Congruences on monoids of transformations preserving the orientation on a finite chain

Transcription

1 Congruences on monoids of transformations preserving the orientation on a finite chain Vítor H Fernandes, Gracinda M S Gomes, Manuel M Jesus 1 Centro de Álgebra da Universidade de Lisboa, Av Prof Gama Pinto, 2, Lisboa, Portugal s: vhf@fctunlpt, ggomes@ciifculpt, mrj@fctunlpt Abstract The main subject of this paper is the description of the congruences on certain monoids of transformations on a finite chain X n with n elements Namely, we consider the monoids OR n and POR n of all full, respectively partial, transformations on X n that preserve or reverse the orientation, as well as their respective submonoids OP n and POP n of all orientation-preserving elements The inverse monoid PORI n of all injective elements of POR n is also considered 2000 Mathematics Subject Classification: 20M20, 20M05, 20M17 Keywords: congruences, orientation-preserving, orientation-reversing, transformations Introduction and preliminaries For n N, let X n be a finite chain with n elements, say X n = {1 < 2 < < n} As usual, we denote by PT n the monoid (under composition of all partial transformations of X n The submonoid of PT n of all full transformations of X n and the (inverse submonoid of all injective partial transformations of X n are denoted by T n and I n, respectively Let a = (a 1, a 2,, a t be a sequence of t (t 0 elements from the chain X n We say that a is cyclic [anti-cyclic] if there exists no more than one index i {1,, t} such that a i > a i+1 [a i < a i+1 ], where a t+1 denotes a 1 Notice that, the sequence a is cyclic [anti-cyclic] if and only if a is empty or there exists i {0, 1,, t 1} such that a i+1 a i+2 a t a 1 a i [a i+1 a i+2 a t a 1 a i ] (the index i {0, 1,, t 1} is unique unless a is constant and t 2 Let s PT n and suppose that Dom(s = {a 1,, a t }, with t 0 and a 1 < < a t We say that s is an orientation-preserving [orientation-reversing] transformation if the sequence of its images (a 1 s,, a t s is cyclic [anti-cyclic] It is easy to show that the product of two orientationpreserving or of two orientation-reversing transformations is orientation-preserving and the product of an orientation-preserving transformation by an orientation-reversing transformation is clearly orientation-reversing 1 This wor was developed within the activities of Centro de Álgebra da Universidade de Lisboa, supported by FCT and FEDER, within project POCTI/MAT/893/2003 Fundamental and Applied Algebra 1

2 Denote by POP n the submonoid of PT n of all orientation-preserving transformations of X n As usual, OP n denotes the monoid POP n T n of all full transformations of X n that preserve the orientation This monoid was considered by Catarino in [3] and by Arthur and Rušuc in [2] The injective counterpart of OP n, ie the inverse monoid POPI n = POP n I n, was studied by the first author in [9, 11] Comprehensiver classes of monoids are obtained when we tae transformations that either preserve or reverse the orientation In this way we get POR n, the submonoid of PT n of all transformations that preserve or reverse the orientation Within T n sits the submonoid OR n = POR n T n and inside I n is PORI n = POR n I n The following diagram, with respect to the inclusion relation and where C n denotes the cyclic group of order n, exposes the relationship between these semigroups: PORn POP n PORI n OR n POPI n OP n C n The study of transformations that respect the orientation is intrinsically associated to the nowledge of the ones that respect the order A transformation s in PT n is called order-preserving if x y implies xs ys, for all x, y Dom(s Denote by PO n the submonoid of PT n of all partial order-preserving transformations of X n The monoid PO n T n of all full transformations of X n that preserve the order is denoted by O n This monoid has been largely studied by several authors (eg see [1, 15, 16, 18] The injective counterpart of O n is the inverse monoid POI n = PO n I n, which is considered, for example, in [5, 7, 8, 10, 12] In this paper, on one hand we aim to describe the Green relations on some of the monoids mentioned above and to use the descriptions obtained to calculate their sizes and rans This type of questions were also considered by Catarino and Higgins [4] for OP n and for OR n ; by Fernandes [9] for POPI n and by the authors [13] for PORI n So, it remains to study the monoids POP n and POR n and that is done in Section 1 On the other hand, we want to describe the congruences of the monoids OP n, POP n, OR n, POR n and PORI n It was proved by Aǐzenštat [1], and later by Lavers and Solomon [18], that the congruences of O n are exactly the Rees congruences A similar result was proved by the first author [10] for the monoid POI n and by the authors [14] for the monoid PO n Fernandes [9] proved that the congruences on POPI n are associated with its maximal subgroups In Section 2, under certain conditions, on an arbitrary finite semigroup we define a class of congruences associated to its maximal subgroups In Section 3, we show that, as in the case of POPI n, all congruences in the monoids referred above are of this type Next, for completion, we recall some notions and fix the notation Let M be a monoid We denote by E(M its set of idempotents Let J be the quasi-order on M defined by u J v if and only if MuM MvM, 2

