# Commutators, centralizers, and strong conciseness in profinite groups

Eloisa Detomi  Dipartimento di Ingegneria dell’Informazione - DEI, Università
di Padova, Via G. Gradenigo 6/B, 35121 Padova, Italy<EMAIL_ADDRESS>,
Marta Morigi  Dipartimento di Matematica, Università di Bologna
Piazza di Porta San Donato 5
40126 Bologna
Italy<EMAIL_ADDRESS>and  Pavel Shumyatsky  Department of Mathematics,
University of Brasilia
Brasilia-DF
70910-900 Brazil<EMAIL_ADDRESS>
###### Abstract.

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 $G$. Shalev
showed that a profinite group with restricted centralizers is virtually
abelian. We take interest in profinite groups with restricted centralizers of
uniform commutators, that is, elements of the form $[x_{1},\dots,x_{k}]$,
where $\pi(x_{1})=\pi(x_{2})=\dots=\pi(x_{k})$. Here $\pi(x)$ denotes the set
of prime divisors of the order of $x\in G$. It is shown that such a group
necessarily has an open nilpotent subgroup. We use this result to deduce that
$\gamma_{k}(G)$ is finite if and only if the cardinality of the set of uniform
$k$-step commutators in $G$ is less than $2^{\aleph_{0}}$.

###### Key words and phrases:

Commutators, centralizers, conciseness

###### 2010 Mathematics Subject Classification:

20E45, 20F24, 20E18

The first and second authors are members of GNSAGA (INDAM), and the third
author was supported by FAPDF and CNPq.

##  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 [34] where he showed that a profinite group
with restricted centralizers is virtually abelian. As usual, we say that a
profinite group has a property virtually if it has an open subgroup with that
property. Recently, profinite groups with restricted centralizers of some
specific elements were considered in [2, 10]. The article [10] handles
profinite groups with restricted centralizers of $w$-values for a multilinear
commutator word $w$. The theorem proved in [10] 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 $w(G)$ generated by all $w$-values is virtually abelian. Recall that
the lower central words $\gamma_{k}$ are recursively defined by

$\gamma_{1}=x_{1},\qquad\gamma_{k}=[\gamma_{k-1}(x_{1},\ldots,x_{k-1}),x_{k}].$

Of course $\gamma_{k}(G)$ is the $k$-th term of the lower central series of
$G$. Thus, if the $\gamma_{k}$-values have restricted centralizers in a
profinite group $G$, then $\gamma_{k}(G)$ is virtually abelian.

In this paper we will show that if the $\gamma_{k}$-values have restricted
centralizers, then the group $G$ is virtually nilpotent. In fact, we will
establish a stronger result.

If $G$ is a profinite group, then $|G|$ denotes its order, which is a Steinitz
number, and $\pi(G)$ denotes the set of prime divisors of $|G|$. Similarly, if
$g$ is an element of $G$, then $|g|$ and $\pi(g)$ respectively denote the
order of the procyclic subgroup generated by $g$ and the set of prime divisors
of $|g|$.

We will say that an element $g$ of a profinite group $G$ is a uniform $k$-step
commutator  (uk-commutator for short) if there are elements
$x_{1},x_{2},\ldots,x_{k}\in G$ such that $\pi(x_{1})=\dots=\pi(x_{k})$ and
$g=[x_{1},x_{2},\ldots,x_{k}]$. When $k=2$, the element $g$ will be referred
to simply as a uniform commutator (such elements were called anti-coprime
commutators in [6, 11]).

######  Theorem 1.1.

Let $G$ be a profinite group in which the centralizers of uniform $k$-step
commutators are either finite or open. Then $G$ is virtually nilpotent and
$\gamma_{k}(G)$ is virtually abelian.

We do not know if there exists a constant, say $C$, depending only on $k$ such
that any group $G$ satisfying the hypothesis of Theorem 1.1 has an open
nilpotent subgroup of class at most $C$. In the case where $k=2$ the
affirmative answer is furnished by the next result. A key tool employed in the
proof was established in [12] using probabilistic arguments.

######  Theorem 1.2.

Let $G$ be a profinite group in which the centralizers of uniform commutators
are either finite or open. Then $G$ has an open subgroup which is nilpotent of
class at most $3$.

A somewhat unexpected by-product of the proof of Theorem 1.1 is related to the
concept of strong conciseness in profinite groups introduced in [5]. A group-
word $w$ is strongly concise if the verbal subgroup $w(G)$ is finite in any
profinite group $G$ in which $w$ takes less than $2^{\aleph_{0}}$ values. A
number of recent results on strong conciseness of group-words can be found in
[5, 3, 22]. The concept of strong conciseness can be applied in a wider
context: suppose $\varphi(G)$ is a subset that can be naturally defined in
every profinite group $G$, then one can ask whether the subgroup generated by
$\varphi(G)$ is finite whenever $|\varphi(G)|<2^{\aleph_{0}}$.

It was shown in [11] that this is the case if $\varphi(G)$ is the set of
u2-commutators; that is, a profinite group $G$ has finite commutator subgroup
$G^{\prime}$ if and only if the cardinality of the set of u2-commutators in
$G$ is less than $2^{\aleph_{0}}$. In view of this, it was natural to ask
whether finite-by-nilpotent profinite groups admit a similar characterisation.
Now we are able to answer the question in the affirmative.

######  Theorem 1.3.

Let $k\geq 1$, and let $G$ be a profinite group. Then $\gamma_{k}(G)$ is
finite if and only if the cardinality of the set of uniform $k$-step
commutators in $G$ is less than $2^{\aleph_{0}}$.

Note that in Theorem 1.3 the order of $\gamma_{k}(G)$ is bounded in terms of
the number of uk-commutators (see Proposition 7.1). It would be interesting to
see if also profinite groups $G$ in which the $k$-th term of the derived
series is finite can be characterised in the same spirit. For now this remains
an open problem.

##  2\. Preliminaries

Results on finite groups often admit a natural interpretation for profinite
groups (see, for example, [29] or [37]). Throughout the paper we use certain
profinite versions of facts on finite groups without explaining in detail how
these can be deduced from the corresponding finite cases. On all such
occasions the deduction can be performed via the routine inverse limit
argument. Every homomorphism of profinite groups considered in this paper is
continuous, and every subgroup of a profinite group is closed, unless
otherwise specified.

If $x$ is an element of a group $G$, we write $x^{G}$ for the conjugacy class
of $x$ in $G$. On the other hand, if $K$ is a subgroup of $G$, then $K^{G}$
denotes the normal closure of $K$ in $G$, that is, the subgroup generated by
all conjugates of $K$ in $G$, with the usual convention that if $G$ is a
topological group then $K^{G}$ is a closed subgroup.

We will denote by $\Delta(G)$ the set of $FC$-elements of $G$, i.e.

$\Delta(G)=\\{x\in G\mid|x^{G}|<\infty\\}.$

Obviously $\Delta(G)$ is a normal abstract subgroup of $G$. Note that if $G$
is a profinite group, $\Delta(G)$ does not need to be closed.

Recall that a pro-$p$ group is an inverse limit of finite $p$-groups, a
pro-$\pi$ group is an inverse limit of finite $\pi$-groups, a pronilpotent
group is an inverse limit of finite nilpotent groups, a prosoluble group is an
inverse limit of finite soluble groups.

