1 Introduction

A group G is said to have restricted centralizers if for each g in G the centralizer \(C_G(g)\) either is finite or has finite index in G. This notion was introduced by Shalev in [13] where he showed that a profinite group with restricted centralizers is virtually abelian. We say that a profinite group has a property virtually if it has an open subgroup with that property. The article [3] handles profinite groups with restricted centralizers of w-values for a multilinear commutator word w. The theorem proved in [3] says that if w is a multilinear commutator word and G is a profinite group in which the centralizer of any w-value is either finite or open, then the verbal subgroup w(G) is virtually abelian. In [1] we study profinite groups in which p-elements have restricted centralizers, that is, groups in which \(C_G(x)\) is either finite or open for any p-element x. The following theorem was proved.

Theorem 1.1

Let p be a prime and G a profinite group in which the centralizer of each p-element is either finite or open. Then G has a normal abelian pro-p subgroup N such that G/N is virtually pro-\(p'\).

The present paper grew out of our desire to determine whether this result can be extended to profinite groups in which the centralizer of each \(\pi \)-element, where \(\pi \) is a fixed set of primes, is either finite or open. As usual, we say that an element x of a profinite group G is a \(\pi \)-element if the order of the image of x in every finite continuous homomorphic image of G is divisible only by primes in \(\pi \) (see [10, Section 2.3] for a formal definition of the order of a profinite group).

It turned out that the techniques used in the proof of Theorem 1.1 were not quite adequate for handling the case of \(\pi \)-elements. The basic difficulty stems from the fact that (pro)finite groups in general do not possess Hall \(\pi \)-subgroups.

In the present paper we develop some new techniques and establish the following theorem about finite groups.

If \(\pi \) is a set of primes and G a finite group, write \(O^{\pi '}(G)\) for the unique smallest normal subgroup M of G such that G/M is a \(\pi '\)-group. The conjugacy class containing an element \(g\in G\) is denoted by \(g^G\).

Theorem 1.2

Let n be a positive integer, \(\pi \) be a set of primes, and G a finite group such that \(|g^G|\le n\) for each \(\pi \)-element \(g\in G\). Let \(H=O^{\pi '}(G)\). Then G has a normal subgroup N such that

  1. 1.

    The index [G : N] is n-bounded;

  2. 2.

    \([H,N]=[H,H]\);

  3. 3.

    The order of [H, N] is n-bounded.

Throughout the article we use the expression “\((a, b,\ldots )\)-bounded” to mean that a quantity is finite and bounded by a certain number depending only on the parameters \(a,b,\ldots \).

The proof of Theorem 1.2 uses some new results related to Neumann’s BFC-theorem [8]. In particular, an important role in the proof is played by a recent probabilistic result from [2]. Theorem 1.2 provides a highly effective tool for handling profinite groups with restricted centralizers of \(\pi \)-elements. Surprisingly, the obtained result is much stronger than Theorem 1.1 even in the case where \(\pi \) consists of a single prime.

Theorem 1.3

Let \(\pi \) be a set of primes and G a profinite group in which the centralizer of each \(\pi \)-element is either finite or open. Then G has an open subgroup of the form \(P\times Q\), where P is an abelian pro-\(\pi \) subgroup and Q is a pro-\(\pi '\) subgroup.

Thus, the improvement over Theorem 1.1 is twofold – the result now covers the case of \(\pi \)-elements and provides additional details clarifying the structure of groups in question. Furthermore, it is easy to see that Theorem 1.3 extends Shalev’s result [13] which can be recovered by considering the case where \(\pi =\pi (G)\) is the set of all prime divisors of the order of G.

We now have several results showing that if the elements of a certain subset X of a profinite group G have restricted centralizers, then the structure of G is very special. This suggests the general line of research whose aim would be to determine which subsets of G have the above property. At present we are not able to provide any insight on the problem. Perhaps one might start with the following question:

Let n be a positive integer. What can be said about a profinite group G such that if \(x\in G\) then \(C_G(x^n)\) is either finite or open?

Proofs of Theorems 1.2 and 1.3 will be given in Sects. 2 and 3 , respectively.

2 Proof of Theorem 1.2

The following lemma is taken from [1]. If \(X\subseteq G\) is a subset of a group G, we write \(\langle X\rangle \) for the subgroup generated by X and \(\langle X^G\rangle \) for the minimal normal subgroup of G containing X.

