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Recall from last time:

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

In particular, for every , is a -invariant subtree of .

**Exercise 21:** If and , then is loxodromic, , and .

(Hint: It is enough to construct an axis for .)

**Lemma 20 (Helly’s Theorem for Trees): **If are closed subtrees and for every , then .

**Proof.** Case : Let where . Let be the center of the triangle with vertices . Then .

General case: Let for . Then by the case. Now, by induction,

**Corollary**: If a finitely generated group acts on a tree with no global fixed point, then there is a (unique) -invariant subtree that is minimal with respect to inclusion, among all -invariant subtrees. Furthermore, is countable.

**Proof.** Let

For any -invariant , it is clear that . Therefore is minimal. Let be a finite generating set for . Suppose that every element of is elliptical, so for all , is elliptical. Then for all ,

Therefore by Lemma 20, a contradiction.

Suppose that are loxodromic and . By Exercise 21, intersects and non-trivially. Thus, is connected.

It remains to show that is -invariant. Let . For any ,

This implies that , and

So we conclude that is -invariant.

**Definition:** If is connected, then the graph of groups carried by is the graph of groups with underlying graph such that the vertex is labelled by , and the edge is labelled (with obvious edge maps). There is a natural map , where and . This map is an injection by the Normal Form Theorem.

**Lemma 21:** If is countable and is finitely generated, then there is a finite subgraph such that (where is the graph of groups carried by ).

**Proof. ** Let be an exhaustion of by finite connected subgraphs. Let denote the graph of groups carried by and set . Since is finitely generated, there is an such that contains each generator of , and since , we conclude that .

**Lemma 22:** If is countable and is finitely generated then there is a finite minimal (wrt inclusion) subgraph that carries .

**Proof.** Let be the Bass-Serre tree of . Let . It’s an easy exercise to check that this is as required.

We will refer to (and , the graph of groups carried by ) as the core of .

**Theorem 17: **If is finitely generated, then decomposes as of a finite graph of groups . The vertex groups of are conjugate into the vertex groups of . The edge groups are likewise.

**Proof.** Let be the Bass-Serre tree of , and set .

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, .

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