It is well-known that a finite-by-prosoluble group is virtually prosoluble,
and a profinite group $G$ that is an extension of a prosoluble group $N$ by a
prosoluble group $G/N$ is prosoluble (see e.g. [23, Lemma 2.2]).

If $G$ is a profinite group and $\pi$ a set of primes, $O_{\pi}(G)$ stands for
the maximal normal pro-$\pi$ subgroup of $G$.

Recall that $\gamma_{\infty}(G)$ stands for the intersection of the terms of
the lower central series of a group $G$. A profinite group $G$ is pronilpotent
if and only if $\gamma_{\infty}(G)=1$. Each profinite group $G$ has a maximal
pronilpotent normal subgroup, its Fitting subgroup $F(G)$. Set $F_{0}(G)=1$
and $F_{i+1}(G)/F_{i}(G)=F(G/F_{i}(G))$ for every integer $i\geq 0$. We say
that the profinite group $G$ has (finite) Fitting height $h=h(G)$ if
$G=F_{h}(G)$ and $h$ is the least integer with this property. Obviously, $G$
has finite Fitting height at most $h$ if, and only if, $G$ is an inverse limit
of finite soluble groups of Fitting height at most $h$. A profinite group $G$
is metapronilpotent if and only if it has Fitting height at most $2$ or,
equivalently, if and only if $\gamma_{\infty}(G)$ is pronilpotent. As usual,
$Z(G)$ denotes the centre of the group $G$.

We mention the following result about nilpotent profinite groups.

######  Lemma 2.1.

Let $G$ be an infinite nilpotent profinite group. Then $Z(G)$ is infinite.

###### Proof.

Assume that $Z(G)$ is finite. Then there exists an open normal subgroup $K$ of
$G$ such that $K\cap Z(G)=1$. Since $G$ is nilpotent, this implies that $K=1$
and so $G$ is finite, a contradiction. ∎

Profinite groups have Sylow $p$-subgroups and satisfy analogues of the Sylow
theorems. Any prosoluble group $G$ has a Sylow basis (a family of pairwise
permutable Sylow $p_{i}$-subgroups $P_{i}$ of $G$, exactly one for each prime
$p_{i}\in\pi(G)$), and any two Sylow bases are conjugate (see [29, Proposition
2.3.9]). The basis normalizer (also known as the system normalizer) of such a
Sylow basis in $G$ is $T=\bigcap_{i}N_{G}(P_{i}).$ If $G$ is a prosoluble
group and $T$ is a basis normalizer in $G,$ then $T$ is pronilpotent and
$G=\gamma_{\infty}(G)T$ (see [28, Lemma 5.6]).

The set of uniform $k$-step commutators of $G$ will be denoted by
$\mathcal{U}_{k}(G)$ and we write $\mathcal{U}_{k}=\mathcal{U}_{k}(G)$ when it
is clear which group we are referring to. Remark that $\mathcal{U}_{k}$ is
symmetric, that is, it is closed under taking inverses, since
$[x,y]^{-1}=[x^{y},y^{-1}]$. Moreover,
$\mathcal{U}_{k}(G/N)=\mathcal{U}_{k}(G)N/N$ whenever $N$ is a normal subgroup
of $G$.

If $x,y$ are elements of a group $G$ and $k$ is a positive integer,
$[x,{}_{k}y]$ is recursively defined by $[x,{}_{1}y]=[x,y]$ and
$[x,{}_{k+1}y]=[[x,{}_{k}y],y]$ for every $k\geq 1$. We will often use the
fact that if $G$ is profinite, then the element $[x,y,y]=[y^{-xy},y]$ is a
uniform commutator. Note that if the profinite group $G$ is pronilpotent, then
it is a direct product of its Sylow subgroups and so every $\gamma_{k}$-value
in $G$ is a uk-commutator.

######  Lemma 2.2.

Let $G$ be a profinite group. Then the set $\mathcal{U}_{k}$ generates
$\gamma_{k}(G)$.

###### Proof.

If $k=1$, then $\mathcal{U}_{k}=G$ and we have nothing to prove. Therefore we
assume that $k\geq 2$. Let $N$ be the subgroup generated by $\mathcal{U}_{k}$.
Obviously, $N\leq\gamma_{k}(G)$ so we only need to show that
$\gamma_{k}(G)\leq N$. Recall that $[x,y,y]=[y^{-xy},y]$ is a uniform
commutator for any $x,y\in G$. It follows that if $\bar{x}=Nx$ and
$\bar{y}=Ny$ are elements of $G/N$, then $[\bar{x},_{k}\bar{y}]=1$, as
$[y^{-xy},{}_{k-1}y]\in\mathcal{U}_{k}$. Since finite Engel groups are
nilpotent (see [30, 12.3.4]), we deduce that $G/N$ is pronilpotent. We
mentioned earlier that in a pronilpotent group every $\gamma_{k}$-value is a
uk-commutator. Therefore $G/N$ is nilpotent of class at most $k-1$, whence
$\gamma_{k}(G)\leq N$. ∎

If $\phi$ is an automorphism of a finite group $H$ of coprime order, that is,
such that $(|\phi|,|H|)=1$, then we say for brevity that $\phi$ is a coprime
automorphism of $H.$ This definition is extended to profinite groups as
follows. We say that $\phi$ is a coprime automorphism of a profinite group $H$
meaning that the procyclic group $\langle\phi\rangle$ faithfully acts on $H$
by continuous automorphisms and $\pi(\langle\phi\rangle)\cap\pi(H)=\emptyset$.
Since the semidirect product $H\langle\phi\rangle$ is also a profinite group,
$\phi$ is a coprime automorphism of $H$ if and only if for every open
$\phi$-invariant normal subgroup $N$ of $H$ the automorphism (of finite order)
induced by $\phi$ on $H/N$ is a coprime automorphism. We will need some
profinite equivalent of well-known results about coprime actions in finite
groups (see e.g. [23]).

######  Lemma 2.3.

If $\phi$ is a coprime automorphism of a profinite group $G,$ then for every
prime $q\in\pi(G)$ there is a $\phi$-invariant Sylow $q$-subgroup of $G.$

######  Lemma 2.4.

[23, Lemma 4.6] Let $\phi$ be a coprime automorphism of a finite nilpotent
group $G$. Then the set of elements of the form $[g,\phi]$, where $g\in G$,
coincides with the set of elements of the form $[g,\phi,\phi]$.

A repeated application of the previous lemma yields the following result for
profinite groups:

######  Lemma 2.5.

Let $\phi$ be a coprime automorphism of a pronilpotent group $G$ and let $k$
be a positive integer. Then the set of elements of the form $[g,\phi]$, where
$g\in G$, coincides with the set of elements of the form $[g,_{k}\phi]$.

###### Proof.

For every positive integer $n$, let $\alpha_{n}$ be the continuous map defined
on $G$ by $x\mapsto[x,_{n}\phi]$. We can repeatedly apply Lemma 2.4 to every
finite image $\bar{G}$ of $G$, which is nilpotent, to get
$\alpha_{1}(\bar{G})=\alpha_{2}(\bar{G})$, and so also
$\alpha_{1}(\bar{G})=\alpha_{k}(\bar{G})$. Since this holds for every finite
image of $G$ and the maps are continuous, we deduce that
$\alpha_{1}(G)=\alpha_{k}(G)$. ∎

##  3\. On groups in which centralizers of word values are either finite or
open

In this section we handle the important particular case of Theorem 1.1 where
the centralizers of all $\gamma_{k}$-values in $G$ are either finite or open.
We will use some results from [10], which were proved for general multilinear
commutator words.

