Do all algebraic integers in some $\mathbb{Z}[\zeta_n]$ occur among the character tables of finite groups?

Solution 1:

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.

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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 $\chi$.