Separability Properties of Nilpotent $\mathbb{Q}[x]$-Powered Groups

TL;DR

This paper investigates conjugacy and subgroup separability in nilpotent groups extended over the polynomial ring f[x], generalizing known theorems and exploring structural properties of these algebraic objects.

Contribution

It generalizes Baumslag's theorem to finitely f[x]-generated, torsion-free nilpotent f[x]-powered groups, establishing new intersection properties for f[x]-isolated subgroups.

Findings

Generalization of Baumslag's theorem to f[x]-powered groups

Establishment of intersection properties for f[x]-isolated subgroups

Extension of techniques from nilpotent groups to f[x]-powered groups

Abstract

In this paper we study conjugacy and subgroup separability properties in the class of nilpotent $\mathbb{Q}[x]$-powered groups. Many of the techniques used to study these properties in the context of ordinary nilpotent groups carry over naturally to this more general class. Among other results, we offer a generalization of a theorem due to G. Baumslag. The generalized version states that if $G$ is a finitely $\mathbb{Q}[x]$-generated $\mathbb{Q}[x]$-torsion-free nilpotent $\mathbb{Q}[x]$-powered group and $H$ is a $\mathbb{Q}[x]$-isolated subgroup of $G,$ then for any prime $\pi \in \mathbb{Q}[x]$, $\bigcap_{i = 1}^{\infty} G^{{\pi}^{i}}H = H.$

Equations

$$\bigcap _{i=1}^{\infty }{G}^{p^i}H=H.$$

$$\left(\genfrac{}{}{0ex}{}{r}{k}\right)=\frac{r(r-1)\cdots (r-k+1)}{k!}\in \mathbb{Q}[x].$$

$$G\times R\to Gdefined\; by(g,\alpha )\mapsto g^{\alpha },$$

$$g_1^{\alpha }\cdots g_n^{\alpha }={\tau }_1(\bar{g}{)}^{\alpha }{\tau }_2(\bar{g}{)}^{\left(\genfrac{}{}{0ex}{}{\alpha }{2}\right)}\cdots {\tau }_{k-1}(\bar{g}{)}^{\left(\genfrac{}{}{0ex}{}{\alpha }{k-1}\right)}{\tau }_k(\bar{g}{)}^{\left(\genfrac{}{}{0ex}{}{\alpha }{k}\right)},$$

$${\tau }_1(\bar{g})=g_1\cdots g_{n}and{\tau }_2\left(\overline{g}\right)={\left(g_1\cdots g_n\right)}^{-2}{g}_1^2\cdots g_n^2.$$

$$\bigcap _{i=1}^{\infty }{G}^{{\pi }^i}H=H.$$

$$[g^{-1}{h}^{-1}g,h]$$

$$H=\bigcap _{X\subset H_i{\le }_{R}G}\{H_i\}.$$

$$(gN{)}^{\alpha }=g^{\alpha }Nfor\; allgN\in G/Nand\alpha \in R,$$

$$1\to H\to G\to G/H\to 1$$

$$G^{\beta }=g{p}_R(g^{\beta }|g\in G).$$

$$G^{{\pi }_1{\pi }_2\cdots {\pi }_n}=1\; but{G}^{{\pi }_1{\pi }_2\cdots \widehat{{\pi }_i}\cdots {\pi }_n}\ne 1\; for\; any\; 1\le i\le n.$$

$$I(H,G)=\{g\in G|g^{\lambda }\in Hfor\; some\; 0\ne \lambda \in R\}.$$

$$H=g{p}_{\mathbb{Q}[x]}(g_1,g_2,\dots ),whereg=g_1^{\pi }and{g}_j=g_{j+1}^{\pi }for\; allj\in \mathbb{N}.$$

$$M=\underset{msummands}{\underbrace{D\oplus \cdots \oplus D}}\oplus T,$$

$$G=G_1{⊵}_R{G}_2{⊵}_R\cdots {⊵}_R{G}_{n+1}=1$$

$$\sum _{i=1}^{n}di{m}_K(K{\otimes }_R{G}_i/G_{i+1}).$$

$$G={\gamma }_{1}G\ge {\gamma }_{2}G\ge \cdots \ge {\gamma }_{c+1}G=1$$

$$H=H_1{\ge }_R{H}_2{\ge }_R\cdots {\ge }_R{H}_c{\ge }_R{H}_{c+1}=1$$

$$\frac{H_i}{H_{i+1}}=\frac{H\cap {\gamma }_{i}G}{H\cap {\gamma }_{i+1}G}=\frac{H\cap {\gamma }_{i}G}{(H\cap {\gamma }_{i}G)\cap {\gamma }_{i+1}G}{\cong }_R\frac{{\gamma }_{i+1}G(H\cap {\gamma }_{i}G)}{{\gamma }_{i+1}G}$$

$$1=N_0{⊴}_R{N}_1{⊴}_R\cdots {⊴}_R{N}_k=N$$

$$1=G_0/N{⊴}_R{G}_1/N{⊴}_R\cdots {⊴}_R{G}_l/N=G/N$$

$$1=N_0{⊴}_R{N}_1{⊴}_R\cdots {⊴}_R{N}_k{⊴}_R{G}_1{⊴}_R\cdots {⊴}_R{G}_l=G.$$

$$S=\{1,f\in \mathbb{Q}[x]|fis\; monic\},$$

$$H_{\sigma }=g{p}_{\mathbb{Q}[x]}\left(a^{\sigma }\right).$$

$$\left(G/H_{\sigma }\right)/\left(N_{\sigma }/H_{\sigma }\right){\cong }_{\mathbb{Q}[x]}G/N_{\sigma }$$

$$g{H}_{\sigma }\sim g{a}^t{H}_{\sigma }for\; each\sigma \in S.$$

$$y_{\sigma }^{-1}g{y}_{\sigma }=g{a}^t{\left(a^{\sigma }\right)}^{u_{\sigma }}.$$

$$\phi :L\to g{p}_{\mathbb{Q}[x]}(a)defined\; by\phi (l)=[l,g].$$

$$L=g{p}_{\mathbb{Q}[x]}(ker\phi ,b)and[b,g]=a^{\alpha }.$$

Full text

††copyright: ©2006: American Mathematical Society

1. Introduction

Let $\mathcal{C}$ be a class of groups. Among the types of separability properties of groups, there are:

• •

Conjugacy $\mathcal{C}$-Separability : A group $G$ is conjugacy $\mathcal{C}$-separable if whenever $g$ and $h$ are not conjugate elements in $G,$ there exists $K\in\mathcal{C}$ and a homomorphism $\varphi$ from $G$ onto $K$ such that $\varphi(g)$ and $\varphi(h)$ are not conjugate elements in $K.$

• •

Residually $\mathcal{C}$ : A group $G$ is residually $\mathcal{C}$ if for every non-identity element $g$ in $G,$ there exists $K\in\mathcal{C}$ and a homomorphism $\varphi$ from $G$ onto $K$ such that $\varphi(g)$ is not the identity.

