. So we may assume that V is not strong quadratic. Now by 1.16 |A| > q. Further 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}))| ≠ 1 ≠ C_{A}(Z(Q_{2})) 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 with 1.12 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 strong quadratic, we get with [Str] that G ≅ ^{3}D_{4}(q) and V is the natural Ω^{+}(8, q^{3})–module. If A ≰ Q_{1}, then we see that |V : C_{V} (A)| ≥ q^{12}, a contradiction. So we have A ≤ Q_{1}. Now as |V : C_{V} (Q_{1})| = |Q_{1}|^{2}, we get with 1.11 |V : C_{V} (A)| = |A|^{2}. As by 1.7 V is defined over GF (q^{3}), we get that |A| > q^{4} and so A containes a root element. But then A also containes some x which is conjugate to some element in G_{1} \ Q_{1}. Then |V : C_{V} (x)| > q^{10}, a contradiction.

So we have that V is not strong quadratic. In particular by 1.16 |A| > q and then A ≰ Q_{1}. Let C_{V} (Q_{1}) be a trivial G_{1}–module and 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}, which shows |A| ≥ q^{4}. As A∩Z(Q_{1}) = 1, we get |A| = q^{4} and C_{V} (A) = C_{V} (t) for t ∈ A\Q_{1}. As |A ∩ Q_{1}| ≤ q, we see that G ≅ ^{3}D_{4}(q) and |A/A ∩ Q_{1}| = q^{3}. But a Sylow 2-subgroup of L_{2}(q^{3}) does not at quadratically on the eight–dimensional module, a contradiction.

So we have with 1.10 that C_{V} (Q_{1}) is a 2F –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, or |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A/A∩Q_{1})| = |A/A ∩ Q_{1}|^{2}. In the latter A ∩ Q_{1} is an offender too and so A ∩ Q_{1} = 1. This again shows G ≅ ^{3}D_{4}(q^{3}). But then there is exactly one such module involved, i.e. [V, G_{1}] ≤ C_{V} (Q_{1}) and so V would be strong quadratic, a contradiction. So we have the former. This then shows that [V, R] involves the eight dimensional module too, where R as before is the root group normal in G_{1}. Now for t ∈ A \ Q_{1} we get |V : C_{V} (t)| ≥ q^{14}, a contradiction.

Lemma 2.5 Let G/Z(G) = G(q) ≅ L̸_{n}(q) be classical of rank at least three and V = V (λ_{n}) be a 2F –module with quadratic offender A. Then (G, V) is one of the theorem.

Proof: By 2.3 we may assume A ≰ Q_{1}. So by 1.10 and 1.8 it is enough to prove that V (λ_{n}) is not a 2F –module for G/Z(G) ≅ Sp(12, q), Ω^{+}(14, q), Ω^{−}(12, q), U_{5}(q) and U_{8}(q). The case G/Z(G) ≅ U_{5}(q) follows from 2.4. As the spin modules for Sp(12, q) and Ω^{−}(12, q) are the same, it is enough to handle G/Z(G) ≅ Sp(12, q), Ω^{+}(14, q) and U_{8}(q). As V (λ_{n}) is not an F –module by 1.1, we get that A ∩ Q_{1} ≠ 1. Let G/Z(G) ≅ U̸_{8}(q). Then we see that A/A ∩ Q_{1} ≤ O_{2}(G_{12}/Q_{1}). Now we see |A ∩ Q_{1}| ≤ q^{3} and |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| ≥ |A/A ∩ Q_{1}|^{2}/q^{2}. Hence |V : C_{V} (A ∩ Q_{1})| ≤ q^{8}, a contradiction, as |V : C_{V} (x)| = q^{16}, or q^{32} for x ∈ Q_{1}^{♯}.

So let now G/Z(G) ≅ U_{8}(q). Now V involves twice V (λ_{3}) and once V (λ_{2}) for G_{1}/Q_{1}. This shows that |V : C_{V} (A/A ∩ Q_{1})| > |A/A ∩ Q_{1}|^{4}, as V (λ_{2}) is not a 2F –module with quadratic offender by 2.1 and V (λ_{3}) is not an F –module by 1.1. So |A ∩ Q_{1}|^{2} > |A|. On the other hand |C_{V} (Q_{1})| = q^{20} and so |V : C_{V} (Q_{1})| ≥ |Q_{1}|^{3.8}. Hence by 1.11 we get |V : C_{V} (A∩Q_{1})| ≥ |A∩Q_{1}|^{3.8}. So we get |A|^{2} ≥ |V : C_{V} (A)| = |V : C_{V} (A∩Q_{1})||C_{V} (A∩Q_{1}) : C_{C_{V} (A∩Q_{1})}(A)| ≥ |A ∩ Q_{1}|^{3.8}|A/A ∩ Q_{1}|. So |A| ≥ |A ∩ Q_{1}|^{2.8} > |A|, a contradiction.

Lemma 2.6 Let G ≅ Ω^{−}(2n, q), n > 3, and V be a an irreducible faithful 2F –module with quadratic offender A. Then V ≅ V (λ_{1}), or n = 4, 5 and V ≅ V (λ_{n}).

