aa r X i v : . [ m a t h . G R ] J a n STRONGLY p -EMBEDDED SUBGROUPS CHRIS PARKER AND GERNOT STROTH Introduction
In this paper we study finite groups which possess a strongly p -embedded subgroup for some prime p . Suppose that p is a prime. Asubgroup H of the finite group G is said to be strongly p -embedded in G if the following two conditions hold.(i) H < G and p divides | H | ; and(ii) if g ∈ G \ H , then p does not divide | H ∩ H g | .One of the most important properties of strongly p -embedded sub-groups is that N G ( X ) ≤ H for any non-trivial p -subgroup X of H .Groups with a strongly 2-embedded subgroup have been classifiedby Bender [2] and Suzuki [15] and their classification forms a pedestalupon which the classification of the finite simple groups stands. Theclassification of groups with a strongly 2-embedded subgroup statesthat if G is a finite group with a strongly 2-embedded subgroup, then O ′ ( G/O ( G )) is a simple rank 1 Lie type group defined in characteristic2 or G has quaternion or cyclic Sylow 2-subgroups. Of course rank1 Lie type groups in characteristic 2 are the building blocks of thegroups of Lie type in characteristic 2. When dealing with groups definedin characteristic p for odd p , strongly p -embedded subgroups play anequally influential role.Assume from here on that p is an odd prime and G is a finite group.If G has cyclic Sylow p -subgroup P , then N G (Ω ( P )) is strongly p -embedded in H . There is thus no prospect of listing all such groups.However, if m p ( G ) ≥
2, then the almost simple groups with a strongly p -embedded subgroup are known as a corollary to the classificationof the finite simple groups. They include the rank 1 Lie type groupsdefined in characteristic p , and there are only additional examples if p ≤
11. See Proposition 2.5 for a complete list. When we examine thelist of simple groups with a strongly p -embedded subgroup H , we seethat either H is a p -local subgroup or that F ∗ ( H ) ∼ = Alt( p ) × Alt( p )with p ≥ +8 (2) with p = 5. Thus the fact that H is strongly p -embedded severely restricts its structure. One major application of Date : October 22, 2018. Chris Parker and Gernot Stroth the results of this paper will be to the investigation of groups of lo-cal characteristic p orchestrated by Meierfrankenfeld, Stellmacher andStroth [13]. In the described application, we have a subgroup H of G generated by all the normalizers of non-trivial p -subgroups containedin a fixed Sylow p -subgroup of G . Generally speaking under such cir-cumstances H will be a Lie type group defined in characteristic p . Wewould like to assert that H = G . Assuming that this is not the case,work in progress by Salarian and Stroth will show that H is strongly p -embedded in G . The conclusion of our second theorem, Theorem 1.2, isthat under mild restrictions on the structure of H , H cannot in fact bestrongly p -embedded. Recall that a K -group is a group in which everycomposition factor is from the list of “known” simple groups. That is,every simple section is either a cyclic group of prime order, an alternat-ing group, a group of Lie type or one of the twenty six sporadic simplegroups. A group is K -proper if all its proper subgroups are K -groups.Obviously K -proper groups are the focus of attention in the proof ofthe classification of the finite simple groups. Theorem 1.1.
Suppose that G is a finite group, p is an odd primeand that H is a strongly p -embedded subgroup of G such that H ∩ K has even order for all non-trivial normal subgroups K of G . Assumethat F ∗ ( H ) = O p ( H ) and m p ( C H ( t )) ≥ for every involution t in H . If N G ( T ) is a K -group for all non-trivial -subgroups T of G , thenthere exists n ≥ such that either F ∗ ( G ) ∼ = PSU ( p n ) or, p = 3 and F ∗ ( G ) ∼ = G (3 n − ) . Theorem 1.2.
Suppose that G is a finite group, p is an odd prime andthat H is a strongly p -embedded subgroup of G . Assume that O p ′ ( H ) = 1 and that m p ( C H ( t )) ≥ for every involution t of H . If N G ( T ) is a K -group for all non-trivial -subgroups T of G and H is a K -group, then F ∗ ( H ) = O p ( H ) . We choose to compartmentalize our theorems in this way so thatwe can be precise about the K -group hypothesis we are invoking. Forexample in the expected application of this work to the project to clas-sify the finite groups of local characteristic p as explained above, H will be a finite quasisimple K -group and this will be known indepen-dently with just a K -group assumption on the normalizers of non-trivial p -subgroups of G . Our two theorems combine to give the following the-orem which is the main result of the paper. Theorem 1.3.
Suppose that G is a finite K -proper group, p is an oddprime and that H is a strongly p -embedded subgroup of G such that H ∩ K is of even order for all non-trivial normal subgroups K of G . Assume that O p ′ ( H ) = 1 and that m p ( C H ( t )) ≥ for every involution t of H . Then there exists n ≥ such that either F ∗ ( G ) ∼ = PSU ( p n ) or, p = 3 and F ∗ ( G ) ∼ = G (3 n − ) . The hypothesis in Theorems 1.1 and 1.3 that | H ∩ K | is even for allnormal subgroups K of G is there to guarantee that we cannot transferall the elements of order 2 from H and finish up with a configurationwhere the strongly p -embedded subgroup has odd order. In particular,we do not want our group G to have G > O ( G ) and H ∩ O ( G ) astrongly p -embedded subgroup of O ( G ) of odd order. For example,taking G ∼ = PGL (27), we see that G has a strongly 3-embedded sub-group of even order, but the normal subgroup of index 2 does not.Recall that our primary application of our theorem is to the projectto classify groups of local characteristic p . With this in mind, we havethe following corollary to Theorem 1.3. Corollary 1.4.
Suppose that p is an odd prime and H is a strongly p -embedded subgroup of G . Then F ∗ ( H ) is not a Lie type group definedin characteristic p of rank at least . In fact, if G and H are as in Corollary 1.4, then Theorem 1.3 can beused to show that if F ∗ ( H ) is a Lie type group in characteristic p ofrank 2, then the only possibility is that F ∗ ( H ) ∼ = PSL ( p ) or PSp ( p ).In work in preparation by the authors, it is shown that the latter casecannot happen. To eliminate the possibility that F ∗ ( H ) ∼ = PSL ( p ), itseems that different techniques need to be developed.We now proceed to describe the contents of the paper and at thesame time give an outline of the proof of our theorems. In Section 2, wepresent the preliminary results that we shall call upon throughout thepaper. We begin with our main characterization theorems. These arethe results which formally identify the groups in Theorem 1.1. Thus westate Aschbacher’s Classical Involution Theorem (Theorem 2.2) whichwill be used to identify the unitary groups and the theorem which isthe culmination of work by Walter, Bombieri, and Thompson (Theo-rem 2.3) to recognize the Ree groups. We then move on to results aboutalmost simple K -groups. The most prominent and important of theseis Proposition 2.5 which describes the structure of the almost simple K -groups of p -rank at least 2 which possess a strongly p -embeddedsubgroup. This is followed by Proposition 2.7 which catalogues thestructure of a strongly p -embedded subgroup in each of the groupslisted in Proposition 2.5. Next a series of corollaries to the propositionsare presented, perhaps the most useful being Corollary 2.8. We then Chris Parker and Gernot Stroth collect some results related to the Thompson Transfer Lemma and re-sults which limit the structure of 2-groups which admit certain typesof automorphisms, see, for example, Lemmas 2.23 and 2.26.Our investigation of groups with a strongly p -embedded subgroupsstarts in earnest in Section 3. We assume that G is a group and that H is a strongly p -embedded subgroup of G and prove some basic prop-erties about such configurations. In particular, easy, but key facts thatare used throughout the work are established such as if K is a subgroupof G which is not contained in H and H ∩ K has order divisible by p , then H ∩ K is strongly p -embedded in K and, of particular impor-tance, if m p ( H ∩ K ) ≥
2, then O p ′ ( K ) ≤ H and F ∗ ( C G ( t ) /O p ′ ( C G ( t )))is an almost simple group of p -rank at least 2 containing a strongly p -embedded subgroup. This is a simple consequence of the propertiesof strongly p -embedded subgroups and coprime action. A further con-sequence of these elementary lemmas is that F ∗ ( G ) is a non-abeliansimple group. Now the hypothesis in the main theorems that H ∩ K has even order immediately implies that H ∩ F ∗ ( G ) has even order.Thus for our investigations we may as well suppose that G = O ( G )and so, in Section 4, where we introduce the main hypothesis Hypothe-sis 4.1, this is included. In Section 4 we start our investigation of groups G with a strongly p -embedded subgroup H such that O p ′ ( H ) = 1, m p ( C H ( t )) ≥ t ∈ H , and G = O ( G ). We also im-pose our K -group hypothesis, that N G ( T ) is a K -group for all 2-groups T of G . We may also suppose that G is not a K -group. Thus H isnot strongly 2-embedded in G and so, as H has even order, there isan involution t ∈ H with C G ( t ) H . Since m p ( C H ( t )) ≥
2, we have O p ′ ( C H ( t )) ≤ H and C H ( t ) is strongly p -embedded in C G ( t ). This al-lows us to use the K -group hypothesis to get hold of the structure of C G ( t ) /O p ′ ( C G ( t )). One stark consequence of the information that weobtain in this section is that for any involution t ∈ H , C G ( t ) H andthe pertinent structural information about C G ( t ) is listed in Lemma 4.4.We close Section 4 with an immediate application of Lemma 4.4 whichstates that if E ( H ) = 1, then E ( H ) is quasisimple. This is Lemma 4.5.Sections 5, 6 and 7, study groups satisfying Hypothesis 4.1 whichhave Q = F ∗ ( H ) = O p ( H ). Since H is strongly p -embedded in G , wethen have H = N G ( Q ). Of course we may also assume that G doesnot have a classical involution. Our objective then in these three sec-tions is to prove that G ∼ = G (3 n − ) for some n ≥
2. To do this weseek to establish the hypothesis of Theorem 2.3. So we need to showthat G has Sylow 2-subgroups which have order 8 and an involution t such that C G ( t ) contains a normal subgroup isomorphic to PSL ( p a )for some a ≥
2. We start this work in Section 5. For t an involution in H , t acts on Q and, of course, either C Q ( t ) = 1 or C Q ( t ) = 1. Inthe former case Lemma 5.2 shows that there is a unique component L t contained in C G ( t ) which is normal and has order divisible by p . In thelatter case, we can only establish that C G ( t ) has a normal 2-component(Lemma 5.3) with the same properties. In both cases, as by hypoth-esis C G ( t ) is a K -group, we may use Proposition 2.5 to describe theisomorphism types of L t /O ( L t ). We then make a careful choice of t .We assume that | C Q ( t ) | is maximal from among all involutions t ∈ H with C Q ( t ) = 1 or that C Q ( t ) = 1. This choice leads to Lemma 5.7which states that a Sylow 2-subgroup of O p ′ ( C G ( t )) is cyclic and that O p ′ ( C G ( t )) has a normal 2-complement. In Section 6 our objective isto prove that L t /O p ′ ( L t ) ∼ = PSL ( p a ) for some a ≥
2. The exact con-clusion being posted in Theorem 6.1. This section deals with all theother possibilities for L t /O p ′ ( L t ) the most resilient case arising when H is strongly 3-embedded and L t /O ( L t ) is a covering group of PSL (4).Section 7 is the most technical of the paper. We begin with an invo-lution t in H such that L t /O p ′ ( L t ) ∼ = PSL ( p a ) and O p ′ ( C G ( t )) hascyclic Sylow 2-subgroups. Arguments about the fusion of involutionsin G centreing around the Thompson Transfer Lemma and the factthat G = O ( G ) eventually establish the structure of C G ( t ) required toinvoke Theorem 2.3.In Section 8, we start to investigate the situation when O p ′ ( H ) = 1and E = E ( H ) = 1. By Lemma 4.5 we already know that E is a qua-sisimple group. We further assume that E is a K -group. In Lemma 8.2we show that O p ( H ) = 1. We remark that our proof of this resultalready requires the K -group hypothesis on H as well as on N G ( T )for non-trivial 2-subgroups T of G . A very useful though easy conse-quence of Lemma 4.4 in conjunction with Propositions 2.5 and 2.7 isthat if C H ( t ) /O p ′ ( C G ( t )) is not soluble, then F ∗ ( C H ( t ) /O p ′ ( C G ( t ))) ∼ =Alt( p ) × Alt( p ) with p ≥ +8 (2) and p = 5 (see Lemma 8.3).Now for most candidates for E , we can select an involution t ∈ E suchthat C E ( t ) is non-soluble and, since we can show that p divides | C E ( t ) | (Lemma 8.7), we finish up only having to consider small Lie rank simplegroups defined over small fields in any significant detail. In these cases,when there is more than one class of involutions in E , it is usually pos-sible to choose a further involution s such that m p ( C H ( s )) <
2. Thusthe meat in this section is contained in dealing with the possibility that E might be a rank 1 Lie type group or that E ∼ = PSL (2 a ) for some a ≥
1. The arguments dispatching these possibilities are presented inLemmas 8.8 and 8.9. Finally we present proofs of Theorems 1.1, 1.2and 1.3 as well as Corollary 1.4 in Section 9.
Chris Parker and Gernot Stroth
Our group theoretical notation is mostly standard and can be foundin [9]. Particularly we use the following notation and conventions. Sup-pose that X is a finite group. We use O ( X ) to represent O ′ ( X ), thelargest normal subgroup of X of odd order. As usual Z ∗ ( X ) is thesubgroup of X which contains O ( X ) and satisfies Z ∗ ( X ) /O ( X ) = Z ( X/O ( X )). For Y ≤ X and x ∈ X , x Y denotes the set of Y -conjugatesof x . If a ∈ x Y , we write a ∼ Y x to indicate that a and x are conjugateby an element of Y . If p is a prime, P ∈ Syl p ( G ) and x ∈ Z ( P ) ,then x is called a p -central element. Our notation for the simple groupsis also standard or self-explanatory. The dihedral group of order n isdenoted by Dih( n ), The semi-dihedral group of order n is denoted bySDih( n ) and the generalized quaternion group of order 2 a is denotedby Q a and is usually referred to as a quaternion group of order 2 a . Theextraspecial 2-groups P of order 2 n are denoted 2 n + if P has anelementary abelian subgroup of order 2 n +1 and are otherwise denotedby 2 n − . If A and B are groups, then A ∗ B represents the centralproduct of A and B , A : B denotes a semidirect product of A and B with undefined action, and A.B denotes a non-split extension of A by B . Acknowledgement.
The first author thanks the Institut f¨ur Mathe-matik, Universit¨at Halle–Wittenberg and the second author for theirhospitality during visits to Halle to carry out research on this project.2.
Preliminaries
In this section we gather together an eclectic collection of preliminaryresults which we will invoke at various places throughout the paper. Webegin with the principal recognition theorems that we shall apply.
Definition 2.1.
Suppose that G is a group, t is an involution in G and C = C G ( t ) . Then t is a classical involution in G provided there existsa subnormal subgroup R of C which satisfies the following conditions. (i) R has non-abelian Sylow -subgroups and t is the unique invo-lution in R ; (ii) r G ∩ C ⊂ N G ( R ) for all -elements r ∈ R ; and (iii) [ R g , R ] ≤ O ′ ( C ) for all g ∈ C \ N G ( R ) . Notice that if G is a group, t is an involution in G and R is a nor-mal subgroup of C G ( t ) containing t which has quaternion Sylow 2-subgroups, then t is a classical involution. The following result is offundamental importance in our work. Theorem 2.2 (Aschbacher) . Suppose that G is a group and F ∗ ( G ) issimple. If G has a classical involution then G is a K -group.Proof. This is the main theorem in [1]. (cid:3)
Theorem 2.3 (Walter, Bombieri, Thompson) . Suppose that G is aperfect group, S ∈ Syl ( G ) and t ∈ G is an involution. If S is abelian oforder and C G ( t ) contains a normal subgroup isomorphic to PSL ( p a ) , p an odd prime, then G ∼ = J or G (3 a ) with a ≥ , a odd.Proof. From the main result in [17] we have that C F ∗ ( G ) ( t ) = h t i × PSL ( q ). The assertion then follows from [16] and [3]. (cid:3) We now move on to properties of K -groups with strongly p -embeddedsubgroups. Definition 2.4.
Suppose that p is a prime, X is a group and P ∈ Syl p ( X ) . Then the p -core of X is defined as follows: Γ P, ( X ) = h N X ( R ) | = R ≤ P i . We say that a proper subgroup M of X is strongly p -embedded ifand only if p divides | M | and Γ P, ( X ) ≤ M for some Sylow p -subgroup P of M . That this definition is equivalent to the one given in theintroduction is an easy application of Sylow’s Theorem (see [9, 17.10]for example). Proposition 2.5.
Suppose that p is a prime, X is a K -group and K = F ∗ ( X ) is simple. Let P ∈ Syl p ( X ) and Q = P ∩ K . If m p ( P ) ≥ and Γ P, ( X ) < X , then Γ Q, ( K ) < K and p and K are as follows. (i) p is arbitrary, a ≥ and K ∼ = PSL ( p a +1 ) , or PSU ( p a ) , B (2 a +1 ) ( p = 2) or G (3 a +1 ) ( p = 3) and X/K is a p ′ -group. (ii) p > , K ∼ = Alt(2 p ) and | X/K | ≤ . (iii) p = 3 , K ∼ = PSL (8) and X ∼ = PSL (8) : 3 . (iv) p = 3 , K ∼ = PSL (4) and X/K is a -group. (v) p = 3 and X = K ∼ = M . (vi) p = 5 and K ∼ = B (32) and X ∼ = B (32) : 5 . (vii) p = 5 , K ∼ = F (2) ′ and | X/K | ≤ . (viii) p = 5 , K ∼ = McL and | X/K | ≤ . (ix) p = 5 , K ∼ = Fi and | X/K | ≤ . (x) p = 11 and X = K ∼ = J .Proof. See [10, 7.6.1] or [8, 24.1]. (cid:3)
Chris Parker and Gernot Stroth
Notation 2.6.
We let E be the set of pairs ( K, p ) in the conclusion ofProposition 2.5 with p odd. Proposition 2.7.
Suppose that X , K , P and Q are as in Proposi-tion 2.5. Let H = Γ Q, ( K ) . Then the following hold. (i) If 2.5 (i) holds, then H = N K ( Q ) is a Borel subgroup of K . Inparticular, H = J Q where J is cyclic and operates irreduciblyon each of Q/ Φ( Q ) and Z ( Q ) . (ii) If 2.5(ii) holds, then H is isomorphic to the subgroup of Sym( p ) ≀ Sym(2) consisting of even permutations. In particular, if p > ,then O p ( H ) = 1 . (iii) If 2.5(iii) holds, then Q is cyclic of order , H = N K ( Q ) isdihedral of order and N X ( P ) ∼ = 3 − . . In particular, N K ( Q ) does not act irreducibly on P/ Φ( P ) . (iv) If 2.5(iv) holds, then Q is elementary abelian of order , H = N K ( Q ) , C K ( Q ) = Z ( Q ) , H/Q ∼ = Q and the complement to Q in H acts irreducibly on Q . (v) If 2.5(v) holds, then Q is elementary abelian of order , H = N K ( Q ) , C H ( Q ) = Q , H/Q ∼ = SDih(16) and a complement to H in K acts irreducibly on Q . (vi) If 2.5(vi) holds, then Q is cyclic of order , H = N K ( Q ) is aFrobenius group of order , and N X ( P ) ∼ = 5 − . . In partic-ular, a complement to Q in N K ( Q ) does not act irreducibly on P/ Φ( P ) . (vii) If 2.5(vii) holds, Q is elementary abelian of order , H = N K ( Q ) , Q = C H ( Q ) , H/Q is isomorphic to a central productof a cyclic group of order with SL (3) and a complement to Q in H acts irreducibly on Q . (viii) If 2.5(viii) holds, then Q is extraspecial of order , H = N K ( Q ) , C K ( Q ) = Z ( Q ) , H/Q isomorphic to a non-abelian ex-tension of a cyclic group of order by a cyclic group of order and a complement to Q in H acts irreducibly on Q/ Φ( Q ) . (ix) If 2.5(ix) holds, then Q has order and H ∼ = Aut(Ω +8 (2)) . Inparticular O ( H ) = 1 . (x) If 2.5(x) holds, then Q is extraspecial of order , H = N K ( Q ) , C K ( Q ) = Z ( Q ) , H/Q is isomorphic to the direct product of acyclic group of order and GL (3) . In particular, a comple-ment to Q in H acts irreducibly on Q/ Φ( Q ) .Proof. This is mostly [10, Theorem 7.6.2], the statement regarding theaction of complements to Q in H are easily deduced and are well-known. (cid:3) Corollary 2.8.
Assume that X , K , P , Q and H are as in Proposi-tion 2.7. If O p ( H ) = 1 , then N H ( P ) acts irreducibly on P/ Φ( P ) , unlesseither (i) p = 3 and X ∼ = PSL (8) : 3 ; or (ii) p = 5 and X ∼ = B (32) : 5 .Proof. The assertion follows directly from Proposition 2.7. (cid:3)
Corollary 2.9.
Assume that X , K , P , Q and H are as in Proposi-tion 2.7 with p odd. Then H contains an involution inverting P/ Φ( P ) unless cases (i) or (ii) of Corollary 2.8 holds or K ∼ = PSL ( p a ) with p a ≡ .Proof. As long as N H ( P ) /C H ( P ) P has a central involution the resultfollows with Corollary 2.8. By Proposition 2.7, if N H ( P ) /C H ( P ) doesnot have a central involution, we must be in case (i) of Proposition 2.5.Furthermore the Borel subgroup must be of odd order. Therefore K ∼ =PSL ( p a ) with p a ≡ (cid:3) Lemma 2.10.
Let F ∗ ( X ) = K ∼ = PSL (4) and t ∈ X be an involutionwhich induces an outer automorphism on L . Then C K ( t ) ∼ = 3 . Q ∼ =PSU (2) , PSL (2) or Alt(5) .Proof.
