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.