finite normal subgroups of strongly verbally closed groups

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. ∎

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

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

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 $M$:

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. ∎

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:

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 $Z(G)$. ∎

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. ∎

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:

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

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  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. ∎

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 $H$:

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]. ∎

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 $H$:

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:

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

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 $Q$):

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:

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

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

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

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

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:

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:

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:

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

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 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 $T$. ∎