Let $G$ be a finite simple group of order $n$. Prove that if $n \geq 3$, then $G$ is isomorphic to to a subgroup of $A_n$, the alternating subgroup of the symmetric group $S_n$.
My idea here was to use the First Isomorphism Theorem. If we could construct a homomorphism $\phi: G \longrightarrow H$, where $H$ is a subgroup of $A_n$ ($n \geq 3$), such that $\phi$ is surjective (so that $im(\phi)$ is exactly $A_n$) and $\ker(\phi) \neq G$ (so that, since $\ker(\phi)$ is normal in $G$ and $G$ is simple, $\ker(\phi)$ would have to be trivial), it would follow that $G/\ker(\phi) = G \cong im(\phi) = H$.
How can one construct such a homomorphism ? Can we use the fact that every finite group is isomorphic to a subgroup of $S_n$, and go from there ?