Equivalently, for some non-empty index set $\Lambda,$ there exists a family $\{N_{\lambda}\ |\ \lambda\in\Lambda\}$ of normal subgroups of $G$ such that $G/N_{\lambda}\in\mathcal{C}$ for all $\lambda\in\Lambda$ and $\cap_{\lambda\in\Lambda}N_{\lambda}=1.$

• •

Subgroup $\mathcal{C}$-Separability : A subgroup $H$ of a group $G$ is called $\mathcal{C}$-separable in $G$ if, for every element $g\in G\setminus H,$ there is a homomorphism $\varphi$ of $G$ onto $K\in\mathcal{C}$ such that $\varphi(g)\notin\varphi(H).$ The group $G$ is said to be separated in a group from $\mathcal{C}$, and $K$ is said to separate $g$ and $H$.

It is clear that $G$ is residually $\mathcal{C}$ if and only if the identity subgroup of $G$ is $\mathcal{C}$-separable. Furthermore, every conjugacy $\mathcal{C}$-separable group is residually $\mathcal{C}.$

Let $\mathcal{F}$ and $\mathcal{F}_{p}$ denote the classes of finite groups and finite $p$-groups ($p$ is a prime) respectively. In the literature, there are several results pertaining to these properties for nilpotent groups. For instance,

(i)  In [4], N. Blackburn proved that every finitely generated nilpotent group is conjugacy $\mathcal{F}$-separable.

(ii)  In [9], E. A. Ivanova proved that if $G$ is a finitely generated torsion-free nilpotent group which is conjugacy $\mathcal{F}_{p}$-separable for some prime $p,$ then $G$ must be abelian.

(iii)  Let $G$ be a finitely generated torsion-free nilpotent group, and let $H$ be an isolated subgroup of $G.$ In [3], G. Baumslag proved that for any prime $p,$

[TABLE]

Equivalently, $H$ is $\mathcal{F}_{p}$-separable in $G$ for any given prime $p.$

If $H$ is the trivial subgroup in (iii), then we get a classical theorem of K. W. Gruenberg [7] which states that finitely generated torsion-free nilpotent groups are residually $\mathcal{F}_{p}$ for any prime $p.$

(iv)   In [2], R. B. J. T. Allenby and R. J. Gregorac prove the following: Let $G$ be a finitely generated nilpotent group and $H\leq G.$ Suppose there is an element $g\in G\setminus H$ such that $g^{n}\notin H$ for any positive integer $n.$ If $p$ is any prime, then $g$ and $H$ can be separated in a finite $p$-group.

Our goal in this paper is to prove results similar to those mentioned above for nilpotent $\mathbb{Q}[x]$-powered groups. The ring $\mathbb{Q}[x]$ is an example of a binomial ring, an integral domain of characteristic zero with unity such that for any $r\in\mathbb{Q}[x]$ and positive integer $k,$

[TABLE]

In this paper, $R$ will always be a binomial ring.

Definition 1.1**.**

[8] Let $G$ be a (locally) nilpotent group which comes equipped with an action by $R,$

[TABLE]

such that for all $g\in G$ and for all $\alpha\in R$, the element $g^{\alpha}\in G$ is uniquely determined. Then $G$ is called a nilpotent $R$-powered group if the following axioms hold:

• •

$g^{1}=g,\,g^{\alpha}g^{\beta}=g^{\alpha+\beta},$ and $(g^{\alpha})^{\beta}=g^{\alpha\beta}$ for all $g\in G$ and $\alpha,\beta\in R.$

• •

$(h^{-1}gh)^{\alpha}=h^{-1}g^{\alpha}h$ for all $g,\ h\in G$ and for all $\alpha\in R.$

• •

(Hall-Petresco Axiom) If $\{g_{1},\,\ldots,\,g_{n}\}\subset G$ and $\alpha\in R,$ then

[TABLE]

where $k$ is the nilpotency class of the group generated by $\{g_{1},\,g_{2},\,\ldots,\,g_{n}\},$ $\bar{g}=(g_{1},\,\ldots,\,g_{n}),$ and $\tau_{i}(\bar{g})$ is the $i$th Hall-Petresco word.

The Hall-Petresco words are obtained by letting \alpha=1,\ 2,\ and so on. For instance,

[TABLE]

$R$-modules and $R$-completions of finitely generated torsion-free nilpotent groups with respect to a Mal’cev basis are examples of nilpotent $R$-powered groups [8]. Categorical notions such as $R$-subgroup, $R$-homomorphism, etc. are defined in the obvious way [8].

Various theorems on residual properties of nilpotent $R$-powered groups have already been proven in some papers by the authors ([11], [14], [15]). A result on the conjugacy separability of nilpotent $\mathbb{Q}[x]$-powered groups can also be found in [11].

The definitions of separability carry over to nilpotent $R$-powered groups in a natural way. We merely state them here without giving additional background definitions. All necessary definitions, notations, and results on nilpotent $R$-powered groups will be provided in the next section.

Definition 1.2**.**

Let $\mathcal{C}$ denote a class of nilpotent $R$-powered groups. A nilpotent $R$-powered group $G$ is conjugacy $\mathcal{C}$-separable if whenever $g$ and $h$ are not conjugate elements in $G,$ there exists $K\in\mathcal{C}$ and an $R$-homomorphism $\varphi$ of $G$ onto $K$ such that $\varphi(g)$ and $\varphi(h)$ are not conjugate in $K.$

From this point on, $\mathcal{FT}$ and $\mathcal{FT_{\pi}}$ will denote the classes of nilpotent $\mathbb{Q}[x]$-powered groups of finite type and finite $\pi$-type for a prime $\pi\in\mathbb{Q}[x]$ respectively.

Theorem 1.1**.**

Let $G$ be a finitely $\mathbb{Q}[x]$-generated nilpotent $\mathbb{Q}[x]$-powered group. Two elements of $G$ are conjugate if and only if they are conjugate in every factor $\mathbb{Q}[x]$-group of $G$ of finite type.

