Groups in which each subgroup is commensurable with a normal subgroup

TL;DR

This paper studies a special class of groups called cn-groups, showing that under certain conditions they are finite-by-abelian-by-finite, and provides detailed descriptions using automorphisms of abelian groups.

Contribution

It characterizes cn-groups with periodic images as finite-by-abelian-by-finite and analyzes their structure via automorphisms of abelian groups.

Findings

Cn-groups with locally finite periodic images are finite-by-abelian-by-finite.

Such groups with bounded index are also finite-by-abelian-by-finite.

Automorphisms of abelian groups help describe the structure of these groups.

Abstract

A group G is a cn-group if for each subgroup H of G there exists a normal subgroup N of G such that the index of both H and N in HN is finite. The class of cn-groups contains properly the classes of core- finite groups and that of groups in which each subgroup has finite index in a normal subgroup. In the present paper it is shown that a cn-group whose periodic images are locally finite is finite-by-abelian-by-finite. Such groups are then described into some details by considering automorphisms of abelian groups. Finally, it is shown that if G is a locally graded group with the property that the above index is bounded independently of H, then G is finite-by-abelian-by-finite.

Full text

In the present paper it is shown that a cn-group whose periodic images are locally finite is finite-by-abelian-by-finite. Such groups are then described into some details by considering automorphisms of abelian groups. Finally, it is shown that if $G$ is a locally graded group with the property that the above index is bounded independently of $H$, then $G$ is finite-by-abelian-by-finite.

to the memory of Jim Wiegold

Carlo Casolo, Ulderico Dardano, Silvana Rinauro

Key words and phrases: locally finite, core-finite, subnormal, inert, cf-group.

2010 Mathematics Subject Classification: Primary 20F24, Secondary 20F18, 20F50, 20E15

1 Introduction and main results

In a celebrated paper, B.H.Neumann [10] showed that for a group $G$ the property that each subgroup $H$ has finite index in a normal subgroup of $G$ (i.e., $|H^{G}:H|$ is finite) is equivalent to the fact that $G$ has finite derived subgroup ($G$ is finite-by-abelian).††margin:

††margin:

A class of groups with a dual property was considered in [1]. A group $G$ is said to be a cf-group (core-finite) if each subgroup $H$ contains a normal subgroup of $G$ with finite index in $H$ (i.e., $|H:H_{G}|$ is finite). As Tarski groups are cf, a complete classification of cf-groups seems to be much difficult. However, in [1] and [12] it has been proved that a cf-group $G$ whose periodic quotients are locally finite is abelian-by-finite and, if $G$ is periodic, there exists an integer $n$ such that $|H:H_{G}|\leq n$ for all $H\leq G$ (say that $G$* is bcf, boundedly cf*) and that a locally graded bcf-group is abelian-by-finite. Furthermore, an easy example of a metabelian (and even hypercentral) group which is cf but not bcf is given. It seems to be a still open question whether every locally graded cf-group is abelian-by-finite. Recall that a group is said to be abelian-by-finite if has an abelian subgroup with finite index and that a group is said to be locally finite (locally graded, respectively) if each non-trivial finitely generated subgroup is finite (has a proper subgroup with finite index, respectively).

With the aim of considering the above properties in a common framework, recall that two subgroups $H$ and $K$ of a group $G$ are said to be commensurable if $H\cap K$ has finite index in both $H$ and $K$. This is an equivalence relation and will be denoted by $\sim$. Clearly, if $H\sim K$, then $(H\cap L)\sim(K\cap L)$ and $HM\sim KM$ for each $L\leq G$ and $M\lhd G$.

Thus, in the present paper we consider the class of cn-groups, that is, groups in which each subgroup is commensurable with a normal subgroup. Into details, for a subgroup $H$ of a group $G$ define $\delta_{G}(H)$ to be the minimum index $|HN:(H\cap N)|$ with $N\lhd G$. Then $G$ is a cn-group if and only if $\delta_{G}(H)$ is finite for all $H\leq G$. Clearly, subgroups and quotients of cn-groups are also cn-groups.

Note that if a subgroup $H$ of a group $G$ is commensurable with a normal subgroup $N$, then $S:=(H\cap N)_{N}$ has finite index in $H$. Thus the class of cn-groups is contained in the class of sbyf-groups, that is, groups in which each subgroup $H$ contains a subnormal subgroup $S$ of $G$ such that the index $|H:S|$ is finite (i.e., $H$ is subnormal-by-finite). It is known that locally finite sbyf-groups are (locally nilpotent)-by-finite (see [7]) and nilpotent-by-Chernikov (see [3]).

The extension of a finite group by a cn-group is easily seen to be a cn-group, see Proposition 1.1 below. Moreover, from Proposition 9 in [4] it follows that for an abelian-by-finite group properties cn  and cf are equivalent. However, for each prime $p$ there is a nilpotent $p$-group with the property cn  which is neither finite-by-abelian nor abelian-by-finite, see Proposition 1.2.

Our main result is the following.

Theorem A *Let G be a cn-group such that every periodic image of $G$ is locally finite. Then $G$ is finite-by-abelian-by-finite. *

Here by a finite-by-abelian-by-finite group we mean a group which has a finite-by-abelian subgroup of finite index. The proof of Theorem A will be given in Sect. 3. The strategy of the proof will be to reduce to the case when $G$ is nilpotent and then to apply techniques of nilpotent groups theory. To this end, in Sect. 2, we will study the action of a cn-group on its abelian sections.

We will consider also bcn*-groups*, that is, groups $G$ for which there is $n\in{\mathbb{N}}$ such that $\delta_{G}(H)\leq n$ for all $H\leq G$ and prove the following theorem.

