# Can finitely generated subgroups of limit groups be detected in free group quotients?

In Henry Wilton's excellent paper "Hall's Theorem for limit groups" (Geom. Funct. Anal. 18, pp. 271–303, 2008 ) he proves the following result (see his paper or here and here for the relevant definitions).

Wilton's Theorem: Limit groups are subgroup separable.

This theorem has me wondering whether, in some cases, it is possible to see this result by passing through free group quotients. To make this precise, say a group $G$ is freely subgroup separable if any finitely generated and infinite-index (see Ian Agol's answer, below) $H \leq G$ is the intersection of subgroups in $\{ H \Delta : G/\Delta \text{ is a free group} \}$.

Question 1: Which limit groups are freely subgroup separable? In particular, are surface groups freely subgroup separable?

If the answer to the above question is "all of them", then this, along with the fact that free groups are subgroup separable, would imply Wilton's Theorem, so I suspect the answer is more complicated than that. In light of this, here is a refinement of Question 1.

Question 2: Given a limit group, $G$, for which finitely generated subgroups $H \leq G$, do we have $H = \cap \{ H \Delta : G/\Delta \text{ is a free group} \}$?

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You could give the subgroups as in Question 2 a name, such as "pro-freely closed". –  YCor Apr 22 '14 at 10:29
For the first part of Question 1, in view of Agol's answer, any group having an infinite index non-abelian surface subgroup is not freely subgroup separable. I think that most limit groups satisfy this, but experts would know better. –  Ashot Minasyan Apr 22 '14 at 10:56

I suspect the answer is no for surface groups, although I don't have an explicit example. Suppose one has a finite simple group $K$, and a surjection $\phi: G=\pi_1(\Sigma_g) \twoheadrightarrow K$, so $H=ker(\phi)$, $\Sigma_g$ a closed surface of genus $g$. Then for any surjection $\psi:G\twoheadrightarrow F$, where $F$ is a free group, one has $\psi(H) \lhd F$, and $K\twoheadrightarrow F/\psi(H)$. Thus, this image is either trivial or $K$, so $H$ is separable iff there exists $\psi$ with $F/\psi(H)=K$.

One could try to find an obstruction to this. For example, suppose we have $H^2(\phi): H^2(K;\mathbb{F}_p) \twoheadrightarrow H^2(G;\mathbb{F}_p)$ a surjection, for some prime $p$. Then this map factors through $H^2(K)=H^2(F/\psi(H)) \to H^2(F) \to H^2(G)$, which is impossible since $H^2(F)=0$. So one could search for a map $H^2(K)\twoheadrightarrow H^2(G)$ with some finite coefficients. I would suggest looking at a paper of Dunfield and Thurston to find such a rep. Of course, if this ansatz works, then you would probably have to assume $H$ should be infinite index.

Addendum: Indeed, I had a look at Dunfield-Thurston's paper, and they prove that there are many representations $\pi_1(\Sigma_2)\to A_5$ which do not extend over a genus 2 handlebody (Example 6.3). If one considers a free quotient of $\pi_1(\Sigma_2)$ of rank $2$, then this map comes from a map to a handlebody (this may be seen by considering the preimages of midpoints of edges of a rank 2 graph). Thus, these maps to $A_5$ do not factor through a free group, and therefore the kernel cannot be separated in a free quotient.

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Dear Ian, thank you very much for this--I will update my question with this soon. –  Khalid Bou-Rabee Apr 18 '14 at 14:43

I believe that Ben McReynolds thought about this for infinite-index subgroups of surface groups---you might ask him if he got anywhere. I had some reasoning similar to Ian's to suggest that it shouldn't be true for finite-index subgroups. I'll try to reconstruct it here later if I can remember it.

I'm certain that it's not true in general that limit groups are freely subgroup separable (even on their infinite-index subgroups), though I don't have an explicit counterexample. Here's my reasoning.

Consider a double $\Gamma =F*_{\mathbb{Z}}F$ for $F$ a free group, and suppose the given splitting is the JSJ. Then $\Gamma$ is a limit group.

Conjecture: The first layer of the Makanin--Razborov diagram for $\Gamma$ over $F$ consists of just two homomorphisms: the 'obvious' retraction $\rho:\Gamma\to F$ and the abelianization map $\eta:\Gamma\to H_1(\Gamma)$. (This just means that every homomorphism $\Gamma\to F$ factors through either $\rho\circ\delta$ or $\eta\circ\delta$, where $\delta$ is some Dehn twist.)

Anyway, if this conjecture is true for even one example---and it surely is, though I don't think a proof is known in any case---then it follows that $\Gamma$ is not `freely subgroup separable'. Indeed, in this case one copy of $F$ in $\Gamma$ can't be separated from any commutator.

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