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Recall from last time:

Theorem 16: Every \gamma \in \mathrm{Isom}\ T is semisimple. If \gamma is loxodromic, then \mathrm{Min}(\gamma) is isometric to \mathbb{R} and \gamma acts on \mathrm{Min}(\gamma) as translation by |\gamma|.

In particular, for every \langle \gamma \rangle, \mathrm{Min}(\langle \gamma \rangle) is a \langle \gamma \rangle-invariant subtree of T.

Exercise 21: If \gamma, \delta \in \mathrm{Isom}\ T and \mathrm{Min}(\gamma) \cap \mathrm{Min}(\delta) = \emptyset, then \gamma \delta is loxodromic, \mathrm{Min}(\gamma \delta) \cap \mathrm{Min}(\gamma) \neq \emptyset, and \mathrm{Min}(\gamma \delta) \cap \mathrm{Min}(\delta) \neq \emptyset.

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

Lemma 20 (Helly’s Theorem for Trees): If S_1, \ldots, S_n \subseteq T are closed subtrees and S_i \cap S_j \neq \emptyset for every i,j, then \bigcap _i S_i \neq \emptyset.

Proof. Case n=3: Let x_i \in S_j \cap S_k where \{ i,j,k \} = \{1, 2, 3\}. Let o be the center of the triangle with vertices x_1, x_2, x_3. Then o \in S_1 \cap S_2 \cap S_3.

General case: Let S_i ' = S_i \cap S_n for 1\leq i \leq n-1. Then S_i' \cap S_j' = S_i \cap S_j \cap S_n \neq \emptyset by the n=3 case. Now, by induction,

\bigcap _{i=1} ^n S_i = \bigcap _{i=1}^{n-1} S_i ' \neq \emptyset .

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

Proof. Let


For any \gamma-invariant T' \subseteq T, it is clear that \mathrm{Axis}(\gamma) \subseteq T'. Therefore T_G is minimal. Let S be a finite generating set for G. Suppose that every element of G is elliptical, so for all s, s' \in S, ss' is elliptical. Then for all s,s' \in S,

\mathrm{Fix}(s) \cap \mathrm{Fix}(s') \neq \emptyset.

Therefore by Lemma 20, \bigcap _{s\in S} \mathrm{Fix}(s) \neq \emptyset, a contradiction.

Suppose that \gamma, \delta \in G are loxodromic and \mathrm{Axis}(\gamma) \cap \mathrm{Axis}(\delta) = \emptyset. By Exercise 21, \mathrm{Axis}(\gamma \delta) intersects \mathrm{Axis}(\gamma) and \mathrm{Axis}(\delta) non-trivially. Thus, T_G is connected.

It remains to show that T_G is G-invariant. Let \gamma, \delta \in G. For any x\in T,

d(x, \delta ^{-1} \gamma \delta x) = d(\delta x, \gamma(\delta x)).

This implies that |\delta^{-1} \gamma \delta | = |\gamma |, and

\delta \mathrm{Axis}(\gamma) = \mathrm{Axis} (\delta^{-1} \gamma \delta).

So we conclude that T_G is G-invariant.

Definition: If \Gamma' \subseteq \Gamma is connected, then the graph of groups carried by \Gamma' is the graph of groups \mathcal{G}' with underlying graph \Gamma' such that the vertex v \in V(\Gamma') \subseteq V(\Gamma) is labelled by G_v, and the edge e \in E(\Gamma') \subset E(\Gamma) is labelled G_e (with obvious edge maps). There is a natural map G' \rightarrow G, where G = \pi_1 (\mathcal{G}) and G' = \pi_1(\mathcal{G}'). This map is an injection by the Normal Form Theorem.

Lemma 21: If \Gamma is countable and G = \pi_1 \mathcal{G} is finitely generated, then there is a finite subgraph \Gamma' \subset \Gamma such that \pi_1 \mathcal{G}' = \pi_1 \mathcal{G} (where \mathcal{G}' is the graph of groups carried by \Gamma').

Proof. Let \Gamma_0 \subseteq \Gamma_1 \subseteq \cdots \subseteq \Gamma be an exhaustion of \Gamma by finite connected subgraphs. Let \mathcal{G}_n denote the graph of groups carried by \Gamma_n and set G_n = \pi_1 (\mathcal{G}_n). Since G is finitely generated, there is an n such that G_n contains each generator of G, and since G_n \leq G, we conclude that G_n = G.

Lemma 22: If \Gamma is countable and G = \pi_1 \mathcal{G} is finitely generated then there is a finite minimal (wrt inclusion) subgraph \Gamma' that carries G.

Proof. Let T be the Bass-Serre tree of \mathcal{G}. Let \Gamma' = G \backslash T_G \subseteq G \backslash T = \Gamma. It’s an easy exercise to check that this is as required.

We will refer to \Gamma' (and \mathcal{G}', the graph of groups carried by \Gamma') as the core of \mathcal{G}.

Theorem 17: If H \leq G = \pi_1 \mathcal{G} is finitely generated, then H decomposes as \pi_1 of a finite graph of groups \mathcal{H}. The vertex groups of \mathcal{H} are conjugate into the vertex groups of \mathcal{G}. The edge groups are likewise.

Proof. Let T be the Bass-Serre tree of \mathcal{G}, and set \mathcal{H} = H \backslash T_H.

Fact: There exists a finitely generated non-Hopf group.  (An example is the Baumslag-Solitar group B(2,3), 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 G is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection F_r \rightarrow G is not separable in F_r.  However, finitely generated subgroups of free groups are separable:

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

This proof is associated with Stallings.

Proof: As usual, let F_r = \pi_1(X) where X is a rose. Let X' \rightarrow X be a covering map with \pi_1 (X') finitely generated.  Let \Delta \subset X' be compact. We need to embed \Delta in an intermediate finite-sheeted covering.

Enlarging \Delta if necessary, we may assume that \Delta is connected and that \pi_1 (\Delta) = \pi_1(X').  Note that we have \Delta \subset X' \rightarrow X. By Theorem 5 (see below), the immersion \Delta \rightarrow X extends to a covering \hat{X} \rightarrow X. Then \pi_1 (X') = \pi_1(\Delta) \subset \pi_1 (\hat{X}). So X' \rightarrow X lifts to a map X' \rightarrow \hat{X}.

The main tool in the proof above is this:

Theorem 5: The immersion \Delta \rightarrow X can be completed to a finite-sheeted covering \hat{X} \rightarrow X into which \Delta embeds:


Proof: Color and orient the edges of X. Any combinatorial map of graphs \Delta \rightarrow X corresponds uniquely to a coloring and orientation on the edges of \Delta.  A combinatorial map is an immersion if and only if at every vertex of \Delta, 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.

Completing <img src='' alt='\Delta' title='\Delta' class='latex' /> to a finite-sheeted covering <img src='' alt='\hat{X} \rightarrow X' title='\hat{X} \rightarrow X' class='latex' />.” width=”467″ height=”283″ /><p class=Completing \Delta to a finite-sheeted covering \hat{X} \rightarrow X.

Let k be the number of vertices of \Delta.  For each color c, let k_c be the number of edges of \Delta colored c. Then there are k-k_c vertices of \Delta missing “arriving” edges colored c, and there are k-k_c vertices of \Delta missing “leaving” edges colored c.  Choose any bijection between these two sets and use this to glue in k-k_c edges colored c. When this is done for all colors, the resulting map \Delta \subset \hat{X} \rightarrow X is clearly a covering.

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

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

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


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