On groups where the twisted conjugacy class of the unit element is a subgroup

TL;DR

This paper investigates groups where the conjugacy class of the identity element under any automorphism forms a subgroup, revealing structural properties like nilpotency and bounds on nilpotency class.

Contribution

It proves that such groups with finitely generated conditions have bounded verbal width in their lower central series and are nilpotent if they satisfy the descending chain condition.

Findings

Lower central series members have finite verbal width bounded by generators and series index.

Groups satisfying the descending chain condition are proven to be nilpotent.

For finite abelian-by-cyclic groups, an upper bound on the nilpotency class is established.

Abstract

We study groups $G$ where the $\varphi$-conjugacy class $[e]_{\varphi}=\{g^{-1}\varphi(g)~|~g\in G\}$ of the unit element is a subgroup of $G$ for every automorphism $\varphi$ of $G$. If $G$ has $n$ generators, then we prove that the $k$-th member of the lower central series has a finite verbal width bounded in terms of $n,k$. Moreover, we prove that if such group $G$ satisfies the descending chain condition for normal subgroups, then $G$ is nilpotent. Finally, if $G$ is a finite abelian-by-cyclic group, we construct a good upper bound of the nilpotency class of $G$.

Equations

$$[e{]}_g=\{[x,g]\text{leavevmode}|\text{leavevmode}x\in G\}$$

$$[e{]}_{g_1}>[e{]}_{[g_1,g_2]}>[e{]}_{[g_1,g_2,g_3]}>\dots$$

$$[xy,z]$$

$$[x,yz]$$

$$[x,y{]}^z$$

$$[x,y]$$

$$[x,y^{-1}]={\left([x,y{]}^{y^{-1}}\right)}^{-1}$$

$$[y_i,h_i]=[x_1,h_i{]}^{a_{1i}}[x_2,h_i{]}^{a_{2i}}\dots [x_n,h_i{]}^{a_{ni}}[x_1^{-1},h_i{]}^{b_{1i}}[x_2^{-1},h_i{]}^{b_{2i}}\dots [x_n^{-1},h_i{]}^{b_{ni}}$$

$$x=\prod _{\begin{array}{c}i=1,\dots ,n\\ j=1,\dots ,s\end{array}}{p}_{ij}^{a_{ij}}{q}_{ij}^{b_{ij}}$$

$$[t,x]=\bigl{[}t,\prod_{i,j}p_{ij}^{a_{ij}}q_{ij}^{b_{ij}}\bigr{]}=\prod_{i,j}\bigl{[}t,p_{ij}^{a_{ij}}\bigr{]}^{c_{ij}}\bigl{[}t,q_{ij}^{b_{ij}}\bigr{]}^{d_{ij}}=\prod_{i,j}\bigl{[}t^{a_{ij}^{-1}},p_{ij}\bigr{]}^{a_{ij}c_{ij}}\bigl{[}t^{b_{ij}^{-1}},q_{ij}\bigr{]}^{b_{ij}d_{ij}}$$

$$[x^{-1},y]={\left([x,y{]}^{x^{-1}}\right)}^{-1}$$

$$[z,q_{ij}]=[z,[x_i^{-1},h_j]]=\left[z,{\left([x_i,h_j{]}^{x_i^{-1}}\right)}^{-1}\right]={\left[z^{x_i},p_{ij}^{-1}\right]}^{x_i^{-1}}={\left({\left[z^{x_i},p_{ij}\right]}^{p_{ij}^{-1}{x}_i^{-1}}\right)}^{-1},$$

$${\gamma }_3(G)=\prod _{\begin{array}{c}i=1,\dots ,n\\ j=1,\dots ,n-1\\ i>j\end{array}}[e{]}_{[x_i,x_j]}.$$

$$\mathrm{wid}({\gamma }_{k+2}(G))\le \frac{n(n-1)}{2}\left(\begin{array}{c}n+k-2\\ k-1\end{array}\right),$$

$${\gamma }_k(G)=[e{]}_{[g_{11},\dots ,g_{1j}]}[e{]}_{[g_{21},\dots ,g_{2j}]}\dots [e{]}_{[g_{N1},\dots ,g_{Nj}]}$$

$$[e{]}_{g_{11}}<\prod _{i=1}^N[e{]}_{[g_{i1},q_i]}<[e{]}_{[g_{11},q_1]}\prod _{i=2}^N[e{]}_{g_{i1}}=[e{]}_{[g_{11},g_{12}]}\prod _{i=2}^N[e{]}_{g_{i1}}.$$

$$h_1=g_1,h_2=[g_1,g_2],h_3=[g_1,g_2,g_3],\dots ,h_k=[g_1,\dots ,g_k],\dots$$

$$[e{]}_{h_m}=[e{]}_{h_n^a}=[e{]}_{h_n}^a=[e{]}_{h_n}.$$

$$[e{]}_{g_1}>[e{]}_{[g_1,g_2]}>[e{]}_{[g_1,g_2,g_3]}>\dots$$

$${\gamma }_k(G)=[e{]}_{[g_{11},\dots ,g_{1j}]}[e{]}_{[g_{21},\dots ,g_{2j}]}\dots [e{]}_{[g_{N1},\dots ,g_{Nj}]},$$

$${\phi }_1:G\to H=G/[e{]}_{[g_{11},\dots ,g_{1j_1}]}.$$

$$e\ne {\phi }_1(a)\in {\phi }_1({\gamma }_k(G))\le [e{]}_{[{\phi }_1(g_{21}),\dots ,{\phi }_1(g_{2j})]}\dots [e{]}_{[{\phi }_1(g_{N1}),\dots ,{\phi }_1(g_{Nj})]}$$

$${\phi }_2:H\to H/[e{]}_{[{\phi }_1(g_{21}),\dots ,{\phi }_1(g_{2j_2})]}$$

$${\phi }_2{\phi }_1({\gamma }_k(G))\le [e{]}_{[{\phi }_2{\phi }_1(g_{31}),\dots ,{\phi }_2{\phi }_1(g_{3j})]}\dots [{\phi }_2{\phi }_1(e){]}_{[{\phi }_2{\phi }_1(g_{N1}),\dots ,{\phi }_2{\phi }_1(g_{Nj})]}$$

$$d(h_1{h}_2)=d(h_1)d(h_2{)}^{h_1}$$

$$d(t^k)=(id+\psi +{\psi }^2+\cdots +{\psi }^{k-1})a$$

$$0=d(e)=d(t^n)=(id+\psi +{\psi }^2+\cdots +{\psi }^{n-1})a.$$

$${\phi }_d:g\mapsto d(\overline{g})g$$

$$(id+\psi +\cdots +{\psi }^{p^n-1})a=0.$$

$$(id+\psi +\cdots +{\psi }^k)b$$

$${\psi }^{p^{n-1}+1}={\psi }^{p^{n-1}}+\psi -id.$$

Full text

Keywords: twisted conjugacy classes, verbal width, (residually) nilpotent groups.

1 Introduction

Conjugacy classes in a group reflect properties of this group. For example, the groups where every conjugacy class is finite (so called FC-groups introduced by Baer in [2]) have very restrictive structure. In particular, B. Neumann discovered that if $G$ is a group in which all conjugacy classes are finite with bounded size, then the derived subgroup $G^{\prime}$ is finite [17]. Another interesting connections between properties of conjugacy classes and properties of groups can be found, for example, in [6, 13].

