Finite subgroups of simple algebraic groups with irreducible centralizers

Martin W. Liebeck; Adam R. Thomas

doi:10.1515/jgth-2017-0002

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Finite subgroups of simple algebraic groups with irreducible centralizers

Published/Copyright: February 14, 2017

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From the journal Journal of Group Theory Volume 20 Issue 5

Abstract

We determine all finite subgroups of simple algebraic groups that have irreducible centralizers – that is, centralizers whose connected component does not lie in a parabolic subgroup.

1 Introduction

Let G be a simple algebraic group over an algebraically closed field. Following Serre [15], a subgroup of G is said to be G-irreducible (or just irreducible if the context is clear) if it is not contained in a proper parabolic subgroup of G. Such subgroups necessarily have finite centralizer in G (see [13, 2.1]). In this paper we address the question of which finite subgroups can arise as such a centralizer. It turns out (see Corollary 4 below) that they form a very restricted collection of soluble groups, together with the alternating and symmetric groups Alt5 and Sym5.

The question is rather straightforward in the case where G is a classical group (see Proposition 3 below). Our main result covers the case where G is of exceptional type.

Theorem 1.

Let G be a simple adjoint algebraic group of exceptional type in characteristic p≥0, and suppose F is a finite subgroup of G such that CG⁢(F)0 is G-irreducible. Then |F| is not divisible by p, and F, CG⁢(F)0 are as in Tables 7–12 in Section 5 at the end of the paper (one G-class of subgroups for each line of the tables).

Remark.

(1) The notation for the subgroups F and CG⁢(F)0 is described at the end of this section; the notation for elements of F is defined in Proposition 2.2.

(2) The theorem covers adjoint types of simple algebraic groups. For other types, the possible finite subgroups F are just preimages of those in the conclusion.

(3) We also cover the case where G=Aut⁡E6=E6⁢.2 (see Table 10 and Section 3.3).

(4) A complete determination of all G-irreducible connected subgroups is carried out in [16].

Every finite subgroup F in Theorem 1 is contained in a maximal such finite subgroup. The list of maximal finite subgroups with irreducible centralizers is recorded in the next result.

Corollary 2.

Let G be a simple adjoint algebraic group of exceptional type, and suppose F is a finite subgroup of G which is maximal subject to the condition that CG⁢(F)0 is G-irreducible. Then F, CG⁢(F)0 are as in Table 1.

Table 1

Maximal finite subgroups F with irreducible centralizer.

For the classical groups we prove the following.

Proposition 3.

Let G be a classical simple algebraic group in characteristic p≥0 with natural module V, and suppose F is a finite subgroup of G such that CG⁢(F)0 is G-irreducible. Then p≠2 and F is an elementary abelian 2-group. Moreover, G≠SLn and the following hold.

If G=Sp2⁢n, then
CG⁢(F)0=∏iSp2⁢ni=∏iSp⁢(Wi),
where ∑ni=n and Wi are the distinct weight spaces of F on V.
If G=SOn, then
CG⁢(F)0=∏iSOni=∏iSO⁢(Wi),
where ni≥3 for all i, ∑ni=n or n-1, and Wi are weight spaces of F.

Remark.

In Section 4 we prove a version of this result covering finite subgroups of Aut⁡G for G of classical type.

Corollary 4.

Let G be a simple adjoint algebraic group, and let D be a proper connected G-irreducible subgroup. Then the finite group CG⁢(D) is either elementary abelian or isomorphic to a subgroup of one of the following groups:

2-1+4,G12,Sym4×2,SL2⁢(3), 32.Dih8,Sym5.

Notation.

Throughout the paper we use the following notation for various finite groups:

Zn, or just ⁢ncyclic group of order ⁢n,ps⁢(p⁢ prime)elementary abelian group of order ⁢ps,Altn,Symnalternating and symmetric groups,Dih2⁢ndihedral group of order ⁢2⁢n,4∘Dih8order 16 central product with centre ⁢Z4,2-1+4extra-special group of order 32 of minus type,Frob20Frobenius group of order ⁢20,G12dicyclic group ⁢〈x,y|x6=1,xy=x-1,y2=x3〉⁢ of order ⁢12.

In the tables in Section 5, and also in the text, we shall sometimes use A¯1 to denote a subgroup A1 of a simple algebraic group G that is generated by long root subgroups; we use the notation A¯2 similarly. Also, Br denotes a natural subgroup of type SO2⁢r+1 in a group of type Dn.

We use the following notation when describing modules for a semisimple algebraic group G. We let L⁢(G) denote the Lie algebra of G. If λ is a dominant weight, then VG⁢(λ) (or simply λ) denotes the rational irreducible G-module with high weight λ. When G is simple, the fundamental dominant weights λi are ordered with respect to the labelling of the Dynkin diagrams as in [3, p. 250]. If V1,…,Vk are G-modules, then V1/…/Vk denotes a module having the same composition factors as V1+⋯+Vk. Finally, when H is a subgroup of G and V is a G-module, we use V↓H for the restriction of V to H.

2 Preliminaries

In this section we collect some preliminary results required in the proof of Theorem 1.

Proposition 2.1.

Let G be a simple adjoint algebraic group in characteristic p. Suppose t∈Aut⁡G∖G is such that t has prime order and CG⁢(t)0 is G-irreducible. Then t, CG⁢(t)0 are given in Table 2.

If G=D4 and t has order 3 with CG⁢(t)=A2, there is an involutory graph automorphism of G that inverts t and acts as a graph automorphism on CG⁢(t).

Table 2

Centralizers of graph automorphisms in simple algebraic groups.

Proof.

The first part follows from [8, Tables 4.3.3, 4.7.1] for p≠2, from [1, Section 8] for G=Dn, p=2, and from [1, 19.9] for G=An,E6, p=2. The last part follows from [9, 2.3.3]. ∎

Proposition 2.2.

Let G be a simple adjoint algebraic group of exceptional type in characteristic p, and let x∈G be a nonidentity element such that CG⁢(x)0 is G-irreducible. Then x and CG⁢(x) are as in Table 3; we label x according to its order, which is not divisible by p.

Proof.

First observe that if p divides the order of x, then CG⁢(x)0 is G-reducible by [2, 2.5]. Hence x is a semisimple element and CG⁢(x)0 is a semisimple subgroup of maximal rank. It follows from [14, 4.5] that this implies the order of x is equal to one of the coefficients in the expression for the highest root in the root system of G; these are at most 6 for G=E8, and at most 4 for the other types. The classes and centralizers of elements of these orders can be found in [5, 3.1, 4.1] and [7, 3.1]. ∎

Table 3

Elements of exceptional groups with irreducible centralizers.

