Let \(G\) act on a set \(X\) and let \(x\in X\). Show that the stabilizer \[ G_x=\{g\in G:g\cdot x=x\} \] is a subgroup of \(G\).
Solution
The identity fixes \(x\), so \(e\in G_x\).
If \(g,h\in G_x\), then \[ (gh^{-1})\cdot x =g\cdot(h^{-1}\cdot x). \] Since \(h\cdot x=x\), applying \(h^{-1}\) gives \(h^{-1}\cdot x=x\). Therefore \[ (gh^{-1})\cdot x=g\cdot x=x, \] so \(gh^{-1}\in G_x\).
By the subgroup criterion, \(G_x\le G\).