If $H$ is a subgroup of $G$ of index $n$, then $g^{n!} \in H \ \forall g \in G$

Question

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.

Answer 1

Do it in two steps.

First you prove that there is normal subgroup $K$ of $G$ that is contained in $H$, and has index a divisor of $n!$

For this, as in the previous answer, you let $G$ act on the left cosets of $H$ by multiplication, regard this action as a homomorphism $G \to S_{n}$, and consider the kernel $K$.

Then you consider the factor group $G/K$, and use the fact that if a group $L$ has order $m$, then $g^{m} = 1$ for each $g \in L$.

Answer 2

Hint Consider the left cosets $X=G/H$ and the group action of $G$ on this set.

Since $X$ has $n$ elements, what can you say about the orbit of $g$ under this action?

Answer 3

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 $[G:H]=n$.