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’.