If Γ has finite in $SL_2(\mathbb{Z})$ , show that there are only finitely many $\Gamma$- inequivalent elliptic points .
Proof . Suppose that $z \in \mathbb{H}$ is an elliptic point for $\Gamma$ . By the definition of a fundamental domain, there exists $\gamma \in SL_2(\mathbb{Z})$ such that $\gamma z$ in $F$ , where $F$ is the standard fundamental domain for $SL_2(\mathbb{Z}) $ . It is easily verified that $\Gamma_{\gamma z}=\gamma \Gamma_z \gamma^{-1}\neq \lbrace \pm I\rbrace$ . Hence $\gamma z$ is an elliptic point for $\Gamma$ and hence also for $SL_2(\mathbb{Z}) $ . We see that z is $SL_2(\mathbb{Z}) $-equivalent to either $i$ or $\rho$ .
Let $\gamma_1,...\gamma_d$ be a complete set of coset representatives for $\Gamma $ in $SL_2(\mathbb{Z}) $ .
We deduce that z is $\Gamma$-equivalent to an element of the finite set $$ \lbrace \gamma_j(i),\gamma_j(\rho):1\leq j\leq d \rbrace$$
I do not understand the last part why we consider a complete set of representatives and at the end the deduction .
Thanks for the help .
The elliptic points of $\Gamma \setminus \Bbb{H}$ are of the form $\Gamma z$,
$SL_2(\Bbb{Z})\Gamma z = SL_2(\Bbb{Z}) z$ is an elliptic point of $SL_2(\Bbb{Z}) \setminus \Bbb{H}$
so $ SL_2(\Bbb{Z}) z = SL_2(\Bbb{Z}) i$ or $ SL_2(\Bbb{Z}) e^{2i \pi / 3}$,
$\gamma z = i$ or $ e^{2i \pi / 3}$ for some $\gamma \in SL_2(\Bbb{Z})$
$\Gamma z = \Gamma \gamma^{-1} i$ or $ =\Gamma \gamma^{-1} e^{2i \pi / 3}$ for some $\gamma \in SL_2(\Bbb{Z})$ ie. $\Gamma \gamma^{-1} \in \Gamma \setminus SL_2(\Bbb{Z})$ and the elliptic points of $\Gamma \setminus \Bbb{H}$ are contained in the finite set $$\bigcup_{\Gamma \gamma^{-1} \in \Gamma \setminus SL_2(\Bbb{Z})} \Gamma \gamma^{-1}i \ \cup \ \Gamma \gamma^{-1} e^{2i \pi / 3}$$