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.

We have the following theorem

Let $G$ be a group, acting on a set $\Omega$ and let $p^m\Bigm||\omega^G|$ wherein $p$ is prime and $\omega \in \Omega$. If $P$ is a $p$-Sylow subgroup of $G$ then, $p^m\Bigm||\omega^P|$.

In this theorem $\omega^G=\{\omega^g|g\in G\}$ is any orbit of $(G|\Omega)$. The proof is easy and it is based of using Orbit-Stabilizer Equation for $G$ and $P$. As theorem takes, I assume that the group $G$ acts on a set $\Omega$. we can prove that for any subgroup $H$ of $G$, $H_\omega$ is a subgroup of $G_\omega$ where $G_\omega$ is the stablizer of $\omega$. I am asking kindly: Can we say that $P_\omega$ is a $p-$ Sylow subgroup of $G_\omega$ when $P$ is a $p-$ Sylow subgroup of $G$? Thanks for your help.

share|improve this question
    
Some conjugate of P works like this, but maybe not P itself. Let G be S3 and P be S2 and w be 1. Then Pw is 1 and is not a Sylow of Gw which is its own Sylow of order 2. –  Jack Schmidt Jun 28 '12 at 15:16
    
I think the abstract groups G where this always works are exactly the nilpotent groups. –  Jack Schmidt Jun 28 '12 at 15:18
    
@JackSchmidt: Is this useful if I know that $(P|\Omega)$ transitively? –  Babak S. Jun 28 '12 at 16:16
    
@JackSchmidt: It seems to me if we assume $(P|\Omega)$ be transitive so, $(G|\Omega)$ will be and then the index of $P_\omega$ in $G_\omega$ is free of $p$. Am I right? –  Babak S. Jun 28 '12 at 17:47
    
Yup. If P is transitive your argument is good. –  Jack Schmidt Jun 28 '12 at 20:23

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.