ON THE CENTER OF MAXIMAL SUBGROUPS OF LOCALLY FINITE SIMPLE GROUPS OF

ALTERNATING TYPE

Neil Flowers Youngstown State University Youngstown, OH 44555 flowers@math.ysu.edu

Ulrich Meierfrankenfeld Michigan State University East Lansing, MI 48824 meier@math.msu.edu

Abstract Let G be a locally finite simple group of alternating type, p a prime, and Z ≤ G be elementary abelian of order p^{2}. We prove that there exists 1 ≠ z ∈ Z with C_{G}(z) ≠ C_{G}(Z).

1 Introduction

In [LS] it was shown that the centers of maximal subgroups of finite simple groups are always cyclic. In this paper we extend this result to locally finite simple groups of alternating type. Note here that by [Me, 4.1] every countable locally finite simple group has maximal subgroups. While our result is of interest in itself, our paper is also meant as an illustration how the structure theory of locally finite simple groups in [Me] can be used. The techniques developed in this paper already have been used in [DM] to shed light on the structure of finite subgroups in locally finite simple groups of 1-type. We also expect the techniques to help answer some of the questions on centralizers in locally finite simple groups posted by B. Hartley in [Har].

A group G is called locally finite if every finite subset of G lies in a finite subgroup of G. A Kegel cover K of a locally finite group G is a set of pairs (H, M) such that H is a finite subgroup of G, M is a maximal normal subgroup of H, and for each finite subgroup K of G there exists (H, M) ∈ K with K ≤ H and K ∩ M = 1. The groups H/M are called the factors of K. We say that K is alternating if all of its factors are alternating groups. For a finite subgroup X of G, K(X) is the set of pairs (H, N) ∈ K with X ≤ H and X ∩ N = 1. A group G is finitary if there exists a field F and a faithful F G-module V such that dim_{F} [V, g] < ∞ for all g ∈ G. A locally finite simple group is of alternating type if it is not finitary and has an alternating Kegel cover. We are now able to state our main theorem

Theorem 1.1 Let G be a locally finite simple group of alternating type, p a prime, and let Z be elementary abelian subgroup of order p^{2}. Then there exists 1 ≠ z ∈ Z such that C_{G}(z) ≠ C_{G}(Z). In particular, if M is a maximal subgroup of G then Z(M) is locally cyclic.

We would like to thank the referee for pointing out numerous misprints.

2 Preliminaries

Lemma 2.1 Let H be a finite group, p a prime, S_{0} a p-subgroup of H, N ⊴ H and T ≤ H minimal with respect to S_{0} ≤ T and H = T N. Let S ∈ Syl_{p}(T) with S_{0} ≤ S. Then

(a) S ∩ N = O_{p}(T ∩ N) ⊴ T.

(b) T ∩ N/O_{p}(T ∩ N) ≤ Φ(T/O_{p}(T ∩ N)).

(c) If O^{p}(H)N/N is a q-group for some prime q ≠ p, then O^{p}(T)O_{p}(T)/O_{p}(T) is a q-group.

Proof Note that S_{0} ≤ N_{T} (N ∩ S) and, by the Frattini argument, T = N_{T} (N ∩ S)(N ∩ T). Thus by the minimality of T we have T = N_{T} (S ∩ N) and so S ∩ N = O_{p}(N ∩ T). So (a) holds. Without loss we may assume T = H and O_{p}(T) = 1. Then N ∩ S = 1 and N is a p′-group. Suppose N ≰ Φ(T). Then there exists a maximal subgroup M of T with N ≰ M. Then T = MN and, since N is a p′-group, M contains a Sylow p-subgroup of H. Thus S_{0} ≤ M^{k} for some k ∈ T. Now as T = NM^{k}, we have T = M^{k} = M by the minimality of T, a contradiction. Therefore (b) holds. For (c), since N is a p′-group, O^{p}(T) is a p′-group. From [Gor, 6.2.2] we conclude that there exists an S-invariant Q ∈ Syl_{q}(O^{p}(T)). Since O^{p}(T)/N is a q-group we get O^{p}(T) = QN and so T = SQN. Hence, by the minimality of T we get T = QS and O^{p}(T) = Q is a q-group.

