Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following is a question from Dummit & Foote.

Prove that if $U$ and $W$ are normal subsets of a Sylow $p$-subgroup $P$ of a finite group $G$ then $U$ is $G$-conjugate to $W$ if and only if $U$ is $N_G(P)$-conjugate to $W$.

Ofcourse G-conjugate means, there exists a $g \in G$ such that $gUg^{-1}=W$, and for $N_G(P)$ conjugate the element is restricted to $N_G(P)$.

The reverse implication is obvious. But I havent been able to prove the implication. D&F gives the hint that $N_P(U)=N_P(W)=P$, but I have not idea how this can be used to get information about $N_G(P)$.

Any help will be appreciated.

Thanks!

share|improve this question
    
Normal subsets!!? :-) –  Babak S. Feb 17 '13 at 6:30
    
@BabakSorouh "Normal subset" means a subset closed under conjugation. –  Ted Feb 17 '13 at 6:34
    
@BabakSorouh: Yes. That's what the question says. The internet defines normal subsets as groupprops.subwiki.org/wiki/Normal_subset –  user23238 Feb 17 '13 at 6:35

1 Answer 1

up vote 1 down vote accepted

Here's a sketch, there are a few details left out for you to fill in:

  1. Assume $gUg^{-1} = W$ for some $g \in G$.
  2. Show that $gPg^{-1} \subseteq N_G(W)$.
  3. Use the Sylow theorems to get that $xgPg^{-1}x^{-1} = P$ for some $x \in N_G(W)$.
  4. Observe that $xg \in N_G(P)$ and $xgUg^{-1}x^{-1} = W$.
share|improve this answer
    
Thanks! I have worked through the steps. But please could you provide me some motivation or insight on the logical flow that lead you to the solution. I am finding it a bit hard to justify the steps. For e.g. Why did you try to look for a subset of $N_G(W)$, and how did you come with $gPg^{-1}$. If it is obvious then i I excessively dumb. –  user23238 Feb 17 '13 at 9:08
    
I think it is impossible for me to have solved this, even if I put a day into it. –  user23238 Feb 17 '13 at 9:11
    
I was plying around with the general formula $xN_G(H)x^{-1} = N_G(xHx^{-1})$ when I got $gPg^{-1} \subseteq N_G(W)$. After that it's a common tactic to use the fact that if you have a tower $P \subseteq H \subseteq G$ and $P$ is a $p$-Sylow subgroup of $G$ then it is also a $p$-Sylow subgroup of $H$. This is the only theorem I could imagine allows to to choose a conjugating element in a smaller subgroup. –  Jim Feb 17 '13 at 16:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.