Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 19H / d

Table of contents
- Solution d

Solution

With class representatives

(0, 0, z)

and

(x, y, 0)

for

(x, y) \neq = (0, 0)

, the complete character table is described by

χ_{a, b} (0, 0, z) = 1, χ_{a, b} (x, y, 0) = ω^{a x + b y},

for

a, b \in F_{p}

, and

χ_{c} (0, 0, z) = p ω^{cz}, χ_{c} (x, y, 0) = 0

for

c \in F_{p}^{\times}

. There are

p^{2} + p - 1

rows, matching the number of conjugacy classes, and

p^{2} \cdot 1^{2} + (p - 1) p^{2} = p^{3} = ∣ G ∣,

so no irreducible characters are missing.

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