# The determinant of finite semigroups of the pseudovariety $\mathsf{ECOM}$

M. H. Shahzamanian  M. H. Shahzamanian
CMUP, Departamento de Matemática, Faculdade de Ciências, Universidade do
Porto, Rua do Campo Alegre s/n, 4169–007 Porto (Portugal).
<EMAIL_ADDRESS>
###### Abstract.

The purpose of this paper is to identify the non-zero semigroup determinant of
a class of finite semigroups in which every two idempotents commute. These
semigroups are a larger class than the semigroups with central idempotents and
also inverse semigroups.

2010 Mathematics Subject Classification. Primary 20M25, 16L60, 16S36.
Keywords and phrases: Frobenius algebra, semigroup determinant, paratrophic
determinant, semigroup algebra.

##  1\. Introduction

In the 1880s, Dedekind introduced the concept of the group determinant of
finite groups and with Frobenius, began to study it in depth. At the same
time, Smith also examined this concept, but in a different way, as outlined in
Smith’s work [14]. This study involved the examination of the determinant of a
$G\times G$ matrix, where the entry at the $(g,h)$ position is $x_{gh}$, with
$G$ being a finite group and the variables $x_{k}$, for all $k$ in $G$.
Additionally, the study has been expanded to include finite semigroups, with
various objectives for researchers [17, 10, 18]. An application of the
semigroup determinant for finite semigroups is in the extension of the
MacWilliams theorem for codes over a finite field to chain rings. The non-zero
semigroup determinant is an essential component in this application. It is
only non-zero when $\mathbb{C}S$ is a Frobenius algebra, which also means that
it is unital, as demonstrated by Theorem 2.1 in [15] or Proposition 18 in
Chapter 16 of [11].

In the paper [15] by Steinberg, he provides a factorization of the semigroup
determinant of commutative semigroups. The semigroup determinant is either
zero or it factors into linear polynomials. Steinberg describes the factors
and their multiplicities explicitly. This work was a continuation of previous
studies on commutative semigroups with Frobenius semigroup algebras by
Ponizovskiĭ [12] and Wenger [16]. Steinberg also showed that the semigroup
determinant of an inverse semigroup can be computed as the semigroup
determinant of a finite groupoid.

In [15], the determining the determinant of idempotent semigroups with central
idempotents, is crucial for understanding the determinant of commutative
semigroups. This current paper investigates a class of finite semigroups in
which every two idempotents commute. This is a larger class than the
semigroups with central idempotents and also inverse semigroups discussed in
the previous paper. The paper defines a partial order relation for this class
of finite semigroups. This relation extends the natural partial ordering of
the idempotents within the semigroup. This partial order relation is crucial
for determining the determinant of these semigroups. The paper then proposes a
conjecture regarding some of these semigroups with zero determinant, and
applies this conjecture to recognize semigroups in this class with a non-zero
determinant. Our recognition is more specific for this class of semigroups. To
investigate small semigroups with a maximum order of 8 or less, the author
uses a combination of techniques such as the package Smallsemi [5] in GAP [8]
as a database, and programs developed in C# to test this conjecture about
these types of semigroups, ensuring that there is no contradiction for these
types of semigroups.

The paper is organized as follows. We begin with a preliminary section on
semigroups and determinant of a semigroup. We present a partial order relation
on the semigroups of the pseudovariety $\mathsf{ECOM}$ and investigate their
properties. We then turn to determine the determinant of finite semigroups of
this pseudovariety. To demonstrate the method, several examples are provided,
and their calculations are performed using programs developed in C#. These
examples are discussed in an appendix section at the end of the paper.

##  2\. Preliminaries

###  2.1. Semigroups

For standard notation and terminology relating to semigroups, we refer the
reader to [1, Chap. 5], [4, Chaps. 1-3] and [13, Appendix A].

Let $S$ a finite semigroup. Let $a,b\in S$. We say that
$a\mathrel{\mathscr{R}}b$ if $aS^{1}=bS^{1}$, $a\mathrel{\mathscr{L}}b$ if
$S^{1}a=S^{1}b$ and $a\mathrel{\mathscr{H}}b$ if $a\mathrel{\mathscr{R}}b$ and
$a\mathrel{\mathscr{L}}b$. Also, we say that $a\mathrel{\mathscr{J}}b$, if
$S^{1}aS^{1}=S^{1}bS^{1}$. The relations
$\mathrel{\mathscr{R}},\mathrel{\mathscr{L}}$, $\mathrel{\mathscr{H}}$ and
$\mathrel{\mathscr{J}}$ are Green relations and all of them are equivalence
relations first introduced by Green [9]. We call $R_{a},L_{a},H_{a}$ and
$J_{a}$, respectively, the
$\mathrel{\mathscr{R}},\mathrel{\mathscr{L}},\mathrel{\mathscr{H}}$ and
$\mathrel{\mathscr{J}}$-class containing $a$.

An element $e$ of $S$ is called idempotent if $e^{2}=e$. The set of all
idempotents of $S$ is denoted by $E(S)$. An idempotent $e$ of $S$ is the
identity of the monoid $eSe$. The group of units $G_{e}$ of $eSe$ is called
the maximal subgroup of $S$ at $e$.

A left ideal of a semigroup $S$ is a non-empty subset $A$ of $S$ such that
$SA\subseteq A$. A right ideal of $S$ is defined similarly, with the condition
$AS\subseteq A$. An ideal of $S$ is a subset of $S$ that is both a left and a
right ideal. An element $s$ of $S$ is called (von Neumann) regular if there
exists an element $t\in S$ such that $sts=s$. Note that an element $s$ is
regular if and only if $s\mathrel{\mathscr{L}}e$, for some $e\in E(S)$, if and
only if $s\mathrel{\mathscr{R}}f$, for some $f\in E(S)$. A
$\mathrel{\mathscr{J}}$-class $J$ is regular if all its elements are regular,
if and only if $J$ has an idempotent, if and only if $J^{2}\cap
J\neq\emptyset$.

A pseudovariety of semigroups is a class of finite semigroups that is closed
under taking subsemigroups, homomorphic images, and finite direct products.
The pseudovariety $\mathsf{S}$ consists of all finite semigroups, while the
pseudovariety $\mathsf{G}$ is the class of all finite groups, $\mathsf{Sl}$
and $\mathsf{Com}$ are the pseudovarieties of all finite, respectively,
semilattices and commutative semigroups. The operator $\mathsf{E}$ associates
a pseudovariety $\mathsf{V}$ to the class of finite semigroups such that the
subsemigroup generated by the idempotents of the semigroup belongs to
$\mathsf{V}$, which can be written as

$\mathsf{EV}=\\{S\in\mathsf{S}\mid\langle E(S)\rangle\in\mathsf{V}\\}.$

If a finite semigroup $S$ is a member of $\mathsf{ECom}$, then the set of
idempotents of $S$ is equal to the subsemigroup generated by the idempotents
of $S$, which is itself equal to the set of idempotents. Therefore, the
pseudovariety $\mathsf{ESl}$ is equal to the pseudovariety $\mathsf{ECom}$.

###  2.2. Incidence Algebras and Möbius Functions

Let $(P,\leq)$ be a finite partially ordered set (poset). The incidence
algebra of $P$ over $\mathbb{C}$, which we denote $\mathbb{C}\llbracket
P\rrbracket$, is the algebra of all functions $f\colon P\times
P\rightarrow\mathbb{C}$ such that

$f(x,y)\neq 0\Rightarrow x\leq y$

equipped with the convolution product

$(f\ast g)(x,y)=\sum\limits_{x\leq z\leq y}f(x,z)g(z,y).$

The convolution identity is the delta function $\delta$ given by

$\delta(x,y)=\begin{cases}1&\text{if}\ x=y\\\ 0&\text{otherwise.}\end{cases}$

The zeta function, denoted as $\zeta_{P}$, of the poset $P$ is an element of
$\mathbb{C}\llbracket P\rrbracket$ that it is equal to 1 when $x\leq y$ and 0
otherwise. It has an inverse function, known as the Möbius function,
represented by $\mu_{P}$. In instances where the poset $P$ is clear from
context, the subscript $P$ will be omitted.

###  2.3. Determinant of a semigroup

For standard notation and terminology relating on finite dimensional algebras,
the reader is referred to references [2] and [3].

A based algebra is a finite dimensional complex algebra $A$ with a
distinguished basis $B$. We often refer to the pair of the algebra and its
basis as $(A,B)$. The multiplication in the algebra is determined by its
structure constants with respect to the basis $B$ defined by the equations

$bb^{\prime}\ =\sum\limits_{b^{\prime\prime}\in
B}c_{b^{\prime\prime},b,b^{\prime}}b^{\prime\prime}$

where $b,b^{\prime}\in B$ and
$c_{b^{\prime\prime},b,b^{\prime}}\in\mathbb{C}$. Let $X_{B}=\\{x_{b}\mid b\in
B\\}$ be a set of variables in bijection with $B$. These structure constants
can be represented in a matrix called the Cayley table, which is a $B\times B$
matrix with elements from the polynomial ring $\mathbb{C}[X_{B}]$. It is
defined as a $B\times B$ matrix over $\mathbb{C}[X_{B}]$ with entries given by

$C(A,B)_{b,b^{\prime}}=\sum\limits_{b^{\prime\prime}\in
B}c_{b^{\prime\prime},b,b^{\prime}}x_{b^{\prime\prime}}.$

The determinant of this matrix, called $\theta_{(A,B)}(X_{B})$, is either
identically zero or a homogeneous polynomial of degree $\left|B\right|$.