Proof: By [Str] and 2.5 it is enough to prove that V is strong quadratic. So assume not. Then by 2.3 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 strong quadratic by 1.12. 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 by 1.12 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 by 1.12 and so V is strong quadratic. So we may assume that C_{Q_{1}}(A) = C_{Q_{1}}(t) for all t ∈ A^{♯}. Hence |A| ≤ q^{n−2}. 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 strong quadratic, we get |[V, a]| > |Q_{2}/Q′_{2}| = q^{4n−8}, a contradiction.

Lemma 2.7 Let G ≅ Sp(2n, q), n ≤ 6, and V ≅ V (λ_{1} + λ_{n}). Then V is not a 2F –module with quadratic offender.

Proof: By [Str] we know that V is not strong quadratic. Let A be a quadratic offender. As |V : C_{V} (Q_{n})| > |Q_{n}|^{2} , we get with 1.11 that A ≰ Q_{n}. But now there is some t ∈ A\Q_{n}, with [t, Q_{n}] contains a root element. As A does not contain root elements, we see with 1.12 that V is strong quadratic if A ∩ Q_{n} ≠ 1. So we have A ∩ Q_{n} = 1 and again with 1.12 C_{Q_{n}}(A) = C_{Q_{n}}(a) for all a ∈ A^{♯}. This shows |A| ≤ q (recall that for n = 3 we have that Q_{n}/U is dual to U, U the natural submodule in Q_{n}, and so the stabilizer of a hyperplane in the natural module becomes the stabilizer of a point in the dual). But as V is not strong quadratic this contradicts 1.16.

Proposition 2.8 Let G/Z(G) = G(q) be classical and V be an irreducible faithful 2F – module with quadratic offender A. Then one of the following holds

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

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

(iii) G(q) ≅ L_{6}(q), or U_{6}(q) and V ≅ V (λ_{3}).

(iv) G(q) ≅ Sp(2n, q), n = 3, 4, 5, and V is the spin module.

(v) G(q) ≅ Ω^{±}(2n, q), n = 4, 5, or Ω^{+}(12, q) and V is the half spin module.

(vi) G(q) ≅ L_{2}(q^{2}) and V ≅ V_{1}^{σ} ⊗ V_{1}, where V_{1} is the natural module and σ the field automorphism of order two.

Proof: By 2.4 we may assume that the rank of G is at least three. By 2.6 we may assume G(q) ≅ Ω̸^{−}(2n, q). By 2.3 and symmetry for G(q) ≅ L_{n}(q) we may assume that A ≰ Q_{1}. By 2.2 we may assume that C_{V} (Q_{1}) is a 2F –module with quadratic offender. So by induction we first assume that C_{V} (Q_{1}) is the natural module. Let V be strong quadratic, we now get with [Str] that V is the natural module in case of L_{n}(q) or V ≅ V (λ_{2}) in which case the assertion follows with 2.1. So V is not strong quadratic. In particular |A∩Q_{1}| ≤ q^{2}. Let first G(q) ≅ Sp(2n, q) or Ω^{+}(2n, q). Then we get A ∩ Q_{1} = 1, or G(q) ≅ Sp(2n, q) and A/A ∩ Q_{1} induces a transvection on Q_{1}. So assume the latter. Then |A| ≤ q^{3}. Choose a ∈ (A ∩ Q_{1})^{♯}. Then ⟨(A ∩ Q_{1})^{C_{G_{1}}(a)}⟩ must not contain a root element by 1.12. So we have |A ∩ Q_{1}| ≤ q. Hence for a as above, we get |V : C_{V} (a)| ≤ q^{4}. Then there is a root element r with [r, G′_{2}] = 1 and |V : C_{V} (r)| ≤ q^{8}. But as |Q_{2}/Z(Q_{2})| ≥ q^{8} and V is not strongly quadratic, we see that |[V, r]| > q^{8}. So in both cases we have |A ∩ Q_{1}| = 1.

Let now G(q) ≅ U_{n}(q). Then by 1.12 |A ∩ Q_{1}| ≤ q. Further [A, C_{Q_{1}}(A ∩ Q_{1})] ≤ A ∩ Q_{1}. This shows that |A/A ∩ Q_{1}| ≤ q and then |A| ≤ q^{2}. Hence for a root element r we get |[V, r]| ≤ q^{8}. As n ≥ 6, we get [V, r] = C_{V} (Q_{1}), a contradiction.

So in any case we have A ∩ Q_{1} = 1. Let a, b ∈ A^{♯} with C_{Q_{1}}(a) ≠ C_{Q_{1}}(b). Then by 1.12 we may assume that there is some x ∈ Q_{1}^{♯} such that ⟨x, a⟩ acts quadratically on V. But then V is strong quadratic or a induces a transvection on Q_{1}. Hence we have that |A| ≤ q^{2} and |Q_{1} : C_{Q_{1}}(A)| = |A|. If C_{Q_{1}}(a) = C_{Q_{1}}(A) for all a ∈ A^{♯}, we also get |Q_{1} : C_{Q_{1}}(A)| = |A|. So if G(q) ≅ Sp(2n, q) or U_{n}(q), we now see that C_{V} (R) involves exactly one nontrivial irreducible module, C_{V} (Q_{1}), where R is the root group normal in G_{1}. But then we see that |C_{V} (R) : C_{V} (Q_{1})| ≤ q, q^{2}, respectively. In both cases we see that C_{V} (Q_{n} ∩ Q_{1}) ≤ [V, R] = C_{V} (Q_{1}) and so G_{1} ∩ G_{n} acts irreducibly on C_{V} (Q_{n}), which is not possible, as Q_{n} ≠ Q_{1} ∩ Q_{n}.

