I have this proof for my problem but I do not understand $g^{m}N=(gN)^{m}=N$, and how come Lagrange plays a role here.
Let G be a finite group and let N be a normal subgroup of G.
Suppose that the order n of N is relatively prime to the index $|G:N|=m$.
Prove that $N=\{a\in G| a^{n}=e\}$.
Proof:
Note: since m and n are relatively primes: $\exists$ s,t such that 1=ms+nt
Note: Since $|G/N|=|G:N|=m$ then $g^{m}N=(gN)^{m}=N,\ \forall g\in G$ by Lagrange Theorem. Thus $g^{m}\in N$
As such, $a= a^{ms+nt}=a^{ms}a^{nt}=a^{ms}=(a^{s})^{m}\in N$. This proves $\{a\in G | a^{n}=e\} \subset N$.
On the other hand if $a\in N$, then we have $a^{n}=e$.
So $N \subset \{a\in G | a^{n}=e\}$.
So we get the equality.
Because $k:=\text{ord}(gN)|m$ (by lagrange) and $(gN)^k = N$, so $(gN)^m = ((gN)^k)^{m/k} = N$ (since $m/k$ is integer).