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).

henrylec151

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<t'-2\delta, then d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma), a contradiction. Similarly t'<t-2\delta. 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.

About these ads