ISOLATED SUBGROUPS IN FINITE GROUPS

ULRICH MEIERFRANKENFELD CHRISTOPHER PARKER PETER ROWLEY

October 19, 2007

1. Introduction

Suppose that p is a prime, P is a finite group and S ∈ Syl_{p}(P). Then P is pminimal if S is not normal in P and S is contained in a unique maximal subgroup of P. Now suppose that G is a finite group and S ∈ Syl_{p}(G). The set of subgroups of G which contain S and are p-minimal will be denoted by P_{G}(S). For P ∈ P_{G}(S) we put

L_{G}(P, S) := ⟨P_{G}(S) \ {P }, N_{G}(S)⟩,

and next introduce isolated p-minimal subgroups.

Definition 1.1. A p-minimal subgroup P in P_{G}(S) is A-isolated where A is a normal p-subgroup of L_{G}(P, S) if A ≰ O_{p}(P). We say that P ∈ P_{G}(S) is an isolated subgroup of G if it is A-isolated for some A.

Clearly P is an isolated subgroup of G if and only if O_{p}(L_{G}(P, S)) ≰ O_{p}(P). As a consequence G ≠ L_{G}(P, S). Thus, as G = ⟨P_{G}(S), N_{G}(S)⟩ (see Lemma 3.1), any minimal generating set of G using subgroups from P_{G}(S), together with N_{G}(S), must include P.

The archetypical p-minimal subgroup is to be found in groups of Lie type defined in characteristic p. If G is such a group and R is a minimal parabolic subgroup of G, then P = O^{p′}(R) is a p-minimal subgroup of G. For such a P, L := L_{G}(P, S) is the unique maximal parabolic subgroup of G not containing R and O_{p}(L) ≰ O_{p}(P), and so in fact P is an isolated subgroup of G. This in itself makes isolated subgroups worthy of study yet there are other compelling reasons for instigating a study of this type of subgroup. Such subgroups arise in the on-going work whose aim is to classify finite K_{p}-proper groups of local characteristic p. We return to this important aspect of isolated subgroups shortly.

Our first theorem includes a statement about a special class of p-minimal subgroups which we call narrow subgroups. Before defining narrow subgroups, we first recall that a group G acts imprimitively on a GF(p)G-module V provided there exists a non-trivial decomposition of V into a direct sum of subspaces W_{1}, …, W_{k} which are permuted by the action of G. A group which does not act imprimitively on a module V is said to act primitively on V.

Definition 1.2. Suppose that P is a p-minimal group, S ∈ Syl_{p}(P) and let M be the unique maximal subgroup of P containing S. Set E := O^{p}(P)F/F where F is the core of M in P. Then we say that P is narrow if either E is a simple group or E is elementary abelian and M acts primitively on E.

Again examples of narrow, isolated subgroups appear naturally in Lie type groups of characteristic p.

Our first theorem shows that finite groups with isolated subgroups have a restricted structure. Its proof consists of an analysis of the over groups of Sylow p-subgroups and relies only on results from classical group theory. This contrasts with the proof of Theorem 1.7 where we are forced to draw upon specific structural properties of certain finite simple groups.

Theorem 1.3. Suppose that p is a prime, G is a finite group, S ∈ Syl_{p}(G) and P ∈ P_{G}(S) is an isolated subgroup of G. Set L := L_{G}(P, S) and Y := ⟨O^{p}(P)^{G}⟩. Then G = Y L, P is an isolated subgroup of Y S and either

(a) Y = O^{p}(P); or

(b) Y/O_{p}(Y) is a central product of quasisimple groups which are transitively permuted by S under conjugation and, if P is narrow, then Y/O_{p}(Y) is quasisimple. Furthermore, Y/O_{p}(Y) has order divisible by p.

Of course, if G is a quasisimple group and G has an isolated subgroup, then Y = G and alternative (b) of Theorem 1.3 holds. If G is a p-minimal group and S ∈ Syl_{p}(G), then L_{G}(G, S) is the unique maximal subgroup of G containing S. Hence, if O_{p}(L_{G}(G, S)) > O_{p}(G), G is an isolated subgroup of G and the possibility in Theorem 1.3(a) trivially holds. As an indication of the limitations of Theorem 1.3, we could take G = H_{1} × H_{2} with P an isolated subgroup of H_{1}S with S ≤ P where S ∈ Syl_{p}(G). Then P is an isolated subgroup of G (see Lemma 3.5) and of course Y ≤ H_{1}. So the direct factor H_{2} is invisible as far as Y is concerned.

For a finite group G and p a prime, let F_{p}^{∗}(G) denote the inverse image in G of F^{∗}(G/O_{p}(G))), the generalized Fitting subgroup of G/O_{p}(G). Then we have the following corollary of Theorem 1.3.

Corollary 1.4. Suppose that G is a finite group and P ∈ P_{G}(S) is an isolated subgroup of G where S ∈ Syl_{p}(G), p a prime.

(a) O^{p}(P) ≤ F_{p}^{∗}(G).

(b) If Y := ⟨O^{p}(P)^{G}⟩ is soluble, then Y = O^{p}(P) and there is a prime t ≠ p such that Y/O_{p}(Y) is a t-group of class at most 2.

Again casting an eye over the Lie type groups in characteristic p we see that every p-minimal subgroup is an isolated subgroup. This leads us to give

Definition 1.5. Let G be a finite group, p a prime and S ∈ Syl_{p}(G). Then G is called completely isolated if P is isolated for all P ∈ P_{G}(S).

For P_{1}, P_{2} ∈ P_{G}(S) write P_{1} ∼ P_{2} whenever ⟨O^{p}(P_{1})^{G}⟩ = ⟨O^{p}(P_{2})^{G}⟩. Clearly ∼ is an equivalence relation on P_{G}(S). Our second theorem describes the structure of completely isolated finite groups.

Theorem 1.6. If G is a completely isolated finite group, then G = F_{p}^{∗}(G)N_{G}(S), where S ∈ Syl_{p}(G). Moreover, F^{∗}(G/O_{p}(G)) is a central product of the subgroups generated by the ∼ equivalence classes of P_{G/O_{p}(G)}(S/O_{p}(G)) and F (G/O_{p}(G)) has class at most two.

For p a prime, we define

L_{1}(p) := {O^{p}(X) | X ≅ H/Z where H is a universal rank 1 Lie type group

defined in characteristic p and Z ≤ Z(H)}.

Observe that some of the groups in L_{1}(p) are smaller than we might have expected. For example, O^{2}(SL_{2}(2)) has order 3 and O^{2}(^{2}B_{2}(2)) has order 5. So, to be explicit we list the members of L_{1}(p). For p ≥ 5, we have

L_{1}(p) = {SL_{2}(p^{a}), PSL_{2}(p^{a}), SU_{3}(p^{a}), PSU_{3}(p^{a}) | a ≥ 1}.

Because of the non-simplicity of SL_{2}(3) and ^{2}G_{2}(3), we have

L_{1}(3) = {Q_{8}, 2^{2},^{2}G_{2}(3)′ ≅ SL_{2}(8)}∪{SL_{2}(3^{a}), PSL_{2}(3^{a}), SU_{3}(3^{a−1}),^{2}G_{2}(3^{2a−1}) | a ≥ 2}

and similarly, because of the non-simplicity of SL_{2}(2), SU_{3}(2) and ^{2}B_{2}(2), we have

L_{1}(2) = {3, 5, 3_{+}^{1+2}, 3^{2}} ∪ {SL_{2}(2^{a}), SU_{3}(2^{a}), PSU_{3}(2^{a}),^{2}B_{2}(2^{2a−1}) | a ≥ 2}.

The groups in L_{1}(p) will be discussed further in Section 3. Notice that if O^{p}(P/O_{p}(P)) ∈ L_{1}(p), S ∈ Syl_{p}(P), and S > O_{p}(P), then P is a narrow unless p = 2 and O^{2}(P/O_{p}(P)) ≅ 3^{2} or 3_{+}^{1+2} in which case we have to be rather more careful. Thus Theorem 1.3 applies to narrow isolated subgroups with the property that O^{p}(P/O_{p}(P)) ∈ L_{1}(p) and tells us that when O_{p}(G) = 1, Y := ⟨O^{p}(P)^{G}⟩ is either quasisimple or equal to O^{p}(P). In the latter case we have that either Y = O^{p}(P) is quasisimple or Y is soluble and O^{p}(P) is isomorphic to one of 3, 5, Q_{8}, 2^{2}, 3_{+}^{1+2} and 3^{2}.

Our next theorem concerns certain isolated subgroups of CK-groups. We say that a finite simple group is a K-group if it is isomorphic to a cyclic group of prime order, an alternating group of degree at least 5, a simple group of Lie type (including the Tits simple group) or one of the 26 sporadic simple groups. A group is a CK-group, if each of its composition factors is a K-group.

Theorem 1.7. Suppose that p is a prime, G is a finite CK-group and P ∈ P_{G}(S). If O^{p}(P/O_{p}(P)) ∈ L_{1}(p), then either O^{p}(P) ⊴ G or O_{p}(Z(L_{G}(P, S))) ≤ O_{p}(P).

In other words Theorem 1.7 is asserting that if P is O_{p}(Z(L_{G}(P, S)))-isolated and O^{p}(P/O_{p}(P)) ∈ L_{1}(p), then O^{p}(P) is normal in G. We note that if we take G = J_{2} (Janko’s second simple group) we have a narrow subgroup P of G with P ≅ PSU_{3}(3) ∈ L_{1}(3) and L_{G}(P, S) ≅ 3.PGL_{2}(9). So O_{3}(P) = 1, O_{3}(L_{G}(P, S)) has order 3 and just misses being in Z(L_{G}(P, S)).

We comment that not every finite group possesses an isolated subgroup. However, as we have seen, if it does then it imposes global constraints on the structure of the group. Another example of this is the following corollary to [MPR, Corollary 1.2]. This asserts that a simple CK-group with an isolated subgroup P satisfying O^{p}(P/O_{p}(P)) ∈ L_{1}(p) with p ≥ 11 must be a Lie type group in characteristic p.

