1 Introduction

All our action are from the right. To save parenthesis we write abc for (ab)c, ab.cd for (ab)(cd), ab.cde.fg for ((ab)(c(de)))(ef) and so on.

2 Cubic and Nearly Quadratic Action

In this section A is a group, 𝔽 is a field and V an 𝔽A-module.

Definition 2.1 [bdefinition cubic] Let V be an 𝔽A-module. Then V is

(a) [a] quadratic 𝔽A-module if [V, A, A] = 0,

(b) [b] Q_{V} (A) is the sum of all quadratic 𝔽A-submodules of V (and so the largest quadratic 𝔽Asubmodule of V.

(c) [b] cubic 𝔽A-module if [V, A, A, A] = 0,

(d) [c] nearly quadratic 𝔽A-module if V is a cubic 𝔽A-module such that

[V, A] + C_{V} (A) = [v𝔽, A] + C_{V} (A) for every v ∈ V \ [V, A] + C_{V} (A).

In the corresponding cases we also say that A acts quadratically, cubically and nearly quadratically on V.

Definition 2.2 [def:imp] A system of imprimitivity for A in V is a set Δ of 𝔽-subspaces of V such that

(i) [1] |Δ| > 1 and Δ^{A} = Δ, and

(ii) [2] V = ⊕_{W ∈Δ}W.

Definition 2.3 [def:semiliner] Let 𝕂 be a field extension of 𝔽 and σ : A → Aut(𝕂) a homomorphism. Then V is a semi-linear 𝕂A-module (with respect to ϕ) provided that vka = v.kσ.a for every k ∈ K, a ∈ A and v ∈ V. Set A_{𝕂} := ker σ and 𝕂_{A} := C_{𝕂}(Aσ).

Lemma 2.4 [quadratic implies nearly] Let V be a quadratic 𝔽A-module. Then V is a nearly quadratic 𝔽A-module.

Proof: Since A is quadratic, [V, A] ≤ C_{V} (A) ≤ [v𝔽, A] + C_{V} (A) for all v ∈ V. □

Lemma 2.5 [trivial property] Let V be a nearly quadratic 𝔽A-module and W be an 𝔽A-subspace of V. Then

(a) [a] Either W ≤ [V, A] + C_{V} (A) or [V, A] ≤ [W, A] + C_{V} (A).

(b) [b] Either Q_{V} (A) = V or Q_{V} (A) = [V, A] + C_{V} (A).

(c) [c] W and V/W are nearly quadratic 𝔽A-modules.

(d) [d] A is quadratic on W or on V/W.

Proof: (a) If W ≰ [V, A] + C_{V} (A) the definition of nearly quadratic implies [V, A] ≤ [W, A] + C_{V} (A).

(b) Since A is cubic, [V, A] + C_{V} (A) ≤ Q_{V} (A). Since A is quadratic on Q_{V} (A) [Q_{V} (A), A] ≤ C_{V} (A). By (a) either [V, A] + C_{V} (A) ≤ Q_{V} (A) or [V, A] ≤ [Q_{V} (A), A] + C_{V} (A). In the first case [V, A] + C_{V} (A) = Q_{V} (A). In the second case [V, A] ≤ C_{V} (A) and so A acts quadratically on V and V = Q_{V} (A).

(c) We first show that \overline{V} := V/W is a nearly quadratic 𝔽A-module. Let v ∈ V with \overline{v} ∉ [\overline{V} , A] + C\overline{V} (A). Since \overline{[V, A] + C_{V} (A)} ≤ [\overline{V} , A] + C\overline{V} (A) we get that v ∉ [V, A] + C_{V} (A). Thus [V, A] ≤ [v𝔽, A] + C_{V} (A) and [\overline{V} , A] ≤ [\overline{v}𝔽, A] + C\overline{V} (A). Hence \overline{V} is nearly quadratic for A.

To show that W is nearly quadratic we may assume by ?? that A is not quadratic on W. Then by (a) [V, A] ≤ [W, A] + C_{V} (A). Thus Q_{V} (A) = [V, A] + C_{V} (A) = [W, A] + C_{V} (A) and so Q_{W} (A) = W ∩ Q_{V} (A) = [W, A] + C_{W} (A). hence if w ∈ W with w ∉ [W, A] + C_{W} (A), then w Q̸_{V} (A) = [V, A] + V_{V} (A0 and so [V, A] ≤ [𝔽w, A] + C_{V} (A) and so [W, A] ≤ [V, A] ∩ W ≤ [w𝔽, A] + C_{W} (A). Thus W is nearly quadratic.

(d) Suppose A is not quadratic on V. Then W ≰ Q_{V} (A) = [V, A] + C_{V} (A) and so by (a) [V, A] ≤ [W, A] + C_{V} (A). Hence [V, A, A] = [W, A, A] ≤ W. So A is quadratic on V/W. □

Lemma 2.6 [belementary cubic prop] Let V be a cubic 𝔽A-module and put A_{0} = C_{A}(Q_{V} (A)) Then the following hold:

(a) [b] A_{0} acts quadratically on V.

(b) [c] For z ∈ ℤ, a ∈ A and v ∈ V with [v, a, a] = 0

z[v, a] = [v, a^{z}] = [zv, a].

(c) [a] If A acts quadratically on V, then A/C_{A}(V) is an elementary abelian char F -group. (Here an elementary abelian m-group is an abelian group all of its non-trivial elements have order m, if m is a prime, and infinite order if m = 0.)

(d) [d] A_{0} ⊴ A, and A/A_{0} and A_{0}/C_{A}(V) are elementary abelian char F -groups.

(e) [e] If char F = 0, then all non-trivial elements in A have infinite order. If char 𝔽 is a prime, then A is a char F -group.

Proof: (a) Since A is cubic, [V, A_{0}] ≤ [V, A] ≤ Q_{V} (A) ≤ C_{V} (A_{0}).

(b): For z = 0 (b) is obvious, and clearly z[v, a] = [zv, a] for every z ∈ ℤ. Moreover, the property [v, a, a] = 0 and the Three Subgroups Lemma give

[v, a^{−1}] = −[v, a] and [v, a^{z}]^{a} = [v, a^{z}].

Now an elementary induction shows (b) for z ≥ 0. Then also the case z < 0 follows using [v, a^{−1}] = −[v, a].

(c) Since [V, A, A] = 0 = [A, V, A] the Three Subgroup Lemma gives [A, A, V] = 0. So A/C_{A}(V) is abelian. Now let a ∈ A and v ∈ V with [v, a] ≠ 0. Let n be a positive integer and put p = char F. By (b) [v, a^{i}] = i[v, a]. If p > 0 we conclude that [v, a^{p}] = 0 and if p = 0, then [v, a^{i}] ≠ 0. So aC_{A}(V) has order p if p > 0 and infinite order if p = 0.

(d): This follows from (c), since A acts quadratically on Q_{V} (A) and A_{0} acts quadratically on V.

(e) follos immediatley from (d). □

Lemma 2.7 [binherit2] Let V be a nearly quadratic 𝔽A-module and Y and X be 𝔽A-submodules of V such that V = X ⊕ Y. Then either V is a quadratic 𝔽A-module, or at least one of the submodules X and Y is centralized by A.

Proof: We may assume that V is not a quadratic 𝔽A-module. Since V = X ⊕ Y, at least one of the summands, say X, is not a quadratic 𝔽A-module. Then by 2.5

[V, A] + C_{V} (A) = [X, A] + C_{V} (A) = Q_{V} (A).

In particular [Y, A, A] ≤ Y ∩ [X, A] = 0, so Y ≤ Q_{V} (A) = [X, A] + C_{V} (A). Now the same argument gives gives [Y, A] ≤ Y ∩ [X, A] = 0. □

Lemma 2.8 [bimprimitivity1] Suppose that V be an 𝔽A-module and Δ is a system of imprimitivity for A in V. Let Δ_{1} be an orbit for A on Δ and Δ_{0} ⊆ Δ_{1}. Then each following conditions immplies that Δ_{0} = Δ_{1}

1. [a] ⨁Δ_{0} ∩ C_{V} (A) ≠ 0.

2. [b] A is cubic and ⨁Δ_{0} ∩ [V, A, A] ≠ 0.

3. [c] A is quadratic and ⨁Δ_{0} ∩ [V, A] ≠ 0.

Proof: Put U = ⨁ Δ_{0}. Cubic action of A gives [V, A, A] ≤ C_{V} (A), so (2) implies (1). Similarly (3) implies (1). Hence, we may assume that C_{U}(A) ≠ 0. For X ∈ Δ let π_{X} be the projection of V onto X. Set

Δ_{2} := {X ∈ Δ_{1} | C_{U}(A)π_{X} ≠ 0}.

Then Δ_{2} ⊆ Δ_{0}, Δ_{1} ≠ ∅ and Δ_{2} is A-invariant. Thus Δ_{2} = Δ_{1} and so also Δ_{0} = Δ_{1} □

Lemma 2.9 [quadratic imprimitivity] Let V be an quadratic 𝔽A-module, Δ be a system of imprimitivity for A in V, Δ_{1} be a non-trivial orbit for A on Δ and let W ∈ Δ_{1}. Then |Δ_{1}| = char 𝔽 = |A/C_{A}(⨁ Δ_{1})| = 2.

Proof: Let W ∈ Δ_{1} and B = N_{A}(W). Then {W } ≠ Δ_{1} and so by ??(??), W ∩ [V, A] = 0. In particular, [W, B] = 0. Let a ∈ A\B. Then 0 ≠ [W, a] ≤ W +W^{a} and so by ??(??), Δ_{1} = {W, W^{a}}, B = C_{A}(Δ_{1}) ⊴ A, B centralizes W + W^{a} and so |A/C_{A}(⨁ Δ_{1})| = 2. 2.6 gives char 𝔽 = 2. □

Lemma 2.10 [bimprimitivity2] Let V be an cubic 𝔽A-module, Δ be a system of imprimitivity for A in V, Δ_{1} be a non-trivial orbit for A on Δ and let W ∈ Δ_{1}

(a) [a] A/C_{A}(⨁ Δ_{1}) is an elementary abelian p-group for some prime p.

(b) [b] p = char 𝔽 ∈ {2, 3}.

(c) [c] One of the following holds:

1. [a] |A/C_{A}(⨁ Δ_{1})| = |Δ_{1}| ≤ 4 and N_{A}(W) = C_{A}(⨁ Δ_{1}) = C_{A}(Δ_{1})

2. [b] p = |Δ_{1}| = 2 and N_{A}(W) = C_{A}(Δ_{1}) acts quadratically on ⨁ Δ_{1}.

Proof: We may assume with loss that V = ⨁ Δ_{1} and V is a faithful A-module. If A is quadratic on V, then 2.9 implies the the lemma holds. Hence we may assume

1^{∘} [1] A is not quadratic on V.

Next we prove

2^{∘} [2] Suppose char 𝔽 = 2. Then A is an elementary abelian 2-group.

Let a ∈ A and suppose that a^{2} ≠ 1. Pick W ∈ Δ_{1} with [W, a^{2}] ≠ 0. Since char 𝔽 = 2 we have a^{2} − 1 = (a − 1)^{2} in End_{𝔽}(V). Thus [W, a, a] = [W, a^{2}] ≤ W + W^{a^{2}}. Hence 2.8(2) implies that Δ_{1} = {W, W^{a^{2}}}. Thus |Δ_{1}| ≤ 2 and so a^{2} acts trivial on Δ_{1}. But then W = W^{a^{2}} and Δ_{1} = {W }, a contradiction.

So a^{2} = 1 for all a ∈ A and (1^{∘}) holds.

3^{∘} [3] Let W ∈ Δ_{1} and a, b ∈ V with [W, b, a] ≠ 0. Then Δ_{1} = {W, W^{b}, W^{a}, W^{ab}}.

Note that [W, b] ≤ W + W^{b} and so also [W, b, a] ≤ W + W^{b} + W^{a} + W^{ab}. Hence 2.8(2) implies that Δ_{1} = {W, W^{b}, W^{a}, W^{ab}}.

4^{∘} [4] |Δ_{1}| ≤ 4

By (1^{∘}) there exists a, b ∈ A and [V, b, a] ≠ 0. Since V = ⊕Δ_{1} there exists W ∈ Δ_{1} with [W, b, a] ≠ 0 and so (4^{∘}) follows from (3^{∘}).

Case 1 [NA(W) acts] Suppose that [W, N_{A}(W)] ≠ 0 for some W ∈ Δ_{1}.

Pick b ∈ B := N_{A}(W) with [W, b]] ≠ 0 and a ∈ A \ B. Since [W, b] ≤ W we get [W, b, a] ≠ 0. Since W^{b}, (3^{∘}) gives Δ_{1} = {W, W^{a}}. Since Δ_{1} ≠ {W }, 2.8(2) gives that [W, B, B] = 0 and so B acts quadratically on V. Note that A acts quadratically on [W, B] + [W^{a}, B] and so by 2.9, p = 2. Thus (c:2) holds. By (2^{∘}), A us elementary abelian and so the Lemma holds in this case.

Case 2 [NA(W) trivial] Suppose that [W, N_{A}(W)] = 0 for all W ∈ Δ_{1}.

Put p = char 𝔽. From the assumption we have that C_{A}(Δ_{1}) = 1. Thus A is finite and so by 2.6(e), p > 0 and A is a p-group. If |Δ_{1}| ≤ 3 we conclude that |Δ_{1}| = p = |A| and the lemma holds. If |Δ_{1}| ≥ 4, then by (4^{∘}), Δ_{1} = 4. So p = 2 and by (2^{∘}), A is elementary abelian. Since A acts tranistively and faithfully on Δ_{1}, this implies |A| = 4 and N_{A}(W) = C_{A}(Δ_{1}) = 1 for W ∈ Δ_{1}. Again the lemma holds. □

Lemma 2.11 [binherit3] Let V be a nearly quadratic 𝔽A-module, and let Δ be a system of imprimitivity for A in V. Then one of the following holds:

1. [1] There exists at most one W ∈ Δ with [W, A] ≠ 0.

2. [2] A acts trivially on Δ and quadratically on V.

3. [3] A acts quadratically on V, char 𝔽 = 2, and |A/C_{A}(W)| ≤ 2 for every W ∈ Δ \ C_{Δ}(A).

4. [4] A does not act quadratically on V, and there exists a unique A-orbit W^{A} ⊆ Δ with [W, A] ≠ 0. Moreover, B := N_{A}(W) acts quadratically on V, B = C_{A}(Δ) and one of the following holds:

1. [1] char 𝔽 = 2, |W^{A}| = 4, dim_{𝔽} W = 1, B = C_{A}(V), and A/C_{A}(V) ≅ C_{2} × C_{2}.

2. [2] char 𝔽 = 3, |W^{A}| = 3, dim_{𝔽} W = 1, B = C_{A}(V), and A/C_{A}(V) ≅ C_{3}.

3. [3] char 𝔽 = 2, |W^{A}| = 2, and C_{A}(W) = C_{A}(V). Moreover, dim_{𝔽} W/C_{W} (B) = 1 and C_{W} (B) = [W, B].

Proof: Suppose first that A acts quadratically on V and non-trivially on Δ. Then 2.9 gives that (2) or (3) holds. So we may assume from now does no act quadratically on W. Pick V_{1} ∈ Δ such that [V_{1}, A] ≠ 0.

Suppose next that that A acts trivially on Δ. Then by 2.7, either V_{1} is the unique element of Δ satisfying [V_{1}, A] ≠ 0. Hence, (1^{∘}) holds.

Suppose finally that A does not act quadratically on V. Pick W ∈ Δ with [W, A, A] ≠ 0 and set

B := N_{A}(W), U := ⨁ W^{A} and U_{0} := ⨁ C_{Δ}(A)E.

By 2.7 [E, A] = 0 for all E ∈ Δ \ W^{A}. It follows that V = U ⊕ U_{0} and [U_{0}, A] = 0. Thus after replacing V by U, we may assume that Δ = W^{A}.

If W^{A} = W then (1^{∘}) holds. Thus we may assume that |W^{A}| ≥ 2. We prove next that

1^{∘} [1] W is not the direct sum of two proper 𝔽B submodules.

Suppose that W = W_{1} ⊕ W_{2} for some proper 𝔽B-submodules W_{1} and W_{2}. Then V = ⨁ W_{1}^{A} ⊕ ⨁ W_{2}^{A} and A act non-trivial on both direct summands. But this contradicts 2.7.

Note that we can apply 2.10. If 2.10(c:1) holds, then [W, B] = 0. Thus (1^{∘}) gives dim_{𝔽} W = 1. Thus the lemma in this case. So suppose that 2.10(c:2) holds and [W, B] ≠ 0. Pick an 𝔽-subspace W_{1} ≤ W with [W, B] ≤ W_{1}, W_{1} + C_{W} (B) = W and W_{1} ∩ C_{W} (B) = [W, B] ∩ C_{W} (B). Also pick an 𝔽-subspace W_{2} ≤ C_{W} (B) with C_{W} (B) = W_{2} ⊕ ([W, B] ∩ C_{W} (B). Then W = W_{1} ⊕ W_{2} and W_{1} and W_{2} are 𝔽B-submodules of W. Thus by (1^{∘}), W_{2} = 0 and so C_{W} (B) ≤ [W, B] By 2.10(c:2) B is quadratic on W. Thus C_{W} (B) = [W, B]. Let a ∈ A\B and w ∈ W \C_{W} (B). Put W_{0} = w𝔽+C_{W} (B) and V_{0} = ⟨W_{0}^{A}⟩. Then V = W + W^{a} and V_{0} = W_{0} + W_{0}^{a}.

By 2.5(b), Q_{V} (A) = [V, A] + C_{V} (A). Since [W_{0}, B] ≠ 0 and so also [V_{0}, B, A] ≠ 0. Thus V_{0} ≰ Q_{V} (A) = [V, A] + C_{V} (A) and so 2.5(a), [V, A] ≤ V_{0} = W_{0} + W_{0}^{a}. Thus

V = W ⊕ W^{a} = W + [V, A] = W + W_{0} + W_{0}^{a} = W + W_{0}^{a}

Hence W^{a} = W_{0}^{a}, W = W_{0} and C_{W} (B) = [W, B] is an 𝔽-hyperplane in W. Thus (4:3) holds. □

Lemma 2.12 [bfield ext] Suppose 𝕂 is a field with 1 ≠ A ≤ Aut(𝕂) and put 𝔼 == C_{𝕂}(A). Suppose that 𝕂 is a cubic 𝔽A-module-module. Then

(a) [a] p := char 𝕂 ∈ {2, 3}.

(b) [b] A is an elemntary abelian p group of order at most two.

(c) [c] dim_{𝔽} 𝕂 = |A| and 𝕂 ≅ 𝔽A and as an 𝔽A-module.

(d) [d] One of the follwing holds:

1. [1] |A| = 2, A acts quadraticillay on 𝕂 and [𝕂, A] = 𝔽.

2. [2] |A| > 2, A does not act quadratically on V and [K, A, A] = 𝔽.

Proof: We consider first the case where A = ⟨σ⟩ for some σ ∈ V. Let V_{1} be the largest quadratic 𝔼⟨σ⟩-submodule of 𝕂 and V_{0} := [V_{1}, σ], so V_{0} ⊆ 𝔼. As V_{0} is an 𝔼-space, we get that V_{0} = 𝔼.