By [10, 2.5.12] K just posseses involutory outer automorphismswhich are field-, graph,- or graph-field automorphisms. A field au-tomorphism centralizes PSL (2), a graph automorphism centralizesPSL (4) ∼ = Alt(5) and a graph-field automorphism centralizes PSU (2). (cid:3) Corollary 2.11.
Assume that X , K , P , Q are as in Proposition 2.5.Set H = Γ P, ( X ) . If O p ′ ( H ) = 1 , then K ∼ = PSL (4) or Fi , | X/K | ≥ and O p ′ ( H ) = Z ( H ) has order .Proof. This is [8, 24.2]. (cid:3)
Lemma 2.12.
Suppose that p is an odd prime and ( K, p ) ∈ E . If twodivides the order of the Schur multiplier of K , then K is isomorphicto PSL ( p a ) for some a , Alt(2 p ) , PSL (4) , or Fi .Proof. For this we just consult [10, Theorem 6.1.4]. (cid:3)
Corollary 2.13.
Assume that p is an odd prime, ( K, p ) ∈ E and H = Γ Q, ( K ) where Q ∈ Syl p ( K ) . If b K is a non-trivial perfect cen-tral extension of K with Z ( K ) a -group and if b H has quaternion orcyclic Sylow -subgroups, then b K ∼ = SL ( p a ) for some a . Chris Parker and Gernot Stroth
Proof.
Suppose that K = PSL ( p a ). Then, by Lemma 2.12, K ∼ =Alt(2 p ), PSL (4) or Fi . In these cases, Proposition 2.7 shows that H does not have cyclic or dihedral Sylow 2-subgroups and so thesegroups do not arise. It is well-known that the Sylow 2-subgroups ofSL ( p a ) are quaternion. (cid:3) Lemma 2.14.
Suppose that p is an odd prime and X is group with F ∗ ( X ) = K where ( K, p ) ∈ E . Let P ∈ Syl p ( X ) and Q = P ∩ K . Set H = Γ Q, ( K ) . If | K : H | is odd, then K ∼ = M .Proof. This follows by inspection of the groups in E and the subgroupscorresponding to H . (cid:3) Lemma 2.15.
Suppose that p is an odd prime and X is group with F ∗ ( X ) = K where ( K, p ) ∈ E . Let P ∈ Syl p ( X ) and Q = P ∩ K . Set H = Γ Q, ( K ) . If | H | is odd, then K ∼ = PSL ( p f ) for some odd f and p ≡ .Proof. This follows by inspection of the groups in E and the subgroupscorresponding to H described in Proposition 2.7. (cid:3) Lemma 2.16.
Suppose that K is a simple K -group and that p is anodd prime. Then m p (Out( K )) ≤ and, if P ≤ Out( K ) has p -rank ,then P O p (Out( K )) .Proof. Since the alternating groups and sporadic simple groups haveouter automorphism groups which are 2-groups (perhaps trivial), wemay suppose that E is a Lie type group. Then the result can be deducedfrom [10, Theorem 2.5.12]. (cid:3) Lemma 2.17.
Assume that G is a group with Z ∗ ( G ) = O p ′ ( G ) . If t ∈ G is an involution, then t G ∩ C G ( t ) = { t } .Proof. Let R ∈ Syl ( C G ( t )) and assume that t G ∩ C G ( t ) = { t } . Then t G ∩ R = { t } and so t ∈ Z ( N G ( R )). Therefore, R ∈ Syl ( G ) andGlauberman’s Z ∗ -Theorem [6] implies that t ∈ Z ∗ ( G ) = O p ′ ( G ) whichis impossible. Hence the lemma holds. (cid:3) The next result is the famous Thompson Transfer Lemma.
Lemma 2.18.
Let G be a group, S ∈ Syl ( G ) , T E S with S = T A , A ∩ T = 1 , A cyclic and non-trivial. If G has no subgroup of index twoand u is the involution in A , then there is some g ∈ G with u g ∈ T and C S ( u g ) ∈ Syl ( C G ( u g )) . In particular | C S ( u ) | ≤ | C S ( u g ) | .Proof. This is [9, (15.16)]. (cid:3)
Lemma 2.19.
Suppose that G is a group and S ∈ Syl ( G ) . Then atleast one of the following hold. (i) G = O ( G ) ; (ii) Ω ( Z ( S )) ≤ Φ( S ) ; or (iii) N G ( S ) acts non-trivially on Ω ( Z ( S )) .Proof. Suppose that (i) and (ii) do not hold. Then there is a maximalsubgroup M of S and an involution t ∈ Ω ( Z ( S )) such that t M .By the Thompson Transfer Lemma 2.18, there is x ∈ G such t x ∈ M and C S ( t x ) ∈ Syl ( C G ( t x )). Since t is 2-central, it follows that t x ∈ Ω ( Z ( S )). Burnside’s Lemma [9, 16.2] now implies that t and t x areconjugate by an element of N G ( S ). Thus (iii) holds. (cid:3) Lemma 2.20.
Suppose that G is a group with G = O ( G ) , t ∈ G isan involution and S ∈ Syl ( C G ( t )) . Assume that S = h y i × S , t ∈ h y i and Z ( S ) is elementary abelian. Then h t i = h y i .Proof. Assume that y has order greater than 2. Then Z ( S ) = h y i × Z ( S ) and so t is the unique involution in Φ( Z ( S )). Therefore h t i is acharacteristic subgroup of S . Hence t ∈ Z ( N G ( S )) and so S ∈ Syl ( G ).As S = h y i × S and G = O ( G ), we may now apply the ThompsonTransfer Lemma 2.18 to see that there exists g ∈ G such that t g ∈ S and C S ( t g ) ∈ Syl ( C G ( t g )). It follows that t g ∈ Z ( S ). Hence t and t g areconjugate in N G ( S ) by Burnside’s Lemma. But we have already notedthat h t i is a characteristic subgroup of S and so we have h t i = h t g i which is a contradiction. Thus h t i = h y i as claimed. (cid:3) Lemma 2.21.
Suppose that G is a group, r is an odd prime and G/O ( G ) ∼ = Dih(2 r ) . Let R ∈ Syl r ( G ) and assume that C O ( G ) ( R ) = 1 .Then N G ( R ) ∼ = Dih(2 r ) and, for a an involution in N G ( R ) , all the in-volutions of G \ O ( G ) are conjugate to a and | C O ( G ) ( a ) | = | O ( G ) | .Proof. Set Q = O ( G ). Since N Q ( R ) = C Q ( R ) = 1, the Frattini ar-gument shows that N G ( R ) ∼ = Dih(2 r ). Let a, b ∈ N G ( R ) be involu-tions with a = b . Assume that c ∈ bQ is an involution. Then h a, c i is a dihedral group of order divisible by r . Thus h a, c i contains aconjugate, R x of R . Since h a, c i ∩ Q ≤ C Q ( R x ) = 1, we have that h a, c i ∼ = Dih(2 r ). Thus a and c are G -conjugate. It follows that everyinvolution in G \ Q is conjugate to a . Assume that a normalizes R x for some x ∈ Q . Then xax − ∈ N G ( R ) and xax − a ∈ Q ∩ N G ( R ) = 1.Thus x ∈ C Q ( a ). It follows that a normalizes exactly | C Q ( a ) | con-jugates of R . Let x ∈ bQ be an involution. Then h a, x i contains aconjugate of R and if y ∈ bQ is an involution with h a, x i = h a, y i ,then x = y . Thus a normalizes at least | Q : C Q ( b ) | conjugates of R .It follows that | Q | ≤ | C Q ( b ) || C Q ( a ) | = | C Q ( b ) | . On the other hand, as C Q ( b ) ∩ C Q ( a ) ≤ C Q ( R ) = 1, we have | C Q ( b ) | ≤ | Q | and this completesthe proof of the lemma. (cid:3) Chris Parker and Gernot Stroth
We recall that if A = Z k × Z k and G = Aut( A ). Then G ∼ = GL ( Z k )and that O ( G ) = (cid:26)(cid:18) a cd b (cid:19) | a, b ∈ Z ∗ k , c, d ∈ Z k (cid:27) , where Z ∗ k denotes the groups of units of Z k . In particular, we havethat | O ( G ) | = 2 k − and G/O ( G ) ∼ = SL (2). Lemma 2.22.
Suppose that k ≥ , A = Z k × Z k and G = Aut( A ) .If H ≤ G ∼ = GL ( Z k ) and H ∼ = Alt(4) , then O ( H ) = (cid:28)(cid:18) k − k − k − (cid:19) , (cid:18) k − k − (cid:19)(cid:29) . In particular, there is exactly one conjugacy class of subgroups of G isomorphic to Alt(4) .Proof.
We first note that t = (cid:18) − −
11 0 (cid:19) is an element of order 3 in G and, since | G | = 3, we may suppose that t ∈ H .We proceed by induction on k . Suppose that k = 2. Then O ( G ) hasorder 16 and is abelian. Since O ( H ) ≤ O ( G ) and Z ( G ) = 1, we thenhave O ( H ) = [ O ( G ) , t ] = (cid:28)(cid:18) (cid:19) , (cid:18) (cid:19)(cid:29) . Now suppose that k >
2. Let B = 2 A ∼ = Z k − × Z k − . Then C G ( B ) = (cid:26)(cid:18) k − a k − c k − d k − b (cid:19) | a, b, c, d ∈ { , } (cid:27) which hasorder 2 . Furthermore, G/C G ( B ) ∼ = Aut( B ). If H ∩ C G ( B ) = O ( H ),then O ( H ) = [ C G ( B ) , t ] = (cid:28)(cid:18) k − k − k − (cid:19) , (cid:18) k − k − (cid:19)(cid:29) . and we are done. Hence HC G ( B ) /C G ( B ) ∼ = Alt(4). By induction, wehave O ( H ) ≤ R = (cid:28)(cid:18) k − k − k − (cid:19) , (cid:18) k − k − (cid:19)(cid:29) C G ( B ).However, an easy calculation then shows that every element of R \ B has order 4 and this contradicts O ( H ) having exponent 2. This provesthe lemma. (cid:3) Lemma 2.23.
Let G be a group, T ≤ G be a -group and V ≤ Z ( T ) be a fours group. Assume that t, ρ ∈ N G ( V ) ∩ N G ( T ) are elements oforder and respectively with [ t, h ρ, V i ] = 1 . If [ V, ρ ] = C T ( t ) = V ,then T is isomorphic to one of the following groups. (i) An elementary abelian group of order 16. (ii)
A homocyclic group of rank , Z n × Z n , n ≥ . (iii) A Sylow -subgroup of PSL (4) . (iv) A Sylow -subgroup of PSU (4) . Proof.
Because C T ( t ) = V and [ t, ρ ] = 1, we have that C T ( ρ ) = 1. If T = V , then (ii) holds with n = 1. So assume that T > V and let W be the preimage of C T/V ( t ). Then, since C T ( ρ ) = 1 and C T ( t ) = V , wehave that | W : V | = 4. In particular, W/V has rank 2 which meansthat | W ′ | ≤
2. Thus, as ρ acts on W ′ , we have that W is abelian andso W is either elementary abelian or homocyclic Z × Z Assume first that W is elementary abelian. Then all involutions in W t are conjugate in h W, t i . As C T ( t ) = V , this means that T = W ,which is (i).Assume next that W is homocyclic. Then, as W/V = C T/V ( t ) is ele-mentary abelian of order 4, T /V satisfies the hypothesis of the lemma.Hence we may assume by induction
T /V is isomorphic to one of thegroups in (i) - (iv).Assume that
T /V is elementary abelian of order 16. If V = Ω ( T ),then, as T /V can be considered as a direct some of two irreducible2-dimensional modules for h ρ i , we have that T = W W , where W is also homocyclic of order 16 and is normalized by ρ . Furthermore, C W ( W ) = C W ( T ) = V . Using Lemma 2.22 shows that the actionof W on W is uniquely determined. As W = [ t, W ], we see that T is uniquely determined and so we have that T is isomorphic to aSylow 2-subgroup of PSU (4). If Ω ( T ) = V , then we have T = W W ,where W is elementary abelian of order 4 and is normalized by ρ . If C W ( W ) = 1, then the action of W on W is uniquely determined asabove. We therefore have that V W and ( V W ) t are elementary abelianof order 16 and the action of W on ( V W ) t is uniquely determined. Itfollows that T is isomorphic to a Sylow 2-subgroup of PSL (4). Hencewe have C W ( W ) = W and consequently T is abelian. Then Ω ( T ) iselementary abelian. But this forces W = Ω ( T ) which is a contradictionas W ∼ = Z × Z .Assume next that T /V is isomorphic to a Sylow 2-subgroup of PSU (4)or PSL (4). Then there is a homocyclic group W ∼ = Z × Z on which h T /W , ρ i acts as Alt(4). Using Lemma 2.22 we get [ W , T ] = V andthis contradicts Z ( T /V ) =
W/V . Hence we finally have
T /V is ho-mocyclic. As T ′ = h [ x, y ] i where T = h x, y i , the fact that C T ′ ( ρ ) = 1implies that T is abelian. As Ω ( T ) = V and C T ( ρ ) = 1, we have that T is homocyclic. (cid:3) Corollary 2.24.
Assume the hypothesis of Lemma 2.23. If the coset
T t contains more than one T -conjugacy class of involutions, then T ishomocyclic and T t contains exactly four T -conjugacy classes of invo-lutions. Chris Parker and Gernot Stroth
Proof.
We consider each of the possibilities for T given in Lemma 2.23.If T is elementary abelian, then C T ( t ) = V has index 4 in T and so T t contains exactly 4 involutions and they are all T -conjugate. Assumenow that T is non-abelian. Let W = C T/V ( t ). We note that | t T | = | T /V | = | T | / x ∈ T and ( xt ) = 1, then t inverts x . As T is non-abelian, then T has exponent 4 and so x ∈ V . We infer that,if x ∈ T is inverted by t , then x ∈ W . It follows that T t containsat most | W | involutions. Since | W | = | T /V | , we have that T t containsexactly one T -conjugacy class of involutions in this case. Finally, if T ishomocyclic, we have that t inverts every element of T . Thus T t consistsof involutions and it follows that
T t contains exactly four T -conjugacyclasses of involutions. (cid:3) Lemma 2.25.
Let t ∈ G be an involution and T be a Sylow -subgroupof C G ( t ) such that T ∼ = h t i × D , where D is dihedral of order or .Assume that there is a fours group V ≤ D , such that C G ( t ) containsa 3-element ρ which acts non-trivially on V . Then there is a Sylow -subgroup R of C G ( V ) which is normalized by h T, ρ i such that R = U h t i where U is isomorphic to one of the groups listed in the conclusion ofLemma 2.23.Proof. By considering a minimal counter example to the lemma, wemay assume that G = N G ( V ) = C G ( V ) T h ρ i . Set E = h V, t i . As G isa counter example to the lemma, we have V Syl ( C G ( V )). Set E = h V, t i . Then E is elementary abelian of order 2 . As T ∈ Syl ( C G ( t )),we have E ∈ Syl ( C G ( E )). Thus C G ( E ) = E × O ′ ( C G ( E )). Assumethat S ∈ Syl ( C G ( V )) is normalized by T . Then W = N S ( E ) > E andsince W = N S ( E ) centralizes V , we deduce that h ρ i W C G ( E ) /C G ( E ) ∼ =Alt(4). In particular, we have | W | = 2 and W ∈ Syl ( N G ( E )). If W ρ = W , then there exists c ∈ O ′ ( C G ( E )) such that ρc normalizes W . Let ρ ∗ be the 3-part of ρc . Then ρ ∗ acts non-trivially on V andcentralizes t . Thus we may as well suppose that ρ normalizes W . Set G = G/V . Then h T i W ∈ Syl ( C N G ( V ) ( t )) and ρ is a 3-element whichacts non-trivially on [ W, ρ ] and centralizes t . Therefore, by induction,there is a subgroup U ≥ V such that U h t i ∈ Syl ( C G ([ W ρ ])) where U is one of the groups listed in the conclusion of Lemma 2.23 and h ρ, T i normalizes U . By considering U h T, ρ i , we have that U is also listed inthe conclusion of Lemma 2.23. We may assume that U h t i ≤ S and, as G is a counter example to the lemma, S = U h t i . Since t is not centralizedby an abelian group of order 16 and U = V , we see that t G ∩ U = ∅ . If U is either elementary abelian or is isomorphic to a Sylow 2-subgroup of PSL (4) or PSU (4), then, by Corollary 2.24, every in-volution in U h t i \ U is conjugate to t by an element of U . Setting X = N S ( U h t i ), we then have U h t i < X = C X ( t ) U h t i = U h t i , a con-tradiction. Thus U is homocyclic of rank 2. Since | U | ≥
16, we havethat U is a characteristic subgroup of U h t i . It follows that X normal-izes Ω ( U ) = [ W ρ ]. Since X centralizes V , we have [[ W ρ ] , X ] ≤ V and thus X ≤ C G ([ W, ρ ]) and this is our final contradiction as U h t i ∈ Syl ( C G ([ W, ρ ])). This concludes the verification of Lemma 2.25. (cid:3)
Lemma 2.26.
Assume that p is an odd prime, W is a -group and that E ≤ Aut( W ) is a non-cyclic abelian p -subgroup. If C W ( e ) contains atmost one involution for each e ∈ E , then | E | = 9 , W ∼ = 2 , − or and C W ( E ) = Z ( W ) .Proof. Let W be a minimal counter example to the claim. Since E isabelian and non-cyclic, W = h C W ( e ) | e ∈ E i . Choose e such that C W ( e ) = C W ( E ). Then, as C W ( e ) contains exactly one involution, C W ( e ) is cyclic or generalized quaternion. Because E normalizes anddoes not centralize C W ( e ), it follows that C W ( e ) is a quaternion groupof order 8 and p = 3. In particular, we have that C W ( E ) = Z ( C W ( e ))has order 2 and if C W ( f ) > C W ( E ) for some f ∈ E , then C W ( f ) is aquaternion group of order 8. We also have that W is non-abelian and Z ( W ) = C W ( E ). We now choose e ∈ E such that C W ( e ) is containedin the second centre of W . Then C W ( e ) is normal in W . Since theautomorphism group of C W ( e ) is isomorphic to Sym(4), we get that W = C W ( e ) C W ( C W ( e )). Set W = C W ( C W ( e )). Then W C W ( e ),for otherwise e would be the trivial automorphism of W . Thus W is a non-trivial 2-group. Suppose that W ≤ C W ( f ) for some f ∈ E .Then, as C W ( f ) is quaternion of order 8 and W is normalized by E , weinfer that C W ( f ) = W and W = C W ( e ) C W ( f ) ∼ = 2 . Furthermore,we have that E is elementary abelian of order 9. Thus we may supposethat every element of E induces a non-trivial automorphism of W . As W < W , we have that | E | = 9 and that W ∼ = 2 , 2 − or 2 by induction. The first two cases immediately deliver W ∼ = 2 − or W ∼ = 2 and the theorem then holds. So suppose that W ∼ = 2 .Then, for each f ∈ E , C W ( f ) is a quaternion group of order 8. But e ∈ E and C W ( e ) ∩ W = Z ( W ) and so we have a contradiction. Thiscompletes the proof of the lemma. (cid:3) Lemma 2.27.
Suppose that X ∼ = Aut(PSL (2 a )) with a ≥ , S ∈ Syl ( X ) and T = S ∩ F ∗ ( X ) . If U ≤ S and U ∼ = T , then U = T . Chris Parker and Gernot Stroth
Proof.
Set K = F ∗ ( X ). Suppose that U ≤ S with U ∼ = T and U = T .Let F and F be the two elementary abelian subgroups of T of order2 a and note that every involution of T is contained in F ∪ F andthat F ∩ F = Z ( T ). Since the group of diagonal outer automorphismsof K has order 3, Out( K ) has abelian Sylow 2-subgroups. Therefore Z ( U ) = Φ( U ) = U ′ ≤ T and U K/K is elementary abelian of order atmost 4. Let E and E be the elementary abelian subgroups of U oforder 2 a . As U = E E , we may suppose that E T . As | U K/K | ≤ | E ∩ T | ≥ a − . Assume that ( E ∩ T ) Z ( T ) = Z ( T ). Then we maysuppose that ( E ∩ T ) Z ( T ) ≤ F . But then F u ≥ ( E ∩ T ) Z ( T ) > Z ( T )for all u ∈ U . Hence U normalizes F and therefore U K/K consists offield outer automorphisms. Hence | E ∩ T | ≥ a − . Now let e ∈ E \ T .Then e centralizes either ( E ∩ T ) Z ( T ) /Z ( T ) = F /Z ( T ) or Z ( T ) ≤ E and e centralizes Z ( T ). In either case e cannot be a field automorphismof K and we have a contradiction. Thus E ∩ T = Z ( T ), | U T /T | = 4and a = 2. But then Z ( U ) = E ∩ T = Z ( T ) and we see that U centralizes Z ( T ). This means that U K/K does not contain non-trivialfield outer automorphisms. It follows that | U K/K | ≤ (cid:3) Basic properties of groups with a strongly p -embeddedsubgroups In this section we reveal the basic structural properties of groupswith strongly p -embedded subgroups. The first lemma is one which wealready alluded to in Section 2 and states the equivalence between thetwo definitions of strongly p -embedded subgroups which we have given. Lemma 3.1.
Assume that G is a group, p is a prime, H ≤ G and S ∈ Syl p ( H ) . Then Γ S, ( G ) ≤ H if and only if p divides | H | and | H ∩ H g | is not divisible by p for all g ∈ G \ H .Proof. See [9, Lemma 17.11]. (cid:3)
The fundamental lemmas which gets us started in the proof of The-orem 1.3 are as follows.
Lemma 3.2.