Table 4

Elements with irreducible centralizers in G=Aut⁡E6.

We shall need a similar result for the group Aut⁡E6=E6⁢.2.

Proposition 2.3.

Let G=Aut⁡E6 and let x∈G∖G′ be such that CG⁢(x)0 is G′-irreducible. Then x and CG′⁢(x) are as in Table 4.

Proof.

If x is an involution, then the result follows from Proposition 2.1. Suppose now that 1≠x2∈G′. Then x2 has order 2 or 3 and CG′⁢(x2)=A1⁢A5 or A23⁢.3, respectively, by Proposition 2.2. In the former case x acts as a graph automorphism on the A5 factor and CG′⁢(x)=A1⁢C3 or A1⁢A3 by [8, Table 4.3.1]. Here A1⁢C3 is not possible since this lies in a subgroup F4 and hence centralizes an involution in G∖G′.

In the case where CG′⁢(x2)=A23⁢.3, the element x2 is of order 3 in CG′⁢(x3). By Proposition 2.2, the latter group must be F4 and CF4⁢(x2)=A2⁢A2. ∎

We also require information on normalizers of certain maximal rank subgroups. The following proposition can be deduced from [4, Tables 7–11] and direct calculation in the Weyl groups of exceptional algebraic groups; many of the results can be found in [12, Chapter 11].

Proposition 2.4.

Let G be a simple algebraic group of exceptional type. Then Table 5 gives the groups NG⁢(M)/M (or NAut⁡G⁢(M)/M for G of type E6) for the given maximal rank subgroups M of G.

Table 5

Normalizers of maximal rank subgroups of G.

Next we have a result about the Spin group Spinn in characteristic p≠2. Recall that the centre of Spinn is 22 if n is divisible by 4 and is Z2 if n is odd. In the former case the quotients of Spinn by the three central subgroups of order two are SOn and the two half-spin groups HSpinn.

Proposition 2.5.

Let G be HSpinn (where 4|n) or Spinn (n odd), in characteristic p≠2. Let 〈t〉=Z⁢(G) (so that G/〈t〉=PSOn) and suppose F is a finite 2-subgroup of G containing t such that CG⁢(F)0 is G-irreducible. Then the preimage of F/〈t〉 in SOn is elementary abelian. Moreover, an element e∈F has order 2 if and only if its preimage in SOn has -1-eigenspace of dimension divisible by 4.

Proof.

If the preimage contains an element e of order greater than 2, then CSOn⁢(e) has a nontrivial normal torus, and hence CG⁢(F)0 cannot be irreducible. The assertion in the last sentence is well known. ∎

In the following statement, by a pure subgroup of G we mean a subgroup all of whose nonidentity elements are G-conjugate.

Proposition 2.6.

Let G=E8 in characteristic p.

If p≠2, then G has two conjugacy classes of subgroups E≅22 such that CG⁢(E)0 is G-irreducible, and one class of pure subgroups E≅23; these are as follows:
E𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠CG⁢(E)0222⁢B3D422⁢A2,2⁢BA12⁢D6232⁢B7A18
Further, G has no pure subgroup 24.
If p≠3, then G has one class of subgroups E≅32 such that CG⁢(E)0 is G-irreducible. For this class, CG⁢(E)=A24.

Proof.

Part (i) follows from [5, 3.7, 3.8]. For part (ii), let E=〈x,y〉<G with E≅32 and CG⁢(E)0 irreducible. Then CG⁢(x)≠A8, so Proposition 2.2 implies that CG⁢(x)=A2⁢E6, and also that CA2⁢E6⁢(y)0=A24, as required. ∎

The next result is taken from [13, Lemma 2.2].

Proposition 2.7.

Suppose G is a classical simple algebraic group in characteristic p≠2, with natural module V. Let X be a semisimple connected subgroup of G. If X is G-irreducible, then either

G=An and X is irreducible on V, or
G=Bn,Cn or Dn and V↓X=V1⊥…⊥Vk with the Vi all nondegenerate, irreducible and inequivalent as X-modules.

3 Proof of Theorem 1

3.1 The case G=E8

We now embark on the proof of Theorem 1 for the case G=E8. Let F be a finite subgroup of G such that CG⁢(F)0 is G-irreducible. Then CG⁢(F)0 is semisimple (see [13, 2.1]). Moreover, CG⁢(E)0 is irreducible for all nontrivial subgroups E of F. Also F is a {2,3,5}-group by Proposition 2.2.

Lemma 3.1.

If F is an elementary abelian 2-group, then F is as in Table 7.

Proof.

Suppose F≅2k. If k≤2, or if k=3 and F is pure, the conclusion follows from Proposition 2.6 (i).

Now assume that k=3 and F is not pure. By considering the 22 subgroups of F, all of which must be as in (i) of Proposition 2.6, we see that one of these, say 〈e1,e2〉, is 2⁢B-pure, so that CG⁢(e1,e2)0=D42. We have CG⁢(e1)/〈e1〉≅PSO16, and consider the preimage of F/〈e1〉 in SO16. This preimage is elementary abelian by Proposition 2.5, so can be diagonalized, and we can take e2=(-18,18). Let e3 be a further element of F that is in class 2⁢A. Then the -1-eigenspace of e3 has dimension 4 or 12, and so the fact that CG⁢(F)0 is G-irreducible means that we can take e3=(-14,14,18), so that CSO16⁢(F)0=SO4⁢SO4⁢SO8, and so CG⁢(E)0=A14⁢D4 as in Table 7.

Next suppose k≥4. Then F is not pure by Proposition 2.6 (i), so F contains a subgroup 〈e1,e2,e3〉≅23 as in the previous paragraph. Arguing as above, we can take a further element e4 of F to be (18,-14,14), so that CG⁢(e1,…,e4)0 equals A18. There is no possible further diagonal involution in F such that CG⁢(F)0 is irreducible, so k=4. ∎

In view of the previous lemma, we assume from now on that F is not an elementary abelian 2-group. Hence if F is a 2-group, it has exponent 4 by Proposition 2.2.

Lemma 3.2.

Suppose F is a 2-group, and has no element in the class 4⁢B. Then one of the following holds:

F≅Z4, generated by an element in class 4⁢A, and CG⁢(F)0=A1⁢A7.
F≅Q8 with elements 2⁢A,4⁢A6, and CG⁢(F)0=A1⁢D4.

In both cases F is as in Table 7.

Proof.

