Show that the stabilizer of an element \(G_x\) is a subgroup of \(G\).
Solution
Let \[ G_x=\{g\in G:g\cdot x=x\}. \] We use the subgroup test.
1. The set \(G_x\) is nonempty.
The identity satisfies \(e\cdot x=x\), so \(e\in G_x\).
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 \[ (gh^{-1})\cdot x=g\cdot(h^{-1}\cdot x)=g\cdot x=x. \] Therefore \(gh^{-1}\in G_x\).
By the subgroup criterion, \(G_x\le G\).