Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 78 / 3 / a / Solution

Use the group representation generated by the displayed matrices. The coordinate reflections generate $({\mathbb{Z}}_2{)}^3$, and the cyclic permutation conjugates ${\kappa }_x$ to ${\kappa }_y$, then to ${\kappa }_z$. Thus the group is the semidirect product $({\mathbb{Z}}_2{)}^3⋊{\mathbb{Z}}_3$, of order $24$. The converted TeX's multiplication sign must not be interpreted as a direct product of commuting factors.

To prove absolute irreducibility, complexify to ${\mathbb{C}}^3$ and let $W$ be a nonzero complex invariant subspace. It is preserved by the coordinate projections $P_j=(I-{\kappa }_j)/2$. For a nonzero $w\in W$, at least one coordinate is nonzero, so $P_{j}w$ is a nonzero multiple of a coordinate vector. Applying the cyclic permutation repeatedly shows that all three coordinate vectors belong to $W$. Hence $W={\mathbb{C}}^3$. Therefore the real representation is absolutely irreducible.

Equivalently, a commuting matrix must be diagonal because it commutes with all three reflections; commuting with $\rho$ forces its diagonal entries equal. This also confirms that every equivariant linearization is a scalar multiple of the identity, the condition needed in the branching calculation.