The actions on the ordered pair $(x,y)$ satisfy
$$
\sigma^4=1,
\qquad
\tau^2=1,
\qquad
\tau\sigma\tau=\sigma^{-1}.
$$
Thus every element of $G$ has the form $\sigma^i$ or $\tau\sigma^i$ for $0\leq i<4$, so $|G|\leq8$. Their actions are
$$
\begin{array}{c|cccccccc}
g&1&\sigma&\sigma^2&\sigma^3&\tau&\tau\sigma&\tau\sigma^2&\tau\sigma^3\\ \hline
(g(x),g(y))
&(x,y)&(y,-x)&(-x,-y)&(-y,x)
&(x,-y)&(-y,-x)&(-x,y)&(y,x).
\end{array}
$$
Because $x,y$ are <algebraically independent elements> over $\mathbb Q$, these eight substitutions are distinct. Hence
$$
\boxed{|G|=8},
$$
and the presentation identifies $G$ with the <dihedral group> $D_8$.

Put
$$
u=x^2+y^2,
\qquad
v=x^2y^2.
$$
Both generators $\sigma$ and $\tau$ fix $u$ and $v$, so for $F=\mathbb Q(u,v)$,
$$
F\subseteq K^G.
$$
The elements $x^2$ and $y^2$ are the two roots of
$$
T^2-uT+v,
$$
so $[F(x^2,y^2):F]\leq2$. Adjoining first $x$ and then $y$ takes at most two quadratic extensions, and the <tower law> gives
$$
[K:F]\leq2\cdot2\cdot2=8.
$$
Part (b) gives $8=|G|\leq[K:K^G]$. Since $F\subseteq K^G\subseteq K$, another use of the tower law gives
$$
8\leq[K:K^G]\leq[K:F]\leq8.
$$
All inequalities are equalities and $[K^G:F]=1$. Therefore
$$
\boxed{K^G=\mathbb Q(x^2+y^2,x^2y^2)},
$$
the <dihedral fixed field of a two-variable rational function field>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 18F / c / Solution

On $K=\mathbb Q(x,y)$, let $\sigma(x,y)=(y,-x)$ and $\tau(x,y)=(x,-y)$. They generate a <dihedral group> of order eight, and
$$
K^{\langle\sigma,\tau\rangle}
=\mathbb Q(x^2+y^2,x^2y^2).
$$
The inclusion from right to left is immediate. In the other direction, $x^2,y^2$ solve a quadratic over the displayed field and adjoining their square roots has degree at most four; the resulting total degree is at most eight, while the <automorphism-count bound for a finite field extension> gives the reverse bound.

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