WEAK CLOSURE AND OLIVER’S p-GROUP CONJECTURE

DAVID J. GREEN AND JUSTIN LYND

Abstract. To date almost all verifications of Oliver’s p-group conjecture have proceeded by verifying a stronger conjecture about weakly closed quadratic subgroups. We construct a group of order 3^{49} which refutes the weakly closed conjecture but satisfies Oliver’s conjecture.

1. Introduction

Chermak has shown that every saturated fusion system has a unique centric linking system [2]. Both Chermak and Oliver in his subsequent proof of the higher limits conjecture [12] invoke the general FF-module theorem [10], which depends on the classification of finite simple groups.

In [11], Oliver formulated a conjecture about p-groups and gave a classificationfree proof that it would imply both the higher limits conjecture and Chermak’s theorem at odd primes. Oliver’s p-group conjecture has been verified in several cases, almost all of which involve verifying the stronger Quadratic Conjecture: Quadratic Conjecture (Conjecture 1.4 of [7]) Let p be a prime, G a finite p-group, and V a faithful 𝔽_{p}G-module. If V is an F -module then there is a quadratic element in Ω_{1}(Z(G)). The Quadratic Conjecture implies Oliver’s Conjecture by [7, Theorem 1.5]. In [7, Proposition 4.5] it is shown that the Quadratic Conjecture in turn follows from another conjecture. Indeed, up till now all verifications of the Quadratic Conjecture have proceeded by verifying this Weakly Closed Conjecture: Weakly Closed Conjecture (Conjecture 4.6 of [7]) Let p be a prime, G a finite p-group, and V a faithful 𝔽_{p}G-module. If there is an elementary abelian subgroup 1 ≠ E ≤ G which is both quadratic on V and weakly closed in C_{G}(E) with respect to G, then there are quadratic elements in Ω_{1}(Z(G)). In this paper we present a counterexample to the Weakly Closed Conjecture which still satisfies the Quadratic Conjecture.

Theorem 1.1. There is a 3-group G of order 3^{49} with the following properties: (1) The Weakly Closed Conjecture fails for one faithful 𝔽_{3}G-module V_{0}.

(2) Every faithful 𝔽_{3}G-module V satisfies the Quadratic Conjecture.

Date: 10 December 2011.

2000 Mathematics Subject Classification. Primary 20D15.

Key words and phrases. p-group; offending subgroup; quadratic offender; p-local finite group.

The group G is constructed in Section 4 and the module V_{0} in Section 5. Part (1) of Theorem 1.1 is proved as Theorem 5.3. Part (2) is proved as Proposition 4.3 using Proposition 3.2.

Remark. It may aid the reader if we recall the exact relationship between the Quadratic Conjecture and the original form of Oliver’s p-group conjecture. Let S be a finite p-group. In [11, Definition 3.1] Oliver defines a characteristic subgroup 𝔛(S) ≤ S, and conjectures that J(S) ≤ 𝔛(S) always holds at odd primes. Here J(S) is the Thompson subgroup generated by the elementary abelian subgroups of greatest order. Section 2 of [7] modifies Oliver’s construction and introduces a second characteristic subgroup Y(S) ≤ S, with Y(S) = S for p = 2 and Y(S) ≤ 𝔛(S) for odd p. Hence the conjecture J(S) ≤ Y(S) is a strengthening of Oliver’s conjecture. Theorem 1.5 of [7] states that the Quadratic Conjecture is equivalent to the conjecture J(S) ≤ Y(S). More specifically, if S is a counterexample to J(S) ≤ Y(S) then G = S/Y(S) and V = Ω_{1}(Z(Y(S))) give a counterexample to the Quadratic Conjecture; and if (G, V) is a counterexample to the Quadratic Conjecture then S = V ⋊ G is a counterexample to J(S) ≤ Y(S).

If p = 3 then the Quadratic Conjecture is equivalent to Oliver’s p-group conjecture, as the definitions of 𝔛(S) and Y(S) coincide for p = 3.

