Inverse Galois problem over Q

Problem. Prove or disprove: for every finite group $G$ there is a finite Galois extension $L/\mathbb Q$ with

$$\operatorname{Gal}(L/\mathbb Q)\cong G .$$

Equivalently, for every finite group $G$ there is a monic irreducible polynomial $f\in\mathbb Q[x]$ whose splitting field over $\mathbb Q$ has Galois group isomorphic to $G$. The two forms agree because every finite Galois extension of $\mathbb Q$ is the splitting field of the minimal polynomial of a primitive element.

A stronger variant is the regular inverse Galois problem: for every finite group $G$, find a Galois extension $L/\mathbb Q(t)$ with $\operatorname{Gal}(L/\mathbb Q(t))\cong G$ and $L\cap\overline{\mathbb Q}=\mathbb Q$. By Hilbert's irreducibility theorem, a regular realization specializes at infinitely many $t_0\in\mathbb Q$ to Galois extensions of $\mathbb Q$ with group $G$. So a positive answer to the regular problem implies a positive answer here. The problem here is the one over $\mathbb Q$ as stated above.

Background

The question goes back to Hilbert's 1892 irreducibility theorem, which he used to realize the symmetric and alternating groups over $\mathbb Q$. Emmy Noether proposed a geometric approach in 1918. If the fixed field $\mathbb Q(x_1,\dots,x_n)^G$ of a permutation action is purely transcendental, then $G$ has a regular realization. Swan showed that this fails for the cyclic group of order $47$, so Noether's approach cannot work for all groups [2].

The most productive general method is rigidity. It uses rational rigid tuples of conjugacy classes to produce $G$-covers of $\mathbb P^1$ defined over $\mathbb Q$, branched over three or more points, hence regular realizations [2, 3]. Thompson used it to realize the Monster group [4]. Other approaches use embedding problems, which proceed by lifting a realization of a quotient of $G$, and arithmetic geometry, including rational points on moduli spaces of covers and Galois representations attached to modular forms [1, 2].