Indecomposable involutive set-theoretic solutions of the Yang-Baxter equation and orthogonal dynamical extensions of cycle sets
aa r X i v : . [ m a t h . QA ] N ov Indecomposable involutive set-theoretic solutions of theYang-Baxter equation and orthogonal dynamicalextensions of cycle sets ⋆ Marco CASTELLI a , Francesco CATINO a , Paola STEFANELLI a a Dipartimento di Matematica e Fisica “Ennio De Giorgi”Universit`a del SalentoVia Provinciale Lecce-Arnesano73100 Lecce (Italy)
Abstract
Employing the algebraic structure of the left brace and the dynamical extensionsof cycle sets, we investigate a class of indecomposable involutive set-theoretic so-lutions of the Yang-Baxter equation having specific imprimitivity blocks. More-over, we study one-generator left braces of multipermutation level 2.
Keywords: cycle set, set-theoretic solution, Yang-Baxterequation, brace
1. Introduction
An intriguing challenge that attracts mathematicians in recent years is pro-viding a complete description of the solutions to the quantum Yang-Baxterequation, a fundamental equation of Theoretical Physics. The interest in thisfield of research has arisen from the paper by Drinfel’d [15], where the authorsuggested the study of set-theoretic solutions. A set-theoretic solution of theYang-Baxter equation is a pair (
X, r ) where X is a non-empty set and r is amap from X × X into itself satisfying the braid relation( r × id X )(id X × r )( r × id X ) = (id X × r )( r × id X )(id X × r ) . Writing the map r as r ( x, y ) = ( λ x ( y ) , ρ y ( x )), for all x, y ∈ X , where λ x and ρ y are maps from X into itself, a solution ( X, r ) is said to be non-degenerate if ⋆ This work was partially supported by the Dipartimento di Matematica e Fisica “EnnioDe Giorgi” - Universit`a del Salento. The authors are members of GNSAGA (INdAM). ∗ Corresponding author
Email addresses: [email protected] (Marco CASTELLI), [email protected] (Francesco CATINO), [email protected] (Paola STEFANELLI)
November 23, 2020 x , ρ y ∈ Sym( X ), for all x, y ∈ X . A systematic investigation of set-theoreticsolutions started in the late ’90s, by Gateva-Ivanova and Van den Bergh [17]and Etingof, Schedler, and Soloviev [16] who focused on the special class ofnon-degenerate involutive solutions, i.e., r = id X × X . For convenience, we callinvolutive non-degenerate solutions simply solutions throughout the paper.A conceptually simple but efficacy strategy for determining all solutions con-sists of constructing solutions starting from smaller ones. To implement thismethod, it is crucial finding solutions which can not be deconstructed into otherones, the so-called indecomposable solutions. Precisely, a solution ( X, r ) is saidto be decomposable if there exists a partition { Y, Z } of X such that r | Y × Y and r | Z × Z are still solutions; otherwise, ( X, r ) is called indecomposable . In particular,Etingof, Schedler, and Soloviev in [16] proved that every decomposable solution(
X, r ) can be obtained through two suitable solutions (
Y, s ) and (
Z, t ), with Y and Z proper subsets of X , s = r | Y × Y , and t = r | Z × Z . Moreover, they showedthat the unique indecomposable solution having a prime number p of elementsis, up to isomorphism, the pair ( Z /p Z , u ) where u ( x, y ) := ( y − , x + 1), for all x, y ∈ Z /p Z .In recent years, some authors used the links between solutions and algebraicstructures to provide new descriptions of indecomposable solutions, as Chouraqui[10], Rump [29], Smoktunowicz and Smoktunowicz [33], and Pinto together withthe first two authors [5]. In particular, Smoktunowicz and Smoktunowicz [33]found a striking connection between these solutions and left braces, a general-ization of radical rings, introduced by Rump in [26]. As reformulated in [7], aset A is said to be a left brace if it is endowed of two operations + and ◦ suchthat ( A, +) is an abelian group, ( A, ◦ ) a group, and a ◦ ( b + c ) + a = a ◦ b + a ◦ c is satisfied, for all a, b, c ∈ A . Later, the link between indecomposable solu-tions and left braces found in [33] has been developed by Rump [29]. Indeed,he showed that a complete description of left braces, on the one hand, and asuitable theory of coverings for solutions, on the other, allows to classify allindecomposable solutions. As a first result of this paper, we show that justTheorem 2 of [29], together with a result of Smocktunowitcz [32, Theorem 19]on left braces with nilpotent multiplicative group, reduces the classification offinite indecomposable solutions with