2. Notation and known results

Let G be a finite p-group and V a faithful right 𝔽_{p}G-module. Our notation for commutators is [x, y] = x^{−1}y^{−1}xy and [x_{1}, …, x_{n−1}, x_{n}] = [[x_{1}, …, x_{n−1}], x_{n}]. We often view V, G as subgroups of the semidirect product V ⋊ G, giving meaning to expressions such as [v, g] and C_{V} (H) for v ∈ V, g ∈ G and H ≤ G. Note that [v, g] = v(g − 1) and C_{V} (H) = V^{H}.

An element g ∈ G is called quadratic if its action on V has minimal polynomial (X − 1)^{2}. An elementary abelian subgroup 1 ≠ E ≤ G is called quadratic if [V, E, E] = 0, or equivalently if every 1 ≠ g ∈ E is quadratic (see Corollary 3.2 of [7]). An elementary abelian subgroup 1 ≠ E ≤ G is called an offender if j_{E}(V) ≥ 1, where

j_{H}(V) = \frac{|H| · |C_{V} (H)|}{|V |}

for any subgroup H ≤ G. If there is an offender then V is called an F -module. Finally the set 𝒫(G, V) of best offenders is defined by

𝒫(G, V) = {E ≤ G | E an offender and j_{F} (V) ≤ j_{E}(V) for all F ≤ E} .

Note that every offender contains a best one. Timmesfeld’s Replacement Theorem ([1, Theorem 2]) Let G be a p-group and V a faithful 𝔽_{p}G-module. Suppose that E ∈ 𝒫(G, V). Then F = C_{E}([V, E]) is a quadratic best offender, and j_{F} (V) = j_{E}(V).

Meierfrankenfeld–Stellmacher Lemma ([9, Lemma 2.6]) Let G be a p-group and V a faithful 𝔽_{p}G-module. For H, K ≤ G one has

j_{H}(V)j_{K}(V) ≤ j_{H∩K}(V)j_{⟨H,K⟩}(V)

with equality if and only if ⟨H, K⟩ = HK and C_{V} (H ∩ K) = C_{V} (H) + C_{V} (K). Normal Abelian Lemma ([6, Theorem 1.5]) Let G be a p-group and V a faithful 𝔽_{p}G-module with no central quadratics. Suppose that A ⊴ G is an abelian normal subgroup. Then no offender lies in A. Finally we require a slight strengthening of [5, Lemma 4.1]. Descent Lemma ([5]) Suppose that p is an odd prime, that G ≠ 1 is a finite p-group, and that V is a faithful 𝔽_{p}G-module. Suppose that a, b ∈ G are such that 1 ≠ c := [a, b] lies in C_{G}(a, b). Suppose further that b is quadratic. Then ⟨b, c⟩ is quadratic too.

Proof. c is quadratic by the statement of [5, Lemma 4.1]. So by [7, Corollary 3.2] it suffices to prove that [V, b, c] = 0. This follows from the assertion βγ = 0 in the proof of [5, Lemma 4.1]. □

3. A technical result on offenders

Hypothesis 3.1. Let G be a finite p-group. Suppose that A, N ⊴ G with A abelian, A ≤ N, [N, N] ≤ A und [N, A] = 1. Suppose that V is a faithful 𝔽_{p}G-module and that there are no quadratic elements in Z(G). Denote by E the set of all offenders E ≤ N such that |E: E ∩ A| = p.

Proposition 3.2. Suppose that Hypothesis 3.1 is satisfied. Then no E ∈ E is weakly closed in C_{G}(E) with respect to G.

Before we prove the proposition we need to establish three lemmas.

Lemma 3.3. Let G be a finite p-group and V a faithful 𝔽_{p}G-module with no central quadratics. Then there is no offender E ≤ G with |E| = p.