Theorem B *Let $G$ be a finite-by-abelian-by-finite group.

i) $G$ is cn if and only if it is finite-by-cf.

ii) $G$ is bcn if and only if it is finite-by-bcf. *

It follows that if the group $G$ is periodic and finite-by-abelian-by-finite, then $G$ is bcn if and only if it is cn. Then we consider non-periodic finite-by-abelian-by-finite bcf-groups in Proposition 3.2.

The more restrictive property bcn remains treatable when we consider the wider class of locally graded groups.

Theorem C *A locally graded bcn-group is finite-by-abelian-by-finite. *

Preliminaries

Our notation is mostly standard. For undefined terminology and basic facts we refer to [11]. If $\Gamma$ is a group acting on a group $G$ and $H\leq G$, we denote $H_{{\Gamma}}:=\cap_{\gamma\in{\Gamma}}H^{\gamma}$ and $H^{{\Gamma}}:={\langle}H^{\gamma}\ |\ \gamma\in{\Gamma}{\rangle}$. We say that $H$ is $\Gamma$-invariant (or a $\Gamma$-subgroup) if $H^{\Gamma}=H$.

We first point out a sufficient condition for a group to be cn (or even bcn) and give examples of non trivial cn-groups.

Proposition 1.1

Let $G$ be a group with a normal series $G_{0}\leq G_{1}\leq G$, where $G_{0}$ and $G/G_{1}$ have finite order, $m$ and $n$ respectively.

If $H\leq G$, then $H$ is commensurable with $H_{1}:=(H\cap G_{1})G_{0}\leq G_{1}$ and $\delta_{G}(H)\leq mn\cdot\delta_{G/G_{0}}(H_{1}/G_{0})$.

In particular, if each subgroup of $G_{1}/G_{0}$ is commensurable with a normal subgroup of $G/G_{0}$, then $G$ is a cn-group. $\square$

Proposition 1.2

For each prime $p$ there is a nilpotent $p$-group with the property bcn, which is neither abelian-by-finite nor finite-by-abelian.

**Proof.  **Consider a sequence $P_{n}$ of isomorphic groups with order $p^{4}$ defined by $P_{n}:={\langle}x_{n},y_{n}\ |\ x_{n}^{p^{3}}=y_{n}^{p}=1,\ x_{n}^{y_{n}}=x_{n}^{1+p^{2}}{\rangle}={\langle}x_{n}{\rangle}\rtimes{\langle}y_{n}{\rangle}$ where clearly $P_{n}^{\prime}={\langle}x_{n}^{p^{2}}{\rangle}$ has order $p$. Let $P:=\mathop{\rm Dr}_{n\in{\mathbb{N}}}P_{n}$ and consider the automorphism $\gamma$ of $P$ such that $x_{n}^{\gamma}=x_{n}^{1+p}$ and $y_{n}^{\gamma}=y_{n}$, for each $n\in{\mathbb{N}}$. Clearly, $\gamma$ has order $p^{2}$, acts as the automorphism $x\mapsto x^{1+p}$ on $P/P^{\prime}$ (which has exponent $p^{2}$) and acts trivially on $P^{\prime}$ (which is elementary abelian). Finally let $N:={\langle}x_{0}^{p^{2}}x_{n}^{p^{2}}\ |\ n\in{\mathbb{N}}{\rangle}$. Then $N$ is a $\gamma$-invariant subgroup of $P^{\prime}$ with index $p$. Thus the $p$-group $G:=(P\rtimes{\langle}\gamma{\rangle})/N$ is a bcn-group by Proposition 1.1 applied to the series $P^{\prime}/N\leq P/N\leq G$.

We have that $G^{\prime}$ is infinite, since for each $n$ we have $x_{n}^{p}=[x_{n},\gamma]\in[P_{n},\gamma]>P_{n}^{\prime}$. Moreover, we have that $gN\in Z(P/N)$ if and only if $\forall i\ [g,P_{i}]\leq N$, and $N\cap P_{i}=1$. Thus $Z(P/N)=Z(P)/N$ where $Z(P)=\mathop{\rm Dr}_{n}{\langle}x_{n}^{p}{\rangle}$ has infinite index in $P$.

If, by contradiction, $G$ is abelian-by-finite, then there is an abelian normal subgroup $A/N$ of $P/N$ with finite index. Then for some $m\in{\mathbb{N}}$ we have $P=AF$, where $F=\mathop{\rm Dr}_{n<m}P_{n}$ is a finite normal sugroup of $P$. Therefore $P/N$ is center-by-finite, a contradiction. $\square$

2 Automorphisms of abelian groups

Recall that an automorphism $\gamma$ of a group $A$ is said to be a power automorphism if $H^{\gamma}=H$ for each subgroup $H\leq A$. ††margin: It is well-known (see [11]) that, if $A$ is an abelian $p$-group, then there exists a $p$-adic integer $\alpha$ such that $a^{\gamma}=a^{\alpha}$ for all $a\in A$. Here $a^{\alpha}$ stands for $a^{n}$, where $n$ is any integer congruent to $\alpha$ modulo the order of $a$. On the other hand, a power automorphism of a non-periodic abelian group is either the identity or the inversion map.

As in [4], if $\Gamma$ is a group acting on an abelian group $A$, we consider the following properties:

 p)    $\forall H\leq A\ \ H=H^{\Gamma}$;

ap)  $\forall H\leq A\ \ |H:H_{\Gamma}|<\infty$;

bp)  $\forall H\leq A\ \ |H^{\Gamma}:H|<\infty$;

cp)  $\forall H\leq A\ \ \exists K=K^{\Gamma}\leq A$ such that $H\sim K$,  ($H$, $K$ are commensurable).

Obviously both ap and bp imply cp. Moreover, from Propositions 8 and 9 in [4] it follows that these three properties are equivalent, provided $A$ is abelian and ${\Gamma}$ is finitely generated, while they are in fact different in the general case even when $A$ and ${\Gamma}$ are elementary abelian $p$-groups. On the other hand, the properties ap and bp have been previously characterized in [6] and [2] respectively, as we are going to recall.