The notion of $\varphi$-conjugacy classes generalizes the notion of conjugacy classes. Let $G$ be a group and $\varphi$ be an automorphism of $G$. Elements $x,y$ of the group $G$ are said to be (twisted) $\varphi$-conjugated if there exists an element $z$ in $G$ such that $x=zy\varphi(z)^{-1}$. In particular, if $\varphi=id$, then we have the usual definition of conjugated elements. The relation of $\varphi$-conjugation is an equivalence relation, equivalence classes of which are called $\varphi$-conjugacy classes. Denote by $[x]_{\varphi}$ the $\varphi$-conjugacy class of the element $x$ and by $R(\varphi)$ the number of these classes. The $\varphi$-conjugacy classes are also called Reidemeister classes, and the number $R(\varphi)$ is called the Reidemeister number of the automorphism $\varphi$. As well as conjugacy classes, twisted conjugacy classes in a group reflect properties of the group itself. For example, the following conjecture (partially proved in [7, Proposition 5.2] and [11, Theorems A, B and C]) describes very strong connection between the Reidemeister number $R(\varphi)$ and the structure of the finitely generated residually finite group.

[8, Conjecture R]****. Let $G$ be a finitely generated residually finite group. If there exists an automorphism $\varphi$ such that $R(\varphi)$ is finite, then $G$ is almost solvable.

Another interesting conjecture states the connection between the twisted conjugacy class $[e]_{\varphi}$ of the unit element $e$ and the structure of the group itself.

[4, Conjecture 1]****. Let $G$ be a group such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$. Then $G$ is nilpotent.

In [4] it is proved that if the class $[e]_{\varphi}$ is a subgroup of $G$ for every inner automorpism $\varphi$, then $G$ belongs to the Kurosh-Chernikov class $\overline{Z}$ and there exists a strictly descending series of normal subgroups in $G$. In particular, if $G$ is a finite group in which the class $[e]_{\varphi}$ is a subgroup for every inner automorphism $\varphi$, then $G$ is nilpotent.

In the present paper we continue to study groups where the twisted conjugacy class $[e]_{\varphi}$ is a subgroup of $G$ for every automorphism $\varphi$. In particular, we study the lower central series of such groups and prove that if $G$ is a group with $n$ generators such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every inner automorphism $\varphi$, then the $k$-th member of the lower central series of $G$ can be decomposed into the product of the finite number (depending on $n$ and $k$) of subgroups of the form $[e]_{\varphi_{i}}$ for different inner automorphisms $\varphi_{i}$ (Theorem 1). This result has several interesting corollaries.

The first important corollary of Theorem 1 is about the verbal width of the members of the lower central series. Let $F_{n}$ be a free group with generators $x_{1}$, $\dots$, $x_{n}$ and $w(x_{1},\dots,x_{n})$ be an element of $F_{n}$. For the elements $g_{1},\dots,g_{n}$ of a group $G$ we use the symbol $w(g_{1},\dots,g_{n})$ to denote the value of $w$ on the elements $g_{1},\dots,g_{n}$. Denote by $w(G)=\langle w(g_{1},\dots,g_{n})\leavevmode\nobreak\ |\leavevmode\nobreak\ g_{1},\dots,g_{n}\in G\rangle$ the subgroup of $G$ defined by the word $w\in F_{n}$. For the element $g\in w(G)$ denote by $l_{w}(g)$ the minimal number $k$ such that $g$ is a product of $k$ elements of the form $w(g_{1},\dots,g_{n})^{\pm 1}$ for $g_{1},\dots,g_{n}\in G$. The supremum of the values $l_{w}(g)$ for $g\in w(G)$ is called the (verbal) width of $w(G)$ and is denote by ${\rm wid}(w(G))$. Width of verbal subgroups in different groups has been studied by various authors (see, for example, [19] and and references therein). It is well known that the commutator width (the word $w$ is a commutator) of a free non-abelian group is infinite [3, 18], but the commutator width of a finitely generated nilpotent group is finite [1]. In the present paper we prove that if $G$ is a group with $n$ generators such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every inner automorphism $\varphi$, then the width of the $k$-th member of the lower central series $\gamma_{k}(G)$ is finite and is bounded in terms of $n,k$.

As an another interesting corollary of Theorem 1, we prove that if $G$ is a finitely generated group which satisfies the descending chain condition for normal subgroups such that $[e]_{\varphi}$ is a subgroup of $G$ for every inner automorphism $\varphi$, then $G$ is nilpotent (Corollary 3). This result generalises the result from [4, Theorem 3].

As we already said, if $G$ is a finite group such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi$, then $G$ is nilpotent [4, Theorem 3]. In the present paper for finite groups of the form $1\to A\to G\to\mathbb{Z}_{n}\to 1$, where $A$ is abelian and $n$ is odd, we construct a good upper bound of the nilpotency class of $G$ which depends only on the number $n$ (Theorem 3).

Several open problems are formulated throughout the paper.

The authors are greatful to Vasiliy Bludov and Evgeny Khukhro for the examples from Section 6.

2 Definitions and known results

We use the following notation. If $x,y$ are element of a group $G$, then $x^{y}=y^{-1}xy$ is conjugacy of $x$ by $y$ and $[x,y]=x^{-1}y^{-1}xy$ is a commutator of $x$ and $y$. We use the symbol $\gamma_{k}(G)$ to denote the $k$-th term of the lower central series: $\gamma_{1}(G)=G$, $\gamma_{k+1}(G)=[\gamma_{k}(G),G]$. The group $G$ is called nilpotent group of nilpotency class $m$ if $\gamma_{m+1}(G)=e$ and $\gamma_{k}(G)\neq e$ for all $k\leq m$. The group $G$ is called residually nilpotent if $\bigcap_{k=1}^{\infty}\gamma_{k}(G)=e$. Every nilpotent group is obviously residually nilpotent.

Let $F_{n}$ be a free group with generators $x_{1}$, $\dots$, $x_{n}$ and $w$ be an element of $F_{n}$. If $G$ is a group and $g_{1}$, $\dots$, $g_{n}$ are elements of $G$, then we use the symbol $w(g_{1},\dots,g_{n})$ to denote the value of $w$ on the elements $g_{1},\dots,g_{n}$. So we have a map $w:G^{n}\to G$ defined by $w$. The subgroup $w(G)=\langle w(g_{1},\dots,g_{n})\leavevmode\nobreak\ |\leavevmode\nobreak\ g_{1},\dots,g_{n}\in G\rangle$ generated by all values of $w$ on $G$ is called a verbal subgroup of $G$ defined by the word $w\in F_{n}$. The subgroup $\gamma_{k}(G)$ is a verbal subgroup defined by the word $\gamma_{k}$ from $F_{k}$: $\gamma_{1}(x_{1})=x_{1}$, $\gamma_{k+1}(x_{1},\dots,x_{k+1})=[\gamma_{k}(x_{1},\dots,x_{k}),x_{k+1}]$. The (verbal) width ${\rm wid}(w(G))$ of the verbal subgroup $w(G)$ defined by the word $w$ is the smalles number $m\in\mathbb{N}\cup\{\infty\}$ such that every element of $w(G)$ is the product of at most $m$ words $w(g_{1},\dots,g_{n})$ and their inverses. The equality $w(G)=v(G)$ for different words $v,w$ in general does not imply the equality ${\rm wid}(w(G))={\rm wid}(v(G))$.

Let $\varphi$ be an automorphism of the group $G$. Elements $x,y$ of $G$ are said to be $\varphi$-conjugated if there exists an element $z$ in $G$ such that $x=z^{-1}y\varphi(z)$. The relation of $\varphi$-conjugation is an equivalence relation, equivalence classes of which are called $\varphi$-conjugacy classes. Denote by $[x]_{\varphi}$ the $\varphi$-conjugacy class of the element $x$. If $\hat{g}:x\mapsto x^{g}$ is an inner automorphism of $G$ induced by the element $g$, then for the sake of simplicity we will write $[x]_{g}$ instead of $[x]_{\hat{g}}$ in order to denote the $\hat{g}$-conjugacy class of the element $x$. In particular, the $\hat{g}$-conjugacy class of the unit element $e$ has the following form.

