To prove this fact, note that there is a natural group action of U(n) on the set of states of H n. This action is continuous and transitive on the pure states. For any state | ψ ψ ⟩ ⟩ , the isotropy group of | ψ ψ ⟩ ⟩ , (defined as the set of elements g of U(n) such that g | ψ ψ ⟩ ⟩ = | ψ ψ ⟩ ⟩ ) is isomorphic to the product group