This establishes that finitely $\mathbb{Q}[x]$-generated nilpotent $\mathbb{Q}[x]$-powered groups are conjugacy $\mathcal{FT}$-separable, thus proving an analogue of (i). Our proof mimics the one given by G. Baumslag for Theorem 1.3 of [3].

Our next theorem is similar to (ii).

Theorem 1.2**.**

Let $G$ be a finitely $\mathbb{Q}[x]$-generated $\mathbb{Q}[x]$-torsion-free nilpotent $\mathbb{Q}[x]$-powered group. If $G$ is conjugacy $\mathcal{FT_{\pi}}$-separable for some prime $\pi\in\mathbb{Q}[x],$ then $G$ is an abelian $\mathbb{Q}[x]$-group.

The proof mimics the one given by E. A. Ivanova in [9].

Definition 1.3**.**

Let $G$ be a nilpotent $R$-powered group, and let $\mathcal{C}$ be a class of nilpotent $R$-powered groups. An $R$-subgroup $H$ of $G$ is called $\mathcal{C}$-separable in $G$ if, for every element $g\in G\setminus H,$ there is an $R$-homomorphism $\varphi$ of $G$ onto $K\in\mathcal{C}$ such that $\varphi(g)\notin\varphi(H).$ The group $G$ is said to be separated in a group from $\mathcal{C}$, and $K$ is said to separate $g$ and $H$.

We prove the following analogue of (iii):

Theorem 1.3**.**

Let $G$ be a finitely $\mathbb{Q}[x]$-generated $\mathbb{Q}[x]$-torsion-free nilpotent $\mathbb{Q}[x]$-powered group, and let $H$ be a $\mathbb{Q}[x]$-isolated subgroup of $G.$ For any prime $\pi\in\mathbb{Q}[x]$,

[TABLE]

Equivalently, $H$ is $\mathcal{FT_{\pi}}$-separable for any given prime $\pi\in\mathbb{Q}[x].$

Another result on $\mathcal{FT_{\pi}}$-separability is:

Theorem 1.4**.**

Let $G$ be a finitely $\mathbb{Q}[x]$-generated nilpotent $\mathbb{Q}[x]$-powered group and $H\leq_{\mathbb{Q}[x]}G.$ Suppose there is an element $g\in G\setminus H$ such that $g^{\lambda}\notin H$ for any $0\neq\lambda\in\mathbb{Q}[x].$ If $\pi$ is any prime in $\mathbb{Q}[x],$ then $g$ and $H$ can be separated in a nilpotent $\mathbb{Q}[x]$-powered group of finite $\pi$-type.

The proof is similar to the one given for Theorem 1 (a) in [2].

2. Preliminaries

In this section, we discuss some terminology and results on nilpotent $R$-powered groups which will be used in this paper. We remind the reader that $R$ will always be a binomial ring. We will usually denote a group identity element, a multiplicative identity ring element, and the trivial subgroup of a group by $1.$

Let $G$ is a nilpotent $R$-powered group. If $[g,\ h]=1$ for $g,\ h\in G$, the Hall-Petresco axiom implies that $(gh)^{\alpha}=g^{\alpha}h^{\alpha}$ for any $\alpha\in R.$ In particular, every abelian $R$-group is an $R$-module.

Proposition 2.1**.**

Suppose $G$ is a nilpotent $R$-powered group. Let $g,\ h\in G$ such that $[g,\ h]\in Z(G).$ Then $[g,\ h]^{\alpha}=[g,\ h^{\alpha}]$ for all $\alpha\in R$.

Proof.

Since $[g,\ h]\in Z(G),$ we have

[TABLE]

By the previous paragraph, $(g^{-1}h^{-1}gh)^{\alpha}=(g^{-1}h^{-1}g)^{\alpha}h^{\alpha}$ for any $\alpha\in R.$ Since $(g^{-1}h^{-1}g)^{\alpha}=g^{-1}h^{-\alpha}g,$ the result follows. ∎

Let $G$ be a nilpotent $R$-powered group. If $H$ is an $R$-subgroup or a normal $R$-subgroup of $G,$ then we write $H\leq_{R}G\hbox{ and }H\unlhd_{R}G$ respectively. We say that $H$ is $R$-generated by $X=\{x_{1},\,\ldots,\,x_{j}\}\subseteq G$ if

[TABLE]

In this case, we write $H=gp_{R}(x_{1},\,\ldots,\,x_{j}).$

If $G$ is a nilpotent $R$-powered group and $N\unlhd_{R}G,$ then the $R$-action on $G$ induces an $R$-action on $G/N,$

[TABLE]

which turns $G/N$ into a nilpotent $R$-powered group.

It is not hard to show using the Hall-Petresco Axiom that if $G$ is a nilpotent $R$-powered group with $H\leq_{R}G$ and $N\unlhd_{R}G,$ then $HN\leq_{R}G$ and $HN=gp_{R}(H,\ N).$

The usual isomorphism theorems carry over to nilpotent $R$-powered groups in the obvious way. If $G_{1}$ and $G_{2}$ are nilpotent $R$-powered groups, then the kernel of an $R$-homomorphism $\varphi:G_{1}\rightarrow G_{2},$ abbreviated as $ker\ \varphi,$ is a normal $R$-subgroup of $G_{1}.$ If $G_{1}$ and $G_{2}$ happen to be $R$-isomorphic, then we write $G_{1}\cong_{R}G_{2}.$

The upper and lower central subgroups of a nilpotent $R$-powered group $G$ are $R$-subgroups of $G.$ We denote them by $\zeta_{i}G$ and $\gamma_{i}G$ respectively. The center of $G$ will be written as $Z(G).$

Theorem 2.2**.**

[10]** If $R$ is noetherian, then every $R$-subgroup of a finitely $R$-generated nilpotent $R$-powered group is finitely $R$-generated.

Definition 2.1**.**

[16] If $G$ is a nilpotent $R$-powered group, then $g\in G$ is an $R$-torsion element if $g^{\alpha}=1$ for some $0\neq\alpha\in R.$ If every element of $G$ is $R$-torsion, then $G$ is called an $R$-torsion group. We say that $G$ is $R$-torsion-free if the only $R$-torsion element of $G$ is $1.$

Definition 2.2**.**

[11] Let $\pi\in R$ be a prime. A finitely $R$-generated nilpotent $R$-powered group $G$ is of

1. (1)

finite type if it is an $R$-torsion group. 2. (2)

finite $\pi$-type if for each $g\in G,$ there exists a positive integer $k$ such that $g^{\pi^{k}}=1.$

The next theorem and lemma are proven by the authors in [12].