3 for all u, v M Denote by J u the J-class of an element u M As usual, a partial order relation J is defined on the set M/J by setting J u J J v if and only if u J v, for all u, v M For u, v M, we write u < J v and also J u < J J v if and only if u J v and (u, v J The Rees congruence ρ I on M associated to an ideal I of M is defined by (u, v ρ I if and only if u = v or u, v I, for all u, v M For convenience, we admit the empty set as an ideal In what follows the identity congruence will be denoted by 1 and the universal congruence by ω The ran of M is, by definition, the minimum of the set { X : X M and X generates M} For more details, see eg [17] A subset C of the chain X m is said to be convex if x, y C and x z y imply that z C An equivalence ρ on X m is convex if its classes are convex We say that ρ is of weight if X m /ρ = Clearly, the number of convex equivalences of weight on X m is ( m 1 1 Now let G be a cyclic group of order n It is well nown that there exists a one-to-one correspondence between the subgroups of G and the (positive divisors of n Since G is abelian, all subgroups are normal, and so there is a one-to-one correspondence between the congruences of G and the (positive divisors of n These correspondences are, in fact, lattice isomorphisms The dihedral group D n of order 2n (n 3 can be defined by the group presentation and its proper normal subgroups are: x, y x n = 1, y 2 = 1, yx = x 1 y (1 x 2, y, x 2, xy and x n p, with p a divisor of n, when n is even; (2 x n p, with p a divisor of n, when n is odd See [6] for more details The following concept will be used in Section 3 Let (P 1, 1 and (P 2, 2 be two disjoint posets The ordinal sum of P 1 and P 2 (in this order is the poset P 1 P 2 with universe P 1 P 2 and partial order defined by: for all x, y P 1 P 2, we have x y if and only if x P 1 and y P 2 ; or x, y P 1 and x 1 y; or x, y P 2 and x 2 y Observe that this operator on posets is associative but not commutative 1 The monoids POP n and POR n In this section we describe the Green relations and calculate the sizes and the rans of the monoids POP n and POR n We show that their structure is similar to the one of the monoids POPI n, PORI n, PT n and I n In particular, in all of them, the J-classes are the sets of all elements with the same ran and form a chain, with respect to the partial order J Notice also that all these monoids are regular In what follows, we must have in mind that an element of POR n is either in POP n or it reverses the orientation Denote by Por n the set of all orientation-reversing partial transformations of X n Clearly, POR n = POP n Por n In view of the next lemma, we have POP n Por n = {s POP n : Im(s 2} Lemma 11 [4] Let a be a cyclic [anti-cyclic] sequence Then a is (also anti-cyclic [cyclic] if and only if a has no more than two distinct values 3

4 It is easy to show that E(POR n = E(POP n Let us consider the permutation of order two ( 1 2 n 1 n h = n n Clearly, h 2 = 1 and h is an orientation-reversing full transformation We showed in [13] that POPI n together with h form a set of generators of PORI n Similarly, by just noticing that, given an orientation-reversing transformation s, the product sh is an orientation-preserving transformation, it follows that POR n is generated by POP n {h} We prove that POR n is regular, using the fact that POP n is already nown to be regular [9] It remains to show that all the elements of Por n are regular Let s be an orientation-reversing transformation Then sh POP n and so there exists s POP n such that (shs (sh = sh Thus, multiplying on the right by h, we obtain s(hs s = s and so s is a regular element of POR n Next consider the following permutation g = ( 1 2 n 1 n 2 3 n 1 which is an element of POP n such that g n = 1 As in [9, Proposition 31], it is a routine matter to prove the following (non unique factorisation of an element of POP n : Proposition 12 Let s POP n Then there exist i {0, 1,, n 1} and u PO n such that s = g i u As an immediate consequence of this proposition, we have: Corollary 13 The monoid POP n is generated by PO n {g}, Corollary 14 Let s POR n Then there exist i {0, 1,, n 1}, j {0, 1} and u PO n such that s = g i uh j Notice that, with the notation of the last corollary, we can always tae: (1 j = 0, if s POP n ; (2 u O n, if s OR n Therefore, wherever in this paper we tae such a factorisation of an element s of POR n, we will consider j and u as above Denote by M n either the monoid POR n or the monoid POP n Proposition 15 Let s and t be elements of M n Then: (1 s R t if and only if Ker(s = Ker(t; (2 s L t if and only if Im(s = Im(t; (3 s J t if and only if Im(s Im(t 4

