= 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 surjection
$$
G\longrightarrow\operatorname{Gal}(F/\mathbb Q)
$$
with normal kernel $H=\operatorname{Gal}(L/F)$. 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 $H\to\langle\xi\rangle$: 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 gives
$$
\boxed{H\text{ cyclic and normal},\qquad G/H\cong(\mathbb Z/n\mathbb Z)^\times.}
$$
There is no need to assume that $|H|=n$; the intersection with the cyclotomic <field> can reduce its order.
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