Lemma 2.1

Let i, j be positive integers and G a group having a subgroup K such that \(|x^G|\le i\) for each \(x\in K\). Suppose that \(|K|\le j\). Then \(\langle K^G\rangle \) has finite (i, j)-bounded order.

If K is a subgroup of a finite group G, we denote by

$$\begin{aligned} Pr(K,G)=\frac{|\{(x,y) \in K\times G : [x,y]=1\}|}{|K||G|} \end{aligned}$$

the relative commutativity degree of K in G, that is, the probability that a random element of G commutes with a random element of K. Note that

$$\begin{aligned} Pr(K,G)=\frac{\sum _{x\in K}|C_G(x)|}{|K||G|}. \end{aligned}$$

It follows that if \(|x^G|\le n\) for each \(x\in K\), then \(Pr(K,G)\ge \frac{1}{n}\).

The next result was obtained in [2, Proposition 1.2]. In the case where \(K=G\) this is a well known theorem, due to P. M. Neumann [9].

Proposition 2.2

Let \(\epsilon >0\), and let G be a finite group having a subgroup K such that \(Pr(K,G)\ge \epsilon \). Then there is a normal subgroup \(T\le G\) and a subgroup \(B\le K\) such that the indexes [G : T] and [K : B], and the order of the commutator subgroup [T, B] are \(\epsilon \)-bounded.

We will now embark on the proof of Theorem 1.2.

Assume the hypothesis of Theorem 1.2. Let X be the set of all \(\pi \)-elements of G. Clearly, \(H=\langle X\rangle \). Given an element \(g\in H\), we write l(g) for the minimal number l with the property that g can be written as a product of l elements of X. The following result is straightforward from [4, Lemma 2.1].

Lemma 2.3

Let \(K\le H\) be a subgroup of index m in H, and let \(b\in H\). Then the coset Kb contains an element g such that \(l(g)\le m-1\).

Let m be the maximum of indices of \(C_H(x)\) in H where \(x\in X\). Obviously, we have \(m\le n\).

Lemma 2.4

For any \(x\in X\) the subgroup [H, x] has m-bounded order.

Proof

Take \(x \in X\). Since the index of \(C_H(x)\) in H is at most m, by Lemma 2.3, we can choose elements \(y_1,\ldots ,y_m\) in H such that \(l(y_i)\le m-1\) and the subgroup [H, x] is generated by the commutators \([y_i,x]\), for \(i=1,\ldots ,m\). For any such i write \(y_i=y_{i1}\ldots y_{i(m-1)}\), with \(y_{ij}\in X\). Using standard commutator identities we can rewrite \([y_i,x]\) as a product of conjugates in H of the commutators \([y_{ij},x]\). Let \(\{h_1,\ldots ,h_s\}\) be the conjugates in H of all elements from the set \(\{x,y_{ij} \mid 1\le i,j \le m\}.\) Note that the number s here is m-bounded. This follows form the fact that \(C_H(x)\) has index at most m in H for each \(x\in X\). Put \(T=\langle h_1,\ldots ,h_s \rangle \). Since [H, x] is contained in the commutator subgroup \(T'\), it is sufficient to show that \(T'\) has m-bounded order. Observe that the centre Z(T) has index at most \(m^s\) in T, since the index of \(C_H(h_i)\) is at most m in H for any \(i=1,\ldots ,s\). Thus, by Schur’s theorem [11, 10.1.4], we conclude that the order of \(T'\) is m-bounded, as desired. \(\square \)

Select \(a\in X\) such that \(|a^H|=m\). Choose \(b_1,\ldots ,b_m\) in H such that \(l(b_i)\le m-1\) and \(a^H=\{a^{b_i}; i=1,\ldots ,m\}\). The existence of the elements \(b_i\) is guaranteed by Lemma 2.3. Set \(U=C_G(\langle b_1,\ldots ,b_m\rangle )\). Note that the index of U in G is n-bounded. Indeed, since \(l(b_i)\le m-1\) we can write \(b_i=b_{i1}\ldots b_{i(m-1)}\), where \(b_{ij}\in X\) and \(i=1,\ldots ,m\). By the hypothesis the index of \(C_G(b_{ij})\) in G is at most n for any such element \(b_{ij}\). Thus, \([G:U]\le n^{(m-1)m}\).

The next result is somewhat analogous to [14, Lemma 4.5].

Lemma 2.5

If \(u\in U\) and \(ua\in X\), then \([H,u]\le [H,a]\).

