The splitting field of $x^n-2$ over $\mathbb{Q}$ is $K.L$ where
$K=\mathbb{Q}(\zeta_n)$ and $L=\mathbb{Q}(\sqrt[n]{2})$, so the order of the Galois group
is
$$
[K.L:\mathbb{Q}] = \frac{[K:\mathbb{Q}]\cdot[L:\mathbb{Q}]}{[K\cap L:\mathbb{Q}]}
= \frac{n \phi(n)}{[K\cap L:\mathbb{Q}]}.
$$
It remains to compute $m:=[K\cap L:\mathbb{Q}]$.

First show that $K\cap L = \mathbb{Q}(\sqrt[m]{2})$. For this, note that the norm
$N_{L/(K\cap L)}(\sqrt[n]{2})$ is in $K\cap L$. This norm is the product
of the $n/m$ conjugates of $\sqrt[n]{2}$ over $K$, so it is the product of $n/m$ of the conjugates of $\sqrt[n]{2}$ over $\mathbb{Q}$, and each of these conjugates has the form
$\zeta_n^i \sqrt[n]{2}$. Hence the norm has the form $\zeta_n^i \sqrt[n]{2}^{n/m}
= \zeta_n^i \sqrt[m]{2}$.
Since this is in $K\cap L$, and $\zeta_n^i\in K$,
it follows that $\sqrt[m]{2}\in K$, so $\sqrt[m]{2}\in K\cap L$.
But $[\mathbb{Q}(\sqrt[m]{2}):\mathbb{Q}]=m=[K\cap L:\mathbb{Q}]$,
so indeed $K\cap L=\mathbb{Q}(\sqrt[m]{2})$.

Next, since $\sqrt[m]{2}\in K$, and $K/\mathbb{Q}$ is abelian, it follows that $\mathbb{Q}(\sqrt[m]{2})/\mathbb{Q}$ is abelian and hence is Galois. Since $\zeta_m\sqrt[m]{2}$ is a conjugate of $\sqrt[m]{2}$ over $\mathbb{Q}$, it follows that $\zeta_m\in\mathbb{Q}(\sqrt[m]{2})$, so in particular $\zeta_m\in\mathbb{R}$, whence $m\le 2$.

The final step is to determine when $\sqrt{2}\in\mathbb{Q}(\zeta_n)$. For this, note
that $\sqrt{2}\in\mathbb{Q}(\zeta_8)$ but $\sqrt{2}\notin\mathbb{Q}(\zeta_4)$, and that
$\mathbb{Q}(\zeta_r)\cap\mathbb{Q}(\zeta_s)=\mathbb{Q}(\zeta_{(r,s)})$. Thus
$\sqrt{2}\in\mathbb{Q}(\zeta_n)$ if and only if $8\mid n$. So the conclusion is that
the splitting field of $x^n-2$ over $\mathbb{Q}$ has degree $n\phi(n)$ if $8\nmid n$,
and has degree $n\phi(n)/2$ if $8\mid n$.

**Question:** is there a way to do this without showing that $K\cap L=\mathbb{Q}(\sqrt[m]{2})$, by using from the start that $K\cap L$ is a subfield of $\mathbb{Q}(\sqrt[n]{2})$ which is Galois over $\mathbb{Q}$, and then somehow going directly to the final step?

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