# finite normal subgroups of strongly verbally closed groups

Filipp D. Denissov
Faculty of Mathematics and Mechanics of Moscow State University
Moscow 119991, Leninskie gory, MSU.
Moscow center for Fundamental and Applied Mathematics.
<EMAIL_ADDRESS>
In the recent paper by A. A. Klyachko, V. Yu. Miroshnichenko, and A. Yu.
Olshanskii, it is proven that the center of any finite strongly verbally
closed group is its direct factor. One of the results of the current paper is
the generalization of this nontrivial fact to the case of finite normal
subgroups of any strongly verbally closed groups. It follows from this
generalization that finitely generated nilpotent groups with nonabelian
torsion subgroups are not strongly verbally closed.

## 1\. Introduction

A subgroup $H$ of a group $G$ is called verbally closed [MR14] if any equation
of the form

$w(x_{1},x_{2},\dots,x_{n})=h,\text{ where }w\text{ is an element of the free
group }F(x_{1},\dots,x_{n})\text{ and }h\in H,$

having solutions in $G$ has a solution in $H$. If each system of equations
with coefficients from H

$\\{w_{1}(x_{1},\dots)=1,\dots,w_{m}(x_{1},\dots)=1\\},\text{ where }w_{i}\in
H\ *\ F(x_{1},\dots,x_{n})\text{ (and }*\text{ means the free product),}$

having solutions in $G$ has a solution in $H$, then the subgroup $H$ is called
algebraically closed in $G$. Note that if the subgroup $H$ is algebraically
closed in the group $G$, then it is verbally closed in $G$.

A group $G$ is called strongly verbally closed if it is algebraically closed
in any group containing $G$ as a verbally closed subgroup. Thus, the verbal
closedness (as well as the algebraic closedness) is a property of a subgroup,
while the strong verbal closedness is a property of an abstract group. The
class of strongly verbally closed groups is fairly wide. For example, it
includes

  * —

all abelian groups [Maz18],

  * —

all free groups [KM18],

  * —

all virtually free groups containing no nontrivial finite normal subgroups
[KM18], [KMM18],

  * —

all groups decomposing nontrivially into a free product [Maz19],

  * —

fundamental groups of all connected surfaces except the Klein bottle [Maz18],
[Kly21],

  * —

all finite groups with nonabelian monolith [KMO21],

  * —

inifinite dihedral group [KMM18] and any finite dihedral group whose order is
not divisible by $8$ [KMO21],

  * —

all acylindrically hyperbolic groups with no nontrivial finite normal
subgroups [Bog22].

The class of non-strongly-verbally-closed groups is fairly wide too. Among
such groups are the following:

  * —

the already mentioned fundamental group of the Klein bottle [Kly21],

  * —

the discrete Heisenberg group [KMO21],

  * —

any finite group, whose center is not its direct factor (in particular, any
finite nonabelian nilpotent group) [KMO21], [RKK17], [KM18].

Proving the strong verbal closedness (as well as its absence) of a group is
not easy. In [KMO21], for example, a question is raised:

######  Question 1.

Does there exist a finitely generated nilpotent nonabelian strongly verbally
closed group?

A negative answer to this question would yield a broad generalization of the
last two examples of non-strongly-verbally-closed groups mentioned above. So
far, we managed to give a partial answer to this question. More precisely, we
proved the absence of strong verbal closedness of finitely generated nilpotent
groups with nonabelian torsion subgroups and of some finitely generated
nilpotent nonabelian groups with abelian torsion subgroups.

A property that is stronger than the strong verbal closedness is the property
of being a strong retract [KMO21]. A group $H$ is called a strong retract if
it is a retract of any group $G\geqslant H$ from the variety generated by the
group $H$.

Let us recall some terminology [Neu67]:

  * —

the variety generated by a class of groups $\mathcal{K}$ is the class of all
groups satisfying all identities that hold in all groups from $\mathcal{K}$,

  * —

the variety generated by a group $G$ is designated by
$\operatorname{\mathbf{var}}G$.

This gives rise to the following question from [KMO21]:

######  Question 2.

What is an arbitrary finite strong retract?

In [KMO21] some examples of strong retracts are provided. In the next section,
we describe all the nilpotent strong retracts.

Below we provide a brief list of notation we use.

If $x,y$ are elements of some group, then the symbol $[x,y]$ denotes their
commutator $x^{-1}y^{-1}xy$. The symbol $\operatorname{ord}(x)$ denotes the
order of an element $x$ of a group $G$. The center of a group $G$ is denoted
by $Z(G)$, and its commutator subgroup is denoted by $G^{\prime}$. The
centralizer of a subset $X$ of a group $G$ is denoted by $C(X)$. The symbol
$\langle\langle X\rangle\rangle$ stands for the normal closure of a subset $X$
of a group $G$ (that is the intersection of all normal subgroups of $G$
containing $X$). The free group with a basis $X$ is denoted as $F(X)$ or
$F_{n}$ in case $X$ has $n\in\mathbb{N}$ elements. Identical mapping from $X$
to itself is denoted by $id$. We use the symbol $H\cong G$ to express the fact
that groups $H$ and $G$ are isomorphic. Finally, the symbol $H\leqslant G$
denotes the fact that a group $H$ is a subgroup of $G$. The symbol $H\unlhd G$
denotes the fact that $H$ is a normal subgroup of $G$.

The author is grateful to his supervisor Anton Alexandrovich Klyachko for
formulation of the problem and for valuable remarks during the work.

## 2\. Nilpotent strong retracts

Note that in case when $G$ is an abelian group, $H\leqslant G$ is its retract
if and only if $H$ is a direct summand of $G$. It means that the property of
being a strong retract for the abelian group $G$ is equivalent to the property
of $G$ being a direct summand of any group $H\in\operatorname{\mathbf{var}}G$
containing $G$. For the further discussion, we need the description of all
varieties of abelian groups (see [Fuc70], paragpaph $18$, exercise $7$):

> Varieties of abelian groups are precisely the following classes of groups:
> 1) the class of all abelian groups; 2) the class of all abelian groups of a
> period divising $n\in\mathbb{N}$.

Recall that the period of a group $G$ is the least number $n\in\mathbb{N}$,
such that $x^{n}=1$ for any $x\in G$. If such a number exists, then $G$ is a
group of bounded period.

To begin with, consider the case, when $G$ is not a group of bounded period.
Then, according to the description, $\operatorname{\mathbf{var}}G$ is the
class of all abelian groups. The following is true of divisible abelian groups
(see, for example, [Kur60]):

> If $G$ is a divisible abelian group, and $H$ is an abelian group such that
> $G\leqslant H$, then $G$ is a direct summand of $H$.

