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This website uses cookies to improve your experience. We’ll assume you’re ok with this, but you can opt-out if you wish. Cookie settingsACCEPT Privacy & Cookies Policy Privacy Overview This website uses cookies to improve your experience while you navigate through the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are as essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may have an effect on your browsing experience. Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information.Q: Orbits of an action of $G$ on a set $X$ Let $G$ be a finite group of order $p^\alpha q^\beta$ where $p, q$ are distinct primes and $\alpha,\beta>0$. Let $X$ be a set of size $p^\alpha$ and $G$ acts on $X$ via the action map $G\times X\to X,\, (g,x)\mapsto g.x$. Prove that there exists a subset $Y$ of $X$ such that $|\mathcal{O}_G(Y)|=q^\beta$. $\textbf{My attempt:}$ Since the group $G$ is not cyclic, it is not necessarily true that a non-trivial element always has exactly a single orbit. Hence, $|\mathcal{O}_G(Y)|>1$ for any $Y\subseteq X$. My idea is to prove that $|\mathcal{O}_G(Y)|$ is independent of $Y\subseteq X$ as long as $Y$ is chosen such that $|Y|\geq q^\beta.$ Clearly, $Y eq \emptyset$. Now, a $(g_1,g_2)\in G\times G$ maps $Y\times Y$ to a2fa7ad3d0
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