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 .