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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?

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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.

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