Proof. This will first be proven for
$G$ cyclic.
Let $|G|=n$ . By Dirichlet's theorem on primes in arithmetic progressions, there exists a prime $p$ with $p \equiv 1 \operatorname{mod} n$ . Let $\zeta_p$ denote a primitive $p^{{th}}$ root of unity. Let $L=\mathbb{Q}(\zeta_p)$ . Then $L/\mathbb{Q}$ is Galois with $\operatorname{Gal}(L/\mathbb{Q})$ cyclic of order $p-1$ . Since $n$
divides $p-1$ , there exists a subgroup $H$ of $\operatorname{Gal}(L/\mathbb{Q})$ such that $\displaystyle |H|=\frac{p-1}{n}$ . Since $\operatorname{Gal}(L/\mathbb{Q})$ is cyclic, it is abelian, and $H$ is a normal subgroup of $\operatorname{Gal}(L/\mathbb{Q})$ . Let $K=L^H$ , the subfield of
$L$ fixed by $H$ . Then $K/\mathbb{Q}$ is Galois with $\operatorname{Gal}(K/\mathbb{Q})$ cyclic of order $n$ . Thus, $\operatorname{Gal}(K/\mathbb{Q}) \cong G$ .
Let $p$ and $q$ be distinct primes with $p \equiv 1 \operatorname{mod} n$ and $q \equiv 1 \operatorname{mod} n$ . Then there exist subfields $K_1$ and $K_2$ of $\mathbb{Q}(\zeta_p)$ and $\mathbb{Q}(\zeta_q)$ , respectively, such that $\operatorname{Gal}(K_1/\mathbb{Q}) \cong G$ and $\operatorname{Gal}(K_2/\mathbb{Q}) \cong G$ . Note that $K_1 \cap K_2=\mathbb{Q}$ since $\mathbb{Q} \subseteq K_1 \cap K_2 \subseteq \mathbb{Q}(\zeta_p) \cap \mathbb{Q}(\zeta_q)=\mathbb{Q}$ . Thus, $K_1 \neq K_2$ . Therefore, for every prime $p$ with $p \equiv 1 \operatorname{mod} n$ , there exists a distinct
number field $K$ such that $K/\mathbb{Q}$ is Galois and $\operatorname{Gal}(K/\mathbb{Q}) \cong G$ . The theorem in the cyclic case follows from using the full force of Dirichlet's theorem on primes in arithmetic progressions: There exist infinitely many primes $p$ with $p \equiv 1 \operatorname{mod} n$ .
The general case follows immediately from the above argument, the fundamental theorem of finite abelian groups, and a theorem regarding the Galois group of the compositum of two Galois extensions. 