Proof that orbits partition the set

Proof

We check the three axioms of an equivalence relation, then identify the classes. Let $e$ be the identity of $G$.

1. Reflexivity. By the first axiom of an action, $x = e \cdot x$. So $x \sim x$ for each $x \in X$.
2. Symmetry. Let $x \sim y$, so $y = g \cdot x$ for some $g \in G$. By the two axioms of an action, $g^{-1} \cdot y = g^{-1} \cdot (g \cdot x) = (g^{-1} g) \cdot x = e \cdot x = x$.
3. So $x = g^{-1} \cdot y$ with $g^{-1} \in G$, and $y \sim x$.
4. Transitivity. Let $x \sim y$ and $y \sim z$, so $y = g \cdot x$ and $z = h \cdot y$ with $g, h \in G$. By the second axiom, $z = h \cdot (g \cdot x) = (hg) \cdot x$.
5. Since $hg \in G$, we get $x \sim z$. By steps 1 to 4, $\sim$ is an equivalence relation on $X$.
6. The equivalence class of $x$ is $\set{y \in X : x \sim y} = \set{g \cdot x : g \in G}$. This is the orbit $G \cdot x$. So the equivalence classes are exactly the orbits.
7. The equivalence classes form a partition of $X$. So each point of $X$ lies in exactly one orbit.