TITLE: let $H\subset G$ with $|G:H|=n$ then $\exists~K\leq H$ with $K\unlhd G$ such that $|G:K|\leq n!$ (Dummit Fooote 4.2.8)
QUESTION [6 upvotes]: Question is to prove that
For $H\subset G$ with $|G:H|=n$, $\exists~K\leq H$ with $K\unlhd G$ such that $|G:K|\leq n!$ 
What i have done so far is that : 
$H$ be a subgroup of index $n$ in $G$ and let $\{g_i :1\leq i\leq n\}$ be its coset representatives.\
Consider the action of  $G$ on set of left cosets $G\times \{g_i H:1\leq i \leq n\}\rightarrow  \{g_i H:1\leq i \leq n\}$.
In  other words we have the action  $G\times \{1,2,3,...n\}\rightarrow \{1,2,3,...n\}$.
For each $g\in G$ acting on $\{1,2,3,...n\}$ we have image in $\{1,2,3,...n\}$.
So,each $g\in G$ gives a permutation in $\{1,2,3,...n\}$ So, we have $G\rightarrow S_n$ a homomorphism. 
As $\eta : G\rightarrow S_n$ is a homomorphism, $Ker(\eta)$ would be a normal subgroup of $G$ and by Isomorphism theorem we have $G/Ker(\eta)$ is isomorphis to subgroup of $S_n$.
Set $K=Ker(\eta)$, we see that $K\leq H$ and $G/K\cong M$ where $M\leq S_n$.
As $|S_n|=n!$ we see that $|G/K|\leq n!$ and so, $|G:K|\leq n!$.
Infact $|G:K|$ divides $n!$ which is not asked to prove in the Question.
So, I am wondering whether my approach is fine or i have just proved something more.
Please look at this as just a proof verification Question. 
Thank You.

REPLY [2 votes]: Yes, you are correct. Also, note the following corollary (since you've been thinking about simple groups recently):
Corollary
If $n>1$ and if $G$ is a simple group with a subgroup $H\subseteq G$ of index $n$, then $\left|G\right|\leq n!$. 
The Corollary is a non-simplicity result because its contrapositive gives you conditions under which a group cannot be simple.
Exercise 1
Let $G$ be a simple group and let $p$ be a prime number. If $n_p$ is the number of Sylow $p$-subgroups of $G$, then prove that $\left|G\right|\leq n_p!$ or else $\left|G\right|=p$.
I hope this helps!