nilpotent regular permutation group to theones having prime-power size. Alongside these theoretical results, concrete fam-ilies of indecomposable solutions have been provided by the first two authorstogether with Pinto [5] using Vendramin’s dynamical cocycles [34].Following a different approach, some authors studied indecomposable solutionsfocusing on the associated permutation group , i.e., the group G ( X, r ) generatedby the set { λ x | x ∈ X } . Specifically, one of the main objectives is to describeall indecomposable solutions having a specific permutation group. Actually,2into, Rump and the first author [6] developed a method to construct all in-decomposable solutions having prime-power size and with cyclic permutationgroup. In addition, they constructed, distinguishing the isomorphism classes,the indecomposable ones with cardinality pq (where p and q are prime numbersnot necessarily distinct) and abelian permutation group. Another remarkableresult was recently obtained by Rump, who gave a theoretical classification, viaindecomposable cycle sets and cocyclic left braces, all the solutions (not neces-sarily indecomposable) having cyclic permutation group [30]. Let us recall thata set X with a binary operation · is called a cycle set if each left multiplication σ x : X −→ X, y x · y is bijective and( x · y ) · ( x · z ) = ( y · x ) · ( y · z ) (1)holds, for all x, y, z ∈ X . Rump introduced this structure in [24] precisely toinvestigate solutions. Indeed, he found a one-to-one correspondence betweensolutions and the class of non-degenerate cycle sets. Up until now, the studyof non-degenerate cycle sets has been allowed for obtaining descriptions andconstructions of new interesting families of solutions, as one can see in [28, 3, 4].However, some authors continued the study of indecomposable solutions with-out using cycle sets. Ced´o, Jespers, and Okni´nski [8], following a suggestiondue to Ballester-Bolinches [20], classified indecomposable solutions with primi-tive permutation group, showing that this class of solutions coincides with theindecomposable ones having a prime number of elements and extending thepartial results obtained by Pinto in [23, Chapter 4]. In a recent work, Jedliˇcka,Pilitowska, and Zamojska-Dzienio [18] concretely classified the indecomposablesolutions having abelian permutation group and multipermutation level 2, ex-tending some results contained in [6]. These works naturally open a new per-spective on the investigations of all indecomposable solutions. Indeed, one cantry to describe the indecomposable solutions which have not prime size usingfurther information coming from the imprimitivity blocks. Besides, since all themultipermutation indecomposable solutions provided in [18, 6, 5, 8] have mul-tipermutation level at most 2, except [6, Example 3], a first step in this sensecan be the concrete construction of indecomposable solutions having greatermultipermutation level.In this spirit, the main aim of this paper is to investigate a class of indecom-posable solutions obtained by dynamical cocycles of cycle sets. In particular,we study the dynamical extensions X × α S for which the complete blocks sys-tem B X induced by X has an orthogonal system of blocks. For this reason, wecall this extensions orthogonal . As one can find in [14, Definition 2.9], given animprimitive group G acting on a set X of size mn , for certain positive integers n and m greater than 1, we say that two systems of blocks A = { A , . . . , A m } and B = { B , . . . , B n } of size m and n , respectively, are orthogonal if | A i ∩ B j | = 1,for all i ∈ { , . . . , m } and j ∈ { , . . . , n } (for some remarkable results involving3rthogonal systems of blocks see, for instance, [22, 13]). In this context, wepay particular attention to the constant dynamical extensions. This class ofextensions was recently investigated by a cohomological point of view by Lebedand Vendramin [21]. Following a different approach, using a property of theorthogonal systems of blocks, we show that orthogonal constant dynamical ex-tensions can be constructed by particular semidirect products of cycle sets, aconstruction of cycle sets introduced by Rump [27]. As one can expect, or-thogonal dynamical extensions are also related to the semidirect product of leftbraces. In this regard, the powerful of left braces to check this class of cycle setsbecomes clear in the further description of these extensions which we provide.As an application of our results, we give several instances of