Theorem 2.3**.**

Let $R$ be noetherian and consider the short exact sequence

[TABLE]

in the category of nilpotent $R$-powered groups. Then $G$ is of finite type if and only if $H$ and $G/H$ are both of finite type.

Lemma 2.4**.**

Let $R$ be a PID. If $G$ is a finitely $R$-generated nilpotent $R$-powered group and $Z(G)$ is of finite type, then $G$ is of finite type.

The next lemma will be needed later.

Lemma 2.5**.**

Let $R$ be a PID. If $G$ is a finitely $R$-generated nilpotent $R$-powered group and is not of finite type, then $Z(G)$ has an element which is not $R$-torsion.

Proof.

Assume, on the contrary, that $Z(G)$ is an $R$-torsion group. Since $G$ is finitely $R$-generated, so is $Z(G)$ by Theorem 2.2. Thus, $Z(G)$ is of finite type. By Lemma 2.4, $G$ is also of finite type. ∎

If $G$ is a nilpotent $R$-powered group and $\beta\in R,$ then we define

[TABLE]

It is clear that $G^{\beta}\unlhd_{R}G.$

Theorem 2.6**.**

[13]** If $R$ contains $\mathbb{Q}$ and $G$ is a nilpotent $R$-powered group, then every element of $G^{\beta}$ is a $\beta$th power of an element of $G$ for any $\beta\in R.$

The notion of the exponent of a group has been generalized for the category of nilpotent $R$-powered groups and will be used in the proof of Theorem 1.4.

Definition 2.3**.**

[12] Let $R$ be a PID, and let $G$ be a nilpotent $R$-powered group  of finite type. We say that $G$ has exponent $\pi_{1}\pi_{2}\cdots\pi_{n},$ where the $\pi_{i}$’s are primes (not necessarily distinct) in $R,$ if

[TABLE]

Here, $\pi_{1}\pi_{2}\cdots\widehat{\pi_{i}}\cdots\pi_{n}$ stands for the omission of $\pi_{i}$ in the product $\pi_{1}\pi_{2}\cdots\pi_{n}.$

The $R$-isolator of a nilpotent $R$-powered group plays an important role in what follows.

Definition 2.4**.**

Let $G$ be a nilpotent $R$-powered group.

(i)  An $R$-subgroup $H$ of $G$ is $R$-isolated in $G$ if $g\in G$ and $g^{\lambda}\in H$ for some $0\neq\lambda\in R$ imply $g\in H.$

(ii)  The $R$-isolator of $S\subseteq G,$ denoted by $I(S,\ G),$ is the intersection of all $R$-isolated subgroups of $G$ containing $S.$

Clearly, $I(S,\ G)$ is an $R$-isolated subgroup of $G.$ The next theorem describes the $R$-isolator of an $R$-subgroup and can be found in [16].

Theorem 2.7**.**

[16]** If $G$ is a nilpotent $R$-powered group and $H\leq_{R}G,$ then

[TABLE]

Note that $I(1,\ G)$ is the $R$-subgroup of $G$ consisting of the $R$-torsion elements of $G.$ This is referred to as the $R$-torsion subgroup of $G$ and denoted by $\tau(G).$ It is clear that $\tau(G)\unlhd_{R}G$ and $G/\tau(G)$ is $R$-torsion-free.

We record a simple lemma which will be used in what follows.

Lemma 2.8**.**

Let $G$ be a nilpotent $R$-powered group and $H\unlhd_{R}G.$ Then $G/H$ is $R$-torsion-free if and only if $H$ is $R$-isolated in $G.$

The theory of extraction of roots appears in several papers by the authors (see [12], [13], [14], and [15]).

Definition 2.5**.**

If $G$ is a nilpotent $R$-powered group and $h^{\alpha}=g$ for some $g,\ h\in G$ and $\alpha\in R,$ then $h$ is called an $\alpha$th root of $g.$

Theorem 2.9**.**

[16]** Let $G$ be a nilpotent $R$-powered group. Then $G$ is $R$-torsion-free if and only if every $1\neq g\in G$ has at most one $\alpha$th root for every $0\neq\alpha\in R;$ that is, if $g,\ h\in G$ and $g^{\alpha}=h^{\alpha}$ for some $0\neq\alpha\in R,$ then $g=h.$

Corollary 2.10**.**

If $G$ is an $R$-torsion-free nilpotent $R$-powered group, then $Z(G)$ is $R$-isolated.

Proof.

If $g^{\alpha}\in Z(G)$ for some $g\in G$ and $\alpha\in R,$ then $h^{-1}g^{\alpha}h=g^{\alpha}$ for any $h\in G.$ Thus, $(h^{-1}gh)^{\alpha}=g^{\alpha}.$ By Theorem 2.9, $h^{-1}gh=g,$ and thus, $g\in Z(G).$ ∎

Another result that we will rely on is:

Lemma 2.11**.**

Let $G$ be a non-trivial finitely $\mathbb{Q}[x]$-generated $\mathbb{Q}[x]$-torsion-free nilpotent $\mathbb{Q}[x]$-powered group. For any $1\neq g\in G$ and prime $\pi\in\mathbb{Q}[x],$ there exists $n\in\mathbb{N}$ such that $g$ has no $\pi^{n}$th root in $G.$

Proof.

Assume, on the contrary, that $g$ has $\pi^{n}$th roots in $G$ for every $n\in\mathbb{N}.$ By Theorem 2.9, these $\pi^{n}$th roots must be unique. Thus, we let

[TABLE]

By Theorem 2.2, $H$ is finitely $\mathbb{Q}[x]$-generated. Consequently, there exists $k\in\mathbb{N}$ such that $H=gp_{\mathbb{Q}[x]}(g_{1},\ g_{2},\ \ldots,\ g_{k}).$ By construction of the $g_{j}$’s, this means that $H=gp_{\mathbb{Q}[x]}(g_{k}).$

Now, $H$ should contain all $\pi^{n}$th roots of $g.$ However, this is not the case since $g_{k},$ the $\pi^{k}$th root of $g,$ has no $\pi$th root in $H.$ For if $g_{k}$ had a $\pi$th root in $H,$ then there would exist $\mu\in\mathbb{Q}[x]$ such that $(g_{k}^{\mu})^{\pi}=g_{k};$ that is, $g_{k}^{\mu\pi-1}=1.$ Since $H$ is $\mathbb{Q}[x]$-torsion-free and $g_{k}\neq 1,$ $\mu\pi-1=0.$ It follows that $\pi$ must be a unit. However, $\pi$ is a prime by hypothesis, a contradiction. ∎

Next we discuss some results on the $R$-torsion-free rank of a finitely $R$-generated nilpotent $R$-powered group, where $R$ is noetherian (see [16]). First we recall the notion of torsion-free rank for a module.

