[Math] Every finite abelian group is the Galois group of some finite extension of the rationals

galois-theorygroup-theory

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

Best Answer

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.

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