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?