Proving the stabilizer is a subgroup of the group to prove the Orbit-Stabiliser theorem I have to prove the OS theorem. The OS theorem states that for some group $G$, acting on some set $X$, we get
$$
|G| = |\mathrm{Orb}(x)| \cdot |G_x| $$
To prove this, I said that this can be written as
$$ |\mathrm{Orb}(x)| = \frac{|G|}{|G_x|}$$
In order to prove the RHS, I can say that we can use Lagrange theorem, assuming that the stabiliser is a subgroup of the group G, which I'm pretty sure it is. I don't really know how I'd go about proving this though. Also, I was thinking, would this prove the whole theorem or just the RHS?
I didn't just want to Google a proof, I wanted to try and come up with one myself because it's easier to remember. Unless there is an easier proof?

We can have the claim for $(G\mid\Omega)$ which is transitive and so we get $$|G|=|\Omega||G_{\omega}|$$ To see this, put $\Omega^*=\{G_{\omega}x\mid x\in G\}$and define the following map: $$f:\Omega^*\to\Omega,\;\;f(G_{\omega}x)=\omega^x$$ You can easily verify that $f$ is well-defined and is a bijective. Both $\Omega^*$ and $\Omega$ are finite and then $$|\Omega|=|\Omega^*|=[G:G_{\omega}]$$
No put the condition of being transitive aside, you will have your own claim as others confirmed.
