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Publicly Available Published by De Gruyter July 14, 2018

Groups with bounded centralizer chains and the Borovik–Khukhro conjecture

  • Alexander A. Buturlakin EMAIL logo , Danila O. Revin and Andrey V. Vasil’ev
From the journal Journal of Group Theory

Abstract

Let G be a locally finite group and let F(G) be the Hirsch–Plotkin radical of G. Let S denote the full inverse image of the generalized Fitting subgroup of G/F(G) in G. Assume that there is a number k such that the length of every nested chain of centralizers in G does not exceed k. The Borovik–Khukhro conjecture states, in particular, that under this assumption, the quotient G/S contains an abelian subgroup of finite index bounded in terms of k. We disprove this statement and prove a weak analogue of it.

Introduction

Following [10, 24], the maximal length of nested chains of centralizers of subsets in a group G is called the c-dimension of G (the same number is also known as the length of the centralizer lattice of G, see, e.g., [25]). Here we consider groups of finite c-dimension. The class of such groups is quite wide: it includes, for example, abelian groups, linear groups, torsion-free hyperbolic groups, in particular, free groups. It is closed under taking subgroups, finite direct products, and extensions by finite groups, and is not closed under taking homomorphic images and arbitrary extensions. Obviously, every group of finite c-dimension satisfies the minimal condition on centralizers, therefore, it is a so-called 𝔐C-group.

The minimal condition on centralizers is a useful notion in the class of locally finite groups. Indeed, locally finite 𝔐C-groups satisfy an analogue of the Sylow theorem [4, Theorem B]. Periodic locally soluble 𝔐C-groups satisfy an analogue of the Hall theorem on π-subgroups [5, Theorem 1.6]; moreover, these groups are soluble and contain a nilpotent-by-abelian subgroup of finite index [5, Theorems 2.1 and 2.2]. When we restrict ourselves to the class of locally finite groups of finite c-dimension, it is natural to ask if the aforementioned results can be strengthened. For example, is the derived length of a locally soluble group in this class bounded in terms of the c-dimension? Another natural question is whether or not such a group contains a nilpotent-by-abelian subgroup of finite index bounded in the same terms. The first question has been answered in the affirmative [21, Theorem 1]. The answer to the second question is negative [21, Example 1], but a weaker statement holds: if G is a periodic locally soluble group of c-dimension k, then the quotient group of G by the second Hirsch–Plotkin radical F2(G) contains an abelian subgroup of k-bounded index.

In the same paper [21], E. I. Khukhro posed the following conjecture which he attributed to A.V. Borovik.

Borovik--Khukhro Conjecture.

Let G be a locally finite group of finite c-dimension k. Let S be the full inverse image of F*(G/F(G)) in G. Then:

  1. the number of nonabelian simple composition factors of G is finite and k-bounded,

  2. G/S has an abelian subgroup of finite k-bounded index.

Recall that the Hirsch–Plotkin radical F(G) is the maximal normal locally nilpotent subgroup. The i-th Hirsch–Plotkin radical Fi(G) is defined inductively: F1(G)=F(G) and Fi(G) is the full inverse image of F(G/Fi-1(G)) in G. The layer E(G) of G is the subgroup generated by all components of G, that is, its subnormal quasisimple subgroups. The generalized Fitting subgroup F*(G) is the product of F(G) and E(G).

While part (1) of the conjecture is known to be true [7], the second part turns out to be false.

Theorem 1.

Statement (2) of the Borovik–Khukhro conjecture does not hold.

It is clear that a counterexample proving Theorem 1 should be an infinite series of locally finite groups G of uniformly bounded c-dimensions with unbounded minimal indices of abelian subgroups in the quotients G/S. In fact, we construct infinitely many counterexamples as follows. For every integer n>0, we define an infinite series 𝒢n of finite groups Gn,r, where r runs over an infinite set of primes. The c-dimensions of groups in 𝒢n are bounded in terms of n only. The quotient of G=Gn,r by the corresponding subgroup S (that is, the full inverse image of the generalized Fitting subgroup of Gn,r/F(Gn,r)) is isomorphic to the symplectic group Sp2n(r). This contradicts statement (2) of the Borovik–Khukhro conjecture, because the minimum of the indices of the proper subgroups of G/S=Sp2n(r) increases with r.

