October 30, 2024
An idempotent element of a groupoid is an element \(e\) such that \(e^2 = e\), and a band is an idempotent semigroup, a semigroup in which every element is an idempotent. Every commutative band is a semilattice by taking \(\wedge\) or \(\vee\) to be the operation, and as a semilattice, it is uniquely determined by the sets of all elements less than or equal to each of its elements or analogously the sets of all elements greater than or equal to each of its elements. We generalize these notions.
First we establish a definition for a groupoid or semigroup stabilized with respect to a collection of relations. Then we define the left identity relation, \(\mathcal{LI}\). For any \(a\) and \(b\) in a groupoid \((S,\star)\), \(a \, \mathcal{LI} \, b\) if and only if \(b \star a = a\), meaning \(b\) acts as a left identity of \(a\). There is a similar definition for the right identity relation, \(\mathcal{RI}\). We defined the left zero relation, \(\mathcal{LZ}\), as \(a \, \mathcal{LZ} \, b\) if and only if \(b \star a = b\), meaning \(b\) acts as a left zero of \(a\). There is a similar definition for the right zero relation, \(\mathcal{RZ}\).
For groupoid \((S,\star)\), we label the left identity set of \(a \in S\) as \(\mathcal{LI}(a)=\{b \in S|b \star a = a\}\), and give similar definitions for the right identity set, left zero set, and right zero set of \(a\). These can then be expanded to any subset \(A \subset S\) by taking the intersections of the respective one-sided identity or one-sided zero sets of the elements of \(A\). These sets can be used to identify idempotents, identities, and zeroes, and they can be utilized to determine maximal subgroups, maximal right and left zero subsemigroups, maximal right and left subgroups, and rectangular band subsemigroups.
In this paper, we then identify classes of groupoids and semigroups that are uniquely determined by the one-sided identity and zero sets of their elements, which is called being stabilized with respect to the relations of \(\mathcal{LI}\) and \(\mathcal{RI}\) (or equivalently \(\mathcal{LZ}\) and \(\mathcal{RZ}\)). A right group whose maximal subgroup contains at most \(2\) elements is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\), meaning uniquely determined the one-sided identity sets of its elements (or equivalently \(\mathcal{LZ}\) and \(\mathcal{RZ}\), meaning the one-sided zero sets of its elements). We define a commutative-rectangular band to be a band in which every pair of elements either commute or are generalized inverses of each other. A commutative-rectangular band is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)].
See [1] and [2] for references on fundamental semigroup theory including relations, maximal subgroups, right [left] zero semigroups, rectangular bands, and right groups.
For the purposes of this paper, let a binary relation \(\rho\) be defined for each groupoid [semigroup, monoid, group, or other class of groupoids] \((S,\star)\) such that \(\rho\) is invariant under isomorphisms, i.e. if \((S,\star)\) is isomorphic to \((S',\star')\) by the isomorphism \(f\), then for any \(a,b \in S\), \(a \, \rho(S,\star) \, b\) if and only if \(f(a) \, \rho(S',\star') \, f(b)\).
For example, Green’s relations \(\mathscr{L}\), \(\mathscr{R}\), \(\mathscr{D}\), \(\mathscr{H}\), and \(\mathscr{J}\) are binary relations defined for every semigroup and satisfying this property that each relation is invariant under isomorphisms between semigroups. See [1] and [2] for references and Green’s seminal paper [3].
Definition 1. Let \(\rho_i\), for \(i\) belonging to an index set \(I\), be a binary relation for groupoids and invariant under isomorphisms, and let \((S,\star)\) be a groupoid. Then \((S,\star)\) is a stabilized groupoid with respect to \(\{\rho_i|i \in I\}\) if and only if when groupoid \((S,\star')\) satisfies \(a \, \rho_i(S,\star) \, b\) if and only if \(a \, \rho_i(S,\star') \, b\) for any \(a,b \in S\) and any \(i \in I\), then \(\star=\star'\).
Similarly, we define a stabilized semigroup and a stabilized member of any other class of groupoids.
Remark: In this definition, \(\star=\star'\) is equivalent to \(I:(S,\star) \to (S,\star')\) such that \(I(a)=a\) for any \(a \in S\) is an isomorphism of groupoids.
For example, the trivial group, \(\{e\}\) with \(e \star e = e\), is stabilized with respect to any relation.
Since any binary relation \(\rho\) on a set \(S\) can be identified with \(\rho(a)=\{b \in S|a \, \rho \, b\}\) for all \(a \in S\), we have the following alternate definition.
Definition 2. Let \(\rho_i\), for \(i\) belonging to an index set \(I\), be a binary relation for groupoids invariant under isomorphisms, and let \((S,\star)\) be a groupoid. Then \((S,\star)\) is a stabilized groupoid with respect to \(\{\rho_i|i \in I\}\) if and only if when groupoid \((S,\star')\) satisfies \(\rho(S,\star)(a)=\rho(S,\star')(a)\) for all \(a \in S\) and all \(i \in I\), then \(\star=\star'\).
To simplify notation in later sections, for groupoid \(S\) with \(a \in S\), we use \(\rho(S,\star)(a)=\rho(a)\) and \(\rho(S,\star')(a)=\rho'(a)\).
If \(\rho\) is an equivalence relation, then \((S,\star)\) is a stabilized groupoid with respect to \(\rho\) if and only if when groupoid \((S,\star')\) satisfies the equivalence class of \(a\) in \((S,\star)\) is equal to the equivalence class of \(a\) in \((S,\star')\), then \(\star=\star'\).
Lemma 1. Let \(\rho_i\), for \(i\) belonging to an index set \(I\), be a binary relation for groupoids invariant under isomorphisms, and let \((S,\star)\) be a groupoid. Then \((S,\star)\) is a stabilized groupoid with respect to \(\{\rho_i|i \in I\}\) if and only if when \((S',\star')\) is a groupoid such that there exists a bijective function \(f:S \to S'\) where \(f(\rho(S,\star)(a))=\rho(S',\star')(f(a))\) for all \(a \in S\) and all \(i \in I\), then \(f\) is an isomorphism.
Therefore the definition of stabilized is consistent for isomorphic groupoids [semigroups, etc.].
Lemma 2. Let \(\rho_i\), for \(i\) belonging to an index set \(I\), be a binary relation for groupoids invariant under isomorphisms, and let \((S,\star)\) and \((S',\star')\) be isomorphic groupoids. Then \((S,\star)\) is stabilized with respect to \(\{\rho_i|i \in I\}\) if and only if \((S',\star')\) is stabilized with respect to \(\{\rho_i|i \in I\}\).
For any \(a,b \in S\), \(b \in \rho(S,\star)(a)\) if and only if \(a \in \rho^{-1}(S,\star)(b)\). Therefore \(\rho(S,\star)(a)=\rho(S,\star')(a)\) for all \(a \in S\) if and only if \(\rho^{-1}(S,\star)(a)=\rho^{-1}(S,\star')(a)\) for all \(a \in S\).
Lemma 3. Let \(\rho_i\), for \(i\) belonging to an index set \(I\), be a binary relation for groupoids invariant under isomorphisms, and let \((S,\star)\) be a groupoid. Then \((S,\star)\) is a stabilized groupoid with respect to \(\{\rho_i|i \in I\}\) if and only if \((S,\star)\) is a stabilized groupoid with respect to \(\{\rho_i^{p_i}|i \in I\}\) and \(p_i=\pm 1\).
Since every semigroup is a groupoid, we have the following result.
Lemma 4. Let \(\rho_i\), for \(i\) belonging to an index set \(I\), be a binary relation for groupoids invariant under isomorphisms, and let \((S,\star)\) be a semigroup. If \((S,\star)\) is a stabilized groupoid with respect to \(\{\rho_i|i \in I\}\), then \((S,\star)\) is a stabilized semigroup with respect to \(\{\rho_i|i \in I\}\).
Similarly, if \((S,\star)\) is a groupoid satisfying some conditions and \((S,\star)\) is a stabilized groupoid satisfying some subset of those conditions with respect to \(\{\rho_i|i \in I\}\), then \((S,\star)\) is a stabilized groupoid satisfying the larger set of of conditions with respect to \(\{\rho_i|i \in I\}\).
Recall an element \(a\) of a groupoid \(S\) is an idempotent if and only if \(aa=a\) in \(S\), and a band is an idempotent semigroup, a semigroup in which every element is an idempotent. Any commutative band, \(S\), is a semilattice with \(a \leq b\) if and only if \(ab=ba=a\) for any \(a,b \in S\). Therefore each commutative band is uniquely determined by the partial ordering \(\leq\), which is a binary relation, and equivalently by the sets \(\{b \in S|ab=ba=a\}\) for all \(a \in S\) or \(\{a \in S|ab=ba=a\}\) for all \(b \in S\). A commutative band is a stabilized commutative band with respect to \(\{b \in S|ab=ba=a\}\), and similarly a commutative band is a stabilized commutative band with respect to \(\{a \in S|ab=ba=a\}\). Now we expand the notion of these sets to general groupoids. In doing so, we distinguish between left and right to account for noncommutative cases, and we define one-sided identity and zero sets. We also include some fundamental theory and observations related to one-sided identity and zero sets.
Definition 3. Let \(S\) be a set with a binary operation. For any \(a,b \in S\),
The left identity relation in \(S\) denoted by \(\mathcal{LI}\) is defined as
\(a \, \mathcal{LI} \, b\) if and only if \(ba=a\).
The left zero relation in \(S\) denoted by \(\mathcal{LZ}\) is defined as
\(a \, \mathcal{LZ} \, b\) if and only if \(ba=b\).
Similarly, we define the right identity relation in \(S\), \(\mathcal{RI}\), which is the inverse relation of \(\mathcal{LZ}\), and the right zero relation in \(S\), \(\mathcal{RZ}\), which is the inverse relation of \(\mathcal{LI}\). These relations exist for every nonempty set with a binary operation and are invariant under isomorphisms.
Note that when \(S\) is a band, for any \(a,b \in S\), \(Sa \subset Sb\) if and only if \(a=ab\), and \(aS \subset bS\) if and only if \(a=ba\). Thus when \(S\) is a band, the Green’s relations \(\leq_{\mathscr{L}}\) and \(\leq_{\mathscr{R}}\) coincide with \(\mathcal{RI}\) and \(\mathcal{LI}\), respectively. See [3]. These differ from the Rees order, which is a partial order on all idempotents of a semigroup, such that \(e < f\) if and only if \(ef=fe=e\). See [4].
For any \(a\) belonging to groupoid \(S\), \(\mathcal{LI}(a)=\{b \in S|ba=a\}\) and similarly for the other one-sided identity and zero relations. Now we expand the definition of such sets in such a way that coincides with these relations.
Definition 4. Let \(S\) be a set with a binary operation. For any nonempty \(A \subset S\),
The left identity set of \(A\) in \(S\) denoted by \(\mathcal{LI}(A)\) is defined as \[\mathcal{LI}(A) = \{b \in S|ba = a, \, \text{for every a \in A}\}.\]
The left zero set of \(A\) in \(S\) denoted by \(\mathcal{LZ}(A)\) is defined as \[\mathcal{LZ}(A) = \{b \in S|ba=b, \, \text{for every a \in A}\}.\]
Similarly, we define the right identity set of \(A\) in \(S\) and the right zero set of \(A\) in \(S\). In particular, if \(A=\{a\}\), then \(\mathcal{LI}(\{a\})=\mathcal{LI}(a)\), and for any nonempty \(A \subset S\), \(\mathcal{LI}(A)\) is the intersection of \(\mathcal{LI}(a)\) for any \(a \in A\). If \(A\) is the empty set, then we define the one-sided identity and zero sets of \(\emptyset\) in \(S\) to be equal to the empty set.
Remark: From the definition, it follows
\(\mathcal{LI}^{-1}=\mathcal{RZ}\).
\(\mathcal{RI}^{-1}=\mathcal{LZ}\).
For any \(a,b \in S\),
\(b\) belongs to \(\mathcal{LI}(a)\) if and only if \(a\) belongs to \(\mathcal{RZ}(b)\).
\(b\) belongs to \(\mathcal{RI}(a)\) if and only if \(a\) belongs to \(\mathcal{LZ}(b).\)
For any nonempty \(A,B \subset S\),
\(B\) is contained in \(\mathcal{LI}(A)\) if and only if \(A\) is contained in \(\mathcal{RZ}(B)\).
\(B\) is contained in \(\mathcal{RI}(A)\) if and only if \(A\) is contained in \(\mathcal{LZ}(B)\).
Lemma 5. Let \((S,\star)\) be a groupoid. Then \((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{LI}\) [\(\mathcal{(}RI)\)] if and only if \((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{RZ}\) [\(\mathcal{LZ}\)], and the following are equivalent:
\((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\).
\((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{RZ}\) and \(\mathcal{LZ}\).
\((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{LI}\) and \(\mathcal{LZ}\).
\((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{RZ}\) and \(\mathcal{LZ}\).