The mappings

V_{1} → V_{0} with v_{1} ↦ [v_{1}, σ]

and

𝕂 → V_{1}/V_{0} with v_{1} ↦ [v_{1}, σ] + V_{0}

are 𝔼-module homomorphisms. The kernel of the first one is 𝔼 and that of the second one is V_{1}. It follows that

(∗)  V_{1}/𝔼 ≅ 𝔼 and either 𝕂/V_{1} = 0 or 𝕂/V_{1} ≅ V_{1}/V_{0} ≅ 𝔼.

Thus dim_{𝔼} 𝕂 ≤ 3. Note that 𝕂 over 𝔼 is a finite Galois extention since the fixed field of the Galois group Aut_{𝔼}(𝕂) is 𝔼, so |A| = | Aut 𝔼(𝕂) = dim_{𝔼} 𝕂 ∈ {2, 3}. Since A is cubic we conclude from ??(??) char 𝔼 = |A|. Let k ∈ 𝕂 \ [𝕂, A]. Then ite easy to see that k^{A} is an 𝔼-basis for 𝕂 and so 𝕂 ≅ 𝔼A as an 𝔼A-module.

We now consider the general case. Let 1 ≠ σ ∈ A and put L = C_{𝕂}(σ). Then p = char 𝔼 = |σ| = dim_{𝕂} 𝔼.

Suppose that p = 2 and A acts quadratically on 𝕂 or that p = 3. If p = 2, then L = [𝕂, σ and if p = 3 then L = [𝕂, σ, σ]. So in any case L ≤ C_{𝕂}(A) = 𝔼. Thus dim_{𝔼} 𝕂 =p and so also | Aut_{L}(𝕂)| = p. Tuse A = ⟨σ⟩ and the lemma hold.s

So suppose that p = 2 and A is not quadratic on 𝕂. Then A ≠ ⟨σ⟩ and there exists 1 ≠ μ ∈ Aut(𝕂) with μ ≠ σ. Then [L, μ] ≤ [𝕂, A, A] ≤ C_{𝕂}(A) = 𝔼 ≤ L and so L^{μ} = L. Since Aut_{L}(𝕂) = ⟨σ⟩, [L, μ] ≠ 0. The cyclic case applied to L in place of 𝕂 show that dim_{[E,μ]} L = 2. We conclude that [L^{,}μ] = 𝔼 and so dim_{𝔼} 𝕂 = 4. It follows that A = Aut_{𝔼}(𝕂) and |A| = 4. Since σ^{2} = 1 for all σ ∈ A, A is elementary abelian. Finally if k ∈ 𝕂 \ [𝕂, A], then k^{A} is a 𝔼-basis for 𝕂 and so K ≅ 𝔽A as an 𝔽A-module. □

Lemma 2.13 [u:dimension commutator] Let V a semi-linear, cubic 𝕂A-module. Put with 𝔼 = 𝕂_{A} and suppose that 𝔼 ≠ 𝕂. Then one of the following holds.

1. [a] |A/C_{A}(V)| = char 𝔼 = dim_{𝔼} 𝕂 = 2, A is quadratic on V and V is isomorphic to a direct sum of copies of 𝕂.

2. [b] |A/C_{A}(V)| = char 𝔼 = dim_{𝔼} 𝕂 = 3, A is not quadratic on V and V is isomorphic to a direct sum of copies of 𝕂.

3. [c] |A/C_{A}(V) ≅ C_{2} × C_{2}, A_{𝕂} = C_{A}(V), char 𝔼 = 2, dim_{𝔼} 𝕂 = 4, A is not quadratic on V and V is isomorphic to a direct sum of copies of 𝕂.

4. [d] |A/A_{𝕂}| = char 𝔼 = dim_{𝔼} 𝕂 = 2, A is not quadratic on V, A/C_{A}(V) is elementary abelian and there exists an 𝔼A submodule W of V such that V ≅ W ⊗_{𝔼} 𝕂 as an 𝔼K module, A = C_{A}(W)A_{𝕂} and A_{𝕂} acts quadratically on V and W.

Proof: We may assume that A acts faithfully on V and V ≠ 0.

Case 1 [ak=1] Suppose that A_{𝕂} = 1.

Let B be a subset of C_{V} (A) maximal with respect to being linearly independent over 𝕂. Note that B exists by Zorn’s Lemma. Let U be the 𝕂-span of B. Since A is cubic, C_{V} (A) ≠ 0 and so B ≠ ∅. Let b ∈ B, then b𝕂 is isomorphic to 𝕂 as an 𝔼A-module. So A acts cubicly on 𝕂 and so we can apply 2.12. |A| ≤ 4,A is elementary abelian of order at most 4. Moreover either |A| = 2 and [𝕂, A] = 𝔼 or |A| > 2 and [𝕂, A, A] = 𝔼.

Suppose that U ≠ V. Then V/U has an 𝔼A-submodule isomorphic to 𝔼 and so [V/U, A] ≠ 0 and if |A| > 2 then [V/U, A, A] ≠ 0. So if |A| = 2 we can v ∈ [V, A] with v ∉ U and if |A| > 2 we can choose v ∈ [V, A, A] with v i̸nU. If |A| = 2, then char 𝔼 = 2 and A acts quadratically on V. So in amy case v ∈ C_{V} (A). Since U is a 𝕂-subspace, B ∪ {v} is linearly independent over 𝕂, a contradiction to the maximality of B.

Thus U = V and so V = ⨁_{b∈B} b𝕂 is a direct sum of copies of 𝕂 as an 𝔼A-module. Thus one of (1), (2) and (3) holds.

Case 2 [ak neq 1] Suppose A_{𝕂} ≠ 1.

Note that [V, A_{𝕂}, A_{𝕂}] is a 𝕂-subspace of V centralized by A. Thus [V, A_{𝕂}, A_{𝕂}] = 0. Moreover, A act quadratically on the non-trivial 𝕂-space V and so (Case 1) shows that |A/A_{𝕂}| = 2 and dim_{𝔼} 𝕂 = 2. Let a ∈ A. Since (a − 1)^{2} = a^{2} − 1 in End_{𝔼}(V), [V, a^{2}] = [V, a, a]. Since a^{2} ∈ A_{𝕂} we conclude that [V, a^{2}] is a 𝕂-subspace centralized by A. Thus [V, a^{2}] = 0 and A is elementary abelian. Let a ∈ A \ A_{𝕂}. Put W = C_{V} (a). Then W is an 𝔼-subspace of V. By (Case 1) applied to ⟨a⟩, W = [V, a]. Hence [W, A] = [V, a, A] ≤ C_{V} (A) ≤ W and so W is an 𝔽A-submodule of V. Since a ∈ C_{A}(W), A = C_{A}(W)A_{𝕂}. By the universal property of the tensor product, there exists an 𝔽A-homomorphism ρ : W ⊗_{𝔽} 𝕂 → V with (w ⊗ k)ρ = wk for all w ∈ W and k ∈ 𝕂. By (Case 1) applied to ⟨a⟩, ρ is a bijection. Thus (4) holds in this case. □

Comment:this needs to be moved after ??

Lemma 2.14 [bnearly quadratic fieldext] Let V ba a nearly quadratic 𝔽A-module. Suppose that there exists a field 𝕂 with 𝔽 ≤ 𝕂 such that V is a semi-linear 𝕂A-module and A ≠ A_{𝕂}. Then one of the following holds:

1. [a] [V, A, A] = 0, [V, A_{𝕂}] = 0, and char 𝕂 = 2 = |A/A_{𝕂}|.

2. [b] [V, A, A] ≠ 0, [V, A_{𝕂}] = C_{V} (A_{𝕂}), dim_{𝕂} V/C_{V} (A_{𝕂}) = 1, 𝔽 = 𝕂_{A}, and char 𝕂 = 2 = |A/A_{𝕂}| = dim_{𝔽} 𝕂.

3. [c] [V, A, A] ≠ 0, [V, A_{𝕂}] = 0, 𝔽 = 𝕂_{A}, dim_{𝕂} V = 1, and char F = 3 = |A/A_{𝕂}| = dim_{𝔽} 𝕂.

4. [d] [V, A, A] ≠ 0, [V, A_{𝕂}] = 0, 𝔽 = 𝕂_{A}, dim_{𝕂} V = 1, and char F = 2, A/A_{𝕂} ≅ C_{2} × C_{2} and dim_{𝔽} 𝕂 = 4..

Proof: Without loss V is a faithful 𝔽A-module. Put 𝔼 = C_{𝕂}(A). Since A is cubic on V we can apply ?? If A is quadratic on V we conclude that (1) holds. So we may assume that A is not quadratic.

Suppose first that [V, A_{𝕂}] = 0. Then V is the direct sum of copies of 𝕂. Hence by 2.7, V ≅ 𝕂 as an 𝔽A-module. Thus by 2.12, V ≅ 𝔼A as an 𝔼A. As an 𝔽A-module, 𝔼A is a direct sum of dim_{𝔽} 𝔼-copoies of 𝔽A and so 2.7 gives 𝔽 = 𝔼. Thus (3) or (4) holds.

Suppose finally that [V, A_{𝕂}] ≠ 0. Then by ?? p = |A/A_{𝕂}| = dim_{𝔼} 𝕂 = 2 and there exists an 𝔼A-module W such that W ≅ 𝔼𝕂 as an 𝔼A-module and A = A_{𝕂}C_{A}(W). Hence we can apply 5.1.

Note that [𝕂, A] = 𝔼. By 5.1(d), [𝕂, A] = [z𝔽, A] for some z ∈ 𝕂. Since |A/A_{𝕂}| = 2 this implies that [𝕂, A] is 1-dimensional over 𝔽. Hence 𝔼 = 𝔽. Also by 5.1(d) [W, A] = C_{W} (A) is an 𝔼-hyperplane of V. Since A = A_{𝕂}C_{A}(W) we conclude that [V, A_{𝕂}] = C_{V} (A_{𝕂}) is an 𝕂-hyperplane of V. Thus (2) holds. □

3 Normalizing Tensor Products

Lemma 3.1 [char scalar] Let 𝕂 be division ring and V a 𝕂 space of dimension at least 2 and 𝔽 a subfield of 𝕂 and α ∈ GL_{𝔽}(V) with v𝕂α = v𝕂 for all v ∈ V. Then there exists k ∈ 𝕂 with vα = vk for all v ∈ V. Morever, kf = fk for all f ∈ 𝔽.

Proof: Let 0 ≠ v ∈ V. Then by assumption vα = vk_{v} for a unique k_{v} ∈ 𝕂. Let 0 ≠ w ∈ 𝕂. We will show that k_{v} = k_{w}.

Suppose first that v𝕂 ≠ w𝕂. We have

vk_{v} + wk_{w} = vα + wα = (v + w)α = (v + w)k_{v+w} = vk_{v+w} + wk_{v+w}

Since v and w are linearly independent over 𝕂 we conclude that k_{v} = k_{v+w} = k_{w}.

Suppose next that v𝕂 = w𝕂. Since V is at least two dimensional over 𝕂 ther exists u ∈ V \ v𝕂. Thus by the preceeding case k_{v} = k_{u} = k_{w}.

Put k = k_{v}. We conclude that uα = uk for all u ∈ U. Let f ∈ 𝔽. Since α is 𝔽-linear

vkf = vαf = vfα = vfk

and so kf = fk. □

Lemma 3.2 [char semilinear] Let 𝕂 be divsion ring, V a 𝕂 space of dimensional at least 2 and 𝔽 a subfield of 𝕂. Let α ∈ GL_{𝔽}(V) with v𝕂α = vα𝕂 for all v ∈ V. Then there exists σ ∈ Aut_{𝕂}(𝔽) such that α acts σ-semi-linear on V.

Proof: Let 0 ≠ v ∈ V. Since v𝕂α = vα𝕂 there exists a function σ_{v} : 𝕂 → 𝕂 with

vkα = vα.kσ_{v}

Since α is a bijection, so is σ_{v}. Since α is 𝔽-linear, α |_{𝔽}= id_{𝔽} and σ_{k} is 𝔽-linear.

Let 0 ≠ w ∈ V. We will show that σ_{v} = σ_{w}. Suppose first that v𝕂 ≠ w𝕂. Then also v𝕂α ≠ w𝕂α and so vα𝕂 ≠ wα𝕂. Hence vα and wα are lineraly independent over 𝕂. We compute

(v + w)kα = (v + w)α.kσ_{v+w} = vα.kσ_{v+w} + wα.kσ_{v+w}

and also

(v + w)kα = vkα + wkα = vα.kσ_{v} + wα.kσ_{w}

Since vα and wα are 𝕂-lineary independent we get kσ_{v+w} = kσ_{v} and kσ_{v+w} = kσ_{w}. Thuse σ_{v} = σ_{v+w} = σ_{w}.

Suppose next that v𝕂 = w𝕂. Since V is at least two dimensional over 𝕂 ther exists u ∈ V \ v𝕂. Thus by the preceeding case σ_{v} = σ_{u} = σ_{w}.

Put σ = σ_{v}. We conclude that

ukα = uα.kσ

for all u ∈ V and k ∈ 𝕂.

Hence for k, l ∈ 𝕂, vklα = vkα.lσ = vα.kσ.lσ and so klσ = kσ.lσ. Thus σ ∈ Aut_{𝔽}(𝕂) and α is σ-semi-linear. □

Theorem 3.3 [normalizing pure tensors] Let 𝕂 be a field, 𝔽 a subfield of 𝕂, I a finite set with |I| ≥ 2 and for i ∈ I let V_{i} be at least 2-dimensional 𝕂-space. Put V := ⊗_{𝕂}^{i∈I}V_{i} and P := {⊗v_{i} | v_{i} ∈ V_{i}, i ∈ I}. Let α ∈ GL_{𝔽}(V) with Pα = P. Then there exist σ ∈ Aut_{𝔽}(𝕂), π ∈ Sym(I) and σ-semilinear maps α_{i} : V_{iπ^{−1}} → V_{i} such that

(⊗v_{i})α = ⊗(v_{iπ^{−1}}α_{i})

for all v_{i} ∈ V_{i}, i ∈ I.

Proof: For j ∈ I let V^{j} := ⊗^{j≠i∈I}V_{i} and identify V with V_{j} ⊗_{𝕂} V^{j}. Similarly if v_{i} ∈ I, i ∈ I we put v^{j} = ⊗_{j≠i∈I}v_{j}.

1^{∘} [1] For i ∈ I let 0 ≠ v_{i}, w_{i} ∈ V_{i}. Then 𝕂 ⊗ v_{i} = 𝕂 ⊗ w_{i} if and only if 𝕂v_{i} = 𝕂w_{i} for all i ∈ W_{i}.

Suppose that 𝕂v_{j} ≠ 𝕂w_{j} for some j ∈ I. Then v_{j} ⊗ V^{j} ∩ w_{j} ⊗ V^{j} = 0 and so 𝕂 ⊗ v_{i} ≠ 𝕂 ⊗ w_{i}. The other direction is obvious.

2^{∘} [2] Let 0 ≠ v_{i}, w_{i} ∈ V_{i} with ⊗v_{i} + ⊗w_{j}inP. Then there exists j ∈ J with 𝕂v_{i} = 𝕂w_{i} for all j ≠ i ∈ I.

We may assume that 𝕂v_{j} ≠ 𝕂w_{j} for some j ∈ j. Let z = v + w and choose z_{i} ∈ V_{i} with z = ⊗z_{i}. Let D be a subspace of V_{j} with V_{j} = 𝕂v_{j} ⊗ 𝕂w_{j} ⊗ D. Then z_{k} = kv_{1} + lw_{1} + d for some k, l ∈ 𝕂 and d ∈ D. From z = u = w we conclude that

v_{j} ⊗ kz^{j} + w_{j} ⊗ lz^{j} + d ⊗ z^{j} = v_{j} ⊗ v^{j} + w_{j} ⊗ w^{j}

Since v = v_{j} ⊗ V^{j} ⊕ w_{j} ⊗ V^{j} ⊕ D ⊗ V^{j} we conclude that v^{j} = kz^{j} and w^{j} = lz^{j}. Hence 𝕂v^{j} = 𝕂w^{j} and (2^{∘}) follows from (1^{∘}).

Define v, w ∈ P to be adjacent if v + w ∈ P. Observe that α induces a graph automorphism on P.

3^{∘} [3] Let U be a maximal clique in P. Then there exists j ∈ J and u_{i} ∈ U_{i}, j ≠ i ∈ I with U = u^{j} ⊗ V_{j}.

Let 0 ≠ u ∈ U and u_{i} ∈ I with u = ⊗u_{i}. Observe that for each j ∈ J, u^{j} ⊗ V_{j} ⊆ P and u^{j} ⊗ V_{j} is closed under addition. Thus u^{j} ⊗ V_{j} is a clique in P. Since u𝕂 is a proper subset of u^{j} ⊗ V_{j}, U ⊈ U𝕂. Choose v ∈ U with v ∉ u𝕂. Then by (2^{∘}), v ∈ u^{j}V_{j} for some j ∈ V with v_{j}𝕂 ≠ u_{j}𝕂.

Suppose for a contradiction that there exists w ∈ U with w ∉ u^{j} ⊗ V^{j}. Then w_{i}𝕂 ≠ u_{i}𝕂 = v_{i}𝕂 for some j ≠ i ∈ V. Moreover, either v_{j}𝕂 ≠ w_{j}𝕂 or u_{j}𝕂 ≠ w_{j}𝕂. Since w is adjacent to u and v, this contradicts (2^{∘}). Hence U ⊆ u^{j} ⊗ V_{j} and so by maximality U = u^{j} ⊗ V_{j}.

4^{∘} [5] There exists σ ∈ Aut_{𝔽}(𝕂) such that α is σ-semi-linear.

Let v ∈ P and let Δ(v) be the set of neighbors of v in P. It is easy to see that 𝕂v = ⋂_{w∈Δ(v)} Δ(w). Thus (𝕂v)α = 𝕂(vα). From 3.2 we conclude that the restriction of α to each maximal clique C of σ_{C}-semi-linear, for some σ_{C} ∈ Aut_{𝔽}(𝕂). If C and D are maximal cliques with C ∩ D ≠ ∅, then clearly σ_{C} = σ_{D}. Consider the graph on the maximal clique of P, where C and D are adjacent if C ∩ neq∅. It is readily verifeid that this graph is connected. It follos that σ := σ_{C} = σ_{D} for all maximal cliques. Since V is spanned by P, α is σ-semi-linear.

Fix 0 ≠ x_{i} ∈ V_{i} and 0 ≠ y_{i} ∈ V_{i} with xα = y where x = ⊗x_{i} and y = ⊗y_{i}.

5^{∘} [4] There exists η ∈ Sym(I) with (x^{iη} ⊗ V_{iη})α = y^{i} ⊗ V_{j} for all i ∈ I.

Let U be a maximal clique of ∩P containing x. By (3^{∘}), U = u^{j} ⊗ V_{j} amd since x ∈ U we conclude that U = x^{j} ⊗ V_{j}. Hence thhe maximal cliques containing x are x^{i} ⊗ V_{i} | i ∈ I. Similarly the maximal cliques of P containing y are y^{i} ⊗ V_{i}, i ∈ I. From xα = y we conclude that

{x^{i} ⊗ V_{i} | i ∈ I}α = {y^{i} ⊗ V_{i} | i ∈ I}

This clearly implies (5^{∘}).

