Orthodox semigroups and permutation matchings

TL;DR

This paper characterizes when orthodox semigroups admit permutation matchings to their inverses, showing such permutations can be involutions, with effective conditions for finite cases and applications to class characterization.

Contribution

It provides necessary and sufficient conditions for the existence of permutation matchings in orthodox semigroups, including finite cases, and characterizes certain semigroup classes via these matchings.

Findings

Permutation matchings exist under specific conditions involving Green's relations.

Such permutations can always be chosen as involutions.

Finite orthodox semigroups have an effective criterion for permutation matchings.

Abstract

We determine when an orthodox semigroup S has a permutation that sends each member of S to one of its inverses and show that if such a permutation exists, it may be taken to be an involution. In the case of a finite orthodox semigroup the condition is an effective one involving Green's relations on the combinatorial images of the principal factors of S. We also characterise some classes of semigroups via their permutation matchings.

Equations

$$(a^2=abab=ab)\Rightarrow (a^3=aba=a).$$

$$\varphi :S\setminus \{0\}\to \{(i,j):1\le i,j\le k\}$$

$$(|V_{i,j}|=|V_{j,i}|)\Leftrightarrow(m_{j}n_{i}=m_{i}n_{j})\Leftrightarrow\big{(}\frac{m_{i}}{n_{i}}=\frac{m_{j}}{n_{j}}\big{)}$$

Full text

Dept of Mathematical Sciences, University of Essex

Wivenhoe Park, Colchester, UK, CO4 3SQ

email: peteh@essex.ac.uk

1 Introduction and Background

In [2] the author introduced the topic of *permutation matchings, *which are permutations on a regular semigroup that map each element to one of its inverses. The main results of this note are in Section 3 where we characterise the class of orthodox semigroups that have a permutation matching and show that every orthodox semigroup with a permutation matching possesses an involution matching. In Section 2 we detail how some classes of semigroups may be characterised by their permutation matchings.

Following the text of Howie [4] and the author’s [3] we denote the set of idempotents of a semigroup $S$ by $E(S)$. We shall write $(a,b)\in V(S)$ if $a$ and $b$ are mutual inverses in $S$ and denote this as $b\in V(a)$ so that $V(a)$ is the set of inverses of $a\in S$. We extend the notation for inverses to sets $A$: $V(A)=\bigcup_{a\in A}V(a)$. Standard results on Green’s relations, particularly those stemming from Green’s Lemma, will be assumed (Chapter 2 of [4], specifically Lemma 2.2.1) and indeed basic facts and definitions concerning semigroups that are taken for granted in what follows are all to be found in [3,4] . Recall that a regular semigroup $S$ is orthodox if $E(S)$ forms a subsemigroup of $S$. We say that a semigroup $S$ is *combinatorial *(or aperiodic) if Green’s ${\cal H}$-relation on $S$ is trivial. A completely [math]-simple combinatorial semigroup is known as a [math]-rectangular band. When $S$ is a finite semigroup, we shall write $a^{\omega}$ for the unique idempotent power of $a\in S$. We shall also write $a^{\omega+1}$ to denote $a^{\omega}a$ and let $a^{\omega-1}$ stand for the least positive power $a^{k}$ of $a$ such that $a^{k+1}=a^{\omega}$. Following [1] we say that a member $a\in S$ is co-regular if there exist $b\in S$ such that $a=aba=bab$ and $S$ is co-regular if all of its members have that property. Co-regularity of $a$ is equivalent to saying that $a$ is self-inverse (which is the term we shall henceforth use) for if that is the case then $a$ satisfies $a=a^{3}$ and we may take $b=a$ in order to satisfy the definition of co-regularity. On the other hand given that $a$ is co-regular then we have by the given equations that $ab=abab$ and so

[TABLE]

Let $C=\{A_{i}\}_{i\in I}$ be any finite family of finite sets (perhaps with repetition of sets). A set $\tau\subseteq\bigcup A_{i}$ is a *transversal *of $C$ if there exists a bijection $\phi:\tau\rightarrow C$ such that $t\in\phi(t)$ for all $t\in\tau$. We assume Hall’s Marriage Lemma in the form that $C$ has a transversal if and only if *Hall’s Condition *is satisfied, which says that for all $1\leq k\leq|I|$, the union of any $k$ sets from $C$ has at least $k$ members. For this and related background see, for example, the text [6].

