2-SUBNORMAL QUADRATIC OFFENDERS AND OLIVER’S p-GROUP CONJECTURE

JUSTIN LYND

Abstract. Bob Oliver conjectures that if p is an odd prime and S is a finite p-group, then the Oliver subgroup 𝔛(S) contains the Thompson subgroup J_{e}(S). A positive resolution of this conjecture would give the existence and uniqueness of centric linking systems for fusion systems at odd primes. Using ideas and work of Glauberman, we prove that if p ⩾ 5, G is a finite p-group, and V is an elementary abelian p-group which is an F-module for G, then there exists a quadratic offender which is 2-subnormal (normal in its normal closure) in G. We apply this to show that Oliver’s conjecture holds provided the quotient G = S/𝔛(S) has class at most log_{2}(p−2)+1, or p ⩾ 5 and G is equal to its own Baumann subgroup.

1. Introduction

Let F be a saturated fusion system on a finite p-group S. If F = F_{S}(G) for some finite group G, then Broto, Levi, and Oliver show [BLO03] that one can associate a classifying space to F whose completion at p has the homotopy type of the completion of BG at p. The construction of such a classifying space is based upon a centric linking system associated to F. But for arbitrary F, it is an open question^{1} as to whether an associated centric linking system (and thus a suitable classifying space for F) exists. See the survey article [BLO04] for more details.

In the course of proving the Martino-Priddy conjecture for odd primes [Oli04], Oliver defines a certain subgroup 𝔛(S) of S, now called the Oliver subgroup. He shows that if p is odd and the Thompson subgroup J_{e}(S) is contained in 𝔛(S), then the homological obstruction to the existence and uniqueness of a centric linking system for F vanishes.

In this paper we investigate Oliver’s (purely group-theoretic) conjecture that this inclusion J_{e}(S) ⩽ 𝔛(S) always holds for odd p and prove it for p-groups whose quotient G = S/𝔛(S) has small nilpotence class.

Theorem A. Let p be an odd prime, and suppose S is a finite p-group such that G = S/𝔛(S) has nilpotence class at most log_{2}(p − 2) + 1. Then J_{e}(S) ⩽ 𝔛(S), so Oliver’s conjecture holds for S.

Green, Héthelyi, and Mazza show in [GHM10] that J_{e}(S) ⩽ 𝔛(S) if the quotient G is metabelian, of maximal class, or of p-rank at most p. They also settle the case where G has class at most 4, so the above result is known for p ⩽ 17. To our knowledge though, Theorem A provides the first nonconstant lower bound on the class of a counterexample to Oliver’s conjecture.

Date: December 8, 2011.

1This question has now been resolved in the affirmative by Andrew Chermak [Che11], but Oliver’s p-group conjecture is still open.

The Baumann subgroup (see [MSS03]) of a finite p-group G is defined as Baum(G) = C_{G}(Ω_{1}Z(J_{e}(G))). In particular, Baum(G) ⩾ J_{e}(G). We define in Section 2 the k-th Oliver subgroup for 3 ⩽ k ⩽ p in direct analogy with 𝔛(S), which gives a chain of characteristic, self-centralizing subgroups 𝔛_{3}(S) ⩽ 𝔛_{4}(S) ⩽ ⋯ ⩽ 𝔛_{p}(S) of S. The last one 𝔛_{p}(S) is the original Oliver subgroup 𝔛(S). We prove the following.

Theorem B. Let p ⩾ 5 and 3 ⩽ k ⩽ p. Suppose S is a finite p-group, and set G = S/𝔛_{k}(S). If Baum(G) = G, then J_{e}(S) ⩽ 𝔛_{k}(S). In particular, taking k = p, Oliver’s conjecture holds for S.

Perhaps one reason there has not been a general attack on Oliver’s conjecture to date lies in its resistance to any kind of inductive argument. The strategy for proving Theorem B gives some hint of how such an inductive argument may possibly proceed, at least for p ⩾ 5. See Lemma 5.2 and Section 6 for speculation on this matter.

In general, we take the point of view of Green, Héthelyi, and Lilienthal in [GHL08], where they reformulate Oliver’s conjecture in terms of a statement about representations of the quotient group G over F_{p}. In particular, in a putative counterexample to Oliver’s conjecture, V = Ω_{1}Z(𝔛(S)) is an F-module for G.

Our methods rely on ideas of Glauberman [Gla08] and a modification of Glauberman’s generalization of Thompson’s Replacement Theorem in [Gla97]. For p ⩾ 5 and V an Fmodule for G, the modification gives rise to the existence of 2-subnormal quadratic offenders in G, and we use this to prove Theorems A and B. Recall that a subgroup H of G is 2subnormal if it is normal in a normal subgroup of G, or equivalently, normal in its normal closure in G.