Multilinear commutator words are words which are obtained by nesting
commutators, but using always different variables. More formally, the word
$w(x)=x$ in one variable is a multilinear commutator; if $u$ and $v$ are
multilinear commutators on disjoint sets of variables then the word $w=[u,v]$
is a multilinear commutator, and all multilinear commutators are obtained in
this way.

Clearly, the lower central words $\gamma_{k}$ are particular instances of
multilinear commutator words.

Throughout the article, if $w$ is a word and $G$ is a group, $G_{w}$ will
denote the set of all $w$-values in $G$.

We will need a combinatorial lemma proved in [10].

######  Lemma 3.1.

[10, Lemma 3.4] Let $w=w(x_{1},\dots,x_{n})$ be a multilinear commutator word.
Assume that $T$ is a normal subgroup of a group $G$ and $a_{1},\dots,a_{n}$
are elements of $G$ such that every element in the set
$\\{w(a_{1}t_{1},\dots,a_{n}t_{n})\mid t_{1},\dots,t_{n}\in T\\}$ has at most
$m$ conjugates in $G$. Then every element in $T_{w}$ has at most $m^{2^{n}}$
conjugates in $G$.

We will first consider profinite groups where every $\gamma_{k}$-value is an
$FC$-element. The following result was recently obtained in [35]. Here
$(k,m)$-bounded means bounded by a function of the parameters $k$ and $m$.

######  Theorem 3.2.

Let $k\geq 1$ and $G$ be a group in which $|x^{G}|\leq m$ for any
$\gamma_{k}$-value $x\in G$. Then $G$ has a nilpotent subgroup of
$(k,m)$-bounded index and $(k,m)$-bounded class.

To deal with the case $k=2$ we require a more complicated result from [12]
(see Theorem 1.2 and the preceding comments).

######  Theorem 3.3.

If $G$ is a group in which $|x^{G}|\leq m$ for any commutator $x\in G$, then
$G$ has a subgroup $H$ of nilpotency class at most $4$ such that $[G:H]$ and
$|\gamma_{4}(H)|$ are both finite and $m$-bounded.

######  Proposition 3.4.

Let $k$ be a positive integer and $G$ a profinite group in which every
$\gamma_{k}$-value is an $FC$-element. Then $G$ is virtually nilpotent. If
$k=2$ then $G$ has an open subgroup $H$ of nilpotency class at most $3$.

###### Proof.

For each positive integer $j$ consider the set $\Delta_{j}$ of elements $g\in
G$ such that $|g^{G}|\leq j$. Note that the sets $\Delta_{j}$ are closed (see
for instance [26, Lemma 5]). Set:

$C_{j}=\\{(g_{1},\dots,g_{k})\mid g_{i}\in G{\textrm{ and
}}[g_{1},\dots,g_{k}]\in\Delta_{j}\\}.$

Each set $C_{j}$ is closed in $G\times\cdots\times G$, being the inverse image
of the closed set $\Delta_{j}$ under the continuous map
$(g_{1},\dots,g_{k})\mapsto[g_{1},\dots,g_{k}]$. Moreover the union of the
sets $C_{j}$ is the whole group $G\times\cdots\times G$. By the Baire category
theorem (cf. [21, p. 200]) at least one of the sets $C_{j}$ has nonempty
interior. Hence, there exists a positive integer $m$, some elements $z_{i}\in
G$, and an open normal subgroup $T$ of $G$ such that

$[z_{1}T,\dots,z_{n}T]\subseteq\Delta_{m}.$

By Lemma 3.1, every $\gamma_{k}$-value in elements of $T$ has at most
$m^{2^{k}}$ conjugates in $G$. It follows from Theorem 3.2 that $T$ has a
nilpotent abstract subgroup $B$ of $(k,m)$-bounded index and $(k,m)$-bounded
class. Then the topological closure of $B$ is an open nilpotent subgroup of
$G$. Thus $G$ is virtually nilpotent, as claimed.

If $k=2$, by Theorem 3.3 the subgroup $T$ contains a nilpotent abstract
subgroup $B$ of $m$-bounded index and such that $\gamma_{4}(B)$ has
$m$-bounded order. Then the topological closure $\overline{B}$ of $B$ is an
open subgroup of $G$ such that $\gamma_{4}(\overline{B})$ has finite order.
Choose an open subgroup $K\leq G$ such that $K\cap\gamma_{4}(\overline{B})=1$.
Observe that $K\cap\overline{B}$ is open in $G$ and nilpotent of class at most
$3$, as required. ∎

We will require the following corollary of the main result in [10].

######  Lemma 3.5.

[10, Corollary 1.2] Let $w$ be a multilinear commutator word and $G$ a
profinite group in which the centralizers of $w$-values are either finite or
open. Then $G$ has an open subgroup $T$ such that $w(T)$ is abelian.

######  Proposition 3.6.

Let $k$ be a positive integer and $G$ a profinite group in which the
centralizers of $\gamma_{k}$-values are either finite or open. Then $G$ is
virtually nilpotent. If $k=2$, then $G$ has an open subgroup which is
nilpotent of class at most $3$.

###### Proof.

By Lemma 3.5, $G$ has an open subgroup $T$ such that $\gamma_{k}(T)$ is
abelian. Without loss of generality, we can assume that $G=T$.

If all $\gamma_{k}$-values of $G$ are $FC$-elements, then by Proposition 3.4
we conclude that $G$ is virtually nilpotent. In particular, if $k=2$, then $G$
has an open subgroup which is nilpotent of class at most $3$.

So assume that there exists a $\gamma_{k}$-value whose centralizer in $G$ is
finite. As $\gamma_{k}(G)$ is abelian, we conclude that $\gamma_{k}(G)$ is
finite. Therefore there exists an open normal subgroup $N$ of $G$ such that
$N\cap\gamma_{k}(G)=1$. In particular, $\gamma_{k}(N)=1$ and $N$ is an open
nilpotent subgroup of $G$. If $k=2$ then $N$ is abelian and $G$ is virtually
abelian. This concludes the proof. ∎

##  4\. On groups in which centralizers of uniform commutators are finite

In this section we will prove Theorem 1.1 in the special case where the
elements of $\mathcal{U}_{k}$ have finite centralizers (see Proposition 4.2).
Throughout, it will be assumed that $k\geq 2$. Note that if $G$ is a profinite
group where all $\gamma_{k}$-values have finite centralizers, then $G$ is
either finite or nilpotent of class at most $k-1$ [10, Corollary 1.3].

We will repeatedly use the following observation:

######  Lemma 4.1.

Let $G$ be a profinite group in which the centralizers of uk-commutators are
finite. If $H$ is a pronilpotent subgroup of $G$, then either $H$ is finite or
$H\cap\mathcal{U}_{k}=1$. In the latter case, $H$ is nilpotent of class at
most $k-1$.