Let G(q) = Ω^{+}(2n, q), then |A| ≤ q^{n−1}. The same argument as before also shows that even A ∩ Q_{2} = 1 and further C_{Q_{2}}(a) = C_{Q_{2}}(A) for a ∈ A^{♯}. This now shows |A| ≤ q^{n−2}. So for a ∈ A^{♯} we have |V : C_{V} (a)| ≤ q^{2n−4}. This shows that for a root element r we have |V : C_{V} (r)| ≤ q^{4n−8}. We may assume r ∈ Z(Q_{2}). We have |Q_{2}| = q^{1+(4n−8)}. As V ≅ V̸ (λ_{2}) by 2.1, we get that C_{V} (Q_{2}) is a nontrivial module on which O^{2}(G′_{1} ∩ G_{2}) acts trivially. Hence |C_{V} (Q_{2})| ≥ q^{2} by 1.8. Further by 1.8 we also have C_{V} (Q_{2}) ≤ [V, r]. Now we see that [V, r] has to involve at least two natural modules for Ω^{+}(2n − 4, q). Hence |[V, r]| ≥ q^{2}q^{2(2n−4)} = q^{4n−6}, a contradiction.

So we are left with G(q) ≅ L_{n}(q). Again V is not strong quadratic. So A ∩ Q_{n−1} = 1. Hence by induction, we have C_{V} (Q_{n−1}) is the trivial, the natural module or then V (λ_{n−3})– module. This shows that we have V ≅ V (λ_{n−1}), V (λ_{2}), or V (λ_{1} + λ_{n−1}). But the latter is out by 1.13 and 1.16.

So we may assume that C_{V} (Q_{1}) is not natural. Let first this module correspond to λ_{n}. Then G(q) ≅ Sp(2n, q), Ω^{+}(2n, q) , or U_{6}(q). If V is strong quadratic the assertion follows with 2.5. So we may assume that V is not strong quadratic. As before we see that A ∩ Q_{1} = 1. If A ≤ Q_{n}, we see that A cannot be an F –module offender on C_{V} (Q_{1}). If there is a root group R normal in G_{1}, then by 1.8 we have C_{V} (Q_{1}) ≤ [V, R] and so also V/C_{V} (R) involves a module isomorphic to C_{V} (Q_{1}). Hence there is no root group normal in G_{1}, which shows that G(q) ≅ Ω^{+}(2n, q). Now as above, we get that |A| ≤ q^{n−2} and |[V, r]| ≤ q^{4n−8}. As C_{V} (Q_{2}) involves the half spin module for Ω^{+}(2n − 4, q) and the parabolic in G_{2} not in G_{1} acts nontrivially on C_{V} (Q_{2}), otherwise V ≅ V (λ_{n}), we see that two half spin modules are involved. Now C_{V} (Q_{2}) ≅ Q̸_{2}/Z(Q_{2}) and so [V, Z(Q_{2})] has to involve another nontrivial Ω^{+}(2n − 4, q)–module. This shows |[V, r]| ≥ q^{2^{n−2}+2n−4}. Hence we get n = 4. But in this case we have symmetry and might assume that A ≰ Q_{n} too.

So we have A ≰ Q_{n}. If 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 a 2F –module too. Then we get V ≅ V (λ_{1} + λ_{n}). Now 1.13 and 1.16 yield G(q) ≅ Sp(2n, q) or Ω^{+}(2n, q). If G(q) ≅ Sp(2n, q) then by induction we have n ≤ 6 and we get a contradiction with 2.7. So let G(q) ≅ 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 this is not a 2F –module with quadratic offender, we get A ≤ Q_{n−1}. Now again we see that |A| ≤ q^{n−2} and get the same contradiction as before.

So by induction the only case left is G(q) ≅ L_{n}(q) and C_{V} (Q_{1}) corresponds to λ_{n−2} or λ_{3}, or G(q) ≅ L_{7}(q) or U_{8}(q) and C_{V} (Q_{1}) corresponds to λ_{4}. Assume that G(q) ≅ L_{n}(q) and C_{V} (Q_{1}) corresponds to λ_{n−2}. If V is strong quadratic we get with [Str] that V ≅ V (λ_{n−2}). So let V not be strong quadratic. Then C_{V} (Q_{n−1}) is not the natural module. In particular it is not a 2F –module with quadratic offender. Hence by 1.10 A ≤ Q_{n−1} which contradicts 1.16 and the fact that V is not strong quadratic.

