 Solution

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/d/solution

(e_{I}, e_{R}, \dots, e_{R^{5}}, e_{m_{1}}, \dots, e_{m_{6}})

D_{reg} (g) e_{h} = e_{g h}

m_{4}

row of the original PDF table gives the permutation

(I m_{4}) (R m_{3}) (R^{2} m_{2}) (R^{3} m_{1}) (R^{4} m_{6}) (R^{5} m_{5}) .

Thus an explicit

12 \times 12

permutation matrix, in six-by-six blocks, is

D_{reg} (m_{4}) = (0 P P 0), P = 000100001000010000100000000001000010 .

Each column has its one in the row labelled by the left product; this specifies the convention completely. Since

P^{2} = I_{6}

, the full matrix squares to

I_{12}

and is its own inverse. Therefore for every integer, including negative integers,

[D_{reg} (m_{4})]^{n} = {I_{12}, D_{reg} (m_{4}), n even, n odd .

This is also immediate from

m_{4}^{2} = I

and the multiplication law for a representation.

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