###### Proof.

Since $H$ is a direct product of its Sylow subgroups, it follows that any
$\gamma_{k}$-value of $H$ lies in $\mathcal{U}_{k}$ and, by [10, Corollary
1.3], $H$ is either finite or nilpotent of class at most $k-1$. Assume
$H\cap\mathcal{U}_{k}\neq 1$ and let $y$ be a nontrivial element of
$H\cap\mathcal{U}_{k}$. Since $Z(H)\leq C_{G}(y)$, it follows that $Z(H)$ is
finite. In view of Lemma 2.1 this implies that $H$ is finite. ∎

######  Proposition 4.2.

Let $G$ be a profinite group in which the centralizers of uk-commutators are
finite. Then $G$ is either finite or nilpotent of class at most $k-1$.

###### Proof.

If $G$ is pronilpotent, then the result is immediate from the previous lemma.
Assume that $G$ is not pronilpotent. We want to prove that $G$ is finite. We
will use the fact that the Sylow subgroups of $G$ are either finite or
nilpotent of class at most $k-1$.

The profinite version of Burnside’s theorem [15, Theorem 3.3] says that if
$N_{G}(P)/C_{G}(P)$ is a pro-$p$ group for every nontrivial pro-$p$ subgroup
$P$ of $G$, then $G$ has a normal $p$-complement. Note that a group is
pronilpotent whenever it has normal $p$-complement for every prime $p$. As our
group $G$ is not pronilpotent, there exists a prime $p$ and a pro-$p$ subgroup
$P\leq G$ such that $N_{G}(P)/C_{G}(P)$ contains a nontrivial
$p^{\prime}$-element. So $N_{G}(P)\setminus C_{G}(P)$ contains a
$p^{\prime}$-element $a$ that induces a nontrivial coprime automorphism on
$P$. We deduce from Lemma 2.5 that there exists a nontrivial $p$-element
$y=[x,_{k}a]\in\mathcal{U}_{k}\cap P$. It follows from Lemma 4.1 that the
Sylow $p$-subgroups of $G$ are finite.

Let $N$ be an open normal pro-$p^{\prime}$ subgroup of $G$ intersecting
$C_{G}(y)$ trivially. Then $C_{N}(y)=1$ and $y$ acts coprimely on $N$. Let $q$
be a prime in $\pi(N)$. By Lemma 2.3, there is a $y$-invariant Sylow
$q$-subgroup $Q$ of $N$.

As $C_{N}(y)=1$, the map $x\mapsto[x,y]$ is injective on $Q$. Hence, by Lemma
2.5, $\mathcal{U}_{k}\cap Q\neq 1$. Thus $Q$ is a finite $q$-group by Lemma
4.1, and the map $x\mapsto[x,y]$ is also surjective. Therefore
$Q\subseteq\mathcal{U}_{k}$ by Lemma 2.5.

Choose an element $z\in Q$ of prime order $q$. Since $C_{G}(z)$ is finite,
there exists an open normal $q^{\prime}$-subgroup $K$ of $G$ that has trivial
intersection with $C_{G}(z)$. In particular, $z$ acts coprimely and fixed-
point-freely on $K$. As $z$ has prime order, combining the well-known results
of Thompson and Higman [18, 36] (see also [37, Theorem 2.6.2]) we conclude
that $K$ is nilpotent. Again, by Lemma 2.5, we deduce that
$K\subseteq\mathcal{U}_{k}$, whence $K$ is finite by Lemma 4.1. It follows
that $G$ is finite. ∎

##  5\.  On groups in which uniform commutators are FC-elements

In this section we will prove the following proposition.

######  Proposition 5.1.

Let $G$ be a profinite group such that $\mathcal{U}_{k}\subseteq\Delta(G)$.
Then $G$ is virtually nilpotent.

We will need a technical results about commutators.

######  Lemma 5.2.

Let $x_{1},\dots,x_{n}$ be elements of a group $G$ and let
$y\in\\{x_{1},\dots,x_{n}\\}$. Then $[x_{1},\dots,x_{n}]$ is a product of
$2^{n-1}$ conjugates of $y^{\pm 1}$.

###### Proof.

Using basic commutator identities it can be easily seen that if $v$ is a
product of $t$ conjugates of $y^{\pm 1}$, then both $[v,x]=v^{-1}v^{x}$ and
$[x,v]=v^{-x}v$ are products of $2t$ conjugates of $y^{\pm 1}$. Then, by
induction on $n$, it follows that $[x_{1},\dots,x_{n}]$ is a product of
$2^{n-1}$ conjugates of $y^{\pm 1}$. ∎

Suppose that a set $\Sigma$ is a union of finitely many subsets
$\Sigma_{1},\dots,\Sigma_{m}$. Then $\Sigma$ admits a partition with blocks
$\Omega_{1},\dots,\Omega_{s}$ such that $s\leq 2^{m}$ and each $\Sigma_{i}$ is
a (disjoint) union of some of the blocks $\Omega_{1},\dots,\Omega_{s}$.
Formally, we can take $\Omega_{1},\dots,\Omega_{s}$ to be all the nonempty
sets of the form

$\Gamma_{1}\cap\Gamma_{2}\cap\dots\cap\Gamma_{m},\quad\textrm{ where
}\Gamma_{i}\textrm{ is either }\Sigma_{i}\textrm{ or
}\Sigma\setminus\Sigma_{i}\textrm{ for all }i.$

This will be used in the next lemma.

######  Lemma 5.3.

Let $N$ be a normal pronilpotent subgroup of a profinite group G. Let
$X\subseteq G$ be the set of commutators $[g_{1},...,g_{k}]$ where at least
one of the entries $g_{1},...,g_{k}$ belongs to $N$. Then every element of $X$
is a product of $k$-boundedly many elements in $\mathcal{U}_{k}\cap N$.

###### Proof.

Let us fix an element $[g_{1},\dots,g_{k}]\in X$, with $g_{i}\in N$. The set

$\pi(g_{1})\cup\dots\cup\pi(g_{k})$

admits a finite partition with blocks $\pi_{1},\dots,\pi_{s}$ with the
property that each $\pi(g_{t})$ is a union of some of these blocks. Note that
$s\leq 2^{k}$. So each element $g_{t}$ is a product

$g_{t}=\prod_{j=1}^{s}y_{tj}\textrm{ where }y_{tj}\in\langle
g_{t}\rangle,\,\,\pi(y_{tj})=\pi_{j}\textrm{ whenever }y_{tj}\neq 1.$

Repeatedly using the commutator identity $[ab,c]=[a^{b},c^{b}]\,[b,c]$ we
obtain that $[g_{1},\dots,g_{k}]$ is a product of $k$-boundedly many elements
of the form: $y=[u_{1},\dots,u_{k}]$, where, for each $t$, $u_{t}$ is a
conjugate of some $y_{tj}$. In particular $u_{i}\in N$, and for each
$t,j=1,\dots,k$ the sets $\pi(u_{t})$ and $\pi(u_{j})$ are either equal or
disjoint. So, taking into account that $N$ is normal, we just need to show
that every such element $y=[u_{1},\dots,u_{k}]$ is a product of $k$-boundedly
many elements in $\mathcal{U}_{k}\cap N$.

