Verbal subgroup equals power subgroup in abelian group

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a Abelian group. That is, it states that in a Abelian group (?), every subgroup satisfying the first subgroup property (i.e., Verbal subgroup (?)) must also satisfy the second subgroup property (i.e., Power subgroup (?)). In other words, every verbal subgroup of Abelian group is a power subgroup of Abelian group.
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Statement

Suppose $G$ is an abelian group and $H$ is a verbal subgroup. Then, there exists an integer $n$ such that $H$ is precisely the set of $n^{th}$ multiples in $G$ :

$H=\{nx\mid x\in G\}$ .

Conversely, the set of $n^{th}$ multiples form a verbal subgroup for any integer $n$ .

Proof

Verbal subgroup implies it is the set of nth powers/multiples

Given: A group $G$ , a verbal subgroup $H$ with a collection $W$ of words generating it.

To prove: There exists a natural number $n$ such that $H$ is the set of $n^{th}$ multiples.

Proof:

Let $w\in W$ $w\in W$ . We first prove that there exists an integer $n_{w}$ $n_{w}$ such that an element of $G$ $G$ can be expressed using the word $w$ $w$ if and only if it is a $n_{w}^{th}$ $n_{w}^{th}$ multiple: Note first that by Abelianness, we can replace $w$ $w$ by a word (written additively) of the form $\sum _{i=1}^{r}a_{i}x_{i}$ $\sum _{i=1}^{r}a_{i}x_{i}$ , where $a_{i}$ $a_{i}$ are integers and $x_{i}$ $x_{i}$ are indeterminates. Let $n_{w}$ $n_{w}$ be the gcd of all the $a_{i}$ $a_{i}$ s. We claim this $n_{w}$ $n_{w}$ works.
- There exist integers $m_{i}$ such that $\sum _{i=1}^{r}a_{i}m_{i}=n_{w}$ . Thus, if $y=n_{w}x$ , we get $y=\sum _{i=1}^{r}a_{i}(m_{i}x)$ , so $y$ can be written using the word $w$ .
- Conversely, if $y$ can be written using the word $w$ , then $y=\sum _{i=1}^{r}a_{i}x_{i}=\sum _{i=1}^{r}n_{w}(a_{i}/n_{w})x_{i}=n_{w}\sum _{i=1}^{r}(a_{i}/n_{w})x_{i}$ , which is clearly a multiple of $n_{w}$ .
We now claim that the gcd of all the $n_{w},w\in W$ $n_{w},w\in W$ , is precisely the $n$ $n$ that we seek:
- Note that each element expressible using a word $w\in W$ is a multiple of $n_{w}$ , and thus, a multiple of $n$ . Thus, any element in the subgroup generated by such elements is also a multiple of $n$ .
- Conversely, since $n$ is the gcd of the $n_{w}$ , there exist $w_{1},w_{2},\dots ,w_{s}$ and integers $b_{1},b_{2},\dots ,b_{s}$ such that $\sum b_{i}n_{w_{i}}=n$ . Thus, if $y=nx$ , $y=\sum b_{i}(n_{w_{i}}x)$ , hence $y$ is in the subgroup generated by multiples of $n_{w}$ .