So let C_{V} (Q_{1}) correspond to λ_{3}, or n = 7 and C_{V} (Q_{1}) corresponds to λ_{4}. Then either C_{V} (Q_{n−1}) is a 2F –module or V is strong quadratic. In both cases we get V ≅ V (λ_{3}), or n = 7 and V ≅ V (λ_{4}), which is just the dual of V (λ_{3}). Hence we may assume V = V (λ_{3}) in all cases. Further we may assume that n > 5, otherwise λ_{3} = λ_{n−2} or λ_{n−1}.

Let first n = 7. Now |V | = q^{35}. Again A ≰ Q_{1} and A ≰ Q_{6}. Now for G_{1} we get a series of submodules 1 < V_{1} < V, where V_{1} is the V (λ_{3})–module and V/V_{1} is the V (λ_{2})–module for G_{1}/Q_{1}. By 1.17 we see that |V/V_{1} : C_{V/V_{1}}(A/A ∩ Q_{1})| ≥ |A/A ∩ Q_{1}|. As V (λ_{3}) is not an F –module for L_{6}(q) by 1.1, we see that A∩Q_{1} ≠ 1. So we may assume that A∩Z(Q_{1,6}) ≠ 1. As |V : C_{V} (Q_{1,6})| = q^{25} > q^{22} = |Q_{1,6}|^{2}, we see that A ∩ Q_{1,6} must not be an offender. But by quadratic action we have [V, Z(Q_{1,6}), A] = 1, and so A ≤ Q_{1,6}, a contradiction.

Let now n > 7, then C_{V} (Q_{n−1}) by induction is not a 2F –module with quadratic offender. So A ≤ Q_{n−1}. But |V : C_{V} (Q_{n−1})| = q^{\binom{n−1}{2}} > q^{2(n−1)}, which contradicts 1.11.

So we are left with G(q) ≅ U_{8}(q) and C_{V} (Q_{1}) is the V (λ_{3})–module for G_{1}/Q_{1}. As this module is involved at least twice in V and V (λ_{3}) is not an F –module for U_{6}(q) by 1.1, we see that A ∩ Q_{1} ≠ 1. Let R be the root group normal in G_{1}. Then for r ∈ R^{♯} we get |V : C_{V} (r)| ≤ |A|^{8}. As [V, r] ≥ C_{V} (Q_{1}), we see |[V, r]| ≥ q^{20}. Hence |A| > q^{2}. So we see that either |A ∩ Q_{1}| > q or A contains some a with |Q_{1}/Z(Q_{1}) : C_{Q_{1}/Z(Q_{1})}(a)| ≥ q^{4}. So by 1.12 we now see that V is strong quadratic and then V ≅ V (λ_{4}) by [Str]. Now we get a contradiction with 2.5.

Proposition 2.9 Let G/Z(G) = E_{n}(q), n = 6, 7, 8, and V be an irreducible faithful 2F – module with quadratic offender A. Then n = 6 and V ≅ V (λ_{1}) or V (λ_{6}), or n = 7 and V ≅ V (λ_{7}).

Proof: Suppose false. Let first G/Z(G) = E_{6}(q). If A ≤ Q_{4}, then V is strong quadratic and by [Str] we have V ≅ V (λ_{1}) or V (λ_{6}). So let A ≰ Q_{4}. So we may assume that V is not strong quadratic. Hence we have with 1.12 that A ∩ Q_{4} = 1 and C_{Q_{4}}(a) = C_{Q_{4}}(A) for all a ∈ A^{♯}. If C_{V} (Q_{4}) is nontrivial, then it is a 2F –module by 1.11. If A ≤ Q_{1} or A ≤ Q_{6}, then we now get |A| ≤ q^{5}. Now elements in different root groups in A have different centralizer in Q_{4}, and so |A| ≤ q, which contradicts 1.16. Hence we have A ≰ Q_{1} and A ≰ Q_{6}. By symmetry we may assume that C_{V} (Q_{1}) is nontrivial. So it is the natural module or the half spin module. If we have the half spin module it corresponds to λ_{4} or λ_{2}. So we get that C_{V} (Q_{6}) is the V (λ_{2}) or V (λ_{1} +λ_{4})–module, which both are not 2F –modules with quadratic offender, contradicting 1.10. So C_{V} (Q_{1}) is the natural module. This then tells us that V ≅ V (λ_{6}).

So we may assume that C_{V} (Q_{4}) is a trivial module. Now [V, Z(Q_{4})] involves the 20dimensional L_{6}(q)–module, and so V involves this module twice. But it is not an F –module by 1.1, contradicting A ∩ Q_{4} = 1.

