Every finite abelian group is the Galois group of some finite extension of the rationals

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

The simplest way I know to do this is what Dinesh suggests in the comment, together with my hint above.

1. Show that if $n\geq 1$, then the $n$th cyclotomic field $\mathbb{Q}(\zeta_n)$ (where $\zeta_n$ is a primitive $n$-th root of unity) has Galois group isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$. The structure of $(\mathbb{Z}/n\mathbb{Z})^*$ is well-understood in terms of the prime factorization of $n$.

2. Given a finite abelian group $G$, find an $n$ such that $G$ is a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$.

3. Prove that if $A$ is an abelian group, and $H$ is a subgroup of $A$, then there exists a subgroup $K$ of $A$ such that $A/K\cong H$.

4. Use the Fundamental Theorem of Galois Theory and the points above to obtain the desired result.

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.