In the above lemma, groupoid could also be replaced by semigroup. Let \(S\) be a set with a binary operation. Recall \(S\) is a left [right] zero semigroup if and only if for any \(a,b \in S\), \(ab=a\) [\(ab=b\)].
Lemma 6. Let \(S\) be a set with a binary operation. Let \(e, z, a, b \in S\). Then
\(e\) is an idempotent if and only if \(e\) belongs to \(\mathcal{LI}(e)\) [\(\mathcal{RI}(e)\), \(\mathcal{LZ}(e)\), \(\mathcal{RZ}(e)\)].
\(e\) is a left [right] identity of \(S\) if and only if \(e\) belongs to \(\mathcal{LI}(S)\) [\(\mathcal{RI}(S)\)].
\(e\) is a left [right] identity of \(S\) if and only if \(\mathcal{RZ}(e) = S\) [\(\mathcal{LZ}(e) = S\)].
\(z\) is a left [right] zero of \(S\) if and only if \(z\) belongs to \(\mathcal{LZ}(S)\) [\(\mathcal{RZ}(S)\)].
\(z\) is a left [right] zero of \(S\) if and only if \(\mathcal{RI}(z)=S\) [\(\mathcal{LI}(z)=S\)].
\(\mathcal{RZ}(S)=S\) [\(\mathcal{LI}(S)=S\)] if and only if \(S\) is a right zero semigroup.
\(\mathcal{LZ}(S)=S\) [\(\mathcal{RI}(S)=S\)] if and only if \(S\) is a left zero semigroup.
\(\mathcal{LI}(e) \cap \mathcal{LZ}(e) \neq \emptyset\) [\(\mathcal{RI}(e) \cap \mathcal{RZ}(e) \neq \emptyset\)] if and only if \(e\) is an idempotent, in which case the intersection equals \(\{e\}\).
Let \(S\) be a semigroup. Let \(A \subset S\). Then
If it is nonempty, \(\mathcal{LI}(A)\) [\(\mathcal{RI}(A)\)] is a subsemigroup of \(S\).
If it is nonempty, \(\mathcal{LZ}(A)\) [\(\mathcal{RZ}(A)\)] is a left [right] ideal of \(S\).
\(z\) is a left [right] zero of \(S\) if and only if \(\mathcal{RZ}(z)=\{z\}\) [\(\mathcal{LZ}(z)=\{z\}\)].
If \(a \in \mathcal{LI}(b)\), then \(\mathcal{LI}(a) \subset \mathcal{LI}(b)\) and similarly for \(\mathcal{RI}\), \(\mathcal{LZ}\), and \(\mathcal{RZ}\).
If \(a \in \mathcal{LI}(b)\), then \(\mathcal{RZ}(b) \subset \mathcal{RZ}(a)\) and similarly for \(\mathcal{RI}\), \(\mathcal{LZ}\), \(\mathcal{RZ}\) with their inverses.
Let \(S\) and \(T\) be groupoids with \(A \subset S\) and \(B \subset T\). Then
Lemma 7. Let \((S,\star)\) be a monoid. Then \((S,\star)\) is a stabilized monoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) if and only if \((S,\star)\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\).
Proof. Let \((S,\star)\) be a stabilized monoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). Let \((S,\star')\) be a semigroup such that \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for any \(x \in S\).
Let \(e\) be the identity in \(S\) with respect to \(\star\). As the identity, \(e \in \mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(e \in \mathcal{RI}(x)=\mathcal{RI}'(x)\) for all \(x \in S\). Therefore \(e\) is the identity in \((S,\star')\), and \((S,\star')\) is a monoid.
Since \((S,\star)\) is a stabilized monoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\), then \(\star = \star'\), and \((S,\star)\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\).
Since every monoid is a semigroup, if \((S,\star)\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\), then it is automatically a stabilized monoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). ◻
Similarly, a semigroup with zero is a stabilized semigroup with zero with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) if and only if it is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\).
Every one-sided identity or zero is an idempotent. Therefore every subsemigroup containing a one-sided identity or zero contains an idempotent. For semigroup \(S\), if \(e \in S\) is an idempotent, then the trivial group \(\{e\}\) is a subsemigroup of \(S\), which contains \(e\) as an identity and zero. Thus the existence of such subsemigroups coincides with idempotents. In particular, for a semigroup \(S\) with idempotent \(e\), there exist maximum subsemigroups of \(S\) containing \(e\) as a one-sided identity or one-sided zero.
\(\mathcal{LI}(e)\) [\(\mathcal{RI}(e)\)] is the maximum subsemigroup of \(S\) containing \(e\) as a right [left] zero, and \(\mathcal{LI}(e) \cap \mathcal{RI}(e)\) is maximum subsemigroup of \(S\) containing \(e\) as a zero.
\(\mathcal{LZ}(e)\) [\(\mathcal{RZ}(e)\)] is the maximum subsemigroup of \(S\) containing \(e\) as a right [left] identity with \(\mathcal{LZ}(e)=Se\) [\(\mathcal{RZ}(e)=eS\)], and \(\mathcal{LZ}(e) \cap \mathcal{RZ}(e)\) is the maximum subsemigroup of \(S\) containing \(e\) as the identity and equals \(eSe\).
A rectangular band is a semigroup \(S\) in which \(aba=a\) for any \(a,b \in S\). Also a right group is a left cancellative and right simple semigroup. A left group is a right cancellative and left simple semigroup. For semigroup \(S\), we refer to a right group [left group] subsemigroup of \(S\) as a right [left] subgroup of \(S\).
If a semigroup contains a subgroup, then it contains the subgroup’s identity as an idempotent. If a semigroup contains a left zero [right zero, rectangular band] subsemigroup, then it contains an idempotent because these subsemigroups are bands. Also, if a semigroup contains a right [left] subgroup of \(S\), then it contains an idempotent using the definition as a left [right] cancellative and right [left] simple semigroup. For semigroup \(S\), If \(e \in S\) is an idempotent, then the trivial group \(\{e\}\) is a subsemigroup of \(S\), which is also a group, left zero semigroup, right zero semigroup, rectangular band, right subgroup, and left subgroup. Thus the existence of such subsemigroups coincides with idempotents.
Let \(S\) be a semigroup. For every idempotent \(e \in S\), there exists a maximum subgroup of \(S\) containing \(e\): the subset of \(eSe\), the maximum subsemigroup of \(S\) containing \(e\) as its identity, of units with respect to \(e\). We let \({\boldsymbol{H}}(e)\) be this maximum subgroup of \(S\) containing idempotent \(e\). Any subgroup of \(S\) is a subgroup of the maximum subgroup containing its idempotent identity. For distinct idempotents \(e\) and \(f\), \({\boldsymbol{H}}(e) \cap {\boldsymbol{H}}(f) = \emptyset\). Let \(E\) be the set of idempotents in \(S\), \(\{{\boldsymbol{H}}(e)|e \in E\}\) is the set of all maximal subgroups of \(S\). These are also the \(\mathscr{H}\)-classes of \(S\) containing idempotents.
In the next three sections, we use the notions of identity and zero sets of idempotents to identify maximal right [left] zero subsemigroups, to identify maximal right [left] subgroups , and to identify all rectangular band subsemigroups and their decompositions in terms of identity and zero sets. In the rest of the sections of this paper, we also apply the concept of stabilized semigroups with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)] in the context of the semigroup classes of right [left] zero semigroups, right [left] groups, and rectangular bands. Then we continue by applying this notion to groupoids. Definitions for a groupoid class labeled left-right zero groupoids and for a semigroup class labeled commutative-rectangular bands then arise to satisfy stability with respect to one-sided identity and zero sets.
Lemma 8. Let \(S\) be a semigroup. Then \(\mathcal{LI}(S) \cap \mathcal{RZ}(S) \neq \emptyset\) if and only if \(S\) is a right zero semigroup.
Similarly, \(\mathcal{RI}(S) \cap \mathcal{LZ}(S) \neq \emptyset\) if and only if \(S\) is a left zero semigroup.
Proof. Given there exists \(e \in S\) such that \(ex=x\) and \(xe=e\) for any \(x \in S\), then for any \(a,b \in S\), \(ab=a(eb)=(ae)b=eb=b\), and \(S\) is a right zero semigroup. If \(S\) is a right zero semigroup, then \(\mathcal{LI}(S) \cap \mathcal{RZ}(S) = S\). ◻
Let \({\boldsymbol{R}Z}(e)=\mathcal{LI}(e) \cap \mathcal{RZ}(e)\) and \({\boldsymbol{L}Z}(e)=\mathcal{RI}(e) \cap \mathcal{LZ}(e)\).
Proposition 1. Let \(S\) be a semigroup. Let \(e \in S\). Then \({\boldsymbol{R}Z}(e) \neq \emptyset\) if and only if \(e\) is an idempotent. In this case, \({\boldsymbol{R}Z}(e)\) is the maximum right zero subsemigroup of \(S\) containing \(e\).
Similarly, \({\boldsymbol{L}Z}(e) \neq \emptyset\) if and only if \(e\) is an idempotent. In this case, \({\boldsymbol{L}Z}(e)\) is the maximum left zero subsemigroup of \(S\) containing \(e\).
Proof. If \(a \in S\) such that \(ae=e\) and \(ea=a\), then \(eae=(ea)e=ae=e\) and \(eae=e(ae)=ee\). Therefore \(ee=e\), and \(e\) is an idempotent.
If \(e\) is an idempotent, then \(ee=e\) and \(e \in \mathcal{LI}(e) \cap \mathcal{RZ}(e)={\boldsymbol{R}Z}(e)\), so this intersection is nonempty.
In this case, for any \(a,b \in {\boldsymbol{R}Z}(e)\), \(ab = a(eb) = (ae)b = eb = b.\) Thus the intersection is a right zero semigroup.
If \(R\) is a right zero subsemigroup of \(S\) containing \(e\), then for any \(a \in R\), \(ae=e\) and \(ea=a\), so that \(a \in {\boldsymbol{R}Z}(e)\). Thus \(R \subset {\boldsymbol{R}Z}(e)\), and \({\boldsymbol{R}Z}(e)\) is maximal. ◻
Proposition 2. Let \(S\) be a semigroup, let \(e,f \in S\), and let \(e\) be an idempotent. Then the following are equivalent:
\({\boldsymbol{R}Z}(e) \cap {\boldsymbol{R}Z}(f) \neq \emptyset\).
\(ef=f\) and \(fe=e\).
\({\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\).
In each of these cases, it is implied that \(f\) is also an idempotent by Proposition 1. There is a similar statement for \({\boldsymbol{L}Z}(e)\) and \({\boldsymbol{L}Z}(f)\).
Proof. \((1) \implies (2)\): Let \(a \in {\boldsymbol{R}Z}(e) \cap {\boldsymbol{R}Z}(f)\). Then \(ef = e(af) = (ea)f = af = f\) and \(fe = f(ae) = (fa)e = ae =e\).
\((2) \implies (3)\): Since \(e \in {\boldsymbol{R}Z}(f)\), \(f\) is an idempotent by Proposition 1. For any \(a \in {\boldsymbol{R}Z}(e)\), \(af = a(ef) = (ae)f = ef = f\) and \(fa = f(ea) = (fe)a = ea = a\). Therefore \({\boldsymbol{R}Z}(e) \subset {\boldsymbol{R}Z}(f)\). Similarly, \({\boldsymbol{R}Z}(f) \subset {\boldsymbol{R}Z}(e)\).
\((3) \implies (1)\): Since \(e\) is an idempotent, \(e \in {\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f) \neq \emptyset\), which implies \({\boldsymbol{R}Z}(e) \cap {\boldsymbol{R}Z}(f) = {\boldsymbol{R}Z}(e) \neq \emptyset\). ◻
Lemma 9. Let \(S\) be a semigroup with idempotents \(e\) and \(f\). \(e \, \mathscr{R} \, f\) if and only if \({\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\).
By Proposition 2, \(e \, \mathscr{R} \, f\) is equivalent to \({\boldsymbol{R}Z}(e) \cap {\boldsymbol{R}Z}(f) \neq \emptyset\). Similarly, for idempotents \(e\) and \(f\), \(e \, \mathscr{L} \, f\) if and only if \({\boldsymbol{L}Z}(e)={\boldsymbol{L}Z}(f)\), and this is equivalent to \({\boldsymbol{L}Z}(e) \cap {\boldsymbol{L}Z}(f) \neq \emptyset\) by the dual of Proposition 2.
Proof. If \(e \, \mathscr{R} \, f\), then \(eS^1 = fS^1\). Since \(e\) and \(f\) are idempotents, \(eS^1=eS\) and \(fS^1=fS\). Thus \(ex=f\) for some \(x \in S\), and \(e=fy\) for some \(y \in S\). Then \(eex=ef\), or \(ex=ef\) because \(e\) is an idempotent, and \(f=ef\). Similarly, \(fe=e\). By Proposition 2, \({\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\).
