Put
$$
u=\begin{pmatrix}1&2\\0&1\end{pmatrix},
\qquad
v=\begin{pmatrix}1&0\\2&1\end{pmatrix}.
$$
For every nonzero <integer> $n$,
$$
u^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix},
\qquad
v^n=\begin{pmatrix}1&0\\2n&1\end{pmatrix}.
$$
Let
$$
A=\{(x,y):|x|>|y|\},
\qquad
B=\{(x,y):|x|<|y|\}.
$$
These are disjoint nonempty subsets of $\mathbb R^2$. If $(x,y)\in B$, then
$$
|x+2ny|\geq2|n||y|-|x|>|y|,
$$
so $u^n(B)\subseteq A$. Similarly, if $(x,y)\in A$, then
$$
|2nx+y|\geq2|n||x|-|y|>|x|,
$$
so $v^n(A)\subseteq B$.

The <ping-pong lemma> now identifies the subgroup generated by $u$ and $v$ with
$$
\langle u\rangle*\langle v\rangle
\cong\mathbb Z*\mathbb Z
\cong F_2.
$$
Both generators have infinite order, so this is a <free group> of rank two inside the <special linear group> $\operatorname{SL}_2(\mathbb Z)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 133 / 1 / a / Solution

Every finitely generated <free group> is a <residually finite group>. For $F_2$, the <ping-pong lemma> embeds it into $\operatorname{SL}_2(\mathbb Z)$, and <reduction modulo a prime in an integral matrix group> separates each nonidentity element in a finite quotient.

Residual finiteness of a free group

{wiki}

Suppose nontrivial groups $G_1,G_2$ act on a set and there are disjoint nonempty subsets $X_1,X_2$ such that every nonidentity element of $G_1$ maps $X_2$ into $X_1$, while every nonidentity element of $G_2$ maps $X_1$ into $X_2$. Then the subgroup generated by $G_1$ and $G_2$ is their <free product>.

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 Incoming links: Ping-pong lemma