TITLE: Every finite abelian group is the Galois group of some finite extension of the rationals
QUESTION [21 upvotes]: I'm trying to prove that every finite abelian group is the Galois group of of some finite extension of the rationals. I think I'm almost there. 
Given a finite abelian group $G$, I have constructed field extensions whose Galois groups are the cyclic groups occurring in the direct product of $G$. How do I show that the compositum of these fields has Galois group $G$.
Cheers

REPLY [2 votes]: All that u have to use from the group structure of (Z/nZ)* is that it is direct product of cyclic groups. To find the appropriate $n$ use the dirichlet's theorem from which we can say there are infinitely many primes which are 1(mod n). And the thing about 'final bit' it is just the fundamental theorem of Galois theory. When you see that the $G$ can be seen as a quotient of $(Z/nZ)$* then according to the fundamental theorem there exists an intermediate field say $K$ between $\mathbb{Q}(\zeta_n)$ and $\mathbb{Q}$ such that $Gal(K/Q)$ is $G$. Now the only task remaining for you is showing that $G$ can be seen as a quotient of $(Z/nZ)$* using dirichlet's theorem.
Good luck.