The smallest degree of a nontrivial complex representation of the monster group $ M $ is $ 196883 $. This irrep has Schur indicator $ 1 $, so the image must lie in the orthogonal group. In fact, since $ M $ is perfect the image must lie in the special orthogonal group. Thus $ M $ is a subgroup of the Lie group $ SO(196883) $.
Is $ M $ maximal among the closed subgroups of $ SO(196883) $?
To give a very small example of why I think this might be true consider the simple group $ A_5 $. The smallest degree of a nontrivial complex representation of $ A_5 $ is $ 3 $. And the degree 3 irrep of $ A_5 $ has Schur indicator $ 1 $. So, for the same reasons described above, $ A_5 $ must be a subgroup of the Lie group $ SO(3) $.
And it turns out that $ A_5 $ is maximal among the closed subgroups of $ SO(3) $.
This seems interesting to me since $ SO(3)/A_5 $ is the famous Poincare homology sphere. Perhaps $ SO(196883)/M $ also has some interesting properties. For example maybe the isometry group of $ SO(196883)/M $ acts primitively? https://www.semanticscholar.org/paper/Primitive-actions-and-maximal-subgroups-of-Lie-Golubitsky/aa52e30d07f08aaeb48fee1b044558cd045fe3b3