The Other P(G, V)-Theorem

U. Meierfrankenfeld, B. Stellmacher

October 13, 2009

1 Introduction

Suppose that H is a finite group, T ∈ Syl_{p}(H), p any prime, and V is an elementary abelian normal p-subgroup of H. Then an elementary Frattini argument shows that either H = C_{H}(V)N_{H}(J(T)) or J(T) ≰ C_{H}(V), where J(T) denotes the Thompson subgroup of T.

In this paper we are interested in this second case. One of the questions is how \overline{J(T)} := J(T)C_{H}(V)/C_{H}(V) is embedded in \overline{H} := H/C_{H}(V). The first problem is to find suitable properties of the Thompson subgroup J(T) that can be expressed in terms of \overline{H} only. This is done in the following way.

Recall that J(T) is generated by the elementary abelian subgroups A of maximal order of T. It is evident that BC_{V} (B) is elementary abelian for every subgroup B ≤ A. Hence by the maximality of |A|

|A| ≥ |BC_{V} (B)| = |B||C_{V} (B)||V ∩ B|^{−1} ≥ |B||C_{V} (B)||C_{V} (A)|^{−1}.

This gives rise to the condition

(∗)  |B||C_{V} (B)| ≤ |A||C_{V} (A)| for every B ≤ A.

Note that B := C_{A}(V) yields an important special case of (∗):

(∗∗)  |V/C_{V} (A)| ≤ |A/C_{A}(V)|.

Both conditions (∗) and (∗∗) can be phrased in terms of the factor group \overline{H} and the GF (p)\overline{H}-module V just by replacing A by its image \overline{A} in \overline{H}. Evidently A satisfies (∗) with respect to V and H iff \overline{A} satisfies (∗) with respect to V and \overline{H}.

This consideration gives rise to the following definition in the more general set up of a finite group G and a finite dimensional GF (p)G-module V.

Definition 1.1 Let A be a subgroup of G such that A/C_{A}(V) is an elementary abelian p-group. Then A is an offender of G on V if |V/C_{V} (A)| ≤ |A/C_{A}(V)|; and A is a best offender of G on V if |B||C_{V} (B)| ≤ |A||C_{V} (A)| for every B ≤ A. The normal subgroup of G generated by the best offenders of G on V is denoted by J_{G}(V).

In the literature, at least in the case of a faithful GF (p)G-module V, the set of best offenders is denoted by P(G, V).

The classical result about J_{G}(V) is the following:

The P(G, V )-Theorem. Suppose that V is a faithful finite dimensional GF (p)G-module and that K is a component of G. Then either [J_{G}(V), K] = 1 or K ⊴ J_{G}(V).

This theorem was proved by Timmesfeld [Ti] for the case p = 2. Later Chermak [Ch] gave a proof for arbitrary p. Earlier proofs by Aschbacher [As] and Thompson (unpublished) used a K-group hypothesis.

As we also want to allow certain solvable analogues of components, we need one further definition.

Definition 1.2 A non-trivial subgroup K of J_{G}(V) is a J_{G}(V)-component if K is minimal with respect to K = [K, J_{G}(V)].

The Other P(G, V )-Theorem. Suppose that V is a faithful finite dimensional GF (p)Gmodule and that O_{p}(J_{G}(V)) = 1. Then [E, K] = 1 and [V, E, K] = 0 for any two distinct J_{G}(V)- components E and K.

In the case of a faithful GF (p)G-module with O_{p}(J_{G}(V)) = 1 we also show that every component of J_{G}(V) is a J_{G}(V)-component (see 3.4) and that every J_{G}(V)-component, which is not a component, is isomorphic to SL_{2}(p)′, p = 2 or 3 (see 3.2).

2 Offenders

In this section G is a finite group, p is a prime, and V is a finite dimensional GF (p)G-module. Some of the arguments in this section got their inspiration from [CD].

Definition 2.3 Let A be a subgroup of G.

(a) A acts quadratically on V if [V, A, A] = 0.

(b) j_{A}(V) := \frac{|A||C_{V} (A)|}{|V ||C_{A}(V)|}.

(c) V is a simple GF (p)G-module, if V ≠ 0 and V and 0 are the only G-submodules in V.

