# 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 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.

• It looks fine, except your action should read $(g, g_iH) \mapsto gg_i H$ – Prahlad Vaidyanathan Sep 2 '13 at 6:39
• @PrahladVaidyanathan : yes,yes that is what i mean . Thank You. :) – user87543 Sep 2 '13 at 6:41
• – Derek Holt Sep 2 '13 at 8:09

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!

• Yes, Yes. Just now i realize this. As in original Question $K\unlhd G$ with $|G:K|\leq n!$ for a simple group there is no proper normal subgroup, we see that $K=(1)$ and so $|G:K|=|G|\leq n!$. – user87543 Sep 2 '13 at 6:58
• Ah,Just now i remember seeing this result "If $n>1$ and if $G$ is a simple group with a subgroup $H\subseteq G$ of index $n$, then $|G|≤n!$"in Herstein Topics in algebra long ago. I did not remember which way he proved but i have got frustrated by the way it was proved there. But, Now, accidentally I got the same result :).. Special thanks to Mr.Jyrki Lahtonen :) – user87543 Sep 2 '13 at 7:04