Factorizations

7/25/2019 Factorizations

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Semigroup Forum OF1OF13c 2004 Springer

DOI: 10.1007/s00233-004-0145-x

RESEARCH ARTICLE

On Factorisations and Generators inTransformation Semigroups

Gonca Ayk , Hayrullah Ayk and John M. Howie

Communicated by John B. Fountain

Abstract

It is shown that the classical decomposition of permutations into disjoint

cycles can be extended to more general mappings by means of path-cycles, andan algorithm is given to obtain the decomposition. The device is used to obtaininformation about generating sets for the semigroup of all singular selfmaps ofXn = {1, 2, . . . , n} . Let Tn,r = Sn Kn,r , where Sn is the symmetric groupand Kn,r is the set of maps : Xn Xn such that |im()| r . The smallestnumber of elements ofKn,r which, together with Sn , generate Tn,r is pr(n) ,the number of partitions of n with r terms.

Introduction

The full transformation semigroup TX on a set X, the semigroup analogue ofthe symmetric group, has been much studied over the last fifty years, in boththe finite and infinite cases. Here we are concerned solely with the case whereX=Xn = {1, 2, . . . , n} , and we denote the semigroup TXn of all self maps of

Xn by Tn .The notational difficulties one encounters in analysing maps in Tn are

much greater than those one meets in the study of symmetric groups. Graphsare certainly helpful, but are clumsy on the printed page. Lipscomb (see [7])was the first to develop what one might call a linear notation, and over severalpapers showed the power of his methods. In this paper we describe an alternativeapproach.

In Section 1 we develop a notation for certain primitive elements of Tncalled path-cycles, and describe an algorithm to decompose an arbitrary in Tninto a product of path-cycles. In the following section we use our techniques toobtain information about generators in Tn , and in particular give a new proof ofa theorem in [4], that every map in STn= Tn \ Sn (where Sn is the symmetricgroup) is a composition of 2-paths (idempotents of rank n 1).

The first author is supported by the Scientific and Technical Research Council of Turkey(TUBITAK) with a NATO (A2) Science Fellowship.

The second author is supported by the TUBITAK Basic Science (TBAG) MathematicsWorking Group.


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OF2 Ayk, Ayk, and Howie

IfV is a subsemigroup of a semigroup S, then the relative rank r(S: V)of V in Sis defined by

r(S: V) = min{|G| : G Sand V G= S} .

This has been studied for infinite TX in [2] and [3]. In the final section weconsider a related idea. For 1 r n 1, we consider the semigroupsKn,r={ Tn : |im()| r} , and Tn,r= Kn,rSn , and define thecontingentrank cont(Tn,r) by

cont(Tn,r) = min{|G| : G Kn,r andG Sn= Tn,r} .

It is known that, for Tn,n1 = STn the contingent rank is 1. We show thatcont(Tn,r) = pr(n), the number of partitions of n with r terms.

For undefined terms in semigroup theory, see [5].

1. Path-cycles

Let Xn ={1, . . . , n} and let Tn be the full transformation semigroup on Xn .If, for X= {x1, . . . , xm} Xn , Tn is defined by

x1= x2, . . . , xm1= xm, xm= xr andx = x (x Xn\ X)

where xr X, then is called a path-cycle of length m and period r , or an(m, r)-path-cycle, and is denoted by = [x1, . . . , xm| xr]. Moreover:

if r= m , we say that is an m-pathto xm ;

if m 2 and r= 1, we say that is an m-cycle;

if m= r = 1, we say that is a loop.

An (m, r)-path-cycle with r= 1 is called a properpath-cycle. Notice that thereis exactly one loop, namely the identity map I= [1| 1] .

If = [x1, . . . , xm| xr] is an (m, r)-path-cycle, we define

r1() = {x1, . . . , xm}.

Let and be two path-cycles of Tn . If r1() r1() = then and arecalled disjoint. If r1() r1() consists of one and only one element, then and are called 1-joint.

2. Factorisations

We show that every in Tn is expressible as a product (that is to say, acomposition) of path-cycles, using the following algorithm.


