Is every finite subgroup the integer points of a linear algebraic group?

Question

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 $.

Answer 1

The answer to this question is no. There exist finite groups $\Gamma \subset K$ of a compact connected Lie group $K$, such that for any algebraic $\mathbb Q$-group $G$ with $G(\mathbb R)=K$, the finite group $\Gamma $ cannot be a subgroup of $G(\mathbb Z)$:

We take $K=SU(2)$ and $\Gamma $ to be a Dihedral group of the form $(\mathbb Z/l\mathbb Z)\rtimes {\mathbb Z}/2{\mathbb Z}$ where the nontrivial element of the group ${\mathbb Z}/2{\mathbb Z}$ operates by $x\mapsto -x$ on ${\mathbb Z}/l{\mathbb Z}$. Here $l$ is a large prime.

Suppose $G$ is an algebraic group defined over $\mathbb Q$ such that $G({\mathbb R})=K$, and $\Gamma \subset G(\mathbb Z)$. Then for almost all primes $p$, $\Gamma $ injects into $G({\mathbb F}_p)=G(\mathbb Z/p\mathbb Z)$. Moreover, for almost all primes $p$, the order of $G({\mathbb F}_p)$ is $(p^2-1)(p^2-p)/(p-1)=p(p^2-1)$. Further, $l$ divides this order since $\Gamma $ is a subgroup of $G({\mathbb F}_p)$.

Therefore, for almost all primes $p$, we have $p(p^2-1)\equiv 0 \quad (mod \quad l)$. But by Dirichlet's theorem on primes in arithmetic progressions, the residue class of the generator of the unit group of ${\mathbb Z}/l{\mathbb Z}$ is represented by infinitely many primes $p$. Hence the order of such a $p$ (modulo $l$) is $l-1$. On the other hand $p(p^2-1)$ is divisible by $l$ which means that $l-1\leq 2$, and $l$ cannot be a large prime.

I am pretty sure that a much simpler proof can be found, but this is "a proof".

ADDED later: The proof shows that the "large prime" $l$ need only satisfy $l\geq 5$. Moreover, the gcd of the numbers $p(p^2-1)$ as $p$ varies over primes large enough, is just $24$. Hence the order of $\Gamma $ is $\leq 24$.