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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!

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  • $\begingroup$ How do you know that there exists a prime $p$ such that $n\mid p-1$? You can prove it using Dirichlet's theorem on primes in arithmetic progressions, but this is far from a simple result. $\endgroup$ – Mathmo123 Nov 23 '16 at 11:35
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As pointed out by @Mathmo123, the existence of $p\equiv 1$ mod $n$ appeals to a non elementary result in the style of Dirichlet’s theorem on primes in an arithmetic progression. But note that this special case is an easy consequence of the properties of cyclotomic polynomials (see Washington, chap.2, coroll.2.11). While we are there, I propose the following solution to your problem using only the classical theory of cyclotomic fields.

Let us show that any cyclic group $C_n$ can be realized as the Galois group of a normal extension of $\mathbf Q$. Since $C_n$ is a direct product of cyclic groups $C_{{p_i}^{r_i}}$ (where $p_i$ are distinct primes), let us first consider the problem for $C_{p^r}$. For simplicity, we’ll always assume $p$ odd, the case $p=2$ usually requiring extra-care e.g. about exponents (not difficult, but boring). It is classically known that the cyclotomic extension $\mathbf Q(\zeta_{p^{r+1}}) / \mathbf Q $ has Galois group isomorphic to $(\mathbf Z / p^{r+1})^* \cong (\mathbf Z / (p-1)) \times (\mathbf Z / p^{r})$, hence $\mathbf Q(\zeta_{p^{r+1}})$ is the compositum of $\mathbf Q(\zeta_p)$ and a $C_{p^r}$ -extension of $\mathbf Q$, usually denoted $\mathbf B_{p,r}$. For any prime $q\neq p$, $\mathbf B_{q,s}$ and $\mathbf B_{p,r}$ are linearly disjoint, so by composing such extensions, we get the required $C_n$-extension of $\mathbf Q$. It does not appear that your « square free » hypothesis on $n$ brings any simplification.

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  • $\begingroup$ I suppose the minor simplification would be that your $r_i$ must all be 1 so that one need not look at $\text{B}_{p,r}$. $\endgroup$ – Curious Nov 24 '16 at 19:01
  • $\begingroup$ But you have to look at B_p, 1 , so... $\endgroup$ – nguyen quang do Nov 24 '16 at 20:58
  • $\begingroup$ oh sorry, I overlooked that you took a $\zeta^{r+1}$st primitive root instead of just a $\zeta^{r}$th. $\endgroup$ – Curious Nov 26 '16 at 4:29

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