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Before we can state our next lemma we need to recall some definitions from [Me]. Suppose G is a group and that G/M ≅ Alt(Σ) for some set Σ and some normal subgroup M of G. Let t be a positive integer with t ≤ |Σ|/2. Then G acts t-pseudo naturally on Ω with respect to M if G acts transitively on Ω and if there exists a G-invariant partition Δ for G on Ω such that C_{G}(Δ) = M and the action of G on Δ is isomorphic to the action of G on subsets of size t of Σ. G acts pseudo naturally on Ω with respect to M, if G acts 1-pseudo naturally on Ω. G acts essentially on Ω with respect to M if C_{G}(Ω) ≤ M.

Lemma 2.2 Let p be a prime. Then there exists an integer n (depending only on p) with the following property:

If

(a) H is a finite group,

(b) Z is an elementary abelian subgroup of H of order p^{2},

(c) M ⊴ H with H/M ≅ Alt(Ω),

(d) H acts transitively on a set Σ such that C_{H}(Σ) ≤ M,

(e) Z has no regular orbits on Σ, and

(f) deg_{Ω}(z) ≥ n for all 1 ≠ z ∈ Z,

then Σ is pseudo natural for H with respect to M.

Proof By [Me, 2.14] there exists an integer n such that under the above assumptions Σ is t-pseudo natural for some t ≤ p^{2} − 2. We do assume without loss that n > 3p^{2}. Then there exists an H-invariant partition Δ of Σ such that the action of H on Δ is isomorphic to the action of H on the set of subsets of size t of Ω. If t = 1 we are done. Suppose that t > 1. As Z has no regular orbits on Σ, Z has no regular orbits on Δ. Hence, every subset of size t of Ω is invariant under some nontrivial element of Z. Let Ω_{1} and Ω_{2} be two nontrivial orbits of Z on Ω such that Z acts faithfully on Ω_{1} ∪ Ω_{2}. Choose T ⊆ Ω such that |T | = t and |T ∩Ω_{i}| = 1 for i = 1, 2 (this is possible since n ≥ 3p^{2}). Then there exists 1 ≠ z ∈ Z leaving T invariant. But then z fixes T ∩ Ω_{i} for i = 1, 2. Since Z is abelian and acts transitively on Ω_{i} we conclude that z fixes all elements of Ω_{1} ∪ Ω_{2} contradicting the faithful choice of Ω_{1} ∪ Ω_{2}.

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3 On a Lemma of Alperin

In this section we partially generalize a result of J.L. Alperin [Al] to odd primes. Let B be a group acting on a group A and i a non-negative integer. Define [A, B, 0] = A and inductively, [A, B, i + 1] = [[A, B, i], B].

Theorem 3.1 Let H be a finite group, p and q distinct primes with q ≠ 2, V a faithful GF (p)H-module, Z ≤ H a noncyclic elementary abelian pgroup, Q = [O_{q}(H), Z], and X be the set of all subgroups of index p in Z. For X ∈ X, set Q_{X} = [C_{Q}(X), Z] and suppose that

[V, Z, p] = 0.

Then

(a) [V, X, Q_{X}] = [V, Q_{X}, X] = 0,

(b) Q = ⨁_{X∈X} Q_{X},

(c) [V, Q] = ⨁_{X∈X} [V, Q_{X}].

Proof For (a), let X ∈ X and z ∈ Z \ X. Since [V, Z, p] = 0, we have [V, X, z, p − 1] = 0. Put H = Q_{X}⟨z⟩ and observe that H normalizes [V, X]. Let F be a composition factor for H on [V, X]. Then O_{p}(H/C_{H}(F)) = 1 and so, by the Hall-Higman Theorem B [Gor, 11.1.1], z centralizes F. Since

Q_{X} = [Q_{X}, Z] = [Q_{X}, X⟨z⟩] = [Q_{X}, z],

also Q_{X} centralizes F. As Q_{X} is a p′-group, we get [V, X, Q_{X}] = 0. Since Q_{X} centralizes X, [V, [X, Q_{X}]] = 0 and so by the Three Subgroup Lemma, [V, Q_{X}, X] = 0 and (a) holds.

For (b), let D = N_{Q}([V, Q_{X}]). Then C_{Q}(X) ≤ D and by (a) ⟨X^{D}⟩ centralizes [V, Q_{X}]. Therefore

[V, Q_{X}, [D, X]] = 0.

Now by co-prime action we have V = C_{V} ([D, X]) ⊕ [V, [D, X]] and so since Q_{X} normalizes [D, X] we get

[V, [D, X], Q_{X}] ≤ [V, [D, X]] ∩ [V, Q_{X}] ≤ [V, [D, X]] ∩ C_{V} ([D, X]) = 0.

