Impose the discrete <Z2 symmetry> $\boxed{\Phi\mapsto-\Phi}$. The quadratic and quartic <trace> invariants are unchanged, while $\operatorname{tr}\Phi^3$ changes sign. Invariance therefore forbids a nonzero coefficient of that cubic operator. This is an additional symmetry of the <scalar potential>, not a consequence of <SU(5)> conjugation alone; cubic traces are permitted by the latter.

Solution

Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 7 / a / Solution

 Solution (source code)