It is worth noting that our construction depends on some deep number-theoretical results. It requires some results on Dickson’s conjecture and Schinzel’s hypothesis H on prime values of finite sets of linear forms and irreducible integer-valued polynomials obtained in [1]. We are thankful to participants of the forum on mathoverflow.net [28], especially to T. Tao, and to J. Maynard for the information on various results on these conjectures and related sieve methods. There is a more detailed discussion on this topic after the proof of Theorem 1 in Section 2.

Though the second part of the Borovik–Khukhro conjecture is false, a weaker version of it holds, at least in the case of finite groups.

Theorem 2.

Let G be a finite group of finite c-dimension k. Put G¯=G/F3(G). The group G¯/E(G¯) contains an abelian subgroup of k-bounded index.

The proof of Theorem 2 relies on [21, Theorem 1] and ideas from [7]. The difference from the proof in [7] is that we bound not just the number of nonabelian composition factors but the overall “size” of them.

Theorem 2 implies that, for a finite group G, the quotient of G/F3(G) by the generalized Fitting subgroup contains an abelian subgroup of index bounded in terms of the c-dimension, while the conjecture was that F(G) instead of F3(G) is enough. So the natural question is whether or not one can replace F3(G) by F2(G). Furthermore, Theorem 2 considers only finite groups. Hence it would be interesting to obtain some positive statement for locally finite groups. Since every iterated Hirsch–Plotkin radical is contained in the locally soluble radical, the following seems to be the “weakest” analogue of the conjecture: Let G be a locally finite group of finite c-dimension k, and R the (locally) soluble radical of G; the quotient group of G/R by the generalized Fitting subgroup contains an abelian subgroup of finite index bounded in terms of k. The proof of this statement for finite groups follows from Theorem 2 (see Corollary 2 in Section 3).

Section 1 contains preliminary results on centralizers and some information on simple groups. In Section 2, we give a counterexample to the Borovik–Khukhro conjecture and discuss for which c-dimensions such counterexamples exist. Theorem 2 is proved in Section 3.

1 Preliminaries

The following lemma lists some well-known properties of centralizers.

Lemma 1.1.

Let G be a group and let X, Y be subsets in G. Then:

  1. CG(X)=xXCG(x).

  2. CG(X)=CG(X).

  3. CG(X)CG(Y) if and only if CG(CG(X))CG(CG(Y)); furthermore, CG(X)=CG(Y) if and only if CG(CG(X))=CG(CG(Y)).

Let G be a group and V a normal abelian subgroup of G. Let ¯:GG/V be the natural homomorphism. The conjugation action of G on V induces an action of G/V on V. In the following lemma the centralizer CV(Y¯) for YG is defined with respect to this action, and therefore coincides with CV(Y).

Lemma 1.2.

Let G be a group and V a normal abelian subgroup of G. Suppose that the length of every chain of nested subgroups in G/V is at most l. Then the c-dimension of G is at most 2l. If V is central, then the c-dimension of G is at most l.

Proof.

Suppose that the c-dimension of G is equal to k. Let[1]

(1.1)Y0>>Yk

be a chain of subgroups of G such that

(1.2)CG(Y0)<<CG(Yk).

As above, ¯:GG/V is the natural homomorphism. In view of (1.1) and (1.2), we have the following chains of inclusions:

(1.3)Y¯0Y¯k,
(1.4)CV(Y0)CV(Yk),
(1.5)CG(Y0)¯CG(Yk)¯,
(1.6)CV(Y¯0)CV(Y¯k).

