# Generating infinite index subgroups of a free group

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 $[F : \langle H,a,xay\rangle] = \infty$ ?

• What is the motivation to this question? Though i admit, it's interesting on its own! Dec 31, 2014 at 12:45
• I can't think of a nice argument, so I delete my answer. I do have the feeling such an $a$ should always exist.
– jmc
Dec 31, 2014 at 13:18
• 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 Nielsen-Schreier Theorem and Kurosh's Theorem? (I don't have Lyndon-Schupp now) -- it was about finding paths in that complex which does all the job, and it seems this could suit here Dec 31, 2014 at 13:37
• 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 Dec 31, 2014 at 13:39
• You can always find a one-relator quotient into which H maps injectively and properly and infinite index. Take a to be that relator. Dec 31, 2014 at 14:21