Let us remind that a group $G$ is called divisible if for any $g\in G$ and
$n\in\mathbb{N}$, the equation $x^{n}=g$ has a solution in $G$.

######  Proposition 1.

An abelian group $G$ of unbounded period is a strong retract if and only if it
is divisible.

######  Proof.

Sufficiency follows from the fact provided above. Let $G$ be an abelian group
of unbounded period. Then, as it was noted earlier,
$\operatorname{\mathbf{var}}G$ is the class of all abelian groups. In
particular, $\operatorname{\mathbf{var}}G$ contains a divisible group $H$
containing $G$ [Kur60]. Though, if $G$ is not divisible itself, it is not a
direct summand of $H$ (as direct summands of a divisible group are divisible
themselves [Kur60]), so $G$ is not a strong retract. ∎

Let us move on to abelian groups of bounded period. The first Prüfer theorem
provides a complete description of these groups [Kur60]:

> An abelian group $G$ of bounded period $d$ is a direct sum of primary cyclic
> groups, i.e. $G\cong\bigoplus_{i\in I}\mathbb{Z}_{p_{i}^{k_{i}}}$, where
> $p_{i}$ are prime numbers and $k_{i}$ are natural numbers such that
> $p_{i}^{k_{i}}|d$, $i\in I$ ($I$ is an index set).

We need the following variation of the Zorn’s lemma [Fuc70]:

> Let $M\neq\varnothing$ be a partially ordered set. Suppose that every chain
> in $M$ (a totally ordered subset of $M$) has an upper bound. Then $M$
> contains a maximal element.

Now, we are ready to proceed with our description:

######  Proposition 2.

An abelian group $G$ of bounded period is a strong retract if and only if in
its decomposition into the direct sum of primary cyclic groups, orders of any
distinct direct summands are either equal or coprime:

$G\cong\bigoplus_{i=1}^{m}C_{p_{i}^{k_{i}}}(n_{i}),\text{ where
}C_{p_{i}^{k_{i}}}(n_{i})\text{ is equal to the direct sum of }n_{i}\text{
copies of the group }\mathbb{Z}_{p_{i}^{k_{i}}},$

where all prime numbers $p_{i}$ are distinct, $m,k_{i}\in\mathbb{N}$, and
$n_{i}$ are some cardinal numbers.

######  Proof.

Suppose that $G$ cannot be decomposed into such a direct sum. We may assume
that

$G=\bigoplus_{i=1}^{m}\bigoplus_{j\in I_{i}}\mathbb{Z}_{p_{i}^{k_{j}}},$ (1)

where $m\in\mathbb{N}$, $|I_{i}|=n_{i}$ and among $k_{j}$, $j\in I_{i}$ there
are only finitely many different ones (because $G$ is a group of bounded
period) but there exists $i\in\\{1,\dots,m\\}$ such that for some
$j_{1},j_{2}\in I_{i},$ $k_{j_{1}}\neq k_{j_{2}}$.

Consider the group: $\displaystyle
H=\bigoplus_{i=1}^{m}C_{p_{i}^{s_{i}}}(n_{i})$, where

$s_{i}=\max\\{k_{j}\ |\ \mathbb{Z}_{p_{i}^{k_{j}}}\text{ is a direct summand
in the decomposition (1)}\\},\ i=1,2,\dots,m.$

Since both $G$ and $H$ are of the same period $\prod_{i=1}^{m}s_{i}$, it
follows from the description of abelian varieties that
$H\in\operatorname{\mathbf{var}}G$ .

Consider the injection $f:G\to H$, which works on each direct summand from
$(1)$ as follows: let $i\in\\{1,\dots,m\\}$, $j\in I_{i}$,
$f:\mathbb{Z}_{p_{i}^{k_{j}}}\hookrightarrow\mathbb{Z}_{p_{i}^{s_{i}}}$, where
$\mathbb{Z}_{p_{i}^{s_{i}}}$ is the $j$th summand from the decomposition of
$C_{p_{i}^{s_{i}}}(n_{i})$ into the direct sum. Every direct summand from
$(1)$ is mapped into the corresponding direct summand of the decomposition of
$H$, so that the restriction of $f$ to $\mathbb{Z}_{p_{i}^{k_{j}}}$ is a
natural injection: if $k_{j}=s_{i}$, then it is the identical map; otherwise
it is a mapping to the subgroup of $\mathbb{Z}_{p_{i}^{s_{i}}}$ of the order
$p_{i}^{k_{j}}$. From the uniqueness of the decomposition of an abelian group
of bounded period into the direct sum of primary cyclic groups [Fuc70], it
follows that $f(G)$ is not a direct summand of $H$. Thus, $G$ is not a strong
retract.

Now, suppose that $G$ has the decomposition from the statement of the theorem.
Let $H\in\operatorname{\mathbf{var}}G$ and let $f:G\hookrightarrow H$ be a
monomorphism. As any monomorphism preserves the order of an element, the
$p_{i}$th component of $G$ is mapped into the $p_{i}$th component of $H$ under
$f$, so it suffices to prove the theorem only for the case $G=C_{p^{k}}(n)$,
where $p$ is prime, $k\in\mathbb{N}$, and $n$ is some cardinal number.

Let us show that there exists such $X\leqslant H$ that $H=f(G)\oplus X$. In
Zorn’s lemma, take the set of all subgroups of $H$ having trivial intersection
with $f(G)$ as $M$:

$M=\\{Y\leqslant H\ |\ Y\cap f(G)=\\{0\\}\\}.$

Order on $M$ is introduced as follows: for $X,Y\in M$, $X\leqslant Y$ if $X$
is a subgroup of $Y$. It can be verified directly that this is an order on
$M$. Set $M$ is nonempty: $\\{0\\}\in M$. Any chain $\\{Y_{\alpha}\\}\subseteq
M$ of subgroups having trivial intersection with $f(G)$ is bounded by an
element $Y\in M$, where $Y=\cup_{\alpha}Y_{\alpha}$. Consequently, Zorn’s
lemma is applicable, and $M$ contains a maximal element $X$: $X\leqslant H$,
$X\cap f(G)=\\{0\\}$, and $X$ is not a subgroup of any bigger (relatively to
the order we introduced) subgroup satisfying this property.

