Show that any cyclic group with square free order is the Galois group over $\mathbb{Q}$ of some field extension.
I'm curious because I (believe) know the proof when $G$ is a finite cyclic group of any order, so I'm wondering if perhaps the square-free case is easier. This is off an old qualifying exam, and the hint is to consider $x^n-1$ for suitable $n$.
In any case, an outline of the proof for a cyclic group is as follows: $G \cong \mathbb{Z}_n$ for some $n$. Then, there exists a prime $p$ such that $n \mid p-1$. Writing $p-1=n\cdot m$, we know $\mathbb{Z}_{m}$ is a subgroup of $\mathbb{Z}_{p-1}$. Then one considers $\zeta$ a primitive $p$th root of unity. We know $Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong \mathbb{Z}_{p-1}$. As this is abelian, $\mathbb{Z}_m$ is a normal subgroup of this group, and so if $K$ is the fixed field of $\mathbb{Z}_m$, it has Galois group isomorphic to $Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) / \mathbb{Z}_m \cong \mathbb{Z}_n \cong G$.
Assuming the above is correct, how would square free order change this at all? Thanks!