Pre-prints DM-FCT-UNL 2011 Pré-Publicações DM-FCT-UNL 2011

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1 On the ranks of certain monoids of transformations that preserve a uniform partition Vítor H. Fernandes 1 and Teresa M. Quinteiro January 011 Abstract The purpose of this paper is to compute the ranks of the monoid OR m n of all orientation-preserving or orientationreversing full transformations on a chain with mn elements that preserve a uniform m-partition and of its submonoids OP m n of all orientation-preserving transformations and OD m n of all order-preserving or order-reversing full transformations. These three monoids are natural extensions of O m n the monoid of all order-preserving full transformations on a chain with mn elements that preserve a uniform m-partition. Moreover we also determine the ranks of certain semigroups of orientation-preserving full transformations with restricted ranges. 000 Mathematics subject classification: 0M0 0M10. Keywords: order-preserving transformations orientation-preserving transformations equivalence-preserving transformations. Introduction and preliminaries Let X be a set and denote by T X the monoid under composition of all full transformations on X. For n N let X n be a chain with n elements say X n = {1 < < < n} and denote the monoid T X n simply by T n. We say that a transformation α in T n is order-preserving [order-reversing] if for all x y X n x y implies xα yα [xα yα]. Notice that the product of two order-preserving transformations or of two order-reversing transformations is order-preserving and the product of an order-preserving transformation by an order-reversing transformation is order-reversing. Denote by O n the submonoid of T n whose elements are order-preserving and by OD n the submonoid of T n whose elements are either order-preserving or order-reversing. Next let a = a 1 a... 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. Let α T n. We say that α is an orientation-preserving [orientation-reversing] transformation if the sequence of its images 1α... nα is cyclic [anti-cyclic]. Like in the order case the product of two orientation-preserving or of two orientation-reversing transformations is orientation-preserving and the product of an orientation-preserving transformation by an orientation-reversing transformation is orientation-reversing. Denote by OP n the submonoid of T n whose elements are orientation-preserving and by OR n the submonoid of T n whose elements are either orientation-preserving or orientation-reversing. Semigroups of order-preserving transformations have long been considered in the literature. In 196 Aǐzenštat [] gave a presentation for O n from which it can be deduced that O n has only one non-trivial automorphism for n > 1. Also in 196 Aǐzenštat [1] showed that the non-trivial congruences of O n are exactly the Rees congruences. Some years later in 1971 Howie [16] studied some combinatorial and algebraic properties of O n in particular he showed that O n is generated by idempotents of defect one and has F n idempotents where F n is the n th Fibonacci number. Later in 199 Gomes and Howie [15] revisited the semigroup O n and computed its rank and idempotent rank which are n and n respectively. Recall that the [idempotent] rank of a finite [idempotent generated] monoid is the cardinality of a least-size [idempotent] generating set. More recently Fernandes et al. [11] characterized the endomorphisms of O n. The notion of an orientation-preserving transformation was introduced by McAlister in [0] and independently by Catarino and Higgins in [6]. Several properties of the monoids OP n and OR n have been investigated in these two papers. A presentation for the monoid OP n in terms of n 1 generators was given by Catarino in [5]. A new presentation for OP n in terms of its rank generators was found by Arthur and Ruškuc [4] who also exhibited a presentation for the monoid OR n in terms of 3 its rank generators. Finally regarding the monoid OD n a presentation was given by Fernandes et al. in [9]. Its rank computed in [10] by the same authors is n/ The author gratefully acknowledges support of FCT and PIDDAC within the projects ISFL and PTDC/MAT/69514/006 of CAUL. The author gratefully acknowledges support of ISEL and of FCT and PIDDAC within the projects ISFL and PTDC/MAT/69514/006 of CAUL. 1

