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 $G$.


  • 3
    $\begingroup$ 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 $C_2 \times C_2$. $\endgroup$ – Qiaochu Yuan Apr 13 '12 at 16:49
  • $\begingroup$ Hint. If $A$ is an abelian group, and $H$ is a subgroup of $A$, then there is a subgroup $K$ of $A$ such that $A/K\cong H$. $\endgroup$ – Arturo Magidin Apr 13 '12 at 17:19
  • $\begingroup$ 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). $\endgroup$ – Chris Leary Apr 13 '12 at 17:26
  • 3
    $\begingroup$ 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. $\endgroup$ – Dinesh Apr 13 '12 at 17:55

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.

  • $\begingroup$ @ Arturo: For step 2 my group theory knowledge is lacking. By what result does a finite abelian group occur as a subgroup of $(\Bbb{Z}/n\Bbb{Z})$∗ for some n? Other than that I can complete the argument. ... The steps I was following before can be found here: math.tifr.res.in/~eghate/kw.pdf (See the discussion proceeding theorem 2). I would like to understand this as I see it as a nice example to some ramification theory that I have been studying recently. I am only unsure on the final bit, "Then some Galois theory shows Gal(K/Q) = G, and we are done" Thanks for any help $\endgroup$ – daisy Apr 14 '12 at 10:51
  • 5
    $\begingroup$ @daisy: The fundamental theorem of finitely generated abelian groups tells you that $G$ can be written as a direct product of cyclic groups of certain orders, $G\cong C_{n_1}\times\cdots\times C_{n_k}$ with $n_1|n_2|\cdots|n_k$. All you need to do is find primes $p_i$ such that $n_i|p_i-1$, and there are infinitely many such primes for each $i$ by Dirichlet's Theorem of primes in arithmetic progressions. Then $n=p_1\cdots p_k$ will do. $\endgroup$ – Arturo Magidin Apr 14 '12 at 19:40
  • $\begingroup$ @daisy: The "some Galois theory" is the inclusion reversing correspondence between subgroups and subfields. The linear independence translates into the fact that the subgroups intersect trivially and generate the entire group $\mathrm{Gal}(K/Q)$. This in turn tells you that $\mathrm{Gal}(K/Q)$ is isomorphic to the product of the corresponding subgroups. $\endgroup$ – Arturo Magidin Apr 14 '12 at 22:55
  • 1
    $\begingroup$ @Shengwen, $(\mathbb Z/n\mathbb Z)^*$ is not cyclic, except when $n=2^\varepsilon p^r$ where $\varepsilon=0$ or $1$, and $p$ is an odd prime. Plus the special case when $n=4$. $\endgroup$ – Lubin Feb 18 '13 at 15:36
  • 3
    $\begingroup$ In fact, one does not need step 3, as one may take $G$ to be a quotient group from the beginning: $$G \simeq \prod_{i=1}^k \mathbb Z/n_i\mathbb Z \twoheadleftarrow \prod_{i=1}^k \mathbb Z/(p_i-1)\mathbb Z \simeq \left( \mathbb Z/p_1\ldots p_k\mathbb Z \right)^*.$$ $\endgroup$ – evgeny Nov 13 '16 at 10:54

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.


@Daisy, your question is part of something more general called the Inverse Galois Problem, which goes back all the way to David Hilbert and Emmy Noether: which finite groups can be realized as Galois group over a general field?. See for example the discussion https://mathoverflow.net/questions/6743/the-inverse-galois-problem-what-is-it-good-for. There exists a vast literature on the subject.

  • 4
    $\begingroup$ The abelian case is way much simpler than that..for example its an exercise in Lang's Algebra. $\endgroup$ – Dinesh Apr 13 '12 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.