[TABLE]

The following two lemmas about the twisted conjugacy class $[e]_{g}$ are obvious.

Lemma** 1****.**

If $G$ is a group such that $[e]_{g}$ is a subgroup of $G$ for every $g\in G$, then the element $e\neq g\in G$ does not belong to $[e]_{g}$.

Lemma** 2****.**

Let $G$ be a group such that $[e]_{g}$ is a subgroup of $G$ for every $g\in G$ and $H$ be a normal subgroup of $G$. If $g\in H$, then $[e]_{g}<H$.

The following two lemmas are proved in [4, Propositions 1, 3, 5].

Lemma** 3****.**

Let $G$ be a group and $\varphi$ be an automorphism of $G$. If the class $[e]_{\varphi}$ is a subgroup of $G$, then this subgroup is normal in $G$.

Lemma** 4****.**

Let $\varphi$, $\psi$ be automorphisms of a group $G$ such that $[e]_{\varphi}$, $[e]_{\psi}$ are subgroups of $G$. Then $[e]_{\varphi\psi}\subseteq[e]_{\varphi}[e]_{\psi}$.

Lemma** 5****.**

Let $G$ be a group and $[e]_{g}$ be a subgroup of $G$. Then for every normal subgroup $N$ of $G$ the class $[e]_{\overline{g}}$ is a subgroup of $G/N$, where $\overline{g}=gN\in G/N$.

The word $w$ from $F_{n}=\langle x_{1},\dots,x_{n}\rangle$ is called an outer commutator word if it is obtained from a finite sequence of distinct variables $x_{1},\dots,x_{n}$ using only commutator brackets. For example, the word $[[x_{1},x_{3}],x_{2}]$ is an outer commutator word, while the words $[x_{1},x_{2}][x_{3},x_{4}]$ and $[[x_{1},x_{2}],[x_{1},x_{3}]]$ are not outer commutator words. The following lemma is proved in [4, Lemma 2(1)].

Lemma** 6****.**

Let $G$ be a group such that the twisted conjugacy class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. If $w\in F_{n}$ is an outer commutator word, then $w(g_{1},\dots,g_{n})$ belongs to $[e]_{g_{1}}\cap\dots\cap[e]_{g_{n}}$.

From Lemma 1, Lemma 2 and Lemma 6 follows that

[TABLE]

is a strictly descending chain of normal subgroups if each member of this chain is not trivial. The following conjecture of Bardakov-Nasybullov-Neshchadim is formulated in [4, Conjecture 1] and [15, Problem 18.14].

Conjecture** 1****.**

If $G$ is a group such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every automorphism $\varphi$, then $G$ is nilpotent.

We say that the group $G$ satisfies the descending chain condition for normal subgroups (Min-n condition) if for every chain $H_{1}>H_{2}>H_{3}>\dots$ of normal subgroups there exists a number $n$ such that $H_{n}=H_{n+1}=\dots$ We say that $G$ satisfies the ascending chain condition for normal subgroups (Max-n condition) if for every chain $H_{1}<H_{2}<H_{3}<\dots$ of normal subgroups there exists a number $n$ such that $H_{n}=H_{n+1}=\dots$ The following result from [4, Theorem 3] gives a partial positive solution of Conjecture 1.

Proposition** 1****.**

If a group $G$ satisfies both Min-n and Max-n conditions and for every $g\in G$ the class $[e]_{g}$ is a subgroup of $G$, then $G$ is nilpotent.

From Proposition 1 follows, in particulat, that if $G$ is a finite group such that for every $g\in G$ the class $[e]_{g}$ is a subgroup of $G$, then $G$ is nilpotent.

In the conclusion of this Section we mention that the following commutator identities hold in every group.

[TABLE]

In particular, from this equalities follows that for every element $x$ which belongs to the subgroup of $G$ generated by the elements $x_{1},\dots,x_{n}$ and for every element $y$ which belongs to the subgroup of $G$ generated by the elements $y_{1},\dots,y_{m}$ there exist some elements $a_{1}$, $\dots$, $a_{n}$, $b_{1}$, $\dots$, $b_{n}$, $c_{1}$, $\dots$, $c_{n}$, $d_{1}$, $\dots$, $d_{n}$ in $G$ such that

[TABLE]

3 Structure of the lower central series

Theorem** 1****.**

Let $G$ be a group with generators $x_{1},\dots,x_{n}$ such that the twisted conjugacy class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. Then

$\gamma_{2}(G)=[e]_{x_{1}}[e]_{x_{2}}\dots[e]_{x_{n-1}}$, 2. 2.

if $\gamma_{k}(G)=[e]_{h_{1}}[e]_{h_{2}}\dots[e]_{h_{s}}$ and $p_{ij}=[x_{i},h_{j}]$, then $\gamma_{k+1}(G)=\prod_{i,j}[e]_{p_{ij}}$.

Proof. By Lemma 4 and Lemma 6 for every elements $y,z\in G$ the commutator $[y,z]$ belongs to $[e]_{y}\cap[e]_{z}<[e]_{y}\subseteq[e]_{x_{1}}[e]_{x_{2}}\dots[e]_{x_{n}}[e]_{x_{1}^{-1}}[e]_{x_{2}^{-1}}\dots[e]_{x_{n}^{-1}}$. Since in every group the equality

[TABLE]

holds, the subgroups $[e]_{x_{i}}$ and $[e]_{x_{i}^{-1}}$ coincide, and hence $[y,z]\in[e]_{x_{1}}\dots[e]_{x_{n}}$. From equality (2) and equality $[e]_{x_{i}}=[e]_{x_{i}^{-1}}$ follows that $[e]_{x_{n}}\leq[e]_{x_{1}}\dots[e]_{x_{n-1}}$, therefore every commutator $[x,y]$ belongs to $[e]_{x_{1}}\dots[e]_{x_{n-1}}$. Since by Lemma 3 for every $g,h\in G$ the classes $[e]_{g}$, $[e]_{h}$ are normal subgroups of $G$, we have $[e]_{g}[e]_{h}=[e]_{h}[e]_{g}$ and therefore every product of commutators belongs to $[e]_{x_{1}}\dots[e]_{x_{n-1}}$, what proves the first item of the theorem.

Let $\gamma_{k}(G)=[e]_{h_{1}}\dots[e]_{h_{s}}$, then for an arbitrary element $x\in\gamma_{k}(G)$ there exist some elements $y_{1},\dots,y_{s}$ in $G$ such that $x=[y_{1},h_{1}]\dots[y_{s},h_{s}]$. Using equality (2) for every $i=1,\dots,s$ we have

[TABLE]

for some elements $a_{1i},\dots,a_{ni},b_{1i},\dots,b_{ni}$ from $G$. If we denote by $p_{ij}=[x_{i},h_{j}]$, $q_{ij}=[x_{i}^{-1},h_{j}]$, then we conclude that every element $x$ from $\gamma_{k}(G)$ can be written in the form

[TABLE]

for some elements $a_{ij},b_{ij}\leavevmode\nobreak\ (i=1,\dots,n,j=1,\dots,s)$ of $G$. For an arbitrary element $t$ from $G$ and element $x$ from $\gamma_{k}(G)$ using equalities (1) and (3) we have

[TABLE]

for some elements $c_{ij},d_{ij}\leavevmode\nobreak\ (i=1,\dots,n,j=1,\dots,s)$ from $G$. Therefore since subgroups $[e]_{p_{ij}}$ and $[e]_{q_{ij}}$ are normal in $G$, the element $[t,x]$ belongs to $\prod_{i,j}[e]_{p_{ij}}[e]_{q_{ij}}$ for every $t\in G$, $x\in\gamma_{k}(G)$. Since in every group the equality

[TABLE]

holds, we have

[TABLE]

therefore $[e]_{q_{ij}}=[e]_{p_{ij}}$ and for every elements $t\in G$, $x\in\gamma_{k}(G)$ the commutator $[t,x]$ belongs to $\prod_{i,j}[e]_{p_{ij}}$, and hence $\gamma_{k+1}(G)=\prod_{i,j}[e]_{p_{ij}}$. $\square$

Corollary** 1****.**

Let $G$ be a group with generators $x_{1},\dots,x_{n}$ such that the class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. Then

${\rm wid}(\gamma_{2}(G))\leq n-1$, 2. 2.