Thus by the Three Subgroup Lemma, [V, [[D, X], Q_{X}]] = 0 and as H acts faithfully on V,

[[D, X], Q_{X}] = 1.

By co-prime action D = C_{D}(X)[D, X] and so since C_{D}(X) ≤ C_{Q}(X) ≤ N_{Q}(Q_{X}), we have

Q_{X} ⊴ D.

Suppose D ≠ Q. Then since Q is a q-group, D < N_{Q}(D). Since Q is Z-invariant, each of Q_{X}, D, and N_{Q}(D) is Z-invariant. Let E be minimal Z-invariant with D < E ≤ N_{Q}(D). Then Z acts irreducibly on the q-group E/D. Since Z is noncyclic, Z does not act faithfully on E/D. Thus, there exists Y ∈ X such that [E/D, Y] = 1 and therefore [E, Y] ≤ D. By co-prime action E = C_{E}(Y)[E, Y] = C_{E}(Y)D and so since C_{Q}(X) ≤ D, we have Y ≠ X. Let y ∈ Y \X and put W = [V, Q_{X}] and U = [W, y, p−1]. As we saw above Q_{X} = [Q_{X}, y] and so y normalizes Q_{X} and Q_{X}⟨y⟩ acts on W/⟨U^{Q_{X}} ⟩. By the Hall-Higman Theorem B, y acts trivially on the composition factors of Q_{X}⟨y⟩ on W/⟨U^{Q_{X}} ⟩ and so Q_{X} acts trivially on W/⟨U^{Q_{X}} ⟩. Hence W = [W, Q_{X}] ≤ ⟨U^{Q_{X}} ⟩ ≤ W, and therefore W = ⟨U^{Q_{X}} ⟩. In particular, as D normalizes W and Q_{X} ≤ D, we have

W = ⟨W^{D}⟩ = ⟨U^{Q_{X}D}⟩ = ⟨U^{D}⟩.

Let e ∈ C_{E}(y) and z ∈ Z. Then

U^{e} = [W^{e}, y, p − 1] ≤ [V, Z, p − 1] ≤ C_{V} (Z) ≤ C_{V} (z).

It follows that U^{e} ≤ W^{e} ∩ W^{ez}. As E normalizes D and W^{ez} is D-invariant, we compute

W^{e} = ⟨U^{De}⟩ = ⟨U^{eD}⟩ ≤ ⟨W^{ezD}⟩ = W^{ez}

and so W^{e} = W^{ez}. Therefore Z acts trivially on the set S = {W^{e}|e ∈ C_{E}(y)}. Moreover, since E normalizes D and D normalizes W, D acts trivially on S as well. By co-prime action E = C_{E}(y)[E, y] = C_{E}(y)D and so E acts on S.

Now D[E, Z] is Z-invariant and so D[E, Z] = D or E = D[E, Z] by the minimality of E. In the first case, [E, Z] ≤ D and so by co-prime action we get E = C_{E}(Z)[E, Z] = C_{E}(Z)D ≤ N_{Q}(W) = D, a contradiction. On the other hand, E = D[E, Z] implies E acts trivially on S and again we get E ≤ N_{Q}(W) = D, a contradiction.

We conclude that D = Q and so Q normalizes [V, Q_{X}]. By (a), X centralizes [V, Q_{X}]. Hence also [Q, X] centralizes [V, Q_{X}] and

[V, Q_{X}] ≤ C_{V} ([Q, X]).

Now by co-prime action V = C_{V} ([Q, X]) × [V, [Q, X]] and so

[V, Q_{X}] ∩ [V, [Q, X]] ≤ C_{V} ([Q, X]) ∩ [V, [Q, X]] = 0.

In particular,

Q_{X} ∩ [Q, X] = 1.

Let Y ∈ X \ {X}. Then Q_{Y} = [Q_{Y} , Z] = [Q_{Y} , XY] = [Q_{Y} , X] ≤ [Q, X]. Thus

Q_{X} ∩ ⟨Q_{Y} | Y ∈ X \ {X}⟩ ≤ Q_{X} ∩ [Q, X] = 1.

Since Q_{X} ⊴ Q, we conclude that

Q = [Q, Z] = ⟨Q_{X}|X ∈ X ⟩ = ⨁_{X∈X} Q_{X}.