For j ∈ J define α_{j} : V_{jη} → V_{j} by

(∗)  (x^{jη} ⊗ u)α = y^{j} ⊗ uα_{j}

for all u ∈ V_{jη}. Since α is σ-linear each α_{i} is σ-linear. By the universial property of the tenosr product there exists a unique σ-liner map β : V → V with (⊗v_{i})β = ⊗v_{iη^{−}1}α_{i} for all v_{i} ∈ V_{i}. Observer that Pβ = P. Put δ = β^{−1} ∘ α. Then

6^{∘} [6] Pδ = P, δ is 𝕂-linear and (x^{i} ⊗ u)δ = x^{i} ⊗ u for all i ∈ I, u ∈ V_{i}.

We will next that

7^{∘} [7] Let J ⊆ I. Then (x^{J} ⊗ u)δ = x^{J} ⊗ u for all u ∈ V_{J}.

The proof is by induction on |J|. If |J| ≤ 1, this follows from (6^{∘}). So suppose |J| ≥ 2. Since δ is 𝕂 linar we may assume that u = ⊗_{t∈J}u_{y} for some u_{t} ∈ V_{t}. Let j, k ∈ J with j ≠ k. Put v = x^{J} ⊗ u and w = x^{J\{j,k}} ⊗ u. Then v = u_{j} ⊗ u_{k} ⊗ w. By induction on |J|,

(u_{j} ⊗ x_{k} ⊗ w)δ = u_{j} ⊗ x_{k} ⊗ w and x_{j} ⊗ u_{k} ⊗ w)δ = x_{j} ⊗ u_{k} ⊗ w)

Since v is adjacent to u_{j} ⊗ x_{k} ⊗ w and x_{j} ⊗ u_{k} ⊗ w in P we conclude that also vδ is adjacent to u_{j} ⊗ x_{k} ⊗ w and x_{j} ⊗ u_{k} ⊗ w in P. This implies that vδ𝕂 is either (x_{h} ⊗ x_{k} ⊗ w) or (u_{h} ⊗ u_{k} ⊗ w). In the first case we conlude that vδδ = vδ and so also vδ = b. In either case vδ = vδk_{v} for some k_{v} ∈ 𝕂. Since (u_{j} ⊗ x_{k} ⊗ w)δ = u_{j} ⊗ x_{k} ⊗ w, 3.1 applied to δ restricted to u_{k} ⊗ V_{j} ⊗ w gives that k_{v} = 1. Thus (7^{∘}) holds.

From (7^{∘}) applied with J = I we have δ = id_{V}. Thuse α = β and the Theorem is proved. □

In this section G is a group, 𝕂 is a field, and J is a finite G-set.

Definition 3.4 [udef: tensor decomposition] Let V be a 𝔽G-module. Then a tensor decomposition of V is a tuple (𝕂, V_{i}∈ I, Φ))) where K is field extension of 𝔽, (V_{i}, i ∈ I) is a finite family of 𝕂-spaces, Φ : ⨂_{i∈I} V_{i} → V a 𝔽 linear isomorphism a such that

Φ(⊗_{i∈I}) | v_{i} ∈ V_{i}

is G -invariant.

A tensor decomposition is called proper if |I| > 1 and dim_{𝕂} V_{i} ≥ 2 for all i ∈ I.

Definition 3.5 [def normalize tensor] Let V_{j}, j ∈ J, be (disjoint) finite dimensional 𝕂-spaces of dimension at least 1. We say that G normalizes the tensor product ⊗_{𝕂}^{J}V_{j} (with respect to σ and (ϕ_{j,h} | h ∈ G, j ∈ J)) if there exists a homomorphism σ : G → Aut(𝕂) and a hσ-semi-linear map ϕ_{j,h} : V_{j} → V_{jh} for every h ∈ G and j ∈ J such that

(i) [i] ⊗v_{j(ab)^{−1}}ϕ_{j(ab)^{−1},ab} = ⊗v_{j(ab)^{−1}}ϕ_{j(ab)^{−1},a}ϕ_{jb^{−1},b} for every a, b ∈ G and v_{j} ∈ V_{j},

(ii) [ii] ϕ_{j,1} = id_{V_{j}} for every j ∈ J.

We say that a subgroup A ≤ G strictly normalizes the the tensor product ⊗_{𝕂}^{J}V_{j} if ϕ_{j,a}ϕ_{ja,b} = ϕ_{j,ab} for all j ∈ J and a, b ∈ A.

Lemma 3.6 [induced action] Spuppose that G normalizes the tensor product V := ⊗_{𝕂}^{J}V_{j} with respect to σ and (ϕ_{j,h} | h ∈ G, j ∈ J). For j ∈ J let G_{j} be the stabilizer of j in G. Then the following hold:

(a) [a] G acts semi-linearly on V via

(⊗v_{j})h := ⊗v_{jh^{−1}}ϕ_{jh^{−1},h} (v_{j} ∈ V_{j}, h ∈ G).

(b) [f] Let A ≤ G such that A stricly normalizes the tensor product V. Then A ∩ G_{j} acts semi-

linearly on V_{j} via

v_{j}a := v_{j}ϕ_{i,a} (v_{j} ∈ V_{j}, a ∈ A ∩ G_{j}).

(c) [d] ϕ_{j,h} is bijective for every h ∈ G.

(d) [e] For every g, h ∈ H there exists λ_{j,g,h} ∈ 𝕂^{∗} such that

ϕ_{j,g}ϕ_{jg,h} = ϕ_{j,gh}λ_{j,g,h} and ∏_{j∈J} λ_{j,g,h} = 1.

Proof: Let j ∈ J, v_{j} ∈ V_{j}, k ∈ 𝕂, and a, b ∈ G.

(a): Property 3.5(ii) shows that 1 induces the identity on V. From property 3.5(i) we get

(⊗v_{j})(ab) = ⊗v_{j(ab)^{−1}}ϕ_{j(ab)^{−1},ab} = ⊗v_{j(ab)^{−1}}ϕ_{j(ab)^{−1},a}ϕ_{jb^{−1},b} = (⊗(v_{ja^{−1}})ϕ_{ja^{−1},a})b = ((⊗v_{j})a)b.

This shows that G acts on the abelian group V.

Fix r ∈ J and put w_{j} := v_{j} for j ≠ r and w_{r} := v_{r}k. Then

(⊗v_{j})ka = (⊗w_{j})a = ⊗w_{ja^{−1}}ϕ_{ja^{−1},a} = ⊗v_{j}^{∗},

where v_{j}^{∗} := v_{ja^{−1}}ϕ_{ja^{−1},a} for ja^{−1} ≠ r and v_{j}^{∗} := v_{ja^{−1}}ϕ_{ja^{−1},a}k^{aσ} for ja^{−1} = r. It folllows that (⊗v_{j})ka = (⊗v_{j})ak^{aσ}. Hence G acts semi-linearly on V.

(b): This is a direct consequence of the definition of strictly normalizing.

(c) and (d): For i := j(ab)^{−1} 3.5(i) gives

(∗)  ⊗v_{i}ϕ_{i,ab} = ⊗v_{i}ϕ_{i,a}ϕ_{ia,b},

Hence by 3.5(ii)

⊗v_{i} = ⊗v_{i}ϕ_{i,1} = ⊗v_{i}ϕ_{i,a}ϕ_{ia,a^{−1}}.

This shows that ker ϕ_{i,a} = 0, and (c) follows.

Now let v_{i} ≠ 0. Then (∗) implies that v_{i}ϕ_{i,ab} = v_{i}ϕ_{i,a}ϕ_{ia,b}k for suitable k ∈ 𝕂^{∗}. Since this dependence holds for any 0 ≠ v_{i} ∈ V_{i} and by (c) ϕ_{i,ab}, ϕ_{i,a} and ϕ_{ia,b} are bijective σ-homomorphisms, the scalar k is independent of the choice of v_{i}. This shows that ϕ_{i,ab} = ϕ_{i,a}ϕ_{ia,b}λ_{i,a,b} for λ_{i,a,b} := k. Moreover,

⊗v_{i}ϕ_{i,a}ϕ_{ia,b} = ⊗v_{i}ϕ_{i,ab} = (⊗v_{i}ϕ_{i,a}ϕ_{ia,b}) ∏_{i∈J} λ_{i,a,b},

and so ∏_{i∈J} λ_{i,a,b} = 1. □

Hypothesis 3.7 [hyp tensor] Let G be a group strictly normalizing ⊗_{𝕂}^{J}V_{j} with respect to σ and (ϕ_{j,h} | h ∈ G, j ∈ J), as defined in 3.5.

Notation 3.8 [not tensor] Assume Hyothesis 3.7. Put V := ⊗_{𝕂}^{J}V_{j}, and let 𝔽 := 𝕂_{G} be the fixed field of Gσ in 𝕂. We use the action of G on V, as given in 3.6(a), and the action of G_{j}, as given in 3.6(b). In particular V is an 𝔽G-module and V_{j} is an 𝔽G_{j}-module, and we can use the usual notation [V, B], C_{V} (B), [V_{j}, B], C_{V_{j}}(B), where B ≤ G and B ≤ G_{j}, respectively.

We say that U ≤ G centralizes V_{j} if U fixes j and [V, U] = 0.

Lemma 3.9 [elementary property] Assume Hypothesis 3.7. For j ∈ J let U_{j} be an 𝔽-subspace of V_{j} with dim_{𝔽} U_{j} ≥ 1. Suppose that there exists r ∈ J and B ≤ G such that

B ≰ G_{r} and {⊗u_{j} | u_{j} ∈ U_{j}} ⊆ C_{V} (B).

Then dim_{𝕂}⟨U_{r}⟩_{𝕂} = 1.

Proof: Pick 0 ≠ u_{j} ∈ U_{j} and a ∈ B with ra ≠ r. Then

⊗u_{j} = (⊗u_{j})a = ⊗u_{ja^{−1}}ϕ_{ja^{−1},a},

and so u_{ra} = u_{r}ϕ_{r,a}k for some k ∈ 𝕂^{∗}. Fixing u_{ra} and allowing u_{r} to run through the elements of U_{r} \ {0} shows that U_{r}ϕ_{r,a} ≤ u_{r}ϕ_{r,a}𝕂. By 3.6(c) ϕ_{r,a} is injective, so the result follows. □

Lemma 3.10 [trivial action] Assume Hypothesis 3.7. Suppose that char K = p > 0 and that G is a finite p-group. Let i ∈ J with dim_{𝕂} V_{i} ≥ 2. Then C_{G}(V) ≤ C_{G}(V_{i}).

Proof: Pick h ∈ C_{G}(V) and 0 ≠ v_{j} ∈ V_{j} and set k := ih. Then

⊗v_{j} = (⊗v_{j})h = ⊗v_{jh^{−1}}ϕ_{jh^{−1},h},

so v_{i}ϕ_{i,h} ∈ v_{k}𝕂.

If i ≠ k, then V_{i}ϕ_{i,h} ≤ v_{j}𝕂, and 3.6(c) gives dim_{𝕂} V_{i} = 1, a contradiction. Hence i = k, and h induces scalar multiplication by a fixed scalar k ∈ 𝕂 on V_{i}. On the other hand, since char 𝕂 = p and G is a finite p-group, C_{V_{i}}(G) ≠ 0 and so k = 1. □

Lemma 3.11 [centralizer index] Assume Hypothesis 3.7. Let 𝕂 = 𝔽 and J be a trivial G-set with |J| ≥ 2. Suppose that there exists r ∈ J with [V_{r}, G] + C_{V_{r}}(G) ≠ V_{r}. Then

\frac{dim_{𝕂} V}{dim_{𝕂} V_{r}} ≤ dim_{𝕂} V/C_{V} (G).

Proof: Let I := J \ {r} and W := ⊗_{𝕂}^{I}V_{i}. Then V and V_{r} ⊗_{𝕂} W are isomorphic 𝕂G-modules. Moreover, it is easy to see that G strictly normalizes V_{r} ⊗_{𝕂} W. If |J| ≥ 3, then by induction

\frac{dim_{𝕂} V}{dim_{𝕂} V_{r}} = \frac{dim_{𝕂}(V_{r} ⊗ W)}{dim_{𝕂} V_{r}} ≤ dim_{𝕂}(V_{r} ⊗ W)/C_{V_{r}⊗W} (G) = dim_{𝕂} V/C_{V} (G).

Hence, we may assume that |J| = 2 and

dim_{𝕂} V/C_{V} (G) = dim_{𝕂} V − dim_{K} C_{V} (G) < \frac{dim_{𝕂} V}{dim_{𝕂} V_{r}} = dim_{𝕂} W.

Pick v ∈ V_{r} \ ([V_{r}, G] + C_{V_{r}}(G)). Then dim_{𝕂} v ⊗ W = dim_{𝕂} W and so v ⊗ W ∩ C_{V} (G) ≠ 0. Let w ∈ W such that 0 ≠ v ⊗ w ∈ C_{V} (G), and pick a ∈ G with [v, a] ≠ 0. Then

0 = [v ⊗ w, a] = [v, a] ⊗ [w, a] + v ⊗ [w, a] + [v, a] ⊗ w.

It follows that (v + [v, a]) ⊗ [w, a] = −[v, a] ⊗ w ≠ 0 and thus v + [v, a] ∈ [v, a]𝕂. But this contradicts v ∉ [V_{r}, A]. □

Lemma 3.12 [tensor and imprimitivity] Assume Hypothesis 3.7. Suppose J is a transitive Gset and |J| ≥ 2. Fix r ∈ J and let X_{r} be a proper G_{r}-invariant 𝕂-subspace of V_{r}. For h ∈ G put X_{rh} := X_{r}ϕ_{r,h} and

X := ⊗_{𝕂}^{J}X_{j}, U_{j} := V_{j} ⊗ (⊗_{i≠j}X_{i}), U := ∑_{j∈J} U_{j}, Δ := {U_{j}/X | j ∈ J}.

Then

(a) [a] U and U/X are semi-linear 𝕂G-modules,

(b) [b] Δ is a system of imprimitivity for G in U/X.

(c) [c] G acts transitively on Δ.

Proof: Observe that X_{rh} = X_{rg} for g ∈ G_{r}h since X_{r} is G_{r}-invariant. For h ∈ G

Xh = (⊗X_{j})h = ⊗X_{jh^{−1}}ϕ_{jh^{−1},h} = ⊗X_{j} = X

and similarly

U_{j}h = V_{jh^{−1}}ϕ_{jh^{−1},h} ⊗ (⊗_{ih^{−1}≠jh^{−1}}X_{ih^{−1}}ϕ_{ih^{−1},h} = U_{jh}.

This shows that X and U are G-invariant, so (a) holds, and that G acts on Δ, so (c) holds. As clearly U_{i} ∩ U_{ℓ} = X for i ≠ ℓ, also (b) holds. □

Lemma 3.13 [quadratic and tensor1] Assume Hypothesis 3.7 and in addition:

(i) [i] 𝕂 = 𝔽 and char 𝕂 = p,

(ii) [ii] |J| ≥ 2, and J is a trivial G-set.

(iii) [iii] G acts quadratically on V, and C_{V_{j}}(G) ≠ 0 for j ∈ J.

and V_{j} ≠ C_{V_{j}}(G) ≠ 0 for every j ∈ J can one deduce this from the quadratic action of G on V? .

Then the following hold:

(a) [a] G acts quadratically on V_{j} for every j ∈ J.

(b) [b] For every j ∈ J with [V_{j}, G] ≠ 0 and i ∈ J \ {j} either [V_{i}, G] = 0 or there exists a faithful 2-dimensinal 𝕂G-module W and a trivial 𝕂G-module C such that V_{j} ≅ W^{n} ⊕C for some n ∈ ℕ.

Proof: Let i, j ∈ J with i ≠ j. Then

[V_{i} ⊗ C_{V_{j}}(G), G] = [V_{i}, G] ⊗ C_{V_{j}}(G) ≤ C_{V} (G),

so [V_{i}, G, G] ⊗ C_{V_{i}}(G) = 0 and thus [V_{i}, G, G] = 0. This (a).

Now assume that [V_{j}, G] ≠ 0 ≠ [V_{i}, G], and let a, b ∈ G and x ∈ V_{j}. Since G acts non-trivially on V_{i} there exists y ∈ V_{i} with [y, a] ≠ 0 and [y, b] ≠ 0. Then

[x ⊗ y, a] = [x, a] ⊗ [y, a] + x ⊗ [y, a] + [x, a] ⊗ y

and by (a)

(∗)  0 = [x ⊗ y, a, b] = [x, b] ⊗ [y, a] + [x, a] ⊗ [y, b].

The condition on y shows that C_{V_{j}}(a) = C_{V_{j}}(b) = C_{V_{j}}(G) and that 𝕂[x, a] = 𝕂[x, b]. In other words, for every 1-dimensional subspace U ≤ V_{j} either [U, G] = 0, or C_{G}(U) = 1 and U + [U, G] is 2dimensional. In the latter case it is easy to see that all such 2-dimensional 𝕂-modules are isomorphic. Moreover, x ↦ [x, a] induces an isomorphism from V_{j}/C_{V_{j}}(a) → [V_{j}, a]. Since C_{V_{j}}(a) = C_{V_{j}}(G) and [V_{j}, a] = [V_{j}, G], (b) follows. □

Lemma 3.14 [quadratic and tensor2] Assume Hypothesis 3.7 and in addition:

(i) [i] 𝕂 = 𝔽 and char 𝕂 = p,

(ii) [ii] J is a transitive G-set, and |J| ≥ 2.

(iii) [iii] dim_{𝕂} V_{j} ≥ 2 for j ∈ J.

(iv) [iv] |G| > 2 and V is a a faithful quadratic 𝕂G-module.

Then p = 2, |J| = 2, dim_{𝕂} V_{j} = 2, and [V_{j}, C_{G}(J)] = C_{V_{j}}(C_{G}(J)) is a 1-dimensional subspace of V_{j}.

Proof: Recall from 2.6(e) that G is an elementary abelian p-group. Put B := C_{G}(J) and fix j ∈ J. Then B = C_{G}(j) since G is abelian.

Let X_{j} be an 1-dimensional 𝕂B-subspace of V_{j}. We apply 3.12 with the notation given there. Then Δ is a system of imprimitivity for G in U/X, so we can apply 2.11. Since G acts transitively on J and quadratically on U/X, we are in case (3) of 2.11, so |J| = 2, say J = {1, 2}, and [U_{j}, B] ≤ X. As |G| > 2 we also get that B ≠ 1. Since the 𝕂B-modules U_{j}/X and V_{j}/X_{j} are isomorphic, [V_{j}, B] = X_{j}.

Pick 1 ≠ b ∈ B and a ∈ G \ B, and put C_{1} := C_{V_{1}}(b). Then by the quadratic action of G,

[C_{1} ⊗ V_{2}, b] = C_{1} ⊗ X_{2} ≤ C_{V} (G).

Hence 3.9 shows that dim_{𝕂} C_{1} = dim_{𝕂} X_{2} = 1, so also dim_{𝕂} X_{1} = 1. The quadratic action of b on V_{1} gives C_{1} = X_{1} and dim_{𝕂} V_{1} = 2. □

I no longer understand where to use the next lemma

Lemma 3.15 [commutators] Assume Hypothesis 3.7. Suppose that

(i) [i] G = A, and G is a finite abelian abelian p-group,

(ii) [ii] J := {1, 2}, σ = id, B := C_{G}(J) ≠ G, and [V, B] ≠ 0.

(iii) [iii] there exist non-zero 𝕂B-submodules Z_{1} and Y_{1} of V_{1} such that [Z_{1}, B] = 0 and V_{1} = Z_{1} ⊕ Y_{1}.

Pick a ∈ G \ B and set

Z_{2} := Z_{1}ϕ_{1,a}, Y_{2} := Y_{1}ϕ_{1,a}, Y := Z_{1} ⊗ Y_{2} + Y_{1} ⊗ Z_{2}.