Definition 2.6**.**

Let $D$ be an integral domain with fraction field $K,$ and let $M$ be a $D$-module. The torsion-free rank of $M$ is the $K$-dimension of the $K$-vector space $K\otimes_{D}M,$ denoted by $dim_{K}(K\otimes_{D}M).$

If $M$ is a finitely generated $D$-module, then $K\otimes_{D}M$ is finite dimensional as a $K$-vector space and $dim_{K}(K\otimes_{D}M)$ is the maximal number of $D$-linearly independent elements in $M,$ i.e. the maximal rank of a free $D$-submodule of $M$ (see Theorem 6.11 in [6] for a proof). In the literature, torsion-free rank is usually referred to simply as rank. In particular, if $D$ is a PID and

[TABLE]

where $T$ is the torsion part of $M,$ then the (torsion-free) rank of $M$ is $m.$

Definition 2.7**.**

Let $G$ be a finitely $R$-generated nilpotent $R$-powered group, where $R$ is noetherian, and let $K$ be the fraction field of $R.$ If

[TABLE]

is a subnormal $R$-series, where $G_{i}/G_{i+1}$ is an $R$-module for $i=1,\ 2,\ \ldots,\ n,$ then the $R$-torsion-free rank of $G$ is

[TABLE]

It follows from Theorem 2.2 that each quotient $G_{i}/G_{i+1}$ is a finitely generated $R$-module of finite rank. Thus, the above sum is well-defined and $G$ has finite $R$-torsion-free rank. An application of the Jordan-Hölder theorem shows that the $R$-torsion-free rank of a nilpotent $R$-powered group is independent of the subnormal $R$-series chosen [16].

The next theorem and corollary are extensions of known facts about the rank of a finitely generated $R$-module over a PID.

Theorem 2.12**.**

Let $R$ be a PID, and suppose $G$ is a finitely $R$-generated nilpotent $R$-powered group with $R$-torsion-free rank $n.$

(i)  If $H\leq_{R}G,$ then the $R$-torsion-free rank of $H$ is at most $n.$

(ii)  If $N\unlhd_{R}G$ and $N$ has $R$-torsion-free rank $m,$ then $G/N$ has $R$-torsion-free rank $n-m.$

Proof.

(i)  Suppose $G$ has nilpotency class $c,$ and let

[TABLE]

be the lower central series of $G.$ Assume that $\gamma_{i}G/\gamma_{i+1}G$ has $R$-torsion-free rank $n_{i}$ for $i=1,\ 2,\ \ldots,\ c.$ Since $G$ has $R$-torsion-free rank $n,$ we have $n_{1}+\cdots+n_{c}=n.$ Put $H_{i}=H\cap\gamma_{i}G$ for $i=1,\ 2,\ \ldots,\ c.$ Then

[TABLE]

is a central $R$-series for $H.$ Now,

[TABLE]

for $i=1,\ 2,\ \ldots,\ c.$ Therefore, $H_{i}/H_{i+1}$ is $R$-isomorphic to an $R$-subgroup of $\gamma_{i}G/\gamma_{i+1}G.$ Since $\gamma_{i}G/\gamma_{i+1}G$ is an $R$-module, the $R$-torsion-free rank of $H_{i}/H_{i+1}$ is at most $n_{i}.$ Hence, the $R$-torsion-free rank of $H$ is at most $n.$

(ii)  Let

[TABLE]

and

[TABLE]

be subnormal $R$-series of $N$ and $G/N$ respectively. Clearly, $G$ has a subnormal $R$-series

[TABLE]

Since $(G_{i+1}/N)/(G_{i}/N)\cong_{R}G_{i+1}/G_{i}$ for $i=0,\ 1,\ \ldots,\ l-1,$ the $R$-torsion-free rank of $G$ is the sum of the $R$-torsion-free ranks of $N$ and $G/N.$ ∎

The next corollary follows immediately from Theorem 2.12 (ii).

Corollary 2.13**.**

Let $R$ be a PID, and let $G$ be a finitely $R$-generated $R$-torsion-free nilpotent $R$-powered group. If $N\unlhd_{R}G,$ then $G$ and $N$ have the same $R$-torsion-free rank if and only if $G/N$ is an $R$-torsion group.

Corollary 2.14**.**

Let $R$ be a PID, and let $G$ be a finitely $R$-generated $R$-torsion-free nilpotent $R$-powered group. If the $R$-torsion-free rank of $G$ is 1, then $G$ is an $R$-module.

Proof.

It suffices to show that $G$ is abelian. Since $Z(G)$ is $R$-torsion-free, its $R$-torsion-free rank equals some $m>0.$ By Theorem 2.12 (i), $m$ must equal $1.$ Hence, $G/Z(G)$ is an $R$-torsion group by Corollary 2.13. Now, $Z(G)$ is $R$-isolated by Corollary 2.10. By Lemma 2.8, $G/Z(G)$ must be trivial. Thus, $G$ is abelian. ∎

3. Proofs of Theorems

We use the following notation: If $g$ and $h$ are (are not) conjugate group elements, then we write $g\sim h$ ($g\nsim h$).

Proof of Theorem 1.1: The proof is by induction on the $\mathbb{Q}[x]$-torsion-free rank $r$ of $G.$ If $r=0,$ then $G$ is of finite type and there is nothing to prove.

