I'm reading Serre's Trees and in Example 1.5.3 he claims that $PSL_2(\mathbb{Z})$ is isomorphic to $\mathbb{Z}_2 \ast \mathbb{Z}_3$.
$\mathbb{Z}_2 \ast \mathbb{Z}_3 = \langle x, y | x^2 = y^3 = 1 \rangle$. But does $PSL_2(\mathbb{Z})$ even have elements of order $2$ or $3$ to begin with?
Previously, I thought since all elements of $PSL_2(\mathbb{Z})$ is in the form of $$ \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}. $$ Hence $\mathbb{Z}$ injects in $PSL_2(\mathbb{Z})$. Let $\phi: \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \mapsto n$. $\phi$ is a homomorphism and in fact bijective. Hence $\mathbb{Z} \cong PSL_2(\mathbb{Z})$. Thus $PSL_2(\mathbb{Z})$ cannot have any $2$-torsion or $3$-torsion.
I must be missing something. Any help is appeciated!