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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.

**Lemma 29: ***Suppose is a finite set of infinite degree elvations and is compact. Then for all sufficiently large , there exists an intermediate covering such that*

*(a)* * embeds in *

*(b) every * *descends to an elevation of degree *

*(c) the are pairwise distinct *

**Proof: **We claim that the images of * *never share an infinite ray (a ray is an isometric embedding of *). *Neither do two ends of the same elevation* **. * Let’s claim by passing to the universal cover of* **, *a tree *.*

For each , lift * *to a map . If and share an infinite ray then there exists such that and overlay in an infinite ray. The point is that correspond to cosets and . But this implies that

This implies that . So . A similar argument implies that the two ends of do not overlap in an infinite ray. This proves the claim.

Let be the core of . Enlarging if necessary, we can assume that

(i) ;

(ii) is a connected subgraph;

(iii) for each , for some , ;

(iv) for each , .

For each identifying with so that is identified with and is identified with . Let

For all sufficiently large ,

Now, the restriction of factors through , where . This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required.

**Theorem 19:*** is LERF.*

Recall the set-up from the previous lecture. We built a graph of spaces for .

**Proof: **Let be finitely generated. Let be the corresponding covering space of and let be compact. Because is finitely generated, there exists a subgraph of spaces such that . We can take large enough so that . We can enlarge so that it contains every finite-degree edge space of . Also enlarge so that

for any . For each let and let incident edge map of infinite degree .

Applying lemma 29 to , for some large , set . (Here we use the fact that vertex groups of are finitely generated)

Define as follows:

For each , the edge space is the that the lemma produced from the corresponding .

Now, by construction, can be completed to a graph of spaces so that the map

factors through and embeds. Let be identical to except with +’s and -‘s exchanged. Clearly satisfies Stallings condition, as required.

Agol-Groves-Manning’s Theorem predicts that, for every word-hyperbolic group we can easily construct, every quasiconvex subgroup is separable (otherwise, we would find a non-residually finite hyperbolic group!).

In this section, we use graphs of groups to build new hyperbolic groups:

**Combination Theorem (Bestvina & Feighn):** If is a quasiconvex malnormal subgroup of hyperbolic groups , then is hyperbolic.

Recall: is called a malnormal subgroup of if it satisfies: if , then .

For a proof, see M. Bestvina and M. Feighn, “A combination theorem for negatively curved groups”, J. Differential Geom., 35 (1992), 85–101.

**Example:** Let be free, not a proper power. By Lemma 11, is malnormal, so is hyperbolic. As a special case, if is closed surface of even genus , considered as the connected sum of two copies of the closed surface of genus , then by Seifert-van Kampen Theorem, for some .

**Question:** (a) Which subgroups of are quasiconvex? (b) Which subgroups of are separable?

We will start by trying to answer (b). The following is an outline of the argument: Let be a finite graph so that , let be two copies of . Realize as maps , where . Let be the graph of spaces with vertex spaces , edge space , and edge maps . Then clearly, , and finitely generated subgroups are in correspondence with covering spaces . We can then use similar technique to sections 27 and 28.

Let us now make a few remarks about elevations of loops. Let be a loop in some space , i.e., and . Consider an elevation of :

The conjugacy classes of subgroups of are naturally in bijection with . The degree of the elevation is equal to the degree of the covering map .

**Definition:** Suppose is a covering map and is an intermediate covering space, i.e., factors through , and we have a diagram

If and are elevations of and the diagram commutes, then we say that descends to .

Let be a finite graph, a finitely generated subgroup and a loop. Let be a covering space corresponding to .

**Lemma 29:** Consider a finite collection of elevations of to , each of infinite degree. Let be compact. Then for all sufficiently large , there exists an intermediate, finite-sheeted covering space satisfying: (a) embeds in ; (b) every descends to an elevation of degree exactly ; (c) these are pairwise distinct.

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