Past exam of the algebra course of the University of Oxford / 2008 / A / 4 / b / Solution

Complete reducibility and character orthogonality imply that the squared multiplicities sum to three. The trivial constituent has multiplicity one, so the remaining two units come from two distinct nontrivial constituents, each with multiplicity one. Their degrees add to $4-1=3$, so they are two and one. ThusThe two-dimensional constituent is the complexification of the geometric action on the plane. Its trace is $2$ on the identity, $-2$ on the half-turn, zero on the quarter-turns, and zero on every reflection. Its character norm is $(4+4)/8=1$, proving it irreducible. Alternatively, the span of the two coordinate functions on the vertex set is a two-dimensional invariant subspace of the permutation module with this geometric action. This identifies the requested representation rather than merely its degree.