TITLE: The inverse Galois problem and the Monster
QUESTION [72 upvotes]: I have a slight interest in both the inverse Galois problem and in the Monster group.  I learned some time ago that all of the sporadic simple groups, with the exception of the Mathieu group $M_{23}$, have been proven to be Galois groups over $\mathbb{Q}$.
In particular, the Monster group has been proven to be a Galois group over $\mathbb{Q}$.  What techniques are used to prove such an assertion?
Is proving that $M_{23}$ is also Galois over $\mathbb{Q}$ within reach?  I assume that the same techniques do not apply, for it is a much more manageable group than the Monster.

REPLY [21 votes]: I'll add a brief commment to Arne Semeets's thorough and useful answer.  If I fix three rational conjugacy classes c_0, c_1, c_infty in a finite group G, then there are finitely many isomorphism classes of unramified G-coverings
X -> P^1 - 0,1,infty /Qbar
with the property that the image of tame inertia at 0 (resp 1,infty) lies in the class c_0 (resp c_1,c_infty) of G.
Call the set of such covers H.  What is |H|?  One can check (by comparison with the complex case) that the number of such covers is the number of conjugacy classes of triples (g_0,g_1,g_infty) with g_i in c_i and g_0 g_1 g_infty = 1.  To say that (c_0,c_1,c_infty) is rigid is just to say that there is precisely ONE such triple.
In that case, H consists of just one cover.  But H is evidently preserved by Galois conjugacy.  So this unique cover is defined over Q.  (One has to be slightly more careful when G has nontrivial center, in which case what I've really proved is something more like "there's a cover whose isomorphism class is defined over Q," not quite the same in general as "there's a cover defined over Q."