Non-abelian, simple subgroups of $S_n$ I am trying to prove, as part of a larger theorem, that if $G$ is a non-abelian finite, simple group of order $>2$ and $G$ is a subgroup of $S_n$, then $G$ must be a subgroup of $A_n$.
Any ideas where along which lines I should be thinking? I am allowed to assume basic group theory and Sylow's theorem.
 A: Hint : consider the restriction of the signature to $G$, as a morphism $G\to \mathbb{Z}/2\mathbb{Z}$. Can it be injective ?
A: Another approach. 
Observe that for $\;n=3,4\;$ you can check it directly, as there are no non-abelian simple subgroups of $\;S_n\;$ (the closest thing is a $\;2\,-$ subgroup of order eight in $\;A_4\;$ which of course isn't simple). Suppose thus that $\;n\ge5\;$ :
$$G\rlap{\;\,/}\subset A_n\implies [G:G\cap A_n]=2\implies G\cap A_n\lhd G\implies $$
either $\;G\cap A_n=1\;$ or $\;G\cap A_n=G\;$ as $\;G\;$  is simple. The second relation leads directly to the contradiction $\;G\subset A_n\;$ , whereas the first one leads us either to the fact that we can form the semidirect product $\;A_n\rtimes G\;$ which would have to be equal to $\;S_n\;$ by order considerations and thus $\;G\cong C_2\;$, contradiction, or else, knowing that any subgroup of $\;S_n\;$ is either contained in $\;A_n\;$ or exactly half its elements are even permutations, we'd get $\;G\cap A_n=1\implies G=1\;$
