If $[G:H]=n$, then $g^{n!}\in H$ for all $g\in G$. 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!
 A: In case anyone wanted a nice name for the minimal number, it is basically the obvious guess, though I think it might be a little surprising that the obvious guess is actually right.
Proposition: If G is a group, and H is a subgroup of G, then the exponent of $G/\operatorname{Core}_G(H)$ is the smallest positive integer n such that for all g in G, $g^n \in H$.
Proof: Suppose n is a positive integer such that $g^n \in H$ for all g in G. Let k in G, then $(g^{k^{-1}})^n \in H$ as well. Conjugating by k, $g^n \in H^k$.  In particular, $g^n$ is contained in every conjugate of H, not just H itself.  Hence $g^n$ is contained in $\operatorname{Core}_G(H) = \displaystyle \bigcap_{g\in G}H^g$, the largest G-normal subgroup of H.  In particular, n is a multiple of the exponent of the group $G/\operatorname{Core}_G(H)$.  Conversely, the exponent n of the group $G/\operatorname{Core}_G(H)$ definitely satisfies $g^n \in \operatorname{Core}_G(H) \leq H$. $\qquad \square$
Equivalently, if G is a transitive permutation group, then the smallest positive integer n such that the nth power of every element of G fixes the first point is also the smallest positive integer n such that the nth power of every element of G fixes every point.
A: Take $g\in G.$ Apply the pigeonhole principle to the $n+1$-tuple $g_0,\ldots,g_n,$ inductively defined by $g_0=e,$ and $g_{i+1}=g.g_i.$
You find $g_i$ and $g_j$ in the same left coset of $H$ in $G,$ for some $i,j=0,1,\ldots,n$ with $i<j.$
Consequently $g^{j-i}=g_jg_i^{-1}\in H,$ therefore a fortiori $g^{n!}\in H.$
A: 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) $H \unlhd G$.

A: Let $G$ act on the left cosets of $H$ by left multiplication,
$$g: aH\longmapsto gaH.$$
This gives an action of $G$ on a set of $n$ elements, hence induces a homomorphism $G\to S_n$. In particular, for every $g\in G$, $g^{n!}$ must lie in the kernel, since $|S_n|=n!$. But the kernel is contained in $H$: if $g$ lies in the kernel, then $gaH = aH$ for all $a\in G$, hence in particular $gH=H$, so $g\in H$. Thus, for every $g\in G$, $g^{n!}\in H$.
A: This is an easy question if you start by showing that $g^n \in H$. Consequently, $g$ raised to the power of any multiple of $n$ is also in H (since $[G:H] = n$). Just an idea.