Therefore by the commutator laws

[V, Q] = ∏_{X∈X}[V, Q_{X}].

Moreover, since

[V, Q_{X}] ∩ ∏_{Y ≠X,Y ∈X}[V, Q_{Y}] ≤ [V, Q_{X}] ∩ [V, [Q, X]] = 0,

we get

[V, Q] = ⨁_{X∈X}[V, Q_{X}].

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Lemma 3.2 Let P be a p-group acting on an abelian p-group A. Suppose that C_{A}(Z) = C_{A}(P) for some Z ≤ Z(P).

(a) For a subset X ⊆ P and non-negative integer i define C_{A}(X, i) inductively by C_{A}(X, 0) = 1 and C_{A}(X, i + 1)/C_{A}(X, i) = C_{A/C_{A}(X,i)}(X). Then for all non-negative integers i,

C_{A}(Z, i) = C_{A}(P, i)

(b) If A is elementary abelian and |Z| = p, then [A, P, p] = 1.

Proof For (a) we use induction on i. The result holds for i = 0 and i = 1 by definition and by assumption respectively. Let i ≥ 2 and suppose the result holds for all positive integers less than i. Then since Z ≤ P, C_{A}(P, i) ≤ C_{A}(Z, i). Let W = C_{A}(Z, i)/C_{A}(Z, i − 2). Then [C_{A}(Z, i), Z] ≤ C_{A}(Z, i − 1) = C_{A}(P, i − 1) and so [W, Z, P] = 1. Also, [Z, P, W] = 1 and so by the Three Subgroup Lemma [W, P, Z] = 1. We conclude that

[W, P] ≤ C_{A/C_{A}(Z,i−2)}(Z) = C_{A}(Z, i − 1)/C_{A}(Z, i − 2).

Hence [C_{A}(Z, i), P] ≤ C_{A}(Z, i−1). By induction C_{A}(Z, i−1) = C_{A}(P, i−1) and so C_{A}(Z, i) ≤ C_{A}(P, i). For (b), let 1 ≠ z ∈ Z. Then in End_{ℤ}(A), (z − 1)^{p} ≡ z^{p} − 1( mod p), and so [A, z, p] = A(z − 1)^{p} = 1. Thus C_{A}(z, p) = A. Hence, by (a) A = C_{A}(P, p) and [A, P, p] = 1.

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4 An Abelian Normal Subgroup

Lemma 4.1 Let H be a finite group, p a prime, and R = O^{p}(H). Then one of the following holds

(a) Φ(O_{p}(R)) ≤ Z(R).

(b) There exists an elementary abelian normal p-subgroup A of H with [A, R] ≠ 1.

Proof Let Q = O_{p}(R) and suppose [Φ(Q), R] ≠ 1. Let D ⊴ H minimal with respect to D ≤ Φ(Q) and [D, R] ≠ 1. Put E = C_{D}(R). Then by the minimality of D, D/E is elementary abelian and H acts irreducibly on D/E. Since R ⊴ H, Q ⊴ H and so [D, Q] ⊴ H. Since Q is nilpotent, [D, Q] < D. Thus by the minimality of D, [D, Q] ≤ E. Since E ≤ Z(R), [D, Q, Q] = 1 and therefore [D, Q′] = 1 by the Three Subgroup Lemma. Let d ∈ D and q ∈ Q. Since D/E is elementary, d^{p} ∈ E. Hence, since [D, Q] ≤ E ∩ Q ≤ Z(Q), we get

1 = [d^{p}, q] = [d, q^{p}].

Therefore [D, q^{p}] = 1 and so [D, Φ(Q)] = 1. Now since D ≤ Φ(Q), D is abelian. Let A = Ω_{1}(D). Then A is an elementary abelian p-group and A char D ⊴ H implies A ⊴ H. Also as [D, R] ≠ 1 and R = O^{p}(H), [Gor, 5.2.4] implies [A, R] ≠ 1. □

5 Over-groups for Z in Alt(n)

Lemma 5.1 Let p be a prime, Ω a finite set and m > 1. Suppose Z ≤ Alt(Ω) is elementary abelian of order p^{2} such that Z has at least m · p^{3} regular orbits on Ω. Then there exists a subgroup H ≤ Alt(Ω) such that

(a) Z ≤ H and [Φ(O_{p}(O^{p}(H))), O^{p}(H)] ≠ 1.

