My favourite group as far as the inverse Galois problem goes is . This is not known to be a Galois group over for any , the difficulty of course being that is must correspond to an even Galois representation. A more tractable case is , and this was recently answered by David Zywina here. Here is a more elementary version of that construction. Suppose that is a classical modular form of weight three with coefficients in and quadratic Nebentypus character . Note that there is an isomorphism . For all primes in , one obtains a representation:
with determinant . There are two cases, depending on whether is split or not. If splits, then, assuming is not CM, the image of restricted to the kernel of is for sufficiently large which can be explicitly determined in any specific case. Thus the image of is plus the image of complex conjugation:
Since , there exists an element of square , and hence an element in equal to
Hence the image of contains a scalar element of determinant , and thus it has projective image .
If , then, from the isomorphism , there is an isomorphism where is the Galois conjugate induced by complex conjugation. It follows that the *projective* image of lands in . The image of is thus, for sufficiently large , a subgroup of with projective image containing . We first observe that this implies that contains ). It suffices to show that it contains all the transvections; yet the lift of any transvection in is a transvection of order times a scalar of order prime to , which one can remove by taking an appropriate power. Since the determinant of is , this leaves only the following three possibilities for the image of :
1. The subgroup of of matrices with determinant .
2. The previous subgroup together with the the scalar element with .
3. The group together with .
The third group does not have a non-scalar element of order correponding to complex conjugation, and the first has traces which do not generate . Hence the image must be the second, which has projective image .
To conclude the argument, it suffices to show that there exists such a . Consulting William Stein’s tables, one may take
for a quadratic where . Since , this form does not have CM by or , so is a Galois group for sufficiently large , which one could compute exactly if one wanted. My impression from the notation in William Stein’s tables is that the fixed field of the kernel of is , so this is presumably the same family of examples that arises in Zywina. Other examples (in the range of William’s tables) are as follows:
Note that this argument requires slightly more than pure thought; it was key that there existed a non-CM form with coefficient field , and there is no *a priori* reason why there should exist any such form. For example, suppose one wanted to generalize this argument to to . Then one would want to look for a non-endoscopic Siegel cusp form of weight where (edit) is odd with Hecke eigenvalues in and quadratic Nebentypus character. Possibly such things exist but perhaps they don’t!