Given a subgroup $H $, find a normal subgroup of $G$ with index less than or equal to $n!$

Question

Let $G$ be a group and $H$ a subgroup with finite index $n$. Prove that $G$ has a normal subgroup $N$ such that $N\subseteq H$and $|G: N| \le n!$.

Answer 1

$G$ acts on $G/H$ by left multiplication and this action is sujective, you deduce a morphism $f:G\rightarrow S_n$ which is not trivial, take $N$ the kernel of $f$.

Answer 2

You don't show any self work so I can't know what you've tried and what you know, so I'll give the answer and you find out whether it is true and its properties:

$$N=\bigcap_{g\in G} g^{-1}Hg$$

and it gets even better, since in fact $\;N\lhd G\;$ .