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Suppose $H$ is some group with finite isomorphic subgroups $A$ and $B$, and isomorphism $\psi: A\rightarrow B$. Suppose that the HNN-extension $G=\langle H, t\mid a^t=\psi(b)~\forall~a\in A\rangle$ is hyperbolic. Is $H$ hyperbolic?

My thoughts on this are rather sparse. I know that the opposite is true - that if $H$ is hyperbolic and $A$ and $B$ are finite then $G$ is hyperbolic. Clearly $H$ needs to be a finitely presented non-hyperbolic subgroup of a hyperbolic group. I know that such subgroups exist (discovered by Noel Brady), but that their construction is a serious result.

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    $\begingroup$ It is enough to show that $H$ embeds quasi-isometrically into $G$ which, since $A$ and $B$ are finite seems clear. In fact if you include elements of $A$ in the generating set of $G$ then any element of $H$ has a geodesic representative in the generators of $H$. $\endgroup$
    – Derek Holt
    Commented Dec 12, 2016 at 12:24

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As Derek Holt indicates in comments, this is standard. You should do the following easy exercise.

If $G = A*_CB$ or $G=A*_C$ is hyperbolic and $C$ is quasiconvex, then so is $A$ (and $B$).

Since finite subgroups are trivially quasiconvex, and quasiconvex subgroups are themselves hyperbolic, the result follows.

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  • $\begingroup$ Okay, thanks. I'll do the exercise :-). $\endgroup$
    – user1729
    Commented Dec 12, 2016 at 13:00

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