Let $\tilde{\pi}=\pi(u_{i})$. As $N$ is pronilpotent and normal in $G$, the
subgroup $O_{\tilde{\pi}}(N)$ is normal in $G$. In the sequel, we will use the
following fact: If $a\in O_{\tilde{\pi}}(N)$ and $b$ induces a coprime
automorphism on $O_{\tilde{\pi}}(N)$, then $[a,b]\in\mathcal{U}_{k}$. Indeed,
by Lemma 2.5, there exists $d\in O_{\tilde{\pi}}(N)$ such that
$[a,b]=[d,{}_{k}b]=[b^{-db},{}_{k-1}b]$.

If $\tilde{\pi}=\pi(u_{1})=\dots=\pi(u_{k})$ then $y\in\mathcal{U}_{k}\cap N$.
Otherwise there exists $j$ such that

$\pi(u_{j})\neq\tilde{\pi}.$

Assume that $j>i$. We have that $x=[u_{1},\dots,u_{j-1}]\in
O_{\tilde{\pi}}(N)$, and $u_{j}$ induces a coprime automorphism on
$O_{\tilde{\pi}}(N)$. Therefore,
$[u_{1},\dots,u_{j}]=[x,u_{j}]\in\mathcal{U}_{k}\cap N$ and it follows from
Lemma 5.2, that $y=[u_{1},\dots,u_{k}]$ is a product of $k$-boundedly many
elements in $\mathcal{U}_{k}\cap N$, which are all conjugates of
$[u_{1},\dots,u_{j}]$ or $[u_{1},\dots,u_{j}]^{-1}$.

So we can assume that $j<i$. By Lemma 5.2 the element
$x=[u_{1},\dots,u_{i-1}]$ is the product of $k$-boundedly many conjugates of
$u_{j}$ or $u_{j}^{-1}$, so that $[x,u_{i}]^{-1}=[u_{i},x]$ is the product of
$k$-boundedly elements of the form $[a,b]$, where $a\in O_{\tilde{\pi}}(N)$
and $b$ is a conjugate of $u_{j}$, so it acts coprimely on
$O_{\tilde{\pi}}(N)$. All elements $[a,b]$ belong to $\mathcal{U}_{k}\cap N$,
thus $[x,u_{i}]^{-1}=[u_{1},\dots,u_{i}]^{-1}$ is the product of $k$-boundedly
many elements in $\mathcal{U}_{k}\cap N$. Since $\mathcal{U}_{k}$ is
symmetric, the result follows. ∎

######  Lemma 5.4.

Let $G$ be a metapronilpotent group. Then every ${\gamma_{k}}$-value in $G$ is
a product of $k$-boundedly many elements in $\mathcal{U}_{k}$.

###### Proof.

Write $G=NT$, where $T$ is a system normalizer and $N=\gamma_{\infty}(G)$.
Since $G$ is metapronilpotent, it follows that $N$ and $T$ are pronilpotent.
Let $x$ be a $\gamma_{k}$-value of $G$. Then we can write $x$ in the form

$x=[n_{1}t_{1},\dots,n_{k}t_{k}],$

where $n_{i}\in N$ and $t_{i}\in T$ for all $i=1,\dots,k$. Using the basic
commutators identities $[ab,c]=[a,c]^{b}[b,c]$, $[a,bc]=[a,c][a,b]^{c}$ and
$ab=b^{a^{-1}}a=ba^{b}$ we can write $x$ as

$x=[t_{1},\dots,t_{k}]\prod_{i=1}^{2^{k}-1}[x_{i1},\dots,x_{ik}]$

where all $x_{ij}$ are conjugates of elements in $N\cup T$ and at least one
entry in each $\gamma_{k}$-value $[x_{i1},\dots,x_{ik}]$ belongs to $N$. As
$T$ is pronilpotent, $[t_{1},\dots,t_{k}]\in\mathcal{U}_{k}$. Moreover, by
Lemma 5.3, all $[x_{i1},\dots,x_{ik}]$ are products of $k$-boundedly many
elements in $\mathcal{U}_{k}\cap N$. This concludes the proof. ∎

Any finite group $H$ has a normal series each of whose factors either is
soluble or is a direct product of nonabelian simple groups. The nonsoluble
length of $H$, denoted by $\lambda(H)$, is defined as the minimum number of
nonsoluble factors in a series of this kind. It is easy to see that the
nonsoluble length $\lambda(H)$ is equal to the least positive integer $l$ such
that there is a series of characteristic subgroups

$1=L_{0}\leq R_{0}<L_{1}\leq R_{1}<\dots\leq R_{l}=H$

in which each quotient $L_{i}/R_{i-1}$ is a (nontrivial) direct product of
nonabelian simple groups, and each quotient $R_{i}/L_{i}$ is soluble (possibly
trivial).

It is natural to say that a profinite group $G$ has finite nonprosoluble
length at most $l$ if $G$ has a normal series

$1=L_{0}\leq R_{0}<L_{1}\leq R_{1}<\dots\leq R_{l}=G$

in which each quotient $L_{i}/R_{i-1}$ is a (nontrivial) Cartesian product of
nonabelian finite simple groups, and each quotient $R_{i}/L_{i}$ is prosoluble
(possibly trivial). In particular if, for some positive integer $m,$ all
continuous finite quotients of a profinite group $G$ have nonsoluble length at
most $m,$ then $G$ has finite nonprosoluble length at most $m$ (see e.g. [38,
Lemma 2]).

In the rest of this section $G$ will be a profinite group such that
$\mathcal{U}_{k}\subseteq\Delta(G)$. Of course, in every quotient $G/N$ of $G$
the elements of $\mathcal{U}_{k}(G/N)$ are $FC$-elements.

######  Lemma 5.5.

Suppose $G$ is a direct product of nonabelian finite simple groups. Then $G$
is finite.

###### Proof.

Let $G=\Pi_{i\in I}S_{i}$, where each $S_{i}$ is a nonabelian finite simple
group. In every factor $S_{i}$ choose a nontrivial element
$a_{i}\in\mathcal{U}_{k}(S_{i})$. Note that $a=\prod_{i\in
I}a_{i}\in\mathcal{U}_{k}$. Further, observe that $C_{G}(a)=\prod_{i\in
I}C_{S_{i}}(a_{i})$ has finite index in $G$ if and only if $I$ is finite. This
proves the result. ∎

######  Lemma 5.6.

The group $G$ is virtually prosoluble.

###### Proof.

Suppose that $G$ is not prosoluble. Let $P$ be a Sylow $2$-subgroup of $G$.
Since $\mathcal{U}_{k}(P)=P_{\gamma_{k}}$, it follows from Proposition 3.6
that $P$ is virtually nilpotent. Thus $P$ is soluble, say of derived length
$d$. By [24, Theorem 1.4] every finite image of $G$ has nonsoluble length at
most $d$ and so also $G$ has nonprosoluble length at most $d$.

Let

$1=L_{0}\leq R_{0}<L_{1}\leq R_{1}<\dots\leq R_{s}=G$

be a normal series of finite length in which each section $L_{i}/R_{i-1}$ is a
(nontrivial) direct product of nonabelian finite simple groups, and each
section $R_{i}/L_{i}$ is prosoluble (possibly trivial). By Lemma 5.5, every
section $L_{i}/R_{i-1}$ is finite. Let

$H=\\{x\in G\mid\ [L_{i},x]\leq R_{i-1}\text{ for all }i\\}$

be the centralizer in $G$ of the non-prosoluble sections of the series. This
is an open prosoluble subgroup of $G$. ∎

######  Lemma 5.7.

If $G$ is prosoluble, then $F(G)\neq 1$.

