Two-generated verbally closed subgroups of a free solvable group G are retracts of G

TL;DR

This paper proves that in free solvable groups of finite rank, any two-generated verbally closed subgroup is necessarily a retract of the entire group, revealing a structural property of these subgroups.

Contribution

It establishes that all two-generated verbally closed subgroups in free solvable groups are retracts, a new insight into subgroup structure.

Findings

Every two-generated verbally closed subgroup is a retract.

The result applies to free solvable groups of finite rank.

Provides a structural characterization of such subgroups.

Abstract

We prove that every verbally closed two-generated subgroup of a free solvable group G of a finite rank is a retract of G.

Equations

$$(x_2,x_1,x_1,x_2)=(z_2,z_1,z_1,z_2).$$

$$h=z_1^{k_1}...z_r^{k_r}{h}^{\prime }(z_1,\dots ,z_r),$$

$$h=x_1^{k_1}...x_r^{k_r}{h}^{\prime }(x_1,\dots ,x_r)$$

$$h_1^{k_1}...h_r^{k_r}{h}^{\prime }(h_1,\dots ,h_r)=h_1^{k_1}...h_r^{k_r}=h^{ds}\ne h,$$

$$(f(x_1,...,x_r),g(x_1,...,x_r),g(x_1,...,x_r),f(x_1,...,x_r))=(f,g,g,f)$$

$$(f(h_1,...,h_r),g(h_1,...,,h_r),g(h_1,...,,h_r),f(h,...,h_r))=(f,g,g,f).$$

$$f(h_1,...,h_r)\equiv f^{ϵ}(mod{H}^{\prime }),g(h_1,...,h_r)\equiv g^{\eta }(mod{H}^{\prime })\text{w}hereϵ,\eta \in \{\pm 1\}.$$

$$g\mapsto g(h_1,...,h_r),f\mapsto f(h_1,...,h_r)$$

$$g(h_1,...,h_r)=g^v,f(h_1,...,h_r)=f^v.$$

$$\beta :z_i\mapsto h_i^{v^{-1}},i=1,...,r.$$

$$\beta (g)=\beta (g(z_1,...,z_r))=g(h_1,...,h_r{)}^{v^{-1}}=g,$$

$$\beta (f)=\beta (f(z_1,...,z_r))=f(h_1,...,h_r{)}^{v^{-1}}=f.$$

$$g^{\eta }\mapsto g(h_1,...,h_r),f^{ϵ}\mapsto f(h_1,...,h_r)$$

$$u(x,y)=w(3;2;1;1)(x,y)w(2;2;1;2)(x,y)$$

$$u(g(x_1,...,x_r),f(x_1,...,x_r))=u(g,f)$$

$$g(h_1,...,h_r)=g^t,f(h_1,...,h_r)=f^t.$$

$$\varphi (g)=\varphi (g(z_1,\dots ,z_r))=g(h_1,\dots ,h_r{)}^{t^{-1}}=g,\varphi (f)=\varphi (f(z_1,\dots ,z_r))(f(h_1,\dots ,h_r){)}^{t^{-1}}=f.$$

Full text

1 Introduction

Algebraically closed objects play an important part in modern algebra. In this paper we study verbally closed and algebraically closed subgroups of free solvable groups.

We first recall that a subgroup $H$ is called algebraically closed in a group $G$ if for every finite system of equations $S=\{E_{i}(x_{1},\ldots,x_{n},H)=1\mid i=1,\ldots,m\}$ with constants from $H$ the following holds: if $S$ has a solution in $G$ then it has a solution in $H$. Then we recall that a subgroup $H$ of a group $G$ is called a *retract * of $G$, if there is a homomorphism (termed retraction) $\phi:G\to H$ which is identical on $H$. It is easy to show that every retract of $G$ is algebraically closed in $G$. Furthermore, if $G$ is finitely presented and $H$ is finitely generated then the converse is also true (see [1]). This result still holds for any finitely generated group $G$ which is equationally Noetherian. Recall that a group $G$ is called equationally Noetherian if for any $n$ every system of equations in $n$ variables with coefficients from $G$ is equivalent (has the same solution set in $G$) to some finite subsystem of itself (see [2], [3].

