# Abelian extension of an algebraic number field whose Galois group is isomorphic to a given finite abelian group

Let $K$ be an algebraic number field, i.e. a finite extension of $\mathbb{Q}$. Let $G$ be a finite abelian group. Does there exist a Galois extension of $K$ whose Galois group is isomorphic to $G$? I can prove this if $K$ is $\mathbb{Q}$ by using a special case of Dirichlet's prime number theorem, i.e. there are infinitely many prime numbers $p$ such that $p \equiv 1$ (mod $n$) for a given integer $n \ge 1$. So if $|G|$ is relatively prime to $[K: \mathbb{Q}]$, there exists such an extension, but I have no idea otherwise.

Let $m=[K:\Bbb{Q}]$. By the structure theorem of finite abelian groups there exists integers $d_1\mid d_2\mid \cdots\mid d_k$ such that $$G\cong C_{d_1}\times C_{d_2}\times \cdots\times C_{d_k}.$$ Choose distinct primes $p_1,p_2,\ldots,p_k$ such that $p_i\equiv1\pmod{md_i}$. Let $\zeta_i$ be a complex primitive root of unity of order $p_i$. Consider the extension $K(\zeta_i)/K$. As a splitting field of a cyclotomic polynomial $K(\zeta_i)$ is Galois over $K$. Because any $K$-automorphism of $K(\zeta_i)$ is determined by the image of $\zeta_i$, the group $Gal(K(\zeta_i)/K)$ is isomorphic to a subgroup $G_i$ of $Gal(\Bbb{Q}(\zeta_i)/\Bbb{Q})\cong C_{p_i-1}.$ The degree of the extension $$[K(\zeta_i):K]=\frac{[K(\zeta_i):\Bbb{Q}]}{[K:\Bbb{Q}]}= \frac{[K(\zeta_i):\Bbb{Q}(\zeta_i)]\cdot(p_i-1)}{m}$$ is a multiple of $d_i$. Therefore $d_i\mid |G_i|$. This implies that there is an intermediate field $K_i, K\subseteq K_i\subseteq K(\zeta_i)$ such that $K_i/K$ is cyclic of order $d_i$.
If we can show that the extensions $K_i/K$ are linearly disjoint, then we reach the conclusion that their compositum $\tilde{K}=K_1K_2\cdots K_k$ is a Galois extension with $Gal(\tilde{K}/K)\cong G$. Initially I felt that this follows immediately from the fact that the extensions $\Bbb{Q}(\zeta_i)/\Bbb{Q}, i=1,2,\ldots,$ are linearly disjoint over the rationals. Unfortunately I cannot justify this at the moment. Leaving this as incomplete for now.
• May be simpler to build it without a compositum using a subfield of $K(\zeta)$ with $\zeta$ of order $p_1p_2\cdots p_k$? – Jyrki Lahtonen Nov 25 '15 at 7:30