# proving the sylow p-subgroups are abelian

• Sep 13th 2012, 06:55 PM
hollywood
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
• Sep 15th 2012, 10:23 PM
Deveno
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 S2p+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.
• Sep 15th 2012, 11:24 PM
hollywood
Re: proving the sylow p-subgroups are abelian
Brilliant!!

Thanks,
Hollywood
• Sep 17th 2012, 07:57 PM
johnsomeone
Re: proving the sylow p-subgroups are abelian
Deveno - well done!