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Ayk, Ayk, and Howie OF3

(A) Decomposition Algorithm. Let Tn be any non-identity map,and let x Xn . Then the sequence (x,x,x2, . . .) must have repe-titions, and so there exists a unique kx N such that x, x, . . . , x

kx

are all distinct, but xkx+1 {x, x, . . . , xkx} . Each path-cycle x =[x, x, . . . , xkx | xkx+1] is called a divisor of . The first factor of , denoted by 1 , is one of the longest divisors of . (If there are two ormore divisors with maximum length, choose the one which has the small-est first entry among them, in the natural order of Xn . In applying thisalphabetical rule, assume that each cycle [x1, . . . , xk | x1] is written sothat x1 = min{x1, . . . , xk} .) Then define

(1) :Xn Xn by

x(1) =

x ifx r1(1)x ifx / r1(1) ,

and call it the first residue of . It is clear that

1(1) = .

Now carry out a similar procedure on (1) , obtaining 2 , the secondfactorof , and (2) , thesecond residueof . Continue this procedureuntil

p= [1 | 1] , (p) = [1 | 1] ,

and = 1. . . p .

The integer p is called the path-cycle rank of and is denoted by pcr(). If = Iis the identity map, then we take I= [1| 1] and pcr(I) = 0. It is clearthat, if Sn , the algorithm gives the standard decomposition into disjointcycles.

Example 1. Let

=

1 2 3 4 5 6 7 8 9 104 7 9 6 7 4 10 4 3 7

.

Then, applying the algorithm (A), we obtain

= [1, 4, 6| 4] [2, 7, 10| 7] [3, 9| 3] [5, 7| 7] [8, 4| 4] .

The decomposition of a mapping into path-cycles is not unique. Forexample, the mapping in Example 1 could be expressed as

[8, 4, 6| 4] [5, 7, 10| 7] [1, 4| 4] [2, 7| 7] [3, 9| 3] .

If, however, we apply the algorithm, observing the alphabetical principle in

parentheses, the decomposition is unique. The number of cycles in the decom-position is certainly unique, and we denote it by cycl(). Thedefect def() isdefined by

def() = |Xn\ im()| .


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Theorem 2. Let Tn . Then

(i) pcr() = def() + cycl() ;

(ii) if1 , 2 , . . ., p are the factors of , then, for all i, j in{1, 2, . . . , p} ,such that i < j , the path-cycles i and j are either disjoint or 1-joint.The latter possibility, with r1(i) r1(j) = {x} , can occur only if

(a) both i and j are proper path-cycles; and

(b) x is not the first element of i , and j is a path to x.

Proof. (i) By definition, if the factor i= [x1, . . . , xk | xr] is a proper path-cycle, then x1 / im(). Each i is either a cycle or a proper path-cycle, andthe number of the latter is precisely def().

(ii) For p= 1 there is nothing to prove. For p 2, consider

i = [y1, . . . , yk | yr] andj = [z1, . . . , zl| zs]

(where 1 i < j p). Since, by definition, (i) fixes all the elementsy1, . . . , yk , the only way in which j can avoid being disjoint from i is forj to be an l -path to zl (that is, for s to be equal to l ) with zl {y1, . . . , yk} .If i is not a cycle, then zl im() and y1 / im , and so we cannot havezl= y1 .

Suppose now that i is a cycle. Writing zl as x , we can supposethat we have written i with x as the first element. Then there is a path-cycle [z1, . . . , zl1, x , y2, . . . , yk | x] longer than i . The algorithm makes thisimpossible.

It is clear that disjoint path-cycles commute with each other. Hence weimmediately have the following corollary.

Corollary 3. The factors i and j commute if and only if either they aredisjoint or they are paths to the same x . In particular they commute if at leastone of i and j is a cycle.

Proof. It remains to prove that if i and j are both proper path-cycles,are not disjoint and are not both paths to the same x , then they do notcommute. Suppose that i < j and i = [y1, . . . , yk | yr], with r = 1, andj = [z1, . . . , zl | zl]. (From Theorem 2 we know that j must be a path.)Again from Theorem 2, we must have zl = ys , with 1 < s k . Nowzl1ij = zl1j = zl , while zl1ji = zli = ysi , and this equals zlonly if i is a path to yk , with yk = zl .

Remark. This is not the place to give a full explanation of the Lipscombapproach, but an example may draw attention to the difference. Let

=

1 2 3 4 5 6 7 8 92 1 1 3 3 7 6 9 9

.


