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Last time we proved that if $\mathcal{G}$ is a graph of groups, then $G = \pi_1 \mathcal{G}$ acts on a tree $T$, namely the underlying graph of the universal cover of $X_\mathcal{G}$.  Lemmas 16, 17, and 18 describe the vertex and edge sets of $T$ and the action of $G$.  The following is immediate.

Theorem 14 (Serre): Every graph of groups is developable.

Corollary: If $\mathcal{G}$ is a graph of groups and $v \in V(\Gamma)$, then the map $G_v \rightarrow G = \pi_1 \mathcal{G}$ is injective.

The tree $T$ is called the Bass-Serre Tree of $\mathcal{G}$.  We will usually equip $T$ with a length metric in which each edge has length 1.  The approach we’ve taken is due to Scott-Wall.  Here’s a sample application.

Lemma 19: Let $G = A \ast_C B$.  If $g \in G$ is torsion, then $g$ is conjugate into $A$ or $B$.

Proof. Let $T$ be the Bass-Serre Tree.  Fix a vertex $v \in T$ amd consider $\{g^n v \mid n \in \mathbb{Z} \}$.  This is a finite set.

Exercise 19: There is an $x \in T$ that minimizes $\mathrm{max}_{n \in \mathbb{Z}} d(x, g^n v)$ (when $g$ is torsion).

Such an $x$ is fixed by $g$.  Therefore, $g$ stabilizes a vertex.  The stabilizers of the vertices are precisely the conjugates of $A$ and $B$$\Box$

Our next goal is to understand the elements of $G = \pi_1 \mathcal{G}$ — which are trivial and which not?  We’ll start with a presentation.  If $\Gamma$ is a tree, then $G$ can be thought of as a sequence of amalgamated free products.  If $\Gamma$ is a rose, then $G$ can be thought of as a sequence of HNN-extensions.  In general, fix a maximal tree $\Lambda \subseteq \Gamma$.  This determines a way of decomposing $G$ as a sequence of amalgamated products followed by a sequence of HNN-extensions.  This process is sometimes called “Excision.”  It follows that $G = \pi_1 \mathcal{G}$ has a “presentation” like this:

$G = \langle \{ G_v \mid v \in V(\Gamma) \}, \{t_e \mid e \in E(\Gamma)\} \mid \\ \{t_e \partial_+^e(g)t^{-1}_e = \partial_-^e(g) \mid e \overline{t}(\Gamma), g \in G_e\}, \{t_e = 1 \mid e \in E(\Lambda)\} \rangle$

We can write any $g \in G$ in the following form:

$g = g_0 t^{\epsilon_1}_{e_1} g_1 t^{\epsilon_2}_{e_2} \cdots t^{\epsilon_{n-1}}_{e_{n-1}} g_{n-1}t^{\epsilon_n}_{e_n} g_n$

where $\epsilon_i = \pm 1$, $e_1^{\epsilon_1} \cdots e_n^{\epsilon_n}$ is a loop in $\Gamma$, and $g_i \in G_{v_i}$ where $v_i$ is the terminus of $e_i^{\epsilon_i}$ and the origin of $e_{i+1}^{\epsilon_{i+1}}$.  A priori, we just know that $g = g_0 t^{\epsilon_1}_{e_1} g_1 \cdots t^{\epsilon_n}_{e_n} g_n$.  Suppose, say, that $e_1$ doesn’t originate from $v_0$.  Then insert stable letters corresponding to edges in $\Lambda$ that form a path from $v_0$ to the origin of $e_1$.  Now repeat.

Ping-Pong Lemma

Question. Let G be a group and $a, b \in G$.  When is $\langle a, b \rangle \cong F_2$?

Ping-Pong Lemma: Let G be a group acting on a set X and $a, b \in G$. Assume:

1. a and b have infinite orders.
2. There exist $X_1, X_2 \subseteq X$ such that $X_2 \nsubseteq X_1$ and $a^m X_2\subseteq X_1$, $b^m X_1 \subseteq X_2$ for all $m \neq 0$.

Then $\langle a, b \rangle \cong F_2$.

Proof. Consider $\varphi : F_2 = \langle x, y \rangle \twoheadrightarrow \langle a,b \rangle \leq G$ such that $\varphi (x) = a$ and $\varphi (y) = b$.  Choose a reduced word w for a nontrivial element in $F_2$, i.e., either $w = x^{m_1} y^{n_1} x^{m_2} \cdots$ or $w = y^{m_1} x^{n_1} y^{m_2} \cdots$ and $m_i, n_i \neq 0$.

Case 1: $w = x^{m_1} y^{n_1} \cdots x^{m_k}$.  Then $\varphi (w) X_2 = a^{m_1} b^{n_1} \cdots a^{m_k} X_2$ and $a^{m_k} X_2 \subseteq X_1$, so $\varphi (w) X_2 \subseteq a^{m_1} b^{n_1} \cdots b^{n_{k-1}} X_1$ and $b^{n_{k-1}} X_1 \subseteq X_2$, and so on until $\varphi (w) X_2 \subseteq a^{m_1} X_2 \subseteq X_1$, so $\varphi (w) X_2 \neq X_2$.  Therefore $\varphi (w) \neq 1$.

