Enumerate the elements of a finite group $G$ as follows: $g_1,g_2,\dots,g_n$. Introduce $n$ variables indexed by the elements of $G$: $x_{g_1},\dots,x_{g_n}$.
Consider the matrix $X_G$ with entries $a_{ij}=x_{g_ig_j}$. The determinant $\det(X_G)$ of is now a polynomial of degree $n$, in the commuting variables $x_{g_1},\dots,x_{g_n}$, which is called the Frobenius determinant.
The following theorem, discovered by Dedekind and proved by Frobenius, became the starting point for the development of representation theory.
Theorem. The polynomial $\det(X_G)$ factors into a product of irreducible polynomials $P_j(X)$, over $\mathbb{C}$, with multiplicity equaling the degree of $P_j(X)$; that is, $$\det(X_G)=\prod_{j=1}^rP_j(X)^{\deg P_j}$$ where $r$ is the number of conjugacy classes of $G$.
Question still waiting for an answer. Is the converse true for Latin squares?
Given just the multiplication table for a Latin square, compute the determinant, and assume that it factors into irreducible polynomials with multiplicity equal to the degree of the corresponding polynomial as in the theorem of Frobenius. We know that if it does not factor as in the theorem then the Latin square is not that of a group. If it does factor well, can we conclude that the table, up to a permutation of its rows, is the multiplication table of a group?
Note. I have never seen this question asked in the literature. Have you?
Remark. I have checked for small values of $n\leq6$. The converse holds there.