5 Proof Since M n is a regular submonoid of PT n, conditions (1 and (2 follow immediately from well nown results on regular semigroups (eg see [17] Next we prove condition (3 First, suppose that s J t Then there exist x, y M n such that s = xty Since Im(s Im(ty and Im(ty = Im(ty Im(t, then Im(s Im(t Conversely, let s, t M n be such that Im(s Im(t By Corollary 14, there exist i 1, i 2 {0,, n 1}, j 1, j 2 {0, 1} and u, v PO n such that s = g i 1 uh j 1 and t = g i 2 vh j 2 Thus Im(s = Im(u and Im(t = Im(v, since g i 1, g i 2, h j 1, h j 2 are permutations Hence u J v in PO n (see [15] and so there exist x, y PO n such that u = xvy Then s = g i 1 uh j 1 = g i 1 xvyh j 1 = (g i 1 xg n i 2 g i 2 vh j 2 (h 2 j 2 yh j 1 = (g i 1 xg n i 2 t(h 2 j 2 yh j 1, with g i 1 xg n i 2, h 2 j 2 yh j 1 M n, and so s J t in M n, as required It follows, from condition (3, that M n /J = {J 0 < J J 1 < J < J J n }, where J = {s M n Im(s = }, for all 0 n On the other hand, given an element s M n I n, from conditions (1 and (2 above and from the corresponding descriptions for the monoids POPI n and PORI n ([9, Proposition 24] and [13, Proposition 53], respectively, it follows that the H-class of s in M n I n coincides with its H-class in M n Thus, as for the monoid POPI n ([9, Proposition 26], we have: Proposition 16 Let s POP n be such that 1 Im(s = n Then H s = Moreover, if s is an idempotent then H s is a cyclic group of order Since a transformation s POR n is both orientation-preserving and orientation-reversing if and only if Im(s 2, we have the following: Corollary 17 Let s POR n be such that 1 Im(s = 2 Then H s = Moreover, if s is an idempotent then H s is a cyclic group of order Also, as for the monoid PORI n ([13, Proposition 53], we have: Proposition 18 Let s POR n be such that 3 Im(s = n Then H s = 2 Moreover, if s is an idempotent then H s is a dihedral group of order 2 Let s be an element of POR n with ran, 0 n Suppose that Im(s = {b 1,, b } Then, considering the ernel classes of s, we obtain two types of partitions of the domain of s into intervals: ( (a Dom(s = i=1 P P1 P i with s = ; or b 1 b ( (b Dom(s = +1 i=1 P P1 P i with s = P +1 b 1 b b 1 5

6 Notice that, in the first case, P 1,, P are precisely the ernel classes of s whereas, in the second one, the ernel classes are P 1 P +1, P 2,, P Now let {2,, n} and suppose that s is an element of POP n with ran If s verifies (a then Ker(s is a convex equivalence on Dom(s of weight On the other hand, if s verifies (b then we can associate to s a convex relation of weight + 1 (with classes P 1,, P, P +1 Therefore the number of R-classes of ran with the same domain as s is given by ( ( ( Dom(s 1 Dom(s 1 Dom(s + =, 1 whence the total number of R-classes of ran is equal to n ( n ( j ( j= j As n ( j ( j = n n ( j, we have n ( n ( j ( j= j = n n ( n ( j= j = n n ( n ( j=0 j = n 2 n Since the number of L-classes of ran is, clearly, equal to ( n and, by Proposition 16, each H-class of ran has elements, the monoid POP n has precisely ( n n ( 2 n elements of ran Furthermore, by noticing that the number of transformations of ran 1 of PT n (and so of POP n is equal to (2 n 1n, we conclude the following result: Proposition 19 POP n = 1 + (2 n 1n + n =2 ( n 22 n As there is a natural bijection between POP n and Por n (obtained by simply reversing the sequence of the images, we have POR n = 2 POP n {s POP n Im(s 2}, whence: Proposition 110 POR n = 1 + (2 n 1n + 2 ( n 22 2 n 2 + n =3 2( n 22 n and Naturally, at this point, we would lie to compute the ran of these monoids Let us consider the following elements s 0, s 1,, s n 1 of POI n : ( 2 n 1 n s 0 = 1 n 2 n 1 s i = ( 1 n i 1 n i n i + 2 n 1 n i 1 n i + 1 n i + 2 n for i {1, 2,, n 1} Consider also the elements u 1,, u n 1 of O n defined by u i = ( 1 i 1 i i + 1 n 1 i 1 i + 1 i + 1 n for 1 i n 1 Since PO n = s 0,, s n 1, u 1,, u n 1 (see [15], it follows from Corollary 13 that: Corollary 111 POP n = s 0,, s n 1, u 1,, u n 1, g Also, as g n 1 u i g = u i+1, for 1 i n 2, s 0 = g n 1 (s 1 g n 1 and s i = g i 1 s 1 g n i+1, for 1 i n 1, we get: Corollary 112 POP n = s 1, u 1, g,, 6

