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Let $\mathcal{S}$ be the class of finitely presented torsion-free groups which occur as the subgroup of some hyperbolic group (so for every $G\in \mathcal{S}$ there exists a hyperbolic group $H$ such that $G\hookrightarrow H$). Define $\mathcal{S}_{\mathrm{tf}}$ to be the class of finitely presented torsion-free groups which occur as the subgroup of some torsion-free hyperbolic group. Clearly $\mathcal{S}_{\mathrm{tf}}\subseteq \mathcal{S}$.

My question is: Are $\mathcal{S}_{\mathrm{tf}}$ and $\mathcal{S}$ equal?

That is, when studying the class of finitely presented torsion-free subgroups of hyperbolic groups $\mathcal{S}$, is anything lost by restricting to torsion-free hyperbolic groups?

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    $\begingroup$ Nice question! it also makes sense for finitely generated subgroups, and even arbitrary subgroups. I'm not even sure to be able to answer in the probably easier case of a torsion-free subgroup of a virtually torsion-free hyperbolic groups. At least if this case is doable, then no counterexample to your example is known, since it's unknown whether every hyperbolic f.g. group is virtually torsion-free. $\endgroup$
    – YCor
    Commented Nov 8, 2019 at 13:57
  • $\begingroup$ @YCor Thanks. Yes, the other questions make sense and I would be interested to know their answer also. I only asked one question as I didn't want to make the question too convoluted (and this specific question cropped up in my work). $\endgroup$
    – ADL
    Commented Nov 8, 2019 at 14:05
  • $\begingroup$ I insist but it would be nice to settle the virtually torsion-free case. Indeed, there exist a hyperbolic (indeed virtually free) group $G$ with a torsion-free (normal) subgroup $N$ not contained in any torsion-free finite index subgroup of $G$. Namely, choose a $k$-generated infinite simple group with an element of order 2 (there are many); surject $G=F_k\ast C_2$ onto it with $C_2$ mapped injectively, and let $N$ be the kernel. Then $N$ is torsion-free and the only finite index subgroup of $G$ containing $N$ is $G$ itself, which is not torsion-free. $\endgroup$
    – YCor
    Commented Nov 8, 2019 at 16:14
  • $\begingroup$ The answer is yes for small cancelation groups (Gersten 96): every f.p. subgroup of a small cancelation group is itself hyperbolic. $\endgroup$
    – user6976
    Commented Nov 8, 2019 at 19:04

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