Let e∈F have order 4. By hypothesis, e is in class 4⁢A, so we have CG⁢(e)=A1⁢A7. There is nothing to prove if F≅Z4, so assume |F|>4 and pick f∈∖⁡NF⁢(〈e〉)〈e〉. If f∈A1⁢A7, then CG⁢(e,f) has a normal torus, so ef=e-1 and f induces an involutory graph automorphism on A7 (see Proposition 2.4). Hence CA7⁢(f)0=C4 or D4 by Proposition 2.1. The subgroup C4 lies in a Levi subgroup E6 of CG⁢(A1)=E7 (see the proof of [6, 2.15]), so the irreducibility of CG⁢(F)0 implies that CA7⁢(f)0=D4, hence CG⁢(e,f)0=A1⁢D4. Also we have 〈e,f〉≅Q8, as shown in the proof of [6, 2.15]. Finally, if NF⁢(〈e,f〉)>〈e,f〉, then some element of order 4 in 〈e,f〉 has centralizer in F of order greater than 4, which we have seen to be impossible above. Hence F=〈e,f〉. ∎

Lemma 3.3.

Suppose F is a 2-group and has an element e in the class 4⁢B. If CF⁢(e)≠〈e〉, then F is as in Table 7 (one of the entries 4×2, Dih8×2, 4∘Dih8, Q8×2, 2-1+4).

Proof.

Assume that CF⁢(e)≠〈e〉. Then CF⁢(e) contains a group of order 8 in which e is central, and hence there is an involution e1∈∖⁡CF⁢(e)〈e〉. Diagonalizing the preimage of CF⁢(e2)/〈e2〉 in SO16 (as in the proof of Lemma 3.1), we can take

e=(-16,110),e1=(16,-14,16),

so that 〈e,e1〉≅Z4×Z2 and CG⁢(e,e1)0=A12⁢A32, as in the 4×2 entry in Table 7.

Write E0=〈e,e1〉. If there exists f∈∖⁡(F∩A12⁢A32)E0, then CG⁢(e,e1,f) has a nontrivial normal torus, which is a contradiction. Hence F∩CG⁢(E0)0=E0.

Suppose CF⁢(e)≠E0. If there is no involution in ∖⁡CF⁢(e)E0, then CF⁢(e) contains a subgroup Z4×Z4, which is impossible. So let e2 be an involution in ∖⁡CF⁢(e)E0. As e2 does not centralize e1, we can take

e2=(16,-1,13,-13,13).

Then we have CG⁢(e,e1,e2)0=A3⁢B13 (where each B1 corresponds to a natural subgroup SO3 in SO16), and 〈e,e1,e2〉=〈e〉∘〈e1,e2〉≅4∘Dih8, as in Table 7. If F≠E1:=〈e,e1,e2〉, then there is an element e3∈∖⁡NF⁢(E1)E1, and adjusting by an element of E1 we can take

e3=(-13,13,-1,19).

Then CG⁢(E1,e3)0=B15 and 〈E1,e3〉≅2-1+4, as in Table 7. Finally, there are no possible further elements of F such that CG⁢(F)0 is irreducible.

Now suppose CF⁢(e)=E0 and F≠E0. Pick f∈∖⁡NF⁢(E0)E0. Then f centralizes e2, so we can diagonalize in the usual way; adjusting by an element of E0 and using the fact that CG⁢(E0,f) has no nontrivial normal torus, we can take f∈{e4,e5,e6,e7}, where

e4=(-13,13,14,-1,15),e5=(-13,13,14,-13,13),

e6=(-13,13,-1,13,16),e7=(-1,15,14,-13,13).

Let E2=〈E0,f〉.

If f=e4, then CG⁢(E2)0=B12⁢A¯12⁢B2 and E2=〈e,e4〉×〈e1〉≅Dih8×Z2, as in Table 7. There are no possible further elements of F in this case.

If f=e5, then CG⁢(E2)0=A¯12⁢B14 and E2=〈e,e5〉×〈e1〉≅Q8×Z2, as in Table 7. Any further element of F would centralize e2, and hence would violate the fact that CF⁢(e)=E0.

Finally, if f=e6 or e7, then E2 is D8-conjugate to 〈e,e1,e2〉 or 〈e,e1,e4〉, cases considered previously. ∎

Lemma 3.4.

If F is a 2-group, then it is as in Table 7.

Proof.

Suppose F is a 2-group. In view of the previous two lemmas, we can assume that F contains an element e in the class 4⁢B, and that CF⁢(e)=〈e〉 – and indeed that F has no subgroup Z4×Z2. We can also assume that F≠〈e〉. Hence there exists f∈F such that ef=e-1. As usual we can diagonalize 〈e,f〉, and hence take e=(-16,110) and f∈{f1,f2,f3,f4,f5}, where

f1=(-13,13,-1,19),f2=(-13,13,-13,17),

f3=(-13,13,-15,15),f4=(-1,15,-13,17),

f5=(-1,15,-15,15).

Moreover, the fact that F has no subgroup Z4×Z2 implies that F=〈e,f〉.

It is easily seen that the possibilities for F and CG⁢(F)0 are as follows:

FCG⁢(F)0〈e,f1〉≅Dih8B12⁢B4〈e,f2〉≅Q8B13⁢B3〈e,f3〉≅Dih8B12⁢B22〈e,f4〉≅Dih8B1⁢B2⁢B3〈e,f5〉≅Q8B23

All these possibilities are in Table 7. ∎

Lemma 3.5.

If F is a 3-group, then F=3 or 32 is as in Table 7.

Proof.

Suppose F is a 3-group. It has exponent 3, by Proposition 2.2. If F=3 or 32, it is as in Table 7 by Propositions 2.2 and 2.6 (ii), so assume |F|>9.

If F has an element e with CG⁢(e)=A8, there is an element f∈CF⁢(e)∖〈e〉, and CA8⁢(f) is reducible in A8, a contradiction. Hence all nonidentity elements of F have centralizer A2⁢E6 (see Proposition 2.2). In particular, they have trace 5 on the adjoint module L⁢(G) (see [5, 3.1]).

Let V be a normal subgroup of F with V≅32. Then CG⁢(V)=A24 by Proposition 2.6 (ii). If f∈F∖V, then f acts as a 3-cycle on the four A2 factors, so CG⁢(V,f)0=A¯2⁢A2. On the other hand, since every nonidentity element has trace 5 on L⁢(G), we have

dim⁡CL⁢(G)⁢(V,f)=127⁢(248+26⋅5)=14.

This is a contradiction, showing that |F|>9 is impossible. ∎

From now on, we assume that F is not a 2-group or a 3-group. Let J=Fit⁢(F), the Fitting subgroup of F.

Lemma 3.6.

Suppose J is a nontrivial 2-group. Then F is as in Table 7.

Proof.