Let $S$ be a finite semigroup, and let $\mathbb{C}S$ be a finite dimensional
semigroup $\mathbb{C}$-algebra with a basis $S$. The semigroup
$\mathbb{C}$-algebra $\mathbb{C}S$ consisting of all the formal sums
$\sum\limits_{s\in S}\lambda_{s}s$, where $\lambda_{s}\in\mathbb{C}$ and $s\in
S$, with the multiplication defined by the formula

$(\sum\limits_{s\in S}\lambda_{s}s)\cdot(\sum\limits_{t\in
S}\mu_{t}t)=\sum\limits_{u=st\in S}\lambda_{s}\mu_{t}u.$

If $A=\mathbb{C}S$ and $B=S$, then the Cayley table of $C(S)=C(\mathbb{C}S,S)$
is the $S\times S$ matrix over $\mathbb{C}[X_{S}]$ with
$C(S)_{s,s^{\prime}}=x_{ss^{\prime}}$ where $X_{S}=\\{x_{s}\mid s\in S\\}$ is
a set of variables in bijection with $S$. We denote the determinant
$\mathop{\mathrm{Det}}\nolimits C(\mathbb{C}S,S)$ by $\theta_{S}(X_{S})$ and
call it the (Dedekind-Frobenius) semigroup determinant of $S$. If the
semigroup $S$ is fixed, we often write $X$ instead of $X_{S}$. For more
information on this topic, the reader is referred to references [7], [11,
Chapter16] and [15].

The contracted semigroup algebra of a semigroup $S$ with a zero element 0 on
the complex numbers is defined as $\mathbb{C}_{0}S=\mathbb{C}S/\mathbb{C}0$;
note that $\mathbb{C}0$ is a one-dimensional two-sided ideal. This algebra can
be thought of as having a basis consisting of the non-zero elements of $S$ and
having multiplication that extends that of $S$, but with the zero of the
semigroup being identified with the zero of the algebra. The contracted
semigroup determinant of $S$, denoted by $\widetilde{\theta}_{S}$, is defined
as $\widetilde{C}(S)=C(\widetilde{C}_{0}S,S\setminus\\{0\\})$, where
$\widetilde{X}=X_{S\setminus\\{0\\}}$ if $S$ is understood, and
$\widetilde{C}(S)_{s,t}$ is equal to $x_{st}$ if $st\neq 0$ and 0 otherwise.

According to Proposition 2.7 in [15] (the concept mentioned in [18]), there is
a connection between the contracted semigroup determinant and the semigroup
determinant of a semigroup $S$ with a zero element. There is a
$\mathbb{C}$-algebra isomorphism between the $\mathbb{C}$-algebra
$\mathbb{C}S$ and the product algebra $\mathbb{C}_{0}S\times\mathbb{C}0$,
which sends $s\in S\setminus\\{0\\}$ to $(s,0)$ and $0$ to $(0,0)$. Put
$y_{s}=x_{s}-x_{0}$ for $s\neq 0$ and let $Y=\\{y_{s}\mid s\in
S\setminus\\{0\\}\\}$. Then $\theta_{S}(X)=x_{0}\widetilde{\theta}_{S}(Y)$.
Therefore, $\widetilde{\theta}_{S}(\widetilde{X})$ can be obtained from
$\theta_{S}(X)/x_{0}$ by replacing $x_{0}$ with 0.

##  3\. The relation $\ll$

There are necessary conditions are given for a semigroup $S$ to have a non-
zero $\theta_{S}(X)$, as stated in [15]. If $\theta_{S}(X)$ is not equal to 0,
then the semigroup algebra $\mathbb{C}S$ is a unital algebra, according to
Theorem 2.1 in [15]. Furthermore, if $\theta_{S}(X)$ is not equal to 0, then
$S$ satisfies the condition $S^{2}=S$ and also satisfies the condition
described in ($\star$) ‣ 3 below, as stated in Corollaries 2.2 and 2.4 in
[15]:

  1. ($\star$)

For each element $s$ in $S$, the number of elements of $S$ fixed by $s$ under
left and right multiplication is the same.

Let $S$ be a semigroup. We define the functions $\varphi^{r}$ and
$\varphi^{l}$ from $S$ to the power set of $E(S)$ as follows:

$\varphi^{r}(s)=\\{e\in E(S)\mid se=s\\}\ \text{and}\ \varphi^{l}(s)=\\{e\in
E(S)\mid es=s\\}.$

######  Lemma 3.1.

If $S$ is finite and $\mathbb{C}S$ is a unital algebra, then the subsets
$\varphi^{r}(s)$ and $\varphi^{l}(s)$ are non-empty, for every $s\in S$.

###### Proof.

Let $S=\\{s_{1},\ldots,s_{n}\\}$. Suppose the contrary that
$\varphi^{r}(s)=\emptyset$, for an element $s\in S$. Let
$1_{S}=c_{1}s_{1}+\cdots+c_{n}s_{n}$, for some
$c_{1},\ldots,c_{n}\in\mathbb{C}$, be the identity of the algebra
$\mathbb{C}S$. Since $s1_{S}=s$, there exists an integer $1\leq i\leq n$ such
that $c_{i}\neq 0$ and $ss_{i}=s$. Since $\varphi^{r}(s)=\emptyset$, we have
$s_{i}\not\in E(S)$. If $e\in\varphi^{r}(s_{i})$, then we have
$e\in\varphi^{r}(s)$, a contradiction. Thus, we have
$\varphi^{r}(s_{i})=\emptyset$. Similarly, there is a sequence
$s_{i}=s_{i_{1}},s_{i_{2}},s_{i_{3}},\ldots$ of elements of $S$ such that
$s_{i_{j}}s_{i_{j+1}}=s_{i_{j}}$, $s_{i_{j}}\not\in E(S)$ and
$\varphi^{r}(s_{i_{j}})=\emptyset$. Since $S$ is finite, there exist integers
$j_{1}<j_{2}$ such that $s_{i_{j_{1}}}=s_{i_{j_{2}}}$. Now, as

$s_{i_{j_{1}}}s_{i_{j_{1}+1}}\cdots
s_{i_{j_{2}-1}}s_{i_{j_{2}}}=s_{i_{j_{1}}}s_{i_{j_{1}+1}}\cdots
s_{i_{j_{2}-1}}=s_{i_{j_{1}}},$

we have $s_{i_{j_{1}}}s_{i_{j_{1}}}=s_{i_{j_{1}}}s_{i_{j_{2}}}=s_{i_{j_{1}}}$.
A contradiction with $s_{i_{j_{1}}}\not\in E(S)$.

Similarly, we have $\varphi^{l}(s)\neq\emptyset$. ∎

For the rest of this section, we will assume that $S$ belongs to the
pseudovariety $\mathsf{ECom}$ and $S$ satisfies the following conditions
designated as ($\star\star$):

  1. (1)

$S^{2}=S$;

  2. (2)

$S$ satisfies Condition ($\star$) ‣ 3;

  3. (3)

The semigroup algebra $\mathbb{C}S$ is a unital algebra.

The elements of $E(S)$ commute, so we have $E(S)=\langle E(S)\rangle$. Hence,
$E(S)$ is a meet semilattice with respect to the ordering $e\leq f$, if
$ef=e$. By Lemma 3.1, the subsets $\varphi^{r}(s)$ and $\varphi^{l}(s)$ are
non-empty. Since $E(S)$ is a meet semilattice, $\varphi^{r}(s)$ and
$\varphi^{l}(s)$ have unique minimum elements respect to the ordering $\leq$
which are denoted by $s^{{+}_{r}}$ and $s^{{+}_{l}}$, respectively.

We define the following equivalence relations

  1. (1)

$s\sim_{r}t$ if $s^{{+}_{r}}=t^{{+}_{r}}$;

  2. (2)

$s\sim_{l}t$ if $s^{{+}_{l}}=t^{{+}_{l}}$;

  3. (3)

$s\sim t$ if $s^{{+}_{r}}=t^{{+}_{r}}$ and $s^{{+}_{l}}=t^{{+}_{l}}$,

and write $\widetilde{\mathrel{\mathscr{L}}}_{s}$,
$\widetilde{\mathrel{\mathscr{R}}}_{s}$ and
$\widetilde{\mathrel{\mathscr{H}}}_{s}$ for the equivalence classes of $s$ of
these relations, respectively. Note that $s^{{+}_{r}}=t^{{+}_{r}}$ if and only
if $\varphi^{r}(s)=\varphi^{r}(t)$. Also, we have $s^{{+}_{l}}=t^{{+}_{l}}$ if
and only if $\varphi^{l}(s)=\varphi^{l}(t)$. Then, the class
$\widetilde{\mathrel{\mathscr{L}}}_{s}$ is the
$\widetilde{\mathrel{\mathscr{L}}}$-class of $s$, the class
$\widetilde{\mathrel{\mathscr{R}}}_{s}$ is the
$\widetilde{\mathrel{\mathscr{R}}}$-class of $s$ and the class
$\widetilde{\mathrel{\mathscr{H}}}_{s}$ is the
$\widetilde{\mathrel{\mathscr{H}}}$-class of $s$ in the sense of Fountain et
al. [6].