To shorten statements we define a further property: $\tilde{\textsc{p}}$)  ${\Gamma}$* has p on the factors of a ${\Gamma}$-series $1\leq V\leq D\leq A$ where

m i)   $V$ is free abelian of finite rank,

m ii) $D/V$ is divisible periodic with finite total rank,

m iii) $A/D$ is periodic and has finite $p$-exponent for each prime $p\in\pi(D/V)$. *

Theorem 2.1

[6],[2]*  Let $\Gamma$ be group acting on an abelian group $A$. Then:

a) ${\Gamma}$ has ap  on $A$ if and only if there is a ${\Gamma}$-subgroup $A_{1}$ such that $A/A_{1}$ is finite and ${\Gamma}$ has either p or $\tilde{\textsc{p}}$ on $A_{1}$.

b)   ${\Gamma}$ has bp  on $A$ if and only if there is a ${\Gamma}$-subgroup $A_{0}$ such that $A_{0}$ is  finite and ${\Gamma}$ has either p or $\tilde{\textsc{p}}$  on $A/A_{0}$.*

By next statement we give a characteration of the property cp  along the same lines.

Theorem 2.2

*Let $\Gamma$ be group acting on an abelian group $A$. Then:

c) $\Gamma$ has cp on $A$ if and only if there are ${\Gamma}$-subgroups $A_{0}\leq A_{1}\leq A$ such that $A_{0}$ and $A/A_{1}$ are finite and ${\Gamma}$ has either p or $\tilde{\textsc{p}}$  on $A_{1}/A_{0}$.*

The proof of Theorem 2.2 is at the end of this section. Here we deduce a corollary.

Corollary 2.3

*For a group $\Gamma$ acting on an abelian group $A$, the following are equivalent:

a) $\Gamma$ has ap on $A/A_{0}$ for a finite $\Gamma$-subgroup $A_{0}$ of $A$,

b) $\Gamma$ has bp on a finite index $\Gamma$-subgroup $A_{1}$ of $A$,

c) $\Gamma$ has cp on $A$. $\square$*

Let us state a couple of elementary basic facts.

Proposition 2.4

Let $\Gamma$ be group acting on a locally nilpotent periodic group $A$. Then $\Gamma$ has ap, bp, cp  on $A$, respectively, if and only if ${\Gamma}$ has ap, bp, cp  on finitely many primary components of $A$, respectively, and p on all the other ones.…

**Proof.  **This proof uses the same argument as in Proposition 4.1 in [5]. The sufficiency of the condition is clear once one notes that for each $H\leq A$ it results $H=\mathop{\rm Dr}_{p}(H\cap A_{p})$, where $A_{p}$ denotes the $p$-component of $A$. Concerning necessity, suppose ${\Gamma}$ does not have p on the primary $p$-component $A_{p}$ of $A$ for infinitely many primes $p$. Then for each such $p$ there is $H_{p}\leq A_{p}$ which is not ${\Gamma}$-invariant. We have that the subgroup generated by the $H_{p}$’s is not commensurable to any ${\Gamma}$-subgroup. $\square$

Lemma 2.5

*Let $\Gamma$ be a group acting on an abelian group $A$. If ${\Gamma}$ has cp on $A$, then:

i)  ${\Gamma}$ has p on the largest periodic divisible subgroup of $A$.

ii) if $A$ is torsion-free, then each ${\gamma}\in{\Gamma}$ acts on $A$ by either the identity or the inversion map.*

**Proof.  **Statement ($i$) follows from Lemma 4.3 in [5]. Concerning ($ii$), by Propositions 3.3 and 3.2 of [5] we have that there are coprime non-zero integers $n,m$ such that $a^{m}=(a^{\gamma})^{n}$ for each $a\in A$. Consider $H$ such that $1\neq H:={\langle}a_{0}{\rangle}\leq A$. Then there is a ${\Gamma}$-invariant subgroup $K$ of $A$ which is commensurable with $H$. Thus there is $r\in{\mathbb{N}}$ such that $K^{r}$ is a ${\Gamma}$-invariant nontrivial subgroup of $H$. This forces $mn=\pm 1$. $\square$

Now we prove some lemmas. In the first one we do not require that the group $A$ is abelian.

Lemma 2.6

Let $\Gamma$ be a group acting on a fc-group $A$. If ${\Gamma}$ has cp on $A$, then ${\Gamma}$ has bp on the subgroup $X:=\{\,a\in A\,|\,{\langle}a{\rangle}^{\Gamma}\,{\rm is\ finite}\}$ of $A$.

**Proof.  **Notice that $X$ is the set of elements $a$ of finite order of $A$ such that $|{\Gamma}:C_{{\Gamma}}(a)|$ is finite, so $X$ is a locally finite ${\Gamma}$-subgroup of $A$. For any $H\leq X$ there is $K\leq X$ such that $H\sim K=K^{\Gamma}\leq A$. Then there is a finite subgroup $F\leq X$ such that $H\leq KF$. Thus $H^{\Gamma}\leq KF^{\Gamma}$ and $|H^{\Gamma}:H|\leq|F^{\Gamma}|\cdot|HK:H|$ is finite. $\square$

Lemma 2.7

Let $\Gamma$ be a group acting on a $p$-group $A$ which is the direct product of cyclic groups. If ${\Gamma}$ has cp on $A$, then the subgroup $X:=\{\,a\in A\,|\,{\langle}a{\rangle}^{\Gamma}\,{\rm is\ finite}\}$ has finite index in $A$.

