 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4/18f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The relations are

σ^{n} = τ^{2} = 1

and

τ σ τ = σ^{- 1}

, so the generated group is

D_{2 n}

X Y

and

X^{n} + Y^{n}

are fixed. Conversely

X^{n}, Y^{n}

are roots of

T^{2} - (X^{n} + Y^{n}) T + (X Y)^{n}

, and adjoining

X

then

Y

gives degree at most

2 n

. Artin gives degree exactly

∣ G ∣ = 2 n

, proving

M^{G} = C (X^{n} + Y^{n}, X Y)

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