Lemma 2.4 Let A ≤ G such that A/C_{A}(V) is elementary abelian. Then the following hold:

(a) j_{A}(V) = \frac{|A/C_{A}(V)|}{|V/C_{V} (A)|}.

(b) A is an offender on V iff j_{A}(V) ≥ 1.

(c) If [V, A] = 0, then j_{A}(V) = 1.

(d) A is a best offender iff j_{B}(V) ≤ j_{A}(V) for all B ≤ A.

Proof: (a) is obvious, and (b) and (c) follow from (a).

(d) By (a) j_{B}(V) = j_{BC_{A}(V)}(V) and so we may assume C_{A}(V) ≤ B. Then C_{A}(V) = C_{B}(V) and so j_{B}(V) ≤ j_{A}(V) iff |B||C_{V} (B)| ≤ |A||C_{V} (A)|. □

Lemma 2.5 Let A ≤ G be a best offender on V and B ≤ G be an offender on V. Then the following hold:

(a) A is an offender on V.

(b) B contains a best offender B^{∗} on V with j_{B}(V) ≤ j_{B^{∗}}(V) such that [V, B^{∗}] ≠ 0 or B^{∗} = B.

(c) A is a best offender on every A-submodule of V.

Proof: (a): By 2.4(c),(d) 1 = j_{1}(V) ≤ j_{A}(V), so by 2.4(b) A is an offender on V.

(b): Choose B^{∗} ≤ B such that first j_{B^{∗}}(V) is maximal and then |B^{∗}| is maximal. If [V, B^{∗}] = 0, then j_{B^{∗}}(V) = 1 and thus by 2.4(b) also j_{B}(V) = 1. Now the maximal choice of |B^{∗}| yields B^{∗} = B.

(c): Let W be an A-submodule of V and A_{0} ≤ A. Then

|A_{0}||C_{W} (A_{0}) + C_{V} (A)| ≤ |A_{0}||C_{V} (A_{0})| ≤ |A||C_{V} (A)|

and thus

|A_{0}||C_{W} (A_{0})||C_{V} (A)||C_{W} (A_{0}) ∩ C_{V} (A)|^{−1} ≤ |A||C_{V} (A)|.

Since C_{W} (A_{0}) ∩ C_{V} (A) = C_{W} (A) we get |A_{0}||C_{W} (A_{0})| ≤ |A||C_{W} (A)|. □

Lemma 2.6 Let A and B be subgroups of G. Then

j_{⟨A,B⟩}(V)j_{A∩B}(V) ≥ j_{A}(V)j_{B}(V)

with equality iff C_{V} (A ∩ B) = C_{V} (A) + C_{V} (B) and ⟨A, B⟩ = AB.

Proof: We may assume that C_{G}(V) = 1 since j_{A}(V) = j_{AC_{H}(V)/C_{H}(V)}(V), so |V |j_{A}(V) = |A||C_{V} (A)|. Observe that

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

Then

|V |^{2}j_{⟨A,B⟩}(V)j_{A∩B}(V) =  |⟨A, B⟩||C_{V} (⟨A, B⟩)||A ∩ B||C_{V} (A ∩ B)|
≥ |AB||C_{V} (A) ∩ C_{V} (B)||A ∩ B||C_{V} (A) + C_{V} (B)|
≥  |A||B||C_{V} (A)||C_{V} (B)|
=  |V |^{2}j_{A}(V)j_{B}(V).
□

Fundamental for the investigation of best offenders is a replacement property first proved by Thompson, the Thompson Replacement Theorem, and then generalized by Timmesfeld, the Timmesfeld Replacement Theorem. We will use the following version of the Timmesfeld Replacement Theorem with [KS] as a reference.

Lemma 2.7 Let A be a best offender of G on V and W be a subgroup of V. Then for A^{∗} := C_{A}([W, A]) the following hold:

(a) A^{∗} is a best offender on V with j_{A}(V) = j_{A^{∗}}(V).

(b) C_{V} (A^{∗}) = [W, A] + C_{V} (A).

(c) [W, A^{∗}] ≠ 0 if [W, A] ≠ 0.

(d) [W, A^{∗}, A] = 0.

