I recently read the statement "up to conjugacy there are 4 nontrivial finite subgroups of ${\rm SL}\_2(\mathbb{Z})$." They are generated by 


$$\left(\begin{array}{cc} -1&0 \\\ 0&-1\end{array}\right),
 \left(\begin{array}{cc} -1&-1 \\\ 1&0\end{array}\right),
 \left(\begin{array}{cc} 0&-1 \\\ 1&0\end{array}\right),
 \left(\begin{array}{cc} 0&-1 \\\ 1&1\end{array}\right) $$
and are isomorphic to $\mathbb{Z}\_2$, $\mathbb{Z}\_3$, $\mathbb{Z}\_4$, and $\mathbb{Z}\_6$, respectively. Does someone know a reference for this statement? (Or, is it easy to see?) My attempt at a Google search turned up this statement, but I wasn't able to find a reference.



Finite subgroups of ${\rm SL}_2(\mathbb{Z})$ (reference request)
A finite subgroup will map to a finite subgroup of $PSL\_2(\mathbb Z)$, which is a free product $Z\_2 \* Z\_3$. I believe I have been told that finite subgroups of free products of finite groups are conjugate to subgroups of the factors being free-producted together.