Let $e$ be an idempotent of $\widetilde{\mathrel{\mathscr{L}}}_{s}$. We have
$e=e^{{+}_{r}}=s^{{+}_{r}}$. Then $s^{{+}_{r}}$ is the unique idempotent in
$\widetilde{\mathrel{\mathscr{L}}}_{s}$. Also, $s^{{+}_{l}}$ is the unique
idempotent in $\widetilde{\mathrel{\mathscr{R}}}_{s}$. Note that, it is not
necessarily true that for an idempotent $s\in S$, $s^{{+}_{r}}=s^{{+}_{l}}$
(as demonstrated in Example 5.1).

Let $s$ and $t$ be elements of $S$. We define $s\leq^{r}t$ if $s=te$ and
$s\leq^{l}t$ if $s=et$ with $e\in E(S)$. Also, we define $s\ll t$, if there
exists an element $u$ in $S$ such that

$s\leq^{r}u\leq^{l}t\ \text{or}\ s\leq^{l}u\leq^{r}t.$

######  Proposition 3.2.

The following conditions hold:

  1. (1)

$s\leq^{r}t$ if and only if $s=ts^{{+}_{r}}$. Also, we have $s\leq^{l}t$ if
and only if $s=s^{{+}_{l}}t$.

  2. (2)

if $s\leq^{r}t$ or $s\leq^{l}t$, then $s^{{+}_{l}}\leq t^{{+}_{l}}$ and
$s^{{+}_{r}}\leq t^{{+}_{r}}$.

  3. (3)

$s\ll t$ if and only if $s=s^{{+}_{l}}ts^{{+}_{r}}$.

  4. (4)

if $s\ll t$, then $s^{{+}_{l}}\leq t^{{+}_{l}}$ and $s^{{+}_{r}}\leq
t^{{+}_{r}}$.

  5. (5)

the relation $\ll$ is a partial order extending the ordering $\leq$.

###### Proof.

(1) Suppose that $s\leq^{r}t$. Then, there exists an idempotent $e\in E(S)$
such that $s=te$. Hence, we have $se=s$ and, thus $s^{{+}_{r}}\leq e$.
Therefore, we have $s=ss^{{+}_{r}}=tes^{{+}_{r}}=ts^{{+}_{r}}$. Also, if
$s=ts^{{+}_{r}}$, then we have $s\leq^{r}t$.

Similarly, the proposition holds for the relation $\leq^{l}$.

(2) First, suppose that $s\leq^{r}t$. By part (1), we have $s=ts^{{+}_{r}}$.
Hence, if $e\in\varphi^{l}(t)$ then $e\in\varphi^{l}(s)$ and, thus,
$s^{{+}_{l}}\leq t^{{+}_{l}}$. Also, if $e\in\varphi^{r}(t)$ then
$se=ts^{{+}_{r}}e=tes^{{+}_{r}}=ts^{{+}_{r}}=s$. Thus, $s^{{+}_{r}}\leq
t^{{+}_{r}}$.

Similarly, if $s\leq^{l}t$, then $s^{{+}_{l}}\leq t^{{+}_{l}}$ and
$s^{{+}_{r}}\leq t^{{+}_{r}}$.

(3) First, suppose that $s\ll t$. Then, there exists an elements $u$ in $S$
such that $s\leq^{r}u\leq^{l}t$ or $s\leq^{l}u\leq^{r}t$. By symmetry, we may
assume that $s\leq^{r}u\leq^{l}t$. By parts (1) and (2), since $s\leq^{r}u$,
we have $s=us^{{+}_{r}}$ and $s^{{+}_{l}}\leq u^{{+}_{l}}$. As $u\leq^{l}t$,
by part (1), we have $u=u^{{+}_{l}}t$ and, thus, $s=u^{{+}_{l}}ts^{{+}_{r}}$.
Therefore, it follows that
$s=u^{{+}_{l}}ts^{{+}_{r}}=s^{{+}_{l}}u^{{+}_{l}}ts^{{+}_{r}}=s^{{+}_{l}}ts^{{+}_{r}}$.

Now, suppose that $s=s^{{+}_{l}}ts^{{+}_{r}}$. Hence, we have
$s\leq^{r}s^{{+}_{l}}t\leq^{l}t$ and $s\leq^{l}ts^{{+}_{r}}\leq^{r}t$.
Therefore, $s\ll t$.

(4) By part (3), we have $s=s^{{+}_{l}}ts^{{+}_{r}}$. Now, as
$t=tt^{{+}_{r}}$, we have
$st^{{+}_{r}}=s^{{+}_{l}}ts^{{+}_{r}}t^{{+}_{r}}=s^{{+}_{l}}tt^{{+}_{r}}s^{{+}_{r}}=s^{{+}_{l}}ts^{{+}_{r}}=s$.
Hence, $s^{{+}_{r}}\leq t^{{+}_{r}}$. Similarly, we have $s^{{+}_{l}}\leq
t^{{+}_{l}}$.

(5) Since $s\leq^{r}s$ and $s\leq^{l}s$, we have $s\ll s$. Hence, the relation
$\ll$ is reflexive.

Suppose that $s\ll t$ and $t\ll w$, for some elements $s$, $t$ and $w$ in $S$.
Then, by part (3), we have $s=s^{{+}_{l}}ts^{{+}_{r}}$ and
$t=t^{{+}_{l}}wt^{{+}_{r}}$. By part (4), we have $s^{{+}_{r}}\leq
t^{{+}_{r}}$ and $s^{{+}_{l}}\leq t^{{+}_{l}}$. It follows that
$s=s^{{+}_{l}}t^{{+}_{l}}wt^{{+}_{r}}s^{{+}_{r}}=s^{{+}_{l}}ws^{{+}_{r}}$.
Hence, we have $s\ll w$ and, thus, the relation $\ll$ is transitive.

Now, suppose that $s\ll t$ and $t\ll s$. Then, we have
$s=s^{{+}_{l}}ts^{{+}_{r}}$ and $t=t^{{+}_{l}}st^{{+}_{r}}$. Hence,
$s=s^{{+}_{l}}ts^{{+}_{r}}=s^{{+}_{l}}t^{{+}_{l}}st^{{+}_{r}}s^{{+}_{r}}=t^{{+}_{l}}s^{{+}_{l}}ss^{{+}_{r}}t^{{+}_{r}}=t^{{+}_{l}}st^{{+}_{r}}=t$.

Therefore, $\ll$ is a partial order and it clearly extends the order $\leq$ on
$E(S)$. ∎

If $s^{\prime}\ll s$ and $t^{\prime}\ll t$, it does not mean that
$s^{\prime}t^{\prime}\ll st$. The semigroup’s determinant can be either zero
or non-zero, as shown in Examples 5.2 and 5.3.

For an idempotent $e\in E(S)$, we define the subsets $I^{r}_{e}=\\{s\mid
s^{{+}_{r}}<e\\}$ and $I^{l}_{e}=\\{s\mid s^{{+}_{l}}<e\\}$. Note that if
$s\in I^{r}_{e}$, then we have $e\in\varphi^{r}(s)$. Also, if $s\in
I^{l}_{e}$, then we have $e\in\varphi^{l}(s)$. We have $Se=\\{s\in S\mid
se=s\\}$. Also, if $s\in Se$ and $s\not\in I^{r}_{e}$, then we have
$s^{{+}_{r}}=e$. Hence, we have
$Se=I^{r}_{e}\bigcupdot\widetilde{\mathrel{\mathscr{L}}}_{e}$. For every $f\in
E(I^{r}_{e})$, it is clear that
$\widetilde{\mathrel{\mathscr{L}}}_{f}\subseteq I^{r}_{e}$. Then, we have
$I^{r}_{e}=\bigcupdot\limits_{f\in
E(I^{r}_{e})}\widetilde{\mathrel{\mathscr{L}}}_{f}$. Similarly, we have
$eS=I^{l}_{e}\bigcupdot\widetilde{\mathrel{\mathscr{R}}}_{e}$ and
$I^{l}_{e}=\bigcupdot\limits_{f\in
E(I^{r}_{e})}\widetilde{\mathrel{\mathscr{R}}}_{f}$.

######  Proposition 3.3.

The following properties hold:

  1. (1)

the subsets $I^{r}_{e}$ and $I^{l}_{e}$ are left and right ideal of $S$,
respectively.

  2. (2)

we have $I^{r}_{e}=\emptyset$ if and only if $I^{l}_{e}=\emptyset$.

  3. (3)

we have $I^{r}_{e}=\emptyset$ if and only if
$Se=L_{e}=\widetilde{\mathrel{\mathscr{L}}}_{e}$ where $L_{e}$ is the
$\mathrel{\mathscr{L}}$-class of the element $e$ in $S$.

  4. (4)

we have $I^{l}_{e}=\emptyset$ if and only if
$eS=R_{e}=\widetilde{\mathrel{\mathscr{R}}}_{e}$ where $R_{e}$ is the
$\mathrel{\mathscr{R}}$-class of the element $e$ in $S$.

###### Proof.

(1) Let $s\in I^{r}_{e}$ and $t\in S$. Then, we have $tss^{{+}_{r}}=ts$ and
thus $(ts)^{{+}_{r}}\leq s^{{+}_{r}}<e$. Similarly, $I^{l}_{e}$ is an right
ideal of $S$.

(2) If $I^{r}_{e}\neq\emptyset$ then there exists an idempotent $f\in
I^{r}_{e}$. Then, we have $f<e$. It follows that $I^{l}_{e}\neq\emptyset$.
Similarly, if $I^{l}_{e}\neq\emptyset$, then we have $I^{r}_{e}\neq\emptyset$.

