Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 324 / 2 / v / Solution

The group averageis Hermitian. In its square, every $R\in G$ occurs exactly $|G|$ times among products $PQ$, and therefore ${\Pi }_G^2={\Pi }_G$. Moreover $g{\Pi }_G={\Pi }_G$ for every $g\in G$, so its image lies in the stabilizer subspace $V_G$, while ${\Pi }_G|\psi ⟩=|\psi ⟩$ for every $|\psi ⟩\in V_G$. Thus ${\Pi }_G$ is the orthogonal projector onto $V_G$. If $g_1,\dots ,g_l$ are independent generators, expanding the product chooses each element of $G$ exactly once and gives the stabilizer-projector formula