Let $G$ be a finite group and $N$ a normal subgroup of $G$. Let $\phi:G\rightarrow G/N$ be the homomorphism of $G$ onto $G/N$.
Suppose $X\leq G/N$. Show that $[G:\phi^{-1}(X)]=[G/N:X]$.
$X\leq G/N$ implies that $N\trianglelefteq\phi^{-1}(X)\leq G$ and $[\phi^{-1}(X):N]=|X|$. I don't know what to do next.
Well, that's it, you sould just write explicitly $\phi^{-1}(X)$ to make things clear. By Correspondence Theorem, there esists $K\leq G$ containing $N$ such that $K=\phi^{-1}(X)$. This means that you can write $X=K/N$. As you are dealing with finite groups, by Lagrange,
$$\begin{align} [G:\phi^{-1}(X)]&=[G:K]\\ &=\frac{|G|}{|K|}\\ &=\frac{|G|/|N|}{|K|/|N|}\\ &=\frac{|G/N|}{|X|} \\ &=[G/N:X]. \end{align}$$