${\rm wid}(\gamma_{k}(G))\leq n^{k-2}(n-1)/2$* for $k\geq 3$.*

Proof. The first item of Theorem 1 implies that ${\rm wid}(\gamma_{2}(G))\leq n-1$. From the second item of Theorem 1 we have $\gamma_{3}(G)=\prod_{i,j}[e]_{[x_{i},x_{j}]}$, where $i$ varies from $1$ to $n$ and $j$ varies from $1$ to $n-1$. Since $[e]_{[x_{i},x_{i}]}=e$ and $[e]_{[x_{i},x_{j}]}=[e]_{[x_{i},x_{j}]^{-1}}=[e]_{[x_{j},x_{i}]}$ we can rewrite $\gamma_{3}(G)$ in the form

[TABLE]

Thereforel ${\rm wid}(\gamma_{3}(G))$ is less then or equal to the number of members in this product which is equal to $n(n-1)/2$. From the second item of Theorem 1 it follows that ${\rm wid}(\gamma_{k+1}(G))\leq n\cdot{\rm wid}(\gamma_{k}(G))$, i. e. ${\rm wid}(\gamma_{k}(G))\leq n^{k-2}(n-1)/2$ for $k\geq 3$. $\square$

Remark** 1****.**

If $G$ is generated by two elements, then ${\rm wid}(\gamma_{2}(G))\leq 2^{1}-2^{0}=1$, i. e. every element of $\gamma_{2}(G)$ is a commutator. Moreover ${\rm wid}(\gamma_{3}(G))\leq 2(2-1)/2=1$, i. e. every element from $\gamma_{3}(G)$ has the form $[[x,y],z]$. Since two generated group $G$ such that $[e]_{g}\leq G$ for all $g\in G$ does not have to be abelian or $2$-step nilpotent (see Section 6), the estimations ${\rm wid}(\gamma_{2}(G))\leq n-1$ and ${\rm wid}(\gamma_{3}(G))\leq n(n-1)/2$ cannot be improved.**

Problem** 1****.**

Construct the sharp estimation of ${\rm wid}(\gamma_{k}(G))$ for $k\geq 4$.**

If the group $G$ from Corollary 1 is metabelian, then for every elements $y_{1},\dots,y_{m}$ $(m\geq 3)$ from $G$ and every permutation $\sigma\in{\rm Sym}(\{3,\dots,m\})$ the commutator equality $[y_{1},y_{2},y_{3},\dots,y_{n}]=[y_{1},y_{2},y_{\sigma(3)},\dotsm,y_{\sigma(n)}]$ holds [16, §1.1]. So, in the case of metabelian groups we can strongly improve the estimation for ${\rm wid}(\gamma_{k+2}(G))$ $(k\geq 1)$ in the form

[TABLE]

where $\begin{pmatrix}n+k-2\\ k-1\end{pmatrix}$ is the number of ways to distribute $k-1$ indistinguishable balls to the $n$ distinguishable baskets: $\gamma_{k+2}(G)=\prod_{i,j,i_{3},\dots,i_{k+1}}[e]_{[x_{i},x_{j},x_{i_{3}},\dots,x_{i_{k+1}}]}$ and $[e]_{[x_{i},x_{j},x_{i_{3}},\dots,x_{i_{k+1}}]}$ depents only on $x_{i}$, $x_{j}$ for $i=1,\dots,n$, $j=1,\dots,n-1$ for $i>j$ ($n(n-1)/2$ variants) and the number of each $x_{i}\leavevmode\nobreak\ (i=1,\dots,n)$ ($n$ distinguishable baskets) in the set $\{x_{i_{3}},\dots,x_{i_{k+1}}\}$ ($k-1$ indistinguishable balls).

Another improvement of the estimation of the width of the members of lower central series can be obtained in the case when $\gamma_{k}(G)=\gamma_{k+1}(G)$ for some $k$.

Proposition** 2****.**

Let $G$ be a group with $n$ generators such that $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. Let $k\geq 2$ satisfies $\gamma_{k}(G)=\gamma_{k+1}(G)$. Then for some number $N\leq n^{k-1}(n-1)$ there exist $N$ sequences $\{g_{1i}\}_{i}$, $\{g_{2i}\}_{i}$, $\dots$, $\{g_{Ni}\}_{i}$ such that $g_{11},g_{21},\dots,g_{N1}\in\gamma_{k}(G)$ and for all $j=1,2,\dots$ the following equality holds.

[TABLE]

Proof. By Theorem 1 there exist some elements $h_{1},\dots,h_{s}$ $(s\leq n^{k-2}(n-1))$ from $\gamma_{k-1}(G)$ such that $\gamma_{k}(G)=[e]_{h_{1}}\dots[e]_{h_{s}}$. If we denote by $p_{ij}=[x_{i},h_{j}]$, then by Theorem 1(2) we have $\gamma_{k+1}(G)=\prod_{i,j}[e]_{p_{ij}}$. Let $N=ns\leq n^{k-1}(n-1)$ and $\{g_{11},g_{21},\dots,g_{N1}\}=\{p_{ij}\leavevmode\nobreak\ |\leavevmode\nobreak\ 1\leq i\leq n,1\leq j\leq s\}$ (we just renamed the elements $p_{ij}$ in the way which we need in the formulation of the lemma). All the element $g_{11},g_{21},\dots,g_{N1}$ belong to $\gamma_{k}(G)$.

Since $\gamma_{k}(G)=\gamma_{k+1}(G)$, the element $g_{11}$ belongs to $\gamma_{k+1}(G)=\prod_{i=1}^{N}[e]_{g_{i1}}$. Therefore there exist some elements $(g_{12}:=)q_{1},q_{2},\dots,q_{N}$ such that $g_{11}=\prod_{i=1}^{N}[g_{i1},q_{i}]$. Then by Lemma 4 we have

[TABLE]

Therefore $\gamma_{k}(G)=\gamma_{k+1}(G)=[e]_{[g_{11},g_{12}]}\prod_{i=2}^{N}[e]_{g_{i1}}$. Now taking $g_{21}$ instead of $g_{11}$ and repeating the reasoning above we conclude that $\gamma_{k}(G)=[e]_{[g_{11},g_{12}]}[e]_{[g_{21},g_{22}]}\prod_{i=3}^{N}[e]_{g_{i1}}$. Repeating this idea for all $g_{i1}$ we have $\gamma_{k}(G)=\prod_{i=1}^{N}[e]_{[g_{i1},g_{i2}]}$. Repeating this proccess $k$ time we conclude that $\gamma_{k}(G)=\prod_{i=1}^{N}[e]_{[g_{i1},g_{i2},\dots,g_{ik}]}$ for some elements $g_{ij}(1\leq i\leq N,1\leq j\leq k)$. This fact finishes the proof. $\square$

Corollary** 2****.**

Let $G$ be a group with $n$ generators such that the class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$ and let $k$ satisifies $\gamma_{k}(G)=\gamma_{k+1}(G)$. Then ${\rm wid}(\gamma_{s}(G))\leq n^{k-1}(n-1)$ for all $s$.

