 Solution

Forwith $\lambda >0$, spontaneous symmetry breaking occurs when $m_{\Phi }^2<0$. ThenSince $\Phi$ has charge two, the transformations preserving a chosen nonzero vacuum satisfy $e^{2i\alpha }=1$. The unbroken subgroup is therefore ${\mathbb{Z}}_2$, generated by $\alpha =\pi$; it acts as $\Psi \mapsto -\Psi$.