There is no injective morphism from $\mathbb S_7$ to $\mathbb A_8$

Question

I am trying to show that it doesn't exist an injective group morphism $f:\mathbb S_7 \to \mathbb A_8$. If there is an injective morphism from $\mathbb S_7$to $\mathbb A_8$ then $\mathbb S_7$ is isomorphic to a subgroup of $\mathbb A_8$. With this in mind, if there exists an element $x$ of order $n$ in $\mathbb S_7$, then $f(x)$ has the same order in $\mathbb A_8$. I was trying to arrive at a contradiction but I couldn't find the appropiate element. I would really appreciate some suggestions or hints.

Answer 1

This is probably overkill, but here goes: In view of the orders of $S_7$ and $A_8$, it suffices to show that $A_8$ has no subgroup $H$ of index $4$. If there were such an $H$, then $A_8$ would act transitively on the set of four left cosets of $H$, and the kernel of that action would be a normal subgroup of $A_8$ of index at least $4$ and at most $4!=24$. That contradicts the fact that $A_8$ is simple.