(b) H/O_{p}(H) ≅ Alt(mp^{2}) and Z acts semi-regularly on {1, 2, 3, …, mp^{2}}.

Proof By assumption there exists a Z-invariant subset Σ ⊆ Ω such that |Σ| = mp^{5} and Z acts semi-regularly on Σ. Let n = mp^{2} and let P_{1} be an extra special group of order p^{3}. By letting P_{1} act on itself regularly and Alt(n) act naturally, we see that Alt(Σ) contains a subgroup H^{∗} = P_{1} ≀ Alt(n). Now since p^{2} | n, Alt(n) contains a subgroup Z^{∗} isomorphic to Z with Z^{∗} acting semi-regularly on {1, 2, …, n}. Then Z^{∗} is also semiregular on Σ. Finally we can choose H^{∗} and Z^{∗} so that Z^{∗} is the image of Z under the restriction map from N_{Alt(Ω)}(Σ) to Sym(Σ). Letting H^{∗} fix all the points in Ω \ Σ elementwise, we may view H^{∗} as a subgroup of Alt(Ω). Then since Z^{∗} ≤ H^{∗}, Z normalizes H^{∗}. Let H = H^{∗}Z. Then Z ≤ H and H = H^{∗}C_{H}(Σ). Since H^{∗} centralizes Ω\Σ, we conclude that H/C_{H}(Ω\Σ) ≅ Z/C_{Z}(Ω\Σ) is a p-group. Thus C_{H}(Σ) is a p-group, O_{p}(H) = O_{p}(H^{∗})C_{H}(Σ) and H/O_{p}(H) ≅ H^{∗}/O_{p}(H^{∗}) ≅ H^{∗}/⟨P_{1}^{H^{∗}}⟩ ≅ Alt(n). Thus (b) holds. Let P = ⟨P_{1}^{H}⟩. Then P is a normal p-subgroup of H. We leave it as an exercise to the reader to verify that P′_{1} ≤ Φ(O_{p}(O^{p}(H))). Also O^{p}(H) does not normalize P′_{1} and (a) holds.

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Theorem 5.2 Let m ≥ 5^{3} and let Z ≤ Alt(mp^{2}) be an elementary abelian p-group of order p^{2} acting semi-regularly. Then there exists a prime q with 2 ≠ q ≠ p and a Z-invariant q-subgroup Q of Alt(mp^{2}) such that

[Q_{X}, Q_{Y}] ≠ 1

for all X and Y distinct proper subgroups of Z, where Q_{X} = [C_{Q}(X), Z] and Q_{Y} = [C_{Q}(Y), Z].

Proof. Let q = 3 if p ≠ 3, q = 5 if p = 3, and Q_{1} be an extra special q-group of order q^{3}. Since m ≥ 5^{3} ≥ q^{3}, there exists a Z invariant subset Σ ⊆ {1, 2, …, mp^{2}} with |Σ| = q^{3}p^{2}. Moreover, as in Lemma 5.1, Alt(Σ) contains a subgroup H^{∗} = Q_{1} ≀ Alt(p^{2}), such that Z normalizes H^{∗} and H^{∗} contains the image of Z in Alt(Σ).

Let Q = ⟨Q_{1}^{Z}⟩, X and Y distinct be proper subgroups of Z, 1 ≠ x ∈ X, and 1 ≠ y ∈ Y. Then Q is a Z-invariant q-subgroup of Alt(mp^{2}). Let π be the projection map from Q to Q_{1}. We claim (Q_{X})π = Q_{1}. Clearly (Q_{X})π ≤ Q_{1}. Let q_{1} ∈ Q_{1}. Then q = q_{1}q_{1}^{x} … q_{1}^{x^{p−1}} ∈ C_{Q}(X) and

[q, y] = q^{−1}q^{y} = q_{1}^{−x^{p−1}} … q_{1}^{−x}q_{1}^{−1}q_{1}^{y} … q_{1}^{x^{p−1}y}

Now as Z acts semi-regularly on {1, 2, …, mp^{2}}, it acts semi-regularly on ⟨Q_{1}^{Z}⟩. Therefore each of the 2p factors of [q, y] lie in different conjugates of Q_{1}. In particular, ([q, y])π = q_{1}^{−1}. Since [q, y] ∈ Q_{X} and q_{1} ∈ Q_{1} was chosen arbitrarily, we have Q_{1} ≤ (Q_{X})π and so (Q_{X})π = Q_{1}. By symmetry (Q_{Y})π = Q_{1} and hence ([Q_{X}, Q_{Y}])π = Q′_{1}. Since Q_{1} is non-abelian, we obtain [Q_{X}, Q_{Y}] ≠ 1 and the theorem is proven.

