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 $[F : \langle H,a,xay\rangle] = \infty$ ?
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$\begingroup$ What is the motivation to this question? Though i admit, it's interesting on its own! $\endgroup$– VictorDec 31, 2014 at 12:45
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$\begingroup$ 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, 2014 at 13:18
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$\begingroup$ 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 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 $\endgroup$– VictorDec 31, 2014 at 13:37
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$\begingroup$ 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, 2014 at 13:39
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1$\begingroup$ You can always find a one-relator quotient into which H maps injectively and properly and infinite index. Take a to be that relator. $\endgroup$– Benjamin SteinbergDec 31, 2014 at 14:21
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1 Answer
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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.
answered Dec 31, 2014 at 17:02
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$\begingroup$ 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, 2015 at 11:31
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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, 2015 at 19:54