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The Classification of Primitive Sharp PermutationGroups of Type {0, k}Douglas P. Brozovic aa Department of Mathematics , University of North Texas , Denton , Texas , USAPublished online: 13 Mar 2014.

To cite this article: Douglas P. Brozovic (2014) The Classification of Primitive Sharp Permutation Groups of Type {0, k},Communications in Algebra, 42:7, 3028-3062, DOI: 10.1080/00927872.2013.780061

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Communications in Algebra®, 42: 3028–3062, 2014Copyright © Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927872.2013.780061

THE CLASSIFICATION OF PRIMITIVE SHARPPERMUTATION GROUPS OF TYPE �0� k�

Douglas P. BrozovicDepartment of Mathematics, University of North Texas, Denton, Texas, USA

The purpose in this paper is to complete the classification of primitive sharp permutationgroups of type ��0� k�� n� by proving that no such group can be almost simple.

Key Words: Almost simple group; Primitive permutation group; Sharp permutation group.

2010 Mathematics Subject Classification: 20B10; 20B15; 20E32.

1. INTRODUCTION

Let G be a finite group, � a (virtual) complex character of G and L =���g� � g ∈ G#�. Blichfeldt [8] has shown that

�G�∣∣∣∣∣∏k∈L

���e�− k� �

When equality holds, � is said to be a sharp character of type L. In addition, if � is apermutation character of degree n associated to the G-set X, the group G is calledsharp permutation group of type �L� n� on X.

In Theorem 1.1 of [9] and theMain Theoremof [10], it was shown that a primitivesharp permutation group of type ��0� k�� n� on X is of one of the following forms:

(a) a 2-transitive Frobenius group; or(b) G � ASO3�p

r� in its natural action on X = �GF�pr��3, where p is an oddprime; or

(c) G is an almost simple group. In this case, either Gx is solvable, or is anonsolvable Frobenius group.

In this note we investigate the viability of case (c). In particular, we prove thefollowing theorem.

Theorem 1.1. Let G be a primitive sharp permutation group of type ��0� k�� n� on X.Then G is not an almost simple group.

Received December 30, 2011; Revised February 12, 2013. Communicated by A. Turull.Address correspondence to Dr. Douglas P. Brozovic, Department of Mathematics, University of

North Texas, 1155 Union Circle, #311430, Denton, TX 76203-5017, USA; E-mail: brozovic@unt.edu

3028

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In light of the above remarks, Theorem 1.1 completes the classification ofprimitive sharp permutation groups of type ��0� k�� n�.

Corollary 1.2. Let G be a primitive sharp permutation group of type ��0� k�� n� on X.Then either G is a 2-transitive Frobenius group, or G is isomorphic to ASO3�p

r� in itsnatural action on X = �GF�pr��3, where p is an odd prime.

Sharp permutation groups can be viewed as generalizations of sharplyk-transitive groups. Another family of examples is found in the so-called geometricgroups. The geometric groups were classified between 1980 and 1990 by Deza andCameron [12], Maund [37], and [47] Zil’ber. It should be noted that a geometricgroup may be assumed to be transitive, and with the exception of the sharplyt-transitive groups, geometric groups act imprimitively on the set that affords thegiven geometric action. Every group can be realized as a sharply 1-transitive group,and as a geometric group of rank 1, so the rank 1 case for these classes is irrelevant.

In the late 1970s, Bannai proposed the problem of classifying sharppermutation groups. To date, there is nothing approaching a complete classification.The main exception is the case where L = �k� and �fixX�G�� < k. Sharp groupsof this type must admit a nontrivial proper partition. There are general resultsrelated to finite groups admitting a partition due to Suzuki [41], Iwahori [24], andBaer [3, 4], and for infinite groups due to Franchi [18]. The complete structuraldescription in the case of finite nonsolvable groups is due to Suzuki [41]. In the caseof finite solvable groups, the classification is essentially due to Baer in [3] and [4](see also Section 3.5 of [38] and Proposition 2.6 below). In [19] and [20] Franchiproves additional results related to sharp permutation groups of type L with �L� = 1including a careful analysis of non-abelian p-groups and finite abelian groups ofthis type.

The sharp permutation groups with �L� ≥ 2 were initially studied by Tsuzuku[43], Cameron and Deza [12], [13], Ito and Kiyota [23], Li [32], and Bundy [11].

In the case of primitive sharp permutation groups, the main result of [9] hasbeen partially generalized to arbitrarily large values of �L� by B. Birszki [5]. Inaddition, Birszki has obtained results in [6] and [7] related to the general case ofprimitive sharp permutation groups where L is arbitrary.

Our consideration will be divided according to whether Soc�G� is analternating group, a classical group, an exceptional group of Lie type, or a sporadicsimple group. In each case, Propositions 2.4–2.6 and Corollaries 2.7 and 2.8 areplayed off against the wealth of information on the structure of the maximalsubgroups of almost simple groups. Sometimes, Proposition 2.4 will be used when itquickly eliminates cases. Alternatively, we use the structural constraints imposed byProposition 2.6 and the associated corollaries to eliminate most of the other cases.

It should be noted that while they are comparatively rare, there are examplesof maximal subgroups of almost simple groups that satisfy the inequalitiesfrom Proposition 2.4 and whose structure is compatible with the conclusion ofProposition 2.6 or the associated corollaries. In these cases we will often useelementary arguments to produce elements of the group G that fix a unique memberof X, forcing G to be sharp of type �0� 1� on X, contrary to hypothesis.

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3030 BROZOVIC

2. PRELIMINARY RESULTS

Let G be a primitive permutation group of type ��0� k�� n� on X. Then bydefinition, �fixX�g�� = k for every g ∈ Gx − �1�. We will henceforth write fix�g� inplace of fixX�g� and will indicate a set as a subscript only if it differs from X. Notealso that as fix�Gx� is a block for the G-action on X, the primitivity of G on Xrequires fix�Gx� = �x�.

Definition 2.1. Let M be a group and X a M-set with fix�M� = �. X is said to bea M-set of type k > 0 if every non-identity element of M fixes exactly k points of X.If, in addition,

X = ⋃m∈M

fix�m��

then X is called a pure M-set of type k.

The following proposition is part of Proposition 2.2 of [24].

Proposition 2.2. Let X be a pure M-set of type k. Then k ≥ 2.

We will require the following lemma.

Lemma 2.3. Let G be a primitive sharp permutation group of type ��0� k�� n� on X.If k < 3, then G is a 2-transitive permutation group.

By Proposition 2.1 of [9], the permutation rank of G on X is k+ 1. Inparticular, k = 1 forces G to be a 2-transitive permutation group. Thus we mayassume that k = 2.

Suppose that �fix�Gx�� = 1. Since G is a sharp permutation group of type��0� 2�� n� on X, �Gx� = n− 2. Set X0 = X − fix�Gx�, so by assumption �X0� = n− 1.Observe that if Gx has a regular orbit on X0 then as �Gx� = n− 2 = �X0� − 1, Gx

must fix the remaining point, contrary to �fix�Gx�� = 1. Thus Gx has no regularorbits on X0, and so we have the following situations:

(i) X0 =⋃

g∈GxfixX0

�g�, and(ii) �fixX0

�g�� = 1 for every g ∈ Gx − �e�.

Thus X0 is a pure Gx-space of type 1, contrary to Proposition 2.2.Finally, if �fix�Gx�� = 2, then G acts imprimitively on X, contrary to

assumption.In view of Lemma 2.3, we may assume that k ≥ 3.

Proposition 2.4. Suppose G is a primitive sharp permutation group of type ��0� k�� n�on X, where k ≥ 3.

(a) �X�+12 ≤ �Gx�.

(b) �Gx�2 + 2��X� − 1� ≤ �G�.(c) �Gx� ≤

√�G� − 2��X� − 1�.

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(d) �G� = �Gx���Gx� + k�.(e) �X� = G Gx� > �Gx�.

Proof. Parts (a)–(c) are the content of Proposition 2.6 in [9]. Part (d) follows fromthe fact that G is a sharp permutation group of type ��0� k�� n� on X. Finally, (e) isan immediate consequence of (d). �

Throughout this paper F�K� and F ∗�K� will denote (respectively) the Fittingsubgroup and the Generalized Fitting subgroup of a group K.

Proposition 2.5. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X. Then Gx is either solvable, or is a nonsolvable Frobenius Group.In particular, F ∗�Gx� = F�Gx�. Furthermore, if Gx is nonsolvable and H is a Frobeniuscomplement, then there exists a normal subgroup H0 of H such that the followingstatements hold:

(i) H H0� ≤ 2;(ii) H0 � SL2�5�×M , where gcd��M�� 30� = 1 and every Sylow r-subgroup of M is

cyclic.

Moreover, if H H0� ≤ 2, then Z�H� is a 2-group, M is cyclic, and every element ofH −H0 inverts every element of M .

Proof. That Gx is either solvable or a nonsolvable Frobenius group is Theorem 1.1of [10]. The classification of nonsolvable Frobenius complements is due toZassenhaus [46]. �

As we have seen above, if G is a primitive sharp permutation group of type��0� k�� n� on X with k ≥ 3, then X0 = X − fix�Gx� is a pure Gx-set of type k− 1 ≥ 2.

By assumption, for each x ∈ X and each g ∈ Gx, we have �fix�g�� = k. Fixx ∈ X, set

� = �S ⊂ X �S� = k and S = fix�g� for some g ∈ Gx��

and set

� = �GS � S ∈ ���

where GS is the element-wise stabilizer of the set S.It easily follows that the distinct members � have pairwise trivial intersection

and the union over � covers Gx. By definition, GS �= �1� for each S ∈ �. Moreover,since Gx does not have global fixed points on X0 = X − fix�Gx�, we have that G >GS for every S ∈ �. In particular, the family � forms a so-called nontrivial properpartition of the group Gx.

We shall refer to this as the partition of Gx associated to the Gx-set X. Note alsothat for every h ∈ Gx and every A ∈ �, Ah ∈ �. In particular, the partition � of Gx

is said to be normal.Let M be a group that admits a partition and let H ≤ M . H is called a -

admissible subgroup of M if for each A ∈ , either A ≤ H or A ∩H = �1�.

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Recall that for a group G, the Hughes subgroup of G, denoted by Hp�G�, isdefined by the rule

Hp�G� = �g � o�g� �= p��

The structure of a group that admits a nontrivial, proper normal partitionis highly restricted. The following theorem, due primarily to Baer, describes thesituation when the underlying group has a nontrivial Fitting subgroup.

Proposition 2.6. Let M be a group that admits a nontrivial, proper, normal partition . Suppose further that the fitting subgroup F�M� �= �1�. Then exactly one of thefollowing statements hold:

(I) M has no proper, nontrivial -admissible normal subgroup, in which case M �Sym�4� and is the collection of all maximal cyclic subgroups of M;

(II) M contains a proper, nontrivial -admissible normal subgroup N . In this case,either

(A) M is a Frobenius group with N ≤ F�M�, F�M� the Frobenius Kernel (and theFitting subgroup) of M , and each Frobenius complement is contained in ; or

(B) There exists p � �G� such that every element g ∈ M\N has order p. Moreover,either

(1) M is a p-group; or(2) N = Hp�M� (the Hughes subgroup of M), N ∈ , and M N� = p. In

addition

(i) For each g ∈ M\N , �CN�g�� = p,(ii) �1� �= Op�M� < F�M� = N ,(iii) M is not a Frobenius Group.

Proof. Conclusion (II)(B)(2)(i) is the content of Theorem 8 of [18]. All of the otherparts are essentially due to Baer in [3] and [4].