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The graphical representation is

4 5

3

1

2

6

7

8

9

In the Lipscomb notation this becomes

(4 3 1 (5 3 1 (1 2) (6 7) (8 9 (9) ,

where (4 3 1 , (5 3 1 and (8 9 are (non-maximal) proper paths, while (1 2),(6 7) and (9) are circuits. Our notation gives

[4, 3, 1, 2| 1] [5, 3| 3] [6, 7| 6] [8, 9| 9]

and has the important advantage that each of the factors is a mapping in itsown right. It should of course be pointed out that Lipscombs notation wasdeveloped primarily to cope with partialmaps.

3. Generating sets

It is well known (see [4]) that the full singular transformation semigroup ofdegree n , denoted by STn(= Tn\Sn), is generated by the set of all 2-paths

(idempotents of rank n 1) in Tn , and that Tn is generated by the set of all 2-path-cycles ofTn . In this section we give an alternative proof using path-cycles,and give a new generating set for STn and Tn .

Theorem 4. Let STn . Then there exist proper path-cycles 1, . . . , ksuch that = 1. . . k .

Proof. In general, the factors 1, 2, . . . , p of will include some cycles,but there must be at least one proper path-cycle, since is singular. ByCorollary 3, we may suppose that we have re-adjusted the factors so that inthe product 1. . . p all of the cycles come after the first of the proper path-cycles. Let 1 = [x1, . . . , xm | xr] with xr = x1 . If j = [y1, . . . , yq | y1] is acycle factor of , then 1j =1j , where

j = [x1, y1, . . . , yq | y1] .

Thus each of the cycles j can be replaced by the proper path-cycle j , andthe result is proved.


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Theorem 5. The set of all 2-paths in STn generates STn .

Proof. First observe that, for 2 r < m , we have

[x1, . . . , xr, . . . , xm| xr] = [xm, xr1| xr1] [x1, x2, . . . , xm| xm] , (1)

and, for m 3 ,

[x1, . . . , xm1, xm| xm] = [xm1, xm| xm] [x1, x2, . . . , xm1 | xm1] . (2)

The result now follows inductively from (1), (2) and Theorem 4.

Theorem 6. For eachm in{2, . . . , n} , the semigroup STn can be generatedby its path-cycles of length m .

Proof. We show first that every proper path-cycle of lengthk can be writtenas a product of proper path-cycles of length k+ 1 .

Let = [x1, x2, . . . , xk |xr] with r= 1. If k is odd, define

= [x1, x3, . . . , xk2, xk, x2, x4, . . . , xk1, xk+1 | xr]

and

=

[xk+1, xk1, xk2, . . . , x2, x1, xk |xr2] if r 3[xk+1, xk1, xk2, . . . , x2, x1, xk |xk] if r= 2,

so that = . If k is even, define

= [x1, x3, . . . , xk1, xk+1, x2, x4, . . . , xk2, xk | xr]

and

=[xk, xr2, xr3, . . . , x2, x1, xk+1, xk1, xk2, . . . , xr1 | xr2] ifr 3

[xk, xk+1, xk1, xk2, . . . , x2, x1, |xk+1] ifr = 2,

so that = . Therefore, by induction on k , we can write all proper 2-pathsas a product of proper path-cycles of length m , and so the result follows fromTheorem 5.

We can even generate STn by a subset of the set of all path-cycles oflength m .

Theorem 7. For eachm in{2, . . . , n} , the semigroup STn can be generatedby its m-paths.

Proof. Define

= [xm, xm1, . . . , x4, x3, x1, x2 | x2]

= [x1, x3, x4, . . . , xm1, xm, x2 | x2] .

Then = [x1, x2 | x2], and the result now follows from Theorem 5.


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It is well known that the set of all the 2-cycles of Sn generates Sn , andthat the set of all the 3-cycles of An generates An , the alternating group.SinceSTn is an ideal ofTn , it is also clear that every generating set ofTn must

contain a subset which generates Sn .

Theorem 8. If n m 2 and m is even, then the set of all the m-pathsand m-cycles in Tn generates Tn . If n m 3 and m is odd, then the set ofall the m-paths and m-cycles in Tn generates SAn= STn An .

