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The following question has been asked about a week ago on MathUnderflow (no answers).

Let $F$ be a free group and let $N$ be a normal subgroup of $F$ such that \begin{equation*} \tag{*} F = [F,F] N. \end{equation*} Is it true that $N=F?$ If no, what about the case when $N$ is a verbal subgroup of $F?$

It is easy to see that $(*)$ implies that $$ F = F^{(k)} N $$ where $F^{(k)}$ with $k \ge 1$ is the $k$-th commutant of $F.$ This implies that every solvable $n$-generator group, where $n=\mathrm{rank}(F),$ is a homomorphic image of $N.$

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    $\begingroup$ For everyone's benefit, add a link to the original post when crossposting. $\endgroup$
    – Boris Bukh
    Commented Aug 4, 2015 at 17:42

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The equation $F = [F,F]N$ is equivalent to $G = [G,G]$ for $G:=F/N$. Thus, if $N$ is the kernel of a homomorphism of $F$ onto a perfect group, then the equation will hold.

On the other hand, if $N$ is a proper verbal subgroup of a free group $F$, then $F/N$ is never perfect, so $F = [F,F]N$ cannot hold in this case. The reason that $F/N$ is not perfect is that it is a nontrivial relatively free group, so it has nontrivial abelian quotients.

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