TITLE: If $[G:H]=n$, then $g^{n!}\in H$ for all $g\in G$.
QUESTION [22 upvotes]: I have the following question:

Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$
Then this holds: $g^{n!}\in H$ for all $g\in G$.
Why is this true?

I think, that's not very difficult, but I have no idea at the moment.
Thanks!

REPLY [2 votes]: See also my post here

Proposition. Let $H$ be a non-trivial subgroup of the finite group $G$, with $n = [G:H]$. Assume that $\gcd(|H|,n)=1$. Then the following are equivalent.
  (a) For all $g \in G$: $g^n \in H$.
  (b) $H \unlhd G$.