indecomposable so-lutions. A lot of these are different from those already contained in [5, 6, 30, 18].Specifically, we construct indecomposable cycle sets (and so indecomposablesolutions) having abelian non-cyclic permutation group and multipermutationlevel 3. Moreover, we use the class of latin cycle sets, i.e., the cycle sets that alsoare quasigroups recently studied in [3], to provide a construction of orthogonaldynamical extensions which give rise to indecomposable irretractable solutionsdifferent from the ones obtained in [5].In the last part of this paper, which is self-contained with respect to the pre-vious ones, we consider the class of one-generator left braces, which are closelyrelated to indecomposable solutions (see [33, 31] for more details). Recently,Rump showed several results involving one-generator left braces: in particular,he described one-generator left braces [31, Theorem 2] of multipermutation levelat most 2, giving an answer to [33, Question 5.5]. At this end, he introduceda left brace structure over the group ring Z [ G ], where G is the infinite cyclicgroup. Following a different approach, we study one-generator left braces ofmultipermutation level 2 focusing on the map λ . We show that, if A is anarbitrary one-generator left brace of multipermutation level 2, then the group λ ( A ) is cyclic. As a consequence of this fact, we finally provide a proof of [31,Theorem 2] which does not make use of the structure of the left brace over thegroup ring Z [ G ].
2. Basic results
In this section, we mainly recall some basics on cycle sets and left braces thatare useful throughout the paper. In addition, using some well-known results onleft braces, we describe the structure of uniconnected cycle sets with nilpotentpermutation group.
In [24], Rump found a one-to-one correspondence between solutions and aspecial class of cycle sets, i.e., non-degenerate cycle sets. To illustrate this4orrespondence, let us firstly recall the following definition.
Definition 1 (p. 45, [24]) . A pair ( X, · ) is said to be a cycle set if each leftmultiplication σ x : X −→ X, y x · y is invertible and( x · y ) · ( x · z ) = ( y · x ) · ( y · z ) , for all x, y, z ∈ X . Moreover, a cycle set ( X, · ) is called non-degenerate if thesquaring map x x · x is bijective. Convention.
All the cycle sets are non-degenerate throughout the paper.
Proposition 1 (Propositions 1-2, [24]) . Let ( X, · ) be a cycle set. Then the pair ( X, r ) , where r ( x, y ) := ( σ − x ( y ) , σ − x ( y ) · x ) , for all x, y ∈ X , is a solution of theYang-Baxter equation which we call the associated solution to ( X, · ) . Moreover,this correspondence is one-to-one. Recall that a first useful tool to construct new solutions, introduced in [16],is the so-called retract relation , an equivalence relation on X which we denoteby ∼ r . Precisely, if ( X, r ) is a solution, then x ∼ r y if and only if λ x = λ y , forall x, y ∈ X . In this way, ( X, r ) induces another solution, having the quotient X/ ∼ r as underlying set, which is named retraction of ( X, r ) and is denoted byRet(
X, r ). As one can expect, the retraction of a solution corresponds to theretraction of a non-degenerate cycle set. Specifically, in [24] Rump showed thatthe binary relation ∼ σ on X given by x ∼ σ y : ⇐⇒ σ x = σ y for all x, y ∈ X , is a congruence of ( X, · ) and he proved that the quotient X/ ∼ ,which we denote by Ret( X ), is a cycle set whenever X is non-degenerate andhe called it the retraction of ( X, · ). As the name suggests, if ( X, · ) is the cycleset associated to a solution ( X, r ), then the retraction Ret( X ) is the cycle setassociated to Ret( X, r ). Besides, a cycle set X is said to be irretractable ifRet( X ) = X , otherwise it is called retractable .The following definition is of crucial importance for our scopes. Definition 2.
A cycle set X has multipermutation level n if n is the minimalnon-negative integer such that Ret n ( X ) has cardinality one, whereRet ( X ) := X and Ret i ( X ) := Ret(Ret i − ( X )), for i > . Clearly, a cycle set of multipermutation level n is retractable, but the converseis not necessarily true. 5or a cycle set X , the permutation group generated by the set { σ x | x ∈ X } will be denoted by G ( X ) and we call it the associated permutation group . Ob-viously, in terms of solutions, the associated permutation group is exactly thepermutation group generated by the set { λ x | x ∈ X } .Our attention is mainly posed on cycle sets that are indecomposable. Definition 3.
