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.

I have the following problem:

Let $P$ be a Sylow subgroup of a group $G$. Prove that if $x$ and $y$ are elements of the centralizer of $P$ that are conjugate in $G$, then they are also conjugate in the normalizer of $P$.

Any hints will be welcome. Thanks in advance.

share|improve this question
5  
A beautiful application of the Sylow $C$-theorem (i.e., the fact that any two Sylow $p$-subgroups of a finite group are conjugate) in a "hidden subgroup". I will give you the following hint: if $y=x^g=g^{-1}xg$, then apply the Sylow $C$-theorem to $P$ and $P^g=\{g^{-1}zg:z\in P\}$ $\textbf{in}$ $\textbf{C}_G(y)=\{z\in G:zy=yz\}$. (However, try not to look at my answer below until you are truly stuck.) –  Amitesh Datta Jun 15 '11 at 12:13

1 Answer 1

up vote 7 down vote accepted

The following steps lead to a solution. Please note that I use the following notation: if $z,w\in G$, then $z^w=w^{-1}zw$, i.e., $z^w$ is the $\textit{conjugation}$ of $z$ by $w$. Similarly, if $H$ is a subgroup of $G$, then $H^w=\{z^w:z\in H\}$ is the $\textit{conjugation}$ of $H$ by $w$.

(1) Let $C=\textbf{C}_G(P)$ and $N=\textbf{N}_G(P)$ denote the centralizer and normalizer of $P$ in $G$ respectively. Also, let $x,y\in C$ be conjugate in $G$.

(2) The "trick" is to observe that $P\subseteq \textbf{C}_G(y)$ (the centralizer of $y$ in $G$) and $P\subseteq \textbf{C}_G(x)$. Choose $g\in G$ such that $y=x^g$. Prove that $(\textbf{C}_G(x^g))=(\textbf{C}_G(x))^g$.

(3) Deduce that $P^g\subseteq \textbf{C}_G(y)$. Hence $P$ and $P^g$ are Sylow $p$-subgroups of $\textbf{C}_G(y)$. Use the Sylow $C$-theorem in $\textbf{C}_G(y)$; i.e., use the fact that $P$ and $P^g$ are conjugate in $\textbf{C}_G(y)$.

(4) Conclude that $x$ and $y$ are conjugate in $\textbf{N}_G(P)$.

I believe this result is due to Burnside and is also referred to as $\textbf{Burnside's Fusion Lemma}$.

share|improve this answer

Your Answer

 
discard

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.