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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
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 ;
(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,
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 .