**Proof.  **Assume by contradiction that $A/X$ is infinite.

Let us see, by elementary facts, that there is a sequence $(a_{n})$ of elements of $A$ such that ††margin:

$1$)x ${\langle}a_{n}|n\in{\mathbb{N}}{\rangle}=\mathop{\rm Dr}_{n\in{\mathbb{N}}}{\langle}a_{n}{\rangle}$,

$2$)x$A_{I}/A_{I}\cap X$ is infinite, for each infinite subset $I$ of ${\mathbb{N}}$, where $A_{I}:={\langle}a_{n}|n\in I{\rangle}$.

In fact, if $A/X$ has finite rank, it has a Prüfer subgroup $Q/X$. ††margin: Let $Y$ be a countable subgroup of $A$ such that $Q=YX$. By Kulikov Theorem (see [11]) $Y$ is the direct product of cyclic groups, so that we may choose elements $a_{n}\in Y$ such that ${\langle}a_{n}|n\in{\mathbb{N}}{\rangle}=\mathop{\rm Dr}_{n\in{\mathbb{N}}}{\langle}a_{n}{\rangle}\leq Y$ and $|a_{n}X|<|a_{n+1}X|$. The claim holds. Similarly, if $A/X$ has infinite rank, consider a countably infinite subgroup $Q/X$ of the socle of $A/X$. As above, let $Y$ be a countable subgroup of $A$ such that $Q=YX$. Then we may choose elements $a_{n}\in Y$ which are independent mod $X$ and generate their direct product as claimed.

We claim now that *there are sequences of infinite subsets $I_{n}$, $J_{n}$ of ${\mathbb{N}}$ and ${\Gamma}$-subgroups $K_{n}\leq A$ such that for each $n\in{\mathbb{N}}$:

$3)$ xx $I_{n}\cap J_{n}=\emptyset$ and $I_{n+1}\subseteq J_{n}$

$4)$ xx $K_{n}\sim A_{I_{n}}$

$5)$ xx $(K_{1}\ldots K_{i})\cap(A_{I_{1}}\ldots A_{I_{n}})\leq(A_{I_{1}}\ldots A_{I_{i}})$, $\forall i\leq n$. *

To prove the claim, proceed by induction on $n$. Choose an infinite subset $I_{1}$ of ${\mathbb{N}}$ such that $J_{1}:={\mathbb{N}}\setminus I_{1}$ is infinite. By cp-property there exists $K_{1}=K_{1}^{\Gamma}$ commensurable with $A_{I_{1}}$.

Suppose we have defined $I_{j}$, $J_{j}$ and $K_{j}$ for $1\leq j\leq n$ such that 3-5 hold. Since $(K_{1}\ldots K_{n})\sim(A_{I_{1}}\ldots A_{I_{n}})$, there is $m\in{\mathbb{N}}$ such that

6. xx $(K_{1}K_{2}\ldots K_{n})\cap A_{{\mathbb{N}}}\leq(A_{I_{1}}A_{I_{2}}\ldots A_{I_{n}}){\langle}a_{1},\ldots,a_{m}{\rangle}$.

Let $I_{n+1}$ and $J_{n+1}$ be disjoint infinite subsets of $J_{n}\setminus\{1,\dots,m\}$. By cp-property there exists $K_{n+1}=K_{n+1}^{\Gamma}$ commensurable with $A_{I_{n+1}}$. By the choice of $I_{n+1}$ it follows that

7. xx $(K_{1}\ldots K_{i})\cap(A_{I_{1}}\ldots A_{I_{n+1}})\leq(K_{1}\ldots K_{i})\cap(A_{I_{1}}\ldots A_{I_{n}})\ \ \forall i\leq n$

and so (5) holds for $n+1$, as whished. The claim is now proved.

Note that by (2) and (5) it follows that $A_{I_{n}}/A_{I_{n}}\cap X$ is infinite for each $n\in{\mathbb{N}}$ and that also the following property holds

8. xx $(K_{1}K_{2}\ldots K_{n})\cap\bar{A}\leq(A_{I_{1}}A_{I_{2}}\ldots A_{I_{n}})\ \forall n$, where $\bar{A}:=\mathop{\rm Dr}_{n\in{\mathbb{N}}}A_{I_{n}}$.

Now for each $n\in{\mathbb{N}}$, choose an element $b_{n}\in(A_{I_{n}}\cap K_{n})\setminus X$. Then we have $B:={\langle}b_{n}\ |\ n\in{\mathbb{N}}{\rangle}=\mathop{\rm Dr}_{n}{\langle}b_{n}{\rangle}$, where ${\langle}b_{n}{\rangle}^{\Gamma}$ is infinite and ${\langle}b_{n}{\rangle}^{\Gamma}\leq K_{n}\sim A_{I_{n}}$, so that

9. xx ${\langle}b_{n}{\rangle}^{\Gamma}\cap A_{I_{n}}$ is infinite for each $n$.

Since there exists $B_{0}=B_{0}^{\Gamma}\sim B$, we may take

• $B_{*}:=(B_{0}\cap B)^{\Gamma}=(B_{*}\cap B)^{\Gamma}\leq B^{\Gamma}$ where $B_{*}\sim B$.

Now $B_{*}/(B_{*}\cap B)$ and $B/(B_{*}\cap B)$ are both finite and there is $n\in{\mathbb{N}}$ such that if $B_{n}:={\langle}b_{1},\ldots,b_{n}{\rangle}$ we have

• $(B_{*}\cap B)^{\Gamma}=B_{*}\leq(B_{*}\cap B)B_{n}^{\Gamma}$  and

• $B=(B_{*}\cap B)B_{n}$.

