I want to show that there are no simple groups of order $p^{k}(p+1)$ where $k>0$ and $p$ is a prime number.
So suppose there is such a group. Then if we let $n_{p}$ denote the number of $p$-Sylow subgroups of $G$ we have that $n_{p}=p+1$. Now by letting $G$ act on $Sylow_{P}(G)$ by conjugation we obtain a group homomorphism $G \rightarrow S_{p+1}$. Since $G$ is simple then either $ker(f)$ is trivial or all $G$. Now here's my question: assume $ker(f)=G$ this would imply then that $G$ has a unique $p$-Sylow subgroup no? but then such subgroup is normal which contradicts the fact that $G$ is simple. So the map in fact is injective but then $|G|$ divides $(p+1)!$ which cannot be.
Basically my question is if my argument is correct, namely thta if $ker(f)=g$ implies the existence of a unique $p$-Sylow subgroup which implies such subgroup is normal in $G$ which cannot be. In case this is wrong, how do you argue that $ker(f)$ cannot be all $G$?
Thanks