I am trying to understand constructions of exceptional groups of type $G\_2$ (over rings). In this post, by a model (of type $G\_2$) I mean an affine smooth group scheme over $\mathbb{Z}$ such that the fibres are connected simple algebraic groups of type $G\_2$.


In Gross' paper *[Groups over Z](https://link.springer.com/article/10.1007/s002220050053)* (see Page 272) a model $\mathbb{G}$ is given using Coxeter's integral octonions, and it is mentioned there that $\mathbb{G}(\mathbb{Z})$ is isomorphic to $G\_2(\mathbb{F}\_2)$ as abstract groups. Models are not unique: For example, this $\mathbb{G}$ is a non-split model, and there is also a split one as given in Appendix B of Conrad's paper *[Non-split reductive groups over Z](http://virtualmath1.stanford.edu/%7Econrad/papers/redgpZsmf.pdf)*.


**Question:** Do we have $\mathbb{G}'(\mathbb{Z})\cong G\_2(\mathbb{F}\_2)$ for every model $\mathbb{G}'$ of type $G\_2$?



Does $G_2(\mathbb{Z})$ depend on the choice of an integral model?
The short answer is no, because a Chevalley group has infinitely many $\mathbb{Z}$-points (even Zariski dense by the Borel density theorem).


For the long answer, let me first completely describe all $\mathbb{Q}$-models and $\mathbb{Z}$-models of groups of type $G\_2$. Let $G\_0/\mathbb{Q}$ be the split reductive group of type $G\_2$. (It is both simply connected and adjoint since the $G\_2$ Cartan matrix has determinant $1$.)
This group has a reductive $\mathbb{Z}$-model $\underline{G}\_0$ which is unique up to isomorphism by the theory of Chevalley groups.


Now the $\mathbb{Q}$-forms of $G\_0$ are classified by $\mathrm{H}^1(\mathbb{Q},G\_0)$, bearing in mind that $G\_0$ is adjoint and has no outer automorphisms. By the work of many people, the restriction map $\mathrm{H}^1(\mathbb{Q},G\_0) \rightarrow \mathrm{H}^1(\mathbb{R},G\_0)$ is an isomorphism (for references see Theorems 5.12.24 and 5.12.31 of Poonen's rational points book). By table 1.3 in the Gross' paper you mention, $\mathrm{H}^1(\mathbb{R},G\_0)$ has size $2$, where the nontrivial element is represented by the compact form.


Let's call the corresponding $\mathbb{Q}$-form $G$. It can be constructed as the automorphism group of (a $\mathbb{Q}$-form of) the octonions (as done in the Gross' paper) and its real points $G(\mathbb{R})$ are compact.
Gross constructs a $\mathbb{Z}$-model $\underline{G}$ of $G$, again using octonions. It has the property that $\underline{G}(\mathbb{Z})\simeq \underline{G}\_0(\mathbb{F}\_2)$ as abstract groups. He shows using the mass formula that $\underline{G}$ is the only $\mathbb{Z}$-model of $G$.


**Conclusion** Up to isomorphim, there exist exactly two semisimple groups of type $G\_2$ over $\mathbb{Q}$. Both admit unique models over $\mathbb{Z}$.


This is a very special property of $G\_2$! Usually a semisimple group has infinitely many $\mathbb{Q}$-forms, most of them will not have integral models, and if such integral models exist they might not be unique. (Considering $\mathrm{SL}\_2$ should already give you an idea.)


Back to your question: you are asking whether we have an isomorphism $\underline{G}\_0(\mathbb{Z}) \simeq \underline{G}\_0(\mathbb{F}\_2)$. But the left hand side is not even finite: by a theorem of Borel and Harish-Chandra, $\underline{G}\_0(\mathbb{Z})$ is a lattice in $G\_0(\mathbb{R})$. By the Borel density theorem, this implies that $\underline{G}\_0(\mathbb{Z})$ is Zariski dense in $G\_0$, so in particular $\underline{G}\_0(\mathbb{Z})$ is infinite.


