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.



Hyperbolic HNN-extension with finite associated subgroups implies hyperbolic base group?
As Derek Holt indicates in comments, this is standard. You should do the following easy exercise.



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> If $G = A\*\_CB$ or $G=A\*\_C$ is hyperbolic and $C$ is quasiconvex, then so is $A$ (and $B$).
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Since finite subgroups are trivially quasiconvex, and quasiconvex subgroups are themselves hyperbolic, the result follows.