Proof: Properties (a) and (b) can be found in [KS, 9.2.1 and 9.2.3], (c) is stated there differently, so we give a proof here.

Assume that [W, A^{∗}] = 0. Then by (b) W ≤ [W, A] + C_{V} (A); in particular [W, A] = [W, A, A]. As A/C_{A}(V) is a p-group, this last property gives [W, A] = 0.

Finally, by definition [W, A, A^{∗}] = 0, and [A, A^{∗}, W] = 0 since A/C_{A}(V) is abelian. Hence the Three Subgroups Lemma implies (d). □

Lemma 2.8 Let A and B be quadratic offenders of G on V such that

[A, B] ≤ C_{G}(V) and AC_{G}(V) ∩ BC_{G}(V) = C_{G}(V).

Then ⟨A, B⟩ is a quadratic best offender on V, or there exists a quadratic best offender X ≤ ⟨A, B⟩ with

j_{X}(V) > max{j_{A}(V), j_{B}(V)}.

Proof: Let D := ⟨A, B⟩. Then D/C_{D}(V) is an elementary abelian p-group. We may assume that G is faithful on V, so D is elementary abelian and A ∩ B = 1. From 2.6 we get that j_{D}(V) ≥ j_{A}(V)j_{B}(V) since j_{A∩B}(V) = 1.

Assume first that j_{D}(V) > max{j_{A}(V), j_{B}(V)}. Then j_{D}(V) > 1, and by 2.5(b) there exists a best offender D^{∗} ≤ D with j_{D^{∗}}(V) ≥ j_{D}(V) > 1; in particular [V, D^{∗}] ≠ 0. Now 2.7 gives the desired quadratic best offender X ≤ D^{∗}.

Assume now that j_{D}(V) ≤ max{j_{A}(V), j_{B}(V)}. By 2.6

j_{D}(V) = j_{D}(V)j_{1}(D) ≥ j_{A}(V)j_{B}(V),

so j_{D}(V) = j_{A}(V)j_{B}(V) and again by 2.6 C_{V} (A) + C_{V} (B) = C_{V} (A ∩ B) = V. This yields [V, A] = [C_{V} (B), A] and [V, A, B] = 0, and the quadratic action of AB follows. □

Lemma 2.9 Let A ≤ G and L ≤ G wih L = [L, A]. Then C_{A}(L) ∩ C_{A}([V, L, A]) ≤ C_{A}([V, L]).

Proof: Note that O^{p}(L) = L ≤ ⟨A^{L}⟩ since L = [L, A]. Thus, we may assume that V = [V, L], in particular [V, A, A_{0}] = 0. Let A_{0} := C_{A}(L) ∩ C_{A}([V, L, A]). Then

[V, L, A_{0}] ≤ [V, ⟨A^{L}⟩, A_{0}] = ⟨[V, A, A_{0}]^{L}⟩ = 0.
□

Lemma 2.10 Let A be a best offender of G on V, L ⊴ G, and W a simple L-submodule of V. Suppose that [L, A] ≰ C_{L}(W). Then W is A-invariant.

Proof: Note that [L, A] ≰ C_{G}(W) implies [W, A] ≠ 0. Hence by 2.7 there exists a quadratic best offender A^{∗} ≤ A such that [W, A^{∗}] ≠ 0 and [W, A^{∗}, A] = 0. Then

[W, N_{A^{∗}}(W)] ≤ W ∩ C_{V} (A) ≤ W ∩ W^{b} for every b ∈ A.

Since W ∩ W^{b} is an L-module and W is simple, we conclude that either W is A-invariant or [W, N_{A^{∗}}(W)] = 0. In the first case we are done, so we may assume that [W, N_{A^{∗}}(W)] = 0. In particular

(1)  W ∩ W^{a} = 0 for every a ∈ A^{∗} \ C_{A^{∗}}(W).

Pick c ∈ A^{∗} \ C_{A^{∗}}(W). Then

(2)  U := W ⊕ W^{c} = W ⊕ [W, c] = W^{c} ⊕ [W, c].

and

(3)  [W, c] ≤ U ∩ C_{V} (A) ≤ U ∩ U^{b} for every b ∈ A.