2 Now let ρ be an equivalence relation on a set X and denote by T ρ X the submonoid of T X of all transformations that preserve the equivalence relation ρ i.e. T ρ X = {α T X aα bα ρ for all a b ρ}. This monoid was studied by Huisheng in [18] who determined its regular elements and described its Green s relations. Let m n N. Of particular interest is the submonoid T m n = T ρ X mn of T mn with ρ the equivalence relation on X mn defined by ρ = A 1 A 1 A A A m A m where A i = {i 1n in} for i {1... m}. Notice that the ρ-classes A i with 1 i m form a uniform m-partition of X mn. Regarding the rank of T m n first Huisheng [17] proved that it is at most 6 and later Araújo and Schneider [3] improved this result by showing that for X mn 3 the rank of T m n is precisely 4. The ranks of its partial and partial injective counterparts were determined by the first author together with Cicalò and Schneider [7]. Finally denote by OR m n the submonoid of T m n of all orientation-preserving or orientation-reversing transformations i.e. OR m n = T m n OR mn. Similarly let OD m n = T m n OD mn OP m n = T m n OP mn and O m n = T m n O mn. Example Consider the following transformations of T 1 : α 1 = ; α = ; α 3 = ; α = ; α 5 = ; α = Then we have: α 1 T 3 4 but α 1 OR 3 4 ; α OR 3 4 but α OP 3 4 ; α 3 OD 3 4 but α 3 O 3 4 ; α 4 OP 3 4 but α 4 O 3 4 ; α 5 O 3 4 ; and finally α 6 T 3 4. In [19] Huisheng and Dingyu described the regular elements and the Green relations of O m n. On the other hand in [1] the authors proved that the monoid O m n has rank mn n. A description of the regular elements and a characterization of the Green relations of the monoid OP m n were given by Sun et al. in [1]. The cardinals of the monoids OR m n OP m n OD m n and O m n were determined by the authors in [13]. In this paper we continue the work of [1] and compute the ranks of the monoids OP m n OD m n and OR m n Sections 3 and 4 respectively. In order to help achieving this goal we use the wreath product description of T m n due to Araújo and Schneider [3] that we recall in the beginning of Section. On the other hand since it will be useful to determine the rank of OP m n in Section 1 we find generating sets and the ranks of certain subsemigroups of OP n with restricted ranges. 1 On the semigroups OP nr Let n N and 1 r n. Consider the subsemigroup with restricted range OP nr = {α OP n Imα {1... r}} of OP n. Recall that the ranks and other properties of the subsemigroups of restricted range of PT n T n and I n were studied by Fernandes and Sanwong in [14]. In this section we determine a set of generators of OP nr that we will use in the next section. Moreover we deduce that OP nr has rank equal to n r for r n 1. Notice that OP n1 is a trivial semigroup and OP nn = OP n. Therefore in what follows we consider r n 1. We begin by showing that OP nr is generated by its elements of rank r. Lemma 1.1 For 1 k < r any transformation of OP nr of rank k is a product of elements of OP nr of rank k + 1. I1 I Proof. Let α = I k be an element of OP a 1 a a nr of rank k where I 1 I... I k are the kernel classes of α k in order 1 I 1 and min I i < min I i+1 for i = 1... k 1. Notice that I... I k are intervals and I 1 is an interval if and only if n I k otherwise I 1 is a union of two intervals. Observe also that a 1 a... a k is a k-cycle. On the other hand as k < n there exits j {1... k} such that I j > 1. 1 k j k j + 1 k k + 1 n Consider γ =. Clearly γ OP a j+1 a k a 1 a j a j a nr and γ has rank k. j I Next if j k let β = 1 I j 1 min I j I j \{min I j } I j+1 I k ; if j = 1 and n I k j + 1 k 1 k k k j k let 1 max I1 I β = I k I ; and if j = 1 and n I k k + 1 k k 1 1 let β = 1 I I k I 1 k k 1 k

3 where I 1 and I 1 are intervals such that I 1 I 1 = I 1 and max I 1 < min I 1 notice that we also have max I k < min I 1. Hence in all cases it is a routine matter to check that β is an element of OP nr of rank k + 1 and α = βγ. Now we focus our attention on γ. Let b 1... b k be the k-cycle a j+1... a k a 1... a j. Observe that with this 1 k k + 1 n notation we have γ =. Take b {1... r}\ Imγ. If b b 1 b k b k b k < b < b 1 or b 1 < b k < b k 1 k k + 1 n 1 k k + 1 k + n let γ 1 = and γ 1 k k + 1 k + 1 =. On the other hand if b 1 b k b k b b b i < b < b i+1 or b < b i+1 < b i or b i+1 < b i < b for some i {1... k 1} let 1 i 1 i i + 1 k k + 1 n γ 1 = k i + k k k i k i + 1 k i + 1 and 1 k i k i + 1 k i + k + 1 k + n γ = b i+1 b k b k b 1 b i b b notice that k < n 1 whence k + n. Then in both cases it is easy to show that γ 1 γ OP nr γ 1 and γ have rank k + 1 and γ = γ 1 γ. Therefore we proved that α = βγ 1 γ with β γ 1 and γ elements of OP nr of rank k + 1 as required. From this lemma by induction on the rank of the transformations we may deduce that OP nr is generated by its elements of rank r as announced above. 