By Lemma 3.4, J and CG⁢(J)0 are as in Table 7. Also CF⁢(J)≤J, so F contains an element x of order r=3 or 5 acting nontrivially on J and as a graph automorphism of CG⁢(J)0. By inspection of Table 7, the possibilities for J with these properties are as follows:

JCG⁢(J)0r22D42323A14⁢D43A183,524A183,5Q8A1⁢D43B233B13⁢B334∘Dih8A3⁢B133Q8×2A12⁢B1432-1+4B153,5

For the last five cases, where CG⁢(J)0▷B1r (or B2r), CG⁢(J,x)0 has a factor B1 (or B2) which is a diagonal subgroup of this, and so CG⁢(J,x)0 is a reducible subgroup of D8 in these cases, by Proposition 2.7. Also if CG⁢(J)0=A18, then NG⁢(A18)/A18≅AGL3⁢(2) by Proposition 2.4, so r=3 and x acts as a product of two 3-cycles on the eight A1 factors. We claim that again CG⁢(J,x)0 is reducible. To see this, regard A18 as a subgroup of D8 corresponding to SO44 in SO16. Observe that CJ⁢(x)0=A¯12⁢A12, where each of the last two A1 factors is diagonal in A¯13. There are two possible actions of the subgroup A12<A¯16<D6 on the 12-dimensional natural module, namely (1,1)3 or (2,0)+(1,1)+(0,2)+(0,0)2. In both cases the subgroup A12 is D6-reducible by Proposition 2.7 and hence CJ⁢(x)0 is D8-reducible.

This leaves the following possibilities remaining, all with r=3:

JCG⁢(J)022D4223A14⁢D4Q8A1⁢D4

Suppose J=22, CG⁢(J)0=D42. Then x induces a triality automorphism on both D4 factors (see Proposition 2.4). So by Proposition 2.1, CG⁢(J,x)0=G2⁢G2, A2⁢A2 or A2⁢G2. In the first case G2⁢G2<D7<D8, so is reducible. Hence if F=〈J,x〉, we have the possibilities

F=〈J,x〉≅Alt4,CG⁢(F)0=A2⁢A2⁢ or ⁢A2⁢G2,

both in Table 7. Now assume F≠〈J,x〉. Then F=〈J,x,t〉≅Sym4, where t is an involution inverting x (since Fit⁢(F)≅22). If CG⁢(J,x)0=A2⁢A2, then t acts as a graph automorphism on each A2 factor (see Proposition 2.1), and so CG⁢(F)0=A1⁢A1; and if CG⁢(J,x)0=A2⁢G2, then t acts as a graph automorphism on the A2 factor and centralizes the G2 factor, so CG⁢(F)0=A1⁢G2. Hence we have the possibilities

F=〈J,x,t〉≅Sym4,CG⁢(F)0=A1⁢A1⁢ or ⁢A1⁢G2,

both in Table 7.

Next suppose J=23, CG⁢(J)0=A14⁢D4. Then x acts as a 3-cycle on the A1 factors and as a triality on D4, so CG⁢(J,x)0=A¯1⁢A1⁢G2 or A¯1⁢A1⁢A2 (where A¯1 denotes a fundamental SL2 generated by a root group and its opposite). The first subgroup is reducible as it is contained in a subgroup D7 of D8. So if F=〈J,x〉, we have

F=〈J,x〉≅2×Alt4,CG⁢(F)0=A¯1⁢A1⁢A2,

as in Table 7. Now assume F≠〈J,x〉, and let 〈v〉=Z⁢(〈J,x〉). As F/J is isomorphic to a subgroup of GL3⁢(2) with no nontrivial normal 2-subgroup, we have F/J≅Dih6 and F=〈J,x,t〉 where xt=x-1 and t2=1 or v. Such an element t centralizes both A1 factors of CG⁢(J,x)0, and induces a graph automorphism on the A2 factor, so CG⁢(F)0=A¯1⁢A1⁢A1. This is contained in the above centralizer G2⁢A2 of an Alt4 subgroup, and A¯1⁢A1 centralizes an involution in G2. Hence in fact t2=1 and we have

F=〈J,x,t〉≅2×Sym4,CG⁢(F)0=A¯1⁢A1⁢A1,

as in Table 7.

Finally, suppose J=Q8, CG⁢(J)0=A1⁢D4<A1⁢A7. Then x induces triality on the D4 factor (see [6, 2.15]), so CG⁢(J,x)0=A1⁢G2 or A1⁢A2. The first subgroup is reducible in A1⁢A7, so if F=〈J,x〉, we have

F=〈J,x〉=Q8⁢.3≅SL2⁢(3),CG⁢(F)0=A¯1⁢A2,

as in Table 7. If F≠〈J,x〉, then F has an element t inducing a graph automorphism on the A2 factor, so CG⁢(J,x,t)0=A1⁢A1, which is reducible in A1⁢A7. This completes the proof. ∎

Lemma 3.7.

If |J|3=3, then F is as in Table 7.

Proof.

Assume |J|3=3, and let x∈J be of order 3.

Suppose first that |F|3=3 also. As F has no element of order 15, we have |F|5=1, so F/〈x〉 is a 2-group. The case where |F|=3 is in Table 7, so assume |F|>3.

Suppose CG⁢(x)=A2⁢E6. If t is an involution in CF⁢(x), then CG⁢(x,t) is equal to A2⁢A1⁢A5; moreover CF⁢(x) has no element of order 4 (as F has no element of order 12), and no subgroup V≅22 (as CG⁢(x,V) would have a normal torus). Hence CF⁢(x) has order 3 or 6, and so |F| is 6 or 12.

If |F|=6, then either F≅Z6, CG⁢(F)=A1⁢A2⁢A5, or F=〈x,t〉≅Dih6 with t inducing graph automorphisms on both factors of CG⁢(x)=A2⁢E6 (see Proposition 2.4), in which case CG⁢(F)0=A1⁢F4 or A1⁢C4. All these possibilities are in Table 7.

If |F|=12, then F=〈y,u〉, where y has order 6, yu=y-1 and u2=1 or y3. Then u induces a graph automorphism on the A2, A5 factors of CG⁢(y)=A1⁢A2⁢A5, so CG⁢(F)0=A¯1⁢A1⁢C3 or A¯1⁢A1⁢A3. The subgroup A¯1⁢A1⁢C3 is contained in A1⁢F4 and in A1⁢C4, while the subgroup A¯1⁢A1⁢A3 is contained in neither. Hence u is an involution in the first case, and has order 4 in the second. This gives the possibilities

F=〈y,u〉≅Dih12⁢ or ⁢G12,CG⁢(F)0=A¯1⁢A1⁢C3⁢ or ⁢A¯1⁢A1⁢A3⁢ (resp.),

both in Table 7.