Suppose that G is a group, p is a prime, H is a strongly p -embedded subgroup of G and K ≤ G such that H ∩ K has orderdivisible by p . Then the following statements hold. (i) Syl p ( H ) ⊆ Syl p ( G ) ; (ii) if K H and , then H ∩ K is a strongly p -embedded subgroupof K ; (iii) Syl p ( H ∩ K ) ⊆ Syl p ( K ) ; (iv) T H G ≤ O p ′ ( G ) ; and (v) if m p ( H ) ≥ , then O p ′ ( G ) = T H G .Proof. Let S ∈ Syl p ( H ). Then, as H is strongly p -embedded in G , N G ( S ) ≤ H . Thus S ∈ Syl p ( G ) and (i) holds.Suppose that S is chosen so that S ∩ K ∈ Syl p ( H ∩ K ). Let T bea non-trivial subgroup of S ∩ K . Then, as H is strongly p -embedded, N K ( T ) ≤ N G ( T ) ≤ H , thus N K ( T ) ≤ H ∩ K and so we have that H ∩ K is strongly p -embedded in K . Thus (ii) holds.Part (iii) follows from (i) and (ii).Since, by Lemma 3.1, p does not divide | H ∩ H g | , we have that T H G ≤ O p ′ ( G ). So (iv) holds.Finally, assume that m p ( H ) ≥ A ≤ S is elemen-tary abelian of order p . Since H is strongly p -embedded, for a ∈ A ,we have C O p ′ ( G ) ( a ) ≤ C G ( a ) ≤ H . Therefore, from coprime action, O p ′ ( G ) = h C O p ′ ( G ) ( a ) | a ∈ A i ≤ h C G ( a ) | a ∈ A i ≤ H. Thus (v) follows from (iv). (cid:3)
Lemma 3.3.
Suppose that G is a group, p is a prime and H is astrongly p -embedded subgroup of G . Set G = G/O p ′ ( G ) and assumefurther that H = G . Then (i) H is strongly p -embedded in G . (ii) F ∗ ( G ) is a non-abelian simple group; and (iii) if G is a K -group and m p ( G ) ≥ , then ( F ∗ ( G ) , p ) ∈ E .Proof. Let S ∈ Syl p ( H ). Then S ∈ Syl p ( H ). Choose X a non-trivialsubgroup of S . Then there exists T ≤ S such that X = T . Therefore,as H is strongly p -embedded in G , we have HO p ′ ( G ) ≥ N G ( T ) O p ′ ( G ) = N G ( X )by the Frattini Argument. Hence H ≥ N G ( X ). Since H = G , we con-clude that H is strongly p -embedded in G . Hence (i) holds.Since H is strongly p -embedded in G and O p ′ ( G ) = 1, we have F ( G ) = 1. Assume that E ( G ) is not simple. Let K be an arbitrarycomponent of G and let K be a component with K = K . Then,as O p ′ ( G ) = 1, p divides | K | . Let T = S ∩ K . Then T ∈ Syl p ( K ).Thus, as H is strongly p -embedded in G and [ K , K ] = 1, K ≤ N G ( T ) ≤ H . Since K was chosen arbitrarily, we have that E ( G ) ≤ H .Then Lemma 3.2 (iv) implies that E ( G ) = 1 which is impossible. Thus F ∗ ( G ) = F ( G ) E ( G ) = E ( G ) is a simple group and (ii) holdsFinally (iii), follows directly from the definition of E and Proposi-tion 2.5. Chris Parker and Gernot Stroth (cid:3)
Corollary 3.4. If m p ( H ) ≥ and O p ′ ( H ) = 1 , then F ∗ ( G ) is a non-abelian simple group.Proof. We have m p ( H ) = m p ( G ) ≥
2. Therefore, Lemma 3.2(iv) im-plies O p ′ ( G ) = T H G ≤ O p ′ ( H ) = 1. Then Lemma 3.3(ii) implies that F ∗ ( G ) is a non-abelian simple group as claimed. (cid:3) Lemma 3.5.
Suppose that p is an odd prime and H is strongly p -embedded of G . Assume that for all involutions t ∈ H , p divides | C H ( t ) | .Then for all involutions t ∈ H , t G ∩ H = t H .Proof. Suppose that t ∈ H is an involution. Obviously t H ⊆ t G ∩ H . As-sume that g ∈ G and t g ∈ H . Let X ∈ Syl p ( C H ( t )), Y ∈ Syl p ( C H ( t g ))and note that by assumption X and Y are non-trivial. As H is strongly p -embedded in G , X and Y are Sylow p -subgroups of C G ( t ) and C G ( t g )respectively. Thus X g , Y ∈ Syl p ( C G ( t g )) and so there exists c ∈ C G ( t g )such that Y = X gc . Therefore Y ≤ H ∩ H gc and, since H is strongly p -embedded in G , we get gc ∈ H . But then t g = t gc ∈ t H and we aredone. (cid:3) Lemma 3.6.
Suppose that p is an odd prime and H is strongly p -embedded of G . Assume that t ∈ H is an involution, t G ∩ H = t H and F is a subgroup of H which contains t . If N G ( F ) H , then C G ( t ) H .Proof. Aiming for a contradiction, we assume that C G ( t ) = C H ( t ). Let k ∈ N G ( F ) \ H . Then t k ∈ F ≤ H and, as t G ∩ H = t H , there exists h ∈ H such that t kh = t . But then kh ∈ C G ( t ) = C H ( t ) ≤ H . Hence k ∈ H , which is a contradiction. Thus C G ( t ) H as claimed. (cid:3) Involutions in H In this section we initiate the investigation of groups satisfying thehypotheses of Theorems 1.1 and 1.2. Specifically throughout the re-mainder of this article we assume that the following hypothesis holds.
Hypothesis 4.1. p is an odd prime, H is a strongly p -embedded sub-group in G and H ∩ K has even order for each non-trivial normalsubgroup K of G . Furthermore we assume that the following hold. (i) O p ′ ( H ) = 1 . (ii) m p ( C H ( t )) is at least for each involution t in H . (iii) N G ( T ) is a K -group, for all non-trivial -subgroups T of G . (iv) O ( G ) = G . Because of Hypothesis 4.1 (i) and (ii), we can apply Corollary 3.4 tosee that F ∗ ( G ) is a non-abelian simple group. The Frattini Argumentthen shows G = HF ∗ ( G ). Furthermore, we have that F ∗ ( G ) ∩ H Suppose that K is a subgroup of G , K H , m p ( H ∩ K ) ≥ and K is a K -group. Then the following hold. (i) O p ′ ( K ) ≤ H . (ii) ( F ∗ ( K/O p ′ ( K )) , p ) ∈ E . (iii) Either O p ′ ( H ∩ K ) = O p ′ ( K ) , or ( F ∗ ( K/O p ′ ( K )) , p ) = (PSL (4) , or (Fi , and | O p ′ ( H ∩ K ) /O p ′ ( K ) | = 2 .Proof. As H is strongly p -embedded in G , we can use Lemma 3.2 (ii)and (v) to see that O p ′ ( K ) ≤ O p ′ ( H ∩ K ) ≤ H . Then, as K is a K -groupand m p ( H ∩ K ) ≥ 2, Lemma 3.3(iii) implies ( F ∗ ( K/O p ′ ( K )) , p ) ∈ E .Thus (ii) holds. Part (iii) follows from (ii) and Corollary 2.11. (cid:3) Lemma 4.3. There exists an involution t ∈ H such that C G ( t ) H .Proof. Assume the lemma is false. Set K = F ∗ ( G ). Then K is a non-abelian simple group and C K ( t ) ≤ H ∩ K for all involutions t in H ∩ K .It follows from [9, Lemma 17.13] that K contains a strongly 2-embeddedsubgroup. Therefore, by [2], we have K ∼ = SL (2 a ), PSU (2 a ) or B (2 a )for some a . In particular, K is a K -group and as m p ( G ) ≥ 2, Proposi-tion 2.5 delivers the contradiction. Thus there exists an involution t in H such that C G ( t ) H . (cid:3) We can now present our first significant result. Lemma 4.4. If t ∈ H is an involution, then (i) C G ( t ) H and C H ( t ) is strongly p -embedded in C G ( t ) ; (ii) O p ′ ( C G ( t )) ≤ H ; (iii) ( F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) , p ) ∈ E ; and (iv) O ( C G ( t )) = O ( O p ′ ( C G ( t ))) .Proof. By Lemma 4.3, there exists an involution t ∈ G such that C G ( t ) H . Choose t with | C H ( t ) | maximal. Set K = C G ( t ) andnote that C H ( t ) is strongly p -embedded in C G ( t ) by Lemma 3.2 (ii).Then, by Hypothesis 4.1(ii) and Lemma 4.2(i) and (ii), O p ′ ( K ) ≤ H and ( F ∗ ( K/O p ′ ( K )) , p ) ∈ E . Let T ∈ Syl ( H ∩ K ). We will show that C G ( s ) H , for all involutions s ∈ T . Assume first that T Syl ( K ).Then N G ( T ) H . Hence Hypothesis 4.1 (ii) and Lemmas 3.5 and 3.6combine to give us that C G ( s ) H for all involutions s ∈ T whichis our claim. So assume that T ∈ Syl ( K ). Then, as ( H ∩ K ) /O p ′ ( K )is strongly p -embedded in K/O p ′ ( K ), ( F ∗ ( K/O p ′ ( K ) , p ) ∈ E and | K : Chris Parker and Gernot Stroth H ∩ K | is odd, Lemma 2.14 implies that F ∗ ( K/O p ′ ( K )) = K/O p ′ ( K ) ∼ =M . Let F ≤ T be such that F ∩ O p ′ ( K ) is a Sylow 2–subgroup of O p ′ ( K ) and | F : F ∩ O p ′ ( K ) | = 2. Assume that s ∈ T is an involutionwith C G ( s ) ≤ H . Then, as K/O p ′ ( K ) ∼ = M has exactly one conjugacyclass of involutions, all the involutions of T are G -conjugate to involu-tions in F . Thus ∅ 6 = s G ∩ F ≤ s G ∩ H = s H by Lemma 3.5 and wemay therefore assume that s ∈ F . However, from the structure of M , N K ( F O p ′ ( K )) /O p ′ ( K ) ( H ∩ K ) /O p ′ ( K ) and so we have N G ( F ) H by the Frattini Argument. Thus Lemma 3.6 implies that C G ( s ) H which is a contradiction. Hence C G ( s ) H for all involutions s ∈ T and our claim is proved.Let z be a 2-central involution of H which centralizes t . Then wemay suppose that z ∈ T . It follows that C G ( z ) H . By maximality of | C H ( t ) | we may assume t = z and so T is a Sylow 2-subgroup of H and this proves (i).Suppose that t is an involution in H . Then by Hypothesis 4.1(ii), m p ( C H ( t )) ≥ C G ( t ) H by (i). Hence Lemma 4.2 (i) and (ii)gives parts (ii) and (iii). Finally, as O ( C G ( t )) is soluble, (iii) implies(iv). (cid:3) As a first application of Lemma 4.4, we show that if E ( H ) is non-trivial then it is quasisimple. Lemma 4.5. If E ( H ) = 1 , then E ( H ) is quasisimple.Proof. Set E = E ( H ) and suppose that E is the product of at leasttwo components of H . Assume that L is a component of H andlet L be the product of all the components of H which are dis-tinct from L . Then E = L L and, since O ( H ) ≤ O p ′ ( H ) = 1by Hypothesis 4.1 (i), there exists an involution t ∈ L \ L . Wehave C G ( t ) ≥ L and, as p divides the order of each component in L , we have that L /Z ( L ) is isomorphic to a subnormal section of C H ( t ) /O p ′ ( C H ( t )). Using Lemma 4.4 (ii) and (iii), Proposition 2.7 andthe fact that O ( H ) = 1, we deduce that ( L , p ) is (Ω +8 (2) , p ) , p )or (Alt( p ) × Alt( p ) , p ) where in the latter two cases we have p ≥ O p ( C H ( t ) /O p ′ ( C H ( t ))) = 1 and thus, as O p ( H ) ≤ C G ( t ), we have O p ( H ) = 1. By applying the above argu-ment to L with an involution taken from L , we have that ( L , p )is (Ω +8 (2) , 5) or (Alt( p ) , p ). The possibilities for the isomorphism typeof E when p = 5 are thus Ω +8 (2) × Ω +8 (2), Ω +8 (2) × Alt(5) × Alt(5),Ω +8 (2) × Alt(5), Alt(5) × Alt(5) or Alt(5) × Alt(5) × Alt(5) and, for p > 5, we have E ∼ = Alt( p ) × Alt( p ) × Alt( p ) or Alt( p ) × Alt( p ). Since O p ′ ( H ) = 1 = O p ( H ) we have E = F ∗ ( H ). Therefore, as p ≥ the structure of E and the fact that the given components have noouter automorphisms of order p implies that E ( H ) contains a Sylow p -subgroup of H . Select an involution d ∈ L L which projects non-trivially onto each component as a 2-central involution. Then p doesnot divide C E ( d ) and we have a contradiction to Hypothesis 4.1(ii).Thus E is quasisimple as claimed. (cid:3) Centralizers of involutions in H In this section we work under Hypothesis 4.1 and aim to uncoverthe basic structure of the centralizers of involutions from H under theadditional hypothesis that H has no components and that none ofthe involutions in H are classical involutions (see Definition 2.1). Weformalize the configuration we shall be investigating in the followinghypothesis. Hypothesis 5.1. Hypothesis 4.1 holds, (i) E ( H ) = 1 ; and (ii) G does not contain a classical involution. Assume Hypothesis 5.1. Then, as O p ′ ( H ) = 1 by Hypothesis 4.1 (i),we have that F ∗ ( H ) = O p ( H ). Set Q = O p ( H ) and note that, as H isstrongly p -embedded in G ,we have H = N G ( Q ) and, as Q = F ∗ ( H )), C G ( Q ) ≤ Q . Lemma 5.2. Suppose that s ∈ H is an involution with C Q ( s ) > .Then there exists a normal component L in C G ( s ) such that LO p ′ ( C G ( s )) /O p ′ ( C G ( s )) = F ∗ ( C G ( s ) /O p ′ ( C G ( s ))) . Furthermore, if Z ( L ) has order divisible by , then L/Z ( L ) ∼ = PSL (4) .Proof. Since C Q ( s ) > O p ′ ( C G ( s )) ≤ H by Lemma 4.4(ii), wehave that [ C Q ( s ) , O p ′ ( C G ( s ))] = 1. Hence F ∗ ( C G ( s )) O p ′ ( C G ( s )).Thus, as C G ( s ) /O p ′ ( C G ( s )) is an almost simple group by Lemma 4.4(iii), we have E ( C G ( s )) O p ′ ( C G ( s )) and so there is a component L of C G ( s ) such that LO p ′ ( C G ( s )) /O p ′ ( C G ( s )) = F ∗ ( C G ( s ) /O p ′ ( C G ( s ))).Since L is the unique component of C G ( s ) which has order divisibleby p , we deduce that L is normal in C G ( s ). If 2 divides | Z ( L ) | , thenLemma 2.12, the fact that C Q ( s ) O p ′ ( C G ( s )) /O p ′ ( C G ( s )) is a non-trivialnormal subgroup of C H ( s ) /O p ′ ( C H ( s )) and Proposition 2.7, togetherimply that L ∼ = SL ( p a ) or L/Z ( L ) ∼ = PSL (4). Since s is not a classicalinvolution by Hypothesis 5.1 (ii), we must have L/Z ( L ) ∼ = PSL (4) andthus the lemma is established. (cid:3) Chris Parker and Gernot Stroth Lemma 5.3. Let u ∈ H be an involution with C Q ( u ) = 1 . Thenthere is a -component L u in C G ( u ) such that L u is not contained in O p ′ ( C G ( u )) .Proof. Let W be a Sylow 2-subgroup of O p ′ ( C G ( u )). Then C G ( u ) = N C G ( u ) ( W ) O p ′ ( C G ( u )). Set C G ( u ) = C G ( u ) /O { p, } ′ ( C G ( u )) and let F = F ∗ ( C G ( u )). Since, by Lemma 4.4(iii), C G ( u ) /O p ′ ( C G ( u )) is an almostsimple group, we may assume that F ≤ O p ′ ( C G ( u )) for otherwise therewould be a component of C G ( u ) not contained in O p ′ ( C G ( u )). Notethat, as C H ( u ) is strongly p -embedded in C G ( u ), O p ( F ) = 1. Sup-pose that C C G ( u ) ( W ) O p ′ ( C G ( u )). Then F is not a 2-group andconsequently E ( C G ( u )) = 1. Since W intersects each component of C G ( u ) in a Sylow 2-subgroup, we see that C C G ( u ) ( W ) normalizes eachcomponent of C G ( u ). Therefore, as C C G ( u ) ( W ) F /F is not soluble, theSchreier property of simple groups delivers a contradiction to F beingself-centralizing. Hence C C G ( u ) ( W ) ≤ O p ′ ( C G ( u )).As C Q ( u ) = 1, we have that H = QC H ( u ) and Q is abelian. Assumethat s ∈ W is an involution such that s = u and C Q ( s ) Syl p ( C G ( s )).Since s = u , we have C Q ( s ) > 1. Therefore Lemma 5.2 implies that C G ( s ) has a normal component L s with L s O p ′ ( C G ( s )). Let P ∈ Syl p ( H ∩ C G ( s )). Then, as H ∩ C G ( s ) normalizes C Q ( s ) and uQ ∈ Z ( H/Q ), C Q ( s ) ≤ P and [ P, u ] ≤ C Q ( s ). Since C Q ( s ) < P and C Q ( s )is normal in H ∩ C G ( s ), Corollary 2.8 implies that C Q ( s ) ≤ Φ( P )or ( L s , p ) ∼ = (PSL (8) , 3) or ( B (32) , P, u ] ≤ C Q ( s ) ≤ Φ( P ) and so P ≤ C G ( u ), which contradicts[ C Q ( s ) , u ] = 1. Thus we have the latter situation, and, as P = C Q ( s ),this means that C Q ( s ) is of order either p or p . As u ∈ N C G ( s ) ( u ),we see that C Q ( s ) ≤ L s and so C Q ( s ) is cyclic. As [ P, u ] ≤ P also[ P, us ] ≤ Q and so also C Q ( s ) is not a Sylow p –subgroup of C G ( us ).This implies that also C Q ( us ) is cyclic of order p or p . In particular,since h u, s i acts on Q , we get, using coprime action, that | Q/ Φ( Q ) | = p .But then H/Q is isomorphic to a subgroup of GL ( p ). It follows that m p ( C H ( u )) ≤ m p ( C H ( u )) ≥ 2. We have seen that for any involution s = u in W we have that C Q ( s ) is a Sylow p -subgroup of C G ( s ).As, by Hypothesis 4.1 (ii), m p ( H/Q ) = m p ( C H ( u )) ≥ 2, there isa non-cyclic abelian p -subgroup E contained in C H ( u ). Suppose that e ∈ E and w ∈ C W ( e ) is an involution. Then C Q ( w ) h e i is a p -subgroup of C G ( w ). If w = u , then C Q ( w ) ∈ Syl p ( C G ( w )). Since this is not the case, we see that for all e ∈ E , u is the unique invo-lution in C W ( e ). Therefore using Lemma 2.26 we have p = 3 and W ∼ = 2 , 2 − or 2 . By Lemma 4.4 (iii) F ∗ ( C G ( u ) /O p ′ ( C G ( u )))is a non-abelian simple group and so, since C G ( W ) ≤ O p ′ ( C G ( u )), wehave that N G ( W ) /C G ( W ) is non-soluble. In particular, we have that W is not isomorphic to 2 which has soluble automorphism group.Let s ∈ W be an involution with s = u . Then C W ( s ) ∼ = 2 × − if W ∼ = 2 − and C W ( s ) ∼ = 2 × if W ∼ = 2 . Since u inverts Q , it in-verts C Q ( s ). Thus C Q ( s ) is abelian and C Q ( s ) admits a faithful action of C W ( s ) / h s i . By Lemma 4.4 (iii), ( F ∗ ( C G ( s ) /O p ′ ( C G ( s ))) , p ) ∈ E . Since C Q ( s ) ∈ Syl p ( C G ( s )) and C W ( s ) / h s i is extraspecial of order 2 or 2 ,we have a contradiction to Proposition 2.7 (i), (iii), (iv) and (v). Thisfinally proves that there is a component in C G ( u ) /O { ,p } ′ ( C G ( u )). (cid:3) We let I be the set of involutions t ∈ H with C Q ( t ) > | C Q ( t ) | ≥ | C Q ( s ) | for all involutions s ∈ H . For any involution x ∈ H with C Q ( x ) > 1, we let L x denote the normal component of C G ( x ) described in Lemma 5.2. Lemma 5.4. Suppose that s is an involution in H with C Q ( s ) > .Then C C G ( s ) ( L s C Q ( s )) = C C G ( s ) ( L s ) = O p ′ ( C G ( s )) .Proof. We clearly have [ C Q ( s ) , O p ′ ( C G ( s ))] = 1. Therefore, we alsohave [ h C Q ( s ) C G ( s ) i , O p ′ ( C G ( s ))] = 1. Hence, as L s C Q ( s ) ≤ h C Q ( s ) C G ( s ) i , O p ′ ( C G ( s )) ≤ C C G ( s ) ( L s C Q ( s )) ≤ C C G ( s ) ( L s ). On the other hand, be-cause F ∗ ( C G ( s ) /O p ′ ( C G ( s ))) = L s O p ′ ( C G ( s )) /O p ′ ( C G ( s )), we also have C C G ( s ) ( L s ) ≤ O p ′ ( C G ( s )). This proves the lemma. (cid:3) Lemma 5.5. Let s ∈ H be an involution such that C Q ( s ) > . Then C H ( s ) /O p ′ ( C G ( s )) is p -closed. In particular, F ∗ ( C G ( s ) /O p ′ ( C G ( s ))) isnot isomorphic to Alt(2 p ) , p ≥ or to Fi .Proof. As C Q ( s ) O p ′ ( C G ( s )) /O p ′ ( C G ( s )) is a non-trivial normal p -subgroupof C H ( s ) /O p ′ ( C G ( s )), the assertion follows from Proposition 2.7. (cid:3) Lemma 5.6. Assume that s, t ∈ H are involutions, C Q ( t ) = 1 and t = s ∈ C H ( t ) . Then C G ( h s, t i ) is a p ′ -group. In particular, if P ∈ Syl p ( C G ( t )) , then m ( N C G ( t ) ( P )) ≤ .Proof. Since C Q ( t ) = 1, we have H = QC H ( t ) and C H ( s ) = C Q ( s ) C H ( h s, t i ).Let P ∈ Syl p ( C H ( h s, t i )). Then C Q ( s ) P ∈ Syl p ( C H ( s )) ⊆ Syl p ( C G ( s ))by Lemma 3.2 (i). Since t normalizes C Q ( s ) P and t centralizes P but in-verts C Q ( s ), we must have that C Q ( s ) Φ( C Q ( s ) P ). Assume that P =1. Then C Q ( s )Φ( C Q ( s ) P ) = C Q ( s ) P . In particular, C H ( s ) /O p ′ ( C H ( s ))does not act irreducibly on C Q ( s ) P/ Φ( C Q ( s ) P ). Since C H ( s ) /O p ′ ( C H ( s )) Chris Parker and Gernot Stroth is p -closed by Lemma 5.5, Corollary 2.8 implies that ( L s , p ) is either(PSL (8) , 3) or ( B (32) , t inverts C Q ( s ), we see that C Q ( s ) ≤ L s and then C Q ( s ) is a cyclic group of order dividing p . Since Q = C Q ( st ) C Q ( s ), we have that | Q/ Φ( Q ) | ≤ p . This means that H/Q isisomorphic to a subgroup of GL ( p ) and implies that C H ( t ) has cyclicSylow p -subgroups. This of course contradicts Hypothesis 4.1(ii). Thus C G ( h s, t i ) is a p ′ -group as claimed. (cid:3) Lemma 5.7. Suppose that t ∈ I or C Q ( t ) = 1 . Then O p ′ ( C G ( t )) hascyclic Sylow -subgroups. In particular, O p ′ ( C G ( t )) has a normal -complement.Proof. Assume that F = h t, s i is a fours group in O p ′ ( C G ( t )). Then Q = C Q ( t ) C Q ( s ) C Q ( ts ).Assume further that t ∈ I . Then, as F ≤ O p ′ ( C G ( t )) ≤ H and[ F, C Q ( t )] ≤ [ O p ′ ( C G ( t )) , C Q ( t )] = 1, the maximality of C Q ( t ) impliesthat C Q ( t ) = C Q ( F ) = C Q ( s ). But then Q = C Q ( t ), which contradicts C G ( Q ) ≤ Q . Hence no such F exists and so O p ′ ( C G ( t )) has eitherquaternion or cyclic Sylow 2-subgroups.Assume that C Q ( t ) = 1. Then, by Lemma 5.3, there is a 2-component L in C G ( t ) with LO p ′ ( C G ( t )) /O p ′ ( C G ( t )) = F ∗ ( C G ( t ) /O p ′ ( C G ( t )))).Now we have [ L, F ] ≤ O ( C G ( t )). In particular, C H ( F ) contains aSylow p -subgroup P of L and, as P is non-trivial, this contradictsLemma 5.6. Thus once again, as t is not a classical involution, wehave that O p ′ ( C G ( t )) has cyclic Sylow 2-subgroups and a normal 2-complement.Since, by Hypothesis 5.1 (ii), t is not a classical involution, we de-duce that O p ′ ( C G ( t )) has cyclic Sylow 2-subgroups and a normal 2-complement. (cid:3) Centralizers of involutions with F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) = PSL ( p f )In this section we will prove the following theorem. Theorem 6.1. Assume that Hypotheses 4.1 and 5.1 hold. Then either (i) there exists t ∈ I with F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) ∼ = PSL ( p f ) forsome f > ; or (ii) for all involutions t ∈ H , C Q ( t ) = 1 and F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) ∼ =PSL ( p f ) with f > and p ≡ . We prove Theorem 6.1 via a sequence of lemmas the first of whichshows that the theorem holds if I is empty. We continue to use thenotation of Section 5. In particular, Q = O p ( H ). ( L t /Z ( L t ) , p ) C L t /Z ( L t ) ( s )(M , 3) GL (3)( F (2) ′ , 5) 2 . Frob(20)(McL , 5) 2 . Alt(8)(J , 11) 2 . . Aut(M )(PSU ( p a ) , p ) O p ′ ( C L t /Z ( L t ) ( s )) ∼ = SL ( p a ) Table 1. Centralizers of s in L t /Z ( L t ) Lemma 6.2. If I = ∅ , then Theorem 6.1 (ii) holds.Proof. Suppose that I = ∅ . Then C Q ( t ) = 1 for all involutions t ∈ H and, in particular, m ( H ) = 1. Therefore the Sylow 2-subgroups of H are either cyclic or quaternion. Let t ∈ H be an involution. Then C Q ( t ) = 1. By Lemma 5.3 there is a normal 2-component L of C G ( t )with L O p ′ ( C G ( t )). Note that Lemma 4.4 (ii) and (iv) imply that O ( L ) ≤ H . Since H has cyclic or quaternion Sylow 2-subgroups, sodoes C H ( t ). If t ∈ L , then tO ( L ) ∈ Z ( L/O ( L )) is non-trivial. Therefore,as the Sylow 2-subgroups of L ∩ H are either cyclic or quaternion,Corollary 2.13 implies that L/O ( L ) ∼ = SL ( p f ) for some f . But then G contains a classical involution, and this contradicts Hypothesis 5.1(ii). Hence t L . Since t is the unique involution in C H ( t ), we deducethat | H ∩ L | is odd. Now Lemma 2.15 shows that L/O ( L ) ∼ = PSL ( p f )where f > p ≡ (cid:3) We may now assume that I 6 = ∅ and so we begin to restrict thepossibilities for the structure of C G ( t ), where t ∈ H is an involutionwith C Q ( t ) > 1. By Lemma 4.4 C G ( t ) H and O p ′ ( C G ( t )) ≤ H andby Lemma 5.2, there is a normal component L t of C G ( t ) such that L t O p ′ ( C G ( t )). By Lemma 4.4 (iii) we then have ( L t /Z ( L t ) , p ) ∈ E .Our aim now is to show that there is an involution t ∈ I with L t ∼ =PSL ( p f ). Lemma 6.3. If t ∈ H is an involution with C Q ( t ) > , then ( L t /Z ( L t ) , p ) is one of (PSL ( p a ) , p ) , with p arbitrary and a ≥ , ( G (3 a − ) , with a ≥ , (PSL (8) , , (PSL (4) , or ( B (32) , .Proof. Assume the claim is false. Then, as ( L t /Z ( L t ) , p ) ∈ E and O p ( H ∩ L t ) = 1, we have that ( L t /Z ( L t ) , p ) is one of (PSU ( p a ) , p )with a ≥ 2, (M , , F (2) ′ , 5) or (J , Z ( L t ) has odd order. In particular, in each of thecases we need to consider there is an involution s ∈ L t such that C L t ( s ) has structure as indicated in the Table 1. In particular, wenote that in each case s ∈ O p ′ ( O p ( C L t ( s ))). By Lemma 3.2 (i) and Chris Parker and Gernot Stroth (ii), H ∩ L t is strongly p -embedded in L t . Since in each of the groupsunder consideration as L t , the centralizer of s has order divisible by p we may assume that s ∈ H ∩ L t . Since Q ∩ L t = C Q ( t ) is nor-malized by C H ( t ), Proposition 2.7 implies that we either have that C Q ( t ) ∈ Syl p ( L t ) or p = 11, L t ∼ = J and C Q ( t ) is the centre ofthe Sylow 11-subgroup of H ∩ L t or L t ∼ = PSU ( p a ). In each case,we have C Q ( s ) ≥ C C Q ( t ) ( s ) > 1. Therefore Lemma 5.2 indicates that C G ( s ) has a normal component L s such that L s O p ′ ( C G ( s )). Since[ O p ′ ( C G ( s )) , L s ] = 1, we have that L s = O p ′ ( O p ′ ( C G ( s )) L s ). Hence, weeither have O p ′ ( O p ( C G ( s ))) = L s , or p divides | C G ( s ) /O p ′ ( C G ( s )) L s | .In the latter case, Proposition 2.5 shows that C G ( s ) /O p ′ ( C G ( s )) ∼ =PSL (8) : 3 and p = 3 or C G ( s ) /O p ′ ( C G ( s )) ∼ = B (32) : 5 and p = 5.In this case we have L s = O p ′ ( O p ( C G ( s ))) as well. It follows that inall cases s ∈ L s , as s ∈ O p ′ ( O p ( C L t ( s ))). Therefore Lemma 5.2 impliesthat L s /Z ( L s ) ∼ = PSL (4) and that p = 3. Furthermore, the Sylow3-subgroups of C G ( s ) have order 9 and so after consulting Table 1 weget L t ∼ = M , PSU (3) or PSU (9). If t ∈ O p ′ ( C G ( s )) L s , then either C G ( h t, s i ) would have order coprime to 3 or would contain a compo-nent K such that K/Z ( K ) is isomorphic to PSL (4), from the choiceof s we see that both of these scenarios are impossible. Hence t in-duces an outer automorphism of L s and so with Lemma 2.10 we getthat C L s /Z ( L s ) ( t ) ∼ = 3 . Q , PSL (7) or Alt(5). Since C L s ( t ) projects toa subgroup of C G ( t ) /O p ′ ( C G ( t )) which is normal in the centralizer ofthe image of s in C G ( t ) /O p ′ ( C G ( t )), this then gives us our final contra-diction. (cid:3) Lemma 6.4. If t ∈ I , then L t /Z ( L t ) is not isomorphic to G (3 n +1 ) , n ≥ .Proof. Suppose that L t /Z ( L t ) ∼ = G (3 n +1 ) with n ≥ 1. Note that by[10, Theorem 2.5.12] the outer automorphisms of L t have odd orderand Z ( L t ) = 1. Let R ∈ Syl ( L t O p ′ ( C G ( t ))) = Syl ( C G ( t )) be suchthat R ∩ H ∈ Syl ( C H ( t )) and set T = R ∩ O p ′ ( C G ( t )) and E = T ∩ L t .Then, by Lemma 5.7, T is a cyclic group and, as L t ∼ = G (3 n +1 ), E iselementary abelian of order 8. Since [ L t , O p ′ ( C G ( t ))] = 1, R = T × E is abelian and so R ∈ Syl ( C G ( R )). By Lemma 2.17, t is conjugate in G to an element t x ∈ R \ { t } and so Lemma 2.20 implies that T hasorder 2. Therefore R is elementary abelian of order 16.From the structure of G (3 n +1 ) we have N L t ( E ) /E ∼ = Frob(21) ∼ = N L xt ( E x ) /E x . Therefore N G ( R ) /C G ( R ) contains distinct subgroups iso-morphic to Frob(21) and N G ( R ) has orbits of lengths either 8 and 7 or15 on R . As GL (2) has no subgroups of order 3 · · N G ( R ) to be transitive on R . Hence | t N G ( R ) | = | t G ∩ R | = 8, N G ( R ) /C G ( R ) ∼ = 2 . Frob(21) or PSL (2) and E is the unique nor-mal subgroup of N G ( R ) of order 2 . Let s ∈ E be chosen so that s normalizes a Sylow 3-subgroup D of C H ( t ), then s and t are not G -conjugate and s ∈ H . Furthermore C L t ( s ) ∼ = 2 × PSL (3 n +1 ) and C N G ( R ) ( s ) has non-abelian Sylow 2-subgroups. In particular, as 3 di-vides | C G ( h s, t i ) | , Lemma 5.6 implies that C Q ( s ) = 1. By Lemma 5.2,there is a normal component L s in C G ( s ) with L s O p ′ ( C G ( s )). Wethen have that t induces an automorphism of L s which centralizes asubgroup of L s which is isomorphic to PSL (3 n +1 ). By Lemmas 2.10and 6.3, we see that L s /Z ( L s ) ∼ = G (3 m +1 ) for some m ≥ L s /Z ( L s ) ∼ = PSL (3 m ) for some m ≥ n + 1. In the former case we have C L s ( t ) ∼ = C L t ( s ) and so we infer that L t ∼ = L s (but perhaps s 6∈ I ). Let B ∈ Syl ( C G ( s )) be chosen so that R ≤ B . Then, as Out( L s ) has oddorder and Z ( L s ) = 1, B = ( B ∩ O p ′ ( C G ( s ))) × ( B ∩ L s ) where B ∩ L s is el-ementary abelian of order 8. Notice that t ∈ B and C B ( t ) ≥ h s i ( B ∩ L s ).Since R ∈ Syl ( C G ( t )) and | R | = 2 , we get that R = h s i ( B ∩ L s ). How-ever, we then have N L s ( B ∩ L s ) ≤ N G ( R ) and so N L s ( B ∩ L s ) leavesboth E and B ∩ L s invariant. Hence E = ( B ∩ L s ) and consequently s ∈ E ⊂ L s which is a contradiction. Assume that L s ∼ = PSL (3 m ).Then, by Proposition 2.5 and Lemma 4.4, L s = O ′ ( C G ( s )). Since O ′ ( C N G ( R ) ( s )) C G ( R ) /C G ( R ) ∼ = Alt(4) and | O ′ ( C N G ( R ) ( s )) ∩ R | = 2 or 2 , we see that O ′ ( C N G ( R ) ( s )) contains a non cyclic abelian groupof order at least 8. Hence L s contains a noncyclic abelian group of or-der 8. So we have a contradiction to the Sylow 2-subgroup structure ofPSL (3 m ). This completes the proof of the lemma. (cid:3) Lemma 6.5. If t ∈ I , then ( L t /Z ( L t ) , p ) is not either (PSL (8) , or ( B (32) , .Proof. Assume ( L t /Z ( L t ) , p ) is either (PSL (8) , 3) or ( B (32) , C G ( t ) /O p ′ ( C G ( t )) ∼ = PSL (8) : 3 or B (32) : 5 respectively. In bothcases, there is an involution s ∈ L t which normalizes C Q ( t ). Thus s ∈ H . Let R be a Sylow 2-subgroup of C G ( t ) which contains h s i . Then R = T × ( R ∩ L t ), where T is a Sylow 2-subgroup of O p ′ ( C G ( t )). ByLemma 5.7 we have that T is cyclic. By Lemma 2.17, t is conjugateto some element t x in R \ { t } . Since Z ( R ∩ L t ) is elementary abelian,Lemma 2.20 implies that T = h t i has order 2.Assume that ( L t , p ) = (PSL (8) , R is elementary abelian oforder 2 and N C G ( t ) ( R ) /C C G ( t ) ( R ) ∼ = Frob(21) ∼ = N C G ( t x ) ( R ) /C C G ( t x ) ( R ).Just as in the proof of Lemma 6.4, we get N G ( R ) /C G ( R ) ∼ = 2 . Frob(21)or PSL (2) and the N G ( R )-orbits on R have lengths 8 and 7 with t in theorbit of length 8. Since s normalizes a Sylow 3-subgroup of C H ( t ) andsince s is contained in a unique Sylow 2-subgroup of L t , we can choose Chris Parker and Gernot Stroth P ∈ Syl ( N G ( R ) ∩ H ) so that t ∈ C R ( P ). Let F = h t, r i = C R ( P ).We may choose notation so that | r N G ( R ) | = 7. Since 3 divides C G ( F ),Lemma 5.6 implies that C Q ( r ) > 1. Furthermore, as C N G ( R ) ( r ) /R con-tains a subgroup isomorphic to Alt(4), we have | O ′ ( C N G ( R ) ( r )) | ≥ . L r /Z ( L r ) is isomorphic to oneof PSL (8), PSL (4), PSL (3 m ), m ≥ 2, or G (3 n +1 ), n ≥ 1. If L r = O ′ ( C G ( r )), then L r ≥ O ′ ( C N G ( R ) ( r )). As G (3 n +1 ) and PSL (8) haveSylow 2-subgroups of order 8 and in PSL (3 m ) the normalizer of a fourgroup is isomorphic to either Alt(4) or Sym(4), these possibilities can-not arise. If L r /Z ( L r ) ∼ = PSL (4), then | C Q ( r ) | = 3 . Now in C G ( t ), wesee that | C C G ( t ) ( r ) | = 3 and C C G ( t ) ( r ) O ′ ( C G ( t )) /O ′ ( C G ( t )) is soluble.Since t induces an automorphism of L r , we now deduce a contradic-tion from Lemma 2.10. Thus L r = O ′ ( C G ( r )). Then L r ∼ = PSL (8).Since R is abelian, R ≤ C G ( r ). Let D ∈ Syl ( C G ( r )) with R ≤ D . Then D = ( D ∩ O ′ ( C G ( r ))) × ( D ∩ L r ). Thus C D ( t ) ≥ h r i × ( D ∩ L r ) ≤ Z ( D ).It follows that D ≤ C G ( t ) and so R = D . Since N G ( R ) normalizes aunique subgroup of order 2 in R , it follows that r ∈ R ∩ L t = R ∩ L r which is of course impossible. This shows that L r = PSL (8) and so L t /Z ( L t ) = PSL (8).Assume that ( L t , p ) = ( B (32) , P ∈ Syl ( C H ( t )) and let s ∈ L t be such that | C G ( s ) ∩ P | = 5. Then s ∈ H and | C L t P ( s ) | =2 · 5. Therefore O ′ ( C G ( s )) has a subgroup of order 2 · L s /Z ( L s ) is isomorphic to B (32) or PSL (5 m ) some m ≥ 1. In the latter case, Proposition 2.5implies that L s = O ′ ( C G ( s )). Since L s /Z ( L s ) has 2-rank 2, we havea contradiction. Thus L s ∼ = B (32). Since s ∈ Z ( R ), R ≤ C G ( s ). Let D ∈ Syl ( C G ( s )) with R ≤ D . Then D = ( D ∩ O ′ ( C G ( s ))) × ( D ∩ L s ).As t ∈ Ω ( D ) and Ω ( D ) ≤ ( D ∩ O ′ ( C G ( s ))) Z ( D ∩ L s ), we have that D ∩ L s ≤ R . Therefore R = ( D ∩ L s ) h s i . But then s ∈ R ′ = ( D ∩ L s ) ′ and this is of course impossible. (cid:3) Lemma 6.6. There exists t ∈ I such that ( L t /Z ( L t ) , p ) is (PSL ( p f ) , p ) .Proof. Suppose the lemma is false and let t ∈ I . Then by Proposi-tion 2.5 and Lemmas 5.2, 5.5, 6.3,6.4 and 6.5, we may assume that( L t /Z ( L t ) , p ) is (PSL (4) , R be a Sylow 2-subgroup of C G ( t ) and set T = R ∩ O ′ ( C G ( t )).By Lemma 5.7, T is cyclic. By Proposition 2.7 (iv), ( L t ∩ H ) /Z ( L t ) ∼ =3 : Q and so | C Q ( t ) O ′ ( C H ( t )) /O ′ ( C H ( t )) | = | C Q ( t ) | = 9. We alsoremark that as T is cyclic, Z ( L t ) is cyclic and if Z ( L t ) is non-trivialthen t ∈ Z ( L t ).We remark that the normalizer of an elementary abelian group E of order 2 in L t /Z ( L t ) has shape 2 : Alt(5) ∼ = 2 : SL (4). Further N L t /Z ( L t ) ( E ) acts transitively on E ♯ . Suppose that | Z ( L t ) | = 2. Thenbecause of the transitive action of L t on E ♯ , we have that E lifts to anelementary abelian group F of order 32. In particular involutions liftto involutions. Suppose that N L t ( F ) acts decomposably on F . Then N L t ( F ) /E is a direct product of a group of order two by PSL (4). Butthen a Sylow 2–subgroup splits over Z ( L t ), which is not possible. Hence N L t ( F ) acts indecomposably on F . Assume now that Z ( L t ) is cyclic oforder four. If the preimage F of E would be abelian, then it is of type(4 , , , , | Ω ( F ) | = 32 and | Φ( F ) | = 2. But then N L t ( F )cannot act indecomposably on F / Φ( F ), which contradicts the factjust proved for | Z ( L t ) | = 2. Hence F is nonabelian and so E lifts to aspecial group of symplectic type.Assume that T L t contains an elementary abelian subgroup F of or-der 32. Then we have that | Z ( L t ) | ≤ 2. We select F such that F ∩ H has order exactly 4 and set F ∩ H = h s, t i where s ∈ L t with s = t .As s ∈ L t , s and st invert C Q ( t ). As [ t, Q ] = 1 there is x ∈ { s, st } such that C Q ( x ) = 1. If, additionally, x 6∈ I , then, as | C Q ( t ) | = 9, | C Q ( x ) | = 3 and, by Lemma 5.2 and Corollary 2.8, L x ∼ = PSL (8)and C G ( x ) /O ′ ( C G ( x )) ∼ = PSL (8) : 3. Since C G ( x ) /O ′ ( C G ( x )) ∼ =PSL (8) : 3 and F ≤ C G ( x ), we have that | F ∩ O ′ ( C G ( x )) | ≥ O ′ ( C G ( x )) ≤ H , we get F ∩ O ′ ( C G ( x )) = F ∩ H . But then t ∈ O ′ ( C G ( x )) and as [ O ′ ( C G ( x )) , C Q ( x )] = 1, we have C Q ( x ) ≤ C Q ( t )whereas x inverts C Q ( t ). Hence, if C Q ( x ) = 1, then x ∈ I . Hencewe also have L x /Z ( L x ) ∼ = PSL (4). Let U be a Sylow 2-subgroupof O ′ ( C G ( x )). Then, as x ∈ I , U is cyclic and U commutes with L x . Since T F ∩ O ′ ( C G ( x )) is cyclic and T F is abelian, we have that | T F O ′ ( C G ( x )) /O ′ ( C G ( x )) | = 2 and hence F ≤ U L x by Lemma 2.10.In particular, U and F commute. As T F is a Sylow 2-subgroup of C G ( F ), we have T F = U F and | U | = | T | . Assume that | T | > 2. Then,on the one hand t is the unique involution in Φ( T F ), while on theother hand, x is the unique involution in U F and so, as x = t , wemust have T = h t i has order 2. As N L t ( F ) centralizes only t in F and N L x ( F ) centralizes only x in F , we have N G ( F ) C G ( t ). Of course wealso have that N G ( F ) /C G ( F ) is isomorphic to a subgroup of GL (2).From the structure of L t , we know that N C G ( t ) ( F ) induces orbits oflength 1,15,15 on F ♯ and so t has 31 or 16 N G ( F )-conjugates in F .If all involutions in F are conjugate, we get that N G ( F ) /C G ( F ) hasorder 2 · · · 31 if N C G ( t ) ( F ) induces Alt(5) on F or 2 · · · Chris Parker and Gernot Stroth if N C G ( t ) ( F ) induces Sym(5) on F . As in both cases, by Sylow’s The-orem, the Sylow 31-subgroup would be normal, this contradicts thestructure of the normalizer of a Sylow 31-subgroup of GL (2). So t has16 N G ( F )-conjugates in F . Since x ∈ I , we may argue in a similar wayto see that there are 16 N G ( F )-conjugates of x in F . Recall that x waschosen in { s, st } and s = st in L t , at least one of s and st is not N G ( F )conjugate to t . It follows that either C Q ( s ) = 1 or C Q ( st ) = 1. Assumethat y ∈ { s, st } is such that C Q ( y ) = 1. Then Q = C Q ( t ) C Q ( x ) and, as | C Q ( x ) | = | C Q ( t ) | = 9, Q is elementary abelian of order 3 and H/Q isisomorphic to a subgroup of GL (3). By Lemmas 5.3 and 5.7 there is a2-component L in C G ( y ) and O ′ ( C G ( y )) has a normal 2-complementand a cyclic Sylow 2-subgroup. Since O ( L ) ≤ H and as | O ( L ) | hasorder coprime to both 2 and 3 and divides | GL (3) | , we have that O ( L ) is cyclic of order dividing 65. Since p = 3 and C H ( y ) is strongly3-embedded in C G ( y ), we have that | O ( C H ( y ) /O ( C G ( y ))) | ≥ O ( L ) is cyclic, O ( C H ( y )) = 1, but thisis impossible as H = C H ( y ) Q , Q = O ( H ) and Q ∩ C H ( y ) = 1. Thiscontradiction shows that T L t does not contain an elementary abeliansubgroup of order 32. Therefore L t is a central extension of PSL (4) bya cyclic group of order 4. In particular, t ∈ L t and L t has precisely twoconjugacy classes of involutions.Assume that there is an s ∈ t G ∩ C G ( t ) \ L t O p ′ ( C G ( t )). Then, byLemma 2.10 we may assume that C C Q ( t ) ( s ) = 1 which means that s ∈ H . If C Q ( t ) = C Q ( s ), then, as t ∈ I , we get [ Q, s ] = 1, a con-tradiction. So we have that | C Q ( s ) ∩ C Q ( t ) | = 3. As t ∈ I , we seeby coprime action that | Q | = 3 . Furthermore, we have that Q is ele-mentary abelian. Now choose u ∈ L t an involution such that u inverts C Q ( t ). By Proposition 2.7 (iv), we have that u is a square in H . Inparticular, u acts on Q as an element of PSL (3). Since u ∼ L t ut ,Lemma 3.5 implies that u ∼ H ut . Therefore both C Q ( u ) and C Q ( ut )are non-trivial. Because u and ut act on Q as elements of SL (3), wesee that | C Q ( u ) | = | C Q ( ut ) | = 3 . Finally, we have C Q ( h u, t i ) = 1 andso by coprime action | Q | = 3 , a contradiction. We have proved that t G ∩ C G ( t ) ⊆ L t O p ′ ( C G ( t )). In particular, by Lemma 2.17 all involutionsin C G ( L t ) L t are conjugate to t .Let u ∈ L t be an involution such that uZ ( L t ) /Z ( L t ) ∈ Z ( RZ ( L t ) /Z ( L t )).We have shown that u ∈ t G . Set R = C R ( u ) and note that as h t, u i is a fours group and u Z ( R ), we have | R : R | = 2. Since u ∼ G t ,there exists R ∈ Syl ( C G ( u )) such that R > R . Obviously R isnormal in R . If R ≤ C G ( t ), then Ω ( Z ( R )) = h t, u i and this con-tradicts the fact that | Ω ( Z ( R )) | = 2. Hence R doesn’t centralize t . Let W ∼ = Z ∗ Q ∗ Q be the preimage in L t of an elementaryabelian subgroup of L t /Z ( L t ) of order 16 which is contained in R .Then | W : C W ( u ) | = 2 and C W ( u ) = h u i × V , where V ∼ = Z ∗ Q .Set R = h t G ∩ R i . Then R ≤ R ∩ L t and R normalizes R . As C W ( u ) is generated by involutions, we have that C W ( u ) ≤ R . As C R/Z ( L t ) ( C W ( u ) /Z ( L t )) = W/Z ( L t ), we see that Z ( R ) = Z ( C W ( u )) = h u, Z ( L t ) i . In particular, R normalizes h u, Z ( L t ) i and hence R nor-malizes h t i which is the characteristic subgroup of h u, Z ( L t ) i generatedby squares. Since R C G ( t ) we have our final contradiction. Hencethere exists t ∈ I such that L t /Z ( L t ) ∼ = PSL ( p f ) for some f ≥ (cid:3) Proof of Theorem 6.1. This is a consequence of a combination of allthe lemmas in this section culminating in Lemma 6.6. (cid:3) Centralizers of involutions with F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) ∼ = PSL ( p f )We continue to assume that Hypotheses 4.1 and 5.1 hold. We usethe notation established in the Sections 4, 5 and 6. In particular, Q = O p ( H ). Because of Theorem 6.1, we may assume that the followinghypothesis is satisfied: Hypothesis 7.1. Hypotheses 4.1 and 5.1 hold and either (i) there exists t ∈ I and F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) ∼ = PSL ( p f ) forsome f > ; (ii) for all involutions t ∈ H , C Q ( t ) = 1 and F ∗ ( C G ( t ) /O p ′ ( C G ( t ))) ∼ =PSL ( p f ) with f > and p ≡ . Our objective is to prove that if Hypothesis 7.1 holds, then F ∗ ( G ) ∼ = G (3 a +1 ) for some a ≥ 1. We fix the following notation throughoutthis section. We let t ∈ H be an involution such that either t ∈ I or C Q ( t ) = 1 and such that E ( C G ( t ) /O p ′ ( C G ( t ))) ∼ = PSL ( p f ) with f > C G ( t ), whichwe denote by L , with LO p ′ ( C G ( t )) /O p ′ ( C G ( t )) = F ∗ ( C G ( t ) /O p ′ ( C G ( t ))).Since t is not a classical involution by Hypothesis 5.1 (ii), we have L/O ( L ) ∼ = PSL ( p f ).We fix T ∈ Syl ( O p ′ ( C G ( t ))), S ∈ Syl ( C G ( t )) such that T ≤ S , D = S ∩ L and U ∈ Syl ( G ) with S ≤ U .Obviously t ∈ T and, by Lemma 5.7, we have that T is cyclic and O p ′ ( C G ( t )) = T O ( C G ( t )). Finally we note that F ∗ ( C G ( t ) /O ( C G ( t ))) ∼ = T × L/O ( L ) ∼ = Z k × PSL ( p f )for some k ≥ f > Chris Parker and Gernot Stroth Lemma 7.2. There is no involution in C G ( t ) which induces a non-trivial field automorphism on L/O ( L ) .Proof. Assume that there is an involution x ∈ S which induces a non-trivial field automorphism on L/O ( L ). Then C L/O ( L ) ( x ) ∼ = PGL ( p r )where 2 r = f . In particular, p divides | C L ( x ) | and so, as H is strongly p -embedded, there is a C G ( t )-conjugate s of x contained in C H ( t ). Since C G ( h s, t i ) is not a p ′ -group, Lemma 5.6 implies that C Q ( t ) = 1 = C Q ( s ).In particular, Lemma 5.2 implies that O ( L ) = Z ( L ) = 1 and that C G ( s )has a normal component L s with L s O p ′ ( C G ( s )). Since C C Q ( t ) ( s ) = C C Q ( t ) ( st ) and C G ( Q ) ≤ Q , we may assume that C Q ( s ) C Q ( t ). There-fore, | C Q ( s ) | > p r and so L s = PSL ( p r ). Observe that t induces a non-trivial automorphism on L s which centralizes a subgroup of L s which isisomorphic to PSL ( p r ). Suppose that p r = 3. Then Lemma 6.3 showsthat either ( L s /Z ( L s ) , p ) = (PSL ( p f ) , p ), ( G (3 a − ) , (8) , (4) , 3) or ( B (32) , C L s ( t ), contains PSL ( p r ), we see that( L s /Z ( L s ) , p ) = (PSL ( p f ) , p ) or ( G (3 r ) , C C G ( s ) /O p ′ ( C G ( s )) ( t ) = h t i × PSL (3 r ), while C G ( t ) ∩ C G ( s )involves PGL ( p r ), a contradiction. Thus, if p r > 3, we deduce that L s ∼ = PSL ( p f ) and t induces a field automorphism on L s . Assumenow that p r = 3. Then we have that L t ∼ = PSL (9) and C L t ( s ) ∼ =Sym(4). Furthermore, we have C C G ( t ) ( s ) is soluble. Since L s L t , t does not centralize L s . Thus t induces an automorphism of L s and, as[ C Q ( s ) , O p ′ ( C G ( s ))] = 1, C Q ( s ) ∩ C L t ( s ) > C L t ( s ) ∼ = Sym(4), C L t ( s ) ∩ O p ′ ( C G ( s )) ≤ C C Lt ( s ) ( C Q ( s ) ∩ C L t ( s )) ≤ Z ( C L t ( s )) = 1 . Thus t centralizes a subgroup of C G ( s ) /O p ′ ( C G ( s )) which is isomorphicto Sym(4).Using Lemma 6.3, we have the following possibilities for the isomor-phism type of L s /Z ( L s ): L s /Z ( L s ) ∼ = PSL (3 m ), m ≥ G (3 a +1 ), a ≥ 1, PSL (4) or PSL (8). In the first case, we get | C Q ( s ) | = 3 m and so, as t ∈ I , we must have L s ∼ = PSL (9) as well, with t induc-ing a non-trivial field automorphism on L s . If L s ∼ = G (3 a +1 ), then t acts as an inner automorphism of L s and so t centralizes a subgroupof L s isomorphic to PSL (3 a +1 ). This subgroup would of course haveto be contained in PSL (9) which is impossible. Thus L s is not a Reegroup. Suppose that L s /Z ( L s ) ∼ = PSL (4). We now use the fact that C C G ( s ) /O p ′ ( C G ( s )) ( t ) contains a subgroup isomorphic to Sym(4) and issoluble. If t acts as an inner automorphism of L s , we have C L s ( t ) isa 2-group which is not the case. Thus t acts as an outer automor-phism of L s . Hence Lemma 2.10 indicates that C C G ( s ) /O p ′ ( C G ( s )) ( t ) isnot soluble which is also impossible. Finally, if L s ∼ = PSL (8), we see C C G ( s ) /O p ′ ( C G ( s )) ( t ) ∼ = 2 × Alt(4) which does not contain a subgroup iso-morphic to Sym(4). Hence this case is also impossible. Thus we haveshown that, if L t ∼ = PSL (9), then L s ∼ = PSL (9) and t induces a fieldautomorphism on L s .We have shown that, if L t ∼ = PSL ( p f ) and s induces a non-trivialfield automorphism on L t , then L s ∼ = PSL ( p f ) and t induces a non-trivial field automorphism on L s . It follows that L s ∩ L t ∼ = PGL ( p r ) andthat L s ∩ L t ∩ H = N L s ∩ L t ( C Q ( h s, t i )). Therefore there is an involution u ∈ L s ∩ L t ∩ H , which centralizes h s, t i and inverts C Q ( x ) for x ∈ { s, t } .If C Q ( st ) ≤ C Q ( t ), then u inverts C Q ( st ) while, if C Q ( st ) C Q ( t ), wehave that L st ∼ = PSL ( p f ) by the argument above. Since L s ∩ L t = C L t ( s ) = C L t ( st ), we deduce that L s ∩ L t = L s ∩ L st . In particular, u ∈ L st inverts C Q ( st ) in this case as well. Therefore, in any case wehave that Q = C Q ( t ) C Q ( s ) C Q ( st ) is inverted by u . Hence C Q ( u ) = 1.Let L u be the 2-component in C G ( u ) given by Lemma 5.3.If ( L u ∩ H ) /O ( L u ) is p -closed, then h s, t i normalizes a Sylow p -subgroup of L u ∩ H . Thus p divides | C G ( h u, x i ) | for some x ∈ h s, t i .This of course contradicts Lemma 5.6. Thus ( L u ∩ H ) /O ( L u ) is not p -closed and hence, either p > L u /O ( L u ) ∼ = Alt(2 p ) or 2 . Alt(2 p ), or p = 5 and L u ∼ = Fi or 2 . Fi . Notice that C L t ( u ) is a Dihedral group oforder 2( p f ± C C G ( t ) ( u ) = C C G ( u ) ( t ) is soluble. Hence t inducesa non-trivial automorphism on L u /O ( L u ). If L u /O p ′ ( L u ) ∼ = Alt(2 p ),then by considering the possibilities for t we see that p = 5 and L u /O p ′ ( L u ) ∼ = Alt(10). Furthermore, in this case we have that theimage of t corresponds to a permutation of cycle shape 1 . or 1 . .In each case C L u ( t ) contains a section which is isomorphic to Sym(4).Since T is cyclic, we see that C C G ( t ) ( u ) contains no such section. Soassume that L u /O p ′ ( L u ) ∼ = Fi . In this case we select an involution w of L u which centralizes a non-trivial 5-subgroup of L u (see [4, page 160]to see that this is possible). Then w may be chosen in H and we have C G ( h u, w i ) has order divisible by 5 in contradiction to Lemma 5.6. Thiseliminates all the possibilities for L u /O ( L u ) and so we conclude thatno involution induces a field automorphism on L t as claimed. (cid:3) Lemma 7.3. We have | T | = 2 .Proof. Suppose that | T | > 2. We have that D is a dihedral groupand, setting R = T × D , we have R ∈ Syl ( F ∗ ( C G ( t )) /O ( C G ( t ))). Let v ∈ S denote a PGL -automorphism of L (which may be trivial). Weremark that T is normal in S , R h v i /T is a dihedral group and wenote that we do not know the action of v on T . Finally, let z be aninvolution in C D ( h v i ). Then, by Lemma 7.2, Ω ( S ) ≤ ( T × D ) h v i andΩ ( C S (Ω ( S ))) ≤ ( T × D ) h v i . Chris Parker and Gernot Stroth If Ω ( T ) ≤ C S (Ω ( S )), we see that h t i = Ω (Φ( C S (Ω ( S )))) . On the other hand, if Ω ( T ) does not centralize Ω ( S ), then v is aninvolution and [Ω ( T ) , v ] = 1. In particular Ω ( T ) ≤ Ω ( S ). Set W = { W | W ≤ Ω ( S ) , W ∼ = Z × Z × Z } . Suppose that there is a W ∈ W with W R . Then W ∩ R has or-der 8 and is either elementary abelian or is isomorphic to Z × Z .In the former case, W ∩ D is elementary abelian of order 4 and as R h v i /T is a dihedral group, we have that C Ω ( S ) ( W ∩ D ) ≤ R , whichis a contradiction as W R . Therefore W ∩ R ∼ = Z × Z and wemay suppose that v ∈ W . Since C Ω ( S ) ( v ) = h t, v, z i , this case cannothappen either. Hence every member of W is contained in R and conse-quently R = hWi char S . Thus h t i = Ω (Φ( Z ( R ))) is a characteristicsubgroup of S . We have shown in both cases that h t i char S. Hence S is a Sylow 2-subgroup of G and Burnside’s Lemma implies t G ∩ Z ( S ) = { t } . Since every involution in T L/O ( L ) is conjugate to an involution in Z ( R ) O ( L ) /O ( L ), we have t G ∩ R = { t } . On the other hand, by the Z ∗ -Theorem, t G ∩ S = { t } and so we mayassume that t ∼ G v . As v Φ( C S ( t )), we have that also t Φ( C S ( v )).This shows that [Ω ( T ) , v ] = 1, which means that v ∼ G vt . Further-more, conjugating by elements from D , we see that v ∼ G vz . Hence also v ∼ G vtz . Thus z and tz are the only involutions in Ω ( Z ( C S ( v ))) = h t, v, z i which are not conjugate to t in G . But then N G ( C S ( v )) nor-malizes h z, tz i and consequently N G ( C S ( v )) ≤ C G ( t ). This contradicts t ∼ G v . Thus T = h t i has order 2 as claimed. (cid:3) Lemma 7.4. No element of C G ( t ) induces a non-trivial field automor-phism of even order on L .Proof. Let R = T D . Suppose that some element of S induces a non-trivial field automorphism on L . Then f must be even and so D ∼ =Dih(2 m ) with m ≥ ( S ) = R h v i where v is a perhaps trivial PGL automorphism of L . Thus we haveΩ ( S ) ∼ = Z × Dih(2 m +1 ) if v = 1 and otherwise Ω ( S ) ∼ = Z × Dih(2 m ).Let z ∈ Z ( D ). Then h z i = Z (Ω ( S )) ∩ Ω ( S ) ′ . Hence h z i char S. Set W = h x | x = z, x ∈ Ω ( S ) i . Then W is characteristic in S and W ∼ = Z × Z . Additionally, we have that C Ω ( S ) ( W ) ∼ = Z m × Z if v = 1 and C Ω ( S ) ( W ) ∼ = Z m − × Z if v = 1. As any field automorphism of L/O ( L ) containedin S centralizes the cyclic group of order 4 in D , we get C S ( W ) = Z h y i where Z ≤ D h v i is cyclic and has index 2 and y induces a non-trivialfield automorphism on L/O ( L ) ∼ = PSL ( p f ). By Lemma 7.2, we alsohave t ∈ h y i . We again aim to show that t G ∩ R = { t } . Let y ∈ h y i with y = t and assume that Z is generated by c . Suppose that [ c, y ] = 1.Then we have that [ D, y ] = 1 and so v = 1 which mean that Ω ( S ) = R . This shows that C S ( R ) = h y , z i and implies h t i = Ω (Φ( C S ( R ))) char S. Assume now that [ c, y ] = 1. Then, as C L ( y ) ∼ = PGL ( p r ) where 2 r = f , c has order at least 8. Furthermore either [ c, y ] = z or [ c, y ] = tz .In both cases [ c , y ] = 1. For h ∈ S , set X h = h s | s ∈ C S ( W ) , s = h i . We have that ( cy ) = cy cy = tc [ c, y ]. As c is of order at least fourwe have that cy is of order at least 8. So for h = z or h = tz we havethat X h ⊆ h c , y i . But then X t = h y , c m − i ∼ = Z × Z while X tz = h c m − y i ∼ = Z . Since C S ( W ) is characteristic in S , we conclude that t and tz are notconjugate by an automorphism of S . Since h z i is characteristic in S ,we deduce that h t i char S. Therefore S ∈ Syl ( G ) and, as every involution in LT O ( C G ( t )) /O ( C G ( t ))is conjugate into Z ( R ) O ( C G ( t )) /O ( C G ( t )), Burnside’s Lemma [9, 6.2]implies t G ∩ R = { t } . By the Z ∗ -Theorem, we may assume that t ∼ G v . We have Ω ( Z ( C S ( v ))) = h t, v, z i . Suppose that there is some field automorphism x such that[ x, v ] ∈ R \ D . Then as there are exactly three LT -classes of involu-tions in R \ D , we deduce that v ∼ G vt . Then, using elements of D , Chris Parker and Gernot Stroth we have v ∼ G vz and vt ∼ G vtz . Hence v ∼ G vz ∼ G vt ∼ G tvz . Thisshows that { z, tz } are the only involutions in h t, v, z i , which are notconjugate to t in G . Hence N G ( C S ( v )) ≤ C G ( t ), and this contradicts t ∼ G v . Therefore S = D h v i Y where Y = h y i is cyclic Ω ( Y ) = h t i and Y ∩ D h v i = 1. Then by the Thompson Transfer Lemma 2.18, we havethat t is conjugate to a 2-central involution in D h v i and so t G ∩ R = { t } ,a contradiction. (cid:3) We now summarize what we have discovered about the structure of C G ( t ). By Lemma 7.3, we have that T = h t i and Lemmas 7.2 and 7.4imply that no element of S induces a field automorphism on L/O ( L ).Thus we have F ∗ ( C G ( t )) /O ( C G ( t )) ∼ = h t i PSL ( p f )and C G ( t ) /O p ′ ( C G ( t )) is isomorphic to either PSL ( p f ), PGL ( p f ) orto PSL ( p f ) . S/ h t i is either a dihedral or semi-dihedral group. Lemma 7.5. Assume Hypothesis 7.1. Suppose that Syl ( C G ( t )) ⊆ Syl ( G ) .Then G has elementary abelian Sylow -subgroups.Proof. Aiming for a contradiction, suppose that S ∈ Syl ( C G ( t )) ⊆ Syl ( G ) and that S is not elementary abelian. Then Z ( S ) = h t, z i where h z i = Z ( S ) ∩ D . Since z is a commutator in S and t is not, z and t are not G -conjugate and so, as S ∈ Syl ( G ), Burnside’s Lemma[9, 6.2] implies that Z ( S ) contains representatives from three distinct G -conjugacy classes. By Hypothesis 4.1 (iv), G = O ( G ) and so wemust have that t ∈ Φ( S ) by Lemma 2.19. Since t ∈ Φ( S ), S/D is cyclicof order 4. In particular, Ω ( S ) = T D . Since T L has exactly threeconjugacy classes of involutions with representatives z , t and zt , wededuce that t G ∩ T L = { t } . Therefore t G ∩ S ⊆ t G ∩ T D = { t } . Finallythe Glauberman Z ∗ -Theorem [6] implies that t ∈ Z ∗ ( G ) = 1 and wehave our contradiction. Hence, if S ∈ Syl ( G ), then S is abelian andconsequently is elementary abelian. (cid:3) We recall that U is a Sylow 2-subgroup of G containing S . Lemma 7.6. U has a normal elementary abelian subgroup of order 4.Proof. As U is not dihedral or semi-dihedral, this follows from [9, 10.11]. (cid:3) Lemma 7.7. S is either elementary abelian or S/T ∼ = Dih(8) . Fur-thermore, if S is not abelian, then there is a fours group in S which isnot contained in Z ( S ) but is normal in U . Proof. We may suppose that S is non-abelian. Let h z i = Z ( S ) ∩ D .Since S is non-abelian, Lemma 7.5 implies that U = S . By Lemma 7.6,there exists a normal elementary abelian subgroup V of U of order 4.As G = O ( G ), the Thompson Transfer Lemma 2.18 implies that t isconjugate to some involution s ∈ C G ( V ) such that U contains a Sylow2-subgroup of C G ( s ). Hence we may assume that V ≤ S .Suppose that V = Z ( S ) = h t, z i . Then, as U = S , | U : S | = 2 andso we can write U = S h x i for some x ∈ U . Since z is a commutatorin S and t is not, t and z are not U -conjugate. Therefore, as U = S , t x = tz . In particular, C Q ( t ) = 1 and O ( L ) = 1.Let h z, s i be a fours group in D . Note that, as S/T contains a di-hedral subgroup of