Since CG(Yi)/CV(Yi)CG(Yi)¯ and CG(Yi-1)<GG(Yi) for every i, at least one of the inclusions

CV(Y¯i-1)=CV(Yi-1)CV(Yi)=CV(Y¯i)andCG(Yi-1)¯CG(Yi)¯

is strict. Moreover, if CV(Y¯i-1)<CV(Y¯i), then Y¯i-1>Y¯i. Thus k does not exceed the sum of the numbers of strict inclusions in the chains (1.3) and (1.5) of subgroups of G/V. Hence k2l. Finally, if V is cental, then all inclusions in (1.4) are equalities, and all inclusions in (1.5) are strict. ∎

Remark.

We want to point out that the bound for the c-dimension obtained in Lemma 1.2 is sharp. Let H be a finite group and let

1=H0<H1<<Hl=H

be the longest chain of subgroups. Fix some prime p. For i=0,,l, denote by Vi the permutation module of H over the field of order p given by the action of H on the right cosets by Hi via right multiplications. Put

V=V0V1Vl

and denote by G the natural semidirect product VH. We claim that the c-dimension of G is equal to 2l.

Let ei be a vector of Vi such that CH(ei)=Hi. Let Wi be the subspace of V spanned by ei,ei+1,,el. It is easy to see that CG(Wi)=VHi. Also the centralizer of VHi contains Wi and does not contain ei-1. Therefore we have

CG(Wl)>CG(Wl-1)>>CG(W0)=CG(V)
=CG(VH0)>CG(VH1)>>CG(VHl)=Z(G).

The length of this chain is exactly 2l, i.e. the maximal possible length.

The following lemma is crucial to the proof of Theorem 2.

Lemma 1.3 ([21, Lemma 3]).

If an elementary abelian p-group E of order pn acts faithfully on a finite nilpotent p-group Q, then there exists a series of subgroups

E=E0>E1>E2>>En=1

such that

CQ(E0)<CQ(E1)<<CQ(En).

Lemma 1.4.

The c-dimension of a central product of finite groups H and K is at least the sum of their c-dimensions.

Proof.

Let G be a central product HK. Let

H0<H1<<HmandK0<K1<<Kl

be series of subgroups of H and K such that the series of their centralizers are strict and have maximal lengths. We may assume that both H and K are subgroups of G, and, therefore, all their subgroups are also subgroups of G. We have CG(Hi)=CH(Hi)K and CG(HKi)=Z(H)CK(Ki). Hence the subgroups Hi and HKi provide a series of subgroups with the centralizer chain of the desired length. ∎

Remark.

The c-dimension of HK can be larger than the sum of c-dimensions of H and K. Indeed, let H be the group SL2(3). Then H is a semidirect product of the quaternion group of order 8 and the cyclic group of order 3 which permutes the subgroups of order 4. It is not hard to see that the c-dimension of H is 2. Denote by Q the Sylow 2-subgroup of H. If Q is generated by two elements a and b, then the commutator [a,b] is the generator of the center of Q and, therefore, of H. Let K be a group isomorphic to H and let c and d be the images of a and b under this isomorphism.

Put G=(H×K)/[a,b][c,d] and let ¯ stand for the natural homomorphism from H×K onto G. By definition, G is a central product of H and K. Consider the following series of centralizers:

(1.7)G>CG(a¯)CG(a¯c¯)CG(a¯,c¯)>CG(H¯,c¯)>Z(G).

We have b¯d¯CG(a¯c¯)CG(a¯,c¯). Indeed, a¯b¯d¯=a¯-1 and

[a¯c¯,b¯d¯]=[a¯,b¯][c¯,d¯]=1.

Therefore the third inclusion in (1.7) is strict.

We argue that no element of order 3 centralizes a¯c¯. Let g¯ be an element of G of order 3. Without loss of generality, we may assume that g¯ does not normalize any subgroup of order 4 in H. Moreover, we may assume that a¯g¯=b¯. Suppose that (a¯c¯)g¯=a¯c¯. Then (b¯a¯-1)(c¯g¯c¯-1)=1. Since ba-1 is an element of H of order 4 and c¯g¯c¯-1 is a 2-element of K, their product is an element of order 4 and has nontrivial image in G. Therefore g¯ does not centralize a¯c¯. Thus all inclusions in (1.7) are strict and the c-dimension of HK is at least 5 (one can show that it is exactly 5).