Proof. Recall from [7, Section 3] that we write g ⊥ h if [g, h] = 1 and [V, g, h] = 0. This relation is symmetric. Suppose that E = ⟨x⟩ is an offender. Since x acts nontrivially and j_{E}(V) ≥ 1 it follows that C_{V} (E) = C_{V} (x) has codimension 1 in the 𝔽_{p}-vector space V. Pick y ≠ 1 to lie in the intersection of Ω_{1}(Z(G)) with the normal closure of E. If x ⊥̸ y then [v_{0}, y, x] ≠ 0 for some v_{0} ∈ V. But then v_{0} and [v_{0}, y] are linearly independent in V/C_{V} (x), since v ↦ [v, y] is nilpotent and [v_{0}, y, x] = [v_{0}, x, y] ≠ 0. This contradicts codimension 1. We conclude that x ⊥ y.

Now consider y^{⊥} = {g ∈ G | g ⊥ y}. By [7, Lemma 3.1 (3)] this is a normal subgroup of C_{G}(y) = G. Since x ∈ y^{⊥} it follows by choice of y that y ∈ y^{⊥}. So y is quadratic by [7, Lemma 3.1 (1)], contradicting the assumption that there are no central quadratics. □

Lemma 3.4. Suppose that Hypothesis 3.1 is satisfied. Let E ∈ E and set F = E ∩ A. Then

(1) F ≠ 1

(2) j_{E}(V) = 1, j_{F} (V) = p^{−1} and C_{V} (E) = C_{V} (F).

(3) E ∈ 𝒫(G, V).

(4) The set C_{E}([V, E]) \ A is non-empty. Each x ∈ C_{E}([V, E]) \ A satisfies E = ⟨x, F ⟩ and [V, x] ≤ C_{V} (E).

Proof. By the Normal Abelian Lemma, A contains no offenders.

(1): If F = 1 then |E| = p, which cannot happen by Lemma 3.3.

(2): Follows by definition of j_{E}(V) since E offends but 1 ≠ F ≤ A does not. (3): Any offender E_{1} < E also lies in E. Apply (2). (4): E_{1} = C_{E}([V, E]) is a quadratic offender by Timmesfeld’s Replacement

Theorem. Since E_{1} offends we have E_{1} ≰ A. So E = ⟨x, F ⟩ for any x ∈ E_{1} \ A. Moreover for e ∈ E we have

[v, x, e] = v(x − 1)(e − 1) = v(e − 1)(x − 1) = [v, e, x] = 0 ,

since [x, e] = 1 and [V, E, E_{1}] = 0 by construction of E_{1}. So [V, x] ≤ C_{V} (E). □

Lemma 3.5. Suppose that Hypothesis 3.1 is satisfied and that E, E_{1} ∈ E are such that E ≤ E_{1}, E ∩ A ≠ 1, and E is weakly closed in C_{G}(E) with respect to G. Then there is E_{2} ∈ E such that E_{1} ≤ E_{2} and E_{2} is weakly closed in C_{G}(E_{2}) with respect to G.

Proof. We argue by contradiction and assume that E_{1} is a largest counterexample. Setting F = E ∩ A and F_{1} = E_{1} ∩ A we have F ≠ 1, F ≤ F_{1} and E_{1} = EF_{1}. As E_{1} is not weakly closed there is g ∈ G with E_{1}^{g} ≠ E_{1} and [E_{1}, E_{1}^{g}] = 1. Since E is weakly closed and E ≤ E_{1} it follows that E^{g} = E. Hence F = F^{g} ≤ F_{1}^{g} and F_{1}^{g} ≰ E_{1}. Set E_{3} = E_{1}F_{1}^{g} > E_{1}. Applying the Meierfrankenfeld–Stellmacher Lemma to E_{1}, F_{1}^{g} we have j_{E_{3}}(V)j_{E_{1}∩F_{1}^{g}}(V) ≥ 1 · p^{−1} by Lemma 3.4. So since 1 ≠ F ≤ E_{1} ∩ F_{1}^{g} ≤ A the Normal Abelian Lemma says that j_{E_{1}∩F_{1}^{g}}(V) ≤ p^{−1} and hence E_{3} ∈ E. This contradicts the maximality of E_{1}. □