1 r 1 r r + 1 n Next let g nr = OP 3 r nr. Hence we have: Lemma 1. Let α and β be two elements of OP nr of rank r such that Kerβ = Kerα. Then β = αgnr k for some k {0... r 1}. I1 I Proof. Take α = I r I1 I and β = I r where I a 1 a a r b 1 b b 1 I... I r are the kernel classes of α and r β in order 1 I 1 and min I i < min I i+1 for i = 1... r 1. Then as a 1 a... a r and b 1 b... b r are two r-cycles of {1... r} we have a 1... a r = i r 1... i and b 1... b r = j r 1... j for some 1 i j r. Take k = j i if i j and k = r i + j otherwise. Hence k {0... r 1} and it is a routine matter to prove that β = αgnr k as required. Now notice that if α is an element of OP nr of rank r and α 1 and α are two elements of OP nr such that α = α 1 α then Kerα 1 = Kerα. On the other hand it is clear that the number of distinct kernels of transformations of OP nr of rank r coincides with the number of distinct kernels of transformations of OP n of rank r which is precisely n r see [6]. These observations together with the previous two lemmas prove the following result. Theorem 1.3 For r the semigroup OP nr is generated by any subset of transformations of rank r containing at least one element from each distinct kernel. Furthermore OP nr has rank equal to n r. The rank of the monoid OP m n Let m n. Following [3] we define the wreath product T n T m of T n and T m as being the monoid with underlying set T m T m and multiplication defined by n α 1... α m ; βα 1... α m; β = α 1 α 1β... α m α mβ; ββ for all α 1... α m ; β α 1... α m; β Tn m T m. Let α T m n and let β T m be the quotient map of α by ρ i.e. for all j {1... m} we have A j α A jβ. For each j {1... m} define α j T n by kα j = j 1n + kα jβ 1n 1 for all k {1... n}. Let α = α 1 α... α m ; β T m T m. With this notation the function n ψ : T m n T n T m α α 3

4 is an isomorphism see [3 Lemma.1]. Notice that from 1 we have kα j < lα j if and only if j 1n + kα < j 1n + lα for all 1 k l n and j {1... m}. Furthermore if jβ = j + 1β for some j {1... m 1} then nα j < 1α j+1 if and only if jnα < jn + 1α. Also if mβ = 1β then nα m < 1α 1 if and only if mnα < 1α. Now admit that α is an orientation-preserving transformation. Then 1. 1α mnα; or. r + 1α mnα 1α rα and rα > r + 1α for some r {1... mn 1}. In the first case notice that α is order-preserving clearly α j O n for all j {1... m}. Next suppose that α satisfies the second condition. If r A j \{jn} for some j {1... m} then α j OP n \O n and α i O n for all i {1... m}\{j}. Furthermore Imα A j α whence β is constant. Otherwise i.e. r = jn for some j {1... m 1} it is clear that we have α i O n for all i {1... m}. On the other hand also as a consequence of 1 if inα jnα then iβ jβ for all 1 i j m. In fact suppose that iβ > jβ for some 1 i j m. Then iβ = jβ + t for some t 1 and so iβn = jβn + tn. Hence inα = i 1n+nα = nα i +iβ 1n = nα i +jβ 1n+tn > nα j +jβ 1n = j 1n+nα = jnα as required. Now if α is orientation-preserving then as any subsequence of a cyclic sequence is also cyclic see [8 Proposition.1] the sequence nα nα... mnα is cyclic and so by the above observation the sequence 1β β... mβ is also cyclic i.e. β OP m. Recall that the authors showed in [1 Lemma 1.] that O m n ψ = {α 1... α m ; β O m n O m jβ = j + 1β implies nα j 1α j+1 for all j {1... m 1}}. Considering addition modulo m in particular m + 1 = 1 for OP m n we have: Proposition.1 A m + 1-tuple α 1 α... α m ; β of Tn m T m belongs to OP m n ψ if and only if it satisfies one of the following conditions: 1. β is a non-constant transformation of OP m for all i {1... m} α i O n and for all j {1... m} jβ = j + 1β implies nα j 1α j+1 ;. β is a constant transformation for all i {1... m} α i O n and there exists at most one index j {1... m} such that nα j > 1α j+1 ; 3. β is a constant transformation there exists one index i {1... m} such that α i OP n \ O n and for all j {1... m} \ {i} α j O n and for all j {1... m} nα j 1α j+1. Proof. We will take into consideration several times the observations stated above. First assuming that a m + 1-tuple α 1... α m ; β satisfies 1 or 3 it is just a routine matter to check that if α T m n is such that αψ = α 1... α m ; β then α OP m n. Conversely let α OP m n and take α = αψ = α 1... α m ; β. If α is order-preserving then by α 1... α m ; β O m n O m and for all j {1... m 1} jβ = j + 1β implies nα j 1α j+1. If β is not constant then mβ 1β and so the m + 1-tuple α satisfies 1. Otherwise α satisfies. Next suppose that r + 1α mnα 1α rα and rα > r + 1α for some r {1... mn 1}. If r A j \ {jn} for some j {1... m} it is easy to deduce that α satisfies 3. On the other hand if r = jn for some j {1... m 1} it is easy to show that α satisfies 1 if β is not constant and that α satisfies otherwise. Let α OP m n. For i {1 3} we say that α and αψ are of type i if αψ satisfies the condition i. of the previous proposition. Notice that if α 1... α m ; β = αψ is of type and for all j {1... m} nα j 1α j+1 then α must be a constant transformation. Moreover as clearly the product of m + 1-tuples of types 1 or respectively or 3 cannot be a m + 1-tuple of type 3 respectively 1 then the subset M respectively N of OP m n ψ of all m + 1-tuples of types 1 or respectively or 3 is a submonoid respectively subsemigroup of OP m n ψ. 4

