TITLE: Determinant associated with a group action
QUESTION [11 upvotes]: Let $G$ be a finite group and $S$ be a finite set, with $G$ acting on $S$. I consider indeterminates $x_g$ indexed by $g\in G$ and form the matrix of the group action $A\in M_{S\times S}$. Its entries are indexed by pairs $(s,t)$ of elements of $S$. By definition, $a_{st}$ is the sum of the $x_g$'s over the group elements such that $s^g=t$.

What are the irreducible factors of $\det A$ ?

Without loss of generality, we may assume that the action is transitive, because otherwise $A$ is block diagonal. Just sort out the elements of $S$ by orbits. We may also assume that the action is faithful.
This determinant has an obvious factor $\ell:=\sum_{g\in G}x_g$, because $\ell$ is the sum of the entries of every row (or of every column). What does it mean about the action that the quotient $(\det A)/\ell$ be irreducible ?
Notice that if $S=G$ and $G$ acts by multipication, this determinant is that considered by Dedekind and Frobenius, which led the latter to the theory of representations.

REPLY [5 votes]: I wish to elaborate a bit on Geoffrey Robinson's answer:
Given a representation $D\colon G\to M_{n\times n}(F)$ which is similar to a representation in upper block form
$$ D \sim \begin{pmatrix} 
             D_1 & * \\
                 & D_2
          \end{pmatrix},
$$
then the determinant factors into
$$ \Theta_D(x):= \det\left( \sum_{g} x_g D(g) \right) = \det\left( \sum_g x_g D_1(g) \right) 
                               \det\left( \sum_g x_g D_2(g) \right) .$$
Thus $\Theta_D$ factors accordingly to the composition factors of $D$, even if the characteristic of $F$ divides $|G|$.  
If $D$ is absolutely irreducible, then $\Theta_D$ is irreducible (see here), and inequivalent representations yield different irreducible polynomials. In particular, your $\det(A)/l$ is irreducible over $\mathbb{C}$ iff the action is doubly transitive.  
The coefficients of $\Theta_D(x)$ as polynomial in the $x_g$'s generate the same field as the values of the character of $D$, and of course $\Theta_D(x)$ remains irreducible over that smaller field (for $D$ absolutely irreducible). If $D$ is irreducible over $F$, but not absolutely irreducible, then $D \cong m(D_1 \oplus \dotsb \oplus D_r)$, where the $D_i$ are algebraically conjugate over $F$, and $\Theta_{D_1}(x)\cdot \dotsm \cdot \Theta_{D_r}(x) $ is irreducible over $F$. Thus in your situation, $\det(A)/l$ is irreducible as a polynomial over $\mathbb{Q}$ iff the permutation module 
$$ \mathbb{Q}S \cong \mathbb{1} \oplus( \text{rationally irreducible module}). $$
This is a slight weakening of double transitivity. There is a paper about this situation:

Dixon, John D., Permutation representations and rational irreducibility, Bull. Austral. Math. Soc. 71 (2005), p. 493-503 (link).  

There are only few examples where this occurs and the action is not doubly transitive, but there is, to the best of my knowledge, no classification.