So my task is to prove that if $N$ is finite group and $G$ is a normal subgroup of $N$ and $(|G| , |N : G|) = 1$ , for every subgroup of N (let's call it F) if $|F|$ divides $|G|$, $F$ is a subgroup of $G$.
So my first approach was with Lagrange's theorem. But I don't know how to use the statement that $(|G| , |N : G|) = 1$. Any tips ?
Take $g\in F$ and consider the standard map $\pi\colon N\longrightarrow N/G$. Then the order of $\pi(g)$ diveds both the order of $g$ and the order of $N/G$. But these numbers are coprime. Therefore, the order of $g$ is $1$. In other words, $g\in\ker\pi= G$.