(3) If $Se\neq L_{e}$, then there exists an element $s\in S$ such that
$se\not\in L_{e}$ and, thus, $(se)^{\omega}\neq e$. Hence, we have
$(se)^{\omega}<e$. Since $((se)^{\omega})^{{+}_{r}}=(se)^{\omega}$, we have
$(se)^{\omega}\in I^{r}_{e}$. Then, $I^{r}_{e}\neq\emptyset$. Also, if
$I^{r}_{e}\neq\emptyset$, then there exists an idempotent $f\in I^{r}_{e}$.
So, we have $f<e$ and, thus, $fe=f$. Now, as $f\neq e$, we have $f\in
Se\setminus L_{e}$. Then, $I^{r}_{e}=\emptyset$ if and only if, $Se=L_{e}$.

Now, we prove that $I^{r}_{e}=\emptyset$ if and only if
$Se=\widetilde{\mathrel{\mathscr{L}}}_{e}$. Suppose that
$I^{r}_{e}\neq\emptyset$. Then, again there exists an idempotent $f$ in
$I^{r}_{e}$ and, thus, we have $f<e$. Then $f\in
Se\setminus\widetilde{\mathrel{\mathscr{L}}}_{e}$. Now, suppose that
$Se\setminus\widetilde{\mathrel{\mathscr{L}}}_{e}\neq\emptyset$. Let $se\in
Se\setminus\widetilde{\mathrel{\mathscr{L}}}_{e}$. Then, we have
$(se)^{{+}_{r}}\neq e$ and, thus, we have $(se)^{{+}_{r}}<e$. Thus we have
$I^{r}_{e}\neq\emptyset$.

(4) Similar to (3), the statement holds. ∎

######  Proposition 3.4.

We have $E(I^{r}_{e})=E(I^{l}_{e})$ and
$\left|I^{r}_{e}\right|=\left|I^{l}_{e}\right|$. Moreover, we have
$\left|\widetilde{\mathrel{\mathscr{L}}}_{e}\right|=\left|\widetilde{\mathrel{\mathscr{R}}}_{e}\right|$.

###### Proof.

Let $f\in E(I^{r}_{e})$. Then, we have $f<e$. So $f\in E(I^{l}_{e})$.
Similarly, we have $E(I^{l}_{e})\subseteq E(I^{r}_{e})$.

Let $n$ be the largest integer such that there is a chain
$e_{1}<e_{2}<\cdots<e_{n}=e$ with $e_{1},e_{2},\ldots,e_{n}\in E(I^{r}_{e})$.
We show by induction on $n$ that
$\left|I^{r}_{e}\right|=\left|I^{l}_{e}\right|$. If $n=1$, then we have
$I^{r}_{e}=\emptyset$ and by Proposition 3.3.(2), $I^{r}_{e}=I^{l}_{e}$.
Assume then that $n>1$. Let $f\in E(I^{r}_{e})$. By the assumption of the
induction, we have $\left|I^{r}_{f}\right|=\left|I^{l}_{f}\right|$. Since $S$
satisfies Condition ($\star$) ‣ 3, we have $\left|Sf\right|=\left|fS\right|$.
It follows that
$\left|\widetilde{\mathrel{\mathscr{L}}}_{f}\right|=\left|\widetilde{\mathrel{\mathscr{R}}}_{f}\right|$.
Since $I^{r}_{e}=\bigcupdot\limits_{f\in
E(I^{r}_{e})}\widetilde{\mathrel{\mathscr{L}}}_{f}$ and
$I^{l}_{e}=\bigcupdot\limits_{f\in
E(I^{r}_{e})}\widetilde{\mathrel{\mathscr{R}}}_{f}$, we have
$\left|I^{r}_{e}\right|=\left|I^{l}_{e}\right|$. Also, we have
$\left|\widetilde{\mathrel{\mathscr{L}}}_{e}\right|=\left|\widetilde{\mathrel{\mathscr{R}}}_{e}\right|$.
∎

Note that, we can also conclude that $I^{r}_{e}=\emptyset$ if and only if
$eSe=G_{e}=\widetilde{\mathrel{\mathscr{H}}}_{e}$ where $G_{e}$ is the maximal
subgroup of $S$ at $e$ is the group of units of $eSe$.

It is generally not true that for an idempotent $e\in E(S)$, the subset
$I_{e}^{r}$ is an ideal (see Example 5.4). Also, it is not necessarily true
that for an idempotent $e\in E(S)$, the subset $I_{e}^{r}$ is a subset of
$eSe$ (see Example 5.5).

######  Open Problem 3.5.

Let $S^{\prime}=\\{s\mid s\in S\ \text{and}\ s^{{+}_{r}}\neq s^{{+}_{l}}\\}$
and let $m=\left|S^{\prime}\right|$. In this case, we may write
$S^{\prime}=\\{s_{1},\ldots,s_{m}\\}$. Suppose that $\theta_{S}(X)\neq 0$.
Does exist a permutation $\sigma$ of the symmetric group $S_{m}$ such that
$s_{i}^{{+}_{l}}=s_{\sigma(i)}^{{+}_{r}}$ and
$s_{i}^{{+}_{r}}=s_{\sigma^{-1}(i)}^{{+}_{l}}$, for every $1\leq i\leq m$?

In Examples 5.6 and 5.7, the classes of $\widetilde{\mathrel{\mathscr{R}}}$
and $\widetilde{\mathrel{\mathscr{L}}}$ of semigroups $S_{5}$ and $S_{6}$ with
the non-zero determinant are described. In $S_{5}$, we have
$S^{\prime}=\\{u,t\\}$, $u^{{+}_{l}}=t^{{+}_{r}}$ and
$t^{{+}_{l}}=u^{{+}_{r}}$. Then, $\sigma=(1,2)$. Also, in $S_{6}$, we have
$S^{\prime}=\\{y,z,t\\}$, $y^{{+}_{l}}=z^{{+}_{r}}$, $z^{{+}_{l}}=t^{{+}_{r}}$
and $t^{{+}_{l}}=y^{{+}_{r}}$. Then, $\sigma=(1,2,3)$.

##  4\. The determinant of finite semigroups of the pseudovariety
$\mathsf{ECOM}$

The relationships $\leq^{l}$, $\leq^{r}$ and $\ll$ are all the same when $S$
is central idempotents. They can all be represented by $\leq$, where $s\leq t$
if there exists an idempotent $e\in E(S)$ such that $s=te$. Also, we have
$I^{r}_{e}=I^{l}_{e}$, for an idempotent $e$. We can simplify the notation
$I^{r}_{e}=I^{l}_{e}$ to $I_{e}$. In this case, for an idempotent $e\in E(S)$,
we define $\widetilde{H}^{0}_{e}$ as the Rees quotient $eSe/I_{e}$ if $I_{e}$
is non-empty. If $I_{e}$ is empty, then
$\widetilde{H}^{0}_{e}=G_{e}\cup\\{z\\}$ where $z$ is an adjoined zero. Note
that $\widetilde{H}^{0}_{e}$ forms a monoid with identity $e$ and can be
identified with $\widetilde{H}_{e}\cup\\{z\\}$ where $z$ is a zero and, for
$a,b\in\widetilde{H}_{e}$, we have that

$a\cdot b=\begin{cases}ab,&\text{if}\ ab\in\widetilde{H}_{e}\\\
z,&\text{else}.\end{cases}$

The determinant of $S$ when $S$ is central idempotents, as stated in Theorem
6.6 of [15], is computed as follows:

######  Theorem 4.1.

Let $S$ be a finite idempotent semigroup with central idempotents. For $s\in
S$, put $y_{s}=\sum\limits_{t\leq s}\mu_{S}(t,s)x_{t}$. Then
$\theta_{S}=\prod\limits_{e\in
E(S)}\widetilde{\theta}_{\widetilde{H}^{0}_{e}}(Y_{e})$ where
$Y_{e}=\\{y_{s}\mid s\in\widetilde{H}_{e}\\}$.

Continuing on, our aim is to recognize non-zero determinant of semigroups in
$\mathsf{ECOM}$ which they may not central idempotent, using a similar method
as in [15] to arrive at Theorem 4.1, by utilizing the relation $\ll$. Then,
let $S\in\mathsf{ECom}$ in the this section such that $S$ satisfies Conditions
3.1.

Let $Z\colon\mathbb{C}S\rightarrow\mathbb{C}S$ be a map given by
$Z(s)=\sum\limits_{s^{\prime}\ll s}s^{\prime}$ on $s\in S$.

######  Proposition 4.2.

The mapping $Z$ is bijective.

###### Proof.

Suppose that there are non-zero scalars $c_{1},\ldots,c_{n}$ of $\mathbb{C}$
and pairwise distinct elements $s_{1},\ldots,s_{n}$ of $S$ such that
$Z(c_{1}s_{1}+\ldots+c_{n}s_{n})=0$. Since, by Lemma 3.2.(5), the relation
$\ll$ is partial order, there is an integer $j$ such that $s_{j}\not\ll
s_{i}$, for all $i\in\\{1,\ldots,n\\}\setminus\\{j\\}$. Now, as $c_{j}\neq 0$
and $s_{j}\leq s_{j}$, the term $c_{j}s_{j}$ is included in
$Z(c_{1}s_{1}+\ldots+c_{n}s_{n})$, and therefore
$Z(c_{1}s_{1}+\ldots+c_{n}s_{n})$ is not equal to zero. This contradicts our
assumption. Then, the map $Z$ is one to one.

