# Hyperbolic HNN-extension with finite associated subgroups implies hyperbolic base group?

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.

• 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 $H$. – Derek Holt Dec 12 '16 at 12:24

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