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 🙂
Best Answer
A group is abelian if and only if satisfies the identity $[x_1,x_2]$; that is, if the verbal subgroup determined by $[x_1,x_2]$ (the commutator subgroup) is trivial. So if a set of words $W$ determines a variety all of whose elements are abelian groups, then the set $W'=W\cup\{[x_1,x_2]\}$ determines the same variety: all groups $G$ for which $W(G)=1$ also satisfy $W'(G)=1$, and conversely. Thus, every variety of abelian groups is determined by a set of words $W$ such that $[x_1,x_2]\in W$.
However, the verbal subgroup determined by any word $w$ is equivalent to the verbal subgroup determined by a pair of words, one of which is a commutator word (an element of $[F_{\omega},F_{\omega}]$, where $F_{\omega}$ is the free group on countably many variables), and the other is a power word (a word of the form $x^a$ for some $a\geq 0$). A sketch of this fact can be found in this answer. Thus, we can replace every element of $W$ with a commutator word and a power word. Note that this process does not replace $[x_1,x_2]$ (well, technically it does, but it replaces it with $x^0$ and $[x_1,x_2]$, which is equivalent to $[x_1,x_2]$, since $x^0$ evaluates to $e$ in any group, so it does not contribute to the verbal subgroup).
Since the value of a commutator word in a group $G$ is necessarily contained in $[G,G]$, if $W$ is a set of group words that includes $[x_1,x_2]$, then we can replace $W$ with a new set, $W'$, which consists of $[x_1,x_2]$ and power words; and then $W(G)=W'(G)$ for every group $G$.
Moreover, in any group, the subgroup generated by all $n$th powers and all $m$th powers is equal to the subgroup generated by all $\gcd(n,m)$th powers: clearly the latter contains the former, since very $n$th power is also a $\gcd(n,m)$th power; and if $\gcd(m,n) = \alpha n + \beta m$ with $\alpha,\beta$ integers, then $x^{\gcd(m,n)} = x^{\alpha n + \beta m} = (x^n)^{\alpha}(x^m)^{\beta}$, so $x^{\gcd(m,n)}$ lies in the subgroup generated by all $n$th and $m$th powers.
Thus, if $W$ is a set of words that consists of $[x_1,x_2]$ and a collection of power words, then we can replace $W$ with a set $W'$ that contains only $[x_1,x_2]$ and a single power word $x^a$ for some $a\geq 0$; and then for any group $G$, we have $W(G)=W'(G)$.
Thus, any set of words $W$ that determines a variety of abelian groups is equivalent to a set $W'$ of the form $W'=\{[x_1,x_2],x^n\}$ for some $n\geq 0$; in the sense that for every group $G$, $W(G)=W'(G)$.
Since a variety of groups corresponds to a set of words $W$, and is determined as "all groups for which $W(G)$ is trivial", we conclude that a variety of abelian groups corresponds to a nonnegative integer $n\geq 0$, and is determined as "all abelian groups $G$ for which $g^n = e$ for all $g\in G$." That is, the abelian groups of exponent $n$. Denote this variety by $\mathcal{Ab}_n$. Note that the variety of all abelian groups is $\mathcal{Ab}_0$.
So the varieties of abelian groups are precisely the varieties $\mathcal{Ab}_n$, for some fixed $n\geq 0$.