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.

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    $\begingroup$ 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

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 $\mathbb C$.


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