The normalizer of the normalizer of a p-sylow supgroup

[Niall101]

Im trying to prove N(N(P)) = N(P)

So N(P) = set oh h where h^-1Ph = p

Then N(N(P)) = k where k^-1hk = h

the fact that p is a p sylow subgroup gives me what information? I am unsure.

Thanks in advance!

 

[MathematicalPhysicist]

Well I guess if you can prove that one is a subset of the other, and both of them have the same cardinality, then they are equal.

Do you see how you can prove this?

 

[Niall101]

Not really sorry. Both are subgroups of G yes?

If I let H = the normalizer of P
Can I say P is normal in H and that as all sylow subgroups are conjugate H contains no other sylow subgroups then if N(N(P)) moves P somewhere else then there would be 2 P sylow subgroups in H?

Note: Havent been told that P is normal in HThanks very much for you reply

 

[VeeEight]

If P is the only subgroup of H, then P is normal in H

 

[Niall101]

Thanks! So I can do it this way then with that result.