Sylow Theory Problem: Proving N[N[P]] = N[P] for a Finite Group G and Prime p

[ehrenfest]

[SOLVED] sylow theory problem

Homework Statement

Let G be a finite group and let p be a prime dividing |G|. Let P be a Sylow p-subgroup of G. Show that N[N[P]]=N[P].

Homework Equations

The Attempt at a Solution

Please confirm that this proof is correct.

The Second Sylow Theorem tells us that any two Sylow p-subgroups of a finite group must be conjugate.
The normalizer of P is the subgroup N[P]={$g \in G$ : $gPg^{-1} = P$}. It is obvious that N[P] is the largest subgroup of G in which P is normal.
It follows immediately that P is the only Sylow p-subgroup in N[P].
The normalizer of the normalizer of P is the subgroup N[N[P]] = {$g \in G$ : $gN[P]g^{-1} = N[P]$}. I claim that P is normal in N[N[P]]. If g in N[N[P]], then $gPg^{-1} \subset N[P]$ and since P is the only Sylow p-subgroup in N[P], it follows immediately that gPg^{-1} = P. Thus g is contained in N[P] and N[N[P]] is contained in N[P].

Why did they tell us that p is a prime that divides |G|? Doesn't that follow from the fact that there exists a Sylow-p-subgroup?

 

[NateTG]

Looks kosher.