Proof of Proposition 3.2. We argue by contradiction and suppose that E ∈ E is weakly closed and of the largest possible order. By Lemma 3.4 (4) we have E = ⟨x, F ⟩ with x ∈ N \ A quadratic. Now E ⋬ G by the Normal Abelian Lemma, so since E is weakly closed there is a g ∈ G with [E, E^{g}] ≠ 1. Since E, E^{g} ≤ N and [N, A] = 1 it follows that 1 ≠ [x, x^{g}] ∈ [N, N] ≤ A. Once more using [N, N] ≤ A and [N, A] = 1 we deduce that [E, E^{x^{g}}] = 1. So since E is weakly closed it follows that E^{x^{g}} = E and therefore 1 ≠ [x, x^{g}] ∈ F = A ∩ E. Similarly with E_{1} = E^{g} we have E_{1}^{x} = E_{1}, and hence [x, x^{g}] ∈ F^{g} = A ∩ E_{1}. So F ∩ F^{g} ≠ 1.

We now apply the Meierfrankenfeld–Stellmacher Lemma to E and F^{g}. Since 1 ≠ [x, x^{g}] ∈ E ∩ F^{g} ≤ A we have j_{E∩F^{g}}(V) ≤ p^{−1}. Hence j_{EF^{g}}(V) ≥ 1 and EF^{g} ∈ E. Since E is maximal weakly closed by assumption, Lemma 3.5 gives us EF^{g} = E and hence F^{g} = F. So by Lemma 3.4 (2) we have

C_{V} (E^{g}) = C_{V} (F^{g}) = C_{V} (F) = C_{V} (E) ,

whence by (4) we have [V, x, x^{g}] ≤ [C_{V} (E), x^{g}] = [C_{V} (E^{g}), x^{g}] = 0. And since [v, x^{g}] = [vg^{−1}, x]g we also have [v, x^{g}, e^{g}] = [vg^{−1}, x, e]g = 0, meaning that [V, x^{g}] ≤ C_{V} (E^{g}) and hence [V, x^{g}, x] ≤ [C_{V} (E^{g}), x] = [C_{V} (E), x] = 0. So [V, [x, x^{g}]] = 0, contradicting the fact that V is faithful and [x, x^{g}] ≠ 1. □

4. The group

The construction we describe in this section was inspired by an example of J. L. Alperin, see [8, p. 349] or [3, pp. 324–5].

Let p be an odd prime and Ω a finite set with at least two elements. Write F (Ω) for the free group on Ω, and let N(Ω) ⊴ F (Ω) be the normal subgroup generated by all twofold commutators [g_{1}, g_{2}, g_{3}] and all pth powers g^{p}. Define E(Ω) to be the quotient group E(Ω) = F (Ω)/N(Ω). We shall call E(Ω) the free group of class two and exponent p on the set Ω. Note that if p = 2 or |Ω| = 1 then E(Ω) is abelian and therefore of class one.

Lemma 4.1. Let p be an odd prime and Ω a finite set with at least two elements. The free class two exponent p group E(Ω) has the following properties:

(1) E(Ω) is indeed of class two and exponent p.

(2) E(Ω) is special^{1}: its derived and Frattini subgroups coincide with its centre. This is elementary abelian of rank \binom{|Ω|}{2}, generated by the [x, y] with x, y ∈ Ω.

(3) |E(Ω)| = p^{n} for n = \frac{1}{2} |Ω| (|Ω| + 1).

(4) E(a, b) is extraspecial of order p^{3} and exponent p. The \binom{|Ω|}{2} projection maps E(Ω) ↠ E(a, b) with a, b ∈ Ω detect all elements of E(Ω).

Proof. (1): Let P = ⟨x, y⟩ be the extraspecial group of order p^{3} and exponent p. Recall that P′ = Φ(P) = Z(P) is cyclic of order p, generated by [x, y]. As P has class two and exponent p, taking two distinct elements a, b ∈ Ω induces a group homomorphism f = f_{a,b}: E(Ω) → P with f(a) = x, f(b) = y and f(c) = 1 for all c ∈ Ω \ {a, b}. Then f is surjective and f([a, b]) = [x, y] ≠ 1, so E(Ω) really does have class two and exponent p.