If \({\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\), then by Proposition 2, \(ef=f\) and \(fe=e\). Thus \(efS=fS\) and \(feS=eS\), so \(fS \subset eS\) and \(eS \subset fS\). Therefore \(eS=fS\), which implies \(eS^1=fS^1\) because \(e\) and \(f\) are idempotents, and \(e \, \mathscr{R} \, f\). ◻
Corollary 1. Let \(S\) be a band with element \(e\). \(R_e = {\boldsymbol{R}Z}(e)\) and \(L_e = {\boldsymbol{L}Z}(e)\).
For groupoids \((S,\star)\) and \((S,\star')\) and any \(a \in S\), we denote \(\mathcal{LI}(a)\) as the left identity set of \(a\) in \((S,\star)\) and \(\mathcal{LI}'(a)\) as the left identity set of \(a\) in \((S,\star')\).
Proposition 3. Any right zero semigroup is a stabilized groupoid with respect to \(\mathcal{LI}\) [\(\mathcal{RZ}\)] and a stabilized semigroup with respect to \(\mathcal{LZ}\) [\(\mathcal{RI}\)].
Similarly any left zero semigroup is a stabilized groupoid with respect to \(\mathcal{RI}\) [\(\mathcal{LZ}\)] and a stabilized semigroup with respect to \(\mathcal{RZ}\) [\(\mathcal{LI}\)].
Proof. Let \((S,\star)\) be a right zero semigroup.
Let \((S,\star')\) be a groupoid such that \(\mathcal{LI}(a)=\mathcal{LI}'(a)\) for any \(a \in S\). Since \((S,\star)\) is a right zero semigroup, for any \(a,b \in S\), \(b \star a=a\) and \(\mathcal{LI}(a)=S\). Therefore \(\mathcal{LI}'(a)=S\), and \((S,\star')\) is a right zero semigroup and \(\star=\star'\).
Let \((S,\star')\) be a semigroup such that \(\mathcal{LZ}(a)=\mathcal{LZ}'(a)\) for any \(a \in S\). Then \(\mathcal{LZ}(a)=\{a\}=\mathcal{LZ}'(a)\) for any \(a \in S\). Since \(a \in \mathcal{LZ}'(a)\), \(a\) is an idempotent in \((S,\star')\) and \(a \star' a = a\). For any \(b \in S\), \((b \star' a) \star' a = b \star' (a \star' a) = b \star' a \in \mathcal{LZ}(a)\). Therefore \(b \star' a = a\), and \((S,\star')\) is a right zero semigroup with \(\star=\star'\).
The rest of the cases follow similarly. ◻
If \(S\) is a right zero semigroup, then for any \(a,b \in S\), \(ab=b\) and \(\mathcal{LI}(b)=\{b\}\). We can then determine \(\star\) with one-sided identity sets.
Proposition 4. Let \(S\) be a right zero semigroup. Then for any \(a,b \in S\), \(\{ab\} = \mathcal{LI}(b)\).
A right group is left cancellative and right simple semigroup. Let \(S\) be a right group. For \(a \in S\), there exists \(b \in S\) such that \(ab=a\) because \(S\) is right simple. Then \(abb=ab\), which implies \(bb=b\) because \(S\) is left cancellative. Therefore \(S\) contains the idempotent \(b\). Similarly, a left group contains an idempotent.
The structure of a right group as a product of a maximal subgroup and a right zero subsemigroup is well known. Dually the structure of a left group is a product of a left zero subsemigroup and a maximal subgroup. We include a proof in the context of right [left] subgroups for completeness.
Lemma 10. Let \(S\) be a semigroup, and let \(e \in S\) be an idempotent. Then \({\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\), for some subgroup \({\boldsymbol{H}}'(e) \subset {\boldsymbol{H}}(e)\) and some \({\boldsymbol{R}Z}'(e) \subset {\boldsymbol{R}Z}(e)\) containing \(e\), is a right subgroup of \(S\) containing \(e\).
Similarly, \({\boldsymbol{L}Z}'(e){\boldsymbol{H}}'(e)\), for some subgroup \({\boldsymbol{H}}'(e) \subset {\boldsymbol{H}}(e)\) and some \({\boldsymbol{L}Z}'(e) \subset {\boldsymbol{L}Z}(e)\) containing \(e\), is a left subgroup of \(S\) containing \(e\).
Since \({\boldsymbol{H}}'(e)\) is a subgroup of \({\boldsymbol{H}}(e)\), \({\boldsymbol{H}}'(e)\) contains \(e\) as its identity. Since \(e \in {\boldsymbol{R}Z}'(e) \subset {\boldsymbol{R}Z}(e)\), \({\boldsymbol{R}Z}'(e)\) is a right zero subsemigroup of \(S\) containing \(e\).
Proof. \(e=ee \in {\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\).
For any \(x_1,x_2 \in {\boldsymbol{H}}'(e)\) and \(e_1,e_2 \in {\boldsymbol{R}Z}'(e)\), \[(x_1e_1)(x_2e_2) = x_1e_1(ex_2)e_2 = x_1(e_1e)x_2e_2 = x_1ex_2e_2 = x_1x_2e_2 \in {\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e).\] Therefore \({\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\) is a subsemigroup.
For any \(x_1,x_2,x_3 \in {\boldsymbol{H}}'(e)\) and \(e_1,e_2,e_3 \in {\boldsymbol{R}Z}'(e)\), if \((x_1e_1)(x_2e_2)=(x_1e_1)(x_3e_3)\), then \[\begin{align} x_1x_2e_2 &= x_1x_3e_3 \\ x_1^{-1}x_2e_2 &= x_1^{-1}x_1x_3e_3 &\text{x_1^{-1} \in {\boldsymbol{H}}'(e)} \\ ex_2e_2 &= ex_3e_3 &\text{x_1^{-1}x_1=e} \\ x_2e_2 &= x_3e_3 &\text{ex=x for any x \in {\boldsymbol{H}}'(e).} \end{align}\] Therefore \({\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\) is left cancellative.
For any \(x_1,x_2 \in {\boldsymbol{H}}'(e)\) and \(e_1,e_2 \in {\boldsymbol{R}Z}'(e)\), \[(x_1e_1)(x_1^{-1}x_2e_2)=x_1x_1^{-1}x_2e_2=ex_2e_2=x_2e_2.\] Therefore \({\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\) is right simple, and thus it is a right group. ◻
Proposition 5. Let \(S\) be a semigroup, and let \(e \in S\). Then \({\boldsymbol{R}G}(e) \neq \emptyset\) if and only if \(e\) is an idempotent. In this case, \({\boldsymbol{R}G}(e)={\boldsymbol{H}}(e){\boldsymbol{R}Z}(e)\) is the maximum right subgroup of \(S\) containing \(e\).
Similarly, \({\boldsymbol{L}G}(e) \neq \emptyset\) if and only if \(e\) is an idempotent, and \({\boldsymbol{L}G}(e)={\boldsymbol{L}Z}(e){\boldsymbol{H}}(e)\) is the maximum left subgroup of \(S\) containing \(e\).
Note that \({\boldsymbol{R}Z}(e){\boldsymbol{H}}(e)={\boldsymbol{H}}(e)\) because for any \(e_1 \in {\boldsymbol{R}Z}(e)\) and \(x \in {\boldsymbol{H}}(e)\), \[e_1x=e_1(ex)=(e_1e)x=ex=x.\] Similarly \({\boldsymbol{H}}(e){\boldsymbol{L}Z}(e)={\boldsymbol{H}}(e)\).
Proof. By Proposition 1, \({\boldsymbol{R}Z}(e) \neq \emptyset\) if and only if \(e\) is an idempotent. If \({\boldsymbol{R}Z}(e) = \emptyset\), then \(e\) is not an idempotent and \({\boldsymbol{R}G}(e) = \emptyset\), and if \({\boldsymbol{R}Z}(e) \neq \emptyset\), then \(e\) is an idempotent and \(e=ee \in {\boldsymbol{H}}(e){\boldsymbol{R}Z}(e)={\boldsymbol{R}G}(e)\).
By Lemma 10, \({\boldsymbol{R}G}(e)\) is a right subgroup of \(S\) containing \(e\).
Let \({\boldsymbol{R}G}'(e)\) be a right subgroup of \(S\) containing \(e\). For any \(x \in RG'(e)\), since \(ee=e\), \(eex=ex\), and \(ex=x\) because \(RG'(e)\) is left cancellative. Therefore \(e\) is a left identity of \(RG'(e)\).
Let \(a \in {\boldsymbol{R}G}'(e)\). Since \(a,e \in {\boldsymbol{R}G}'(e)\), \(ae \in {\boldsymbol{R}G}'(e)\), and there exists \(b \in {\boldsymbol{R}G}'(e)\) such that \((ae)b=a\) because \({\boldsymbol{R}G}'(e)\) is right simple. Then we get the following implications. \[\begin{align} a(eb) &= a \\ a(eb)e &= ae &\text{multiply by e} \\ (eb)e &= e &\text{because {\boldsymbol{R}G}'(e) is left cancellative} \\ ebe &= ee &\text{because ee=e} \\ be &= e &\text{because {\boldsymbol{R}G}'(e) is left cancellative.} \end{align}\] Since \(e\) is a left identity of \(RG'(e)\), \(eb=b\). Therefore \(b \in {\boldsymbol{R}Z}(e)\) because \(be=e\) and \(eb=b\).
Since \({\boldsymbol{R}G}'(e)\) is right simple, there exists \(c \in {\boldsymbol{R}G}'(e)\) such that \(ac=e\). Since \(e\) is a left identity for any element of \({\boldsymbol{R}G}'(e)\), \(e(ae)=(ea)e=ae \in eSe\) and \(e(ce)=(ec)e=ce \in eSe\). Also, \[\begin{align} ac &= e \\ aca &= ea &\text{multiply by a}\\ aca &= a &\text{ea=a} \\ acae &= ae &\text{multiply by e} \\ cae &= e &\text{because {\boldsymbol{R}G}'(e) is left cancellative.} \end{align}\] Therefore \[(ae)(ce)=a(ec)e = ace = ee = e,\] and \[(ce)(ae)=c(ea)e=cae=e,\] so \(ae \in {\boldsymbol{H}}(e)\).
Thus \(a=(ae)b \in {\boldsymbol{H}}(e){\boldsymbol{R}Z}(e)={\boldsymbol{R}G}(e)\). This means that \({\boldsymbol{R}G}(e)\) is the maximum right subgroup of \(S\) containing \(e\). ◻
Lemma 11. Let \(S\) be a semigroup, and let \(e \in S\) be an idempotent. Let \({\boldsymbol{H}}'(e)\) and \({\boldsymbol{H}}''(e)\) be subgroups of \({\boldsymbol{H}}(e)\), and let \({\boldsymbol{R}Z}'(e)\) and \({\boldsymbol{R}Z}''(e)\) be subsets of \({\boldsymbol{R}Z}(e)\) containing \(e\). If \({\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)={\boldsymbol{H}}''(e){\boldsymbol{R}Z}''(e)\), then \({\boldsymbol{H}}'(e)={\boldsymbol{H}}''(e)\) and \({\boldsymbol{R}Z}'(e)={\boldsymbol{R}Z}''(e)\).
Proof. \[\begin{align} {\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e) &= {\boldsymbol{H}}''(e){\boldsymbol{R}Z}''(e) & \\ {\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\{e\} &= {\boldsymbol{H}}''(e){\boldsymbol{R}Z}''(e)\{e\} &\text{multiply by \{e\}} \\ {\boldsymbol{H}}'(e)\{e\} &= {\boldsymbol{H}}''(e)\{e\} &\text{because {\boldsymbol{R}Z}(e)\{e\}=\{e\}} \\ {\boldsymbol{H}}'(e) &= {\boldsymbol{H}}''(e) &\text{because e is the identity of {\boldsymbol{H}}(e).} \end{align}\]
For any \(f \in {\boldsymbol{R}Z}'(e)\), \(f=ef \in {\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e) = {\boldsymbol{H}}'(e){\boldsymbol{R}Z}''(e)\). Therefore \(f=xf'\) for some \(x \in {\boldsymbol{H}}'(e)\) and \(f' \in {\boldsymbol{R}Z}(e)''\). \[\begin{align} f &= xf' & \\ fe &= xf'e &\text{multiply by e} \\ e &= xe &\text{fe=f'e=e for any f,f' \in {\boldsymbol{R}Z}(e)} \\ e &= x &\text{e is the identity in {\boldsymbol{H}}(e).} \end{align}\] Thus \(f=ef'=f'\), and \({\boldsymbol{R}Z}'(e) \subset {\boldsymbol{R}Z}''(e)\). Similarly, \({\boldsymbol{R}Z}''(e) \subset {\boldsymbol{R}Z}'(e)\), so \({\boldsymbol{R}Z}'(e)={\boldsymbol{R}Z}''(e)\). ◻
Corollary 2. Let \(S\) be a semigroup, let \(e \in S\) be an idempotent, and let \(T \subset S\). Then \(T\) is a right subgroup of \(S\) containing \(e\) if and only if \(T={\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\) for some unique subgroup \({\boldsymbol{H}}'(e) \subset {\boldsymbol{H}}(e)\) and some unique \({\boldsymbol{R}Z}'(e) \subset {\boldsymbol{R}Z}(e)\) containing \(e\).
