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  "body": "\n## Proof\n\nWe prove that $g \\cdot x = h \\cdot x$ exactly when $g G_x = h G_x$. Let $e$ be the identity of $G$, and let $H = G_x$ be the [[stabiliser|0RGGF8ZS6D]] of $x$.\n\nThe set $H$ is a subgroup of $G$, by the [[subgroup criterion|05N4TQKPH0]]. First, $e \\cdot x = x$, so $e \\in H$. If $a, b \\in H$, then $b^{-1} \\cdot x = b^{-1} \\cdot (b \\cdot x) = e \\cdot x = x$. So $(a b^{-1}) \\cdot x = a \\cdot (b^{-1} \\cdot x) = x$, and $a b^{-1} \\in H$.\n\n1. Let $g, h \\in G$. By the axioms of an [[action|0YSE802YCQ]], $g \\cdot x = h \\cdot x$ implies $(g^{-1} h) \\cdot x = g^{-1} \\cdot (h \\cdot x) = g^{-1} \\cdot (g \\cdot x) = e \\cdot x = x$. So $g^{-1} h \\in H$.\n2. Conversely, if $g^{-1} h \\in H$, then $h \\cdot x = (g g^{-1} h) \\cdot x = g \\cdot ((g^{-1} h) \\cdot x) = g \\cdot x$. So $g \\cdot x = h \\cdot x$ if and only if $g^{-1} h \\in H$.\n3. Next, if $g^{-1} h = k \\in H$, then $h = g k$. Each element of $h H$ is $g k k'$ with $k' \\in H$, and $k k' \\in H$. So $h H \\subseteq g H$.\n4. In the same way, $h^{-1} g = k^{-1} \\in H$ gives $g H \\subseteq h H$. So $g^{-1} h \\in H$ implies $g H = h H$.\n5. Conversely, if $g H = h H$, then $h = h e \\in h H = g H$. So $h = g k$ for some $k \\in H$, and $g^{-1} h = k \\in H$.\n6. By steps 2 to 5, $g \\cdot x = h \\cdot x$ if and only if the [[left cosets|05FD317V2W]] $g H$ and $h H$ are equal.\n7. Define $\\Phi(g H) = g \\cdot x$. By the direction \"equal cosets give equal points\" of step 6, the value does not depend on the representative $g$. So $\\Phi$ is well defined.\n8. By the other direction of step 6, $\\Phi(g H) = \\Phi(h H)$ implies $g H = h H$. So $\\Phi$ is injective.\n9. Each point of the [[orbit|00F2JFA41J]] $G \\cdot x$ has the form $g \\cdot x = \\Phi(g H)$. So $\\Phi$ maps the set of left cosets of $H$ onto $G \\cdot x$.\n10. Hence $\\Phi$ is a bijection from the set of left cosets of $G_x$ to $G \\cdot x$. By the [[definition of the index|04QEBF8S49]], $\\abs{G \\cdot x} = [G : G_x]$.\n\n## Idea\n\nTwo group elements move $x$ to the same point exactly when they differ by an element that fixes $x$. That is the same as lying in the same left coset of the stabiliser. So the points of the orbit match the cosets one to one.\n",
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