For a finite group G, we define a nonnegative integer λ(G) as follows. First, let G be a nonabelian simple group. If G is a group of Lie type, then λ(G) is the minimal Lie rank of groups of Lie type isomorphic to G (the Lie rank is the rank of the corresponding (B,N)-pair [3, p. 249]). If G is an alternating group, then λ(G) is the degree of G (except for the groups Alt5, Alt6 and Alt8 which are isomorphic to groups of Lie type and, therefore, already have assigned values of λ(G)). Put λ(G)=1 for the sporadic groups. If G is an arbitrary finite group, then λ(G) is the sum of λ(S), where S runs over the nonabelian composition factors of G.

Recall that the p-rank of a group G for a given prime p is the largest rank of elementary abelian p-subgroups of G.

Lemma 1.5.

Let r{2,3,5}. There exists a positive constant c such that the r-rank of any nonabelian finite simple group G whose order is divisible by r is at least cλ(G).

Proof.

For the sporadic groups, there is nothing to prove. In the case of alternating groups, this fact is obvious. In the case of groups of Lie type, it follows, for example, from [14, Part I, 10-6]. ∎

The next lemma directly follows from the formulas for the orders of groups of Lie type [3, Table 16.1] and the well-known Zsigmondy theorem [27].

Lemma 1.6.

Let G be a finite simple group of Lie type. The number of distinct prime divisors of the order of G is at least λ(G).

Denote by μ(G) the minimal degree of a faithful permutation representation of a finite group G.

Lemma 1.7 ([19, Theorem 2]).

Let G be a finite group. Let 𝔏 be a class of finite groups closed under taking subgroups, homomorphic images and extensions. Let N be the 𝔏-radical of G, that is, the maximal normal 𝔏-subgroup of G. Then μ(G)μ(G/N).

Lemma 1.8 ([11, Theorem 3.1]).

Let S1,S2,,Sr be finite simple groups. Then μ(S1×S2××Sr)=μ(S1)+μ(S2)++μ(Sr).

Lemma 1.9.

If S is a finite nonabelian simple group, then μ(S)λ(S).

Proof.

For alternating and sporadic groups, the assertion is obvious. If S is a group of Lie type having a faithful permutation representation of degree n, then the number of prime divisors of |S| is less than n and, by Lemma 1.6, is not less than λ(S). ∎

Note that Lemma 1.9 can also be deduced from the information on the values of μ(S) for finite simple groups S of Lie type which are all known (the complete list of these numbers can be found, for example, in [17, Table 4]).

Lemma 1.10.

If G is a subgroup of the symmetric group Symn, then λ(G)<5n4.

Proof.

We proceed by induction on |G|. If the soluble radical R of G is nontrivial, then Lemma 1.7 implies that G/R is also a subgroup of Symn and the inequality follows by induction.

Let R be trivial. If the socle Soc(G) of G is the direct product of nonabelian simple groups S1,S2,,Sl, then G is a subgroup of the semidirect product of Aut(S1)×Aut(S2)××Aut(Sl) and a subgroup of Syml. Lemmas 1.8 and 1.9 imply that λ(Soc(G))n. Since μ(S)5 for every finite simple group S, from Lemma 1.8 it follows that ln5. By the Schreier conjecture, the kernel of the natural homomorphism from G/Soc(G) to Syml is soluble. By the inductive hypothesis, λ(G/Soc(G))5l4n4. Finally, we have

λ(G)=λ(Soc(G))+λ(G/Soc(G))n+n4=5n4.

Lemma 1.11 ([7, Proposition 2.1]).

If G is a finite group of c-dimension k, then the number of nonabelian composition factors of G is less than 5k.

The following lemma is a direct consequence of [21, Theorem 1].

Lemma 1.12.

If a periodic locally soluble group G has finite c-dimension k, then the quotient G/F2(G) has an abelian subgroup of finite k-bounded index. In particular, the order of G/F3(G) is k-bounded.

