= Solution
The <character> is the trace. For the planar representation, the rotation trace and reflection trace give
$$
\boxed{\chi_2(R^k)=2\cos\frac{k\pi}{3},\qquad\chi_2(m_j)=0.}
$$
In the source’s order $(I,R,R^2,R^3,R^4,R^5,m_1,\ldots,m_6)$, 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. Consequently
$$
\boxed{\chi_{\rm reg}(I)=12,\qquad\chi_{\rm reg}(g)=0\quad(g\ne I).}
$$
In particular $\chi_{\rm reg}(m_4)=0$, consistently with the zero diagonal in part (d). The two characters have different identity values because the representation dimensions are two and twelve.
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