7 Finally, since any generating set of POP n must clearly contain a permutation, a non-permutation full transformation and a non-full transformation, we must have: Theorem 113 POP n has ran 3 Next we observe that, given an orientation-reversing partial transformation s, we have sh POP n, whence sh = x 1 x 2 x, for some x 1, x 2,, x {s 1, u 1, g} and N Thus s = sh 2 = x 1 x 2 x h and so we may conclude the following: Corollary 114 POR n = s 1, u 1, g, h Let A be a set of generators of POR n As for POP n, the set A must contain at least one nonpermutation full transformation and one non-full transformation On the other hand, for n 3, the group of units of POR n is the dihedral group D n, which has ran two Hence we must also have two permutations in A We have proved the next result Theorem 115 For n 3 the monoid POR n has ran 4 2 Congruences associated to maximal subgroups In this section we construct a family of congruences associated to maximal subgroups of a J-class that satisfies certain conditions As we will show in Section 3, this family provides a description for the congruences of the monoids we want to consider We start with a simple technical lemma Lemma 21 Let S be a semigroup and let s, t, u S be such that s is regular and s H t Then there exist v 1, v 2 S such that v 1 s = us, v 1 t = ut, sv 2 = su, tv 2 = tu, v 1 s R v 1 R v 1 t and sv 2 L v 2 L tv 2 Proof It is well nown (eg see [17] that ss = tt and s s = t t, for some inverses s of s and t of t Let v 1 = uss = utt and v 2 = s su = t tu Then v 1 s = us, v 1 t = ut, sv 2 = su and tv 2 = tu On the other hand, as ss R s, tt R t, s s L s and t t L t, we obtain v 1 = uss R us = v 1 s, v 1 = utt R ut = v 1 t, v 2 = s su L su = sv 2 and v 2 = t tu L tu = tv 2, as required Let S be a finite semigroup and let J be a J-class of S Denote by B(J the set of all elements s S such that J J J s It is clear that B(J is an ideal of S We associate to J a relation π J on S defined by: for all s, t S, we have s π J t if and only if (a s = t; or (b s, t B(J; or (c s, t J and s H t Lemma 22 [9]The relation π J is a congruence on S 7

8 Assume that J is regular and tae a group H-class H 0 of J Also, suppose that there exists a mapping ε : J H 0 s s which satisfies the following property: given s, t J such that st J, there exist x, y H 0 such that b H t implies sb = x s b (1 a H s implies ãt = ã ty (2 The existence of such a map for the monoid I n (and for some of its submonoids was showed by the first author in [9] To each congruence π on H 0, we associate a relation ρ π on S defined by: given s, t S, we have s ρ π t if and only if s π J t and s, t J implies s π t Theorem 23 The relation ρ π is a congruence on S Proof First, observe that ρ π is an equivalence relation, since H and π are equivalence relations and B(J J = So, it remains to prove that ρ π is compatible with the multiplication Let s, t S be such that s ρ π t and assume that s t Let u S As s π J t and π J is a congruence, we have us π J ut and su π J tu In order to prove that us ρ π ut, suppose that us, ut J Then us, ut B(J and, as B(J is an ideal, s, t B(J Since s t, we must therefore have s, t J and s H t Also we get s π t Now, by Lemma 21, there exists v 1 S such that v 1 s = us, v 1 t = ut and v 1 s R v 1 R v 1 t Hence we have s, v 1, v 1 s J As t H s, it follows from condition (1 that ṽ 1 s = xṽ 1 s and ṽ 1 t = xṽ 1 t, for some x H 0 Thus, as π is a congruence, ũs = ṽ 1 s = xṽ 1 s π xṽ 1 t = ṽ 1 t = ũt and so us ρ π ut Similarly, we prove that su ρ π tu, as required 3 On the congruences of OP n, POP n, OR n, PORI n and POR n The goal of this section is to describe the congruences of the monoids OP n, POP n, OR n, PORI n and POR n We will use a method that generalises the process developed by the first author to describe the congruences of the monoid POPI n [9] In fact, this new technique will also comprise that case Although there are details that differ from one case to the other, we will present the proof in a way that solves the problem simultaneously for all these monoids To prove our main result, Theorem 33, we need to fix some notation and recall some properties of the monoids OP n, OR n, POPI n, PORI n, POP n and POR n presented in [4, 9, 13] or in this paper First, remember that OP n = O n, g, POPI n = POI n, g, POP n = PO n, g, OR n = O n, g, h, PORI n = POI n, g, h and POR n = PO n, g, h Let us fix T {O n, POI n, PO n } and let M be either the monoid T, g or the monoid T, g, h Both T and M are regular monoids (moreover, if T = POI n then M and T are inverse monoids 8