Proof. Let = [x1, x2, . . . , xm| x1] Sn . Then define

= [x1, . . . , xm, xm+1, xm+2 | x1]

= [x1, xm+1, xm, . . . , x2, xm+2 | x1]

= [x1, x3. . . , xm, xm+2, x2, x4, . . . , xm1, xm+1 | x1]

so that

=2 if m is even

if m is odd.

Thus every m-cycle can be written as a product of (m+2) -cycles. In particular,by repetition, every 2-cycle is a product of ((k 1)2 + 2)-cycles, and every 3-cycle is a product of ((k 1)2+ 3)-cycles (k Z+ ). The results follow from thefacts that Sn is generated by its 2-cycles and An is generated by its 3-cycles.

4. Ranks

For 1 r n , let

Kn,r = { Tn : |im()| r} , Tn,r = Kn,r Sn.

Thus Kn,n= Tn,n= Tn,n1 = Tn , and Kn,n1 = STn .It is evident from the well known inequality

def() max{def(), def()} (, Tn) (3)

(see [4]) that both Kn,r and Tn,r are subsemigroups of Tn .

For each Tn, we denote the smallest subsemigroup of Tn containingboth Sn and by . It is well known (see [8]) that [x, y | y]= Tn forall distinct x , y in Xn .

Before stating the next theorem, it is convenient to prove a lemma thatwill be useful both now and later. Each in Tn determines an equivalenceker(). This in turn determines an (arithmetical) partition of n , consisting ofthe sizes of the classes of ker(), and we denote this partition by part() Thus,

for example, if=

1 2 3 4 52 5 2 5 3

,

then part() = (2, 2, 1).


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Lemma 9. Let , Tn . The following statements are equivalent:

(i) there exist , in the symmetric group Sn such that = ;

(ii) |im()|=|im()| and part() = part() .

Proof. (i) (ii). Since = and = 11 , we have J, andso |im()|=|im()| . Also, since R , it follows that ker() = ker(), andso certainly part() = part(). Next, observe that, for all x , y in Xn ,

(x, y) ker() if and only if (x1, y1) ker() = ker() .

Thus ker() is simply a translate of ker() = ker(), and it follows thatpart() = part().

(ii) (i). We may suppose, in a standard notation, that

=

A1 A2 . . . Akb1 b2 . . . bk

, =

C1 C2 . . . C kd1 d2 . . . dk

,

where |Ai|=|Ci|(i= 1, 2, . . . , k). Let be the permutation mapping Ci ontoAi for i= 1, 2, . . . , k . Let map bi to di(i= 1, 2, . . . , k) and map Xn \ im()bijectively onto Xn\ im(). Then = .

Theorem 10. Let , Tn .

(i) = if and only if there exist , Sn such that = .

(ii) = Tn if and only if def() = 1 .

Proof. (i) It is clear that= implies that = . Suppose nowthat = . We observe first that

|im()|=|im()| , (4)

for if, for example, |im()|< |im()| , it would follow that / , for contains only elements of rank not exceeding rank ().

Next, we shall show that part() = part(), and from this it will followby Lemma 9 that = for some , in Sn . Since , it followsthat for some m and some 1, 2, . . . , m+1 in Sn ,

= 12 . . . mm+1.

It is clear from Lemma 9 that the required result holds if m= 1. If m 2 wewrite = m+1 , where

= 12 . . . m.


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We may suppose inductively that part() = part(). Then

|im()| |im()| |im()| ,

and so, by (4), |im()| = |im()| . Suppose that , and belong to theJ-class Jr = { Tn : |im()| = r} . Then the elements and m+1 arenon-zero elements in the completely 0-simple principal factor

Pn,r= Kn,r/(Kn,r1 Kn,r2 . . . Kn,1) .

By [5, Lemma 3.3.5], = m+1 R within Pn,r , and so certainly within Tn .Hence ker() = ker(), and it follows that

part() = part() = part() .

(ii) () If = [x1, x2, . . . , xm| xr] is a path-cycle, define

= [x1, xm, xm1, . . . , x2 | x1] Sn. (5)

It is clear that if is a cycle then = 1 . If is a proper path-cycle, then

= [x1, xr |xr] , = [xm, xr1 | xr1] and = . (6)

If def() = 1 and has factors 1, . . . , p , then all the factors are cyclesexcept one. By Corollary 3 we may assume that we have rearranged the factorsso that the unique proper factor is

p = [x1, . . . , xm| xr] .