Case 2: $w = y^{m_1} x^{n_1} \cdots y^{m_k}$.  Then, by Case 1, $\varphi (xwx^{-1}) \neq 1$, so $\varphi (x) \varphi (w) \varphi (x^{-1}) \neq 1$.  Therefore $\varphi (w) \neq 1$.

Case 3: $w = x^{m_1} y^{n_1} \cdots y^{n_k}$.  (similar to above)

Case 4: $w = y^{m_1} x^{n_1} \cdots x^{n_k}$.  (similar to above)

Free Groups Are Linear

Theorem 3. $F_2$ is linear.

Proof. $\mathrm{SL}_2\mathbb{R}$ acts on $\mathbb{R}^2$ by linear transformations.  Let $a = \left(\begin{array}{cc} 1&2 \\ 0&1 \\ \end{array}\right)$ and $b = \left(\begin{array}{cc} 1&0 \\ 2&1 \\ \end{array}\right)$.  Then $a^m = \left(\begin{array}{cc} 1&2m \\ 0&1 \\ \end{array}\right)$ and $b^m = \left(\begin{array}{cc} 1&0 \\ 2m&1 \\ \end{array}\right)$. Let $X_1 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| > |y|\right\}$ and $X_2 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| < |y|\right\}$.  Then $a^m X_2 \subseteq X_1, b^m X_1 \subseteq X_2$ for all $m \neq 0$, so $\langle a, b \rangle \cong F_2$ by the Ping-Pong Lemma.

Corollary 1. Finitely generated free groups are linear, hence residually finite.

Proof. The case of $\mathbb{Z}$ is obvious; otherwise, this follows from $F_n \hookrightarrow F_2$ as proved in Exercise 1.

Separability

Definition. Let G be a group.  The profinite topology on G is the coarsest topology such that every homomorphism from G to a finite group (equipped with the discrete topology) is continuous.

Definition. A subgroup H is separable in G if H is closed in the profinite topology of G.

Exercise 3. Let G be a group. $X \subseteq G$ is separable if and only if for all $g \in G \smallsetminus X$, there exists a homomorphism to a  finite group $q : G \rightarrow Q$ such that $q(g) \not \in q(X)$.  Note that if X = {1}, this is equivalent to: G is RF if and only if {1} is separable.

Hint. For the “if” direction, let $g \in G$ and consider $q^{-1} \circ q(g)$.  For the other direction, use the definition of subbase and that $p : G \rightarrow P, q: G \rightarrow Q \Rightarrow \text{ker}(p \times q) = \text{ker}(p) \cap \text{ker}(q)$.

Definition. Let G be a group.

1. G is Extended RF (ERF) if any subgroup of G is separable.
2. G is Locally ERF (LERF, subgroup separable) if any finitely generated subgroup is separable.

Lemma 3. Let G be a group. A subgroup H of G is separable if and only if for all $g \in G \smallsetminus H$, there exists a finite-index subgroup $K \leq G$ such that $H \leq K \leq G$ and $g \not \in K$.

Proof. In the “only if” direction, by the previous exercise, for all $g \in G \smallsetminus H$, there exists a homomorphism to a  finite group $q : G \rightarrow Q$ such that $q(g) \not \in q(H)$. Then $g \not \in q^{-1} \circ q(H) = K$.  Conversely, let $g \in G \smallsetminus H$. By hypothesis, there exists a finite-index subgroup $K \leq G$ such that $H \leq K \leq G$ and $g \not \in K$.  Let $L = \cap_{h \in G} K^h$. Note that this is a finite number of intersections ($|G/N_G (K)|$, to be precise). There exists a finite quotient $q : G \rightarrow G/L$.  Then  $q^{-1} \circ q(H) = H \cdot L$.  Therefore, $g \not \in q^{-1} \circ q(H)$, i.e., $q(g) \not \in q(H)$, and the lemma follows by the previous exercise.

Scott’s Criterion (1978). Let X be a Hausdorff topological space and $G = \pi_1 (X,x)$.  Let $X' \rightarrow X$ be a covering and $H = \pi_1 (X',x')$. Then H is separable in G if and only if for any compact $\Delta \subseteq X'$, there exists and intermediate finite-sheeted cover $\hat{X} \rightarrow X$ such that $X' \rightarrow \hat{X}$ embeds $\Delta$ into $\hat{X}$.

Exercise 4. Let $H \leq G$ be a separable subgroup.

1. If $G' \leq G$, then $G' \cap H$ is separable in G’.
2. If $G \leq G'$ has finite index, then $H$ is separable in G’.