Here I am asking for an analogue of [Generating infinite index subgroups of a free group](https://mathoverflow.net/questions/191855/generating-infinite-index-subgroups-of-a-free-group)


Let $F$ be a nonabelian finitely generated free group, let $H \leq F$ be a finitely generated subgroup of infinite index, and let $x \in F$. Must there be some $y \in F$ such that $[F : \langle H, yxy^{-1}\rangle] =\infty$ ?



Conjugates and infinite index subgroups of free groups
The proof is like before. Let $\Gamma$ be the Stallings graph for $H$. Choose a word $w$ labeling a path from the base point to a vertex q where some letter or its inverse, call it a, cannot be read. If $axa^{-1}$ is reduced, then by sowing an edge a at q followed by x as a loop, we obtain a Stallings graph of an infinite index subgroup containing H and $yxy^{-1}$ where $y=wa$. Else choose b different than a or its inverse and sow at q a pair of edges labeled ab followed by $x$ as a loop. This results in a Stallings graph of an infinite index subgroup containing H and $yxy^{-1}$ with y=wab.