Isolated subgroups, as mentioned earlier, also surface in the project to classify finite K_{p}-proper groups of local characteristic p. That is finite groups all of whose p-local subgroups H satisfy F^{∗}(H) = O_{p}(H) and have all simple sections from the list of finite simple groups itemized above. Let G to be such a group and let C be a maximal p-local subgroup containing N_{G}(Ω_{1}(Z(S))) where S ∈ Syl_{p}(G). Then in this context the P̃!-Theorem [MMPS] asserts that in the presence of certain additional conditions there is an isolated subgroup P̃ of G contained in P_{C}(S). Moreover, O^{p}(P̃/O_{p}(P̃)) ∈ L_{1}(p). Thus all of our theorems apply to tell us that Y := ⟨O^{p}(P̃)^{C}⟩O_{p}(C)/O_{p}(C) is a quasisimple group. In particular, as C is a CK-group the results of [MPR] which determine the finite simple K-groups possessing an isolated subgroup P with O^{p}(P/O_{p}(P)) ∈ L_{1}(p), can be applied to limit the possibilities for Y. For an account of the genealogy of this (still in progress) programme as well as an overview of its aims see [MSS].

Briefly this paper is organized as follows. In Section 2 we gather some elementary background results that we require in this paper. The following section contains a number of general properties of p-minimal groups ranging from their generational properties, their behaviour under quotients to the important compendium of structural properties given in Lemma 3.2. Towards the end of Section 3, in Lemmas 3.6 and 3.7, detailed information about p-minimal isolated subgroups with O^{p}(P/O_{p}(P)) ∈ L_{1}(p) is listed. These two results plus Corollary 3.9 are needed in our proof of Theorem 1.7. We begin Section 4 by noting a number of basic facts about isolated subgroups – Lemma 4.3 being a result we use frequently. After establishing further preparatory results such as Lemmas 4.4, 4.9 and 4.10, we then prove Theorems 1.3 and 1.6. The remainder of this section gives various consequences of Theorem 1.3 which will be of use in [MPR]. Our final section is devoted to a proof of Theorem 1.7.

If X is a collection of subsets of G, then ⋂ X means the intersection of all the subsets in X. Assume that H is a subgroup of G. Then H^{G} will denote the set of all G-conjugates of H. Hence ⋂ H^{G} is just the core of H in G. For p a fixed prime, we sometimes write Q_{G} for O_{p}(G), the largest normal p-subgroup of G. Also we denote the preimage of F (G/Q_{G}) by F_{p}(G) and the preimage of Φ(G/Q_{G}) by Φ_{p}(G). The remainder of our group theoretic notation is standard and can be found in [As] or [KS]. Atlas [Atlas] names and conventions will be followed with a few exceptions which we now note. We shall use Sym(n) and Alt(n) to denote, respectively, the symmetric and alternating group of degree n. And Dih(n), SDih(n) stand, respectively, for the dihedral and semidihedral group of order n (so n is even ).

The genesis of the work in this paper and [MPR] occurred while the authors were participants in the RiP (Research in Pairs) programme at the Mathematische Forschungsinstitut Oberwolfach. The completion of these papers was further supported by an LMS scheme 4 grant and funding from the Manchester Institute of Mathematical Sciences – the authors express their gratitude to all these organizations.

2. Preliminary Results

This section contains an assorted collection of background results. The first two concern subnormal subgroups while Lemma 2.3 is an elementary criterion for a psubgroup of a group G to be contained in O_{p}(G). Lemmas 2.4 and 2.6 look at the Frattini subgroup and Lemma 2.7 examines Φ_{p}(O^{p}(G)). And we end with Proposition 2.8 which gives a property of finite simple K-groups with abelian Sylow psubgroups.

Our first lemma records a well-known property of components.

Lemma 2.1. Let G be a finite group, K a component of G and N a subnormal subgroup of G. Then either K ≤ N or [K, N] = 1.

Proof. [As, 31.4]. □

In the proof of Lemma 3.2 we shall need the following result.

Lemma 2.2. Suppose that G is a finite group, M is a maximal subgroup of G and N is a subnormal subgroup of G with N ≤ M. Then N ≤ ⋂ M^{G}.

Proof. We may assume that N is maximal, under inclusion, with the property that N is subnormal in G and N ≤ M. Let N = N_{0} ⊴ N_{1} ⊴ … ⊴ N_{k} = G be a subnormal chain from N to G. By Wielandt’s Subnormality Theorem [KS, 6.7.1], ⟨N^{M}⟩ ⊴ ⊴G. Because of the maximality of N, ⟨N^{M}⟩ = N and so N ⊴M. Thus N ⊴⟨M, N_{1}⟩. Also, by the maximality of N, N_{1} ≰ M, whence N ⊴ G, and this proves the lemma. □

Lemma 2.3. Suppose that G is a finite group, p is a prime, and A is a p-subgroup of G for which [A, G] ≤ N_{G}(A). Then A ≤ O_{p}(G) and [A, G] is a p-group.

Proof. Since

A ≤ O_{p}([A, G]A) = O_{p}(⟨A^{G}⟩) ⊴ G,

A ≤ O_{p}(G) and [A, G] is a p-group. □

Lemma 2.4. Suppose that G is a finite group and N is a normal subgroup of G. Then Φ(N) ≤ Φ(G).

Proof. Assume that M is a maximal subgroup of G and that Φ(N) is not contained in M. Then, as M is maximal in G and Φ(N) is normal in G, G = Φ(N)M. Thus

N = N ∩ G = N ∩ Φ(N)M = Φ(N)(N ∩ M).

Thus N = N ∩M and so M ≥ N ≥ Φ(N) which is not the case. Hence Φ(N) ≤ Φ(G) as claimed. □

Lemma 2.5. Suppose that G is a finite group. Then

(a) Φ_{p}(G) is the intersection of all the maximal subgroups of G which contain a Sylow p-subgroup of G;

(b) O_{p}(G) ∈ Syl_{p}(Φ_{p}(G)); and

(c) if O_{p}(O^{p}(G)) = 1, then Φ_{p}(G) is nilpotent.

Proof. Let F := Φ_{p}(G) and H be the intersection of all the maximal subgroups of G which contain a Sylow p-subgroup of G. Obviously F ≤ H and H is normal in G. Let T ∈ Syl_{p}(H). Then the Frattini Argument gives G = N_{G}(T)H. Since T has p′ index in H, we infer that N_{G}(T) contains a Sylow p-subgroup of G. Thus N_{G}(T) = G. Hence T ≤ O_{p}(G) and so as O_{p}(G) ≤ F, we have O_{p}(G) = T. Assume that F < H. Then there exists a maximal subgroup M of G containing T which does not contain H. But then G = MH and, as M ∩ H ≥ T, we have that M contains a Sylow p-subgroup of G which delivers the contradiction H ≤ M. Hence F = H and O_{p}(G) ∈ Syl_{p}(F), so proving (a) and (b). Suppose that O_{p}(O^{p}(G)) = 1. Then [O_{p}(G), F ∩ O^{p}(G)] = 1 and, since F = O_{p}(G)(F ∩ O^{p}(G)) by (b), we get that F is nilpotent. Hence (c) holds. □

Lemma 2.6. Suppose G is a finite group with H and K subgroups of G. If G = HK and H ∩ K = [H, K] = 1, then Φ(G) = Φ(H)Φ(K).

Proof. Clearly we may assume H ≠ 1 ≠ K. Now every maximal subgroup M of G containing H, as H ∩ K = 1, must be of the form HM_{1} where M_{1} is a maximal subgroup of K. Since a similar statement holds for maximal subgroups of G containing K, intersecting over all maximal subgroups of G which contain either H or K shows that Φ(G) ≤ Φ(H)Φ(K). Lemma 2.4 gives the reverse inclusion and so Φ(G) = Φ(H)Φ(K). □

Lemma 2.7. Suppose that G is a finite group and p is a prime. Then Φ_{p}(O^{p}(G)) = O^{p}(G) ∩ Φ_{p}(G).

Proof. Suppose that G is a minimal counterexample to the statement in the lemma. Set X := O^{p}(G), F := Φ_{p}(G), F_{X} := F ∩ X and let S ∈ Syl_{p}(G). By the minimality of G, Q_{X} = 1.

Using Lemma 2.4, we get Φ_{p}(X) = Φ(X) ≤ Φ(G) ∩ X ≤ F_{X}. Hence F_{X} ≰ Φ(X). Assume that M is a maximal subgroup of X and that M does not contain F_{X}.

Then, as M is maximal in X and F_{X} is normal in X, X = F_{X}M. Since F_{X} ≅ F_{X}Q_{G}/Q_{G} ≤ Φ_{p}(G)/Q_{G} = Φ(G/Q_{G}), F_{X} is nilpotent of p′-order. Hence N_{F_{X}}(F_{X} ∩ M) > F_{X} ∩ M which together with the maximality of M in X implies that F_{X} ∩ M is a normal subgroup of X. Furthermore, because of the maximality of M in X, F_{X}/(F_{X} ∩ M) has no non-trivial proper X-invariant subgroups. Hence F_{X}/(F_{X} ∩ M) is an elementary abelian t-group for some prime t with t ≠ p. Let R = ⋂(F_{X} ∩ M)^{G}. Assume that R > 1. Then by induction we have O^{p}(G/R)∩Φ_{p}(G/R) = Φ_{p}(O^{p}(G/R)). Furthermore, as R ≤ Φ_{p}(G), Φ_{p}(G/R) ≥ Φ_{p}(G)/R. Hence, using X/R = O^{p}(G/R), we have

F_{X}/R = Φ_{p}(G)/R ∩ X/R ≤ Φ_{p}(G/R) ∩ O^{p}(G/R) = Φ_{p}(X/R).

