Size of orbit of $x$ divides the order of $G$

Question

I'm having a bit of trouble "closing" this question (finishing it):

Suppose that $G$ acts on $X$ and let $x \in X$. Let stab($x$) = {$g$ | $gx = x$}. Show that the size of the orbit of $x$ divides the order of $G$. (Hint: LaGrange's Theorem and the fact that stab($x$) is a subgroup of $G$).

Here's what I got so far:

I'm thinking that the orbit of $x$ $\in$ stab($x$), which is a subgroup of $G$. But I have no basis for this line of logic, and am having some trouble figuring out what the orbit of $x$ is. I know that the orbit of $x$ is all the possible values that can be reached by the actions of $G$. So wouldn't that mean the orbit of $x$ is just all of $X$? Any help with this problem would be great. Thank you for your help.

EDIT: Is the orbit of $x$ the order of stab($x$)? If so, how would I go about proving it?

Answer 1

The stabilizer subgroup consists of the elements of $G$ that fix $x$. Any time you have a subgroup of a group, you can consider the cosets of that subgroup. If $H$ is the stabilizer subgroup of $x$, what can you say about the action of the cosets on $x$? (Note: none of this requires $H$ to be normal, btw.) For example, if $g \not \in H$, so that $gx \ne x$, what can you say about the action on $x$ of elements in the coset $gH$? Well, I'll leave it to you to show this, but each element does the same thing (i.e., sends $x$ to the same point in the orbit.)

This allows us to establish a correspondence between the cosets of $H$ and elements of the orbit of $x$: $gH \mapsto gx$. As a second exercise, you can show that this correspondence is bijective (i.e., the cardinality of $G/H$ coincides with that of the orbit $\mathcal{O}_x$. Your desired result follows from the fact that $|G| = |G/H| \cdot |H|$.

Answer 2

since $G$ act on $X$, by Orbit-Stabilizer Theorem, we have $$|Orb(x)|=|G: Stab_G(x)|.$$ Thus $|Orb(x)|=\frac{|G|}{|Stab(x)|}$. Clearly, $\frac{|G|}{|Stab(x)|} \mid |G|$. So $|Orb(x)| \mid |G|$.