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

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.

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.
• I don't understand two points: 1) why or how would $\mathbb A_8$ act transitively on the four left cosets of $H$? 2) I see that the kernel can't be just the identity element since the order of $\mathbb A_8$ is $\dfrac{8!}{2}$ and the order of the image is less than or equal to $24$, but why the kernel can't be all $\mathbb A_8$? Maybe I am not understanding the underlying action. Commented Jul 9, 2015 at 2:13
• @user156441 The transitive action is as follows. An element $g$ of $A_8$ sends a left coset $qH$ of $H$ to the left coset $(gq)H$. This is transitive because, given any two left cosets, say $qH$ and $rH$, there is some $g$ sending the first to the second, namely $g=rq^{-1}$. If the kernel of the action were all of $A_8$, that would mean every element of $A_8$ sends every left coset to itself, which is certainly not the case (by transitivity, or because the only way $g$ can fix the coset $H$ is to be a member of $H$). Commented Jul 9, 2015 at 2:18
• So, I guess you've basically proven that if a simple group $G$ has a subgroup $H$, then $|G| \leq [G:H]!$ Commented Jul 9, 2015 at 2:20
• @AlexZorn Right (except for $H=G$ of course). I expect this general form of the result is in some standard sources, but I'm away from home and can't easily check. Commented Jul 9, 2015 at 2:26