We provide a brief sketch using Schmidt’s presentation [38] of the relevantresults of Baer.

Suppose first that M has no nontrivial, proper -admissible normal subgroups.Then Conclusion (I) follows from a theorem of Baer [4] (Theorem 3.5.9, p. 149of [38]).

Thus we may suppose that there exists a proper, nontrivial, normal subgroupN of M that is -admissible.

Suppose first that there exists A ∈ for which A is not a subgroup of N andCN�A� = �1�. In this case, Baer shows that M is a Frobenius group with Frobeniuscomplement A and Frobenius Kernel F�M�, with N ≤ F�M� (see Theorem 3.5.7(a)and its proof, pp. 148–149 of [38]). Since the partition is normal, all of theFrobenius complements are contained in , and so conclusion (II)(A) holds.

Thus we may assume that for every B ∈ with B �= N , that CN�G� �= �1�.Since N is a proper subgroup of M , it cannot contain all of the subgroups of , sothere are subgroups B ∈ for which B � N .

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In this case, Baer [3] proves (see also Theorem 3.5.7(b) and its proof, pp. 148–149 of [38]) the existence of a prime divisor p of �G� such that every element g ∈M\N has order p. Moreover, either M is a p-group, or N = Hp�M� (the Hughessubgroup of M), N ∈ , M N� = p. In particular, either (II)(A) holds, or the firstpart of (II)(B) holds.

Suppose M is not a p-group. By Theorem 8 of [18], we have that �CN�g�� = pfor every element g ∈ M\N , and so conclusion (II)(B)(2)(i) holds. Note also thatp � N , M is not a p-group, and every element of M\N has order p. So we must have�1� �= Op�M� < N .

Now N is not a self normalizing -admissible subgroup of M , and so byLemma 3.4.5 of [38], N is nilpotent and so N ≤ F�M�. Since F�M� is divisible byat least 2 distinct primes, if follows from Lemma 3.5.2 of [38] that F�M� ≤ C ∈ ,and so F�M� < M . As M N� = p, we have that N = F�M�, and so conclusion(II)(B)(2)(ii) holds.

Finally, observe that M = N g� for any g ∈ M\N and N = F�M�. Since ButCN�g� �= �1�, M cannot be a Frobenius group, and we are done. �

The following two corollaries are essentially immediate consequences ofProposition 2.6.

Corollary 2.7. Let G be a primitive sharp permutation group of type ��0� k�� n� on X.Suppose further that F�Gx� = Op�Gx� for x ∈ X. Then exactly one of the followingstatements hold:

(a) Gx is a p-group; or(b) Gx � Sym�4�; or(c) Gx is a Frobenius group with Frobenius Kernel Op�Gx�.

Corollary 2.8. Let G be a primitive sharp permutation group of type ��0� k�� n� on X.Suppose further that Gx is neither a p-group nor isomorphic to Sym�4�. Then either

(a) Gx is a Frobenius group; or(b) Gx = F�Gx� g� and g� � �p for every g ∈ Gx\F�Gx�. Moreover,

�1� < Op�Gx� < F�Gx�

and �CF�Gx��g�� = p for every g ∈ Gx\F�Gx�. In particular, Op′�Gx� g� is a

Frobenius Group no nontrivial p′-element of F�Gx� is centralized by any element ofGx\F�Gx�.

In either case, no r-power element of Gx − F�Gx� centralizes a nontrivial s-powerelement of F�Gx�, where r and s are distinct primes.

Finally, let S � G ≤ Aut�G� be an almost simple group, and Gx a maximalsubgroup of G. Recall from [16] that Gx is said to be trivial (or a triviality) if S ≤ Gx.Observe that by Proposition 2.5 Gx cannot be a triviality.

3. ALTERNATING GROUPS

We treat the possibility Soc�G� = Alt�n� (n ≥ 5) in the following proposition.

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Proposition 3.1. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X. Then Soc�G� is not an alternating group.

Proof. Suppose first that Soc�G� � Alt�6�. If G � Alt�6�, then from [16] we seethat the permutation characters associated to the distinct maximal subgroups of Gare either rank 2 or 3. By Proposition 2.1 [9], the permutation rank of G on X isk+ 1 and so k ≤ 2. Thus Lemma 2.3 forces G to act on X as a 2-transitive Frobeniusgroup, contrary to the (almost) simplicity of G.

Next suppose G Soc�G�� = 2, so G is isomorphic to one of PGL2�9�, M10,or Sym�6�. From Proposition 2.4(a) and (c), we have

G Gx�+ 12

≤ �Gx� ≤√�G� − 2��X� − 1��

Examining the list of maximal subgroups from p. 4 of [16] for each of the threepossibilities for G, the inequality above is satisfied only if G is isomorphic to eitherM10 or PGL2�9� and Gx is a maximal subgroup of order 20. In either case Gx mustbe the normalizer of a Sylow 5-subgroup of G. It is easy to see that if � ∈ Gx is anyelement of order 5, then � fixes only the trivial right Gx-coset of G. Whence k = 1,and as above, Lemma 2.3 forces G to act as a 2-transitive Frobenius group on X, acontradiction to the almost-simplicity of G.

Thus we may assume that G � Aut�Alt�6�� � P�L2�9�, and as before, theinequality above is satisfied only if Gx ∈ Syl2�G�. We consider this configurationbelow. Suppose G induces a sharp permutation group of type ��0� k�� n� on X =Gx\G (the right cosets of Gx in G). Then �X� = 45 and so by Proposition 2.4(d), k =�X� − �Gx� = 13. In particular, every non-identity element of Gx must fix 13 distinctmembers of X.

Set G � P�L2�9�, G0 = Soc�G�, H � PGL2�9�, and let � denote the set ofinvolutions of G0. Observe that ��� = 45, and � is the unique class of involutionsof G0. From this, we conclude that Syl2�G� = �CG��� � � ∈ ��, and Syl2�H� =�CH��� � � ∈ ��.

Set Gx = CG��� for � ∈ � fixed, and let T = H ∩Gx. Then T ∈ Syl2�H� andT � D16. Choose g ∈ T of order 8. Observe that T is the unique Sylow 2-subgroupof H containing g. Indeed, if U is another, then U = CH��� for some � ∈ � . Butg� � U , g� � T , and U � T � D16. This forces �� � ∈ g� � �8, whence � = �.

Now, if g is contained in another Sylow 2-subgroup of G, say W = CG���,then g ∈ W ∩H ∈ Syl2�H�, whence W ∩H = T . Thus CH��� = CH��� for �� � ∈ � ,and so � = �. Thus g is contained in a unique G-conjugate of Gx and so g fixes aunique right coset of Gx in G. Again, we have k = 1 and obtain a contradiction asabove.

For the remainder of the argument, we may assume that Soc�G� � Alt�n� andAlt�n� � G ≤ Sym�n�, n ≥ 5, n �= 6.

By Proposition 2.5, the socle of the maximal subgroup Gx of G cannot benonsolvable. According to [33], a solvable maximal subgroup M ≤ G must satisfyone of the following:

(a) M acts intransitively on X;(b) M acts imprimitively on X;

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PRIMITIVE SHARP PERMUTATION GROUPS 3035

(c) M is of wreathed type and is primitive on X;(d) M � M ≤ AGLt�p�, where pt = n.

If (a) or (b) hold, then Proposition 2.5 yields an embedding of Gx intoa subgroup of the form Sym�a�× Sym�b� or Sym�a� Sym�b� with substantialconstraints on a and b. It either case, it is a simple matter to verify that �G� < �Gx�2,contrary to Proposition 2.4(b).

If case (c) holds, then one similarly shows that Gx embeds in a subgroup of theform Sym�a� Sym�b� with ab = n. Proposition 2.5 forces 1 < a� b ≤ 4. As in theimprimitive case, this yields 8 possibilities. For each of these, one easily verifiesthat Proposition 2.4(e) fails.

Finally, suppose (d) holds, and so Gx is an affine primitive maximal subgroupof G. Let AGLt�p� � M ≤ Sym�pt� so Gx = M ∩G.

Then M Gx� ≤ 2, and so the solvability of Gx forces the solvability ofAGLt�p�. In particular, either t = 1, or t = 2 and p = 2 or p = 3.

If t = 2, then as 5 ≤ n = pt, we must have p = 3 and M � AGL2�3�. Thus eitherGx � AGL2�3� orGx � ASL2�3�. In particular, F�Gx� = O3�Gx� < Gx, and there existelements of order 3 outside of F�Gx�. In particular, F�Gx� is not a Frobenius group,and it is clearly neither symmetric nor a 3-group, contrary to Corollary 2.7.

Hence t = 1, and �X� = n = p < �Gx�, contrary Proposition 2.4(e) fails.Henceforth we assume that Gx is nonsolvable, so by Proposition 2.5 Gx is a

nonsolvable Frobenius group. From [16], we may assume pt > 7. Now Gx � Ept Lwhere L is subgroup of GLt�p� acting irreducibly on Ept .

It follows that F�Gx� = Ept is the Frobenius Kernel of Gx, and so L is aFrobenius complement. Thus �L� ≤ pt − 1 and �Gx� ≤ pt�pt − 1�. Since �pt − 2�! ≥2�pt�2 for all pt ≥ 8, we have

G Gx�+ 12

>�pt − 2�!

2≥ �pt�2 > �Gx��

contrary to Proposition 2.4(a). This completes the proof. �

4. GROUPS OF LIE TYPE

Our approach will be to separate the analysis of classical groups andexceptional groups. However, the following proposition applies to all of the simplegroups of Lie type, and so we present it here.

If G0 is a simple group of Lie type and G0 � G ≤ Aut�G0�, we say M is aparabolic subgroup of G if M is an overgroup of some Borel subgroup of G0 withG0 � M .

Proposition 4.1. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X. If Soc�G� is a simple group of Lie type, then Gx ∩ Soc�G� is nota parabolic subgroup of Soc�G�.

Proof. Suppose not. Then Gx ∩ Soc�G� is a parabolic subgroup of the simplegroup of Lie type Soc�G�.

Throughout this argument, we set G0 = Soc�G�, P0 = Gx ∩G0, and we assumethat G0 is a simple group of Lie type in characteristic p. Since P0 is a parabolic

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subgroup of G0, there is a Borel subgroup B0 of G0 with B0 ⊆ P0. Fix a split torusH0 of B0. The pair �B0� H0� determines a root system, , a base � (of simple roots),and a set of positive roots +. In particular, Xr � r ∈ +� = U0 = Op�B0� is a Sylowp-subgroup of G0 (Xr the root subgroup associated to r) and B0 = U0 H0 is a Levidecomposition.

It is well known (see Section 8.3 of [14]) that the lattice of proper parabolicsubgroups between B0 and G0 corresponds bijectively to the lattice of proper subsetsof � or, equivalently, to sub-diagrams of the associated Dynkin diagram for G0.Indeed, there is a proper (possibly empty) subset J ⊂ � such that

P0 = H0� Xr� X−s � r ∈ +� s ∈ J��

Let L0 be the unique subgroup of P0 containing H0 for which P0 = Op�P0� L0 is aLevi decomposition of P0. Note that the Lie Rank of Op′�L0� is �J �.

Since P0 is a parabolic subgroup of G0, NG0�P0� = P0. As Gx is a maximal

subgroup of G which does not contain G0, Gx covers G/G0. Note also that P0 � Gx

and since Gx is maximal in G, P0 is a maximal Gx-invariant parabolic subgroup G0

with Gx = NG�P0�.

