 Solution

Let $a=\sqrt[n]{2}$ and $F=\mathbb{Q}(\xi )$. The roots of $X^n-2$ are $a{\xi }^j$, so $L=F(a)$. This splitting extension is separable in characteristic zero, hence Galois over $\mathbb{Q}$. The subextension $F/\mathbb{Q}$ is itself Galois by part (iii), so restriction gives a surjectionwith normal kernel $H=\mathrm{Gal}(L/F)$. Surjectivity follows from the extension theorem for embeddings and normality of $L/\mathbb{Q}$.

For $\sigma \in H$, write $\sigma (a)=a{\xi }^{j(\sigma )}$. This defines an injective homomorphism $H\to ⟨\xi ⟩$: injectivity holds because $L=F(a)$, and multiplicativity holds because $H$ fixes all roots of unity in $F$. A subgroup of a cyclic group is cyclic. The first isomorphism theorem therefore givesThere is no need to assume that $|H|=n$; the intersection with the cyclotomic field can reduce its order.