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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 .

**Fact:** There exists a finitely generated non-Hopf group. (An example is the Baumslag-Solitar group , although we cannot prove it yet.) So, by Lemma 5, there is a finitely generated non-residually finite group. Thus, free groups are not ERF: if is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection is not separable in . However, finitely generated subgroups of free groups are separable:

**Marshall Hall’s Theorem (1949):** is LERF.

This proof is associated with Stallings.

**Proof:** As usual, let where is a rose. Let be a covering map with finitely generated. Let be compact. We need to embed in an intermediate finite-sheeted covering.

Enlarging if necessary, we may assume that is connected and that . Note that we have . By Theorem 5 (see below), the immersion extends to a covering . Then . So lifts to a map .

The main tool in the proof above is this:

**Theorem 5:** The immersion can be completed to a finite-sheeted covering into which embeds:

**Proof: ** Color and orient the edges of . Any combinatorial map of graphs corresponds uniquely to a coloring and orientation on the edges of . A combinatorial map is an immersion if and only if at every vertex of , we see each color arriving at most once and leaving at most once. Likewise, it’s a covering map if and only if at each vertex, we see each color arriving exactly once and leaving exactly once.

Let be the number of vertices of . For each color , let be the number of edges of colored . Then there are vertices of missing “arriving” edges colored , and there are vertices of missing “leaving” edges colored . Choose any bijection between these two sets and use this to glue in edges colored . When this is done for all colors, the resulting map is clearly a covering.

Note that the proof in fact gives us more. For instance:

**Exercise 6:** If is a finitely generated subgroup of , then is a free factor of a finite-index subgroup of .

**Exercise 7 (Greenberg’s Theorem):** If and is finitely generated, then is of finite index in .

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