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  "body": "\n## Proof\n\nWe prove the two inclusions between $G_{g \\cdot x}$ and $g G_x g^{-1}$. Let $e$ be the identity of $G$, and let $y = g \\cdot x$.\n\n1. Let $h \\in G$. By the axioms of an [[action|0YSE802YCQ]], $(g^{-1} h g) \\cdot x = g^{-1} \\cdot (h \\cdot (g \\cdot x)) = g^{-1} \\cdot (h \\cdot y)$.\n2. Suppose $h \\in G_y$, the [[stabiliser|0RGGF8ZS6D]] of $y$. Then step 1 gives $(g^{-1} h g) \\cdot x = g^{-1} \\cdot y = g^{-1} \\cdot (g \\cdot x) = e \\cdot x = x$.\n3. So $g^{-1} h g \\in G_x$, and $h = g (g^{-1} h g) g^{-1} \\in g G_x g^{-1}$. Thus $G_y \\subseteq g G_x g^{-1}$.\n4. Conversely, let $k \\in G_x$ and $h = g k g^{-1}$. Then $h \\cdot y = (g k g^{-1}) \\cdot (g \\cdot x) = (g k) \\cdot x = g \\cdot (k \\cdot x) = g \\cdot x = y$.\n5. So $h \\in G_y$, and $g G_x g^{-1} \\subseteq G_y$.\n6. By steps 3 and 5, $G_{g \\cdot x} = g G_x g^{-1}$, a [[conjugate subgroup|0D6WHTTH7T]] of $G_x$.\n\n## Idea\n\nTo fix $g \\cdot x$, an element may first move it back to $x$ with $g^{-1}$, fix $x$, and then move it forward again with $g$. So the stabiliser of $g \\cdot x$ is the stabiliser of $x$ seen through the relabelling $g$.\n",
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