Similarly, \(T\) is a left subgroup of \(S\) containing \(e\) if and only if \(T={\boldsymbol{L}Z}'(e){\boldsymbol{H}}'(e)\) for some unique subgroup \({\boldsymbol{H}}'(e) \subset {\boldsymbol{H}}(e)\) and some unique \({\boldsymbol{L}Z}'(e) \subset {\boldsymbol{L}Z}(e)\) containing \(e\).
Proof. Every right subgroup \(T\) of \(S\) containing idempotent \(e\) is the maximum right subgroup of itself containing \(e\), so that \(T={\boldsymbol{H}}'(e){\boldsymbol{R}Z}'(e)\) for \({\boldsymbol{H}}'(e)\) as the maximum subgroup of \(T\) containing \(e\) and maximum right zero subsemigroup of \(T\) containing \(e\). Since \(T\) is a subsemigroup of \(S\), \({\boldsymbol{H}}'(e)\) is a subgroup of the maximum subgroup of \(S\) containing \(e\), \({\boldsymbol{H}}(e)\), and \({\boldsymbol{R}Z}'(e)\) with \(e \in {\boldsymbol{R}Z}'(e)\) is a right zero subsemigroup of the maximum right subsemigroup of \(S\) containing \(e\), \({\boldsymbol{R}Z}(e)\). By the preceding lemma, \({\boldsymbol{H}}'(e)\) and \({\boldsymbol{R}Z}'(e)\) are unique.
By Lemma 10, if \(T={\boldsymbol{L}Z}'(e){\boldsymbol{H}}'(e)\) for some subgroup \({\boldsymbol{H}}'(e) \subset {\boldsymbol{H}}(e)\) and some \({\boldsymbol{L}Z}'(e) \subset {\boldsymbol{L}Z}(e)\) containing \(e\), then \(T\) is a right subgroup of \(S\). ◻
If \(S\) is a right [left] group, then it is its own maximal right [left] subsemigroup.
Corollary 3. If \(S\) is a right [left] group, then \(S={\boldsymbol{H}}(e){\boldsymbol{R}Z}(e)\) [\(S={\boldsymbol{L}Z}(e){\boldsymbol{H}}(e)\)].
Let \(E\) be the set of idempotents in semigroup \(S\). Every right [left] subgroup contains an idempotent, and every idempotent in \(S\) is contained in a maximal right [left] subgroup. Then \(\{{\boldsymbol{R}G}(e)|e \in E\}\) is the set of all maximal right subgroups in \(S\), \(\{{\boldsymbol{L}G}(e)|e \in E\}\) is the set of all maximal left subgroups in \(S\). For these sets and \(e,f \in E\), either \({\boldsymbol{R}G}(e)={\boldsymbol{R}G}(f)\) or \({\boldsymbol{R}G}(e) \cap {\boldsymbol{R}G}(f)= \emptyset\), and either \({\boldsymbol{L}G}(e)={\boldsymbol{L}G}(f)\) or \({\boldsymbol{L}G}(e) \cap {\boldsymbol{L}G}(f)= \emptyset\), as seen in the following theorem.
Proposition 6. Let \(S\) be a semigroup, let \(e,f \in S\), and let \(e\) be an idempotent. Then the following are equivalent:
\({\boldsymbol{R}G}(e) \cap {\boldsymbol{R}G}(f) \neq \emptyset\).
\(ef=f\) and \(fe=e\).
\({\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\).
\({\boldsymbol{R}G}(e)={\boldsymbol{R}G}(f)\).
In these cases, it is implied that \(f\) is also an idempotent. There are similar equivalent statements for \({\boldsymbol{L}G}(e)\) and \({\boldsymbol{L}G}(f)\).
Proof. \((1) \implies (2)\): Let \({\boldsymbol{R}G}(e) \cap {\boldsymbol{R}G}(f) \neq \emptyset\). Since \({\boldsymbol{R}G}(f) \neq \emptyset\), then \(f\) is an idempotent by Proposition 5. Let \(x \in {\boldsymbol{H}}(e)\), \(e' \in {\boldsymbol{R}Z}(e)\), \(y \in {\boldsymbol{H}}(f)\), and \(f' \in {\boldsymbol{R}Z}(f)\) such that \(xe'=yf'\) be the element in \({\boldsymbol{R}G}(e) \cap {\boldsymbol{R}G}(f)\).
Therefore \[\begin{align} xe'e &= yf'e &\text{multiply by e} \\ xe &= yf'e &\text{e'e=e} \\ x &= yf'e &\text{xe=x in {\boldsymbol{H}}(e)} \\ fx &= fyf'e &\text{multiply by f} \\ fx &= yf'e &\text{fy=y in {\boldsymbol{H}}(f)} \\ fx &= x &\text{x=yf'e,} \end{align}\] and similarly \(ey=y\).
Since \(fx=x\) and there exists \(x^{-1} \in {\boldsymbol{H}}(e)\), \(fxx^{-1}=xx^{-1}\), or \(fe=e\). Similarly, \(ey=y\) implies \(ef=f\).
\((2) \implies (3)\): By Proposition 2, \(ef=f\) and \(fe=e\) implies \({\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\).
\((3) \implies (4)\): By Proposition 1, since \(e\) is an idempotent, \({\boldsymbol{R}Z}(e) \neq \emptyset\), and thus \({\boldsymbol{R}Z}(f)={\boldsymbol{R}Z}(e) \neq \emptyset\), so that \(f\) is an idempotent.
Let \(x \in {\boldsymbol{H}}(e)\) and \(e' \in {\boldsymbol{R}Z}(e)={\boldsymbol{R}Z}(f)\), so that \(xe' \in {\boldsymbol{H}}(e){\boldsymbol{R}Z}(e)={\boldsymbol{R}G}(e)\). Then \(xe'=x(fe')=(xf)e'\) because \(e' \in {\boldsymbol{R}Z}(f)\).
Also since \(x \in {\boldsymbol{H}}(e)\) and \(f \in {\boldsymbol{R}Z}(f)={\boldsymbol{R}Z}(e)\), \(fx=f(ex)=(fe)x=ex=x\). Therefore \(f(xf)=(fx)f=xf \in fSf\).
Since \(x \in {\boldsymbol{H}}(e)\), there exists \(x^{-1} \in {\boldsymbol{H}}(e)\), and similarly to \(x\), \(fx^{-1}=x^{-1}\) and \(x^{-1}f \in fSf\). Then \[(xf)(x^{-1}f)=x(fx^{-1})f=xx^{-1}f=ef=f,\] and \[(x^{-1}f)(xf)=x^{-1}(fx)f=x^{-1}xf=ef=f,\] so \(xf \in {\boldsymbol{H}}(f)\).
Therefore \(xe'=(xf)e' \in {\boldsymbol{H}}(f){\boldsymbol{R}Z}(f)\), and \({\boldsymbol{H}}(e){\boldsymbol{R}Z}(e) \subset {\boldsymbol{H}}(f){\boldsymbol{R}Z}(f)\). Similarly \({\boldsymbol{R}G}(f) \subset {\boldsymbol{R}G}(e)\), so that \({\boldsymbol{R}G}(e)={\boldsymbol{R}G}(f)\).
\((4) \implies (1)\): Given \({\boldsymbol{R}G}(e)={\boldsymbol{R}G}(f)\), since \(e=ee \in {\boldsymbol{H}}(e){\boldsymbol{R}Z}(e)={\boldsymbol{R}G}(e)={\boldsymbol{R}G}(f)\), \({\boldsymbol{R}G}(e) \cap {\boldsymbol{R}G}(f) \neq \emptyset\). ◻
Proposition 7. Let \(G\) be a group. Then \(G\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)] if and only if \(|G| \leq 3\).
Proof. For any group \(G\) with identity \(e\), \(\mathcal{LI}(x)=\{e\}=\mathcal{RI}(x)\) for any \(x \in G\).
The trivial group \((G,\star)\) is the only semigroup on the one-element set \(G\).
If \(|G|=2\), then \((G,\star)=\{e,a\}\) with identity \(e\) and \(a^2 = e\). If \((G,\star')\) is a semigroup with \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for any \(x \in G\), then \(e\) is an identity in \((G,\star')\) and \(a \star' a \neq a\) because \(a\) is not an idempotent in either \((G,\star)\) or \((G,\star')\). Therefore \(a \star' a = e\), and \(\star' = \star\).
If \(|G|=3\), then \((G,\star)=\{e,a,b\}\) with identity \(e\), \(a^2=b\), \(b^2=b\), and \(ab=ba=e\). If \((G,\star')\) is a semigroup with \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for any \(x \in G\), then \(e\) is an identity in \((G,\star')\). Since \(ab=ba=e\), then \(a\) and \(b\) are not left or right identities or zeroes of each other, and \(a \star' b = b \star' a = e\). Since \(a\) is not an idempotent in \((G,\star)\) or \((G,\star')\), \(a \star' a = b\) or \(a \star' a = e\). If \(a \star' a = e\), then \((a \star' a) \star' b = e \star' b = b\) and \(a \star' (a \star' b) = a \star' e = a\), which is a contradiction. Therefore \(a \star' a = b\), and similarly, \(b \star' b = a\). Hence \(\star' = \star\).
If \(|G|=n>3\), let \(G=\{e,a_1,\ldots,a_{n-1}\}\). Define \((G,\star)\) to be the cyclic group such that \(a_1^k = a_k\) for \(k=1,\ldots,n-1\) and \(a_1^n=e\). Define \((G,\star')\) to be the cyclic group such that \(a_1^k = a_k\) for \(k=1,\ldots,n-3\), \(a_1^{n-2}=a_{n-1}\), \(a_1^{n-1}=a_{n-2}\), and \(a_1^n=e\). Then \((G,\star)\) and \((G,\star')\) are both groups with \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for any \(x \in G\) but \(\star \neq \star'\).
If \(G\) is infinite, there are multiple non-isomorphic group operations on \(G\). ◻
Theorem 8. Let \(S\) be a right group in which every maximal subgroup \(H\) satisfies \(|H| \leq 2\). Then \(S\) is stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)].
Proof. If \(S\) is a right group with \(|H|=1\), then \(S\) is a right zero semigroup, which is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)]. Now assume \(|H|=2\).
Let \((S,\star)\) be a right group such that \(S=H \star E\) with subgroup \(H=\{e,a\}\), such that \(e\) is its identity, whereby \({\boldsymbol{H}}(e)=H\) and \({\boldsymbol{R}Z}(e)=E\). Let \((S,\star')\) be a semigroup such that \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for any \(x \in S\). \(S = E \cup a \star E\). To simplify notation, we will also represent \(\star\) using concatenation, so \(S=E \cup aE\). If a statement is true for an element in \((S,\star)\) and \((S,\star')\), we may simply say the statement is true for that element in \(S\).
Since \((S,\star)\) is a right group, for any \(f \in E\), \(\mathcal{LI}(f)=E\), \(\mathcal{LI}(af)=E\), \(\mathcal{RI}(f)=\{f\}\), and \(\mathcal{RI}(af)=\{f\}\). Then for any \(f \in E\), \(\mathcal{LI}'(f)=E\), \(\mathcal{LI}'(af)=E\), \(\mathcal{RI}'(f)=\{f\}\), and \(\mathcal{RI}'(af)=\{f\}\)
The idempotent sets of \((S,\star)\) and \((S,\star')\) must be equal because \(f\) is an idempotent if and only if \(f \in \mathcal{LI}(f),\mathcal{RI}(f)\). Also, for any \(f \in E\), \(f\) is a left identity of \((S,\star)\), so \(f\) is a left identity of \((S,\star')\) because \(f \in \mathcal{LI}(x)=\mathcal{LI}'(x)\) for all \(x \in S\). Thus \(E\) is a right zero semigroup for both \((S,\star)\) and \((S,\star')\), and \(e_1 \star e_2 = e_1 \star' e_2\) for any \(e_1,e_2 \in E\).
Now consider \(ae_1 \in S\) for any \(e_1 \in E\). Then \((ae_1) \star e_2 = a(e_1e_2) = ae_2\) for any \(e_2 \in E\). Assume \((ae_1) \star' e_2 = f\) for some \(f \in E\). Then for any \(b \in S\), \(((ae_1) \star' e_2) \star' b = ae_1 \star' (e_2 \star' b) = ae_1 \star' b\) because \(e_2\) is a left identity, and \(((ae_1) \star' e_2) \star' b = f \star' b = b\) because \(f\) is a left identity. Therefore \(ae_1 \star' b = b\) for any \(b \in S\). This is a contradiction because \(ae_1\) is not a left identity of \((S,\star)\), so it is not a left identity of \((S,\star')\). Thus \(ae_1 \star' e_2 = af\) for some \(f \in E\).
Since \(e_2\) is an idempotent in \(S\), \((ae_1 \star' e_2) \star' e_2 = ae_1 \star' (e_2 \star' e_2) = ae_1 \star' e_2\), which implies \(af \star' e_2 = af\) and \(e_2 \in \mathcal{RI}'(af)=\mathcal{RI}(af)=\{f\}\). Therefore \(f=e_2\).
