Verbal subgroup equals power subgroup in abelian group

This article gives a proof/explanation of the equivalence of multiple definitions for the term verbal subgroup of abelian group
View a complete list of pages giving proofs of equivalence of definitions

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 $n$ .

Related facts

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 $n_{w}$ .