# Subgroups of hyperbolic groups

Let $G$ be a finitely generated hyperbolic group, and let $H \leq G$ be a subgroup whose profinite completion is finitely generated. Must $H$ be finitely generated?

In view of Ian Agol's answer, I am ready to assume that $G$ is residually finite.

• In this paper (Trans AMS, 1988: ams.org/journals/tran/1988-308-02/S0002-9947-1988-0951631-0), D. Long shows (Section 3) that any cocompact lattice in $SL_2(\mathbf{C})$ has a finite index subgroup (thus finitely generated and residually finite) with a profinitely dense infinitely generated subgroup. So the answer is no even if $G$ is assumed RF. – YCor Nov 22 '15 at 19:25
• @YCor I fail to see how these facts imply that the profinite completion of that infinitely generated subgroup is finitely generated... – Pablo Nov 22 '15 at 19:48
• Ah, OK, I interpreted "profinite completion of a subgroup" as "profinite closure" of this subgroup, so I indeed don't answer your question. Actually $G$ is not really part of the data, and the question can be rephrased as: let $H$ be a group that is isomorphic to a subgroup of a (residually finite) hyperbolic group. Assume that $H$ has a f.g. profinite completion, does it imply that $H$ is f.g.? – YCor Nov 23 '15 at 1:14
• @YCor You are right! Is it reasonable to ask a new question with this formulation? – Pablo Nov 23 '15 at 6:00
• Of course not a new question (it would be a duplicate). You could edit this one (since it's the same question), but it's not necessary. – YCor Nov 23 '15 at 8:55

If there's a non-residually finite hyperbolic group, then the answer is no. By a result of Kapovich-Wise, in this case there is a hyperbolic group $H$ whose profinite completion is trivial. Then consider the group $H\ast \mathbb{Z}$. There is a subgroup $\underset{\mathbb{Z}}{\ast}\ H$ which is the kernel of the map $H\ast \mathbb{Z}\to \mathbb{Z}$, which is infinitely generated but has trivial profinite completion (which is finitely generated).

• @IanAgol if on the other hand, I assume that $G$ is residually finite. Is my conjecture correct? Must $H$ be finitely generated? – Pablo Nov 22 '15 at 16:41

Let $A$ be a finitely generated group, and let $\beta \colon A \to A$ be an injective homomorphism which is not surjective. Freely construct a group $G$ generated by $A$ and some formal element $t$ such that the equality $tat^{-1} = \beta(a)$ holds in $G$ for each $a \in A$. $G$ is called the strict ascending HNN extension of $(A, \beta)$. Set $$H := \bigcup_{n=0}^{\infty} t^{-n}At^n$$ where the union is taken in $G$. $H$ is a strictly ascending union of finitely generated subgroups of $G$ which are all isomorphic to $A$. It follows that $H$ is a subgroup of $G$ which is not finitely generated (if it were, the union could not be strictly ascending). On the other hand, every finite image of $H$ is clearly a finite image of one of the groups in the union (which is isomorphic to $A$) and can thus be generated by $d(A)$ elements. It follows that the profinite completion of $H$ is finitely generated.

To answer my question it suffices to choose $A$ and $\alpha$ in a way that will make $G$ hyperbolic. Take $A$ to be the free group on $x,y$ and define $\beta$ by $\beta(x) = xy$ and $\beta(y) = yx$. It is easy to see that $\beta$ is injective but not surjective. In this case, $G$ is the Sapir group, and its hyperbolicity is established in Theorem 4.1 of http://arxiv.org/pdf/1302.5370.pdf

If we want an answer to the extended question, i.e. with $G$ residually finite, then we can take $A$ to be the free group on $x,y$ and define $\beta$ by $\beta(x) = xy^{-1}x^2y$ and $\beta(y) = yx^{-1}y^2x$. Again, it is easy to see that $\beta$ is injective but not surjective. By Theorem 4.2 of http://arxiv.org/pdf/1302.5370.pdf the resulting $G$ is hyperbolic and linear over $\mathbb{Z}$, and thus, residually finite.

• I believe the results of Hagen and Wise apply to show that the Sapir group is residually finite: arxiv.org/abs/1311.2084 – Ian Agol Nov 25 '15 at 14:38
• @IanAgol it is possible. See my edit - for a slightly different group it is possible to prove linearity over $\mathbb{Z}$. – Pablo Nov 25 '15 at 17:12
• I don't think Hagen--Wise's results apply directly to the Sapir group (though perhaps they could be extended), but this wonderful result of Borisov and Sapir certainly does apply: arxiv.org/abs/math/0309121 . – HJRW Nov 26 '15 at 10:33