I asked the following question (https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con) on math.stackexchange.com and received no answers, so I thought I would ask it here. I've asked several people in my department who were all stumped by the question.
The question is: why is every finite subgroup of $\operatorname{GL}_n(\mathbb{Q})$ conjugate to a finite subgroup of $\operatorname{GL}_n(\mathbb{Z})$?
Note that at least for $n=2$ the question of isomorphism is much easier, since one can (with some effort) work out exactly which finite groups can be subgroups of $\operatorname{GL}_2(\mathbb{Q})$. Further, there are isomorphic finite subgroups of $\operatorname{GL}_2(\mathbb{Q})$ that are not conjugate to each other. For example, the group generated by $-I_{2 \times 2}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are both isomorphic to $C_2$, but they cannot be conjugate to each other because the eigenvalues of the two generators are different.
If there is a relatively simple proof, that would be ideal, but a reference with a potentially long proof is fine as well.
Thanks for any assistance.
Best Answer
This argument is fairly standard, but it is quicker to repeat it than to find a reference: Let $G$ be a finite subgroup of $GL_n(\mathbb{Q})$. Set $\Lambda = \sum_{g \in G} g \cdot \mathbb{Z}^n \subset \mathbb{Q}^n$. Then $\Lambda$ is a finitely generated torsion free abelian group, hence isomorphic to $\mathbb{Z}^r$ for some $r$. Since $\mathbb{Z}^n \subseteq \Lambda \subset \mathbb{Q}^n$, we have $r=n$. Clearly, $g \Lambda = \Lambda$ for all $g \in G$.
Let $h \in GL_n(\mathbb{Q})$ take the standard basis to a basis of $\Lambda$, so $h \mathbb{Z}^n = \Lambda$ and $\mathbb{Z}^n = h^{-1} \Lambda$. Then $h^{-1} g h$ takes $h^{-1} \Lambda = \mathbb{Z}^n$ to itself for all $g \in G$, so $h^{-1} G h \subset GL_n(\mathbb{Z})$.