Claim 1. We may assume that one of the following statements hold:

(a) G0 � P�+8 �q� and G induces a graph or graph-field automorphism of G0 of

order 3; or(b) G induces a graph or graph-field automorphism of G0 of order 2, P0 is a

submaximal parabolic subgroup of G0 such that the Lie rank of Op′�L0� is 2 lessthan that of G0; or

(c) P0 is a maximal parabolic subgroup of G0.

Proof. First, note that according to Theorem 30 of Steinberg [40] (see alsoTheorem 2.5.1 [22]), every automorphism of G0 can be expressed as a product ����,where � is an inner automorphism of G0 and where �, �, and � are standard (relativeto the fixed H0-root system) diagonal, field, and graph automorphisms of G0. Inparticular, the standard field and diagonal automorphisms stabilize every subgroupin the lattice of subgroups of G0 containing B0. Thus, if no element of G inducesa graph or graph-field automorphism of G0, then Gx (recall Gx = NG�P0�) muststabilize every parabolic subgroup of G0 containing P0. As P0 is a maximal Gx-invariant subgroup of G0, P0 must then be a maximal subgroup of G0.

It follows that if neither (a) nor (c) hold, G must induce a graph or graph-fieldautomorphism of G0 of order 2 that stabilizes the nonmaximal parabolic subgroupP0 of G0. For the remainder of the proof of this claim, we assume that (a) and (c)do not hold.

As P0 is a nonmaximal parabolic subgroup of G0, choose P a maximalparabolic of G0 with P0 ⊂ P, and let � ∈ Gx induce a graph or graph-fieldautomorphism of G0 (since Gx covers G/G0) of order 2. Since P0 is a maximalGx-invariant parabolic subgroup of G0 and as P is stable under every standarddiagonal and/or field automorphism induced by G, P cannot be stable under � but isstabilized by �2. In particular, P� ∩ P must be a Gx-invariant parabolic subgroup of

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G0 containing P0, whence P0 = P� ∩ P. Thus P0 is a submaximal parabolic subgroupof G0, by which we mean a parabolic obtained from a set J with �J � = ��� − 2,and by the remarks above, the Lie rank of Op′�L0� is ��� − 2, and (b) holds. ThusClaim 1 holds.

Since P0 � Gx, we have Op�Gx� �= �1� and so by Corollary 3.1.4 p. 94 of [22]

F ∗�P0� = Op�P0� and F ∗�Gx� = Op�Gx��

Since Gx admits a nontrivial normal partition, Corollary 2.7 implies that exactly oneof the following statements holds:

(i) Gx is a Frobenius group with Frobenius Kernel Op�Gx�; or(ii) Gx is a p-group; or(iii) Gx � Sym�4�.

Suppose first that Claim 1(a) holds. By Lemma 1.6.1 p. 193 of [29], P0 isnot a Borel subgroup of G0. Thus the Lie rank of Op′�L0� is nonzero, and so L0

contains the image of a fundamental SL2�q� of G0 (Xr�X−r�). In particular, sinceG0 � P�+

8 �q�, it follows that Gx �= Op�Gx� and Gx � Sym�4�, so (i) must hold. Onthe other hand, p divides �L0�, and so P0 cannot be a Frobenius group with kernelF�P0� = Op�P0� = Ru�P0�. It follows that Gx cannot be a Frobenius group withkernel F�Gx�, a contradiction.

Henceforth, we suppose that either (b) or (c) of Claim 1 hold.

Claim 2. Gx is not Frobenius group.

Proof. Suppose not, and let Gx = F�Gx� K. By the same argument as in case (a)above, p cannot divide �L0�. Thus the Lie rank of Op′�L0� is 0 and P0 is a Borelsubgroup of G0, whence Gx is solvable and the Lie rank of G0 is at most 2. Weconsider the two possible values for the Lie rank of G0 separately.

Suppose first that G0 is has Lie rank 1. Then G0 must be isomorphic to oneof the groups PSL2�q�, PSU3�q�,

2B2�22m+1� with m > 0 (since 2B2�2� is solvable), or

2G2�32m+1� with m > 0 (since 2G2�3�

′ � PSL2�8�). We treat each case separately.Suppose G0 � PSL2�q�. Let g ∈ Ru�P0� be a non-identity element of order p.

Now Ru�P0� = F�Gx� ∩G0 and Gx is a Frobenius group. Thus CG�g� ⊆ F�Gx� andso CG0

�g� ⊆ Ru�P0�. From the maximality of Gx in G, we have Gx = NG�Ru�P0��

and so if g fixes the coset Gxh, then g ∈ NG�Ru�P0��h = NG�Ru�P0�

h�. It follows thatg ∈ NG0

�Ru�P0�h� and so g ∈ Ru�P0� ∩ Ru�P0�

h. But Ru�P0� is abelian, so Ru�P0�h ⊆

CG0�g� ⊆ Ru�P0�, whence h normalizes Ru�P0� and h ∈ Gx = NG�Ru�P0��. Therefore

g fixes a unique right coset of Gx in G, contrary to G a sharp permutation group oftype ��0� k�� n� on X with k > 2.

Suppose next that G0 � PSU3�q�. Since G0 is simple q > 2. Using Table 3.5.Bp. 71 of [30], it is easy to show that �G�

�Gx �2 < 1, and so �X� < �Gx�, contrary toProposition 2.4(b).

If G0 � 2B2�22m+1� (m > 0), then P0 is a Borel subgroup of G0. Since P0 is

maximal in G0, we have G Gx� = G0 P0�. Setting 22m+1 = q > 8, from [42] we

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have

�G��Gx�2

= q2 + 1q2�q − 1�G G0�

≤ q2 + 1q2�q − 1�

�

Since q ≥ 8, q2+1q2�q−1� < 1, whence �G� < �Gx�2, contrary to Proposition 2.4(b).

Finally, suppose that G0 � 2G2�q� with q = 32m+1 ≥ 27. Now P0 is a Borelsubgroup of G0, and as q ≥ 27, and from [27] we easily obtain �G�

�Gx �2 = q3+1q3�q−1� < 1,

contrary to Proposition 2.4(b).Now we treat the case in which the Lie rank of G0 is 2. Since P0 is a Borel

subgroup of G0, Claim 1(b) holds and so G0 is untwisted and G induces a graph orgraph-field automorphism of G0 of order 2. Recall that such graph or graph-fieldautomorphisms exist for groups of type G2 or B2 only if the field of definitions haveorder 32m+1 and 22m+1, respectively (see Sections 12.2, 12.3 of [14]). Therefore, G0 isisomorphic to one of G2�3

2m+1�, PSp4�22m+1�, or PSL3�q�.

Suppose G0 � G2�32m+1�, m ≥ 0. By the remarks above Gx = NG�P0�, P0 a

Borel subgroup of G0, and Gx covers G/G0. In particular, �Gx� = �P0� · G G0�, andfrom [27] we have

�G��Gx�2

= q6�q6 − 1��q2 − 1�G G0�

q12�q − 1�4G G0�2

= �q6 − 1��q + 1��G0�q6�q − 1�3�G� <

q + 1�q − 1�3

< 1

for all q ≥ 3, contrary to Proposition 2.4(b).Next suppose G0 � Sp4�2

2m+1�, and so p = 2. Since Sp4�2�′ � Alt�6�,

Proposition 3.1 allows us to assume m > 0. In this case there exists a prime r �= 2for which the r-rank of a split torus of G0 is strictly greater than 1, contrary to Gx

a Frobenius group.Finally, suppose G0 � PSL3�q�. Suppose first that q > 2. Then L0 is a split

torus of G0. In particular, if q − 1 �= 3, then there is an element of L0 that centralizesa central element of Ru�P0�. Thus Ru�P0� L0 is not a Frobenius group and so Gx isnot a Frobenius, a contradiction.

Thus we may assume 3 = q − 1 and G0 � PSL3�4�. In this case we easily verifythat G Gx� < �Gx�, contrary to Lemma 2.4(e). This final contradiction completesthe proof of Claim 2.

Claim 3. Gx is not a p-group.

Proof. Suppose not. Then P0 is also a p-group, it must also be a Borel subgroupof G0 with trivial Levi factor L0 and so the field of definition of G0 must be GF�2�.As above, the Lie rank of G0 is at most 2. As none of the rank 1 groups over GF�2�are simple, it follows that G0 has Lie rank 2 and so Claim 1(b) holds.

Thus G0 is untwisted and admits a graph or graph field automorphism oforder 2 induced by G. Therefore, either G0 � PSL3�2� or Sp4�2�

′ � Alt�6�. Thelatter case is prohibited by Proposition 3.1 and so we may assume G0 � PSL3�2�and G � Aut�PSL3�2��. Let t ∈ Ru�P0� � D8 have order 4, and suppose Gxh isfixed by t. Then as we have seen above, t ∈ Gh

x = NG�Ru�P0��h = NG�Ru�P0�

h�, sot ∈ NG0

�Ru�P0�h� = Ru�P0�

h, whence t ∈ Ru�P0� ∩ Ru�P0�h. But t2 �= e is the unique

central involution of both Ru�P0� and Ru�P0�h, and so Ru�P0�

h ⊆ CG0�t2� ⊆ Ru�P0�.

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Thus h ∈ NG�Ru�P0�� = Gx and t fixes a unique element of X, contrary to G a sharppermutation group of type ��0� k�� n� on X with k > 2. This completes the proof ofClaim 3.

Claim 4. Gx � Sym�4�.

Proof. Suppose not. If G0 �= G, then �P0�2 = 4 and G0 contains a self centralizingSylow 2-subgroup of order 4. Thus Theorem 2.1 p. 421 of [21] implies G0 � PSL2�q�with q ≡ 3� 5 mod 8. Thus G0 � PSL2�5� � Alt�5�, contrary to Proposition 3.1.

Thus we may assume G = G0 and Gx � Sym�4� is a maximal parabolicsubgroup of G. Since Gx must be a parabolic subgroup of G, the definingcharacteristic of G is 2. On the other hand, G has a dihedral Sylow 2-subgroup andis not isomorphic to Alt�7� and so G � PSL2�q� with q odd (see p. 462 of [21]).Thus by Proposition 2.9.1 of [30] we must have G � PSL2�7� � GL3�2�. But thenG Gx� = 7 < 24 = �Gx�, contrary to Propositions 2.4(e). Thus Claim 4 holds.

From Claims 1–4, we obtain a contradiction to Corollary 2.7, and the proofis complete. �

4.1. Classical Groups

The goal of this section is a proof of the following proposition.

Proposition 4.2. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X. Then Soc�G� is not a simple classical group.

We will divide our analysis according to the collections �i�G�, 1 ≤ i ≤ 8from [30]. The reader is warned that the definition of these subgroup classes differsslightly from those introduced by Aschbacher in [1]. According to Aschbacher’stheorem, if the maximal subgroup Gx is not in one of these classes, then Gx is almostsimple, contrary to Proposition 2.5.

The main results in [30] assume the following conditions:

(a) If Soc�G� � P�+8 �q�, then G does not induce a triality automorphims; and

(b) If Soc�G� � PSp4�2k�, then G does not induce a graph (or graph field)

automorphism of order 2.

For this reason, we rule out these two cases with the following two lemmas.

Lemma 4.3. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X. If Soc�G� � P�+

8 �q� then G does not induce a triality automorphismon Soc�G�.

Proof. Suppose, by way of contradiction, that G induces a triality automorphismon Soc�G�.

Throughout this argument, we set G0 = Soc�G�, and set M = Gx. Moreover,for any subgroup K of G, we set K0 = K ∩G0.