Proof. By Corollary 1 for $s\leq k+1$ we have ${\rm wid}(\gamma_{s}(G))\leq n^{s-1}-n^{s-2}\leq n^{k}-n^{k-1}$ and we only need to estimate the width ${\rm wid}(\gamma_{s}(G))$ for $s>k+1$. From Proposition 2 it follows that ${\rm wid}(\gamma_{s}(G))\leq N\leq n^{k-1}(n-1)$ for $s>k+1$. $\square$

Remark** 2****.**

In Corollary 2 we suppose that $\gamma_{k}(G)=\gamma_{k+1}(G)$ but when we write ${\rm wid}(\gamma_{k+1}(G))$ we also specify the generating set of $\gamma_{k}(G)=\gamma_{k+1}(G)$ and the equality $\gamma_{k}(G)=\gamma_{k+1}(G)$ does not imply the equality ${\rm wid}(\gamma_{k}(G))={\rm wid}(\gamma_{k+1}(G))$.**

4 Structure of the group

At first we prove one simple general necessary condition which must be satisfied by an infinite group $G$ where the class $[e]_{g}$ is a subgroup of $G$ for every $g$.

Proposition** 3****.**

Let $G$ be an infinite group such that $[e]_{g}$ is a subgroup of $G$ for every $g$. Then $G$ has infinitely many conjugacy classes.

Proof. Suppose that $g$ has $s<\infty$ conjugacy classes, and let $g_{1},g_{2},\dots$ be elements of $G$. Consider the sequence of elements

[TABLE]

Since there are only $s$ conjugacy classes in the group $G$, there exist two numbers $m<n\leq s+1$ such that $h_{m}$ and $h_{n}$ are conjugated, i. e. $h_{m}=h_{n}^{a}$ for some $a\in G$. Therefore

[TABLE]

The element $h_{n}=[g_{1},\dots,g_{n}]=[h_{m},g_{m+1},\dots,g_{n}]$ belongs to $[e]_{h_{m}}$. If $h_{n}\neq e$, then by Lemma 1 it does not belong to $[e]_{h_{n}}=[e]_{h_{m}}$. This contradiction says that $h_{n}=e$. Therefore for $r\geq s+1$ we have $h_{r}=[h_{n},g_{n+1},\dots,g_{r}]=e$ and since $g_{1},g_{2},\dots$ is an arbitrary sequence of elements from $G$, then $G$ is nilpotent of nilpotency class less then or equal to $s$. Since a nilpotent group has finitely many conjugacy classes if and only if it is finite, $G$ must be a finite group, what contradicts the condition of the proposition. $\square$

If $G$ is a groups such that the class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$, then for elements $g_{1},g_{2},\dots$ from $G$ if $[e]_{[g_{1},\dots,g_{k}]}$ is not trivial, then

[TABLE]

is a strictly descending chain of normal subgroups. Then it is reasonable to ask about the structure of $\cap_{k}[e]_{[g_{1},\dots,g_{k}]}$.

Problem** 2****.**

Let $G$ be a group such that the class $[e]_{g}$ is a subgroup of $G$ for every $g$. Is it true that $\cap_{k}[e]_{[g_{1},\dots,g_{k}]}=e$ for every elements $g_{1},g_{2},\dots$ from $G$?**

A positive solution of Problem 2 would emply the fact that the lower central series of a group is strictly descending.

Theorem** 2****.**

Let $G$ be a finitely generated group such that the class $[e]_{g}$ is a subgroup of $G$ for every $g$. If for every sequence of elements $g_{1},g_{2},\dots$ the intersection $\cap_{k}[e]_{[g_{1},\dots,g_{k}]}$ is trivial, then either $\gamma_{k}(G)=e$ or $\gamma_{k}(G)\neq\gamma_{k+1}(G)$.

Proof. Suppose that $\gamma_{k}(G)=\gamma_{k+1}(G)$, from Proposition 2 we have

[TABLE]

for an infinite set of elements elements $g_{ij}$ and every $j>1$. Suppose that in $\gamma_{k}(G)$ there exists an element $a\neq e$. Since $\cap_{j}[e]_{[g_{11},\dots,g_{1j}]}=e$, for $j_{1}$ big enough the element $a$ does not belong to $[e]_{[g_{11},\dots,g_{1j_{1}}]}$. Consider the natural homomorphism

[TABLE]

Since $a$ does not belong to $[e]_{[g_{11},\dots,g_{1j_{1}}]}$, then by Lemma 5

[TABLE]

for $j>j_{1}$. Using the same argument we see that there exists a homomorphism

[TABLE]

such that $e\neq\varphi_{2}\varphi_{1}(a)\in\varphi_{2}\varphi_{1}(\gamma_{k}(G))$ and

[TABLE]

for $j>j_{2}>j_{1}$. Repeating this process $N-1$ times we conclude that there exist a homomorphism $\varphi=\varphi_{N-1}\dots\varphi_{1}$ from $G$ to $G/H$ for some subgroup $H$ such that $e\neq\varphi(a)\in\varphi(\gamma_{k}(G))\leq[e]_{[\varphi(g_{N1}),\dots,\varphi(g_{Nj})]}$ for $j$ big enough. The equality $\cap_{k}[e]_{[g_{N1},\dots,g_{Nk}]}=e$ implies $\varphi(a)=e$ what contradicts the choice of $a$, i. e. in $\gamma_{k}(G)$ there are no nontrivial elements. $\square$

Corollary** 3****.**

Let $G$ be a finitely generated group such that $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. If $G$ satisfies Min-n condition, then $G$ is nilpotent.

Proof. From Min-n condition follows that for every sequence of elements $\{g_{i}\}_{i}$ the intersection $\cap_{k}[e]_{[g_{1},\dots,g_{k}]}$ is trivial. If $\gamma_{k}(G)=\gamma_{k+1}(G)$ for some $k$, then by Theorem 2 we have $\gamma_{k}(G)=e$, i. e. $G$ is nilpotent. If $\gamma_{k+1}(G)\neq\gamma_{k}(G)$ for all $k$, then $\gamma_{1}(G)>\gamma_{2}(G)>\dots$ is a strictly descending chain of normal subgroups, therefore by Min-n condition $\gamma_{k}(G)=e$ for some $k$. $\square$

Corollary 3 generalizes the result from [4, Theorem 3] which we recalled in Proposition 1: we change Max-n condition by a weaker condition which says that $G$ is a finitely generated group.

5 Finite abelian-by-cyclic groups

Let a group $H$ acts on a group $A$, i. e. there exists a homomorphism from $H$ to ${\rm Aut}\leavevmode\nobreak\ A$ which maps an element $h$ from $H$ to the automorphism $\hat{h}:a\mapsto a^{h}$ of the group $A$. The map $d:H\to A$ is called a derivation if for every elements $h_{1},h_{2}\in H$ the equality

[TABLE]

holds. The set of all derivations $H\to A$ is denoted by the symbol $Der(H,A)$.

Let $H=\mathbb{Z}_{n}=\langle t\rangle$ be a cyclic group of order $n$ and $A$ be an abelian group (with the operation written additively). Denote by $\psi$ the automorphism $a\mapsto a^{t}$. If $d$ is a derivation from $H$ to $A$ such that $d(t)=a$, then using equality (4) we conclude that

[TABLE]

for all $k=1,\dots,n$. So, in order to define a derivation from $\mathbb{Z}_{n}=\langle t\rangle$ to $A$ it is enough to define only the valude $d(t)=a$. The map $d(t)=a$ can be extended to the derivation $\mathbb{Z}_{n}=\langle t\rangle\to A$ if an only if

[TABLE]

In particular, for every $b\in A$ the map $d(t)=(id-\psi)b$ can be extended to the derivation from $Der(\mathbb{Z}_{n},A)$. Such derivations are called principal derivations [5, Chapters II, III].

