proving the sylow p-subgroups are abelian

Assume that $\displaystyle p$ is an odd prime and $\displaystyle G$ is a finite simple group with exactly $\displaystyle 2p+1$ Sylow $\displaystyle p$-subgroups. Prove that the Sylow $\displaystyle p$-subgroups are abelian.

I haven't made much progress: It's easy if $\displaystyle p^2$ doesn't divide the order of $\displaystyle G$, since then the Sylow $\displaystyle p$-subgroups have order $\displaystyle p$ and are therefore cyclic.

Thanks,

Hollywood

Re: proving the sylow p-subgroups are abelian

i figured it out:

let G act on the sylow p-subgroups by conjugation.

this gives a homomorphism of G into S_{2p+1}.

since G is simple, and the action is transitive, the kernel of this action must be trivial.

therefore |G| divides (2p+1)!.

now we can ask, what is the highest possible power of p that divides (2p+1)! ?

well, the only integers 2 ≤ n ≤ 2p+1 that p divides are p and 2p. this means that the highest possible power of p that divides (2p+1)! is p^{2}.

thus the sylow p-subgroups of G have either order p, or p^{2}, and are abelian in either case.

Re: proving the sylow p-subgroups are abelian

Brilliant!!

Thanks,

Hollywood

Re: proving the sylow p-subgroups are abelian