**Definitions 1.1 **Let $S$ be a regular semigroup and let $F=\{f\in T_{S}:f(a)\in V(a)\}$. We call $F$ the set of *inverse matchings *of $S$. We call $f\in F$ a *permutation matching *if $f$ is a permutation of $S$; more particularly $f$ is an involution matching if $f^{2}=\iota$, the identity mapping.

We shall often denote a matching simply by $(\cdot)^{\prime}$, so that the image of $a$ is $a^{\prime}$. When discussing an involution $f$ we may sometimes write $a\leftrightarrow b$ to indicate that $f(a)=b$ and $f(b)=a$. In general the inverse $f^{-1}$ of a permutation matching $f$ is also a permutation matching, and so if a semigroup has a unique permutation matching $f$, then $f$ must necessarily be an involution matching. Permutation matchings are not however closed under composition of permutations.

When using the following theorem we work with the family of subsets of $S$ given by $V=\{V(a)\}_{a\in S}$. The members of $V$ may have repeated elements—for example $S$ is a rectangular band if and only if $V(a)=S$ for all $a\in S$. However, for the purposes of the next result we consider the members of $V$ to be marked by the letter $a$, so that $V(a)$ is an unambiguous member of $V$ (strictly, we are using the pairs $\{a,V(a)\},$ $(a\in S)$). We summarise some results of [2].

**Proposition 1.2 **For a finite regular semigroup $S$ the following are equivalent:

(i) $S$ has a permutation matching;

(ii) $S$ is a transversal of $V=\{V(a)\}_{a\in S}$;

(iii) $|A|\leq|V(A)|$ for all $A\subseteq S$.

**Remark 1.3 **In general, a finite regular semigroup may have no permutation matchings. A minimal example is given by the $7$-element orthodox [math]-rectangular band $B=\{(i,j):1\leq i\leq 2,\,1\leq j\leq 3\}\cup\{0\}$, where $E(B)=\{(1,2),(1,3),(2,1)\}\cup\{0\}$. Then $V\{(2,2),(2,3)\}=\{(1,1)\}$ and so Hall’s Condition is violated, and by Proposition 1.2, $B$ has no permutation matching. The class of finite regular semigroups with a permutation matching is closed under the taking of direct products but not of homomorphic images or regular subsemigroups [2, Proposition 1.5.]

**Theorem 1.4 [see 2, Theorem 1.6] **The following are equivalent for a finite regular semigroup $S$:

(i) $S$ has a permutation matching;

(ii) $S$ has a permutation matching that preserves the ${\cal H}$-relation (meaning that $\alpha{\cal H}\beta\Rightarrow\alpha^{\prime}{\cal H}\beta^{\prime}$);

(iii) each principal factor $D_{a}\cup\{0\}$ $(a\in S)$ has a permutation matching;

(iv) each [math]-rectangular band B=D_{a}\cup\{0\}/{\cal H}$$(a\in S) has a permutation matching.

Remarks 1.5 We may re-write the proof of Theorem 1.4 as given in [2], replacing the word ‘permutation’ by ‘involution’ to recover the implications ((i) $\Leftrightarrow$ (iii)) $\Leftarrow$ ((ii) $\Leftrightarrow$ (iv)). However the missing forward implication has not been proved and remains an open question.

Proof (i) $\Leftrightarrow$ (iii) Suppose that ${}^{\prime}:S\mapsto S$ is an involution matching of $S$ and let $D$ denote a ${\cal D}$-class of $S$. For each $a\in D$ we have $V(a)\subseteq D$ so that ${}^{\prime}|_{D}$ is an involution of $D$; extending this by $0\mapsto 0$ gives an involution matching for the principal factor $D\cup\{0\}$. Conversely, if each principal factor has an involution matching then the union of these matchings over the set of ${\cal D}$-classes yields an involution matching of $S$.