Let $F(X)$ be the free group of countably infinite rank with basis $X=\{x_{1},x_{2},...,x_{n},...\}.$ For $w=w(x_{1},\ldots,x_{n})\in F(X)$ and a group $G$ by $w[G]$ we denote the set of all $w$-elements in $G$, i.e., $w[G]=\{w(g_{1},\ldots,g_{n})\mid g_{1},\ldots,g_{n}\in G\}$. The verbal subgroup $w(G)$ is the subgroup of $G$ generated by $w[G]$. The $w$-width $l_{w}(g)=l_{w,G}(g)$ of an element $g\in w(G)$ is the minimal natural number $n$ such that $g$ is a product of $n$ $w$-elements in $G$ or their inverses; the width of $w(G)$ is the supremum of widths of its elements. The first question about $w$-width goes back to the Ore’s paper [4] where he asked whether the $[x,y]$-width (the commutator width) of every element in a non-abelian finite simple group is equal to $1$ (Ore Conjecture). The conjecture was established in [5]. Observing paper [6] and monographs [7], [8] present results about verbal width in groups.

Two important questions arise naturally for an extension $H\leq G$ and a given word $w\in F(X)$:

• •

when it is true that $w[H]=w[G]\cap H$?

• •

when $l_{w,G}(h)=l_{w,H}(h)$ for a given $h\in w(H)$?

To approach these questions Mysnikov and the author introduced in [1] a new notion of verbally closed subgroups.

Definition 1.1**.**

A subgroup $H$ of $G$ is called verbally closed if for any word $w\in F(X)$ and element $h\in H$ equation $w(x_{1},\ldots,x_{n})=h$ has a solution in $G$ if and only if it has a solution in $H$, i.e., $w[H]=w[G]\cap H$ for every $w\in F(X)$.

Theorem 1.2**.**

[1*]**.

Let $F$ be a free group of a finite rank. Then for a subgroup $H$ of $F$ the following conditions are equivalent:*

1. a)

$H$* is a retract of $F$.* 2. b)

$H$* is a verbally closed subgroup of $F$.* 3. c)

$H$* is an algebraically closed subgroup of $F$.*

This result clarifies the nature of verbally or algebraically closed subgroups in $F$. Surprisingly, the ”weak” verbal closure operator in this case is as strong as the standard one.

Similar result is true for any free nilpotent group $N$ of a finite rank.

Theorem 1.3**.**

[9*]**.

Let $N$ be a free nilpotent group of a finite rank and class $c$. Then for a subgroup $H$ of $N$ the following conditions are equivalent:*

1. a)

$H$* is a retract of $N$.* 2. b)

$H$* is a verbally closed subgroup of $N$.* 3. c)

$H$* is an algebraically closed subgroup of $N$.* 4. d)

$H$* is a free factor of the free group $N$ in the variety ${\cal N}_{c}$ of all nilpotent groups of the class $\leq c.$*

2 Preliminaries

In this section we collect some known or simple facts on verbally or algebraically closed subgroups.

Proposition 2.1**.**

[1]. Let $H\leq G$ be a group extension. Then the following holds:

If $H$ is a retract of $G$ then $H$ is algebraically closed in $G.$

• 2)

Suppose $G$ is finitely presented and $H$ is finitely generated. Then $H$ is algebraically closed in $G$ if and only if $H$ is a retract.

• 3)

Suppose $G$ is finitely generated relative to $H$ and $H$ is equationally Noetherian. Then $H$ is algebraically closed in $G$ if and only if $H$ is a retract.

Further in the paper $M_{r}$ denotes a free metabelian group of rank $r$ and $M_{rk}$ denotes a free metabelian nilpotent of class $k$ group of rank $r.$

The following statement was proved in [10] (see also [8]).

Lemma 2.2**.**

Let $G$ denotes $M_{2}$ or $M_{rk},k\geq 4,$ with basis $z_{1},z_{2}$. Let tuple $(g_{1},g_{2})$ is a solution of the equation

[TABLE]

Then $g_{i}\equiv z_{i}^{\pm 1}(mod\ G^{\prime}),i=1,2.$

Lemma 2.3** ([9], Lemma 1.1).**

Let $G$ be a group and let $N$ be a verbal subgroup of $G.$ If $H$ is a verbally closed subgroup of $G$ then its image $H_{N}$ is verbally closed in $G_{N}=G/N$.