A cycle set ( X, · ) is said to be indecomposable if the permutationgroup G ( X ) acts transitively on X . Moreover, X is said to be uniconnected if G ( X ) acts regularly on X , i.e., G ( X ) acts freely and transitively on X .Note that a solution ( X, r ) is indecomposable if and only if the associated cycleset ( X, · ) is indecomposable. In the rest of the paper, we will study indecom-posable solutions by their associated cycle sets. In every case, all the resultsinvolving cycle sets can be translated in terms of solutions by Proposition 1.A particular family of indecomposable cycle sets that has been considered bysome authors (see, for example, [16, 6]) is that of cycle sets having cyclic per-mutation group. In [6], a method to construct all the indecomposable cycle setsof prime-power size with cyclic permutation group and multipermutation levelgreater than 1 was developed. Theorem 2 (Theorems 8 - 9, [6]) . Let p be a prime number, X := { , . . . , p k − } , n ∈ N \ { } , j , . . . , j n ∈ N ∪ { } such that j n = 0 , j = k , j i < j i − , forevery i ∈ { , . . . , n } , and { f i } i ∈{ ,...,n − } a set of maps such that f i : Z /p j i Z −→ { , . . . , p j i − /p j i − } f i (0) = 0 , for every i ∈ { , . . . , n − } , and the map ϕ i : { , . . . , p j i − } −→ { , . . . , p j i − − } l p j n − f n − ( l ) + · · · + p j i f i ( l ) is injective, for every i ∈ { , . . . , n − } . Moreover, set ψ := (0 . . . p k − and σ i := ψ p jn − f n − ( i )+ ··· + p j f ( i ) , (2) for every i ∈ X . Define K j,i := j + 1 + p n − f n − ( i ) + · · · + p j f ( i ) and Q j,i := p j n − f n − ( i ) + · · · + p j f ( i ) + p j n − f n − ( K j,i ) + · · · + p j f ( K j,i ) , for every i, j ∈ X , and suppose that Q i,j ≡ Q j,i ( mod p k ) for every i, j ∈{ , . . . , p k − } .Then X is an indecomposable cycle set of level n and cyclic permutation group < ψ > such that | Ret i ( X ) | = p j i , for every i ∈ { , . . . , n } .Conversely, every indecomposable cycle set with cyclic permutation group arisesin this way. .2. Solutions of the Yang-Baxter equation and left braces At first, we introduce the following definition that, as observed in [7], is equiv-alent to the original introduced by Rump in [26].
Definition 4 ([7], Definition 1) . A set A endowed of two operations + and ◦ is said to be a left brace if ( A, +) is an abelian group, ( A, ◦ ) a group, and a ◦ ( b + c ) + a = a ◦ b + a ◦ c, for all a, b, c ∈ A .Given a left brace A and a ∈ A , let us denote by λ a : A −→ A the map from A into itself defined by λ a ( b ) := − a + a ◦ b, (3)for all b ∈ A . As shown in [26, Proposition 2] and [7, Lemma 1], these mapshave special properties. We recall them in the following proposition. Proposition 3.
Let A be a left brace. Then, the following are satisfied:1) λ a ∈ Aut( A, +) , for every a ∈ A ;2) the map λ : A −→ Aut( A, +) , a λ a is a group homomorphism from ( A, ◦ ) into Aut( A, +) . The map λ is of crucial importance to construct solutions of the Yang-Baxterequation using left braces, as one can see in the following proposition. Proposition 4 (Lemma 2, [7]) . Let A be a left brace and r : A × A → A × A the map given by r ( a, b ) := ( λ a ( b ) , λ − λ a ( b ) ( a )) , for all a, b ∈ A . Then, ( A, r ) is a solution of the Yang-Baxter equation whichwe call the associated solution to the left brace A . For the following definition, we refer the reader to [26, p. 160] and [7, Definition3].
Definition 5.
