The argument below claiming that d is a rational integer is incorrect.
Indeed, the character table of the cyclic group of order 3 has
determinant a square root of -27, which is pure imaginary. And the
character table of the cyclic group or order 5 has determinant a square
root of 5^5, which is real but not rational. (The argument given just
shows that the only conjugate d can have over the rationals is -d. That
doesn't mean d is necessarily rational.)
Also, to add a little to the original question, d^2 is actually
divisible by the square of the group order. As noted below, d^2 is
plus or minus the product of the orders of the centralizers associated
with the all different conjugacy classes. One of these classes is the
identity, also noted below. This gives one factor of d^2 which is the
group order. The centralizer of a non-trivial element in the center of
a Sylow p-group has order divisible by the p-part of the group order.
Accumulating these p-parts for all the different prime divisors p of the
group order gives a second factor of d^2 which is the group order.
Frank Luebeck wrote:
>The following questions of Stefan Kohl already got answered on another
>mailing list (By M. Isaacs and B. Howlett). Here is a short summary
>and an additional remark for the GAP-forum recipients.
>> 1. Does the character table of a group with n
>> conjugacy classes only contain
>> character values which are algebraic of degree
>> strictly smaller than n ?
>> (Clearly, this is not a consequence of the fact that
>> character values of a group G are sums of
>> Exponent(G)'th roots of unity)
>This follows from the following stronger statement:
>Let g in G be of order m. Let d be the degree of the field generated
>by all character values on g over the rationals. Then d is the number
>of conjugacy classes of G represented by powers g^k of g with
>Proof: This field is a subfield of K_m = Q[zeta_m], zeta_m a primitive
>m-th root of unity. If sigma_k is the automorphism of K_m over Q
>mapping zeta_m to zeta_m^k and chi is a character of G then
>chi(g)^sigma_k = chi(g^k). The result now follows from basic Galois
>Note that this degree d can be found in GAP without knowing the values
>of irreducible characters:
>or you can read it of from the power map information in the character
>> 2. Let d be the 'determinant' of the character table
>> of a group G of order n
>> (in GAP : d := DeterminantMat(List(Irr(G),ValuesOfClassFunction))).
>> - Is d always different from zero ?
>Yes, the irreducible characters are a *basis* of complex class functions.
>> - Is d^2 always an integer which is divisible by n ?
>> (Obviously, d is determined up to the sign by the group G,
>> hence d^2 is uniquely determined by G)
>Even d is an integer: It is integral since all matrix entries are
>integral and it is rational since applying any field automorphism of
>an algebraic closure of the rationals to the table induces a
>permutation of the rows. (This also give a direct proof of question
>1). If C is the matrix of character values and A is the complex
>conjugate transpose of C then the "second orthogonality relations" say
>that AC is diagonal with entries the centralizer orders of the
>correspondig classes. (Of course, det(AC)=d^2 and the trivial element
>has centralizer order |G|.)
>With best regards,
>%%% Dr. Frank Lübeck, Lehrstuhl D für Mathematik, Templergraben 64, %%%
>%%% 52064 Aachen, Germany %%%
>%%% E-mail: Frank.Luebeck@Math.RWTH-Aachen.De %%%
>%%% Tel: +49-241-80-4549 %%%