Show that the stabilizer of an element need not be a normal subgroup.
Solution
Consider the natural action of \(S_3\) on \(X=\{1,2,3\}\).
1. The stabilizer of \(1\) is\[ (S_3)_1=\{1,(23)\}. \]
A permutation fixes \(1\) exactly when it permutes only \(2\) and \(3\). Hence the only possibilities are the identity and \((23)\).
2. This stabilizer is not normal in \(S_3\).
Conjugating its nonidentity element by \((12)\) gives \[ (12)(23)(12)^{-1}=(13), \] and \((13)\notin\{1,(23)\}\). Therefore \[ (12)(S_3)_1(12)^{-1}\neq(S_3)_1, \] so \((S_3)_1\) is not normal in \(S_3\).
3. Thus a point stabilizer need not be a normal subgroup.
The action above supplies the required counterexample.