(ii) $\Rightarrow$(iv) The given involution matching $(\cdot)^{\prime}$ of $S$, when restricted to a ${\cal D}$-class $D$, defines an involution matching on the principal factor $D\cup\{0\}$. In addition we are told that $(\cdot)^{\prime}$ preserves the ${\cal H}$-relation and so $(\cdot)^{\prime}$ induces an involution matching on $B$.

(iv) $\Rightarrow$ (ii) Let $D$ be an arbitrary ${\cal D}$-class of $S$. The given involution matching of $B$ induces an involution on the set of ${\cal H}$-classes within $D$ in such a way that $H_{1}\leftrightarrow H_{2}$ then implies that each $a\in H_{1}$ has a unique inverse $a^{\prime}\in H_{2}$. The mapping defined by $a\leftrightarrow a^{\prime}$ then defines an involution of $H_{1}\cup H_{2}$. Taking the union of these involutions over all such pairs $(H_{1},H_{2})$ then gives a required involution matching of $D$. (Note that $H_{1}=H_{2}$ is possible, in which case $H_{1}$ is a group and the involution $a\leftrightarrow a^{\prime}$ is the unique involution matching of that group.).

(ii) $\Rightarrow$(i) as a special case.

**Theorem 1.6 **[3, Theorem 1.4.18] For a regular semigroup $S$, $V(E^{n})=E^{n+1}$. In particular for an orthodox semigroup, $V(E)=E^{2}=E$.

**Theorem 1.7 **[4, see Theorems 6.2.4 and 6.2.5] Let $S$ be a semigroup. Then $S$ is orthodox if and only if $\{V(a):a\in S\}$ forms a partition of $S$. Let $\gamma$ be the relation on the semigroup $S$ whereby $a\gamma b$ if and only if $V(a)=V(b)$. The relation $\gamma$ is an equivalence relation for any semigroup $S$. If $S$ is orthodox then $\gamma$ is the least inverse semigroup congruence on $S$.

We close this section with several observations based on these results that will be invoked in Section 3.

**Corollary 1.8 **For the orthodox semigroup $S$, the set of inverse sets $I=\{V(a):a\in S\}$ equals the set of classes of the least inverse congruence $\gamma$ of $S$. Moreover for any $a\in S$, $V(V(a))=a\gamma$.

*Proof *For any semigroup $S$ the relation $\gamma$ whereby $a\gamma b$ if $V(a)=V(b)$ is an equivalence relation. If $(a,b)\in V(S)$ then $a\gamma\subseteq V(b)$. It follows that if $S$ is regular, every $\gamma$-class is contained in $V(b)$ for some $b\in S$. If we assume further that $S$ is orthodox we have by Theorem 1.7 that the classes of inverses $I$ also partition $S$. Since each $\gamma$-class is contained in some class of inverses it follows that each class of inverses $V(a)$ is the union of $\gamma$ classes. However if $b,c\in V(a)$ then $a\in V(b)\cap V(c)$ whence $V(b)=V(c)$, which is to say $b\gamma c$. Hence each class of inverses $V(a)$ consists of exactly one $\gamma$ class. Therefore the two partitions of $S$ are identical: $\{a\gamma:\,a\in S\}=\{V(a):\,a\in S\}$.

For the second statement take any $b\in V(a)$. Then since $a\in V(b)$ it follows that $a\in V(V(a))$. For any $c\in a\gamma$ we also have $c\in V(b)$ and so it follows that $a\gamma\subseteq V(V(a))$. Indeed, since $b\in V(a)$ was arbitrary, it follows that $a\gamma\subseteq\cap_{b\in V(a)}V(b)$. However since $S$ is orthodox, the distinct classes in this intersection are pairwise disjoint. Since $a\gamma\neq\emptyset$ it follows that all members of this intersection are equal and so $V(V(a))$ is itself a single class of inverses. Since $a\gamma$ is a member of $I$ that is contained in $V(V(a))$, it follows that $V(V(a))=a\gamma$, as required.

Remark 1.9: It follows from Theorem 1.7 and Corollary 1.8 that for any member $a$ of an orthodox semigroup $S$, either $a\gamma\cap V(a)=\emptyset$ or $a\gamma=V(a)$, the latter occurring exactly when $a=a^{3}$. In this case all members $b\in V(a)=a\gamma$ are self-inverse for it follows that $b\gamma=a\gamma$ and so $b\in V(b)$ and $b=b^{3}$ also.