The possible shapes of maximal subgroups M in G are given in Table Iof [29]. By Proposition 2.5, F ∗�M� = F�M� and M has, at most, one nonsolvable

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composition factor which is necessarily isomorphic to Alt�5�. Moreover, byProposition 4.1, M0 cannot be a parabolic subgroup of G0. Combining these facts,we reduce to 5 possible subgroup configurations corresponding to rows 26, 55–57,and 61 of Table I from [29]. We treat each case separately.

If M is of type N1 (as in row 26 of Table I of [29]), then q = 2 and

M0 � ��3 ×GU3�2����2�

In this case G/G0 � �3 or Sym�3�. Evidently, we have F�M0� = O3�M0�. Moreover,M is neither a p-group nor isomorphic to Sym�4�. However, there are 2-powerelements of M0 − F�M� that centralize nontrivial 3-power elements of F�M�, contraryto Corollary 2.8.

Suppose next that M is of type I�2 (as in rows 55 and 56 of Table I of [29]).Then M0 = M ∩G0 is the stabilizer of a decomposition V = V1 ⊥ V2 ⊥ V3 ⊥ V4,where each Vi is a 2-dimensional orthogonal space of type �. So M0 = S0 J whereS0 is the kernel of the permutation action of M0 on � = �V1� V2� V3� V4� and J �Sym�4�, with J acting naturally on �. Note also that S0 contains a normal subgroupwhose corresponding subgroup in O+

8 �q� is

D2�q−�� ×D2�q−�� ×D2�q−�� ×D2�q−��

and with J acting naturally on each component of the direct product.Obviously, M is neither a p-group, nor isomorphic to Sym�4�, so Corollary 2.8

applies.Let r be the largest prime divisor of q − �. If r > 3, then it is easy to check

that Or�M� = Or�M0� �= �1�. But O3�M� ∩ J = �1�, together with the action of J onS0 yield an element t ∈ M − F�M� of order 3 (it is an element of J ) that centralizesan element of order r in Or�M�, contrary to Corollary 2.8.

Suppose next that r = 2, so q − � is a power of 2. Then O2�M0� �= �1� and isnot a Sylow 2-subgroup of M0. Moreover, O3�M0� = �1�. According to row 55 ofTable I in [29], q − 1 �= 2. In particular, arguing as above we see that there is a 3-element of M − F�M� (it is an element of J ) that centralizes a 2-power element ofO2�M�, contrary to Corollary 2.8.

Thus we may assume that 3 is the largest prime dividing q − �. In this casethere is an involution in M − F�M� (it is an element of J ) that must centralize anelement of O3�M� of order 3, a contradiction as before.

Suppose next that M is of type I+4 (as in rows 57 of Table I of [29]). Then M0

contains a normal subgroup W0 with

W0 � �+4 �q� ��+

4 �q� � SL2�q� � SL2�q� � SL2�q� � SL2�q�

and so as F ∗�M� = F�M� (by Proposition 2.5), we have q = 2 or 3. The case q = 2is excluded by the restrictions in Row 57, Table I of [29], an so

S � SL2�3� � SL2�3� � SL2�3� � SL2�3��

It follows that O3�W0� = O3�M0� = �1� and O2�M0� = F�M0� and, as above, weobtain 3 power elements outside of F�M� that centralize 3′ elements of F�M�,contrary to Corollary 2.8.

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Finally, we suppose M is of type N3 (as in row 61 of Table I of [29]). Then

M0 � �D 2d �q

2+1� ×D 2d �q

2+1�� ��2�2�

where d = gcd�2� q − 1�. Write M0 = D1 ×D2 where Di = Ci ti�, with Ci cyclicof order q2+1

dand ti and involution inverting every element of Ci, for i = 1� 2.

By Zsigmondy’s theorem, there exists r a prime divisor of q4 − 1 that does notdivide qi − 1 for i < 4. It follows from the order of G0 that M0 is the normalizerin G0 of a Sylow r-subgroup of G0–indeed, M = NG�Or�M0�� and Or�M0� ∈Sylr �G0�. Evidently, r �= 2 and Or�G0� = Or�C1�× Or�C2�. Since t1� C2� = �1� andt1 �∈ O2�M� ∩G0, we have an element of order 2 centralizing a nontrivial element ofOr�M�. But M is neither a p-group nor a symmetric group of degree 4, contrary toCorollary 2.8. This completes the proof. �

Lemma 4.4. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X. If Soc�G� � PSp4�2

m�, then G does not induce a graph or graph-fieldautomorphism on Soc�G�.

Proof. Suppose, by way of contradiction, that G induces a graph or graph fieldautomorphism on Soc�G�.

As before, we set G0 = Soc�G�, and set M = Gx. Moreover, for any subgroupK of G, we set K0 = K ∩G0.

According to Section 14 of [1], M = NG�Z�, and one of the followingstatements hold:

�1: Z ∈ Syl2�G0�;�2: q > 2 and Z ≤ Y , Y of type O+

4 �2m� (a �8 subgroup of Sp4�2

m�) and Z stabilizesa nondegenerate 2-dimensional subspace;

�3: Z � �q2+1;�4: Z is a subgroup of type �5�G0�;�5: The field of definition of G0 is GF2m, m odd and Z is generated by an

involutory outer automorphism (the centralizer of a graph or graph-fieldautomorphism of G0).

The �1 case is prohibited by Proposition 4.1. In the �3 case, if r is aZsigmondy prime divisor of q4 − 1, then R = Or�M0� ∈ Sylr �G0� (as in the proof ofLemma 4.3) and M = NG�R�. Since R is cyclic, the usual argument shows that agenerator of R fixes only the trivial right coset M , contrary to the fact that G issharp of type ��0� k�� n� on the right coset space M\G with k > 2.

In the �4 case Z corresponds to the centralizer in G0 of a field automorphismof prime order. Thus Sp4�2� embeds in M0, which is prohibited by Proposition 2.5.

In the �5 case Z corresponds to the centralizer in G0 of a graph or graph-fieldautomorphism of order 2. Since q = 2m, with m odd, 19.5 of [2] forces all involutionsof G−G0 are conjugate. Moreover, for such an involution t, Z = CG0

�t� � Sz�2m�.As q > 2, it follows that F ∗�M� �= F�M�, contrary to Proposition 2.5.

It remains to treat the case �2. Since q > 2, let r� (� = ±1) denote thelargest prime divisor of q − � and let R� ∈ Sylr� �G0�. According to the proof of14.3 in [1], there are two G-conjugacy classes of type �2 with representatives

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NG�R+� and NG�R−�. Observe that M0 = NG0�R�� corresponds to a subgroup

of O+4 �q� that stabilizes and orthogonal decomposition of V = V1�� ⊥ V2�� with

Vi�� a 2-dimensional orthogonal space of � type. In particular, M0 contains anormal subgroup isomorphic to D2�q−�� ×D2�q−�� and the usual argument yields aninvolution t � O2�M� that centralizes a nontrivial r�-element of Or�

�M�, contrary toCorollary 2.8. This completes the proof. �

Now we proceed case by case through the various classes within each of theclassical groups.

Lemma 4.5. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �1

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0.

By Proposition 4.1 M0 cannot be a parabolic subgroup of G0 (maximal orotherwise).

Suppose that G0 is conjugate to one of PSpn�q� or P�+n �q�, or P�−

n �q�.According to Tables 3.5.C, E, and F, pp. 72–74 of [30], all possible subgroups oftype �1 must have nonsolvable socles, contrary to Proposition 2.5. Thus we mayassume that G0 is either linear, unitary, or odd dimensional orthogonal. We handleeach case separately.

Case 1. G0 � PSLn�q�According to [30], M0 must be of type GLm�q�×GLn−m�q� with 1 ≤ m < n−

m. Moreover, as M0 is never maximal in G0 (such subgroups are Levi factors ofmaximal parabolic subgroups of G0) we must have G �= G0 and G must inducea graph or graph-field automorphism of order 2 that stabilizes M0. In addition,Proposition 2.5 requires F ∗�M� = F�M�, and so it follows that m = 1, n = 3 andq = 2� 3.

If G0 � PSL3�2�, then M0 � Sym�3� and M � Sym�3�× �2. In particular,M = NG�T� where T ∈ Syl3�G�. Let t be an element of T of order 3. Then Mh isfixed by t if and only if t ∈ T ∩ Th. Thus T = Th, h normalizes T and Mh = M . Sot fixes a unique right coset of M in G, contrary to G a sharp permutation group oftype ��0� k�� n� on X with k > 2.

On the other hand, if G0 � PSL3�3�, then M0 � GL2�3� and M � GL2�3� �2.In this case F�M� = O2�M� and so by Corollary 2.7, M must be a Frobenius groupwith Frobenius kernel O2�G�, an obvious contraction (M centralizes the uniquecentral involution of M0).

Case 2. G0 � PSUn�q�Then according to [30], M0 must be of type GUm�q�×GUn−m�q� with 1 ≤ m <

n−m. As before, Proposition 2.5 requires F ∗�M� = F�M� and so m− n < 4. Thusm = 1� 2 and n ≤ 5.

Suppose first that m = 2. Then n−m = 3 and F ∗�M� = F�M� forces q = 2.We then have M0 � �SU2�2�× SU3�2�� �3 and M M0� = G G0� ≤ 2. EvidentlyO3�M0� = F�M0�. However SU3�2� � 31+2 Q8 and so a 2-power element of M0 −F�M� centralizes a nontrivial element of O3�M�, contrary to Corollary 2.8.

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Suppose m = 1. Then either n = 3 and q = 3 or n = 4 and q = 2. Suppose firstthat n = 4 and G0 � PSU4�2�. Then M0 � GU3�2� and arguing as in the m = 2 caseyields a contradiction.

Finally, suppose m = 1, n = 3 and q = 3. Then M0 � SL2�3� �4 and thecentral involution of SL2�3� is M-stable. In particular, M cannot satisfy theconclusion of Proposition 2.6, a contradiction.

Case 3. G0 � �n�q�,with q odd and n ≥ 7In this case, Proposition 2.5 forces M0 to be of type O3�3�×O+

4 �3�. In this caseG G0� ≤ 2 and a Sylow 3-subgroup of G has 3-rank 3. Now F�M0� = O2�M0� andso there are 3-power elements of M0 − F�M� which centralize elements in O2�M�,contrary to Corollary 2.8. This completes the proof. �

Lemma 4.6. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X where Soc�G� is a simple classical group. Then Gx is not of type �2

Proof. Suppose not. As before, we set G0 = Soc�G�, and set M = Gx. Moreover,for any subgroup K of G, we set K0 = K ∩G0.

Case 1. G0 � PSLn�q�.According to Table 3.5.A p. 70 of [30], M0 must have the shape

GLm�q� Sym�t��

where mt = n. Let J ≤ GLn�q� with J � Sym�t� and which permutes thecomponents of

GLm�q�× · · · ×GLm�q�

naturally, and J the corresponding subgroup of M . By Proposition 2.5, we havem ≤ 2, t ≤ 4.

Suppose first that m = 2. Since F ∗�M� = F�M� (by Proposition 2.5), we haveq ≤ 3. When q = 2, the fact that PSL4�2� � Alt�8� together with Proposition 3.1allow us to assume 2 < t ≤ 4. For either choice of t, we have O3�M0� = F�M0�and there exist involutions in M0 − F�M� centralizing nontrivial 3-power elementsof F�M�, contrary to Corollary 2.8. If q = 3, then O2�M0� = F�M0� and there existelements of order 3 in M0 − F�M� centralizing nontrivial 2-power elements of F�M�,contrary to Corollary 2.8.

