Normal subgroup of index a divisor of n! [duplicate]

Question

I’ve found this exercise on “basic algebra”:

Show that if a finite group $G$ has a subgroup $H$ of index $n$ then $H$ contains a normal subgroup of $G$ of index a divisor of n!.

My attempt: I’ve used the action of $G$ on left cosets of $\frac{G}{H}$. This action is transitive so every stabilizer,that is a subgroup of $G$,has index $n$. Now I consider $$K=\cap Stab (r_iH).$$ So K is a subgroup of $G$ and $K \subset H$ , because $ stab (1H)=H$. $K=\{g \in G| rgr^{(-1)} \in H\}$ so K is a normal subgroup of $H$, because for every $g \in K$ and for every $h \in H$, $hgh^{(-1)} \in H$.

I am pretty sure that this is the normal subgroup searched, because the hint of the book says “consider the action of $G$ on $ \frac{G}{H}$ by left translations. But I am not able to try that the index of $K$ in $G$ is a divisor of n!

Answer 1

Whenever a group $G$ acts on a set with $n$ elements, this gives us a homomorphism $\phi:G\to S_n$. Each element $g\in G$ will permute the $n$ elements, and $\phi(g)$ is the permutation you get from letting $g$ act on the elements.

Note that $\ker(\phi)$ is the intersection of all the stabilizers of the action. So the intersection of all the stabilizers will be a normal subgroup of $G$.

Also note that the index of $\ker(\phi)$ is $|\mathrm{im}(\phi)|$. Since $\mathrm{im}(\phi)$ is a subgroup of $S_n$, the index of $\ker(\phi)$ divides $n!$.

Answer 2

What you have defined is a homomorphism $\varphi: G \rightarrow S_{G/H}$, where $S_{G/H}$ is the group of permutations of the set $G/H$. And the group $K$ is simply the kernel of $\varphi$.

Therefore the quotient group $G/K$ is isomorphic to the image of $\varphi$, which is a subgroup of $S_{G/H}$. Since the set $G/H$ has $n$ elements, the group $S_{G/H}$ has $n!$ elements, and hence the image of $\varphi$, being a subgroup, has cardinal dividing $n!$.