Then Y is G-invariant. Moreover, G does not act quadratically on Y, or

dim_{𝕂} Z_{1} = dim_{𝕂}[V_{1}, B] = 1 = dim_{𝕂} V_{1}/C_{V_{1}}(B) = 1 and dim_{𝕂} V_{1} = 3.

Proof: Note that |G/B| = 2 and a^{2} ∈ B. Since G strictly normalizes V, we have

ϕ_{1,a}ϕ_{2,a} = ϕ_{1,a}ϕ_{1a,a} = ϕ_{1,a^{2}}.

It follows that Y_{2}ϕ_{2,a} = Y_{1}ϕ_{1,a^{2}} = Y_{1} and similarly Z_{2}ϕ_{2,a} = Z_{1}. This shows that (Z_{1} ⊗ Y_{2})a = Y_{1} ⊗ Z_{2}, and Y is A-invariant.

To prove the remaining claim we may assume by induction on |G| that B = ⟨b⟩. Observe that [V_{1}, B] = [Y_{1}, B] and [Y_{1} ⊗ Z_{2}, B] = [Y_{1}, B] ⊗ Z_{2}. Assume now that G acts quadratically on Y. Then [Y_{1}, B] ⊗ Z_{2} ≤ C_{Y} (G) and so by 3.9 dim_{𝕂} Z_{1} = dim_{𝕂} Z_{2} = dim_{𝕂}[V_{1}, B] = 1. On the other hand B = ⟨b⟩ and

dim_{𝕂} V_{1}/C_{V_{1}}(B) = dim_{𝕂} V_{1}/C_{V_{1}}(b) = dim_{𝕂}[V_{1}, b] = 1,

so dim_{𝕂} V_{1} = 3. □

4 Tensor Decompositions of Homogeneous Modules

Lemma 4.1 [simple tensor] Let 𝔽 be a finite field of characteristic p, and let D and E be subgroups of G such that [D, E] = 1. Suppose that V is a homogeneous 𝔽D-module and X is a simple 𝔽Dsubmodule of V. Then the following hold, where Y := Hom_{𝔽D}(X, V) and 𝕂 := End_{𝔽D}(X):

(a) [a] 𝕂 is a finite field and Y is a right 𝕂E-module via

αk : x ↦ (xk)α and αe : x ↦ (xα)e (x ∈ X, α ∈ Y, k ∈ 𝕂, e ∈ E).

(b) [b] X is an absolutely simple 𝕂D-module.^{1}

(c) [d] There exists an 𝔽(D × E)-module isomorphism Φ : X ⊗_{𝕂} Y → V with (x ⊗ α)Φ = xα for all x ∈ X and α ∈ Y.

(d) [c] For 𝕂-subspaces Z ≤ Y the map Z ↦ (X ⊗ Z)Φ is an E-invariant bijection between the 𝕂-subspaces Z of Y and the 𝔽D-submodules of V with inverse U ↦ Hom_{𝔽D}(X, U), U an 𝔽Dsubspace of V.

(e) [e] V is a simple 𝔽DE-module iff Y is a simple 𝕂E-module.

(f) [f] End_{𝔽D}(V) ≅ End_{𝕂}(Y) and 𝕂 ≅ Z(End_{𝔽D}(V)).

Proof: (a): By Schur’s Lemma 𝕂 is a divisionring. As dim_{𝔽} X is finite, also dim_{𝔽} K is finite. Now the finiteness of 𝔽 shows that 𝕂 is finite, so by Wedderburn’s Theorem 𝕂 is a finite field. The remaining assertions in (a) are readily verified.

(b): Since 𝔽 and 𝕂 are commutative we have 𝔽 ≤ 𝕂 and 𝕂 ≤ End_{𝕂D}(X), so 𝕂 ≤ End_{𝕂D}(X) ≤ End_{𝔽D}(X) = 𝕂.

(c): Define

Φ : X ⊗_{𝕂} Y → V with x ⊗ α ↦ xα.

If x ⊗ α = x′ ⊗ α′, then there exist k, h ∈ 𝕂 with x′ = xk, α′ = αh and kh = 1, so

x′α′ = (xk)(αh) = (xkh)α = xα.

Hence Φ is a map, and it is evident that this map is 𝔽-linear.

Let (d, e) ∈ D × E. Then

((x ⊗ α)(d, e))Φ = (xd ⊗ αe)Φ = (xd)(αe) = (xdα)e = (xαd)e = (xα)(de) = (x ⊗ α)Φ · (de),

so Φ is 𝔽(D × E)-linear.

(d): Let W be an 𝔽D-submodule of V. Since V is homogenous as an 𝔽D-module there exist simple 𝔽D-submodules X_{1}, … X_{n} with W = ⨁_{i=1}^{n} X_{i}, X_{i} ≅ X as an 𝔽D module. Choose α_{i} ∈ Y with Xα_{i} = X_{i} and set Z := ∑_{i=1}^{n} α_{i}𝕂 ≤ Y. Then (X ⊗ α_{i})Φ = X_{i} and so (X ⊗_{𝕂} Z)Φ = W. Moreover,

Z ≤ Hom_{𝔽D}(X, W) ≅ ⨁_{i=1}^{n} Hom_{𝔽D}(X, X_{i}) ≅ ⨁_{i=1}^{n} Hom_{𝔽D}(X, X) ≅ 𝕂^{n}.

It is evident that the elements α_{1}, …, α_{n} are linearly independent in Y. Hence dim_{𝕂} Z = n, so Z ≅ 𝕂^{n} and Z = Hom_{𝔽D}(X, W). This shows that Z ↦ (X ⊗ Z)Φ gives a a bijection between the 𝕂-subspaces of Y and the 𝔽D-submodules of V. Because of (c) this bijection is E-invariant.

1An 𝔽G-module V is absolutely simple if End_{𝔽G}(V) = 𝔽.

(e): This is a direct consequence of (d) since a 𝕂-subspace of Y is E-invariant if and only if the corresponding 𝔽D-submodule of V is E-invariant.

(f): Let x ∈ X, α ∈ Y and β ∈ End_{𝔽D}(V). Then the map

αβ : x ↦ (xα)β

is in Y, and Y is an End_{𝔽D}(V)-module. Moreover, for k ∈ 𝕂

x((αβ)k) = xk(αβ) = (xkα)β = x(αk)β = x((αk)β),

so End_{𝔽D}(V) acts 𝕂-lineary on Y. Hence we obtain a ring homomorphism End_{𝔽D}(V) → End_{𝕂} Y. Observe that X ⊗_{𝕂} Y is an End_{𝕂}(Y)-module via

(x ⊗ α)δ = x ⊗ αδ (x ∈ X, α ∈ Y, δ ∈ End_{𝕂}(Y)).

So by (c) V is an End_{𝕂}(Y)-module with ((x ⊗ α)Φδ := (x ⊗ α)δ)Φ. Since Φ is an 𝔽(D × E)-module homomorphism this action of End_{𝕂}(Y) on V is 𝔽D-linear. Hence, we obtain a ring homomorphism End_{𝕂}(Y) → End_{𝔽D}(V) which is inverse to the one from the previous paragraph. Thus the first statement in (f) holds. Since Z(End_{𝕂}(Y)) ≅ 𝕂, the second statement follows from the first. □

Lemma 4.2 [pure tensors] Let 𝔽 be a finite field of characteristic p and (D_{i}, i ∈ I) be a finite family of subgroups of G with H = ⟨D_{i} | i ∈ I⟩ and [D_{i}, D_{j}] = 1 for all i, j ∈ I. Suppose that V is a simple 𝔽G-module and put 𝕂 := End_{𝔽G}(V). Then the following hold:

(a) [a] For every i ∈ I there exists an absolutely simple 𝕂D_{i}-module V_{i} isomorphic to a 𝕂D_{i}- submodule of V, and there exists a 𝕂(×_{i∈I} D_{i})-isomorphism Φ : ⊗_{𝕂}^{I}V_{i} → V (where D_{i} acts trivially on V_{j} for i ≠ j).

(b) [b] For 0 ≠ v ∈ V the following two statements are equivalent:

1. [1] For all i ∈ I, v𝕂D_{i} is a simple 𝕂D_{i}-submodule of V.

2. [2] There exist 0 ≠ v_{i} ∈ V_{i} with v = (⊗v_{i})Φ.

Proof: Replacing 𝔽 be 𝕂 we may assume that V is an absolutely simple 𝔽G-module and so 𝔽 = 𝕂. By induction on |J| we may also assume that |J| = 2, say J = {1, 2}. Let V_{1} be a simple 𝕂D_{1}- submodule of V and put V_{2} := Hom_{𝕂D_{1}}(V_{1}, V).

(a): Let 𝔼 := End_{𝕂D_{1}}(V_{1}). By 4.1 there exists a 𝕂(D_{1} × D_{2})-isomorphism

Φ : V_{1} ⊗_{𝔼} V_{2} → V with (w ⊗ α)Φ = wα (w ∈ V_{1}, α ∈ V_{2}).

It follows that End_{𝕂(D_{1}×D_{2})}(V_{1} ⊗_{𝔼} V_{2}) ≅ End_{𝕂G}(V). Since V is absolutely simple and 𝕂 ≤ 𝔼 ≤ End_{𝕂(D_{1}×D_{2})}(V_{1} ⊗_{𝔼} V_{2}), we get that 𝕂 = 𝔼, and V_{1} is an absolutely simple 𝕂 D_{1}-module.

Let 0 ≠ v_{1} ∈ V_{1}. Then, again by 4.1, V_{2} is isomorphic to the 𝕂D_{2}-submodule (v_{1} ⊗ V_{2})Φ of V, and V_{2} is a simple 𝕂D_{2}-module since V is a simple 𝕂G-module. So by symmetry also V_{2} is an absolutely simple 𝕂D_{2}-module.

(b): Let 0 ≠ v ∈ V. Suppose first that (b:1) holds. Since V is a homogeneous 𝕂D_{1}-module, there exists an 𝕂D_{1}-isomorphism α ∈ V_{2} such that V_{1}α = v𝕂D_{1}. Put v_{1} = vα^{−1}. Then (v_{1} ⊗ α)Φ = v.

Suppose next that (b:2) holds. Then v𝕂D_{1} = (v_{1} ⊗ v_{2})Φ𝕂D_{1} = (V_{1} ⊗ v_{2})Φ ≅ V_{1} as 𝕂D_{1}-module and similarly v𝕂D_{2} ≅ V_{2} as 𝕂D_{2}-module. Thus (b:1) holds. □

Definition 4.3 [def:tensor decomposition] A tensor decomposition of V for G is tuple

(Φ, ⊗_{𝕂}^{J}V_{j}, σ, (ϕ_{j,h} | j ∈ J, h ∈ G)),

where 𝕂 is a field extension of 𝔽, and V_{j}, j ∈ J, are finite dimensional 𝕂-vectorspaces satisfying

(i) [i] G normalizes the tensor product ⊗_{𝕂}^{J}V_{j} with respect to σ and (ϕ_{j,h} | j ∈ J, h ∈ G),

(ii) [ii] Φ : ⨂_{𝕂}^{J} V_{j} → V is an 𝔽-isomorphism such that

(⊗v_{j})Φg = (⊗ v_{jg^{−1}}ϕ_{jg^{−1},g})Φ (g ∈ G, v_{j} ∈ V_{j}, j ∈ J).

The tensor decomposition is called strict if ϕ_{j,a}ϕ_{ja,b} = ϕ_{j,ab} for every j ∈ J and a, b ∈ G.

Lemma 4.4 [tensor decomposition for homogenous] Let 𝔽 be a finite field of characteristic p, G be a finite group and I a finite G-set. Further let T be a p-subgroup of G and (D_{i}, i ∈ I) be a family of subgroups of G with D := ⟨D_{i} | i ∈ I⟩. Suppose that

(i) [i] D_{i}^{h} = D_{ih} and [D_{i}, D_{j}] = 1 for all i ≠ j ∈ I, h ∈ H, and

(ii) [ii] V is a homogeneous 𝔽D-module.

Put 𝕂 := Z(End_{𝔽D}(V)), J := I ⊎ {0}, and view J has a G-set with G fixing 0. Then there exist absolutely simple 𝕂 D_{i}-modules V_{i}, i ∈ I, a finite dimensional 𝕂-vectorspace V_{0} and a tensor decomposition (Φ, ⊗_{𝕂}^{J}V_{j}, σ, (ϕ_{j,g}, j ∈ J, g ∈ H)) such that the following hold:

(a) [a] V_{j} is an absolutely simple 𝕂 D_{j}-module for j ≠ 0, and V_{0} is a trivial 𝕂D-module. Moreover, every simple D_{j}-submodule of V is isomorphic to V_{j} as a 𝕂D_{j}-module.

(b) [b] Φ : ⊗_{𝕂}^{J}V_{j} → V is a 𝕂(×_{i∈I} D_{i})-module isomorphism (where D_{i} acts trivially on V_{j} for j ≠ i).

(c) [c] The tensor decomposition restricted to T is strict.

(d) [d] For fixed i ∈ I the map ϕ_{i} : d ↦ ϕ_{i,d} is a homomorphism from D_{i} into GL_{𝔽}(V_{i}).

Proof: To simplify notation we assume without loss that V is a faithful 𝔽G-module and that G is subgroup of GL_{𝔽}(V). Let D_{0} := GL_{𝔽D}(V), and let R_{j} be the subring of End_{𝔽}(V) spanned by 𝕂 and D_{j}, j ∈ J. By 4.1(f),(e) (with D_{0} in place of E) V is a simple 𝔽DD_{0}-module and so V is an absolutely simple 𝕂DD_{0}-module. Thus 4.2 implies:

1^{∘} [1] There exist absolutely simple 𝕂D_{j}-modules V_{j} and a 𝕂(×_{i∈J} D_{j})-isomorphism

Φ : ⊗_{𝕂}^{J}V_{j} → V.

The number of elements of the form (⊗v_{j})Φ, 0 ≠ v_{j} ∈ V_{j}, is not divisible by p and so we can choose 0 ≠ w_{j} ∈ V_{j}, j ∈ J, such that w := (⊗w_{j})Φ is centralized by T.

Let g ∈ G. Then for j ∈ J, wR_{j} is a simple R_{j}-module. Since g normalizes 𝕂 and D_{j}^{g} = D_{jg} we have R_{j}^{g} = R_{jg} and so wR_{j}g is a simple R_{jg}-module. Also wR_{j}g = wgR_{jg} and so for j ∈ J, wgR_{jg} is a simple R_{jg}-module. Thus by 4.2(b), there exist 0 ≠ u_{j} ∈ V_{j}, j ∈ J, with wg = (⊗u_{j})Φ. If g ∈ T then wg = w and so we can choose u_{j} = w_{j}.

Let v_{j} ∈ V_{j}. Since V_{j} is a simple R_{j}-module there exists r_{j} ∈ R_{j} with v_{j} = w_{j}r_{j}. Next we show:

2^{∘} [2] Let r_{j}, s_{j} ∈ R_{j} with w_{j}r_{j} = w_{j}s_{j}. Then u_{jg}r_{j}^{g} = u_{jg}s_{j}^{g}.

Put t_{j} = r_{j} − s_{j}. Then w_{j}t_{j} = 0 and since Φ is R_{j}-linear wt_{j} = 0. Thus wgt_{j}^{g} = wt_{j}g = 0 and since Φ is also R_{jg}-linear, (⊗^{k∈J}u_{k})t_{j}^{g} = 0. Since t_{j}^{g} ∈ R_{jg} this implies u_{jg}t_{j}^{g} = 0 and so u_{jg}r_{j}^{g} = u_{jg}s_{j}^{g}.

We now define

ϕ_{j,g} : V_{j} → V_{jg} with v_{j} → u_{jg}r_{j}^{g}, where r_{j} ∈ R_{j} and v_{j} = w_{j}r_{j}.

For g = 1 we have chosen w_{j} = u_{j}, so we get in this case that v_{j} = u_{j}r_{j} = u_{j1}r_{j}^{1} and v_{j}ϕ_{j,1} = v_{j}. Hence ϕ_{j,1} = id_{V_{j}}.

Using (2^{∘}) we get

3^{∘} [3] ϕ_{j,g} is independent from the choice of r_{j}.

Clearly ϕ_{j,g} is a homomorphism of the additive group V_{j}. Next we define a homomorphism σ : G → Aut(𝕂). Observe that for fixed g ∈ G,

g^{σ} : 𝕂 → 𝕂 with k ↦ k^{g}

is an element of Aut_{𝔽}(𝕂) and so g ↦ g^{σ} defines the desired homomorphism σ.

Let k ∈ 𝕂. Since v_{j}k = w_{j}r_{j}k = w_{j}(r_{j}k) and r_{j}k ∈ R_{j}, the definition of ϕ_{j,g} shows that

(v_{j}k)ϕ_{j,g} = u_{jg}(r_{j}k)^{g} = u_{jg}r_{j}^{g}k^{g} = v_{j}ϕ_{j,g}k^{g^{σ}}.

Hence ϕ_{i,g} is g^{σ}-semi-linear.

To verify 4.3(ii) we compute

(∗)  \begin{aligned}(⊗v_{jg^{−1}}ϕ_{jg^{−1},g})Φ & = (⊗u_{j}r_{jg^{−1}}^{g})Φ = (⊗u_{j})Φ ∏ r_{jg^{−1}}^{g} = wg(∏ r_{j})^{g} = w(∏ r_{j})g \\ & = (⊗w_{j})Φ(∏ r_{j})g = (⊗w_{j}r_{j})Φg = (⊗v_{j})Φg. \end{aligned}

This is 4.3(ii).

Let g, h ∈ G, v := ⊗v_{j}, and v_{j}^{∗} := v_{jg^{−1}}ϕ_{jg^{−1},g}. Since v(gh) = (vg)h we get from 4.3(ii):

v(gh) = (⊗v_{j(gh)^{−1}}ϕ_{j(gh)^{−1},gh})ϕ =
(vg)h = (⊗v_{jg^{−1}}ϕ_{jg^{−1},g})Φh = (⊗v_{j}^{∗})Φh
= (⊗v_{jh^{−1}}^{∗}ϕ_{jh^{−1},h})Φ = (⊗v_{jh^{−1}g^{−1}}ϕ_{jh^{−1}g^{−1},g}ϕ_{jh^{−1},h})Φ.

Since Φ is bijective, this implies:

(∗∗)  ⊗(v_{jh^{−1}g^{−1}}ϕ_{jh^{−1}g^{−1},g}ϕ_{jh^{−1},h}) = ⊗(v_{j(gh)^{−1}}ϕ_{j(gh)^{−1},gh}).

Hence H normalizes the tensorproduct ⊗_{𝕂}^{J}V_{j} with respect to σ and (ϕ_{j,h}/j ∈ J, h ∈ H).

Observe that by 3.6(b) V_{j} is a semi-linear 𝕂D-module (and thus an 𝔽D-module) if λ_{j,a,b} = 1 for a, b ∈ D. Hence to prove (c) and (d) it suffices to show that the u_{j}, j ∈ J, can be chosen such that λ_{r,a,b} = 1 for a, b ∈ T and a, b ∈ D_{r}, r fixed.

If a ∈ T, we have already seen that we can choose w_{j} = u_{j} and thus

w_{j}ϕ_{ja,a}ϕ_{ja,b} = w_{ja}ϕ_{ja,b} = w_{jab} = w_{j}ϕ_{j,ab},

and λ_{k,a,b} = 1.