Henceforth, we suppose m = 1.Suppose now that t = 4. Since PSL4�2� � Alt�8�, by Proposition 3.1 we may

assume q > 2.According to p. 69 of [16], there are no maximal subgroups of type GL1�3�

Sym�4� in G where PSL4�3� = G0 � G ≤ Aut�G0�, so q �= 3.If q = 5, then O2�M0� = F�M0�, and there exist elements of order 2 and 3 in

M0 − F�M�, contrary to Corollary 2.8.Thus, we may assume that q − 1 > 4, and so either there exists an odd prime

divisor r of q − 1, or q − 1 = 2s with s > 2.

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Let � be a generator of the multiplicative group of �q, and let a denote theimage of

� 0 0 00 � 0 00 0 � 00 0 0 ���−3

in G0. By the restrictions on q, a is a nontrivial element of F�M�. If q − 1 is nota power of 2, then a is centralized by an involution of J . In particular, there isan r-element of F�M� centralized by a r ′-element of M − F�M�, a contradiction asbefore. On the other hand, if q − 1 = 2s, there is an element of order 3 in J ∩ �M −F�M�� that centralizes a, contrary to Corollary 2.8.

Henceforth, we may assume that t = 3 and m = 1. The familiar boundingarguments using Proposition 2.4 yield a contradiction unless q = 2. If q = 2, thenG � Aut�PSL3�2�� and M � D12. As M is the normalizer of a Sylow 3-subgroupof G, the usual argument yields an element of M of order 3 fixing a unique rightcoset of M in G, contrary to G a sharp permutation group of type ��0� k�� n� on Xwith k > 2.

Case 2. G0 � PSUn�q�.According to Table 3.5.B, p. 71 of [30], M0 must have the shape

GUm�q� Sym�t��

where mt = n, or

GL n2�q2���2�

In the later case, F ∗�M� �= F�M�, contrary to Proposition 2.5.Let J ≤ GUn�q� with J � Sym�t� and which permutes the components of

GUm�q�× · · · ×GUm�q�

naturally, and J the corresponding subgroup of M . By Proposition 2.5, we havem ≤ 3, t ≤ 4.

If m = 1, then t = 3 or 4. The t = 3 case is eliminating by the standardbounding arguments using Proposition 2.4. The t = 4 case is eliminated along thesame lines as in Case 1 above.

If m = 2, then q ≤ 3, and this is handled as in Case 1 above.Finally, if m = 3 then q = 2 and t = 2� 3 or 4. In any case, there is an

involution outside of O2�M� that centralizes a nontrivial element of O3�M�, contraryto Corollary 2.8.

Case 3. G0 � PSpn�q�.By Lemma 4.4 we may assume that G does not induce a graph or graph field

automorphism on G0.According to Table 3.5.C, p. 72 of [30], M0 must have the shape

Spm�q� Sym�t��

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where mt = n and m even, or

GL n2�q���2�

where q is odd.Suppose first that M0 has shape Spm�q� Sym�t�. As usual, Proposition 2.5

requires m = 2 and q ≤ 3.Suppose q = 2. As PSp4�2� � Alt�6�, Proposition 3.1 allows us to assume 2 <

t ≤ 4. As in Case 1 above, F�M0� = O3�M0� there is an element of order 2 in M0 −F�M� centralizing a nontrivial element of F�M�, contrary to Proposition 2.8. Ifq = 3, (as in Case 1 above) there are elements of order 3 in M0 − F�M� centralizingnontrivial 2-power elements in F�M�, contrary to Corollary 2.8.

On the other hand, if M0 has shape GL n2�q���2, Proposition 2.5 requires

q = 3 and n = 4. In particular, F�M0� = O2�M0�. Observe that since F�M0� is acharacteristic subgroup of M0, and M0 � M , we have F�M� ∩M0 = F�M0� andso G G0� = M M0� = F�M� F�M0�� ≤ 2. Since G G0� ≤ 2, we have F�M� =O2�M�. But there exist elements of order divisible by 2 in M0 − F�M�, and so Mcannot be a Frobenius group. Moreover, M is clearly not a 2-group, nor is itisomorphic to Sym�4�, contrary to Corollary 2.7.

Case 4. G0 � �n�q�, with q odd and n ≥ 7.According to Table 3.5.D of [30], M0 must have the shape

Om�q� Sym�t��

where mt = n and m > 1 odd, or

O1�q� Sym�n��

The latter is impossible by Proposition 2.5. The former shape requires m = 3, q = 3(since q odd), and t = 3 or 4. The corresponding configurations can be eliminatedas above.

Case 5. G0 � P�+n �q�.

By Lemma 4.3, we may assume G does not induce a triality automorphism onG0. From Table 3.5.E p. 73 of [30], together with the fact that n ≥ 8 and F ∗�M� =F�M�, M0 must have the shape

O�m�q� Sym�t�

or

O1�q� Sym�n��

The latter is impossible by Proposition 2.5. The former shape requires m ≤ 4, and1 < t ≤ 4. As before, the corresponding configurations are ruled out as above.

Case 6. G0 � P�−n �q�. This can be handled as above.

This completes the proof. �

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Lemma 4.7. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �3

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0.

Suppose first that G0 isomorphic to one of P�+n �q�, P�

−n �q�, or �n�q�,with

q odd and n ≥ 7. According to Tables 3.5.D–3.5.F, pp. 72–74 of [30], all possiblesubgroups of type �3 must have nonsolvable socles, contrary to Proposition 2.5.Thus we may assume that G0 is either linear, unitary, or symplectic. We handle eachcase separately.

Case 1. G0 � PSLn�q�.According to Table 3.5.A, p. 70 of [30], M0 has shape GL n

r�qr�: �r where r is

a prime divisor of n. Since F ∗�M� = F�M� by Proposition 2.5, it follows that n = r.Suppose first that r ≥ 3. If, in addition, �r� q� �= �3� 4�, then Zsigmondy’s

theorem forces the existence of a prime divisor t of prm − 1 (here q = pm) such thatt does not divide pi − 1 for any i < rm. Let T ∈ Sylt�G0�. Then T is characteristicin M0 and so M = NG�T�. But T is cyclic, so the usual argument shows that for anygenerator � of T , � fixes a unique right coset of M in G, contrary to G a sharppermutation group of type ��0� k�� n� on X, with k ≥ 3.

Next suppose G0 � PSL3�4�. Then M0 � �7 �3. In particular, if S ∈ Syl7�G0�,then S is a characteristic, cyclic subgroup of M0 and M = NG�S�. As in the previouscase, this forces any generator of S to fix a unique right coset of M in G, acontradiction as before.

Finally, suppose r = 2. By Proposition 3.1, we may assume G0 � Alt�5� soq > 5. If p2m − 1 admits a primitive prime divisor, then we can argue as above toproduce an element with precisely one fixed point in X, which is impossible. Thus,by Zsigmondy’s theorem, either p = 2 and m = 3, or m = 1 and p+ 1 = 2v for somev > 2.

If p = 2 and m = 3, then G0 � PSL2�8�. It follows that M0 contains a cyclic,normal Sylow 3-subgroup T . The usual argument forces a generator of T to fix aunique right coset of M in G, a contradiction as above.

Thus, we may assume G0 � PSL2�p� where p > 5 and p+ 1 = 2v for somev > 2. Then M0 � Dp+1, and as p+ 1 ≥ 8, M0 has a unique cyclic subgroup ofindex 2, which we denote by C0. Then C0 is characteristic in M0 and M = NG�C0�.Let g be any generator of C0 and suppose Mh is fixed by g. This forces g ∈ NG�C

h0 �

so g ∈ NG0�Ch

0 � = Mh0 . But this forces g ∈ Ch

0 and so C0 = Ch0 , whence h ∈ M , and g

fixes a unique right coset of M in G, a contradiction as above.

Case 2. G0 � PSUn�q�.By Table 3.5.B, p. 71 of [30], M0 must have shape GU n

r�qr�, with r an odd

prime. As in the previous case, Proposition 2.5 forces n = r.Set q = pm. Note that �r� q� �= �3� 2�, since G0 is simple. Now

M0 � � qr+1�q+1� gcd�r�q+1�

In particular, by Zsigmondy’s theorem, there exists a primitive prime divisor s ofp2r − 1. It follows that s divides prm + 1 but does not divide pm + 1. If T is a Sylow

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s-subgroup of M0, then T is a cyclic characteristic subgroup of M0, and so the usualargument yields a nontrivial element of T fixing exactly one right coset of M in G,a contradiction, as above.

Case 3. G0 � PSpn�q�.By Lemma 4.4 we may assume that G does not induce a graph or graph-field

automorphism on G0. By Table 3.5.C, p. 72 of [30] together with Proposition 2.5,we have n = 4, q = 3, and M0 has the shape GU2�3�.

According to Table 3.5.D, p. 72 of [30] (see also Proposition 4.3.7, p. 118of [30]),

M0 � �GU2�3�/�2���2

In this case, F�M0� = O2�M0� as G G0� ≤ 2, and by the same argument as inCase C of Lemma 4.6 we have F�M� = O2�M�. However, there are elements oforder 3 in M0 − F�M� that centralizes a nontrivial involution in F�M�, contrary toProposition 2.8. This completes the proof. �

Lemma 4.8. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �4

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0.

Suppose first that G0 is conjugate to one of PSLn�q�, P�−n �q�, or �n�q� with q

odd and n ≥ 7. According to Tables 3.5.A, D, and F, pp. 70–74 of [30], all possiblesubgroups of type �4 in these groups must have nonsolvable socles, contrary toProposition 2.5. Thus we may assume that G0 is either unitary, or symplectic, or isa +-type orthogonal group. We handle each case separately.

Case 1. G0 � PSUn�q�.By Table 3.5.B, p. 71 of [30] together with Proposition 2.5, G0 � PSU6�2� and

M0 must have shape GU2�2� �GU3�2�. In fact, By Proposition 4.4.10 of [30],

M0 � PSU2�2�× PSU3�2��

In particular, there exists a 2-power element in M0 − F�M� that centralizes anontrivial element in O3�M�, contrary to Corollary 2.8.

Case 2. G0 � PSpn�q�.By Lemma 4.4 we may assume that G does not induce a graph or graph field

automorphism on G0. By Table 3.5.C, p. 72 of [30] together with Proposition 2.5,we have either

M0 � PSp2�3�× PO3�3�

or

M0 � �PSp2�3�× PO+4 �3����2�

In either case, F�M0� = O2�M0�, and there is a nontrivial 3-element of M0 − F�M�centralizing a nontrivial 2-element of O2�M�, contrary to Corollary 2.8.

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Case 3. G0 � P�+n �q�.

By Lemma 4.3 we may assume G does not induce a triality automorphism onG0. From Table 3.5.E, p. 73 of [30] together with Proposition 2.5, we conclude thatG0 � P�+

12�3� and

M0 � P�+4 �3�× SO3�3��

But as before, this is incompatible with Corollary 2.8. This completes the proof. �

Lemma 4.9. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �5.

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0.

Suppose first that G0 is conjugate to one of PSpn�q�, P�±n �q�, or �n�q�,with

q odd and n ≥ 7. According to Tables 3.5.C–3.5.F, pp. 72–74 of [30], all possiblesubgroups of type �5 in these groups must have nonsolvable socles, contrary toProposition 2.5. Thus we may assume that G0 is either a linear group or a unitarygroup. We handle the two cases separately.

Case 1. G0 � PSLn�q�.By Table 3.5.A, p. 70 of [30] and Proposition 2.5, we have G0 � PSL2�p

r�where r is a prime and p ∈ �2� 3�.