Proof.

Suppose that equation $w(x_{1},...,x_{n})=h$ ($h\in H_{N}$) has a solution $(g_{1},...,g_{n})$ in $G_{N}$. Let $(g_{1}^{\prime},...,g_{n}^{\prime})$ be a preimage of this solution in $G.$ Then $(g_{1}^{\prime},...,g_{n}^{\prime})$ is a solution of the equation $w(x_{1},...,x_{n})=f^{\prime}$ in $G.$ Here $f^{\prime}=w(g_{1}^{\prime},...,g_{n}^{\prime})$ is a preimage of $h$ in $G.$ There also is a preimage $h^{\prime}$ of $h$ in $H.$ Then $f^{\prime}=h^{\prime}v,v\in N$. Let $v=v(x_{n+1},...,x_{n+m})$ be a corresponding to $N$ word that has value 1 in $G/N.$ The equation $w(x_{1},...,x_{n})v(x_{n+1},...,x_{n+m})^{-1}=h^{\prime}$ is solvable in $G$. Then it has a solution in $H.$ The image of the $n$ first components of this solution will be a solution of $w(x_{1},...,x_{n})=h$ in $H_{N}.$ ∎

Lemma 2.4**.**

Let $H=gp(g,f)$ be a two-generated verbally closed noncyclic subgroup of a free solvable group $S_{rd}(r,d\geq 2$). Then the image $\bar{H}$ in the abelianization $A_{r}=S_{rd}/S_{rd}^{\prime}$ (that is a free abelian group of rank $r$) is a direct factor of rank $2$, and $H$ is a free solvable group of rank $2$ and class $d.$

Proof.

By Lemma 2.3 $\bar{H}$ is verbally closed in $A_{r}$. By Theorem 1.3 $\bar{H}$ is a direct factor of $A_{r}.$ Obviously, $\bar{H}$ is non-trivial. We need to prove that $\bar{H}$ is not cyclic. By induction on $d$, we can assume that the image of $H$ in $S_{rd-1}=S_{rd}/S_{rd}^{(d-1)}$ is generated by the image $g^{\prime}$ of $g$ (so it is cyclic) and that $f\in S_{rd}^{(d-1)}$. Let $f(z_{1},...,z_{r})$ be an expression of $f$ in the free generators $z_{1},...,z_{r}$ of $S_{rd}.$ The equation $f(x_{1},...,x_{r})=f$ is solvable in $S_{rd}$. Hence, it has a solution $h_{1},...,h_{r}$ in $H.$ Then we can write the components in the form $h_{i}=g^{t_{i}}f^{\alpha_{i}},t_{i}\in\mathbb{Z},\alpha_{i}\in\mathbb{Z}[g^{\prime}],i=1,...,x_{r}.$ Then $f(h_{1},...,h_{r})=f(g^{t_{1}},...,g^{t_{r}})f^{\delta}=f^{\delta},$ where $\delta=\sum_{i=1}^{r}\alpha_{i}d_{i}(f)$ and $d_{i}(f)$ is a value of $i$-th partial Fox derivation in $\mathbb{Z}[g^{\prime}].$ Details about Fox derivatives see in [8]. It is easy to compute $\delta$ directly without Fox derivatives. We can collect exponents of $f$ from the expression $f(h_{1},...,h_{r}).$ Note that $\delta$ belongs to the fundamental ideal of $\mathbb{Z}[g^{\prime}]$, because $f(x_{1},...,x_{r})$ is a commutator word. Then $f^{1-\delta}=1$ and $1-\delta\not=0.$ This is impossible because $S_{rd}^{(d-1)}$ has no module torsion (see [8]).

By the G. Baumslag’s result [11] $H=gp(g,f)$ is free solvable group of rank $2$ and class $d$ with basis $g,f.$ Recall that Baumslag proved in [11] the following statement: A subgroup H of a free solvable group $S$ is itself a free solvable group if and only if there exists a set $Y$ of generators of $H$ which freely generates, modulo some term of the derived series of $S$, a free abelian group. He also gave an obvious formula of the solvability class of $H.$ See also [12]. ∎

3 Description of verbally and algebraically closed cyclic subgroups of free solvable groups

We start with the case of a cyclic subgroup.