2 Counterexamples to the Borovik–Khukhro conjecture

Denote by Ω(n) the number of prime divisors of a positive integer n counting multiplicities, i.e. if n=p1α1pmαm for primes p1,,pm, then

Ω(n)=α1++αm.

For positive integers n and M, we set

πn,M={r:r is an odd prime and Ω(rn-1)M}.

It is clear that the following inclusions hold for every n:

(2.1)πn,1πn,2πn,3.

A crucial ingredient in constructing our counterexamples is the following number-theoretical statement.

Proposition 2.1 ([1, Theorem C]).

For every positive integer n, there exists a positive integer M such that πn,M is infinite.

Given n, put

Mn=min{M:πn,M is infinite}andπn=πn,Mn.

Thus πn is the first infinite set in the chain (2.1).

Fix a positive integer n and an odd prime r. Let Rn,r be an extra-special group of order r2n+1 and exponent r. It is known that Aut(Rn,r) is split over Inn(Rn,r) (see [16, p. 404]) and the image in Out(Rn,r) of the centralizer in Aut(Rn,r) of Z(Rn,r) is isomorphic to the symplectic group Sp2n(r) (see [3, Exercise 8.5, p. 116]). Hence Aut(Rn,r) contains a subgroup An,rSp2n(r) that centralizes Z(Rn,r). We can form the natural semidirect product

Xn,r=Rn,rAn,r.

Take a prime p with p1(modr). It is known (see [22, p. 151]) that the extra-special group Rn,r has a faithful irreducible representation of degree rn over the field 𝔽p of order p. Moreover, this representation extends to a faithful representation of Xn,r (see [2, (3A) and (3B)]). Let Vn,r be a faithful irreducible 𝔽pXn,r-module of dimension rn corresponding to this representation and form the natural semidirect product

Gn,r=Vn,rXn,r.

The next assertion lists some basic properties of Gn,r which can be easily deduced from the definition of this group.

Proposition 2.2.

The following statements hold:

  1. F(Gn,r)=Vn,r,

  2. F*(Gn,r/Vn,r)=F(Gn,r/Vn,r)Rn,r,

  3. Gn,r/(Rn,rVn,r)An,rSp2n(r).

For a positive integer n, put

n¯=2lcm(1,2,,n)and𝒢n={Gn,r:rπn¯}.

Observe that by definition the set 𝒢n is infinite.

Theorem 3.

For an arbitrary positive integer n, the c-dimension of every group in 𝒢n does not exceed

(2.2)2((n+1)2+nMn¯).

Proof.

Take G=Gn,r𝒢n. By definition we have rπn¯ and Ω(rn¯-1)Mn¯. The Lagrange theorem implies that the length of every chain of nested subgroups in Xn,r does not exceed l=Ω(|Xn,r|). Since

|Xn,r|=|Rn,r||An,r|=r2n+1|Sp2n(r)|=r(n+1)2i=1n(r2i-1)

and r2i-1 divides rn¯-1 for every i=1,,n, we have

l=(n+1)2+i=1nΩ(r2i-1)(n+1)2+nΩ(rn¯-1)(n+1)2+nMn¯.

Now the theorem follows from Lemma 1.2. ∎

Now we prove Theorem 1. Assume that statement (2) of the Borovik–Khukhro conjecture is true. Fix a positive integer n and consider the set 𝒢n. Theorem 3 implies that the c-dimensions of groups in 𝒢n are bounded by the number given in (2.2). Hence every group Gn,r/(Rn,rVn,r)Sp2n(r) for rπn¯ contains an abelian subgroup, and therefore a normal abelian subgroup, of index bounded in terms of n. But there is a unique maximal proper normal subgroup of Sp2n(r), namely its center of order 2, and the index of these subgroups increases with r. This contradiction completes the proof of Theorem 1.

