 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2018/1/c/solution

Put

α = e^{2 πi /3}

, so

α^{3} = 1

and

1 + α + α^{2} = 0

. Since the group is Abelian, every conjugacy class is a singleton. The character table is

χ_{0} χ_{1} χ_{2} 0111 11 α α^{2} 21 α^{2} α .

⟨ χ_{j}, χ_{k} ⟩ = \frac{1}{3} \sum_{g = 0}^{2} \overline{χ_{j} (g)} χ_{k} (g) = \frac{1}{3} (1 + α^{k - j} + α^{2 (k - j)}) .

For

j = k

, all three summands are one, giving unit norm. For

k - j \equiv 1 (mod 3)

the sum is

1 + α + α^{2} = 0

; for

k - j \equiv 2 (mod 3)

it is

1 + α^{2} + α = 0

. This checks every distinct pair, proving

⟨ χ_{j}, χ_{k} ⟩ = δ_{jk} .

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