Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 10 / e / Solution

The character is the trace. For the planar representation, the rotation trace and reflection trace giveIn the source’s order $(I,R,R^2,R^3,R^4,R^5,m_1,\dots ,m_6)$, these values are $(2,1,-1,-2,-1,1,0,0,0,0,0,0)$.

In the regular representation, the trace counts basis vectors fixed by left multiplication. The equation $gh=h$ forces $g=I$, so a nonidentity element fixes none, while the identity fixes all 12. ConsequentlyIn particular ${\chi }_{\mathrm{reg}}(m_4)=0$, consistently with the zero diagonal in part (d). The two characters have different identity values because the representation dimensions are two and twelve.