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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 $\Delta$ 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 $F_r$.