Assume that [W, c] is L-invariant. Then [W, c, A] = 0 implies that also [W, c, [L, A]] = 0. The decomposition (2) shows that [W, [L, A]] = 0, a contradiction. Thus we have

(4)  [W, c] is not L-invariant.

Let b ∈ A. Since U ∩U^{b} is an L-module and W is simple, we get from (3) that either U ∩U^{b} = [W, c] or U = U^{b}. The first case contradicts (4), so U is A-invariant. But then, as [W, A, c] = [W, c, A] = 0, we get that

C_{U}(A^{∗}) = C_{U}(c) = [W, c] and |U/C_{U}(A^{∗})| = |W |.

By 2.5(c) A^{∗} is a best offender on U, so |W | = |U/C_{U}(A^{∗})| ≤ |A^{∗}/C_{A^{∗}}(U)|. Observe that |U^{♯}| = |W |^{2} − 1 = (|W | + 1)|W^{♯}|. On the other hand, by (1) any two distinct A^{∗}-conjugates of W intersect trivially, so there are at most |W | + 1 such conjugates. We conclude that |U/C_{U}(A^{∗})| = |W | = |A^{∗}|, and the A^{∗}-conjugates of W together with [W, c] form a partition of U. Since all these conjugates of W are L-invariant, also [W, c] is L-invariant. This contradicts (4). □

3 J_{G}(V)-Components

As in the last section G is a finite group and V is a finite dimensional GF (p)G-module.

Lemma 3.1 Suppose that V is a faithful GF (p)G-module. Let L ⊴ J_{G}(V) with O_{p}(L) = 1 and [L, J_{G}(V)] ≠ 1. Then L contains a J_{G}(V)-component of G.

Proof: Let J := J_{G}(V), and let K ⊴ J be minimal with respect to K ≤ L and [K, J] ≠ 1. It suffices to show that K = [K, J].

Assume that K ≠ [K, J]. Then the minimality of K gives [K, J, J] = 1 and thus [K, K, K] = 1. Hence K is nilpotent. Again the minimality of K shows that K is an r-group, r a prime. Moreover r ≠ p since O_{p}(L) = 1. As J is generated by p-elements we get that J = O^{r}(J). Thus, [K, J, J] = 1 implies [K, J] = 1, a contradiction. □

In the next lemma we will use a result of Glauberman [Gl] as it is stated in [KS].

Lemma 3.2 Suppose that V is a faithful GF (p)G-module. Let R be a J_{G}(V)-component with O_{p}(R) = 1 that is not a component of J_{G}(V). Then the following hold:

(a) p = 2 or 3.

(b) R ≅ SL_{2}(p)′ and |[V, R]| = p^{2}.

(c) For every J_{G}(V)-component K with K ≠ R, [R, K] = 1 and [V, R, K] = 0.

Proof: Set J := J_{G}(V) and J̃:= J/Z(J). By 3.1 R̃ is a minimal normal subgroup of J̃. Hence, either R̃ is the product of components of J̃ or R̃ is an elementary abelian q-group.

Assume first that R̃ is the product of components. Then an elementary argument shows that R′ is the product of components of J (see [KS, 6.5.1]). By the P(G, V)-Theorem each of these components is normal in J, so by 3.1 R is a component, which is not the case.

Assume now that R̃ is a q-group. Then R is nilpotent and thus also R is a q-group. Moreover q ≠ p since O_{p}(R) = 1, so R ≤ O_{p′}(G). Pick T_{0} ∈ Syl_{p}(C_{J}(R)), and set

J_{0} := J_{N_{J} (T_{0})}(V), W := C_{V} (T_{0}) and \overline{J}_{0} := J_{0}/C_{J_{0}}(W).

Observe that J = J_{0}C_{G}(R), so R is also a J_{J_{0}}(V)-component. By the P × Q-Lemma R acts faithfully on W. Also C_{J_{0}}(R)/J_{0} ∩ T_{0} is a p′-group. Let X be the inverse image of C_{\overline{J}_{0}}(\overline{R}) in J_{0}. Then [R, X] ≤ C_{R}(W), and the faithful action of R gives X ≤ C_{J_{0}}(R), so \overline{X} is a p′-group.