For any \(e_1,e_2 \in E\), \(ae_1 \star ae_2 = e_2\). Assume \(ae_1 \star' ae_2 = af\) for some \(f \in E\). Since \(ae_2 \star' e_2 = ae_2\), \((ae_1 \star' ae_2) \star' e_2 = ae_1 \star' (ae_2 \star' e_2) = ae_1 \star' ae_2\), which implies \((ae_1 \star' ae_2) \star' e_2 = af \star' e_2 = af\). Also by above, \(af \star' e_2 = ae_2\), so \(af=ae_2\). Therefore \(ae_1\) is a left identity of \(ae_2\) in \((S,\star')\) and hence \((S,\star)\), which is a contradiction. Thus \(ae_1 \star' ae_2 = f\) for some \(f \in E\). Since \(ae_2 \star' e_2 = ae_2\), \((ae_1 \star' ae_2) \star' e_2 = ae_1 \star' (ae_2 \star' e_2) = ae_1 \star' ae_2\), which implies \(f \star' e_2 = f\). Since \(E\) is a right zero subsemigroup, \(f=e_2\), and \(ae_1 \star' ae_2 = e_2\).
Therefore \(\star = \star'\), and \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). ◻
If \((S,\star)\) is a group containing three elements, then \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)]. However, if \((S,\star)\) is a right group, in which each maximal group contains three elements and the idempotent set contains two or more elements, then \((S,\star)\) is not stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). For example, consider the right group \((S,\star)\) such that \(H=\{e,a,b\}\) with identity \(e\) and \(E=\{e,f\}\). Then \(S=\{e,a,b,f,af,bf\}\), and the following Cayley table gives a binary operation \(\star'\) on \(S\) with equivalent one-sided identity and zero sets, but \(\star' \neq \star\).
\(\begin{array}{c|ccccccc} \star & e & a & b & f & af & bf \\ \hline e & e & a & b & f & af & bf \\ a & a & b & e & af & bf & f \\ b & b & e & a & bf & f & af \\ f & e & a & b & f & af & bf \\ af & a & b & e & af & bf & f \\ bf & b & e & a & bf & f & af \\ \end{array}\) \(\begin{array}{c|ccccccc} \star' & e & a & b & f & af & bf \\ \hline e & e & a & b & f & af & bf \\ a & a & b & e & bf & f & af \\ b & b & e & a & af & bf & f \\ f & e & a & b & f & af & bf \\ af & b & e & a & af & bf & f \\ bf & a & b & e & bf & f & af \\ \end{array}\)
For \((S,\star)\) with \(H=\{e,a,b\}\) and \(|E|>2\), \((S,\star')\) can be generated by extending the definition above and defining \(\star'\) as the above Cayley table for every pair of distinct idempotents.
Corollary 4. A right group \(S\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) if and only if \(|H| \leq 2\) or \(|H|=3\) and \(|E|=1\).
A rectangular band is a semigroup \(S\) in which \(aba=a\) for any \(a,b \in S\). There are several equivalent definitions, and the structure of a rectangular band as a product of a left zero and right zero subsemigroups is well known. We develop some equivalent definitions in the context of one-sided identity and zero sets, and we describe all rectangular band subsemigroups containing a particular idempotent, which reiterates this structure.
Lemma 12. Let \(S\) be a semigroup. Then the following conditions are equivalent:
\(S\) is a rectangular band;
\(S\) is a band and for any \(a,b \in S\), \(ab=b\) if and only if \(ba=a\).
\(S\) is a band and for any \(a,b \in S\), \(ab=a\) if and only if \(ba=b\).
Proof. \((1) \implies (2)\): If \(S\) is a rectangular band, then \(S\) is a band. For any \(a,b \in S\) with \(ab=b\), \(aba=(ab)a=ba\). Since \(S\) is a rectangular band, \(aba=a\), and \(ba=a\). Similarly, for any \(a,b \in S\) with \(ba=a\), \(bab=(ba)b=ab\). Since \(S\) is a rectangular band, \(bab=b\), and \(ab=b\).
\((2) \implies (1)\): \(S\) is a band, and for any \(a,b \in S\), \(ab=b\) if and only if \(ba=a\). For any \(a,b \in S\), \(a(ab)=(aa)b=ab\), which implies \((ab)a=a\). Therefore \(S\) is a rectangular band.
The proofs for \((1) \implies (3)\) and \((3) \implies (1)\) are similar. ◻
Conditions \((2)\) and \((3)\) can then be explicitly expressed in terms of identity and zero sets.
Corollary 5. Let \(S\) be a semigroup. Then the following conditions are equivalent:
\(S\) is a rectangular band;
For any \(a \in S\), \(\mathcal{LI}(a)=\mathcal{RZ}(a) \neq \emptyset\).
For any \(a \in S\), \(\mathcal{RI}(a)=\mathcal{LZ}(a) \neq \emptyset\).
Proof. Given \((2)\) from Lemma 12, \(S\) is a band and for any \(a,b \in S\), \(ab=b\) if and only if \(ba=a\). Since \(S\) is a band, \(aa=a \in \mathcal{LI}(a),\mathcal{RZ}(a)\), and \(\mathcal{LI}(a),\mathcal{RZ}(a) \neq \emptyset\). Since for any \(b \in S\), \(ab=b\) if and only if \(ba=a\), \(\mathcal{RZ}(a)=\mathcal{LI}(a)\) by definition. Therefore \((2)\) from Corollary 5 is implied.
Given \((2)\) from Corollary 5, for any \(a \in S\), \(\mathcal{LI}(a)=\mathcal{RZ}(a) \neq \emptyset\). For any \(a \in S\), since \(\mathcal{LI}(a)=\mathcal{RZ}(a) \neq \emptyset\), \(\mathcal{LI}(a) \cap \mathcal{RZ}(a) \neq \emptyset\), and by Proposition 1, \(a\) is an idempotent, so that \(S\) is a band. Since \(\mathcal{LI}(a)=\mathcal{RZ}(a)\) for any \(a \in S\), then for any \(b \in S\), \(ba=a\) if and only if \(ab=b\) by definition. Therefore \((2)\) from Lemma 12 is implied. ◻
Lemma 13. Let \(S\) be a semigroup, let \(e\) be an idempotent, and let \(T \subset S\). If \(T=T_LT_R\) for some sets \(T_L\) and \(T_R\) such that \(T_L \subset {\boldsymbol{L}Z}(e)\), \(T_R \subset {\boldsymbol{R}Z}(e)\), and \(T_RT_L=\{e\}\), then \(T_L\) and \(T_R\) are unique.
Proof. Let \(T=T_LT_R\) and \(T=T_L'T_R'\) such that \(T_L,T_L' \subset {\boldsymbol{L}Z}(e)\), \(T_R,T_R' \subset {\boldsymbol{R}Z}(e)\), \(T_RT_L=\{e\}\), and \(T_R'T_L'=\{e\}\).
Since \(T_RT_L = \{e\}\) and \(T_R'T_L'=\{e\}\), \(T_L,T_R,T_L',T_R'\) must all be nonempty. For any nonempty \(A,B\) subsets of a left zero semigroup, \(AB=A\), and for any nonempty \(A,B\) subsets of a right zero semigroup, \(AB=B\).
\[\begin{align} T_LT_R &= T_L'T_R' \\ T_LT_RT_R &= T_L'T_R'T_R &\text{multiply by T_R on the right} \\ T_LT_R &= T_L'T_R &\text{T_R,T_R' \subset {\boldsymbol{R}Z}(e)} \\ T_LT_RT_L &= T_L'T_RT_L &\text{multiply by T_L on the right} \\ T_L\{e\} &= T_L'\{e\} &\text{T_RT_L=\{e\}} \\ T_L &= T_L' &\text{T_L,T_L',\{e\} \subset {\boldsymbol{L}Z}(e).} \end{align}\]
Similarly, \(T_R=T_R'\). ◻
Proposition 9. Let \(S\) be a semigroup, and let \(e\) be an idempotent. \(T\) is a rectangular band subsemigroup of \(S\) containing \(e\) if and only if \(T=T_LT_R\) for some sets \(T_L\) and \(T_R\) such that \(e \in T_L \subset {\boldsymbol{L}Z}(e)\), \(e \in T_R \subset {\boldsymbol{R}Z}(e)\), and \(T_RT_L = \{e\}\). In this case, \(T_L\) and \(T_R\) are unique.
Proof. If \(T\) is a rectangular band subsemigroup of \(S\) containing \(e\), then for any \(x \in T\), \[x=xex=x(ee)x=(xe)(ex) \in (Te)(eT).\] For any \(x,y \in T\), \[(xe)(ey)=xey \in T\] because \(T\) is a subsemigroup containing \(x,e,y\). Therefore \(T=(Te)(eT)\).
\(Te \subset Se = \mathcal{LZ}(e)\) and \(eT \subset eS = \mathcal{RZ}(e)\). For any \(x \in T\), \(e(xe)=exe=e\), so \(Te \subset \mathcal{RI}(e)\), and \(Te \subset {\boldsymbol{L}Z}(e)\). For any \(x \in T\), \((ex)e=exe=e\), so \(eT \subset \mathcal{LI}(e)\), and \(eT \subset {\boldsymbol{R}Z}(e)\). For any \(x,y \in T\), \((ex)(ye)=e(xy)e=e\) because \(T\) is a rectangular band subsemigroup with \(xy \in T\). Therefore \((eT)(Te)=\{e\}\).
If \(T=T_LT_R\) for some sets \(T_L\) and \(T_R\) such that \(e \in T_L \subset {\boldsymbol{L}Z}(e)\), \(e \in T_R \subset {\boldsymbol{R}Z}(e)\), and \(T_RT_L = \{e\}\), then \(ee=e \in T\).
For any \(a,c \in T_L\) and \(b,d \in T_R\), \(ab,cd \in T\), and \[(ab)(cd)=a(bc)d=aed=(ae)d=ad \in T.\] Thus \(T\) is a subsemigroup of \(S\) containing \(e\).
For any \(a,c \in T_L\) and \(b,d \in T_R\), \(ab,cd \in T\), and \[(ab)(cd)(ab)=a(bc)(da)b=a(ee)b=aeb=a(eb)=ab.\] Therefore \(T\) is a rectangular band subsemigroup of \(S\) containing \(e\).
\(T_L=Te\) and \(T_R=eT\) are unique by the previous lemma. ◻
Let \(S\) be a semigroup. For any \(a \in S\), define \(V(a)\) to be the set of all generalized inverses of \(a\): \(V(a)=\{b \in S|aba=a, \, bab=b\}\). If \(S\) is a band, then \(V(a)\) is the maximum rectangular band subsemigroup containing \(a\).
Corollary 6. Let \(S\) be a rectangular band. For any \(e \in S\), \(S=(Se)(eS)\) such that \(Se\) is a maximal left zero subsemigroup of \(S\) and \(eS\) is a maximal right zero subsemigroup of \(S\).
Proof. If \(S\) is a rectangular band and \(e \in S\), then \(e\) is an idempotent. By Proposition 1 and its dual, \({\boldsymbol{L}Z}(e)\) is the maximum left zero subsemigroup of \(S\) containing \(e\) and \({\boldsymbol{R}Z}(e)\) is the maximum right zero subsemigroup of \(S\) containing \(e\).
For any \(a \in S\), \((ae)e=a(ee)=ae\) and \(e(ae)=e\). Therefore \(ae \in {\boldsymbol{L}Z}(e)\) and \(Se \subset {\boldsymbol{L}Z}(e)\). For any \(b \in {\boldsymbol{L}Z}(e)\), \(b=be \in Se\), so \({\boldsymbol{L}Z}(e) \subset Se\). Thus \({\boldsymbol{L}Z}(e)=Se\). Similarly, \({\boldsymbol{R}Z}(e)=eS\).
Since \(S\) is a rectangular band, \((eS)(Se)=\{e\}\) and \(S=(Se)(eS)\). ◻
Proposition 10. Let \(S\) be a band. For any \(e \in S\), \({\boldsymbol{R}B}(e)={\boldsymbol{L}Z}(e){\boldsymbol{R}Z}(e)\) is the maximum rectangular band subsemigroup of \(S\) containing \(e\). In this case, \(V(e)={\boldsymbol{R}B}(e)\).
