The strategy is to construct a bijection between $\operatorname{Orb}_G(x)$ and the [set](/page/Set) of left cosets $(G : \operatorname{Stab}_G(x))$, then apply [Lagrange's Theorem](/theorems/782) to conclude. The bijection sends $g(x)$ to the coset $g\operatorname{Stab}_G(x)$; the key insight is that $g(x) = h(x)$ precisely when $g$ and $h$ lie in the same coset of the stabiliser. **Step 1: $\operatorname{Stab}_G(x)$ is a subgroup.** We verify the subgroup axioms. The identity fixes $x$, so $e \in \operatorname{Stab}_G(x)$. If $g, h \in \operatorname{Stab}_G(x)$, then $(gh)(x) = g(h(x)) = g(x) = x$, so $gh \in \operatorname{Stab}_G(x)$. If $g \in \operatorname{Stab}_G(x)$, then $x = e(x) = (g^{-1}g)(x) = g^{-1}(g(x)) = g^{-1}(x)$, so $g^{-1} \in \operatorname{Stab}_G(x)$. **Step 2: Construct the bijection.** Define: \begin{align*} \vartheta : \operatorname{Orb}_G(x) &\to (G : \operatorname{Stab}_G(x)) \\ g(x) &\mapsto g\operatorname{Stab}_G(x). \end{align*} [claim:Orbit Stabiliser Bijection] $\vartheta$ is a well-defined bijection. [/claim] [proof] *Well-defined:* Suppose $g(x) = h(x)$. Then $h^{-1}g(x) = x$, so $h^{-1}g \in \operatorname{Stab}_G(x)$. By the [Equal Cosets Criterion](/theorems/786), $g\operatorname{Stab}_G(x) = h\operatorname{Stab}_G(x)$. *Injective:* Suppose $g\operatorname{Stab}_G(x) = h\operatorname{Stab}_G(x)$. By the [Equal Cosets Criterion](/theorems/786), $h^{-1}g \in \operatorname{Stab}_G(x)$, so $h^{-1}g(x) = x$, giving $g(x) = h(x)$. *Surjective:* Given $g\operatorname{Stab}_G(x)$, the element $g(x) \in \operatorname{Orb}_G(x)$ satisfies $\vartheta(g(x)) = g\operatorname{Stab}_G(x)$. [/proof] **Step 3: Count.** Since $\vartheta$ is a bijection, $|\operatorname{Orb}_G(x)| = |G : \operatorname{Stab}_G(x)|$. By [Lagrange's Theorem](/theorems/782): \begin{align*} |G| = |G : \operatorname{Stab}_G(x)| \cdot |\operatorname{Stab}_G(x)| = |\operatorname{Orb}_G(x)| \cdot |\operatorname{Stab}_G(x)|. \end{align*}