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  "body": "\n## Proof\n\nWe check the three axioms of an equivalence relation, then identify the classes. Let $e$ be the identity of $G$.\n\n1. Reflexivity. By the first axiom of an [[action|0YSE802YCQ]], $x = e \\cdot x$. So $x \\sim x$ for each $x \\in X$.\n2. Symmetry. Let $x \\sim y$, so $y = g \\cdot x$ for some $g \\in G$. By the two axioms of an action, $g^{-1} \\cdot y = g^{-1} \\cdot (g \\cdot x) = (g^{-1} g) \\cdot x = e \\cdot x = x$.\n3. So $x = g^{-1} \\cdot y$ with $g^{-1} \\in G$, and $y \\sim x$.\n4. 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, $z = h \\cdot (g \\cdot x) = (hg) \\cdot x$.\n5. Since $hg \\in G$, we get $x \\sim z$. By steps 1 to 4, $\\sim$ is an [[equivalence relation|0SGTMBEWRF]] on $X$.\n6. The [[equivalence class|082RMRSY26]] of $x$ is $\\set{y \\in X : x \\sim y} = \\set{g \\cdot x : g \\in G}$. This is the [[orbit|00F2JFA41J]] $G \\cdot x$. So the equivalence classes are exactly the orbits.\n7. The [[equivalence classes form a partition|01HT2CH8BH]] of $X$. So each point of $X$ lies in exactly one orbit.\n\n## Idea\n\nThe group axioms are exactly what is needed: the identity gives reflexivity, inverses give symmetry, and closure under products gives transitivity. Equivalence classes always partition a set, so the orbits do too.\n",
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