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Suppose $G$ is a linear algebraic group over $\mathbb{C}$, defined over $\mathbb{Z}$ (for example, $SL(n, \mathbb{C})$ is defined by $\det x = 1,$ which visibly has integer coefficients). Let $H$ be an algebraic subgroup of $G.$ Is it always true that some conjugate of $H$ is defined over $\mathbb{Z}?$ This sounds like it should be (if true) be a totally soft fact, but what do I know...

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Is a finite cyclic subgroup of $\text{SL}_2(\mathbb{C})$ defined over $\mathbb{Z}$? – Qiaochu Yuan Mar 10 '12 at 2:44
@Qiaochu, sure. – Mariano Suárez-Alvarez Mar 10 '12 at 3:17
I think there exist continuous families of nilpotent complex Lie algebras of dimension $7$. The general such Lie algebra would give rise to a unipotent algebraic group which cannot be defined over $\mathbb{Z}$. If you embed such a group in $SL_n$ for some $n$ you would get an algebraic subgroup $H$ such that no conjugate is defined over $\mathbb{Z}$. – ulrich Mar 10 '12 at 6:29
Igor, take $G={\mathbb C}^2$ with obvious integral structure and the subgroup given by equation $z=\sqrt{2}w$. Since $G$ is an abelian group, no conjugation will help you. Thus, you probably want to assume that $G$ is semisimple and $H$ is too. Then the answer could be positive. – Misha Mar 10 '12 at 11:20
It is not particularly easy but could be found for instance in Steinberg's "Lectures on Chevalley Groups" (published in Russian but not in English, I think). One can trace this result to rationality of characters of algebraic tori. – Misha Mar 10 '12 at 17:25

As some of the comments indicate, arbitrary Zariski-closed subgroups of for instance $SL(n, \mathbb{C})$ can be defined by all kinds of polynomial equations over $\mathbb{C}$. This makes it very unlikely that much can be said in general about integral forms of conjugate subgroups.

The whole subject of $\mathbb{Z}$-forms for linear algebraic groups strikes me as quite delicate. In the case of simple (or more generally reductive) groups the best insight originates with Chevalley, though it has taken a long time to reach a clear definitive treatment: this is explained by Lusztig in a paper published in JAMS 22 (2009), with an arXiv version available here. (As noted in some of the comments, there is relevant background in Steinberg's lectures on Chevalley groups from 1967-68, sold for many years in mimeographed format by the Yale math department but never formally typeset and "published" in English in spite of some efforts by people over the years. These notes are still quite useful, but lack for instance an index. See Steinberg's UCLA homepage cited above here.)

Even with explicit information about existence of integral forms, it would require some further argument if you start with an arbitrary (even simple) algebraic group having such a form and then consider all its closed reductive subgroups. Unless the big group is of general or special linear type to begin with, I'm not convinced there is enough motivation to get into this (?)

ADDED: Questions about algebraic groups over $\mathbb{Z}$ get rather sophisticated even when you consider only reductive groups. This is developed in a short conference paper:

"Non-split reductive groups over $\mathbb{Z}$" by Brian Conrad and Benedict Gross (from Luminy last September), online at the conference page here.

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