5 Let M = Mψ 1. Hence clearly M is the submonoid of OP m n whose elements are the order-preserving transformations and so in particular M contains O m n and the transformations α OP m n such that jn + 1α mnα 1α jnα and jnα > jn + 1α for some j {1... m 1}. 1 n 1 n Recall that being g n the n-cycle OP 3 n 1 n each element s OP n admits a factorization s = gnu j with 0 j n 1 and u O n which is unique unless s is constant see [6]. Now consider the permutations of {1... mn} 1 mn 1 mn g = g mn = OP 3 mn 1 mn and f = g n 1 n n + 1 mn n mn n + 1 mn = OP n + 1 n n + 1 mn 1 n m n. Let α M \ O m n and take j {1... m 1} such that jnα > jn + 1α. Then as jn + 1α mnα 1α jnα it is clear that f j α O m n. Thus we have: Lemma. Each element α M admits a factorization α = f j γ with 0 j m 1 and γ O m n which is unique unless α is constant. In particular the monoid M is generated by O m n and f. Notice that the uniqueness stated in the previous lemma follows immediately from the fact that f is a power of g and from Catarino and Higgins s result mentioned above. Now let N = Nψ 1. Clearly N is the subsemigroup of OP m n whose elements are the transformations α OP m n such that Imα A j for some j {1... m}. Next we justify the study made in the previous section by considering OP mnn which is a subsemigroup of N. For j {1... m} let ν j = 1 γ... γ m ; β j where γ = = γ m = 1 n 1 m and β n n j =. Clearly ν j j j N for all j {1... m}. Next let α = α 1... α m ; β j N with j {1... m}. Then γ = α 1... α m ; β 1 OP mnn ψ and α = γ ν j. On the other hand noticing that fψ = ; g m we also have αfψ m j+1 = γ i.e. α = γfψ j 1. Thus being ν j the element of N such that ν j ψ = ν j with j {1... m} we have: Lemma.3 The semigroup N is generated by OP mnn {ν... ν m }. Moreover every element of N is a product of an element of OP mnn by a power of f. Next for j {1... n 1} let 1 n j n j + 1 n n + 1 p j = j + 1 j + n m 1n m 1n + 1 mn j mn j + 1 mn OP j + 1 mnn. Notice that p i 1 n i n i + 1 n n + 1 m 1n m 1n + 1 mn 1 mn 1 = i + 1 n 1 i i i i i i + 1 for i {1... n 1} and p n 1 n n + 1 m 1n m 1n + 1 mn 1 mn 1 = 1 n n n n n 1 is a right identity of OP mnn. Lemma.4 Any transformation of OP mnn is a product of elements of M {p j 1 j }. Proof. By Theorem 1.3 it suffices to consider only transformations of OP mnn with rank n. Let γ be such a transformation. step 1. Let i = 1γ and α = γp n i+1 1. Then 1α = 1γp n i+1 1 = ip n i+1 1 = 1 and γ = αp n+i 1 1. If α M then γ satisfies the statement of the lemma. 5

6 Therefore suppose that α M. Hence mnα = 1 otherwise mnα = n whence α O mn and so α M. Let r {1... mn} be the least integer such that {r... mn}α = {1}. As α also has rank n and 1α = 1 then r n + 1. Thus r = t 1n + k + 1 for some t {... m} and k {1... n 1} notice that if k = 0 then α M. Let j = t 1nα 1 notice that 0 j n 1. If j = 0 then 1 t 1n t 1n + 1 tn 1 tn tn + 1 mn α = 1 1 n whence αp 1 = 1 t 1n t 1n + 1 tn 1 tn tn + 1 mn M n n 1 n 1 n n n and so as γ = αp1 n+i 1 = αp 1 p i 1 in this case γ also satisfies the statement of the lemma. Otherwise let β T mn be defined by mn j + 1 xα if 1 x t 1n xβ = xα j if t 1n + 1 x t 1n + k n if t 1n + k + 1 x mn. Then β M and α = βp j. step. Now in order to disregard the transformations p l with l > for a given j {1... n 1} we repeat step 1 considering in particular γ = p j. As 1p j = j + 1 we take α j = p j p n j = p j p n j 1 n j n j + 1 n n = 1 n j n j + 1 n j + 1 n j + 1 m 1n m 1n + 1 mn j mn j + 1 mn 1 mn. n j + 1 n j + 1 n j + 1 n j + n 1 Notice that α j M. Now by step 1 there exists β j M such that α j = β j p n j+1 1 = β j p n j. Thus p j = α j p j 1 = β j p n j p j 1 for some β j M. Finally by noticing that < j n 1 implies 1 n j we may deduce that any transformation of OP mnn with rank n is a product of elements of M {p j 1 j } as required. Now let 1 i 1n i 1n + 1 i 1n + i 1n + 3 in in + 1 mn c i = O 1 i 1n i 1n + 1 i 1n + 1 i 1n + in 1 in + 1 mn m n and 1 i 1n i 1n + 1 i 1n + j 1 i 1n + j i 1n + j + 1 in b ij = 1 i 1n i 1n + 1 i 1n + j 1 i 1n + j + 1 i 1n + j + 1 in in + 1 mn O in + 1 mn m n for i {1... m} and j {1... n 1}; and 1 i 1n i 1n + 1 i 1n + in s i = 1 i 1n i 1n + 1 i 1n + 1 i 1n + 1 in + 1 in + i + 1n i + 1n + 1 mn O i 1n + 1 i 1n + in i + 1n + 1 mn m n and 1 i 1n i 1n + 1 in j + 1 in j + in t ij = 1 i 1n in + 1 in + 1 in + in + j in + 1 in + j in + j + 1 i + 1n i + 1n + 1 mn O in + j in + j in + j + 1 i + 1n i + 1n + 1 mn m n for i {1... m 1} and j {1... n}. The authors proved in [1 Proposition.5] that the set {c i b ij s k t kl 1 i m 1 j n 1 1 k m 1 l n} is a generating set of the monoid O m n. On the other hand it is a routine matter to show that: 6

7 1. c i = f m i+1 c 1 f i 1 for i m;. b ij = f m i+1 b 1j f i 1 for i m and 1 j n 1; 3. s i = f m i+1 s 1 f i 1 for i m 1; and 4. t ij = f m i+1 t 1j f i 1 for i m 1 and j n. These observations combined with the previous three lemmas allow us to deduce the following result. Proposition.5 The set A = {f c 1 b b 1 s 1 t 1... t 1n p 1... p } is a generating set with n+ +1 elements of the monoid OP m n. Example.1 The monoid OP 3 4 is generated by the following transformations: f = ; c = ; b 11 = ; s = ; b 1 = ; t = ; b 13 = ; t = ; p 1 = ; t = ; p = The next series of lemmas will allow us to conclude that A is a least size generating set of OP m n for m > and contains a least size generating set of OP n. Lemma.6 Any generating set of OP m n contains at least a non identity