In the second case, let d ∈ D_{r}, and observe that d induces the identity on V_{j} for j ≠ k. It follows that λ_{j,a,b} = 1 for j ≠ r and thus also λ_{r,a,b} = 1 since by 3.6(d) ∏_{j∈J} λ_{j,a,b} = 1. □

5 Tensor Products and Nearly Quadratic Modules

In this section p is a prime, 𝕂 is a field of characteristic p, A is a finite elementary abelian p-group, J is a finite A-set. Moreover A strictly normalizes the tensor product ⊗_{𝕂}^{J}V_{j} with respect to σ and (ϕ_{j,h} | j ∈ J, h ∈ A), as defined in 3.5. The fixed field of A in 𝕂 is denoted by 𝕂_{A}.

Lemma 5.1 [nearly quadratic and tensor] Let 𝕂_{A} = 𝕂, J = {1, 2}, J be a trivial A-set, and dim_{𝕂} V_{j} ≥ 2 for j ∈ J. Suppose that there exists a subfield 𝔽 ≤ 𝕂 such that V is a nearly quadratic 𝔽A-module but not a quadratic 𝔽A-module. Then the following hold for j ∈ J:

(a) [z] A acts quadratically and non-trivially on V_{j}.

(b) [a] [V_{j}, A] is a 𝕂-hyperplane of V_{j}.

(c) [b] [𝔽z, A] = [V_{j}, A] for all z ∈ V_{j} \ [V_{j}, A].

(d) [c] Q_{V} (A) = [V_{1}, A] ⊗ V_{2} + V_{1} ⊗ [V_{2}, A] is a 𝕂-hyperplane of V.

(e) [d] One of the following holds:

1. [a] C_{V} (A) = [V_{1}, A] ⊗ [V_{2}, A] and if 𝔽 = 𝔽_{p} then A = C_{A}(V_{1})C_{A}(V_{2}).

2. [b] dim_{𝕂} C_{V} (A) = 2, char F ≠ 2, dim_{𝕂} V_{1} = 2 = dim_{𝕂} V_{2}, and V_{1} and V_{2} are isomorphic as 𝕂A-modules.

(f) [e] If F = 𝔽_{p} then A induces on Q_{𝕂}(V_{j}, [V_{j}, A]) on V_{j}.

Proof: As A is not quadratic on V, by 2.5 Q_{V} (A) = [V, A] + C_{V} (A) is the largest quadratic 𝔽A-submodule of V and V ≠ Q_{V} (A). But then Q_{V} (A) is a 𝕂-space, so Q_{V} (A) is the also the largest quadratic 𝕂A-submodule of V.

Put U := C_{V_{1}}(A) and W := C_{V_{2}}(A). Observe that for v_{1} ∈ V_{1} and v_{2} ∈ V_{2}

(∗)  [v_{1} ⊗ v_{2}, a] = [v_{1}, a] ⊗ [v_{2}, a] + v_{1} ⊗ [v_{2}, a] + [v_{1}, a] ⊗ v_{2}.

To simplify notation and since the hypothesis is symmetric in V_{1} and V_{2} we show the claims for j = 1.

1^{∘} [0] A acts non-trivially on V_{1}.

Assume that [V_{1}, A] = 0. Then dim_{𝕂} V_{1} ≥ 2 implies that V_{1} = Z_{1} ⊕ Z_{2} with Z_{1} ≠ 0 ≠ Z_{2}. Thus V = (Z_{1} ⊗ V_{2}) ⊕ (Z_{2} ⊗ V_{2}), and by 2.7 A centralizes one of these submodules, say Z_{1} ⊗ V_{2}. But then (∗) shows that A also centralizes V_{2}, a contradiction since A is not quadratic on V.

2^{∘} [1] A is quadratic on V_{1}, and (a) holds.

Suppose A is not quadratic V_{1}. By (1^{∘}) [V_{2}, A] ≠ 0. On the other hand, since char K = p and A is a p-group, [V_{2}, A] < V_{2}. Let v_{1} ∈ V_{1} and a, b ∈ A with [v_{1}, a, b] ≠ 0. Then A is not quadratic on \overline{V} = V/V_{1} ⊗ [V_{2}, A] since by (∗)

[\overline{v_{1} ⊗ v_{2}}, a, b] = \overline{[v_{1}, a, b] ⊗ v_{2}} ≠ 0 (0 ≠ v_{2} ∈ V_{2} \ [V_{2}, A]).

Similarly A is not quadratic on [V_{1} ⊗ [V_{2}, A]/V_{1} ⊗ [V_{2}, A, A]. But this contradicts 2.5. This shows that A acts quadratically on V_{1}, and (a) follows from (1^{∘}).

3^{∘} [1-2] U = [V_{1}, A] and W = [V_{2}, A].

By (2^{∘}) [V_{1}, A] ≤ U. Suppose [V_{1}, A] ≠ U. Then there exit 𝕂A-submodules X_{1} and X_{2} in V_{1} such that

X_{1} ≤ U, [X, A] ≤ X_{2} and V_{1} = X_{1} ⊕ X_{2}.

Thus V = (X_{1} ⊗ V_{2}) ⊕ (X_{2} ⊗ V_{2}), and again by 2.7 A centralizes (X_{1} ⊗ V_{2}) or (X_{2} ⊗ V_{2}). In the first case (∗) shows that [V_{2}, A] = 0 which contradicts (1^{∘}). In the second case A also centralizes [V_{1}, A] ⊗ V_{2} and [V_{1}, A, A] = 0. Hence again (∗) yields [V_{2}, A] = 0, a contradiction. 4^{∘} [2] U ⊗ V_{2} + V_{1} ⊗ W ≤ Q_{V} (A).

By (∗) A acts quadratically on U ⊗ V_{2} + V_{1} ⊗ W, so (4^{∘}) follows.

Since V ≠ Q_{V} (A) we can choose x ∈ V_{1} and y ∈ V_{2} with x ⊗ y ∉ Q_{V} (A). By (4^{∘}) x ∉ U and y ∉ W. 5^{∘} [3] Q_{V} (A) = x ⊗ W + U ⊗ y + C_{V} (A).

By the definition of nearly quadratic we have Q_{V} (A) = [𝔽(x ⊗ y), A] + C_{V} (A). So (5^{∘}) follows from (∗), (3^{∘}) and (4^{∘}). 6^{∘} [3-5] Let t ∈ V_{1} such that t ⊗ y ∈ Q_{V} (A). Then t ∈ U.

By (5^{∘}) there exist u ∈ U, w ∈ W and c ∈ C_{V} (A) such that t ⊗ y = x ⊗ w + u ⊗ y + c. For a ∈ A, (∗) yields

[t ⊗ y, a] = [t, a] ⊗ [y, a] + [t, a] ⊗ y + t ⊗ [y, a] = [x, a] ⊗ w + u ⊗ [y, a].

Hence (3^{∘}) gives [t, a] ⊗ y ∈ V_{1} ⊗ W. Since y ∉ W we get [t, a] = 0 and thus t ∈ C_{V_{1}}(A) = U. 7^{∘} [4] dim_{𝕂} V_{1}/U = 1.

We split the proof of (7^{∘}) into three cases. Case 1 [case 1] dim_{𝕂} U ≥ 2.

In this case we can choose two distinct 1-dimensional 𝕂-subspaces U_{1} and U_{2} in U. Since U_{1} ⊗ V_{2} ∩ U_{2} ⊗ V_{2} = 0 there exists i ∈ {1, 2} such that [V, A, A] ≰ U_{i} ⊗ V_{2}. As the 𝕂A-modules V/(U_{i} ⊗ V_{2}) and (V_{1}/U_{i}) ⊗ V_{2} are isomorphic, induction on dim V_{1} (together with (3^{∘})) shows that dim_{𝕂}(V_{1}/U_{i})/[V_{1}/U_{i}, A] = 1 and thus also dim_{𝕂} V_{1}/[V_{1}, A] = 1 since U_{i} ≤ [V_{1}, A]. Hence again (3^{∘}) gives (7^{∘}). Case 2 [case 2] dim_{𝕂} W ≥ 2.

Similar to the previous case we choose two distinct 1-dimensinal 𝕂A subspaces W_{1} and W_{2} in W and i ∈ {1, 2} such that [V, A, A] ≰ X ⊗ W_{i}. Thus A acts not quadratically on X ⊗ (Y/W_{i}) and again by induction (7^{∘}) holds. Case 3 [case 3] dim_{𝕂} U = 1 and dim_{𝕂} W = 1.

We may assume that dim_{𝕂} V_{1} ≥ 3. By (4^{∘}), V_{1} ⊗ W ≤ Q_{V} (A) and by (5^{∘}), dim_{𝕂} Q_{V} (A)/(U ⊗ y + C_{V} (A)) ≤ 1. Hence R := V_{1} ⊗ W ∩ (U ⊗ y + C_{V} (A)) contains a hyperplane of V_{1} ⊗ W. Moreover U ⊗ W ≤ R since U ⊗ W ≤ C_{V} (A). As dim_{𝕂} V_{1} ≥ 3, there exists t ∈ V_{1} \ U and w ∈ W such that t ⊗ W ≤ R. As W is 1-dimensional, also t ⊗ W ≤ R.

From (6^{∘}) we get that t ⊗ y ∉ Q_{V} (A), so (5^{∘}) applies with t in place of x; i.e., Q_{V} (A) = t⊗W +U ⊗y +C_{V} (A). But then the choice of t yields Q_{V} (A) = U ⊗y +C_{V} (A), and V_{1} ⊗W ∩C_{V} (A) is a hyperplane of V_{1} ⊗ W containing U ⊗ W. It follows that C_{V_{1}}(A) = U is a hyperplane of V_{1} contradicting dim_{𝕂} U = 1 and dim_{K} V_{1} ≥ 3.

8^{∘} [6*] (b) and (d) hold.

Claim (b) follows from (7^{∘}) and (3^{∘}). In particular, U ⊗ V_{2} + V_{1} ⊗ W is a 𝕂-hyperplane of V. Since Q_{V} (A) is the largest quadratic 𝕂A-submodule of V, we get (d).

9^{∘} [5] Suppose that C_{V} (A) = U ⊗ W. Then (c), (e:1) and (f) hold.

To prove (c) let z ∈ V_{1} \ U. According to (6^{∘}) we may assume that z = x. From (4^{∘}), (5^{∘}) we get

U ⊗ y ≤ Q_{V} (A) = [𝔽(x ⊗ y), A] + U ⊗ W ⊆ 𝔽{[x, a] | a ∈ A} ⊗ y + x ⊗ W + U ⊗ W.

Since y ∉ W we get that U ⊗ y = [𝔽x, A] ⊗ y and U = [𝔽y, A]. This is (c).

The first part of (e:1) is true by assumption, so we can assume that 𝔽 = 𝔽_{p}. Then

U ⊗ y + x ⊗ W ≤ [𝔽x ⊗ y, A] + U ⊗ W = {[x, a] ⊗ y + x ⊗ [y, a] | a ∈ A} + U ⊗ W.

Hence, for every u ∈ U and w ∈ W, there exists a ∈ A with [x, a] = u and [y, a] = w. The particular case when u = 0 (or w = 0) gives (e:1). Moreover, (f) follows. define or quote Q_{𝕂}(V_{j}, [V_{j}, A])

10^{∘} [6] Suppose that C_{V} (A) ≠ U ⊗ W. Then (c), (e:2) and (f) hold.

Since by (5^{∘}) U ⊗ W < C_{V} (A) ≤ U ⊗ W + U ⊗ y + x ⊗ W, we conclude that there exist u ∈ U and w ∈ W with 0 ≠ u ⊗ y − x ⊗ w ∈ C_{V} (A). Hence for all a ∈ A

(∗∗)  u ⊗ [y, a] = [x, a] ⊗ w.

Suppose that u = 0. By the choice of x, there exists a ∈ A with [x, a] ≠ 0, so w = 0, which contradicts u ⊗ y − x ⊗ w ≠ 0. Hence u ≠ 0 and similarly also w ≠ 0. Then (∗∗) implies that [x, a] ∈ 𝕂u for all a ∈ A. By (3^{∘}) and (7^{∘}), U is 1-dimensional and V_{1} 2-dimensional, and similarly dim_{𝕂} W = 1 and dim_{𝕂} V_{2} = 2.

For a ∈ A define λ(a), μ(a) ∈ 𝕂 by [x, a] =: λ(a)u and [y, a] =: μ(a)w. Then by (∗∗), λ(a) = μ(a) and so V_{1} and V_{2} are isomorphic as 𝔽a-spaces. Moreover,

C_{V} (A) = U ⊗ W + 𝕂(u ⊗ y − x ⊗ w).

Let L be the 𝔽-subspace of 𝕂 spanned by {λ(a) | a ∈ A}. Then

Q_{V} (A) = [𝔽x ⊗ y, A] + C_{V} (A) = L(u ⊗ y + x ⊗ w) + 𝕂(u ⊗ y − x ⊗ w) + U ⊗ W.

Let k ∈ 𝕂. Then kx ⊗ w ∈ Q_{V} (A), so there exists ℓ ∈ L and s ∈ 𝕂 with

kx ⊗ w = ℓ(u ⊗ y + x ⊗ w) + s(u ⊗ y − x ⊗ w).

This implies s = −l and k = 2l. Since k ∈ 𝕂 was arbitray we conclude that char 𝔽 ≠ 2 and L = 𝕂. Thus (e:2) and (c) holds.

If 𝔽 = 𝔽_{p}, then 𝕂 = L = {λ(a) | a ∈ A} and so also (f) is proved. □

Lemma 5.2 [nearly quadratic and tensor decompositions] Suppose that

(i) [i] |A| > 2, |J| ≥ 2, and dim_{𝕂} V_{j} ≥ 2 for j ∈ J,

(ii) [ii] there exists 𝔽 ≤ 𝕂_{A} such that V is a faithful nearly quadratic 𝔽A-module.

Then A acts 𝕂-linearly on V, and one of the following holds, where B := C_{A}(I):

1. [2] A is quadratic on V, and there exists j ∈ J such that A centralizes V_{i} for all i ∈ J \ {j}.

2. [3] p = 2, A is quadratic on V, and there exists an A-invariant subset J_{0} in J with |J_{0}| = 2 such that A centralizes V_{i} for all i ∈ J \ J_{0}. Moreover one of the following holds:

1. [1] A acts trivially on J_{0}, and there exists a faithful 2-dimemsional 𝕂A-module W, a trivial 𝕂A-modules C_{j} and a postive integer k_{j} such that V_{j} ≅ W^{k_{j}} ⊕ C_{j} for j ∈ J_{0}.

2. [2] A acts non-trivially on J_{0}, dim_{𝕂} V_{j} = 2 and C_{B}(V_{j}) = C_{B}(V) for all j ∈ J.

3. [4] |J| = 2, A is not quadratic on V and for j ∈ J, [V_{j}, B] = C_{V_{j}}(B) is a 𝕂-hyperplane of V_{j}. Moreover one of the following holds:

1. [1] A acts trivially on J, and [V_{j}, A] = [𝔽v_{j}, A] for v_{j} \ [V_{j}, A], j ∈ J.

2. [2] A acts non-trivially on J, p = 2, 𝔽 = 𝕂, and C_{B}(V_{j}) = C_{B}(V) for all j ∈ J.

Proof: The proof is by induction on |A| and |J|. Note that dim_{𝕂} V ≥ 4 since |J| ≥ 2 and dim_{𝕂} V_{j} ≥ 2. First we show:

1^{∘} [11] A acts 𝕂-linearly on V.

Let A_{0} := ker σ and assume that A ≠ A_{0}. Then 2.14 yields p = 2, [V, A_{0}] = C_{V} (A_{0}), and dim_{𝕂} V/C_{V} (A_{0}) = 1. If |A_{0}| = 2, then dim_{𝕂}[V, A_{0}] = 1 and so dim_{K} V = 2, which contradicts dim_{𝕂} V ≥ 4. If |A_{0}| > 2, then we can apply induction with A_{0} in place of A since A_{0} acts quadratically on V. Hence one of the cases (1) or (2) holds for A_{0}. In both cases |A_{0}/A_{0} ∩ B| ≤ 2, so [V_{r}, A_{0} ∩ B] ≠ 0 for some r ∈ J. Hence 3.11 shows that C_{V} (A_{0} ∩ B) is not a 𝕂-hyperlane of V. This contradiction shows that A acts 𝕂-linearly on V.

Case 1 [non-transitive] A is not transitive on J.

Let L be an orbit of A on J. We choose L in such a way that |L| is minimal and that A centralizes V_{j} for j ∈ L if this is possible. Put I := J \ L, V_{L} := ⨂_{j∈L} V_{j} and V_{I} = ⊗_{i∈I}V_{i}. Then A stricly normalizes the tensor product V_{L} ⊗_{κ} V_{I} ≅ V, so by (1^{∘}) A induces 𝕂-linear transformations on V_{L} and V_{I}.

2^{∘} [22] A acts quadratically on V_{L} and V_{I}.

If A acts quadratically on V, then by 3.13(a) A also acts quadratically on V_{L} and V_{I}. In the other case 5.1 shows that A acts quadratically on V_{L} and V_{I}.

3^{∘} [33] Suppose that A centralizes V_{L}. Then case (1) or (2) holds.

Note that by our choice of L and 3.10, |L| = 1. Moreover, the faithful action of A on V shows that A acts faithfully on V_{I}.

If |I| = 1, then (1) holds. If |I| > 1 then by induction on J we see that (1) or (2) holds for V_{I} since A is quadratic on V_{I}. But then the same case also holds for V.

4^{∘} [44] Suppose that A acts non-trivially on V_{L} and V_{I}. Then (2:1) or (3:1) holds.

By our choice of L, A does not centralize any V_{j} for j ∈ J. Assume first that A acts quadratically on V. If |L| = |I| = 1, then 3.13 shows that (2:1) holds. If |I| > 1, then by induction |I| = 2 and (2:1) or (3:1) holds for V_{I}. how do you continue? The reference you used does not apply to the quadratic case.

Assume now that A is not quadratic on V. Then by 5.1(b) (with V_{1} = V_{L} and V_{2} = V_{I}), C_{V_{I}} (A) is a 𝕂-hyperplanes of V_{I}. Hence by 3.11 (applied to V_{I}) |I| = 1, and thus by our choice of L also |L| = 1. Now 5.1(c) gives (3:1).

Case 2 [transitive] A is transitive on I.

Let X_{1} be a 1-dimensional 𝕂B-submodule of V_{1}. We apply 3.12 (with A in place of G) and use the notation introduced there.

5^{∘} [55] Either |J| ≥ 3 and 2.11 (4:1) or (4:2) holds for Δ and A, or |J| = 2 and 2.11 (3) or (4:3) holds for Δ and A. In particular [V_{1}, B, B] ≤ X_{1}.

This follows from 2.11 using the transitivity of A on Δ.

6^{∘} [66] |J| = 2, in particular B ≠ 1.

If |J| ≥ 3, then by (5^{∘}) A is not quadratic on U/X. But A is also not quadratic on V/U (how does one calculate this?) , which contradicts 2.5. Now |A/B| = |J| = 2 and thus B ≠ 1 since |A| ≥ 3.

7^{∘} [66.5] Suppose that A is quadratic on V. Then (2:2) holds.

This follows from (6^{∘}) and 3.14.

8^{∘} [77] Suppose that C_{V_{1}}(B) ≠ X_{1}. Then (2:2) or (3:2) holds.

