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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.

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2 Answers 2

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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 $m>1$.

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  • $\begingroup$ All the irreducible representations of $S_n\ltimes G$ are obtained as follows: start with a one-dimensional representation $\chi$ of $G$, extend it trivially to $S\ltimes G$, where $S$ is the stabiliser of $\chi$ in $S_n$, take any irreducible representation $\rho$ of $S$, viewed as a representation of $S\ltimes G$. Then ${\rm Ind}^{S_n\ltimes G}\chi\otimes \rho$ is an irreducible representation, and they all arise in this way. Is the representation you have constructed somewhere among these? I don't quite see that right now. $\endgroup$
    – Alex B.
    Commented Mar 6, 2013 at 10:31
  • $\begingroup$ So here is the problem, I think: suppose that among the $m$-th roots of unity, the first two are 1, and the remaining ones are distinct primitive $m$-th roots of unity. Then the subgroup whose $S_n$-component is trivial and whose $G$-component is of the form $(x,y,1,1,\ldots, 1)\in (\mathbb{Z}/m\mathbb{Z})^n$ is the kernel of the representation you are constructing. But this subgroup is not normal. So, unless I am mistaken, what you have constructed is not a representation. $\endgroup$
    – Alex B.
    Commented Mar 6, 2013 at 10:38
  • $\begingroup$ But where is the mistake in his proof? He has provided an explicit construction of an irreducible imprimitive subgroup of ${\rm GL}(n,{\mathbb C})$ that is isomorphic to the wreath product of a cyclic group of order $m$ by $S_n$. $\endgroup$
    – Derek Holt
    Commented Mar 6, 2013 at 11:00
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    $\begingroup$ Alex, I think you have misunderstood Bob's construction. In your notation, $\chi$ is the representation sending each generator to a primitive $m$th root of unity (the same one), so your $S$ is $S_n$, $\rho$ is the natural $n$-dimensional representation of $S$, and Bob's your uncle. $\endgroup$
    – user30035
    Commented Mar 6, 2013 at 11:03
  • $\begingroup$ The image of $S_n \ltimes G$ in $\mathrm{GL}(n, \mathbb{C})$ is the set of all monomial matrices having $m$-th roots of unity as nonzero entries. (By monomial I mean: having exactly one nonzero entry in each row and column.) In fact, every $z$ that can be written as a sum of $n$ or fewer $m$-th roots of unity occurs as trace of such a matrix. $\endgroup$
    – Frieder Ladisch
    Commented Mar 6, 2013 at 14:20
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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.

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