Let $F$ be a free group, and let $A,B \leq F$ be two subgroups such that $AB$ contains a nontrivial normal subgroup of $F$. Must either $A$ or $B$ contain a nontrivial normal subgroup of $F$?


What if $AB = F$? 



Products of subgroups of a free group
The answer is negative even if $A$ is finitely generated. Here is a simple construction.


Let $F$ be the free group on $\{x,y,z\}$ and let $A=\langle x,y \rangle$. Then there is a natural retraction $\rho:F \to A$ with kernel $N:=\ker \rho=\langle z^F\rangle$. Choose a subgroup $H \leqslant A$ freely generated by an infinite countable set $\{a\_1,a\_2,\dots\}$ (e.g., $a\_i=x^i y x^{-i}$). Now fix some enumeration $N=\{f\_1,f\_2,\dots\}$ and set 
$$B=\langle a\_if\_i \mid i=1,2,\dots\rangle .$$


Clearly $F=A B$, as $F=AN$, so it remains to show that $B$ does not contain non-trivial normal subgroups of $F$ (this is of course true for $A$ as it is even malnormal in $F$). To this end, it's enough to show that $zBz^{-1}\cap B=\{1\}$. Indeed, suppose that $zbz^{-1}=b'$ for some $b,b' \in B \setminus\{1\}$. 


Note that since the elements $a\_i$ freely generate $H$ and $\rho(a\_if\_i)=a\_i$, the elements $a\_if\_i$ freely generate $B$, and the restriction of $\rho$ to $B$ induces an isomorphism of $B$ with $H$. In particular $\rho$ is injective on $B$. Hence $zbz^{-1}=b'$ implies $\rho(b)=\rho(zbz^{-1})=\rho(b')$, yielding $b=b'$, i.e., $z$ must centralize $b$ in $F$. But centralizers of non-trivial elements in $F$ are cyclic, so $z^n=b^m\in N\cap B$ for some $m,n\in \mathbb Z\setminus\{0\}$, contradicting to the fact that $\rho$ is injective on $B$. Therefore $zBz^{-1}\cap B=\{1\}$, as claimed.


