$\DeclareMathOperator\perm{perm}$Let $G$ be a finite group. Define the determinant $\det(G)$ of $G$ as the determinant of the character table of $G$ over $\mathbb{C}$ and define the permanent $\perm(G)$ of $G$ as the permanent of the character table of $G$ over $\mathbb{C}$. Note that due to the properties of the determinant and the permanent, this definition just depends on $G$ and not on the ordering of the conjugacy classes etc.

I'm not experienced with character theory but did some experiments with GAP on this and found nothing related in the literature, which motivates the following questions (sorry, in case they are trivial). Of course finite groups are dangerous and it is tested just for all finite groups of order at most $n \leq 30$ and some other cases, which might not be too good evidence for a question on finite groups.

Question 1: Are $\perm(G)$ and $\det(G)^2$ always integers?

I was able to prove this for cyclic groups. Since the character table of the direct product of groups is given by their Kroenecker product, one can conclude that $\det(G)^2$ is also an integer for all abelian groups $G$. Maybe there is a formula for the permanent of the Kroenecker product of matrices to conclduge that $\perm(G)$ is also an integer for all abelian groups or even better a more direct proof that question 1 is true at least for abelian groups. Note that $\det(G)$ is in general not an integer, even for cyclic groups.

Now call a finite group permfect in case $\perm(G)=0$.

Question 2: Is it true that all finite groups of order $n$ are permfect if and only if $n=4r+2$ for some $r \geq 2$?

Being permfect could be seen as having a high symmetry. It seems symmetric groups are permfect and for the alternating groups I only found $A_6$ to be permfect yet.

Permfectis a great term. $\endgroup$ – Brian Hopkins Jun 25 at 13:03