I am trying to understand an argument in a group theory, which is not my strong suit (I mostly work with semigroups and rarely delve into the actual structure of groups).
Let $\boldsymbol{\mathcal{V}}$ be a non-Abelian variety of groups such that every solvable group in $\boldsymbol{\mathcal{V}}$ is Abelian. Apparently, the relatively free group on two generators, $F_2$, has a perfect commutator subgroup; that is $F_2^{(1)} = F_2^{(2)}$. This is presented as an obvious fact, but I have no idea of how to prove it. I would like to know why it is true. For my purposes, I can assume that the variety has bounded exponent, but this doesn't seem to be needed.
Thanks!
In fact, this holds for every group $G\in\mathcal{V}$.
Let $G\in\mathcal{V}$, and let $K=G/G’’$. This is solvable, since $K’’=G’’/G’’$ is trivial. By assumption, this means that $K$ is abelian, and hence that $K’=\{e\}$. But $K’=G’/G’’$, hence $G’=G’’$.
(The fact that $K’’=G’’/G’’$ and $K’=G’/G’’$ follows because for any variety $\mathfrak{W}$ and any surjective group homomorphism $H\to M$, we have that $\mathfrak{W}(H)$ maps onto $\mathfrak{W}(M)$. Thus, $G’$ maps onto $K’$, and $G’’$ onto $K’’$; using the variety of abelian groups, and of solvable groups of length at most $2$.)