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Let $F$ be a free group and let $V$ be another verbal subgroup of $F$ such that $$ F = [F,F] V. $$ Is it true that $V=F?$ More generally, if $N$ is a normal (or even characteristic) subgroup of $F$ which also supplements $[F,F],$ $$ F=[F,F] N $$ is it true that $N=F?$

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    $\begingroup$ The question has been answered by Keith Kearnes on MathOverflow: if $V$ is a nontrivial verbal subgroup, then $F/V,$ which a relatively free group, has abelian quotients, hence can't be perfect, hence $F=[F,F]V$ is impossible. $\endgroup$ – P.H. Aug 5 '15 at 7:37

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