Corollary 1.10 For an orthodox semigroup $S$ the mapping $V:V(a)\mapsto V(V(a))$ is an involution on the set $I=\{V(a):a\in S\}$. Moreover the fixed points of this involution are exactly the classes $V(a)$ where $a=a^{3}$.

*Proof *By Corollary 1.8, any member of $I$ has the form $a\gamma$ $(a\in S)$ and conversely each class $a\gamma\in I$. Hence $V(V(a))=a\gamma$ shows that $V$ is a mapping from $I$ into $I$. Moreover $V(V(a\gamma))=V(V(a))=a\gamma$ shows that $V$ does indeed define an involution on $I$. By Remark 1.9, $V(a\gamma)=a\gamma$ if and only if $a$ is self-inverse, which occurs if and only if every member of $a\gamma$ is self-inverse. Hence $V(a)$ is fixed by our involution if and only if $a=a^{3}$.

2 Characterisation of classes of regular semigroups by their permutation

matchings

In this section we show how classes of finite regular semigroups can be characterised by their permutation matchings. We begin with *completely regular *semigroups, which are those that are unions of groups.

**Theorem 2.1 **Let $S$ denote a finite regular semigroup.

(i) $S$ is completely regular if and only if the function $f(x)=x^{\omega-1}$ $(x\in S)$ is a permutation matching;

(ii) $S$ is completely simple if and only if $f(x)=x^{\omega-1}(xyx)^{\omega}$ $(x,y\in S)$ is a permutation matching;

(iii) $S$ is a group if and only if $f(x)=y^{\omega}x^{\omega-1}y^{\omega}$ $(x,y\in S)$ is a permutation matching;

(iv) $S$ satisfies $x=x^{k+2}$ $(k\geq 1)$ if and only if $f(x)=x^{k}$ $(x\in S)$ is a permutation matching; in particular $S$ is co-regular if and only if the identity function is a permutation matching;

(v) $S$ is a rectangular band if and only if every permutation of $S$ is a permutation matching.

Remark: In (ii) and (iii) the function $f$ is in general a function of the two variables $x$ and $y$ but, under the hypothesis of each part, the output of $f$ depends on $x$ alone. We record proofs for just the first two statements, the others being straightforward and similar.

*Proof *(i) Given that $S$ is a union of groups, for any $x\in S$ we have $xx^{\omega-1}x=x^{\omega}x=x$ as $x^{\omega}$ is the identity of the group $H_{x}$; similarly we obtain $x^{\omega-1}xx^{\omega-1}=x^{\omega-1}x^{\omega}=x^{\omega-1}$. Since $x$ and $y=x^{\omega-1}$ each have no other inverse in the group $H_{x}=H_{y}$ it follows that $f(x)$ is a permutation matching, indeed $f(x)$ is an involution matching. Conversely, given that $f(x)$ is a permutation matching of $S$ we have that $(x,x^{\omega-1})$ are mutual inverses in $H_{x}$, which is therefore a group, and so $S$ is a union of groups.

(ii) Given that $S$ is completely simple it follows that for any $x,y\in S$ we have that $(xyx)^{\omega}=x^{\omega}$, the identity element of the group $H_{x}$. Hence $f(x)=x^{\omega-1}(xyx)^{\omega}=x^{\omega-1}$ and as in (i) we have that $x$ and $x^{\omega-1}$ are mutually inverse in the group $H_{x}$ and $f(x)$ is an involution matching. Conversely suppose that $f(x)$ is a permutation matching. If $S$ had more than one ${\cal D}$-class then there would exist ${\cal D}$-classes, $D_{1}$ and $D_{2}$ such that $D_{1}<_{{\cal{\cal D}}}D_{2}$. However if we then take $x\in D_{2}$ and $y\in D_{1}$ we get that $f(x)\not\in D_{2}$, contradicting our assumption that $f(x)\in V(x)$. Hence $S$ is completely simple.