Theorem C. Suppose p ⩾ 5. Let V be an elementary abelian p-group which is an F-module for the p-group G. Then there exists a quadratic offender on V in G which is normal in its normal closure in G.

The organization of the paper is as follows. In Section 2, we recap definitions and outline the reformulation of Oliver’s conjecture due to Green, et al. Section 3 contains a couple of elementary lemmas needed for Section 4, where we prove Theorem C. We use Theorem C in Section 5 to produce in a counterexample an elementary abelian subgroup generated by quadratic elements, and we leverage this subgroup to prove Theorems A and B. In the final Section we record an observation which could allow for a inductive approach to the conjecture.

2. Definitions and preliminaries

Let p be an odd prime. We outline in this section the module-theoretic version of Oliver’s conjecture from [GHL08].

Definition 2.1. Let 3 ⩽ k ⩽ p. The k-th Oliver subgroup of S is the largest subgroup 𝔛_{k}(S) such that there exists a series

1 = Q_{0} < Q_{1} < ⋯ < Q_{n−1} < Q_{n} = 𝔛_{k}(S)

with each Q_{i} normal in S and with the property that for all 1 ⩽ i ⩽ n,

[Ω_{1}(C_{S}(Q_{i−1})), Q_{i}; k − 1] = 1.

Taking k = p in the definition, we get Oliver’s original definition, and so 𝔛_{p}(S) = 𝔛(S). For each k, 𝔛_{k}(S) is a characteristic subgroup of S. Furthermore, as is clear from the definition, 𝔛_{k}(S) contains every normal subgroup of S of nilpotence class at most k − 2. In particular it contains every normal abelian subgroup of S, and so C_{S}(𝔛_{k}(S)) = Z(𝔛_{k}(S)) (that is, 𝔛_{k}(S) is self-centralizing). A recent preprint from Green, Héthelyi and Mazza [GHM11] independently calls attention to 𝔛_{3}(S), which is their Y(S).

Recall the Thompson subgroup is the subgroup J_{e}(S) of S generated by the elementary abelian subgroups of maximum order. We will keep the e subscript throughout, since in Section 4, we will be working with (not necessarily elementary) abelian subgroups of maximum order. In [Oli04, Conjecture 3.9], Bob Oliver poses the following.

Conjecture 2.2 (Oliver). Let S be a finite p-group with p odd. Then J_{e}(S) ⩽ 𝔛(S).

Perhaps the most effective way of viewing these Oliver subgroups is by way of the action of the quotient group G := S/𝔛_{k}(S) on V := Ω_{1}Z(𝔛_{k}(S)). If the k-th Oliver subgroup is a proper subgroup of S, then for each element z ∈ S such that z𝔛_{k}(S) is an element of order p in Z(G), the subgroup ⟨z⟩𝔛_{k}(S) is normal in S and properly contains 𝔛_{k}(S). Thus by maximality of the Oliver subgroup,

[V, 𝔛_{k}(S)⟨z⟩; k − 1] ≠ 1.

Since 𝔛_{k}(S) and V commute, this means that [V, z; k − 1] ≠ 1. In other words, z𝔛_{k}(S) acts on V with minimum polynomial of degree at least k.

This motivates the following definition.

Definition 2.3. Suppose G is a finite p-group, and V is an elementary abelian p-group on which G acts. Then V is said to be a PS-module of degree k if for G if each 1 ≠ z ∈ Ω_{1}Z(G) has minimum polynomial of degree at least k. We drop the degree qualifier if k = p, and say V is a PS-module if it is a PS-module of degree p.

Thus, whenever 𝔛_{k}(S) is a proper subgroup of S, V is a PS-module of degree k for G. Conversely, if G is any finite p-group and V is a PS-module of degree k for G, then 𝔛_{k}(V ⋊ G) = V.

Let us examine what happens if Oliver’s conjecture fails. If J_{e}(S) is not contained in 𝔛_{k}(S), then there is an elementary abelian subgroup A ⩽ S of maximum order not contained in 𝔛_{k}(S). Setting E = A𝔛_{k}(S)/𝔛_{k}(S) ⩽ G, this means by maximality of A, that |E||C_{V} (E)| has order at least |V |. This leads us to the next definition and to the reformulation of Oliver’s conjecture due to Green, Héthelyi, and Lilienthal in [GHL08, Theorem 1.2].

Definition 2.4. Let G be a finite p-group and V an elementary abelian p-group on which G acts faithfully. Then V is an F-module for G if there exists a nontrivial elementary abelian subgroup E of G such that |E||C_{V} (E)| ⩾ |V |. In this case, we call E an offender and say that E offends on V.

Theorem 2.5. Let p be an odd prime. Oliver’s conjecture holds if and only if for every finite p-group G, no PS-module for G is an F-module.