9 and, for the partial order J, the quotients T/J and M/J are chains More precisely, for S {T, M}, we have S/J = {J S 0 < J J S 1 < J < J J S n }, if T {POI n, PO n }, and S/J = {J S 1 < J < J J S n }, if T = O n, where J S denotes the J-class of S of the elements of ran, for suitably defined Since S/J is a chain, the sets I S = {s S Im(s }, with 0 n, together with the empty set (if necessary, constitute all the ideals of S (see [10] Observe also that T is an aperiodic monoid (ie T has only trivial subgroups; the H-classes of ran of T, g have precisely elements, for 1 n; and the H-classes of ran of T, g, h have precisely 2 elements, for 3 n, and elements, for = 1, 2 For a J-class J M of M (necessarily regular, since M is regular, with 1 n, we want to find a particular group H-class H in J M and a mapping ε : J M H satisfying the conditions of Theorem 23 Notice that, we have B(J M = J 0 M J 1 M J 1 M or B(J M = J 1 M J 1 M Given s PT n with Dom(s = {i 1 < < i }, where 1 n, define s T n by, for every x X n, (i 1 s, if 1 x i 1 (xs = (i j s, if i j 1 < x i j and 2 j (i s, if i < x n It is clear that s and s have the same ran Moreover: (a If s PO n then s O n ; (b If s POP n then s OP n ; and (c If s POR n then s OR n Fix 1 n and consider the following elements of I n (which are permutations of {1,, }: ( ( ( ɛ =, γ = and η = Let J be the J-class of M of the elements of ran If T {POI n, PO n }, tae the following elements: e = ɛ, g = γ and h = η When T = O n, consider the following full transformations of X n : e = ɛ, g = γ and h = η Notice that h g i = g i h, for 1 i Denote by H the H-class of M of the idempotent e and observe that: (a If M = T, g, then H is the cyclic group of order generated by g ; and (b If 3 and M = T, g, h, then H is the dihedral group of order 2 generated by g and h 9

10 Let s J Suppose that {a 1 < < a } is the transversal of the ernel of s formed by the minimum element of each ernel class Let Im(s = {b 1 < < b } and tae the injective partial order-preserving transformations ( ( 1 b1 b σ L =, σ a 1 a R = 1 and σ L = Define s L, s R, s L, s R T by: ( a1 a 1 (, σ R 1 = b 1 b (a s L = σ L, s R = σ R, s L = σ L and s R = σ R, if T {POI n, PO n }; (b s L = σ L, s R = σ R, s L = σ L and s R = σ R, if T = O n Clearly, s L R e L s R and s L s L = e = s R s R Now let b p1,, b p {b 1,, b } be such that b pl = a l s, for 1 l There exists i {0,, 1} such that b pi+1 < < b p < b p1 < < b pi if s is orientation-preserving, or b pi+1 > > b p > b p1 > > b pi if s is orientation-reversing It can be proved in a routine manner that s L ss R = g i if s is orientation-preserving, or s L ss R = g i h if s is orientationreversing Next let a q1,, a q {a 1,, a } be such that b l = a ql s, for 1 l, and consider the following injective partial transformation: ( σ b1 b = a q1 a q Define ŝ M by: (a ŝ = σ, if T {POI n, PO n }; (b ŝ = σ, if T = O n Clearly, ŝ is an inverse of s and it is easy to show that s = sŝs L s Lss R s R and s Lsŝs L = e Next consider the mapping ε : J H s s = s L ss R Observe that, given s, t J such that s H t, we have s L = t L, s R = t R, s L = t L and s R = t R Moreover, since sŝ and tˆt are idempotents of J with the same ernel and the same image, we also have sŝ = tˆt Lemma 31 Let s, t M be such that s, t, st J Then there exist l 1, l 2 {0, 1, 1} such that (1 b R t implies sb = g l 1 s b; (2 a L s implies ãt = ã tg l 2 10

