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.


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 –  ramanujan_dirac 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. –  ramanujan_dirac 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. –  ramanujan_dirac 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


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.