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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]={[itex]g \in G[/itex] : [itex]gPg^{-1} = P[/itex]}. 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]] = {[itex]g \in G[/itex] : [itex]gN[P]g^{-1} = N[P][/itex]}. I claim that P is normal in N[N[P]]. If g in N[N[P]], then [itex] gPg^{-1} \subset N[P] [/itex] 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?

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# Sylow theory problem

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