Let us denote by $\mathrm{SL}(2,\mathbb{Z})$ the set of $(2 \times 2)$-matrices whose determinant is $1$, and by $\mathrm{PSL}(2,\mathbb{Z})$ the quotient $\mathrm{SL}(2,\mathbb{Z}) / \{ \pm \mathrm{Id} \}$. Our aim here is to prove that $\mathrm{PSL}(2,\mathbb{Z})$ is a free product:

Theorem: $\mathrm{PSL}(2,\mathbb{Z}) \simeq \mathbb{Z}_2 \ast \mathbb{Z}_3$.

As noticed in the previous note Some SQ-universal groups, as a corollary we get that $\mathrm{SL}(2,\mathbb{Z})$ is SQ-universal, that is every countable group is embeddable into a quotient of $\mathrm{SL}(2, \mathbb{Z})$.

Usually, the theorem above is proved thanks to a natural action of $\mathrm{PSL}(2,\mathbb{Z})$ on the hyperbolic plane by Möbius transformations. However, as noticed by Roger Alperin in his article published in The American Mathematical Monthly, it is sufficient to make $\mathrm{PSL}(2,\mathbb{Z})$ act on the set of irrational numbers via:

$\displaystyle \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \cdot r = \frac{a r + b }{ c r + d}$.

Notice that $cr+d \neq 0$ for all $c,d \in \mathbb{Z}$ precisely because $r$ is irrational. The argument below may be thought of as an application of ping-pong lemma (that we already met in the note Free groups acting on the circle in order to find free subgroups).

First, it is a classical exercice to prove that $\mathrm{SL}(2,\mathbb{Z})$ is generated by the two following matrices:

$A= \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$  and  $B= \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right)$.

For convenience, let $C= AB$. Then $B$ and $C$ also generate $\mathrm{SL}(2, \mathbb{Z})$, and consequently their images in $\mathrm{PSL}(2,\mathbb{Z})$ generate it; furthermore, $B$ (resp. $C$) has order two (resp. order three) in $\mathrm{PSL}(2, \mathbb{Z})$. Notice that

$\displaystyle B=B^{-1} : z \mapsto - \frac{1}{z}$,    $\displaystyle C : z \mapsto 1- \frac{1}{z}$    and    $\displaystyle C^{-1} : z \mapsto \frac{1}{1-z}$.

Now, let $\mathcal{P}$ and $\mathcal{N}$ denote the set of positive and negative irrationals respectively. Clearly,

$B( \mathcal{P} ) \subset \mathcal{N}$  and  $C^{ \pm 1} ( \mathcal{N} ) \subset \mathcal{P}$.

To conclude, we want to prove that, for any alternating word $w$ from $\langle B \rangle$ and $\langle C \rangle$, $w \neq 1$ in $\mathrm{PSL}(2,\mathbb{Z})$.

Case 1: $w$ has odd length. Then either $w$ begins and ends with a $B$, hence $w( \mathcal{P} ) \subset \mathcal{N}$, or $w$ begins and ends with a $C^{\pm 1}$, hence $w( \mathcal{N} ) \subset \mathcal{P}$. In particular, we deduce that $w \neq 1$ in $\mathrm{PSL} (2, \mathbb{Z})$.

Case 2: $w$ has even length. Without loss of generality, we may suppose that $w$ begins with a $C^{\pm 1}$; otherwise, just conjugate $w$ by $B$. Then either $w$ begins with a $C$ and ends with a $B$, hence

$w( \mathcal{P} ) \subset C(\mathcal{N}) \subset \{ r \ \text{irrationals} \mid r> 1 \}$,

or $w$ begins with a $C^{-1}$ and ends with a $B$, hence

$w( \mathcal{P} ) \subset C^{-1} ( \mathcal{N} ) \subset \{ r \ \text{irrationals} \mid r< 1 \}$.

In either case, we deduce that $w(1) \neq 1$, so $w \neq 1$ in $\mathrm{PSL}(2, \mathbb{Z})$. The proof is complete.