Let $n \in \mathbb N^*$, $p(x) = x^n-1$ and $F$ be the splitting field of $p(x)$ over the rationals. Show that the Galois group of $F$ over $\mathbb Q$ is abelian.
Let us set $\omega\triangleq\mathrm{cis}(2\pi/n)$. This proof is trivial for $n$ equal to $1$ and $2$, so let's start with odd $n\neq1$.
$p(x) = (x-1)q(x)\,,$
where $q$ is seen to be irreducible in $\mathbb Q[x]$ by plotting its roots on the trigonometric circle. $F$ is actually the splitting field of $q(x)$:
$F = \mathbb Q(\omega,\dots,\omega^{n-1})\,.$
The Galois group is the set of automorphisms in $F$, which are the morphisms from $F$ to $F$ that map $\omega$ to a root of $q$:
$\mathrm{Gal}(F) = \{\omega\overset{\large\sim}\longmapsto\omega^k\,:\,k = 1,\dots,n-1\}\,.$
Therefore, it has order $n-1$.
For even $n\neq2$,
$p(x) = (x+1)(x-1)r(x)\,,$
$F = \mathbb Q(\omega,\dots,\omega^{\frac n2-1},\omega^{\frac n2+1},\dots,\omega^{n-1})$
and
$\begin{align} \mathrm{Gal}(F) &= \{\omega\overset{\large\sim}\longmapsto\omega^k\,:\,k = 1,\dots,n-1;\,k\neq n/2\}=\\ &= \{\omega\overset{\large\sim}\longmapsto\pm\omega^k\,:\,k = 1,\dots,n/2-1\}\,, \end{align}$
meaning $|Gal(F)| = n-2$.
Now what? Why does $\mathrm{Gal}(F)$ have to be abelian in both cases?
One idea I had was to prove it's cyclic instead, which immediately implies commutativity. I think the automorphism that maps $\omega$ to $\omega^{-2}$ always has the order of the group. Should I try that or am I missing an easier path?