Theorem 1 implies that there exists a positive integer k such that the second part of the Borovik–Khukhro conjecture fails for the finite groups of c-dimension k. On the other hand, the description of the finite groups of c-dimension 2 (see [25, Theorem 9.3.12]) yields the full version of the conjecture for these groups. So it is natural to ask what is the minimal value of the c-dimension for which the Borovik–Khukhro conjecture is false. We already know that 3 is a lower bound for this number and we do not know any better lower bound. Let us obtain an upper bound by bounding M2.

Recall that M2 is the smallest M such that there exist an infinite number of prime numbers r satisfying Ω(r2-1)M.

Since r2-1 is divisible by 24 for r>3, it is clear that Ω(r2-1)6 for sufficiently large r, and therefore M26. There is a question attributed to P. Neumann [26, p. 316] asking whether or not the number of prime numbers r such that the order of PSL2(r) is a product of six prime numbers is infinite. Since

Ω(r2-1)=Ω(|PSL2(r)|),

the affirmative answer to this question would imply that M2=6 and the c-dimensions of groups in 𝒢1 would not exceed 2(22+6)=20.

The affirmative answer to Neumann’s question would follow from the validity of Dickson’s conjecture (the k-tuple conjecture) [9]. This conjecture states that for a finite set of integral linear forms a1n+b1,,akn+bk with ai>0, there are infinitely many positive integers n such that all these forms are simultaneously prime unless there are fixed divisors of their product. Equivalently,

Ω(i=1k(ain+bi))=kfor infinitely many integers n.

If Dickson’s conjecture is valid for the triplet 12n+1, 6n+1 and n, then there are infinitely many primes r of the form r=12n+1 such that n and 6n+1 are also primes and

Ω(r2-1)=Ω(12n(12n+2))=Ω(233n(6n+1))=6.

Both Dickson’s and Neumann’s conjectures provide an exact value of M2, but for our purposes a partial result might be satisfactory. There are well developed sieve methods that allow to obtain partial results for this kind of statement (see [18] for details). For example, the main result of [23] is very close to what is required. In this paper, J. Maynard proved that for any triplet of the forms ain+bi, i=1,2,3, without fixed divisors of their product, there are infinitely many positive integers n such that

Ω(i=13(ain+bi))7.

In particular, there are infinite many n such that the product n(6n+1)(12n+1) has at most seven prime divisors, and, therefore, there are infinitely many integers m such that Ω(m(m2-1))11. Unfortunately, one cannot guarantee the primality of m in this result. Nevertheless, in a private letter, J. Maynard expressed confidence that one can show that there are infinitely many primes r such that Ω(r(r2-1))11, or equivalently Ω(r2-1)10, by slightly changing his proof. This would mean that the c-dimensions of groups in 𝒢1 do not exceed 28.

Proposition 2.1, which is also proved in [1] by using sieve methods, has the following refinement [1, Corollary 4.2]: there are infinitely many primes r such that Ω(r2-1)21. This statement implies that M221, and therefore the following is a consequence of Theorem 3.

Corollary 1.

The c-dimensions of groups in 𝒢1 do not exceed 50.

It is clear that the number p in the construction of the module V1,r can be chosen so that p would not divide |X1,r|. Then one can apply the description of subgroups in SL2(r)Sp2(r)A1,r, the character table of this group and the results of [20] to calculate the character of the representation of X1,r corresponding to the module V1,r and use it to calculate the precise value of the c-dimension of G1,r.

Remark (June 2018).

According to a recent result by D. R. Heath-Brown [6, Theorem A.1], there are infinitely many primes r with Ω(r2-1)11. This result allows to improve the bound in Corollary 1 to 30.

3 Proof of Theorem 2

Throughout the rest of the paper G denotes a finite group and k denotes the c-dimension of G. The function λ was defined in Section 1. We start by bounding the value of λ on the layer E(G) of G.

Proposition 3.1.

There exists a universal constant b such that λ(E(G))bk for every G.

Proof.

