Let $G$ be a finite group and $H$ a subgroup of $G$ of index $n$, i.e., $[G:H]=n$. Prove that $$\forall g \in G,\; g^{n!} \in H.$$
This is a question I've had in a past exam for Group Theory and I'm really struggling to come up with a solution. There were two hints given in the question, namely that $\# G = n \cdot \# H$ (which is just from Lagrange's Theorem), and we were told to recall that $\#S_{n} = n!$.
I've thought about using Cayley's theorem, that every group is isomorphic to a subgroup of $S_{n}$, but can't figure out how to get the desired result.
Best Answer
Alternatively, one can argue by contradiction.
So suppose there is a $g\in G$ with the property $g^{n!}\notin H$.
Then neither of the elements $g,g^{2},...,g^{n}$ can belong to $H$. This implies that $$H, Hg, Hg^{2},...,Hg^{n}$$ are distinct Right-cosets. But this violates the assumption $[G:H]=n$.