If $H \lhd G$ and $[G:H] = n,$ then $a^n \in H$ for all $a \in G$

Question

If $H \lhd G$ and $[G:H] = n,$ then $a^n \in H$ for all $a ∈ G.$

I know that $H$ is a normal subgroup of $G$ with the index of $G:H = n.$ This means that $|G|/|H|=n,$ so $|G|=|H|n. $

I attempted to say that $a^{|G|} = a^{|H|}n = e,$ the identity, but have gotten stuck here, and have also tried to interpret the question in the quotient group $G/H.$

If someone could provide a thorough proof I'd appreciate it.

Answer 1

There is a stronger statement:

$$n=[G:H]=|G/H|$$ In other words, there is a homomorphism $\phi:G \to G/H$, and for any $\phi(a) \in G/H$, we have that $\phi(a^n)=\phi(a)^n=1$ which occurs if and only if $a^n \in \ker \phi$ or, $a^n \in H$.