I have read on wikipedia that a Gromov hyperbolic group which is solvable is elementary (i.e. virtually cyclic). Where can I find a proof of this fact?


There is a proof of a similar fact in Bridson-Haefliger that if a solvable group $\Gamma$ acts properly and cocompactly on a CAT(0) space, then it is virtually abelian. Is the proof similar to this one?



Gromov hyperbolic groups which are solvable are elementary
I do not know an exact reference. I think it is a folklore. Here is a proof using basic properties of hyperbolic groups which can be found in any book on hyperbolic groups. Let $G$ be solvable and hyperbolic. 
Then it has an Abelian normal subgroup $H$, the last nontrivial member of the derived series. If $H$ is finite, then $G/H$ is quasi-isometric to $G$, so it is hyperbolic and we can proceed by induction on the solvability class. So $H$ is infinite. It cannot contain an infinite locally finite subgroup because all finite subgroups of a hyperbolic group have uniformly bounded orders. $H$ cannot contain a subgroup isomorphic to $\mathbb Z^2$ because $G$ is hyperbolic. Hence $H$ is virtually cyclic and contains a characteristic infinite cyclic subgroup $H\_0$. Then $H\_0$ is normal in $G$. So $G$ has a homomorphism into $Out(\mathbb Z)$ which is a group of order 2. Hence $G$ has a subgroup $N$ of index at most 2 which centralizes $H\_0$. A centralizer of an infinite cyclic subgroup in a hyperbolic group is virtually cyclic. Hence $G$ is virtually cyclic. 


PS. I was talking about Gromov hyperbolic, hence finitely generated discrete groups. The (solvable) group of $2 \times 2 $ upper triangular matrices with determinant 1, with Riemannian metric is quasi-isometric to the hyperbolic plane, hence is hyperbolic itself.


PPS. Another proof is based on Druţu, Cornelia; Sapir, Mark Tree-graded spaces and asymptotic cones of groups. With an appendix by Denis Osin and Mark Sapir. Topology 44 (2005), no. 5. If $G$ is solvable, it satisfies a non-trivial law. Hence if $G$ is finitely generated but not virtually cyclic, its asymptotic cones do not contain cut points. But if $G$ is hyperbolic then all asymptotic cones are trees where every point is a cut-point. Hence $G$ is virtually cyclic.


This proof is of course an overkill, but it works for every group satisfying a non-trivial law, not just solvable groups.
Also "hyperbolic " can be replaced by "relatively hyperbolic", "acylindrically hyperbolic" and even by "lacunary hyperbolic".


PPPS Of course the easiest proof is using the fact that every nonelementary hyperbolic group contains a noncyclic free subgroup (a solvable group cannot contain such a subgroup). I do not remember who proved it first. Gromov, probably. Maybe Olshanskiy or Delzant.