There exists a 1-dimensional 𝕂-subspace X′_{1} of V_{1} different from X_{1}. Hence by (5^{∘}) [V_{j}, B, B] ≤ X_{1} ∩ X′_{1} = 0, so B acts quadratically on V_{1}. By (6^{∘}) |J| = 2, say J = {1, 2}, |A/B| = 2 and B ≠ 1.

Assume first that C_{V_{1}}(b) ≠ [V_{1}, B]. Then there exists a 𝕂B-submodule Z_{1} ≤ C_{V_{1}}(B) with Z_{1} ∩ [V_{1}, B] = 0. Hence, there also exists a 𝕂B-submodule Y_{1} ≤ V_{1} with X_{1} ≤ Y_{1} and V_{1} = Y_{1} ⊕ Z_{1}. Pick a ∈ A \ B and put

Z_{2} := Z_{1}ϕ_{1,a}, Y_{2} := Y_{1}ϕ_{1,a}, Y := Z_{1} ⊗ Y_{2} + Y_{1} ⊗ Z_{2}, D := Y_{1} ⊗ Y_{2}.

Then Y, D and Z_{1} ⊗ Z_{2} are 𝕂A-submodules of V and

V = (Z_{1} ⊕ Y_{1}) ⊗ (Z_{2} ⊕ Y_{2}) = (Z_{1} ⊗ Z_{2}) ⊕ (Z_{1} ⊗ Y_{2}) ⊕ (Y_{1} ⊗ Z_{2}) ⊕ (Y_{1} ⊗ Y_{2}) = (Z_{1} ⊗ Z_{2}) ⊕ Y ⊕ D.

As dim_{𝕂} Y_{1} > 1, 3.9 implies that A neither centralizes Y nor D. Hence, 2.7 shows that A is quadratic on Y ⊕ D. On the other hand [Z_{1} ⊗ Z_{2}, B] = 0 and a acts as an involution on Z_{1} ⊗ Z_{2}. It follows that A acts quadratically on V, and (7^{∘}) implies (8^{∘}).

Assume now that C_{V_{1}}(B) = [V_{1}, B]. We apply (5^{∘}). If 2.11 (4:3) holds, then (3:2) follows. If 2.11 (3) holds, then [V_{1}, B] = X_{1} and so C_{V_{1}}(B) = X_{1}, a contradiction.

9^{∘} [88] Suppose that C_{V_{1}}(B) = X_{1}. Then (2:2) holds.

If dim_{𝕂} V_{1} = 2, then (2:2) follows. Hence we may assume that dim_{𝕂} V_{1} ≥ 3. Fix 1 ≠ b ∈ B and a ∈ A \ B. Then dim_{𝕂} V_{1}/C_{V_{1}}(b) = dim_{𝕂} X_{1} = 1, so there exists x, y, z ∈ V_{1} such that for Y := ⟨x, y, z⟩

dim_{𝕂} Y = 3, X_{1} := x𝕂, [z, b] = 0, [y, b] = x.

Since A strictly normalizes V_{1} ⊗ V_{2} we get for v_{1} ∈ V_{1}

v_{1}ϕ_{1,a}ϕ_{2,b} = v_{1}ϕ_{1,ab} = v_{1}ϕ_{1,ba} = ϕ_{1,b}ϕ_{1,a}.

Hence the map

V_{1} → V_{2} with v_{1} ↦ v_{1}ϕ_{1,a} =: v′_{1}

is a 𝕂B-module isomorphism. It is easy to calculate that

x ⊗ x′ ∈ C_{V} (A), y ⊗ y′ ∈ C_{V} (a) and [(y ⊗ z′), a] = y ⊗ z′ + z ⊗ y′.

This shows that

[y ⊗ z′, a, b] = [y ⊗ z′ + z ⊗ y′, b] = x ⊗ z′ + z ⊗ x′ ≠ 0,

and for Y′ := Y ϕ_{1,a}

(∗)  Q_{V} (A) = 𝔽[x ⊗ z′, A] + C_{V} (A) ≤ 𝔽(y ⊗ z′ + z ⊗ y′) + Y ⊗ x′ + x ⊗ Y′ + C_{V} (A).

If y ⊗ y′ ∉ Q_{V} (A), then a centralizes Q_{V} (A) since y ⊗ y′ ∈ C_{V} (a). But then [V, A, a] = 0, which contradicts [y ⊗ z′, a, b] ≠ 0. Thus we have y ⊗ y′ ∈ Q_{V} (A), so (∗) shows that there exist u, w ∈ Y, t ∈ 𝔽 and c ∈ C_{V} (A) such that

y ⊗ y′ = x ⊗ u′ + w ⊗ x′ + t(y ⊗ z′ + z ⊗ y′) + c.

This gives

x ⊗ x + x ⊗ y′ + y ⊗ x′ = [y ⊗ y′, b] = x ⊗ [u′, b] + [w, b] ⊗ x′ + t(x ⊗ z′ + z ⊗ x)

and

x ⊗ y′ + y ⊗ x′ ≡ t(x ⊗ z′ + z ⊗ x) mod (x ⊗ x)𝕂.

But then x, y, z are not linearly independent in V_{1} which contradicts dim_{𝕂} Y = 3. This contradiction shows (9^{∘}) and completes the proof of 5.2. □

6 The Nearly Quadratic Subgroup Theorem

Lemma 6.1 [simple slm] Let 𝔽 be a field, V a finite dimensional 𝔽-vector space, and M ≤ SL_{𝔽}(V). Suppose that V is M-reduced define M-reduced and G = ⟨C_{SL_{𝔽}(V)}(V/U)^{G}⟩ for some 1-dimensional subspace U of V. Then G = SL_{𝔽}(V).

Proof: We may assume that that dim_{𝔽} V > 1 since otherwise M = SL_{𝔽}(V) = 1. Let P be the set of 1-dimensional subspaces of V, and let L(X) := C_{SL_{𝔽}(V)}(V/X) for X ∈ P and

P(M) := {X ∈ P | L(X) ≤ M}.

As SL_{𝔽}(V) = ⟨L(X) | X ∈ P⟩, it suffices to show that P(M) = P.

Since [V, L(U)] = U we get [V, M] = ∑_{U∈P(M)} U. If [V, M] ≠ V, then 1 ≠ C_{L(U)}([V, M]) ≤ C_{M}([V, M])) ∩ C_{M}(V/[V, M]). But then V is not M-reduced, a contradiction. Thus V = [V, K] = ∑_{U∈P(M)} U.

Let U_{1}, U_{2} ∈ P(M). Then L(U_{1}) acts tranistively on the 1-dimensional subspaces of U_{1} + U_{2} unequal to U_{1}. Hence P(M) contains all the 1-dimensional subspaces of U_{1} + U_{2}. Since V = ∑_{U∈P(M)} U we conclude that ∑_{U∈P(M)} U = ⋃_{U∈P(M)} U, and P(M) contains all the 1-dimensional subspaces of V. □

Remark. Let L(U) be defined as in the previous lemma. Then U = C_{V} (L(U)) and by the dimension formula

dim U + dim[V^{∗}, L(U)] = dim V,

where V^{∗} is the dual vector space. Hence passing to the dual space V^{∗} transforms 6.1 into a statement about subgroups M ≤ SL_{𝔽}(V) satisfying O_{p}(M) = 1 and M = ⟨C_{SL_{𝔽}(V)}(U)^{M}⟩ for some hyperplane U ≤ V. We will refer to this version as the “dual version of 6.1”.

Theorem 6.2 [nearly quadratic decomposition] Let 𝔽 be field, H a group and V be a faithful semisimple 𝔽H-module. Let Q be the set of nearly quadratic, but not quadratic subgroups of H. Suppose that H = ⟨Q⟩. Then there exists a partition (Q_{i})_{i∈I} of Q such that

(a) [a] H = ⨁_{i∈I} H_{i}, where H_{i} = ⟨Q_{i}⟩.

(b) [b] V = C_{V} (H) ⊕ ⨁_{i∈I}[V, H_{i}].

(c) [c] For each i ∈ I, [V, H_{i}] is a faithful simple 𝔽H_{i}-module.

Proof: Since V is semisimple, V = ⊕_{j∈J}V_{j} for some simple 𝔽H-submodules V_{j}, j ∈ J. Let I = {i ∈ J | [V_{i}, H] ≠ 0}. Then clearly

V = C_{V} (H) ⊕ ⨁_{i∈I} V_{i}

For i ∈ I let Q_{i} = {A ∈ Q | [V_{i}, A] ≠ 0}. Then by 2.7 each A ∈ Q is contained in a unique Q_{i} and so (Q_{i})_{i∈I} is a partition of Q. So (a) holds. Observe that H_{i} centralizes V_{j} for all i ∈ I and j ∈ J with i ≠ j. In particular V_{i} is a faithful H_{i}-module and H_{i} ∩ ⟨H_{j} | i ≠ j ∈ I⟩ = 1. Thus (b) holds. Since V_{i} is a simple 𝔽H module, we conclude that V_{i} is a simple 𝔽H. Since 0 ≠ [V, H_{i}] ≤ V_{i} we get V_{i} = [V, H_{i}] and so (c) holds. □

Lemma 6.3 [nh(a)] Let 𝕂 be a finite field of charteristic 2, 𝔽 ≤ 𝕂 a subfield with |𝔽| ≥ 4, V a 3-dimensional 𝕂-space, L ≤ GL_{𝕂}(V) such that L ≅ SL_{2}(𝔽) and V ≅ W ⊗_{𝔽} 𝕂, where W is a natural Ω_{3}(𝔽)-module for 𝔽L. Let A ∈ Syl_{2}(L). Put H = N_{GL_{𝕂}(V)}(A), B = C_{SL_{𝕂}(V)}(C_{V} (A)), Z = Z(GL_{𝕂}(V)), V_{1} = [C_{V} (A), H ∩ L] and V_{2} = C_{V} (L). Then C_{V} (A) = V_{1} ⊕ V_{2} and

(a) [a] if |𝔽| > 4, then V_{1} and V_{2} are H-invariant and H = (H ∩ L)ZB.

(b) [b] if 𝔽| = 3, then V_{1}^{H} = {V_{1}, V_{2}} and N_{H}(V_{1}) = (H ∩ L)ZB.

Proof: Let q be the L-invariant quadratic form on W and s the corresponding bilinear form. Let (w_{1}, w_{2}, w_{3}) be an 𝔽-basis for W such that w_{2} ∈ C_{W} (L), (w_{1}, w_{3})s = 1, w_{1}q = 0, w_{2}q = 1 and w_{3}q = 0 and w_{1} ∈ C_{V} (A). Let a ∈ A then w_{3}aq = 0 and so w_{3}a = w_{1}λ^{2} + w_{2}λ + w_{3} for some λ ∈ 𝔽. We denote the element of A by a_{λ}. Let v_{i} be the image of w_{i} ⊗ 1 in V under the isomorphism from W ⊗ 𝔽κ to V. Then (v_{1}, v_{2}, v_{3}) is an 𝕂-basis for V and the matric of a_{λ} with respect to this basis is

a_{λ} ↔ (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ λ^{2} & λ & 1 \end{matrix})

Note that B ie abelian and A ≤ B. Thus ZB ≤ H. Since ZB acts transitively on V \ C_{V} (A) we have H = C_{H}(v_{3})ZB. Let h ∈ C_{H}(v_{3}). Then

h ↔ (\begin{matrix}a & b & 0 \\ c & d & 0 \\ 0 & 0 & 1 \end{matrix})

for some a, b, c, d ∈ 𝕂 with ad − cb ≠ 0. Let λ ∈ 𝔽. Since h ∈ N_{H}(A), a_{λ}h = ha_{μ} for some μ ∈ 𝔽. We have

a_{λ}h ↔ (\begin{matrix}a & b & 0 \\ c & d & 0 \\ λ^{2}a + λc & λ^{2}b + λd & 1 \end{matrix}) and ha_{μ} ↔ (\begin{matrix}a & b & 0 \\ c & d & 0 \\ μ^{2} & μ & 1 \end{matrix})

Hence

λ^{2}b + λd = μ and λ^{2}a + λc = μ^{2} = λ^{4}b^{2} + λ^{2}d^{2}

Thus

λc + λ^{2}(a + d^{2}) + λ^{4}b^{2} = 0 for all λ ∈ 𝔽

Suppose that |𝔽| > 4 and consider the polynomial f = cx + (a + d^{2})x^{2} + b^{2}x^{4}. The each λ ∈ 𝔽 is a root of f. Since deg f ≤ 4 < |𝔽| we conclude that f is the zero polynomial. Hence c = 0, d = 0 and a = d^{2}. From μ = λ^{2}b + λd = λd we conclude that d ∈ 𝔽. Moreover,

h ↔ (\begin{matrix}d^{2} & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix}d & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & d^{−1} \end{matrix}) (\begin{matrix}d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d \end{matrix})

And so h ∈ (H ∩ L)Z. Thus H = (H ∩ L)BZ and (b) holds.

Suppose next that |F | = 4. Then λ^{4} = λ for all λ in 𝔽. Thus λ(c + b^{2}) + λ^{2}(a + d^{2}) = 0. Hence f = (c + b^{2})x + (a + d^{2})x^{2} is the zero polynomial and so c = b^{2} and a = d^{2}. From λ^{2}b + λd = μ ∈ 𝔽 for all λ ∈ 𝔽 we conclude that b, d ∈ 𝔽. Moreover, 0 ≠ ad − bc = a^{3} − b^{3}. Since u^{3} = 1 for all 0 ≠ u ∈ 𝔽 we have a = 0 or b = 0. If a = 0, then v_{1}h = v_{2}b ∈ V_{2} and if b = 0 then v_{1}h = v_{1}a ∈ V_{1}. Thus V_{1}^{H} = {V_{1}, V_{2}}. Also of b = 0 then as above h ∈ (L ∩ H)ZB and so (b) holds. □

Proposition 6.4 [char slm] Let H be a finite group, 𝕂 a finite field and V faithful finite dimensional 𝕂H-module. Put

H = {A ≤ H | C_{V} (A) = [V, H] and dim V/C_{V} (H) = 1}

Suppose that H = ⟨H⟩. Then C_{V} (H) is the unique maximal 𝕂H-submodule of V and C_{H}(V/C_{V} (H)) = O_{p}(H). Moreover, put \overline{V} /V/C_{V} (H), H̃ = H/C_{H}(\overline{V}) and n = dim_{𝕂}(\overline{V}). Then one of the following holds

1. [a] p = 2, n = 2 and H̃ ≅ D_{2m} for some odd integer m with m > 3

2. [b] p = 3, n = 2, 𝔽_{9} ≤ 𝕂 and H̃ ≅ SL_{2}(5).

3. [c] p = 2, n = 2 and H̃ ≅ 3. Alt(6).

4. [d] H̃ ≅ SL_{n}(𝔽) for some subfield 𝔽 of 𝕂. Moreover, \overline{V} ≅ W ⊗_{𝔽} 𝕂 for a natural 𝔽H̃-module W.

Proof: The proof is by induction on n. We first show that

1^{∘} [1] C_{V} (H) is the unique maximal 𝕂H-submodule.

Let U be a 𝕂H-submodule of V with U ≰ C_{V} (H). let A ∈ H with [U, A] ≠ 0. Since dim_{𝕂} V/C_{V} (A) = 1 we have V = U + C_{V} (A). Thus C_{V} (A) = [V, A] = [U, A] ≤ U and so V = U. (1^{∘}) holds.

2^{∘} [3] \overline{V} is a simple 𝕂H-module and O_{p}(H) = C_{H}(\overline{V}).

Teh first statement follows immediatley from (1^{∘}). In particular, O_{p}(H) ≤ C_{H}(\overline{V}). Since C_{H}(\overline{V}) centalizes C_{V} (H) and V/C_{V} (H) we get that C_{H}(\overline{V}) is a p-group and so (2^{∘}) holds.

3^{∘} [2] Let A be then set of maximal elements of V wit respect to inclusion. Then H = ⟨A⟩, each A in H is weakly closed and H acts transitively on A.

Since each B ∈ H is contained in some A ∈ A, H = ⟨A⟩. Let A, B ∈ A such that B normalizes A. By (1^{∘}) applied to AB in place of H, C_{V} (A) ≤ C_{V} (AB). Thus C_{V} (A) = C_{V} (B) and so AB ∈ H. By maximality of A and B we get A = AB = B. Thus A is weakly closed in H and any Sylow p-subgroup of H contains a unique member of A. So (3^{∘}) holds.

If n = 1, then (d) holds with 𝕂 = 𝔽. If n = 2 the propostion follows from Dickson’s List [?, II.8.27]. Replacing V be \overline{V} and H by H̃ we may assume from now on that

4^{∘} [3.5] V is a faithful simple 𝕂H module and n ≥ 3.

For D ≤ H let A_{H} = {A ≤∈ A | A ≤ L}. Since C_{V} (H) = 0 and dim V/C_{V} (A) = 1 for all A ∈ A, the exists L ≤ H with L = ⟨A_{L}⟩ and dim C_{V} (L) = 1. Let L be the set of such subgroups of H.

5^{∘} [4] Suppose that O_{p}(L) = 1 for some L ∈ L. Then n = 3, p = 2, L ≅ SL_{2}(𝔽) for some subfield 𝔽 of 𝕂 and A_{L} = Syl_{p}(L).

By induction the theorem holds for L in place of H. By (1^{∘}) applied to L, V is indecomposable as 𝕂L-module. In particular, O_{p′}(L). We conclude that (d) hold for L. Moreover, if n = 3, then p = 2. Let A ∈ A_{L} and put P^{∗} = N_{L}(C_{V} (A)) and P = C_{P^{∗}}(V/C_{V} (A)). Note that P^{∗} acts simply on O_{p}(P^{∗}). Since A is weakly closed in H and A‘O_{p}(P^{∗}) we conclude that A ⊴ P^{∗} and A = O_{p}(P^{∗}). If n = 2 we conclude (5^{∘}) holds. Su suppose that n > 3. Then A = O_{p}(P^{∗}) is natural module for P/O_{p}(P^{∗}) ≅ SL_{n}(𝔽). Let x ∈ V \ C_{V} (A). Then [x, A] ≅ A/C_{A}(x) ≅ A as an F_{p}P -module. Thus [x, A] is a nontrivial simple module for P. Since this holds for any such x, [V, A] is a sum of non -trivial simple P module and so C_{[V,A]}(P) = 0. But this is a cntradiction to C_{V} (L) ≤ C_{V} (A) = [V, A].

6^{∘} [5] Suppose L ∈ L with O_{p}(L) = 1 and let A ∈ A_{L}. Then

(a) [5:a] Let L ≤ R ∈ L. Then L = R.

(b) [5:b] Let L_{−}(A) = {R ∈ L | A ≤ R, O_{p}(R) = 1}. If |A| = 4, then |L_{−}(A) ≤ 2 and if |A| > 4, then L_{−}(A) = {L}.

(c) [5:c] There exists R ∈ L with O_{p}(R) ≠ 1.

Let L ≤ R ∈ L. Since A ≤ O_{p}(R) ≤ O_{p}(L) ≤ 1 and A is weakly closed, [A, O_{p}(R)] = 1. On the other hand since V/C_{V} (R) is a simple 𝕂R-module, O_{p}(R) is as an 𝔽_{p}R-module the direct sum of copies of V/C_{V} (R). So [O_{p}(R), A] = 1, implies O_{p}(R) = 1. So by (5^{∘}), both L and R are isomorphic to SL_{2}(|A|) and so L = R. Thus (a) holds.