Also, since $Z(\sum\limits_{t\ll s}\mu_{S}(t,s)t)=s$, we can conclude that $Z$
is surjective. ∎

For $s$ and $t$ in $S$, one recursively defines two sequences $s_{n}$ and
$t_{n}$ by

$s_{0}=s,t_{0}=t$

and

$s_{n+1}=s_{n}t_{n}^{{+}_{l}},t_{n+1}=s_{n}^{{+}_{r}}t_{n}.$

######  Proposition 4.3.

We have $s_{n+1}=st_{n}^{{+}_{l}}$ and $t_{n+1}=s_{n}^{{+}_{r}}t$. Also, we
have $s_{n+1}^{{+}_{r}},t_{n+1}^{{+}_{l}}\leq
s_{n}^{{+}_{r}},t_{n}^{{+}_{l}}$.

###### Proof.

First, we prove that $s_{n+1}^{{+}_{r}},t_{n+1}^{{+}_{l}}\leq
s_{n}^{{+}_{r}},t_{n}^{{+}_{l}}$. Since

$s_{n+1}s_{n}^{{+}_{r}}=s_{n}t_{n}^{{+}_{l}}s_{n}^{{+}_{r}}=s_{n}s_{n}^{{+}_{r}}t_{n}^{{+}_{l}}=s_{n}t_{n}^{{+}_{l}}=s_{n+1},$

we have $s_{n+1}^{{+}_{r}}\leq s_{n}^{{+}_{r}}$. Also, as

$s_{n+1}t_{n}^{{+}_{l}}=s_{n}t_{n}^{{+}_{l}}t_{n}^{{+}_{l}}=s_{n}t_{n}^{{+}_{l}}=s_{n+1},$

we have $s_{n+1}^{{+}_{r}}\leq t_{n}^{{+}_{l}}$. Similarly, we have
$t_{n+1}^{{+}_{l}}\leq s_{n}^{{+}_{r}},t_{n}^{{+}_{l}}$.

Let us show that $s_{n+1}=st_{n}^{{+}_{l}}$ and $t_{n+1}=s_{n}^{{+}_{r}}t$ by
induction on $n$. The case $n=0$ is clear. Assume then that $n>0$, and that
the result holds for smaller values of $n$. We have
$s_{n+1}=s_{n}t_{n}^{{+}_{l}}$. By hypothesis of induction, we have
$s_{n}=st_{n-1}^{{+}_{l}}$. It follows that
$s_{n+1}=st_{n-1}^{{+}_{l}}t_{n}^{{+}_{l}}$. Now, as $t_{n}^{{+}_{l}}\leq
t_{n-1}^{{+}_{l}}$, we have $s_{n+1}=st_{n}^{{+}_{l}}$. Similarly, we have
$t_{n+1}=s_{n}^{{+}_{r}}t$. ∎

######  Proposition 4.4.

We have $s_{n}t_{n}=st$, for every integer $n\geq 0$. Moreover, we have
$s_{n}et_{n}=set$, for every idempotent $e$.

###### Proof.

By Proposition 4.3, we have $s_{n}=st_{n-1}^{{+}_{l}}$ and
$t_{n}=s_{n-1}^{{+}_{r}}t$.

Let us show that $st_{m}^{{+}_{l}}s_{m}^{{+}_{r}}t=st$ by induction on $m\leq
n-1$. First assume that $m=0$. Hence, we have

$st_{0}^{{+}_{l}}s_{0}^{{+}_{r}}t=ss_{0}^{{+}_{r}}t_{0}^{{+}_{l}}t=st.$

Assume then that $m>0$, and that the result holds for smaller values of $m$.
By hypothesis of induction and Proposition 4.3, we have

$\displaystyle st_{m}^{{+}_{l}}s_{m}^{{+}_{r}}t$
$\displaystyle=st_{m-1}^{{+}_{l}}t_{m}^{{+}_{l}}s_{m}^{{+}_{r}}s_{m-1}^{{+}_{r}}t=s_{m}t_{m}^{{+}_{l}}s_{m}^{{+}_{r}}t_{m}$
$\displaystyle=s_{m}s_{m}^{{+}_{r}}t_{m}^{{+}_{l}}t_{m}=s_{m}t_{m}=st_{m-1}^{{+}_{l}}s_{m-1}^{{+}_{r}}t$
$\displaystyle=st.$

Therefore, we have $s_{n}t_{n}=st_{n-1}^{{+}_{l}}s_{n-1}^{{+}_{r}}t=st$.

Similar as above, we have $s_{n}et_{n}=set$. ∎

Since $S$ is finite, by Proposition 4.3, there exists an integer $n_{s,t}$
such that if $n_{s,t}\neq 0$, then
$t_{n_{s,t}}^{{+}_{l}}=s_{n_{s,t}}^{{+}_{r}}$ and $t_{n_{s,t}-1}^{{+}_{l}}\neq
s_{n_{s,t}-1}^{{+}_{r}}$, otherwise $t_{0}^{{+}_{l}}=s_{0}^{{+}_{r}}$. We
denote the elements $t_{n_{s,t}}^{{+}_{l}}$ and $s_{n_{s,t}}^{{+}_{r}}$ by
$\epsilon_{s,t}$. Then, we have $s_{n_{s,t}+1}=s_{n_{s,t}}=s\epsilon_{s,t}$,
$t_{n_{s,t}+1}=t_{n_{s,t}}=\epsilon_{s,t}t$. It follows that
$(s\epsilon_{s,t})^{{+}_{r}}=(\epsilon_{s,t}t)^{{+}_{l}}=\epsilon_{s,t}$.

Let $E^{st}=\\{e\in E(S)\mid set=st\\}$. Since the idempotents $s^{{+}_{r}}$
and $t^{{+}_{l}}$ are in the set $E^{st}$, $E^{st}$ is non empty. By
Propositions 4.3 and 4.4, we have $s\epsilon_{s,t}t=st$ and
$\epsilon_{s,t}\leq t^{{+}_{l}},s^{{+}_{r}}$. It is possible that the subset
$E^{st}$ has more than one minimal element (see Example 5.8).

We define the multiplication
$\sharp\colon\mathbb{C}S\times\mathbb{C}S\rightarrow\mathbb{C}S$ as follows:

$s\sharp t=\begin{cases}st,&\text{if}\
s^{{+}_{l}}=(st)^{{+}_{l}},t^{{+}_{r}}=(st)^{{+}_{r}}\ \text{and}\
s^{{+}_{r}}=t^{{+}_{l}};\\\ 0,&\text{otherwise},\end{cases}$

for every $s,t\in S$. The multiplication operation $\sharp$ is not always
associative (as demonstrated in Example 5.9). Also, we define the
multiplication
$\sharp\limits^{e}\colon\mathbb{C}S\times\mathbb{C}S\rightarrow\mathbb{C}S$,
for some $e\in E(S)$, as follows:

$s\sharp^{e}t=\begin{cases}st&\text{if}\
s^{{+}_{l}}=(st)^{{+}_{l}},t^{{+}_{r}}=(st)^{{+}_{r}}\ \text{and}\
s^{{+}_{r}}=t^{{+}_{l}}=e;\\\ 0&\text{otherwise},\end{cases}$

According to Proposition 4.2, the mapping $Z$ is a one-to-one correspondence.
We define the following multiplication on $\mathbb{C}S\times\mathbb{C}S$.

$Z(s)\boldsymbol{*}Z(t)=\sum\limits_{\begin{subarray}{c}s^{\prime}\ll s,\\\
t^{\prime}\ll
t\end{subarray}}s^{\prime}\stackrel{{\scriptstyle\epsilon_{{s^{\prime}}^{{+}_{l}}s,t{t^{\prime}}^{{+}_{r}}}}}{{\sharp}}t^{\prime}$

Since $Z$ is bijective, for every $u,u^{\prime}\in S$, there exist elements
$s_{1}$, $\ldots$, $s_{n_{u}}$, $t_{1}$, $\ldots$, $t_{n_{u^{\prime}}}$ and
scalars $c_{1},\ldots,c_{n_{u}}$, $d_{1},\ldots,d_{n_{u^{\prime}}}$ such that
$u=c_{1}Z(s_{1})+\cdots+c_{n_{u}}Z(s_{n_{u}})$ and
$u^{\prime}=d_{1}Z(t_{1})+\cdots+d_{n_{u^{\prime}}}Z(t_{n_{u^{\prime}}})$.
Then

$u\boldsymbol{*}u^{\prime}=\prod\limits
c_{i}d_{j}Z(s_{i})\boldsymbol{*}Z(t_{j}).$

######  Proposition 4.5.

We have

$Z(s)\boldsymbol{*}Z(t)=Z(st)$

for every $s,t\in S$.