11 Proof Suppose that s reverses the orientation and t preserves the orientation Let ( P1 P s = i P i+1 P P +1 s 1 s i s i+1 s s 1 and ( Q1 Q t = j Q j+1 Q Q +1 t 1 t j t j+1 t t 1 with possibly P +1 = or Q +1 = Then Im(s = {s i+1 > > s > s 1 > > s i } and Im(t = {t j+1 < < t < t 1 < < t j }, for some 0 i, j 1 Hence we have s = g i h and t = g j As s, t, st J, then Im(s is a transversal of Ker(t and we have two possible situations: (a If s i Q 1,, s 1 Q i, s Q i+1,, s i+1 Q, then ( P1 P st = i P i+1 P P +1 t i t 1 t t i+1 t i Since (P i j st = {t j+1 } if 0 j i 1 and (P i j+ st = {t j+1 } if i j 1, it follows that st = g i+j h, whence st = s t (b If s i Q 2,, s 1 Q i+1, s Q i+2,, s i+1 Q +1, then ( P1 P st = i+1 P i+2 P P +1 t i+1 t 1 t t i+2 t i+1 and so (P i j+1 st = {t j+1 } if 0 j i and (P +i j+1 st = {t j+1 } if i + 1 j 1 Therefore, in both cases, st = g +j i 1 h and so st = g 1 s t = s tg If s preserves the orientation or t reverses the orientation, it is a routine matter to show that, in the situation analogous to (a, we always have st = s t On the other hand, the situation analogous to (b can be summarised by the following table: s t l 1 l 2 orientation-preserving orientation-preserving 1 1 orientation-reversing orientation-preserving 1 1 orientation-preserving orientation-reversing 1 1 orientation-reversing orientation-reversing 1 1 Finally, as elements R-related have the same ernel and elements L-related have the same image, it is clear that l 1 does not depend of the element of the R-class of t (s fixed and l 2 does not depend of the element of the L-class of s (t fixed, as required The next proposition follows from this lemma and Theorem 23 Proposition 32 Let {1,, n} and let π be a congruence on H Then ρ π is a congruence on M 11

12 Notice that, if π is the universal congruence of H, then the relation ρ π is the congruence π J of M On the other hand, if π is the identity congruence of H, then the relation ρ π is the Rees congruence of M associated to the ideal I 1 M Thus, for = 1, the relation ρ π is the identity congruence of M and, for 2 n, there exist s B(J = I 1 M and t J, whence (s, t ρ π and so ρ π is not the universal congruence of M At this point, we can state our main result Theorem 33 The congruences of M {OP n, POPI n, POP n, OR n, PORI n, POR n } are exactly the congruences ρ π, where π is a congruence on H, for {1,, n}, and the universal congruence Denote by Con(S the lattice of the congruences on a semigroup S Recall that Con(T is formed only by the Rees congruences of T {O n, POI n, PO n } (see [1, 10, 14] On the other hand, it is clear that Con(OP 1 = Con(OR 1 = {1} and Con(M = {1, ω} if M {POPI 1, PORI 1, POP 1, POR 1 } To prove Theorem 33 we start by establishing some auxiliary results Let c 1,, c n T n be the constant mappings such that Im(c i = {i}, for all 1 i n Notice that, if s, t T n are such that c i s = c i t, for all 1 i n, then we must have s = t The version of this property for partial transformation is the following: let c 1,, c n PT n be the n partial identities of ran one such that Dom(c i = Im(c i = {i}, for all 1 i n Then, given s, t PT n such that c i s = c i t, for all 1 i n, we must also have s = t In what follows, c 1,, c n denote the constant mappings of T n if T = O n, and the partial identities of ran one of PT n if T {PO n, POI n } In both cases c 1,, c n T Moreover, for all 1 i n and s M, we have c i s T In fact, c i s is either a constant map of image {(is} or the empty map Let ρ be a congruence on M and consider ρ = ρ (T T Then ρ is a Rees congruence of T and so ρ = ρ I T 1, for some 1 n + 1 This notation will be used in the sequel Lemma 34 If = 1 then ρ = 1 Proof First notice that = 1 if and only if ρ = 1 Let s, t M be such that s ρ t Then c i s ρ c i t and, since c i s, c i t T, we have c i s ρ c i t, whence c i s = c i t, for all 1 i n Thus s = t and so ρ = 1, as required From now on, consider 2 Lemma 35 ρ I M 1 ρ Proof It suffices to show that s ρ t, for all s, t I 1 M Let s, t IM 1 (1 If s, t T then s, t I 1 T and so s ρ t, whence s ρ t (2 If s M \T and t T then, by Corollary 14, there exist i {0, 1,, n 1}, j {0, 1} and u T such that s = g i uh j Since s J u, we get g n i sh 2 j = u I 1 T As c 1 I 1 T, we have u ρ c 1, whence u ρ c 1 Then s = g i uh j ρ g i c 1 h j On the other hand, since g i c 1 h j I 1 T (in fact, gi c 1 h j is a constant map, we also have g i c 1 h j ρ t Hence g i c 1 h j ρ t and so s ρ t 12

