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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?

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At the positive side, if $G$ acts irreducibly on $\mathbf{C}^n$ and $k$ is an arbitrary subfield of $\mathbf{C}$, then the answer is yes (allowing some field extension $k'$ of degree dividing $n$). This even works assuming that $G$ is a multiplicative submonoid of $M_n(\mathbf{C})$ (keeping the irreducibility assumption).

See for instance Proposition 2.2 in H. Bass, Groups of integral representation type. Pacific J. Math. 86, Number 1 (1980), 15-51. (ProjectEuclid link, unrestricted access)

Robert Israel's simple example shows that some assumption such as irreducibility has to be done.

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Here's one easy example. Let $G$ be generated by $\pmatrix{1 & x\cr 0 & 1}$ for $x$ in some finite set $X$ of complex numbers. All eigenvalues are $1$, so we can take $k = \mathbb Q$. If $G$ is conjugate by $S$ to a subgroup of $GL(2,\mathbb Q)$, then the members of $X$ are in the field generated by the matrix elements of $S$, and we can choose $X$ so that this is impossible (e.g. take more than $4$ numbers that are algebraically independent).

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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.

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    $\begingroup$ $k$ cyclotomic (including $k=\mathbf{Q}$) doesn't mean that eigenvalues have finite order... $\endgroup$
    – YCor
    Apr 2, 2019 at 14:22
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    $\begingroup$ and also, that all elements have only eigenvalues of finite order doesn't imply being finite: just take the cyclic subgroup generated by a nontrivial unipotent element. $\endgroup$
    – YCor
    Apr 2, 2019 at 14:26
  • $\begingroup$ @YCor: You are right, I was careless. I will re-edit or delete. Schur's theorem is of course correct, but the eigenvalues being roots of unity does not give periodicity, as you say. And I did not say what I meant in the first part either. $\endgroup$ Apr 2, 2019 at 14:42
  • $\begingroup$ If I'm not wrong, the fact that every finite subgroup is conjugate into the algebraic closure of $\mathbf{Q}$ is immediate from basic theory (which basically works over an arbitrary algebraically closed field of characteristic zero, and in particular by counting, every irreducible is defined over the algebraics). $\endgroup$
    – YCor
    Apr 2, 2019 at 15:03
  • $\begingroup$ That is true, but it does not a priori get you into a representation over a cyclotomic field. It does get you into some number field. The content of Schur's theorem is that finitely generated periodic linear groups over complex numbers are in fact finite. $\endgroup$ Apr 2, 2019 at 16:25

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