How can one convert rational matrices into integer ones?

Question

Let $G$ be a finite subgroup of $GL_n(\Bbb{Q})$. I want to show the existence of a matrix $A\in GL_n(\Bbb{Q})$ with the property that $AGA^{-1} \subseteq M_n(\Bbb{Z})$.

Answer 1

There is a standard trick here, which works for any principal ideal domain. You are looking for a change of $\mathbb{Q}$-basis which makes the matrices representing group elements integral. Let $V$ be the underlying $\mathbb{Q}G$-module, with $\mathbb{Q}$-basis $\{v_{i} : 1 \leq i \leq n \}$. Let $M$ be the $\mathbb{Z}$-submodule of $V$ generated by the vectors in $\{ v_{i}g : 1 \leq i \leq n, g \in G \}$. Then $M$ is in fact a $\mathbb{Z}G$-submodule of $V$ by its construction, because $mg \in M$ whenever $m \in M$ and $g \in G$. Now $M$ is a finitely generated torsion-free $\mathbb{Z}$-module, so has a $\mathbb{Z}$-basis. Furthermore $M \otimes_{\mathbb{Z}}\mathbb{Q} = V$, so this $\mathbb{Z}$-basis has $n$ elements, say it is $\{ m_{i} : 1 \leq i \leq n \}$. Then $m_{i}g $ is a $\mathbb{Z}$-linar combination of $\{m_{j} : 1 \leq j \leq n \}$ for each $i$ and each $g \in G$. Hence the matrices representing the elements of $G$ with respect to the $\mathbb{Q}$-basis $\{ m_{i} : 1 \leq i \leq n \}$ of $V$ are all integral, as required.