# Sylow theory problem

1. Mar 4, 2008

### ehrenfest

[SOLVED] sylow theory problem

1. The problem statement, all variables and given/known data
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].

2. Relevant equations

3. 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?

Last edited: Mar 4, 2008
2. Mar 4, 2008

### NateTG

Looks kosher.