Since $b_{n}\in K_{n}$ for each $n$, we have $B_{n}\leq\bar{K}_{n}:=K_{1}K_{2}\ldots K_{{n}}$ and

• $B^{\Gamma}=(B_{*}\cap B)^{\Gamma}B_{n}^{\Gamma}\leq(B_{*}\cap B)B_{n}^{\Gamma}\leq(B_{*}\cap B)\bar{K}_{n}\leq B\bar{K}_{n}$, so that

• $B^{\Gamma}\cap\bar{A}\leq B\bar{K}_{n}\cap\bar{A}=B(\bar{K}_{n}\cap\bar{A})\leq BA_{I_{1}}A_{I_{2}}\ldots A_{I_{n}}$ by (8) above.

Thus

${\langle}b_{n+1}{\rangle}^{\Gamma}\cap A_{I_{n+1}}\leq B^{\Gamma}\cap A_{I_{n+1}}\ \leq(BA_{I_{1}}A_{I_{2}}\ldots A_{I_{n}})\cap A_{I_{n+1}}={\langle}b_{n+1}{\rangle}$ is finite, a contradiction with (9). $\square$

Lemma 2.8

Let $\Gamma$ be a group acting on an abelian periodic reduced group $A$. If ${\Gamma}$ has cp on $A$, then there are ${\Gamma}$-subgroups $A_{0}\leq A_{1}\leq A$ such that $A_{0}$ and $A/A_{1}$ are finite and ${\Gamma}$ has p on $A_{1}/A_{0}$.

**Proof.  **By Proposition 2.4 it is enough to consider the case when $A$ is a $p$-group. If $A$ is the direct product of cyclic groups, by Lemma 2.7 we have that $A_{1}:=\{\,a\in A\,|\,{\langle}a{\rangle}^{\Gamma}\,{\rm is\ finite}\}$ has finite index in $A$. Further, by Lemma 2.6, ${\Gamma}$ has bp on $A_{1}$. Then the statement follows from Theorem 2.1.

Let $A$ be any reduced $p$-group and $B_{*}$ be a basic subgroup of $A$. Then there is $B=B^{\Gamma}\sim B_{*}$. Since $A/B_{*}$ is divisible, then the divisible radical of $A/B$ has finite index. Thus we may assume that $A/B$ is divisible. By Kulikov Theorem (see [11]), also $B$ is a direct product of cyclic groups, therefore by the above there are $\Gamma$-subgroups $B_{0}\leq B_{1}\leq B$ such that $B_{0}$ and $B/B_{1}$ are finite and ${\Gamma}$ has p on $B_{1}/B_{0}$. We may assume $B_{0}=1$. Also, since $A/B_{1}$ is finite-by divisible, it is divisible-by-finite and we may assume it is divisible.

Let ${\gamma}\in{\Gamma}$ and $\alpha$ be a $p$-adic integer such that $x^{\gamma}=x^{\alpha}$ for all $x\in B_{1}$. Consider the endomorphism ${\gamma}-\alpha$ of $A$ and note that $B_{1}\leq\ker({\gamma}-\alpha)$. Thus $A/\ker({\gamma}-\alpha)\simeq\mbox{\rm im}({\gamma}-\alpha)$ is both divisible and reduced, hence trivial. It follows ${\gamma}=\alpha$ on the whole $A$.. $\square$

**Proof of Theorem 2.2 ** For the sufficiency of the condition note that for any subgroup $H\leq A$ we have $H\sim H\cap A_{1}$ and the latter is in turn commensurable with a ${\Gamma}$-subgroup since ${\Gamma}$ has bp on $A_{1}$ by Theorem 2.1.

Concerning necessity, we first prove the statement when $A$ is periodic. Let $A=D\times R_{1}$, where $D$ is divisible and $R_{1}$ is reduced. Then there is a subgroup $R=R^{\Gamma}\sim R_{1}$. Thus $DR$ and $D\cap R$ are ${\Gamma}$-subgroups of $A$ with finite index and order respectively. Then we can assume $A=D\times R$. Let $X:=\{\,a\in A\,|\,{\langle}a{\rangle}^{\Gamma}\,{\rm is\ finite}\}$. Clearly $D\leq X$, as ${\Gamma}$ has p on $D$ by Lemma 2.5. On the other hand, $X\cap R$ has finite index in $R$ by Lemma 2.8. It follows $A/X$ is finite and by Lemma 2.6 and Theorem 2.1 the statement holds.

In the non-periodic case, note that if $V_{0}$ is a free subgroup of $A$ such that $A/V_{0}$ is periodic, then there is $V_{1}=V_{1}^{\Gamma}\sim V_{0}$. Let $n:=|V_{1}/(V_{0}\cap V_{1})|$. Thus by applying Lemma 2.5 to the ${\Gamma}$-subgroup $V:=V_{1}^{n}$ we have

• there is a free abelian ${\Gamma}$-subgroup $V$ such that $A/V$ is periodic and each ${\gamma}\in\Gamma$ acts on $V$ by either the identity or the inversion map.

Suppose that $V$ has finite rank. Consider now the action of ${\Gamma}$ on the periodic group $A/V$ and apply the above. Then there is a series $V\leq A_{0}\leq A_{1}\leq A$ such that $A_{0}/V$ and $A/A_{1}$ are finite and ${\Gamma}$ has either p or $\tilde{\textsc{p}}$ on $A_{1}/A_{0}$. Since $A_{0}$ has finite torsion subgroup $T$ we can factor out $T$ and assume $A_{0}=V$. Then ${\Gamma}$ has either p or $\tilde{\textsc{p}}$ on $A_{1}$ as straightforward verification shows.

Suppose finally that $V$ has infinite rank. Let $V_{2}\leq V$ be such that $V/V_{2}$ is divisible periodic and its $p$-component has infinite rank for each prime $p$. We may assume $V:=V_{2}$. By the above case when $A$ is periodic, there is a ${\Gamma}$-series $V\leq A_{0}\leq A_{1}\leq A$ such that $A_{0}/V$ and $A/A_{1}$ are finite and ${\Gamma}$ has p on $A_{1}/A_{0}$. We may factor out the torsion subgroup of $A_{0}$, as it is finite, and assume $A_{0}=V$.

