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Last time we saw that if acts on a tree, then has the structure of a graph of groups (remeber stabilizers). Such a is called *developable.*

**Exercise 18:** Show that .

**Theorem 13 (Scott-Wall): **Let be a graph of spaces. The universal cover of is itself a graph of spaces. Also, each vertex space (resp. edge space) is the universal cover of a vertex space of (.edge space).

**Sketch of Proof:** For any , let , where is the set of edges that are the edges incident to . It should be noted that is a deformation retract of . Also, recall that the edge maps are injective. From covering space theory we know that given a map and a covering space of that lifts to a map if and only if . It therefore follows that . So is built from by attaching covering spaces of edge spaces. Because is injective, these covering spaces of edge spaces really are universal covers. By iteratively gluing together copies of the we can construct a simply connected cover of .

Given we have constructed the universal cover . The underlying graph, of is a tree because is a surjection. We now need to check that the action of on is interesting.

Let be a graph of groups and let . Fix a base point and a choice of lift . Let . The space is a universal cover of by Theorem 13, and so the lift of to the universal cover of at is contained in . Therefore the preimages of that are contained in correspond to the elements of .

Now consider . If is lifted at then the terminus of this lift is not in , but in some other component of the preimage of . Call the component where the lift terminates . If are such that both have lifts that terminate in then . We have just proved the following lemma.

**Lemma 16:** Let be a graph of groups and let be the underlying graph of the universal cover . For any vertex the set of vertices of lying above is in bijection with and acts by left translation.

We can also prove the following two lemmas in a similar fashion.

**Lemma 17: **For any the set of edges of that lie above is in bijection with and acts by left translation.

**Lemma 18:** If adjoins a vertex then for any lying above the set of edges of adjoining lying above is in bijection wiht and acts by left translation.

We begin with a lemma that among other things demonstrates that most free groups are non-trivial

**Lemma 1:** If are sets then the map , induced by inclusion has a left inverse. Consequently, is injective.

**Proof:** Let and be the roses with petals indexed by and respectively. In particular we have that and . If we view as a subspace of in the obvious way then there exists such that (One such map is obtained by crushing the petals in ). This map induces a function and this map can easily be seen to be the desired left inverse. **QED**

The next thing that we would like to do is examine exactly what do elements of free groups “look like.” Informally, elements look like words in the elements of where only the obvious cancellations by inverses are allowed. More formally, we will identify the elements of with their images under the map (This is not the same from Lemma 1). Next we consider products of the form , where and . Next, we prove a theorem that formalizes the previous intuition about free groups

**Normal Form Theorem:** Given a free group the following 2 statements are true

- Every element of is of the form .
- If then is either the empty word or there exists such that and

**Proof:** We will prove the theorem when , but the proof generalizes to sets of all cardinality. Let be the rose with 2 petals and wedge point and so we have that . Next, let be the universal cover of based at . Because acts freely on and has only one vertex we see that the vertices of are in bijection with the elements of . We next observe 2 facts. First, from lemma 1 we see that since is simply connected that it must be a tree. Second, since each vertex of looks locally like we see that every vertex of is 4-valent. From covering space theory we know that if and is a vertex of then the lift of to that starts at ends at .

We proceed by induction on , which is the minimum number of edges between and . Suppose that . Since is a tree with all 4-valent vertices we see that there exists such that and . Next, observe that there exists such that when is lifted at ends at . Therefore we see that and since the group action is free this implies that which proves 1.

For the second proposition, suppose that and that the second proposition is not true. From covering space theory we know that the lift of at is a loop based at . First, since is not the empty word we see that it will not lift to the constant path. Finally, since contains no obvious cancellations we see that must contain a loop based at that is not the constant loop, which contradicts the fact that is a tree. Thus 2 must hold. **QED**

As a consequence of the previous theorem we see that it is possible to solve the word problem of whether it is algorithmically possible to decide if a group element is the identity.

**Definition:** A group is finitely generated if and only if there exists a surjection from to for some finite .

**Exercise 2: **

- Show that if then has finitely many subgroups of index .
- Deduce that the same holds for any finitely generated group .

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