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

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!

• Use a single dollar sign $ for inline symbols, like $x$ for$x$instead of $$x$$ for $$x$$ – Shaun Dec 11 '19 at 19:29 • Yes,i’ve Done. Thank you! – Francesco_Trig Dec 11 '19 at 19:33 • "The left cosets of$\frac{G}{H}$" should be "the left cosets of$H$in$G$" (it's not the cosets of the quotient, which may not even be defined). The hint in the book is suggesting that you use the action to define a homomorphism into$S_n$, and look at the kernel of that. It's the same thing you are doing, as your$K\$ is exactly the kernel of that action. – Arturo Magidin Dec 11 '19 at 19:34

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 $$n!$$.
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 $$n!$$.