Verbal subgroup equals power subgroup in abelian group

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

 * Verbal upper-hook verbal equals verbal in abelian group
 * Intersection of two verbal subgroups of abelian group is verbal in both and in whole group

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:


 * 1) Let $$w \in W$$. We first prove that there exists an integer $$n_w$$ such that an element of $$G$$ can be expressed using the word $$w$$ if and only if it is a $$n_w^{th}$$ multiple: Note first that by Abelianness, we can replace $$w$$ by a word (written additively) of the form $$\sum_{i=1}^r a_ix_i$$, where $$a_i$$ are integers and $$x_i$$ are indeterminates. Let $$n_w$$ be the gcd of all the $$a_i$$s. We claim this $$n_w$$ works.
 * 2) * There exist integers $$m_i$$ such that $$\sum_{i=1}^r a_im_i = n_w$$. Thus, if $$y = n_wx$$, we get $$y = \sum_{i=1}^r a_i(m_ix)$$, so $$y$$ can be written using the word $$w$$.
 * 3) * Conversely, if $$y$$ can be written using the word $$w$$, then $$y = \sum_{i=1}^r a_ix_i = \sum_{i=1}^r n_w(a_i/n_w)x_i = n_w \sum_{i=1}^r (a_i/n_w) x_i$$, which is clearly a multiple of $$n_w$$.
 * 4) We now claim that the gcd of all the $$n_w, w \in W$$, is precisely the $$n$$ that we seek:
 * 5) * 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 $$n$$.
 * 6) * 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 $$n_w$$.

Set of nth powers is a verbal subgroup
Since $$G$$ is Abelian, the subset is a subgroup. Moreover, since $$nx = x + x + \dots + x$$ is itself a word, we see that it is a verbal subgroup.