Prove that $PSL(2,\mathbb{Z})$ is free product of $C_2$ and $C_3$

Solution 1:

Let $\text{PSL}(2,\mathbb{Z})$ act on the set of irrational numbers via $$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \cdot z \mapsto \frac{az+b}{cz+d}.$$

• Show that $A=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right)$ and $B= \left( \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right)$ generate $\text{PSL}(2,\mathbb{Z})$.

• For convenience, let $C=AB$. Notice that $B$ is an order-two element and $C$ an order-three element. Therefore, it defines an epimorphism $\mathbb{Z}_2 \ast \mathbb{Z}_3 \twoheadrightarrow \text{PSL}(2,\mathbb{Z})$.

• Let $w$ be a non-trivial alternating word written over $\{ B \}$ and $\{ C, C^{-1} \}$. Now, according to what $w$ looks like, find an irrational $z$ such that $w(z) \neq z$. In particular, $w \neq 1$.

• Conclude that our morphism $\mathbb{Z}_2 \ast \mathbb{Z}_3 \to \text{PSL}(2,\mathbb{Z})$ turns out to be an isomorphism.

More information and references can be found in An elementary application of ping-pong lemma.