Lemma 3.1**.**

Let $S$ be a free solvable group of a finite rank $r$ and let $H=gp(h)$ be a cyclic subgroup of $S$ generated by a non-trivial element $h\in S$. Then the following conditions are equivalent:

$H$* is verbally closed in $S$;*

• 2)

$H$* is a retract of $S$;*

• 3)

the image of $h$ in the abelianization $A_{r}=S/S^{\prime}$ (that is free abelian group of rank $r$) is primitive.

Proof.

Let $\{z_{1},...,z_{r}\}$ be a basis of $S$. The element $h$ can be expressed uniquely in the form

[TABLE]

where $k_{1},...,k_{r}\in\mathbb{Z}$ and $h^{\prime}(z_{1},\ldots,z_{r})$ is a product of commutators of words in $z_{1},\ldots,z_{r}$ (a commutator word). Then $h^{\prime}\in S^{\prime}.$

To show that 1) $\rightarrow$ 3) assume that $h$ has a non primitive image in $A_{r}$, i.e., either $h\in S^{\prime}$ or $gcd(k_{1},...,k_{n})=d>1.$

We first suppose that $h\in S^{\prime}$, so $k_{1}=\ldots=k_{r}=0$. Replacing each $z_{i}$ by a variable $x_{i}$ in (2) one gets an equation $h=x_{1}^{k_{1}}...x_{r}^{k_{r}}h^{\prime}(x_{1},\ldots,x_{r})$, with $h$ as a constant from $H$, which has a solution in $S$. However, this equation does not have a solution in $H$, since $H$ is abelian, so $h^{\prime}(h_{1},\ldots,h_{r})=1$ for any $h_{1},\ldots h_{r}\in H$. This shows that $H$ is not verbally closed in $S$ - contradiction. So $h\not\in S^{\prime}$. Then in this case $gcd(k_{1},...,k_{r})=d>1.$ The equation

[TABLE]

still has a solution in $S$, but for any $h_{1},\ldots h_{r}\in H$ one has

[TABLE]

for some $s\in\mathbb{Z}$. Hence, the equation does not have a solution in $H$, so $H$ is not verbally closed - contradiction.

To show that 3) $\rightarrow$ 2) assume that $h$ is primitive in the abelianization of $S$. Then there are integers $l_{1},...,l_{r}$ such that $k_{1}l_{1}+...k_{r}l_{r}=1.$ Now we define a homomorphism $\varphi:S\rightarrow H=gp(h)$ by putting $\varphi(z_{i})=h^{l_{i}}$ for $i=1,...,r.$ Since $H$ is abelian $\varphi(h^{\prime})=1$, so $\varphi(h)=h$ and $\varphi$ is a retraction. Hence $H$ is a retract, as claimed.

2. $\rightarrow$ 1) follows from Proposition 2.1, statement 1). ∎

4 Description of verbally and algebraically closed two-generated subgroups of free metabelian groups

Further in the Section, $z_{1},...,z_{r}$ is denoted a basis of group $G$ that is $M_{r}$ or $M_{rk}.$

Theorem 4.1**.**

Let $H=gp(g,f)$ be a two-generated subgroup of $G$ that is $M_{r}$ or $M_{rk}$, $r\geq 2,k\geq 4.$ Then the following conditions are equivalent:

$H$* is a retract of $G$.* 2. 2)

$H$* is a algebraically closed subgroup of $Gr$.* 3. 3)

$H$* is an verbally closed subgroup of $G$.*

Proof.

The implications 1) $\rightarrow$ 2) $\rightarrow$ 3) are obvious. We are to prove that 3) $\rightarrow$ 1). By Lemma 2.3 the image of $H$ in the abelianization $A_{r}=G/G^{\prime}$ is verbally closed. By Lemma 2.4 this image is a direct factor of $A_{r}$ of rank 2. We can assume that $g=uz_{1},uϵG^{\prime}$ and $f=vz_{2},vϵG^{\prime}$. Let $g(z_{1},...,z_{r})$ and $f(z_{1},...,z_{r})$ be expressions of $g$ and $f$ respectively. Then an equation

[TABLE]

has a solution in $G.$ Thus it has a solution in $H.$ It means that there are elements $h_{1},...,h_{r}$ in $H$ such that

[TABLE]