Proof

Assume that \(u\in U\) and \(ua\in X\). For each \(i=1,\ldots ,m\) we have \((ua)^{b_i}=ua^{b_i}\), since u belongs to U. We know that \(ua\in X\) so taking into account the hypothesis on the cardinality of the conjugacy class of ua in H, we deduce that \((ua)^H\) consists exactly of the elements \(ua^{b_i}\), for \(i=1,\ldots ,m\). Thus, given an arbitrary element \(h\in H\), there exists \(b\in \{b_1,\ldots ,b_m\}\) such that \((ua)^h=ua^b\) and so \(u^ha^h=ua^b\). It follows that \([u,h]=a^ba^{-h}\in [H,a]\), and the result holds. \(\square \)

Lemma 2.6

The order of the commutator subgroup of H is n-bounded.

Proof

Let \(U_0\) be the maximal normal subgroup of G contained in U. Recall that, by the remark made before Lemma 2.5, U has n-bounded index in G. It follows that the index \([G:U_0]\) is n-bounded as well.

By the hypothesis a has at most n conjugates in G, say \(\{a^{g_1},\ldots ,a^{g_n}\}\). Let T be the normal closure in G of the subgroup [H, a]. Note that the subgroups \([H,a^{g_i}]\) are normal in H, therefore \(T=[H,a^{g_1}]\ldots [H,a^{g_n}]\). By Lemma 2.4 each of the subgroups \([H,a^{g_i}]\) has n-bounded order. We conclude that the order of T is n-bounded.

Let \(Y=Xa^{-1}\cap U\). Note that for any \(y\in Y\) the product ya belongs to X. Therefore, by Lemma 2.5, for any \(y\in Y\), the subgroup [H, y] is contained in [H, a]. Thus,

$$\begin{aligned}{}[H,Y]\le T. \end{aligned}$$
(1)

Observe that for any \(u\in U_0\) the commutator \([u,a^{-1}]=a^ua^{-1}\) lies in Y and so

$$\begin{aligned} \left[ H,\left[ U_0,a^{-1}\right] \right] \le [H,Y]. \end{aligned}$$
(2)

Since \([U_0,a^{-1}]=[U_0,a]\), we deduce from (1) and (2) that

$$\begin{aligned} \left[ H,\left[ U_0,a\right] \right] \le T. \end{aligned}$$
(3)

Since T has n-bounded order, it is sufficient to show that the derived group of the quotient H/T has finite n-bounded order. We pass now to the quotient G/T and for the sake of simplicity the images of G, \(H,U,U_0,X\) and Y will be denoted by the same symbols. Note that by (1) the set Y becomes central in H modulo T. Containment (3) shows that \([U_0,a]\le Z(H)\). This implies that if \(b\in U_0\) is a \(\pi \)-element, then \([b,a] \in Z(H)\) and the subgroup \(\langle a,b\rangle \) is nilpotent. Thus the product ba is a \(\pi \)-element too and so \(b\in Y\). Hence, all \(\pi \)-elements of \(U_0\) are contained in Y and, in view of (1), we deduce that they are contained in Z(H).

Next we consider the quotient G/Z(H). Since the image of \(U_0\) in G/Z(H) is a \(\pi '\)-group and has n-bounded index in G, we deduce that the order of any \(\pi \)-subgroup in G/Z(H) is n-bounded. In particular, there is an n-bounded constant C such that for every \(p\in \pi \) the order of the Sylow p-subgroup of G/Z(H) is at most C. Because of Lemma 2.1 for any \(p\in \pi \) each Sylow p-subgroup of G/Z(H) is contained in a normal subgroup of n-bounded order. We deduce that all such Sylow subgroups of G/Z(H) are contained in a normal subgroup of n-bounded order. Since H is generated by \(\pi \)-elements, it follows that the order of H/Z(H) is n-bounded. Thus, in view of Schur’s theorem [11, 10.1.4], we conclude that \(|H'|\) is n-bounded, as desired. \(\square \)

We will now complete the proof of Theorem 1.2.

Proof

Assume first that H is abelian. In this case the set X of \(\pi \)-elements is a subgroup, that is, \(X=H\). By the hypothesis we have \(|x^G|\le n\) for any element \(x\in H\) and so the relative commutativity degree Pr(H, G) of H in G is at least \(\frac{1}{n}\). Thus, by virtue of Proposition 2.2, there is a normal subgroup \(T\le G\) and a subgroup \(B\le H\) such that the indexes [G : T] and [H : B], and the order of the commutator subgroup [T, B] are n-bounded.