Suppose that $r>0.$ Let $g$ and $h$ be elements of $G$ which are conjugate in every factor $\mathbb{Q}[x]$-group of finite type. We claim that $g\sim h.$ Assume, on the contrary, that $g\nsim h.$ Since $r>0,$ $G$ is not of finite type. By Lemma 2.5, there exists $a\in Z(G)$ that is not a $\mathbb{Q}[x]$-torsion element. Let

[TABLE]

and for each $\sigma\in S,$ define

[TABLE]

For each $\sigma\in S$, observe that $H_{\sigma}\unlhd_{\mathbb{Q}[x]}G.$ By Theorem 2.12, the $\mathbb{Q}[x]$-torsion-free rank of each $G/H_{\sigma}$ is less than $r.$

Suppose that $gH_{\sigma}\nsim hH_{\sigma}$ in $G/H_{\sigma}$ for some $\sigma\in S.$ By induction and a $\mathbb{Q}[x]$-Isomorphism Theorem, there exists a normal $\mathbb{Q}[x]$-subgroup $N_{\sigma}/H_{\sigma}$ of $G/H_{\sigma}$ such that

[TABLE]

is of finite type and the images of $g$ and $h$ under the natural $\mathbb{Q}[x]$-homomorphism from $G$ onto $G/N_{\sigma}$ are not conjugate in $G/N_{\sigma}.$ This contradicts the assumption that the images of $g$ and $h$ are conjugate in every factor $\mathbb{Q}[x]$-group of $G$ of finite type. Therefore, $gH_{\sigma}\sim hH_{\sigma}$ for all $\sigma\in S.$ In particular, $gH_{1}\sim hH_{1}$ in $G/H_{1}.$ Thus, since $a\in Z(G),$ we can find a non-zero element $t\in\mathbb{Q}[x]$ such that $h\sim ga^{t}.$ Since $gH_{\sigma}\sim hH_{\sigma},$ we have

[TABLE]

Thus, for each $\sigma\in S,$ we can find $y_{\sigma}\in G$ and $u_{\sigma}\in\mathbb{Q}[x]$ such that

[TABLE]

Let $Y$ be the set of those $y_{\sigma}\in G$ arising in (1), and put $L=gp_{\mathbb{Q}[x]}(g,\ a,\ Y).$ Clearly, $\left[y,\ g\right]\in gp_{\mathbb{Q}[x]}(a)$ for all $y\in Y$ by (1). Moreover, $gp_{\mathbb{Q}[x]}(a)\leq_{\mathbb{Q}[x]}Z(L)$ since $a\in Z(G).$ It follows from the commutator calculus and the axioms that $[l,\ g]\in gp_{\mathbb{Q}[x]}(a)$ for all $l\in L.$ Thus, we obtain a $\mathbb{Q}[x]$-homomorphism

[TABLE]

Since $\mathbb{Q}[x]$ is a PID, $L/ker\ \varphi$ is $\mathbb{Q}[x]$-cyclic. Thus, there exists an element $b\in L\setminus ker\ \varphi$ and $\alpha\in\mathbb{Q}[x]$ such that

[TABLE]

Note that $\alpha\neq 0$ because $b\notin Ker\ \varphi.$ In fact, $b$ and $\alpha$ can be chosen so that $\alpha\in S.$

Next, we find all of the conjugates of $g$ in $L.$ By (2), such a conjugate has the form $(b^{\mu}k)^{-1}g(b^{\mu}k),$ where $k\in ker\ \varphi$ and $\mu\in\mathbb{Q}[x].$ Since $[g,\ b]=a^{-\alpha}\in Z(L)$ and $ker\ \varphi$ is the centralizer of $g$ in $L,$

[TABLE]

by Proposition 2.1. It follows that the set of conjugates of $g$ in $L$ is

[TABLE]

Now, $g\nsim ga^{t}$ in $L$ because $g\nsim ga^{t}$ in $G.$ Thus $ga^{t}\notin C,$ and so, $\alpha\mu\neq t$ for all $\mu\in\mathbb{Q}[x].$ On the other hand, (1) yields

[TABLE]

because $\alpha\in S.$ This means that $ga^{t}\left(a^{\alpha}\right)^{u_{\alpha}}\in C,$ so $t+\alpha u_{\alpha}=\alpha\lambda$ for some $\lambda\in\mathbb{Q}[x].$ Hence, $t=\alpha(\lambda-u_{\alpha}),$ a contradiction since $\lambda-u_{\alpha}\in\mathbb{Q}[x]$. Theorem 1.1 is proven.

Proof of Theorem 1.2: Suppose that $G$ has nilpotency class $c,$ and assume that $G$ is not abelian. In this case, $c>1.$ Thus, there exist elements $z_{2}\in\zeta_{2}G\setminus Z(G)$ and $h\in G$ such that $[z_{2},\ h]=z_{1}$ for some $1\neq z_{1}\in Z(G).$ Let $\alpha$ be a prime in $\mathbb{Q}[x]$ different from $\pi.$ Since $\mathbb{Q}[x]$ is noetherian and $G$ is finitely $\mathbb{Q}[x]$-generated, $Z(G)$ is also finitely $\mathbb{Q}[x]$-generated by Theorem 2.2. Hence, $Z(G)$ is a finitely $\mathbb{Q}[x]$-generated $\mathbb{Q}[x]$-torsion-free abelian $\mathbb{Q}[x]$-group. By Lemma 2.11, there exists $n\in\mathbb{N}$ such that $z_{1}$ has no $\alpha^{n}$th root in $Z(G).$ We claim that $z_{2}^{\alpha^{n}}\nsim z_{2}^{\alpha^{n}}z_{1}$ in $G,$ whereas $\varphi\left(z_{2}^{\alpha^{n}}\right)\sim\varphi\left(z_{2}^{\alpha^{n}}z_{1}\right)$ in $\varphi(G)$ for every $\mathbb{Q}[x]$-homomorphism $\varphi$ from $G$ onto a nilpotent $\mathbb{Q}[x]$-powered group of finite $\pi$-type. Once this is shown, the theorem will be proved.

First we show that $z_{2}^{\alpha^{n}}\nsim z_{2}^{\alpha^{n}}z_{1}$ in $G.$ If this were not the case, then there would exist $g\in G$ such that $g^{-1}z_{2}^{\alpha^{n}}g=z_{2}^{\alpha^{n}}z_{1}.$ Now, since $z_{2}\in\zeta_{2}G\setminus Z(G),$ there exists a central element $z$ such that $[z_{2},\ g]=z.$ Thus,

[TABLE]

Therefore, $z_{1}=z^{\alpha^{n}}.$ This contradicts the fact that $z_{1}$ has no $\alpha^{n}$th root in $Z(G).$