(2): Write G = E(Ω). Then Φ(G) = G′ since the exponent is p, and G′ ≤ Z(G) since the class is two. And as all commutators are central we have [g, hk] = [g, h]·[g, k] and [gh, k] = [g, k][h, k]. So G′ is generated by the [a, b] with a, b ∈ Ω. If g ∈ G \ G′ then f_{a,b}(g) ∈ P is noncentral for some a, b ∈ Ω, so G′ = Z(G) since f_{a,b} is surjective. Moreover as f_{a,b} kills all commutators except [a, b] = [b, a]^{−1} it follows that the \binom{|Ω|}{2} commutators [a, b] are linearly independent in the 𝔽_{p}-vector space which is the elementary abelian p-group G′. Part (3) follows since G/Φ(G) is elementary abelian of rank |Ω|. Finally (4) now follows by the proof of (2). □

1In the sense of [4, p. 183].

From now on we restrict ourselves to the prime three. Set H = E(Ω) for Ω = {x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}, z_{1}, z_{2}, z_{3}}. Define G to be the semidirect product G = H ⋊Q, where Q = C_{3} ≀ C_{3} = ⟨σ_{x}, σ_{y}, σ_{z}, τ⟩ is the the Sylow 3-subgroup of S_{9}, with the following action on H:

• The x-cycle σ_{x}: x_{i}^{σ_{x}} = x_{i+1}, y_{i}^{σ_{x}} = y_{i}, z_{i}^{σ_{x}} = z_{i};

• The analogous y- and z-cycles σ_{y} and σ_{z};
• The top-level cycle x_{i}^{τ} = y_{i}, y_{i}^{τ} = z_{i}, z_{i}^{τ} = x_{i}.

Since |H| = 3^{45} and |Q| = 3^{4} we have |G| = 3^{49}.

Lemma 4.2. Let G = H ⋊ Q be this 3-group.

(1) Z(G) lies in H′ and is elementary abelian of rank two, generated by the Q-orbit products of [x_{1}, x_{2}] (length 9) and [x_{1}, y_{1}] (length 27).

(2) Suppose that V is a faithful 𝔽_{3}G-module with no central quadratics. Then every quadratic element lies in H.

Proof. Since H is special we have Z(H) = H′. The diagonal action of Q partitions Ω^{2} into five orbits: three of length nine with representatives (x_{1}, x_{1}), (x_{1}, x_{2}) and (x_{2}, x_{1}); and two of length 27 with representatives (x_{1}, y_{1}) and (y_{1}, x_{1}). Since [x, x] = 1 and [y, x] = [x, y]^{−1}, this means that Z(G) ∩ H is the rank two elementary abelian described in (1).

(2): Set K = ⟨H, σ_{x}, σ_{y}, σ_{z}⟩. If g ∈ G \ K then g ∈ Kτ ∪ Kτ^{2}, so commuting g with ∏_{i,j}[x_{i}, y_{j}][x_{i}, z_{j}] yields the Q-orbit product of [x_{1}, y_{1}]. This is central, so g and ∏_{i,j}[x_{i}, y_{j}][x_{i}, z_{j}] are non-quadratic by the Descent Lemma. And if k ∈ K \ H then without loss of generality k ∈ H⟨σ_{y}, σ_{z}⟩σ_{x} ∪ H⟨σ_{y}, σ_{z}⟩σ_{x}^{2}. So its commutator with ∏_{i}[x_{1}, y_{i}]^{−1}[x_{1}, z_{i}]^{−1}[x_{2}, y_{i}][x_{2}, z_{i}] yields ∏_{i,j}[x_{i}, y_{j}][x_{i}, z_{j}], which is non-quadratic. Hence k is non-quadratic by the Descent Lemma. □

Proposition 4.3. Let G = H ⋊ Q be the 3-group constructed above. Every faithful 𝔽_{3}G-module V satisfies the Quadratic Conjecture.

Proof. Suppose not. Then V has no central quadratics and yet there is an offender E ≤ G. By [7, Theorem 4.5] we may assume that E is quadratic and weakly closed in C_{G}(E) with respect to G. So E ≤ H by Lemma 4.2 (2).

