I was reading Dummit and Foote and they talked about the study of Class Field Theory, which studies abelian extensions of an arbitrary finite extension of $\mathbb{Q}$ (and related, a theorem that says any abelian $G$ may be realized as the Galois group of some subfield of a cyclotomic extension). So my question is, is there any group that cannot be realized as a Galois group over $\mathbb{Q}$? I know that from above $G$ must be nonabelian, and that if we don't require $\mathbb{Q}$ as our base field specifically, using the Fundamental Theorem of Galois Theory the answer is "no" via realizing $G < S_n$ for some $n$ and then taking the extension $K/K^G$, where $K = \mathbb{Q}[x_1, ... , x_n]$. But if we require the base field to be $\mathbb{Q}$?
Is there an example of a group $G$ which cannot be a Galois group over $\mathbb{Q}$?
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galois-theory
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class-field-theory
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7Pretty big open problem. Look up inverse galois theory. Maybe I should add the mention that this only concerns finite groups though (right, finite extensions)... – 2017-02-02
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0But for many groups, including the solvable groups, the groups $A_n$ and the groups $S_n$, it is known that they can be the galois group of a polynomial in $\mathbb Q[x]$ – 2017-02-02
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1Editing the tags, because [tag:finite-fields] applies only when the fields are finite as sets. Here you only ask about extensions of $\Bbb{Q}$, so they are all infinite, because there are infinitely many rational numbers. – 2017-02-03
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1In the same vein as your question, the following approach by J. Minac and his collaborators (*) could be of interest. Let G_F be the absolute Galois group of a number field F containing a primitive q-th root of 1, q being a power of a prime p. The authors study the q-central descending series of G_F using the Bloch-Kato-Milnor conjecture (now a theorem of Voevodsky) on a canonical isomorphism between the cohomology algebra H*(G_F, Z/qZ) and the Milnor K-theory algebra K*(F) mod q. – 2017-02-04
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1(continued) One surprising result is that the third quotient G[3] in the descending filtration determines the cohomology algebra. This puts constraints on G_F , which allow to construct examples of profinite groups which cannot be absolute Galois groups of number fields. See e.g. Chebolu-Efrat-Minac, Math. Annalen 352 (2012) and Efrat-Minac, Amer. J. of Math. 133 (2011) . – 2017-02-04
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0What about ${\mathbb Z}$? – 2017-02-24
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0Related: https://mathoverflow.net/questions/80359/which-small-finite-simple-groups-are-not-yet-known-to-be-galois-groups-over-q – 2018-07-04
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0There is a remark in _Topics in Galois Theory_ (second edition, p. 16, just before §2.2), by J.-P. Serre, which says that the $p$-adic Lie group $\Bbb Z_p \times \Bbb Z_p$ is not a Galois group over $\Bbb Q$. – 2018-07-08