Let $p>2$ be a prime, $G$ be a finite $p$-group, $A$ be a normal abelian subgroup of $G$ and $H=G/A$. For $g\in G$ denote by $\overline{g}=gA$ the image of $g$ under the natural homomorphism $G\to H$. If $d\in Der(H,A)$, then the map

[TABLE]

is an automorphism of $G$ (see [9, Satz 2] for details).

Following [14] for a finite abelian $p$-group $A$ denote by $\Omega_{k}(A)=\{a\in A\leavevmode\nobreak\ |\leavevmode\nobreak\ a^{p^{k}}=e\}$.

Lemma** 7****.**

Let $p>2$ be a prime, $A$ be a finite abelian $p$-group and $G$ be an extension of $A$ by $\mathbb{Z}_{p^{n}}=\langle t\rangle$. If the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$, then for every $a\in\Omega_{1}(A)$ the map $d(t)=a$ can be extended to the derivation $d\in Der(\mathbb{Z}_{p^{n}},A)$.

Proof. Denote by $\psi$ the automorphism $\psi:a\mapsto a^{t}$ of the group $A$. Since $\psi^{p^{n}}=id$, the order of $\psi$ is a power of $p$. Using the additive form for the operation in abelian group $A$, by equality (6) the map $d(t)=a$ can be extended to derivation $d:\mathbb{Z}_{p^{n}}\to A$ if an only if

[TABLE]

So, we need to prove that for every element $a\in\Omega_{1}(A)$ the equality (7) holds.

Let $c$ be an arbitrary element of the group $A$ and $b=(id-\psi)c$. The map $d(t)=b$ can be extended to the derivation $d:\mathbb{Z}_{p^{n}}\to A$ and by the condition of the lemma, for the automorphism $\varphi:g\mapsto d(\overline{g})g$ the class $[e]_{\varphi}$ is a subgroup of $G$. This subgroup consists of the elements $g\varphi(g)^{-1}=d(\overline{g})^{-1}$. Since $[e]_{\varphi}$ is a subgroup, we have $[e]_{\varphi}=[e]_{\varphi}^{-1}=d(\mathbb{Z}_{p^{n}})$ and by equality (5) this subgroup consists of elements

[TABLE]

for all $k=1,\dots,p^{n}$. Since by Lemma 3 the class $[e]_{\varphi}$ is a normal subgroup of $G$, the map $\psi$ maps $[e]_{\varphi}$ to itself. Therefore since $b$ belongs to $[e]_{\varphi}$, the elements $\psi^{k}(b)$ belong to $[e]_{\varphi}$ for all $k=1,\dots,p^{n}$. Since the order of $[e]_{\varphi}=d(\mathbb{Z}_{p^{n}})$ is less then or equal to $p^{n}$ and all elements $\psi^{k}(b)\leavevmode\nobreak\ (k=1,\dots,p^{n})$ are not trivial, all the elements $\psi^{k}(b)\leavevmode\nobreak\ (k=1,\dots,p^{n})$ cannot be distinct. Therefore $\psi^{r}(b)=\psi^{s}(b)$ for some $0\leq r\neq s\leq p^{n}-1$, i. e. $\psi^{k}(b)=b$ for some $0<k=r-s\leq p^{n}-1$. Since the order of $\psi$ is a power of $p$, we can assume that $k$ is a power of $p$, i. e. for some $0\leq k\leq n-1$ we have $\psi^{p^{k}}(b)=b$. Acting on this equality by $\psi^{p^{k}(p^{n-k-1}-1)}$ we have $\psi^{p^{n-1}}(b)=b$ for all $b=(id-\psi)c$. Since $c$ is an arbitrary element of $A$ we have the equality $\psi^{p^{n-1}}(id-\psi)=(id-\psi)$ which emplies the following equality

[TABLE]

Using simple induction on the pair $(r,s)$ in lexicographic order, this equality implies the equality

[TABLE]

for all $r=0,\dots,p-1$ and $s=0,\dots,p^{n-1}-1$. Therefore we have

[TABLE]

This equality emplies that $(id+\psi+\dots+\psi^{p^{n}-1})a=0$ for every element $a$ of order $p$ (i. e. $a\in\Omega_{1}(A)$), so, the map $d(t)=a$ can be extended to the derivation $d:\mathbb{Z}_{p^{n}}\to A$. $\square$

Lemma** 8****.**

Let $p>2$ be a prime, $A$ be a finite abelian $p$-group and $G$ be an extension of $A$ by $\mathbb{Z}_{p^{n}}=\langle t\rangle$. If the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$, then $a^{t}=a$ for every $a\in\Omega_{1}(A)$.

Proof. Denote by $\psi$ the automorphism $\psi:a\mapsto a^{t}$ of the group $A$. Since $\psi^{p^{n}}=id$, the order of $\psi$ is a power of $p$. Let $a\in\Omega_{1}(A)$ be an arbitrary element. Then by Lemma 7 the map $d(t)=a$ defines a derivation. For the automorphism $\varphi:g\mapsto d(\overline{g})g$ the class $[e]_{\varphi}$ is a subgroup of $G$ which coincides with $d(\mathbb{Z}_{p^{n}})$ and consists of the elements

[TABLE]

for $k=1,\dots,p^{n}$. Similarly to Lemma 7 we have $\psi^{p^{m}}(a)=a$ for some $0\leq m\leq n-1$. If $m=0$, then there is nothing to prove. Suppose that $\psi(a)\neq a$ and let $m\neq 0$ be a minimal number such that $\psi^{p^{m}}(a)=a$.

Suppose that $(id+\psi+\dots+\psi^{p^{m}-1})a=0$. In this case using the equality (8) the group $d(\mathbb{Z}_{p^{n}})$ consists of the elements

[TABLE]

for $k=1,\dots,p^{m}$, i. e. it contains only $p^{m}$ elements. This group contains $a$ and it is fixed under the action of $\psi$. Therefore $\psi^{r}(a)$ belongs to $d(\mathbb{Z}_{p^{n}})$ for all $r=1,\dots,p^{m}$. Since $d(\mathbb{Z}_{p^{n}})$ contains only $p^{m}$ elements and all the elements $\psi^{r}(a)$ are not trivial, there exists some numbers $0\leq r\neq s<p^{m}$ such that $\psi^{r}(a)=\psi^{s}(a)$, i. e. $\psi^{k}(a)=a$ for some $0<k=r-s<p^{m}$. Since the order of $\psi$ is a power of $p$, we can assume that $k$ is a power of $p$. It contradicts the fact that $m$ is a minimal number such that $\psi^{p^{m}}(a)=a$. Therefore $(id+\psi+\dots+\psi^{p^{m}-1})a\neq 0$.

