Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 15E / e / Solution

Both operators map linear combinations of $|{\psi }_g⟩$ and $|{\psi }_b⟩$ to linear combinations of the same two vectors: $R_g$ merely changes one coefficient, whilehas $|\Omega ⟩\in P$. Therefore each leaves $P$ invariant.