Since the c-dimension of a subgroup does not exceed the c-dimension of the group, we may assume that G=E(G). By Lemma 1.4, we may also assume that G is quasisimple. Let ¯ be the natural homomorphism from G to G/Z(G). Observe that if G¯ is a sporadic group, or a group of Lie type of bounded Lie rank, or an alternating group of bounded degree, then we can choose b large enough to make the proposition trivial. So we may assume that G¯ is either an alternating group of sufficiently large degree, or a classical group of sufficiently large Lie rank.

Suppose that r is a prime divisor of the order of G¯ satisfying the following two properties. First, it is coprime to the order of the center Z(G). Second, there exists a strict nested chain of subsets of r-elements such that its length l is bounded from below by a linear function of λ(G) (not depending on G) and the centralizers of distinct sets are distinct. It is clear that every subset of r-elements of G¯ can be presented in the form M¯ for some subset M of G also consisting of r-elements. Let M0M1MlG be a chain of subsets of r-elements such that

CG¯(M¯0)>CG¯(M¯1)>>CG¯(M¯l).

By the properties of coprime action, CG¯(Mi¯)=CG(Mi)¯. Hence

CG(M0)>CG(M1)>>CG(Ml).

Therefore l is at most the c-dimension of G. Since it is bounded from below by a linear function of λ(G), the desired inequality follows. Thus it remains to determine such a prime r for every G.

Let us consider the condition that r does not divide the order of Z(G). Every prime divisor of the order of Z(G) is a prime divisor of the order of the Schur multiplier M(G¯) of G¯. The orders of Schur multipliers of all finite simple groups are known and can be found in, for example, [15, Section 6.1]. Due to these results, if M(G¯) is not a {2,3}-group, then G¯ is isomorphic to An(q) or An2(q) and every prime divisor of the order of M(G¯) divides 6(q2-1). So r should be chosen subject to these restrictions.

Now we consider the second condition on r, that is, a linear lower bound on the maximal length of a chain of centralizers of r-elements. If G is an alternating group Altn, then we can take r to be 5. For 1in5, we can choose nested sets Mi consisting of i disjoint 5-cycles. Their centralizers form a strict chain whose length exceeds n5-1. By the definition of λ, this gives λ(G)5-1 as a lower bound for l.

Let G¯ be a classical group over a field of order q. According to [8, Propositions 7–12], the group G¯ contains a subgroup isomorphic to a central product of quasisimple subgroups H1,H2,,Hs, where every Hi is a group of Lie type A3 or A32 over a field of order q. Moreover, the number s of these factors is at least λ(G)-64. By the Zsigmondy theorem, there exists a prime r dividing the order of each Hi, but not 6(q2-1). For each i=1,,s, let hiHi be an element of order r. Then hi is not central in Hi. Therefore

CG¯(h1)>CG¯(h1,h2)>>CG¯(h1,h2,,hs),

and we have found a strict series of centralizers of subsets of r-elements whose length is at least λ(G)-64-1. The proposition is proved. ∎

Proposition 3.2.

There exists a universal constant d such that λ(G)dk for every G.

Proof.

Let R be the soluble radical of G. If P is a Sylow subgroup of R, then G/RNG(P)/(RNG(P)). So nonabelian composition factors of NG(P) and G coincide. Furthermore, the c-dimension of NG(P) as a subgroup of G is at most k. Therefore we may assume that R is the Fitting subgroup of G.

Put G¯=G/R. Let l be the number of nonabelian composition factors of the socle L¯ of G¯ (note that L¯ is the direct product of nonabelian simple groups). The quotient G¯/L¯ is an extension of a soluble group by a subgroup of the symmetric group Syml. Lemma 1.11 implies that l<5k. Hence λ(G¯/L¯)<25k4 by Lemma 1.10. Thus it is sufficient to show that λ(L¯)dk for some d. In particular, we may assume that G coincides with the full inverse image of L¯ in G.