Again let $V_{2}\leq V$ be such that $V/V_{2}$ is divisible periodic and its $p$-component has infinite rank for each prime $p$. Let ${\gamma}\in{\Gamma}$ and, for each prime $p$, let $\alpha_{p}$ be a $p$-adic integer such that $x^{\gamma}=x^{\alpha_{p}}$ for all $x$ in the $p$-component of $A_{1}/V$. Let $\epsilon=\pm 1$ be such that $x^{\gamma}=x^{\epsilon}$ for all $x\in V$. By Lemma 2.5, $\gamma$ has p on the maximum divisible subgroup $D_{p}/V_{2}$ of the $p$-component of $A_{1}/V_{2}$. Thus $\alpha_{p}=\epsilon$ on $D_{p}/V_{2}$. Therefore $x^{\gamma}=x^{\epsilon}$ for all $x\in V$ and for all $x\in A_{1}/V$. We claim that $a^{\gamma}=a^{\epsilon}$ for each $a\in A_{1}$. To see this, for any $a\in A_{1}$ consider $n\in{\mathbb{N}}$ such that $a^{n}\in V$. Then there is $v\in V$ such that $a^{\gamma}=a^{\epsilon}v$. Hence $a^{n\epsilon}=(a^{n})^{\gamma}=(a^{\gamma})^{n}=(a^{\epsilon}v)^{n}=a^{n\epsilon}v^{n}$. Thus $v^{n}=1$. Therefore, as $V$ is torsion-free, we have $v=1$, as whished. $\square$

3 Proofs of the Theorems

Recall that locally finite cf-groups are abelian-by-finite and bcf (see [1]).

Proof of Theorem B It follows from Proposition 1.1 and Proposition 3.1 below. $\square$

Proposition 3.1

*Let $G$ be an abelian-by-finite group.

i)  if $G$ is cn, then $G$ is cf;

ii) if $G$ is bcn, then $G$ is bcf.*

**Proof.  **Let $A$ be a normal abelian subgroup with finite index $r$. Then each $H\leq A$ has at most $r$ conjugates in $G$. If $\delta_{G}(H)\leq n<\infty$ then for each $g\in G$ we have $|H:(H\cap H^{g})|\leq 2\delta_{G}(H)\leq 2n$ hence $|H/H_{G}|\leq(2n)^{r}$. More generally, if $H$ is any subgroup of $G$, then $|H/H_{G}|\leq r(2n)^{r}$. $\square$

Let us characterize bcf-groups among abelian-by-finite cf-groups.

Proposition 3.2

*Let $G$ be a non-periodic group with an abelian normal subgroup $A$ with finite index. Then the following are equivalent:

i) $G$ is a bcf-group;

ii) $G$ is a cf-group and there is $B\leq A$ such that $B$ has finite exponent, $B\lhd G$ and each $g\in G$ acts by conjugation on $A/B$ by either the identity or the inversion map.*

**Proof.  **Let $T$ be the torsion subgroup of $A$. By Lemma 2.5, for each $g\in G$ there $\epsilon_{g}=\pm 1$ such that ${\gamma}$ acts on $A/T$ as the automorphism $x\mapsto x^{\epsilon_{g}}$. Then the equivalence of (i) and (ii) holds with $B:=\langle A^{g-\epsilon_{g}}\ |\ g\in G\rangle$, by Theorem 3 of [4]. $\square$

To prove Theorem A, our first step is a reduction to nilpotent groups.

Lemma 3.3

A soluble $p$-group $G$ with the property cn is nilpotent-by-finite.

**Proof.  **By Theorem 2.2, one may refine the derived series of $G$ to a finite $G$-series $\cal S$ such that $G$ has p on each infinite factor of $\cal S$. Recall that a $p$-group of power automorphisms of an abelian $p$-group is finite (see [11]). Then the stability group $S\leq G$ of the series $\cal S$, that is, the intersection of the centralizers in $G$ of the factors of the series, has finite index in $G$. On the other hand, by a theorem of Ph.Hall, $S$ is nilpotent. $\square$

We recall now an elementary property of nilpotent groups.

Lemma 3.4

Let $G$ be a nilpotent group with class $c$. If $G^{\prime}$ has finite exponent $e$, then $G/Z(G)$ has finite exponent dividing $e^{c}$.

**Proof.  **Argue by induction on $c$, the statement being clear for $c=1$. Assume $c>1$ and that $G/Z$ has exponent dividing $e^{c-1}$, where $Z/\gamma_{c}(G):=Z(G/\gamma_{c}(G))$. Then for all $g,x\in G$ we have $[g^{e^{c-1}},x]\in\gamma_{c}(G)\leq G^{\prime}\cap Z(G).$ Therefore $1=[g^{e^{c-1}},x]^{e}=[g^{e^{c}},x],$ and $g^{e^{c}}\in Z(G)$, as claimed. $\square$

Next lemma follows easily from Lemma 6 in [9].

Lemma 3.5

Let $G$ be a nilpotent p-group and $N$ a normal subgroup such that $G/N$ is an infinite elementary abelian group. If $H$ and $U$ are finite subgroup of $G$ such that  ${H}\cap U=1$, there exists a subgroup $V$ of $G$ such that $U\leq V$, ${H}\cap V=1$ and $VN/N$ is infinite. $\square$

We deduce a technical lemma which is a tool for our pourpose.

Lemma 3.6

Let $G$ be a nilpotent p-group and $N$ be a normal ††margin: subgroup such that $G/N$ is an infinite elementary abelian group. If $N$ contains the fc-center of $G$ and $G^{\prime}$ is abelian with finite exponent, then there are subgroups $H$, $U$ of $G$ such that $H\cap U=1$, with injective maps $n\mapsto h_{n}\in H$ and $(i,n)\mapsto u_{i,n}\in[G,h_{i}^{-1}h_{n}]\cap U$, where $i,n\in{\mathbb{N}}$, $i<n$.