Since H is a normal \(\pi \)-subgroup and [G : H] is a \(\pi '\)-number, by the Schur–Zassenhaus Theorem [5, Theorem 6.2.1] the subgroup H admits a complement L in G such that \(G=HL\) and L is a \(\pi '\)-subgroup. Set \(T_0=T\cap L\). Observe that the index \([L:T_0]\) is n-bounded since it is at most the index of T in G. Thus we deduce that the index of \(HT_0\) is n-bounded in G, as well.

We claim that the order of \([H,T_0]\) is n-bounded. Indeed, the \(\pi '\)-subgroup \(T_0\) acts coprimely on the the abelian \(\pi \)-subgroup \(B_1=B[B,T_0]\), and so we have \(B_1=C_{B_1}(T_0)\times [B_1,T_0]\) ( [7, Corollary 1.6.5]). Note that \([B_1,T_0]=[B,T_0]\). Since the oder of \([B,T_0]\) is n-bounded (being at most the order of [T, B]), we deduce that the index \([B_1:C_{B_1}(T_0)]\) is n-bounded. In combination with the fact that [H : B] is n-bounded, we obtain that the index \([H:C_{B_1}(T_0)]\) is n-bounded and so in particular \([H:C_{H}(T_0)]\) is n-bounded. Since \(T_0\) acts coprimely on the abelian normal \(\pi \)-subgroup H, we have \(H=C_{H}(T_0)\times [H,T_0]\). Thus we obtain that the order of the commutator subgroup \([H,T_0]\) is n-bounded, as claimed. Let \(T_1=C_{T_0}([H,T_0])\) and remark that the index \([T_0:T_1]\) of \(T_1\) in \(T_0\) is n-bounded too. Set \(N=HT_1\). From the fact that the indexes \([T_0:T_1]\) and \([G:HT_0]\) are both n-bounded, we deduce that the index of N in G is n-bounded, as well.

Note that N is normal in G since the image of N in \(G/H\cong L\) is isomorphic to \(T_1\) which is normal in L. Furthermore, we have \([H,T_1,T_1]=1\), since \(T_1=C_{T_0}([H,T_0])\). Hence by the standard properties of coprime actions we have \([H,T_1]=1\) ( [7, Corollary 1.6.4]). Therefore \([H,N]=1\). This proves the theorem in the particular case where H is abelian.

In the general case, in view of Lemma 2.6, the commutator subgroup [H, H] is of n-bounded order. We pass to the quotient \({\overline{G}}=G/[H,H]\). The above argument shows that \({\overline{G}}\) has a normal subgroup \({\overline{N}}\) of n-bounded index such that \({\overline{H}}\le Z({\overline{N}})\). Here \(Z({\overline{N}})\) stands for the centre of \({\overline{N}}\). Let N be the inverse image of \({\overline{N}}\). We have \([H,N]=[H,H]\) and so N has the required properties. The proof is now complete. \(\square \)

3 Proof of Theorem 1.3

We will require the following result taken from [1, Lemma 4.1].

Lemma 3.1

Let G be a locally nilpotent group containing an element with finite centralizer. Suppose that G is residually finite. Then G is finite.

Profinite groups have Sylow p-subgroups and satisfy analogues of the Sylow theorems. Prosoluble groups satisfy analogues of the theorems on Hall \(\pi \)-subgroups. We refer the reader to the corresponding chapters in [10, Ch. 2] and [15, Ch. 2].

Recall that an automorphism \(\phi \) of a group G is called fixed-point-free if \(C_G(\phi )=1\), that is, the fixed-point subgroup is trivial. It is a well-known corollary of the classification of finite simple groups that if G is a finite group admitting a fixed-point-free automorphism, then G is soluble (see for example [12] for a short proof). A continuous automorphism \(\phi \) of a profinite group G is coprime if for any open \(\phi \)-invariant normal subgroup N of G the order of the automorphism of G/N induced by \(\phi \) is coprime to the order of G/N. It follows that if a profinite group G admits a coprime fixed-point-free automorphism, then G is prosoluble. This will be used in the proof of Theorem 1.3.

Proof of Theorem 1.3

Recall that \(\pi \) is a set of primes and G is a profinite group in which the centralizer of every \(\pi \)-element is either finite or open. We wish to show that G has an open subgroup of the form \(P\times Q\), where P is an abelian pro-\(\pi \) subgroup and Q is a pro-\(\pi '\) subgroup.