But we know that M has index a power of t in X, so M/R ≥ Φ_{p}(X/R) by Lemma 2.5(a). Thus F_{X}/R ≤ Φ_{p}(X/R) ≤ M/R, which is a contradiction to our choice of M. Therefore, R = 1 and, in particular, F_{X} is an elementary abelian t-group. Notice that M/(F_{X} ∩ M) is a complement to F_{X}/(F_{X} ∩ M) in X/(F_{X} ∩ M). Choose K of minimal order in M such that X = KF_{X}. Then K ∩ F_{X} is normalized by both F_{X} and K and so is normal in X. Since K ≤ M, K ∩ F_{X} ≠ F_{X}. We claim that K is a complement to F_{X} in X. Suppose that K ∩ F_{X} > 1. Then, as R = 1, there exists g ∈ G such that K^{g} ∩ F_{X} ≰ F_{X} ∩ M. It follows that K^{g}(F_{X} ∩ M) = K^{g}(F_{X} ∩ K^{g})(F_{X} ∩ M) = K^{g}F_{X} = X as (F_{X} ∩ M)(F_{X} ∩ K^{g}) is normal in X. Therefore, as K^{g}/(K^{g} ∩ (F_{X} ∩ M)) ≅ X/(F_{X} ∩ M) and X/(F_{X} ∩ M) splits over F_{X}/(F_{X} ∩ M), K^{g}/(K^{g} ∩ F_{X} ∩ M) splits over (K^{g} ∩ F_{X})/(K^{g} ∩ F_{X} ∩ M) and thus K contains a proper subgroup K_{0} such that K = K_{0}(K ∩ F_{X}). But then K_{0}F_{X} = KF_{X} = X and we have a contradiction to the minimal choice of K. Therefore, X splits over F_{X}. By [As, 17.7] the total number of complements to F_{X} in X is |F_{X}/C_{F_{X}}(X)||H^{1}(X, F_{X})| = t^{a} for some integer a. Since S permutes (by conjugation) the complements to F_{X} in X and S is a p-group, we get that there exists a complement K to F_{X} in X which is normalized by S. So KS is a subgroup of G and F_{X}KS = XS = G. Since KS ≥ Q_{G}, we infer that KS = G. As F_{X} is a p′-group, we get F_{X} = 1 and we have a contradiction to F_{X} ≰ Φ(X). This concludes the proof of Lemma 2.7 □

Proposition 2.8. Suppose that p is a prime, G is a finite non-abelian simple K-group and S is a Sylow p-subgroup of G. If S is abelian, then N_{G}(S) acts irreducibly on Ω_{1}(S).

Proof. See [GL, 12-1,pg 158] or [GLS3, 7.8.1]. □

3. p-minimal groups

The first result of this section is elementary to prove but is never-the-less a fundamental result for our work on p-minimal subgroups.

Lemma 3.1. For G a finite group, p a prime and S a Sylow p-subgroup of G, G = ⟨P_{G}(S)⟩N_{G}(S).

Proof. This is also proved in [PPS, 1.3]. By induction on |G|. If S is contained in a unique maximal subgroup of G, then either S is normal in G or G ∈ P_{G}(S) and the lemma holds. Otherwise we can choose different maximal subgroups M_{1} and M_{2} of G which contain S. Since G = ⟨M_{1}, M_{2}⟩, applying induction to M_{1} and M_{2} gives G = ⟨P_{G}(S)⟩N_{G}(S). □

We now move on to study the structure of p-minimal groups – here is the main structural result for such groups.

Lemma 3.2. Suppose that p is a prime, P is p-minimal and S ∈ Syl_{p}(P). Let M be the (unique) maximal subgroup of P containing S and set F := ⋂ M^{P}. Then the following hold.

(a) Q_{P} ∈ Syl_{p}(F).

(b) F = Φ_{p}(P) and, in particular, if O_{p}(O^{p}(P)) = 1, then F is nilpotent.

(c) If N is a subnormal subgroup of P contained in M, then N ∩ S ≤ Q_{P}.

(d) If O^{p}(P) is p-closed, then P is a {t, p}-group for some prime t ≠ p.

(e) For N ⊴ P, either O^{p}(P) ≤ N or N ≤ F. In particular, P = O^{p′}(P).

(f) O^{p}(P)/(F ∩ O^{p}(P)) is a minimal normal subgroup of P/(F ∩ O^{p}(P)).

(g) If P is soluble, then O^{p}(P) is p-closed and P is a {t, p}-group for some prime t ≠ p.

(h) O^{p}(P) = [O^{p}(P), P].

Proof. Parts (a) and (b) follow immediately from Lemma 2.5(a) and (b). (c) From Lemma 2.2, N ≤ ⋂ M^{P} = F. Hence by (a), N ∩ S ≤ Q_{P}.

(d) For r a prime divisor of |O^{p}(P)| with r ≠ p, O^{p}(P) possesses a Sylow rsubgroup S_{r} such that O_{p}(O^{p}(P))S_{r} is S-invariant. If (d) were false, then we would get SS_{r} ≤ M for all prime divisors r of |O^{p}(P)|, which then forces M = P. Thus (d) holds.

(e) Clearly S ≤ SN ≤ P and so either SN = P or SN ≤ M. The former alternative yields O^{p}(P) ≤ N and the latter N ≤ ⋂ M^{P} = F, as required. If O^{p′}(P) ≤ F, then by (a) Q_{P} = S a contradiction. Thus O^{p′}(P) ≰ F and so P = SO^{p}(P) ≤ O^{p′}(P).

(f) Note that O^{p}(P) > F ∩ O^{p}(P), as O^{p}(P) ≤ F would give F S = P whereas F S ≤ M. Then (f) follows immediately from (e).

(g) Suppose that P is soluble. Hence, by (f), O^{p}(P)/(F ∩ O^{p}(P)) is a t-group, t a prime with t ≠ p. Since F is p-closed by (a), F ∩ O^{p}(P) is also p-closed and so O^{p}(P) is p-closed. Now (d) gives the result.

(h) As [O^{p}(P), P]⊴P, applying (e) gives either O^{p}(P) ≤ [O^{p}(P), P] or [O^{p}(P), P] ≤ F. So to prove (e) we must show that [O^{p}(P), P] ≤ F cannot occur. Suppose that [O^{p}(P), P] ≤ F. Then [O^{p}(P), M] ≤ F and, as O^{p}(P)M = P, we infer that M ⊴ P. Hence M = F and so, by (a), S = Q_{P}. But then S⊴P contrary to P being p-minimal. Thus [O^{p}(P), P] ≤ F cannot hold. □

It is of course important to know about the quotients of a p-minimal group P. For N a normal subgroup of P, having O^{p}(P) ≤ N is the kiss of death as far as P/N being a p-minimal group (because a p-group is not p-minimal). However, as we see next, this is the only bad case.

Lemma 3.3. Suppose that p is a prime, P is a p-minimal group and let N be a normal subgroup of P with O^{p}(P) ≰ N. Put \overline{P} := P/N. Then

(a) \overline{P} is a p-minimal group;

(b) P/Φ_{p}(P) ≅ \overline{P} /Φ_{p}(\overline{P}); and

(c) if P is narrow, then so is \overline{P}.

Proof. Let S ∈ Syl_{p}(P) and let M be the unique maximal subgroup of P containing S. By Lemma 3.2(e), as O^{p}(P) ≰ N, N ≤ M and hence \overline{M} is the unique maximal subgroup of \overline{P} containing \overline{S}. Suppose that \overline{S} ⊴ \overline{P}. Then, by the Frattini argument, P = N_{P} (S)N ≤ M, which is impossible. Therefore \overline{S} is not normal in \overline{P} and so \overline{P} is a p-minimal group. Clearly (b) follows from Lemma 3.2(b) and (c) follows from (a) and (b). □

Lemma 3.4. P_{G}(S) = P_{O^{p′}(G)}(S) and O^{p′}(G) = ⟨P_{G}(S)⟩.

Proof. Set X := O^{p′}(G). The inclusion P_{X}(S) ⊆ P_{G}(S) is obvious. Assume that P ∈ P_{G}(S). Then, by Lemma 3.2(e), P = O^{p′}(P) ≤ P ∩ X. Thus P_{G}(S) ⊆ P_{X}(S) and so P_{G}(S) = P_{X}(S). Since ⟨P_{X}(S)⟩ = ⟨P_{G}(S)⟩ is normalized by N_{G}(S), Lemma 3.1 implies that ⟨P_{X}(S)⟩ is a normal subgroup of G which contains S. Thus ⟨P_{X}(S)⟩ = O^{p′}(G). □

We next see that p-minimal subgroups are nicely located within central products.

Lemma 3.5. Assume that G has subgroups H_{1} and H_{2} such that [H_{1}, H_{2}] = 1 and G = H_{1}H_{2}. Then

P_{G}(S) = P_{H_{1}S}(S) ∪ P_{H_{2}S}(S).

Proof. For i = 1, 2, set S_{i} := S ∩ H_{i} and let P ∈ P_{G}(S). Then, by Lemma 3.2(e), P = ⟨S_{1}^{P} ⟩⟨S_{2}^{P} ⟩. Let M be the unique maximal subgroup of P containing S. We may assume ⟨S_{1}^{P} ⟩ ≰ M. So, using Lemma 3.2(e), O^{p}(P) ≤ ⟨S_{1}^{P} ⟩ ≤ H_{1} and P ≤ H_{1}S. Thus P ∈ P_{H_{1}S}(S). □

We continue this section by illuminating, in the next two lemmas, the structure of narrow groups with O^{p}(P/Q_{P}) ∈ L_{1}(p).

Lemma 3.6. Suppose that p is a prime, P is a narrow group and O^{p}(\overline{P}) ∈ L_{1}(p) where \overline{P} = P/Q_{P}. Then exactly one of the following holds.

(a) p^{a} ≥ 4 and \overline{P} is isomorphic to SL_{2}(p^{a}) or PSL_{2}(p^{a}) perhaps extended by field automorphisms of order a power of p.

(b) p^{a} ≥ 3 and \overline{P} is isomorphic to SU_{3}(p^{a}) or PSU_{3}(p^{a}) perhaps extended by field automorphisms of order a power of p.

(c) p = 2, a > 1 and \overline{P} is isomorphic to ^{2}B_{2}(2^{a}).

(d) p = 3, a > 1 and \overline{P} is isomorphic to ^{2}G_{2}(3^{a}) perhaps extended by field automorphisms of order a power of 3.

(e) p = 2, O^{2}(\overline{P}) ≅ 3 and \overline{P} ≅ SL_{2}(2) ≅ Sym(3).

(f) p = 3, O^{3}(\overline{P}) ≅ Q_{8} or 2^{2} and, respectively, \overline{P} ≅ SL_{2}(3) ≅ 2. Alt(4) or PSL_{2}(3) ≅ Alt(4).

(g) p = 2, O^{2}(\overline{P}) ≅ 3_{+}^{1+2} or 3^{2} and P/O^{2}(P)Q_{P} ≅ SDih(16), 8 or Q_{8}.

(h) p = 2, O^{2}(\overline{P}) ≅ 5 and \overline{P} ≅ Dih(10) or ^{2}B_{2}(2).

