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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 $A_{12}$.