Show that $a^m$ is in $H$ for every $a$ in $G$

Question

Let $H$ be a normal subgroup of $G$, and let $m=(G:H)$. Show that $a^m \in H$ for every $a \in G$.

I have been thinking about this question for a few days but I get something informal.

What am I missing?

1. There are $m$ $H$-cosets.
2. If we consider the cosets of $e$, $a$, $a^2$, $\cdots$, $a^{m-1}$, we get $m$ distinct cosets(actually I claim this without a valid proof, any hints?) Then the coset of $a^m$ should be one the above coset. (Then why must it be $eH$?)

Thanks in advance.

Answer 1

let xH be any element of G/H, then (xH)^m=H as |G:H|=n this implies that x^nH=H then x^n belongs to H for all x in G.

Answer 2

Hint.

Do you know the case $H=1$?

Does the general case reduce to this case?

Answer 3

Hint: look at $G/H$, being a group of order $m$. What do you know about orders of group elements and the order of the group they live in? Joseph-Louis Lagrange could help.