Let A be a left brace. A subset I of A is said to be a left ideal if it is a subgroup of the multiplicative group and λ a ( I ) ⊆ I , for every a ∈ A .Moreover, a left ideal is an ideal if it is a normal subgroup of the multiplicativegroup.Given an ideal I , it holds that the structure A/I is a left brace called the quotientleft brace of A modulo I .More in general, examples of left ideals can be found easily in a finite left brace,as the following proposition shows (see [1, p. 11]).7 roposition 5. Let A be a finite left brace and p a prime number dividing | A | .Then, the p -Sylow subgroup A p of the additive group ( A, +) is a left ideal of A . In [26], Rump introduced the special notion of the socle of a left brace that,in the terms of [7, Section 4], is the following.
Definition 6.
Let A be a left brace. Then, the setSoc( A ) := { a ∈ A | ∀ b ∈ A a + b = a ◦ b } is named socle of A .Clearly, Soc( A ) := { a ∈ A | λ a = id A } . Moreover, we have that Soc( A ) is anideal of A . In [26], Rump defined the socle series of A by setting Soc ( A ) := { } and Soc n ( A ) := { a | ∀ b ∈ A a + b − a ◦ b ∈ Soc n − ( A ) } , for every n ∈ N . Obviously, Soc ( A ) coincides with Soc( A ).As highlighted in [26, Proposition 7] and [7, Lemma 3], the socle of a left brace A has a special linkage with the solution associated to A . Proposition 6.
Let A be a left brace and ( A, r ) the solution associated to A .If ( A/ Soc( A ) , r ) is the solution associated to the left brace A/ Soc( A ) , then ( A/ Soc( A ) , r ) coincides with the retraction Ret(
A, r ) of ( A, r ) . Definition 7.
A left brace A has finite multipermutation level if the sequence S n defined as S := A and S n +1 = S n / Soc( S n ) for n ≥ Proposition 7.
Let A be a left brace. Then, A has finite multipermutationlevel if and only if A admits an s -series, i.e., there exists a positive integer n such that A = Soc n ( A ) . In particular, if n is the smallest positive integer such that A = Soc n ( A ), then A is said to be a left brace of multipermutation level n . In this subsection, we use some well-known results involving left braces todetermine the structure of uniconnected cycle sets with nilpotent permutationgroup. As a consequence, we use this result and the ones contained in [6] toprovide a method for constructing all the indecomposable cycle sets with cyclicpermutation group.Below we recall two results, the first due to Smocktunowitcz [32, Theorem 19]and the second to Rump [29], which are of crucial importance for our purposes.8 emma 8 (Theorem 19, [32]) . Let A be a finite left brace such that ( A, ◦ ) isnilpotent and suppose that | A | = p n · · · p n m m . Then, A and Q i A p i are isomorphicas left braces. Before introducing Rump’s Theorem on uniconnected cycle sets, recall that if A is a left brace, by Proposition 3, the maps λ a in (3) determines an action of( A, ◦ ) on ( A, +). According to [26, 29], a subset X of A which is a union oforbits with respect to such an action and generating the additive group ( A, +)is called cycle base . Moreover, if a cycle base is a single orbit then is said to bea transitive cycle base . Lemma 9 (Theorem 2, [29]) . Let A be a left brace, X a transitive cycle baseand g ∈ X . Define on A the binary operation • a • b := ( λ a ( g )) − ◦ b, for all a, b ∈ A . Then, ( A, • ) is a uniconnected cycle set and G ( A ) ∼ = ( A, ◦ ) .Conversely, every uniconnected cycle set can be constructed in this way. If A is a left brace and ( A, • ) is the uniconnected cycle set constructed as in theprevious lemma, from now on we refer to ( A, • ) as the cycle set associated to theleft brace A . Now, we provide the main result of this section, namely we showthat the classification of finite indecomposable cycle sets with a nilpotent regularpermutation group reduces to the classification of the ones having prime-powersize. Theorem 10.
Let m, n , ..., n m be natural numbers, p , ..., p m distinct primenumbers and ( X, · ) an uniconnected cycle set of cardinality p n · · · p n m m and withnilpotent permutation group G ( X ) . Then, there exist m indecomposable cyclesets ( Y , · ) , . . . , ( Y m , · m ) such that | Y i | = p n i i and ( X, · ) ∼ = ( Y , · ) × · · · × ( Y m , · m ) . Proof.
