If $HIf $H$ is a subgroup of a finite group $G$, and if $|G|=m|H|$, then for all $g\in G$ we have $g^{m!}\in H$.
The question suggests I adapt the proof of Lagrange's Theorem in the book (Groups and Symmetry, MA Armstrong). Below is my attempt.
Assume $H$ is a proper subgroup of $G$ and let $g\in G-H$. The map $g \mapsto gh$ for $h\in H$ is a bijection between $H$ and $gH$ and so $|H|=|gH|$. We can do this only so many times since there are only finitely many elements in $G-H$.
Am I on the right track?
 A: Let G act on the set of left cosets of H via left multiplication. This action affords a homomorphism from G to $S_{m}$ with kernel K being contained in H. Let G be an arbitrary element of G, then gK is an element of the quotient group G/K whose order divides m!. Hence $(gK)^{m!} = K$, which implies that $g^{m!}$ is in K and hence it is in H. 
A: Not sure if your last statement will lead you to the conclusion you want to arrive at.
Idea: (which can be generalized easily)
Suppose $|G|=3|H|$, then consider $g \in G$.


*

*If $g \in H$, then for sure $g^{3!} \in H$.

*If $g \not\in H$, then $gH$ is a coset different from $H$. Then ask yourself, where is $g^2$? It cannot be in $gH$, so it must be in $H$, hence $(g^{2})^{3}=g^{3!} \in H$.

*If $g^2 \not\in H$, then $H,gH,g^2H$ are distinct cosets. Ask where is $g^3$?


Hope this will help you generalize.
A: You're on the right track. Here is a formalization of your argument using group theory.
The set of left cosets $G/H$ has $m$ elements. Therefore its group of permutation $P$ has order $m!$. The map $f_g:G/H\to G/H:kH\mapsto ghK$ is a bijection, therefore $f_g \in P$. So we have that $f_g^{m!}=Id_P$. 
It follows that $f_g^{m!}(H)=g^{m!}H=H$. Therefore $g^{m!}\in H$. 
