The normalizer of the normalizer of a p-sylow supgroup

1. Oct 29, 2009

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.

2. Oct 29, 2009

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?

3. Oct 29, 2009

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 H

Thanks very much for you reply

Last edited: Oct 29, 2009
4. Oct 29, 2009

VeeEight

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

5. Nov 2, 2009

Niall101

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