13 (3 Finally, suppose that s, t M \ T By Corollary 14, there exist i 1, i 2 {0, 1,, n 1}, j 1, j 2 {0, 1} and u, v T such that s = g i 1 uh j 1 and t = g i 2 vh j 2 Since s ρ I M t, it follows that 1 u ρ I M 1 g n i 1+i 2 vh 2 j 1+j 2 If g n i 1+i 2 vh 2 j 1+j 2 M \ T then, by (2, we have u ρ g n i 1+i 2 vh 2 j 1+j 2 On the other hand, if g n i 1+i 2 vh 2 j 1+j 2 T then, by (1, again we have u ρ g n i 1+i 2 vh 2 j 1+j 2 Hence s = g i 1 uh j 1 ρ g i 2 vh j 2 = t, as required Lemma 36 Let s, t M be such that s ρ t Then Im(s if and only if Im(t Proof It suffices to show that Im(s implies Im(t So, suppose that Im(s (1 If s, t T then s ρ t Since s I 1 T, we have s = t, whence Im(t (2 Next consider s T and t M \ T If t I 1 M then t ρ I 1 M c 1 and so t ρ c 1, by Lemma 35 Hence s ρ c 1 By (1, we obtain s = c 1 and this is a contradiction, for c 1 has ran one Therefore Im(t (3 Finally, if s M \ T, by Corollary 14, there exist i {0, 1,, n 1}, j {0, 1} and u T such that s = g i uh j Then g n i sh 2 j = u T and u ρ g n i th 2 j Since u J s and Im(s, by (1 or (2, we deduce that Im(t = Im(g n i th 2 j, as required Lemma 37 Let s M Then there exists an inverse s M of s such that s s E(T Proof Let s M By Corollary 14, there exist i {0, 1,, n 1}, j {0, 1} and u T such that s = g i uh j Let u T be an inverse of u and consider s = h 2 j u g n i Then s is an inverse of s and s s = h 2 j u g n i g i uh j = h 2 j u uh j E(T, as required Lemma 38 Let t M and let D be a transversal of Ker(t Then there exists an inverse t M of t such that Im(t = D Proof Consider the injective partial transformation ξ defined by Dom(ξ = Im(t and, for all x Dom(ξ, (xξ is the unique element in D (xt 1 Define t by: (a t = ξ, if T {POI n, PO n }; (b t = ξ, if T = O n It is a routine matter to show that t M and t is an inverse of t with image D, as required Lemma 39 Let s, t M be such that s ρ t and Im(s Then s H t Proof Let s and t be inverses of s and t, respectively, such that s s, t t T, by Lemma 37 As s ρ t, then s st t ρ s tt t = s t ρ s s Since s s, s st t T, we have s s ρ s st t Now, as Im(s s = Im(s, it follows that s s = s st t and so s = (st t Similarly, as Im(t, by Lemma 36, we also have t = (ts s, whence s L t Next let D be a transversal of Ker(t By Lemma 38, there exists an inverse t of t such that Im(t = D As Im(t t = Im(t and t t ρ t s, by the argument above, it follows that t t L t s Since t t L t L s, we get t s L s and so D = Im(t contains a transversal of Ker(s As s and t are L-related, the transversals of Ker(s and Ker(t have the same number of elements, whence D must also be a transversal of Ker(s We conclude that any transversal of Ker(t is a transversal of Ker(s and vice-versa Thus Ker(s = Ker(t and so s R t Therefore s H t, as required 13