From $X\cap f(G)=\\{0\\}$ it follows that $f(G)+X=f(G)\oplus X$. It remains to
prove that $H=f(G)+X$. Let $h\in H$. There exists such $k\in\mathbb{N}$ that
$kh\in f(G)+X$. Indeed, otherwise $\langle h\rangle\cap(f(G)+X)=\\{0\\}$,
which means that $(\langle h\rangle+X)\cap f(G)=\\{0\\}$, leading to a
contradiction with the maximality of $X$.

Let $s$ be the least of such numbers $k$. Without loss of generality, assume
that $s$ is prime or that $s=1$ (otherwise, take a power of $h$ instead of
$h$). Two cases are possible:

1) $s=p$. Then, $ph=f(g)+x$ for some $g\in G$, $x\in X$. If $g=pg_{1}$,
$g_{1}\in G$ ($g_{1}$ may be equal to zero), then $ph-f(pg_{1})=x$. However,
from $h-f(g_{1})\not\in X$ (as $h\not\in f(G)+X$) it can be obtained that
$(X+\langle h-f(g_{1})\rangle)\cap f(G)=\\{0\\}$, which leads to a
contradiction with the maximality of $X$. Consequently, $g\neq pg_{1}$ for any
$g_{1}\in G$. As $g\neq 0$, $\operatorname{ord}(g)=p^{k}$. Though,
$\operatorname{ord}(ph)=p^{r}<p^{k}$, so $p^{r}(ph)=0=p^{r}(f(g))+p^{r}x$. As
the sum $f(G)+X$ is direct, $p^{r}f(g)=p^{r}x=0$, which means that $p^{r}g=0$,
which is impossible.

2) $s\neq p$. For abelian groups of period $p$, the mapping $g\mapsto sg$ is
an automorphism, so, as $sh=f(g)+x$ for some $g\in G$, $x\in X$, there exist
such $g_{1}\in G$, $x_{1}\in X$ that $g=sg_{1}$, $x=sx_{1}$. Thus,
$s(h-f(g_{1})-x_{1})=0$. No nontrivial element of $H$ has the order of $s$, so
$h=f(g_{1})+x_{1}$.

As a result, $H=f(G)\oplus X$, and $G$ is a strong retract. ∎

######  Proposition 3.

The center of a strong retract is its direct factor.

######  Proof.

Let $G$ be a strong retract. The center of any group is a normal subgroup, so
it suffices to prove that $Z(G)$ is a retract of $G$. Consider the central
product of $G$ with its copy $\widetilde{G}$ with joined center:

$K=G\underset{Z(G)=Z(\tilde{G})}{\times}\tilde{G}=(G\times\tilde{G})/\\{(g,g^{-1})|g\in
Z(G)\\}\in\operatorname{\mathbf{var}}G.$

The group $\tilde{G}$ is isomorphic to the group $G$, so it is a strong
retract too. Let $\rho$ be a retraction of $K$ to its subgroup $\tilde{G}$.
From the fact that in the group $K$, the group $G$ commutes with the group
$\tilde{G}$, we obtain $\rho(G)\leqslant Z(G)$. By definition of the
retraction, $\rho(g)=g$ is true for any element $g\in Z(G)$. Thus, the
restriction of $\rho$ to the subgroup $G$ of the group $K$ is the desired
retraction to $Z(G)$. ∎

The following simple proposition shows that consideration of nilpotent groups
does not yield any new strong retracts:

######  Proposition 4.

Nilpotent strong retract is an abelian group.

######  Proof.

Any nontrivial normal subgroup of a nilpotent group intersects the center of
this group nontrivially (see [KM79]). From this fact and from the proposition
$3$, we obtain that any nilpotent strong retract is equal to its center. ∎

As a result, we proved the following theorem:

######  Nilpotent-strong-retract theorem.

Nilpotent strong retracts are precisely divisible abelian groups and abelian
groups of bounded period in whose decomposition into the direct sum of primary
cyclic groups, orders of any distinct direct summands are either equal or
coprime.

In the next paragraph we show that many nilpotent groups are not even strongly
verbally closed.

## 3\. Finite normal subgroups of strongly verbally closed groups

We say that a group presentation $\langle X\ |\ R\rangle$ is finitely
presented over a group presentation $\langle Y\ |\ S\rangle$, if there exist
such finite sets $A$ and $B$ that $\langle X\ |\ R\rangle\cong\langle
X^{\prime}\ |\ R^{\prime}\rangle$, where $X^{\prime}=Y\cup A$,
$R^{\prime}=S\cup B$.

The following lemma reveals that this definition is, in fact, a group property
(which means it does not depend on the choice of a group presentation), so it
makes sense to speak about the finite presentability of one group over the
other group:

######  Lemma 1.

Suppose that a group presentation $\langle X\ |\ R\rangle$ is finitely
presented over a group presentation $\langle Y\ |\ S\rangle$ and $\langle Y\
|\ S\rangle\cong\langle Y^{\prime}\ |\ S^{\prime}\rangle$. Then $\langle X\ |\
R\rangle$ is finitely presented over $\langle Y^{\prime}\ |\
S^{\prime}\rangle$.

######  Proof.

We may assume that $X=Y\cup A$ and $R=S\cup B$ for some finite sets $A$ and
$B$. It is known (see, for example, [LS15]) that groups defined by group
presentations $\langle Y\ |\ S\rangle$ and $\langle Y^{\prime}\ |\
S^{\prime}\rangle$ are isomorphic if and only if presentation $\langle
Y^{\prime}\ |\ S^{\prime}\rangle$ is obtained from presentation $\langle Y\ |\
S\rangle$ by applying a finite number of Tietze transformations:

  * —

adding to the set $S$ an arbitrary set $T\subseteq\langle\langle
S\rangle\rangle\unlhd F(Y)$ of its consequences,

  * —

adding to the set $Y$ an arbitrary set $\widetilde{Y}$ while ading to $S$ a
set $\\{\widetilde{y}=w_{\widetilde{y}}\ |\
\widetilde{y}\in\widetilde{Y},w_{\widetilde{y}}\in F(Y)\\}$,

and their inverses. It is sufficient to prove the lemma only for the case,
when $\langle Y^{\prime}\ |\ S^{\prime}\rangle$ is obtained from $\langle Y\
|\ S\rangle$ by applying one Tietze transformation. One can easily verify that
in case of the first transformation, $X^{\prime}=X$ and $R^{\prime}=R\cup T$,
while in case of the second transformation, $X^{\prime}=X\cup\widetilde{Y}$
and $R^{\prime}=R\cup\\{\widetilde{y}=w_{\widetilde{y}}\ |\
\widetilde{y}\in\widetilde{Y},w_{\widetilde{y}}\in F(Y)\\}$ provide the
desired group presentation. ∎

By virtue of Lemma $1$, the following definition may be introduced:

A group $G$ is finitely presented over a group $H$, if there exists such a
presentation of $G$ that it is finitely presented over any presentation of
$H$.