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6 The Regular Case

Theorem 6.1 Let F be a finite group, Z an elementary abelian subgroup of order p^{2}, and N ⊴ F with F/N ≅ Alt(Ω). Suppose that Z has at least 5^{3} · p^{3} regular orbits on Ω. Then C_{F} (z) ≠ C_{F} (Z) for some 1 ≠ z ∈ Z.

Proof Let H ≤ F/N be given by 5.1 with m = 5^{3}. Let H^{∗} and N^{∗} be the pre-images of H and O_{p}(H) in F, and S ∈ Syl_{p}(H^{∗}) with Z ≤ S. Also let T ≤ H^{∗} be minimal with respect to S ≤ T and H^{∗} = N^{∗}T. Set P = S∩N^{∗} = S∩N^{∗}∩T. By Lemma 2.1(a), P ⊴T. Since P is a Sylow p-subgroup of N^{∗} and N^{∗}/N is a p-group, N^{∗} = P N. Thus H^{∗} = T N^{∗} = T P N = T N. Note that O_{p}(O^{p}(H)) = (P ∩ O^{p}(T))N/N. As [Φ(O_{p}(O^{p}(H)), O^{p}(H)] ≠ 1 we conclude that [Φ(O_{p}(O^{p}(T))), O^{p}(T)] ≠ 1. Hence, by Theorem 4.1 there exists A ⊴ T elementary abelian with [A, O^{p}(T)] ≠ 1. It follows that there exists a composition factor D for T on A with [D, T] ≠ 1. Let \overline{T} = T/C_{T} (D). Then D is a faithful irreducible GF(p)\overline{T} -module and O_{p}(\overline{T}) = 1. Suppose that C_{T} (D) ≰ N^{∗}. Then since H^{∗}/N^{∗} is simple, H^{∗} = N^{∗}C_{T} (D). Hence, by the minimality of T, we have T = C_{T} (D)S and so \overline{T} is a p-group. It follows that \overline{T} = 1, contradicting the choice of D. Since O_{p}(\overline{T}) = 1, Φ(\overline{T}) is a p′ group. By Lemma 2.1, \overline{N^{∗}} ∩ \overline{T} ≤ Φ(\overline{T}). Since \overline{T} /\overline{N^{∗}} ∩ \overline{T} is simple we conclude that \overline{N^{∗}} ∩ \overline{T} = Φ(\overline{T}) and so \overline{T} /Φ(\overline{T}) ≅ Alt(mp^{2}). Let Q be the Zinvariant q-subgroup of \overline{T} /Φ(\overline{T}) given by Theorem 5.2. Since the pre-image of Q in \overline{T} is a p′-group, the pre-image of Q contains a Z-invariant Sylow q-subgroup Q^{∗}. As [Q_{X}, Q_{Y}] ≠ 1 for all distinct proper subgroups X and Y of Z, [Q_{X}^{∗}, Q_{Y}^{∗}] ≠ 1. Suppose that C_{A}(z) = C_{A}(Z) for all 1 ≠ z ∈ Z.

Then by Lemma 3.2 [A, Z, p] = 1 and also [D, Z, p] = 0, where we view D as a vector space over GF(p). Thus Theorem 3.1 (b) implies [Q_{X}^{∗}, Q_{Y}^{∗}] = 1, a contradiction. Therefore C_{A}(z) ≠ C_{A}(Z) for some 1 ≠ z ∈ Z and the result of the theorem follows.

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7 A Bifurcation Lemma

Theorem 7.1 Let G be a non-finitary, locally finite simple group with alternating Kegel cover K and let E ≤ G be a finite subgroup of G. Then one of the following holds

(a) For all positive integers t

K_{reg}^{t}(E) = {(H_{K}, M_{K}) ∈ K(E)| E has at least t regular orbits on Ω_{K}}

is a Kegel cover for G.

(b)

K_{nat}(E) = {(H_{K}, M_{K}) ∈ K(E)| E has no regular orbits on Ω_{K}}

is a Kegel cover for G.