**Theorem 2.2 **An inverse semigroup $S$ has a unique permutation matching, which is an involution. Conversely, any orthodox semigroup $S$ with a unique permutation matching is an inverse semigroup. However, there exists a $5$-element combinatorial semigroup that is not an inverse semigroup, which has a unique permutation matching.

Proof For an inverse semigroup $S$, clearly the mapping whereby $a\mapsto a^{-1}$ is the unique permutation matching of $S$ and is an involution. Conversely, suppose that $S$ is orthodox and has a unique permutation matching $f$. If $S$ is not inverse then there exists two distinct idempotents $e,f\in E=E(S)$ such that $e{\cal G}f$, where ${\cal G}$ denotes either of Green’s relations ${\cal L}$ or ${\cal R}$. Since $S$ is orthodox, it follows by Theorem 1.6 that $f(E)=E$ and $f(S\setminus E)=S\setminus E$. However these equations now allow us to construct two distinct permutation matchings $g,h$ of $S$ as follows. Put $g|_{(S\setminus E)}=h_{|(S\setminus E)}=f_{|(S\setminus E)}$ and put $g_{|E}=\iota_{|E}$ the identity mapping on $E$, while $h$ is identical to $g$ except that $h(e)=f$ and $h(f)=e$. This contradicts the uniqueness of matchings and so it follows that if $S$ is orthodox with a unique permutation matching then $S$ is inverse and that matching is the standard involution by inverses.

Next consider the $5$-element combinatorial semigroup $S=D\cup\{0\}$ where the $2\times 2$ ${\cal D}$-class $D=\{(1,1),(1,2),(2,1),(2,2)\}$ consists of elements that are all idempotent except $(1,2)$. This semigroup is regular but not orthodox as the idempotent $(2,1)$ is the unique inverse of the non-idempotent $(1,2).$ However $S$ has a unique permutation matching $f$: for any permutation matching $f$ we necessarily have $(1,2)\mapsto(2,1)$; we can now complete an involution matching $f$ by saying that under $f$, $(2,1)\mapsto(1,2)$ while the three idempotents $0,\,(1,1)$, and $(2,2)$ are fixed by $f$. What is more, by inspection we see there is no alternative to this definition for $f$. Therefore $S$ possesses a unique permutation matching but $S$ is not an inverse semigroup.

**Remark 2.3 **We observe that if $S$ is any semigroup that is not inverse but has a unique permutation matching $f$, then $S$ contains the ${\cal D}$-class structure of $D$ as given in the example of the previous proof. To see this, note that by the argument of the proof of Theorem 2.2, the uniqueness of $f$ ensures that $f(E)\not\neq E$ so there exists $e\in E=E(S)$ and $a\in S\setminus E$ such that $(e,a)\in V$. Hence $(e,a)\not\in{\cal G}$ in $S$ and so $ea=f\in E$ with $e{\cal R}f$ and $ae=g\in E$ with $e{\cal L}g$ and $e,f,g$ are pairwise distinct. Then $gf=ae\cdot ea=aea=a$ and so $\{e,f,g,a\}$ has the ${\cal D}$-class structure of $D$ above, meaning that $e,f,g\in E$, $a\not\in E$ and $a{\cal R}g{\cal L}e{\cal R}f{\cal L}a$ and $gf=a,\,fg\not\in D$, $(a,e)\in V(S)$.

3 Orthodox semigroups

Theorem 3.1 An orthodox semigroup $S$ has a permutation matching if and only if $|V(V(a)|=|V(a)|$ for all $a\in S$. If orthodox $S$ has a permutation matching then in fact $S$ possesses an involution matching.

Proof We prove that the condition that $S$ has a permutation matching implies the stated condition on cardinalities of sets of inverses and that that condition in turn allows the construction of an involution matching for $S$. From this follows the claims in the statement of the theorem.

