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Last time: **Theorem 21** (Groves–Manning–Osin): If is hyperbolic rel then there exists a finite subset such that if then

(a) is injective;

(b) is hyperbolic rel .

**Theorem 22** (Gromov, Olshanshkii, Delzant): If is hyperbolic relative to the infinite cyclic then there is a such that for all there exists a hyperbolic such that for each .

The proof is an easy application of Groves–Manning–Osin.

**Definition**: If (infinite cyclic) is malnormal then we say are **independent**. A group G is **omnipotent** if for every independent there exists a such that for all there exists a homomorphism $\phi$ from to a finite group such that for all .

Omnipotence strengthens residual finiteness for torsionfree groups.

**Exercise 29**: If every hyperbolic group is residually finite then every hyperbolic group is omnipotent.

We’ll finish off by talking about a similar theorem of Agol–Groves–Manning. I’m going to seem a little cavalier about torsion. This is OK. In fact, if every hyperbolic group is residually finite then every hyperbolic group is virtually torsionfree.

**Theorem 22** (Agol–Groves–Manning): If every hyperbolic group is residually finite then every quasi-convex subgroup of any hyperbolic group is separable.

Let . The idea is to Dehn fill to get a new hyperbolic group in which the image is finite and . If we could do this, we would be done by residual finiteness. This works if is malnormal. But it probably isn’t. Fortunately, we can quantify how far is from being malnormal:

**Definition**: The **height** of is the maximal such that there are distinct cosets such that the intersection

is infinite.

H is height iff is finite. In a torsionfree group, is height iff is malnormal.

**Theorem 23** (Gitik, Mitra, Rips, Sageev): A quasiconvex subgroup of a hyperbolic group has finite height.

Agol, Groves and Manning are able to prove:

**Theorem 24**: Let be a (torsionfree) residually finite hyperbolic group, and a quasiconvex subgroup of height . Let . Then is an epimorphism to a hyperbolic group such that

(i) is quasiconvex in ;

(ii) ;

(iii) has height .

The idea of the proof of Theorem 24 is to Dehn fill a finite index subgroup of a maximal infinite intersection of conjugates of . Theorem 22 is an easy consequence.

Our goal is to understand the abelian subgroups of a hyperbolic group; e.g., can a hyperbolic group contain as a subgroup? We already know that it cannot be a quasiconvex subgroup, but it may be possible for to be “twisted” in some manner.

**Theorem 10.** The intersection of two quasiconvex subgroups is quasiconvex.

**Pf.** As usual, we work in a -hyperbolic . Let be quasiconvex subgroups each with the same corresponding constant . Let , and let . Our goal is therefore to show that is in a bounded neighborhood of . Let be the (or more precisely, a particular) closest element of to , and suppose that . Let be such that and let be such that ; such elements exist since and are quasiconvex. Let be such that . We sketch the situation below.

Consider the geodesic triangle with vertices , , and . Because this triangle is -slim, for each there exists some such that .

Likewise, for each there exists such that . Next, let and . Then .

Suppose . Then by the Pigeonhole Principle there are integers and with such that and . Now, we can use this information to find a closer element of .

Consider . By the triangle inequality,

.

All that remains is to prove that . But since ,

.

Playing this same game with , we get , and hence we have found our contradiction.

For a group , recall that is the **center** of .

**Corollary.** For any (where is -hyperbolic) of infinite order, the subgroup generated by is quasiconvex. Equivalently, the map sending is a quasigeodesic (which is a sensible statement to make given that is quasi-isometric to ).

**Pf.** By Theorem 9, we deduce that is quasiconvex, and so is finitely generated. Let be a finite generating set for . Notice

where is the centralizer in of , so is quasiconvex by Theorem 10 and thus is a finitely generated abelian group containing . By Exercise 16 (below), we deduce that is quasi-isometrically embedded in , and hence is quasiconvex in .

**Exercise 16.** Cyclic subgroups of finitely generated abelian groups are quasi-isometrically embedded.

**Lemma 10.** Suppose the order of is infinite. If is conjugate to , then .

**Pf.** Suppose . An easy induction on shows . Therefore, applying the triangle inequality,

.

But is a -quasi-isometric embedding, so

.

This is impossible unless (eventually the exponential growth dominates). Similarly, reversing the roles of and in the above argument implies that , so .

**Theorem 8: **Let be a -hyperbolic group with respect to . If are conjugate then there exists such that

where depends only on .

**Proof:** We work in . Let be such that . Let be such that . We want to find a bound on .

Let . By Lemma 9,

Also

So . Thus . Suppose that . By the Pigeonhole Principle there exist integers such that . It follows that one can find a shorter conjugating element by cutting out the section of between and .

Recall, for , is the *centralizer* of .

**Theorem 9:** If is -hyperbolic with respect to and , then is quasi-convex in .

**Proof: **Again we work in . Let , . We need to prove that is in a bounded neighborhood .

Just as in the proof of Theorem 8,

Well, and are conjugate. By Theorem 8 there exists such that

But so that and .

**Exercise 15: **Prove that

is not hyperbolic for any Anosov .

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