If $G$ is a group, $H$ is a normal subgroup of $G$, $[G:H]=k,{\rm ord}_G(a)=n,a\in G,|\langle a\rangle\cap H|=m,$ prove: The term of $\langle a\rangle$ in $G/H$ is $\frac{n}{m}$.
I have some thoughts but to be honest I do not think I solve it completely.
First, we can prove that: $H\cap\langle a\rangle\unlhd G\cap\langle a\rangle$. It's easy and trivial.
Then attention to $[(\langle a\rangle\cap G)\backslash(\langle a\rangle\cap H]=[(\langle a\rangle\cap G):(\langle a\rangle\cap H]=\frac{|\langle a\rangle\cap G|}{|\langle a\rangle\cap H|}$.
So we can prove that $\langle a\rangle \cap (G\backslash H)= (\langle a \rangle \cap G)\backslash(\langle a \rangle\cap H)$.
But I do not know how to solve it.