Next we show that the images of $z_{2}^{\alpha^{n}}$ and $z_{2}^{\alpha^{n}}z_{1}$ are conjugate under every $\mathbb{Q}[x]$-homomorphism from $G$ onto an arbitrary nilpotent $\mathbb{Q}[x]$-powered group of finite $\pi$-type. Let $N\unlhd_{\mathbb{Q}[x]}G$ such that $G/N$ is of finite $\pi$-type and $(z_{1}N)^{\pi^{m}}=N$ for some nonnegative integer $m;$ that is, $z_{1}^{\pi^{m}}\in N.$ Since $\alpha$ and $\pi$ are distinct primes, $\alpha^{n}$ and $\pi^{m}$ are coprime. Thus, there exist $\beta,\ \gamma\in\mathbb{Q}[x]$ such that $\beta\alpha^{n}+\gamma\pi^{m}=1.$ Now, $[z_{2},\ h]=z_{1}$ and $z_{1}\in Z(G)$ imply that $h^{-\beta}z_{2}h^{\beta}=z_{2}z_{1}^{\beta}$ by Proposition 2.1. Therefore,

[TABLE]

Since $z_{1}^{\pi^{m}}\in N,$ we have $z_{2}^{\alpha^{n}}z_{1}N=h^{-\beta}z_{2}^{\alpha^{n}}h^{\beta}N$ in $G/N.$ And so, $z_{2}^{\alpha^{n}}z_{1}N\sim z_{2}^{\alpha^{n}}N$ in $G/N.$ This completes the proof of Theorem 1.2.

Proof of Theorem 1.3: This proof invokes the following lemma:

Lemma 3.1**.**

Let $G$ be a finitely $\mathbb{Q}[x]$-generated $\mathbb{Q}[x]$-torsion-free nilpotent $\mathbb{Q}[x]$-powered group. Suppose that $Y\leq_{\mathbb{Q}[x]}Z(G),$ and let $H$ be a $\mathbb{Q}[x]$-isolated subgroup of $G.$ If $I(HY,\ G)=G,$ then $H\unlhd_{\mathbb{Q}[x]}G.$

Proof.

The proof is by induction on the $\mathbb{Q}[x]$-torsion-free rank $r$ of $G$. If $r=0,$ then $G$ is a $\mathbb{Q}[x]$-torsion group, contrary to assumption. If $r=1$, then $G$ is a $\mathbb{Q}[x]$-module by Corollary 2.14 and the result is trivial. Suppose then that $r>1.$ We first prove that $Z(H)\leq_{\mathbb{Q}[x]}Z(G).$ By Theorem 2.9, every element of $G$ has at most one $\alpha$th root for every $0\neq\alpha\in\mathbb{Q}[x].$ It follows that the centralizer of every non-empty subset of $G$ is $\mathbb{Q}[x]$-isolated in $G$ (see Lemma 4.8 in [14]). In particular, $Z(H)$ is $\mathbb{Q}[x]$-isolated in $G$ because $Z(H)=H\cap C_{H}(G),$ where $C_{H}(G)$ denotes the centralizer of $H$ in $G,$ and $H$ is $\mathbb{Q}[x]$-isolated in $G$ by hypothesis. We will use these observations freely in the remainder of the proof.

Pick any element $g\in G=I(HY,\ G).$ By Theorem 2.7, there exists $0\neq\alpha\in\mathbb{Q}[x]$ such that $g^{\alpha}\in HY.$ We assert that $g^{\alpha}$ centralizes $Z(H).$ Suppose that $h\in Z(H).$ Since $g^{\alpha}\in HY\leq_{\mathbb{Q}[x]}HZ(G),$ there exist elements $k\in H$ and $z\in Z(G)$ such that $g^{\alpha}=kz.$ Now, $[h,\ k]=1$ because $h\in Z(H),$ and so $[h,\ g^{\alpha}]=1$ because $z\in Z(G).$ Thus, $g^{\alpha}$ centralizes $Z(H)$ as asserted. Therefore, $g$ also centralizes $Z(H)$ because the centralizer of $Z(H)$ is $\mathbb{Q}[x]$-isolated in $G.$ Since $g$ is an arbitrary element of $G,$ we have that $Z(H)\leq_{\mathbb{Q}[x]}Z(G)$ as claimed.

The fact that $Z(H)\leq_{\mathbb{Q}[x]}Z(G)$ immediately gives $Z(H)\unlhd_{\mathbb{Q}[x]}G.$ Clearly, $G/Z(H)$ is finitely $\mathbb{Q}[x]$-generated. It is also $\mathbb{Q}[x]$-torsion-free by Lemma 2.8. Since $G$ has $\mathbb{Q}[x]$-torsion-free rank $r,$ $G/Z(H)$ has $\mathbb{Q}[x]$-torsion-free rank less than $r$ by Theorem 2.12 (i) and Corollary 2.13. We check that the remaining hypotheses of the lemma hold for $G/Z(H).$ Indeed, $YZ(H)/Z(H)\leq_{\mathbb{Q}[x]}Z(G/Z(H))$ since $Y\leq_{\mathbb{Q}[x]}Z(G).$ Moreover, since $H$ is $\mathbb{Q}[x]$-isolated in $G$, it is easy to see that $H/Z(H)$ must be $\mathbb{Q}[x]$-isolated in $G/Z(H).$ It remains to show that $G/Z(H)\leq_{\mathbb{Q}[x]}I(HY/Z(H),\ G/Z(H))$ (the reverse inclusion is trivially true). We begin by noting that $Z(H)\unlhd_{\mathbb{Q}[x]}HY$ since $Y\leq_{\mathbb{Q}[x]}Z(G).$ Suppose $gZ(H)\in G/Z(H)$ for some $g\in G.$ Since $G=I(HY,\ G),$ $g^{\beta}\in HY$ for some $0\neq\beta\in\mathbb{Q}[x].$ Thus, $g^{\beta}Z(H)\in HY/Z(H);$ that is, $(gZ(H))^{\beta}\in HY/Z(H).$ And so, $gZ(H)\in I(HY/Z(H),\ G/Z(H))$ as required. By induction, $H/Z(H)\unlhd_{\mathbb{Q}[x]}G/Z(H),$ and consequently, $H\unlhd_{\mathbb{Q}[x]}G.$ ∎

We now prove Theorem 1.3 by induction on the nilpotency class $c$ of $G.$ If $c=1$, then $G$ is a free $\mathbb{Q}[x]$-module of finite rank. By Lemma 2.8, $G/H$ is $\mathbb{Q}[x]$-torsion-free, and thus, a free $\mathbb{Q}[x]$-module as well. Hence, $G=H\times K$ for some $K\leq_{\mathbb{Q}[x]}G.$

We claim that

[TABLE]

To begin with, observe that $HK^{\pi^{i}}=H\times K^{\pi^{i}}$ because $H\cap K^{\pi^{i}}=1.$ We show that $HK^{\pi^{i}}=G^{\pi^{i}}H.$ Clearly, $HK^{\pi^{i}}\leq_{\mathbb{Q}[x]}G^{\pi^{i}}H.$ We assert that $G^{\pi^{i}}H\leq_{\mathbb{Q}[x]}HK^{\pi^{i}}.$ If $g^{\pi^{i}}h$ is a generator of $G^{\pi^{i}}H,$ then

[TABLE]

for some $h_{1}\in H$ and $k_{1}\in K.$ Thus, $g^{\pi^{i}}h\in HK^{\pi^{i}},$ proving the assertion and the claim.

