Simple group of order $660$ is isomorphic to a subgroup of $A_{12}$ 
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!
 A: 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 $A_n$.
A: This is a comment made into an answer.
Using the Sylow theorems, if $G$ is a simple group of order $660=60\times 11$, the number of Sylow-$11$ subgroups $n_{11}$ has $n_{11}\equiv 1~\mathrm{mod}~11$, $n_{11}|60$ and $n_{11}\neq1$ by simplicity, so $n_{11}=12$. Again, by Sylow's theorem,  $G$ acts transitively on the $12$-element-set of its $11$-Sylows by conjugation, and so we get a non trivial morphism $G\rightarrow S_{12}$. By simplicity of $G$, this is an embedding.
Now apply the signature morphism $\epsilon:S_{12}\rightarrow\lbrace-1,+1\rbrace$:
$$G\hookrightarrow S_{12}\rightarrow \lbrace-1,+1\rbrace$$
The kernel is a normal subgroup of $G$. Since $G$ is simple, it is either $G$ or $1$, and by cardinality reasons it has to be $G$. Thus the embedding $G\hookrightarrow S_{12}$ maps $G$ into $\mathrm{Ker}(\epsilon)=A_{12}$, and $G$ is isomorphic to a subgroup of $A_{12}$.
