I asked the following question (http://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.