First appearing in John Thompson’s N-group paper as obstructions to factorizations in 2-constrained groups, the properties of F-modules are by now well-known. In particular, as proved by Thompson in his Replacement Theorem, if there exists an offender E ⩽ G on V, then one may choose E to act quadratically, that is, satisfying [V, E, E] = 1 but not centralizing V. In particular, the nonidentity elements of E are quadratic elements, that is, they have quadratic minimum polynomial on V. Recall that because of the identity (x − 1)^{p} = x^{p} − 1 in characteristic p, a quadratic element is of order p provided G is faithful on V.

When 3 ⩽ k ⩽ p and J_{e}(S) is not contained in 𝔛_{k}(S), the existence of quadratic offenders in G = S/𝔛_{k}(S) on V = Ω_{1}Z(𝔛_{k}(S)) portends the presence of quadratic elements in Z(G) and leads the way to a possible contradiction. Indeed, many of the proofs for special classes of p-groups have gone this way. So it is quite possible that if Oliver’s original conjecture holds at all, it holds in the stronger form: J_{e}(S) ⩽ 𝔛_{3}(S) for all S.

In Sections 3 and 4, where the focus is on abelian subgroups of S of maximal order, we follow Huppert and Blackburn’s treatment [HB82, pp. 19–21] of Thompson’s Replacement Theorem. Then in the proof of Theorem B, we use inside the quotient group G a version [GLS96, Theorem 25.2] of Thompson Replacement stated in terms of elementary abelian subgroups. Here we state the following well-known preliminary lemma to Thompson Replacement from [HB82] for use in Section 4.

Lemma 2.6. Suppose that S is a p-group and A is an abelian subgroup of S. Let v be an element of S for which N = [v, A] is abelian, and put A^{∗} = NC_{A}(N). Then

(a) A^{∗} is an abelian subgroup and |A ∩ M| ⩽ |A^{∗} ∩ M| for every normal subgroup M of S. In particular |A| ⩽ |A^{∗}|.

(b) If also 1 = Z_{0} ⩽ Z_{1} ⩽ ⋯ ⩽ Z_{n} = S is a central series of S such that |A ∩ Z_{i}| = |A^{∗} ∩ Z_{i}| for all i = 1, …, n, then A^{∗} = A.

3. Product subgroups

Let V be an elementary abelian p-group. Suppose G is a p-group which acts faithfully on V, and set S = V ⋊ G. Let π : S → G denote the canonical projection.

In this section we make a definition and collect a couple of elementary lemmas needed for carrying out a modification of Glauberman’s generalization of Thompson Replacement in the next section.

Definition 3.1. Let B be a subgroup of S. Define B_{×} to be the subgroup (B ∩ V)π(B) of S. We say B is a product subgroup if B = B_{×}.

The following Lemma is clear from the definitions, but is included here for reference.

Lemma 3.2. Let B be a subgroup of S. Then (a) (B_{×})_{×} = B_{×}.

(b) |B| = |B_{×}|.

(c) B = B_{×} if and only if π(B) ⩽ B.

(d) If B ⩽ V or V ⩽ B, then B = B_{×}.

(e) If B is abelian, then B_{×} is abelian.

Lemma 3.3. Let A and M be subgroups of S and suppose M ⩽ V or V ⩽ M. Then (A ∩ M)_{×} = A_{×} ∩ M.

Proof. Since π(A ∩ M) ⩽ π(A) ∩ π(M), we always have (A ∩ M)_{×} ⩽ A_{×} ∩ M_{×}, and by (d) of Lemma 3.2, the right-hand side is A_{×} ∩ M, so it suffices to show the reverse inclusion.

If M ⩽ V then A_{×} ∩ M = A_{×} ∩ V ∩ M = A ∩ V ∩ M = (A ∩ M)_{×}, so we may assume V ⩽ M.

Let s ∈ A_{×} ∩ M, and write s = uh with u ∈ A ∩ M ∩ V and h ∈ π(A) ∩ π(M) ⩽ M, the inclusion by (c) of Lemma 3.2. For s to be in (A ∩ M)_{×}, it is enough to show that h ∈ π(A ∩ M). But as h ∈ π(A), there exists v ∈ V ⩽ M with vh ∈ A. So vh ∈ M as well. Therefore h ∈ π(A ∩ M), completing the proof. □

4. Offenders

In this section, we shall investigate properties of offending subgroups of G, using ideas [Gla08] and results [Gla97] of Glauberman, and work toward the proof of Theorem C.

Given an elementary abelian p-group V and a p-group G acting faithfully on V, we form the semidirect product S = V ⋊ G, as before.

Denote by A(S) the set of abelian subgroups of S of maximum order, and set

A_{×}(S) = {A ∈ A(S) | A = A_{×} and A ⩽̸ V }.