Let $b=(id+\psi+\dots+\psi^{p^{m}-1})a\neq 0$ and $H$ be a group generated by $b$. From the equality $\psi^{p^{m}}(a)=a$ follows that

[TABLE]

for all $r=1,\dots,p^{n-m}$. Therefore the group $H$ is a subgroup of $[e]_{\varphi}$. Since the element $b=(id+\psi+\dots+\psi^{p^{m}-1})a$ is fixed under $\psi$, the group $H$ is a normal subgroup of $G$ which does not contain $a$. The quotient group $[e]_{\varphi}/H$ consists of the elements

[TABLE]

for $k=1,\dots,p^{m}$. This subgroup contains $a+H$ and $\psi([e]_{\varphi}/H)=[e]_{\varphi}/H$. Therefore the elements $\psi^{r}(a)+H$ belong to $[e]_{\varphi}/H$ for all $r=0,\dots,p^{m}-1$. Since the group $[e]_{\varphi}/H$ contains less then or equal to $p^{m}$ elements and all the elements $\psi^{r}(a)+H$ are not trivial in $[e]_{\varphi}/H$, there exist some numbers $0\leq r\neq s<p^{m}$ such that $\psi^{r}(a)+H=\psi^{s}(a)+H$. So, $\psi^{k}(a)+H=a+H$ for some $0<k=r-s\leq p^{m}-1$. Since the order of $\psi$ is a power of $p$, we can assume that $k=p^{l}$ for some $0\leq l\leq m-1$. So, for some number $r$ we have $\psi^{p^{l}}(a)=a+r(id+\psi+\dots+\psi^{p^{m}-1})a$. Since $(id+\psi+\dots+\psi^{p^{m}-1})a$ is fixed under the action of $\psi$, then for all $s$ we have $\psi^{sp^{l}}(a)=a+sr(id+\psi+\dots+\psi^{p^{m}-1})a$. Therefore

[TABLE]

what contradicts the fact that $m$ is a minimal number such that $\psi^{p^{m}}(a)=a$. So, $\psi(a)$ must be equal to $a$. $\square$

Corollary** 4****.**

Let $p>2$ be a prime, $A$ be a finite abelian $p$-group and $G$ be an extension of $A$ by $\mathbb{Z}_{p^{n}}=\langle t\rangle$. If the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$, then $a^{t}=a\theta(a)$, where $\theta$ is a homomorphism from $A$ to $\Omega_{n}(A)$ such that $\theta(\Omega_{k}(A))\leq\Omega_{k-1}(A)$ for all $k\leq n$.

Proof. Denote by $\psi$ the automorphism $\psi:a\mapsto a^{t}$ of the group $A$. Let $b$ be an arbitrary element of the group $A$ and $a=(id-\psi)b$. The map $d(t)=a$ defines a derivation from $Der(\mathbb{Z}_{p^{n}},A)$ and for the automorphism $\varphi:g\mapsto d(\overline{g})g$ the class $[e]_{\varphi}=d(\mathbb{Z}_{p^{n}})$ is a subgroup of $G$ which contains $a$. Since the order of $[e]_{\varphi}=d(\mathbb{Z}_{p^{n}})$ is less then or equal to $p^{n}$, then the order of $a$ is less then or equal to $p^{n}$, i. e. for every $b$ the element $(id-\psi)b$ belongs to $\Omega_{n}(A)$. So, for every $b\in A$ we have $b^{t}=\psi(b)=b\theta(b)$, where $\theta$ is the map from $A$ to $\Omega_{n}(A)$. For arbitrary elements $a,b\in A$ we have $\theta(ab)ab=\psi(ab)=\psi(a)\psi(b)=a\theta(a)b\theta(b)=ab\theta(a)\theta(b)$, therefore $\theta$ is a homomorphism.

If $a$ is an element of $\Omega_{k}(A)$, then $a^{p^{k-1}}$ is an element from $\Omega_{1}(A)$, and then by Lemma 8 we have $\theta(a)^{p^{k-1}}=\theta(a^{p^{k-1}})=e$, so, $\theta(a)$ belongs to $\Omega_{k-1}(A)$. $\square$

Theorem** 3****.**

Let $n$ be an odd integer with prime decomposition $n=p_{1}^{n_{1}}\dots p_{m}^{n_{m}}$ and $A$ be a finite abelian group. If $G$ is an extension of $A$ by $\mathbb{Z}_{n}$ such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$, then $G$ is nilpotent of nilpotency class at most ${\rm max}\{n_{1},\dots,n_{m}\}+1$.

Proof. Since $G$ is a finite group such that the class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$, by Corollary 3 the group $G$ is nilpotent. By Burnside-Wielandt theorem [12, Theorem 17.1.4] the group $G$ is the direct product of its Sylow subgroups $G_{1},\dots,G_{k}$. We can assume that $k=m$ (otherwise rewrite $n=p_{1}^{n_{1}}\dots p_{m}^{n_{m}}p_{m+1}^{0}\dots p_{k}^{0}$).

Since $G$ is an extension of the abelian group $A$ by the cyclic group $\mathbb{Z}_{n}$, the group $G_{i}$ must be an extension of the abelian $p_{i}$-group $A_{i}$ by the cyclic group $\mathbb{Z}_{p_{i}^{n_{i}}}=\langle t_{i}\rangle$. Denote by $\psi_{i}$ the automorphism $a\mapsto a^{t_{i}}$ of the group $A_{i}$. By Corollary 4 for every $a\in A_{i}$ we have $\psi_{i}(a)=a\theta_{i}(a)$. Using simple induction on $r\geq 2$ it is easy to see that

[TABLE]

and by Corollary 4 we have $\gamma_{n_{i}+2}(G_{i})=e$, i. e. $G_{i}$ is a nilpotent group of nilpotency class at most $n_{i}+1$ and $G$ is nilpotent group of nilpotency class at most ${\rm max}\{n_{1},\dots,n_{m}\}+1$. $\square$

Remark** 3****.**

In Lemma 7, Lemma 8, Corollary 4 and Theorem 3 we require that the class $[e]_{\varphi}$ is a subgroup of $G$ for every automorphism $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$. However in proofs we use only inner automorphisms and automorphisms of the form $g\mapsto d(\overline{g})g$ for $d\in Der(\mathbb{Z}_{n},A)$. So, in all of this statements it is enough to require that the class $[e]_{\varphi}$ is a subgroup of $G$ only for these specific automorphisms.**

If the group $A$ in Theorem 3 is elementary abelian, then by Corollary 4 the group $G$ must be abelian. So we have the following problem.

Problem** 3****.**

Let $p$ be an odd prime, $A$ be a finite abelian group and $G$ be an extension of $A$ by $\mathbb{Z}_{p}$ such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$. Is it true that $G$ is abelian?**

6 Examples and counterexamples

Conjecture 1 says that if $G$ is a group such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every $\varphi\in{\rm Aut}\leavevmode\nobreak\ G$, then $G$ is nilpotent. Example 1 and Example 2 below show that it is not sufficient to consider only inner automorphisms to prove Conjecture 1. Also this examples give answers to some questions from [4].

We thank E. Khukhro, who noticed that in the group $G(p,n)$ from Example 1 the twisted conjugacy class of the unit element is a subgroup for every inner automorphism and showed us this example. We also thank V. Bludov, who demonstrates us Example 2 in details.

Example 1. Let $n>2$ be a positive integer and $p>2$ be a prime. Denote by $G(p,n)$ the semidirect product of two cyclic groups $\mathbb{Z}_{p^{n}}\rtimes\mathbb{Z}_{p^{n-1}}$ with generators $x=(1,0)$, $y=(0,1)$ and relation $yxy^{-1}=x^{p+1}$. The order of $G(p,n)$ is $p^{2n-1}$, therefore $G(p,n)$ is nilpotent. Since

[TABLE]

the nilpotency class of $G(p,n)$ is greater then or equal to $n-1$. Using simple induction on $r$ we see that in $G(p,n)$ the following equality holds for all positive integers $a,s$.

[TABLE]

If $\varphi$ is inner automorphism of $G$, then

[TABLE]

From [10, Corollary 4.4 (1b)] it follows that $a\in\mathbb{Z}_{p^{n}}^{*}$ ($a$ is an invertible element of the ring $\mathbb{Z}_{p^{n}}$) and $b\equiv 0\leavevmode\nobreak\ ({\rm mod}\leavevmode\nobreak\ p)$.

