Let $G=C_n$ be the cyclic group and $f_n$ the characteristic polynomial of its character table (over $\mathbb{C}$) in the ordering so that the character table is given by the discrete Fourier transform matrix (https://en.wikipedia.org/wiki/DFT_matrix) without the factor.

Question 1: For which $n$ is $f_n$ irreducible (over the smallest field extension of $\mathbb{Q}$ containing its coefficients)?

I did input this into GAP using the command IrrDixonSchneider to obtain the character table (I hope this is does not change much with respect to irreducibility) and tested for irreducibility with GAP. GAP seems to use another numbering for the character table, so lets call $g_n$ the characteristic polynomial according to GAP (maybe someone can clarify the ordering which GAP uses to get the character table of the cyclic group. I try to see what it is but Im not sure at the moment).

For $n \leq 23$ it was true that the $n$ such that $g_n$ is irreducible coincides with the sequence https://oeis.org/A280862 , which are those $n$ such that $a_n \psi_n = \varphi_n$, where $\varphi_n$ is the Euler phi function (https://oeis.org/A000010), $\psi_n$ is the reduced totient function (https://oeis.org/A002322) and $a_n$ is the greatest common divisor of all $(d-1)$'s, where the $d$'s are the positive divisors of $n$ (https://oeis.org/A258409).

Question 2: Is this true?

Question 3: Does being irreducible depend on the ordering used to obtain the character table here (or even for a general group)?

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    $\begingroup$ What is the meaning of the characteristic polynomial of the character table? There's no canonical ordering on the elements of $C_n$, nor its character; they are not in natural bijection, so there is no way to view the character table as naturally the matrix of an endomorphism of a vector space. $\endgroup$ Jun 30, 2020 at 21:24
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    $\begingroup$ Isn't the matrix you are looking at more or less the DFT matrix? $\endgroup$ Jun 30, 2020 at 21:25
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    $\begingroup$ Well, you can think of the character table as the matrix $[\omega^{(i-1)(j-1)}]: 1 \leq i,j \leq n,$ where $\omega$ is a complex primitive $n$-th root of unity. Admittedly, reordering the rows and columns will lead to a different characteristic polynomial. $\endgroup$ Jun 30, 2020 at 21:42
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    $\begingroup$ I'm not sure what the relationship between the characteristic polynomials is after permuting rows and columns. $\endgroup$ Jun 30, 2020 at 21:54
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    $\begingroup$ @GeoffRobinson, in this case, choosing a different primitive $n$th root of unity $\omega$ just conjugates your matrix (right?), so it doesn't change the characteristic polynomial. $\endgroup$
    – LSpice
    Jun 30, 2020 at 22:07

1 Answer 1


I think a strong pointer to the answer (for the cyclic group of order $n>2 $, where the character table $C$ is chosen so that the $(i,j)$-entry is $\omega^{(i-1)(j-1)}$ for a fixed complex primitive $n$-th root of unity $\omega$) is as follows:

We have $C\overline{C}^{T} = nI,$ but (with this labelling) we note that $C$ is also symmetric. Hence we have $C\overline{C} = nI.$ As noted in my answer to MO363691 (and also previously noted by Denis Serre in his answer to his own question MO78050), we have $\overline{C} = PC$ where $P$ is a permutation matrix of order $2$ (the number of $2$-cycles in the associated permutation is $\frac{n-1}{2}$ if $n$ is odd, and $\frac{n-2}{2}$ if $n$ is even- this is the number of complex conjugate pairs of linear characters of the cyclic group of order $n$ which are not real-valued).

Now we have $CPC = nI$, so that $P = nC^{-2}$ and (since $P^{2} = I$), we have $C^{2} = nP$. Now the eigenvalues of $C$ are all in the set $\{ \sqrt{n},-\sqrt{n}, i\sqrt{ n}, -i\sqrt{n} \}.$ Since $P$ has the eigenvalue $-1$ with multiplicity $\lfloor \frac{n-1}{2}\rfloor$, we see that $\lfloor \frac{n-1}{2}\rfloor$ of the eigenvalues of $C$ are pure imaginary of absolute value $\sqrt{n}$ . If $n$ is odd, then $\frac{n+1}{2}$ of the eigenvalues of $C$ are real of absolute value $\sqrt{n}$, while if $n$ is even, $\frac{n+2}{2}$ of the eigenvalues are real of absolute value $\sqrt{n}$.

Furthermore, as in the answers to MO3639691 and MO78050 that det(C) is real if $n \equiv 1$ or $2$ (mod $4$), and det(C) is pure imaginary if $n \equiv 0$ or $3$ (mod $4$).

Let $F$ denote the field generated by the coefficients of the characteristic polynomial of $C.$ Then the minimum polynomial $p(x)$ of $C$ in $F[x]$ is certainly a divisor of $(x^{2}-n)(x^{2}+n)$ from our knowledge of the eigenvalues of $C$. Hence any irreducible factor of $p(x)$ has degree $1$ or $2$. Since the characteristic polynomial of $C$ has degree $n >2,$ the characteristic polynomial of $C$ is definitely not irreducible in $F[x].$


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