5
$\begingroup$

Is anyone aware of a result (or a counterexample) along the following lines: let $G$ be an algebraic group over $\mathbf Z$. Let $H$ be a finite group such that $H$ occurs as a subgroup of $G(\bar{\mathbf F}_p)$ for all but finitely many primes $p$. Then $H$ also occurs as a subgroup of $G(\mathbf C)$.

I only really need this for $H$ solvable and $G = \mathrm{GSp}(4)$ if that helps at all.

$\endgroup$
  • 1
    $\begingroup$ as angelo says, you just need infinitely many $p$. An alternative argument is to observe that $H$ embeds in a suitable nonprincipal ultraproduct of the $G(\overline{\mathbf{F}_p})$, which is $G$ of the same ultraproduct of the $\overline{\mathbf{F}_p}$, which is, by a cardinality argument, isomorphic to $\mathbf{C}$. $\endgroup$ – YCor Jun 12 '13 at 18:57
10
$\begingroup$

The functor of injective homomorphisms from $H$ to $G$ is represented by a scheme of finite type over $\mathbb Z$ (a locally closed subscheme of the product $G^H$). If this has points over $\overline{\mathbb F}_p$ for infinitely many $p$, then it must dominate $\mathop{\rm Spec}\mathbb Z$, so it has points over $\mathbb C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.