Let G be a group.

Prove that G has no P-Sylow subgroup 'H' for which [G:N(H)]=2.

For every subgroup H in G, the Normalizer is defined as N(H):={ $\displaystyle g\in G: gH=Hg $ }

I need to find a way to prove this. If |G| is infinity, then for every prime p and natural m, we get that |G| is never $\displaystyle m*p^n$ , therefore there is no p-Sylow subgroups anyway (Is this right? I can't think of any other way to 'take care' of the infinite case).

Thanks