I'm struggling with the definition of an arithmetic subgroup of an algebraic group defined over $\mathbb{Q}$.
In Wikipedia you can read:



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> If $\mathrm G$ is an algebraic subgroup of $\mathrm{GL}\_n(\mathbb{Q})$ for some $n$ then we can define an arithmetic subgroup of $\mathrm G(\mathbb{Q})$ as the group of integer points $\Gamma = \mathrm{GL}\_n(\mathbb{Z}) \cap \mathrm G(\mathbb{Q}).$ [...] The subgroup defined above can change when we take different embeddings $\mathrm G \to \mathrm{GL}\_n(\mathbb{Q}).$   
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> Thus a better notion is to take for definition of an arithmetic subgroup of $\mathrm G(\mathbb{Q})$ any group $\Lambda$ which is commensurable [...] to a group $\Gamma$ defined as above (with respect to any embedding into $ \mathrm{GL}\_n$). With this definition, to the algebraic group $\mathrm G$ is associated a collection of "discrete" subgroups all commensurable to each other.
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This definition imply that for two embeddings $\mathrm G \to \mathrm{GL}\_n(\mathbb{Q})$ the groups $\mathrm{GL}\_n(\mathbb{Z}) \cap \mathrm G(\mathbb{Q})$ are commensurable. Why is it true?



Definition of an arithmetic subgroup of an algebraic group
Write the coordinates $a\_{ij}$ of one embedding into $GL\_n$ as polynomial functions, defined over $\mathbb Q$, in the coordinates $b\_{ij}$ of a different embedding into $GL\_n$. We can do this because, by definition, embeddings give an isomorphism of algebraic varieties. (I guess we should allow the inverse of the determinant as one of our coordinates. I will still write them as $a\_{ij}, b\_{ij}$ for notational simplicity.)


Let $N$ be the greatest common denominator of all rational numbers appearing as coefficients of these polynomials.


Now if $b\_{ij} = \delta\_{ij}$ for all $i,j$ then the group element goes to the identity under the second embedding, so it is the identity, so $a\_{ij} =\delta\_{ij}$ for all $i,j$. It follows, by elementary algebra, that if $b\_{ij}$ is congruent to $\delta\_{ij}$ modulo $N$ for all $i,j$, then $a\_{ij}$ is an integer for all $i,j$.


Since matrices congruent to the identity mod $N$ are a finite index subgroup, this shows the matrices with $a\_{ij}$ integral contain a finite index subgroup of the matrices with $b\_{ij}$-integral. Handling the other direction symmetrically, they are commensurable.