Next suppose that CG⁢(x)=A8. Then CF⁢(x)=〈x〉, so F=〈x,t〉≅Dih6, where t induces a graph automorphism on A8, giving CG⁢(F)0=B4. This completes the case where |F|3=3.

Finally, suppose |F|3=32, and let 〈x,y〉 be a Sylow 3-subgroup of F. Again, |F|5=1. If F=〈x,y〉, then CG⁢(F)0=A24 by Proposition 2.6 (ii), as in Table 7. Otherwise, as CF⁢(J)≤J, there must be a subgroup V≅22 of J such that y acts nontrivially on V. But then CG⁢(x,V)0=CA2⁢E6⁢(V)0 is reducible, a contradiction. ∎

Lemma 3.8.

If |J|3=32, then F is as in Table 7.

Proof.

Let V≅32 be a Sylow 3-subgroup of J (also of F, by Lemma 3.5). By Propositions 2.6 (ii) and 2.4, CG⁢(V)=A24 and NG⁢(A24)/A24≅GL2⁢(3). There is no involution in CF⁢(V), so J=Fit⁢(F)=V and F/J is a nontrivial 2-subgroup of GL2⁢(3).

Suppose first that |F/J|=2. There are two classes of involutions in GL2⁢(3), with representatives i=-I and t=(-1001). Then i induces a graph automorphism on each A2 factor of CG⁢(V), so CG⁢(V,i)0=A14; and t fixes two A2 factors, inducing a graph automorphism on one of them, so CG⁢(V,t)0=A1⁢A¯2⁢A2. Both these groups F=32⁢.2 are in Table 7.

If F/J≅22, we can take F=〈V,i,t〉 and so CG⁢(F)0=A12⁢A1, as in Table 7.

Next suppose F/J≅Z4. There is one class of elements of order 4 in GL2⁢(3), with representative u=(01-10); this swaps two pairs of A2 factors, and squares to i. Hence CG⁢(F)0=CG⁢(V,u)0=A12, as in Table 7.

If F/J≅Dih8, we can take F=〈V,u,t〉, and again CG⁢(F)0=A12.

Now suppose F/J≅Q8. Then F/J acts transitively on the four A2 factors, and contains i, so CG⁢(F)0=A1, a diagonal subgroup of A14<A24. We claim that CG⁢(F)0 is reducible. To see this, observe that A14 centralizes the involution i; this involution corresponds to w0, the longest element of the Weyl group of G, and so CG⁢(i)=D8. Now it is easy to check that CG⁢(F)0=A1 is reducible in this D8. Indeed, A14 acts as (1,1,1,1) on the natural module for D8 and hence a diagonal subgroup A1 acts as 4+23+02. Thus F/J≅Q8 is impossible, and we have now covered all possibilities for F/J. ∎

Lemma 3.9.

If |J|5≥5, then F is as in Table 7.

Proof.

Suppose |J|5≥5, and let x∈J have order 5. As CG⁢(x)=A42 by Proposition 2.2, there is no element of order 5 in CF⁢(x)∖〈x〉, and so 〈x〉 is a Sylow 5-subgroup of F.

As F has no element of order 10 or 15, we have CF⁢(x)=〈x〉, and |F|=10 or 20. By Proposition 2.4, NG⁢(A42)/A42=〈t〉≅Z4, where t interchanges the two A4 factors and t2 induces a graph automorphism on both. Hence F is either Dih10 or Frob20, and CG⁢(F)0=B2⁢B2 or B2, respectively, as in Table 7. ∎

Lemmas 3.6–3.9 cover all cases where the Fitting subgroup J is nontrivial.

Lemma 3.10.

Suppose J=Fit⁢(F)=1. Then F=Alt5 or Sym5 is as in Table 7.

Proof.

In this case S:=soc⁢(F) is a direct product of nonabelian simple groups. As 52 does not divide |F|, in fact S is simple. Proposition 1.2 of [10] shows that S≅Alt5 or Alt6.

Suppose S≅Alt5. Then S has subgroups D≅Dih10 and A≅Alt4, and by what we have already proved, these subgroups are in Table 7. Hence the involutions in S are in the class 2⁢B (since those in A are in this class). If the elements of order 3 in S are in class 3⁢A (with centralizer A8), then from [5, 3.1] we see that the traces of the elements in S of orders 2,3,5 on L⁢(G) are -8,-4,-2, respectively, and hence

dim⁡CL⁢(G)⁢(S)=160⁢(248-8⋅15-4⋅20-2⋅24)=0,

which is a contradiction. It follows that the elements of order 3 in S are in the class 3⁢B, with centralizer A2⁢E6 and trace 5, so that

dim⁡CL⁢(G)⁢(S)=160⁢(248-8⋅15+5⋅20-2⋅24)=3.

Since CG⁢(D)0=B2⁢B2<A4⁢A4, it follows that CG⁢(S)0=A1, embedded diagonally and irreducibly in A4⁢A4. Also CG⁢(A1)=Sym5 by [10, 1.5]. Hence we have F=Alt5 or Sym5 and CG⁢(F)0=A1, as in Table 7.

Finally, suppose S≅Alt6 and choose a subgroup T<S with T≅Alt5. By the above, CG⁢(T)0=A1 and so CG⁢(S)0 must also be A1. But as observed before, CG⁢(A1)=Sym5, a contradiction. ∎

We have now established that F and CG⁢(F)0 must be as in Table 7. To complete the proof of Theorem 1, we need to establish that all these examples exist. This is proved in the following lemma.

Lemma 3.11.

Let F and CG⁢(F)0 be as in Table 7. Then CG⁢(F)0 is G-irreducible.

Proof.

Any subgroup containing a G-irreducible subgroup is itself G-irreducible. Thus we need only consider the subgroups CG⁢(F)0 for which F is maximal. These subgroups are given in Table 1; let X be such a subgroup CG⁢(F)0.

Firstly, if X has maximal rank then X is clearly G-irreducible. For the subgroups not of maximal rank we use the fact that a subgroup with no trivial composition factors on L⁢(G) is necessarily G-irreducible (since the Lie algebra of the centre of a Levi subgroup gives a trivial composition factor). It thus remains to show X has no trivial composition factors on L⁢(E8). We find the composition factors of X on L⁢(G) by restriction from a maximal rank overgroup Y, as given in the last column of Table 7. The restrictions L⁢(G)↓Y are given in [12, Lemma 11.2, 11.3] for all of the maximal rank overgroups Y except for A1⁢A7, A14⁢D4 and A24. The latter subgroups are contained in A1⁢E7, D42 and A2⁢E6, respectively, and it is straightforward to compute their composition factors on L⁢(G).

We finish the proof with two examples of how to calculate the composition factors of L⁢(G)↓X from those of a maximal rank overgroup Y. The others all follow similarly and in each case there are no trivial composition factors.