Suppose first that p = 2, so by Proposition 3.1, r > 2. Then byProposition 4.5.3, p. 141 of [30], M0 � Sym�3�. Note that there exists a fieldautomorphism � of G0 of order r such that Aut�G0� � G0 �� and M =CAut�G0�

���. In particular, if G0 �= G, then there will be an involution of M − F�M�that centralizes a nontrivial element of Or�M�, contrary to Corollary 2.8.

Thus we may assume G = G0 and M = M0 � Sym�3�. However, in this case itis simple to verify that GM�+1

2 > �M�, contrary to Proposition 2.4(a).Next suppose that q = 3r . By Proposition 4.5.3, p. 141 of [30] M0 embeds

in PGL2�3� � Sym�4� and S � M0 with S � Alt�4�. In either case, it is astraightforward matter to show GM�+1

2 > �M�, contrary to Proposition 2.4(a).

Case 2. G0 � PSUn�q�.By Table 3.5.B, p. 71 of [30] together with Proposition 2.5, we have either

(i) G0 � PSU3�2r �, r an odd prime and M0 � PGU3�2� or PSU3�2�; or

(ii) G0 � PSU4�3� and M0 � PSO+4 �3� �2.

Suppose (i) holds. Observe that 3 � 2r + 1 for every odd prime r. In addition,9 � 2r + 1 for r an odd prime only when r = 3. Setting

c = 2r + 1

lcm[3� 2r+1

gcd�3�2r+1�

] �

we have c = 3 if r = 3 and c = 1 for every odd prime r > 3. By Proposition 4.5.3p. 141 of [30] we have M0 � PSU3�2� if r > 3 and M0 � PGU 3�2� if r = 3.

In the first case, it is easy to show GM�+12 > �M�, contrary to Proposition 2.4(a).

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Thus we may assume G0 � PSU3�8� and M0 � PGU 3�2� � ASL2�3�. NowO3�M0� = F�M0� = F�M� ∩M0 and there exist elements of order 2 and 3 in M0 −F�M0�. In particular, M cannot satisfy Corollary 2.8, a contradiction.

Next we suppose (ii) holds. In this case, M0 � �Alt�4�×Alt�4�� �2, and sothere is a nontrivial element of order 3 in M0 − F�M� centralizing a nontrivialelement of O2�M�, contrary to Corollary 2.8. This completes the proof. �

Lemma 4.10. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �6.

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0.

By Propositions 4.6.8, 4.6.9, and Tables 3.5.A–F, pp. 70–74 of [30], togetherwith Proposition 2.5, we may assume that G0 is either linear or unitary. We handleeach case individually.

Case 1. G0 � PSLn�q�.From Propositions 4.6.5–4.6.7 of [30], together with Proposition 2.5, one of

the following statements must hold:

(i) G0 � PSL2�p� and M0 � Alt�4� or Sym�4�, p an odd prime;(ii) G0 � PSL3�p

e�, M0 � PSU3�2�, p > 3;(iii) G0 � PSLrm�p

e� and M0 � Z2mr �Sp2m�r�.

Suppose first that (i) holds. By Proposition 3.1, we may assume p > 5. FromProposition 4.6.7 p. 153 of [30] we have that p ≡ 3� 5 �mod 8� if M0 � Alt�4� andp ≡ 1� 7 �mod 8� if M0 � Sym�4�. In either case, a routine argument yields GM�+1

2 >�M�, contrary to Proposition 2.4(a).

Next we suppose (ii) holds. Here, e is the smallest positive odd integerfor which pe ≡ 1 �mod 3�. If p ∈ �2� 5�, then p ≡ −1 �mod 3�, and so pe ≡ −1�mod 3� for every e (since e is odd). Thus we may assume p ≥ 7. Again, a routineargument yields GM�+1

2 > �M�, contrary to Proposition 2.4(a).Finally, suppose (iii) holds. Since r must by odd (By Proposition 4.6.5 pp. 151–

152 of [30]), Proposition 2.5 forces m = 1 and r = 3. In particular, M0 � Z23�Sp2�3�

and G0 � PSL3�pe�, e odd and minimal subject to pe ≡ 1 �mod 3�. In addition,

since we have already treated (ii) above, Proposition 4.6.5 of [30] allows us to assumepe �≡ 4� 7 �mod 9�. As in (ii) above, p > 5. Indeed, a simple computation allows usto assume pe ≥ 19. With this assumption, it is trivial to verify that GM�+1

2 > �M�,contrary to Proposition 2.4(a).

Case 2. G0 � PSUn�q�.From Proposition 2.5 together with Propositions 4.6.5, 4.6.6 pp. 151–152 and

Table 3.5.B p. 71 [30], we may assume that one of the two cases holds:

(a) G0 � PSU3�pf �, M0 � PSU3�2�, e = 2f the smallest even integer for which pe ≡

1 �mod 3�. In addition, pe ≡ 2� 5 �mod 9�;(b) G0 � PSL3�p

f � and M0 � Z23 · Sp2�3�.

However, the argument in this case is essentially the same as in Case 1. In particular,we verify GM�+1

2 > �M�, contrary to Proposition 2.4(a). This completes the proof. �

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Lemma 4.11. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �7.

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0. According to Definition (b) p. 156of [30], the socle of M must be nonsolvable, contrary to Proposition 2.5. Thiscompletes the proof. �

Lemma 4.12. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X with Soc�G� a simple classical group. Then Gx is not of type �8.

Proof. Suppose not. As before, we set G0 = Soc�G� and M = Gx. Moreover, forany subgroup K of G, we set K0 = K ∩G0. According to Tables 3.5.A–F pp. 70–74of [30], such subgroups occur only if G0 is linear or symplectic. In the symplecticcase, Proposition 3.1 together with the fact that Sp4�2� � Sym�6� allows us toassume the field of definition has cardinality at least 4.

By Proposition 4.8.6, p. 168 of [30], F ∗�M0� will be nonsolvable, contrary toProposition 2.5. Thus we need consider only the linear groups.

Suppose, G0 � PSLn�q�. From Propositions 4.8.3–4.8.5 and Table 4.8.A,pp. 165–168 of [30], we are left with the following possible configurations:

(1) G0 � PSL4�3� and M0 � PSO+4 �3� �2;

(2) G0 � PSL3�3� and M0 � S03�3� � Sym�4�; or(3) G0 � PSL3�4� and M0 � PSU3�2�.

If (1) holds then we obtain a 3-element of M0 − F�M� that centralizes a 2-power element in F�M�, contrary to Corollary 2.8.

If (2) holds, it is simple to show that GM�+12 > �M�, contrary to

Proposition 2.4(a).Thus we may assume that (3) holds. Observe that M0 is a Frobenius

group. If G0 = G, then it is straightforward to verify GM�+12 > �M�, contrary to

Proposition 2.4(a).Thuswemay assumeG �= G0. By Proposition 4.8.5 p. 168 of [30], G M� = G0

M0� = 280. Moreover, since G is a sharp permutation of type ��0� k�� n� on X withk ≥ 3, Proposition 2.4(d) forces k = 8�35− 9G G0��, whence 2 ≤ G G0� ≤ 3.

If G G0� = 3. Then by [16], M � ASL2�3�. However, this group (as wehave seen in the proof of Proposition 3.1, for example), is not compatible withProposition 2.6.

On the other hand, if G G0� = 2, then by [16] either M � PSU3�2�× �2 orM � PSU3�2� �2. In either case M = NG�P� where P ∈ Syl3�G�.

Let � ∈ P − �1� We claim that � fixes a unique right coset of M in G. Supposenot, and let � fix Mg �= M . Then � ∈ Mg = NG�P

g�, whence � ∈ P ∩ Pg. Thus C0 =CG0

��� contains distinct Sylow 3-subgroups P and Pg, and so [16] forces C0 to becontained in a maximal subgroup of G0 isomorphic to Alt�6�. But C0 containsdistinct Sylow 3-subgroups of Alt�6� and so by [16], C0 = Alt�6� which contradictsthe simplicity of Alt�6�. Thus the claim is proved. As before, this contradicts thefact that G is a sharp permutation group of type ��0� k�� n� on X with k > 2. Thiscompletes the proof. �

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Proof of Proposition 4.2. Suppose, by way of contradiction, that G is an almostsimple, primitive sharp permutation group of type ��0� k�� n� on X, whose socle is asimple classical group. Then Gx is a maximal subgroup of G, and by Proposition 2.5,Gx is either solvable or is a nonsolvable Frobenius group.

By Lemma 4.4, we may assume that G does not induce a graph or graph fieldautomorphism on G0 when G0 is a symplectic group. Similarly, Lemma 4.3 allowsus to assume G does not induce a triality automorphism on G0 when G0 � P�+

8 �q�.Thus by Aschbacher’s theorem [1], together with Proposition 2.5, we see that

M must be a subgroup from one of the modified Aschbacher classes �1 through �8.However, this is impossible by Lemmas 4.5–4.12. This completes the proof. �

4.2. Exceptional Groups

In this section we consider the case in which Soc�G� is an exceptional groupof Lie type. As G2�2� � Aut�PSU3�3�� and

2G2�3� � Aut�PSL2�8��, Proposition 4.2implies that we may ignore these groups in the analysis which follows. The goal ofthis section is a proof of the following proposition.

Proposition 4.13. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X. Then Soc�G� is not a simple exceptional group of Lie type.

If G is a primitive sharp permutation group of type ��0� k�� n� on X, thenProposition 2.5 forces Gx to be either solvable, or a nonsolvable Frobenius group.In particular, F ∗�Gx� is solvable, and so Gx is a local maximal subgroup.

According to Theorem 1 of [15], a maximal subgroup M of G must be of oneof the following types:

(a) M = CG��� where � is a field, graph, or graph-field automorphism of G0 ofprime order s; or

(b) M is a parabolic subgroup of G; or(c) M is a subgroup of maximal rank as in Table 5.1 or 5.2 of [34]; or(d) M = NG�E� where E is an elementary abelian r-group as in Table 1 of [15].

Lemma 4.14. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X, where Soc�G� is a simple exceptional group of Lie type. Thenthe maximal subgroup Gx is not the centralizer of a field, graph, or graph-fieldautomorphism of G0 of prime order.

Proof. Suppose not. Throughout this argument, we set G0 = Soc�G� and writeM = Gx, and for any subgroup K of G, K0 = K ∩G0.

Recall that by Proposition 4.2, we may assumeG is not of typeG2�2� or2G2�3�.

According to the remarks following Table 1 of [15], one of the followingstatements holds:

(i) M0 is the centralizer of a field automorphism, and involves a group of Lie typedefined over a maximal subfield of �q with the same Dynkin Diagram as G0; or

(ii) M0 is the centralizer of a graph-field automorphism, and involves a twistedversion of G0; or

(iii) G0 = E6�q� or2E6�q� and M0 involves a group of type C4�q� (q odd) sor F4�q�.

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Observe that if (ii) or (iii) holds, then F ∗�M� is certainly nonsolvable, contrary toProposition 2.5. Thus (i) must hold. Again, the fact that F ∗�M� is solvable forcesG0 �2B2�q� (q ≥ 8) andM0 � 2B2�2�, where q

1r = 2, r an odd prime. Then for all q ≥ 8,

G Gx�+ 12�Gx�

>�G�

2�Gx�2≥ q2�q2 + 1��q − 1�

800 log2�q�> 1�

contrary to Proposition 2.4(a). This completes the proof. �

Lemma 4.15. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X, where Soc�G� is a simple exceptional group of Lie type. Then themaximal subgroup Gx is not a parabolic subgroup of G.

