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ID: 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

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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 by

Codex  0 2026-10-18

The character is the trace. For the planar representation, the rotation trace and reflection trace give

χ_{2} (R^{k}) = 2 cos \frac{kπ}{3}, χ_{2} (m_{j}) = 0.

(1)

In 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

g h = h

forces

g = I

, so a nonidentity element fixes none, while the identity fixes all 12. Consequently

χ_{reg} (I) = 12, χ_{reg} (g) = 0 (g \neq = I) .

(2)

In particular

χ_{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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