element of rank mn and n distinct elements of rank mn 1. Proof. Let X be a generating set of OP m n. Notice that the group of units of OP m n is clearly generated by the permutation f = g n which has order m. Hence X must contain a non identity element of rank mn. Now let ξ 1... ξ k k 1 be all the elements of X of rank mn 1. Then any element of OP m n of rank mn 1 is of the form αξ j f i for some j {1... k} i {0... m 1} and a product α OP m n. As the elements of rank mn 1 of the above form can not have more than mk distinct images and on the other hand we have precisely mn possible distinct images for an element of OP m n of rank mn 1 we deduce that mn mk and so k n as required. Lemma.7 For m > any generating set of OP m n contains at least n distinct elements of rank m 1n. Proof. Let T j = {α OP m n rankα = m 1n and knα = kn + 1α = i 1n + j for some 1 i k m} for j {1... n}. Clearly T 1... T n are n two by two disjoint non-empty subsets of elements of rank m 1n of OP m n. Let j {1... n} and take α T j. Let i k {1... m} be such that knα = kn + 1α = i 1n + j. Suppose that α = α α for some α α OP m n and take αψ = α 1... α m ; β α ψ = α 1... α m; β and α ψ = α 1... α m; β. Notice that α α and α are elements of OP m n of type 1 whence α l α l α l O n for l {1... m}. Also observe that nα k = j = 1α k+1 Imα k = {1... j} Imα k+1 = {j... n} and α l = 1 for l {1... m} \ {k k + 1}. Moreover we have α l = α l α lβ for l {1... m} and on the other hand β = β β from which follows that rankβ = m 1 or rankβ = m 1 since rankβ = m 1. Next our goal is to show that α T j or α T j. We consider two cases: rankβ = m 1 or rankβ = m. First admit that β has rank m 1. Then α has rank m 1n and so Kerα = Kerα. Hence Kerα k = Kerα k and Kerα k+1 = Kerα k+1 from which follows that Imα k = Imα k = j and Imα k+1 = Imα k+1 = n j + 1. Thus nα k j and 1α k+1 j since 1α k nα k and 1α k+1 nα k+1. On the other hand the equality Kerα = Kerα also implies that knα = kn + 1α whence nα k = 1α k+1 and so nα k = 1α k+1 = j. Then knα = kn + 1α = kβ 1n + j and finally we conclude that α T j. 7

8 Secondly suppose that β has rank m i.e. β is a power of g m. Then β must have rank m 1 and so α has rank m 1n. As α k α kβ = α k then {1... j} = Imα k Imα kβ and so nα kβ j. Similarly as α k+1 α k+1β = α k+1 then {j... n} = Imα k+1 Imα k+1β and so 1α k+1β j. Now by noticing that β is a power a g m we have k + 1β = kβ + 1 and so kβ + 1β = k + 1β β = k + 1β = kβ = kβ β. Hence j nα kβ 1α kβ +1 = 1α k+1β j i.e. nα kβ = 1α kβ +1 = j from which follows that kβ nα = kβ n + 1α = kβ 1n + j. Thus α T j as required. Now by induction on k it is clear that to write an element of T j as a product of k elements of OP m n we must have a factor belonging to T j for all 1 j n. This fact proves the lemma. The next lemma helps us to find the least number of elements of rank n required on a generating set of OP m n. Lemma.8 Let α 1... α m ; β OP m n ψ be such that iβ = jβ for some 1 i < j m. Then Imα i + Imα j n +. Moreover if Imα i + Imα j = n + then α 1... α m ; β is of type 3 and: 1. α i OP n \ O n or α j OP n \ O n ;. Imα i Imα j = A iβ and so α 1... α m ; βψ 1 is a transformation of OP m n of rank n; 3. Imα k = 1 for k {1... m} \ {i j}. Proof. We begin by proving that Imα i + Imα j n +. First suppose that α i α j O n. Then as iβ = jβ we have 1α i nα i 1α j nα j or 1α j nα j 1α i nα i whence Imα i Imα j has at least Imα i + Imα j 1 distinct elements notice that we may have nα i = 1α j or nα j = 1α i. As Imα i Imα j A iβ it follows that Imα i + Imα j n + 1. Next suppose that α i OP n \ O n. Then α j O n and we have t + 1α i nα i 1α j nα j 1α i tα i for some 1 t m 1 whence Imα i Imα j has at least Imα i + Imα j distinct elements notice that we may have nα i = 1α j and nα j = 1α i and so Imα i + Imα j n +. Finally as the case α j OP n \ O n is similar to the previous one we proved that Imα i + Imα j n + for all cases. Now in order to prove the second part of the lemma admit that Imα i + Imα j = n +. Notice that by the first part of the proof α i OP n \ O n or α j OP n \ O n and so α 1... α m ; β must be of type 3. On the other hand as n Imα i Imα j Imα i + Imα j = n we have Imα i Imα j = A iβ. Suppose that α i OP n \ O n and let t {1... m 1} be as above. Let k {i j 1}. Let l {1... m} \ {i... j}. Then with the obvious adaptation if k or l does not exist we have t + 1α i nα i 1α k nα k 1α j nα j 1α l nα l 1α i tα i. Hence nα i = 1α j and nα j = 1α i otherwise Imα i Imα j would have at least Imα i + Imα j 1 = n + 1 distinct elements which is a contradiction. Thus nα i = 1α k = = nα k = 1α j if k exists and nα j = 1α l = = nα l = 1α i if l exists. Anyway we proved that Imα k = 1 for k {1... m} \ {i j}. Similarly if α j OP n \ O n we have Imα k = 1 for k {1... m} \ {i j} as required. Lemma.9 Any generating set of OP m n contains at least elements of rank n. Proof. For 1 i define P i = {γ 1... γ m ; λ N Imγ k = n i + 1 and Imγ l = i + 1 for some 1 k l m such that k l}. Notice that p i ψ P i and by Lemma.8 all elements of P i are of type 3 and all elements of P i ψ 1 have rank n for 1 i. Moreover P 1... P are two by two disjoint subsets of OP m nψ. In fact suppose there exists γ 1... γ m ; λ P i P j for some 1 i < j. Let 1 k l m with k l be such that Imγ k = n i + 1 and Imγ l = i + 1. Then by Lemma.8 we have Imγ t = 1 for t {1... m} \ {k l}. Hence Imγ k = n j + 1 or Imγ k = j + 1. If Imγ k = n j + 1 then i = j which is a contradiction. On the other hand if Imγ k = j + 1 then n = i + j < + n + n = n which is again a contradiction. Therefore P i P j = for 1 i < j. 8

