I have the following question:
Let $n\geq5$, and suppose that $G$ is a simple subgroup of $S_{n+1}$ of index $k$. Show that if $k\leq2n+2$, then $G=A_{n+1}$ or $G$ is isomorphic to $A_n$.
I have managed to show by using the simplicity of $G$ that $G$ must be a subgroup of $A_{n+1}$. From here, we see that if $G=A_{n+1}$ then we are done. Else, we would like to say something along the lines of $G$ being a subgroup of $A_n$. The problem is that there are $n+1$ subgroups of $A_{n+1}$ that are isomorphic to $A_n$, and as such it would be strange for me to write $G$ as a subgroup of $A_n$. Alternatively, I am trying to show that $G$ can be embedded into $A_n$, so that the statement would follow rather easily. However, I am unable to show why $G$ can be embedded into $A_n$.
Is there a way to show that there must exist some injective homomorphism from $G$ to $A_n$, or is there a way to show that all elements of $G$ must somehow fix an element of $\{1,2,\cdots,n+1\}$?
Well $|A_{n+1}:G| = k/2 \le n+1$ and if $k/2>1$ then, by simplicity of $A_{n+1}$, the permutation action of $A_{n+1}$ on the $k/2$ cosets of $G$ in $A_{n+1}$ is faithful, and hence $A_{n+1}$ embeds into $A_{k/2}$. So we must have $k/2 = n+1$ and, since $G$ is a point stabilizer in the action on cosets, we have $G \cong A_n$.
If $n+1 = 6$ then there is an outer automorphism of $A_6$ that maps $A_5$ to a transitive subgroup of $A_6$ (i.e. ${\rm PSL}(2,5)$), and so $G$ might not fix a point. For $n+1>6$ there is no such outer automorphisms, and all subgroups of $A_{n+1}$ isomorphic to $A_n$ must fix a point.