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Our goal is to understand the structure of subgroups of graphs of groups. This will enable us to prove things like: is LERF.

**Exercise 22**. A group is coherent if every fg subgroup is fp. Prove that of a finite graph of groups with coherent vertex groups and cyclic edge group is coherent.

Recall, that if is fg then H has an induced graph of groups structure. where T is the Bass-Serre tree of .

Topologically, defines a covering space , which inherits a graph of spaces structure with underlying graph .

is a finite connected subgraph and the image of in is a connected subcomplex such that . has the structure of a graph of space with underlying graph

There is an explicit algebraic description of , which follows immediately from Lemmas 16,17,18.

**Lemma 23. **(a) The vertices of are in bijection with .

The vertex is labelled by: , well-defined up to conjugation in .

(b) The edges of are in bijection with . The vertex is labelled with .

(c) The edges of adjoining the vertex corresponding to are in bijection with .

Now we will try to understand this topologically. In particular, we will understand the edge maps of a covering space of a graph of spaces.

Let be a continuous map and a covering map. If makes the diagram commutes, then is a *lift* of .

**Lemma 24**. Fix basepoint , , and mapping to . There is a lift with if and only if . Furthermore, if this lift exists it is unique.

It may be impossible to lift at . But it is possible if we pass to a covering space. Eg., if is the universal cover. Intuitively, an elevation is a minimal lift.

**Definition**: Let be as above. A based connected covering space together with a based map such that

commutes is an *elevation* (of at ) if whenever the diagram

commutes and is a covering map of degree larger than 1 then the composition does not lift to at .

The unbased covering map or equivalently the conjugacy class of the subgroup is called the *degree* of the elevation .

In this lecture, we will use the Normal Form Theorem to understand and construct groups. In particular, we will construct a group that is not residually finite (RF). Note that in the case when is an HNN-extension, then the Normal Formal Theorem is called Britton’s Lemma.

**Example:** The (p,q)-Baumslag-Solitar group has the following presentation:

For example .

Consider . We will show that this group is non-Hopfian (i.e., that there exists a homomorphism that is a surjection with nontrivial kernel), hence is not RF. Consider defined by and . First, we show this is a homomorphism, then a surjection, then that it has nontrivial kernel.

Homomorphism: Hence, the map is a homomorphism.

Epimorphism: To see that is a surjection, we check that . This is obvious for . Further and so . But then .

Nontrivial kernel: Consider . Then . So we just need to show that is nontrivial. We use Britton’s Lemma (Normal Form Theorem). What Britton’s Lemma says is that there will be that can be removed from our word using the relation if it is trivial. But this can’t be done with , so we must have as required.

**Remark:** can never embed in a word-hyperbolic group by a previous Lemma.

**Exercise 20:** Show that , where is the dyadic rationals and we have acting via multiplication by 2. One can deduce that is linear, hence RF by Selberg’s Lemma.

**Theorem 15 (Higman-Neumann-Neumann)****:** Every countable group embeds in a 3-generator group.

**Proof: **Let . Consider . This has some nice free subgroups, unlike : Let . Let . By the Normal Form Theorem, (basically since there will always be some g’s between the s’s whenever you multiply any two elements together). Let . Again, . Since $latex\Sigma_{1}$ and are both countable, .

Consider the HNN-extension $\Gamma = (G \ast \langle s \rangle)\ast_{F_{\Sigma_{1}} \tilde F_{\Sigma_{2}}}$ and let t be the stable letter. For every , . But is generated by , , and : , so or . So by induction . for all n. But by construction is generated by .

**Remark: **In fact 3 can be replaced by 2.

**Isometries of Trees**

Before we saw that graphs of groups correspond to groups acting on trees. As such, we now turn to isometries of trees. Let be a tree and let . The translation length of is defined to be . Let .

**Definition: **If , is called *semisimple*. If is semisimple, and , is called elliptic. If , it is called loxodromic.

**Theorem 16: **Every is semisimple. If is loxodromic, is isometric to , and acts on as translation by .

Notation: If is elliptic, we write . If is loxodromic$, we write . Note that is connected since if we have two points fixed by any path between them must be fixed, for otherwise we would not be in a tree.

**Proof of Theorem 16:** Consider any and the triangle with vertices (i.e., ). Let , and let be the midpoint of .

Case 1:

In this case we have . So , and is elliptic.

Case 2:

Let . Now . Therefore is isometric to and acts as translation by . Furthermore, . Therefore unless is on the line just constructed, .

**Normal form theorem for graphs of groups.** Let be a graph of groups and .

- Any can be written as

as before. - If , this expression includes `backtracking’, meaning that for some , with , and furthermore that if , then .

Similar to the case of the free group, the proof boils down to the fact that the Bass–Serre tree is a tree.

**Proof.** To simplify notation, set , so

.

Fix base points in the vertex spaces , which are chosen to coincide when the vertices do. Then is a loop in based at , and is a path, crossing the corresponding edge space, from to . This allows us to consider as a loop in based at . (We may assume by adding letters from a maximal tree.)

Consider the universal covering and fix a base point over in . Let be the lift of based at and its image in the Bass–Serre tree . We now analyze and closely.

Choose adjoining and so that the edge traversed by when lifted at corresponds to the coset .

Then lifts to a path in which terminates at . Similarly, lifts at to a path across the edge to the vertex space terminating at . Therefore, lifts at to a path which crosses the edge space and ends at .

Then, lifts at to a path in ending at , and lifts at to a path across the edge into the vertex space , and terminating at . Thus lifts at to a path which crosses , through , across , and ending at

.

We continue this process until we have explicitly constructed . By hypothesis, , so and are both loops in and , respectively. Since is a tree, must backtrack.

This implies that and that . That is, by Lemma 18,

.

Therefore, we have found a backtracking, and can accordingly shorten . This proves the theorem.

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