Proof. Since \(S\) is a band and \(e \in S\), \(e\) is an idempotent, and \(ee=e \in {\boldsymbol{R}Z}(e),{\boldsymbol{L}Z}(e)\). For any \(a \in {\boldsymbol{R}Z}(e)\), \(ea=a\) and \(ae=e\), and for any \(b \in {\boldsymbol{L}Z}(e)\), \(eb=e\) and \(be=b\). Therefore \[\begin{align} ab &= eabe &\text{a=ea,b=be} \\ &= ebabae &\text{e=eb,e=ae} \\ &=ebae &\text{S is a band, so baba=ba} \\ &=ee &\text{eb=e,ae=e} \\ &=e &\text{ee=e.} \end{align}\] This implies \({\boldsymbol{R}Z}(e){\boldsymbol{L}Z}(e)=\{e\}\), and by Proposition 9, \({\boldsymbol{R}B}(e)={\boldsymbol{L}Z}(e){\boldsymbol{R}Z}(e)\) is a rectangular band subsemigroup of \(S\) containing \(e\). Also by Proposition 9, any rectangular band subsemigroup of \(S\) containing \(e\) is of the form \(T_LT_R\) for \(T_L \subset {\boldsymbol{L}Z}(e)\) and \(T_R \subset {\boldsymbol{R}Z}(e)\), and \(T_LT_R \subset {\boldsymbol{L}Z}(e){\boldsymbol{R}Z}(e)={\boldsymbol{R}B}(e)\). Thus \({\boldsymbol{R}B}(e)\) is the maximum rectangular band subsemigroup containing \(e\).
Since every element of a rectangular band is a generalized inverse of every other element of a rectangular band, \({\boldsymbol{R}B}(e) \subset V(e)\). For any \(a \in V(e)\), \(a\) is also an idempotent because \(S\) is a band, and \(aea=a\) and \(eae=e\), so that \(\langle a,e \rangle = \{a,e,ae,ea\}\). Also, \(\{a,e,ae,ea\}=T_LT_R\) such that \(e \in T_L \subset {\boldsymbol{L}Z}(e)\), \(e \in T_R \subset {\boldsymbol{R}Z}(e)\), and \(T_RT_L=\{e\}\) for \(T_L=\{e,ae\}\) and \(T_R=\{e,ea\}\), so \(\langle a,e \rangle\) is a rectangular band subsemigroup containing \(e\) and a subset of the maximum \({\boldsymbol{R}B}(e)\). Therefore \(a \in {\boldsymbol{R}B}(e)\), and \(V(e)={\boldsymbol{R}B}(e)\). ◻
Proposition 11. Let \(S\) be a band. For any \(a,b \in S\), the following are equivalent:
\(aba=a\) and \(bab=b\).
\({\boldsymbol{R}B}(a)={\boldsymbol{R}B}(b)\).
\({\boldsymbol{R}B}(a) \cap {\boldsymbol{R}B}(b) \neq \emptyset\).
\({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) \neq \emptyset\) [\({\boldsymbol{L}Z}(a) \cap {\boldsymbol{R}Z}(b) \neq \emptyset\)].
\({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) = \{ab\}\) [\({\boldsymbol{L}Z}(a) \cap {\boldsymbol{R}Z}(b) = \{ba\}\)].
Proof. \((1) \implies (2)\): By Proposition 10, \({\boldsymbol{R}B}(a)=V(a)\) is the maximum rectangular subsemigroup of \(S\) containing \(a\) and \(b\), respectively. Since \(a\) is an idempotent, \(aaa=a \in {\boldsymbol{R}B}(a)\). Since \(aba=a\) and \(bab=b\), \(b \in V(a)={\boldsymbol{R}B}(a)\). For any \(c \in {\boldsymbol{R}B}(a)\), \(bcb=b\) and \(cbc=c\) because \({\boldsymbol{R}B}(a)\) is a rectangular band. Therefore \(c \in V(b)={\boldsymbol{R}B}(b)\), and \({\boldsymbol{R}B}(a) \subset {\boldsymbol{R}B}(b)\). Similarly, \({\boldsymbol{R}B}(b) \subset {\boldsymbol{R}B}(a)\), so \({\boldsymbol{R}B}(a)={\boldsymbol{R}B}(b)\).
\((2) \implies (3)\): Since \(a\) is an idempotent, \(aaa=a \in {\boldsymbol{R}B}(a) \neq \emptyset\). If \({\boldsymbol{R}B}(a)={\boldsymbol{R}B}(b)\), then \({\boldsymbol{R}B}(a) \cap {\boldsymbol{R}B}(b) = {\boldsymbol{R}B}(a) \neq \emptyset\).
\((3) \implies (1)\): There exists \(c \in {\boldsymbol{R}B}(a) \cap {\boldsymbol{R}B}(b)\). By Proposition 10, \({\boldsymbol{R}B}(a)=V(a)\), \({\boldsymbol{R}B}(b)=V(b)\), and \({\boldsymbol{R}B}(c)=V(c)\). Since \(c\) is an idempotent, \(ccc=c \in {\boldsymbol{R}B}(c)\), and \(a,b \in {\boldsymbol{R}B}(c)\). As a rectangular band subsemigroup, \(ab,ba \in {\boldsymbol{R}B}(c)\). Therefore \[\begin{align} aba &= (aca)b(aca) &\text{a=aca} \\ &= aca &\text{cabac=c in rectangular band {\boldsymbol{R}B}(c)} \\ &= a &\text{aca=a,} \end{align}\] and \[\begin{align} bab &= (bcb)a(bcb) &\text{b=bcb} \\ &= bcb &\text{cbabx=c in rectangular band {\boldsymbol{R}B}(c)} \\ &= b &\text{bcb=b.} \end{align}\]
\((1) \implies (4)\): Since \(aba=a\) and \(bab=b\), \(ab \in \mathcal{LI}(a)\) and \(ab \in \mathcal{RI}(b)\). Since \(S\) is a band and \(a\) and \(b\) are idempotents, \(aab=ab\) and \(abb=ab\), so that \(ab \in \mathcal{RZ}(a)\) and \(ab \in \mathcal{LZ}(b)\). Therefore \(ab \in {\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) \neq \emptyset\).
\((4) \implies (5)\): Let \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b)\) be nonempty. For any \(c \in {\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b)\), \(ca=a\), \(ac=c\), \(cb=c\), and \(bc=b\). Therefore \(cabc=(ca)(bc)=ab\).
By Proposition 10, \({\boldsymbol{R}B}(a)=V(a)\), \({\boldsymbol{R}B}(b)=V(b)\), and \({\boldsymbol{R}B}(c)=V(c)\). Since \(ca=a\) and \(ac=c\), \(aca=a\) and \(cac=c\), so that \(a \in V(c)={\boldsymbol{R}B}(c)\), and since \(cb=c\) and \(bc=b\), \(cbc=c\) and \(bcb=b\), so that \(b \in V(c)={\boldsymbol{R}B}(c)\). Since \({\boldsymbol{R}B}(c)\) is a rectangular band subsemigroup of \(S\), \(ab \in {\boldsymbol{R}B}(c)\) and \(cabc=c\).
Therefore \(ab=c\), and \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b)=\{ab\}\).
\((5) \implies (1)\): Let \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b)=\{ab\}\). Then \(ab \in \mathcal{LI}(a)\), so that \((ab)a=a\), and \(ab \in \mathcal{RI}(b)\), so that \(b(ab)=b\). ◻
Proposition 12. Any rectangular band is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)].
This theorem is a special case of Theorem 20
Proof. Let \((S,\star)\) be a rectangular band.
Let \((S,\star')\) be a semigroup such that \(\mathcal{LI}(a)=\mathcal{LI}'(a)\) and \(\mathcal{RI}(a)=\mathcal{RI}'(a)\) for any \(a \in S\). Since \(\mathcal{LI}^{-1}=\mathcal{RZ}\) and \(\mathcal{RI}^{-1}=\mathcal{LZ}\), \(\mathcal{LZ}(a)=\mathcal{LZ}'(a)\) and \(\mathcal{RZ}(a)=\mathcal{RZ}'(a)\) for any \(a \in S\).
Since \((S,\star)\) is a rectangular band, then by Corollary 5, \(\mathcal{LI}(a)=\mathcal{RZ}(a) \neq \emptyset\) and \(\mathcal{RI}(a)=\mathcal{LZ}(a) \neq \emptyset\) for any \(a \in S\).
Therefore \(\mathcal{LI}'(a)=\mathcal{RZ}'(a) \neq \emptyset\) and \(\mathcal{RI}'(a)=\mathcal{LZ}'(a) \neq \emptyset\) for any \(a \in S\), and by Corollary 5, \((S,\star')\) is a rectangular band.
For any \(a,b \in S\), since \((S,\star)\) is a band, \(a \star (a \star b) = (a \star a) \star b = a \star b\) and \((a \star b) \star b = a \star (b \star b) = a \star b\). Therefore, \(a \in \mathcal{LI}(a \star b)=\mathcal{LI}'(a \star b)\) and \(b \in \mathcal{RI}(a \star b)=\mathcal{RI}'(a \star b)\). Since \((S,\star')\) is a rectangular band, for any \(a,b,c \in S\), \(a \star' c \star' b = a \star' b\). Then \[a \star b = a \star' (a \star b) \star' b = a \star' b.\] Thus \(\star=\star'\). ◻
Proposition 13. Let \(S\) be a rectangular band. For any \(a,b \in S\), \(\{ab\} = \mathcal{RZ}(a) \cap \mathcal{LZ}(b)\).
This theorem is a special case of Theorem 21.
Proof. For any \(a,b \in S\), since \(S\) is a band, \(aab=ab\) and \(abb=ab\). Therefore \(ab \in \mathcal{RZ}(a) \cap \mathcal{LZ}(b)\). Also, for any \(c \in \mathcal{RZ}(a) \cap \mathcal{LZ}(b)\), \(acb=c\), and since \(S\) is a rectangular band, \(acb=ab\). ◻
Lemma 14. In a band \(S\) with elements \(a\) and \(b\), \(V(a)=D_a\) and \(a \, \mathscr{D} \, b\) if and only if \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) \neq \emptyset\). When \(a \, \mathscr{D} \, b\), \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) = \{ab\}\).
Proof. From [5], every inverse of an element is contained in its \(\mathscr{D}\)-class, and there exists an inverse of \(a\) in \(H_b\) if and only if \(R_a \cap L_b\) and \(R_b \cap L_a\) contain idempotents. In a band, for any \(a \in S\), \(V(a)=D_a\).
Let \(a,b \in S\) such that \(a \, \mathscr{D} \, b\). Then there exists \(c \in S\) such that \(a \, \mathscr{R} \, c\) and \(c \, \mathscr{L} \, b\). By Lemma 9 and its dual, \({\boldsymbol{R}Z}(a)={\boldsymbol{R}Z}(c)\) and \({\boldsymbol{L}Z}(c)={\boldsymbol{L}Z}(b)\). Since \(c\) is an idempotent, \(c \in {\boldsymbol{R}Z}(c) \cap {\boldsymbol{L}Z}(c)={\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b)\), and \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) \neq \emptyset\).
Let \(a,b \in S\) such that \(c \in {\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) \neq \emptyset\). Since \(c\) is an idempotent, \(c \in {\boldsymbol{R}Z}(c) \cap {\boldsymbol{L}Z}(c)\). Thus \(c \in {\boldsymbol{R}Z}(a) \cap {\boldsymbol{R}Z}(c)\) and \(c \in {\boldsymbol{L}Z}(c) \cap {\boldsymbol{L}Z}(b)\). Since these are nonempty, by Proposition 2 and Lemma 9 and their duals, \(a \, \mathscr{R} \, c\) and \(c \, \mathscr{L} \, b\), which implies \(a \, \mathscr{D} \, b\).
When \(a \, \mathscr{D} \, b\), \(a\) and \(b\) are generalized inverses, so that \(aba=a\) and \(bab=b\), which implies \(ab \in \mathcal{LI}(a) \cap \mathcal{RI}(b)\). Since \(a\) and \(b\) are idempotents, \(aab=ab\) and \(abb=ab\), which implies \(ab \in \mathcal{RZ} \cap \mathcal{LZ}(b)\). Therefore \(ab \in \mathcal{RZ}(a) \cap \mathcal{LI}(a) \cap \mathcal{LZ}(b) \cap \mathcal{RI}(b) = {\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b)\). By Corollary 1, \({\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) = R_a \cap L_b\). Since \(S\) is a band, for any \(e \in S\), \(H_e = {\boldsymbol{H}}(e) = \{e\}\), so \(R_a \cap L_b = H_{ab} = \{ab\}\). ◻
We develop some equivalent definitions for commutative bands in the context of one-sided identity and zero sets.
Lemma 15. Let \(S\) be a semigroup. \(S\) is a commutative band if and only if \(S\) is a band and for any \(a,b \in S\), \(ab=a\) if and only if \(ba=a\).
Proof. If \(S\) is a commutative band, then \(S\) is a band and for any \(a,b \in S\), \(ab=a\) implies \(ba=ab=a\) and \(ba=a\) implies \(ab=ba=a\).
If \(S\) is a band and for any \(a,b \in S\), \(ab=a\) if and only if \(ba=a\), then for any \(x,y \in S\), \(x(xy)=(xx)y=xy\) implies \((xy)x=xy\), and \((yx)x=y(xx)=yx\) implies \(x(yx)=yx\). Therefore \(xy=xyx=yx\). ◻
We can then explicitly express the definition of a commutative band in terms of identity and zero sets.
