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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?

Edit. Sorry for this question a bit too naive. I found the answer 10' after posting it: $$\Delta=\prod_c\frac{|G|}{|c|},$$ where the product is taken over the conjugacy class.

Proof. 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 $$\frac{|G|}{|c|}=|{\mathcal Z}(a)|,\qquad a\in c.$$

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 $G={\mathbb Z}/n{\mathbb Z}$, $\Delta=n^2$.\Delta=n^n$. Is there a close formula for$\Delta$for a general group? Is it always an integer?A square? Edit. Sorry for this question a bit too naive. I found the answer 10' after posting it: $$\Delta=\prod_c\frac{|G|}{|c|},$$ where the product is taken over the conjugacy class. Proof. 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 $$\frac{|G|}{|c|}=|{\mathcal Z}(a)|,\qquad a\in c.$$