The values of irreducible characters of a finite groups are always sums of roots of unity; do all sums of roots of unity (i.e. algebraic integers in the maximal abelian extension of $\mathbb{Q}$) actually appear among the character tables of finite groups?

Obviously roots of unity themselves appear, as can be seen from the character tables of cyclic groups. But then there seems no easy way to get the sum and product of these numbers to appear in another character table, as $\oplus$ and $\otimes$ don't always yield irreducible representations.

Are there substitute constructions which allow you to show that all these numbers actually do appear somewhere?


The following construction answers your question in the affirmative. Assume that $n>1$. Let $\mu_n=\langle \zeta_n\rangle$ consist of the complex solutions of the equation $z^n=1$. Fix a (large) integer $M$. Consider the group $G(M,n)$ of $M\times M$ matrices that have a single non-zero entry from $\mu_n$ on each row and column, i.e. monomial matrices with non-zero entries constrained to come from $\mu_n$.

I claim that the natural action of $G(M,n)$ on $V=\mathbf{C}^M$ gives an irreducible representation. This follows easily from the fact that the symmetric group $S_M$ is a subgroup of $G(M,n)$. We know that the only non-trivial $S_M$-invariant subspaces are the one spanned by the all one vector, and its complement the zero-sum subspace. Neither of these is invariant under the action of all of $G(M,n)$, so $V$ is an irreducible $G(M,n)$ module.

Thus we get as values of the character $\chi_V$ of $G(M,n)$ all the sums of any $M$ elements of $\mu_n$ as $\chi_V(d)$ for some diagonal matrix $d\in G(M,n)$. Your claim follows from this by varying the integer parameter $M$. There are several ways of getting the negative integers to appear as coefficients of the powers $\zeta_n^j$. We can either go to $\mu_{2n}$ or use the relations determined by the cyclotomic polynomial.


Adding a brief description of the representation $V$ in the context of this answer by Alex B. If $\chi$ is the character of the abelian group $A=\mu_n^M$ that maps any element to its last component, then under the action of $S_M$ on the characters of $A$ clearly $\text{Stab}_{S_M}(\chi)$ is the point stabilizer of the last index $M$ (i.e. the usual copy of $S_{M-1}\le S_M$). Therefore $\chi$ extends to a character of the group $S_\chi=A\rtimes S_{M-1}$. The group $S_\chi$ is of index $M$ in the group $G(M,n)=A\rtimes S_M$. The representation $V$ then arises as the induced representation $\text{Ind}_{G(M,n)/S_\chi}(\chi)$. The irreducibility of $V$ (that was clear in our context) then also follows from the general theory. Note: the details of the interpretation of $V$ as an induced module depend on the choice of $\chi$.

  • $\begingroup$ If I remember my basics correctly, the group $G(M,n)$ is the wreath product of $\mu_n$ and $S_M$. $\endgroup$ – Jyrki Lahtonen Oct 18 '11 at 13:18
  • $\begingroup$ This is very nice. I wonder if this has anything to do with the fact that $G(M,n) = \mu_n^M \rtimes S_M$. $\endgroup$ – Joel Cohen Oct 18 '11 at 13:25
  • $\begingroup$ Great construction! $\endgroup$ – Will Oct 18 '11 at 13:37
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    $\begingroup$ The character values of the natural representation of G(M,n) are the sums of k members of μ(n) where 0 ≤ k ≤ M and k ≠ M−1, so you get fairly good coverage by a single finite group. I think Jyrki only mentioned k = M (only the diagonal matrices). G(M,n) is called the monomial group and is indeed the wreath product of μn and Sym(M). $\endgroup$ – Jack Schmidt Oct 18 '11 at 14:07
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    $\begingroup$ Very nice indeed! In fact, it's easy to see from the definition of induction, that your representations are inductions of non-trivial characters of $\mu_n^M$ to $G(M,n)$. The fact that (most) such inductions are irreducible follows from general facts about irreducible characters of semi-direct products by abelian groups. $\endgroup$ – Alex B. Oct 18 '11 at 14:46

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