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Codex  0 Created 2026-09-23 Updated 2026-09-25

(r, b)

, where

r = a^{2}

is a nonzero quadratic residue modulo

11

. The square subgroup is

Q = ⟨ 4 ⟩ = {1, 4, 5, 9, 3} ≅ C_{5} .

(1)

Matrix multiplication becomes

(r, b) (s, c) = (rs, b + rc), (r, b)^{- 1} = (r^{- 1}, - r^{- 1} b) .

(2)

Thus

H

is the affine semidirect product of cyclic groups of orders eleven and five

H ≅ (F_{11}, +) ⋊ Q ≅ C_{11} ⋊ C_{5} .

(3)

There are five choices for

r

and eleven for

b

, so

∣ H ∣ = 55

. It is nonabelian, since

(4, 0) (1, 1) = (4, 4) \neq = (4, 1) = (1, 1) (4, 0) .

(4)

Let

N = {(1, b) : b \in F_{11}} ≅ C_{11} .

(5)

Conjugation by the complement gives

(r, 0) (1, b) (r, 0)^{- 1} = (1, r b) .

(6)

Hence the ten nonidentity translations split into the two orbits

T_{Q} = {(1, b) : b \in Q}, T_{N} = {(1, b) : b \in 2 Q},

(7)

each of size five. For

r \neq = 1

(1, t) (r, b) (1, t)^{- 1} = (r, b + (1 - r) t) .

(8)

Since

1 - r \neq = 0

, varying

t

gives every

b \in F_{11}

. Conjugation cannot change

r

in the abelian quotient

H / N

. Therefore, with

C_{k} = {(4^{k}, b) : b \in F_{11}}, 1 \leq k \leq 4,

(9)

the seven conjugacy classes are

{1}, T_{Q}, T_{N}, C_{1}, C_{2}, C_{3}, C_{4},

(10)

of sizes

1, 5, 5, 11, 11, 11, 11

, agreeing with the conjugacy classes in the affine semidirect product of orders eleven and five.

The commutators with

(4, 0)

generate every translation because multiplication by

4 - 1 = 3

is invertible in

F_{11}

. Hence the commutator subgroup is

N

, and the abelianization is

H^{ab} ≅ H / N ≅ C_{5} .

(11)

Put

ζ = e^{2 πi /5}

. The one-dimensional characters factor through the abelianization, giving five characters

χ_{j} (4^{k}, b) = ζ^{jk}, 0 \leq j \leq 4.

(12)

For the remaining characters, put

ω = e^{2 πi /11}

and define a character of

N

θ_{m} (1, b) = ω^{mb} .

(13)

The complement has two free orbits on the nontrivial

θ_{m}

, indexed by

m \in Q

and

m \in 2 Q

. By induction from an abelian normal subgroup with a free character orbit, the characters

ψ_{Q} = Ind_{N}^{H} θ_{1}, ψ_{N} = Ind_{N}^{H} θ_{2}

(14)

are irreducible of degree five and vanish outside

N

Set

η = \sum_{q \in Q} ω^{q} = \frac{- 1 + i 11}{2} .

(15)

The quadratic periods modulo eleven give the nonsquare sum

\overline{η} = (- 1 - i 11) /2

. Multiplication by a square preserves

Q

and multiplication by a nonsquare exchanges the two square classes, so the complete character table is

class size χ_{0} χ_{1} χ_{2} χ_{3} χ_{4} ψ_{Q} ψ_{N} 111111155 T_{Q} 511111 η \overline{η} T_{N} 511111 \overline{η} η C_{1} 111 ζ ζ^{2} ζ^{3} ζ^{4} 00 C_{2} 111 ζ^{2} ζ^{4} ζ ζ^{3} 00 C_{3} 111 ζ^{3} ζ ζ^{4} ζ^{2} 00 C_{4} 111 ζ^{4} ζ^{3} ζ^{2} ζ 00 .

(16)

Finally,

5 \cdot 1^{2} + 2 \cdot 5^{2} = 55 = ∣ H ∣,

(17)

and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the irreducible characters of the affine semidirect product of orders eleven and five.

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 18I / d by