9 It follows that P 1 ψ 1... P ψ 1 are two by two disjoint subsets of OP m n of elements of rank n. Now let i {1... } and take γ = γ 1... γ m ; λ P i. Let 1 k l m with k l be such that Imγ k = n i + 1 and Imγ l = i + 1. Suppose that γ = αψα ψ for some α α OP m n and take αψ = α 1... α m ; β and α ψ = α 1... α m; β. Notice that γ j = α j α jβ for 1 j m. Moreover as γ is of type 3 then either α or α is of type 3. Next our goal is to show that αψ P i or α ψ P i. We begin by observing that if kβ = lβ then by Lemma.8 we have Imα k + Imα l n + and so as n i + 1 = Imγ k = Imα k α kβ Imα k and i + 1 = Imγ l = Imα l α lβ Imα l it follows that Imα k = n i + 1 and Imα l = i + 1. We consider two cases. First if α is of type 3 in particular we have αψ N as β is a constant transformation we have kβ = lβ and so by the above observation we may deduce immediately that αψ P i. On the other hand admit that α is not of type 3. Then kβ lβ. In fact if kβ = lβ then Imα k + Imα l = n i i + 1 = n + and so by Lemma.8 α must be of type 3 which is a contradiction. Also notice that α must be of type 3 and so β is a constant transformation. In particular kββ = lββ. Hence by Lemma.8 we have Imα kβ + Imα lβ n +. Moreover since Imγ k = Imα k α kβ Imα kβ and Imγ l = Imα l α lβ Imα lβ we have n i + 1 = Imγ k Imα kβ and i + 1 = Imγ l Imα lβ. Thus we obtain precisely Imα kβ = n i + 1 and Imα lβ = i + 1 which proves that α ψ P i. Now by induction on k it is easy to show that to write an element of P i ψ 1 as a product of k elements of OP m n we must have a factor that belongs to P i ψ 1 for all 1 i. This fact proves the lemma. Now for m > from the previous lemmas we deduce immediately that A is a least size generating set of OP m n. On the other and regarding OP n it is a routine matter to show that: 1. s 1 = b 11 b 1 b 1 fp n 1 ;. t 1j = fc j 1 1 p j 1 f for j t 1j = c j 1 b 11 p n j p j 1 1 f for + j n ; and 4. t 1 = b 11 p 1 f and t 1n = fc 1 fp n 1 f. Hence from these equalities and Lemmas.6 and.9 it follows that {f c 1 b b 1 p 1... p } is a least size generating set of OP n. Therefore we have proved: Theorem.10 The rank of OP m n is equal to n for m > and equal to n for m =. 3 The rank of the monoid OD m n Consider the reflexion 1 n 1 mn h =. mn mn 1 1 Observe that h is a permutation of order two of X mn and h OD m n. Moreover given α T mn we have α = h α = hhα and α is an order-reversing transformation if and only if hα respectively αh is an order-preserving transformation. Thus clearly the monoid OD m n is generated by O m n {h}. As recalled in Section the authors proved in [1] that C = {c i b ij s k t kl 1 i m 1 j n 1 1 k m 1 l n} is a generating set with mn n elements of the monoid O m n. Hence C {h} generates OD m n. In order to reduce the number of generators consider 1 i 1n i 1n + 1 in j + 1 in j + in s ij = 1 i 1n i 1n + 1 in j + 1 in j + 1 in j + 1 in + 1 in + in + j i + 1n i + 1n + 1 mn O in j + 1 in j + in in i + 1n + 1 mn m n for i {1... m 1} and j {1... n}. Notice that we have s i = s in for i {1... m 1}. Finally if m is odd consider also the transformation u j O m n of rank mn 1 whose image is {1... mn} \ { m 1 n + j} and whose kernel is defined by the partition {{1}... { m 1 n + n m 1 j} { n + n m 1 j + 1 n + n m 1 j + } { n + n j + 3}... {mn}} for each 1 j n. 9

10 Example 3.1 For m = 3 and n = 5 we have: u 1 = u = u 3 = The proofs of the equalities stated in the next lemma are routine. Lemma 3.1 The following identities hold: 1. c i = hb m i+1 b m i+1n b m i+11 h for 1 i m;. b i1 = hb m i+1 b m i+1n b m i+1 c m i+1 h for 1 i m; 3. b ij = hb m i+1n j b m i+1n j 1 b m i+1 c m i+1 b m i+1 b m i+1n b m i+1n j+1 h for 1 i m and j n 1; 4. b m+1 j = u n j+1 u j for m odd and 1 j n ; 5. b m+1 n j = hu n j+1 u j+1 h for m odd and 1 j n n 1; 6. t ij = hs m ij h for 1 i m 1 and 1 j n; 7. t ij = c n j i hb b j hs in j+1hs m i h with b l = b m in j+l 1b m in j+l b m il for l j 1 i m 1 and j n 1. Now let A be the set {c i b ij s kl s m r s t h 1 i m 1 j 1 k m n 1 l n +1 r 1 t m 1} if m is even and the set {c i b ij u k s il s t h 1 i m 1 n 1 j n 1 1 k l n 1 1 t m 1} if m is odd. By using the equalities of Lemma 3.1 it is not difficult to show that any element of C is a product of elements of A. Thus it follows that: Proposition 3. The set A generates the monoid OD m n. Furthermore A has mn + m 1n + 1 