Let $G$ be a group and $N\trianglelefteq G$ with $[G:N]=4$. Show that there is an $N'\trianglelefteq G$ with $[G:N']=2$.
Proof. We have $|G/N|=4$, so there must exist a $M\trianglelefteq G/N$ of order 2. This corresponds to a $N'\trianglelefteq G$ with $N\subseteq N'$. We now have $4=[G:N]=[G:N']\cdot[N':N]$. How can I obtain $[N':N]=2$ from $|M|=2$?