###### Proof.

Assume by contradiction that $F(G)=1$ and let $x$ be a nontrivial element in
$\mathcal{U}_{k}$. Then $K=\langle x^{G}\rangle$ is generated by finitely many
conjugates of $x$ and so its centralizer has finite index in $G$. It follows
that the centre $Z(K)$ of $K$ has finite index in $K$. Moreover $Z(K)\leq
F(G)=1$. Therefore $Z(K)=1$ and consequently $K$ is finite. As $G$ is
prosoluble, $K$ is soluble and $F(K)\neq 1$. As $F(K)\leq F(G)$, this
contradicts the assumption that $F(G)=1$. ∎

Now we are ready to prove Proposition 5.1.

###### Proof of Proposition 5.1.

By Lemma 5.6 we may assume that $G$ is prosoluble and so $F(G)\neq 1$ by Lemma
5.7. If $F(G)=G$, then all $\gamma_{k}$-values of $G$ are contained in
$\mathcal{U}_{k}$ and the result is immediate from Proposition 3.6.

Assume that $G$ is not pronilpotent and suppose first that $G=F_{2}(G)$. Then
$G=NT$ with $N=\gamma_{\infty}(G)$ and $T$ a system normalizer. In this
situation $N$ and $T$ are pronilpotent and in view of Lemma 5.4 every
$\gamma_{k}$-value in $G$ is a product of finitely many elements from
$\mathcal{U}_{k}$. Thus all $\gamma_{k}$-values belong to the $FC$-centre
$\Delta(G)$ of $G$ and it follows from Proposition 3.6 that $G$ is virtually
nilpotent.

This proves that $F_{2}(G)$ is virtually nilpotent, whence $F_{2}(G)/F(G)$ is
finite. Let $\bar{G}=G/F(G)$, and let $\bar{K}$ be an open normal subgroup of
$\bar{G}$ such that $\bar{K}\cap F(\bar{G})=1$. Then $F(\bar{K})=1$ which,
because of Lemma 5.7, implies that $\bar{K}=1$. Hence $\bar{G}$ is finite and
$G$ is virtually nilpotent. This concludes the proof. ∎

##  6\. Proof of Theorem 1

We start with the case where $G$ is a metapronilpotent group.

######  Lemma 6.1.

Let $G$ be a metapronilpotent group in which the centralizers of uk-
commutators are either finite or open. Then $G$ is virtually nilpotent.

###### Proof.

Write $G=NT$, where $T$ is a system normalizer and $N=\gamma_{\infty}(G)$.
Since $G$ is metapronilpotent, it follows that $N$ and $T$ are pronilpotent.
Hence, by Proposition 3.6 both $N$ and $T$ are virtually nilpotent. Let
$N_{0}\leq N$ and $T_{0}\leq T$ be nilpotent open subgroups of $N$ and $T$
respectively. Remark that $N_{0}$ can be chosen characteristic in $G$. Notice
that $L=N_{0}T_{0}$ is open in $G$ and it is sufficient to show that $L$ is
virtually nilpotent. Obviously, $L$ is soluble. Arguing by induction on the
derived length of $L$ we can assume that $L^{\prime}$ is virtually nilpotent.
Then $L$ has an open subgroup whose commutator subgroup is nilpotent and so
without loss of generality we can assume that $L^{\prime}$ is nilpotent.

If $L^{\prime}$ is finite, then $L$ is virtually abelian. Assume that
$L^{\prime}$ is infinite. Lemma 2.1 says that $Z(L^{\prime})$ is infinite,
too. Observe that every element of $\mathcal{U}_{k}(L)$ centralizes
$Z(L^{\prime})$. Hence, every element of $\mathcal{U}_{k}(L)$ is an
$FC$-element, and therefore $L$ is virtually nilpotent by Proposition 5.1. ∎

Recall that a group is locally finite if every finite subset is contained in a
finite subgroup.

######  Lemma 6.2.

Let $G$ be a profinite group and $H$ a locally finite normal abstract subgroup
of $G$. Let $\hat{H}$ be the closure of $H$. Suppose that $C_{H}(a)=1$ for
some torsion element $a\in G.$ Then $C_{G/\hat{H}}(a)$ is the image of
$C_{G}(a)$ in $G/\hat{H}.$

###### Proof.

Let $h\in H$. Then $J=\langle h\rangle^{\langle a\rangle}$ is a finite
$a$-invariant subgroup of $H$. Consider the map $f:J\mapsto J$ defined by:
$x\mapsto[x,a]$. Since $C_{H}(a)=1$, the map is injective and hence
surjective. So $h=[x,a]$ for some $x\in J$. This holds for every $h\in H$ and
therefore the natural extension of $f$ to $\hat{H}$ is also surjective.

Clearly, $C_{G/\hat{H}}(a)$ contains the image of $C_{G}(a)$ in $G/\hat{H}$.
Conversely, if $x\hat{H}\in C_{G/\hat{H}}(a)$, then $[x,a]\in\hat{H}$, whence
$[x,a]=[h,a]$ for some $h\in\hat{H}$. We deduce that $xh^{-1}\in C_{G}(a)$ and
$x\in C_{G}(a)\hat{H}$. Thus $C_{G/\hat{H}}(a)=C_{G}(a)\hat{H}/\hat{H}$, as
required. ∎

Now we are ready to prove the result on virtually nilpotency.

######  Proposition 6.3.

Let $G$ be a profinite group in which the centralizers of uniform $k$-step
commutators are either finite or open. Then $G$ is virtually nilpotent.

###### Proof.

Suppose that an element $x\in\mathcal{U}_{k}$ has infinite order. Then the
centralizer $C=C_{G}(x)$ is infinite and therefore open. Every uk-commutator
lying in $C$ is centralized by $x$, which has infinite order. Therefore every
uk-commutator in $C$ is an $FC$-element. Apply Proposition 5.1 to deduce that
$C$ is virtually nilpotent, and the same holds for $G$, as $C$ is open in $G$.
Hence, without loss of generality, we will assume that every element of
$\mathcal{U}_{k}$ has finite order.

Let $Y\subseteq\mathcal{U}_{k}$ be the set of elements of $\mathcal{U}_{k}$
that have open centralizers, and let $H$ be the abstract subgroup generated by
$Y$. Note that $H$ is contained in the $FC$-centre of $G$. In particular, $H$
is locally finite. If $Y=\mathcal{U}_{k}$, the result is immediate from
Proposition 5.1 so assume that $Y\neq\mathcal{U}_{k}$.