(i) p = 3 and \overline{P} is isomorphic to ^{2}G_{2}(3)′ ≅ SL_{2}(8) or ^{2}G_{2}(3) ≅ SL_{2}(8) : 3.

Proof. The structure of O^{p}(\overline{P}) follows directly from the structure of the rank one Lie type groups and then the rest follows from the structure of the automorphism groups of O^{p}(\overline{P}) and the fact that P = O^{p′}(P) is p-minimal. However, we note in detail the special case of \overline{P} ≅ SU_{3}(2) or PSU_{3}(2). In these cases we have that O^{2}(P) ≅ 3_{+}^{1+2} or 3^{2} respectively. Thus Out(O^{2}(\overline{P})) ≅ GL_{2}(3) and so X := P/O^{2}(P)Q_{P} is isomorphic to a subgroup of SDih(16) (which is isomorphic to a Sylow 2-subgroup of GL_{2}(3)). Since \overline{P} is 2-minimal, we have that X acts irreducibly on O^{2}(\overline{P})/Φ(O^{2}(\overline{P})) ≅ 3^{2}. Thus X ≅ SDih(16), 8, Q_{8}, Dih(8) or 4. However, the latter two groups act imprimitively on O^{2}(\overline{P})/Φ(O^{2}(\overline{P})) and so these two possibilities do not arise in (g). □

Lemma 3.7. Suppose that p is a prime, P is narrow and that O^{p}(\overline{P}) ∈ L_{1}(p) where \overline{P} := P/Q_{P}. Let \overline{S} ∈ Syl_{p}(\overline{P}), \overline{M} be the unique maximal subgroup of \overline{P} containing \overline{S} and \overline{R} := N_{O^{p}(\overline{P})}(\overline{S} ∩ O^{p}(\overline{P})).

(a) If \overline{P} has abelian Sylow p-subgroups, then either (i) \overline{P} ≅ PSL_{2}(p^{a}) or SL_{2}(p^{a}) for some a ≥ 1; (ii) p = 3 and \overline{P} ≅ ^{2}G_{2}(3)′ ≅ SL_{2}(8); (iii) p = 2 and \overline{P} ≅ ^{2}B_{2}(2) or Dih(10); or (iv) p = 2 and \overline{P} ≅ 3_{+}^{1+2} : 8 or 3^{2} : 8.

(b) If \overline{P} has cyclic Sylow p-subgroups, then \overline{P} ≅ PSL_{2}(p), SL_{2}(p), ^{2}G_{2}(3)′, ^{2}B_{2}(2), Dih(10), 3_{+}^{1+2} : 8, and 3^{2} : 8.

(c) If \overline{D} ≤ \overline{S} is normal in \overline{M} with \overline{D} abelian but not elementary abelian, then \overline{D} is cyclic and either \overline{P} is soluble, or p = 3 and \overline{P}′ ≅ ^{2}G_{2}(3)′.

(d) \overline{R} is soluble.

(e) Either C\overline{S}(\overline{R}) = 1 or \overline{P} is soluble.

(f) If \overline{R} normalizes a non-trivial cyclic subgroup of \overline{S}, then either |\overline{S}| ≤ p^{3}, or p = 2 and \overline{S} ≅ SDih(16).

(g) If P is soluble, then p ≤ 3.

Proof. Suppose first that \overline{P} is soluble. Then Lemma 3.6 implies that p = 2 or 3 and gives an explicit description of \overline{P}. With this information it is easy to verify all the claims in the lemma in this case. So assume that \overline{P} is not soluble. Let \overline{X} = O^{p}(\overline{P}) and \overline{T} = \overline{S} ∩ \overline{X}. If \overline{X} = ^{2}G_{2}(3)′ ≅ SL_{2}(8). Then \overline{P} has cyclic Sylow 3-subgroups if and only if \overline{P} = \overline{X}. In this case we have \overline{S} is cyclic of order 9 and is inverted by \overline{R}. Thus (a), (b), (c) and (d) hold in this case. Since \overline{S} is either cyclic or extraspecial, and \overline{R} inverts \overline{T}, C\overline{S}(\overline{R}) = 1. Hence (e) is true and, as |\overline{S}| ≤ 3^{3}, (f) holds as well. So we may assume that \overline{X} ≅ ^{2̸}G_{2}(3)′. Hence we now consider the cases when \overline{X} is not soluble and

\overline{X} ∈ \{SL_{2}(p^{a}), PSL_{2}(p^{a}), SU_{3}(p^{a}), PSU_{3}(p^{a}),^{2}B_{2}(2^{2a+1}),^{2}G_{2}(3^{2a+1}) | a ≥ 1\}.

In particular, we note that p divides |\overline{X}|. If \overline{P} has abelian Sylow p-subgroups, then so does \overline{X}. Thus the candidates for the groups \overline{P} with abelian Sylow p-subgroups are as listed in (a) and (b). Now for (a) we note that if \overline{P} ≠ \overline{X}, then, by Lemma 3.6 (a) and (i), \overline{P} includes field automorphisms of SL_{2}(p^{a}). Since the field automorphism of SL_{2}(p^{a}) of order p centralizes a subgroup of \overline{X} which has Sylow p-subgroups of order p^{a/p}, we have \overline{S} is abelian if and only if \overline{P} = \overline{X}. Hence (a) holds and (b) follows from (a).

Suppose that \overline{D} ≤ \overline{S} where \overline{D} is abelian and is normalized by \overline{M} = \overline{R}\overline{S}. We will prove (c) by showing that \overline{D} is elementary abelian. Since [\overline{R}, \overline{D}] ≤ \overline{R} ∩ \overline{D} ≤ \overline{T}, we infer that \overline{D} ≤ \overline{T} as in all the cases we are considering the field automorphisms of \overline{X} do not centralize \overline{R}/\overline{T}. If \overline{X} ≅ SL_{2}(p^{a}) or PSL_{2}(p^{a}), we get \overline{D} is elementary abelian. Now suppose that \overline{X} ≅ SU_{3}(p^{a}) or PSU_{3}(p^{a}). We use [H, II 10.12] to deduce the required facts about the structure of \overline{X}. If \overline{D} ≰ Z(\overline{T}), then, as \overline{R} acts irreducibly on \overline{T} /Z(\overline{T}), we have \overline{T} = \overline{D}Z(\overline{T}). Since Z(\overline{T}) = Φ(\overline{T}), this means \overline{D} = \overline{T} is not abelian which is impossible. Thus \overline{D} ≤ Z(\overline{T}) is elementary abelian. In the case when \overline{X} = ^{2}B_{2}(2^{2a+1}), a similar argument to the one above shows that \overline{D} is elementary abelian . Suppose that \overline{X} ≅ ^{2}G_{2}(3^{2a+1}) with a ≥ 1. We use [HB3, 13.2] or [W] to extract facts about the structure of \overline{T}. This time \overline{R} acts irreducibly on \overline{T} /Φ(\overline{T}), Φ(\overline{T})/Z(\overline{T}) and on Z(\overline{T}) and each of these groups has order 3^{2a+1}. Furthermore Φ(\overline{T}) is elementary abelian. If \overline{D} ≰ Φ(\overline{T}), then \overline{T} = \overline{D}Φ(\overline{T}) and so \overline{D} = \overline{T}, again a contradiction. Thus \overline{D} ≤ Φ(\overline{T}) and so \overline{D} is elementary abelian. This completes the proof of (c). In particular, we note from the above proof that the irreducible action of \overline{R} implies that \overline{D} ≥ Z(\overline{T}) for all abelian subgroups of \overline{S} normalized by \overline{R}. Thus if \overline{R} normalizes a non-trivial cyclic subgroup \overline{D} of \overline{S}, we must have Ω_{1}(\overline{D}) = Z(\overline{T}) has order p. It follows that |\overline{S}| = |\overline{T} | ≤ p^{3} and thus (f) holds.

We know from the above sources that \overline{R}/\overline{T} is a cyclic group. Thus \overline{R} is soluble and (d) holds.

Finally for part (e) we cite [DS, (5.1)(e)]. □

For a p-group S, p a prime, we use J(S) to denote the elementary abelian version of the Thompson subgroup of S. That is, J(S) := ⟨𝔄_{e}(S)⟩ where 𝔄_{e}(S) is defined to be the set of elementary abelian subgroups of S which have maximal rank.

Theorem 3.8. Suppose that p is a prime, P is a p-minimal group, C_{P} (Q_{P}) ≤ Q_{P}, S ∈ Syl_{p}(P) and P > ⟨C_{P} (Ω_{1}(Z(S))), N_{P} (J(S))⟩. Let V := Ω_{1}(Z(Q_{P})) and \overline{P} := P/C_{P} (V). Then there exist subgroups E_{1}, …, E_{n} of P such that

(a) S acts transitively on {E_{1}, …, E_{n}} by conjugation;

(b) [\overline{E_{i}}, \overline{E_{j}}] = 1 for i ≠ j;

(c) either \overline{E_{i}} ≅ SL_{2}(p^{n}) for some n ≥ 1 or p = 2 and \overline{E_{i}} ≅ Sym(2^{m} + 1) for some m ≥ 1; and

(d) V/C_{V} (E_{i}) is a corresponding natural \overline{E}_{i}-module and, for i ≠ j, V/C_{V} (E_{i}) is centralized by E_{j}.

Proof. This is the C^{∗∗}(G, T)-Theorem from [BHS]. □

Corollary 3.9. Suppose that P is as in Theorem 3.8 (and use the notation there). Let M be the unique maximal subgroup of P containing S, and set E := ⟨E_{i} | 1 ≤ i ≤ n⟩. Assume there is a non-trivial subgroup A of V which is centralized by M and such that A ∩ C_{V} (E) = 1. Then p = 2, \overline{E_{i}} ≅ Sym(2^{m} + 1) and AC_{V} (E)/C_{V} (E) = C_{V/C_{V} (E)}(S) has order 2.

