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?

Evidence for:

**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.

Evidence against:

**Theorem (Agol-Groves-Manning).** If every hyperbolic group is residually finite, then every quasi-convex subgroup of every hyperbolic group is separable.

## 1 comment

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4 March 2009 at 6.07 am

Henry WiltonThanks, Sungmo! I made a few cosmetic changes.