Now we consider the case $G=M_{r}.$ By the Baumslag’s theorem (see above) (4) is valid in the free metabelian group $H\simeq M_{2}$. By Lemma 2.2

[TABLE]

Recall that an element $u$ of a group $G$ is said to be test element if any endomorphism $\alpha:G\rightarrow G$ for which $\alpha(g)=g$ is an automorphism of $G.$ Timoshenko proved in [13] that every non-trivial element of $M_{2}^{\prime}$ is a test element for $M_{2}.$ Hence the following map

[TABLE]

defines an automorphism of $H.$ Let $\epsilon,\eta=1.$ Then this automorphism is identical $(mod\ H^{\prime}).$ By Bachmuth’s theorem [14] it is an inner automorphism (see also [17]). Then there is an element $v\in H$ for which

[TABLE]

It follows that $(f,g,g,f)^{v}=(f,g,g,f).$ Then $v\in H^{\prime}$ because the centralizer of any non-trivial element of the commutant $M_{r}^{\prime}$ of $M_{r}$ is equal to $M_{r}^{\prime}$. This statement was proved by Mal’cev in [12].

Define an endomorphism of $M_{r}$ by the map

[TABLE]

The image $\beta(M_{r})$ lies in $H.$ Moreover,

[TABLE]

[TABLE]

Thus $\beta$ is identical on $H$ and so $\beta$ is a retraction and $H$ is a retract.

Now let $(\epsilon,\eta)\not=(1,1).$ As above the following map

[TABLE]

defines an automorphism of $H.$ A composition of this automorphism with itself is identical $(mod\ H^{\prime})$ and we can finish our proof as above.

Let $G=M_{rk}.$ This group is free in a nilpotent variety ${\cal N}$. Then a pair of elements $g,f\in H$ which induce a pair of free generators $\bar{g},\bar{f}$ in $A_{r}=M_{rk}/M_{rk}^{\prime}$, generate a free ${\cal N}$-factor in $M_{rk}$. This factor is obviously a retract of $M_{rk}.$ ∎

5 Description of verbally and algebraically closed two-generated subgroups of free solvable groups

To prove the main result of this section in the case of group $S_{r3}$ we will use a concrete example of a test element of $S_{23}$, that was constructed in [15]. The proof in the general case of group $S_{rd}$ will be slightly different.

Example 5.1**.**

Let $x,y$ be a basis of $S_{23}.$ For any pair of positive integers $k$ and $l$ we denote by $z(k;l)(x,y)$ the element $(y,x;k,y;l-1)$, and by $w(k;l;m;n)(x,y)$ we denote the element $(z(k;l)(x,y),z(m;n)(x,y)).$ Then

[TABLE]

is a test element of $S_{23}$.

The following lemma was proved in [15].

Lemma 5.2**.**

Let $\phi$ be an endomorphism of $S_{23}$ for which $\phi(u)\equiv u(mod\ \gamma_{8}S_{23})$ where $u$ is the element in Example 5.1. Then $\phi$ is an automorphism identical modulo $S_{23}^{\prime}.$ In this case $\phi$ is inner by theorem proved in [16] (see also [17]).

The following more general result was proved in [18]

Lemma 5.3**.**

Let $S_{2d}$ be a free solvable group of rank $2$ and class $d\geq 2$ with basis $z_{1},z_{2},$ and let $v\in S_{2d}^{(d-1)},v\not=1.$ Then there is a positive number $m$ such that element $u=v^{(1-z_{1}^{m})(1-z_{2}^{m})}$ is a test element. Moreover, for every automorphism $\phi$ if $\phi(u)=u$ then $\varphi=\phi^{2}$ is an inner automorphism.

Theorem 5.4**.**

Let $H=gp(g,f)$ be a two-generated subgroup of $S_{rd},r,d\geq 2.$ Then the following conditions are equivalent:

$H$* is a retract of $S_{rd}$.* 2. 2)

$H$* is a algebraically closed subgroup of $S_{rd}$.* 3. 3)

$H$* is an verbally closed subgroup of $S_{rd}$.*

Proof.

The implications 1) $\rightarrow$ 2) $\rightarrow$ 3) are obvious. We are to prove that 3) $\rightarrow$ 1).

