Verbal Subgroups of $\mathbb{Z}^2$

Question

So I want to try to show that all the verbal subgroups of an abelian group $G$ are of the form $G(X^n), n \geq 1$. I want to start with $\mathbb{Z}$ and $\mathbb{Z}^n$. So I worked with $\mathbb{Z}$, I know that all the subgroups of $\mathbb{Z}$ are $m\mathbb{Z} = $ $\langle {m} \rangle$ $ = \mathbb{Z}(X^m)$, so then all the subgroups of $Z$ are verbal.

Now if I look at $\mathbb{Z}^2$, the subgroups of this are:

$(0,m) \mathbb{Z} = \langle (0,m) \rangle$

$(m,0) \mathbb{Z} = \langle (m,0) \rangle$

$(a,b) \mathbb{Z} = \langle (a,b) \rangle$ $ a,b \in \mathbb{Z}-0$

$(a,b) \mathbb{Z} + (c,d) \mathbb{Z} = \langle (a,b) \rangle + \langle (c,d) \rangle$

Now I know that a verbal subgroup must be fully characteristic, so i know the first 3 subgroups are not verbal under the endomorphism $f(x,y) = (y,x)$. In the third case if $a = b$ then it fails under the endomorphism $g(x,y) = (x+1,y+2)$.

The last case, I'm not really sure about, I could show it fails in general with $f$ as above but perhaps in certain cases it would work? But then how do I show it is verbal , I'm not even sure how to begin to write any of these subgroups as $\mathbb{Z}(X^n)$, it doesn't seem like you can, at least not as easily as we did for $\mathbb{Z}$.

Am I missing a subgroup of $\mathbb{Z}^2$? How could I extend this to a general $\mathbb{Z^n}$?

Edit: Definition of verbal subgroup. Let $W$ be a set of words, then the verbal subgroup $G(W)$ of a group G, is the subgroup of G generated by the words $w \in W$, written with elements of G.

i.e. $G(X^2) = \langle g^2 | g \in G\rangle$ or $G(X^2, XYX) = \langle g^2, ghg | g,h \in G\rangle$

Answer 1

I wouldn't bother with $\mathbb{Z}^n$, and instead start at full generality.

First off, any word in $x_1,\cdots,x_m$ can be simplified to the form $x_1^{a_1}\cdots x_m^{a_m}$ because $G$ is abelian. Second, we can set all but one of the $x_i$s to be $e$ to say the verbal subgroup contains any of the powers $x^{a_i}$, and conversely if it contains any of those powers then of course it contains the word $x_1^{a_1}\cdots x_m^{a_m}$. Therefore, the verbal subgroup is WLOG generated by a bunch of powers $x^{a_1},\cdots,x^{a_m}$. Now, we can take these themselves to powers and multiply them together to get any $x^{a_1b_1+\cdots+a_mb_m}$ (for given choice of $x$). This in turn means the verbal subgroup contains all powers $x^g$ where $g=\gcd(a_1,\cdots,a_m)$, but conversely all $a_i$s are multiples of $g$ so...

Answer 2

Any word $w(x_1,\ldots,x_n)$ is equivalent to a power word $x^a$ and a commutator word $c$; that is, the verbal subgroup determined by $w$ is equal to the verbal subgroup determined by $x^a$ and $c$.

Indeed, this can be verified through the collection process. Consider an arbitrary word $w$. The first occurrence of $x_1$ will look something like $$w_1 x_1^a w_2$$ where $w_1$ and $w_2$ are words, and $w_1$ does not contain any instances of $x_1$. But we can rewrite this as $$w_1x_1^aw_2 = x_1^aw_1[w_1,x_1^a]w_2 = x_1^aw_1w_2[w_1,x_1^a][[w_1,x_1^a],w_2] = x_1^aw_1w_2c,$$ where $c\in [F_n,F_n]$, the commutator subgroup of the free group in $n$ letters. Then we can do the same thing with the next occurence of $x_1$, which will be in $w_2$, and so on until we move the last original occurrence of $x_1$ all the way to the left. We end up with a word of the form $$x_1^{a_1}w_3c_1$$ where $w_3$ is a word that does not contain $x_1$, and $c_1$ is an element of the commutator subgroup of the free group $F_n$. We can then repeat the process with $x_2$, $x_3$, etc. In the end, we obtain that the original word is equal to the word $$w(x_1,\ldots,x_n)=x_1^{a_1}\cdots x_n^{a_n}c$$ where $c\in [F_n,F_n]$.

Among the elements in $w(G)$ (the verbal subgroup of $G$ determined by $w$) are the elements $w(g,e,\ldots,e)$, $w(e,g,e,\ldots,e),\ldots,w(e,\ldots,e,g)$. For each of these words we have that $c$ evaluates to the trivial element, since the value lies in the subgroup $\langle g\rangle$ which is abelian. Thus, the verbal subgroup $w(G)$ includes all values of the form $g^{a_1},\ldots,g^{a_n}$.

If $d=\gcd(a_1,\ldots,a_n)$, then we know that $d=\alpha_1a_1+\cdots+\alpha_na_n$ for integers $\alpha_1,\ldots,\alpha_n$, hence the elements $g^d$ are also all in the verbal subgroup. And clearly with the elements $g^d$ we also obtain all elements $g^{a_1},\ldots,g^{a_n}$.

Thus, the verbal subgroup determined by the words $x_1^d$ and $c$ (the commutator word from the rewritten $w$) is equal to $w(G)$: because the latter is generated by the values $g_1^{a_1}\cdots g_n^{a_n}c(g_1,\ldots,g_n)$, and since we also have $g_1^{a_1},\ldots,g_n^{a_n}\in w(G)$, we also have $c(g_1,\ldots,g_n)$ separately. Thus, $w(G)$ is contained in the verbal subgroup $W(G)$ with $W=\{x_1^d,c\}$, and conversely.

From this we conclude that any verbal subgroup is equal to the verbal subgroup determined by a single power word $x^a$, plus a set of commutator words $\{c_i\}_{i\in I}$. This holds in any group.

(This argument can be found near the beginning of Hanna Neumman’s book, Varieties of groups, and is due to Bernhard Neumman.)

Now if $G$ is abelian, then all commutator words evaluate to the trivial element. Thus, any verbal subgroup of an abelian group is of the form $G^n = \{g^n\mid g\in G\}$ for some positive integer $n$ (technically, the subgroup generated by these elements, but in an abelian group, the set of $n$th powers is already a subgroup).

In particular, this holds for $\mathbb{Z}^2$.