Proof. As A is centralized by S and S is transitive on {E_{1}, …, E_{n}}, A ∩ C_{V} (E_{1}) ≤ A ∩ C_{V} (E) = 1. Therefore, |A| = |AC_{V} (E_{1})/C_{V} (E_{1})|. Now M ∩ E_{1} is the unique maximal subgroup of E_{1} containing S ∩ E_{1} ∈ Syl_{p}(E_{1}) and so C_{V/C_{V} (E_{1})}(M ∩ E_{1}) = 1 whenever V/C_{V} (E_{1}) is a natural module for \overline{E_{1}} ≅ SL_{2}(p^{n}) with p^{n} ≠ 2. Since AC_{V} (E_{1})/C_{V} (E_{1}) ≤ C_{V/C_{V} (E_{1})}(M ∩ E_{1}), we infer that p = 2, \overline{E_{1}} ≅ Sym(2^{m} + 1) and then that |AC_{V} (E_{1})/C_{V} (E_{1})| = 2 by direct calculation in the natural Sym(2^{m} + 1)- module. □

4. Finite groups with isolated subgroups

Throughout this section G is assumed to be a finite group, p a prime and S ∈ Syl_{p}(G). Our first three results give some elementary properties of isolated subgroups, Lemma 4.2(a) having already been mentioned in Section 1.

Lemma 4.1. Suppose that P ∈ P_{G}(S) is an A-isolated subgroup of G. Then P is an A-isolated subgroup of O^{p′}(G).

Proof. Set X := O^{p′}(G). Suppose that P ∈ P_{G}(S) is A-isolated. Then, by Lemma 3.4, P ∈ P_{X}(S), A is normal in L_{X}(P, S) and, of course, A ≰ Q_{P}. Thus P is A-isolated in X. □

Lemma 4.2. Let P ∈ P_{G}(S) be an isolated subgroup of G. Then

(a) P ≰ L_{G}(P, S) and L_{G}(P, S) ≠ G; and

(b) N_{G}(S) ≤ N_{G}(P).

Proof. Set L := L_{G}(P, S). If P ≤ L, then Q_{L} ≤ Q_{P}, contrary to P being an isolated subgroup. So P ≰ L, and (a) holds. For g ∈ N_{G}(S), P^{g} ∈ P_{G}(S) and, as N_{G}(S) ≤ L and P ≰ L, P^{g} ≰ L. Since P^{g} is also p-minimal, P = P^{g} and the lemma holds. □

The next lemma is the initial structural result about groups which possess an isolated subgroup.

Lemma 4.3. Assume that P ∈ P_{G}(S) is an A-isolated subgroup, and set L := L_{G}(P, S).

(a) Suppose that S ≤ H ≤ G and H ≰ L. Then P ≤ H, L_{H}(P, S) = L ∩ H is a maximal subgroup of H and P is an A-isolated subgroup of H.

(b) P ∩ L is the unique maximal subgroup of P containing S.

(c) L is a maximal subgroup of G.

Proof. From Lemma 3.1, H = ⟨P_{H}(S)⟩N_{H}(S) and so, as H ≰ L, P ∈ P_{H}(S). Thus P ≤ H. By Lemma 4.2(a) and the definition of L, P_{H}(S) \ {P } = P_{L∩H}(S). Since N_{H}(S) ≤ L ∩ H, using Lemma 3.1 gives

L_{H}(P, S) = ⟨P_{H}(S) \ {P }⟩N_{H}(S) = L ∩ H.

If L ∩ H < K ≤ H, then clearly K ≰ L and so, as S ≤ K, P ≤ K. Hence P_{K}(S) = P_{H}(S). Therefore, by Lemma 3.1,

K = ⟨P_{K}(S)⟩N_{K}(S) = ⟨P_{H}(S)⟩N_{H}(S) = H.

Thus L ∩ H is a maximal subgroup of H. Since A is a normal p-subgroup of L ∩ H = L_{H}(P, S), P is an A-isolated subgroup of H and this proves (a). Parts (b) and (c) follow from (a) by taking, respectively, H = P and H = G. □

We now obtain further restrictions on the structure of P.

Lemma 4.4. Suppose that P ∈ P_{G}(S) is an A-isolated subgroup of G and M is the unique maximal subgroup of P containing S. Put X := O^{p}(P) and F := ⋂ M^{P}. Then (a) XA = ⟨A^{P} ⟩;

(b) [XA, X ∩ F] ≤ O_{p}(X);

(c) X/O_{p}(X) is either a central product of quasisimple groups transitively permuted by S or there is a prime t such that X/O_{p}(X) is a t-group of class at most 2; and

(d) Φ_{p}(X) = X ∩ F.

Proof. Since F is p-closed by Lemma 3.2(a) and A ≰ Q_{P}, Lemma 3.2(e) implies that O^{p}(P) ≤ ⟨A^{P} ⟩. So (a) holds.

Set L := L_{G}(P, S). Then (F ∩ X)S ≤ M = P ∩ L by Lemma 4.3(b). Thus F ∩ X normalizes A. Hence, by Lemma 3.2(a),

[A, F ∩ X] ≤ Q_{L} ∩ F ∩ X ≤ Q_{P} ∩ X = O_{p}(X).

Hence (b) follows from (a).

From Lemma 3.2(f) X/(X ∩F) is a minimal normal subgroup of P/(X ∩F). Hence X/(X ∩ F) is isomorphic to a direct product of simple groups. If X/(X ∩ F) is an elementary abelian t-group (t a prime), then X is p-closed, whence X/O_{p}(X) is a t-group by Lemma 3.2(d). Since [X, X ∩ F] ≤ O_{p}(X), X/O_{p}(X) will also have class at most two. In the case when X/(X ∩ F) is a direct product of non abelian simple groups, [X, X ∩ F] ≤ O_{p}(X) and P being p-minimal force X/(X ∩ F) to be a central product of quasisimple groups transitively permuted by S. So (c) holds.

Finally, combining Lemmas 2.7 and 3.2(b) gives

Φ_{p}(X) = X ∩ Φ_{p}(P) = X ∩ F.
□

Lemma 4.5. Suppose that P ∈ P_{G}(S) is an A-isolated subgroup of G.

(a) If A ≤ T ⊴ S, then N_{G}(T) ≤ L_{G}(P, S).

(b) If S ≤ H ≤ G and Q_{H} ≤ Q_{P}, then P ≤ H.

Proof. Put L := L_{G}(P, S). Assume that A ≤ T ⊴ S and N_{G}(T) ≰ L. Since S ≤ N_{G}(T), applying Lemma 4.3(a) yields P ≤ N_{G}(T). But then

A ≤ T ≤ O_{p}(N_{G}(T)) ≤ Q_{P} ,

which is not the case. Thus N_{G}(T) ≤ L, and we have (a). Now assume that S ≤ H ≤ G with Q_{H} ≤ Q_{P}. If we have H ≤ L then

A ≤ Q_{L} ≤ Q_{H} ≤ Q_{P} ,

contrary to P being an A-isolated subgroup. Therefore H ≰ L and, again by Lemma 4.3(a), P ≤ H, so proving the lemma. □

As we shall see in Lemma 4.6 the property of having an isolated subgroup is preserved when taking certain quotients.

Lemma 4.6. Suppose that P ∈ P_{G}(S) and let N be a normal subgroup of G with O^{p}(P) ≰ N. Put L := L_{G}(P, S), Y := ⟨O^{p}(P)^{G}⟩ and \overline{G} := G/N. Then the following hold.

(a) P ∩ N ≤ Φ_{p}(P).

(b) G = LY.

(c) If P is an isolated subgroup of G, then N ≤ ⋂ L^{G} and

⋂ L^{G} = C_{G}(⟨Q_{L}^{G}⟩/Q_{G}) = C_{L}(Y Q_{G}/Q_{G}).

(d) If P is an A-isolated subgroup of G, then \overline{P} is an \overline{A}-isolated subgroup of \overline{G}.

Proof. Part (a) follows immediately from Lemmas 3.2(b) and 3.2(e), and part (b) follows from Lemma 3.1 as LY contains N_{G}(S) and all the p-minimal subgroups containing S.

Set Z := ⋂ L^{G}. Since O^{p}(P) ≰ N, P ≰ NS and so NS ≤ L by Lemma 4.3(a). Hence N ≤ Z. Put J := C_{G}(⟨Q_{L}^{G}⟩/Q_{G}) and K := C_{L}(Y Q_{G}/Q_{G}). Because P is an isolated subgroup of G, Lemma 4.4(a) gives O^{p}(P) ≤ ⟨Q_{L}^{P}⟩ and so Y ≤ ⟨Q_{L}^{G}⟩. Since Q_{L} ⊴ JS, P ≰ JS and so JS ≤ L. Therefore J ≤ K. Because Z ≤ L, Z normalizes Q_{L} and so [Q_{L}, Z] ≤ Q_{L} ∩ Z = Q_{Z} ≤ Q_{G}. So we have Z ≤ J ≤ K. Since Q_{G} ≤ K, Y normalizes K and so K ⊴ ⟨Y, L⟩ = G. So, as K ≤ L, K ≤ Z. Thus Z = J = K and (c) holds.

Since \overline{P} = P N/N ≅ P/(P ∩ N), Lemma 3.3(a) implies that \overline{P} ∈ P\overline{G}(\overline{S}). By (c) we also have \overline{P} ≰ \overline{L}. Now suppose that \overline{R} ∈ P\overline{G}(\overline{S}) with \overline{R} ≰ \overline{L}. Then, as N ≤ L by (c), R ≰ L. Since R ≥ NS ≥ S, we infer from Lemma 4.3(a) that P ≤ R and that, as \overline{R} is p-minimal, L ∩ R is the unique maximal subgroup of R containing NS. But P ≥ S and so P N ≥ NS. Since P ≰ L, we deduce that P N = R and that \overline{R} = \overline{P}. Hence, as N\overline{G}(\overline{S}) = \overline{N_{G}(S)} ≤ \overline{L}, \overline{L} = L\overline{G}(\overline{P} , \overline{S}). Finally assume that \overline{A} ≤ Q\overline{P}. Let K denote the inverse image in G of Q\overline{P}. Evidently ⟨A^{P} ⟩ ≤ K. By Lemma 4.4(a), O^{p}(P) ≤ ⟨A^{P} ⟩. So O^{p}(P) ≤ K and consequently, as K/N is a p-group, we obtain O^{p}(P) ≤ N whereas O^{p}(P) ≰ N. Therefore \overline{P} is \overline{A}-isolated in \overline{G} and so (d) holds. □

Lemma 4.7. Suppose that P ∈ P_{G}(S) is an A-isolated subgroup of G. If O^{p}(G) is p-closed, then G = L_{G}(P, S)P and ⟨A^{G}⟩ = ⟨A^{P} ⟩.

