Existence of finite indexed normal subgroup for a given finite indexed subgroup. [duplicate]

Prove that if H has finite index n then there is a normal subgroup N of G with $N \subset H $ and [G:N]$ \le $n!.

I tried to solve the problem but could not done exactly. Since [G:H]=n , Let A={$a_i H$:i=1,2,..,n}. Consider the left multiplication group action on A .Then kernal of the action is intersection N= $\cap a_iHa_i^{-1}$ i=1,2..,n. Then N is the largest normal subgroup of G contained in H. Now it is clear that $[G:a_iHa_i^{-1}]$ =n for all i.{by orbit stabliser theorem}.Now how to find [G:N]? yes I got it. By Caley's theorem G/N is isomorphic to a subgroup of $ S_{A}$ .Therefone [G:N] divides n!.

Solution 1:

Define a homomorphism $F$ from $G$ to the group $\{f: G/H \to G/H| f \mbox{ is bijective}\}$ and look at its kernel ($N$ is to be related with it).