###### Proof.

Let $s^{\prime}\ll s$ and $t^{\prime}\ll t$. The first step is to prove that
$s^{\prime}t^{\prime}\ll st$, if
$s^{\prime}\stackrel{{\scriptstyle\epsilon_{{s^{\prime}}^{{+}_{l}}s,t{t^{\prime}}^{{+}_{r}}}}}{{\sharp}}t^{\prime}\neq
0$. It is clear that the following condition holds:

(1)
${s^{\prime}}^{{+}_{l}}=(s^{\prime}t^{\prime})^{{+}_{l}},{t^{\prime}}^{{+}_{r}}=(s^{\prime}t^{\prime})^{{+}_{r}}\
\text{and}\ {s^{\prime}}^{{+}_{r}}={t^{\prime}}^{{+}_{l}}.$

Since $s^{\prime}\ll s$ and $t^{\prime}\ll t$, we have
$s^{\prime}={s^{\prime}}^{{+}_{l}}s{s^{\prime}}^{{+}_{r}}$ and
$t^{\prime}={t^{\prime}}^{{+}_{l}}t{t^{\prime}}^{{+}_{r}}$. As
$s^{\prime}\stackrel{{\scriptstyle\epsilon_{{s^{\prime}}^{{+}_{l}}s,t{t^{\prime}}^{{+}_{r}}}}}{{\sharp}}t^{\prime}\neq
0$, we have
${s^{\prime}}^{{+}_{r}}={t^{\prime}}^{{+}_{l}}=\epsilon_{{s^{\prime}}^{{+}_{l}}s,t{t^{\prime}}^{{+}_{r}}}$.
Hence, we have
$s^{\prime}t^{\prime}={s^{\prime}}^{{+}_{l}}s{s^{\prime}}^{{+}_{r}}{t^{\prime}}^{{+}_{l}}t{t^{\prime}}^{{+}_{r}}={s^{\prime}}^{{+}_{l}}s\epsilon_{{s^{\prime}}^{{+}_{l}}s,t{t^{\prime}}^{{+}_{r}}}t{t^{\prime}}^{{+}_{r}}={s^{\prime}}^{{+}_{l}}st{t^{\prime}}^{{+}_{r}}$.
As ${s^{\prime}}^{{+}_{l}}=(s^{\prime}t^{\prime})^{{+}_{l}}$ and
${t^{\prime}}^{{+}_{r}}=(s^{\prime}t^{\prime})^{{+}_{r}}$, we have
$s^{\prime}t^{\prime}\ll st$.

Secondly, we need to show that if $u\ll st$, for some $u$ in $S$, there exists
a unique pair $(s^{\ast},t^{\ast})$ such that $s^{\ast}\ll s$, $t^{\ast}\ll
t$,
$s^{\ast}\stackrel{{\scriptstyle\epsilon_{{s^{\ast}}^{{+}_{l}}s,t{t^{\ast}}^{{+}_{r}}}}}{{\sharp}}t^{\ast}\neq
0$, and $s^{\ast}t^{\ast}=u$. Since $u\ll st$, we have
$u=u^{{+}_{l}}stu^{{+}_{r}}$. Let $s^{\prime}=u^{{+}_{l}}s$ and
$t^{\prime}=tu^{{+}_{r}}$. It is easily follows that
${s^{\prime}}^{{+}_{l}}=u^{{+}_{l}}$ and ${t^{\prime}}^{{+}_{r}}=u^{{+}_{r}}$.
Let $s^{\prime\prime}=s^{\prime}\epsilon_{s^{\prime},t^{\prime}}$ and
$t^{\prime\prime}=\epsilon_{s^{\prime},t^{\prime}}t^{\prime}$. Clearly, we
have $s^{\prime\prime}\ll s$, $t^{\prime\prime}\ll t$ and
${s^{\prime\prime}}^{{+}_{r}}={t^{\prime\prime}}^{{+}_{l}}=\epsilon_{s^{\prime},t^{\prime}}$.
As
$s^{\prime}\epsilon_{s^{\prime},t^{\prime}}t^{\prime}=s^{\prime}t^{\prime}=u$,
we have ${s^{\prime\prime}}^{{+}_{l}}=u^{{+}_{l}}$ and
${t^{\prime\prime}}^{{+}_{r}}=u^{{+}_{r}}$. Then, we have
$s^{\prime\prime}\stackrel{{\scriptstyle\epsilon_{{s^{\prime\prime}}^{{+}_{l}}s=s^{\prime},t{t^{\prime\prime}}^{{+}_{r}}=t^{\prime}}}}{{\sharp}}t^{\prime\prime}=s^{\prime\prime}t^{\prime\prime}=s^{\prime}\epsilon_{s^{\prime},t^{\prime}}t^{\prime}=u$.

There is then our desired pair. Now, we prove the uniqueness of this
existence. Let $s_{1},s_{2},t_{1},t_{2}\in S$ such that $s_{1},s_{2}\ll s$,
$t_{1},t_{2}\ll t$, $s_{1}t_{1}=s_{2}t_{2}=u$, and
$s_{1}\stackrel{{\scriptstyle\epsilon_{{s_{1}}^{{+}_{l}}s,t{t_{1}}^{{+}_{r}}}}}{{\sharp}}t_{1},s_{2}\stackrel{{\scriptstyle\epsilon_{{s_{2}}^{{+}_{l}}s,t{t_{2}}^{{+}_{r}}}}}{{\sharp}}t_{2}\neq
0$. Then, the pairs $(s_{1},t_{1})$ and $(s_{2},t_{2})$ satisfy Condition (1)
and, thus, we have $s_{1}^{{+}_{l}}=s_{2}^{{+}_{l}}=u^{{+}_{l}}$ and
$t_{1}^{{+}_{r}}=t_{2}^{{+}_{r}}=u^{{+}_{r}}$. Also, we have
$s_{1}^{{+}_{r}}=s_{2}^{{+}_{r}}=t_{1}^{{+}_{l}}=t_{2}^{{+}_{l}}=\epsilon_{u^{{+}_{l}}s,tu^{{+}_{r}}}$.
Therefore, we have
$s_{1}=u^{{+}_{l}}s\epsilon_{u^{{+}_{l}}s,tu^{{+}_{r}}}=s_{2}$ and
$t_{1}=\epsilon_{u^{{+}_{l}}s,tu^{{+}_{r}}}su^{{+}_{r}}=t_{2}$.

The result follows. ∎

######  Theorem 4.6.

The mapping $Z$ is an isomorphism of $K$-algebras.

###### Proof.

By Propositions 4.2 and 4.5, the result follows. ∎

We consider the following conjuncture where the determinant of a semigroup
that fulfills the conjuncture condition is equal to zero.

######  Conjecture 4.7.

Suppose that there exist elements $u$ and $u^{\prime}$ such that
$u^{{+}_{r}}\neq{u^{\prime}}^{{+}_{l}}$ and $u\boldsymbol{*}u^{\prime}\neq 0$.
We have $\theta_{S}(X)=0$.

Since we are interested in investigating semigroups with non-zero determinant,
based on our conjecture, we impose the conditions that the following Theorem
holds, as stated below:

  1. ($\diamond$)

if $u^{{+}_{r}}\neq{u^{\prime}}^{{+}_{l}}$, then
$u\boldsymbol{*}u^{\prime}=0$, for every $u,u^{\prime}\in S$.

Let $X_{e}=\\{x_{s}\in X\mid
s\in\widetilde{\mathrel{\mathscr{L}}}_{e}\widetilde{\mathrel{\mathscr{R}}}_{e}\\}$.
Let $\theta_{e}(X_{e})$ be the determinant of the submatrix
$\widetilde{\mathrel{\mathscr{L}}}_{e}\times\widetilde{\mathrel{\mathscr{R}}}_{e}$
of the Cayley table $(Z(S),\boldsymbol{*})$, for every idempotent $e\in S$.

######  Theorem 4.8.

Let $S\in\mathsf{ECOM}$. Suppose that the condition ($\diamond$) holds for the
semigroup $S$. For $s\in S$, put $y_{s}=\sum\limits_{t\ll
s}\mu_{S}(t,s)x_{t}$. The determinant of $S$ is non-zero if and only if
$\widetilde{\theta}_{e}(Y_{e})\neq 0$, for every idempotent $e$, where
$Y_{e}=\\{y_{s}\mid
s\in\widetilde{\mathrel{\mathscr{L}}}_{e}\widetilde{\mathrel{\mathscr{R}}}_{e}\\}$.

###### Proof.

Since $\mathbb{C}S\cong Z(\mathbb{C}S)$, by Theorem 4.6, we can take the set

$X=\\{x_{1},\ldots,x_{\left|S\right|}\\}$

as the base of $Z(\mathbb{C}S)$. If
$s\in\widetilde{\mathrel{\mathscr{L}}}_{e}$ and
$t\in\widetilde{\mathrel{\mathscr{R}}}_{e^{\prime}}$, for some distinct
idempotents $e$ and $e^{\prime}$, by Condition ($\diamond$), then we have
$x_{s}\boldsymbol{*}x_{t}=0$. Let $P$ be a matrix that by rearranging and
shifting the rows and columns of $C(X)$ so that the elements of the subset
$\widetilde{\mathrel{\mathscr{L}}}_{e}$ being adjacent rows and the elements
of the subset $\widetilde{\mathrel{\mathscr{R}}}_{e}$ being adjacent columns
for every idempotent $e\in E(S)$. Then, the determinant of the matrix $P$ is
equal to $\prod\limits_{e\in E(S)}\widetilde{\theta}_{e}(X_{e})$. Hence, the
determinant of $P$ is non-zero if and only if
$\widetilde{\theta}_{e}(X_{e})\neq 0$, for every idempotent $e$. This latter
conclusion also applies to $C(X)$. Now, as $\mathbb{C}S\cong Z(\mathbb{C}S)$
and $Z$ is unimodular, by [15, Proposition 2.6], we have $\theta_{S}(X)\neq 0$
if and only if $\widetilde{\theta}_{e}(Y_{e})\neq 0$, for every idempotent
$e$, where $Y_{e}=\\{y_{s}\mid
s\in\widetilde{\mathrel{\mathscr{L}}}_{e}\widetilde{\mathrel{\mathscr{R}}}_{e}\\}$.
∎

Example 5.10 utilizes Theorem 4.8 to demonstrate that the determinant of the
semigroup $S_{9}$ is non-zero.

