TITLE: Simple group of order $660$ is isomorphic to a subgroup of $A_{12}$
QUESTION [13 upvotes]: Prove that the simple group of order $660$ is isomorphic to a subgroup of the alternating group of degree $12$. 

I have managed to show that it must be isomorphic to a subgroup of $S_{12}$ (through a group action on the set of Sylow $11$-subgroups). Any suggestions are appreciated!

REPLY [7 votes]: Let $G$ be our simple group of order $660$, contained in $S_{12}$. Since $A_{12}$ is normal in $S_{12}$, the subgroup $A_{12} \cap G$ is normal in $G$. Since $G$ is simple and any subgroup of $S_{12}$ with more than two elements  contains an even permutation, we get $A_{12} \cap G = G$.
To generalize, suppose $G$ is a simple group with $|G| > 2$. With the same idea as before, you can show that if $G$ is isomorphic to a subgroup $H$ of $S_n$, then $H$ is a subgroup of $A_n$.