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Nov 8, 2019 at 19:04 comment added user6976 The answer is yes for small cancelation groups (Gersten 96): every f.p. subgroup of a small cancelation group is itself hyperbolic.
Nov 8, 2019 at 16:14 comment added YCor 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.
Nov 8, 2019 at 14:05 comment added ADL @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).
Nov 8, 2019 at 13:57 comment added YCor 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.
Nov 8, 2019 at 13:54 history edited YCor CC BY-SA 4.0
romanized the abbreviation "tf"
Nov 8, 2019 at 13:49 history asked ADL CC BY-SA 4.0