Consider the following open problem in which the determinant of a semigroup
satisfying the conjuncture condition is zero.

######  Open Problem 4.9.

Suppose that there exists an idempotent $e$ such that $st=set$, $s\sharp t\neq
0$ and $e<s^{{+}_{r}}$. We have $\theta_{S}(X)=0$.

## Acknowledgments

The author was partially supported by CMUP, which is financed by national
funds through FCT – Fundação para a Ciência e a Tecnologia, I.P., under the
project UIDB/00144/2020. The author also acknowledges FCT support through a
contract based on the “Lei do Emprego Científico” (DL 57/2016).

##  5\. Appendix A

In the Appendix section, we provide examples of the semigroups that the paper
discusses. We present detailed information for each semigroup, including its
Cayley table and contracted semigroup determinants. In the examples provided,
we exclude the element zero from the rows and columns of the Cayley table. If
the multiplication of two non-zero elements in the Cayley table results in
zero, we represent it with a dot. Additionally, we note the contracted
semigroup determinants.

######  Example 5.1.

Let $S_{1}$ be the semigroup with the following Cayley table.

$S_{1}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$
---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | $z$ | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | $y$ | . | $u$ | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | $y$ | $z$ | $u$ | $t$

$\widetilde{\theta}_{S_{1}}(X_{S_{1}})=-x_{y}x_{z}x_{y}x_{z}$.
The partitions of $S_{1}$ respect the equivalence relations $\sim_{r}$,
$\sim_{l}$ and $\sim$ are as follows:

$S_{1}/\sim_{r}=\\{\\{0\\},\\{y,t\\},\\{z,u\\}\\},S_{1}/\sim_{l}=\\{\\{0\\},\\{y,u\\},\\{z,t\\}\\}\
\text{and}$ $S_{1}/\sim=\\{\\{0\\},\\{y\\},\\{z\\},\\{u\\},\\{t\\}\\}.$

We have $y^{{+}_{r}}=t$ and $y^{{+}_{l}}=u$.

######  Example 5.2.

Let $S_{2}$ be the semigroup with the following Cayley table.

$S_{2}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$ | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | . | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | $y$ | $y$ | $y$ | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | . | $z$ | . | $z$ | $t$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | . | . | $z$ | $y$ | $y$ | $t$ | $w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | . | $u$ | $y$ | $u$ | $v$ | $v$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$ | $y$ | $z$ | $u$ | $t$ | $w$ | $v$ | $q$

$\widetilde{\theta}_{S_{2}}(X_{S_{2}})=x_{y}x_{z}x_{y}x_{z}x_{z}x_{y}x_{z}$.
We have $t\ll w$, $u\ll w$ and $z=tu\not\ll ww=y$.

######  Example 5.3.

Let $S_{3}$ be the semigroup with the following Cayley table.

$S_{3}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{a}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{b}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{s^{\prime}}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t^{\prime}}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{s}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{e}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{f}}$
---|---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{a}}$ | . | . | . | . | . | . | $a$ | $a$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{b}}$ | . | . | . | . | . | . | $b$ | $b$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{s^{\prime}}}$ | . | . | . | $a$ | . | $a$ | $s^{\prime}$ | $s^{\prime}$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t^{\prime}}}$ | . | . | . | . | . | . | $t^{\prime}$ | $t^{\prime}$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{s}}$ | . | . | . | $a$ | . | $b$ | $s^{\prime}$ | $s$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | . | . | . | . | . | $t$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{e}}$ | $a$ | $b$ | $s^{\prime}$ | $t^{\prime}$ | $s$ | $t^{\prime}$ | $e$ | $e$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{f}}$ | $a$ | $b$ | $s^{\prime}$ | $t^{\prime}$ | $s$ | $t$ | $e$ | $f$

$\theta_{S_{3}}(X_{S_{3}})=0$.
We have $s^{\prime}\ll s$, $t^{\prime}\ll t$, $s^{\prime}t^{\prime}=a$, $st=b$
and $a\not\ll b$.

######  Example 5.4.

In the semigroup $S_{1}$ (Example 5.1), we have $I^{r}_{t}=\\{0,z,u\\}$ and
$I^{r}_{t}$ is not an ideal.

######  Example 5.5.

Let $S_{4}$ be the semigroup with the following Cayley table.

$S_{4}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$
---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | $z$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | $u$ | $u$ | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | $z$ | . | $t$ | $t$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | $y$ | $z$ | . | $t$ | $w$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | . | $z$ | $u$ | $t$ | $t$ | $v$

$\widetilde{\theta}_{S_{4}}(X_{S_{4}})=-x_{y}x_{z}x_{u}x_{z}x_{y}x_{u}$.
We have $wS_{4}w=\\{0,z,t,w\\}$, $vS_{4}v=\\{0,u,t,v\\}$,
$I^{r}_{w}=\\{0,u,t\\}$, $I^{l}_{v}=\\{0,z,t\\}$, $I^{r}_{w}\not\subset
wS_{4}w$ and $I^{l}_{v}\not\subset vS_{4}v$.

######  Example 5.6.

Let $S_{5}$ be the semigroup with the following Cayley table.

$S_{5}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$
---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | $z$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | $y$ | $u$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | . | $z$ | . | . | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | . | $z$ | . | $t$ | $w$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | . | $u$ | . | . | $v$

The classes $\widetilde{\mathrel{\mathscr{R}}}$ and
$\widetilde{\mathrel{\mathscr{L}}}$ of $S_{5}$, for idempotents of $S_{5}$,
are as follows:
$wS_{5}$ $S_{5}w$ $\mathrel{\mathscr{R}}_{w}$ $w$ $\ \ ////\ \ $
$\mathrel{\mathscr{L}}_{w}$ $w$
$\widetilde{\mathrel{\mathscr{R}}}_{w}\setminus\mathrel{\mathscr{R}}_{w}$ $z,\
t$ $\ \ ////\ \ $
$\widetilde{\mathrel{\mathscr{L}}}_{w}\setminus\mathrel{\mathscr{L}}_{w}$ $z,\
u$ $I^{l}_{w}$ $0$ $\ \ ////\ \ $ $I^{r}_{w}$ $0$ $vS_{5}$ $S_{5}v$
$\mathrel{\mathscr{R}}_{v}$ $v$ $\ \ ////\ \ $ $\mathrel{\mathscr{L}}_{v}$ $v$
$\widetilde{\mathrel{\mathscr{R}}}_{v}\setminus\mathrel{\mathscr{R}}_{v}$ $y,\
u$ $\ \ ////\ \ $
$\widetilde{\mathrel{\mathscr{L}}}_{v}\setminus\mathrel{\mathscr{L}}_{v}$ $y,\
t$ $I^{l}_{v}$ $0$ $\ \ ////\ \ $ $I^{r}_{v}$ $0$
$\widetilde{\theta}_{S_{5}}(X_{S_{5}})=x_{y}x_{z}x_{y}x_{z}x_{z}x_{y}$.

######  Example 5.7.

Let $S_{6}$ be the semigroup with the following Cayley table.

$S_{6}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$
---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | $z$ | $z$ | $z$ | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | $y$ | . | $u$ | $u$ | $u$ | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | $y$ | . | $u$ | $u$ | $t$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | $y$ | $z$ | $u$ | $u$ | $w$ | $w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | $z$ | $u$ | $t$ | $w$ | $v$

The classes $\widetilde{\mathrel{\mathscr{R}}}$ and
$\widetilde{\mathrel{\mathscr{L}}}$ of $S_{6}$, for idempotents of $S_{6}$,
are as follows:
$uS_{6}$ $S_{6}u$ $\mathrel{\mathscr{R}}_{u}$ $u$ $\ \ ////\ \ $
$\mathrel{\mathscr{L}}_{u}$ $u$
$\widetilde{\mathrel{\mathscr{R}}}_{u}\setminus\mathrel{\mathscr{R}}_{u}$ $y$
$\ \ ////\ \ $
$\widetilde{\mathrel{\mathscr{L}}}_{u}\setminus\mathrel{\mathscr{L}}_{u}$ $z$
$I^{l}_{u}$ $0$ $\ \ ////\ \ $ $I^{r}_{u}$ $0$ $wS_{6}$ $S_{6}w$
$\mathrel{\mathscr{R}}_{w}$ $w$ $\ \ ////\ \ $ $\mathrel{\mathscr{L}}_{w}$ $w$
$\widetilde{\mathrel{\mathscr{R}}}_{w}\setminus\mathrel{\mathscr{R}}_{w}$ $z$
$\ \ ////\ \ $
$\widetilde{\mathrel{\mathscr{L}}}_{w}\setminus\mathrel{\mathscr{L}}_{w}$ $t$
$I^{l}_{w}$ $0,\ y,\ u$ $\ \ ////\ \ $ $I^{r}_{w}$ $0,\ z,\ u$ $vS_{6}$
$S_{6}v$ $\mathrel{\mathscr{R}}_{v}$ $v$ $\ \ ////\ \ $
$\mathrel{\mathscr{L}}_{v}$ $v$
$\widetilde{\mathrel{\mathscr{R}}}_{v}\setminus\mathrel{\mathscr{R}}_{v}$ $t$
$\ \ ////\ \ $
$\widetilde{\mathrel{\mathscr{L}}}_{v}\setminus\mathrel{\mathscr{L}}_{v}$ $y$
$I^{l}_{v}$ $0,\ y,\ z,\ u,\ w$ $\ \ ////\ \ $ $I^{r}_{v}$ $0,\ z,\ u,\ t,\ w$

$\widetilde{\theta}_{S_{6}}(X_{S_{6}})=x_{y}x_{z}x_{y}(x_{t}x_{z}(x_{u}-x_{t})-(x_{u}-x_{t})x_{z}x_{u})$.

