Finally, we are in a position to prove that a hyperbolic group has no subgroup isomorphic to $\mathbb Z^2$.

Theorem 11. Let $\gamma \in \Gamma$ with $o(\gamma)=\infty$. Then $|C(\gamma):\langle \gamma \rangle|<\infty$.

Proof. By Lemma 10, we can assume that $\gamma$ is not conjugate to any element of length $\leq 4\delta$ by replacing $\gamma$ with a power of itself. Suppose $g\in C(\gamma)$. We need to bound $d(g, \gamma)$. Replacing $g$ with $\gamma^{-r}g$ for some $r$, we may assume that $d(1,g)=d(g,\langle \gamma \rangle)$. We will be done if we can bound $l(g)$.

Suppose $l(g)>2(l(\gamma)+2\delta)$. By dividing into triangles, we see that any geodesic rectangle is $2\delta$-slim, in the same way that triangles are $\delta$-slim.

Because the rectangle with vertices $1, \gamma, g\gamma, g$ is $2\delta$-slim, there exists $g_t, g_{t'} \in [1,g]$ such that $d(g_t, \gamma g_{t'}) \leq 2\delta$.

If $t, then $d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma)$, a contradiction. Similarly $t'. So $|t-t'|<2\delta$. Therefore, $d(g_t, \gamma g_t)<4\delta$.

But $l(g_t^{-1}\gamma g_t)=d(g_t, \gamma g_t)<4\delta$. This is a contradiction since we assumed that $\gamma$ is not conjugate to anything so short. Therefore $l(g)\leq2(l(\gamma)+2\delta)$. Thus $|C(\gamma):\langle \gamma \rangle|<\infty$.

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 $\Gamma$ be a torsion-free hyperbolic group. Whenever $\gamma \in \Gamma$ is not a proper power, then $\langle \gamma \rangle$ is malnormal.

Definition. A subgroup $H$ of a group $G$ is malnormal if for all $g\in G$ $gHg^{-1} \cap H \neq 1$ , then $g\in H$.

Remark. By Theorem 11, if $\Gamma$ is hyperbolic and torsion-free, centralizers are cyclic.

Proof of Lemma. Suppose $g \langle \gamma \rangle g^{-1}\cap \langle \gamma \rangle \neq 1$.

Therefore for some $p, q \neq 0$, $g\gamma^{p}g^{-1}=\gamma^q$.

By Lemma 10, $|p|=|q|$. Therefore $g^2\gamma^pg^{-2}=\gamma^p$. Thus $g^2\in C(\gamma^p)= \langle \gamma \rangle$. Therefore $g \in \langle \gamma \rangle$.

Exercise 17. Prove that if $x, y, z \in \Gamma$ where $\Gamma$ is hyperbolic and torsion-free and $xy=yx$ and $yz=zy$ and $y\neq 1$, then $xz=zx$. That is, $\Gamma$ 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.