[proofplan] We verify the subgroup axioms directly for the stabiliser $G_x$. The identity element fixes $x$ by the identity axiom for a [group action](/page/Group%20Action). Closure under multiplication follows from the compatibility axiom of the action, and closure under inverses follows by applying $g^{-1}$ to the equality $g \cdot x = x$. [/proofplan]

[step:Fix a point and verify that the identity belongs to the stabiliser] Fix $x \in X$. By the identity axiom for the left action $\alpha$, we have $e \cdot x = x$. Hence $e \in G_x$, so $G_x$ is nonempty. [/step]

[step:Show that the product of two elements fixing $x$ also fixes $x$]Let $g,h \in G_x$. By definition of $G_x$, we have $g \cdot x = x$ and $h \cdot x = x$. Using the compatibility axiom for the left action, \begin{align*} (gh) \cdot x = g \cdot (h \cdot x) = g \cdot x = x. \end{align*} Therefore $gh \in G_x$.[/step]

[guided]We must prove that $G_x$ is closed under the group operation inherited from $G$. Take arbitrary elements $g,h \in G_x$. The meaning of this membership is exactly that both elements fix the chosen point $x$: \begin{align*} g \cdot x = x, \qquad h \cdot x = x. \end{align*} Now consider the product $gh \in G$. To decide whether $gh$ lies in $G_x$, we must check whether $(gh) \cdot x = x$. The defining compatibility axiom for a left group action says that acting by a product is the same as acting first by the right factor and then by the left factor: \begin{align*} (gh) \cdot x = g \cdot (h \cdot x). \end{align*} Since $h$ fixes $x$, this becomes \begin{align*} g \cdot (h \cdot x) = g \cdot x. \end{align*} Since $g$ also fixes $x$, we conclude \begin{align*} (gh) \cdot x = x. \end{align*} Thus $gh \in G_x$, proving closure under multiplication.[/guided]

[step:Show that the inverse of an element fixing $x$ also fixes $x$] Let $g \in G_x$. Then $g \cdot x = x$. Acting by $g^{-1}$ on both sides and using the compatibility axiom gives \begin{align*} g^{-1} \cdot x = g^{-1} \cdot (g \cdot x) = (g^{-1}g) \cdot x = e \cdot x = x. \end{align*} Hence $g^{-1} \in G_x$. [/step]

[step:Conclude that the stabiliser is a subgroup] We have shown that $G_x$ contains the identity element $e$, is closed under multiplication, and is closed under inverses. Therefore $G_x$ is a subgroup of $G$. [/step]