######  Lemma 2.

Suppose that $G$ contains a subgroup $H$ and a finite normal subgroup $N$ such
that $G/N$ is finitely presented over $H/(H\cap N)$. Then $G$ is finitely
presented over $H$.

Proof (with minor changes) replicates the proof of the Hall theorem [Hal54]
about preservation of finite presentability of a group under extensions (see
also [Rob82]).

Let $G$ be a group, $H=\langle X\ |\ R\rangle\leqslant G$, and $N=\langle Y\
|\ S\rangle\ \unlhd\ G$ be its finite subgroup, where $Y$ and $S$ are finite
sets. By condition of the lemma, the group $G/N$ is finitely presented over
$H/(H\cap N)=\langle X\ |\ R\cup C\rangle$, where $\langle\langle
C\rangle\rangle=H\cap N$ and the set $C$ is finite. Consequently

$G/N\cong\langle X\cup A\ |\ R\cup C\cup B\rangle,$

where sets $A$ and $B$ are finite.

Let us construct a presentation of the group $G$. As the set of generators,
take $\overline{X}\cup\overline{A}\cup\overline{Y}$, where sets
$\overline{X}$, $\overline{A}$, $\overline{Y}$ are in one-to-one
correspondence with sets $X$, $A$, $Y$ respectively. The sets $R$, $S$, $C$,
and $B$ are in correspondence with the sets $\overline{R}$, $\overline{S}$,
$\overline{C}$, and $\overline{B}$ respectively. As the set of defining
relations, take the union of the following sets: $\overline{R}$,
$\overline{S}$, $\overline{C}_{1}=\\{cw_{c}^{-1}\ |\ c\in\overline{C},w_{c}\in
F(\overline{Y})\\}$, $\overline{B}_{1}=\\{bw_{b}^{-1}\ |\
b\in\overline{B},w_{b}\in F(\overline{Y})\\}$ ($c\in\overline{C}$ and
$b\in\overline{B}$ are considered as words from $F(\overline{X})$ and from
$F(\overline{X}\cup\overline{A})$ respectively),
$\overline{T}=\\{a^{-1}yaw_{a,y}^{-1},\ aya^{-1}v_{a,y}^{-1}\ |\
a\in\overline{A},y\in\overline{Y},\ w_{a,y},v_{a,y}\in F(\overline{Y})\\}$:

$\widetilde{G}=\langle\overline{X}\cup\overline{A}\cup\overline{Y}\ |\
\overline{R}\cup\overline{S}\cup\overline{C}_{1}\cup\overline{B}_{1}\cup\overline{T}\rangle.$

Consider a surjective homomorphism $\theta:\widetilde{G}\to G$, defined with
the following bijections $\overline{X}\to X$, $\overline{A}\to A$,
$\overline{Y}\to Y$ on the generators (defining relations are mapped into true
identities under such a map on generators, so such a homomorphism exists). The
restriction $\theta|_{K}:K\to N$ on the subgroup
$K=\langle\overline{Y}\rangle\leqslant\widetilde{G}$ is an isomorphism as all
the relations in the alphabet $\overline{Y}$ in $\widetilde{G}$ are
consequences of the defining relations $\overline{S}$. Besides,
$K\unlhd\widetilde{G}$.

Homomorphism $\widetilde{\theta}:\widetilde{G}/K\to G/N$ generated by
$\theta$, is an isomorphism too. Now, let $g\in\ker\theta$. Then
$gK\in\ker\widetilde{\theta}$, but $\widetilde{\theta}$ is an isomorphism, so
$g\in K$. Finally, $\theta|_{K}$ is an isomorphism, so $g=1$. ∎

The following lemma provides a criterion for algebraic closedness of a
subgroup $H$ of a group $G$ in case, when $G$ is finitely presented over $H$
(for similar propositions, refer to [MR14]):

######  Lemma 3.

Suppose that $H=\langle X\ |\ R\rangle$ is a subgroup of $G$ and $G$ is
finitely presented over $H$. The subgroup $H$ is algebraically closed in $G$
if and only if $H$ is a retract of $G$.

######  Proof.

Suppose $H$ is algebraically closed in $G$ and $A=\\{a_{1},\dots,a_{m}\\}$,
$B=\\{s_{1},\dots,s_{n}\\}$ are the sets from the definition of finite
presentability of $G$ over $H$. The relations $s_{i}(a_{1},\dots,a_{m},X)=1$,
$i=1,\dots,n$ are corresponded to a system of equations with coefficients from
$H$:

$\begin{cases}s_{1}(t_{1},\dots,t_{m},X)=1\\\ \ \ \ \ \ \ \ \ \ \ \dots\\\
s_{n}(t_{1},\dots,t_{m},X)=1\\\ \end{cases}$

which, by condition, has a solution $t_{1}=a_{1},\dots,t_{m}=a_{m}$. By virtue
of algebraic closedness of $H$ in $G$, this system has a solution
$t_{1}=h_{1},\dots,t_{m}=h_{m}$ in $H$. Mapping
$X\sqcup\\{a_{1},\dots,a_{m}\\}\to H,$ $x\in X\mapsto x$, $a_{i}\mapsto h_{i}$
extends to a surjective homomorphism $\varphi:G\to H$, as defining relations
of $G$ are mapped into true identities under such a mapping of generators
(note that $R$ is the set of words in the alphabet $X$).

This homomorphism is the desired retraction: let $h\in H$,
$h=v(x_{1},\dots,x_{r})$, $x_{i}\in X$. Applying to this word the homomorphism
$\varphi$, we get: $\varphi(h)=v(\varphi(x_{1}),\dots,\varphi(x_{r}))=h$.