######  Example 5.8.

Let $S_{7}$ be the semigroup defined as follows.

$\displaystyle S_{7}=\langle 0,s,t,e,f,g,h,1\mid\ $ $\displaystyle
se=s,ft=t,e^{2}=e,f^{2}=f,g^{2}=g,h^{2}=h,$ $\displaystyle
ef=fe,gh=hg=0,ge=eg=g,gf=fg=g,$ $\displaystyle he=eh=h,hf=fh=h,sgt=sht=st,$
$\displaystyle 0\ \text{is the zero of}\ S_{7}\ \text{and}\ 1\ \text{is the
identity of}\ S_{7}\rangle,$

$\theta_{S_{7}}(X_{S_{7}})=0$.
We have $s^{{+}_{r}}=e$, $t^{{+}_{l}}=f$, $E^{st}=\\{g,h,ef\\}$ and $g,h$ are
minimal in $E^{st}$. Also, we have
$(st)^{{+}_{r}}=t^{{+}_{r}}=(et)^{{+}_{r}}=(ht)^{{+}_{r}}=(gt)^{{+}_{r}}=1$
and
$(st)^{{+}_{l}}=s^{{+}_{l}}=(sf)^{{+}_{l}}=(sh)^{{+}_{l}}=(sg)^{{+}_{l}}=1$.

######  Example 5.9.

Let $S_{8}$ be the semigroup with the following Cayley table.

$S_{8}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | $z$ | $z$ | $z$ | $y$ | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | $z$ | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | $z$ | . | $y$ | $y$ | $y$ | $u$ | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | $z$ | . | $y$ | $y$ | $y$ | $u$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | $z$ | . | $y$ | $y$ | $y$ | $w$ | $w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | $z$ | $u$ | $t$ | $u$ | $v$ | $v$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$ | $y$ | $z$ | $u$ | $t$ | $w$ | $v$ | $q$

$\theta_{S_{8}}(X_{S_{8}})=0$.
We have $(u\ast t)\ast w\neq u\ast(t\ast w)$, since $u\ast t=0$ and
$u\ast(t\ast w)=utw$.

######  Example 5.10.

Let $S_{9}$ be the semigroup with the following Cayley table.

$S_{9}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$ | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | . | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | $y$ | $y$ | . | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | . | $z$ | . | $z$ | $t$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | . | . | $z$ | $y$ | $y$ | $t$ | $w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | . | $u$ | . | $u$ | $v$ | $v$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$ | $y$ | $z$ | $u$ | $t$ | $w$ | $v$ | $q$

The Cayley table of the semigroup $Z(S_{9})$ with the multiplication
$\boldsymbol{*}$ is as follows:

$(Z(S_{9}),\boldsymbol{*})$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u+t+w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v+q}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$ | $y$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | . | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | $y$ | $y$ | . | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | . | $z$ | . | $z$ | $t$ | $t$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u+t+w}}$ | . | . | $z$ | $y$ | $y$ | $t$ | $u+t+w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | . | $u$ | . | $u$ | $v$ | $v$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v+q}}$ | $y$ | $z$ | $u$ | $t$ | $u+t+w$ | $v$ | $v+q$

By Theorem 4.6, the Cayley table on the left side can be used for $Z(S_{9})$.

| $(Z(S_{9}),\boldsymbol{*})$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | . | . | . | $y$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | . | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | $y$ | . | . | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | . | $z$ | . | . | $t$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | . | . | . | . | $-z$ | . | $w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | . | $u$ | . | . | $v$ | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$ | . | $z$ | . | $t$ | $w$ | . | $q$
| $P$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | $y$ | . | . | . | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | $z$ | $t$ | . | . | . | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | $u$ | $v$ | . | . | . | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | . | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | . | $y$ | . | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w}}$ | . | . | . | . | . | $-z$ | $w$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q}}$ | . | . | . | $z$ | $t$ | $w$ | $q$

The Cayley table $P$ is on the right side that by rearranging and shifting the
rows and columns of $Z(S_{9})$ so that the elements of the subset
$\widetilde{\mathrel{\mathscr{L}}}_{e}$ being adjacent rows and the elements
of the subset $\widetilde{\mathrel{\mathscr{R}}}_{e}$ being adjacent columns
for every idempotent $e\in E(S_{9})$. Then, it easily may recognize that the
determinant of the matrix $P$ is non-zero and equal to
$(x_{y}x_{z}x_{y})(x_{z}x_{y}x_{z}x_{z})$. It follows that the determinant of
$\widetilde{\theta}_{S_{9}}(X_{S_{9}})$ is non-zero. Additionally, finding the
determinant $\widetilde{\theta}_{e}(Y_{e})\neq 0$ from the given Cayley table
results in $\widetilde{\theta}_{S_{9}}(X_{S_{9}})$ being not equal to 0.

$Z^{-1}(P)$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w-u-t}}$ | ${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q-v}}$
---|---|---|---|---|---|---|---
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{y}}$ | . | . | $y$ | . | . | . | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{t}}$ | . | $z$ | $t$ | . | . | . | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{v}}$ | $y$ | $u$ | $v$ | . | . | . | .
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{z}}$ | . | . | . | . | . | . | $z$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{u}}$ | . | . | . | . | $y$ | . | $u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{w-u-t}}$ | . | . | . | . | . | $-z$ | $w-t-u$
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{q-v}}$ | . | . | . | $z$ | $t$ | $w-u-t$ | $q-v$

## References

  * [1] J. Almeida.  Finite semigroups and universal algebra, volume 3 of Series in Algebra.  World Scientific Publishing Co., Inc., River Edge, NJ, 1994.  Translated from the 1992 Portuguese original and revised by the author.
  * [2] I. Assem, D. Simson, and A. Skowroński.  Elements of the representation theory of associative algebras. Vol. 1, volume 65 of London Mathematical Society Student Texts.  Cambridge University Press, Cambridge, 2006.  Techniques of representation theory.
  * [3] D.J. Benson.  Representations and cohomology. I, volume 30 of Cambridge Studies in Advanced Mathematics.  Cambridge University Press, Cambridge, second edition, 1998.  Basic representation theory of finite groups and associative algebras.
  * [4] A.H. Clifford and G.B. Preston.  The algebraic theory of semigroups. Vol. I.  Mathematical Surveys, No. 7. American Mathematical Society, Providence, R.I., 1961.
  * [5] A. Distler and J. Mitchell.  Smallsemi, a library of small semigroups, Version 0.6.13.  https://gap-packages.github.io/smallsemi/, Feb 2022.  GAP package.
  * [6] J. Fountain, G.M.S. Gomes, and V. Gould.  Enlargements, semiabundancy and unipotent monoids.  Comm. Algebra, 27(2):595–614, 1999.
  * [7] F.G. Frobenius.  Theorie der hyperkomplexen größen.  Sitzungsber. Press. Akad. Wissen., Leipzig, 24:504–537, 1903.
  * [8] The GAP Group.  GAP – Groups, Algorithms, and Programming, Version 4.12.2, 2022\.
  * [9] J.A. Green.  On the structure of semigroups.  Ann. of Math. (2), 54:163–172, 1951.
  * [10] B. Lindström.  Determinants on semilattices.  Proc. Amer. Math. Soc., 20:207–208, 1969.
  * [11] J. Okniński.  Semigroup algebras, volume 138 of Monographs and Textbooks in Pure and Applied Mathematics.  Marcel Dekker, Inc., New York, 1991.
  * [12] I.S. Ponizovskiĭ.  The Frobeniusness of the semigroup algebra of a finite commutative semigroup.  Izv. Akad. Nauk SSSR Ser. Mat., 32:820–836, 1968.
  * [13] J. Rhodes and B. Steinberg.  The $q$-theory of finite semigroups.  Springer Monographs in Mathematics. Springer, New York, 2009.
  * [14] H.J.S. Smith.  On the Value of a Certain Arithmetical Determinant.  Proc. Lond. Math. Soc., 7:208–212, 1875/76.
  * [15] B. Steinberg.  Factoring the Dedekind-Frobenius determinant of a semigroup.  J. Algebra, 605:1–36, 2022.
  * [16] R. Wenger.  Some semigroups having quasi-Frobenius algebras. II.  Canadian J. Math., 21:615–624, 1969.
  * [17] H.S. Wilf.  Hadamard determinants, Möbius functions, and the chromatic number of a graph.  Bull. Amer. Math. Soc., 74:960–964, 1968.
  * [18] J.A. Wood.  Factoring the semigroup determinant of a finite commutative chain ring.  In Coding theory, cryptography and related areas (Guanajuato, 1998), pages 249–259. Springer, Berlin, 2000.