For the first example, let X=B23 so p≠2 and X is contained in the maximal rank overgroup D8. From [12, Lemma 11.2],

L⁢(G)↓D8=V⁢(λ2)+V⁢(λ7),

the sum of the exterior square of the natural module for D8 and a spin module. To find the restriction of the spin module VD8⁢(λ7) to X, we consider the chain of subgroups X<B2⁢D5<B2⁢B5<D8. By [12, Lemma 11.15 (ii)],

VD8⁢(λ7)↓B2⁢B5=01⊗λ5.

Also,

VB5⁢(λ5)↓D5=λ4+λ5

and

VD5⁢(λi)↓B22=01⊗01 for i=4,5.

Therefore,

L⁢(G)↓X=⋀2(10⊗00⊗00+00⊗10⊗00+00⊗00⊗10+0)+(01⊗01⊗01)2

and this has no trivial composition factors.

For the second example, let X=A1⁢D4. Here p=3 and X is contained in a maximal rank subgroup A1⁢A7. Then using the restriction L⁢(G)↓A1⁢E7 given in [12, Lemma 11.2] we find

L⁢(G)↓A1⁢A7=2⊗0+1⊗λ2+1⊗λ6+0⊗(λ1+λ7)+0⊗λ4.

It is sufficient to show there are no trivial composition factors for D4 acting on VA7⁢(λ) for λ=λ1+λ7 and λ4. By weight considerations, the first module restricts to D4 as V⁢(2⁢λ1)+V⁢(λ2) and the second as V⁢(2⁢λ3)+V⁢(2⁢λ4). Hence L⁢(G)↓X has no trivial composition factors. ∎

This completes the proof of Theorem 1 for G=E8.

3.2 The case G=E7

In this section we prove Theorem 1 for G=E7, of adjoint type. Let F be a finite subgroup of G such that CG⁢(F)0 is G-irreducible. As before, CG⁢(F)0 is semisimple and CG⁢(E)0 is G-irreducible for all nontrivial subgroups E of F. Also F is a {2,3}-group by Proposition 2.2.

Lemma 3.12.

If F is an elementary abelian 2-group, then F is as in Table 8.

Proof.

We may suppose that |F|>2. If F has an element e in the class 2⁢B, then any further element f∈∖⁡F〈e〉 must lie in ∖⁡CG⁢(e)CG⁢(e)0=∖⁡A7⁢.2A7, and hence F=〈e,f〉≅22; moreover CG⁢(F)0=D4, as in the proof of Lemma 3.2. Hence F is as in Table 8.

So now suppose that F is 2⁢A-pure. Let 1≠e∈F and e1∈∖⁡F〈e〉. Then CG⁢(e) is equal to A1⁢D6, and diagonalizing in SO12 as in Lemma 3.1, we can take e1=(-14,18). Hence CG⁢(e,e1)0=A13⁢D4. If there is an element e2∈∖⁡E〈e,e1〉, then we can take e2=(14,-14,14), and so CG⁢(e,e1,e2)0=A17. Both these possibilities are in Table 7, and there are no further possible elements in F. ∎

Lemma 3.13.

If F is a 2-group containing an element of order 4, then F is as in Table 8.

Proof.

Let e∈F of order 4. By Proposition 2.2 we have CG⁢(e)0=A1⁢A32. Suppose F≠〈e〉, so there exists f∈F such that ef=e-1. Now CG⁢(e2)=A1⁢D6, and diagonalizing in SO12 as in Lemma 3.1, we may take e=(-16,16) and f∈{f1,f2}, where f1=(-1,15,-13,13), f2=(-13,13,-13,13). If f=f1, then CG⁢(e,f)0=A¯1⁢B12⁢B2 and 〈e,f〉≅Dih8. And if f=f2, then we have CG⁢(e,f)0=A¯1⁢B14 and 〈e,f〉≅Q8. Both possibilities are in Table 8. Finally, there are no possible further elements of F, as can be seen by diagonalizing in the usual way. ∎

In view of the previous two lemmas we assume from this point that F contains an element x of order 3. Let J be the Fitting subgroup of F. Note that F does not contain an element of order 6 by Proposition 2.2. Therefore J is a 2-group or a 3-group.

Lemma 3.14.

If J is a 3-group, then F and CG⁢(F)0 are as given in Table 8.

Proof.

Suppose |J|=3. If |F|=3, then by Proposition 2.2 we have that CG⁢(F) equals A2⁢A5. Otherwise F≅Dih6 and CG⁢(F)0=A1⁢C3 or A1⁢A3.

Finally, |J|>3 is impossible because the centralizer of an element of order 3 in A2⁢A5 is not A2⁢A5-irreducible. ∎

We may now assume that J is a 2-group. By Lemmas 3.12 and 3.13, J is as in Table 8 and the action of x shows that the only possibilities are J≅22, 23 or Q8.

Lemma 3.15.

If J≅22, then F and CG⁢(F)0 are as given in Table 8.

Proof.

Suppose CG⁢(J)0=A13⁢D4. By Proposition 2.4,

NG⁢(A13⁢D4)/A13⁢D4≅Sym3

acting simultaneously on both the A13 and the D4 factors. Therefore CG⁢(J,x)0 equals A1⁢A2 or A1⁢G2 with 〈J,x〉≅Alt4. The subgroup A1⁢G2 is A1⁢D6-reducible by Proposition 2.7, and therefore does not appear in Table 8. If F≠〈J,x〉, then we must have F≅Sym4 with CG⁢(F)0=A1⁢A1.

Now suppose CG⁢(J)0=D4<A7. By [6, Lemma 2.15], we have

NG⁢(D4)/(D4×CG⁢(D4))≅Sym3.

Therefore CG⁢(J,x)0=A2 or G2. The subgroup G2 is A7-reducible and therefore does not appear in Table 8. If F≠〈J,x〉, then F≅Sym4 with CG⁢(F)0=A1. ∎

Lemma 3.16.

There are no possible subgroups F with J≅23 or Q8.

Proof.

Suppose J≅23 so CG⁢(J)0=A17. By Proposition 2.4 we have

NG⁢(A17)/A17≅GL3⁢(2).

The element x∈F therefore acts as a product of two disjoint 3-cycles on the seven A1 factors. But the centralizer CG⁢(J,x)0 is then A1⁢D6-reducible by an argument in the first paragraph of the proof of Lemma 3.6.

Finally, if J≅Q8 and CG⁢(J)0=A1⁢B14, then CG⁢(J,x)0=A1⁢B1⁢B1 which is clearly A1⁢D6-reducible. ∎

The proof of Theorem 1 for G=E7 is now complete, apart from showing that all the subgroups CG⁢(F)0 in Table 8 are G-irreducible. This is proved in similar fashion to Lemma 3.11.