Corollary 7. Let \(S\) be a semigroup. Then the following conditions are equivalent:
\(S\) is a commutative band;
For any \(a \in S\), \(a \in \mathcal{LI}(a)=\mathcal{RI}(a)\).
For any \(a \in S\), \(a \in \mathcal{LZ}(a)=\mathcal{RZ}(a)\).
Proof. \((1) \implies (2)\): If \(S\) is a commutative band, then for any \(a \in S\), \(aa=a \in \mathcal{LI}(a),\mathcal{RI}(a)\). For any \(b \in \mathcal{LI}(a)\), \(ba=a\), so \(ab=ba=a\) and \(b \in \mathcal{RI}(a)\). Similarly \(\mathcal{RI}(a) \subset \mathcal{LI}(a)\).
\((2) \implies (1)\): If for any \(a \in S\), \(a \in \mathcal{LI}(a)=\mathcal{RI}(a)\), then \(aa=a\), and \(S\) is a band. For any \(a,b \in S\), \(b \in \mathcal{LI}(a)\) if and only if \(b \in \mathcal{RI}(a)\), so \(ab=a\) if and only if \(ba=a\). By Lemma 15, \(S\) is a commutative band.
Similarly, \((1)\) if and only if \((3)\). ◻
Proposition 14. Every commutative band is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)].
This theorem is a special case of Theorem 20.
Proof. Let \((S,\star)\) be a commutative band.
Let \((S,\star')\) be a semigroup such that \(\mathcal{LI}(a)=\mathcal{LI}'(a)\) and \(\mathcal{RI}(a)=\mathcal{RI}'(a)\) for any \(a \in S\). Since \((S,\star)\) is a commutative band, \(a \in \mathcal{LI}(a)=\mathcal{RI}(a)\) for any \(a \in S\) by Corollary 7. Then \(a \in \mathcal{LI}'(a)=\mathcal{RI}'(a)\) for any \(a \in S\). Therefore \((S,\star')\) is also a commutative band.
Let \(a,b \in S\). Since \(a\) and \(b\) are idempotents in \((S,\star)\) and \((S,\star')\), \(a \star (a \star b) = (a \star a) \star b = a \star b\) and \((a \star b) \star b = a \star (b \star b) = a \star b\). Therefore \(a \star' (a \star b) \star' b = a \star b\), and since \((S,\star')\) is commutative, \((a \star' b) \star' (a \star b) = a \star b\), which implies \((a \star' b) \star (a \star b) = a \star b\) because \(\mathcal{LI}(a \star b) = \mathcal{LI}'(a \star b)\). Similarly, \((a \star' b) \star (a \star b) = a \star' b\). Thus \(a \star b = a \star' b\), and \(\star=\star'\). ◻
Lemma 16. Let \(S\) be a groupoid. For any idempotents \(a,b \in S\), if \(\mathcal{LI}(a)=\mathcal{LI}(b)\) and \(\mathcal{RI}(a)=\mathcal{RI}(b)\), then \(a=b\).
There is a similar statement for \(\mathcal{LZ}\) and \(\mathcal{RZ}\).
Proof. Since \(a\) and \(b\) are idempotents, \(a,b \in \mathcal{LI}(a)=\mathcal{LI}(b), \mathcal{RI}(a)=\mathcal{RI}(b)\). Therefore \(ab = b\) because \(a \in \mathcal{LI}(b)\), and \(ab = a\) because \(b \in \mathcal{RI}(a)\). Thus \(a=b\). ◻
Proposition 15. Let \(S\) be a band, and let \(a,b \in S\). Then \(a\) and \(b\) commute if and only if \(ab \in \mathcal{LZ}(a) \cap \mathcal{RZ}(a) \cap \mathcal{LZ}(b) \cap \mathcal{RZ}(b)\). In this case, \(ab\) is the unique element \(c \in S\) such that \(\mathcal{LI}(c) = \displaystyle\cap_{x \in U} \mathcal{LI}(x)\) and \(\mathcal{RI}(c)=\displaystyle\cap_{x \in U} \mathcal{RI}(x)\) for \(U = \mathcal{LZ}(a) \cap \mathcal{RZ}(a) \cap \mathcal{LZ}(b) \cap \mathcal{RZ}(b)\).
Proof. Since \(S\) is a band, \(a\) and \(b\) are idempotents, and \(aab=ab\) and \(abb=ab\). Therefore \(ab \in \mathcal{RZ}(a) \cap \mathcal{LZ}(b)\).
If \(ab=ba\), then \(aba=aab=ab\) and \(bab=abb=ab\), and \(ab \in \mathcal{LZ}(a) \cap \mathcal{RZ}(b)\).
If \(ab \in \mathcal{LZ}(a) \cap \mathcal{RZ}(b)\), then \(aba=ab\) and \(bab=ab\). Therefore \(baba=bab\), which implies \(ba=ab\) because \(S\) is a band and \(bab=ab\).
For any \(x \in U=\mathcal{LZ}(a) \cap \mathcal{RZ}(a) \cap \mathcal{LZ}(b) \cap \mathcal{RZ}(b)\), \(xa=x\), \(ax=x\), \(xb=x\), and \(bx=x\). Thus \(abx=ax=x\) and \(xab=xb=x\), so that \(ab \in \mathcal{LI}(x)\) and \(ab \in \mathcal{RI}(x)\). Hence \(\mathcal{LI}(ab) = \displaystyle\cap_{x \in U} \mathcal{LI}(x)\), and \(\mathcal{RI}(ab) = \displaystyle\cap_{x \in U} \mathcal{RI}(x)\). By Lemma 16, \(ab\) must be the unique element of \(S\) satisfying these one-sided identity sets. ◻
Corollary 8. Let \(S\) be a commutative band. For any \(a,b \in S\), \(ab\) is the unique element \(c \in S\) such that \[\mathcal{LI}(c) = \displaystyle\cap_{x \in \mathcal{RZ}(a) \cap \mathcal{LZ}(b)} \mathcal{LI}(x) = \displaystyle\cap_{x \in \mathcal{RZ}(a) \cap \mathcal{LZ}} \mathcal{RI}(x) = \mathcal{RI}(c).\]
This corollary is a special case of Theorem 21.
Proposition 16. Let \((S,\star)\) be a groupoid. If \(|S| \leq 2\), then \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) [\(\mathcal{RI}\), \(\mathcal{LZ}\), \(\mathcal{RZ}\)].
Proof. If \(|S|=1\), then \((S,\star)\) is the trivial group, \(S=\{e\}\) with \(e \star e = e\), and \((S,\star)\) is stabilized with respect to any binary relation and \(\mathcal{LI}\) in particular.
If \(|S|=2\), then \(S=\{a,b\}\). For any \(x \in S\), \(x \star a = a\) if and only if \(x \in \mathcal{LI}(a)\), and \(x \star a = b\) if and only if \(x \notin \mathcal{LI}(a)\). Thus \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\). ◻
Proposition 17. Let \((S,\star)\) be a groupoid with \(|S|=3\). Then \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)] if and only if every \(x \in S\) is an idempotent.
Proof. Let \((S,\star)\) be a groupoid with \(|S|=3\).
Let \((S,\star)\) be stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). For any \(a \in S\), if \(a\) is not an idempotent, then \(a \star a \neq a\), and \(a \notin \mathcal{LI}(a), \mathcal{RI}(a)\). Let \((S,\star')\) be a groupoid with \(\star'\) defined such that \(x \star' y = x \star y\) for every \(x,y \in S\) except when \(x=y=a\), then let \(a \star' a = b\) for \(b \in S\) such that \(b \neq a\) and \(b \neq a \star a\). Since \(|S|=3\), such a \(b\) exists. Then \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for all \(x \in S\). However \(\star \neq \star'\), which is a contradiction because \((S,\star)\) is stabilized. Therefore \(a \in S\) is an idempotent.
Let \((S,\star)\) be an idempotent groupoid in which \(x \star x = x\) for all \(x \in S\), so that \(x \in \mathcal{LI}(x),\mathcal{RI}(x)\) for all \(x \in S\). Let \((S,\star')\) be a groupoid such that \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for all \(x \in S\). Then \((S,\star')\) is an idempotent groupoid because \(x \in \mathcal{LI}'(x),\mathcal{RI}'(x)\) and \(x \star' x = x\), for all \(x \in S\).
For any \(x,y \in S\) with \(x \neq y\), if \(x \star y = x\), then \(y \in \mathcal{RI}(x)=\mathcal{RI}'(x)\), and \(x \star' y = x\). If \(x \star y = y\), then \(x \in \mathcal{LI}(y)=\mathcal{LI}'(y)\), and \(x \star' y = y\). If \(x \star y = z\) for \(z \neq x\) and \(z \neq y\), then \(y \notin \mathcal{RI}(x)=\mathcal{RI}'(x)\) and \(x \notin \mathcal{RI}(y)=\mathcal{RI}'(y)\). Since \(|S|=3\), this implies that \(x \star' y = z\). Since \(|S|=3\), these are all possible cases for \(x \star y\). Therefore \(x \star y = x \star' y\) for all \(x,y \in S\), and \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). ◻
Definition 5. \(S\) is a left-right zero groupoid if and only if for any \(a,b \in S\), \(a \star b = a\) or \(a \star b = b\).
Essentially, \(S\) is a left-right zero groupoid if and only if for any \(a,b \in S\), either \(a\) is a left [right] identity of \(b\) or \(a\) is a left [right] zero of \(b\). If \((S,\star)\) is a left-right zero groupoid, then for any \(a \in S\), \(a \star a = a\), and \((S,\star)\) is an idempotent groupoid. If \((S,\star)\) is a semigroup and a left-right zero groupoid, then we call \((S,\star)\) a left-right zero semigroup, and then \((S,\star)\) is a band.
Proposition 18. Let \((S,\star)\) be a groupoid with \(|S|>3\). Then \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)] if and only if \(S\) is a left-right zero groupoid.
Proof. Let \((S,\star)\) be a groupoid with \(|S|>3\).
Let \((S,\star)\) be stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). For any \(a,b \in S\), if \(a \star b = c\) with \(c \neq a\) and \(c \neq b\), then \(a \notin \mathcal{LI}(b)\) and \(b \notin \mathcal{RI}(a)\). Let \((S,\star')\) be a groupoid with \(\star'\) defined such that \(x \star' y = x \star y\) for every \(x,y \in S\) except when \(x=a\) and \(y=b\), then let \(a \star' b = d\) for \(d \in S\) such that \(d \neq a\), \(d \neq b\), and \(d \neq a \star b\). Since \(|S|>3\), such a \(d\) exists. Then \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for all \(x \in S\). However \(\star \neq \star'\), which is a contradiction because \((S,\star)\) is stabilized with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). Therefore \(a \star b = a\) or \(a \star b = b\) for all \(a,b \in S\).
Let \((S,\star)\) be a groupoid such that \(a \star b = a\) or \(a \star b = b\) for all \(a,b \in S\). Let \((S,\star')\) be a groupoid such that \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for all \(x \in S\). For any \(a,b \in S\), \(a \star b = a\) or \(a \star b = b\). If \(a \star b = a\), then \(b \in \mathcal{RI}(a)=\mathcal{RI}'(a)\), and \(a \star' b = a\). If \(a \star b = b\), then \(a \in \mathcal{LI}(b)=\mathcal{LI}'(b)\), and \(a \star' b = b\). Therefore \(a \star' b = a \star b\) for any \(a,b \in S\), and \((S,\star)\) is a stabilized groupoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). ◻
Proposition 19. Let \(S\) be a left-right zero groupoid. Then \(S\) is a semigroup if and only if for any \(a,b \in S\), \(\mathcal{LI}(a) \subset \mathcal{LI}(b)\) if \(a \in \mathcal{LI}(b)\) and \(\mathcal{RI}(a) \subset \mathcal{RI}(b)\) if \(a \in \mathcal{RI}(b)\).
There is a similar statement involving \(\mathcal{LZ}\) and \(\mathcal{RZ}\).
Proof. Let \(S\) be a semigroup. For any \(a,b \in S\), if \(a \in \mathcal{LI}(b)\), \(ab = b\), and for any \(x \in \mathcal{LI}(a)\) with \(xa = a\), \(xb = x(ab) = (xa)b = ab = b\). Therefore \(\mathcal{LI}(a) \subset \mathcal{LI}(b)\). Similarly, if \(a \in \mathcal{RI}(b)\), then \(\mathcal{RI}(a) \subset \mathcal{RI}(b)\).
Let \(S\) be a left-right zero groupoid satisfying the property of the theorem. Then for any \(a,b,c \in S\), consider the eight cases generated by \(ab = a\) or \(ab = b\), \(ac = a\) or \(ac = c\), and \(bc = b\) or \(bc = c\). The only cases that satisfy the property also satisfy \((ab)c = a(bc)\). ◻
If \((S,\star)\) is a semigroup and a stabilized groupoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\), then \((S,\star)\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\). However, if the semigroup \((S,\star)\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\), then \((S,\star)\) is not necessarily a stabilized groupoid with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\).
