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?

You should show that the identity of $G$ fixes $x$, and that if $g,h\in G$ both fix $x$, then $gh$ fixes $x$. This will establish that $G_x$ is a subgroup of $G$.
This won't finish the proof, although it will let you say that $\frac{\lvert G\rvert}{\lvert G_x\rvert}$ is the number of cosets of $G_x$ in $G$ (left or right, there's the same number either way). You still need to find a bijection between these cosets and the orbit of $x$ under $G$.
