Show that if \(G \actson X\) is a group action, then the stabilizer \(G_x\) of a point is a subgroup.
Solution
Fix \(x\in X\) and write \[ G_x=\{g\in G:g\cdot x=x\}. \]
1. The identity belongs to \(G_x\).
By the identity axiom for a group action, \(e\cdot x=x\).
2. If \(g,h\in G_x\), then \(gh^{-1}\in G_x\).
Since \(h\cdot x=x\), applying \(h^{-1}\) gives \(h^{-1}\cdot x=x\). Hence \[ (gh^{-1})\cdot x =g\cdot(h^{-1}\cdot x) =g\cdot x =x. \] Thus \(gh^{-1}\in G_x\).
3. Therefore \(G_x\le G\).
By 1, \(G_x\) is nonempty, and by 2 it satisfies the one-step subgroup criterion. Hence it is a subgroup of \(G\).