PlanetMath: Galois groups of finite abelian extensions of $\mathbb{Q}$

Proof. This will first be proven for

Let $\vert G\vert=n$ . By Dirichlet's theorem on primes in arithmetic progressions, there exists a prime with $p \equiv 1 \operatorname{mod} n$ . Let $\zeta_p$ denote a primitive $% latex2html id marker 353 $ p^{\text{th}}$$ root of unity. Let $% latex2html id marker 355 $ L=\mathbb{Q}(\zeta_p)$$ . Then $% latex2html id marker 357 $ L/\mathbb{Q}$$ is Galois with $% latex2html id marker 359 $ \operatorname{Gal}(L/\mathbb{Q})$$ cyclic of order . Since divides , there exists a subgroup of $% latex2html id marker 369 $ \operatorname{Gal}(L/\mathbb{Q})$$ such that $\displaystyle \vert H\vert=\frac{p-1}{n}$ . Since $% latex2html id marker 373 $ \operatorname{Gal}(L/\mathbb{Q})$$ is cyclic, it is abelian, and is a normal subgroup of $% latex2html id marker 377 $ \operatorname{Gal}(L/\mathbb{Q})$$ . Let , the subfield of fixed by . Then $% latex2html id marker 385 $ K/\mathbb{Q}$$ is Galois with $% latex2html id marker 387 $ \operatorname{Gal}(K/\mathbb{Q})$$ cyclic of order . Thus, $% latex2html id marker 391 $ \operatorname{Gal}(K/\mathbb{Q}) \cong G$$ .

Let and be distinct primes with $p \equiv 1 \operatorname{mod} n$ and $q \equiv 1 \operatorname{mod} n$ . Then there exist subfields and of $% latex2html id marker 405 $ \mathbb{Q}(\zeta_p)$$ and $% latex2html id marker 407 $ \mathbb{Q}(\zeta_q)$$ , respectively, such that $% latex2html id marker 409 $ \operatorname{Gal}(K_1/\mathbb{Q}) \cong G$$ and $% latex2html id marker 411 $ \operatorname{Gal}(K_2/\mathbb{Q}) \cong G$$ . Note that $% latex2html id marker 413 $ K_1 \cap K_2=\mathbb{Q}$$ since $% latex2html id marker 415 $ \mathbb{Q} \subseteq K_1 \cap K_2 \subseteq \mathbb{Q}(\zeta_p) \cap \mathbb{Q}(\zeta_q)=\mathbb{Q}$$ . Thus, $K_1 \neq K_2$ . Therefore, for every prime with $p \equiv 1 \operatorname{mod} n$ , there exists a distinct number field such that $% latex2html id marker 425 $ K/\mathbb{Q}$$ is Galois and $% latex2html id marker 427 $ \operatorname{Gal}(K/\mathbb{Q}) \cong G$$ . 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 with $p \equiv 1 \operatorname{mod} n$ .