Let G/Z(G) ≅ E_{7}(q). If V is strong quadratic then by [Str] V ≅ V (λ_{7}). So we may assume that V is not strong quadratic. Now if A ≤ Q_{1}, then |A| ≤ q, contradicting 1.16. So we have A ≰ Q_{1} and |A ∩ Q_{1}| ≤ q. As there are no elements in Ω^{+}(12, q) which centralize codimension two in the half spin modules, we even get A ∩ Q_{1} = 1. As the half spin module is not an F –module for Ω^{+}(12, q), we see that this module is not involved in [V, Z(Q_{1})]. So we have C_{V} (Q_{1}) is an F –module and then it has to be the natural one. Now as A has to be an F –offender on [V, Z(Q_{1})], we see that A ≰ Q_{7}. As V is not strong quadratic we get with 1.10 that V ≅ V (λ_{1} + λ_{7}). This shows with 1.10 A ≤ Q_{4}. As Z(Q_{4}) is the natural module and there is no quadratic group containing a root subgroup properly, we see that |A| ≤ q^{7}, otherwise there is b ∈ Q_{4} with [b, A] ≠ 1 and C_{A}(b) ≠ 1. So there is some root element r ∈ [b, A] and then by 1.12 ⟨r, C_{A}(b)⟩ acts quadratically on V. Now as [V, Z(Q_{1})] has to contain another nontrivial module besides C_{V} (Q_{1}), we see that |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| ≤ q^{3}. This now shows |A| ≤ q^{3}, but then we have |V : C_{V} (A)| ≤ q^{6}, which contradicts the fact the V involves at least four nontrivial G_{1}/Q_{1}–modules and G_{1} cannot induce transvections on some module.

Let finally G(q) ≅ E_{8}(q). As there is no strong quadratic module, we see that A∩Q_{8} = 1. As there are at least two nontrivial E_{7}(q)–modules involved in V and E_{7}(q) does not possess an F –module, we get a contradiction.

Lemma 2.10 If V is an irreducible faithful 2F –module with quadratic offender A for the group G ≅ F_{4}(q), then V ≅ V (λ_{1}) or V (λ_{4}).

Proof: By [Str] and 2.1 we may assume that V is not strong quadratic. Let A ≤ Q_{1}, the we get A ≤ Z(Q_{1}). Now if A ≤ Q_{4} too, then we get A ≤ Z(Q_{1}) ∩ Z(Q_{4}), so |A| ≤ q^{2}. As A cannot contain any root element, we even see |A| ≤ q, which contradicts 1.16. So we may assume A ≰ Q_{1}. Let A ∩ Z(Q_{1}) ≠ 1. Then there is x ∈ Q_{1} with [x, A] ≰ Z(Q_{1}). So by 1.12 there is a quadratic group B in Q_{1} with B ≰ Z(Q_{1}) but B ∩ Z(Q_{1}) ≠ 1. Hence again by 1.12 we get a subgroup acting strongly quadratic. So we have A ∩ Z(Q_{1}) = 1. This first with 1.12 shows |A ∩ Q_{1}| ≤ q. If A ∩ Q_{1} ≠ 1, then we choose x ∈ Z(Q_{1}) and proceed as before. So we have A ∩ Q_{1} = 1. Now further we see that C_{Z(Q_{1})}(a) = C_{Z(Q_{1})}(A) for all a ∈ A^{♯}. Hence |A| ≤ q^{3}. As |A| > q by 1.16, we see that A ≰ Q_{4}. But now we also get A ∩ Q_{4} = 1. So we get that both [V, Q′_{i}], i = 1, 4, are F –modules. This shows that we may assume that C_{V} (Q_{1}) is the natural module and |A| = q^{3}. Now [V, Q′_{1}]/C_{V} (Q_{1}) is a trivial module and so C_{Q_{1}}([V, Q′_{1}]) covers Q_{1}/Z(Q_{1}) which shows that V is strong quadratic.

Lemma 2.11 The group G/Z(G) ≅ ^{2}E_{6}(q) does not possess a 2F –module with quadratic offender.

Proof: Suppose false. Let A be a quadratic offender. Assume first that V is not strong quadratic. Let A ≤ Q_{1}. By 1.16, we have |A| > q. Now we see with 1.12 that V is strong quadratic, a contradiction. So we have A ≰ Q_{1}. Further |A ∩ Q_{1}| ≤ q. Let |A/A ∩ Q_{1}| ≤ q. Then A ∩ Q_{1} ≠ 1. Let 1 ≠ a ∈ A ∩ Q_{1}, then |V : C_{V} (a)| ≤ q^{4}, so we get |V : C_{V} (Z(Q_{1}))| ≤ q^{8}, and so V is strong quadratic, a contradiction. So we have |A/A ∩ Q_{1}| > q. As |[Q_{1}/Z(Q_{1}), a]| > q^{2} for all a ∈ A \ A ∩ Q_{1}, we get that A ∩ Q_{1} = 1.

Now 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 C_{V} (Q_{1}) is a nontrivial module not isomorphic to Q_{1}/Z(Q_{1}). Then by 1.10 and 1.8 we get that C_{V} (Q_{1}) is the natural U_{6}(q)–module. As V is not strong quadratic we have that [V, Z(Q_{1})] involves a further nontrivial module. As Q_{1}/Z(Q_{1}) is not an F –module for U_{6}(q), we get a contradiction in the former. In the latter by 1.1 we may assume that V involves four natural modules. So we get for the natural module V_{1} that |V_{1} : C_{V_{1}}(A)| ≤ q^{4}. But then we see that |A| ≤ q^{4} and so |V : C_{V} (A)| ≤ q^{8}. Now even |V_{1} : C_{V_{1}}(A)| ≤ q^{2} and so |A| ≤ q. But then |V : C_{V} (A)| ≤ q^{2}, a contradiction.

