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?


1 Answer 1


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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