Conjugates and infinite index subgroups of free groups

Here I am asking for an analogue of 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 $[F : \langle H, yxy^{-1}\rangle] =\infty$ ?

• This can be answered the same way as your other one. Choose y to not be readable at the base and not cancel too much of x. Jan 7, 2015 at 13:22

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.
• We may assume that x has at least 2 letters. Otherwise replace it by $cxc^{-1}$ for some other letter $c$. But if x has more than one letter then the loop I sowed in labeled by x has missing letters at the end point of the first edge. Jan 7, 2015 at 15:47