5
$\begingroup$

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.

$\endgroup$
7
  • 2
    $\begingroup$ 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. $\endgroup$
    – YCor
    Nov 22, 2015 at 19:25
  • 2
    $\begingroup$ @YCor I fail to see how these facts imply that the profinite completion of that infinitely generated subgroup is finitely generated... $\endgroup$
    – Pablo
    Nov 22, 2015 at 19:48
  • 2
    $\begingroup$ 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.? $\endgroup$
    – YCor
    Nov 23, 2015 at 1:14
  • $\begingroup$ @YCor You are right! Is it reasonable to ask a new question with this formulation? $\endgroup$
    – Pablo
    Nov 23, 2015 at 6:00
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Nov 23, 2015 at 8:55

2 Answers 2

8
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ @IanAgol if on the other hand, I assume that $G$ is residually finite. Is my conjecture correct? Must $H$ be finitely generated? $\endgroup$
    – Pablo
    Nov 22, 2015 at 16:41
5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I believe the results of Hagen and Wise apply to show that the Sapir group is residually finite: arxiv.org/abs/1311.2084 $\endgroup$
    – Ian Agol
    Nov 25, 2015 at 14:38
  • $\begingroup$ @IanAgol it is possible. See my edit - for a slightly different group it is possible to prove linearity over $\mathbb{Z}$. $\endgroup$
    – Pablo
    Nov 25, 2015 at 17:12
  • $\begingroup$ 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 . $\endgroup$
    – HJRW
    Nov 26, 2015 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.