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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've had a lot of trouble with this question, as well as my advisor.

Let x and y be two elements of $Z(P)$, where $P$ is a Sylow $p$-subgroup of $G$. If $x$ and $y$ are conjugate in $G$, prove that $x$ is conjugate to $y$ in $N_G(P)$, the normalizer of $P$ in $G$.

share|cite|improve this question
up vote 5 down vote accepted

Hint: Use Sylow's theorem in $C_G(y)$, the centralizer of $y$ in $G$.

If $y = x^g$ and $x,y$ are in $Z(P)$, then $C_G(y)$ contains both $P$ (because $y \in Z(P)$) and $P^g$ (since $y=x^g \in Z(P^g)$). By Sylow's theorem in $C_G(y)$, $P^g = P^z$ for some $z \in C_G(y)$, so $gz^{-1} \in N_G(P)$ and $x^{gz^{-1}} = y^{z^{-1}} = y$.

share|cite|improve this answer
    
See also math.stackexchange.com/questions/45520/… – Jack Schmidt Aug 13 '12 at 15:37
    
Above, $\,z\in C_G(y)\,$, right? – DonAntonio Aug 13 '12 at 15:48
    
@DonAntonio: yes. I added it to the answer. – Jack Schmidt Aug 13 '12 at 15:52

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.