Algebraic closedness of a subgroup $H$ of a group $G$ follows from retractness
of $H$ in $G$ for every group $G$ [MR14]. ∎

######  Approximation lemma.

Let $C$ be a finite elementary abelian $p$-group (where $p$ is a prime
number). For any $k\in\mathbb{N}$, there exists $t\geqslant k$ such that the
direct product $P=\mbox{\Large$\times$}_{i=1}^{t}C_{i}$ of copies $C_{i}$ of
$C$ contains a subgroup $R$ invariant with respect to the diagonal action on
$P$ of the endomorphism algebra $\operatorname{End}C$ with the following
properties:

  1. 1)

$R\subseteq\bigcup\ker\rho_{j}$, where $\rho_{j}:P\to C_{j}$, $j=1,\dots,t$
are the natural projections,

  2. 2)

But $R\cdot\mbox{\Large$\times$}_{j\not\in J}C_{j}=P$ for any subset
$J\subseteq\\{1,\dots,t\\}$ of cardinality $|J|=k$,

  3. 3)

Moreover, each such $J$ is contained in a set $J^{\prime}\supseteq J$ such
that $P=R\times(\mbox{\Large$\times$}_{j\not\in J^{\prime}}C_{j})$; and there
exist integers $n_{ij}\in\mathbb{Z}$ such that the projection
$\pi:P\to\mbox{\Large$\times$}_{j\not\in J^{\prime}}C_{j}$ with the kernel $R$
acts as: $C_{i}\ni c_{i}\mapsto\prod_{j\not\in J^{\prime}}c_{j}^{n_{ij}}$,
where $c_{j}\in C_{j}$ is the element corresponding to $c_{i}$ under the
isomorphism $C_{i}\cong C\cong C_{j}$.

The following theorem provides a generalization of the result from [KMO21]
about the center of a finite strongly verbally closed group. The proof is also
analogical to the proof of that theorem, with the exception of some nuances.

######  Finite-normal-subgroup theorem.

Let $H$ be a strongly verbally closed group. For any finite normal subgroup
$T$ of $H$, for any abelian subgroup $A$ of $T$, normal in $H$, it is true
that $Z(C_{T}(A))$ is a direct factor of $C_{T}(A)$, and some complement is
normal in $H$. Here $C_{T}(X)=C(X)\cap T$.

######  Proof.

Let $H$ be such a group, and let $L=C_{T}(A)$. It suffices, for each prime
$p$, to find a homomorphism $\psi_{p}:L\to Z(L)$ commuting with the action
$H\curvearrowright L$ by conjugations (this action is well-defined as $L\unlhd
H$) and injective on the $p$-component of the center $Z_{p}(L)$ of $L$. Then
the homomorphism $\psi:L\to Z(L),$
$x\mapsto\displaystyle\prod_{p}\pi_{p}(\psi_{p}(x))$, where $\pi_{p}:Z(L)\to
Z_{p}(L)$ is the projection on the $p$-component, is injective on $Z(L)$, so
its kernel is the desired complement $D$ (normality of $D$ in $H$ follows from
the fact that $\psi$ commutes with $H\curvearrowright L$).

Suppose that there are no such homomorphisms for some prime number $p$, i.e.
every homomorphism $f:L\to Z(L)$ commuting with the action $H\curvearrowright
L$ is not injective on $Z_{p}(L)$. Then it is not injective on the maximal
elementary abelian $p$-subgroup $C\leqslant Z_{p}(L)$ (it is finite as $L$ is
finite). Indeed, if $x\neq 1\in Z_{p}(L)$ is an element such that $f(x)=1$,
then, raising it to the appropriate power $d$, we get $f(x^{d})=1$ and
$x^{d}\in C$, $x^{d}\neq 1$.

Choose $t$ by the approximation lemma applied to $C$ (for some $k$ to be
specified later) and consider the fibered product of $t$ copies of the group
$H$:

$Q=\\{(h_{1},\dots,h_{t})\ |\ h_{1}L=\dots=h_{t}L\\}\leqslant H^{t}.$

First of all, let us show that the subgroup $R\leqslant C^{t}\leqslant Q$ from
the approximation lemma is normal in $Q$. Subgroup $R$ is invariant under the
diagonal action of automorphisms
$\operatorname{Aut}C\leqslant\operatorname{End}C$. It remains to show that $Q$
acts by conjugations on $P=C^{t}$ diagonally. It follows from the lemma:

######  Lemma 4.

Let $G$ be a group, and $N\unlhd G$. If $xC(N)=yC(N)$ for some $x,y\in G$,
then $x$ and $y$ act on $N$ (by conjugations) identically.

######  Proof.

From $xC(N)=yC(N)$ it follows that for some $c\in C(N)$, $x=yc$. Then for
$n\in N$, we have:

$x^{-1}nx=c^{-1}(y^{-1}ny)c=y^{-1}ny.$

The last identity is true, as (due to normality) $y^{-1}ny\in N$ and $c\in
C(N)$. ∎

Let $q=(q_{1},\dots,q_{t})\in Q$, $p=(p_{1},\dots,p_{t})\in P$. As $q_{1}L=$
$q_{2}L=$ $\dots=q_{t}L$, then (according to Lemma $4$) $q^{-1}pq=$
$\tilde{q}^{-1}p\tilde{q}$, where $\tilde{q}=(q_{1},\dots,q_{1})$. It means
that the conjugation action of $Q$ on $P$ is diagonal. On the other hand,
diagonal action by conjugations induces an endomorphism of $C^{t}$ (due to
normality of $C\unlhd H$), and $R$ is invariant with respect to the diagonal
action of such endomorphisms, leading to normality of $R$ in $Q$.

Put $G=Q/R$. First, let us show that $H$ embeds into $G$. The group $H$ embeds
into $Q$ diagonally: $h\mapsto(h,\dots,h)$, $h\in H$. This homomorphism serves
as embedding into $G$ as well, as all projections of any nontrivial diagonal
element of $Q$ are nontrivial (and $R$ is contained in the union of the
kernels of these projections).

Now, let us prove the verbal closedness of this diagonal subgroup (denote it
as $H$ too) in $G$. Consider an equation

$w(x_{1},\dots,x_{n})=(h,\dots,h)$

having a solution in $G$ and let $\tilde{x}_{1},\dots,\tilde{x}_{n}$ be a
preimage (in $Q$) of a solution $x_{1},\dots,x_{n}$. Then (in $Q$):

$w(\tilde{x}_{1},\dots,\tilde{x}_{n})=(hc_{1},\dots,hc_{t}),$

where $(c_{1},\dots,c_{t})\in R$. By the property $1)$ of the approximation
lemma, $c_{i}=1$ for some $i$. It means that in $H$ (the group itself)
$w(\tilde{x}_{1}^{i},\dots,\tilde{x}_{n}^{i})=h$, where $\tilde{x}_{j}^{i}$ is
the $i$th coordinate of the vector $\tilde{x}_{j}$, $j=1,\dots,n$.

Let us take $y_{j}=(\tilde{x}_{j}^{i},\dots,\tilde{x}_{j}^{i})$,
$j=1,\dots,n$. Then in $H\leqslant G$ the following is true:

$w(y_{1},\dots,y_{n})=(h,\dots,h),$

which proves the verbal closedness of $H$ in $G$.

