 Solution

The relations ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$ give the dihedral group; its $2n$ substitutions $z\mapsto {\zeta }_n^{j}z$ and $z\mapsto {\zeta }_n^j/z$ are distinct. The element $w=z^n+z^{-n}$ is fixed. Since $z$ satisfies $X^{2n}-w{X}^n+1=0$, the extension has degree at most $2n$, while the automorphism group gives degree at least $2n$. Hence $L^G=\mathbb{C}(w)$.