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 $AB = F$?

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    $\begingroup$ Is $AB$ a subgroup or just the set of all pairwise products? In the former case, the answer is no. $\endgroup$ Apr 6 '15 at 9:52
  • $\begingroup$ @AlexDegtyarev I am talking about $AB = \{ab : a \in A, \ b \in B\}$. Do you have an answer in case that $AB$ happens to be a subgroup? $\endgroup$
    – Pablo
    Apr 6 '15 at 10:05
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    $\begingroup$ To disambiguate, I would call $AB$ a `double coset'. $\endgroup$
    – HJRW
    Apr 6 '15 at 10:20
  • $\begingroup$ @Pablo, it turns out that I downvoted your question by accident. Please perform some edit of the question so that I can undo my vote. I am sorry for that. $\endgroup$ Apr 6 '15 at 13:27
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    $\begingroup$ @BenjaminSteinberg: it's not assumed in the question $\endgroup$
    – YCor
    Apr 6 '15 at 17:17

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.

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    $\begingroup$ In fact, if one chooses $H$ so that it is malnormal in $A$, then $B$ will be malnormal in $F$. Thus $F$ is a product of two proper malnormal subgroups. $\endgroup$ Apr 10 '15 at 9:22
  • $\begingroup$ Is there a nontrivial normal subgroup of $F$ disjoint from $B$? $\endgroup$
    – Pablo
    Apr 11 '15 at 14:47
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    $\begingroup$ It is shown above that $N$, the kernel of $\rho$, is disjoint from $B$. $\endgroup$ Apr 11 '15 at 15:02

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