## A fact regarding conjugate normal subsets of a Sylow subgroup in a finite group

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 $W$. Deduce that two elements in the center of $P$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$. (A subset $U$ is normal in $G$ if $N_G(U) = G$.)

Let $U,W \subseteq P$ with $N_P(U) = P$ and $N_P(W) = P$.

The $(\Leftarrow)$ direction is immediate. To see $(\Rightarrow)$, suppose $U$ is $G$-conjugate to $W$; then there exists $g \in G$ such that $g^{-1}Ug = W$ and $gWg^{-1} = U$. Note that $N_P(g^{-1}Ug) = g^{-1}N_{gPg^{-1}}(U)g$. Now $N_G(P) = N_G(N_P(W))$ $N_G(N_P(g^{-1}Ug))$ $= N_G(g^{-1}N_{gPg^{-1}}(U)g)$ $= g^{-1}N_G(N_{gPg^{-1}}(U))g$. (@@@)

Now suppose $x,y \in Z(P)$. In this case, $N_P(\{x\}) = C_P(\{x\}) = P$ and $N_P(\{y\}) = C_P(\{y\}) = P$, so that $\{x\}$ and $\{y\}$ are normal in $P$. By the above argument we have that $x$ and $y$ are $G$-conjugate if and only if they are $N_G(P)$-conjugate.

• nhanttruong  On July 5, 2011 at 10:37 pm

In the second paragraph, you said $N_G(N_P(g^{-1}Ug)) = N_G(g^{-}N_P(U)g).$ I can see that $N_G(N_P(g^{-1}Ug)) \supset N_G(g^{-}N_P(U)g),$ which is all you need for this proof. But how did you get the other inclusion?

• nbloomf  On July 6, 2011 at 6:20 am

Good question. I’m not convinced now that that is true.

I’ve got another proof that appears to work, though.

Thanks!

• nhanttruong  On July 6, 2011 at 3:16 pm

I have a problem showing $g^-N_P(U)g = N_G(g^-Ug).$ $\subset$ is clear, but for the new proof you need $\supset.$ It’s the same problem I had with showing the other inclusion in my previous comment.

Your original proof didn’t work because the equalities might not have been true. But, the supersets were obvious, and they were sufficient. Here’s the same proof with equalities replaced by supersets.

Let $U = g^- Wg.$ Then $N_G(P) = N_G(N_P(U)) = N_G(N_P(g^-Wg)) \supset N_G(g^-N_P(W)g) \supset g^-N_G(N_P(W)) = g^-N_G(P)g,$ so $g\in N_G(N_G(P)) = N_G(P),$ because P is sylow p. Therefore U and W are conjugate in $N_G(P).$

• nbloomf  On July 7, 2011 at 12:09 pm

Aha- my proof only shows that $g^{-1}N_P(U)g = N_{g^{-1}Pg}(g^{-1}Ug)$, from which the inequality follows but not equality. That’s what I get for trying to think at 6 AM. 🙂

Thanks!

• nbloomf  On July 7, 2011 at 1:48 pm

Okay… now I’m not so convinced the proof works. I don’t have the time at the moment to dig into it, so I’ll mark this as ‘incomplete’ and come back to it later.

• nhanttruong  On July 6, 2011 at 3:18 pm

Ah, typo. The second last term in the long chain should have been $g^-N_G(N_P(W))g.$ They really should let you preview comments.