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?

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    $\begingroup$ Take a look at Bridson-Haefliger, III.$\Gamma$.3.20. $\endgroup$
    – Steve D
    Oct 7, 2019 at 21:33

1 Answer 1


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.

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    $\begingroup$ Gromov proved everything in the 1987 book. He proved results for an arbitrary action by isometries on a geodesic hyperbolic space, one has the 5-chotomy (a) bd orbits (b) horocyclic: no loxodromic, unbd orbits, (c) axial: there's a loxo and all have the same endpoints (d) focal: not axial, exists loxo and they have common endpoint (e) general type: there are 2 loxo with 4 distinct endpoints. He proved that (e) implies a discrete nonabelian free subgroup. Also (b) excludes cobounded action. Eventually he proved that (d) is impossible for a discrete hyperbolic group acting on itself. $\endgroup$
    – YCor
    Oct 8, 2019 at 10:04
  • $\begingroup$ The Gromov reference is Remark 3.1.A(b) of Gromov, M. "Hyperbolic groups", Essays in group theory, pp. 75–263, Math. Sci. Res. Inst. Publ., 8, Springer, New York, 1987 $\endgroup$
    – HJRW
    Oct 16, 2019 at 11:08
  • $\begingroup$ Gromov did not prove PPS. $\endgroup$
    – user6976
    Oct 16, 2019 at 11:45

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