Rrerku

Let $G$ be a finitely generated subgroup of $GL(n,mathbbC)$. Assume that there exists a number field $k$ (i.e. a finite extension of $mathbbQ$) 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?

Here's one easy example. Let $G$ be generated by $pmatrix1 & xcr 0 & 1$
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).

At the positive side, if $G$ acts irreducibly on $mathbfC^n$ and $k$ is an arbitrary subfield of $mathbfC$, 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 $M_n(mathbfC)$ (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.