Definition 6. \(S\) is a commutative-rectangular band if and only if \(S\) is a band with the property that for any \(a,b \in S\), either \(a\) and \(b\) commute or \(a\) and \(b\) are generalized inverses of each other.
Examples of commutative-rectangular bands include commutative bands, rectangular bands, and left-right zero semigroups.
Lemma 17. Let \(S\) be a band. \(S\) has the property that for any \(x \in S\), \(\mathcal{LI}(x) \cup \mathcal{LZ}(x)=\mathcal{RI}(x) \cup \mathcal{RZ}(x)\), if and only if \(S\) is a commutative-rectangular band.
Proof. Let \(S\) be a band with the property that for any \(x \in S\), \(\mathcal{LI}(x) \cup \mathcal{LZ}(x)=\mathcal{RI}(x) \cup \mathcal{RZ}(x)\). Since \(a(ab)=(aa)b=ab\), \((ab)a=a\) or \((ab)a=ab\). Since \((ab)b=a(bb)=ab\), \(b(ab)=b\) or \(b(ab)=ab\). Similarly, \((ba)b=b\) or \((ba)b=ba\), and \(a(ba)=a\) or \(a(ba)=ba\). In every one of the 16 cases created by these possibilities, either \(a\) and \(b\) commute or \(a\) and \(b\) are generalized inverses.
Let \(S\) be a band with the property that for any \(a,b \in S\), either \(ab=ba\) or \(aba=a\) and \(bab=b\). For any \(x \in S\), if \(y \in \mathcal{LI}(x) \cup \mathcal{LZ}(x)\), then \(yx=x\) or \(yx=y\). If \(xy=yx\), then \(xy=x\) or \(xy=y\) and \(y \in \mathcal{RI}(x) \cup \mathcal{RZ}(x)\). If \(xyx=x\) and \(yxy=y\), then \(y=yxy=xy\) or \(x=xyx=xy\) and \(y \in \mathcal{RI}(x) \cup \mathcal{RZ}(x)\). Therefore \(\mathcal{LI}(x) \cup \mathcal{LZ}(x) \subset \mathcal{RI}(x) \cup \mathcal{RZ}(x)\). Similarly, \(\mathcal{RI}(x) \cup \mathcal{RZ}(x) \subset \mathcal{LI}(x) \cup \mathcal{LZ}(x)\), and \(\mathcal{LI}(x) \cup \mathcal{LZ}(x)=\mathcal{RI}(x) \cup \mathcal{RZ}(x)\). ◻
The property of semigroup \(S\) being a band could have been included in the left side property as \(x \in \mathcal{LI}(x) \cup \mathcal{LZ}(x)=\mathcal{RI}(x) \cup \mathcal{RZ}(x)\) for any \(x \in S\) and in the right side property as \(aa=a\) and either \(ab=ba\) or \(aba=a\) and \(bab=b\) for any \(a,b \in S\).
A commutative-rectangular band is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)], which can be proven directly using the definition of a stabilized semigroup as in the proof of Theorem 20 or by explicitly calculating its products using the one-sided identity and zero sets as in Theorem 21 and its proof.
Theorem 20. Every commutative-rectangular band is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)].
Proof. Let \((S,\star)\) be band such that for any \(x \in S\), \(\mathcal{LI}(x) \cup \mathcal{LZ}(x)=\mathcal{RI}(x) \cup \mathcal{RZ}(x)\). Since \((S,\star)\) is a band, \(x \in \mathcal{LI}(x) \cup \mathcal{LZ}(x)=\mathcal{RI}(x) \cup \mathcal{RZ}(x)\) for any \(x \in S\). Let \((S,\star')\) be a semigroup such that \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) and \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) for any \(x \in S\). Since \(\mathcal{LI}^{-1}=\mathcal{RZ}\) and \(\mathcal{RI}^{-1}=\mathcal{LZ}\), \(\mathcal{RZ}(x)=\mathcal{RZ}'(x)\) and \(\mathcal{LZ}(x)=\mathcal{LZ}'(x)\) for any \(x \in S\). Therefore \(x \in \mathcal{LI}'(x) \cup \mathcal{LZ}'(x)=\mathcal{RI}'(x) \cup \mathcal{RZ}'(x)\) for any \(x \in S\), and \(S\) is a band with the property that \(ab=ba\) or \(aba=a\) and \(bab=b\) for any \(a,b \in S\) by Lemma 17.
Let \(a,b \in S\). Then \(a\) and \(b\) are idempotents in \((S,\star)\) and are either generalized inverses or commute in \((S,\star)\), and \(a\) and \(b\) are idempotents in \((S,\star')\) and are either generalized inverses or commute in \((S,\star')\).
Since \((S,\star)\) is a band, \(a \star (a \star b) = (a \star a) \star b = a \star b\), which implies \(a \star' (a \star b) = a \star b\) because \(\mathcal{LI}(x)=\mathcal{LI}'(x)\) (\(\mathcal{RZ}(x)=\mathcal{RZ}'(x)\)) for any \(x \in S\). Since \((S,\star)\) is a band, \((a \star b) \star b = a \star (b \star b) = a \star b\), which implies \((a \star b) \star' b = a \star b\) because \(\mathcal{RI}(x)=\mathcal{RI}'(x)\) (\(\mathcal{LZ}(x)=\mathcal{LZ}'(x)\)) for any \(x \in S\). Similarly since \((S,\star')\) is a band, \(a \star (a \star' b) = a \star' b\) and \((a \star' b) \star b = a \star' b\).
If \(a \star b \star a = a\), \[\begin{align} (a \star b) \star a &= a &\text{associativity of \star} \\ (a \star b) \star' a &= a &\text{\mathcal{LI}(x)=\mathcal{LI}'(x) (\mathcal{RZ}(x)=\mathcal{RZ}'(x)) for any x \in S} \\ (a \star b) \star' (a \star' b) &= a \star' b &\text{\star' b both sides, associativity of \star'} \\ (a \star b) \star (a \star' b) &= a \star' b &\text{\mathcal{LI}(x)=\mathcal{LI}'(x) (\mathcal{RZ}(x)=\mathcal{RZ}'(x)) for any x \in S.} \end{align}\] If \(b \star a \star b = b\), \[\begin{align} b \star (a \star b) &= b &\text{associativity of \star} \\ b \star' (a \star b) &= b &\text{\mathcal{RI}(x)=\mathcal{RI}'(x) (\mathcal{LZ}(x)=\mathcal{LZ}'(x)) for any x \in S} \\ (a \star' b) \star' (a \star b) &= a \star' b &\text{a \star' both sides, associativity of \star'} \\ (a \star' b) \star (a \star b) &= a \star' b &\text{\mathcal{RI}(x)=\mathcal{RI}'(x) (\mathcal{LZ}(x)=\mathcal{LZ}'(x) for any x \in S.} \end{align}\] If \(a\) and \(b\) are generalized inverses of each other in \((S,\star)\), then \(a \star b\) and \(a \star' b\) commute in \((S,\star)\) and \((S,\star')\), and their product in \((S,\star)\) and \((S,\star')\) is \(a \star' b\). Similarly, if \(a\) and \(b\) are generalized inverses of each other in \((S,\star')\), then \(a \star b\) and \(a \star' b\) commute in \((S,\star)\) and \((S,\star')\), and their product in \((S,\star)\) and \((S,\star')\) is \(a \star b\). Therefore, if \(a\) and \(b\) are generalized inverses of each other in \((S,\star)\) and \((S,\star')\), then \(a \star b = a \star' b\).
If \(a \star b = b \star a\), then \(a \star b \star a = a \star b\) and \(b \star a \star b = a \star b\). \[\begin{align} (a \star b) \star a &= a \star b &\text{associativity of \star} \\ (a \star b) \star' a &= a \star b &\text{\mathcal{RI}(x)=\mathcal{RI}'(x) (\mathcal{LZ}(x)=\mathcal{LZ}'(x)) for any x \in S} \\ (a \star b) \star' (a \star' b) &= (a \star b) \star' b &\text{\star' b both sides, associativity of \star'} \\ (a \star b) \star' (a \star' b) &= a \star b &\text{(a \star b) \star' b = a \star b} \\ (a \star b) \star (a \star' b) &= a \star b &\text{\mathcal{RI}(x)=\mathcal{RI}'(x) (\mathcal{LZ}(x)=\mathcal{LZ}'(x)) for any x \in S.} \end{align}\] Also, \[\begin{align} b \star (a \star b) &= a \star b &\text{associativity of \star} \\ b \star' (a \star b) &= a \star b &\text{\mathcal{LI}(x)=\mathcal{LI}'(x) (\mathcal{RZ}(x)=\mathcal{RZ}'(x)) for any x \in S} \\ (a \star' b) \star' (a \star b) &= a \star' (a \star b) &\text{a \star' both sides, associativity of \star'} \\ (a \star' b) \star' (a \star b) &= a \star b &\text{a \star' (a \star b) = a \star b} \\ (a \star' b) \star (a \star b) &= a \star b &\text{\mathcal{LI}(x)=\mathcal{LI}'(x) (\mathcal{RZ}(x)=\mathcal{RZ}'(x)) for any x \in S.} \end{align}\] If \(a\) and \(b\) commute in \((S,\star)\), then \(a \star b\) and \(a \star' b\) commute in \((S,\star)\) and \((S,\star')\), and their product in \((S,\star)\) and \((S,\star')\) is \(a \star b\). Similarly, if \(a\) and \(b\) commute in \((S,\star')\), then \(a \star b\) and \(a \star' b\) commute in \((S,\star)\) and \((S,\star')\), and their product in \((S,\star)\) and \((S,\star')\) is \(a \star' b\). Therefore, if \(a\) and \(b\) commute in both \((S,\star)\) and \((S,\star')\), then \(a \star b = a \star' b\).
Consider the case where \(a\) and \(b\) are generalized inverses of each other in \((S,\star)\) and commute in \((S,\star')\). Then \((a \star b) \star' a = a\) because \(a \star b \star a = a\), and \((a \star b) \star (a \star' b) = (a \star b) \star' (a \star' b) = a \star' b\) because \(a\) and \(b\) are generalized inverses of each other in \((S,\star)\). \[\begin{align} (a \star b) \star' b &= a \star b &\text{(S,\star) is a band, \mathcal{RI}(x)=\mathcal{RI}'(x) (\mathcal{LZ}(x)=\mathcal{LZ}'(x)) for any x \in S} \\ (a \star b) \star' (b \star' a) &= (a \star b) \star' a &\text{\star' a both sides, associativity of \star'} \\ (a \star b) \star' (a \star' b) &= a &\text{a and b commute in (S,\star'), (a \star b) \star' a=a} \\ (a \star b) \star (a \star' b) &= a &\text{(a \star b) \star (a \star' b) = (a \star b) \star' (a \star' b)} \\ (a \star b) \star ((a \star' b) \star b) &= a \star b &\text{\star b both sides, associativity of \star} \\ (a \star b) \star (a \star' b) &= a \star b &\text{(a \star' b) \star b = a \star' b.} \end{align}\] Therefore \(a \star b = a \star' b\).
The case where \(a\) and \(b\) commute in \((S,\star)\) and are generalized inverses of each other in \((S,\star')\) is implied to be true from the previous case by switching the roles of \(\star\) and \(\star'\). Therefore \(S\) is a stabilized semigroup with respect to \(\mathcal{LI}\) and \(\mathcal{RI}\) [\(\mathcal{LZ}\) and \(\mathcal{RZ}\)]. ◻
Theorem 21. Let \(S\) be a commutative-rectangular band. For any \(a,b \in S\), \(ab\) is the unique element \(c \in S\) such that for \(T=\mathcal{RZ}(a) \cap \mathcal{LI}(a) \cap \mathcal{LZ}(b) \cap \mathcal{RI}(b)\) and \(U=\mathcal{RZ}(a) \cap \mathcal{LZ}(a) \cap \mathcal{LZ}(b) \cap \mathcal{RZ}(b)\),
if \(T \neq \emptyset\), then \(\{c\} = T\), and
if \(T = \emptyset\), then \(\mathcal{LI}(c) = \displaystyle\cap_{y \in U} \mathcal{LI}(y)\) and \(\mathcal{RI}(c) = \displaystyle\cap_{y \in U} \mathcal{RI}(y)\).
Proof. Since \(S\) is a commutative-rectangular band, either \(a,b \in S\) are generalized inverses of each other or they commute. If \(a\) and \(b\) are generalized inverses of each other, then by Proposition 10 (or Lemma 14), \(\{ab\} = {\boldsymbol{R}Z}(a) \cap {\boldsymbol{L}Z}(b) = T\).
If \(a\) and \(b\) are not generalized inverse of each other, then \(T = \emptyset\) by Proposition 10 (or Lemma 14), and \(a\) and \(b\) commute. By Proposition 15, \(ab\) is the unique element \(c \in S\) such that \(\mathcal{LI}(c) = \displaystyle\cap_{y \in U} \mathcal{LI}(y)\) and \(\mathcal{RI}(c) = \displaystyle\cap_{y \in U} \mathcal{RI}(y)\). ◻