3.3 The case G=Aut⁡E6

Let G=Aut⁡E6=E6⁢.2, and let F be a finite subgroup of G such that CG′⁢(F)0 is G′-irreducible.

Lemma 3.17.

If F has an element x of order 4, then F=〈x〉 and CG′⁢(F)0 is equal to A1⁢A3.

Proof.

By Proposition 2.3, we have CG⁢(x)0=A1⁢A3. By Proposition 2.7, A1⁢A3 contains no proper A1⁢A5-irreducible connected subgroups and therefore F=〈x〉, as claimed. ∎

We now assume that F has no element of order 4.

Lemma 3.18.

If F is an elementary abelian 2-group then it appears in Table 10.

Proof.

If |F|=2, then F and CG⁢(F)0 are as in Table 10 by Proposition 2.2. Now suppose F=〈t,u〉≅22. Then CG⁢(t)0=A1⁢A5, F4 or C4 and therefore CG⁢(F)0=A1⁢A3, A1⁢C3, B4 or C22. The A1⁢A3 case is ruled out by Lemma 3.17. The B4 and C22 subgroups are both contained in D5-parabolic subgroups. Therefore CG⁢(F)0=A1⁢C3. Finally, if F has a further involution v, then CG⁢(F)0 is equal to A12⁢C2, which by Proposition 2.7 is A1⁢A5-reducible. ∎

We now let J be the Fitting subgroup of F. Since F is a {2,3}-group, J is nontrivial.

Lemma 3.19.

If J is not a 2-group or a 3-group, then F is as in Table 10.

Proof.

Under the assumptions of the lemma, J has an element x of order 6. Then CG⁢(x)0=A2⁢A2 by Proposition 2.3. If F≠〈x〉, then there exists an element t∈F∖J inverting x with t2∈J. Since F has no element of order 4, we have t2=1 and 〈J,t〉≅Dih12. The element t induces a graph automorphism on both A2 factors and so CG⁢(J,t)0=A1⁢A1. Since the two factors are non-conjugate, there is no element swapping them and hence F=〈J,t〉. ∎

Lemma 3.20.

If J is a 2-group, then F=J.

Proof.

The possibilities for the 2-group J are in Table 10, from which we see that no element of order 3 can act as a graph automorphism on CG⁢(J)0. ∎

Lemma 3.21.

If J is a 3-group, then F is as in Table 10.

Proof.

Suppose J=〈x〉≅Z3. If F=J, then CG⁢(F)0=A23 by Proposition 2.2. Otherwise, F=〈J,t〉≅Dih6, where t is an involution in NG⁢(A23)/A23≅2×S3 by Proposition 2.4. This gives two possibilities for t. If t is the central involution, then t induces a graph automorphism on each factor A2 and CG⁢(J,t)0=A13. If t is not central, then CG⁢(J,t)0=A1⁢A1.

Now suppose |J|>3 and let x,y∈J with 〈x,y〉≅32. Then CG⁢(x)=A23⁢.3 and y cyclically permutes the three A2 factors, so CG⁢(x,y)0 is a diagonal subgroup A2. However, the elements in class 3⁢A have trace -3 on L⁢(G) and so

dim⁡CL⁢(G)⁢(S)=19⁢(78-8⋅3)=6,

a contradiction. ∎

Finally, we need to prove that all the subgroups CG⁢(F)0 in Table 10 are G′-irreducible.

Lemma 3.22.

Let F and CG⁢(F)0 be as in Table 9. Then CG⁢(F)0 is G′-irreducible.

Proof.

This is proved in a similar fashion to Lemma 3.11 for most of the subgroups. Specifically, all of the subgroups have no trivial composition factors on L⁢(G) except for A1⁢A5, A¯1⁢C3 and A¯1⁢A3 when p=3, all of which have exactly one trivial composition factor. There are no Levi subgroups of G′ containing a subgroup of type A1⁢A5 or A1⁢C3. Hence both are G′-irreducible.

Now consider X=A1⁢A3. Assume X is G′-reducible and choose a minimal parabolic subgroup P containing X. By [11, Theorem 1], X is contained in a Levi subgroup L of P and by minimality X is L-irreducible. Hence L=D5⁢T1 or A1⁢A3⁢T2, where Ti denotes a central torus of rank i. The second possibility is ruled out since X has only one trivial composition factor on L⁢(G). So X is an irreducible subgroup of L′=D5. The A1 factor of X is generated by root groups of D5 and so CD5⁢(A1)0=A1⁢A3. Thus CD5⁢(X)0 contains a subgroup A1, contradicting the L-irreducibility of X. Hence X is G′-irreducible, as required. ∎

This completes the proof of Theorem 1 for G=Aut⁡E6.

3.4 The case G=F4

Let G=F4, and let F be a finite subgroup of G such that CG⁢(F)0 is G-irreducible.

Lemma 3.23.

If F is an elementary abelian 2-group, then F and CG⁢(F)0 are given in Table 11.

Proof.

If F≅Z2, then CG⁢(F)=B4 or A1⁢C3 by Proposition 2.2. Now suppose F=〈t,u〉≅22. Then u∈CG⁢(t)=B4 or A1⁢C3. Therefore CG⁢(F)0=D4 or A12⁢B2. Now suppose |F|>4. A 22 subgroup of F must contain a 2⁢A involution, say t, with centralizer B4. Then B4/〈t〉≅SO9 and the image of F in SO9 is elementary abelian by Proposition 2.5. Since CG⁢(F)0 is G-irreducible, it follows that the image is 〈u,v〉≅22 with u=(-18,1) and v=(-14,15). Therefore CSO9⁢(F)0=SO4⁢SO4, and so CG⁢(F)0=A14. ∎

Lemma 3.24.

If F is a 2-group containing an element x of order 4, then F and CG⁢(F)0 are given in Table 11.

Proof.

By Proposition 2.2, CG⁢(x)=A1⁢A3. Now suppose |F|>4. Since A1⁢A3 is not contained in A1⁢C3, it follows that CG⁢(x2)=B4. Therefore F/〈x2〉 is elementary abelian in SO9. Diagonalizing as before, we may assume x=(13,-16). Since CG⁢(F)0 is G-irreducible, it must be the case that |F/〈x2〉|=4 and a further involution in F is either u1=(-16,13) or u2=(15,-14). In the first case the order of x⁢u1 is 4 and hence F≅Q8 with CG⁢(F)0=B13. In the second case the order of x⁢u2 is 2 and hence F≅Dih8 with CG⁢(F)0=B1⁢B2. ∎

Now let J be the Fitting subgroup of F. Since F has no element of order 6, it follows that J is either a 2-group or a 3-group.