The twisted cojugacy class $[e]_{\varphi}$ of the unit element $e$ consists of the elements $(x^{r}y^{s})\varphi(x^{r}y^{s})^{-1}=(x^{r}y^{s})((x^{a}y^{b})^{r}(x^{c}y)^{s})^{-1}$ for $r=0,\dots,p^{n}-1$, $s=0,\dots,p^{n-1}-1$. Using equality (9) and direct calculations we conclude that the class $[e]_{\varphi}$ consists of the elements

[TABLE]

for $r=0,\dots,p^{n}-1$, $s=0,\dots,p^{n-1}-1$. Since $b\equiv 0\leavevmode\nobreak\ ({\rm mod}\leavevmode\nobreak\ p)$, we have $b=kp$ and $b\left(1+(p+1)+\dots+(p+1)^{s-1}\right)=kp\left(1+(p+1)+\dots+(p+1)^{s-1}\right)=k[(p+1)^{s}-1]$. Since the multiplicative order of $p+1$ in $\mathbb{Z}_{p^{n}}$ is equal to $p^{n-1}$ [20, Chapter IV §3.5], the set $\{(p+1)^{s}-1\leavevmode\nobreak\ |\leavevmode\nobreak\ s=0,\dots,p^{n-1}-1\}$ consists of $p^{n-1}$ different elements from $\mathbb{Z}_{p^{n}}$ each of which is divisible by $p$. Since in $\mathbb{Z}_{p^{n}}$ there are only $p^{n-1}$ elements divisible by $p$, the set $\{(p+1)^{s}-1\leavevmode\nobreak\ |\leavevmode\nobreak\ s=0,\dots,p^{n-1}-1\}$ contains all this elements. Therefore the class $[e]_{\varphi}$ consists of the elements

[TABLE]

for $r=0,\dots,p^{n}-1$, $z$ is divisible by $p$. This set of elements obviously forms a subgroup of $G(p,n)$.

So, $G(p,n)$ is a nilpotent group of nilpotency class greater then or equal to $n-1$ such that the twisted conjugacy class of the unit element is a subgroup of $G(p,n)$ for every inner automorphism of $G(p,n)$. Denote by $p_{n}$ the $n$-th prime and by $G$ the infinite direct product

[TABLE]

Then $G$ is non-nilpotent residually nilpotent group such that the class $[e]_{\varphi}$ is a subgroup of $G$ for every inner automorphism $\varphi$ of $G$.

Remark** 4****.**

If $\varphi$ is not inner automorphism of $G(p,n)$, then $[e]_{\varphi}$ is not neccessarily a subgroup of $G(p,n)$. For example, by [10, Corollary 4.4 (1b)] the map

[TABLE]

induces an automorphism of $G(3,3)$. However, the class $[e]_{\varphi}$ contains $3$ elements $(0,0)$, $(0,3)$, $(9,6)$ which do not form a subgroup. So, group $G$ does not give a counter example to Conjecture 1.**

The group $G(p,n)$ for $p>2$, $n>3$ gives a negative answer to the following problems from [4].

[4, Problem 2]****. Let $G$ be a finite group such that the class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. Can we assert that $G$ is a $2$-step nilpotent group?

[4, Problem 3]****. Let $G$ be an extension of an abelian group by a cyclic (or abelian) group. Can we assert that for some $g\in G$ the class $[e]_{g}$ is not a subgroup of $G$?

Also using groups $G(p,n)$ from Example 1 we can answer the following question from [4].

[4, Problem 7]****. Let $G$ be a group such that $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. Let $N$ be a normal subgroup of $G$ and denote by $[e]_{g,N}=\{[h,g]\leavevmode\nobreak\ |\leavevmode\nobreak\ h\in N\}$.

Is it true that $[e]_{g,N}$ is a subgroup of $N$? 2. 2.

How are $[e]_{g,N}$ and $[e]_{g}\cap N$ related?

Let $G=G(3,3)=\langle x,y\leavevmode\nobreak\ |\leavevmode\nobreak\ x^{27}=1,y^{9}=1,yxy^{-1}=x^{4}\rangle$ and $N=\langle x^{3},y^{3}\rangle$. Then $[e]_{x,N}$ contains only three elements $e$, $x^{15}$ and $x^{18}$ which do not form a subgroup. It gives negative answer to the first item of the question.

Let $G=G(p,n)=\langle x,y\leavevmode\nobreak\ |\leavevmode\nobreak\ x^{p^{n}}=1,y^{p^{n-1}}=1,yxy^{-1}=x^{p+1}\rangle$. If $N=G$, then $[e]_{g,N}=[e]_{g}$, i. e. $[e]_{g,N}$ and $[e]_{g}\cap N$ can coincide. If $N=\langle x\rangle=\mathbb{Z}_{p^{n}}$, then $[e]_{x^{p^{k}},N}=e$ but from equality (10) follows that $[e]_{x^{p^{k}}}\cap N=[e]_{x^{p^{k}}}=\langle x^{p^{k+1}}\rangle$, therefore for $[e]_{g,N}=e$ the group $[e]_{g}\cap N$ can be arbitrary. Therefore there are no essential relations between $[e]_{g,N}$ and $[e]_{g}\cap N$.

Example 2. Let $\mathbb{F}$ be a field, $T$ be a variable and $\mathbb{F}(T)$ be the field of rational function over $\mathbb{F}$ with variable $T$. Denote by $P=\mathbb{F}(T)^{+}$ the additive group of the field $\mathbb{F}(T)$, by $D=\{d(T)\in\mathbb{F}(T)\leavevmode\nobreak\ |\leavevmode\nobreak\ d(1)=0\}$ and by $H=\{h(T)\in\mathbb{F}(T)\leavevmode\nobreak\ |\leavevmode\nobreak\ h(1)=1\}$. The group $D$ is a subgroup of $P$ and the group $H$ is a subgroup of the multiplicative group $\mathbb{F}(T)^{*}$ of the field $\mathbb{F}(T)$. Denote by $G$ the group G=\left\{{\footnotesize\begin{pmatrix}1&f\\ 0&h\end{pmatrix}}\leavevmode\nobreak\ \biggl{|}\leavevmode\nobreak\ f\in\mathbb{F}(T),h\in H\right\}. For elements $u={\footnotesize\begin{pmatrix}1&f\\ 0&h\end{pmatrix}}$ and $v={\footnotesize\begin{pmatrix}1&a\\ 0&b\end{pmatrix}}$ we have

[TABLE]

If $b\neq 1$, then denoting by $h=1$, $f=c(b-1)^{-1}$ we have

[TABLE]

i. e. $[e]_{v}$ is a subgroup of $G$. If $b=1$, then denoting by $h=1-d$ for $d\in D$ we have

[TABLE]

i. e. $[e]_{v}$ is again a subgroup of $G$. So the twisted conjugacy class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$.

From equality (11) follows that the center $Z(G)$ is trivial. So, Example 2 gives a negative answer to the following problem from [15, Problem 18.15].

[15, Problem 18.15]****. Let $G$ be a group with trivial center. Is it true that there exists an element $g\in G$ such that the class $[e]_{g}$ is not a subgroup of $G$.

Since $Z(G)$ is trivial, the group $G$ is not nilpotent. However, it is residually nilpotent.

Problem** 4****.**

Let $G$ be a finitely generated group such that the class $[e]_{g}$ is a subgroup of $G$ for every $g\in G$. Is it true that $G$ is residually nilpotent?**

If the group $G$ from Problem 4 is residually finite, then by Corollary 3 it is also residually nilpotent. In particular, if $G$ is finitely generated metabelian or finitely generated linear, then $G$ is residually finite and therefore residually nilpotent.

The positive solution of Problem 4 implies the positive solution of Problem 2. However, the positive solution of Problem 2 together with Theorem 2 gives only a hope to have a positive solution of Problem 4.