Let $\chi$ be a character of G. Let $\phi$ be the representation of G which affords the character $\chi$. We assume that the representation space V is over $\mathbb{C}$. Show that for a suitable basis of V, the entries for $\phi_{g}$ for all g$\in$G lie in some subfield F $\subseteq$ $\mathbb{C}$ such that [F:$\mathbb{Q}$] is finite.
My attempt: If G is Abelian then we can find a basis such that $\phi_g$ are diagonal matrices with eigen values in their diagonal for all g$\in$G. But we know eigen values are $n^{th}$ roots of unity( where $|G|$ =n) and hence we find a basis such that the entries come from $\mathbb{Q}[\zeta_{n}]$ which is a finite extension over $\mathbb{Q}$. But I cannot do the general case. Thanks in advance for the help!!!!
Recently on math.SE, we solved this exercise in which we were asked to prove that every representation over $\mathbb C$ is equivalent to a representation over $\bar{\mathbb Q}$, up to a change of basis.
Once you have picked your basis so that the matrix entries are all in $\bar{\mathbb Q}$, you're guaranteed that the field generated by the matrix entries is a finite extension over $\mathbb Q$, because every entry is algebraic over $\mathbb Q$ and there are only finitely many entries.