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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

  • 3
    You can't show that because it isn't necessarily true. For example, the compositum of a quadratic extension $K$ with itself still has Galois group $C_2$ rather than $C_2 \times C_2$.2012-04-13
  • 0
    **Hint.** If $A$ is an abelian group, and $H$ is a subgroup of $A$, then there is a subgroup $K$ of $A$ such that $A/K\cong H$.2012-04-13
  • 0
    I date myself with this one, but you might find Theorem 5 and its first corollary of interest ( in the 1965 edition of Lang's Algebra).2012-04-13
  • 3
    Why don't you use the fact that $Gal( \mathbb{Q}(\zeta_n) / \mathbb{Q})$ is isomorphic to the group of units of $\mathbb{Z}/n\mathbb{Z}$ ? find an appropriate $n$ such that the finite abelian group $G$ is a quotient of $\mathbb{Z}/n\mathbb{Z}$ and use the galois correspondence to prove the existence of such field.2012-04-13

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