Proof Suppose (a) does not hold. Then there exists a positive integer t such that Kreg^{t}(E) is not a Kegel cover. Then K(E) \ Kreg^{t}(E) is a Kegel cover and we may assume that for all (H_{K}, M_{K}) ∈ K, E ≤ H_{K} and E has less than t regular orbits on Ω_{K}. Let F ≤ G be a finite subgroup with E ≤ F and |F/E| = r > t. Suppose there exists some (H_{K}, M_{K}) ∈ K(F) such that F has a regular orbit s^{F} on Ω_{K}, for some s ∈ Ω_{K}. Then each s^{x_{i}E}, where {x_{i}}_{i=1}^{r} is a transversal for E in F, is a regular orbit for E on Ω_{K}, contrary to our assumptions. Therefore F has no regular orbits on Ω_{K} for all (H_{K}, M_{K}) ∈ K. From [Me, 3.4] we conclude that there exists a Kegel cover J ⊆ K for G such that for all (H_{J}, M_{J}), (H_{K}, M_{K}) ∈ J with H_{J} ≤ H_{K}, all essential orbits for H_{J} on Ω_{K} are pseudo natural with respect to M_{J}. Without loss we may assume K = J.

Pick (H_{K}, M_{K}) ∈ K so that the number d of regular orbits of E on Ω_{K} is maximal. If d = 0 then (b) holds. Suppose d > 0. Let R be the unique minimal subnormal supplement to M_{K} in H_{K} and 1 ≠ r ∈ R. Then by Hall’s Finitary Lemma ( [Ha, 3.13]) there exists (H_{J}, M_{J}) ∈ K(H_{K}) such that deg_{Ω_{J}} (r) > |H_{K}|, where deg_{Ω_{J}} (r) is the number of elements of Ω_{J} moved by r. Hence there exists at least two orbits Ω_{1} and Ω_{2} for H_{K} on Ω_{J} which are not fixed elementwise by r. In particular, Ω_{i} is an essential orbit for H_{K} and therefore pseudo natural for i = 1, 2. So there exists an H_{K}-invariant partition Δ_{i} of Ω_{i} such that Δ_{i} and Ω_{K} are isomorphic as H_{K}-sets. As E has d regular orbits on Ω_{K}, it has d regular orbits on Δ_{i} for i = 1, 2. But then E has at least 2d regular orbits on Ω_{J}, contradicting the maximal choice of d.

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8 The Proof of the Main Theorem

Let G be a locally finite simple group of alternating type and suppose Z is an elementary abelian subgroup of order p^{2} such that C_{G}(Z) = C_{G}(z) for all 1 ≠ z ∈ Z. Let K be an alternating Kegel cover for G and let (H_{K}, M_{K}) ∈ K with Z ≤ H_{K}. Then by Theorem 6.1 Z has less than 5^{3}p^{3} regular orbits on Ω_{K}. Therefore, by Theorem 7.1 we may assume that K = Knat(Z), that is, for all (H_{K}, M_{K}) ∈ K, Z ≤ H_{K} and Z has no regular orbits on Ω_{K}.

Let n be chosen as in Lemma 2.2. By Hall’s Finitary Lemma ( [Ha, 3.13]) there exists (H_{K}, M_{K}) ∈ K such that deg_{Ω_{K}}(z) ≥ (p + 1)(n + 2) for all 1 ≠ z ∈ Z. Also let Δ be the union of the nontrivial orbits of Z on Ω_{K} and let X be the set of nontrivial proper subgroups of Z. As none of the orbits are regular, we have

Δ = ⋃_{X∈X} Fix_{Δ}(X)

where Fix_{Δ}(X) is the subset of Δ fixed elementwise by X. Since |Δ| ≥ (p + 1)(n + 2) and Z has p + 1 proper subgroups, Δ_{2} = Fix_{Δ}(X_{1}) has at least n + 2 elements for some X_{1} ∈ X. Let 1 ≠ x_{1} ∈ X_{1}. Then |Δ \ Δ_{2}| ≥ deg_{Ω_{K}}(x_{1}) ≥ (p + 1)(n + 2). As above there exists X_{2} ∈ X \ {X_{1}} so that Δ_{1} = Fix_{Δ}(X_{2}) has at least n+2 elements. Note that an element in Δ_{1} ∩Δ_{2} is fixed by X_{1}X_{2} = Z. But Z has no fixed points on Δ and so Δ_{1} and Δ_{2} are disjoint.

