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?
Best Answer
If $A$ is the character table and $A^\ast$ is its conjugate transpose, then the orthogonality relations tell us that $A A^\ast = \text{diag}\{|C_G(g)|\} $, where the enties run over a fixed choice of elements of $G$, one from each conjugacy class. Thus $|\Delta| = \det A A^\ast = \prod |C_G(g)|$ is an integer. On the other hand, $\Delta$ must be rational. This follows from the fact that the action of $\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$ permutes the columns of $A$, hence fixes $\Delta = (\det A)^2$. Thus $\Delta=\pm |\Delta|$ is an integer.