Let $a\in\mathcal{U}_{k}\setminus Y$. As the centralizer $C_{G}(a)$ is finite
and $H$ is residually finite, there exists a normal subgroup $K$ of finite
index in $H$ such that $C_{K}(a)=1$. Let $\hat{H}$ and $\hat{K}$ be the
topological closures of $H$ and $K$, respectively. It follows from Lemma 6.2
that $C_{G/\hat{K}}(a)=C_{G}(a)\hat{K}/\hat{K}$ and therefore
$C_{G/\hat{K}}(a)$ is finite. Since $\hat{H}/\hat{K}$ is finite, also
$C_{G/\hat{H}}(a)$ is finite (see e.g. [34, Lemma 2.1]). Thus every nontrivial
element of $\mathcal{U}_{k}$ has finite centralizer in $G/\hat{H}$. In view of
Proposition 4.2 we conclude that $G/\hat{H}$ is virtually nilpotent.

Let us now examine the action of $a$ on $H$. Again, $K$ is a finite index
$a$-invariant subgroup of $H$ such that $C_{K}(a)=1$. A well-known corollary
of the classification of finite simple groups is that a finite group admitting
a fixed-point-free automorphism is soluble (see e.g. [32]). Thus $K$ is
locally soluble. Recall that a Carter subgroup is a self-normalizing nilpotent
subgroup and note that $\langle a\rangle$ is a Carter subgroup in every finite
subgroup $T$ of $K\langle a\rangle$ such that $a\in T$. The main result in
Dade’s paper [4] implies that the Fitting height $h(T)$ of $T$ is bounded by a
function depending only on the order of $a$. We deduce that $K$ has a
characteristic series of finite length all of whose factors are locally
nilpotent. Therefore $\hat{K}$ has a finite characteristic series with
pronilpotent factors, that is, $\hat{K}$ has finite Fitting height. As
$\hat{K}$ is open in $\hat{H}$ and $G/\hat{H}$ is virtually nilpotent, we
conclude that also $G$ has an open prosoluble subgroup whose Fitting height is
finite. So, without loss of generality we can assume that $G$ is prosoluble
and $h(G)$ is finite.

If $h(G)\leq 2$, Lemma 6.1 implies that $G$ is virtually nilpotent. Assume
that $h(G)>2$ and argue by induction of $h(G)$. It follows that $G/F(G)$ is
virtually nilpotent. Hence, $G$ has an open subgroup $M$ such that $h(M)\leq
2$. In view of Lemma 6.1 $M$ (and therefore $G$) is virtually nilpotent. The
proof is now complete. ∎

Now Theorem 1.2 follows.

###### Proof of Theorem 1.2.

Let $G$ be a profinite group in which the centralizers of elements of
$\mathcal{U}_{2}$ are either finite or open. It follows from Proposition 6.3
that $G$ is virtually nilpotent. So we can assume that $G$ is nilpotent. In
that case every commutator in $G$ is a uniform commutator. In view of
Proposition 3.6 we conclude that $G$ has an open subgroup which is nilpotent
of class at most $3$. ∎

It remains to prove the part of Theorem 1.1 which states that $\gamma_{k}(G)$
is virtually abelian.

Let G be a group and $w=w(x_{1},\dots,x_{n})$ a word. The marginal subgroup
$w^{*}(G)$ of $G$ corresponding to the word $w$ is defined as the set of all
$x\in G$ such that

$w(g_{1},\dots,xg_{i},\dots,g_{n})=w(g_{1},\dots,g_{i}x,\dots,g_{n})=w(g_{1},\dots,g_{i},\dots,g_{n})$

for all $g_{1},\dots,g_{n}\in G$ and $1\leq i\leq n$. It is well known that
$w^{*}(G)$ is a characteristic subgroup of $G$ and $[w^{*}(G),w(G)]=1$. Note
that marginal subgroups in profinite groups are closed.

Let $S$ be a subset of a group $G$. Following [10] define the $w^{*}$-residual
of $S$ of $G$ to be the intersection of all normal subgroups $N$ such that
$SN/N$ is contained in the marginal subgroup $w^{*}(G/N)$.

For multilinear commutator words the $w^{*}$-residual of a normal subgroup has
the following properties.

######  Lemma 6.4.

[10, Lemma 4.1] Let $w$ be a multilinear commutator word, $G$ a group and $N$
a normal subgroup of $G$. Then the $w^{*}$-residual of $N$ in $G$ is the
subgroup generated by the elements $w(g_{1},\dots,g_{n})$ where at least one
of $g_{1},\dots,g_{n}$ belongs to $N$.

######  Lemma 6.5.

[10, Lemma 4.2] Let $w$ be a multilinear commutator word, $G$ a profinite
group and $N$ an open normal subgroup of $G$. Then the $w^{*}$-residual of $N$
is open in $w(G)$.

The following result is a particular case of [10, Proposition 4.5]:

######  Proposition 6.6.

Let $G$ be a profinite group and $N$ a normal subgroup of $G$. Let $H$ be the
topological closure of $\Delta(G)$ in $G$. Fix $i\in\\{1,\dots,k\\}$ and
consider the set $X_{i}=\\{[g_{1},\dots,g_{k}]\mid g_{i}\in N,g_{j}\in G\\}$.
If

$X_{i}\subseteq\Delta(G),$

then $[H,\langle X_{i}\rangle]$ is finite.

Following the lines of [10, Theorem 4.3], we have:

######  Theorem 6.7.

Assume that $G$ is a profinite group in which the centralizers of uk-
commutators are either finite or open and $N$ is an infinite normal nilpotent
subgroup of $G$. Then the $\gamma_{k}^{*}$-residual of $N$ has finite
commutator subgroup.

###### Proof.

For $i=1,\dots n$, let $X_{i}$ be the set of $\gamma_{k}$-values
$[g_{1},...,g_{k}]$ such that $g_{i}$ belongs to $N$. It follows from Lemma
5.3 that every element in $X_{i}$ is a product of $k$-boundedly many elements
in $\mathcal{U}_{k}\cap N$. As $N$ is infinite, the center $Z(N)$ of $N$ is
infinite as well, thus every element in $\mathcal{U}_{k}\cap N$ is an
$FC$-element. Therefore also the set $X_{i}$ consists of $FC$-elements. It
follows from Proposition 6.6 that $[H,\langle X_{i}\rangle]$ is finite for
every $i$. By Lemma 6.4, the $\gamma_{k}^{*}$-residual of $N$ is the subgroup
$R$ generated by the set $X=X_{1}\cup\dots\cup X_{k}$. Thus
$[H,R]=\prod_{i=1}^{k}[H,\langle X_{i}\rangle]$ is finite. Finally, note that
$R\leq H$ and so $R^{\prime}\leq[H,R]$ is also finite. ∎

Now Theorem 1.1 follows.

###### Proof of Theorem 1.1.

Assume that $G$ is a profinite group in which the centralizers of uniform
$k$-step commutators are either finite or open. We proved in Proposition 6.3
that $G$ is virtually nilpotent. Now we will show that $\gamma_{k}(G)$ is
abelian-by-finite. Let $N$ be an open nilpotent subgroup of $G$. Of course we
can assume that $N$ is infinite. By Theorem 6.7, the $\gamma_{k}^{*}$-residual
$R$ of $N$ has finite commutator subgroup, thus it is virtually abelian.
Moreover, by Lemma 6.5 $R$ is open in $\gamma_{k}(G)$. Thus $\gamma_{k}(G)$ is
virtually abelian and the proof is complete. ∎

##  7\. Strong conciseness of uniform commutators

A word $w$ is said to be concise in a class of groups $\mathcal{X}$ if $w(G)$
is finite whenever the set of $w$-values in $G$ is finite for a group
$G\in\mathcal{X}$. In the sixties Hall raised the problem whether all words
are concise, but in 1989 S. Ivanov [19] solved the problem in the negative
(see also [27, p. 439]). On the other hand, the problem for residually finite
groups remains open (cf. Segal [33, p. 15] or Jaikin-Zapirain [20]). In recent
years several positive results with respect to this problem were obtained (see
[1, 17, 14, 7, 8, 9]).

