Using the fact that $A_n$ is simple for $n \geq 5$, prove that the only index 6 subgroups of $A_6$ and $S_6$ is $A_5$ and $S_5$ respectively I have seen many proofs involving the Sylow theorems but I am not allowed to use that. Also, I was told that this is equivalent of saying "$A_5$ is the only simple group of order 60", but I fail to see the connection as well.

Let $H$ be a subgroup of order $60$ in $G = A_6$. Consider $G$ acts on the cosets $[G:H]$.
The kernel of this action is trivial, otherwise the kernel is a non-trivial normal subgroup of the simple group $G=A_6$. Hence this action is faithful. Note that the stabilizer of $H\in[G:H]$ is the whole $H$. $H$ acts on the remaining $5$ cosets.
Let $\varphi:H\to S_5$ represent this action. $\ker \varphi = 1$, since the action is faithful. Therefore, $\varphi(H)$ is a subgroup of order $60$ in $S_5$. The only subgroup of order $60$ in $S_5$ up to isomorphism is $A_5$. (This uses the uniqueness of $A_5$ as a simple group of order $60$.) So we are done.