14 Lemma 310 Let s, t M be such that s t and s H t Im(zs = Im(zt = Im(s 1 and (zs, zt L Then there exists z T such that Proof First, notice that, s and t have the same image and the same ernel Next let D be a transversal of Ker(s As s t, there exists i D such that is it Let τ be the partial identity with domain D \ {i} and define z by: (a z = τ, if T {POI n, PO n }; (b z = τ, if T = O n Clearly, z T and Im(zs = Im(s 1 = Im(t 1 = Im(zt On the other hand, as Im(zs = Im(s\{is} and Im(zt = Im(t\{it} = Im(s\{it} and is it, we have Im(zs Im(zt and so (zs, zt L, as required Finally, we can prove Theorem 33 Proof of Theorem 33 Let ρ be a congruence of M Let 1 n + 1 be such that ρ = ρ (T T = ρ I T By Lemma 34, we have ρ = 1, for = 1 Thus we can consider 2 On the 1 other hand, ρ I M ρ, by Lemma 35, and so if = n + 1 the relation ρ is the universal congruence 1 on M Hence, in what follows, we can also assume n Let s, t M be such that s ρ t and Im(s > By Lemma 39, we have s H t Suppose that s t and let m = Im(s By Lemma 310, there exists z T such that Im(zs = Im(zt = m 1 and (zs, zt L On the other hand, as m 1 and zs ρ zt, by Lemma 39, we have (zs, zt H, which is a contradiction Thus s = t Now let π be the congruence of H induced by ρ, ie π = ρ (H H As ρ I M ρ, to 1 prove that ρ = ρ π it remains to show that, for all s, t J M, we have s ρ t if and only if s ρ π t Tae s, t J M First, suppose that s ρ t By Lemma 39, we have s H t and so s π J t Moreover, s L = t L and s R = t R, whence s = s L ss R ρ s L ts R = t L tt R = t and so s π t Thus s ρ π t Conversely, assume that s ρ π t Then s H t and s π t Hence s L = t L, s R = t R, s L = t L, s R = t R and s ρ t Now consider the inverses ŝ and ˆt of s and t, respectively Then sŝ = tˆt and so as required s = sŝs L(s L ss R s R ρ sŝs L(s L ts R s R = tˆtt L(t L tt R t R = t, Let {1,, n} Let π 1, π 2 Con(H If π 1 π 2, it is easy to show that ρ π1 ρ π2 On the other hand, it is clear that given 1, 2 {1,, n} such that 1 < 2, π 1 Con(H 1 and π 2 Con(H 2, we have ρ π1 ρ π2 Denote by D the lattice of the congruences of the group H, for 1 n By Theorem 33, we have the following description of Con(M Theorem 311 The lattice of the congruences of the monoid M is isomorphic to the ordinal sum of lattices D 1 D 2 D n D 1 Example 312 Consider the monoid POR 6 Applying the last result, we get the following Hasse diagram for Con(POR 6 : 14

15 ω 1 References [1] AYa Aĭzenštat, Homomorphisms of semigroups of endomorphisms of ordered sets, Uch Zap, Leningr Gos Pedagog Inst 238 (1962, (Russian [2] RE Arthur and N Rušuc, Presentations for two extensions of the monoid of orderpreserving mappings on a finite chain, Southeast Asian Bull Math 24 (2000, 1-7 [3] PM Catarino, Monoids of orientation-preserving transformations of a finite chain and their presentations, Semigroups and Applications, eds JM Howie and N Rušuc, World Scientific, (1998, [4] PM Catarino and PM Higgins, The monoid of orientation-preserving mappings on a chain, Semigroup Forum 58 (1999, [5] DF Cowan and NR Reilly, Partial cross-sections of symmetric inverse semigroups, Int J Algebra Comput 5 ( [6] D Dummit and R Foote, Abstract Algebra, Second Edition, John Wiley & Sons, Inc,

16 [7] VH Fernandes, Semigroups of order-preserving mappings on a finite chain: a new class of divisors, Semigroup Forum 54 (1997, [8] VH Fernandes, Normally ordered inverse semigoups, Semigroup Forum 58 ( [9] VH Fernandes, The monoid of all injective orientation preserving partial transformations on a finite chain, Comm Algebra 28 (2000, [10] VH Fernandes, The monoid of all injective order preserving partial transformations on a finite chain, Semigroup Forum 62 (2001, [11] VH Fernandes, A division theorem for the pseudovariety generated by semigroups of orientation preserving transformations on a finite chain, Comm Algebra 29 ( [12] VH Fernandes, Semigroups of order-preserving mappings on a finite chain: another class of divisors, Izvestiya VUZ Matematia 3 (478 ( (Russian [13] VH Fernandes, GMS Gomes and MM Jesus, Presentations for some monoids of injective partial transformations on a finite chain, Southeast Asian Bulletin of Mathematics 28 (2004, [14] VH Fernandes, GMS Gomes and MM Jesus, Congruences on monoids of order-preserving or order-reversing transformations on a finite chain, Glasgow Mathematical Journal 47 (2005, [15] GMS Gomes and JM Howie, On the rans of certain semigroups of order-preserving transformations, Semigroup Forum 45 (1992, [16] JM Howie, Product of idempotents in certain semigroups of transformations, Proc Edinburgh Math Soc17 (1971, [17] JM Howie, Fundamentals of Semigroup Theory, Oxford University Press, 1995 [18] T Lavers and A Solomon, The Endomorphisms of a Finite Chain Form a Rees Congruence Semigroup, Semigroup Forum 59 No (2 (1999, First and Third authors second address: Departamento de Matemática, Faculdade de Ciências e Tecnologia, Universidade Nova de Lisboa, Monte da Caparica, Caparica, Portugal Second author s second address: Departamento de Matemática, Faculdade de Ciências, Universidade de Lisboa, Lisboa, Portugal 16