Proof. We set L := L_{G}(P, S), and prove that G = LP. Because of Lemma 4.6(d), without loss of generality we may assume that Q_{G} = 1. So O^{p}(G) is a p′-group, and hence O^{p}(P) is also a p′-group. Therefore, by Lemma 3.2(d), P is a {t, p}-group for some prime t ≠ p. For each prime r dividing |O^{p}(G)|, O^{p}(G) has an S-invariant Sylow r-subgroup, denoted S_{r}, and we may select S_{t} so that P ≤ SS_{t}. Therefore P = (P ∩ S_{t})S. If r is a prime divisor of |O^{p}(G)| with r ≠ t, then evidently P ≰ SS_{r} and consequently SS_{r} ≤ L by Lemma 4.3(a). Thus O^{p}(G) = (O^{p}(G) ∩ L)S_{t}. Since SS_{t} ≰ L, Lemma 4.3(a) implies that SS_{t} ∩ L = S(S_{t} ∩ L) is a maximal subgroup of SS_{t}. Now N_{S_{t}}(S_{t} ∩ L) is S-invariant and so SN_{S_{t}}(S_{t} ∩ L) is a subgroup of SS_{t}. Thus S_{t} ∩ L ⊴ S_{t} and S acts irreducibly on S_{t}/(S_{t} ∩ L). As S_{t} ∩ P ≰ S_{t} ∩ L, this gives S_{t} = (S_{t} ∩ L)(S_{t} ∩ P). Hence we have

G = O^{p}(G)S = (O^{p}(G) ∩ L)S_{t}S
= (O^{p}(G) ∩ L)(S_{t} ∩ L)(S_{t} ∩ P)S = LP.

Finally it follows that ⟨A^{G}⟩ = ⟨A^{LP} ⟩ = ⟨A^{P} ⟩ and this completes the proof of Lemma 4.7. □

Corollary 4.8. Suppose that P ∈ P_{G}(S) is an A-isolated subgroup of G. If Y := ⟨O^{p}(P)^{G}⟩ is p-closed, then Y = O^{p}(P).

Proof. We have that P is an A-isolated subgroup of H := SY. Since O^{p}(H) ≤ Y and Y is p-closed, O^{p}(H) is p-closed. Therefore as A is normalized by L := L_{G}(P, S), Lemmas 4.6(b) and 4.7 imply

⟨A^{P} ⟩ = ⟨A^{SY} ⟩ = ⟨A^{LY} ⟩ = ⟨A^{G}⟩.

By Lemma 4.4(a), O^{p}(P) is a characteristic subgroup of ⟨A^{P} ⟩ and so O^{p}(P) is normal in G. Thus Y = ⟨O^{p}(P)^{G}⟩ = O^{p}(P), as claimed. □

Lemma 4.9. Suppose that P ∈ P_{G}(S) is an A-isolated subgroup of G, and set L := L_{G}(P, S), Z := ⋂ L^{G} and Y := ⟨O^{p}(P)^{G}⟩.

(a) If N ⊴ G and N ≰ L, then Y ≤ N.

(b) If N is a proper characteristic subgroup of Y, then N ≤ Z and [Y, N] ≤ Q_{G}.

(c) Either Y = O^{p}(P) and there exists a prime t ≠ p such that Y Q_{G}/Q_{G} is a t-group of class at most 2, or Y Q_{G}/Q_{G} is a central product of quasisimple groups.

(d) Assume that Y ≠ O^{p}(P), and let K ≤ Y be such that KQ_{G}/Q_{G} is a component in Y Q_{G}/Q_{G}. Then Y Q_{G}/Q_{G} = ⟨K^{S}⟩Q_{G}/Q_{G} and K ∩ P ≰ L.

(e) Y ≤ F_{p}^{∗}(G).

Proof. (a) Since N ⊴ G and N ≰ L, S ≤ NS ≤ G and NS ≰ L. Thus P ≤ NS by Lemma 4.3(a) and so O^{p}(P) ≤ O^{p}(NS) ≤ N. Therefore Y = ⟨O^{p}(P)^{G}⟩ ≤ N.

(b) Because Y ≰ N, part (a) implies that N ≤ L. Hence N ≤ ⋂ L^{G} = Z. Thus, by Lemma 4.6(c), [N, Y] ≤ Q_{G}.

For the proof of parts (c) and (d), by Lemma 4.6(d), we may suppose without loss of generality that Q_{G} = 1.

(c) Suppose that E(Y) ≠ Y. Then [E(Y), Y] = 1 by (b) and so E(Y) = 1. Hence F^{∗}(Y) = F (Y). Since C_{Y} (F^{∗}(Y)) ≤ F^{∗}(Y) using (b) again we see that Y = F^{∗}(Y) = F (Y). So Y is nilpotent and in particular is p-closed. Applying Corollary 4.8 gives Y = O^{p}(P). Therefore either Y = E(Y) or Y = O^{p}(P). Finally, in the latter case, Lemma 4.4(c) implies that Y is a t-group of class at most 2 for some prime t ≠ p.

(d) Assume that Y ≠ O^{p}(P) and let K be a component of Y. Then K ⊴ Y by (c). Hence, as G = Y L,

⟨K^{G}⟩ = ⟨K^{Y L}⟩ = ⟨K^{L}⟩.

Since [Z, Y] = 1 by Lemma 4.6(c), we infer that K ≰ L. Therefore ⟨K^{S}⟩ ≥ O^{p}(P) by Lemma 4.3(a). If K_{1} is a component of Y which is not contained in ⟨K^{S}⟩, then using (b) we get

O^{p}(P) ≤ ⟨K^{S}⟩ ∩ ⟨K_{1}^{S}⟩ ≤ Z(Y) ≤ Z ≤ L,

which is impossible. Therefore Y = ⟨K^{S}⟩ and the first part of (d) holds.

Assume that K ∩P ≤ L, and argue for a contradiction. So K ∩P ≤ L∩P which is the unique maximal subgroup of P containing S. Since (K ∩ P) ⊴ ⊴P, Lemma 3.2(c) gives K ∩ S = (K ∩ P) ∩ S ≤ Q_{P}. Hence, as Y = ⟨K^{S}⟩, Y ∩ S = Y ∩ Q_{P} ⊴ P and, in particular, O^{p}(P) is p-closed. Set R := N_{Y} (Y ∩ S)S. Then O^{p}(R) is p-closed and R ≥ P. Therefore, R = (R ∩ L)P and ⟨A^{R}⟩ = ⟨A^{P} ⟩ ≤ P, using Lemma 4.7. Suppose that A ≤ N_{G}(K). Then

[A, K ∩ R] ≤ K ∩ ⟨A^{R}⟩ ≤ K ∩ P ≤ L

which implies that [A, K ∩R] normalizes A. It follows from Lemma 2.3 that [A, K ∩R] is a p-group. Consequently, as K ∩ R = N_{K}(K ∩ S) is p-closed,

[A, K ∩ R] ≤ Q_{P} ∩ K.

Thus [A, R] = ⟨[A, K ∩ R]^{S}⟩ ≤ Q_{P}. But then ⟨A^{R}⟩ = ⟨A^{P} ⟩ ≥ O^{p}(P) is a p-group, a contradiction. Hence we conclude that A ≰ N_{G}(K). So Q_{L} ≰ N_{G}(K). Since [K ∩ L, Q_{L}] is a p-group we infer that (K ∩ L)Z(Y)/Z(Y) is a p-group which yields that K ∩ L is nilpotent. Hence K ∩ S = O_{p}(K ∩ L) and therefore Y ∩ S = O_{p}(Y ∩ L). Thus

Y ∩ Q_{P} = Y ∩ S ⊴ ⟨P, Y ∩ L⟩ = Y,

and we conclude that Y is p-closed. Now a final application of Corollary 4.8 gives Y = O^{p}(Y) = O^{p}(P). By assumption Y ≠ O^{p}(P) and so this is the desired contradiction. Therefore K ∩ P ≰ L.

Part (e) follows from (c). □

Lemma 4.10. Suppose that P ∈ P_{G}(S) is A-isolated and narrow and set Y := ⟨O^{p}(P)^{G}⟩. Then either Y = O^{p}(P) or Y Q_{G}/Q_{G} is quasisimple.

Proof. As usual we set L := L_{G}(P, S). We assume that Y ≠ O^{p}(P) and aim to show that Y Q_{G}/Q_{G} is quasisimple. Because of Lemma 4.6(d) there is no loss in assuming that ⋂ L^{G} = 1. By Lemma 4.9(c), Z(Y)Q_{Y} < Y and so, by Lemma 4.9(b), Z(Y)Q_{Y} ≤ ⋂ L^{G} = 1. Thus we seek to show that Y is simple. Put F := Φ_{p}(P) and X := O^{p}(P). By Lemma 4.9(c) and 4.9(d), Y is semisimple and for any component K of Y, Y = ⟨K^{S}⟩ and K ∩ P ≰ L. In particular, if K is normalized by S, then Y = ⟨K^{S}⟩ = K is simple and we are done. Hence we assume that K is not normalized by S and look for a contradiction.

Suppose first that XF/F is a non-abelian simple group. Observe that (K∩P)F/F ∩ XF/F ≠ 1. For if (K ∩ P)F/F ∩ XF/F = 1, then (K ∩ P)F/F is a p-group and so, as a result, (K ∩ P)F ≤ F S ≤ L whereas K ∩ P ≰ L. Then, as X ≤ Y, (K ∩ P)F/F is normalized by XF/F and so (K ∩ P)F ≥ XF. Now selecting s ∈ S such that K^{s} ≠ K, we have

X = [X, X] ≤ [XF, XF] ≤ [(K ∩ P)F, (K^{s} ∩ P)F] ≤ [K, K^{s}]F = F,

which is a contradiction as X is not contained in F. Therefore, XF/F is an elementary abelian t-group for some prime t ≠ p and P is a {t, p}-group. Moreover, by Lemma 3.2(f), SF is a maximal subgroup of P. We have that O^{p}(K ∩ P) ≤ O^{p}(P) = X and so, as K ∩ P ≰ L, X = ⟨O^{p}(K ∩ P)^{S}⟩. Set D := O^{p}(K ∩ P) and note that, as Y is semisimple,

X = ×_{T ∈D^{S}}T.

By Lemma 4.4(d), Φ_{p}(X) = X ∩ F. On the other hand, by Lemma 2.6, Φ_{p}(X) = ×_{T ∈D^{S}}Φ_{p}(T) and so we conclude that

XF/F ≅ X/(X ∩ F) = ×_{T ∈D^{S}}T/Φ_{p}(T).

