If you want to read question directly, please go to down.
Firstly, let's give the definitions of $\Gamma(N)$, $\Gamma_1(N)$ ve $\Gamma_0(N)$: $$\Gamma(N) : = \left \{\left[ \begin{array}{cc}a & b \\c & d \end{array} \right] \in SL_2(\mathbb Z) \; : \;\left[ \begin{array}{cc}a & b \\c & d \end{array} \right]\equiv \left[ \begin{array}{cc}1 & 0 \\0 & 1 \end{array} \right] \mod N \right \},$$$$\Gamma_1(N) : = \left \{ \left[ \begin{array}{cc}a & b \\c & d \end{array} \right] \in SL_2(\mathbb Z) \; : \;\left[ \begin{array}{cc}a& b \\c & d \end{array} \right] \equiv\left[ \begin{array}{cc}1 & * \\0 & 1 \end{array} \right] \mod N \right \}, $$$$\Gamma_0(N) : = \left \{\left[ \begin{array}{cc}a & b \\c & d \end{array} \right] \in SL_2(\mathbb Z) \; : \;\left[ \begin{array}{cc}a & b \\c & d \end{array} \right]\equiv\left[ \begin{array}{cc}* & * \\0 & * \end{array} \right]\mod N \right \}.$$ 1) Obviously, $$\Gamma(n) \subset \Gamma_1(N)\subset \Gamma_0(N) \subset SL_2(\mathbb Z)$$ satisfies.
2) Let the function $$\Gamma_1(N) \to \mathbb Z/N\mathbb Z$$ to be $$\left[ \begin{array}{cc}a & b \\c & d \end{array} \right] \to b \mod N.$$ Since this onto function is an homomorphism and the kernel is $\Gamma(N)$, we will have $\Gamma(N) \lhd \Gamma_1(N)$ olur ve $$\Gamma_1(N)/\Gamma(N)\stackrel{\sim}{\to}\mathbb Z/N\mathbb Z$$ and $$[\Gamma_1(N): \Gamma(N)]=N.$$
3) Let the function $$\Gamma_0(N) \to \big(\mathbb Z/N\mathbb Z\big)^*$$ to be $$\left[ \begin{array}{cc}a & b \\c & d \end{array} \right] \to d \mod N.$$ Since this onto function is an homomorphism and the kernel is $\Gamma_1(N)$, we will have $\Gamma_1(N) \lhd \Gamma_0(N)$ and $$\Gamma_0(N)/\Gamma_1(N)\stackrel{\sim}{\to}\big(\mathbb Z/N\mathbb Z\big)^*$$ and so $$[\Gamma_0(N): \Gamma_1(N)]=\phi(N).$$
Question: How can we show the equality $$[SL_2(\mathbb Z): \Gamma_0(N)]=N\prod_{p\mid N}\left(1+\frac1p\right)?$$
Then this will give us $$[SL_2(\mathbb Z): \Gamma(N)]=N^3\prod_{p\mid N}\left(1-\frac1{p^2}\right).$$