Past exam of the groups and representations course of the University of Oxford / 2022 / 1 / a / Solution

The matrices obey $Q^5=P^2=I$ and $PQP=Q^{-1}$. Moving every $P$ to the right therefore reduces every product to $Q^k$ or $Q^{k}P$, with $0\le k<5$. The five diagonal matrices $Q^k$ are distinct because $\alpha$ is a primitive fifth root of unity; the five $Q^{k}P$ are distinct and off-diagonal. ThusThis is the dihedral group of the regular pentagon, with $Q$ a rotation and $P$ a reflection.

Conjugation by $P$ interchanges $Q^k$ and $Q^{-k}$, while conjugation by a power of $Q$ fixes rotations. AlsoSince multiplication by $2$ permutes the residues modulo $5$, all five reflections form one conjugacy class. The full list of conjugacy classes isTo prove the representation is irreducible over $\mathbb{C}$, any proper nonzero invariant subspace would be a line invariant under $Q$. Its two distinct eigenvalues force such a line to be one of its coordinate eigenspaces. But $P$ exchanges those two lines, so neither is $G$-invariant. Hence the given complex representation is irreducible.