Galois Group of $x^n-2$ Let $n \in \mathbb{N}$, then the order of the Galois Group
of $x^n-2$ coincide with $n \phi(n)$ for $n\in \{ 1 , \dots , 36 \}$
except for $n=\{ 8, 16, 24, 32 \}$ where this order is $\frac{ n \phi (n)}{2}$
and is easy to prove that for $p$ prime we have that the order
of the Galois Group of $x^p-2$ is $p(p-1)$.
What is the order of the Galois Group of $x^n-2 : n\in \mathbb{N}$ is general?
Thanks in advance.
 A: 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?
A: See this stackexchange question., or this worked exercise.
