I'm 4 years late, but see Stegeman, Math Annalen 261, 51-54 (1982) link.springer.com/article/10.1007/BF01456409. It is natural to conjecture you can take $C = 4/\pi^2$ (which is what you get from arithmetic progressions), and Stegeman shows $C = 4/\pi^3$ is acceptable.

It's a "noun adjunct" (en.wikipedia.org/wiki/Noun_adjunct), like "surface" in "surface group". It's the same in "novelty gift". But I agree "novel" is semantically better, unless the maximal subgroup is somehow amusing.

Replace $y$ with $-y$ to get rid of the $(-1)^n$.

The very special property of the generating set $S$ you chose is that it's normal (conjugation-invariant). In general if $S$ normal then it's true (though I think you're missing a $1/d_i$ factor). This is a simple consequence of Schur's lemma. Consult any first course in representation theory.

What does it even mean for a groupoid to embed in a group? $M_{13}$ may be defined in terms of permutation of 13 things, but that doesn't make it a subgroupoid (again, what even is that?) of $S_{13}$, because there are peculiar rules (en.wikipedia.org/wiki/Mathieu_groupoid#Construction).

No. For example you could take $I = J = \{n : \sqrt{2} n \in (1/3, 2/3) \pmod 1\}$.

I missed the restriction that $y$ should be in a subfield.

@PabloSpiga I said "more generally", because the situation you describe has needless restrictions. Read the first paragraph of my answer with $q$ replaced by $q^2$ and assume $n$ is prime if you want.