A word $w$ is called boundedly concise in a class of groups $\mathcal{X}$ if
whenever the set of its values is finite of size at most $m$ in a group
$G\in\mathcal{X}$, it always follows that the subgroup $w(G)$ is finite of
order bounded by a function of $m$ and $w$. In [13] it is shown that every
word which is concise in the class of all groups is actually boundedly
concise. There is a conjecture that every word which is concise in residually
finite groups is boundedly concise (cf. [14]) but this probably will remain
open for some time.

On the other hand, the multilinear commutator words and words implying virtual
nilpotency are known to have this property [7]. Recall that a word $w$ is said
to imply virtual nilpotency if every finitely generated metabelian group $G$
where $w$ is a law has a nilpotent subgroup of finite index. It follows from
Gruenberg’s result [16] that the Engel words imply virtual nilpotency, so they
are boundedly concise.

The main result in [13] states that if $w$ is a multilinear commutator word
and the set of $w$-values in a group $G$ has size at most $m$, then the verbal
subgroup $w(G)$ is finite of order bounded by a function of $m$, independently
of $w$. We will show that in the class of profinite groups the set
$\mathcal{U}_{k}$ has the same property.

Recall that the $i$-th centre $Z_{i}(G)$ of a group $G$ is defined inductively
by $Z_{0}(G)=1$ and $Z_{i}(G)/Z_{i-1}(G)=Z(G/Z_{i-1}(G))$ for $i\geq 1$. The
last term of the upper central series of a finite group $G$ will be denoted by
$Z_{\infty}(G)$. A classical result, due to Baer, states that if $Z_{i}(G)$
has finite index $t$ in $G$, then $\gamma_{i+1}(G)$ is finite, and its order
is bounded by a function of $i$ and $t$ (see the proof of [30, 14.5.1]).
Similarly, if $G$ is a finite group such that $[G:Z_{\infty}(G)]=t$, then the
order of $\gamma_{\infty}(G)$ is $t$-bounded (see [25]).

######  Proposition 7.1.

Let $G$ be a profinite group in which $|\mathcal{U}_{k}(G)|\leq m$ for some
positive integer $m$. Then $|\gamma_{k}(G)|$ is $m$-bounded.

###### Proof.

As $\gamma_{k}(G)$ is generated by the set $\mathcal{U}_{k}(G)$ on which $G$
acts by conjugation, it follows that $[G:C_{G}(\gamma_{k}(G)]\leq m!$. Thus
$Z(\gamma_{k}(G))$ has $m$-bounded index in $\gamma_{k}(G)$ and, by Schur
Theorem, $\gamma_{k}(G)^{\prime}$ has $m$-bounded order (see [31, 4.12]). We
may pass to the quotient $G/\gamma_{k}(G)^{\prime}$ and, without loss of
generality, assume that $\gamma_{k}(G)$ is abelian. Now we will prove that
$|\gamma_{k}(G)|$ is finite and $m$-bounded by induction on $m$, the case
$m=1$ being trivial. If $G$ is pronilpotent, then every $\gamma_{k}$-value is
contained in $\mathcal{U}_{k}(G)$, so we can conclude by [13, Theorem A].
Suppose that $G$ is not pronilpotent. Then there exists a Sylow $p$-subgroup
$P$ of $\gamma_{k}(G)$, for some prime $p$, and a $p^{\prime}$-element $a$
such that $[P,a]\neq 1$. Since $P$ is abelian $[zy,a]=[y,a][z,a]$ for each
$y,z\in P$, thus $[P,a]=\\{[y,a]\mid y\in P\\}$. It follows from Lemma 2.5
that $[P,a]\subseteq\mathcal{U}_{k}(G)$ and so it has order at most $m$.
Choose a nontrivial element $x\in[P,a]$. Then $x$ has order at most $m$ and
$[G:C_{G}(x)]\leq m$. Hence $|\langle x^{G}\rangle|$ has order at most
$m^{m}$. Now we can pass to the quotient $G/\langle x^{G}\rangle$ and conclude
by induction on $m$. ∎

As mentioned in the introduction Theorem 1.1 enables us to prove that if $G$
is a profinite group such that the cardinality of the set $\mathcal{U}_{k}(G)$
is less than $2^{\aleph_{0}}$, then $\gamma_{k}(G)$ is finite. Our proof
relies on the fact that multilinear commutator words are strongly concise [5].

The next lemma will be useful.

######  Lemma 7.2.

[5, Lemma 2.2] Let $G$ be a profinite group and let $x\in G$. If the conjugacy
class $x^{G}$ contains less than $2^{\aleph_{0}}$ elements, then it is finite.

###### Proof of Theorem 1.3.

It is enough to prove that if $G$ is a profinite group such that the
cardinality of the set of uk-commutators in $G$ is less than $2^{\aleph_{0}}$
then $\gamma_{k}(G)$ is finite. Under this assumption, the conjugacy class
$x^{G}$ of every uk-commutator $x$ is finite, by Lemma 7.2. Thus all uk-
commutators are $FC$-elements and $G$ is virtually nilpotent, by Proposition
5.1. Let $N$ be an open nilpotent normal subgroup of $G$. If $g\in G$, Lemma
5.4 implies that the cardinality of the set of $\gamma_{k}$-values in
$N\langle g\rangle$ is less than $2^{\aleph_{0}}$. Taking into account that
multilinear commutator words are strongly concise we conclude that
$\gamma_{k}(N\langle g\rangle)$ is finite. Choose a transversal
$g_{1},\dots,g_{s}$ of $N$ in $G$. As each $\gamma_{k}(N\langle g_{i}\rangle)$
is normalized by $N$, its normal closure $N_{i}$ is finite. Thus we can pass
to the quotient over the finite normal subgroup $N_{1}\cdots N_{s}$ and assume
that $\gamma_{k}(N\langle g_{i}\rangle)=1$ for $i=1,\dots,s$. Now, if $x\in
N$, the commutator $[x,{}_{k-1}g]$ is trivial for each $g\in G$, that is, $x$
is a right Engel element. Therefore for every finite quotient $\bar{G}$ of $G$
the image $\bar{N}$ of $N$ is contained in $Z_{\infty}(\bar{G})$ (see for
instance [30, 12.3.7]). It follows that $\gamma_{\infty}(\bar{G})$ has
$s$-bounded order. As this happens for every finite quotient of $G$,
$\gamma_{\infty}(G)$ is finite. Without loss of generality, we can pass to the
quotient over $\gamma_{\infty}(G)$ and assume that $G$ is pronilpotent. In
this case, every $\gamma_{k}$-value is a uk-commutator. Since multilinear
commutator words are strongly concise, it follows that $\gamma_{k}(G)$ is
finite. This concludes the proof. ∎

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