Now observe that G satisfies Hypothesis 3.1 with N = H, A = H′. Since E is a weakly closed offender, we will be done by Proposition 3.2 once we can show that E ∈ E. Certainly E ≰ H′, by the Normal Abelian Lemma. And observe that by construction of H, we have C_{H}(g) = ⟨g, H′⟩ for all g ∈ H \ H′. So since E is elementary abelian we have |E: E ∩ A| ≤ p. So E ∈ E, as claimed. □

5. Constructing the module V_{0}

As above we restrict our attention to the case p = 3. Recall that E(a, b) is extraspecial. We first construct a useful representation of E(a, b) × E(c, d).

Lemma 5.1. Consider the representation V_{1}(a, b) of E(a, b) in GL_{3}(𝔽_{p}) given by

a ↦ (\begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix})  b ↦ (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix})

Then c := [a, b] acts as the matrix (\begin{matrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}). This representation is faithful,

and the elementary abelian subgroups ⟨a, c⟩ and ⟨b, c⟩ are quadratic.

Proof. Since c acts nontrivially the representation is faithful. Every matrix in

⟨a, c⟩ is of the form (\begin{matrix}1 & ∗ & ∗ \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) and is therefore quadratic. Similarly, every

matrix in ⟨b, c⟩ has the form (\begin{matrix}1 & 0 & ∗ \\ 0 & 1 & ∗ \\ 0 & 0 & 1 \end{matrix}). □

Lemma 5.2. Consider the representation V_{2}(a_{1}, a_{2}; b_{1}, b_{2}) of E(a_{1}, a_{2})×E(b_{1}, b_{2}) in GL_{9}(𝔽_{p}) given by

V_{2}(a_{1}, a_{2}; b_{1}, b_{2}) := V_{1}(a_{1}, a_{2}) ⊗_{𝔽_{p}} V_{1}(b_{1}, b_{2}) .

Then V_{2} is faithful; ⟨a_{i}, [a_{1}, a_{2}]⟩ and ⟨b_{i}, [b_{1}, b_{2}]⟩ are quadratic for i = 1, 2; and [a_{1}, a_{2}][b_{1}, b_{2}] does not act quadratically.

Proof. Let e_{1}, e_{2}, e_{3} be the standard basis of 𝔽_{p}^{3}. The centre of E(a_{1}, a_{2}) × E(b_{1}, b_{2}) is elementary abelian of rank two generated by [a_{1}, a_{2}] and [b_{1}, b_{2}]. Since [a_{1}, a_{2}]^{r}[b_{1}, b_{2}]^{s} sends e_{3} ⊗ e_{3} to e_{3} ⊗ e_{3} + re_{1} ⊗ e_{3} + se_{3} ⊗ e_{1} + rse_{1} ⊗ e_{1}, the representation is faithful. The subgroups ⟨a_{i}, [a_{1}, a_{2}]⟩ and ⟨b_{i}, [b_{1}, b_{2}]⟩ are quadratic by Lemma 5.1, since each only operates on one factor of V_{2}. For g = [a_{1}, a_{2}][b_{1}, b_{2}] we have [e_{3} ⊗ e_{3}, g, g] = (e_{3} ⊗ e_{3})(g − 1)^{2} and

(e_{3} ⊗ e_{3})(g − 1)^{2} = (e_{1} ⊗ e_{3} + e_{3} ⊗ e_{1} + e_{1} ⊗ e_{1})(g − 1) = 2e_{1} ⊗ e_{1} ≠ 0 .

So g is not quadratic. □

We now use several copies of V_{2} to construct a faithful representation of G. Write F_{4}(Ω) for the subset of Ω^{4} consisting of those 4-tuples whose components are pairwise distinct. Note that Q acts on F_{4}(Ω) via its diagonal action on Ω^{4}. For each (a, b, c, d) ∈ F_{4}(Ω) we may view V_{2}(a, b; c, d) as a representation of H, with each generator in Ω \ {a, b, c, d} acting as the identity. So the direct sum

V_{0} = ⨁_{(a,b,c,d)∈F_{4}(Ω)} V_{2}(a, b; c, d) ,

is a representation of H, as each summand is; and as Q permutes the summands, V_{0} is a representation of G too.

