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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    $\begingroup$ I would highly doubt one can get the e.g. icosahedral subgroup of $SO(3)$ this way. But maybe I’m misunderstanding the question… $\endgroup$ Mar 4, 2022 at 19:18
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    $\begingroup$ You're not misunderstanding the question Sam that's actually the finite subgroup I'm most interested in I've even specifically asked in the past if there is a form of $ SO_3 $ with icosahedral as the group of integer points math.stackexchange.com/questions/4287901/… $\endgroup$ Mar 4, 2022 at 19:22
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    $\begingroup$ @YCor But then $G(\mathbb Z)$ is not well-defined (as you note). So we must be working with some kind of group scheme $G$ over $\mathbb Z$ whose generic fiber $G_{\mathbb Q}$ is a reductive algebraic group satisfying $G_{\mathbb Q}(\mathbb R) = K$. What conditions to put on finite primes is indeed not obvious but from my reading that's the only thing that's unclear. $\endgroup$
    – Will Sawin
    Mar 4, 2022 at 20:12
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    $\begingroup$ @WillSawin For a Zariski-closed subgroup $G$ of $\mathrm{GL}_n(\mathbf{C})$, $G(\mathbf{Z})$ is well-defined as $G\cap \mathrm{GL}_n(\mathbf{Z})$. $\endgroup$
    – YCor
    Mar 4, 2022 at 20:23
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    $\begingroup$ @LSpice yes of course. But I mean all this needs no knowledge of group schemes over rings to be defined (the OP doesn't seem familiar with this language, let alone the inherent difficulties of dealing with such general group schemes). So I'd understand the question as whether there is $n$ and a continuous representation $f:K\to\mathrm{GL}_n(\mathbf{C})$ such that $f^{-1}(\mathrm{GL}_n(\mathbf{Z}))$ is a given finite subgroup of $K$. $\endgroup$
    – YCor
    Mar 4, 2022 at 20:27

1 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 $\leq 24$.

  • $\begingroup$ What necessary condition does this give for a finite group to be the integer points of an algebraic $ \mathbb{Q} $ group? That is, what about the subgroup $ \Gamma $ do you use in this proof other than that it has order $ 2 \ell $ for some large prime $ \ell $? Also small note $ K=SU_2 $ doesn't contain dihedral subgroups or even any nonabelian groups of order $ 2 \ell $ for $ \ell $ prime. I suppose you wanted to take $ K=SO_3(\mathbb{R}) $. Or your proof probably would work with the order $ 4 \ell $ dicyclic subgroups of $ SU_2 $. $\endgroup$ Mar 8, 2022 at 13:29
  • $\begingroup$ The diagonal subgroup of SU(2) contains $\mathbb Z/l\mathbb Z$, and the Weyl group element gives the extra Z/2Z$. So, the Dihedral group is indeed contained in SU(2). $\endgroup$ Mar 8, 2022 at 13:31
  • $\begingroup$ I cannot offhand, think of a condition, but the order of $\Gamma$ should divide the gcd of $p(p^2-1)$ for all but finitely many $p$. $\endgroup$ Mar 8, 2022 at 13:33
  • $\begingroup$ you are right of course, $ SU_2 $ contains $ O_2(\mathbb{R}) $ so certainly contains dihedral subgroups, I was a bit hasty on that one sorry. $\endgroup$ Mar 8, 2022 at 13:33
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    $\begingroup$ This is probably another silly question but is there a reason that you use the dihedral group instead of just a cyclic group of prime order $ \ell $? $\endgroup$ Mar 8, 2022 at 13:44

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