To this end suppose that $S$ has a permutation matching $(\cdot)^{\prime}$. By Corollaries 1.8 and 1.10 the mapping whereby $V(a)\mapsto V(V(a))$ is an involution on the set $I=\{V(a):a\in S\}=\{a\gamma:a\in S\}$. For any $a\gamma\in I$ we have $(a\gamma)^{\prime}\subseteq V(a)$ and since $(\cdot)^{\prime}$ is one-to-one, it follows that $|a\gamma|\leq|V(a)|$. Similarly $(V(a))^{\prime}\subseteq V(V(a))=a\gamma$ by Corollary 1.10, and so $|V(a)|\leq|a\gamma|$. It follows that $|a\gamma|=|V(a)|$. Since any pair of mutual images under $V:I\rightarrow I$ has the form $(a\gamma,V(a\gamma)=V(a))$ it follows that the members of each pair of mutual images under the involution $V:I\rightarrow I$ are equi-cardinal, which is to say that $|V(V(a)|=|V(a)|$ for all $a\in S$.

Let the classes of $I=\{V(a):a\in S\}$ that are fixed by the involution $V$ of $I$ be $\{I_{i}\}_{i\in R}$ for some index set $R$, and write the remaining classes of $I$ in ordered pairs $\{(J_{i},K_{i})\}_{i\in T},$ for some index set $T$, where $J_{i}\leftrightarrow K_{i}$ under the involution $V$. Recall that $J_{i}\cap K_{i}=\emptyset$ and each pair $(a,b)\in J_{i}\times K_{i}$ is a pair of mutual inverses. The stated condition on cardinalities of inverse sets says that $|J_{i}|=|K_{i}|$ for each $i\in T$. We may now define an involution matching $f$ on $S$ as follows. Since the members of each pair of elements of any $I_{i}$ are mutually inverse, we may take $f|_{I_{i}}$ to be any involution mapping of $I_{i}$ (for example, the identity mapping on $I_{i}$). Since $J_{i}$ and $K_{i}$ are equi-cardinal, for each $J_{i}$ we may let $f|_{J_{i}}$ be any bijection of $J_{i}$ onto $K_{i}$ and let the $f|_{K_{i}}$ be the corresponding inverse bijection. The mapping $f:S\rightarrow S$ is then the union of all these restriction mappings on the members of $I$. By construction $f$ is an involution that maps each member of $S$ to one of its inverses and so $f$ is an involution matching of $S$, as required.

We next use Theorem 3.1 to determine when a finite orthodox semigroup $S$ has a permutation matching and to effectively find all such matchings when they exist. The result hinges on the special case where $S$ is a finite orthodox [math]-rectangular band so to this end let $S$ denote a finite $m\times n$ orthodox [math]-rectangular band (meaning the number of ${\cal R}$- and ${\cal L}$-classes in $D=S\setminus\{0\}$ is $m$ and $n$ respectively) with band of idempotents $E(S)=B$. The non-zero members of $S$ can therefore be taken to be the set of ordered pairs $S\setminus\{0\}=\{(i,j):\,1\leq i\leq m,\,1\leq j\leq n\}$. This structure for $S$ is assumed in Lemmas 3.2, 3.3 and 3.4.

**Lemma 3.2 **The set of inverses $V(e)$ of $e\in B$ is a maximal rectangular subband of $S$.

*Proof *Consider the minimum inverse congruence $\gamma$ on $S$. Then $(e\gamma)^{2}=e^{2}\gamma=e\gamma$ and so from Remark 1.9 we obtain $(V(e))^{2}=V(e)$, which is to say that $V(e)$ is a subsemigroup of $S$ and by Theorem 1.6 it follows that $V(e)$ is a band. For any $a$ such that $(a,e)\in V$ we have that $e\in V(e)\cap V(a)$ and so by Theorem 1.7 we infer that $V(a)=V(e)$. It follows that for any $a,b\in V(e)$ we have $b\in V(a)$ also and so $aba=a$. Therefore $V(e)$ is a rectangular subband of $S$.

Next let $V(e)\subseteq U$, where $U$ is a rectangular subband of $S$. Let $e{\cal G}f$ in $U$, whence $e{\cal G}f$ in $S$. Then $f\in V(e)\cap V(f)$ so that $V(e)=V(f)$. Take any $u\in U$. Since $U$ consists of a single ${\cal D}$-class and $V(e)\subseteq U$ it follows that there exists $f\in U$ such that, in $U$, $e{\cal R}f{\cal L}u$. By the previous argument this gives first that $f\in V(e)$ and then in turn $u\in V(e)$. Therefore $U\subseteq V(e)$ and so we conclude that $U=V(e)$, which is to say that $V(e)$ is indeed a maximal rectangular subband of $S$.