So we have that V is strong quadratic and then by [Str] we have V ≅ V (λ_{6}). In particular |V | = q^{54} and by 1.7 V is defined over GF (q^{2}). As a G_{1}–module we have the following series 1 < V_{1} < V_{2} < V, where V_{1} and V/V_{2} are natural modules and V_{2}/V_{1} is the V (λ_{2})–module for U_{6}(q).

If A ≤ Q_{1}, then we will just consider offender B ∈ P_{2}(Q_{1}, V) with |C_{V} (B)||B| maximal. If Z(Q_{1}) ∩ B = 1. Then choose x ∈ Q_{1} \ C_{Q_{1}}(B). By 1.11 ⟨B, B^{x}⟩ is also an offender. So we may assume that Z(Q_{1}) ∩ B ≠ 1. As |B| ≤ q^{11}, we get |C_{V} (Z(Q_{1})) : C_{C_{V} (Z(Q_{1}))}(B/B ∩ Z(Q_{1}))| ≤ |B/B ∩ Z(Q_{1})|. So by 1.11 we get q^{30} = |C_{V} (Z(Q_{1})) : C_{V} (Q_{1})| ≤ |Q_{1}/Z(Q_{1})| = q^{20}, a contradiction.

So we have A ≰ Q_{1}. In particular by quadratic action A ∩ Z(Q_{1}) = 1, and so |A| ≤ q^{12}. Then |V : C_{V} (A)| ≤ q^{24}. Suppose first that |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| ≥ q^{6}. As by 2.1 V_{2}/V_{1} is not a 2F –module with quadratic offender, we get that |A|^{2} > q^{12}|A/A ∩ Q_{1}|^{2}. This shows |A ∩ Q_{1}| > q^{6} and further |V : C_{V} (A ∩ Q_{1})| ≤ q^{18}. If |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = q^{4}, we get |A/A ∩ Q_{1}| ≤ q^{4}. Hence |A ∩ Q_{1}| > q^{4}. Let finally |C_{V} (Q_{1}) : C_{C_{V} (Q_{1})}(A)| = q^{2}, then |A/A ∩ Q_{1}| ≤ q. Then we see that for a ∈ A \ A ∩ Q_{1} we have |V : C_{V} (a)| ≥ q^{12}, as in the V (λ_{2})–module for U_{6}(q) we get an index q^{8}. Hence again |A ∩ Q_{1}| > q^{4}. So in both cases there is some x with |A ∩ Q_{1}| = q^{4+x}, and |V : C_{V} (A ∩ Q_{1})| ≤ q^{12+2x}, q^{8+2x}, respectively. Now set A_{1} = A ∩ Q_{1}. Then we have |V : C_{V} (A_{1})| ≤ q^{a} and |A_{1}| ≥ q^{b}, where b = 6, 4 + x, 4 + x and a = 18, 12 + 2x, 8 + 2x. Now choose y ∈ Q_{1} with [A_{1}, y] ≠ 1. Set A_{2} = A_{1} ∩ A_{1}^{y} and B_{1} = ⟨A_{1}, A_{1}^{y}⟩. Now we get

|V : C_{V} (B_{1})| ≤ q^{2a}/|V : C_{V} (A_{2})|

By 1.11 we have that |A_{1}|^{1.5} ≤ |C_{V} (Z(Q_{1})) : C_{C_{V} (Z(Q_{1}))}(A_{1})| ≤ |V : C_{V} (B_{1})|/q^{12}. This shows

q^{2a−(1.5b+12)} ≥ |V : C_{V} (A_{2})| = q^{t}

Here we get t ≤ 15, 8 + 2.5x, 2.5x − 2. Now we play the same game with A_{2}. Choose z ∈ Q_{1} with [A_{2}, z] ≠ 1. Set B_{2} = ⟨A_{2}, A_{2}^{z}⟩ and A_{3} = A_{2} ∩ A_{2}^{z}. Then we get

|V : C_{V} (B_{2})| ≤ q^{2t}/|V : C_{V} (A_{3})|

and

|A_{2}|^{1.5} ≤ |C_{V} (Z(Q_{1})) : C_{C_{V} (Z(Q_{1}))}(A_{2})| ≤ |V : C_{V} (B_{2})|/q^{12}

As |A_{1} : A_{2}| ≤ q, we have |A_{2}| ≥ q^{b−1}. Now

q^{2t−(1.5b+10.5)} ≥ |V : C_{V} (A_{3})| = q^{s}

where s = 10.5, 3.5x − 0.5, 3.5x − 20.5. As A_{3} ≠ 1, we see that s ≥ 12. So we have the second case with x > 3. Now we may apply the procedure once more and get some A_{4} ≠ 1 with |V : C_{V} (A_{4})| = q^{r}, where r ≤ 2s − (1.5b + 9) = 5.5x − 16. Once more we get A_{5} with |V : C_{V} (A_{5})| ≤ q^{2r−(1.5b+7.5)} ≤ q^{9.5x−45.5}. As |A_{5}| ≥ q^{4}, we can proceed twice and get A_{7} with |V : C_{V} (A_{7})| ≤ q^{33.5x−217.5}. But then x > 6, a contradiction, as q^{10} ≥ |A∩Q_{1}| = q^{4+x}.