**Proof.  **Let ††margin: us show that for each $n\in N$ there is an $(n+1)$-uple $v_{n}:=(h_{n},u_{0,n},u_{1,n},\dots,u_{n-1,n})$ of elements of $G$ such that:

$\{h_{1},\dots,h_{n}\}$ is linearly independent modulo $N$; 2. 2.

$u_{i,n}\in[G,h_{i}^{-1}h_{n}]$   $\forall i\in\{0,\dots,n-1\}$; 3. 3.

$\{u_{j,h}\ |\ 0\leq j<k\leq n\}$ is ${\mathbb{Z}}$-independent in $G^{\prime}$; 4. 4.

$H_{n}\cap U_{n}=1$, where $H_{n}:={\langle}h_{1},\dots,h_{n}{\rangle}$ and $U_{n}:={\langle}u_{j,h}\ |\ 0\leq j<k\leq n{\rangle}$.

Then the statement is true for $H:=\bigcup\limits_{n\in{\mathbb{N}}}H_{n}$ and $U:=\bigcup\limits_{n\in{\mathbb{N}}}U_{n}$.

Let $h_{0}:=1$ and choose $h_{1}\in G\setminus N$. Since $N$ contains the fc-center $F$ of $G$, we have that $h_{1}$ has an infinite numbers of coniugates in $G$, hence $[G,h_{1}]$ is infinite and residually finite. Thus we may choose $u_{0,1}\in[G,h_{1}]$ such that ${\langle}u_{0,1}{\rangle}\cap{\langle}h_{1}{\rangle}=1$.

Assume then that we have defined $v_{i}$ for $i\leq n$, that is, we have elements $h_{0},\dots,h_{n}$, $u_{j,k}$, with $0\leq j<k\leq n$ such that conditions 1-4 hold. To define an adequate $v_{n+1}$, note that by Lemma 3.5 we have that there exists $V_{n}\leq G$ such that $H_{n}\leq V_{n}$, $U_{n}\cap V_{n}=1$ and $V_{n}N/N$ is infinite. Then choose

$i)\phantom{xxx}h_{n+1}\in V_{n}\setminus NU_{n}H_{n}.$

Note that $h_{n+1}\not\in FH_{n}\leq NH_{n}$, so that $\{h_{1},\dots,h_{n+1}\}$ is independent mod $F$. In particular $\forall i\in\{0,\dots,n\}$, $h_{i}^{-1}h_{n+1}\not\in F$, hence also $[G,h_{i}^{-1}h_{n+1}]$ is infinite. Since $G^{\prime}$ is residually finite, we may recursively choose $u_{0,n+1},\dots,u_{n,n+1}$ such that $\forall i\in\{0,\dots,n\}$

$ii)$ xxx $\ u_{i,n}\in[G,h_{i}^{-1}h_{n}]$

$iii)$ xxx ${\langle}u_{i,n+1}{\rangle}\cap U_{n}{\langle}u_{h,n+1}\ |\ 0\leq h<i{\rangle}H_{n+1}=1$

Then properties 1-3 hold for $v_{n+1}$. Finally suppose there are $h\in H_{n}$, $u\in U_{n}$, $s,t_{0},\dots,t_{n}\in{\mathbb{Z}}$ such that

$iv)$ xxx $a=hh_{n+1}^{s}=uu_{0,n+1}^{t_{1}}\cdots u_{n,n+1}^{t_{n}}\in H_{n+1}\cap U_{n+1}.$

Then from $(iii)$ it follows $u_{n,n+1}^{t_{n}}=\ldots=u_{0,n+1}^{t_{1}}=1$. Hence $a=hh_{n+1}^{s}=u\in V_{n}\cap U_{n}=1$ and 4 holds. $\square$

Lemma 3.7

Let $G$ be a nilpotent $p$-group. If $G$ is cn, then $G^{\prime}$ has finite exponent.

**Proof.  **If, by contradiction, $G^{\prime}$ has infinite exponent, then the same happens to the abelian group $G^{\prime}/\gamma_{3}(G)$ and there is $N$ such that $G^{\prime}\geq N\geq\gamma_{3}(G)$ and $G^{\prime}/N$ is a Prüfer group. We may assume $N=1$, that is, $G^{\prime}$ itself is a Prüfer group and $G^{\prime}\leq Z(G)$. Let us show that for any $H\leq G$ we have $|H^{G}:H|<\infty$, hence $G^{\prime}$ is finite, a contradiction. In fact we have that, by the cn-property, there is $K\lhd G$ such that $K\sim H$. Thus $H$ has finite index in $HK$ and we can also assume $H=HK$, that is, $H/H_{G}$ is finite. Thus, we can assume $H_{G}=1$ and $H\cap G^{\prime}=1$, that is, $H$ is finite with order $p^{n}$ and $HG^{\prime}$ is an abelian Chernikov group. It follows that $H$ is contained in the $n$-th socle $S$ of $HG^{\prime}\lhd G$, where $S$ is finite and normal in $G$, as whished. $\square$

Lemma 3.8

Let $G$ be a nilpotent $p$-group. If $G$ is cn, then $G$ is finite-by-abelian-by-finite.

**Proof.  **Let $G$ be a counterexample. Then both $G^{\prime}$ and $G/Z(G)$ are infinite. However, they have finite exponent by Lemmas 3.7 and 3.4. Moreover, by Lemma 2.6, the fc-center $F$ of $G$ is finite-by-abelian. Thus $F$ has infinite index in $G$. On the other hand, $G/F$ has finite exponent, since $F\geq Z(G)$.

