The Orbit Stabilizer Theorem, is this simple proof true?

Question

So I came up with this simple proof of the OST, and was wondering if it is valid or not. Let $G$ be the set of permutations of the set $S$ and let $|S|=n$. Now $\mathrm{Stab}_G(i)$ is simply all the functions that keep $i$ in place. So $i$ maps to $i$ and the rest can map to whatever. Hence $|\mathrm{Stab}_G(i)|=(n-1)!$. Now $\mathrm{Orb}_G(i)$ is where $i$ can map to, well it can map to anything else in $S$, so $|\mathrm{Orb}_G(i)|=n$. We also know that $|G|=n!=|\mathrm{Stab}_G(i)|\cdot|\mathrm{Orb}_G(i)|$.

Answer 1

This proof is fine for when the group is $S_n$, the symmetric group on $n$ letters, and when the set that $S_n$ is acting on is $\{1,\dots,n\}$. Keep in mind that the orbit-stabilizer theorem is more general and applies to any group $G$ acting on a finite set $X$.

Answer 2

It is true, but it doesn't always happen. By Cayley's theorem, every group is isomorphic to a group of permutations, but that doesn't mean it is finite i.e. it could be something like $\mathbb{S}(\mathbb{R})$, the set of bijective functions. For example you coul have groups of order 33 that act on sets of order 11.
In conclusion, it doesn't happen only for your case, but for any action.