By Lemma 9, there exists a left brace ( A, + , ◦ ) such that the associatedcycle set ( A, • ) is isomorphic to ( X, · ) and G ( X ) ∼ = ( A, ◦ ).Now, since ( A, ◦ ) is nilpotent, by Lemma 8 we have that ( A, + , ◦ ) ∼ = Q i ( A p i , + , ◦ ),where A p i is the p i -Sylow of ( A, ◦ ). Hence, by a routine computation, it followsthat ( A, • ) is isomorphic to the direct product Q i ( A p i , • ), where ( A p i , • ) is theuniconnected cycle set associated to ( A p i , + , ◦ ). Setting ( Y i , · i ) := ( A p i , • ), forevery i ∈ { , . . . , m } , the claim follows.As a first consequence of the previous theorem, we obtain an alternative proofof [30, Proposition 4]. 9 orollary 11 (Proposition 4, [30]) . Let ( X, · ) be an indecomposable cycle setshaving cyclic permutation group G ( X ) . Then, < σ x > = G ( X ) , for every x ∈ X .Proof. If X has prime-power size, the claim follows by [6, Lemma 4-5], otherwise,the claim follows by [6, Lemma 4-5] and Theorem 10.In the following corollary, we specialize our result to indecomposable cyclesets X with cyclic permutation group G ( X ). Observe that, by this hypothesistogether with that of transitivity on X , we have that G ( X ) is a nilpotent groupthat acts regularly on X , i.e., X is uniconnected. Corollary 12.
Let m, n , ..., n m be natural numbers and p , ..., p m distinct primenumbers. Suppose that ( X, · ) is an indecomposable cycle set of cardinality p n · · · p n m m and with cyclic permutation group G ( X ) . Then, there exist m indecomposablecycle sets ( Y , · ) , . . . , ( Y m , · m ) such that | Y i | = p n i i and ( X, · ) ∼ = ( Y , · ) × · · · × ( Y m , · m ) . Moreover, ( Y i , · i ) is a cycle set of multipermutation level or a cycle set con-structed as in Theorem 2.Proof. Since X is uniconnected, by the previous theorem, there exist m inde-composable cycle sets ( Y , · ) , . . . , ( Y m , · m ) such that | Y i | = p n i i and( X, · ) ∼ = ( Y , · ) × · · · × ( Y m , · m ) . Since G ( X ) is cyclic so G ( Y i ) is, hence ( Y i , · i ) is a cycle set constructed as inTheorem 2.We highlight that to classify concretely finite indecomposable cycle sets withnilpotent regular permutation group, by Theorem 10, partial results have beenobtained by Theorem 2. However, these results do not consider all cycle setsbelonging to this family, as one can see in the following example. Example 1.
Let X := { , , , , , , , } be the cycle set given by σ = σ := (1 4)(2 8)(3 7)(5 6) σ = σ := (1 7 6 2)(3 4 8 5) σ = σ := (1 5)(2 3)(4 6)(7 8) σ = σ := (1 2 6 7)(3 5 8 4) . Then, G ( X ) is isomorphic to the dihedral group of order 8 and acts transitivelyon X .We conclude this section by showing that the classification of left braces hasa great impact on indecomposable cycle sets (and hence on indecomposablesolutions) also in the infinite case. To this end, we recall the following resultdue to Ced´o, Smocktunowitcz, and Vendramin.10 emma 13 ([9], Theorem 5.5) . Let A be a left brace such that ( A, ◦ ) ∼ = ( Z , +) .Then, a + b = a ◦ b , for all a, b ∈ A , i.e., A is a trivial left brace. As a consequence of the previous lemma we provide the following significanttheorem. One can note that it is essentially contained in the recent preprint byJedliˇcka, Pilitowska, and Zamojska-Dzienio [18, Proposition 6.2], however herewe give an alternative proof of their result.
Theorem 14.