Definition 4.1. Let

S: 1 = Z_{0} ⩽ Z_{1} ⩽ ⋯ ⩽ Z_{n} = S

be a central series of S. For abelian subgroups A and B of S, we say that A ⩽_{S} B if

|A| = |B| and |A ∩ Z_{i}| ⩽ |B ∩ Z_{i}| for all 1 ⩽ i ⩽ n.

If A ⩽_{S} B and furthermore |A ∩ Z_{i}| < |B ∩ Z_{i}| for some i, we say that A <_{S} B. Let C be a collection of abelian subgroups of S. We say that A is maximal in C with respect to S if A ∈ C and there does not exist B ∈ C with A <_{S} B.

As a consequence of Thompson Replacement, there exists A ∈ A(S) such that for any abelian subgroup B of S which is normalized by A, we have A in turn is normalized by B. The Corollary to [Gla97, Theorem 1] extends this statement for odd primes to include B of nilpotence class at most 2 by choosing A to be maximal in A(S) with respect to any fixed central series of S. Furthermore, for p ⩾ 5, Glauberman obtains with the same choice of A that (∗) A is normalized by any subgroup B of S (not necessarily normalized by A) for which A ⊴ ⟨A^{B}⟩ and [A, u; 3] = 1 for every u ∈ B.

To get a statement modulo V about offenders, we must take A not contained in V, and make sure the replacement still lies outside V. To do this, we modify Glauberman’s proof of his Theorem 1 in a couple of ways to obtain a version of this last result (∗) for p ⩾ 5. Firstly, because of difficulties retaining the commutator condition [A, u; 3] = 1 after mapping u into G, we restrict the discussion and replacement process only to elements of A_{×}(S) and require that B be a product subgroup of S. Secondly, we require that A is maximal with respect to a central series which contains our distinguished subgroup V as a member. Since we are just concerned with what happens modulo V, this is harmless and facilitates verification that our replacement subgroup A^{∗} truly is greater than A in the above ordering.

We begin by showing that A_{×}(S) is nonempty and proving a product subgroup version of a consequence of Thompson Replacement.

Lemma 4.2. The following hold.

(a) If V is an F-module for G, then A_{×}(S) is not empty.

(b) Let S be a central series of S. Suppose A is maximal in A_{×}(S) with respect to S. Then V normalizes A, and so [V, A, A] = 1.

Proof. Suppose V is not an element of A(S). If A ∈ A(S), then in this case |A_{×}| = |A| > |V |, and A_{×} is a product subgroup by Lemma 3.2. Thus, A_{×} ∈ A_{×}(S).

Suppose that V ∈ A(S). By assumption S = V ⋊ G and V is an F-module for G. Choose a nontrivial elementary abelian subgroup E ⩽ G such that |E||C_{V} (E)| ⩾ |V |. Then A := C_{V} (E)E ∈ A(S) by maximality of |V |, and A is a product subgroup. Since E ≠ 1, A ∈ A_{×}(S), proving (a).

Suppose A is maximal in A_{×}(S) with respect to S. Let E = A ∩ G, and note that E ≠ 1 by definition. Set D = A ∩ V = C_{V} (A).

Suppose V does not normalize A and let M = N_{V} (A), which is a proper subgroup of V. Then M and V are normal subgroups of V A. Thus, 1 ≠ V/M ⊴ V A/M and so V/M ∩ Z(V A/M) ≠ 1. Let v ∈ V − M such that v maps into Z(V A/M) in the quotient V A/M. Then N := [v, A] ⩽ M is abelian and normalizes A. Furthermore, N is not contained in D = C_{V} (A) < A since v does not normalize A.

Thus D < ND ⩽ M < V. Set A^{∗} = NC_{A}(N). Since [D, N] = 1 and A = D × E, we get A^{∗} = NDC_{E}(N) so that A^{∗} is a product subgroup. Then as A^{∗}∩V = ND > D = A∩V, we have that A^{∗} ≠ A. By Lemma 2.6(a), A^{∗} is abelian and |A^{∗}| ⩾ |A|. First, this means that A^{∗} ∈ A(S). Second we get A^{∗} ∩ G ≠ 1 and therefore A^{∗} ∈ A_{×}(S), since A^{∗} ∩ V = ND is proper in V and |V | ⩽ |A| = |A^{∗}|. Also, A ⩽_{S} A^{∗} by Lemma 2.6(a). But |A∩Z_{i}| < |A^{∗}∩Z_{i}| for some i by Lemma 2.6(b). This contradicts the maximality of A, and shows that V does indeed normalize A. Thus we also have [V, A, A] ⩽ [A, A] = 1.

□

We now assume for the remainder of this section that p ⩾ 5.

