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

$\begingroup$ What is the motivation to this question? Though i admit, it's interesting on its own! $\endgroup$– VictorDec 31 '14 at 12:45

$\begingroup$ I can't think of a nice argument, so I delete my answer. I do have the feeling such an $a$ should always exist. $\endgroup$– jmcDec 31 '14 at 13:18

$\begingroup$ Perhaps Johan is right. My brain is damaged by semigroup theory, so may be the following does not make any sense: what if using those complexes related to any finite presentation and from which one gets a short proof of NielsenSchreier Theorem and Kurosh's Theorem? (I don't have LyndonSchupp now)  it was about finding paths in that complex which does all the job, and it seems this could suit here $\endgroup$– VictorDec 31 '14 at 13:37

$\begingroup$ Another thing, this time even more vague, what if viewing this situation as inside the free inverse monoid  we can work with elements there as the corresponding Munn trees  do some combinatorial things to glue trees as we need  and then project back onto the free group $\endgroup$– VictorDec 31 '14 at 13:39

1$\begingroup$ You can always find a onerelator quotient into which H maps injectively and properly and infinite index. Take a to be that relator. $\endgroup$– Benjamin SteinbergDec 31 '14 at 14:21
Taking the finite subgraph of the Schreier graph contains the Stallings graph of H and x,y^{1} read from the base point. Going to a conjugate we may assume some letter is not read at the base point. Choose a word a not readable on this finite graph and beginning with the letter not readable at the base and ending with the inverse of that letter. Then we can sow a at the base point and have a Stallings graph. Now sow a from the end of x to the end of y^{1} and do Stallings folding. Because a cannot be read on the original graph the folds should not result in a covering space. I'll try later to write details later.

$\begingroup$ For which finitely generated groups that are not free could your argument (maybe after a slight modification) work? Which features of the free group are really necessary here? $\endgroup$– PabloJan 6 '15 at 11:31

1$\begingroup$ I am not sure how easily this would go beyond virtually free. Maybe for nice enough subgroups of free products. You need to have good control over sewing in loops. Maybe mccammond wise perimeter conditions help. $\endgroup$ Jan 6 '15 at 19:54