Let $U\leqslant L$. We use the following denotion:

$U_{i}:=\\{(1,\dots,1,u,1,\dots,1)\ |\ u\in U\\}\leqslant Q,\
i=1,\dots,t\text{ (coordinate }u\text{ stands on the }i\text{th place)}.$

It remains to prove that $H$ is not algebraically closed in $G$.

######  Lemma 5.

The group $Q$ is finitely presented over its subgroup $H$.

######  Proof.

According to Lemma $2$, it is sufficient to show that
$Q/(L_{1}\times\dots\times L_{t})$ is finitely presented over $H/\tilde{L}$,
where $\tilde{L}=\\{(l,\dots,l)\ |\ l\in L\\}$. However,
$Q=H\cdot(L_{1}\times\dots\times L_{t})$, so the statement we prove follows
from this fact (see [KM79], theorem $4.2.4$):

> Suppose that $G$ is a group, $F$ is its subgroup, and $K$ is its normal
> subgroup. Then $(K\cdot F)/K\cong F/(F\cap K)$.

Thus, the group $Q/(L_{1}\times\dots\times L_{t})$ is not just finitely
presented over $H/\tilde{L}$ but is isomorphic to it. ∎

From Lemma $3$ and Lemma $5$, it follows that it suffices to show that $H$ is
not a retract of $G$. Let $\rho:G\to H$ be a hypothetical retraction, and let
$\hat{\rho}:Q\to H$ be its composition with the natural epimorphism $Q\to
Q/R=G$. Henceforth, all subgroups and centralizers we refer to relate to $Q$.

Let us verify that $\hat{\rho}(L_{i})\leqslant C_{T}(C_{T}(L))\leqslant L$ for
every $i$. First, prove the left inclusion. Let $h\in C_{T}(L)$. Then, $h$
commutes with every element from $L$; consequently, $h$, as an element of $Q$,
commutes with $L_{i}$. Applying the retraction $\hat{\rho}$ to this identity,
we get that $\hat{\rho}(h)$ $(=h)$ commutes with the subgroup
$\hat{\rho}(L_{i})$, which (by definition of the centralizer) proves the
inclusion. The second inclusion follows from the fact that $L=C_{T}(A)=$
$C(A)\cap T$, which means that

$C_{T}(C_{T}(L))\leqslant C_{T}(A\cap T)=C_{T}(A)=L.$

The first inclusion here is true as $C(L)\geqslant A$. The following equality
is true as $A\leqslant T$.

On the other hand, for $i\neq j$, the mutual commutator subgroup
$[L_{i},L_{j}]$ is trivial (as in case $i$ and $j$ are different, $L_{i}$ and
$L_{j}$ are contained in different components of the fibered product). It
means that the image of this mutual commutator subgroup is trivial too:
$[\hat{\rho}(L_{i}),\hat{\rho}(L_{j})]=\\{1\\}$. Consequently,
$[L_{i},\prod_{j\neq i}L_{j}]=\\{1\\}$ and $[\hat{\rho}(L_{i}),\prod_{j\neq
i}\hat{\rho}(L_{j})]=\\{1\\}$. If $\hat{\rho}(L_{i})=\hat{\rho}(L_{l})$ for
some $i\neq l$, then (by the virtue of well-known commutator identities)
$[\hat{\rho}(L_{i}),\prod_{j}\hat{\rho}(L_{j})]=\\{1\\}$, which means that
$\hat{\rho}(L_{i})\leqslant C_{T}(L)$ (as
$L=\hat{\rho}(L)\leqslant\prod_{j}\hat{\rho}(L_{j})$).

Thereby, if for some different $i$ and $j$,
$\hat{\rho}(L_{i})=\hat{\rho}(L_{j})$, then $\hat{\rho}(L_{i})\leqslant
C_{T}(L)$. From here and from the inclusion we proved earlier, we get
$\hat{\rho}(L_{i})\leqslant L\cap C_{T}(L)=$ $Z(L)$.

Let us take $k$ in the approximation lemma to be the number of all subgroups
of $T$, and let $J$ be the set of all exclusive numbers $i$, namely such that
for any $l\neq i$, $\hat{\rho}(L_{i})\neq\hat{\rho}(L_{l})$. Since among
$\hat{\rho}(L_{i})\leqslant T$ there are no more than $k$ different subgroups,
$|J|\leqslant k$. Thus, from the property $3)$ of the approximation lemma, we
have a decomposition:

$\mbox{\Large$\times$}_{i=1}^{t}C_{i}=R\times(\mbox{\Large$\times$}_{i\in
I}C_{i}),$

where $I\subseteq\\{1,\dots,t\\}\setminus J$ is some set of non-exclusive
elements. Again, according to the property $3)$ of the approximation lemma,
the projection
$\pi:\mbox{\Large$\times$}_{i=1}^{t}C_{i}\to\mbox{\Large$\times$}_{i\in
I}C_{i}$ onto the second factor of this decomposition is defined by an integer
matrix $(n_{ij})$, namely, for $c_{i}\in C_{i}$, $\pi:c_{i}\mapsto\prod_{j\in
I}c_{j}^{n_{ij}}$, where $c_{j}$ are elements corresponding to $c_{i}$ under
the isomorphism $C_{i}\cong C\cong C_{j}$.

This means that the restriction of $\pi$ to $C=\\{(c,\dots,c)\ |\ c\in
C\leqslant H\\}$ is defined by formula:

$\hat{\pi}:(c,\dots,c)\mapsto\prod_{j\in I}c_{j}^{m_{j}},\
m_{j}=\sum_{i}n_{ij}.$

Here $c_{j}$ are elements corresponding to $c$ under the isomorphism $C\cong
C_{j}$.