Then $N:=FG^{p}G^{\prime}$ has infinite index in $G$, otherwise the abelian group $G/FG^{\prime}$ has finite rank and finite exponent, hence it is finite. This implies that the nilpotent group $G/F$ is finite, a contradiction.

If $G^{\prime}$ is abelian we are in a condition to apply Lemma 3.6 and get infinite elements and subgroups $h_{n}\in H$, $u_{i,n}\in U$ as in that statement. By cn-property there is $K$ such that $H\sim K\lhd G$. So that the set $\{h_{n}(H\cap K)\ /\ n\in{\mathbb{N}}\}$ is finite. Hence there is $i\in{\mathbb{N}}$ and an infinite set $I\subseteq{\mathbb{N}}\setminus\{1,\ldots,i\}$ such that for each $n\in I$ we have $h_{i}^{-1}h_{n}\in H\cap K$ and $u_{i,n}\in U\cap[G,H\cap K]\leq U\cap K$. Therefore $U\cap K$ is infinite, in contradiction with $U\cap K\sim U\cap H=1$.

For the general case, proceed by induction on the nilpotency class $c>1$ of $G$ and assume that the statement is true for $G/Z(G)$ and even that this is finite-by-abelian. Then there is a subgroup $L\leq G$ such that $G/L$ is abelian and $L/Z(G)$ is finite. Thus $L^{\prime}$ is finite and, by the above, $G/L^{\prime}$ is finite-by-abelian-by-finite, a contradiction. $\square$

Let us consider now non-periodic cn-groups.

Lemma 3.9

Let $G$ be a cn-group and $A=A(G)$ its subgroup generated by all infinite cyclic normal subgroups. Then $G/A$ is periodic, $A$ is abelian and each $g\in G$ acts on $A$ by either the identity or the inversion map, hence $|G/C_{G}(A)|\leq 2$.

**Proof.  **For any $x\in G$ there is $N\lhd G$ which is commensurable with ${\langle}x{\rangle}$. Then $n:=|N:(N\cap{\langle}x{\rangle})|$ is finite. Thus $N^{n!}\leq{\langle}x{\rangle}$ where $N^{n!}\lhd G$. Hence $G/A$ is periodic.

It is clear that $A$ is abelian. Let $g\in G$. If ${\langle}a{\rangle}\lhd G$ and $a$ has infinite order, then there is $\epsilon_{a}=\pm 1$ such that $a^{g}=a^{\epsilon_{a}}$. On the other hand, by Lemma 2.5, there is $\epsilon=\pm 1$ such that for each $a\in A$ there is a periodic element $t_{a}\in A$ such that $a^{g}=a^{\epsilon}t_{a}$. It follows $a^{\epsilon_{a}-\epsilon}=t$. Therefore $\epsilon_{a}=\epsilon$ is independent of $a$, as wished. $\square$

Proof of Theorem A. Recall from the Introduction that all subgroups of $G$ are subnormal-by-finite. If $G$ is periodic, then, by above quoted results in [7] and [3] respectively, we may assume that $G$ is locally nilpotent and soluble. Then, by Proposition 2.4, only finitely many primary components are non-abelian. Thus we may assume $G$ is a $p$-group and apply Lemma 3.3 and Lemma 3.8. It follows that $G$ is finite-by-abelian-by-finite.

To treat the general case, consider $A=A(G)$ as in Lemma 3.9. We may assume $A$ is central in $G$. Let $V$ be a torsion-free subgroup of $A$ such that $A/V$ is periodic. Then $G/V$ is locally finite and we may apply the above. Thus there is a series $V\leq G_{0}\leq G_{1}\leq G$ such that $G$ acts trivially on $V$, $G_{1}/G_{0}$ is abelian, while $G_{0}/V$ and $G/G_{1}$ are finite. Then we can assume $G=G_{1}$ and note that the stabilizer $S$ of the series has finite index. Since $S$ is nilpotent (by Ph.Hall Theorem) we can assume that $G=S$ is nilpotent. If $T$ is the torsion subgroup of $G$, then $VT/T$ is contained in the center of $G/T$. Since all factors of the upper central series of $G/T$ are torsion-free we have $G/T$ is abelian. Thus $G^{\prime}\leq T\cap G_{0}$ is finite. $\square$

Proof of Theorem C. If the statement is false, by Theorem A we may assume there is a counterexample $G$ periodic and not locally finite. Also we may assume $G$ is finitely generated and infinite. Let $R$ be the locally finite radical of $G$. By Theorem A again, $R$ is finite-by-abelian-by-finite. By Theorem B(i), there is a finite subgroup $G_{0}\lhd G$ such that $R/G_{0}$ is abelian-by-finite. We may assume $G_{0}=1$, so that $R$ is abelian-by-finite.

We claim that $\bar{G}:=G/R$ has finite exponent at most $(n+1)!$ where $n$ is such that $n\geq\delta_{G}(H)$ for each $H\leq G$ . In fact, for each $x\in\bar{G}$, there is $\bar{N}\lhd\bar{G}$ such that $|\bar{N}:(\bar{N}\cap{\langle}x{\rangle})|\leq n$. Thus $\bar{N}^{n!}\leq{\langle}x{\rangle}$ and $\bar{N}^{n!}\lhd G$. Hence $\bar{N}^{n!}=1$ and $x^{n\cdot n!}=1$.

By the positive answer (for all exponents) to the Restricted Burnside Problem, there is a positive integer $k$ such that every finite image of $\bar{G}$ has order at most $k$. Since $\bar{G}$ is finitely generated, this means that the finite residual $\bar{K}$ of $\bar{G}$ has finite index and is finitely generated as well. Since also $\bar{G}$ is locally graded (see [8]), we have $\bar{K}=1$ and $\bar{G}$ is finite. Therefore $G$ is abelian-by-finite, a contradiction. $\square$