# G a group, H a subgroup of finite index. Proof that G is hyperbolic $\Leftrightarrow$ H is hyperbolic?

I found the the following claim here:

If $G$ is a group and $G_0 \subset G$ is a subgroup of finite index, then $G$ is hyperbolic if and only if $G_0$ is hyperbolic.

Why is this true? Can anyone provide a proof or reference to a proof? Also, is this a well-known result?

• Try googling quasi-isometry/quasi-isometric. The Cayley graphs of $G$ and $G_0$ with any finite generating sets are quasi-isometric, which means roughly that there is a map between them which is an isometry up to multiplicative and additive constants. You can prove that if two metric spaces are quasi-isometric, then one has uniformly slim/thin triangles if and only if the other one does. This is all straighforward. – Derek Holt Feb 14 '17 at 20:01

If $\Gamma$ is a Cayley graph of $G$ then $G$ acts properly discontinuously and cocompactly on $\Gamma$, hence $\Gamma$ is quasi-isometric to $G$ (by the Milnor-Svarc lemma).
Now consider the restricted action of the subgroup $G_0$ on $\Gamma$. Proper discontinuity is inherited under restriction of the action of $G$ to any subgroup (pretty obvious), hence the action of $G_0$ on $\Gamma$ is properly discontinuous. Cocompactness is inherited under restriction of the action of $G$ to any finite index subgroup, hence the action of $G_0$ on $\Gamma$ is cocompact (proof: if $K \subset \Gamma$ is a finite subgraph whose $G$-translates cover $\Gamma$, and if $g_1,..,g_N$ are left coset representatives of $G_0$ in $G$, then $(g_1 \cdot K) \cup \cdots \cup (g_N \cdot K)$ is a finite subgraph whose $G_0$-translates cover $\Gamma$).
Again (by the Milnor-Svarc lemma) $\Gamma$ is quasi-isometric to $G_0$.
Since quasi-isometry is an equivalence relation, it follows that $G$ is quasi-isometric to $G_0$. Since hyperbolicity is a quasi-isometry invariant, it follows that $G$ is hyperbolic if and only if $G_0$ is hyperbolic.