Definition 4.3. Given a subgroup A of S, let B_{A} be the set of all B ⩽ S such that B = B_{×}, A ⊴ ⟨A^{B}⟩, and [A, u; 3] = 1 for all u ∈ B.

Theorem 4.4. Suppose S_{V} is a central series of S which passes through V and A is maximal in A_{×}(S) with respect to S_{V}. Then A is normalized by every member of B_{A}.

Proof. Suppose not, and choose B ∈ B_{A} such that B does not normalize A. By Lemma 4.2(b), V normalizes A. As B = B_{×} = (B ∩ V)(B ∩ G), there exists b ∈ B ∩ G which does not normalize A. Set T = ⟨A, b⟩ and Â = ⟨A^{T} ⟩. Under this setup, Glauberman shows [Gla97, p.320] (see the last paragraph of Step 1 there) that

Â = ⟨ A, A^{b}, A^{b^{2}} ⟩ and Â has class at most 3.

Since p ⩾ 5 and the class of Â is at most 3, we view Â as a Lie ring by the Lazard corespondence [Gla97, Theorem 3.4]. Let α be the map on Â which is conjugation by b, and let δ = log(α). By [Gla97, pp. 320–323], we have that δ^{3} = 0,

δ = (α − 1) − \frac{1}{2}(α − 1)^{2} is a derivation of Â,
Â = A + δ(A) + δ^{2}(A), and
A ⩾ Z(Â).

Set

A_{1} = {a_{1} − 2δ(a_{2}) | a_{1}, a_{2} ∈ A and δ(a_{1}) − δ^{2}(a_{2}) ∈ Z(Â)},

and

A^{∗} = A_{1} + δ^{2}(A).

Then Glauberman shows by way of a Lie ring theoretic result [Gla97, Theorem 3.2] that as groups, A^{∗} is abelian, |A| = |A^{∗}|, and A <_{S} A^{∗} for every central series S of S. Set A_{×}^{∗} = (A^{∗})_{×}. Then by Lemma 3.2(b), |A_{×}^{∗}| = |A^{∗}| = |A|, and by Lemma 4.5 below, A^{∗} ⩽̸ V so A_{×}^{∗} ⩽̸ V. Thus, A_{×}^{∗} ∈ A_{×}(S). Since every member M of S_{V} satisfies M ⩽ V or V ⩽ M, we have by Lemma 3.3 that A^{∗} ⩽_{S_{V}} A_{×}^{∗}. Therefore, A <_{S_{V}} A_{×}^{∗}. This contradicts the assumption that A is maximal in A_{×}(S) with respect to S_{V} and completes the proof. □

Lemma 4.5. Let A, δ, Â and A^{∗} be as in the proof of Theorem 4.4, with Â and its subgroups viewed as Lie rings. Let D = A ∩ G, and set D̂ = D + δ(D) + δ^{2}(D). Then A^{∗} ∩ D̂ ≠ {0}, and therefore A^{∗} ∩ G ≠ 1.

Proof. Note that for the proof, we do not need knowledge that D̂ is closed under the Lie bracket, but only that D̂ is a abelian group (as a subspace of a Lie ring).

Recall that A^{∗} = A_{1} + δ^{2}(A) where

A_{1} = { a_{1} − 2δ(a_{2}) | a_{1}, a_{2} ∈ A and δ(a_{1}) − δ^{2}(a_{2}) ∈ Z(Â)}.

As D = A ∩ G is nontrivial, we have that D̂ is nontrivial. Since δ^{2}(D) is contained in A^{∗}, the moment δ^{2}(a) ≠ 0 for some a ∈ D, we are finished. So assume that δ^{2}(D) = 0.

Let a be a nonzero element of D. Suppose first that δ(a) = 0. Let a_{2} = 0. Then δ(a) − δ^{2}(a_{2}) = 0 ∈ Z(Â), and so a = a − 2δ(a_{2}) ∈ A_{1} ⩽ A^{∗}, and a ∈ D̂ as well.

Suppose δ(a) ≠ 0. If a ∈ Z(Â), then δ(a) ∈ Z(Â) as well, by [Gla97, Theorem 3.2(b)]. Taking a_{2} = 0 again, we have δ(a) − δ^{2}(a_{2}) ∈ Z(Â), giving a ∈ A^{∗} ∩ D̂. Suppose a ∉ Z(Â). By [Gla97, Theorem 3.2(c)], either δ^{2}(a) ∈ A^{∗} − A or δ(a) ∈ A^{∗} − A. The former cannot hold as δ^{2}(a) = 0, and the latter gives δ(a) ∈ A^{∗} ∩ D̂.