elements. Next we aim to show that the rank of OD m n is precisely mn + m 1n + 1. Let U be a generating set of OD m n. First notice that as h is the unique non-identity permutation in OD m n we must have h U. On the other hand recall that in the proof of [9 Theorem 1.5] Fernandes et al. showed that any generating set of the monoid OD n for n has at least n elements of rank n 1. A similar argument allow us to conclude that U must have at least mn elements of rank mn 1. In fact take K i = {1... mn} \ {i} for 1 i mn and let ξ 1... ξ k be all the elements of U of rank mn 1. Then k 1 and for all 1 i k there exists 1 l i mn such that Imξ i = K li. Now given an element α OD m n of rank mn 1 we have α = ξξ i or α = ξξ i h for some ξ OD m n and 1 i k. Hence Imα = Imξ i = K li or Imα = Imξ i h = K mn li+1. As we have mn possible distinct images for a transformation of OD m n of rank mn 1 the set {K l1... K lk K mn l K mn lk +1} has at least mn elements. It follows that k mn and so k mn. Therefore we have proved: Lemma 3.3 Any generating set of OD m n contains h and at least mn distinct elements of rank mn 1. Regarding generators of rank m 1n we have: 10

11 Lemma 3.4 Any generating set of OD m n contains at least Proof. For 1 i m 1 and 1 j n consider m 1n distinct elements of rank m 1n. Q ij = {α O m n rankα = m 1n and inα = in + 1α = k 1n + j for some 1 k m}. The authors proved in [1 Theorem.6] that the family {Q ij 1 i m 1 1 j n} consists on m 1n two by two disjoint non-empty subsets of O m n such that given α 1 α O m n if α 1 α Q ij then α 1 Q ij or α Q ij for 1 i m 1 and 1 j n. On the other hand given α T mn it is easy to show that α Q ij if and only if hαh Q m in j+1 and consequently hα Q ij if and only if αh Q m in j+1 3 for 1 i m 1 and 1 j n. Next for 1 i m 1 and 1 j n define T ij = {α OD m n α Q ij Q m in j+1 or hα Q ij Q m in j+1 }. Observe that clearly T ij = T m in j+1 for 1 i m 1 and 1 j n. Moreover if 1 i i m 1 and 1 j j n are such that T ij T i j then i j = i j or i j = m i n j + 1. In fact suppose that there exists α T ij T i j. If α O m n then α Q ij Q m in j+1 Q i j Q m i n j +1. On the other hand if α O m n then hα Q ij Q m in j+1 Q i j Q m i n j +1. Then for both cases Q ij Q i j or Q ij Q m i n j +1 or Q m in j+1 Q i j or Q m in j+1 Q m i n j +1 from which follows that i j = i j or i j = m i n j + 1 regarding that {Q ij 1 i m 1 1 j n} has m 1n two by two disjoint elements. Therefore we may deduce that the family {T ij 1 i m 1 1 j n} consists on m 1n two by two disjoint non-empty subsets of OD m n. Now by proving that any generating set of OD m n contains an element of T ij for all 1 i m 1 and 1 j n the proof of the lemma follows. To accomplish this aim we show that for 1 i m 1 and 1 j n given α 1... α k OD m n k N such that α 1 α k T ij we have α t T ij for some t {1... k}. Furthermore in order to prove this last statement by induction on k it suffices to consider k =. First notice that given α T mn it follows from 3 that {α hαh αh hα} T ij or {α hαh αh hα} T ij = 4 for 1 i m 1 and 1 j n. Hence let 1 i m 1 and 1 j n and let α 1 α OD m n be such that α 1 α T ij. Next we show that α 1 T ij or α T ij by considering four cases which finishes the proof. case 1. If α 1 α O m n then α 1 α Q ij Q m in j+1 and so by the observation at the start of the proof we have α 1 Q ij Q m in j+1 or α Q ij Q m in j+1 whence α 1 T ij or α T ij. case. If α 1 O m n and α O m n then α 1 hhα = α 1 α T ij and α 1 h hα O m n and so by case 1 α 1 h T ij or hα T ij. Hence by 4 α 1 T ij or α T ij. case 3. If α 1 O m n and α O m n then hα 1 O m n and α O m n and by 4 hα 1 α = hα 1 α T ij. Hence by case 1 hα 1 T ij or α T ij and so again by 4 α 1 T ij or α T ij. case 4. Finally if α 1 O m n and α O m n then α 1 O m n and α h O m n and by 4 α 1 α h = α 1 α h T ij. Thus by case 1 α 1 T ij or α h T ij and so once again by 4 α 1 T ij or α T ij as required. Now from Proposition 3. and Lemmas 3.3 and 3.4 the main result of this section follows immediately. Theorem 3.5 The rank of OD m n is mn + m 1n The rank of the monoid OR m n As for OD m n if α T mn then α is an orientation-reversing transformation if and only if hα respectively αh is an orientation-preserving transformation. Hence as α = h α = hhα it is clear that the monoid OR m n is generated by OP m n {h}. From Section recall that {f c 1 b b 1 s 1 t 1... t 1n p 1... p } is a generating set with n elements of the monoid OP m n. Furthermore for m = the set {f c 1 b b 1 p 1... p } generates OP n and has just n elements. Now for 1 j n let v j be the transformation of O m n of rank mn 1 whose image is {1... mn} \ {j} and whose kernel is defined by the partition {{1}... { n j} { n j + 1 n j + } { n j + 3}... {mn}}. 11