Let V_{1} = [C_{V} (A), N_{L}(A)] and V_{2} = C_{V} (L). Let R ∈ L_{−}(A). Since N_{R}(A) = O^{2}(N_{R}(A)), 6.3 implies that N_{R}(A) normalizes V_{1} and V_{2}. Since 𝕂N_{R}(A) has exactly proper submodule in C_{V} (A) we conclude that V_{1} = C_{V} (R) or V_{2} = C_{V} (R). In the first case ⟨R, L⟩ ∈ L and so by (a), R = L. So suppose V_{2} = C_{V} (R), then V_{1} = [C_{V} (A), N_{R}(A)]. Put P = N_{H}(A) ∩ N_{H}(V_{1}). We conclude that |P/C_{P} (V_{1})| ≥ |A|−1. On the other hand since P ≤ SL_{κ}(V_{1}), 6.3 implies that P ≤ Z(SL_{𝕂}(V)C_{P} (A) and so P/C_{P} (A)|leq3. Thus |A| = 4. Moreover by (a), R is unique in L with C_{V} (R) = V_{1} and so (b) holds.

Suppose that O_{p}(R) = 1 for all R ∈ L. Let B ∈ A with B ≰ L. Then R := ⟨A, B⟩ ∈ L. Since R ≠ L, (b) gives that |A| = 4 and R is unique detemined. Hence A = A_{L} ∪ A_{R}. Since R ≅ L ≅ SL_{2}(4) we conclude that |A| = 1 + 4 + 4 = 9. Then m = |L|. The there exists 5m pairs (B, P) with B ≤ A, P ∈ L and B ≤ P. On the otherhand each of the 9 members of A is contained in exactly 2 memembrs of L and so 5m = 2 · 9 = 18, a contradiction. Thus (c) holds.

7^{∘} [6] Put H^{∗} = {D ≤ H | [V, D] = C_{V} (D), H^{∗} =< A^{∗}} and let A^{∗} the set of maximal elemensts of H^{∗} under inclusion. Let L ∈ L with O_{p}(L) ≠ 1 and A ∈ L.

(a) [a] H = H^{∗}.

(b) [b] O_{p}(L) ∈ A^{∗} and |O_{p}(L)| = |A|.

(c) [c] L/O_{p}(L) ≅ SL_{n−1}(𝔽) for some subfield 𝔽 of 𝕂.

(d) [d] O_{p}(L) is a natural module for 𝔽_{p}SL_{n−1}(𝔽), |A| = |𝔽|^{n−1} and |A ∩ O_{p}(L)| = |𝔽|.

Observe first that by (6^{∘})(c) there exists L ∈ L with O_{p}(L) ≠ 1. Let B be a simple 𝔽_{p}L-submodule of O_{p}(L). Then B is dual to an simple q F_{p}L-quotient of V/C_{V} (L). By (1^{∘}) V/C_{V} (L) is a simple 𝕂L-module and C_{V} (B) = C_{V} (O_{p}(L)) = C_{V} (L) = [V, B] = [V, O_{p}(B)]. So both B and O_{p}(L) are in H^{∗}. Let D^{∗} ≤ A^{∗} with O_{p}(L) ≤ D^{∗} . Put 𝔽 = End_{L}(B). It follows that [B, A] = C_{B}(A) is 1−dimensional over 𝔽. In particular, |O_{p}(L)/C_{O_{p}(L)}(A) ≥ |B/C_{B}(A)| ≥ |A/C_{A}(B)|. Observe that C_{A}(B) = C_{A}(V/C_{V} (L) = A ∩ O_{p}(L) and since A is weakly closed C_{B}(A) = [B, A] ≤ A. Thus

|A| ≤ |O_{p}(L)|

Since V is a simple 𝕂H-module and 1 ≠ H^{∗} ⊴ H, V = [V, H^{∗}] and so C_{V^{∗}}(H^{∗}) = 0, where V^{∗} is the dual of V has a 𝕂H-module. So we can apply the results the proved about H and V to H^{∗} and V^{∗}. Put

L^{∗} = {L^{∗} ≤ H^{∗} | L^{∗} = ⟨A_{L^{∗}}^{∗}⟩, dim_{𝕂}[V, L^{∗}] = 1}

By (6^{∘})(c) there exists L^{∗} with O_{p}(L^{∗}) ≠ 1. By the result proved in (7^{∘}) so far ( applied to L^{∗} and V^{∗}), O_{p}(L^{∗}) ∈ A and |D^{∗}| ≤ O_{p}(L^{∗}). Let D ∈∈ A with O_{p}(L^{∗}) ≤ D. Then

|A| ≤ |O_{p}(L)| ≤ D^{∗} | ≤ |O_{p}(L^{∗})| ≤ |D| = |A|

It follows that |A| = |D^{∗}, O_{p}(L) = D^{∗} and O_{p}(L^{∗}) = D. From |A| = |O_{p}(L)| we get B = O_{p}(L) and |B/C_{B}(A)| = |A/C_{B}(A). Since the theorem holds for L and V/C_{V} (L) in place of H and V we conclude that L/O_{p}(L) ≅ SL_{n−1}(𝔽) and B = O_{p}(L) is natural module for L. Thus O_{p}(L)| = |𝔽^{n−1}|. Since D ∈ A and H acts tranistively on caA, H = ⟨A⟩ = ⟨D^{H}⟩ ≤ H^{∗} and so H = H^{∗}. Thus all parts of (7^{∘}) are proved.

8^{∘} [6.5] Let A ∈ A and put L_{A}^{∗} := ⟨A_{N_{H}(A)}^{∗}⟩. Then L_{A}^{∗} ∈ L^{∗} and A = O_{p}(L_{A}^{∗}).

By (6^{∘})(c) there exists R ∈ L^{∗}. By (7^{∘}), O_{p}(R) ∈ A and so we may assume that O_{p}(R) = A. Thus A ≤ R ≤ L_{A}^{∗}. Since L_{A}^{∗} normalizes [V, A] we have [V, L_{A}^{∗}] = [V, A] and so L_{A}^{∗} ∈ L^{∗}. Thus by (7^{∘}), |R| = |L_{A}^{∗} and so R = L_{A}^{∗} and A = O_{p}(L_{A}^{∗}).

9^{∘} [7] Let A, B ∈ A with A ≠ B. Then exactly one of the follwing holds.

1. [7:a] There exists D ∈ A^{∗} with D ≤ N_{H}(A) and B ≤ N_{H}(D).

2. [7:b] ⟨A, B⟩ ∈ L and O_{p}(⟨A, B⟩) = 1.

Pick L ∈ L with ⟨A, B⟩ ≤ L.

Suppose that D := O_{p}(L) ≠ 1. Then by (7^{∘}), D ∈ A^{∗}. Clearly B ≤ N_{H}(D) and since AD is a p-group and A is weakly closed, D ≤ N_{H}(D). Thus (2) holds.

Suppose that O_{p}(L) = 1. Then by (5^{∘}), L = ⟨A, B⟩ and so (2) holds.

Suppose for a contradiction that (1) and (2) holds. Since A is weakly closed, A ≤ N_{H}(D) and so L ≤ N_{H}(D). Since O_{p}(L) = 1, [A, D] ≤ A ∩ D ≤ L ∩ D ≤ O_{p}(L) = 1. But C_{H}(D) = Z(H)D and so A ≤ D and A = 1, a contradiction.

Case 1 [op(l)= 1] Suppose that O_{p}(L) = 1 for some L ∈ L. Then (c) holds.

By (5^{∘}) n = 3, p = 2 and L ≅ SL_{2}(𝔽) subfield 𝔽 of 𝕂 with |𝔽| = |A|. Let AinA_{L}. By (6^{∘}) there exists R ∈ L with O_{p}(L) ≠ 1 and by (3^{∘}) we can choose R such that A ≤ R. Suppose that |A ∩ O_{p}(R)| > 2. Then then action of L on V shows that [V, A ∩ O_{p}(LR] = [V, A]. But this is a contradiction since [V, O_{p}(R)] is 1-dimensional and [V, A] is a hyperplane. By (7^{∘}), R/O_{p}(R) ≅ SL_{2}(𝔼) for some subfield 𝔼 of 𝕂. Morever, |A| = |𝔼|^{2} and |A ∩ O_{p}(R)| = |𝔼|. Thus |𝔼| = 2 and |𝔽| = |A| = 4.

By (8^{∘}), N_{H}(A) ≠ N_{H}(C_{V} (L)). So N_{H}(A) does not normalizes L. Thus (b) implies that |L_{−}(A)| = 2. Each T ∈ L_{−}(A) is isomorphic to SL_{2}(4) and so contains four elemenst of A other than A. So there exists eight elemensts of A fulfilling (9^{∘})(2).

By (8^{∘}) and (9^{∘}), L_{A}^{∗} ≅ Sym(4) and so there exists exactly three D ∈ A^{∗} with D ≤ N_{H}(A). Similarly there exists three eleents B ∈ A with B leqN_{H}(D), two of whcig are different from A. If B ≠ A, then ⟨A, B⟩ = L_{D} and so D = O_{p}(⟨A, B⟩) is uniquely determined by A and B. Thus there are 6 = 3·3 elements of A fulfilling (9^{∘})(1). This shows that |A| = 1 + 6 + 8 = 15. Since C_{H}(C_{V} (A)) = A, we conclude from 6.3 that N_{H}(A) = Z(H)L_{A}^{∗}. Thus |N_{H}(A)/Z(H)| = 24 and |H/Z(H)| = 24 · 15 = 360. Since |LZ(H)/Z(H)| = |L| = 60 we have |H/LZ(H)| = 6 and so H/Z(H) ≅ Alt(6). the elements of order three in L_{A}^{∗} acts fixed-point freely on C_{V} (A). Thus N_{L}(A) ≤ L_{A}^{∗}. Hence Z(H) ≠ 1 and |Z(H)| = 3. Therefore H ∼ 3. Alt(6) and (c) holds in this case.

Case 2 [opl neq 1 ] Suppose that O_{p}(L) ≠ 1 for all L ∈ L. Then (d) holds.

Let 𝔽 be as in (7^{∘}). By (7^{∘})(??), |𝔽| and so also 𝔽 is independent of the chooice of L ∈ L. Let

WW = {x ∈ nV | x ∈ C_{V} (A) for some A ∈ A}

Let W♯ = W \ {0}. For x ∈ W♯ let A_{x} ∈ A with x ∈ C_{V} (A_{x}) and observe that A_{x} is uniquely determined by x. Define the relation ’∼’ on W^{♯} by x ∼ y if either x𝔽 = y𝔽 or y ∈ [x, A_{y}]. Since x𝔽 = x𝔽, ∼ is reflexive.

10^{∘} [8] Let x, y ∈ W^{♯} with x ∼ y. Then y ∼ x, x𝔽 + y𝔽 ⊆ W. If x𝔽 ≠ y𝔽, then x𝔽 + y𝔽 is a natural 𝔽SL_{2}(|F)-module for ⟨A_{x}, A_{y}⟩.

If x𝔽 = y𝔽 then clearly y ∼ x and x𝔽 + y𝔽 ⊆ W. So suppose y ∈ [x, A_{y}]. Put R = ⟨A_{x}, A_{y}⟩ and pick L ∈ L with R ≤ L. Let V_{1} = x𝕂 + y𝕂 and observe that R normalizes V_{1}. From (9^{∘}) we conclude that R/C_{R}(V_{1}) ≅ SL_{2}(𝔽) and V_{1} ≅ W_{1} ⊗_{𝔽} 𝕂 for some natural 𝔽SL_{2}(𝔽)-module W_{1} for R. Let z ∈ {x, y}. Since z ∈ C_{V_{1}}(A_{z}), x ↔ w_{z} ⊗k_{z} for some w_{z} ∈ W_{1} and k_{z} ∈ 𝕂. Since y ∈ [x, A_{y}] and since W ⊗ k_{x} is invariant under A_{y} we conclude that w_{y} ⊗ k_{y} ∈ W ⊗ k_{x}. Thus x𝔽 + y𝔽 ↔ W ⊗ k_{x}. Thus x𝔽+y𝔽 is invarinat under under R,x𝔽+y𝔽 is natural SL_{2}(𝔽)-module for R, R acts tranistively on (x𝔽 + y𝔽)^{♯}, x𝔽 + y𝔽 ⊆ W and x ∈ [y, A_{x}]. So (10^{∘}) holds.

11^{∘} [9] ∼ is an equivalence relation on W.

We already proved that ∼ is reflexive and symmetic. To show that ∼ is tranistive, let x, y, z ∈ W^{♯} wirth x ∼ y and y ∼ z. If x𝔽 = y ∼ 𝔽 and y𝔽 = z𝔽, then x𝔽 = y ∼ 𝔽. If x𝔽 = y𝔽 and y𝔽 ≠ z𝔽, then by x ∈ y𝔽 ∈ y𝔽 + z𝔽 and by (10^{∘}), [x, A_{z}] = z𝔽 and so x ∼ z. So we may assume that x𝔽 ≠ y𝔽 and similarly that y𝔽 ≠ z𝔽.

Put V_{1} = x𝕂 + y𝕂, R = ⟨A_{x}, A_{y}⟩. and N = N_{H}(V_{1}). Then ⟨A_{N}⟩ ≤ L for some L ∈ L and from (7^{∘}) gives that ⟨A_{N}⟩ = R. By (10^{∘}) ⟨y^{R}⟩ = x𝔽 + y𝔽.

Suppose first that V_{1} = z𝕂+y𝕂. Then by summetry R = ⟨A_{N}⟩ = {A_{z}, A_{x}⟩ and ⟨y^{R}⟩ = z𝔽+y𝔽. Hence by (10^{∘}), x ∼ z.

Suppose next that z ∉ V_{1}. Since [z, A_{y}] = y𝔽 and [x, A_{y}] = y𝔽 we conclude that A_{y} = C_{A_{y}}(x)C_{A_{y}}(z) and so

[z + V_{1}, C_{A_{y}}(V_{1})] = [z, C_{A_{y}}(x)] = [z, A_{y}] = y𝔽

By (10^{∘}) there exists g ∈ R with y^{g} = x. Since [V, R] ≤ V_{1}, (z + V_{1})^{g} = z + V_{1} and conjugating the precceding line by g gices

[z + V_{1}, C_{A_{x}}(v_{1}) = x𝔽

Hence x ∈ [z, A_{x}] and x ∼ z. Thus (11^{∘}) holds.

12^{∘} [10] Let W be an equivallence class of ∼. Then W is an 𝔽-subspace of V, H normalizes W, H induces SL_{𝔽}(W) on V and V ≅ W ⊗_{𝔽} 𝕂 and an 𝕂H-module.

It follows from (10^{∘}) that W is an 𝔽-subspace of V. Let x ∈ W^{♯}, A ∈ A and y ∈ [x, A]^{♯}. Then A_{y} = A and x ∼ y. So y ∈ W and H normalizes W. If w ∈ W \ x𝔽, then [w, A_{y}] = x𝔽. Hence by 6.1, H induces SL_{𝔽}(W) on W. By the univerrsal property of the tensor product there exists a unique 𝔽-linear map α : W ⊗_{𝔽} 𝕂 → V with (w ⊗ k)α = wk for all w ∈ W, k ∈ 𝕂. Clearly this map is a 𝕂H-homorphism. Since both W ⊗_{𝔽} 𝕂 and V are simple 𝕂H-modules, α is an isomorphism. Hence (12^{∘}) holds.

(12^{∘}) completes the proof for (??). □

Theorem 6.5 [nearly quadratic module theorem]char sl Let H be a finite group, and V a faithful simple 𝔽_{p}H-module. Suppose that H is generated by elementary abelian, nearly quadratic, but not quadratic subgroups of H.

Let W a simple 𝔽_{p} F^{∗}(H)-submodule of V and

𝕂 = End_{F^{∗}(H)}(W).

Then H, V, W, 𝕂 and H/C_{H}(𝕂) are as in the following table:

H  V  W 𝕂 H/C_{H}(𝕂)
(C_{2} ≀ Sym(n))′  𝔽_{3}^{n}  𝔽_{3} 𝔽_{3}  −
SL_{n}(𝔽_{2}) ≀ Sym(m) (𝔽_{2}^{n})^{m} 𝔽_{2}^{n} 𝔽_{2}  −
SL_{2}(𝔽_{2}) × SL_{2}(𝔽_{2}) 𝔽_{2}^{2} ⊗ 𝔽_{2}^{2} 𝔽_{4} 𝔽_{4}  −
Frob(39)  𝔽_{27}  V 𝔽_{27}  C_{3}
ΓGL_{n}(𝔽_{4})  𝔽_{4}^{n}  V 𝔽_{4}  C_{2}
ΓSL_{n}(𝔽_{4})  𝔽_{4}^{n}  V 𝔽_{4}  C_{2}
3. Sym(6)  𝔽_{4}^{3}  V 𝔽_{4}  C_{2}
SL_{n}(𝕂) ∘ SL_{m}(𝕂) 𝕂^{n} ⊗ 𝕂^{m} V any  1
(C_{2} ≀ Sym(4))′  𝔽_{3}^{4}  V 𝔽_{3}  1
F^{∗}(H) quasisimple  ?  V ?  1

Proof: Fix A ∈ Q and put K = K_{A} = [F^{∗}(H), A]. Since K is subnormal in H and V is a simple 𝔽H-module,

1^{∘} [-1] I = I_{A} ⊕ C_{I}(K).

Moreover, C_{G}(I) ∩ F ≤ O_{p}(G) = 1 and thus C_{G}(I) = 1, since F contains its centralizer. We have shown:

2^{∘} [0] I is a faithful semisimple 𝔽_{p}G- and 𝔽_{p}F -module.

Let Δ := {W_{1}, …, W_{r}} be the set Wedderburn components of F on I, and put

Δ_{A} := {W ∈ Δ | [W, K] ≠ 0}.

Observe that Δ is a system of imprimitivity if r > 1.

From now on we assume that A is not quadratic on I. Then by 2.7 C_{I}(K) = C_{I}(KA). Pick W ∈ Δ_{A}, and put

N := N_{G}(W), Ñ := N/C_{G}(W), A_{W} := N_{A}(W).

By 4.1 𝕂 is a finite field. We now devide the proof in seversl cases.

Case 1 [w neq i] The case I_{A} ≠ W.

Then |Δ_{A}| > 1; in particular Δ is a system of imprimitivity. Hence 2.11 applies. Since A is not quadratic on I, we are in case 2.11(4).

Clearly |W | > 2 since C_{I}(F) = 0 by (2^{∘}). Hence 2.11(4) shows that either

(i) [i] p = 3, A_{W} = 1, |W | = |A| = |Δ_{A}| = 3, or

(ii) [ii] p = 2, |A/A_{W} | = |Δ_{A}| = 2, C_{W} (A_{W}) = [W, A_{W}], and |W/C_{W} (A_{W})| = 2.

Suppose that (i) holds. Then |I_{A}| = 3^{3}, and K is an elementary abelian 2-group. Since KA is a subgroup of GL_{3}(3) we get that KA ≅ A_{4} ≅ Ω_{3}(3), and I_{A} is the orthogonal module for KA. Hence (??) holds.

Suppose that (ii) holds. Then |A_{W} | = |[w, A_{W}]| = |[W, A_{W}]| for w ∈ W \ C_{W} (A_{W}) and thus |A_{W} | = |C_{W} (A_{W})|. If [F̃, Ã_{W}] = 1, then F centralizes W/C_{W} (A_{W}) and so F centralizes W since W is a Wedderburn component. This is impossible since C_{I}(F) = 1.