Then (as $i\in I$ are non-exclusive, we have $\hat{\rho}(L_{i})\leqslant
Z(L)$), consider the composition:

$\Psi:C\leqslant Q\to Z(L),\ c\overset{\pi}{\mapsto}\prod_{j\in
I}c_{j}^{m_{j}}\overset{\hat{\rho}}{\mapsto}\prod_{j\in
I}\hat{\rho}(c_{j}^{m_{j}}).$

It extends to a homomorphism $\Phi:L\to Z(L)$ defined by the similar formula:

$\Phi:g\mapsto\prod_{j\in I}\hat{\rho}(g_{j}^{m_{j}}),$

where $g\in L$ and $g_{j}\in L_{j}$ are elements corresponding to $g\in L$.
Obviously, it is an extension of $\Psi$ and a homomorphism, as for $j\in I$,
$\hat{\rho}(L_{j})\leqslant Z(L)$ and the group $Z(L)$ is abelian. This
homomorphism commutes with the conjugation action of $H$ on $L$. Indeed, let
$g\in H$ and let $\mathfrak{g}$ be the action of $g$ on $L$ by conjugation,
namely, for $x\in L$, $\mathfrak{g}(x)=g^{-1}xg$. Let us show that
$\Phi\circ\mathfrak{g}=\mathfrak{g}\circ\Phi$. Let $h\in L$. Then
$\Phi(\mathfrak{g}(h))=$ $\prod_{j\in I}\hat{\rho}(g^{-1}h_{j}^{m_{j}}g)=$
$\prod_{j\in I}g^{-1}\hat{\rho}(h_{j}^{m_{j}})g=$ $\mathfrak{g}(\Phi(h))$.
Penult identity is true, as $\hat{\rho}$ is a retraction on $H$, so it acts
identically on $H$ itself. By assumption we made in the beginning, the kernel
of this homomorphism has nontrivial intersection with $C$: $\ker\Phi\cap
C\neq\\{1\\}$, so the restriction $\Psi=\Phi|_{C}$ has a nontrivial kernel
too.

On the other hand, $\Psi$ is the identical mapping, since
$\Psi=\hat{\rho}|_{C}\circ\pi|_{C}=$ $\hat{\rho}|_{C}\circ\hat{\pi}=$
$\hat{\rho}|_{C}$ (the last identity is true as $\hat{\pi}$ is a projection
<<forgetting>> the $R$ coordinate, and $\hat{\rho}(R)=\\{1\\}$ is a
composition of the natural homomorphism to the quotient group and of the
retraction to $H$) and $\hat{\rho}|_{C}=id$, as $\hat{\rho}$ is the retraction
from $Q$ to $H$, so it acts trivially on $C$. The obtained contradiction
completes the proof. ∎

Let us provide some corollaries of this theorem:

######  Corollary 1.

Finitely generated nilpotent groups with nonabelian torsion subgroups are not
strongly verbally closed.

######  Proof.

Let us take the torsion subgroup of such group as $T$ from the theorem, and
the center of this torsion subgroup as $A\neq\\{1\\}$. Since $T$ is nilpotent
and nonabelian, every nontrivial normal subgroup of $T$ has a nontrivial
intersection with $A$ [KM79], so $A$ is not a direct factor of $T$. ∎

######  Corollary 2 [KMO21].

A finite group, whose center is not a direct factor is not strongly verbally
closed.

This theorem does not cover the case of finitely generated nilpotent
nonabelian groups with abelian torsion subgroups, and it is still unknown
whether there are strongly verbally closed groups among such groups. So far,
we can provide only a partial answer to this question (see the first
proposition of the following paragraph).

## 4\. Nilpotent non-strongly-verbally-closed groups

Let us remind that the discrete Heisenberg group is the free nilpotent group
of the nilpotency class two with two free generators. It can be easily
verified that this group admits a faithful representation in the group of
upper triangular matrices of size $3$ by $3$.

######  Proposition 5.

Let $H$ be the discrete Heisenberg group with $a$ and $b$ being its free
generators and $N$ being its subgroup:

$N=\langle\langle a^{\alpha},[a,b]^{n}\rangle\rangle,\ \alpha,n\geqslant 0.$

The group $G=H/N$ is strongly verbally closed if and only if
$\text{gcd}(\alpha,n)=1$.

######  Proof.

Let $T(G)$ be the torsion subgroup of $G$. The center of the group $H$ is
equal to its commutator subgroup, and it is isomorphic to the infinite cyclic
group. As it was said earlier (refer to the proof of the Proposition $4$), any
nontrivial normal subgroup of a nilpotent group intersects its center
nontrivially. Thus, if $T(G)=\\{1\\}$, then either $G=H$ is the discrete
Heisenberg group, or $G$ is abelian. Non-strong-verbal-closedness of $H$ was
proved in [KMO21], and the strong verbal closedness of abelian groups was
proved in [Maz18]. The case of $T(G)=\\{1\\}$ corresponds to the case of
$\alpha=0$, $n=0$ and $\alpha=0$, $n=1$.

If $\text{gcd}(\alpha,n)=1$, then, once again, $G$ is abelian, since
$[a,b]^{\alpha}=[a^{\alpha},b]\in N$; consequently, it is strongly verbally
closed.

Consider the case, when $\text{gcd}(\alpha,n)=d\neq 1$. Without loss of
generality, we may assume that $\alpha$ and $n$ are the least of non-negative
numbers such that $a^{\alpha}\in N$, $[a,b]^{n}\in N$. Consider the central
product of $G$ with its copy $\tilde{G}$ with joined commutator subgroup:

$K=G\underset{G^{\prime}=\tilde{G}^{\prime}}{\times}\tilde{G}=(G\times\tilde{G})/\\{(c,c^{-1})|c\in
G^{\prime}\\}.$

The group $G$ is not algebraically closed in $K$, since $G$ is not a retract
of $K$. Indeed, let $\rho$ be a hypothetical retraction. The group $G$
commutes with $\tilde{G}$ in $K$, so $\rho(\tilde{G})\leqslant Z(G)$ and
$\rho(\tilde{G}^{\prime})=\\{1\\}$, which leads to a contradiction with the
definition of retraction. However, $G$ is verbally closed in $K$. Let $w\in
F(t_{1},\dots,t_{s})$ be some word and

$w((h_{1}N,h_{1}^{\prime}N),\dots,(h_{s}N,h_{s}^{\prime}N))=(hN,N)$

for some $hN,h_{i}N\in G$, $h_{i}^{\prime}N\in\tilde{G}$. Then, for some
$cN\in G^{\prime}$, the following holds:

$\begin{cases}w(h_{1}^{\prime},\dots,h_{s}^{\prime})N=cN\\\
w(h_{1},\dots,h_{s})N=hc^{-1}N\\\ \end{cases}$

By an automorphism of the free group, the word $w$ can be reduced to a normal
form [KMO21]: $w(t_{1},\dots,t_{s})=t_{1}^{m}w^{\prime}(t_{1},\dots,t_{s})$,
where $m\in\mathbb{N}$, $w^{\prime}\in F_{s}^{\prime}$. From the first
equation, we get $cN\in G^{\prime}\ \cap\ \varphi(G^{s})$, where
$\varphi:G^{s}\to G,$ $(g_{1},\dots,g_{s})$ $\mapsto w(g_{1},\dots,g_{s})$ is
a verbal mapping. This means that for some $w_{1},w_{2}\in N$, in $H$ it is
true that:

$\begin{cases}w(h_{1}^{\prime},\dots,h_{s}^{\prime})=cw_{1}\\\
w(h_{1},\dots,h_{s})=hc^{-1}w_{2}\\\ \end{cases}$

Let us show that in $G$ the identity $(aN)^{x}=[aN,bN]^{z}$ doesn’t hold for
$x\not\in\alpha\mathbb{Z}$. Converse would mean that in the discrete
Heisenberg group the following holds:

$a^{x}[a,b]^{-z}=b^{-k}a^{-l}a^{\alpha t}[a,b]^{ns}a^{l}b^{k}$

for some $k,l,t,s\in\mathbb{Z}$. After some reductions, we get: $a^{x-\alpha
t}=[a,b]^{ns+z+\alpha tk}$. In $H$ it is possible only if $x=\alpha
t\in\alpha\mathbb{Z}$. We obtained a contradiction. Thus,
$h_{1}^{\prime}=[a,b]^{\gamma}$ for some $\gamma\in\mathbb{Z}$, and,
consequently, $cw_{1}\in H^{\prime}$. Since for any verbal mapping $\varphi$
in the discrete Heisenberg group (see [KMO21])

$\text{for any }g\in\varphi(H^{s}),\text{ it is true that
}g(\varphi(H^{s})\cap H^{\prime})\subseteq\varphi(H^{s}),$

for some $g_{1},\dots,g_{s}\in H$, $w(g_{1},\dots,g_{s})=$
$w(h_{1},\dots,h_{s})cw_{1}$. It means that:

$w(g_{1},\dots,g_{s})=hw_{3}$

for some $w_{3}\in W$, and in $G$:

$w(g_{1},\dots,g_{s})N=hN,$

which proves verbal closedness of $G$ in $K$. ∎

At last, let us prove that the higher dimensional Heisenberg groups over any
field are not strongly verbally closed:

The Heisenberg group of dimension $2n+1$ over a field $K$, where
$n\in\mathbb{N}$ is the group of upper triangular matrices of the kind

$H_{n}(K)=\Bigg{\\{}T(\bar{a},\bar{b},c)=\begin{pmatrix}1&\bar{a}&c\\\
0&I_{n}&\bar{b}\\\ 0&0&1\\\
\end{pmatrix}\Bigg{|}\bar{a},(\bar{b})^{\intercal}\in K^{n},c\in K\Bigg{\\}},$

where $I_{n}$ is the identity matrix of size $n$.

######  Proposition 6.

The group $H_{n}(K)$ is not strongly verbally closed.

######  Proof.

Consider the central product of $H_{n}(K)$ with its copy $\tilde{H}_{n}(K)$
with joined commutator subgroup:

$G=H_{n}(K)\underset{H_{n}(K)^{\prime}=\tilde{H}_{n}(K)^{\prime}}{\times}\tilde{H}_{n}(K).$

Denote with the symbol $H$ the first factor of this central product. Let us
show that $H$ is not algebraically closed in $G$. The group $H$ is linear,
and, consequently, it is equationally noetherian [BMR99], so it is
algebraically closed in $G$ if and only if it is a retract of every finitely
generated over $H$ subgroup of $G$ [KMM18]. In particular, of such a subgroup
of $G$:

$\bar{H}=\langle
H,(1,h_{1}),\dots,(1,h_{n}),(1,g_{1}),\dots,(1,g_{n})\rangle\text{, where
}h_{i}=\begin{pmatrix}1&\bar{a}_{i}&0\\\ 0&I_{n}&0\\\ 0&0&1\\\
\end{pmatrix},g_{i}=\begin{pmatrix}1&0&0\\\ 0&I_{n}&\bar{b}_{i}\\\ 0&0&1\\\
\end{pmatrix},$

where $\bar{a}_{i}=(0,\dots,1,\dots,0)=(\bar{b}_{i})^{\intercal}$, (unit is on
the $i$th place). Thus, $N=\langle h_{1},\dots,h_{n},g_{1},\dots,g_{n}\rangle$
is a subgroup of $\widetilde{H}_{n}(K)$, isomorphic to the discrete Heisenberg
group of the dimension $(2n+1)$. Let $\rho$ be a hypothetical retraction.
Since in $G$ the group $H$ commutes with $N$, we get that
$\rho(N^{\prime})=\\{1\\}$, which leads to a contradiction with the definition
of retraction.

Nevertheless, subgroup $H$ is verbally closed in $G$: let $w\in F_{s}$ be some
word (without loss of generality, this word is in the normal form we
established earlier), and let $\varphi:H^{s}\to H$ be the verbal mapping
associated with this word. Suppose that for some $h_{i},h\in H$,
$h_{i}^{\prime}\in\tilde{H}$, $c\in H^{\prime}$:

$\begin{cases}w(h_{1}^{\prime},\dots,h_{s}^{\prime})=c\\\
w(h_{1},\dots,h_{s})=hc^{-1}\\\ \end{cases}$

In general, on matrices $g_{i}=T(\bar{a}_{i},\bar{b}_{i},c_{i})$,
$i=1,\dots,s$, the mapping $\varphi$ acts like that:

$\varphi(g_{1},\dots,g_{s})=\begin{pmatrix}1&m\bar{a}_{1}&mc_{1}+f(\bar{a}_{1},\dots,\bar{a}_{s};\bar{b}_{1},\dots,\bar{b}_{s})\\\
0&I_{n}&m\bar{b}_{1}\\\ 0&0&1\\\ \end{pmatrix},$

where $f:(K^{n})^{s}\times(K^{n})^{s}\to K$ is some function linear in every
argument. The image of $f$ is either trivial or is equal to $K$, which leads
to:

$\varphi(H_{n}(K)^{s})=\begin{cases}\\{1\\},&\text{ if }m=0\text{ and the
image of }f\text{ is trivial}\\\ (H_{n}(K))^{\prime},&\text{ if }m=0\text{ and
the image of }f\text{ equals }K\\\ H_{n}(K),&\text{ if }m\neq 0\end{cases}$

Then $\varphi(H_{n}(K)^{s})\cap(H_{n}(K))^{\prime}\leqslant H_{n}(K)$ and for
every element $h\in\varphi(H_{n}(K)^{s})$ it is true that

$h(\varphi(H_{n}(K)^{s})\cap(H_{n}(K))^{\prime})\subseteq\varphi(H_{n}(K)^{s}),$

whence verbal closedness follows. ∎

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