Let $G$ be a finite group, and let $\Omega$ be a transitive $G$-space. Assume 1 $\neq H \unlhd G$ and that |$\Omega$| = $p$ where $p$ is prime, and $G \leq Sym(\Omega)$.

Deduce that then $H$ must act transitively on $\Omega$.

So far I have been able to deduce that $p$ divides the order of $G$ as the action is transitive, and the intersection of all stabilizers must be trivial as the action is faithful. However I'm not sure how to deduce that $H$ acts transitively.


The key point is if X is an orbit of H, so is $X^g$. (Check this).

  • $\begingroup$ What is $X^{g}$? $\endgroup$ – user130075 Feb 22 '14 at 14:38
  • $\begingroup$ We write the action of $g$ on $x$ top left as $x^g$. And $X^g=\{x^g|x\in X\}$. $\endgroup$ – Wei Zhou Feb 23 '14 at 1:36
  • $\begingroup$ @WeiZhou what actually the constraint $|\Omega|=p$ gives us here? $\endgroup$ – user112072 Mar 16 '14 at 5:52
  • $\begingroup$ @user112072 Once it is shown that $X^g$ is $H$-orbit if $X$ is a $H$-orbit, one can also show that $G$ acts transitively of the set of $H$-orbits. Therefore all $G$-orbits must of the form $g_iX$. As orbits form a partition of $\vert \Omega \vert$, we get that $\vert \Omega \vert = t \vert X \vert$, where $t$ is the number of orbits. As the size of $\Omega$ is prime, we must have $X$ has size 1 or $p$. If $\vert X \vert = 1$, then every element of $H$ fixes every element of $X$, which can only happen if $H = 1$, contradiction. SO $\vert X \vert = p$, and so $H$ is transitive. $\endgroup$ – user130075 May 11 '14 at 12:01

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