TITLE: Find all the varieties of abelian groups (in the sense of universal algebra)
QUESTION [6 upvotes]: This is Exercise 2.3.6 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
(Note: Here "abelian variety" is not, at least a priori, in the sense given in this Wikipedia article. See below for details.)
The Details:
These are essentially the same as in the previous exercise of Robinson's book.
Since definitions vary, on page 15, ibid., paraphrased, it states that

A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:
(i) $xN=Nx$ for all $x\in G$.
(ii) $x^{-1}Nx=N$ for all $x\in G$.
(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.

On page 56, ibid.,

Let $F$ be a free group on a countably infinite set $\{x_1,x_2,\dots\}$ and let $W$ be a nonempty subset of $F$. If $w=x_{i_1}^{l_1}\dots x_{i_r}^{l_r}\in W$ and $g_1,\dots, g_r$ are elements of a group $G$, we define the value of the word $w$ at $(g_1,\dots,g_r)$ to be $w(g_1,\dots,g_r)=g_1^{l_1}\dots g_{r}^{l_r}$. The subgroup of $G$ generated by all values in $G$ of words in $W$ is called the verbal subgroup of $G$ determined by $W$,
$$W(G)=\langle w(g_1,g_2,\dots) \mid g_i\in G, w\in W\rangle.$$

On page 57, ibid.,

If $W$ is a set of words in $x_1, x_2, \dots$ and $G$ is any group, a normal subgroup $N$ is said to be $W$-marginal in $G$ if
$$w(g_1,\dots, g_{i-1}, g_ia, g_{i+1},\dots, g_r)=w(g_1,\dots, g_{i-1}, g_i, g_{i+1},\dots, g_r)$$
for all $g_i\in G, a\in N$ and all $w(x_1,x_2,\dots,x_r)$ in $W$. This is equivalent to the requirement: $g_i\equiv h_i \mod N, (1\le i\le r)$, always implies that $w(g_1,\dots, g_r)=w(h_1,\dots, h_r)$.
[The] $W$-marginal subgroups of $G$ generate a normal subgroup which is also $W$-marginal. This is called the $W$-marginal of $G$ and is written $$W^*(G).$$

On page 58, ibid.,

If $W$ is a set of words in $x_1, x_2, \dots $, the class of all groups $G$ such that $W(G)=1$, or equivalently $W^*(G)=G$, is called the variety $\mathfrak{B}(W)$ determined by $W$.

The Question:

A variety is said to be abelian if all its members are abelian. Find all the abelian varieties.

Thoughts:
It occurred to me to consider two extremes first: where the variety $\mathfrak{W}$ is just the class containing only (the isomorphic copies of) the trivial group and where the variety $\mathfrak{W}$ is the class of all abelian groups. I guess I could get at least one of these extremes by considering
$$\mathfrak{W}=\mathfrak{B}(\{\varepsilon\}),\tag{1}$$
where $\varepsilon$ is the empty word; but something tells me that $(1)$ includes any & all groups, not just abelian ones; I don't know - I'm not particularly confident in calculating it.
Another thought I have is whether an abelian variety might correspond to
$$\mathfrak{W}=\mathfrak{B}(\{ [x_1, x_2]\}),\tag{2}$$
where $[x_1,x_2]=x_1^{-1}x_2^{-1}x_1x_2$ is the commutator of the abstract symbols $x_1,x_2$. Again, I don't know.
Please help :)

REPLY [3 votes]: The following argument is basically the same as Arturo's above, but somewhat streamlined in presentation.
Suppose I have some word $$p=x_1^{z_1}x_2^{z_2}...x_n^{z_n}$$ with $z_i\in\mathbb{Z}$; I claim that there is some $k$ such that the abelian groups in which $p$ evaluates to $1$ on all inputs are exactly those groups where every element has order dividing $k$.
First, since we're only looking at abelian groups we can assume WLOG that $x_i\not=x_j$ whenever $1\le i<j\le n$. Let $k=gcd\{\vert z_1\vert ,...,\vert z_n\vert\}$. Clearly if $G$ is an abelian group where every element has order dividing $k$ we get $p(g_1,...,g_n)=1_G$ for all $g_1,...,g_n\in G$. In the opposite direction, let $g\in G$; I claim that $g^{\vert z_i\vert}=1_G$ for each $1\le i\le n$.
To see this, fix such an $i$ and consider the variable assignment $\nu$ defined by $$x_u=\begin{cases}
g & \mbox{if $u=i$,}\\
1_G & \mbox{otherwise.}
\end{cases}$$ If $p$ always evaluates to $1_G$ in $G$, we must have $p[\nu]=1_G$. But $p[\nu]=g$.

Strictly speaking we're not done; the above classifies the varieties determined by a single word. But each variety is an intersection of such single-word varieties, and any intersection of "$\mathcal{Ab}_m$-classes" is again an "$\mathcal{Ab}_m$-class," using Arturo's notation here. (Note that this latter fact is just a rephrasing of the Noetherianness of $\mathbb{Z}$!)