If p is odd, let Ω_{i} = Δ_{i} for i = 1, 2; if p = 2, let Ω_{i} be a Z-invariant subset of Δ_{i} such that |Δ_{i} \ Ω_{i}| ≤ 2 and 4 divides |Ω_{i}|. Then |Ω_{i}| ≥ n and X_{i} has no fixed points on Ω_{i} for each i.

Let Ω = Ω_{1} ∪ Ω_{2}, H^{∗} = {h ∈ N_{H_{K}}(Ω)|h is even on Ω} and M^{∗} = C_{H^{∗}}(Ω). Then Z = X_{1}X_{2} ≤ H^{∗}. Choose H ≤ H^{∗} minimal so that Z ≤ H and H^{∗} = HM^{∗}, and let M = M^{∗} ∩ H. Then H/M ≅ Alt(Ω). Let R = H′. Suppose N ⊴ H with N ≰ M. Then, since H/M is simple, H = MN. Now by the minimality of H, H = NZ and so R ≤ N. Let A_{i} = C_{H}(Ω_{3−i}) for i = 1, 2. Then X_{i} ≤ A_{i} and A_{i}/M ≅ Alt(Ω_{i}). Let B_{i} = ⟨X_{i}^{A_{1}A_{2}}⟩ for each i. Then as X_{i} ≰ M and A_{i}/M is simple, A_{i} = B_{i}M for each i. Now R ⊴ H and B_{i} ⊴ A_{i} implies that B_{i} ∩ R ⊴ A_{i}. If B_{i} ∩ R ≤ M, then, as B′_{i} ≤ R, B′_{i} ≤ M. Hence, we get that A_{i}/M is abelian, a contradiction. Therefore B_{i} ∩ R ≰ M and so A_{i} = (B_{i} ∩ R)M.

Suppose [B_{1}, B_{2} ∩ R] = 1. Then [X_{1}, B_{2} ∩ R] = 1 and, as C_{G}(X_{1}) = C_{G}(Z) = C_{G}(X_{2}), [X_{2}, B_{2} ∩ R] = 1. Also since B_{2} ∩ R ⊴ A_{1}A_{2} and B_{2} = ⟨X_{2}^{A_{1}A_{2}}⟩, it follows that [B_{2}, B_{2} ∩ R] = 1. Thus [A_{2}, A_{2}] = [B_{2}M, (B_{2} ∩ R)M] ≤ M, a contradiction since A_{2}/M is not abelian. Therefore [B_{1}, B_{2} ∩ R] ≠ 1.

Let D = [B_{1}, B_{2} ∩ R] and (H_{J}, M_{J}) ∈ K with H ≤ H_{J} and H ∩ M_{J} = 1. Since D ≠ 1, there exists an orbit Σ for H on Ω_{J} on which D acts nontrivially. Suppose C_{H}(Σ) ≰ M. Then as seen above D ≤ R ≤ C_{H}(Σ), a contradiction. Therefore C_{H}(Σ) ≤ M. Note that for all 1 ≠ z ∈ Z, deg_{Ω}(z) ≥ min(|Ω_{1}|, |Ω_{2}|) ≥ n. Also Z has no regular orbits on Ω_{J} and so Lemma 2.2 implies that Σ is pseudo natural for H with respect to M. That is, there exists a H-invariant partition Γ of Σ such that Γ and Ω are isomorphic H-sets. Let Γ_{i} be the images of Ω_{i} under the above isomorphism for i = 1, 2. Also let

Σ_{i} = ⋃_{γ∈Γ_{i}} γ

for each i. Then Σ = Σ_{1}∪Σ_{2}. Let σ ∈ Σ_{3−i}. Since Z has no regular orbits on Σ, C_{Z}(σ) ≠ 1. Let γ ∈ Γ_{3−i} with σ ∈ γ. Then C_{Z}(σ) ≤ N_{Z}(γ) = X_{i}. Thus C_{Z}(σ) = X_{i} and X_{i} fixes Σ_{3−i} elementwise. Therefore, as A_{1}A_{2} normalizes Σ_{i} for each i, B_{i} = ⟨X_{i}^{A_{1}A_{2}}⟩ fixes Σ_{3−i} elementwise for each i. Hence B_{1}∩B_{2} acts trivially on Σ = Σ_{1} ∪ Σ_{2}. Since D ≤ [B_{1}, B_{2}] ≤ B_{1} ∩ B_{2}, we get D acts trivially on Σ, contradicting the choice of Σ.

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