12 Example 4.1 For m = 3 and n = 5 we have: v 1 = v = v 3 = It is a routine matter to prove the following lemma. Lemma 4.1 The following equalities hold: 1. b 1j = v n j+1 v j for 1 j n ;. b 1n j = f m 1 hv n j+1 v j+1 f m 1 h for 1 j n n 1; 3. c 1 = hfb 1 b 1n b 1 b 11 f m 1 h; 4. t 1n = f m hs 1 f m h; 5. t 1j = c n j 1 hf b 1n j+1 b 1n j b 1 b 1n j+ b 1n j+1 b 13 b 1 b 1n b 1j t 1n j+1 s 1 f m h for j n 1. Therefore it is easy to prove that: Proposition 4. The set {f s 1 t 1... t 1 n p 1... p v 1... v n h} has n + elements and generates OR m n. Furthermore for m = the set {f p 1... p v 1... v n h} has n + + elements and generates OR n. + In what follows we show that the first and second sets of the last result are a least size generating set of OR m n for m > and of OP n respectively. First notice that any generating set OR m n must contain two distinct permutations of X mn one preserving the orientation and another reversing the orientation. Next we consider transformations of rank mn 1. Lemma 4.3 Any generating set of OR m n contains at least n distinct elements of rank mn 1. Proof. For each 1 t mn let K t = {1... mn}\{t}. Let U be a generating set of OR m n and let ξ 1... ξ k k 1 be all the elements of U of rank mn 1. Then for 1 j k we have that Imξ j = K lj for some 1 l j mn. For 1 j k and 1 i m 1 define l ik+j as being the element of X mn congruent modulo mn with l j + in. Now take a transformation γ OR m n of rank mn 1. Then γ = αξ j f i or γ = αξ j f i h for some j {1... k} i {0... m 1} and α OR m n. Hence Imγ = K lik+j or Imγ = K mn lik+j +1. As we have precisely mn possible distinct images for a transformation of OR m n of rank mn 1 the set {K l1... K lmk K mn l K mn lmk +1} has at least mn distinct elements. Thus mk mn and so k n as required. For the transformations of rank m 1n we have: Lemma 4.4 For m > any generating set of OR m n contains at least n distinct elements of rank m 1n. Proof. This proof is similar to Lemma 3.4 and so we omit some details. For j {1... n} consider T j = {α OP m n rankα = m 1n and knα = kn + 1α = i 1n + j for some 1 i k m}. Recall that in the proof of Lemma.7 we showed that T 1... T n are n two by two disjoint subsets of OP m n such that given α 1 α OP m n if α 1 α T j then α 1 T j or α T j for 1 j n. Moreover it is easy to show that given α T mn we have α T j if and only if hαh T n j+1 and consequently hα T j if and only if αh T n j+1 for 1 j n. Define U j = {α OR m n α T j T n j+1 or hα T j T n j+1 } 1

13 for 1 j n. First observe that clearly U j = U n j+1 for 1 j n. Also it is easy to show that if j j {1... n} are such that U j U j then j {j n j + 1}. It follows that U 1... U n are n two by two disjoint non-empty subsets of OR m n. Secondly notice that given α T mn it is also easy to show that {α hαh αh hα} U j or {α hαh αh hα} U j = for 1 j n. Hence it is a routine matter to prove that for 1 j n and α 1 α OR m n such that α 1 α U j we have α 1 U j or α U j. It follows by induction on k that to write an element of U j as a product of k elements of OR m n we must have a factor that belongs to U j for 1 j n which proves the lemma. Next we deal with transformations of OR m n of rank n. As for OP m n we aim to show that in order to generate OR m n at least distinct transformations of rank n are required. We begin with an observation for which we need to introduce notation first. For each n N denote by h n the reflexion 1 n 1 n permutation of X n n 1 1 n. Observe that with this notation we have h = h mn and moreover hψ = h n h n... h n ; h m. Furthermore being α T m n and αψ = α 1 α... α m ; β we obtain Notice that clearly hαhψ = h n α m h n h n α m 1 h n... h n α 1 h n ; h m βh m. 5 Imh m βh m = Imβ and Imh n α i h n = Imα i 6 for 1 i m. Now recall the two by two disjoint subsets of OP m nψ P i = {γ 1... γ m ; λ N Imγ k = n i + 1 and Imγ l = i + 1 for some 1 k l m such that k l} with 1 i considered in the proof of Lemma.9. Given α T mn from 5 and 6 it follows immediately that αψ P i if and only if hαhψ P i and consequently hαψ P i if and only if αhψ P i 7 for 1 i. Next following the same strategy of Lemmas 3.4 and 4.4 we define Q i = {α OR m n αψ P i or hαψ P i } for 1 i. First observe that as P 1 ψ 1... P ψ 1 are two by two disjoint subsets of transformations of rank n of OP m n it is clear that also Q 1... Q are two by two disjoint subsets of transformations of rank n of OR m n. On the other hand from 7 we also deduce that {α hαh αh hα} Q i or {α hαh αh hα} Q i = 8 for α T mn and 1 i. Now recall we proved in Lemma.9 that α 1α P i ψ 1 implies α 1 P i ψ 1 or α P i ψ 1 for α 1 α OP m n and 1 i. Hence by using properly the property 8 it is easy to show also that given α 1 α OR m n if α 1 α Q i then α 1 Q i or α Q i for 1 i. Thus by induction on k it follows that to write an element of Q i as a product of k elements of OR m n we must have a factor that belongs to Q i for 1 i. Therefore we have proved that: Lemma 4.5 Any generating set of OR m n contains at least distinct elements of rank n. Finally it follows our main objective of this section. Theorem 4.6 The rank of OR m n is equal to n + + for m > and equal to n + + for m =. 13

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