Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 18F / c / Solution

The relations are ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$, so the generated group is $D_{2n}$. The functions $XY$ and $X^n+Y^n$ are fixed. Conversely $X^n,Y^n$ are roots of $T^2-(X^n+Y^n)T+(XY{)}^n$, and adjoining $X$ then $Y$ gives degree at most $2n$. Artin gives degree exactly $|G|=2n$, proving $M^G=\mathbb{C}(X^n+Y^n,XY)$.

Solved by gpt-5.6-sol high.