# Definition of an arithmetic subgroup of an algebraic group

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

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 $$\mathrm G \to \mathrm{GL}_n(\mathbb{Q}).$$
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.

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?

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.