Inverse Galois Problem
All finite groups do occur as Galois groups. It is easy to construct field extensions with any given finite group as Galois group, as long as one does not also specify the ground field.
For that, choose a field K and a finite group G. Cayley's theorem says that G is (up to isomorphism) a subgroup of the symmetric group S on the elements of G. Choose indeterminates {xα}, one for each element α of G, and adjoin them to K to get the field F = K({xα}). Contained within F is the field L of symmetric rational functions in the {xα}. The Galois group of F/L is S, by a basic result of Emil Artin. G acts on F by restriction of action of S. If the fixed field of this action is M, then, by the fundamental theorem of Galois theory, the Galois group of F/M is G.
It is an open problem to prove the existence of a field extension of the rational field Q with a given finite group as Galois group. Hilbert played a part in solving the problem for all symmetric and alternating groups. Igor Shafarevich proved that every solvable finite group is the Galois group of some extension of Q. Various people have solved the inverse Galois problem for selected non-abelian simple groups. Existence of solutions has been shown for all but possibly one (Mathieu group M23) of the 26 sporadic simple groups. There is even a polynomial with integral coefficients whose Galois group is the Monster group.
Read more about this topic: Galois Theory
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