This shows that A^{∗} ∩ D̂ ≠ {0}. However, recall that δ is a polynomial in α, which is an automorphism induced by conjugation by an element of G. It follows that D̂ ⩽ G, and A^{∗} ∩ G is nontrivial as claimed. □

The following is basically the content of [Gla97, Theorem 4.1]. However, since we have restricted the collections from which A and B are chosen in proving Theorem 4.4, the statement and argument have to be modified. We present a complete proof in our case for the convenience of the reader.

Theorem 4.6. Suppose A and B are abelian product subgroups of S, both normalized by V. Assume that A is normalized by every member of B_{A} and B is normalized by every member of B_{B}. Then A and B normalize each other.

Proof. We proceed by induction on the order of G. If G = 1, then S = V, and the statement is trivial. Suppose that G > 1.

Assume first that A ∩ G = G. Then G is abelian and so the conjugation action of S is trivial modulo V. Hence ⟨A^{B}⟩ ⩽ V A, so A is normal in ⟨A^{B}⟩ because V normalizes A. As [S, S] ⩽ V and V normalizes B, we also have [A, u, u, u] ⩽ [V, u, u] ⩽ [B, u] = 1 for all u ∈ B. So B ∈ B_{A}. Because G is abelian we still have ⟨B^{A}⟩ ⩽ V B, and the same argument applies to give A ∈ B_{B}. Thus A and B normalize each other. By symmetry, the same conclusion holds in case B ∩ G = G.

So we may assume that A∩G and B ∩G are proper in G. Let M_{0} be a maximal subgroup of G containing A ∩ G, and set M = V M_{0}. Then M ⊴ S, so

A ⩽ ⟨A^{B}⟩ ⩽ ⟨M^{S}⟩ = M.

If x ∈ B, then A^{x} = A^{π(x)}, since V normalizes A. Hence A^{x} is a product subgroup. Furthermore, it is easy to see that A^{x} is normalized by every member of B_{A^{x}} = (B_{A})^{x}. By the inductive hypothesis, A^{x} normalizes A. It follows that A ⊴ ⟨A^{B}⟩, and

[B, A; 3] ⩽ [⟨A^{B}⟩, A, A] ⩽ [A, A] = 1.

By symmetry [A, B; 3] = 1. This means that B ∈ B_{A}. Hence, B normalizes A by hypothesis. By symmetry, A normalizes B. □

Now we are in a position to prove Theorem C.

Theorem C. Suppose p ⩾ 5. Let V be an elementary abelian p-group which is an F-module for the p-group G. Then there exists a quadratic offender on V in G which is normal in its normal closure in G.

Proof. Let S = V ⋊ G. By Lemma 4.2(a), A_{×}(S) is nonempty. Let S be a central series which passes through V and let A be an element of A_{×}(S) maximal in A_{×}(S) with respect to S. Set E := A ∩ G ≠ 1. As A ∈ A(S), we have |V | ⩽ |A|. Since A is a product subgroup, |V | ⩽ |A| = |A ∩ V ||A ∩ G| = |C_{V} (E)||E|. So E offends on V. By Lemma 4.2(b), V normalizes A, and E acts quadratically on V. Let B be a conjugate of A. Then B is a product subgroup because V normalizes A. Thus, B is maximal in A_{×}(S) with respect to S. By Theorem 4.4, B is normalized by each element in B_{B}. By Theorem 4.6, A is normalized by its S-conjugates. Therefore E is normalized by its G-conjugates. This is another way of saying that E ⊴ ⟨E^{G}⟩. □

5. Oliver’s conjecture

In this section we leverage the existence of a 2-subnormal quadratic offender guaranteed in Theorem C to prove Theorems A and B. First we state the key lemma of Green, Héthelyi, and Lilienthal in [GHL08, Lemma 4.1].

Lemma 5.1. Suppose V is an elementary abelian p-group and G is a p-group acting faithfully on V. Let a be an element of G acting quadratically on V. Let x ∈ G and suppose c := [a, x] ≠ 1 centralizes both a and x. Then c acts quadratically on V.

The previous Lemma allows one, under certain conditions, to begin with a quadratic element and descend the central series of G, finding quadratics further and further down. This process stops when at some point, and having some quadratic element a, one can find no element x ∈ G for which [a, x] ≠ 1 centralizes a, and [a, x, x] = 1. The proof of the next Lemma uses the existence of a 2-subnormal offender to produce normal abelian subgroups of G which act as a container for descent, and identifies a normal abelian subgroup W at the bottom when the descent process terminates.

Lemma 5.2. Suppose p ⩾ 5 and 3 ⩽ k ⩽ p. Suppose S is a finite p-group such that J_{e}(S) is not contained in 𝔛_{k}(S). Set G = S/𝔛_{k}(S), which acts faithfully on V = Ω_{1}Z(𝔛_{k}(S)). Then there exists a normal elementary abelian subgroup W = ⟨a^{G}⟩ of G, generated by the G-conjugates of a quadratic element, such that for each x ∈ G, either [W, x] = 1 or [W, x, x] ≠ 1.

