Let $G$ be a finite simple group and $p$ divides the order of $G$. If $G$ has exactly $n (>1)$ sylow $p$-subgroups, then $G$ is isomporphic to a subgroup of $A_n$.
Thanks in advance!
Let $G$ be a finite simple group and $p$ divides the order of $G$. If $G$ has exactly $n (>1)$ sylow $p$-subgroups, then $G$ is isomporphic to a subgroup of $A_n$.
Thanks in advance!
Your mistake: $G$ need not be a simple normal subgroup of $S_k$. $S_k$ may have many simple subgroups. $A_k$ is just the only normal simple subgroup.
Strong hint / sketch answer:
First we must assume $G$ is not cyclic of prime order otherwise $G$ is simple with one $p$ subgroup but is not $A_1$ (which I guess is the trivial group?).
So $G$ is a non-abelian simple group and therefore has at least $2$ sylow $p$-subgroups.
Let $G$ act on the set of Sylow $p$-subgroups of $G$ by conjugation. This is a map $G\to S_n$. Now you need the map to be injective (this follows from the fact $G$ is simple) and you need $G\le A_n$. Well if $G$ is not in $A_n$ then $G\cap A_n$ is properly contained in $G$...