So we have shown

Proposition 2.12 Let G/Z(G) = G(q) be of Lie type, V be an irreducible faithful 2F – module with quadratic offending group. Then one of the following holds

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

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

(iii) G(q) ≅ L_{6}(q) or U_{6}(q) and V ≅ V (λ_{3}).

(iv) G(q) ≅ Sp(2n, q), n = 3, 4, 5, and V is the spin module.

(v) G(q) ≅ Ω^{±}(2n, q), n = 4, 5, or Ω^{+}(12, q) and V is the half spin module.

(vi) G(q) ≅ L_{2}(q^{2}) and V ≅ V_{1}^{σ} ⊗ V_{1}, where V_{1} is the natural module and σ the field automorphism of order two.

(vii) G(q) ≅ E_{6}(q) and V ≅ V (λ_{1}) or V (λ_{6}).

(viii) G(q) ≅ E_{7}(q) and V ≅ V (λ_{7}).

(ix) G(q) ≅ F_{4}(q) and V ≅ V (λ_{1}) or V (λ_{4}).

3 2F -modules for alternating, sporadic and Lie type groups in odd characteristic

Throughout this chapter we will assume that 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 a 2F -module and A ∈ P_{2}(G, V) an offending subgroup which acts quadratically.

Lemma 3.1 Let G = A_{n} or G = 3A_{m}, m = 6, 7 , and V be an irreducible faithful 2F – module with quadratic offender A, 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) G ≅ 3A_{6} and |V | = 2^{6}.

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

(v) G ≅ A_{9} and V is the eight dimensional spin module.

Proof: Let first n ≥ 9. Suppose |A| = 2. Then |V : C_{V} (A)| ≤ 4. Now we see that there is some involution b ≠ a with [[V, a], b] = 1. So we have a quadratic fours group. The same applies for |A| ≥ 4. So we assume that we have a quadratic fours group. By [MeiStr1] we have that V is either the natural module or the spin module. So assume V to be the spin module. Let a ∈ A^{♯}, where we set A = ⟨a, b⟩ in the first case. 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 of order 2^{m} by A_{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 [MeiStr1, (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)⟩. We will handle this case later.

So let [A, O^{2′}(C_{C_{G}(a)}(K′))] = 1. If [A, K′] ≠ 1, then [K′, [V, a]] = 1. If k ≥ 4, then K′ contains some x ∼ (12)(34). This again contradicts [MeiStr1, (4.3)]. Let k ≤ 3. As [A, O^{2′}(C_{C_{G}(a)}(K′))] = 1 and m > 2 this is impossible. 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^{♯}. We may assume a = (12)(34) ∈ A. Now A ≥ ⟨a, b⟩, b = (13)(24), or (12)(56). Let [b, K′] ≠ 1. Then b = (12)(56) and so K′ contains no involutions by [MeiStr1, (4.3)]. This shows k ≤ 3 and so A ≤ A_{7}, a contradiction.

So we have seen that |A| ≤ 8 and so |V : C_{V} (a)| ≤ 2^{6} for all a ∈ A^{♯}. Furthermore as 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^{12}. Let n > 9, then U = ⟨ω_{1}, ω_{2}⟩, ω_{1} = (1, 2, 3, 4, 5), ω_{2} = (6, 7, 8, 9, 10) act on V. As ⟨ω_{1}⟩, ⟨ω_{2}⟩ act fixed point freely and all other subgroups of order five in U are conjugate in G, we get |V | ≥ 2^{16}, a contradiction. So we have n = 9.

The cases of n = 5, 6, 8 were done before, as G is a group of Lie type in characteristic two. So let n = 7. If |A| = 4, then by [MeiStr1] |V | = 16, which is (ii). So assume |A| = 2 and there is no quadratic fours group. Now we may embedd A into a subgroup U ≅ Σ_{5}. Further V involves exactly one nontrivial U–module. So there is a submodule V_{1} which is the permutation module or an extension of a trivial module by the L_{2}(4)–module, with V_{1} ≥ [V, A]. But now we see that there is a fours group centralizing [V, A].

Let now G ≅ 3A_{6}. As G is generated by four involutions or three fours groups, we get |V | ≤ 2^{12}. Now the assertion follows with [JLPW, page 4].

Let G ≅ 3A_{7}. Then we have a subgroup U ≅ 3A_{6} and so again |V | ≤ 2^{12}. This time we get with [JLPW, page 13] |V | = 2^{12}. Further we get |A| = 4. And so |V : C_{V} (A)| = 16. This shows that the two 3A_{6}–modules involved in V have to be isomorphic. Let V_{1} be a 3A_{6}–submodule. Then |C_{V} (A)| = 16. There is another conjugay class of fours groups in A_{7}. If C is in this class then we get |C_{V_{1}}(C)| = 4. Let B be the fours group in A_{7} which centralizes an element of order three in A_{7}. Then we have that U ≅ 3^{1+2} acts on C_{V} (B). This shows that |C_{V} (B)| = 2^{6}. In particular |C_{V_{1}}(B)| = 16. So B ∼ A, a contradiction.