Proof. This follows from Proposition 4.1. �

Lemma 4.16. Let G be an almost simple, primitive sharp permutation group of type��0� k�� n� on X, where Soc�G� is a simple exceptional group of Lie type. Then themaximal subgroup Gx is not a subgroup of maximal rank.

Proof. Suppose not. Throughout this argument, we set G0 = Soc�G� and writeM = Gx, and for any subgroup K of G, K0 = K ∩G0.

As usual, we treat the various possibilities for G0 individually.Since M is maximal in G and is not a triviality, G0M = G, and so we have

G M�+ 12�M� ≥ �G�

2�M�2 = �G0�2�M0�2G G0�

� (1)

In particular, we will obtain a contradiction to Proposition 2.4(a) whenever

�G0�2�M0�2G G0�

> 1 (2)

is shown to hold.

Case 1. G0 � 2B2�22a+1�, a > 0.

According to [15] and [42], M0 is maximal in G0 and one of the followingstatements must hold:

(a) M0 � �q−1 �2; or(b) M0 � �

q+√

2q+1 �4; or

(c) M0 � �q−√

2q+1 �4.

Suppose first that (a) holds. Then

�G0�2�M0�2G G0�

≥ q2�q2 + 1�8�q − 1� log2�q�

> 1

for all q ≥ 8. Thus inequality (2) holds, a contraction.

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Suppose (b) holds. Two simple calculus arguments yield q�q −√2q + 1� > q +√

2q + 1 on 8��� and q�q − 1� > 32 log2�q� on 32���. So for q ≥ 32, we have

�G0�2�M0�2G G0�

≥ q2�q − 1��q −√2q + 1�

32�q +√2q + 1� log2�q�

> 1�

a contradiction as before. If q = 8, then we have

�G0�2�M0�2G G0�

≥ 2 · 5 · 713 · 3 > 1�

a contradiction.Finally, if (c) holds, then q ≥ 8 forces

�G0�2�M0�2G G0�

≥ q2�q − 1��q +√2q + 1�

32�q −√2q + 1� log2�q�

> 1�

a contradiction, as before.

Case 2. G0 � 2G2�32m+1�. As 2G2�3�

′ � SL2�8�, this case is treated byProposition 4.2 along with the other classical groups. Thus we may assume q =32m+1 ≥ 27. From [15] and [27], together with Proposition 2.5, one of the followingstatements must hold:

(a) M0 � �V ×Dq+12� �3, V a group of order 4; or

(b) M0 � �q+√

3q+1 �6; or

(c) M0 � �q−√

3q+1 �6.

Suppose (a) holds. Then

G M�+ 12

>�G0�2�M0�

= q3�q2 − q + 1��q − 1�12

�

But as q ≥ 27, q2 − q + 1 > 72 and q3�q − 1� > �q + 1� log3�q�, and so we have

G M�+ 12

> �M��

contrary to Proposition 2.4(a).If (b) holds, then we may easily verify that for q ≥ 27,

�G0�2�M0�2G G0�

≥ q3�q2 − 1��q −√3q + 1�

72�q +√3q + 1� log3�q�

> 1�

a contradiction.Finally, if (c) holds, then we may argue as above to obtain a contradiction to

Proposition 2.4(a).

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Case 3. G0 � 3D4�q�. By the Main Theorems of [34] (Tables 5.1, 5.2) and [28],together with Proposition 2.5, one of the following statements must hold:

(a) q = 2 and M0 � SU3�2� · �3 · �2; or(b) M0 � ��q2+q+1�

2 · SL2�3�; or(c) M0 � ��q2−q+1�

2 · SL2�3�; or(d) M0 � �q4−q2+1 · �4.

If (a) holds, set S = SU3�2�. Then F�M� ∩ S = F�S� = O3�S�, and there areelements in S − F�M� of order 4 and 6. In particular, M cannot satisfy Corollary 2.8,a contradiction.

If (b) holds, then

�G0�2�M0�2G G0�

≥ q12�q8 + q4 + 1��q6 − 1��q2 − 1�6�q2 + q + 1�4 log3�q� · �24�2

≥ q12�q8 + q4 + 1��q6 − 1��q2 − 1�2733�q3�4 log3�q�

≥ �q8 + q4 + 1��q6 − 1��q + 1�2733

> 1�

a contradiction. The configurations of type (c) and (d) can be handled in a similarmanner using inequality (2) above.

Case 4. G0 � G2�q�. By Proposition 4.2, we may assume q > 2. Then theMain Theorem of [34] (Tables 5.1, 5.2) together with Proposition 2.5 forces one ofthe following statements to hold:

(a) G = G0 = G2�3� and M0 � �SL2�3� ∗ SL2�3�� · �2; or(b) M0 � ��q−1�

2 ·D12, q > 3 and G induces a graph automorphism on G0; or(c) M0 � ��q+1�

2 ·D12, q > 3 and G induces a graph automorphism on G0; or(d) M0 � �q2−q+1 · �6, q > 3 and G induces a graph automorphism on G0; or(e) M0 � �q2+q+1 · �6, q > 3 and G induces a graph automorphism on G0.

If (a) holds, then F�M� = O2�M�, and there are elements of order 3 in M0 −F�M� centralizing nontrivial elements of O2�M�, contrary to Corollary 2.8.

Note that if any of (b)–(e) hold, then q = 3m (since a graph automorphism ofis induced by G on G0). Moreover, such graph automorphisms have even order, andso 2 divides G G0�. Evidently, F�M� ∩M0 ≤ F�M0�. On the other hand, F�M0� is acharacteristic subgroup of M0 which, in turn, in normal in M , and so F�M0� ≤ F�M�.Thus F�M� ∩M0 = F�M0� and so

G G0� = M M0� = F�M� F�M0���

In particular, 2 � �F�M��, whence O2�M� �= �1�. Note also that according to [27], ineach of the cases (b)-(e) above, M0 is the normalizer of a maximal torus T0 of order�q − 1�2, �q + 1�2, q2 − q + 1, or q2 + q + 1 in case (b), (c), (d), and (e), respectively.In particular, T0 is self-centralizing in G0 (in each case) and so F�M0� = T0.

We will handle the configuration from (b). The rest can be handled inessentially the same manner. To this end, Suppose (b) holds, and set T0 � ��q−1�

2, so

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T0 = F�M0�. Now M0/T0 contains elements of order 6, and so M0 − F�M� containselements of order divisible by 6. Thus Corollary 2.8 forces M to be a Frobeniusgroup. But as we have seen O2�M� �= �1�, so M − F�M� can contain no elementswhose order is divisible by 6, a contradiction.

Case 5. G0 � 2F4�q�′, q = 22m+1. Recall that 2F4�q� is simple unless q = 2. By

the Main Theorems of [34] (Tables 5.1, 5.2) and [36] together with Proposition 2.5,one of the following statements must hold:

(a) G0 � 2F4�2�′ and M0 � SU3�2� · �2; or

(b) M0 � ��q+1�2 GL2�3�, q > 2; or

(c) M0 � ��q+√

2q+1�2 ��4 ∗GL2�3��; or

(d) M0 � ��q−√

2q+1�2 ��4 ∗GL2�3��, q > 2; or

(e) M0 � �q2+

√2q3+q+

√2q+1

�12; or

(f) M0 � �q2−

√2q3+q−

√2q+1

�12, q > 2.

If (a) holds, then we obtain a contradiction to Corollary 2.8 as in Case 3part (a) above.

From [36] each group M0 in (b)–(f) are the normalizers of a maximal torus T0

of G0 of the appropriate order (as in Case 4 above). As before, T0 is self-centralizingin G0 (in each case) and so F�M0� = T0.

Suppose first that one of (b)–(d) hold. Evidently, none of these groups arep-groups nor isomorphic to Sym�4�, and in all cases there exists B ≤ M0 with B ∩F�M� = �1� and B � GL2�3�. The 2-rank of B is 2, and so M cannot be a Frobeniusgroup. On the other hand, B contains elements of order 6 and so M0 − F�M� doesas well. Thus M cannot satisfy the conclusion of Corollary 2.8, a contradiction.

If (e) or (f) hold, it is straightforward to verify that inequality (2) holds, acontradiction as before.

Case 6. G0 � F4�q�. As before, the Main Theorem of [34] (Tables 5.1, 5.2)together with Proposition 2.5 forces one of the following statements to hold:

(a) q = 2 and M0 � �SU3�2� ∗ SU3�2�� · �3 · �2; or(b) M0 � ��q−1�

4 ·W�F4�, q = 2a �= 2� 4, W�F4� the Weyl group of type F4, and Ginduces a graph automorphism on G0; or

(c) M0 � ��q+1�4 ·W�F4�, q = 2a > 2, W�F4� the Weyl group of type F4, and G

induces a graph automorphism on G0; or(d) M0 � ��q2+q+1�

2 · ��3 × SL2�3��, q = 2a and G induces a graph automorphismon G0; or

(e) M0 � ��q2−q+1�2 · ��3 × SL2�3��, q = 2a > 2 and G induces a graph automor-

phism on G0; or(f) M0 � ��q2+1�

2 · ��4 ∗GL2�3��, q = 2a > 2 and G induces a graph automorphismon G0; or

(g) M0 � �q4−q2+1 · �12, q = 2a > 2 and G induces a graph automorphism on G0.

If case (a) holds, then F�M0� = O3�M0�, and so there are 2-power elements inM0 − F�M� centralizing 3-power elements of F�M�, contrary to Corollary 2.8.

As in the arguments for (b)–(e) of Case 4 above, if (b)–(g) hold, then asG induces a graph automorphism on G0, q = 22m+1, and in each case M0 is the

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normalizer in G0 of a maximal torus T0 with F�M0� = T0. Moreover, 2 � F�M� F�M0�� and so O2�M� �= �1�.

Observe that G0/T0 is isomorphic to one of W�F4� (if (b) or (c) hold), �3 ×SL2�3� (if (d) or (e) hold), �4 ∗GL2�3�� (if (f) holds), or �12 (if (g) holds). One canverify that in each case, G0/T0 contains an element of order 6, and so M0 − F�M�contains an element whose order is divisible by 6. In particular, Corollary 2.8 forcesM to be a Frobenius group. But O2�M� �= �1�, and M0 − F�M� contains elements ofeven order, a contradiction.

Observe that if G0 is isomorphic to one of 2E6�q�, E6�q�, or E7�q�, thenInndiag�G0� can properly contain G0. In this case, a bit more care is required inthe use of Tables 5.1 and 5.2 of [34]. Recall G0 � G ≤ Aut�G0�, M a subgroupof G of maximal rank which is maximal in G. Set G1 = Inndiag�G0�, and let G0

denote a simple algebraic group over an algebraically closed field � of positivecharacteristic p, and � an endomorphism of G0 such that �G0�� = G1. The maintheorem of [34] describes subgroup M as the normalizer in G of subgroup D�, andTables 5.1 and 5.2 of [34] give the structure of NG1

�D��.For each of cases 7, 8, and 9, we set M1 = NG1

�D��. For each of these cases itis easy to check that the following inequality holds:

G M�+ 12�M� >

�G�2�M�2 ≥ �G0�

2�M1�2Aut�G0� G1�(3)

In particular, one need only verify that

�G0�2�M1�2Aut�G0� G1�

≥ 1 (4)

to obtain a contradiction to Proposition 2.4(a) whenever the inequality (4) holds.

Case 7. G0 � 2E6�q�. Then the Main Theorem of [34] (Tables 5.1, 5.2)together with Proposition 2.5 forces one of the following statements to hold (recallthe definition of M1 in the paragraph above):

(a) q = 2 and M1 � �3 · �PSU3�2��3 · T · Sym�3�, where �T � = 9; or

(b) M1 � ��q2−q+1�3 · S · SL2�3�, S an extraspecial group of order 27 and q > 2.