order at least 8, N SL ( h z, s i ) /T ∼ = Sym(4) and that N SL ( h z, s i ) normalizes E = h s, t, z i . Since S/ h t i is either dihedral orsemi-dihedral, E ∈ Syl ( C G ( E )). By considering the action of N L ( E )on E , we see that tz ∼ G ts ∼ G tsz and z ∼ G zs ∼ G z and by assump-tion we have t x = tz . So the involutions in E are partitioned into twosets t G ∩ E of size 4 and z G ∩ E of size 3. Suppose that | E ∩ L x | = 4,then N ( SL ) x ( E ∩ L x ) ∼ = Sym(4). Then we must have E ∩ L x = h z, s i .Since tz is centralized by L x , we infer that N G ( E ) /C G ( E ) ∼ = Sym(4).Let R ∈ Syl ( N G ( E )). Then R/E ∼ = Dih(8). We claim U containsno subgroup R which is isomorphic to R . Assume this is false and let F be the subgroup of R with R /F ∼ = Dih(8). Since S has a cyclicsubgroup of index 4, R has a cyclic subgroup C of index 8 with C ≤ S .Since | R | = 2 , | C | ≥ 8. We conclude that CF/F is cyclic of order 4, Z ( CF ) has order 2 and | C | = 8.If F ≤ S , then, as the 2-rank of S is 3, we have that h t, z i = Z ( S ) ≤ F and as C ≤ S , we have a contradiction to | Z ( CF ) | = 2. Thus F S and ( R ∩ S ) / ( F ∩ S ) ∼ = R /F ∼ = Dih(8). Since C is inverted in R ∩ S , we have that [ R ∩ S, C ] has order 4. Because, | F ∩ S | = 2 ,[ R ∩ S, C ] centralizes F ∩ S and so the structure of S indicates that F ∩ S ≤ Z ( S ) = h t, z i . But then F ∩ S ≤ Z ( R ∩ S ) which hasorder 2 and we have a contradiction. This contradiction arose from theassumption that R was isomorphic to a subgroup of U and in turn thisfollowed from the supposition that | E ∩ L x | = 4. Hence we must have | E ∩ L x | ≤ x does not normalize E and EL x /O ( L x ) ∼ = PGL ( p f ).Therefore S = h t i × D where D is a dihedral group and x can bechosen so that S = h E, E x i and D = h r, s, s x i . In particular, D isnormalized by x and, as all the involutions in D are contained in L ∪ L x (and consequently G -conjugate to z ), t is not G -conjugate toan element of D . Chris Parker and Gernot Stroth Let u ∈ N E x L ( C Q ( t )), be an involution conjugate into D . Then u ∈ H and, as u is conjugate to z , C G ( u ) has Sylow 2-subgroups isomorphicto U . By Lemmas 5.2 and 5.3 there is a 2-component M in C G ( u ).Since | Z ( U ) | = 2, we infer that | Z ( M/O ( M )) | ≥ 2. It follows fromLemma 2.12 that U/O p ′ ( U ) ∼ = PSL ( p a ), Alt(2 p ), PSL (4) with p = 3 orFi with p = 5. In the first case, we have that u is a classical involutionswhich is impossible. In the remaining cases, we have incompatible Sylow2-structure, as the sectional 2-rank of U is 3 whereas the sectional 2-rank of the possible components is at least 4.Thus we have shown that there is a fours group different from h t, z i which is normal in S . This shows that S/T is dihedral of order atmost 8. If S/T is abelian, then S = DT is elementary abelian. Thiscompletes the proof of Lemma 7.7. (cid:3) Lemma 7.8. U = S is elementary abelian.Proof. Let V = h r, s i be a fours group in D . Then there is some elementof order three in L , which acts non-trivially on V and, by Lemma 7.7, V is normal in S . By Lemma 2.25, we have that C G ( V ) has a Sylow2-subgroup, which is an extension of B by h t i where B ∼ = Z n × Z n ,a Sylow 2-subgroup of PSL (4), a Sylow 2-subgroup of SU (4) or iselementary abelian of order 16. We write S = h S ∩ L t , t, y i where, if S is abelian, y = 1 and, if S is non-abelian, y ∈ T . Set U = B h t, y i .Assume first that U = U ∈ Syl ( G ). If B = V , then U = U = S andwe are done by Lemma 7.5. So we suppose further B = V and seek acontradiction. Then t is not G -conjugate to any involution in B , as anysuch involution centralizes an abelian group of order 16. In particular,since G = O ( G ), the Thompson Transfer Lemma 2.18 implies that U = BT and U = h y i B . Thus y is an involution and, using theThompson Transfer Lemma again, we have that t is G -conjugate tosome element y in By and to y ∈ Byt . For i = 1 , 2, we have that y i ∈ N C G ( t ) ( V ) and [ y i , V ] = 1. As N G ( V ) /C G ( V ) ∼ = Sym(3), y i B inverts some element ρB of order three. As y i are conjugate to t , we seethat C B ( y i ) does not contain an elementary abelian subgroup of order8. Hence if B is elementary abelian or a Sylow 2-subgroup of PSL (4),we see that all involutions in By i are conjugate. In the other caseswe have that V = Ω ( B ) and so C B ( ρ ) = 1. Therefore we may applyLemma 2.21 to get that all the involutions in h y i , ρ i B \ B are conjugateto y i again. In all cases we have that | C B ( y i ) | = | B | . In particular, wemay suppose that y = y and that y = yt . Since y is conjugate to t , | C G ( y ) | = 2 . As h y, t i is abelian, it follows that | C B ( y ) | ≤ . Thus | B | = | C B ( y ) | ≤ and so B is either elementary abelian of order16 or is isomorphic to Z × Z . We summarize the conclusions about fusion of involutions in U as follows: all the involutions in By ∪ Byt are G -conjugate to t and t G ∩ B is empty. The coset Bt may or maynot contain G -conjugates of z , where z ∈ C V ( y ) . Since z and t arenot G -conjugate, we have that D = V , LS ∼ = Z × PGL ( p f ) and S ∼ = Z × Dih(8). By considering N U ( S ) = S , we see that t and tz are G -conjugate. Therefore z G ∩ S = V .Suppose that B is elementary abelian. Then | C B ( y ) | = 2 and C B ( y )contains no conjugates of t . Since y is conjugate to t , we see that C B ( y ) consists of conjugates of z . It follows that V and C B ( y ) areconjugate in G . Since | C B ( U ) | = 2, B is the Thompson subgroup of U . Therefore C B ( y ) and V are conjugate in N G ( B ). Hence yC G ( B ), tC G ( B ) and, by an argument similar to the one above, tyC G ( B ) areconjugate in N G ( B ) /C G ( B ). In particular, as tC G ( B ) commutes withan element ρC G ( B ) of order 3 which acts fixed-point-freely on B , yC G ( B ) centralizes ρ y C G ( B ) and tyC G ( B ) centralizes ρ ty C G ( B ) where ρ y C G ( B ) and ρ ty C G ( B ) both have order 3 and both act fixed-point-freely on B . The isomorphism between GL (2) and Alt(8) maps suchelements of order 3 to 3-cycles. Up to conjugacy, T h ρ i C G ( B ) /C G ( B ) = h (1 , , , (1 , , , (1 , , i where t = (4 , , 7) and, say, y =(1 , , y and areinverted by t either move 3 or 8. In the former case we see that N G ( B ) /C G ( B ) contains a subgroup isomorphic to Sym(5) and this con-tradicts the fact that | N G ( B ) /C G ( B ) | = 4. Thus ρ and ρ y commute.Since ρ and ρ y cannot both commute with ρ ty , we have a contradiction.Hence B is not elementary abelian.So we have shown that B ∼ = Z × Z . Suppose that H contains aconjugate z g of z . Then U g ≤ C G ( z g ). By Lemmas 5.2 and 5.3, C G ( z g )has a normal 2-component L r . Since U g L r has Sylow 2-subgroup U g and Z ( U g ) = h z g i , we infer that Z ( L r /O ( L r )) has order divisible by 2. ThusLemma 2.12 implies that L r /O p ′ ( L r ) is isomorphic to one of PSL ( p e ),Alt(2 p ), PSL (4) or Fi . Since | U g /Z ( U g ) | = 2 , we can only have L r /O p ′ ( L r ) ∼ = PSL ( p e ). But then z g is a classical involution which isimpossible by hypothesis. Thus H does not contain G -conjugates of z . Itfollows that H ∩ L has odd order and so Corollary 2.9 implies that p f ∼ =3 (mod 4). On the other hand, L h y i /O ( L ) ∼ = PGL ( p f ) and so L h y i∩ H has order divisible by 2. Hence we may assume that H ∩ L h t, y i contains E = h t, y i and E ∈ Syl ( C H ( t )). Since all the involutions in E areconjugate to t , we have E ∈ Syl ( H ). Furthermore, using Lemma 3.5,we have N H ( E ) /O ( N H ( E )) ∼ = Alt(4). Since all the involutions in E areconjugate, we also have C Q ( t ) = 1. Thus Lemma 5.2 implies that L isa component and [ L, O p ′ ( C G ( t ))] = [ L, O ( C G ( t ))] = 1. Chris Parker and Gernot Stroth In C G ( t ), E normalizes exactly two p -subgroups one of which is C Q ( t ). Since C G ( E ) acts on the set of p -subgroups of C G ( t ) whichare normalized by E , we deduce that | C G ( E ) : C H ( E ) | ≤ 2. Notingthat C C G ( t ) ( E ) has order divisible by 2 , we now have that | C G ( E ) : C H ( E ) | = 2 . As y C G ( t ) yt , we deduce that | N G ( E ) /C G ( E ) | = 3 and O ( N G ( E )) ≤ H . In particular, N G ( E ) has elementary abelian Sylow2-subgroups of order 8. Let E ∈ Syl ( N G ( E )) be chosen so that E = E h z i = h t, y, z i ≤ S where z ∈ Z ( S ) . By the Frattini Argument, N N G ( E ) ( E ) C G ( E ) = N G ( E ) and so there exists an element ρ ∈ N G ( E ) of 3-power orderwhich normalizes but does not centralize E and additionally ρ ∈ C G ( E ). Because t , y , ty , tz , zy and zty are pairwise conjugate and t is not conjugate to z , we have that z G ∩ E = { z } . Thus ρ centralizes z . Since t inverts B , the preimage of C B/ h z i ( t ) in B is the subgroup X = { f ∈ B | f ∈ h z i} ∼ = Z × Z . Now y acts non-trivially onΩ ( B ) = V and, as y has order 2, a short calculation shows that y normalizes every subgroup of X of order 4. It follows that C B/ h z i ( t ) = C B/ h z i ( E ) = X/ h z i . Hence N U ( E ) = E X and N U ( E ) /E is the four group of Aut( E )which centralizes E / h z i . This means that N U ( E ) is extraspecial oforder 32 of +-type. Since y and ty are not conjugate in C G ( t ), we have N G ( E ) /C G ( E ) ∼ = Alt(4). In particular, N U ( E ) ∈ Syl ( N G ( E )). Let F ∈ Syl ( N G ( E )) be chosen so that it is normalized by ρ . Note that F has index 2 in a Sylow 2-subgroup U of G and that the subgroup of U corresponding to B intersects F in a subgroup isomorphic to Z × Z .Also C G ( F ) = h z i C H ( F ) and ρ ∈ C H ( F ). Since F is extraspecial of+-type and order 32, N G ( F ) /F C G ( F ) is isomorphic to a subgroup ofO +4 (2) ∼ = Sym(3) ≀ Sym(2). As N G ( F ) /F C G ( F ) has Sylow 2-subgroupsof order two and, as these subgroups consist of a non-trivial elementwhich centralizes Z × Z , we see that N G ( F ) /F C G ( F ) ∼ = Sym(3) or( Z × Z ) : 2. In both cases O ( N G ( F ) /F C G ( F )) is inverted. Supposethe latter. We now have that the 9 involutions of F O ( R ) /O ( R ) areconjugate in R/O ( R ). Since V ≤ F , we get that all the involutions in F are conjugate to z , but t ∈ F and, as z and t are not conjugate, wehave a contradiction. We have that U = F h u i where u ∈ B has order 4.And N G ( F ) = h u, ρ i F C G ( F ), C F ( u ) = B ∩ F ∼ = Z × Z . As t inverts B all elements in Bt are involutions. Hence there are involutions in B h t i \ F . Choose such an involution w . Then w centralizes Ω ( B ). So Z ( B ∩ F ) h w i ) = Ω ( B ). In particular, B ∩ F is not centralized by w , but B ∩ F is normalized by h w i . It follows that W = ( B ∩ F ) h w i ∼ =Dih(8) × 2. Now W = [ F, w ] h w i and so W is normal in U = F h w i . SinceAut( W ) is a 2-group, N G ( W ) = C G ( W ) U . Set C G ( z ) = C G ( z ) / h z i .Then, as N G ( W ) = C G ( W ) U , we have C F ( w ) h w i is a Sylow 2-subgroupof C C G ( z ) ( w ). In particular, w is not C G ( z )-conjugate to a subgroupof F . Therefore, the Thompson Transfer Lemma 2.18, implies that C G ( z ) has a subgroup R of index two with elementary abelian Sylow 2-subgroup F of order 16. Recall that ρ acts fixed point freely on V ≤ F and so [ F , ρ ] = F . Set e R = R/O ( R ) h z i .Suppose that E ( e R ) = 1 and recall that R is a K -group by hypoth-esis. Then, as e R has elementary abelian Sylow 2-subgroups of order16 which admit a fixed-point-free element of order 3, we deduce that R/O ( R ) h z i has either one or two components isomorphic to PSL ( l )for some prime l ≡ , (16). Since F ∈ Syl ( R ) and N G ( F ) /F C G ( F ) ∼ = Sym(3),we see that just one component PSL ( l ), l ≡ , F is quaternion of order eigth, contradicting [ F, ρ ] = F . This contradiction finally shows that E ( e R ) = 1. In particular, C G ( z ) = O ( C G ( z )) N G ( F ) and, as h z i ∈ Syl ( C C G ( z ) ( ρ )), h z i is a Sylow2-subgroup of C G ( ρ ). Therefore C G ( ρ ) has a normal 2-complement. Weclaim that Syl p ( C G ( ρ )) = ∅ . We first investigate ρ which we assumefor now is non-trivial. Recall that ρ ∈ H and ρ ∈ C G ( E ) ≤ C G ( t ).Thus ρ normalizes C Q ( t ). Since C L ( z ) ∩ H = 1, and ρ ≤ C G ( z ),we have ρ L . We have C C G ( t ) ( L ) = O p ′ ( C G ( t )). If L h ρ i is iso-morphic to a direct product A × L (necessarily with A ≤ H ), then A h ρ i ∩ L ≤ H ∩ C L ( z ) = 1, which is impossible unless ρ ∈ A . So inthis case we have C Q ( t ) ≤ C Q ( ρ ). So suppose that L h ρ i is not iso-morphic to a direct product. Hence ρ induces an outer automorphismof L . But then, as ρ is a 3-element, ρ induces a field automorphism on L and C C Q ( t ) ( ρ ) = 1. Hence in each case we have C Q ( ρ ) = 1. Since E ∈ Syl ( C H ( t )), E commutes with ρ and E inverts C Q ( t ), we havethat, setting J = C Q ( ρ ), J = C J ( t ) × C J ( ty ) × C J ( y ). Hence ρ central-izes a diagonal element in this decomposition of J and so C J ( ρ ) = 1.This then proves our claim. Since h z i is a Sylow 2-subgroup of C G ( ρ )and C G ( ρ ) has a normal 2-complement, we now see that h z i is con-tained in a conjugate of H . But then h z i is conjugate to an element of E and this means that z and t are G -conjugate. We have proved thatthis is not the case and so at this stage we deduce U = U . Chris Parker and Gernot Stroth Assume B = V and U = U . If B is not elementary abelian oforder 16, then we claim that V is a characteristic subgroup of U . Set U = BT . If U = U , then V = Z ( U ) and we are done. Thus y = 1and | U : U | = 2. Suppose that α ∈ Aut( U ) with V = α ( V ). Then U = α ( U ). Hence | U : U ∩ α ( U ) | = 2. Assume that B is homocyclic.Since B ∩ α ( B ) is centralized by Bα ( B ) and t inverts B , we either havethat tB Bα ( B ) or B ∩ α ( B ) is elementary abelian. In the lattercase we have that α ( V ) = Ω ( α ( B )) = Ω ( B ∩ α ( B )) = Ω ( B ) = V ,which we supposed was not the case. Therefore, Bα ( B ) = U and so | B : B ∩ α ( B ) | = 2 and again we have α ( V ) = V . So we may assumethat B is non-abelian. In particular, we have that | B | = 2 . Then B ∩ α ( U ) has order 2 . Now U /α ( B ) is abelian and so V = B ′ ≤ α ( U )and α ( V ) = α ( B ) ′ ≤ B . Thus Z ( B ∩ α ( U )) ≥ V α ( V ) > V , thiscontradicts the structure of B as the centre of every subgroup of B ofindex 2 is V . Thus V is a characteristic subgroup of U and our claimis proved.Set U = N U ( U ). As BT ∈ Syl ( C G ( V )), we deduce that | U : U | =2 and that U = BT . In particular, we have y = 1 and S is elementaryabelian of order 2 . Since C U ( t ) = S ≤ U , there must be at leasttwo BT -conjugacy classes of involutions in Bt . It follows from Corol-lary 2.24 that B is homocyclic and there are four BT -conjugacy classesof involutions in Bt . Recall that there is an element ρ ∈ N L ( V ) of or-der 3 and that ρ normalizes B , centralizes t and C B ( ρ ) = 1. Thus Bt contains two B h t, ρ i classes of involutions. Now U does not centralize t and so we deduce h U , ρ i induces a transitive action on the four BT -conjugacy classes of involutions in Bt . It follows that | N G ( U ) : U | isdivisible by 4. But this contradicts BT ∈ Syl ( C G ( V )) and V beingnormalized by N G ( U ).So we have shown that B is elementary abelian of order 16. Note that B is characteristic in U . By Corollary 2.24, all the involutions in Bt areconjugate in BT . Therefore, as U < U , we deduce that U = BS > BT and so, in particular, | U | = 2 . By Lemma 7.7, there is a non-centralfours group X of S which is normal in U . Since U ∈ Syl ( N G ( V )) and U > U , X = V . We have that X ≤ S ≤ BS and so X normalizes B and B normalizes X . Therefore [ B, X ] ≤ B ∩ X ≤ B ∩ S = V .Since X = V , we infer that [ B, X ] = B ∩ X = h z i = Z ( S ). However, | [ B, x ] | = 2 for all x ∈ BS \ B and so we have a contradiction.So we are left with the case that B = V and U > U = S = h V, t, y i .Let U = N U ( S ) > S . We claim that V T is normal in U . This isobviously true if y = 1. So suppose that S is non-abelian. Then, byLemma 7.7, there is a fours group X ≤ S which is normal in U andis not contained in Z ( S ). It follows that T X is an elementary abelian group of order 8. Since S contains exactly two such subgroups, we de-duce that both T X and T V are normal in U . Let E = V T . Since U does not centralize t and since E is normalized by an element of order 3in L , we see that | t G ∩ E | ≥ 4. Since Z ( U ) ∩ E = 1, we get | t G ∩ E | = 4.Therefore, | N G ( E ) /C G ( E ) | = 12 or 24. As N G ( E ) /C G ( E ) is a subgroupof PSL (2), this means N G ( E ) /C G ( E ) ∼ = Alt(4) or Sym(4). In bothcases C E ( O ( N G ( E ) /C G ( E ))) is non-trivial and normal in N G ( E ) andso it must be V . But E = BT ∈ Syl ( C G ( V )) and we have a contradic-tion. This is our final contradiction and so Lemma 7.8 is proved. (cid:3) Lemma 7.9. L is a component of C G ( t ) .Proof. From Lemma 7.8, we have G has elementary abelian Sylow 2-subgroups of order 8 and O ′ ( C G ( t ) /O ( C G ( t ))) ∼ = 2 × PSL ( p f ) where f ≥ 3. Assume that the lemma is false. By Lemma 5.2, we have C Q ( t ) =1. Let P ∈ Syl p ( H ∩ L ). If [ P, O p ′ ( C G ( t ))] = 1, then P centralizes F ( O p ′ ( C G ( t ))) and so F ∗ ( C G ( t )) O ( C G ( t )). Therefore, C G ( t ) has acomponent and of course it must be L , which is a contradiction. Thus,as T ∈ Syl ( O p ′ ( C G ( t ))), there exists an odd prime r such that r = p , r divides | F ( O ( C G ( t ))) | and P acts faithfully on R = O r ( O ( C G ( t ))).Let C be a critical subgroup of R . Then, as r is odd, we may assumethat C is of exponent r (see [7, Theorem 5.3.13]). Choose