This is part $3$ of a $4$ part question involving proving properties of the group action.
Part $1$ (and links to other parts)
The question asks me to show that the set of orbits $\mathcal{O}_{G}(X) = \{G \cdot x \mid x \in X\}$ is a partition of the set $X$. given the action $G \times X \to X$
We declare the relation $x \sim y$ iff $g \cdot x = y$ for some $g \in G$ is an equivalence relation on $X$. Let $x,y,z \in X$ then we have:
- $x \sim x$ since by definition of the action $e \cdot x = x, \forall x \in X$ where $e \in G$ is the identity.
- If $x \sim y \implies g \cdot x = y \implies g^{-1} \cdot (g \cdot x) = (g^{-1}g) \cdot x = e \cdot x = x = g^{-1} \cdot y \implies y \sim x$
- $x \sim y$ and $y \sim z$ imply that there exists $g,g' \in G$ such that: $g \cdot x = y$ and $g' \cdot y = z$. Then we see that: $$z = g' \cdot y = g' \cdot (g \cdot x) = (g'g) \cdot x \implies x \sim z$$ By the definition of an action.
Hence $\sim$ is an equivalence relation. The equivalence class of $x$ is $\{ y \in X \mid x \sim y \} = \{ y \in X \mid g \cdot x = y, \ g \in G \}$ Thus we see the equivalence class of $x$ is equivalent to the orbit of $x$ and hence by the theorem that says the set of equivalence classes $X/\sim$ is a partition of $X$ we see that the set of orbits $\mathcal{O}_{G}(X)$ is a partition of $X$. $\square$
Is this correct? I am not entirely confident in my association of the (set of) equivalence classes and the (set of) orbits so any help/clarification would be greatly appreciated.