Now, $K$ is finitely $\mathbb{Q}[x]$-generated by Theorem 2.2, as well as $\mathbb{Q}[x]$-torsion-free. Thus, $\bigcap_{i=1}^{\infty}K^{{\pi}^{i}}=1$ by Corollary 5.3 in [14]. And so,

[TABLE]

This proves the theorem when $c=1.$

Suppose that $c>1.$ Put

[TABLE]

We need to show that $L=H.$ By Corollary 3.3 in [14], $G/Z$ is $\mathbb{Q}[x]$-torsion-free because $G$ is $\mathbb{Q}[x]$-torsion-free. Since $I$ is $\mathbb{Q}[x]$-isolated in $G,$ $I/Z$ is $\mathbb{Q}[x]$-isolated in $G/Z.$ By induction,

[TABLE]

We claim that $\bigcap_{i=1}^{\infty}G^{{\pi}^{i}}I=I.$ Clearly, $\bigcap_{i=1}^{\infty}G^{{\pi}^{i}}I\geq_{\mathbb{Q}[x]}I.$ Let $x\in\bigcap_{i=1}^{\infty}G^{{\pi}^{i}}I.$ Then

[TABLE]

for all $i\geq 1$. It follows from (4) that $x\in I$, establishing our claim. Now, $\bigcap_{i=1}^{\infty}G^{{\pi}^{i}}H\leq_{\mathbb{Q}[x]}I.$ It follows that $H\leq_{\mathbb{Q}[x]}L\leq_{\mathbb{Q}[x]}I.$ Now, $I$ is finitely $\mathbb{Q}[x]$-generated by Theorem 2.2 and $\mathbb{Q}[x]$-torsion-free, and $H$ is $\mathbb{Q}[x]$-isolated in $I.$ By Lemma 3.1, $H\unlhd_{\mathbb{Q}[x]}I$. Moreover, $I/H$ is $\mathbb{Q}[x]$-torsion-free by Lemma 2.8 because $H$ is $\mathbb{Q}[x]$-isolated in $I.$

We show that $L=H.$ Consider the factor $\mathbb{Q}[x]$-group $L/H.$ Let $\ell H\in L/H,$ and let $i$ be a positive integer. By Theorem 2.6, every element of $G^{{\pi}^{i}}$ is a $\pi^{i}$th power of an element of $G.$ Thus, there exist $h\in H$ and $g_{i}\in G$ such that

[TABLE]

Hence, $g_{i}^{\pi^{i}}=\ell h^{-1}\in I.$ Since $I$ is $\mathbb{Q}[x]$-isolated in $G,$ $g_{i}\in I.$ This implies that $\ell H=(g_{i}H)^{\pi^{i}}$ with $g_{i}H\in I/H.$ It follows that $\ell H$ has a $\pi^{i}$th root in $I/H$ for every $i=1,\ 2,\ \ldots.$ If $\ell H\neq H$, this would contradict Lemma 2.11. Consequently, $\ell H=H,$ and thus, $L=H$ as desired. This finishes the proof of Theorem 1.3.

Proof of Theorem 1.4: Let $c$ be the nilpotency class of $G.$ We can assume, without loss of generality, that $G$ is $\mathbb{Q}[x]$-torsion-free. To see why, put $T=\tau(G)$ and fix a prime $\pi\in\mathbb{Q}[x].$ Since $T$ is a (normal) $\mathbb{Q}[x]$-subgroup of $G,$ it is finitely $\mathbb{Q}[x]$-generated by Theorem 2.2, and thus, of finite type. Let $\sigma\in\mathbb{Q}[x]$ be the exponent of $T,$ and assume that $(gT)^{\alpha}\in HT/T$ for some $0\neq\alpha\in\mathbb{Q}[x].$ There exist $h\in H$ and $t\in T$ such that $g^{\alpha}=ht.$ We claim that there exist non-zero distinct elements $\beta$ and $\mu$ of $\mathbb{Q}[x]$ and $t^{\prime}\in T$ such that $g^{\alpha\beta}=h^{\beta}t^{\prime}$ and $g^{\alpha\mu}=h^{\mu}t^{\prime}.$ Choose any $0\neq\beta\in\mathbb{Q}[x],$ and let $\mu=\beta+\sigma.$ Put $\tau_{i}=\tau_{i}(h,\ t)$ for $i\in\mathbb{N}.$ Since $T\unlhd_{\mathbb{Q}[x]}G,$ $\tau_{i}\in T$ for $i\geq 2$ (see the proof of Theorem 4.16 in [1] for details). By the Hall-Petresco Axiom,

[TABLE]

with $t^{\prime}\in T$ and

[TABLE]

Since $T$ has exponent $\sigma,$ $t^{\beta+\sigma}=t^{\beta}$ and $\tau_{i}^{-\binom{\beta+\sigma}{i}}=\tau_{i}^{-\binom{\beta}{i}}$ for $2\leq i\leq c.$ This proves the claim.

Now, $\alpha\beta-\alpha\mu\neq 0$ and $g^{\alpha\beta-\alpha\mu}=h^{\beta-\mu}\in H,$ contrary to the hypothesis. This means that $(gT)^{\alpha}\notin HT/T$ for any $0\neq\alpha\in\mathbb{Q}[x].$ Hence, $g^{\alpha}\notin HT,$ and thus, $g\notin T.$ Consequently, we might as well assume that $G$ is $\mathbb{Q}[x]$-torsion-free as mentioned in the beginning of the proof.

Since $I(H,\ G)$ is a $\mathbb{Q}[x]$-isolated subgroup of $G,$

[TABLE]

by Theorem 1.3. Now,

[TABLE]

because $g\notin I(H,\ G)$ by hypothesis. Furthermore,

[TABLE]

since $H\leq_{\mathbb{Q}[x]}I(H,\ G).$ And so, there exists an integer $j$ such that $G/G^{\pi^{j}}$ is a nilpotent $\mathbb{Q}[x]$-powered group of finite $\pi$-type separating $g$ and $H.$