Character table entries and sums of roots of unity It is well-known that the entries of the character table of a finite group are sums of roots of unity. 
Question: Is the converse true? Explicitly, given $z\in \mathbb{Z}[\mu_\infty]$, can I find a finite group $G$ and irreducible  character $\chi$ with $z=\chi(g)$ for some $g\in G$?
It's certainly true if I dropped the irreducibility; in this case one can take $G$ to be abelian, and just hit each summand of $z$ one at a time and take a direct sum. Also noteworthy is that if one has a single $G$ one can compose with a surjection to $G$, so it will be true for infinitely-many groups if it is for one. 
A related question (possibly less trivial): If it's YES, is there a natural "minimal" subclass of finite groups that suffice for this purpose?
If it's "NO," are the obstructions completely understood? 
This is partially meant as a (hopefully) easier relative to a previous question: A Realization Problem for Character Tables. I apologize in advance if it is trivial (one way or the other), as I fear it may be. 
 A: I think that a clearer way to construct the desired example is not to let the acting group be $S_n$, but take the cyclic group $C = C_n$ instead. Thus, start with the group $G = (C_m)^n$, as before, and let $C$ act by cycling the direct factors of $G$. Let $\lambda$ be a linear character of $G$ whose kernel is exactly the product of all but the first cyclic factor of $G$ (so the restriction of $\lambda$ to the first factor of $G$ is faithful). Only the identity of $C$ stabilizes $\lambda$, so working in the semidirect product $\Gamma = GC$, the stabilizer of $\lambda$ is $G$, and it follows that the induced character $\chi = \lambda^\Gamma$ is irreducible.
Now the restriction $\chi_G$ is a sum of $m$ linear characters $\lambda_i$, where the kernel of $\lambda_i$ is the product of all but the $i$th cyclic factor of $G$. Now given any list of $n$ $m$th roots of unity $\varepsilon_i$, we can choose an element $g$ of $G$ such that $\lambda_i(g) = \varepsilon_i$, and thus $\chi(g) = \sum\varepsilon_i$, as wanted.
A: I think you can take your idea and make it into an irreducible representation.
Take $m\in\mathbb{Z}$ large enough that all the summands are $m$th roots of unity.  Let $n$ be the number of such summands.  Represent $G=(\mathbb{Z}/m\mathbb{Z})^n$ on $\mathbb{C}^n$ by diagonal matrices, with the $i$th generator of the group acting by multiplication by the primitive $m$th root of unity on the $i$th coordinate.  Your $z$ appears as a value of the character, but it's not irreducible.  
To make it irreducible, you just need to intertwine the coordinates.  So let the permutation group $S_n$ act by automorphisms on $G$ by permuting the generators.  Likewise, represent it on $\mathbb{C}^n$ by permuting the coordinates.  This defines a covariant representation, and hence a representation of the semidirect product $S_n\ltimes G$.  It should be irreducible for $m>1$.
