The notes I'm using to study group theory make a remark that another appropriate name for the "orbit stabiliser theorem" is the "first isomorphism theorem for group actions". For reference, here are the two theorems:
First isomorphism theorem: let $\phi: G \to H$ be a group homomorphism with kernel $K$. Then $G / K \cong \text{Image}(\phi)$. The isomorphism is given by $\psi: G / K \to \text{Image}(\phi)$ with $\psi(gK) = \phi (g)$.
Orbit stabiliser theorem: let $G$ be a group acting on a set $X$. Let $x \in X$. Then the map $\phi: G / \text{Stab} (x) \to \text{Orb} (x)$ where $g \: \text{Stab} (x) \mapsto gx$ is a bijection. Also, if $G$ is finite, then $|G| = | \text{Stab} (x) | | \text{Orb} (x) |$.
I can see why the orbit stabiliser theorem is like the first isomorphism theorem, especially when considering that the stabiliser fixes points (i.e. for the stabiliser, the group action is the same as the identity action), so the stabiliser is a kind of analogue to the kernel. Similarly, the orbit is a bit like the image of a map.
What I want is to understand how they're analogous in a more rigorous way, and maybe if any results can be derived for group actions in much the same way as the second and third isomorphism theorems and the correspondence theorem can be derived from the first isomorphism theorem.
Best Answer
Welcome to MSE, this is a great question!
Say that $\phi : G \to H$ is a group homomorphism.
Then there is a group action $G \curvearrowright H$ defined by $g \cdot h = \phi(g) h$, using the existing multiplication in $H$. If it's not obvious, checking that this really is a group action whenever $\phi$ is a group homomorphism is a worthwhile exercise.
Now let's look at the orbit and stabilizer of the identity element $e_H$. We can compute
Again, if these equalities aren't obvious, they make for great exercises!
But what does the orbit stabilizer theorem tell us in this case? It tells us that the map $\overline{\phi} : G \big / \text{stab}_G(e_H) \to \text{orb}_G(e_H)$ (that is, $\overline{\phi} : G \big / \text{ker}(\phi) \to \text{im}(\phi)$) given by $g \ \text{ker}(\phi) \mapsto g \cdot e_H$ is a bijection. But it's not hard to see that this is the same function as $g \ \text{ker}(\phi) \mapsto \phi(g)$, which is the bijection from the first isomorphism theorem!
I hope this helps ^_^