Solution
= Solution
For
$$
V(\Phi)=m_\Phi^2|\Phi|^2+\lambda|\Phi|^4
$$
with $\lambda>0$, spontaneous symmetry breaking occurs when $m_\Phi^2<0$. Then
$$
|\langle\Phi\rangle|^2=-\frac{m_\Phi^2}{2\lambda}.
$$
Since $\Phi$ 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 $\Psi\mapsto-\Psi$.