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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 $G$?

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Thanks

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You are (mostly) correct. If ker(f) = G, then the image of G in the symmetric group is just the identity, so G does not move any of its Sylow p-subgroups around. However, G acts transitively on its Sylow p-subgroups (they are all conjugate) and the identity is not transitive unless p+1=1, which is silly.

If ker(f) = 1, then G embeds in the symmetric group on p+1 points, so its order divides (p+1)!. This is not a contradiction when k=1.

Indeed, taking p=2, the non-abelian group of order six has order p(p+1) and has exactly p+1 Sylow p-subgroups, and the homomorphism f is injective.

Of course a group of order p(p+1) is also not simple, but probably for a different reason.

“Now here’s my question: assume ker(f)=G this would imply then that G has a unique p-Sylow subgroup no? “

yes, the group action (conjugation on sylow subgroups) is transitive.

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