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ehrenfest
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[SOLVED] sylow theory problem
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].
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?
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?
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