Denote the pairwise distinct maximal rectangular subbands formed by the sets of inverses $V(e)$ $(e\in B\setminus\{0\})$ by $V(e_{1}),V(e_{2}),\cdots,V(e_{k})$ say for suitable fixed representatives $e_{i}$ of each of the rectangular subbands $V(e).$

**Lemma 3.3 **Each ${\cal G}$-class of $D=S\setminus\{0\}$ meets exactly one of the maximal rectangular subbands $V(e)$.

*Proof *By symmetry it is enough to take the case ${\cal G}={\cal R}$ so let $a\in D$, consider the class $R_{a}$ and take any $e\in B\cap R_{a}$ so that $R_{a}$ meets the maximal rectangular subband $V(e)$. Suppose that $V(e_{i})$ is a maximal rectangular subband of $D$ such that $R_{a}\cap V(e_{i})\neq\emptyset$. Then any $f\in R_{a}\cap V(e_{i})$ is idempotent, $e{\cal R}f$ and so $f\in V(e)$. However we then have $f\in V(e_{i})\cap V(e)$ so that $V(e_{i})=V(e)$, thus completing the proof.

We may now arrange the ‘egg-box’ diagram of $D$ so that the maximal rectangular subbands, $U_{1}=V(e_{1}),U_{2}=V(e_{2}),\cdots$ form a leading diagonal of the diagram as follows. We may list the ${\cal{\cal L}}$-classes of $D$ (and similarly the ${\cal R}$-classes) by first listing all the ${\cal{\cal L}}$-classes of $D$ that meet $U_{1}$, then of $U_{2}$ and so on through to those of $U_{k}$, as, by Lemma 3.3, these orderings of the ${\cal L}$- and the ${\cal R}$-classes are well-defined. Let $U$ denote the set of members of $D$ consisting of all the intersections of the ${\cal L}$-classes in $S$ that meet $U_{i}$ with the ${\cal R}$-classes in $S$ that meet $U_{i}$. Clearly $U_{i}\subseteq U$ but the reverse inclusion is also true for take any $u\in U$ so that $\{u\}=R_{f}\cap L_{g}$ say where $f,g\in U_{i}=V(e_{i})$. Since $V(e_{i})$ is a subsemigroup of $S$ it follows that $fg\in U_{i}$; in particular $fg\neq 0$ so that $u=fg\in U_{i}$. Therefore the union of intersections of the ${\cal L}$- and ${\cal R}$-classes of $U_{i}$ forms the maximal rectangular subband $U_{i}$ itself.

Let the number of ${\cal L}$- and ${\cal R}$-classes of $V(e_{i})$ be denoted by $m_{i}$ and $n_{i}$ respectively $(1\leq i\leq k)$. Define a mapping

[TABLE]

where $\phi(a)=(i,j)$ if $V(e_{i})$ and $V(e_{j})$ are the maximal rectangular subbands of $S$ that meet $R_{a}$ and $L_{a}$ respectively.

**Lemma 3.4 **Let $a\in S\setminus\{0\}$ with $\phi(a)=(i,j)$. Then

(i) $V(a)=\{b\in S:\,\phi(b)=(j,i)\}=V(\{a\in S:\,\phi(a)=(i,j)\})$;

(ii) Let the set defined in (i) be denoted by $V_{i,j}$. Then $V(V_{i,j})=V_{j,i}$.

(iii) $|V(a)|=m_{j}n_{i}$.

