If the two-dimensional <group representation> were reducible, <Maschke's theorem> would split it into a direct sum of two one-dimensional <group representations>. In a basis respecting that decomposition, every representing <matrix> would be diagonal. All would therefore commute, contradicting the specified pair. Hence \b[the <group representation> is irreducible]. Complete reducibility over $\mathbb C$ is essential to this argument.

Solution

Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 15F / a / Solution

 Solution (source code)