Question about an application Ping-Pong Lemma The following is an application of the Ping-Pong Lemma.But I am not sure whether I understand the this application well.

Let $A=\begin{bmatrix} 1 & 2\\  0 & 1 \end{bmatrix}$,
$B=\begin{bmatrix} 1 & 0\\  2 & 1 \end{bmatrix}$ Then, the subgroup by
the matrices $A$ and $B$ is a free of rank 2 which is isomorhpic to
the subgroup of $SL(2,\mathbb{Z})$(a special linear group).

I understand the first part of the proof is the case where the group is generated only one generator either $A$ or $B$. Clearly, $H_1=<A>\le SL(2,\mathbb{Z}), ~H_2=<B>\le SL(2,\mathbb{Z})$
Then, when considering the first part of the proof, I think that the subgroup generated by the matrices $A,B$ which is denoted by $<A,B>$ and whose elements consist of
$A^{m_1}B^{n_1}... A^{m_k}B^{n_l} (\in  <A,B>=H$ )where $m_1,...m_k, n_1,...n_j$ are positive integers.
Then, If I can apply Ping-Pong Lemma to this case(just following the wiki), so that $<A,B>\underset{Ping-Pong}{=}<A><B>$, or $H\underset{Ping-Pong}{=}H_1H_2$
From here, I think that $H_1H_2 = \left \{ X*Y : X\in H_1, Y \in H_2 \right \}$ (of course, the operation $*$ means the multiplication of matrix)  so that this means every generated element by $A$,$B$ are elements of $H_1H_2$ and  $H_1H_2$ is a subgroup of $SL(2,\mathbb{Z})$. Hence, every subgroup generated by two matricies $A$ and $B$ is a free of rank 2 which is isomorphic to the subgroup of $SL(2,\mathbb{Z})$. Am I on the right track?
 A: Yes, you are on the right track.
For this application, we get two groups $H_1 = \langle A \rangle$ and $H_2 = \langle B \rangle$. To check that the group $H$ generated by $H_1$ and $H_2$ is free, we want to use the ping pong lemma as you suggested. To do so, we need to find a set $X$ that $H$ acts on with the following property:
There exist disjoint subsets $X_1, X_2 \subset X$ so that for any $h_1 \in H_1$ we have $h_1(X_2) \subset X_1$ and for any $h_2 \in H_2$ we have $h_2(X_1) \subset X_2$.
I think you are supposed to think that $h_1$ and $h_2$ are ``ping pong players'' and the ping pong table is $X$ with two sides $X_1$ and $X_2$. The ball is an element $x \in X_1$. Then $h_2$ acts on $x$ by pushing it into $X_2$ and then $h_1$ acts on $h_2(x)$ pushing it back to $X_1$. Do you get the analogy?
Anyway, in this case, $H \subset \mathrm{SL}_2(\mathbb{Z})$ which naturally acts on $\mathbb{R}^2$ by matrix multiplication. So we can let $X=\mathbb{R}^2$ and $X_1 = \{ (x,y) : |x| > |y| \}$ and $X_2 = \{ (x,y) : |x| < |y| \}$. You can check that these sets have the correct property with respect to $H_1$ and $H_2$ so that we can apply the ping pong lemma.
