Finally, we are in a position to prove that a hyperbolic group has no subgroup isomorphic to .
Theorem 11. Let with . Then .
Proof. By Lemma 10, we can assume that is not conjugate to any element of length by replacing with a power of itself. Suppose . We need to bound .
Replacing with for some , we may assume that . We will be done if we can bound .
Suppose . By dividing into triangles, we see that any geodesic rectangle is -slim, in the same way that triangles are -slim.
Because the rectangle with vertices is -slim, there exists such that .
If , then , a contradiction. Similarly . So . Therefore, .
But . This is a contradiction since we assumed that is not conjugate to anything so short. Therefore . Thus .
An element of a group is torsion if its order is finite.
A group is torsion if every element is torsion.
A group is torsion-free if no nontrivial elements are torsion.
Corollary. Every non-trivial abelian subgroup of a hyperbolic group is virtually cyclic.
Lemma 11. Let be a torsion-free hyperbolic group. Whenever is not a proper power, then is malnormal.
Definition. A subgroup of a group is malnormal if for all , , then .
Remark. By Theorem 11, if is hyperbolic and torsion-free, centralizers are cyclic.
Proof of Lemma. Suppose .
Therefore for some , .
By Lemma 10, . Therefore . Thus . Therefore .
Exercise 17. Prove that if where is hyperbolic and torsion-free and and and , then . That is, is commutative transitive.
We now turn briefly to a fundamental open question about hyperbolic groups. This question is a theme of the course.
Question. Is every word-hyperbolic group residually finite?
The fundamental groups of hyperbolic manifolds are linear, so residually finite by Selberg’s Lemma.
What about for negative curved manifolds?
Theorem (Sela). Every torsion-free hyperbolic group is Hopfian.
Theorem (I. Kapovich-Wise). If every nontrivial hyperbolic group has a proper finite-index subgroup, then every hyperbolic group is residually finite.
Theorem (Agol-Groves-Manning). If every hyperbolic group is residually finite, then every quasi-convex subgroup of every hyperbolic group is separable.