Lemma 3.25.

If J is a 2-group, then F and CG⁢(F)0 are given in Table 11.

Proof.

By the previous two lemmas we may assume that F is not a 2-group, hence contains an element x of order 3. The only possibilities for J are 22, 23 or Q8, with CG⁢(J)0=D4, A14 or B13, respectively. The last two cases are ruled out since any proper diagonal connected subgroup of A14 or B13 such that each projection involves no nontrivial field automorphisms is not B4-irreducible by Proposition 2.7.

Hence J≅22 and CG⁢(J)0=D4. If F=〈J,x〉≅Alt4, then CG⁢(F)0=A2 or G2; and G2 is not possible since it is contained in a Levi subgroup of type B3. And if F≠〈J,x〉, then F≅Sym4 and CG⁢(F)0=A1. ∎

Lemma 3.26.

If J is a 3-group, then F and CG⁢(F)0 are given in Table 11.

Proof.

Let J=〈x〉, so CG⁢(x)=A2⁢A2. Proposition 2.4 gives

NG⁢(A2⁢A2)/A2⁢A2=〈t〉≅Z2,

where t acts as a graph automorphism on each factor. Hence F=〈x,t〉≅Dih6 with CG⁢(F)0=A1⁢A1. ∎

As before, the fact that all the subgroups CG⁢(F)0 in Table 11 are G-irreducible is proved in similar fashion to Lemma 3.11; in particular they all have no trivial composition factors on L⁢(G).

This completes the proof of Theorem 1 for G=F4.

3.5 The case G=G2

Lemma 3.27.

Let F be a finite subgroup of G=G2 such that CG⁢(F)0 is G-irreducible. Then F and CG⁢(F)0 are as in Table 12.

Proof.

By Proposition 2.2, nonidentity elements of F have order 2 or 3. If F is a 2-group, then F contains an involution t with CG⁢(t)=A1⁢A1; the centralizer of an involution in A1⁢A1 is reducible and therefore F=〈t〉. Similarly, if F is a 3-group, then F=〈u〉≅3 and CG⁢(F)=A2. The only remaining possibility is F=〈t,u〉≅Dih6. Since NG⁢(A2)/A2≅2, such an example exists and CG⁢(F)0=A1. ∎

This completes the proof of Theorem 1.

4 Proof of Proposition 3

In this section we prove the following generalisation of Proposition 3.

Proposition 4.1.

Let G be a classical simple adjoint algebraic group in characteristic p≥0 with natural module V, and let H=Aut⁡G. Suppose F is a finite subgroup of H such that CG⁢(F)0 is G-irreducible. Then F is an elementary abelian 2-group (or a group of order 3 or 6 in the case where G=D4), and one of the following holds.

G=PSLn, F∩G=1, |F|=2 and
1. if n is even, then CG⁢(F)0=PSpn or PSOn(p≠2),
2. if n is odd, then p≠2 and CG⁢(F)0=PSOn.
G=H=PSp2⁢n, p≠2, and taking preimages in Sp2⁢n,
CSp2⁢n⁢(F)0=∏iSp2⁢ni=∏iSp⁢(Wi),
where ∑ni=n and Wi are the distinct weight spaces of F on V.
G=PSOn(n≠8), H=POn, p≠2, and taking preimages in On,
COn⁢(F)0=∏iSOni=∏iSO⁢(Wi),
where ni≥3 for all i, ∑ni=n or n-1, and Wi are weight spaces of F.
G=PSO2⁢n(n≠4), p=2, F∩G=1, |F|=2 and CG⁢(F)0=SO2⁢n-1.
G=D4=PSO8, H=D4.Sym3, and F, CG⁢(F)0 are as in Table 6.

Proof.

First suppose G=PSLn. If F∩G≠1, then CG⁢(F∩G)0 is reducible, so F∩G=1. Hence |F|=2 and now the conclusion in part (i) follows from Proposition 2.1. Similarly, if G=Dn=PSO2⁢n with n≠4 and p=2 (so that H=G⁢.2), then F∩G=1, |F|=2 and CG⁢(F)0=Bn-1 by Proposition 2.1, as in (iv).

Now suppose G=PSp2⁢n. Then H=G and the centralizer in G of any element of order greater than 2 is reducible. Hence F is an elementary abelian 2-group and p≠2. The preimage F^ of F in Spn must also be elementary abelian, and if we let Wi(1≤i≤k) be the weight spaces of F^ on V, then V=W1⊥⋯⊥Wk and CSp2⁢n⁢(F^)=∏Sp⁢(Wi), as in conclusion (ii).

A similar proof applies when G=PSOn with n≠8 and p≠2, giving (iii).

It remains to handle G=D4=PSO8. Here H=G.Sym3. If F≤PO8=G⁢.2, then the above proof shows that F=2 or 22 is as in Table 6. Now suppose 3 divides |F|, so that F contains an element x of order 3 inducing a triality automorphism on G. By Proposition 2.1, CG⁢(x)=G2 or A2, with p≠3 in the latter case.

If there is an element y∈CF⁢(x)∖〈x〉, then y∈G2 or A2 has irreducible centralizer, which forces y to be an involution in G2. So in this case F=〈x,y〉≅Z6 and CG⁢(F)=CG2⁢(y)=A¯1⁢A1, as in Table 6. This subgroup has composition factors of dimensions 1, 3 and 4 on V, so is G-irreducible.

We may now suppose that CF⁢(x)=〈x〉 and F≠〈x〉. This implies that

F=〈x,t〉≅Dih6.

If CG⁢(x)=G2, then t must centralize G2, so that CG⁢(F)=G2. If CG⁢(x)=A2, then t induces a graph automorphism on A2 (see Proposition 2.1), so CG⁢(F)=A1 and p≠2, as in Table 6. This completes the proof. ∎

Table 6

G=Aut⁡D4: finite subgroups F with irreducible centralizer.

5 Tables of results

This section consists of the tables referred to in Theorem 1.

Table 7

G=E8: finite subgroups F with irreducible centralizer.

Table 8

G=E7 (adjoint): finite subgroups F with irreducible centralizer.

Table 9

G=E6: finite subgroups F with irreducible centralizer.

Table 10

G=Aut⁡E6: finite subgroups F with irreducible centralizer.

Table 11

G=F4: finite subgroups F with irreducible centralizer.

Table 12

G=G2: finite subgroups F with irreducible centralizer.

Communicated by Gunter Malle

References

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Received: 2016-6-10

Revised: 2016-12-19

Published Online: 2017-2-14

Published in Print: 2017-9-1

Articles in the same Issue

https://doi.org/10.1515/jgth-2017-0002