Proof. Note J_{e}(V ⋊G) is not contained in 𝔛_{k}(V ⋊G) = V, and so we assume that S = V ⋊G. Furthermore, V is both a PS-module of degree k and an F-module for G. Choose by Theorem C a quadratic offender E ⩽ G which is normalized by its conjugates in G. Let N be the normal closure ⟨E^{G}⟩ of E in G, so that E ⊴ N. Then E ∩ Ω_{1}Z(N) ≠ 1, and hence there are quadratic elements in Ω_{1}Z(N). For each i ⩾ 0, set W_{i} = [Ω_{1}Z(N), G; i]. Let j be the greatest integer such that there exists a quadratic element in W_{j}. Choose such a quadratic element a ∈ W_{j}. Note that W_{j} is not in the center of G as Z(G) has no quadratic elements. Set W = ⟨a^{G}⟩. Then W is a normal elementary abelian subgroup of G generated by quadratic elements.

Let x ∈ G. Suppose that x does not centralize W, but that [W, x, x] = 1. Then there exists a G-conjugate b of a such that [b, x] ≠ 1, but [b, x, x] = 1. Setting c = [b, x], we have that c centralizes both b and x. Since b is quadratic, c is as well by Lemma 5.1. Thus c ∈ W_{j+1} and quadratic. This contradicts the maximality of j. Therefore, for every x ∈ G, we have either [W, x] = 1 or [W, x, x] ≠ 1. □

We have recently learned that Green, Héthelyi, and Mazza have independently obtained a similar subgroup (generated by last quadratics in the terminology of their paper) as the W in Lemma 5.2 under only the assumption that G has quadratic elements on V. See Section 8 of their paper [GHM11].

Theorem B. Suppose p ⩾ 5 and 3 ⩽ k ⩽ p. Let S be a finite p-group, and set G = S/𝔛_{k}(S). If Baum(G) = G, then J_{e}(S) ⩽ 𝔛_{k}(S). In particular, Oliver’s conjecture holds in this case.

Proof. Suppose not and as in the proof of Lemma 5.2, let S = V ⋊ G be a counterexample. Let W = ⟨a^{G}⟩ be as in Lemma 5.2. Let A_{e}(G) denote the set of elementary abelian subgroups of G of maximum order.

By Thompson Replacement [GLS96, Theorem 25.2], there exists A^{∗} ∈ A_{e}(G) acting quadratically on W or else [W, A] = 1 for all A ∈ A_{e}(G). The first possibility is ruled out by Lemma 5.2. It follows that [W, A] = 1, whence W ⩽ A for all A ∈ A_{e}(G). So W ⩽ Ω_{1}Z(J_{e}(G)) = Ω_{1}Z(G) as Baum(G) = G. Since W contains quadratic elements and V is a PS-module of degree k ⩾ 3 for G, this is a contradiction. □

We need the following basic lemma for the proof of Theorem A.

Lemma 5.3. Suppose G is a p-subgroup of GL(n, p). Assume that a and b are commuting elements of G whose minimum polynomials on V have degree at most k and l, respectively. Then ab has minimum polynomial of degree at most k +l−1. In particular, if a_{1}, …, a_{k} are commuting quadratic elements, then a_{1} ⋯ a_{k} has minimum polynomial of degree at most k + 1.

Proof. We have ab − 1 = (a − 1) + (b − 1) + (a − 1)(b − 1). Raise the right hand side to the k + l − 1st power, and distribute. Since a and b commute, each term is of the form (a − 1)^{i}(b − 1)^{j} with i + j ⩾ k + l − 1. As (a − 1)^{k} = 0 and (b − 1)^{l} = 0, one of (a − 1)^{i} or (b − 1)^{j} must be zero in the respective term. Therefore, (ab − 1)^{k+l−1} = 0. The last statement follows by induction on k. □

Theorem A. Let p be an odd prime, and suppose S is a finite p-group such that S/𝔛(S) has nilpotence class at most log_{2}(p − 2) + 1. Then J_{e}(S) ⩽ 𝔛(S), so Oliver’s conjecture holds for S.

Proof. The conjecture has been shown in [GHM10, Theorem 5.2] to hold when S/𝔛(S) has nilpotence class at most 4, so the result is known for small primes. Therefore, we may assume that p ⩾ 5.

Suppose not and choose a counterexample S = V ⋊ G with V both a PS-module and an F-module for G. Let W = ⟨a^{G}⟩ be the normal elementary abelian subgroup of G guaranteed by Lemma 5.2.

