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?


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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