Is the group of integer points on a finite-type group scheme over Z finitely presented? 
Let $G$ be a group scheme of finite type over $\mathbf{Z}$.  Must $G(\mathbf{Z})$ be finitely presented?

(The question is inspired by a not yet successful attempt to answer a question of Brian Conrad.)
A few special cases:


*

*If $G$ is the Néron model of an abelian variety over $\mathbf{Q}$, then a positive answer amounts to the Mordell-Weil theorem (combine with restriction of scalars to get the full Mordell-Weil theorem).

*If $G$ is the restriction of scalars of $\mathbf{G}_m$ from the ring of integers of a number field down to $\mathbf{Z}$, then a positive answer follows from Dirichlet's unit theorem.
 A: It follows from Theorem 6.12 of Borel and Harsh-Chandra, "Arithmetic subgroups of algebraic groups", that $G(\mathbb{Z})$ is finitely generated if $G$ is affine. Perhaps one can combine this with Chevalley's theorem to deduce finite generation in the general (not necessarily affine) case.
EDIT  (Added idea for proof of general case; see also Torsten Ekedahl's comment below)
EDIT (Proof completed (assuming $G$ is separated) and simplfied using comments of BCnrd below)
As discussed in the comments below, we may assume $G$ is flat and we may also assume it is connected. By Chevalley's theorem, there is an affine subgroup scheme $H_{\mathbb{Q}}$ of the generic fibre $G_\mathbb{Q}$ of $G$
such that the quotient $G_{\mathbb{Q}}/H_\mathbb{Q}$ is an abelian variety. Let $H$ be the Zariski closure of $H_{\mathbb{Q}}$ in $G$ with the reduced induced structure. Then $H$ is a closed subgroup scheme of $G$. By a theorem of Raynaud (see comment of BCnrd below for the reference) $H$ is also affine.
We have an incusion of groups 
$G(\mathbb{Z})/H(\mathbb{Z}) \subset G(\mathbb{Q})/H(\mathbb{Q}) \subset (G_{\mathbb{Q}}/H_{\mathbb{Q}})(\mathbb{Q})$.
Since $G_{\mathbb{Q}}/H_{\mathbb{Q}}$ is an abelian variety, by the Mordell-Weil theorem $(G_{\mathbb{Q}}/H_{\mathbb{Q}})(\mathbb{Q})$ is a finitely generated abelian group, hence so is $G(\mathbb{Z})/H(\mathbb{Z})$. Since $H(\mathbb{Z})$ is finitely generated by the Borel-Harish-Chandra theorem, it follows that $G(\mathbb{Z})$ is finitely generated.
