# Normal subsets of a Sylow p subgroup are conjugate if and only if they are $N_G(P)$ conjugate.

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!

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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. 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 $xgUg^{-1}x^{-1} = W$.
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 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