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I'm not very familiar with the theory of semigroups, so sorry if this is well-known or easy.

Let $G=\{x_1,...,x_n\}$ be a finite semigroup with $n$ elements. Let $T$ be a field and $K=T(y_1,...,y_n)$ be the function field in the variables $y_i$ over $T$. Let $f:G \rightarrow K$ be the function with $f(x_i)=y_i$. Define the $n \times n$-matrix $F_G=F$ with entries $F_{i,j}=f(x_i x_j)$.

More or less this is the "multiplication table" of $G$ with values in the field $K$.

Question 1: Has this matrix been studied in the literature before? When is $F_G$ invertible for a given $G$?

(in general $F$ is not invertible for general semigroups) The number of semigroups with invertible $F_G$ starts for $n \geq 1$ with 1,2,6,24,113,660.

Question 2: If $G$ is an inverse semigroup, then is $F_G$ an invertible matrix over $K$?

(see https://en.wikipedia.org/wiki/Inverse_semigroup for the definition of inverse semigroup)

The question has a positive answer for inverse semigroups with at most 6 elements.

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  • $\begingroup$ Is it maybe more generally true that if you have a matrix whose entries are given by generators of a function field, the only way the determinant can become $0$ is if two rows or columns are already identical? $\endgroup$ Commented Aug 15 at 7:33
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    $\begingroup$ @AchimKrause I think I have a counterexample for that. Take the matrix with first row [y1,y1,y1], second row [y2,y2,y2] and third row [y1,y2,y3]. Then no two rows or columns are equal but the first two rows are linear dependent (they are both multiples of [1,1,1]). $\endgroup$ Commented Aug 15 at 7:37
  • $\begingroup$ Ah yes of course! $\endgroup$ Commented Aug 15 at 7:44

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I wrote a paper about this. The determinant is nonzero precisely when the semigroup algebra is unital and Frobenius. The semigroup need not be a monoid. This is Frobenius’s paratrophic matrix with respect to the semigroup basis. See Benjamin Steinberg, Factoring the Dedekind-Frobenius determinant of a semigroup, Journal of Algebra, Volume 605, 2022, Pages 1-36. The values of the variables that give a nonzero determinant give you the functionals defining a Frobenius form. This essentially goes back to the work of Frobenius on the group determinant and in the paper where he introduced what are now called Frobenius algebras.

In particular it is always invertible for an inverse semigroup because the algebra is Frobenius. My paper focuses on the complex field but this is not necessary.

In my paper I construct 3-nilpotent semigroups with adjoined identity for any $n\times n$ $0/1$-matrix $A$ so that the determinant of the multiplication table is $-\det(A)x(y-x)^{n+2}$ for certain elements $x,y$. In particular the characteristic of the field matters.

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