Proving a Finite Group is Not Simple

[tom.young84]

I found this problem, and I was wondering if I'm on the right approach.

Let G be a finite group on a finiste set X with m elelements. Suppose there exist a g$$\in$$G and x$$\in$$X such that ^gx not equal to x. Suppose the order of G does not divide m!. Prove that G is not simple.

Would it suffice to show that an isomorphism "f" exists from G to X? Then we just need to prove two cases about the Ker(f). We need to show that Ker(f) can't just be the identity because then it would be an infinite group being isomorphic to a finite group. If the Ker(f)=G, then some stuff. Sorry for the informality, I'm not actually sure what happens if Ker(f)=G.

 

[rasmhop]

Would it suffice to show that an isomorphism "f" exists from G to X?

What would that mean? X isn't necessarily a group, just a set.

Instead try to consider the permutation representation,
$$\varphi : G \to S_X$$
afforded by the group action. We are told that for some g, $\varphi(g)\not= 1$ which tells you $\ker \varphi \not= G$. If $\ker\varphi=1$, then $\varphi$ is an embedding so what is the order of $\varphi(G)$ and how does it relate to $|S_X|$? Use this to show |G| divides $|S_X| = m!$ which is a contradiction.