Let = p. . . 1 . Then, by (6) and again by Corollary 3,

= pp = [x1, xr |xr] .

Hence == Tn .

() Suppose that def() 2. From (3) we deduce that no with1 def()< def() can belong to .

Theorem 11. Let STn with |im()| = r , and let m = n r . Thenthere exists Sn such that

im() = {m+ 1, m+ 2, . . . , n} .

Proof. If im() = {m + 1, m + 2, . . . , n} , take = [1| 1], the identity map.Otherwise, since {m+ 1, . . . , n} contains r elements and |im()| = r , thereexist 1 i1 < < ik m such that {i1, . . . , ik}= im() {1, . . . , m} . This


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implies that im() {m + 1, . . . , n} contains r k elements. Hence there existm+ 1 j1 < < jk n such that j1, . . . , jk / im(). Let

= [i1, j1 | i1] . . . [ik, jk | ik] ; (7)

then it is easy to check that im() = {m+ 1, m+ 2, . . . , n} .

Theorem 12. Let STn with |im()| = r , and let m = n r . Thenthere exist , Sn and i = [i, xi | xi] with xi m + 1 (1 i m) suchthat = 1. . . m .

Proof. From Theorem 11, there exists such that

im() = {m+ 1, m+ 2, . . . , n} .

From Corollary 3 we deduce that =1. . . m , where is the product of all

the cycle factors of , and i= [i, yi2, . . . , yiqi |yiri ] with yiri =i (1 i m ).Let= 12. . . m, = m. . . 21. (8)

Then, noting that Sn , we see that

= = ()1= ()1m. . . 2(11)2. . . m

= ()1(m. . . 2)[1, y1r1 |y1r1 ](2. . . m)

= ()1[1, y1r1 |y1r1 ](m. . . 22. . . m)

= . . .

= ()1[1, y1r1 |y1r1 ] . . . [m, ymrm |ymrm ] .

The result follows, with = ()1 and xi= yiri .

We have already encountered arithmetical partitions, but now we need torecord some notations. For n, r Z+ with n r , let Pr(n) denote the set ofinteger solutions of the equation

x1+x2+ +xr = n with x1 x2 xr 1.

If (n1, n2, . . . , nr) is a solution of the equation, then it is called a partition ofnwith r terms. (See [1].) Let pr(n) = |Pr(n)| .

For n, r Z+ with n r , let m = n r and s = min{2m, n} . If(n1, n2, . . . , nr) is a partition of n , then we define

(n1,n2,...,nr) = [1, x1 | x1] [2, x2| x2] . . . [m, xm| xm] STn,

where

m+ 1 = x1 = x2 = = xn11


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m+ 2 = xn1 =xn1+1 = = xn1+n22...

s 1 = x(n1++nr2)r+3 = x(n1++nr2)r+4 = = x(n1++nr1)r+1s = x(n1++nr1)r+2 = x(n1++nr1)r+3 = = x(n1++nr)r

and, with these notations, define

Xn,r = { (n1,n2,...,nr) : (n1, n2, . . . , nr) is a partition of n }.

For example, consider (4, 2, 1, 1), which is a partition of 8 with 4 terms; then

(4,2,1,1) = [1, 5| 5][2, 5| 5][3, 5| 5][4, 6| 6] X8,4.

Notice that the two 1s contribute nothing to (4,2,1,1) . Indeed

(1,1,...,1) = [1| 1] .

Lemma 13. Let STn with |im()|= r . Then, with the above notations,there exists (n1,n2,...,nr) Xn,r such that =(n1,n2,...,nr).

Proof. Let STn with|im()|= r , and let m = nr . Then, by Theorem12, there exist , Sn and 2-paths 1, . . . , m such that = 1. . . m

1 ,where i= [i, xi| xi] and xi m + 1 (1 i m ). Then =1. . . m .In addition, by Corollary 3, we can use the commuting of disjoint paths tore-order the i so that

m+ 1 x1 = = xm1 < xm1+1 = = xm1+m2 < xm1+m2+1

= = xm1++mk1 < xm1++mk1+1 = = xm1++mk n , (9)

with m1+ +mk = m . Since there are k 1 proper inequalities in (9), wehave that

n m1+ +mk (m+ 1) + (k 1)

and so k n m= r .

