 Solution

For odd prime $n$, every subgroup is one of $\{1\}$, $⟨\sigma ⟩$, a reflection subgroup $H_j=⟨{\sigma }^j\tau ⟩$ for $0\le j<n$, or $G$. Indeed, a subgroup containing a nonidentity rotation contains all rotations; with any reflection it is then $G$. A subgroup with no nonidentity rotation contains at most one reflection, since the product of distinct reflections is a nonidentity rotation.

The Galois correspondence gives the following fixed fields. The trivial subgroup corresponds to $L=\mathbb{C}(z)$, the rotation subgroup to $\mathbb{C}(z^n)$, and $G$ to the base $\mathbb{C}(w)$. Since composition of the stated automorphisms gives $({\sigma }^j\tau )(z)={\zeta }_n^{-j}/z$, the reflection subgroup corresponds toThis rational invariant has degree two, hence generates the fixed field of its order-two subgroup, by the same degree argument as part (i). These fields have degrees respectively $2n,2,n,1$ over $\mathbb{C}(w)$.

An intermediate field is Galois over the base exactly when its corresponding subgroup is normal in $G$. The trivial subgroup, the rotations and $G$ are normal; no reflection subgroup is normal because $n$ is odd and greater than two. Hence the Galois intermediate extensions and groups areThe $n$ reflection fixed fields are not Galois over the base.