If G is a group and N is a normal subgroup of G, show that if a in G has finite order o(a), then Na in G/N has finite order m, where m|o(a). (Prove this by using the homomorphism of G onto G/N.)
Let $\displaystyle \pi : G\to G/N$ be the natural projection homomorphism i.e. $\displaystyle g\mapsto gN$.
Let $\displaystyle a\in G$ then order of $\displaystyle \pi (a)$ must divide order of $\displaystyle a$ by the properties of homomorphism.
Thus, we see that $\displaystyle m | \text{o}(a)$.