Case of group $S_{r3}.$ Let $z_{1},...,z_{r}$ be a basis of $S_{r3}$. By Theorem 4.1 and Lemmas 2.2 and 2.3 we can assume that $g=vz_{1},v\in S_{r3}^{\prime},$ and $f=wz_{2},w\in S_{r3}^{\prime}$. Let $g(z_{1},...,z_{r})$ and $f(z_{1},...,z_{r})$ be two expressions of $g$ and $f$ respectively. Let $u$ be the element described by Example 5.1. Then equation

[TABLE]

has a solution in $S_{r3}$. Hence it has a solution $h_{1},...,h_{r}$ in $H.$ The map $z_{i}\mapsto h_{i},i=1,...,r,$ defines an endomorphism $S_{r3}$ with the image in $H.$ It maps $g=g(z_{1},...,z_{r})$ to $g(h_{1},...,h_{r}),$ and $f=f(z_{1},...,z_{r})$ to $f(h_{1},...,h_{r}).$ By Lemma 5.2 the map $g\mapsto g(h_{1},...,h_{r}),f\mapsto f(h_{1},...,h_{r})$ defines an inner automorphism of $H.$ Then there is $t\in H$ for which

[TABLE]

It follows that $u(g,f)^{t}=u(g,f).$ In any group $S_{rd},r,d\geq 2,$ the centralizer of any non-trivial element $y\in S_{rd}^{(d-1)}$ coincides with $S_{rd}^{(d-1)}$ [12]. Hence $t\in H^{(2)}.$ Let $\phi$ be an endomorphism of $S_{r3}$ defined by the map: $z_{i}\mapsto h_{i}^{t^{-1}},i=1,...,r.$ The image $\phi(S_{r3})$ lies in $H$, and

[TABLE]

Hence $\phi$ is a retraction, and $H$ is a retract.

Case of group $S_{rd},d\geq 2$. Now $u$ is defined in Lemma 5.3. We repeat all the arguments as above until we get the map $g\mapsto g(h_{1},...,h_{r}),f\mapsto f(h_{1},...,h_{r})$ which defines an automorphism $\phi$ of $H.$ Then $\phi(S_{rd})\subseteq H.$ By Lemma 5.3 $\phi^{2}$ is an inner automorphism of $H.$ For some $t\in H$ we have equalities (13). Then we finish proof as in the previous case.

∎

6 Open problems

Problem 6.1**.**

Is it true that for any subgroup $H$ of any free solvable group $S_{rd}$ of rank $r\geq 2$ and class $d\geq 2$ the following conditions are equivalent:

$H$ is a retract of $S_{rd}$, 2. 2)

$H$ is a algebraically closed subgroup of $S_{rd}$, 3. 3)

$H$ is an verbally closed subgroup of $S_{rd}$?

It is likely that the answer to this question is negative. By Timoshenko’s theorem (see [18] or [19]) the test rank of $S_{rd},r,d\geq 2,$ is $r-1.$ Recall, that test rank of a group $G$ is the minimal number of elements of $G$ such that every endomorphism fixing any of this elements is automorphism. In the general case $H$ does not contain a test element. Hence, the methods of this paper do not work completely.

Problem 6.2**.**

Is it true that for any two-generated subgroup $H$ of any free polynilpotent group $P$ of rank $r\geq 2$ and class $(1,c_{1},...,c_{l}),l\geq 1,$ the following conditions are equivalent:

$H$ is a retract of $P$, 2. 2)

$H$ is a algebraically closed subgroup of $P$, 3. 3)

$H$ is an verbally closed subgroup of $P$?

By Timoshenko’s theorem [18] (see also [19]) any free solvable group $S_{2d},d\geq 2,$ contains test elements, all belong to $S_{rd}^{(d-1)}.$ For the affimative solution to this problem sufficient to prove that any $S_{2d}$ contains a test element with properties as in Lemma 5.2. Note that the reference to the statement in [16] (or [17]) in Lemma 5.2 works in the general case. Namely, any automorphism of $S_{2d},d\geq 2$ identical $mod\ S_{2d}^{\prime},d\geq 2,$ is inner.

Note, that by Gupta and the second author’s theorem [20] (see also [19]) any free polynilpotent group of class $(1,c_{1},...,c_{l})$ for $l\geq 1$ has a test element. We need in analog of Lemma 2.4 to solve this problem affirnatively.