Cross Posting this from MSE since it's been there for almost a month and it got a couple upvotes but no answers. MSE link Is every finite subgroup the integer points of a linear algebraic group?

Let $ K $ be a compact connected Lie group. For every finite subgroup $ \Gamma $ of $ K $ does there exist a linear algebraic group $ G $ such that the integer points are $$ G_\mathbb{Z} \cong \Gamma $$ and the real points are $$ G_\mathbb{R} \cong K. $$

I'm interested in this because sometimes the integer points are cool like $$ \operatorname{SO}_3(\mathbb{Z}) \cong S_4. $$

EDIT: Here is an attempt to clarify what I am looking for.

Consider 3 by 3 matrices with complex entries. For a $ 3 \times 3 $ complex matrix the conditions $$ I=MM^T $$ and $$ det(M)=1 $$ are polynomial in the entries of $ M $. The polynomials defining these conditions all have integer coefficients. The subset of matrices that satisfy these two constraints is the Lie group $ SO_3(\mathbb{C}) $. Now if we restrict the entries to be real then we get exactly the group $ SO_3(\mathbb{R}) $ which is a compact connected Lie group. Finally, if we restrict the entries to be integers we get $ SO_3(\mathbb{Z}) $ which is a finite group with 24 elements isomorphic to the symmetric group $ S_4 $.

So what I was really interested in was the idea that for any compact connected lie group $ K $ and finite subgroup $ \Gamma $ we can find a (finite collection of integer coefficient) polynomial constraints on the entries of a square matrix such that the complex matrices satisfying those constraints form a Lie group, the matrices satisfying those constraints and having real entries form a compact connected Lie group, and finally the matrices satisfying those constraints and having integer entries form a finite group isomorphic to $ \Gamma $.

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