Theorem 5.3. G and V_{0} have the following properties:

(1) V_{0} is faithful and there are no quadratic elements in the centre.

(2) The subgroup

E := ⟨x_{1}, [x_{1}, a] | a ∈ Ω}⟩ ≤ G

is elementary abelian of rank 9. It is weakly closed in C_{G}(E) with respect to G.

(3) E has quadratic action on V_{0}. Hence (G, V_{0}, E) is a counterexample to the Weakly Closed Conjecture.

Proof. (1): If the representation has a kernel then this meets Z(G). Pick 1 ≠ g ∈ Z(G). Replacing g by a power if necessary we may assume by Lemma 4.2 (1) that g is one of the following: the Q-orbit product c_{1} of [x_{1}, x_{2}]; the Q-orbit product c_{2} of [x_{1}, y_{1}]; or c_{1}c_{2}^{r} for r = ±1. If g = c_{1} then by Lemma 5.2 its action on V_{2}(x_{1}, x_{2}; y_{1}, y_{2}) is neither trivial nor quadratic, for the image of c_{1} in E(x_{1}, x_{2})×E(y_{1}, y_{2}) is [x_{1}, x_{2}][y_{1}, y_{2}]. Similarly the action of c_{2} on V_{2}(x_{1}, y_{1}; x_{2}, y_{2}) is neither trivial nor quadratic. Finally g = c_{1}c_{2}^{r} has image [x_{1}, x_{2}][y_{1}, z_{1}]^{r} in E(x_{1}, x_{2}) × E(y_{1}, z_{1}), and so for the action of g on V_{2}(x_{1}, x_{2}; y_{1}, z_{1}) we have

[e_{3}⊗e_{3}, g, g] = (e_{3}⊗e_{3})(g−1)^{2} = (re_{3}⊗e_{1}+e_{1}⊗e_{3}+re_{1}⊗e_{1})(g−1) = 2re_{1}⊗e_{1} ≠ 0 ,

so the action of g is neither trivial nor quadratic.

(2) From Lemma 4.1 it follows that E is elementary abelian of rank nine. Weakly closed: We have [E, ⟨H, σ_{y}, σ_{z}⟩] ≤ E, and if g ∈ G \ ⟨H, σ_{y}, σ_{z}⟩ then x_{1}^{g} ∈ H′a for some a ∈ Ω \ {x_{1}}, whence [x_{1}, x_{1}^{g}] ≠ 1.

(3) x_{1} quadratic: If no a_{i} is x_{1}, then x_{1} acts trivially on V_{2}(a_{1}, a_{2}; a_{3}, a_{4}). If x_{1} is amongst the a_{i}, then x_{1} acts quadratically by Lemma 5.2. In particular the action on V_{2}(x_{1}, x_{2}; y_{1}, y_{2}) is nontrivial.

E quadratic: We have E = ⟨x_{1}⟩ × F for F = E ∩ H′. Since [x_{1}, g][x_{1}, h] = [x_{1}, gh] for g, h ∈ H it follows that F = {[x_{1}, g] | g ∈ H}. So for each 1 ≠ e ∈ E there is an a ∈ Ω \ x_{1} with e ∈ F_{a} = ⟨x_{1}, [x_{1}, a]⟩. Since [x_{1}, a] ≠ 1 and x_{1} is quadratic, it follows from the Descent Lemma that F_{a} is quadratic too. So e is quadratic. □

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http://www.abdn.ac.uk/~mth192/papers/o/oliver/ch-limz.dvi.

Department of Mathematics, Friedrich-Schiller-Universität Jena, 07737 Jena, Germany

E-mail address: david.green@uni-jena.de

Department of Mathematics, The Ohio State University, Columbus, OH 43210

E-mail address: jlynd@math.ohio-state.edu