Since P is narrow, it must be that D^{S} = {D} and so X = D. Hence K is normalized by S, a contradiction. □

It is now a simple matter to deduce Theorems 1.3 and 1.6 and Corollary 1.4.

Proof of Theorem 1.3. Combining Corollary 4.8, Lemma 4.9(c), (d) and (e) together with Lemma 4.10 yields Theorem 1.3. For part (b) we note that p divides |Y/Q_{Y} | else Y would be p-closed and so, by part (a), Y = O^{p}(P). □

Proof of Corollary 1.4. This follows from Theorem 1.3 and Lemma 4.9(c). □

Proof of Theorem 1.6. By Lemma 4.9(e), ⟨P_{G}(S)⟩ ≤ F_{p}^{∗}(G) and hence, by Lemma 3.1, G = F_{p}^{∗}(G)N_{G}(S). The next lemma completes the proof of Theorem 1.6. □

Lemma 4.11. Suppose that Q_{G} = 1 and P_{1}, P_{2} ∈ P_{G}(S) are both isolated subgroups of G. Then either ⟨O^{p}(P_{1})^{G}⟩ = ⟨O^{p}(P_{2})^{G}⟩ or [⟨O^{p}(P_{1})^{G}⟩, ⟨O^{p}(P_{2})^{G}⟩] = 1.

Proof. For i = 1, 2, set Y_{i} = ⟨O^{p}(P_{i})^{G}⟩ and L_{i} = L_{G}(P_{i}, S). If Y_{1} and Y_{2} are products of components of G then Theorem 1.3 guarantees that either Y_{1} = Y_{2} or [Y_{1}, Y_{2}] = 1. So suppose that Y_{1} = O^{p}(P_{1}) and without loss that P_{1} ≠ P_{2}. Then P_{1} ≤ L_{2} and so

[Q_{L_{2}}, Y_{1}] = [Q_{L_{2}}, O^{p}(P_{1})] ≤ O_{p}(O^{p}(P_{1})) ≤ Q_{G} = 1.

Since O^{p}(P_{2}) ≤ ⟨Q_{L_{2}}^{P_{2}}⟩ by Lemma 4.4(a), we get Y_{2} ≤ ⟨Q_{L_{2}}^{G}⟩ and conclude that [Y_{1}, Y_{2}] = 1. □

In the companion paper [MPR] we will investigate specific simple groups with an eye to showing that they have or do not have an isolated narrow subgroup. The remaining results in this section will be applied to proper subgroups of such groups. And in all these results for P ∈ P_{G}(S) an isolated subgroup of G we set L := L_{G}(P, S) and Y := ⟨O^{p}(P)^{G}⟩.

Lemma 4.12. Suppose that P ∈ P_{G}(S) is an isolated subgroup in G. Then either P is soluble or O^{p}(P)Q_{P} /Q_{P} is a central product of quasisimple groups. If, additionally, P is narrow, then either P is soluble or O^{p}(P)Q_{P} /Q_{P} is quasisimple.

Proof. Let M := L ∩ P. Then, by Lemma 4.3(b), M is the unique maximal subgroup of P containing S. Put F := ⋂ M^{P}. Then F is normal in P and contained in L. Therefore [Q_{L}, F] ≤ Q_{L} ∩ F ≤ S ∩ F = Q_{P}. Thus F/Q_{P} is centralized by ⟨Q_{L}^{P}⟩ ≥ O^{p}(P) and so F Q_{P} /Q_{P} ≤ Z(O^{p}(P)Q_{P} /Q_{P}). Since P ∈ P_{G}(S), we have that either P is soluble or O^{p}(P)Q_{P} /Q_{P} is a central product of quasisimple groups. If P is narrow, the latter case implies that O^{p}(P)Q_{P} /Q_{P} is quasisimple. □

Lemma 4.13. Suppose that O^{p}(G) is quasisimple and set \overline{G} := G/Z(O^{p}(G)). If P ∈ P_{G}(S) is an A-isolated subgroup of G, then \overline{P} is an \overline{A}-isolated subgroup of \overline{G}.

Proof. Set Z := Z(O^{p}(G)) and L := L_{G}(P, S). If O^{p}(P) ≰ Z, then the result follows from Lemma 4.6(d). So assume O^{p}(P) ≤ Z. Then [O^{p}(P), O^{p}(L)] = 1 which, as G = ⟨L, O^{p}(P)⟩, means that O^{p}(L) ⊴ G. But O^{p}(L) ≤ O^{p}(G) and O^{p}(G) being quasisimple implies that O^{p}(L) ≤ Z or O^{p}(L) = O^{p}(G). Both possibilities are impossible as, in the first case, we get that G = SZ which is soluble and, in the second case, we get P = O^{p}(P)S ≤ O^{p}(G)S = O^{p}(L)S = L. □

Lemma 4.14. Suppose that Q_{G} = 1, F (G) = C_{G}(E(G)) and G operates transitively by conjugation on the set of components of G. If P ∈ P_{G}(S) is narrow and isolated in G, then E(G) = Y is quasisimple.

Proof. Assume that P is a narrow isolated subgroup of G. Suppose first that Y is soluble. Then Corollary 1.4(b) implies that Y = O^{p}(P). As G = LY, we have that Q_{L}Y ⊴ G and this subgroup is also soluble. Hence Q_{L}Y centralizes E(G). But then Q_{L}Y ≤ C_{G}(E(G)) which is nilpotent by assumption. Thus Q_{L}Y is nilpotent and this contradicts Q_{L} ≰ Q_{P}. Thus Y is not soluble and so Lemma 4.10 implies that Y is a component of G. Since G acts transitively on the set of components of G, we obtain Y = E(G) and this proves the lemma. □

Lemma 4.15. Suppose that Q_{G} = 1, N is a normal subgroup of G with N soluble and C_{G}(N) ≤ N. If P ∈ P_{G}(S) is isolated in G, then O^{p}(P) ⊴ G and P is soluble.

Proof. Suppose that N ≤ L. Then O^{p}(P) ≰ N and hence Y ≰ N. However, Lemma 4.6(c) implies that Y ≤ C_{G}(N) ≤ N, which is a contradiction. Therefore N ≰ L. Hence P ≤ NS by Lemma 4.3(a). But then O^{p}(P) ≤ N and, as N is soluble, Lemma 4.9(c) gives the result. □

Lemma 4.16 exploits the fact that, as observed in Lemma 4.2(b), an isolated pminimal subgroup is normalized by N_{G}(S).

Lemma 4.16. Suppose that P ∈ P_{G}(S) is an isolated subgroup of G. Assume that one of the following holds.

(a) N_{G}(S) acts irreducibly on S.

(b) N_{G}(S) = L.

(c) N_{G}(S) is contained in a unique maximal subgroup of G. Then P ⊴ G.

Proof. Suppose (a) holds. Since N_{G}(S) ≤ L, we get that Q_{L} = S and so (b) holds. Now suppose (b) holds. Then, by Lemma 4.3(c), N_{G}(S) is a maximal subgroup of G and hence (c) holds. So we may assume that (c) holds. Since N_{G}(S) ≤ L, we get that L is the unique maximal subgroup of G containing N_{G}(S). Because N_{G}(S)P ≤ N_{G}(P) and P ≰ L, we must have G = N_{G}(P) and so P ⊴ G. □

ISOLATED SUBGROUPS IN FINITE GROUPS 19

Lemma 4.17. Suppose that G is an almost simple group, and set X := F^{∗}(G). If P ∈ P_{G}(S) is A-isolated, then P ∈ P_{XS}(S) is A-isolated. Proof. Note that if H ≥ XS, then O_{p}(H) ≤ C_{G}(X) = C_{G}(F^{∗}(G)) = 1. Thus L_{G}(P, S) ≱ XS, whence P ≤ XS. □

5. The Centre of L_{G}(P, S)

The main purpose of this section is to prove Theorem 1.7. Theorem 5.1. Suppose that p is a prime and G a finite CK-group. If P ∈ P_{G}(S) with O^{p}(P/Q_{P}) ∈ L_{1}(p), then either O^{p}(P) ⊴ G or O_{p}(Z(L_{G}(P, S))) ≤ Q_{P}. Proof. Suppose the theorem is false and let G be a minimal counterexample. Again set L := L_{G}(P, S) and Y := ⟨O^{p}(P)^{G}⟩. So we have O^{p}(P) is not normal in G and D := O_{p}(Z(L)) ≰ Q_{P}. Thus P is D-isolated. The minimality of G gives 5.1.1. Q_{G} = 1.

Thus, by Theorem 1.3, 5.1.2. Y is a central product of quasisimple simple groups each of which has order divisible by p.

Since Q_{P} ≤ L, we have [Q_{P} , D] = 1 and so, by Lemma 4.4(a), 5.1.3. [Q_{P} , O^{p}(P)] = 1.

We next show that 5.1.4. P is soluble and O^{p}(P) is a p′-group.

Since P ∩ L centralizes DQ_{P} /Q_{P}, [DS, (5.1)e] implies that P is soluble. Hence (5.1.4) now follows from (5.1.3). 5.1.5. D ∩ Q_{P} = 1.

By (5.1.3) D ∩ Q_{P} is normalized by ⟨O^{p}(P), L⟩ = G. Thus (5.1.1) gives (5.1.5). Pick 1 ≠ a ∈ D of order p and set A := ⟨a⟩. Then, by (5.1.5), A ≰ Q_{P}. Since L is

a maximal subgroup of G, (5.1.1) implies 5.1.6. N_{G}(A) = L = C_{G}(A). 5.1.7. A is not weakly closed in G (with respect to S).

Suppose that A is weakly closed in G. If there exists g ∈ G such that a^{g} ∈ S, then A^{g} ≤ S whence A = A^{g}. So g ∈ N_{G}(A) and hence, by (5.1.6), a^{g} = a. Therefore a is weakly closed in G and, in particular, in AY (with respect to S ∩ AY). Now, as G is a CK-group, [GLS3, Remark 7.8.3] implies that [A, Y] is a p′-group. Hence by (5.1.2), [A, Y] ≤ Z(Y) and so the Three Subgroup Lemma and (5.1.2) once more give [A, Y] = 1. Hence, using (5.1.6), O^{p}(P) ≤ Y ≤ L, a contradiction and so (5.1.7) holds.

