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Last time, we used the following lemma without justification, so let’s prove it now.

**Lemma 30. **Let be a graph of groups with finite and finitely generated. If is finitely generated for every edge , then is finitely generated for every .

This is not completely trivial: it is certainly possible for finitely generated group to have subgroups that are not finitely generated. Indeed, recall the proof that every countable group embeds in the free group on three generators.

**Pf.** Let be a finite generating set for , and for each let be a finite generating set for the edge group . By the Normal Form Theorem, every can be written in the form

where each is a stable letter and each for some . For a fixed , let

,

where for each adjoining the plus or minus is chosen so that . It is clear then that is contained in . To see that is finite, note that since is finite, the first union is finite; and since is finite there can be only finitely many edges adjoining a given vertex, so the second union is finite. Hence it remains to prove that generates .

Let . Because generates , we have

where . Each has a normal form as above, so we get an expression of the form

, or .

By the Normal Form Theorem, the expression on the right can then be simplified. After the simplification process, we have no stable letters left, and every is either contained in or is a product of elements of the incident edge groups, and in both cases lie in .

Remember that Theorem 19 said is LERF, answering our question (b). But in fact, we get more from the proof of Theorem 19.

**Definition.** Recall that is a **retract** if the inclusion has a left inverse . Similarly, we call a **virtual retract** if is a retract of a finite index subgroup of .

For instance, Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a virtual retract.

**Theorem 20.** Every finitely generated subgroup of is a virtual retract.

**Pf.** Consider the setup of the proof of Theorem 19. We start with a subgroup and end up with a finitely sheeted covering space . The graph of spaces is built using the “obvious” bijection between elevations to and elevations to . Thus the identification extends to a topological retraction . Now, we shall build a map . We build it at each vertex space, one at a time. From the proof of Lemma 29, we see that the core of each is a topological retract of the corresponding . Furthermore, we can choose the retraction so that for each long loop of degree that we added is mapped to a null-homotopic loop in . This allows you to piece together the map into a retraction .

Exercise 14 asserted that a (virtual) retract is quasi-isometrically embedded, and so Theorem 20 has the following corollary:

**Corollary.** Every finitely generated subgroup of is quasi-convex.

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 .

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