How to prove that O(Ng) | O(g)

Question

I have this exercise:

Let $G$ be a group and $N$ a normal subgroup of $G$. Show that for all $Ng\in G/N$, $$o(Ng)\mid o(g).$$

For now, without using the canonic homomorphism $\tau \left(g\right)=Ng$.

Sorry but don't have any directions or ideas, that's new for me.

Any suggestions?

Answer 1

$$(Ng)^{\operatorname{ord}(g)} = Ng^{\operatorname{ord}(g)} = Ne = N$$