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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.

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    Yup. If P is transitive your argument is good.2012-06-28

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