Hence we have [F̃, Ã_{W}] ≠ 1. Let W_{0} be a simple F -submodule of W. Then |W_{0}| > 2, so |W/C_{W} (A_{W})| = 2 shows that W_{0} is invariant. On the other hand [W_{0}, [F, A_{W}]] ≠ 0 sinc W is homogeneous. It follows that [W_{0}, A_{W}] ≠ 0 and W = W_{0}C_{W} (A_{W}). Hence W = W_{0}, again since W is homogeneous. Let E := ⟨A_{W}^{F} ⟩. Then O_{p}(Ẽ) = 1 since W_{0} is a simple F -module, so the dual version of 6.1 implies that Ẽ ≅ SL_{𝔽_{2}}(W). Pick a ∈ A \ A_{W}, so I_{A} = W ⊕ W^{a}.

Assume first that n := dim_{𝔽_{2}} W > 2. Then there exists a component K_{1} ⊴ K such that F = K_{1}C_{F} (W) and C_{F} (W) = C_{F} (K_{1}). Suppose that K_{1} is A-invariant, then K_{1}A_{W} also induce SL_{𝔽_{2}}(W^{a}) on W^{a}. But then C_{F} (W) = C_{F} (I_{A}) W and W^{a} are isomorphic F -modules since A_{W} centralizes a hyperplane in both. This contradicts the fact that W and W^{a} are distinct Wedderburn components. Thus, we have that K = K_{1} × K_{1}^{a}, and (??) follows.

Assume now that n = 2, so |I| = 2^{4} and Ẽ ≅ C_{3}. As W and W^{a} are non-isomorphic F -modules, we get that K ≅ C_{3} × C_{3}, and |A| = 4. Now it is easy to see that (??) holds.

Case 2 [homogeneous] The case I_{A} = W and A not 𝕂-linear on W.

We apply 2.14. Since A is not quadratic on I, we are either in case 2.14(2) or case (3). If 2.14(3) holds, then (??) follows.

Assume now that 2.14(2) holds. Then

(∗)  |𝕂| = 4, [I, A_{𝕂}] = C_{I}(A_{𝕂}), |A/A_{𝕂}| = 2, and |I/C_{I}(A_{𝕂})| = 4.

In particular A_{𝕂} acts quadratically on I_{A} and A_{𝕂} ≠ 1. This last property excludes the case dim_{𝕂} I_{A} = 1.

Suppose that dim_{𝕂} I_{A} = 2. Then KA ≅ X ≤ ΓL_{2}(4). If F̃ is abelian, then K embedds into 𝕂^{∗} and so [K, A_{𝕂}] = 1. But then [F, A_{𝕂}, A_{𝕂}] = 1, which is impossible since O_{2}(H) = 1 and A_{𝕂} ≠ 1. Since all nilpotent subgroups of ΓL_{2}(4) are abelian, F̃ is not nilpotent and so F̃′ ≅ SL_{2}(4). Thus either X = ΓL_{2}(4) or |ΓL_{2}(4)/X| = 3, and (??) follows.

Suppose that dim_{𝕂} I_{A} =: n ≥ 3. By Baer’s theorem there exists x ∈ K such that R := ⟨A_{𝕂}, A_{𝕂}^{x}⟩ is not a 2-group. It follows that \overline{I}_{A} := I_{A}/C_{I_{A}}(R) has order 4^{2}, and R/C_{R}(\overline{I}_{A}) ≅ SL_{2}(4). Assume first that A_{0} := C_{A}(\overline{I}_{A}) ≠ 1. Then

[I_{A}, A_{0}] = [C_{I_{A}}(a), A_{0}] ≤ C_{I_{A}}(A)

and so [I_{A}, A, A_{0}] = 0. But then I_{A}/C_{I_{A}}(A_{0})| = 2, which contradicts the fact that C_{I_{A}}(A_{0}) is a 𝕂-space.

We have shown that A_{0} = 1, so |A_{𝕂}| = 4 and |I_{A}| = 4^{3} since C_{I_{A}} = [I_{A}, A_{𝕂}]. show now that KA ≅ 3S_{6} . Hence (??) holds.

Case 3 [klinear] The case I_{A} = W and A 𝕂-linear on W.

Let K be the set of all X ⊴ F such that [I_{A}, X] ≠ 0 and either X is a component of G or X = O_{q}(G) with X′ ≰ C_{F} (W), q a prime. We first prove:

3^{∘} [2] There exists D ∈ K such that [D, A] ≠ 1.

Suppose that [D, A] = 1 for every D ∈ K. Then K̃ ≤ Z(F̃), and [K, A] ≤ C_{F} (I_{A}) since A acts 𝕂-linearly on I_{A}. Hence (1^{∘}) implies [K, A] = 1 and [F, A, A] = 1. This contradicts O_{p}(G) = 1.

Let D_{1}, …, D_{r} be the elements of the set K and put J := {1, …, r}. Since N acts on K by conjugation, J is an N-set with respect to the induced action; i.e., D_{ig} := D_{i}^{g} for g ∈ N. Hence, the module W, J and K satisfy the conditions (i) and (ii) of 4.4 (with 𝔽 = 𝔽_{p} and N in place of G). Thus there exists a tensor decomposition

(Φ, ⊗_{𝕂}^{J_{0}}V_{j}, σ, , (ϕ_{i,g} | j ∈ J, g ∈ N))

where J_{0} := {0} ∪ J, V_{j} is a simple 𝕂D_{j}-module for j ∈ J and V_{0} is a trivial 𝕂⟨D_{1}, …, D_{r}⟩-module. Moreover

Φ : ⊗_{𝕂}^{J}V_{j} → V

is a 𝕂(× D_{i})-module isomomorphism satisfying

(⊗v_{j})Φg = (⊗v_{jg^{−1}}ϕ_{jg^{−1},g})ϕ,

and the decomposition restricted to A is strict. In particular A strictly normalizes the tensor product ⊗_{𝕂}^{J_{0}}V_{j}. We use the action of A and N on ⊗_{𝕂}^{J_{0}}V_{j}, as described in 3.6. Recall that for every j ∈ J there exists a 𝕂D_{j}-submodule I_{j} ≤ W isomorphic to V_{j}.

As A acts nearly quadratically but not quadratically on W, it also acts (via Φ) nearly quadratically but not quadratically on ⊗_{𝕂}^{J_{0}}V_{j}. Clearly dim_{𝕂} V_{j} = dim_{𝕂} I_{j} ≥ 2 for j ≥ 1 by our choice of K_{I}. If dim_{𝕂} V_{0} = 1, then [V_{0}, A] = 0 and A acts nearly quadratic but not quadratic on ⊗_{j∈J}V_{j}. Thus we are allowed to apply 5.2 with V := ⊗_{𝕂}^{J_{0}}V_{j} if dim_{𝕂} V_{0} ≥ 2, and to ⊗_{j∈J}V_{j} if dim_{𝕂} V_{0} = 1 and |J| ≥ 2. In all these cases (3) of 5.2 holds since A is not quadratic on the corresponding module. We get one of the following cases:

Case 4 [7] dim_{𝕂} V_{0} ≥ 2, |J| = 1, and [V_{j}, A] = C_{V_{j}}(A) = [v_{j}, A] is a 𝕂-hyperplane of V_{j} for v_{j} ∈ V_{J} \ [V_{j}, A] and j ∈ J_{0}.

Case 5 [8] dim_{𝕂} V_{0} = 1, |J| = 2, J is a trivial A-set, and [V_{j}, A] = C_{V_{j}}(A) = [v_{j}, A] is a 𝕂-hyperplane of V_{j} for v_{j} ∈ V_{J} \ [V_{j}, A] and j ∈ J.

Case 6 [9] dim_{𝕂} V_{0} = 1, |J| = 2, J is a non-trivial A-set, 𝕂 = 𝔽_{2}, and [V_{j}, B] = C_{V_{j}}(B) is a 𝕂-hyperplane of V_{j} for j ∈ J, where B := C_{A}(J).

Case 7 [10] dim_{𝕂} V_{0} = 1 and |J| = 1.

We first show:

4^{∘} [12] Suppose that (Case 6) holds. Then (??) holds.

Assume (Case 6). Then C_{V_{1}}(B) = [V_{1}, B] and |V_{1}/C_{V_{1}}(B)| = 2 since 𝕂 = 𝔽_{2}. This shows that |B/C_{B}(V_{1})| = \frac{1}{2}|V_{1}|. Thus the dual version of 6.1 shows that V_{1} is natural SL_{n}(2)-module for ⟨B^{D_{1}}⟩. Going back to N via Φ we conclude that D̃_{1} ≅ SL_{n}(2), and n > 2 since D̃_{1} is not abelian. In addition B̃ = NÃ(D̃_{1}) and |B̃| = \frac{1}{2}|V_{1}|. This shows that B ≤ D_{1}D_{2}C_{F} (W). As I = W ⊕ C_{I}(KA), we also get that C_{D_{j}}(W) = 1, so D_{j} ≅ SL_{n}(2). Now (??) follows.

5^{∘} [13] Suppose that (Case 4) or (Case 5) holds. Then (??) or (??) holds.

Let I_{1} be a simple 𝕂D_{1}-submodule of W isomorphic to V_{1}. In both cases A normalizes I_{1} and C_{I_{1}}(A) = [I_{1}, A] is a 𝕂-hyperplane of I_{1} with [I_{1}, A] = [v, A] for v ∈ I_{1} \ [I_{1}, A].

Since I_{1} is a simple 𝕂D_{1}A-module, O_{p}(D_{1}A) = 1. Hence the dual version of 6.1 implies that I_{1} is a natural SL_{n_{1}}(q)-module for D_{1}A, where n_{1} = dim_{𝕂} I_{1} and q = |𝕂|. Since D_{1} acts faithfully on I_{1} and is not abelian we conclude that D_{1} ≅ SL_{n_{1}}(q) or |𝕂| = 3 and D_{1} ≅ SL_{2}(3)′ ≅ Q_{8}.

Suppose first that D_{1} ≅ SL_{n_{1}}(q). Then there exists a p-group A_{1} ≤ D_{1} with [A_{1}, A] = 1 such that for A_{2} := C_{AA_{1}}(I_{1})

AA_{1} = AA_{2} = A_{1}A_{2} and C_{AA_{1}}(W) ≤ C_{KA}(I) = 1.

Put R_{2} := ⟨A_{2}^{N}⟩. As |J| ≤ 2, N_{N}(D_{1}) has index at most 2 in N and so R_{2} is subnormal in N; in particular O_{p}(R_{2}) ≤ O_{p}(N). As F ≤ N and O_{p}(F) = 1, we get that O_{p}(N) = 1 and thus also O_{p}(R_{2}) = 1. Moreover, since R_{2} centralizes I_{1}, R_{2} acts faithfully on I_{2}, where I_{2} is isomorphic to V_{2} if |J| = 2 and to V_{0} if |J| = 1.Another application of the dual version of 6.1, this time to R_{2} and I_{2}, shows that R_{2} ≅ SL_{n_{2}}(q) and I_{2} is a natural SL_{n_{2}}(q) module for R_{2}. Hence (??) holds.

Assume now that D_{1} ≅ SL_{2}(3)′ ≅ Q_{8}. Then by the definition of K, K = O_{2}(H). If |J| > 1, then D_{2} ≠ O_{2}(H) and so the previous paragraph applies with D_{2} in place of D_{1}. This gives (??). Thus we may assume that |J| = 1 and F = O_{2}(H)C_{F} (A).

Note that |A| = 3. It follows that |[V_{0}, A]| = 3 and thus |V_{0}| = 3^{2}. We get that |W | = 3^{4}. It is now easy to see that (??) holds.

6^{∘} [14] Suppose that (Case 7) holds. Then D_{1} = KC_{D_{1}}(A), and I_{A} is a simple 𝔽_{p}KA-module.

The first part of (6^{∘}) follows from |J| = 1. Moreover I_{A} is a simple 𝕂F A-module. Let I_{1} be a simple 𝔽_{p}KA-submodule of I_{A}. Since I_{A} = ⟨I_{1}^{C_{K}(A)}⟩, I_{A} is a direct sum of A-submodules isomorphic to I_{1}. Hence 2.7 shows that I_{A} = I_{1}.

7^{∘} [14-0] Suppose that (Case 7) holds and D_{1} is not solvable. Then (??) holds.

Since D_{1} is not solvable, D_{1} is a component of G. It follows that D_{1} = [F, A] = K since |J| = 1. This gives (??).

Case 8 [solvable] Suppose that (Case 7) holds and D_{1} is solvable. Then (??) holds.

There exists a prime r such that D_{1} = O_{r}(G) and K ≤ O_{r}(G). We choose a normal subgroup R of F A that is minimal with [R, A] ≠ 1, so R ≤ K.

8^{∘} [14.5] R = [R, KA], and either R is elementary abelian, or C_{R}(A) = Z(R) and A acts simply on R/Z(R).

Observe that F = KC_{F} (A) and [R, F] = [R, D_{1}] < R. Both, [R, F] and [R, KA] are normal subgroups of F. The minimality of R gives [R, F, A] = 1 and either R = [R, KA] or [R, KA, A] = 1. In the second case [R, A] = [R, A, A] = 1, a contradiction.

As [R, F] ≤ C_{R}(A), C_{R}(A) is a normal subgroup of F A and thus also C_{R}(C_{R}(A)) is a normal subgroup of F A. The P × Q-Lemma and the minimality of R yield C_{R}(A) ≤ Z(R). If R is nonabelian, the minimality of R yields C_{R}(A) = Z(R). Since every subgroup of R containing C_{R}(A) is normalized by F, we also get that A acts simply on R/Z(R). If R is abelian, then Ω_{1}(R) = R since R = [R, A] × C_{R}(A), so R is elementary abelian.

9^{∘} [15] 1 ≠ C_{R}(K) embedds into 𝕂; in particular |𝕂| > 2.

Since R is a normal subgroup of O_{r}(F), C_{R}(F) ≠ 1, and by the definition of 𝕂, C_{R}(F) is isomorphic to a subgroup of 𝕂^{∗}, so |𝕂| > 2.

10^{∘} [16] Suppose that I_{A} is not a homogeneous 𝕂R-module. Then (??) holds.

Let Δ be the set of Wedderburn components for R on I. Then Δ is a system of imprimitivity for KA on I_{A}. We apply 2.11.

Let E ∈ Δ. By (6^{∘}) I_{A} is a simple 𝕂KA-module. Hence KA acts transitively on Δ and N_{KA}(E) acts simply on E. In particular [E, R] ≠ 0 since C_{I_{A}}(R) = 0, and by (8^{∘}), [E, A] ≠ 0 if A ≤ N_{KA}(E). This excludes the case 2.11(4:1). The transitivity on Δ and the non-quadratic action of A on I_{A} excludes the cases 2.11(1^{∘}), 2.11(2), and 2.11(3). Moreover, if 2.11(4:3) holds, then |E/C_{E}(N_{A}(E))| = 2 and so |𝕂| = 2, which contradicts (9^{∘}).

Thus we are left with case 2.11(4:2). In this case p = 3 = |E| = |A| and thus r = 2, and A acts as a cycle of length 3 on Δ. On the other hand, KA acts transitively on Δ, so KA/C_{KA}(Δ) ≅ Alt(Δ). As r = 2, the solvability of K gives |Δ| = 4 and |I_{A}| = 3^{4}, so F A acts as a subgroup of C_{2} ≀ Sym(4) on I_{A}. Since by (8^{∘}) R = [R, KA] we conclude that R = [R, KA] ≅ C_{2}^{3} and KA ≅ 2^{3} : Alt(4). This shows that K ≅ Q_{8} ∘ Q_{8}, and (??) holds.

We may assume from now on that I_{A} is a homogeneous 𝕂R-module. By 4.1 𝔼 := Z(End_{R}(I_{A})) is a field, so 𝕂 ≤ 𝔼, and K\overline{A} acts 𝔼-semi-linearly on I_{A}.

11^{∘} [17] A acts 𝔼-linearly on I_{A}, and R is not abelian.

Suppose A does not act 𝔼-linearly on I_{A}. Since A is not quadratic on I_{A}, 2.14(2) or (3) holds. In both cases dim_{𝔽_{p}} 𝔼 = p and so | Aut(𝔼)| = p. Since F/C_{F} (I_{A}) is an r-group, 𝔼 ≤ End_{F} (I_{A}) = 𝕂, and A acts 𝔼-linearly on I, a contradiction.

Thus A is 𝔼-linear. If R is abelian, then R ≤ End_{R}(I_{A}) = 𝔼 and [R, A] = 1, a contradiction.

12^{∘} [18] Suppose R acts homogeneously but not simply on I_{A}. Then (??) holds.

By 4.4 there exists a tensor decomposition (Φ, V_{0} ⊗_{𝔼} V_{1}, σ, (ϕ_{j,h} | j = 0, 1, h ∈ F A)) such that I_{A} ≅ V_{0} ⊗_{𝔼} V_{1} as an 𝔼R-module, and V_{1} is isomorphic to a R-submodule of I_{A}. Since R is not abelian, dim_{𝔼} V_{1} > 1, and since I_{A} is not a simple R-module, dim_{𝔼} V_{0} > 1. Moreover, by (11^{∘}) A acts 𝔼-linearly on V_{0} ⊗_{𝔼} V_{1}. Hence one of the two cases in 5.2(3) holds.

In both cases the dual version of 6.1 shows that V_{i} is a natural SL_{m_{i}}(𝔼)-module for KA. Since R is solvable, p = 2 or 3 and m_{1} = m_{2} = 2, and since R is non-abelian, p = 3, R ≅ Q_{8} and |I_{A}| = 3^{4}. Now (??) follows

13^{∘} [19] I_{A} is no a simple 𝔼R-module.

Assume that I_{A} is a simple 𝔼R-module. By (11^{∘}) R is non-abelian, and by (8^{∘}) R/Z(R) is a simple A-module. In particular |A/C_{A}(R)| = p and C_{A}(R) = C_{A}(I_{A}) = 1 since I_{A} is a simple R-module, so |A| = p. As A is is cubic on I_{A}, the Hall-Higman Theorem B [Gor, 11.1.1] shows that p = 3 and so |R/Z(R)| = 4 and [R, A] ≅ Q_{8}. Since Q_{8} has a unique faithful simple module over 𝔼 and this module has dimension 2, there exists a 2-dimensional 𝔼[R, A]-submodule I_{0} in I_{A}. On the other hand, R = Z(R)[R, A] and Z(R) embeds in 𝔼. Thus R normalizes I_{0} and I_{0} = I_{A}. But then A acts quadratically on I_{A}, a contradiction. □

Theorem 6.6 [components] Let G be a finite group with O_{p}(G) = 1, K a component of G and V be a faithful finite dimensional 𝔽_{p}G-module. Suppose that there exists an elementary abelian psubgroup A ≤ G with |A/C_{A}(K)| > 2 acting nearly quadratically on V. Then |A/N_{A}(K)| ≤ 2 and either A ≤ N_{G}(K), or p = 2 and K ≅ SL_{n}(2^{m}).

Proof: This follows from [Ch] if A is quadratic on F_{V} (G) and from ?? if A is not quadratic on F_{V} (G). □

References

[Ch] A. Chermak, Quadratic action and the P_{G}(V)-theorem in arbitrary characteristic, J. Group Theory 2 (1999), 1 – 13.

[Gor] D. Gorenstein, Finite Groups, Chelsea (1980) New York.