Lemma 3.2 Let G/Z(G) be sporadic and V be an irreducible faithful 2F –module with quadratic offender A. Then G ≅ 3M_{22} and V is the 12-dimensional module comming from the embedding into SU_{6}(2).

Proof: Let first |A| = 2. As any sporadic group but M_{11} contains an elementary abelian subgroup of order eight, we get a quadratic fours group. So we may assume that G ≅ M_{11} or there is a quadratic fours group. So assume the latter. Then by [MeiStr1] we just have to handle the cases E(G) ≅ M_{12}, M_{24}, 3Suz, 3M_{22}, J_{2}, Co_{1}, Co_{2}. Furthermore V is known. In all cases besides G ≅ 3M_{22} we have |A| ≤ 4.

Let G ≅ M_{12}. There is a subgroup U in G , U ≅ M_{10}. This shows |[V, a]| ≥ 16 for all a ∈ A^{♯}. So we have |A| = 4 and C_{V} (a) = C_{V} (A) for all a ∈ A^{♯}. But the centralizer of a is a maximal subgroup of G, a contradiction.

Let G ≅ M_{24}. If t is a non 2-central involution, then it inverts an element of order 11 in G, so |[V, t]| ≥ 32. Hence A just consists of 2–central involutions and so we have that |A| = 4 and C_{V} (a) = C_{V} (A) for all a ∈ A^{♯}. Now we get that U ≅ E_{16}A_{8} acts on C_{V} (A). Hence [C_{V} (A), O_{2}(U)] = 1 and so V would be an F –module, contradicting 1.1.

Let G ≅ 3Suz. If A contains a 2-central involution a, we see that [[a, V], C_{G}(a)′] = 1. Now any fours group containing a is quadratic, which contradicts [MeiStr1]. So we have A just consists of non 2–central involutions. But then U ≅ SL_{3}(4) would centralize [V, a], which contradicts the fact that an element of order three in Z(U) acts fixed point freely on V.

Let G ≅ Co_{1} or Co_{2}. As 3Suz embedds into Co_{1} and the modules are the same and up to conjugation the quadratic subgroups are the same, we get a contradiction in that case as before. So assume that G ≅ Co_{2}. Then A consists of 2–central involutions and for a ∈ A^{♯}, we have [[V, a], C_{G}(a)] = 1. But now any fours group containig a would be quadratic, contradicting [MeiStr1].

Let G ≅ J_{2}. Then V is the G_{2}(4)–module. Hence for the 2-central involutions z we have |[V, z]| = 16, while for the non 2–central ones r we have |[V, r]| = 64. So |A| = 16 and C_{V} (A) = C_{V} (a). But C_{G}(a) is a maximal subgroup in G, a contradiction.

Let finally G ≅ 3M_{22}. Then the assertion follows with [MeiStr1].

So we are left with G ≅ M_{11} and |A| = 2. Let A ≤ T, T a Sylow 2–subgroup of G. Then T acts transitively on an elementary abelian subgroup U of order 9. This shows |[V, U]| ≥ 2^{8}. As A inverts U, we get |[V, A]| ≥ 2^{4}, a contradiction.

Lemma 3.3 Let V be an irreducible faithful 2F –module with quadratic offender A. If G is some covering group of a group of Lie type in odd characteristic then it is a group of Lie type in even characteristic too, or G ≅ 3 · U_{4}(3) and V is the 12–dimensional module, comming from the embedding into SU_{6}(2).

Proof: Suppose false. Assume first that we have a quadratic fours group. So by [MeiStr2] we get G ≅ 3 · U_{4}(3) and |V | = 2^{12}.

So we may assume |A| = 2 and there are no quadratic fours groups. In particular for A ≤ T, T a Sylow 2–subgroup, we have that C_{T} (A) has a cyclic subgroup U of index at most two. Now C_{T} (U) = U, and so by [Go, 5.4.8] we have that T is dihedral or semidihedral. Hence we have that G/Z(G) ≅ L_{2}(q), L_{3}(q), or U_{3}(q), q odd.

Let G ≅ U_{3}(q) or L_{3}(q). Then A acts on a special group U of order q^{3}, centralizing the center and inverting U/Z(U). As [[V, Z(U)], A] ≠ 1, we see that q = 3. So we have G ≅ L_{3}(3), as U_{3}(3) ≅ G_{2}(2)′. But now there is also an involution t ∈ G which inverts Z(U). As all involutions are conjugate, we get that |[V, Z(U)]| = 16 and so U ≤ GL(4, 2), a contradiction.

So it remains G ≅ L_{2}(q). Then G is generated by three conjugates of A. Hence we have |V | ≤ 2^{6}, and so G ≤ GL(6, 2). Now it is easy to see that G ≅ L_{2}(5), L_{2}(7), or L_{2}(9), which all are also Lie groups in characteristic two.

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