Let X be the set of \(\pi \)-elements in G. Consider first the case where the conjugacy class \(x^G\) is finite for any \(x\in X\). For each integer \(i\ge 1\) set

$$\begin{aligned} S_i=\{x\in X;\ |x^G|\le i\}. \end{aligned}$$

The sets \(S_i\) are closed. Thus, we have countably many sets which cover the closed set X. By the Baire Category Theorem [6, Theorem 34] at least one of these sets has non-empty interior. It follows that there is a positive integer k, an open normal subgroup M, and an element \(a\in X\) such that all elements in \(X\cap aM\) are contained in \(S_k\).

Note that \(\langle a^G\rangle \) has finite commutator subgroup, which we will denote by T. Indeed, the subgroup \(\langle a^G\rangle \) is generated by finitely many elements whose centralizer is open. This implies that the centre of \(\langle a^G\rangle \) has finite index in \(\langle a^G\rangle \), and by Schur’s theorem [11, 10.1.4], we conclude that T is finite, as claimed.

Let \(x\in X\cap M\). Note that the product ax is not necessarily in X. On the other hand, ax is a \(\pi \)-element modulo T. This is because \(\langle a^G\rangle \) becomes an abelian normal \(\pi \)-subgroup modulo T and the image of ax in the quotient \(G/\langle a^G\rangle \) is a \(\pi \)-element. In other words, there are \(y\in X\cap aM\) and \(t\in T\) such that \(ax=ty\). Observe that t has an open centralizer in G since \(t\in T\). In fact \([G:C_G(t)]\le |T|\). From the equality \(ax=ty\) deduce that \(|x^G|\le k^2|T|\). This happens for any \(x\in X\cap M\). Using a routine inverse limit argument in combination with Theorem 1.2 we obtain that M has an open normal subgroup N such that the index [M : N] and the order of [H, N] are finite. Here H stands for the subgroup generated by all \(\pi \)-elements of M. Choose an open normal subgroup U in G such that \(U\cap [H,N]=1\). Then \(U\cap M\) is an open normal subgroup of the form \(P\times Q\), where P is an abelian pro-\(\pi \) subgroup and Q is a pro-\(\pi '\) subgroup. This proves the theorem in the case where all \(\pi \)-elements of G have open centralizers.

Assume now that G has a \(\pi \)-element, say b, of infinite order. Since the procyclic subgroup \(\langle b\rangle \) is contained in the centralizer \(C_G(b)\), it follows that \(C_G(b)\) is open in G. This implies that all elements of \(X\cap C_G(b)\) have open centralizers (because they centralize the procyclic subgroup \(\langle b\rangle \)). In view of the above \(C_G(b)\) has an open subgroup of the form \(P\times Q\), where P is an abelian pro-\(\pi \) subgroup and Q is a pro-\(\pi '\) subgroup and we are done.

We will therefore assume that G is infinite while all \(\pi \)-elements of G have finite orders and there is at least one \(\pi \)-element, say d, such that \(C_G(d)\) is finite. The element d is a product of finitely many \(\pi \)-elements of prime power order. At least one of these elements must have finite centralizer. So without loss of generality we can assume that d is a p-element for a prime \(p\in \pi \).

Let \(P_0\) be a Sylow p-subgroup of G containing d. Since \(P_0\) is torsion, we deduce from Zelmanov’s theorem [16] that \(P_0\) is locally nilpotent. The centralizer \(C_G(d)\) is finite and so in view of Lemma 3.1 the subgroup \(P_0\) is finite. Choose an open normal pro-\(p'\) subgroup L such that \(L\cap C_G(d)=1\). Note that any finite homomorphic image of L admits a coprime fixed-point-free automorphism (induced by the coprime action of d on L). Hence L is prosoluble. Let K be a Hall \(\pi \)-subgroup of L. Since any element in K has restricted centralizer, Shalev’s result [13] shows that K is virtually abelian. We therefore can choose an open normal subgroup J in L such that \(J\cap K\) is abelian. If \(J\cap K\) is finite then G is virtually pro-\(\pi '\) and we are done. If \(J\cap K\) is infinite, then all \(\pi \)-elements of J have infinite centralizers. This yields that all \(\pi \)-elements of J have open centralizers in J and in view of the first part of the proof, J has an open normal subgroup of the form \(P\times Q\), where P is an abelian pro-\(\pi \) subgroup and Q is a pro-\(\pi '\) subgroup. This establishes the theorem. \(\square \)