We have shown that C_{\overline{J}_{0}}(O_{p′}(\overline{J}_{0})) ≤ C\overline{J_{0}}(\overline{R}) = \overline{X} ≤ O_{p′}(\overline{J}_{0}). In addition, by 2.5(c) \overline{J}_{0} is generated by best offenders on W. Hence W and \overline{J}_{0} satisfy the hypothesis of Theorem 9.3.7 in [KS]. It follows that either

p = 2, R ≅ C_{3} and |[W, R]| = 4, or
p = 3, R ≅ Q_{8} and |[W, R]| = 9.

Let A ≤ J_{0} be a best offender on V such that [R, A] ≠ 1. Clearly R = [R, A] and |A/C_{A}(R)| = p. Moreover, according to 2.7 there exists a best offender A^{∗} ≤ A such that [V, R, A] ≤ C_{V} (A^{∗}) and [V, R, A^{∗}] ≠ 1. Hence 2.9 shows that [R, A^{∗}] ≠ 1, and we may assume that A acts quadratically on [V, R]. But then again 2.9 shows that that C_{A}(R) = C_{A}([V, R]) and |A/C_{A}([V, R])| = p. As A is an offender on [V, R] by 2.5(c), we get |[V, R]/C_{[V,R]}(A)| = p. This gives [V, R] = [W, R], and (a) and (b) hold.

Now let K be any other J_{G}(V)-component. As we have seen above K is either a component or has a structure as R. In the first case the fact that GL_{2}(p) is solvable for p ≤ 3 shows that [V, R, K] = 0 and so also [R, K] = 1. In the second case we can choose a best offender B with K = [K, B] such that ⟨A, B⟩ is a p-group. Then C_{⟨A,B⟩}(RK) is a normal p-subgroup of RK⟨A, B⟩ and so centralizes [V, RK]. Hence the above result from [KS] applies to RK⟨A, B⟩/C_{RK⟨A,B⟩}([V, RK]). Then

[R, K] ≤ C_{R∩K}([V, RK]) = 1

and [V, R, K] = 0. Hence (c) also holds in this case. □

Lemma 3.3 Suppose that V is a faithful GF (p)G-module and that K is a J_{G}(V)-component with O_{p}(K) = 1. Then there exists a best offender A of G such that [K, A] = K and A is quadratic on [V, K].

Proof: By 3.1 there exists a best offender B such that [K, B] ≠ 1. Hence 3.2 gives [K, B] = K and thus [V, K, B] ≠ 0. Now 2.7 with W := [V, K] gives a best offender A ≤ B satisfying [W, B, A] = 0 and [W, A] ≠ 0. The first property shows that A is quadratic on W. It remains to prove K = [K, A].

Assume that [K, A] ≠ K. Then again 3.2 yields [K, A] = 1 and thus by 2.9 W = [W, K] ≤ C_{W} (A), a contradiction. □

Lemma 3.4 Suppose that V is a faithful GF (p)G-module and that K is a component of J_{G}(V) with O_{p}(K) = 1. Then K is a J_{G}(V)-component.

Proof: Let J := J_{G}(V). By the P(G, V)-Theorem K is normal in J, so K = [K, J]. Hence, either K is a J_{G}(V)-component, or there exists a J_{G}(V)-component R < K. Suppose that the second case holds. Then R ≤ Z(K) since K is a component. Thus R is a J_{G}(V)-component that is not a component. Hence 3.2 implies that |[V, R]| = p^{2}. But then GL([V, R]) is solvable and as above [V, R, K] = 0; particular [V, R, R] = 0. This is impossible since R is a non-trivial p′-group. □

Lemma 3.5 Suppose that V is a faithful GF (p)G-module and that K is a J_{G}(V)-component with O_{p}(K) = 1. Then there exists a best offender A in G such that [K, A] = K and A is quadratic on [V, K].

Proof: By 3.3 there exists a best offender A such that [K, A] ≠ 1 and A is quadratic on [V, A]. It remains to prove K = [K, A].

If K is a component, then [K, A] ≠ 1 implies [K, A] = K. If K is not a component, then by 3.2(a),(b) R ≅ C_{3} or Q_{8} and again [K, A] ≠ 1 implies [K, A] = K. □

Lemma 3.6 Suppose that V be a faithful GF (p)G-module and O_{p}(J_{G}(V)) = 1. Let K and L be two distinct components of J_{G}(V). Then

[V, K, L] = [V, L, K] = 0.

