The following question has been [asked](https://math.stackexchange.com/questions/1377795/free-groups-normal-supplements-of-the-commutator-subgroup) 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 $N.$ 



Commutator subgroups as normal supplmements
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. 


