Finite Group Theory--Sylow SubgroupsDate: 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/ |
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