If x1 = = xm1 = m + 1, then define 1 = [1| 1]. Otherwise, define1 = [m+ 1, x1 | m+ 1]. Since 1 is disjoint from j for all j m1+ 1, andsince 1 is a transposition, it follows from Corollary 3 that

1(12. . . m)1 = (111)(121) . . . (1m11)m1+1 m.

Now, for i= 1, . . . , m1

1i1 = [m+ 1, x1 | m+ 1] [i, x1 | x1] [m+ 1, x1| m+ 1] = [i, m+ 1| m+ 1] ,


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and so

1(1. . . m)1= [1, m+ 1| m+ 1] . . . [m1, m+ 1| m+ 1]m1+1. . . m.

Then we consider xm1+1 = = xm1+m2 m+ 2, and similarly define

2 =

[1 | 1] if xm1+1 = = xm1+m2 =m+ 2[m+ 2, xm1+1 | m+ 2] otherwise.

Continuing, we eventually obtain a permutation = 12. . . k (= k. . . 1 ,since the transpositions i are disjoint) such that

(1) = [1, m+ 1| m+ 1] . . . [m1, m+ 1| m+ 1]

[m1+ 1, m+ 2| m+ 2] . . . [m1+m2, m+ 2| m+ 2]

...

[m1+ +mk1+ 1, m+k| m+k] . . . [m, m+k| m+k] .

Notice that, if k= r , then

(m1+ 1) + + (mk+ 1) = m+k= m+r = n ,

and so (m1+1, m2+1, . . . , mk+1) is a partition ofn . Thus 1 Xn,r . If

k=r , then we consider the partition (m1 + 1, m2 + 1, . . . , mk+ 1, nk+1, . . . , nr)of n where nk+1 = = nr = 1, and so 1 Xn,r . The required resultthen follows from Theorem 10.

Recall now that

cont(Tn,r) = min{|G| : G Kn,r andG Sn= Tn,r} .

We have shown that {p : p Pr(n)} Sn generates { Tn : |im()|= r} ,and it follows from [6, Lemma 4] that it generates the whole of Tn,r . Hencecont(Tn,r) pr(n). In effect, we have proved half of the following result.

Theorem 14. cont(Tn,r) = pr(n) .

Proof. Let Jr . As in the proof of Theorem 10, we can show thatpart() = part() for every in Jr . It follows that a set G in Km,rsuch that G Sn= Tn,r must contain, for each partition p in Pr(n), at leastone element such that part() = p . Hence cont() pr(n), and our proofis complete.

References

[1] Hardy, G. H. and E. M. Wright, An Introduction to the Theory of Num-bers, Fifth Edition, Oxford University Press, 1979.


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[2] Higgins, P. M., J. M. Howie and N. Ruskuc,Generators and factorisationsof transformation semigroups, Proc. Royal Soc. Edinburgh A 128 (1998),13551368.

[3] Higgins, P. M., J. M. Howie, J. D. Mitchell and N. Ruskuc, Countableversus uncountable ranks in semigroups of transformations and relations,Proc. Edinburgh Math. Soc. 46 (2003), 531534.

[4] Howie, J. M., The subsemigroup generated by the idempotents of a fulltransformation semigroup, J. London Math. Soc. 41 (1966), 707716.

[5] Howie, J. M., Fundamentals of Semigroup Theory, Oxford UniversityPress, 1995.

[6] Howie, J. M., and R. B. McFadden, Idempotent rank in finite full transfor-mation semigroups, Proc. Royal Soc. Edinburgh A 114 (1990), 161-167.

[7] Lipscomb, S., Symmetric Inverse Semigroups, Math. Surveys and Mono-graphs, Vol. 46, American Math. Soc., Providence, 1996.

[8] Vorobev, N. N., On symmetric associative systems, Leningrad. Gos. Ped.Inst. Uch. Zap. 89 (1953), 161166 (Russian).

Department of MathematicsCukurova UniversityAdana, Turkey

Department of MathematicsCukurova UniversityAdana, Turkey

[email protected]

Mathematical Institute

University of St AndrewsNorth Haugh

St Andrews, Fife KY16 9SS, [email protected]

Received October 21, 2003and in final form July 30, 2004Online publication December 2, 2004

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