TITLE: Finitely generated matrix groups whose eigenvalues are all algebraic
QUESTION [15 upvotes]: Let $G$ be a finitely generated subgroup of $GL(n,\mathbb{C})$.  Assume that there exists a number field $k$ (i.e. a finite extension of $\mathbb{Q}$) such that for all $g \in G$, the eigenvalues of $g$ all lie in $k$.  This implies that $g$ is conjugate to an element of $GL(n,k)$.
Question: must it be the case that some conjugate of $G$ lies in $GL(n,k)$?  Or at least $GL(n,k')$ for some finite extension $k'$ of $k$?  If this is not true, what kinds of assumptions can I put on $G$ to ensure that it is?

REPLY [2 votes]: Re-edited following YCor's comments: A nice theorem of Schur, building on earlier work of Jordan and Burnside, states that any finitely generated periodic subgroup $G$ of ${\rm GL}(n,\mathbb{C})$ is finite ( this is Theorem 36.2 of the 1962 edition of Curtis and Reiner)-and hence is completely reducible.
Hence the answer to your question is "yes" , if every eigenvalue of every element of $G$ is a root of unity and every element of $G$ is semisimple.  
In that case, once we know that $G$ is finite, then a Theorem of Brauer (which makes use of his induction theorem) asserts that every finite subgroup $X$ of ${\rm GL}(n,\mathbb{C})$ is conjugate within ${\rm GL}(n,\mathbb{C})$ to a subgroup of ${\rm GL}(n,\mathbb{Q}[\omega]),$ where $\omega$ is a primitive complex $|G|$-th root of unity.