Thread: proving the sylow p-subgroups are abelian

Assume that is an odd prime and is a finite simple group with exactly Sylow -subgroups. Prove that the Sylow -subgroups are abelian.

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

Thanks,
Hollywood

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 p².

thus the sylow p-subgroups of G have either order p, or p², and are abelian in either case.

Brilliant!!

Thanks,
Hollywood

Deveno - well done!