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Finite Group Theory--Sylow Subgroups

Date: 07/09/2004 at 19:32:02
From: Alesia
Subject: Finite Group Theory- Sylow subgroups

Hi Dr. Math,

I hope you can help me with this since this problem's been bugging me
for a while.  It seems as if it should be straightfoward, but I'm
having trouble cracking it.

Let U and W be SUBSETS of a Sylow p-subgroup P (of a group G) such
that U and W are normal in P, i.e., the normalizer in P of U is P (N_P 
(U) = P) and N_P (W) = P.

Show that U is conjugate to W in G iff U is conjugate to W is the 
normalizer of P. 

I think that if we assume that there does exist an element g in G 
which conjugates U to W, we know that g is not an element of P.  But,
does this same g have to be in the normalizer of P, or does the 
existence of such a g only guarantee that an element of the normalizer
of P also conjugates the two subsets?  

Is there some underlying property of being normal as a subset that
tells us that if we move U to W, then we must move P to P under this
action of conjugation?  

I tried looking at the subgroups generated by U and W, at the cosets
they form, and at some of the permutation representations afforded by
conjugation by various subgroups of G.  But, I still can't get this
problem.  Am I just missing something which is obviously true, or is
this much tougher than I thought? 



Date: 07/10/2004 at 06:25:39
From: Doctor Jacques
Subject: Re: Finite Group Theory- Sylow subgroups

Hi Alesia,

First, note that the "if" part is obvious--only the "only if" part 
requires some thought.

Let us write x^g for the conjugate of x by g, i.e., g^(-1)xg.  Let us 
also simply write N(X) for N_G(X).

We assume that there is a g in G such that U^g = W, and we must show 
that there is an element h in N(P) such that U^h = W (as you suspect, 
we do not need to prove that g = h).

As U is a normal subset of P, U^g = W is a normal subset of P^g = Q, 
because automorphisms preserve normal subsets.

The normalizer N(W) is a subgroup of G and contains P (by hypothesis) 
and Q.  This means that P and Q are Sylow subgroups of N(W).  As Sylow 
subgroups are conjugate to each other, there is an element z in N(W) 
such that:

  Q^z = P
  P^(gz) = P

and this shows that gz is in N(P).  On the other hand, as z is in 
N(W), we also know that W^z = W.

Can you continue from here?  Please feel free to write back if you 
require further assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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