Let be the set of nonabelian composition factors of CG(R). For a prime p, denote by p the set of nonabelian composition factors of G/CG(Op(R)) whose order is divisible by p (here Op(R) stands for the maximal p-subgroup of R). By the Feit–Thompson theorem [12] and the Thompson–Glauberman theorem [13, Chapter II, Corollary 7.3], the order of every finite nonabelian simple group is divisible by 2 and is not coprime to 15. Since R=Op(R)Oq(R) for distinct primes p and q, we have CG(Op(R))CG(Oq(R))=CG(R). Therefore each nonabelian composition factor of G is contained in 235. So to prove the proposition, it is sufficient to bound the sum of values of λ in each of these four sets by a linear function of k.

Put K=CG(R). Since K¯=KR/R is normal in G¯, it is a direct product of elements of . Let S¯ be in and let S be the full inverse image of S¯ in K. Then S(), that is, the least term in the derived series of S, is a perfect central extension of S¯ which is normal in K, so it is a component of K. Therefore all elements of are composition factors of the layer E(K). Since E(K) is a subgroup of E(G), the number λ(K) is bounded by bk for some constant b due to Proposition 3.1.

Put Kp=G/CG(Op(R)) for p{2,3,5}. It is clear that Kp is an extension of a direct product of some nonabelian composition factors of G by a nilpotent p-group. By the definition, Kp acts faithfully on Op(R), and so does its Sylow p-subgroup P. The group P is the direct product of Sylow p-subgroups of elements of p. By Lemma 1.5, P contains an elementary abelian p-group whose rank is bounded from below by cSpλ(S) for some positive constant c. It follows from Lemma 1.3 that cSpλ(S)k. Finally,

λ(G)3kc+bk

as required. ∎

Now we are ready to complete the proof of Theorem 2. Put G¯=G/F3(G). Let F¯* be the generalized Fitting subgroup of G¯. The quotient G¯/F¯* is isomorphic to a subgroup of the group of outer automorphisms Out(F¯*) of F*. The latter is a subgroup of the direct product of Out(F(G¯)) and Out(E(G¯)). By Lemma 1.12, the order of F(G¯) and, therefore, the order of Out(F(G¯)) is bounded in terms of k. Hence it is sufficient to prove that Out(E(G¯)) contains an abelian subgroup of k-bounded index.

If E(G¯) is a product of components Q1,Q2,,Qs, then Out(E(G)) is a subgroup of a semidirect product of Out(Q1)×Out(Q2)××Out(Qs) and some subgroup of Syms. According to Lemma 1.11, the number s is bounded in terms of k, therefore, it is sufficient to prove that the direct product of the outer automorphism groups of the components has an abelian subgroup of k-bounded index. The group Out(Qi) is a subgroup of the outer automorphism group of the corresponding simple group Si=Qi/Z(Qi) (see [15, Corollary 5.1.4]). If Si is an alternating or sporadic group, then the order of Out(Si) is at most 4. If Si is a group of Lie type, then the index of the cyclic subgroup of the field automorphisms in Out(Si) is bounded in terms of the Lie rank of Si (see [15, Theorems 2.5.1 and 2.5.12]). Due to Proposition 3.2, the Lie rank of every Si is bounded in terms of k. Define Ai to be trivial if Si is not a group of Lie type, and the group of the field automorphisms of Si otherwise. The index of the direct product A1××As in Out(S1)××Out(Ss) is k-bounded by the preceding remarks. This completes the proof of the theorem.

Observe that since the group G¯/Soc(G¯) is a homomorphic image of the quotient group of G/F3(G) by its layer, the following is an immediate consequence of Theorem 2.

Corollary 2.

Let G be a finite group of c-dimension k and let R be its soluble radical. Put G¯=G/R. Then the group G¯/Soc(G¯) contains an abelian subgroup of k-bounded index.


Communicated by Evgenii I. Khukhro


Funding statement: The research was supported by RSF (project N 14-21-00065).

Acknowledgements

We thank E. I. Khukhro for drawing our attention to the subject. We also want to thank the referee for useful suggestions towards improving the exposition.

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Received: 2017-08-04
Revised: 2018-06-10
Published Online: 2018-07-14
Published in Print: 2018-11-01

© 2018 Walter de Gruyter GmbH, Berlin/Boston

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