Is it true that for every finitelty generated subgroup $H$ of infinite index in a free group $F$ on the two letters $\{x,y\}$, there exists a finite index subgroup $K$ of $H$, such that the normal subgroup $N$ of $F$ generated by $K$ is of infinite index in $F$ ?

The normal subgroup $N$ of $F$ generated by $K$ is given by $$N = \langle \bigcup_{g \in F} g^{-1}Kg \rangle.$$


This is true, though the proof uses some heavy machinery! Theorem A.1 of Agol--Groves--Manning's appendix to Agol's proof of the Virtual Haken conjecture states:

Let $G$ be a hyperbolic group, let $H\leq G$ be a quasi-convex virtually special subgroup. For any $g\in G-H$, there is a hyperbolic group $\mathcal{G}$ and a homomorphism $\phi:G\to\mathcal{G}$ such that $\phi(g)\notin\phi(H)$ and $\phi(H)$ is finite.

Furthermore, if you examine the proof carefully, you obtain the following addendum.

If $H$ has infinite index in $G$ then $\mathcal{G}$ can be taken to be infinite.

These statements use Groves--Manning/Osin's combinatorial Dehn filling technology, and Wise's Malnormal Special Quotient Theorem. We don't need to remember all the definitions, except to recall that finitely generated free groups are hyperbolic and virtually special, and that all finitely generated subgroups of free groups are quasiconvex.

Therefore, taking $F=G$, we obtain an infinite, hyperbolic quotient $\phi:F\to\mathcal{G}$ such that $\phi(H)$ is finite. Taking $K=\ker\phi|_H$, we see that $K$ has finite index in $H$ and that $\ker\phi$ contains the normal closure of $K$. In particular, the normal closure of $K$ has infinite index in $F$.

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  • $\begingroup$ Awesome! At the end you mean 'is contained in' instead of 'contains' right? $\endgroup$ – Pablo Mar 6 '17 at 11:31
  • $\begingroup$ @Pablo -- right; corrected. $\endgroup$ – HJRW Mar 6 '17 at 11:35
  • $\begingroup$ And in the statement of the theorem, you mean maybe $\phi(g) \notin \phi(H)$ instead of $\phi(g) \neq \phi(H)$? $\endgroup$ – Pablo Mar 6 '17 at 11:36
  • $\begingroup$ But for this application you don't use some 'smart choice' of $g$ right? any $g$ should work? $\endgroup$ – Pablo Mar 6 '17 at 11:38
  • 1
    $\begingroup$ @Pablo -- yes, the choice of $g$ is irrelevant for this application. Perhaps it's also worth mentioning that if $H$ happens to be malnormal then the proof is much easier -- only combinatorial Dehn filling is required, and we can get away without the MSQT. $\endgroup$ – HJRW Mar 6 '17 at 11:47

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