Let $F$ be a free group, and let $H,K \leq F$ be finitely generated subgroups of infinite index in $F$. Is it possible that for the set of products we have $HK = F$ ?



Is a free group a product of f.g subgroups of infinite index?
It's actually easier to prove the stronger statement that there are infinitely many double cosets $H\backslash F/K$. 


First, note that if $F'$ is a subgroup of finite index in $F$, with $H'$ and $K'$ its intersections with $H$ and $K$ respectively, then the natural map $H'\backslash F'/K'\to H\backslash F/K$ is finite-to-one. So we may pass to finite-index subgroups.


EDIT: (I was a little glib in translating from topology to group theory before. Here's a corrected version of the final paragraph.)


Therefore, by Marshall Hall's theorem, we may take $F$ to be the fundamental group of a graph $X$ and $H$ and $K$ to be carried by embedded subgraphs $Y$ and $Z$, say. But now it's easy. Indeed, let $a$ be a based loop not contained in $Y$ and $b$ a based loop not contained in $Z$. Then the double cosets $Ha^mb^nK$ are all distinct.


