Let $\Gamma_0(m)$ be the subgroup of $SL(2,\mathbb{Z})$ which is defined as follows: $$\Gamma_0(m)=\left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \in SL(2, \mathbb{Z} : c \equiv 0 \mod{m} \right\}$$ Moreover, let $$C(m) = \left\{ \left( \begin{array}{cc} a & b \\ 0 & d \end{array} \right): ad=m, a>0, 0 \leq b < d, \gcd(a,b,d) = 1 \right\}$$ and $\sigma_0 = \left( \begin{array}{cc} m & 0 \\ 0 & 1 \end{array} \right) \in C(m)$.
I need to show that $$\Gamma_0(m) = (\sigma_0^{-1} SL(2,\mathbb{Z}) \sigma_0) \cap SL(2, \mathbb{Z})$$ and more generally we have that For $\sigma \in C(m)$, the set $$(\sigma_0^{-1} SL(2,\mathbb{Z}) \sigma) \cap SL(2, \mathbb{Z})$$ is a right coset of $\Gamma_0(m)$ in $SL(2, \mathbb{Z})$.
The missing step in the theorem I am reading seems to be that I need to show (or know) that there is a one-to-one correspondence between the right cosets of $\Gamma_0(m)$ and elements of $C(m)$.