By (5.1.7) there exists a p-subgroup R of G with A ≤ R and N_{G}(R) ≰ N_{G}(A) = L. So there also exists a subgroup H of G with H ≰ L and A ≤ Q_{H}. Choose such an H with |H ∩ L|_{p} maximal and then with |H| minimal. Let T ∈ Syl_{p}(H ∩ L). Since A ⊴ L we may assume that T ≤ S. If T = S, then, as H ≰ L, P ≤ H and then A ≤ Q_{H} ≤ Q_{P}, a contradiction. Therefore,

5.1.8. S > T.

Next we show

5.1.9. Let U be a p-group with T ≤ U. Then U ≤ N_{G}(U) ≤ L. In particular, T ∈ Syl_{p}(H) and H ≠ N_{H}(T).

We have A ≤ T ≤ O_{p}(N_{G}(U ∩ L)) and either U ∩ L > T or, by (5.1.8),

|N_{G}(U ∩ L) ∩ L|_{p} > |U ∩ L| = |T |.

Hence |N_{G}(U ∩ L) ∩ L|_{p} > |T | = |H ∩ L|_{p} and so, by the maximal choice of H, N_{G}(U ∩ L) ≤ L. In particular, U = U ∩ L and (5.1.9) holds.

5.1.10. L ∩ H is the unique maximal subgroup of H containing T. In particular, H ∈ P_{H}(T).

Indeed let T ≤ M < H. Then A ≤ Q_{H} ≤ Q_{M}, |M ∩ L|_{p} = |T | = |H ∩ L|_{p} and thus the minimal choice of H implies M ≤ L. So M ≤ L ∩ H. By (5.1.9), T ⋬ H and so H ∈ P_{H}(T).

5.1.11. C_{G}(O^{p}(H)) ∩ Z(S) = 1.

Suppose that E := C_{G}(O^{p}(H)) ∩ Z(S) > 1 and put C := C_{G}(E). Then, by (5.1.1), G > C ≥ ⟨O^{p}(H), S⟩. Note that S ≤ C and C ≰ L. Therefore, by Lemma 4.3(a), P ≤ C. Also L ∩ C = L_{C}(P, S) and D ≤ Z(L ∩ C). Thus O_{p}(Z(L ∩ C)) ≰ Q_{P} and hence O^{p}(P) ⊴ C by induction. We now have C = (L ∩ C)O^{p}(P). So

⟨A^{O^{p}(H)}⟩ ≤ ⟨A^{C}⟩ = ⟨A^{(L∩C)O^{p}(P)}⟩ = ⟨A^{O^{p}(P)}⟩ ≤ O^{p}(P)A.

Since also ⟨A^{O^{p}(H)}⟩ ≤ Q_{H}, we get that ⟨A^{O^{p}(H)}⟩ ≤ O^{p}(P)A ∩ Q_{H} = A by (5.1.4). But then, by (5.1.6), O^{p}(H) ≤ N_{G}(A) = L, a contradiction.

By the maximal choice of H, T ∈ Syl_{p}(N_{G}(Q_{H})). Hence Z(S) ≤ T ≤ H. Since H ∈ P_{H}(T) by (5.1.10), Lemma 3.2(a) and (e) imply that either C_{H}(Q_{H}) ≥ O^{p}(H) or Q_{H} ∈ Syl_{p}(C_{H}(Q_{H})) and C_{H}(Q_{H}) is p-closed. In the former case, O^{p}(H) ≤ C_{G}(A) = L which is not the case. Thus, as Z(S) ≤ C_{H}(Q_{H}), we have

5.1.12. C_{H}(Q_{H}) ≤ Q_{H} and Z(S) ≤ Q_{H}.

5.1.13. p = 2 and Ω_{1}(Z(S)) = A has order 2.

Since A ≤ Z(T) and H is p-minimal by (5.1.10), we have C_{H}(Ω_{1}(Z(T))) ≤ H ∩ L. Also, since T < S, N_{S}(J(T)) > T and so the maximal choice of H shows that J(T) is not normal in H. It follows that H > H∩L ≥ ⟨N_{H}(J(T)), C_{H}(Ω_{1}(Z(T)))⟩. Therefore, since Ω_{1}(Z(S)) ≤ Ω_{1}(Z(Q_{H})) and H ∩ L centralizes A, we infer from Corollary 3.9, (5.1.11) and (5.1.12) that p = 2 and |A| ≤ |Ω_{1}(Z(S))| = 2.

5.1.14. Q_{P} = 1.

Since A ∩ Q_{P} = 1 by (5.1.5), (5.1.13) implies Q_{P} ∩ Ω_{1}(Z(S)) = 1. Hence also Q_{P} = 1.

5.1.15. Q_{H} is a fours group, T ≅ Dih(8), H/C_{H}(Q_{H}) ≅ Sym(3), S ≅ SDih(16) and O^{2}(P) ≅ 3^{2} or 3_{+}^{1+2}.

From (5.1.4), (5.1.14) and Lemma 3.6, |S| ≤ 2^{4}. Note that 1 < A < Q_{H} < T < S. Thus |S| = 2^{4}, |Q_{H}| = 4 and |T | = 8. Since A ⋬ H, Q_{H} ≅ 2^{2} and since Q_{H} ⋬ S, T = ⟨Q_{H}^{S}⟩ ≅ Dih(8). Together with Lemma 3.6 we infer that (5.1.15) holds.

5.1.16. G has a unique conjugacy class of involutions.

By (5.1.15) all involutions in S are conjugate in S to an involution in the fours group Q_{H}. Since H acts transitively by conjugation on Q_{H}^{♯}, (5.1.16) follows.

From (5.1.15) and (5.1.16) either G is simple or G′ is simple and T ∈ Syl_{2}(G′). Suppose first that G is simple. Then, since S ≅ SDih(16), G ≅ Mat_{11} or G ≅ PSL_{3}^{ϵ}(q) with q + ϵ ≡ 4 (mod 8) by [ABG]. In the first case there exists S ≤ K ≤ G with K ≅ Mat_{10}. So K ∈ P_{G}(S), but neither P = K nor K ≤ L, a contradiction. If ϵ = +, then L = C_{G}(A) ≅ GL_{2}(q) is not a maximal subgroup of G. So ϵ = −. In this case, as the Sylow 3-subgroups of PSL_{3}^{−}(q) are abelian, we have O^{2}(P) = 3^{2} and P ≅ 3^{2} : SDih(16). Since the Sylow 3-subgroups of G are not cyclic, we infer that 3 divides q + 1 and that |Z(SL_{3}^{−}(q))| = 3. It follows that the preimage J of O^{2}(P) in SL_{3}^{−}(q) is extraspecial and that N_{SL_{3}^{−}(q)}(J)/J ≅ Q_{8}. This contradicts S ≤ N_{G}(O^{2}(P)) and |S| = 2^{4}. Therefore G ≅ P̸SL_{3}^{−}(q). So G is not simple.

Hence G′ is simple and by [GW], G′ ≅ Alt(7) or G′ ≅ PSL_{2}(q) with q ≅ 7, 9 (mod 16). The former possibility is ruled out as a Sylow 2-subgroup of Sym(7) is not isomorphic to SDih(16). Hence G′ ≅ PSL_{2}(q). If q ≠ 3^{a}, then G has cyclic Sylow 3-subgroups and this contradicts O^{2}(P) ≤ G′. Thus q = 3^{a}. In this case P ∩ G′ normalizes a 3-subgroup and hence a Sylow 3-subgroup U of G′. But N_{G′}(U)/U is cyclic, T ≤ G′ and P ≥ T ≅ Dih(8). This contradiction completes the proof of Theorem 5.1. □

Our final observation turns out to be very useful in [MPR].

Corollary 5.2. Let p be a prime and G be a finite CK-group with Q_{G} = 1. Suppose that P ∈ P_{G}(S) is an isolated subgroup of G with O^{p}(P/Q_{P}) ∈ L_{1}(p) and that O^{p}(P) is not normal G. Then the following hold.

(a) If [Q_{P} , O^{p}(P)] = 1, then C_{Q_{L}}(O^{p}(L)) = 1.

(b) Q_{L} ∩ O^{p}(G) ≠ 1.

(c) L ∩ O^{p}(G) is a p-local subgroup of O^{p}(G).

Proof. Put D := O_{p}(Z(L)) where L := L_{G}(P, S). By Theorem 5.1, D ≤ Q_{P} and so [D, O^{p}(P)] = 1 and D ⊴⟨O^{p}(P), L⟩ = G. Thus D = 1 and, since C_{C_{Q_{L}}(O^{p}(L))}(S) ≤ D,

(a) holds.

For (b), suppose that Q_{L} ∩ O^{p}(G) = 1. Then

[Q_{L}, O_{p}(O^{p}(P))] ≤ Q_{L} ∩ O^{p}(P) ≤ Q_{L} ∩ O^{p}(G) = 1.

Therefore [⟨Q_{L}^{P}⟩, O_{p}(O^{p}(P))] = 1. Since O^{p}(P) ≤ ⟨Q_{L}^{P}⟩ by Lemma 4.4(a), we get [O^{p}(P), O_{p}(O^{p}(P))] = 1. Now

[Q_{P} , O^{p}(P)] ≤ Q_{P} ∩ O^{p}(P) = O_{p}(O^{p}(P)),

and so [Q_{P} , O^{p}(P), O^{p}(P)] = 1. Hence [Q_{P} , O^{p}(P)] = 1. Thus C_{Q_{L}}(O^{p}(L)) = 1 by part (a), Further

[Q_{L}, O^{p}(L)] ≤ Q_{L} ∩ O^{p}(L) ≤ Q_{L} ∩ O^{p}(G) = 1

and consequently Q_{L} ≤ C_{Q_{L}}(O^{p}(L)) = 1, a contradiction. Hence Q_{L} ∩ O^{p}(G) ≠ 1 and (b) holds. □

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U. Meierfrankenfeld, Department of Mathematics, Michigan State University, East Lansing, MI 48824-1027, USA

E-mail address: meier@math.msu.edu

C.W. Parker, School of Mathematics, University of Birmingham, Edgbaston, Birmingham B15 2TT, UK

E-mail address: C.W.Parker@bham.ac.uk

P.J. Rowley, School of Mathematics, The University of Manchester, PO Box 88, Manchester M60 1QD, UK

E-mail address: peter.j.rowley@manchester.ac.uk