Proof (i) and (ii). In general, $b\in V(a)$ if and only if there exist $e,f\in E$ such that $a{\cal R}e{\cal L}b{\cal R}f{\cal{\cal L}}a$. Since $\phi(a)=(i,j)$ this implies that $e\in V(e_{i}),\,f\in V(e_{j})$, whence $\phi(b)=(j,i)$. Conversely if $\phi(b)=(j,i)$ then for some $e\in V(e_{j})$ and $f\in V(e_{i})$ we have $f{\cal L}b{\cal R}e$. Let $\{g\}=L_{f}\cap R_{a}$ and $\{h\}=R_{e}\cap L_{a}$. Now $a{\cal R}k$ for some $k\in V(e_{i})$ so that $g=kf\in V(e_{i})$ and similarly $h\in V(e_{j})$. Therefore we have $a{\cal R}g{\cal L}b{\cal R}h{\cal L}a$ and since $g,h\in E(S)$ we conclude that $b\in V(a)$. It now follows that for any $a_{1},a_{2}\in D$, $V(a_{1})=V(a_{2})$ if and only if $\phi(a_{1})=\phi(a_{2})$, whence the second equality in (i) now follows, while (ii) is a re-statement of the second equality in (i).

(iii) By (i) we have $|V(a)|=|\{b\in S:\,\phi(b)=(j,i)\}|$. To say that $\phi(b)=(j,i)$ means that $R_{b}\cap V(e_{j})\neq\emptyset$ and $L_{b}\cap V(e_{i})\neq\emptyset$. This in turn gives $m_{j}$ choices for $R_{b}$ and $n_{i}$ choices for $L_{b}$. Since $b$ is determined by such a pair of choices, which can be made independently of each other, it follows that $|V(a)|=m_{j}n_{i}$, as claimed.

**Definition 3.5 **Let $U_{1}$ and $U_{2}$ be finite rectangular bands, let $m_{i}$ (resp. $n_{i})$ be the respective number of ${\cal R}$-classes and ${\cal L}$-classes of $U_{i}$ $(i=1,2)$. We shall say that $U_{1}$ and $U_{2}$ are *similar *if $\frac{m_{1}}{n_{1}}=\frac{m_{2}}{n_{2}}$.

**Theorem 3.6 **Let $S$ be a finite orthodox [math]-rectangular band. Then $S$ has a permutation matching if and only if the maximal rectangular subbands of $D=S\setminus\{0\}$ are pairwise similar.

Proof Using the notation of Lemma 3.4, we note that the sets $V_{i,j}$ partition $D$ into blocks and, by Lemma 3.4(i) and (ii), two members $a,b\in D$ lie in the same block $V_{i,j}$ if and only if $V(a)=V(b)$. Suppose that $S$ has a permutation matching $f$. It follows from Lemma 3.4(ii) that $f_{|V_{i,j}}$ is a bijection onto $V_{j,i}$ and in particular that $|V_{i,j}|=|V_{j,i}|$ for any pair $(i,j)$ $(1\leq i,j\leq k)$. Hence from Lemma 3.4(iii) we now infer:

[TABLE]

which is equivalent to the condition that the members of each pair of maximal rectangular subbands $U_{i}$ and $U_{j}$ of $D$ are similar.

Conversely, given that the members of each such pair of rectangular subbands in $D$ are similar, it follows that (1) holds for each pair $(i,j)$. It now follows from Lemma 3.4 and Theorem 3.1 that $S$ has a permutation matching. Indeed all such permutation matchings may be constructed as the union of any chosen bijections ${}^{\prime}:V_{i,j}\rightarrow V_{j,i}$ $(1\leq i,j\leq k)$. The involution matchings correspond to the case where the members of each pair of bijections between sets $V_{i,j}$ and $V_{j,i}$ are mutual inverses.

**Theorem 3.7 **Let $S$ be a finite orthodox semigroup. Then $S$ has a permutation matching if and only if for each [math]-rectangular band D_{a}\cup\{0\}/{\cal H}$$(a\in S) the maximal rectangular subbands are pairwise similar.

*Proof *This follows from Theorem 1.4 (i) $\Leftrightarrow$(iv) and Theorem 3.6.

It remains an open question as to whether there exists a finite regular semigroup (necessarily non-orthodox) that has a permutation matching but no involution matching. In particular, it was shown in [2, Theorem 2.12] that any finite full transformation semigroup possesses a permutation matching but, as yet, no involution matching has been identified for $T_{n}$. In [5] it was shown that $T_{n}$ is covered by its inverse subsemigroups (also see [3, Chapter 6.2]) but that fact does not in itself immediately yield an involution matching for $T_{n}$.