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

###### Theorem.

Let $G$ be a finite abelian group with $|G|>1$. Then there exist infinitely many number fields $K$ with $K/\mathbb{Q}$ Galois and $\operatorname{Gal}(K/\mathbb{Q})\cong G$.

###### 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 $p^{\text{th}}$ root of unity. Let $L=\mathbb{Q}(\zeta_{p})$. Then $L/\mathbb{Q}$ is Galois with $\operatorname{Gal}(L/\mathbb{Q})$ cyclic of order (http://planetmath.org/OrderGroup) $p-1$. 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 (http://planetmath.org/FixedField) 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 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 , the fundamental theorem of finite abelian groups (http://planetmath.org/FundamentalTheoremOfFinitelyGeneratedAbelianGroups), and a theorem regarding the Galois group of the compositum of two Galois extensions. ∎

Title Galois groups of finite abelian extensions of $\mathbb{Q}$ GaloisGroupsOfFiniteAbelianExtensionsOfmathbbQ 2013-03-22 16:18:40 2013-03-22 16:18:40 Wkbj79 (1863) Wkbj79 (1863) 11 Wkbj79 (1863) Theorem msc 11R32 msc 11N13 msc 11R20 msc 12F10 AbelianNumberField