$G$ is a finite group of permutations of the set $X$. Suppose $G$ acts transitively on $X$. Then $X$ is a finite set, and $|X|$ divides $|G|$.

So since $G$ acts transitively on $X$, there exists $x\in X$ such that $Gx = X$ for all $x \in X$. I don't really understand what is going on here with the concept of $X$ being a "single $G$-orbit" as my book says.

I think what is happening is that we take a single element from $X$ and we compose this with each element from the group of permutations $G$, then we get that the set of elements in $Gx$ is just $X$. We do this with each element of $X$. But if that is the case then there is some correspondence between $Gx$ and $X$ where $Gx$ is finite because $G$ is finite and therefore $X$ is finite.

I'm stuck moving forward from the transitive definition. Any help is appreciated.


Pick $x\in X$, by the orbit stabilizer theorem you have that the size of the orbit of $x$ is $[G: G_x]$. Since $G$ is finite this is just $\frac{|G|}{|G_x|}$, notice that the the orbit of $x$ is all of $X$, so we are done.

  • $\begingroup$ So we are saying that because $G$ acts transitively on set $X$, the orbit of $x$ is just $X$ by definition and then we use stabilizer theorem to relate order of the orbit of $x$ with the order of $X$? $\endgroup$ – TfwBear Oct 18 '16 at 1:10
  • $\begingroup$ couldn't have said it better myself. :) $\endgroup$ – Jorge Fernández Hidalgo Oct 18 '16 at 1:12

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