D ≤ C minimal so that C C G ( t ) ( D ) ≤ O p ′ ( C G ( t )) and D is normal in C G ( t ).Then, as Φ( D ) < D , C C G ( t ) (Φ( D )) covers O ′ ( C G ( t ) /O p ′ ( C G ( t ))) ∼ =PSL ( p f ). In particular, as P acts faithfully on D , P acts faithfullyon D/ Φ( D ). Since P is elementary abelian of order p f and N L ( P ) hasorbits of length ( p f − / (2( p − P , wehave | D/ Φ( D ) | ≥ r ( p f − / ( p − . Because f ≥ 3, ( p f − / ( p − ≥ | D/ Φ( D ) | ≥ r . In particular, as D is normal in C G ( t ) and F ∗ ( C G ( t ) /O ( C G ( t ))) is generated by three involutions, we have that | D/ Φ( D ) : C D/ Φ( D ) ( x ) | > r for all involutions x ∈ C G ( t ).On the other hand, as the Sylow 2-subgroups of C G ( t ) / h t i have order4 with all involutions conjugate, we also have | C D/ Φ( D ) ( x ) | > r for allinvolutions x ∈ C G ( t ). Therefore, for involutions x ∈ C G ( t ), C D ( x ) con-tains an elementary abelian subgroup of order r . Let W ≤ D be an el-ementary abelian subgroup of order r centralizing some involution s ∈ L \ H . Then, as W ≤ O p ′ ( C G ( t )) ≤ H and C G ( Q ) ≤ Q , W acts faith-fully on Q . Since t inverts Q , Q is abelian and so there is a non-trivial w ∈ W such that | C Ω ( Q ) ( w ) | ≥ p . Then, as s ∈ C G ( w ), C G ( w ) H and m p ( C H ( w )) ≥ 2. Hence Lemma 3.2 (v) implies that O p ′ ( C G ( w )) ≤ Chris Parker and Gernot Stroth H and Lemma 3.3 (ii) implies that F ∗ ( C G ( w ) /O p ′ ( C G ( w ))) is a non-abelian simple group. Since C Q ( w ) = 1 and | [ C Q ( w ) , O p ′ ( C G ( w ))] | = 1,we have that F ∗ ( C G ( w )) O p ′ ( C G ( w )) and so C G ( w ) has a normalcomponent L w with L w O p ′ ( C G ( w )). Since L w has abelian Sylow2-subgroups of order at most 8, we may apply [17] to deduce that L w is a K -group. Hence L w ∼ = PSL ( p b ), b ≥ G (3 c +1 ), c ≥ (8). Now, as D ≤ H , C D ( w ) O p ′ ( C G ( w )) /O p ′ ( C G ( w )) normalizes C Q ( w ) O p ′ ( C G ( w )) /O p ′ ( C G ( w )) and is an r -group of exponent r . It fol-lows from Proposition 2.7 that | C D ( w ) : C D ( w ) ∩ O p ′ ( C G ( w )) | ≤ r .Recall that t centralizes w and so there is an involution x ∈ L w which centralizes t . We have O p ′ ( C G ( w )) ≤ C G ( L w ) ≤ C G ( x ) and so | C D ( w ) : C D ( w ) ∩ C G ( w ) | ≤ r . Since | D/ Φ( D ) : C D/ Φ( D ) ( w ) | > r , weinfer that w Z ( W ) ≥ Φ( D ). Because Φ( D ) ∩ O p ′ ( C G ( t )) centralizes C Q ( w ), we have Φ( D ) ∩ O p ′ ( C G ( t )) = 1 for otherwise w could have beenchosen in Φ( D ) which we know is not possible. On the other hand,as h w i Φ( D ) is elementary abelian and [ D, Φ( D )] = 1, we know that | D : C D ( w ) | ≤ | Φ( D ) | . As Φ( D ) ≤ C D ( w ) and Φ( D ) ∩ O p ′ ( C G ( w )) = 1, | C D ( w ) : Φ( D )( C D ( w ) ∩ O p ′ ( C G ( w ))) | ≤ r / | Φ( D ) | . Finally we have | D/C D ( w ) | ≤ | Φ( D ) | and | C D ( w ) : C C D ( w ) ( x ) | ≤ r / | Φ( D ) | which onceagain gives | D/ Φ( D ) : C D/ Φ( D ) ( x ) | ≤ r our final contradiction. Thus L is a component of C G ( w ). (cid:3) Theorem 7.10. Assume Hypothesis 7.1. Then F ∗ ( G ) ∼ = G (3 a +1 ) .Proof. By Lemma 7.8 the Sylow 2-subgroups of G are elementary abelianof order 8 and, by Lemma 7.9, L ∼ = PSL ( p f ) is normal componentof C G ( t ). Therefore Theorem 2.3 gives F ∗ ( G ) ∼ = G (3 a +1 ) for some a ≥ (cid:3) Components in H From Lemma 4.5, we know that if E ( H ) = 1, then E ( H ) is qua-sisimple. The objective of this section is to prove that E ( H ) = 1. Thusthroughout this section we assume the following hypothesis. Hypothesis 8.1. Hypothesis 4.1 holds as well as (i) E ( H ) is a quasisimple K -group; and (ii) G does not contain a classical involution. Set E = E ( H ). So E is a quasisimple group. Since O p ′ ( G ) = 1, wehave that F ( H ) is a p -group. In particular, E contains involutions.If t ∈ H is an involution, we know by Lemma 4.4 that C G ( t ) H , O p ′ ( C G ( t )) ≤ H and ( C G ( t ) /O p ′ ( C G ( t )) , p ) ∈ E . Whenever t isan involution from H , we use the following bar notation C G ( t ) = C G ( t ) /O p ′ ( C G ( t )). Lemma 8.2. E = F ∗ ( H ) .Proof. Set Q = O p ( H ) and assume that Q E ( H ). Of course [ E, Q ] =1. Suppose that t is an involution in E . Then, by Lemma 4.4, C G ( t ) H , O p ′ ( C G ( t )) ≤ H and ( F ∗ ( C G ( t )) , p ) ∈ E . Since Q and O p ′ ( C H ( t ))commute, we have that F ∗ ( C G ( t )) O p ′ ( C G ( t )). As F ∗ ( C G ( t )) is al-most simple, C G ( t ) has a unique component L t which has order divisibleby p and we have L t = F ∗ ( C G ( t )). Note that, as Q = 1, Proposition 2.7implies that C H ( t ) is p -closed and soluble.Assume that p divides | C E ( t ) /Z ( E ) | . Then O p ( C H ( t )) contains non-trivial normal subgroups Q and O p ( C E ( t )) and they commute. Thiscontradicts the structure of C H ( t ) given in Proposition 2.7. Therefore, C E ( t ) ≤ O p ′ ( C H ( t )) for all involutions t ∈ E . Suppose that C E ( t ) O p ′ ( C G ( t )). Then using Corollary 2.9 and noting that C H ( t ) is soluble,we get | C E ( t ) | = 2 and ( L t , p ) = (PSL (4) , Q is a normal3-subgroup of C H ( t ) and so Proposition 2.7(iv) implies that N L t ( Q ) /Q is a non-abelian 2-group. Suppose that s ∈ C H ( E ) is an involution.Then EO ′ ( C G ( s )) /O ′ ( C G ( s )) is a quasisimple normal subgroup of C H ( s ) /O ′ ( C G ( s )) which by Lemma 4.4(iii) is strongly 3-embedded in C G ( s ) /O ′ ( C G ( s )). However, as ( C G ( s ) /O ′ ( C G ( s )) , ∈ E , we have acontradiction to Proposition 2.7 as, when p = 3, C H ( t ) must be sol-uble. Thus C H ( E ) has odd order and, in particular, N L t ( Q ) /Q mapsisomorphically into H/C H ( E ) and so we deduce that Out( E ) has non-abelian Sylow 2-subgroups. Now E is a K -group and since Out( E ) hasnon-abelian Sylow 2-subgroups, we infer that E is a Lie type groupof Lie rank at least 4 defined over a field of characteristic r . Butthese groups all possess an involution s with C E ( s ) involving PSL ( r ).Since 3 divides | PSL ( r ) | , this contradicts C E ( s ) ≤ O ′ ( C H ( s )). Thus( L t , p ) = (PSL (4) , 3) and we conclude that C E ( t ) ≤ O p ′ ( C G ( t )) for allinvolutions t ∈ E .Suppose now that T ∈ Syl ( E ) and t ∈ Z ( T ) is an involution. Then T ≤ C E ( t ) ≤ O p ′ ( C G ( t )) ≤ C G ( L t ) . Assume that s ∈ T is an involution. Then C E ( s ) ≤ O p ′ ( C G ( s )) andthus[ C E ( s ) , h Q C G ( s ) i ] ≤ [ O p ′ ( C G ( s )) , h Q C G ( s ) i ] = h [ O p ′ ( C G ( s )) , Q ] C G ( s ) i = 1 . As L t ≤ C G ( s ), we see h Q C G ( s ) i ≥ h Q L t i = L t Q . Thus L t centralizes C E ( s ). It follows that L t centralizes K = h C E ( x ) | x ∈ T, x = 1 i .If p divides | K | , then we have the contradiction L t ≤ H as H isstrongly p -embedded. Hence p does not divide | K | and so K = E . Chris Parker and Gernot Stroth It follows from [9, 17.13] that N E ( K ) is a strongly 2-embedded sub-group of E . Hence E ∼ = SL (2 a ), PSU (2 a ) or B (2 a − ) for some a ≥ K ≤ N G ( T ) by [2]. Since L t centralizes T and L t is nor-mal in C G ( t ), we see that L t is a characteristic subgroup of C G ( T ).Hence N G ( T ) normalizes L t . In particular, N G ( T ) = N H ( T ) L t by theFrattini Argument. By Lemmas 3.2(v) and 3.3 (ii) and (iii), we have F ∗ ( N G ( T ) /O p ′ ( N G ( T ))) ∼ = L t /Z ( L t ). Thus using Corollary 2.9 and thefact that L t /Z ( L t ) = PSL (4), we have O p ′ ( N G ( T )) = O p ′ ( N H ( T )). If p divides | N E ( T ) | , then N E ( T ) O p ′ ( N G ( T )) /O p ′ ( N G ( T )) commutes with QO p ′ ( N G ( T )) /O p ′ ( N G ( T )) and as before we have a contradiction tothe structure of L t via Proposition 2.7. Hence N E ( T ) ≤ O p ′ ( N H ( T )) = O p ′ ( N G ( T )). In particular, L t centralizes N E ( T ). Let J be a comple-ment to T in N E ( T ) and set M = N E ( J ). Then E = h N E ( T ) , M i .Since L t centralizes N E ( T ), L t ≤ C G ( J ). Therefore, Lemma 3.3 (ii) and(iii) imply that L t is the unique component in N G ( J ). In particular, M normalizes L t and so E = h N E ( T ) , M i normalizes L t as well. Butagain by Lemma 3.3 (ii) and (iii), we have that F ∗ ( L t E/O p ′ ( L t E )) = L t O p ′ ( L t E ) /O p ′ ( L t E ) which means that E ≤ O p ′ ( L t E ) and contradicts O p ′ ( H ) = 1. This contradiction shows that Q ≤ E and so F ∗ ( H ) = E is quasisimple. (cid:3) Because of Lemma 8.2, we may now assume that F ∗ ( H ) /Z ( F ∗ ( H ))is a simple group. We will now proceed to investigate the possibilitiesfor H using the fact that H is a K -group. We begin with two lemmaswhich will be used frequently. Lemma 8.3. If t ∈ H is an involution and C H ( t ) is not soluble, theneither (i) ( F ∗ ( C G ( t )) , p ) = (Fi , and C H ( t ) ∼ = Aut(Ω +8 (2)) ; or (ii) ( F ∗ ( C G ( t )) , p ) = (Alt(2 p ) , p ) and F ∗ ( C H ( t )) ∼ = Alt( p ) × Alt( p ) with p ≥ .Moreover, in case (ii), the components of C H ( t ) are not normal in C H ( t ) .Proof. This follows from Lemma 4.4(iii) and Proposition 2.7. (cid:3) Lemma 8.4. Assume that t ∈ H is an involution and | C H ( t ) | p = | C E ( t ) | p . Then ( C G ( t ) , p ) = (PSL (8) : 3 , or ( B (32) : 5 , . Inparticular, C G ( t ) has extraspecial Sylow p -subgroups of order p .Proof. Suppose that | C E ( t ) | p = | C H ( t ) | p . Then C H ( t ) > C E ( t ) and,as C H ( t ) /C E ( t ) is soluble and has order divisible by p , Proposition 2.7implies that C H ( t ) is p -closed. Let T ∈ Syl p ( C H ( t )) and let T = T ∩ E . Then T > T and T is normal in C H ( t ). It follows from Corollary 2.8that either T ≤ Φ( T ) or ( C G ( t ) , p ) = (PSL (8) : 3 , 3) or ( B (32) :5 , ≤ m p ( T /T ) ≤ m p ( H/E ). Since the p -rank of Out( E ) is at most 2, we have that T /T has p -rank 2 byLemma 2.16. But then the structure of Out( E ) in Lemma 2.16 showsthat C H ( t ) cannot act irreducibly on T / Φ( T ) and Corollary 2.8 onceagain gives p ∈ { , } and the structure of C G ( t ). (cid:3) Lemma 8.5. E/Z ( E ) is not an alternating group of degree n ≥ .Proof. Suppose that E/Z ( E ) ∼ = Alt( n ) for some n ≥ 5. Assume that Z ( E ) = 1. Let n ≥ 9. Let t correspond to a product of two transpo-sitions in E . Then C H ( t ) contains a normal subgroup isomorphic toAlt( n − 4) with n − ≥ 5. Thus C H ( t ) contains such a normal sub-group and this contradicts Lemma 8.3. So n < 9. But then C E ( t ) is a { , } –group with cyclic Sylow 3-group and so m r ( C H ( t )) < r . This contradicts Hypothesis 4.1(ii).This contradiction shows that Z ( E ) = 1. Since Z ( E ) is a p -group,this means E ∼ = 3 . Alt(6) or 3 . Alt(7) with p = 3. The first possibil-ity fails as the centralizers of involutions in Alt(6) have order 8. Thus E ∼ = 3 . Alt(7). In this case, C H ( t ) has a normal Sylow 3-subgroup oforder 9 and C H ( t ) does not act irreducibly on O ( C H ( t )), this contra-dicts Corollary 2.8. Hence we have shown E/Z ( E ) is not an alternatinggroup. (cid:3) Next we show that E/Z ( E ) cannot be a sporadic simple group. Lemma 8.6. E/Z ( E ) is not a sporadic simple group.Proof. We use Lemmas 4.4 and 8.3 for all the sporadic groups. We firstobserve that the outer automorphism group of a sporadic simple grouphas order dividing 2 [10, Table 5.3]. Hence E has index at most 2 in H . In Table 2, for each possibility for E/Z ( E ), we give the structureof the centralizer of some involution t in E (recall O ( E ) = 1). Foreach case, except E ∼ = 3 . M with p = 3, we see that either p does notdivide | C E ( t ) | or that C H ( t ) /O p ′ ( C H ( t )) has a simple section which isnot isomorphic to Alt( p ) or Ω +8 (2). Assume that E ∼ = 3 . M and p = 3.Then C H ( t ), with t as in Table 2, is soluble and has Sylow 3-subgroupsof order 9. However, C H ( t ) does not act irreducibly on O ( C H ( t )) andthis then contradicts Corollary 2.8. Thus this possibility cannot occureither. Therefore E/Z ( E ) is not a sporadic simple group. (cid:3) We now begin our investigation of the case when E/Z ( E ) is a groupof Lie type. Chris Parker and Gernot Stroth E/Z ( E ) Involution C E/Z ( E ) ( t ) E/Z ( E ) Involution C E/Z ( E ) ( t )M A GL (3) M A × Sym(5)M A . Sym(4) M A : PSL (2)M A . PSL (2) J A × Alt(5)J A − . Alt(5) J A − . Alt(5)J A . . M . A . PSp (2)Co A . PSp (2) Co C . M HS 2 A ∗ . Sym(5) McL 2 A . Alt(8)Suz 2 A − . Ω − (2) He 2 A . PSL (4) . A . Alt(11) Ru 2 A . Sym(5)O’N 2 A . PSL (4) . A . PSU (2)Fi A . Fi Fi ′ A . Fi . A . HS . A . Alt(9)B 2 A . E (2) . A . B Table 2. Centralizers of certain involutions in sporadicsimple groups. Lemma 8.7. Suppose that E/Z ( E ) is a simple group of Lie type. Then p divides | C E ( t ) | for all involutions t ∈ H .Proof. Assume that E/Z ( E ) is a Lie type group defined in charac-teristic r and that C E ( t ) is a p ′ -group. Then, as m p ( C H ( t )) ≥ | C E ( t ) | p = | C H ( t ) | p and hence Lemma 8.4 implies that p ∈ { , } and C H ( t ) has extraspecial Sylow p -subgroups. It follows that H/E andhence Out( E ) has non-abelian Sylow p -subgroups. In particular, wesee that E must admit diagonal automorphisms of order p . This showsthat E/Z ( E ) is PSL n ( q ) or PSU n ( q ) and that p divides n . Since H/E has non-abelian Sylow p -subgroups, we further infer that p divides n .Thus n ≥ 9. But then the canonical form of t shows that the central-izer of t has a section isomorphic to PSL ( r ) or PSU ( r ). Both thesegroups have order divisible by 3 and 5, a contradiction. Thus p divides | C E ( t ) | . (cid:3) Lemma 8.8. E/Z ( E ) is not a rank Lie type group.Proof. Suppose that E/Z ( E ) is defined in characteristic 2. So E/Z ( E ) ∼ =PSL (2 a ), PSU (2 a ) or B (2 a − ) for some a ≥ 2. In the first andthird case, the centralizer of an involution in C E ( t ) is a 2-group andso m p (Out( E )) ≥ 2. Using [10, Theorem 2.5.12] we see that Out( E )is cyclic, which is a contradiction. Therefore E/Z ( E ) ∼ = PSU (2 a )for some a ≥ 2. Let t ∈ E be an involution and S ∈ Syl p ( C H ( t )).Then C E ( t ) is 2-closed, contains a Sylow 2-subgroup T of E and T has a cyclic complement in C E ( t ) of order dividing q + 1. In partic-ular, | C E ( t ) | p = | C H ( t ) | p and so S is extraspecial of order p and( C G ( t ) , p ) = (PSL (8) : 3 , 3) or ( B (32) : 5 , 5) by Lemma 8.4. SinceOut( E ) has abelian Sylow p -subgroups, p must divide q + 1. Further-more, the Sylow p -subgroups of GU (2 a ) are abelian and so SE mustinvolve field automorphisms of E . In particular, we have p divides a and Z ( T ) is not centralized by S . From the structure of C G ( t ), T ≤ O ( C H ( t )) ≤ O p ′ ( C H ( t )) = O p ′ ( C G ( t )). Suppose that T g = T for some g ∈ C G ( t ). Then T g ≤ O ( C H ( t )) and T g E/E is a non-trivial 2-groupof outer automorphisms of E . It follows that T g E/E is cyclic and actsnon-trivially on the cyclic group C E ( t ) /T . However T g ≤ O ( C H ( t ))and so this is impossible. Thus T is normalized by C G ( t ). Since Z ( T )is centralized by C E ( t ), L = h C E ( t ) C G ( t ) i centralizes Z ( T ). Since p divides | C E ( t ) | , the structure of C G ( t ) shows that L ≥ O p ( C G ( t )). Sup-pose that 2 a = 8. Then, as | C E ( t ) /T | = (2 a + 1) / (2 a + 1 , ω ∈ C E ( t ), o ( ω ) = r , r a Zsygmondi prime divid-ing 2 a + 1. Note that r does not divide | C G ( t ) | and that ω acts ir-reducibly on T /Z ( T ). Thus C G ( t ) acts on T /Z ( T ) irreducibly. Since | T | = 2 a with | Z ( T ) | = 2 a , using [14, Lemma 2.7.3] we have that C G ( t ) /C G ( T ) T embeds into SL (2 a ) : a . Since C G ( T ) ≤ O p ′ ( C G ( t )) wehave that C G ( t ) /C G ( T ) T involves SL (8) : 3 when p = 3 and involves B (32) : 5 when p = 5 and these groups must be sections of SL (2 a ) : a .Furthermore, r divides O p ′ ( C G ( t ) /C G ( T ) T ) which is thus non-trivial.The structure of SL (2 a ) now implies that C G ( t ) /C G ( T ) T is soluble,a contradiction. So we are left with 2 a = 8. Let U be a hyperplane of Z ( T ). Then T /U is extraspecial of order 2 . Since L centralizes Z ( T ), L acts on T /U . Furthermore, L/C L ( T ) has a section isomorphic to SL (8)when p = 3 or to B (32) when p = 5. Since Out( T /U ) is isomorphicto O ± (2), this is also impossible. Hence we have shown that E is not arank 1 group defined in characteristic 2.Assume that E/Z ( E ) is PSL ( r a ) with r odd. Since PSL (9) ∼ =Alt(6), Lemma 8.5 implies that r a = 9. Thus the Schur multiplierof E has order 2 and so E ∼ = PSL ( r a ). Let t be an involution in E .Then C E ( t ) is a dihedral group and hence we have | C H ( t ) | p > | C E ( t ) | p and Lemma 8.4 implies that either p = 3 and C G ( t ) ∼ = PSL (8) : 3or p = 5 and C G ( t ) ∼ = B (32). Let S ∈ Syl p ( C H ( t )). Then S isextraspecial of order p . Since the Sylow p -subgroups of Out( E ) arecyclic, we must have S ∩ E is cyclic of order p . In particular, as C E ( t ) is a dihedral group, we have Q = O p ( C E ( t )) > 1. Thus, be-cause O p ′ ( C G ( t )) = O p ′ ( C H ( t )), [ Q, O p ′ ( C G ( t ))] = [ Q, O p ′ ( C H ( t ))] = 1.Therefore L = h Q C G ( t ) i centralizes O p ′ ( C G ( t )) and so, as the Schur Chris Parker and Gernot Stroth multipliers of PSL (8) and B (32) are both trivial and Q is invertedin N C G ( t ) ( Q ), we have that L is a normal component of C G ( t ). Thus C G ( t ) = ( L × O p ′ ( C H ( t ))) S . Let T ∈ Syl ( C H ( t )). Then as E hasone conjugacy class of involutions, T ∈ Syl ( H ) and T normalizes Q .Thus we can write T = ( T ∩ L ) × ( T ∩ O p ′ ( C G ( t ))). Let T L = T ∩ L .Then T L is cyclic of order p − T L ≤ Z ( T ). In par-ticular, Z ( T ) ≥ h t i T L , t T L and | Z ( T ) | ≥ 