As W ⊴ G, there exists an integer k > 0 such that 1 ≠ [W, G; k] ⩽ Ω_{1}Z(G). Hence, there exists a G-conjugate b of a, and elements x_{1}, …, x_{k} of G, such that

1 ≠ [b, x_{1}, …, x_{k}] ∈ Ω_{1}Z(G).

Applying (k times) the standard identity [x, y] = x^{−1}x^{y} one sees that

z := [b, x_{1}, …, x_{k}] = ∏_{j=0}^{k} ∏_{(i_{1},...,i_{j})∈X_{j}}(b^{(−1)^{j+1}})^{x_{i_{1}}···x_{i_{j}}} ,

where X_{j} is the set of strictly increasing sequences of length j from {1, …, k}. It follows that z is a product of ∑_{j=0}^{k} \binom{k}{j} = 2^{k} commuting conjugates of b and b^{−1}. By Lemma 5.3, the degree of the minimum polynomial of z on V is at most 2^{k} + 1. On the other hand, the degree of the minimum polynomial of z is p, because V is a PS-module for G. Thus p < 2^{k} + 2, and hence log_{2}(p − 2) < k. By definition of k, we have that the class of G is at least k + 1. Therefore the class of G is strictly larger than log_{2}(p − 2) + 1, and this completes the proof. □

6. Speculation regarding an inductive approach

Lemma 5.2 applied with k = 3 gives some hint as to how an inductive approach to the conjecture might proceed at least for p ⩾ 5. (In view of the subgroup generated by last quadratics in [GHM11, Section 8], the approach applies for all odd p.) We record here an observation that forms the main motivation for this idea.

Proposition 6.1. Suppose that p ⩾ 5 and S is a p-group such that J_{e}(S) is not contained in 𝔛_{3}(S). Let G = S/𝔛_{3}(S). Then 𝔛_{3}(G) is a proper subgroup of G.

Proof. As J_{e}(S) is not contained in 𝔛_{3}(S), 𝔛_{3}(S) is a proper subgroup of S. Let W be as in Lemma 5.2. Thus W is a normal elementary abelian subgroup of G generated by quadratic elements. Set X := C_{G}(W), and note that X is a proper normal subgroup of G. We will show that 𝔛_{3}(G) ⩽ X, and this will settle the claim. Suppose to the contrary that 𝔛_{3}(G) is not contained in X. Let 1 = Q_{0} < Q_{1} < ⋯ < Q_{n} = 𝔛_{3}(G) be a series of normal subgroups of G such that

[Ω_{1}(C_{G}(Q_{i−1})), Q_{i}, Q_{i}] = 1

for each i. Let j be the greatest integer such that Q_{j−1} ⩽ X. Let x ∈ Q_{j} − X. Then as W ⩽ Ω_{1}Z(X) ⩽ Ω_{1}(C_{G}(Q_{j−1})), we have

[W, x, x] ⩽ [Ω_{1}(C_{G}(Q_{j−1})), x, x] = 1,

a contradiction to Lemma 5.2 because x ∉ X. Therefore, 𝔛_{3}(G) ⩽ X, and 𝔛_{3}(G) is a proper subgroup of G. □

For an inductive approach, we would choose S to be minimal subject to the condition that J_{e}(S) is not contained in 𝔛_{3}(S). We would then like to show that J_{e}(G) is not contained in 𝔛_{3}(G) to contradict the minimality of S. By the proof of Proposition 6.1, we have that 𝔛_{3}(S) ⩽ C_{G}(W). However, the content of the proof of Theorem B is that J_{e}(G) is also contained in C_{G}(W). In fact, one can see this explicitly in the example of Green, Héthelyi, and Lilienthal in Section 5 of their first paper [GHL08]. In this case G ≅ C_{3} ≀ C_{3} acting on an 8-dimensional module V, and we have that X = W = J_{e}(G) = 𝔛_{3}(G) is the base subgroup of G. Note that G is not a counterexample to Oliver’s conjecture, as can be seen from subsequent work ([GHM10, Theorem 1.1] or [GHM11]). But this example shows that taken alone, the existence of a normal elementary abelian subgroup of G generated by quadratic elements is not enough to carry out this inductive argument.

Acknowledgements

I would like to thank Ian Leary for introducing me to Oliver’s conjecture over lunch some time ago. George Glauberman provided me with the starting point for these ideas, and I am grateful to him for his encouragement while writing up the details. Theorem A was proven on the beautiful Isle of Skye at the Conference on Algebraic Topology, Group Theory and Representation Theory in June 2009; thanks to the organizers for providing such a great setting for mathematics. I am indebted to the referee for a very careful reading of this paper and many helpful comments on style and substance which greatly improved the exposition. Finally, I would like to thank David Green, László Héthelyi, and Nadia Mazza for keeping me up to date on their work, and my thesis advisor Ronald Solomon for his constant encouragement and support.

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

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