Ping-Pong Lemma

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\section*{Example:}
\qquad Let
$A = \begin{bmatrix}
1 & 2 \\
0 & 1
\end{bmatrix}$
and
$B = \begin{bmatrix}
1 & 0 \\
2 & 1
\end{bmatrix}$
freely generated a subgroup of $SL_2(\mathbb{Z}).$\\
\vspace{0.3cm}

\noindent Let $H_1:= \Big{\langle}\begin{bmatrix}
1 & 2 \\
0 & 1
\end{bmatrix}\Big{\rangle} $
$= \Big{\{}\begin{bmatrix}
1 & 2n \\
0 & 1
\end{bmatrix}$ $\setminus \  n \in \mathbb{Z}\Big{\}}$ and
$H_2:= \Big{\langle}\begin{bmatrix}
1 & 0 \\
2 & 1
\end{bmatrix}\Big{\rangle} $
$= \Big{\{}\begin{bmatrix}
1 & 0\\
2n & 1
\end{bmatrix}$ $\setminus \  n \in \mathbb{Z}\Big{\}}.$ \\
\vspace{0.3cm}

\noindent Consider the action of $SL_2(\mathbb{Z}).$ on the projectiv $SL_2(\mathbb{Z}) \cap \mathbb{P}(\mathbb{R}^2)$ and the disjoint sets of $\mathbb{P}(\mathbb{R}^2)$
$$X_1:= \{ [x:y]\ \setminus \ |y| \geq |x| \}$$
$$X_2:= \{ [x:y]\ \setminus \ |y| \leq |x| \}$$
We whish to prove that
$\Big{(}H_1\setminus \Big{ \{ }$$\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$
$\Big{\}} \Big{)} \cdot \mathit{X}_2 \subseteq \mathit{X}_2$
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\vspace{0.3cm}

\noindent In fact: $\begin{bmatrix}
1 & 2n \\
0 & 1
\end{bmatrix}$
$ \begin{bmatrix}
x \\
y
\end{bmatrix}$ =
$ \begin{bmatrix}
x+2ny \\
y
\end{bmatrix} $ and it is verified that:
\begin{align*}
|x+2ny| & \geq |2n| |y| - |x|\\
& \geq |y| + (|y| - |x|)\\
& \geq |y|
\end{align*}
Similarly $\Big{(}H_2 \setminus \Big{ \{ }$$\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}$
$\Big{\}} \Big{)} \cdot \mathit{X}_1 \subseteq \mathit{X}_2$\\

\noindent So by Ping - Pong Lemma\\

$$\Big{\langle}\begin{bmatrix}
1 & 2 \\
0 & 1
\end{bmatrix},\  \begin{bmatrix}
1 & 2 \\
0 & 1
\end{bmatrix}\Big{\rangle}\ \cong\   H_1 \ast H_2\  \cong\  \mathbb{Z} \ast \mathbb{Z}\  \cong\  \mathit{F}_2 $$
\subsection*{Proposition}
$\Big{\langle}\overline{\begin{bmatrix}
i & 2 \\
0 & i
\end{bmatrix}},\  \overline{\begin{bmatrix}
0 & i \\
-i & 0
\end{bmatrix}}\Big{\rangle}\ \cong\   \mathbb{Z} \ast \mathbb{Z}\ \big{/}2\  \mathbb{Z}\ $, where $\overline{g} \in \mathrm{PSL}_2(\mathbb{R}) \cong \mathrm{SL}_2(\mathbb{R}) \big{/}\{ \pm I \} $, for $g \in \mathrm{SL}_2(\mathbb{R}).$
\subsubsection*{Proof:}
Consider the continuous action on the hyperbolic metric space $\mathbb{H}$, $$\alpha : \mathrm{SL}_2(\mathbb{R}) \times \mathbb{H} \to \mathbb{H}$$
defined by the rule:
$$\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \cdot z = \frac{az+b}{cz+d}$$
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Observe that this action factors through $\mathrm{PSL}_2(\mathbb{R}).$ Let $H_1 = \Big{\langle}\overline{\begin{bmatrix}
0 & i \\
-i & 0
\end{bmatrix}}\Big{\rangle}$ and $H_2 = \Big{\langle}\overline{\begin{bmatrix}
i & 2 \\
0 & i
\end{bmatrix}}\Big{\rangle}$. Observe that $\begin{bmatrix}
0 & i \\
-i & 0
\end{bmatrix} \cdot z = - \frac{1}{z}$, and $\begin{bmatrix}
i & 2n \\
0 & i
\end{bmatrix} \cdot z = z - 2ni$, and so:\\

$$\Big{(} H_1 \setminus \Big{ \{ } \overline{
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}} \Big{ \} } \cdot \mathit{X_2} \subseteq \mathit{X_1} $$
$$\Big{(} H_2 \setminus \Big{ \{ } \overline{
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}} \Big{ \} } \cdot \mathit{X_1} \subseteq \mathit{X_2} $$
Therefore by Ping-Pong Lemma:
$$\Big{\langle}\overline{\begin{bmatrix}
i & 2 \\
0 & i
\end{bmatrix}},\  \overline{\begin{bmatrix}
0 & i \\
-i & 0
\end{bmatrix}}\Big{\rangle}\ \cong\   \mathbb{Z} \ast \mathbb{Z}\ \big{/}2\  \mathbb{Z}\ $$



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