TITLE: determinant of the table of characters
QUESTION [18 upvotes]: I am certain that the answer to this question exists somewhere. It might be a classical exercise.
Let $G$ be a finite group. Its table of characters is a square matrix, whose rows are indexed by the conjugacy classes and the columns are indexed by the irreducible characters. It is well defined, up to the order of rows and columns. In particular, its determinant if well-defined up to the sign. Let us define $\Delta$ to be the square of this determinant (this is well-defined). Because the characters form a basis of the space of class functions, we know that $\Delta\ne0$. When $G={\mathbb Z}/n{\mathbb Z}$, $\Delta=n^n$.

Is there a close formula for $\Delta$ for a general group? Is it always an integer?

REPLY [9 votes]: I found the following answer after posting it:
$$\Delta=\epsilon\prod_c\frac{|G|}{|c|},\qquad\epsilon=(-1)^m,$$
where the product is taken over the conjugacy class. And $m$ is the number of pairs of complex conjugate irreducible characters.
Proof. On the one hand, the complex conjugate of the table is itself, up to $m$ transpositions of rows. This is because the conjugate of an IC is an IC. Therefore 
$$\overline{\det(TC)}=\epsilon\det(TC)$$
($TC$ stands for ``table of characters''.)
Hence $\det(TC)$ is real if $m$ is even, pure imaginary if $m$ is odd. hence $\Delta$ is real and its sign is $\epsilon$.
Now the characters form a unitary basis. Because a unitary matrix has a unit determinant, we may compute $|\Delta|$ by taking any unitary basis. Take $\phi_c(g)$ to be $0$ if $g\not\in c$ and $|G|^{1/2}/|c|^{1/2}$ if $g\in c$. In particular $|\Delta|$ is an integer because $$\frac{|G|}{|c|}=|{\mathcal Z}(a)|,\qquad a\in c.$$
Another Proof: Let $D$ be the diagonal matrix whose diagonal entries are the cardinals of the congugacy classes. We may assume that the first rows of $TC$ are the real characters and the $2m$ last ones are the pairs of complex conjugate characters. Then the $(i,j)$-entry of $M:=(TC)D(TC)^T$ is $|G|\langle\overline{\chi_i},\chi_j\rangle$. From the orthogonality relations, we see that $M={\rm diag}(1,\ldots,1,J,\ldots,J)$ where 
$$J=\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}.$$
The number of blocks $J$ is precisely $m$. Now take the determinant; we obtain $\Delta\det D=(-1)^m|G|^r$ where $r\times r$ is the size of $TC$. Hence the formula.