In either case, it is a simple matter to verify that the inequality (4) holds, contraryto Proposition 2.4(a).

Case 8. G0 � E6�q�. As usual, the Main Theorem of [34] (Tables 5.1, 5.2) andProposition 2.5 forces

M1 � ��q2+q+1�3 · S · SL2�3�

where S is an extraspecial group of order 27. But it is a simple matter to show thatinequality (4) holds, contrary to Proposition 2.4(a).

Case 9. G0 � E7�q�. By Tables 5.1 and 5.2 of [34], M1 is always isnonsolvable, whence M1 ∩G0 = M0 is nonsolvable, contrary to hypothesis.

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Case 10. G0 � E8�q�. The Main Theorem of [34] (Tables 5.1, 5.2) andProposition 2.5 force one of the following statements to hold:

(a) q = 2 and M0 � T1 · �PSU3�2��4 · T2 ·GL2�3� where Ti = 9 for i = 1� 2; or

(b) M0 � �q8+q7−q5−q4−q3+q+1 · �30; or(c) M0 � ��q4−q2+1�

2 · ��12 ∗GL2�3��; or(d) M0 � �q8−q7+q5−q4+q3−q+1 · �30.

Since Inndiag�E8�q�� = E8�q�, we need show that inequality (2) holds for eachof the above cases. But this is trivial. This final case completes the proof. �

Proof of Proposition 4.13. Suppose, by way of contradiction, that G is an almostsimple, primitive sharp permutation group of type ��0� k�� n� on X, whose socle is asimple exceptional group of Lie type. Then Gx is a maximal subgroup of G, and byProposition 2.5, Gx is either solvable or is a nonsolvable Frobenius group. Thus byTheorem 2 of [35], together with Table 1 from [15], Gx must be the centralizer of agraph, field, or graph-field automorphism of the socle of G, a maximal parabolic, ora subgroup of maximal rank. However, these possibilities are ruled out, respectively,by Lemmas 4.14, 4.15, 4.16. This completes the proof of Proposition 4.13.

5. SPORADIC GROUPS

The goal of this section is a proof of the following proposition.

Proposition 5.1. Let G be an almost simple, primitive sharp permutation group oftype ��0� k�� n� on X. Then Soc�G� is not a sporadic simple group.

Proof. By Lemma 2.3 we may assume that k ≥ 3. As before, we set G0 = Soc�G�.For this argument, we will refer to a class of maximal subgroups as compatible

if a representative subgroup satisfies the conclusion of Proposition 2.5. That is, if itis either solvable, or is a nonsolvable Frobenius group.

The Mathieu group M11: From p. 18 of [16] and Proposition 2.5, we have G = G0

and either M � M9 �2 � �23 SD16 or M � M8 Sym�3� � GL2�3�. In either case,

F�M� is a p-group but M fails to satisfy Corollary 2.7, a contradiction.

The Mathieu group M12: From p. 31 of [16], we have G G0� ≤ 2. If G = G0, fromp. 33 of [16] there are 5 compatible classes with representatives �2

3 GL2�3� (twoclasses), 21+4

+ Sym�3�, 42 D12, and Alt�4�× Sym�3�. The smallest three groupsfail Proposition 2.4(a). For the other two, we have F�M� = O3�M�. But this groupevidently fails Corollary 2.7, a contradiction. On the other hand, if G = Aut�G0�,then there are 4 compatible classes with representatives 21+4

+ �Sym�3���2�, 42 �D12��2�, 3

1+2+ D8, and Sym�4�× Sym�3�. All such groups fail Proposition 2.4(a).

The Janko group J1: From p. 36 of [16], we have G = G0. In this there are 5compatible classes. Four of these have representatives that are Frobenius groups oforder 168, 114, 110, and 42, and the other has shape D6 ×D10. In any case, all ofthese groups fail Proposition 2.4(a).

The Mathieu group M22: From p. 39 of [16], we have G G0� ≤ 2. In addition,whether G is simple or not, there are no compatible classes.

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The Janko group J2: From p. 42 of [16], we have G G0� ≤ 2. If G is simple, there aretwo compatible classes with representative shape 22+4 ��3 × Sym�3�� and �2

5: D12.The former fails Proposition 2.4(e), and the latter Proposition 2.4(a). If G is notsimple, there are two compatible classes with representative shapes

22+4 ��3 × Sym�3����2 and �25 �Z4 × Sym�3���

The former fails Proposition 2.4(e), and the latter Proposition 2.4(a).

The Mathieu group M23: From p. 71 of [16], we have G = G0. In this case, G = G0

has one compatible class of maximal subgroups with representative shape �23 �11.These groups fail Proposition 2.4(a).

The Higman–Sims group HS: From p. 80 of [16], we have G G0� ≤ 2. In thesimple case, there are no compatible classes. If G is not simple, there is onecompatible class with representative shape 51+2

+ 25�. However, all such groups failProposition 2.4(e).

The Janko group J3: From p. 82 of [16], we have G G0� ≤ 2. If G is simple, thereare two compatible classes with representative shapes 32+1+2 8 and 22+4 ��3 ×Sym�3�), both of which fail Proposition 2.4(a). If G is not simple, there are threecompatible classes with representative shapes 32+1+2 8�2, 22+4 �Sym�3�× Sym�3�,and �19 �18. In all cases, the groups fail Proposition 2.4(a).

The Mathieu group M24: From p. 94 of [16], we have G = G0. From p. 96 of [16],there are no compatible classes.

The McLaughln group McL: From p. 100 of [16], we have G G0� ≤ 2. If G issimple, then the one compatible class with representative shape 51+2

+ 3 8 andwhich fails Proposition 2.4(a). If G is not simple, there are two compatible classeswith representative shapes 51+2

+ 3 8�2 and 22+4 �Sym�3�× Sym�3��, both of whichfail Proposition 2.4(a).

The Held group He: From p. 104 of [16], we have G G0� ≤ 2. If G is simple thereare 2 compatible classes with representative shapes 71+2

+ ��3 × Sym�3�� and 52 4�Alt�4�, both of which fail Proposition 2.4(a). If G is not simple, there are threecompatible classes with representative shapes 24+4 �Sym�3�× Sym�3����2, 71+2

+ ��6 × Sym�3��, and 52 4�Sym�4�, each of which fail Proposition 2.4(a).

The Rudvalis group Ru: From p. 126 of [16], we have G = G0. In this case,there is compatible class with representative shape 51+2

+ 25� and which failsProposition 2.4(a). Note that there is a maximal subgroup with shape �2

5 4Sym�5� � �2

5 GL2�5�, with F�M� = O5�M�. This group fails Corollary 2.7.

The Suzuki group Suz: From p. 131 of [16], we have G G0� ≤ 2. If G is simple,there is one compatible class with representative shape 32+4 2��Alt�4�× 22��2 andwhich fails Proposition 2.4(a). If G is not simple, there is one compatible class withrepresentative shape 32+4 2��Sym�4�×D8� and which also fails Proposition 2.4(a).

The O’Nan group O′N : From p. 132 of [16], we have G G0� ≤ 2. If G is simple,there is one compatible class with representative shape 34 21+4

− �D10. If G is notsimple, there are two compatible classes with representative shapes 34 21+4

− �D10�2and 71+2

+ ��3 ×D16�. All of these groups fail Proposition 2.4(a).

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The Conway group Co3: From p. 134 of [16], we have G = G0. In this case, thereis one compatible class with representative shape 22�27�32��Sym�3�, and this failsProposition 2.4(a).

The Conway Co2: From p. 154 of [16], we have G = G0. In this case, thereis one compatible class with representative shape 51+2

+ 4�Sym�4� which failsProposition 2.4(a).

The Fischer group Fi22: From p. 162 of [16], we have G G0� ≤ 2. If G is simple,there is one compatible class with representative shape 31+6

+ 23+4 32 2 whichfails Proposition 2.4(a). If G is not simple, there is one compatible class withrepresentative shape 31+6

+ 23+4 32 2�2 which fails Proposition 2.4(a).

The Harada–Norton group HN : From p. 166 of [16] we have G G0� ≤ 2. If G issimple there are two compatible classes with representative shapes 51+4

+ 21+4− �5�4 and

34 2��Alt�4�×Alt�4���4, both of which fail Proposition 2.4(a). If G is not simple,there are again two compatible classes with representative shapes 51+4

+ 21+4− �5�4�2

and 34 2��Sym�4�× Sym�4���2, both of which fail Proposition 2.4(a).

The Lyons group Ly: From p. 174 of [16], we have G = G0. In this case, there are twocompatible classes with representative shapes �67 �22 and �37 �18, all of whichfail Proposition 2.4(a). Note that there is a third class that, but this group cannotbe a nonsolvable Frobenius group (since O3�M� �= �1�).

The Thompson group Th: From p. 177 of [16], we have G = G0. In thiscase, there are 5 compatible classes with representative shapes 39��GL2�3�,32�37��GL2�3�, 51+2

+ 4��Sym�4�, 72 ��3 ×GL2�3��, and �31 �15. All of thesegroups fail Proposition 2.4(a).

The Fischer group Fi23: From p. 177 of [16], we have G = G0. In this case, there isone compatible class with representative shape 31+8

+ �21+6− �31+2

+ �GL2�3�. This subgroupfails Proposition 2.4(e).

The Conway group Co1: From p. 183 of [16], we have G = G0. According to [45],there are 2 compatible classes with representative shapes

33+4 2��Sym�4�× Sym�4�� and �27 ��3 ×GL2�3���

both of which fail Proposition 2.4(a). Note that there is one more of shape 52 4Alt�5�, which cannot be a nonsolvable Frobenius group.

The Janko group J4: From p. 190 of [16], we have G = G0. In this case, there arefour compatible classes with representative shapes 111+2

+ �5× 2Sym�4��, �29 �28,�43 �14, and �37 �12. All of these groups fail Proposition 2.4(a).

The Fischer group Fi′24: From p. 207 of [16], we have G G0� ≤ 2. When G issimple, there is one compatible class with representative shape �29 �14 which failsProposition 2.4(a). When G is not simple there are two compatible classes withrepresentative shapes 71+2

+ ��6 × Sym�3���2, both of which fall Proposition 2.4(a).

The Baby Monster �: From p. 217 of [16], we have G = G0. From [44] and [45],there are 30 classes of maximal subgroups of B. Two of these classes are compatible

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with representative shapes 311��Sym�4�×GL2�3�� or �47 �23, both of which failProposition 2.4(a).

The Monster : From p. 220 of [16], we have G = G0. According to [26], if Y isa maximal subgroup of the Monster that is not listed in [45] or not isomorphic toPSL2�41�, then Soc�Y� involves PSL2�13�, Sz�8�, PSU3�4�, or PSU3�8�. In particular,such maximal subgroups (if they exist), do not satisfy Proposition 2.5. In particular,the list given in [45] will suffice for our needs. Of these, there are two compatibleclasses with representatives �41 �40 and 131+2

+ ��3 × 4�Sym�4��. But both of thesefail Proposition 2.4(a). This completes the proof. �

By Propositions 3.1, 4.13, 4.2, and 5.1, the proof of Theorem 1.1 is nowcomplete.

ACKNOWLEDGMENTS

The author wishes to express his thanks to Professors Ronald Solomon and J.Matthew Douglass for valuable conversations relating to Proposition 4.1. Finally,the author would like thank the referee for numerous valuable suggestions that havegreatly improved the presentation of these results.

REFERENCES

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