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The inverse Galois problem asks whether every finite group appears as the Galois group of a Galois extension of the rational numbers.

Is anything known about the anologous problem, where the extensions are not required to be Galois? In other words, for a finite group $G$, does there exist a finite field extension $K$ of $\mathbb{Q}$ such that $\mathrm{Aut}(K/\mathbb{Q})=G$?

Is this suspected to be as difficult as the inverse Galois problem or easier?

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The answer to this is positive. The first correct proof, it seems, was given in

Michael D. Fried. A note on automorphism groups of algebraic number fields. Proc. Amer. Math. Soc., 80(3):386–388, 1980.

For a generalization to Hilbertian fields and some history see for example

F. Legrand and E. Paran. Automorphism groups over Hilbertian fields. Journal of Algebra Volume 503, 2018. journal website

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  • $\begingroup$ I have just yesterday arrived at a proof for slighly variant statement, true for Q. It is as follows: Given a finite group $G$, say of order $n$ for which Inverse Galois Problem is true over Q and any prime $p$, there exists infinitely many extensions of Q of degree $np$ which are non-Galois, but with Aut(K/Q) = G. Is this an interesting result? I am currently working to see if it is true when the prime $p$ is replaced by any positive integer. $\endgroup$
    – P Vanchinathan
    Commented Sep 6 at 4:11
  • $\begingroup$ @PVanchinathan: one should check the references I gave whether the methods there maybe already give this or something similar (without the assumption that the inverse Galois problem is true for $G$) $\endgroup$
    – Arno Fehm
    Commented Sep 6 at 12:06

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