4. Assume that E hasnon-abelian Sylow 2-subgroups. Then T = h t i T L and T L ∩ E = 1 forotherwise | Z ( T ) ∩ E | ≥ 4. Let f ∈ Ω ( T L ) ♯ . Then f centralizes a Sylow2-subgroup of E and so we have that f induces a non-trivial a fieldautomorphism of E . Thus, since r a = 3 , C H ( f ) contains a component F isomorphic to PSL ( r a/ ) and a is even. As a is even, we have that C E ( t ) is of order r a − p divides r a − 1. In particular, p divides | PSL ( r a/ ) | = r a/ ( r a − / 2. HencePSL ( r a/ ) is isomorphic to a subgroup of C H ( f ) /O p ′ ( C G ( f )) whichis consequently not soluble. Now Lemma 8.3 delivers a contradiction.It follows that E has abelian Sylow 2-subgroups and, as | Z ( T ) | ≥ H = E ∼ = PSL ( r ) where r ≡ , T = h t i T L .Therefore T ≤ E and T L has order 2 = p − 1. Therefore p = 3 and L ∼ = PSL (8). Let R ∈ Syl ( C G ( t )) such that R ≥ T . Then R is ele-mentary abelian of order 16. Since R ∈ Syl ( C G ( t )), R ∈ Syl ( C G ( R )).Note that N L ( R ) /R is cyclic of order 7. Thus N L ( R ) induces orbits oflength 1, 7 and 7 on the non-trivial elements of R . Since the non-trivialelements of T are all conjugate, T L ≤ L , and te L where e ∈ T L ,we infer that all the involutions in R are conjugate to t . In particular, R ∈ Syl ( C G ( r )) for all r ∈ R ♯ . Therefore N G ( R ) acts transitively on R ♯ and N G ( R ) /C G ( R ) is a subgroup of GL (2) divisible by 3 . . (2) and so we have ourfinal contradiction to this configuration. Hence E/Z ( E ) = PSL ( r a )for odd r .So we are left with E/Z ( E ) ∼ = PSU ( r a ) or G ( r a ), where r = 3, a > (3)), no involution centralizes a groupof order p , we get that r a = 3. Therefore in all the cases, there is aninvolution t such that E ( C E ( t ) / h t i ) ∼ = PSL ( r a ) is a non-abelian simplegroup. If p divides | PSL ( r a ) | , then Lemmas 8.3 gives a contradiction.So we have that p does not divide | PSL ( r a ) | . As C E ( t ) = GU ( r a ) or h t i × PSL ( r a ), we get that C E ( t ) is a p ′ -group and this contradictsLemma 8.7. (cid:3) Lemma 8.9. E/Z ( E ) = PSL (2 a ) for a ≥ .Proof. Suppose that E/Z ( E ) ∼ = PSL (2 a ), a ≥ 1. Then E has exactlyone conjugacy class of involutions. If a = 1, then C E ( t ) is dihedral of order 8 and Out( E ) has order 2, so H does not satisfy our Hypothe-sis 4.1(ii). So we may assume that a ≥ t ∈ E be an involution. Then T = O ( C E ( t )) ∈ Syl ( C E ( t )).Furthermore, a complement to T in C E ( t ) is cyclic of order dividing2 a − 1. Hence, again by Hypothesis 4.1 (ii), | C H ( t ) | p > | C E ( t ) | p . ThusLemma 8.4 implies that p ∈ { , } and C G ( t ) has extraspecial Sylow p -subgroups. Since Out( E ) has abelian Sylow p -subgroups, we have | C E ( t ) | p > p divides 2 a − O ( C H ( t ) /O p ′ ( C G ( t )) ∩ F ∗ ( C G ( t ) /O p ′ ( C G ( t )))) = 1. Inparticular, T is normal in O p ′ ( C G ( t )) and, using Lemma 2.27, we nowsee that T is normal in C G ( t ).By considering T as the subgroup of lower unitriangular matrices inSL (2 a ), we see that T contains exactly two elementary abelian sub-groups F and F of order 2 a (all the elements of T outside these twosubgroups have order 4). Thus, as T is normal in C G ( t ), C G ( t ) per-mutes F and F and consequently C G ( t ) has a subgroup of index atmost 2 which normalizes F . It follows that N G ( F ) H and that m p ( N G ( F )) = m p ( C G ( t )) ≥ 2. By Lemmas 3.2 and 3.3 (ii), we havethat L = F ∗ ( N G ( F ) /O p ′ ( N G ( F ))) is a non-abelian simple group. Since N E ( F ) /O ( N E ( F )) has a section isomorphic to SL (2 a ) with a ≥ p divides | SL (2 a ) | = (2 a − a (2 a + 1), we get with Proposi-tion 2.7 that F ∗ ( N G ( F ) /O p ′ ( N G ( F ))) ∼ = Alt(2 p ) or Fi . Furthermore,we have N H ( F ) /O p ′ ( N G ( F )) is isomorphic to (Alt( p ) × Alt( p )) : 2with no normal components or Ω +8 (2), neither of these has a normalsubgroup isomorphic to PSL (2 a ) and so we have a contradiction. (cid:3) The next two lemmas are needed to finally dispatch the Lie typegroups as possibilities for E . Lemma 8.10. Suppose that K is a simple group of Lie type defined incharacteristic and p is an odd prime. Let t ∈ K be an involution ina long root subgroup of K . If p divides | C K ( t ) | , then either (i) p divides | O ′ ( C K ( t )) | ; or (ii) K is isomorphic to one of PSL (2 a ) , PSU (2 a ) , B (2 a +1 ) or PSL (2 a ) for some a ≥ .Proof. As C K ( t ) = O ′ ( C K ( t )) B , where B is contained in a Borel sub-group of K , we may assume that p divides the order of a Borel sub-group. If the Lie rank of K is at least two and K = PSL (2 a ), we havethat t is centralized by a minimal parabolic subgroup and so p dividesthe order of this minimal parabolic. This is the assertion. (cid:3) Lemma 8.11. Suppose that K is a simple group of Lie type defined incharacteristic r , r odd, and of Lie rank at least . Let t be an involution Chris Parker and Gernot Stroth in a fundamental subgroup of K and assume that p is an odd prime. If p divides | C K ( t ) | , then p divides | O ′ ( C K ( t )) | .Proof. We have that C K ( t ) = L L S , where L is the fundamentalsubgroup, L = C C K ( t ) ( L ) and S is a Sylow 2-subgroup with t ∈ Z ( S ).If L L = O ′ ( L L ), we are done. So we may assume that r = 3. If L = O ′ ( L ) and L = 1, then, as 3 divides the order of L , wealso are done. Hence we have p = 3 and L is a central product ofgroups isomorphic to SL (3) or PSL (3). This now shows that K isisomorphic to one of the following PSL (3), PSp (3), PΩ ± (3), Ω (3),PΩ +8 (3) or G (3). But in all cases we have some GL (3) involved, so 3divides | O ′ ( C K ( t )) | . (cid:3) For X a Lie type group of Lie rank at least two in odd character-istic r we list O r ′ ( C X ( t )) for t a classical involution in Table 4. Theinformation is taken from [10, Table 4.5.1, Theorem 4.5.5]. Lemma 8.12. E/Z ( E ) is not a simple Lie type group defined in char-acteristic .Proof. Suppose that E/Z ( E ) is a Lie type group defined in characteris-tic 2. By Lemma 8.8 we have that the Lie rank of E is at least two. Fur-ther E/Z ( E ) = PSL (2 a ), a ≥ S ∈ Syl ( E ) andlet t be an involution in the centre of the long root group X ρ containedin Z ( S ). Then C E ( t ) is a subgroup of N G ( X ρ ) and t is centralized bythe subgroup K t of the Levi complement of N G ( X ρ ) which is generatedby root subgroups. These subgroups are given in Table 3 and we notethat K t is a characteristic subgroups of C E ( t ) which in turn is normalin C H ( t ). By Lemmas 8.10 and 8.7, we have that p divides | O ′ ( C E ( t )) | .Suppose that K t /O p ′ ( K t ) is not soluble. Then, by Lemma 8.3, we havethat F ∗ ( C H ( t )) ∼ = Alt( p ) × Alt( p ) with p ≥ +8 (2) with p = 5.Since K t is normal in C H ( t ), we infer that either K t ∼ = Alt( p ) × Alt( p )and p ≥ +8 (2) and p = 5. As Alt( p ), p ≥ 5, is isomorphic to aLie type group defined in characteristic 2 if and only if p = 5, we nowhave that p = 5 and K t ∼ = Alt(5) × Alt(5) or to Ω +8 (2). Using Table 3 weget that E/Z ( E ) ∼ = Ω +12 (2). Furthermore, we have Z ( E ) = O ( E ) = 1.Therefore there is a further involution s ∈ E which has centralizer C with C/O ( C ) ∼ = Sp (2). Now using Lemma 8.3, we have a contra-diction. Hence we must assume that C E ( t ) is soluble. By Lemmas 8.5and 8.8, E/Z ( E ) = PSp (2) ′ ∼ = Alt(6) or to G (2) ′ ∼ = PSU (3). Thenusing Table 3 again, we have E/Z ( E ) is one of PSU (2), PSU (2), F (2) ′ , PSL (2) and Ω +8 (2). In all the cases we check that | C H ( t ) | isonly divisible by p when p = 3 and E/Z ( E ) ∼ = PSU (2), PSU (2) or Simple group Levi Factor of Simple group Levi FactorPSL n ( q ) SL n − ( q ) PSU n ( q ) SU n − ( q )PSp n ( q ) ′ Sp n − ( q ) Ω ± n ( q ) SL ( q ) ∗ SO ± n − ( q )G ( q ) ′ SL ( q ) D ( q ) SL ( q )F ( q ) Sp ( q ) F ( q ) B ( q ) E ( q ) SU ( q ) E ( q ) SL ( q ) / Z ( q − , E ( q ) SO +12 ( q ) E ( q ) E ( q ) Table 3. Subgroups generated by root elements in the Levi complements of centralizers of involu-tions in the centre of long root subgroups in Lie type groups of rank at least 2 defined over a field oforder q = 2 a .Simple group X O r ′ ( C X ( t )) Simple group X O r ′ ( C X ( x ))PSL n ( r a ) SL ( r a ) ∗ SL n − ( r a ) PSU n ( r a ) SL ( r a ) ∗ SU n − ( r a )PSp n ( r a ) SL ( r a ) ∗ Sp n − ( r a ) Ω ± n ( r a ) (SL ( r a ) ∗ SL ( r a )) ∗ SO ± n − ( r a )G ( r a ) SL ( r a ) ∗ SL ( r a ) D ( r a ) SL ( r a ) ∗ SL ( r a )F ( r a ) SL ( r a ) ∗ Sp ( r a ) E ( r a ) SL ( r a ) ∗ SU ( r a )E ( r a ) SL ( r a ) ∗ SL ( r a ) E ( r a ) SL ( r a ) × Ω +12 ( r a )E ( r a ) SL ( r a ) × E ( r a ) Table 4. The group generated by the r -elements in the centralizer of a classical involution t in Lietype groups of rank at least 2 and odd characteristic r . Chris Parker and Gernot Stroth Ω +8 (2). Suppose that E/Z ( E ) ∼ = Ω +8 (2). Then E ∼ = Ω +8 (2) and thereis an involution s ∈ E such that C E ( s ) /O ( C E ( s )) has a normal sub-group isomorphic to Sp (2) ′ ∼ = Alt(6). This violates Lemma 8.3. If E ∼ = PSU (2), then there is an involution s ∈ E , which is not central-ized by an elementary abelian group of order 9 contrary to Hypothe-sis 4.1 (ii). So we have E ∼ = PSU (2). In this case C H ( t ) ∼ = 3 . SL (3)or C H ( t ) ∼ = 3 . GL (3). Now Proposition 2.7 shows that this struc-ture is impossible. Thus E/Z ( E ) is not a Lie type group defined incharacteristic 2. (cid:3) Lemma 8.13. E/Z ( E ) is not a Lie type group in odd characteristic.Proof. Suppose that E/Z ( E ) is a Lie type group defined in charac-teristic r , r an odd prime. By Lemma 8.8, we may assume that theLie rank of E is at least two. Let t be classical involution in t in E .By Lemmas 8.7 and 8.11, p divides | O ′ ( C E ( t )) | . Let K be a sub-normal subgroup of C E ( t ) containing t with K ∼ = SL ( r a ). Assumethat p divides | K | and that r a = 3. Then Lemma 8.3 implies that K ∼ = Alt( p ) or Ω +8 (2). We conclude that p = r a = 5 and that F ∗ ( C G ( t )) ∼ = Alt(10). In particular, K is not normal in C H ( t ). Us-ing Table 4 now shows that E ∼ = PSL (5), PSp (5), PSU (5), or G (5).The last case immediately fails, as Out(G (5)) = 1 means E = H and K is normal in C H ( t ). If E ∼ = PSp (5), then there is an invo-lution s ∈ E with O ′ ( C E ( s )) ∼ = PSL (5), which contradicts Hypoth-esis 4.1 (ii). If E ∼ = PSU (5), then there is an involution s ∈ E with F ∗ ( C E ( x ) /O ′ ( C E ( x ))) ∼ = PSL (25) and this contradicts Lemma 8.3. Sowe must have that E ∼ = PSL (5). Since O ′ ( C G ( t )) ≤ H is normalizedby C E ( t ), the structure of Aut( E ) shows that O ′ ( C G ( t )) ∩ E = h t i . Itfollows that C G ( t ) /O p ′ ( C G ( t )) has a subgroup isomorphic to C E ( t ) / h t i .Since the subgroup M of even permutations of Sym(5) ≀ Sym(2) has M/F ∗ ( M ) cyclic of order 4 and C E ( t ) /F ∗ ( C E ( t )) is a fours group, wesee that C G ( t ) /O p ′ ( C G ( t )) ∼ = Sym(10). But then H > E . In partic-ular, there are involutions s ∈ H \ E with F ∗ ( C H ( s )) ∼ = PSL (5) incontradiction to Lemma 8.3.So we may assume next that p does not divide the order of SL ( r a )or r a = 3. Suppose that in the latter case we also have p = 3. Thenin both cases p does not divide r a ± 1. Hence p divides the order ofone of the factors listed in Table 4. Since p does not divide | K | , C E ( t )has a component which is a group of Lie type in characteristic r . Butthese components are not isomorphic to either Alt( p ) or Ω +8 (2), andthis contradicts Lemmas 4.4 and 8.3.So we have that p = 3 = r a . Suppose that C E ( t ) has a compo-nent F , then 3 divides | F | . Therefore Lemmas 4.4, 8.3 and Table 4 give a contradiction. So we have that C E ( t ) is soluble. Then referenceonce again to Table 4 gives E/Z ( E ) is isomorphic to one of PSL (3),PSL (3), PSp (3), PSU (3), Ω (3), PΩ +8 (3), G (3). If E ∼ = Ω (3) orPΩ +8 (3), there is an involution s with F ∗ ( C E ( s )) ∼ = SO − (3) and, if E ∼ = PSL (3) there is an involution s ∈ E F ∗ ( C E ( s )) ∼ = PSL (9). Thusin these cases we obtain a contradiction via Lemma 8.3. If E ∼ = PSL (3)or PSp (3), then there is an involution s ∈ E whose centralizer is notdivisible by 9, in the first case | C E ( s ) | = 2 · 3, while in the secondcase s = t and C E ( t ) ∼ = GL (3). This contradicts Hypothesis 4.1 (ii).So we have that E/Z ( E ) ∼ = PSU (3) or G (3). Assume that H = E .Then, by Lemma 3.2 (v), O ′ ( C G ( t )) ≤ H and Corollary 2.11 gives O ( C E ( t )) = O ′ ( C G ( t )) which is extraspecial 2 . It follows that C G ( t ) /C C G ( t ) ( O ( C E ( t )) is soluble. But C G ( t ) is an almost simple groupand so we have that O ( C E ( t )) is centralized by a Sylow 3-subgroupof C E ( t ) something which is impossible. Thus E = H . Hence thereexists an involution s ∈ H \ E . If E/Z ( E ) ∼ = G (3), C E ( s ) ∼ = G (3). If E/Z ( E ) ∼ = PSU (3), then C E ( s ) has a section isomorphic to PSU (3),PSp (3) or Ω − (3). In all these cases we get a contradiction to Lemma8.3 applied to C G ( s ). (cid:3) Proofs of the Main Theorems In this final section we assemble the proofs of our main theorems. Werefer the reader to the introduction for their statements and continuethe notation of the previous sections. Proof of Theorem 1.1. Let G = O ( G ). Then H = H ∩ G has evenorder and H is strongly p -embedded in G by Lemma 3.2(ii) and (iv).As H is normal in G and as H has even order by hypothesis, wehave that G and H together satisfy Hypothesis 4.1. If G containsa classical involution, then Theorem 2.2 implies that G is a K -groupand then Proposition 2.5 implies that F ∗ ( G ) = F ∗ ( G ) ∼ = PSU ( p a ) forsome a ≥ 2. Thus we may suppose that G has no classical involutions.Since F ∗ ( G ) = O p ( G ), Hypothesis 5.1 is satisfied. Therefore Theo-rems 6.1 and 7.10 together imply that F ∗ ( G ) = F ∗ ( G ) ∼ = G (3 n − )for some n ≥ (cid:3) Proof of Theorem 1.2. Suppose that F ∗ ( H ) = O p ( H ). Then as O p ′ ( H ) =1, we have E = E ( H ) = 1. Let G = O ( G ). Then G ≥ E ( H ) andso G satisfies hypothesis 4.1. Therefore Lemma 4.5 implies that E ( H )is a quasisimple group. Again we may as well assume that G does notcontain a classical involution. In particular, Hypothesis 8.1 is satisfied.By Lemma 8.2, we have O p ( H ) = O p ′ ( H ) = 1. Now we use the fact Chris Parker and Gernot Stroth that E is a K -group and use Lemmas 8.5, 8.6, 8.12 and 8.13 to delivera contradiction. Hence we conclude that F ∗ ( H ) = O p ( H ). (cid:3) Proof of Theorem 1.3. This is merely a combination of Theorems 1.1and 1.2. (cid:3) Finally we prove Corollary 1.4. For this we require the followingproposition about centralizers of involutions in Lie type groups definedin characteristic p . Proposition 9.1. Let p be an odd prime, X be a finite group and set K = F ∗ ( X ) . Suppose that K is a simple group of Lie type defined incharacteristic p . Then either m p ( C X ( x )) ≥ for all involutions x ∈ X or one of the following holds: (i) K ∼ = PSL ( p n ) with n ≥ ; (ii) K ∼ = PSL ( p ) or PSU ( p ) ; (iii) K ∼ = PSp ( p ) ; or (iv) K ∼ = G (3) ′ ∼ = PSL (8) .Proof. We may suppose that K = PSL ( p n ). Let x ∈ X be an invo-lution and set K x = O p ′ ( C K ( x )). Note that since the Lie type groupsare generated by their (perhaps twisted) root subgroups the only Lietype groups L with m p ( L ) = 1 are L ∼ = SL ( p ) for arbitrary p and L ∼ = G (3) ′ ∼ = PSL (8) with p = 3. Assume that q = p n and K hastype d Σ( q ) (using notation as in [10, Chapter 4]). Then, by [10, Defini-tion 2.5.13], x induces either an inner-diagonal, graph, field or graph-field automorphism on K . If x induces a field automorphism, then [10,Proposition 4.9.1] gives that K x ∼ = d Σ( q / ). Since m p ( C X ( x )) ≤ 1, wededuce that K x ∼ = PSL ( p ) and consequently K ∼ = PSL ( p ) which is acontradiction. If x induces a graph-field automorphism of K , then [10,Proposition 4.9.1] implies K x ∼ = Σ( q / ). Thus Σ( q / ) is PSL ( p ) orSL ( p ) and this is impossible. Therefore x is contained in the groupgenerated by inner-diagonal and graph automorphisms of K . For eachof the possibilities for K and x , the structure of K x is described in [10,Table 4.5.1] and we avail ourselves of this information. Thus, if K hastype A ǫn ( q ), then the only possibility is that n = 2 and q = p and so (ii)holds. Similarly if K has type B m ( q ) or C m ( q ), we see again that m = 2and q = p and thus (iii) holds. The groups of type D ǫm ( q ) only need tobe considered for m ≥ 4, and in these cases m p ( K x ) ≥ 2. The same istrue from the groups of type G ( q ) and D ( q ). For K ∼ = G (3 n +1 ) ′ ,the only possibility is that X ∼ = G (3) and K x ∼ = PSL (3) which gives(iv). The remaining exceptional groups are all easily seen to violate m p ( K x ) ≤ (cid:3) Proof of Corollary 1.4. If F ∗ ( H ) is a Lie type group defined in char-acteristic p and of rank at least 3, then Proposition 9.1 implies that m p ( C H ( t )) ≥ t ∈ H . Thus Theorem 1.2 impliesthat H cannot be strongly p -embedded in any groups G . (cid:3) References [1] M. Aschbacher, A characterization of Chevalley groups over fields of odd order,I, II, Ann. of Math. 106 (1977), 353 - 468, Correction, Ann. of Math. 111 (1980),411 - 414.[2] H. 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Tokyo 21, (1974), 133-159. Chris Parker, School of Mathematics, University of Birmingham,Edgbaston, Birmingham B15 2TT, United Kingdom E-mail address : [email protected] Gernot Stroth, Institut f¨ur Mathematik, Universit¨at Halle - Wit-tenberg, Theordor Lieser Str. 5, 06099 Halle, Germany E-mail address ::