Proof: We apply 3.4 and 3.5. Then there exist best offenders A and B such that [K, A] = K and [L, B] = L, and A and B act quadratically on [V, K] and [V, L], respectively. Moreover since K and L are normal in J_{G}(V), we may assume that ⟨A, B⟩ is a p-group.

By way of contradiction we assume that [V, K, L] ≠ 0. Set G_{0} := KL⟨A, B⟩. Then there exists a G_{0}-submodule W of V that is minimal with respect to [W, K, L] ≠ 0. Since [K, L] = 1 also [W, L, K] ≠ 0, and the situation is symmetric in K and L.

Suppose that W ≠ [W, K]. Then the minimality of W gives [W, K, K, L] = 0. On the other hand [W, K] = [W, K, K], so [W, K, L] = 0, a contradiction. Hence we have W = [W, K] and by symmetry also W = [W, L]. Thus A and B are quadratic on W, and by 2.5(c) A and B are quadratic best offenders on W.

Let U be a maximal GF (p)G_{0}-submodule of W and set Ŵ := W/U. Then Ŵ is a simple GF (p)G_{0}-module. By 2.9 [W, C_{A}(K)] = 0 and similarly [W, C_{B}(L)] = 0. Thus we have

(∗)  C_{A}(Ŵ) = C_{A}(K) = C_{A}(W) and C_{B}(Ŵ) = C_{B}(L) = C_{B}(W).

Since A and B are quadratic best offenders on W, (∗) shows that A and B are also quadratic offenders on Ŵ. Hence there exist quadratic best offenders Â and B̂ on Ŵ in ⟨A, B⟩ such that [K, Â] = K and [L, B̂] = L. In addition, we choose Â such that jÂ(Ŵ) is maximal with that property.

Let X̂ be a simple K-submodule of Ŵ. Assume that [K, B̂] ≠ 1. Then 2.10 shows that X̂ is normalized by every L-conjugate of B̂, so L normalizes X̂. As X̂ is a simple K-module, Schur’s Lemma shows that End_{K}(X̂) is a finite division ring and then Wedderburn’s Theorem that L/C_{L}(X̂) is cyclic. This shows that [X̂, L] = 0 since L is perfect. Hence CŴ (L) is a non-trivial G_{0}-submodule of Ŵ. As Ŵ is a simple G_{0}-module, we conclude that [Ŵ , L] = 0, a contradiction.

We have shown that [K, B̂] = 1 and similarly [L, Â] = 1. Observe that C_{⟨A,B⟩}(KL) is a normal p-subgroup of G_{0} and hence centralizes Ŵ, so

[Â, B̂] ≤ C_{G_{0}}(Ŵ) and ÂC_{G_{0}}(Ŵ) ∩ B̂C_{G_{0}}(Ŵ) = C_{G_{0}}(Ŵ).

Now 2.8 and the maximality of jÂ(Ŵ) show that ⟨Â, B̂⟩ is a quadratic best offender on Ŵ. But the above argument with ⟨Â, B̂⟩ in place of B̂ implies that [K, ⟨Â, B̂⟩] = 1, which contradicts [K, Â] = K.

□

The proof of The Other P(G, V)-Theorem. Let E and K be distinct J_{G}(V)-components. Then by 3.2(b) and 3.6 [V, E, K] = 0 = [V, K, E]. So by the Three Subgroups Lemma, [E, K] = 1.

References

[As] M. Aschbacher, GF (2)-Representations of finite groups, Am. J. Math. 104 (1982), 683 – 771.

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

[CD] A. Chermak, A. Delgado, A measuring argument for finite groups, Proc. AMS 107 (1989), 907 – 914.

[Gl] G. Glauberman, Failure of factorization in p-solvable groups, Quart. J. Math. Oxford 24 (1973), 71 – 77.

[KS] H. Kurzweil, B. Stellmacher, Theorie der endlichen Gruppen. Eine Einführung. Springer-Verlag, Berlin, (1998) 341pp

[Ti] F. Timmesfeld, A remark on Thompson’s replacement theorem and a consequence, Arch. Math. 38 (1982), 491 – 499.