[Math] Order and index of a normal subgroup $N$ are relatively prime

Question

Let $N$ be a normal subgroup of a finite group $G$. Assume that the order of $N$ and the index of $N$ in $G$ are relatively prime.
Prove that if $g\in G$ satisfies $o(g)\mid o(N)$, then $g\in N$.

I tried using some properties from number theory, but didn't get anywhere.
Any idea?

Answer 1

Note that $(gN)^{o(g)}=(gN)^{|G/N|}=eN$, and $o(g)$ and $|G/N|$ are relatively prime. There exist integers $m$ and $n$ such that $1=m\cdot o(g)+n|G/N|$. So $gN=(gN)^{m\cdot o(g)+n|G/N|}=eN$ and thus $g\in N$.