Let ( X, · ) be an infinite indecomposable cycle set such that G ( X ) ∼ =( Z , +) . Then, up to isomorphism, ( X, · ) is the cycle set on Z given by a · b := b + 1 , for all a, b ∈ Z .Proof. By Lemma 9, there exists a left brace A such that the associated cycleset ( A, • ) is isomorphic to ( X, · ) and G ( A ) ∼ = ( A, ◦ ), hence ( A, ◦ ) ∼ = ( Z , +). Bythe previous lemma, A is the trivial left brace and, consequently, λ a = id A ,for every a ∈ A . In this way, the unique transitive cycle bases of A are { } and {− } . By Lemma 9, since a • b = ( λ a ( g )) − ◦ b , for some element g of atransitive cycle base, we obtain that a • b = b + 1, for all a, b ∈ A , if g = − a • b = b −
1, for all a, b ∈ A , if g = 1. Therefore, the claim follows.
3. Orthogonal dynamical extensions of cycle sets
In this section, we consider a particular type of dynamical extensions whichwe call orthogonal. Specifically, we focus on those that are constant, and weshow that, for these extensions, one can simplify the cocycle-condition (4). Thisresult allows for adapting [5, Theorem 7] to characterize them.Let us begin by recalling some suitable notions (see for instance [12]). If G isa finite transitive group acting on a set X , then G is said to be imprimitive ifthere exists a subset B of X , B = X , with at least two elements such that, foreach permutation g of G , either g ( B ) = B or g ( B ) ∩ B = ∅ . The subsets g ( B )of X are called blocks and the set { g ( B ) } g ∈ G is said to be a system of blocks .In this way, the action of G on X induces an action on a system of blocks in anatural way.Let G be an imprimitive group acting on a set X of size mn , for some nat-ural numbers n and m greater than 1, and let A = { A , . . . , A m } and B = { B , . . . , B n } be two systems of blocks of size m and n . Referring to [14], wesay that two systems of blocks A and B are orthogonal if | A i ∩ B j | = 1, forall i ∈ { , . . . , m } and j ∈ { , . . . , n } . The following classical lemma involvingimprimitive groups is useful for our purpose. Lemma 15 (Lemma 2.2, [13]) . Let G be an imprimitive group acting on a set X and let A and B be two orthogonal systems of blocks. Then, the action of G on X is equivalent to the action of G on A×B given by g ( A, B ) := ( g ( A ) , g ( B )) ,for all A ∈ A , B ∈ B . X is a cycle set, S a set and α : X × X × S −→ Sym( S ), α ( x, y, s ) α ( x,y ) ( s, − ) a function such that α ( x · y,x · z ) ( α ( x,y ) ( r, s ) , α ( x,z ) ( r, t )) = α ( y · x,y · z ) ( α ( y,x ) ( s, r ) , α ( y,z ) ( s, t )) , (4)for all x, y, z ∈ X and r, s, t ∈ S , then α is said to be a dynamical cocycle andthe operation · given by( x, s ) · ( y, t ) := ( x · y, α ( x,y ) ( s, t )) , for all x, y ∈ X and s, t ∈ S makes X × S into a cycle set which we denote by X × α S and we call dynamical extension of X by α . A dynamical cocycle is saidto be constant if α ( x,y ) ( s, − ) = α ( x,y ) ( t, − ), for all x, y ∈ X and s, t ∈ S . In thiscase, X × α S is said to be a constant dynamical extension .Moreover, a dynamical extension X × α S is called indecomposable if X × α S is an indecomposable cycle set: by [5, Theorem 7], this happens if and only if X is an indecomposable cycle set and the subgroup of G ( X × S ) generated by { h | ∀ s ∈ S h ( y, s ) ∈ { y } × S } acts transitively on { y } × S , for some y ∈ X .In this case, G ( X × S ) acts imprimitively and the set {{ x } × S } x ∈ X is a systemof blocks which we call the system of blocks induced by X and we denote it by B X . Definition 8.
Given a finite indecomposable cycle set X and X × α S a (con-stant) indecomposable dynamical extension of X by α , we say that X × α S is a (constant) orthogonal dynamical extension of X by α if the permutation group G ( X × S ) has a system of blocks A which is orthogonal to B X .Let us observe that one can find examples of orthogonal dynamical extensionsin [5], [6], and [18]. Below, we recall some of them. Examples 2.
1) If X and Y are indecomposable cycle sets of multipermutation level 1 andhaving coprime cardinalities, the direct product X × Y coincides with theorthogonal dynamical extension of X by α , where α : X × X × Y → Sym( Y )is the function given by α ( x,y ) ( s, − ) := σ s , for all x, y ∈ X and s ∈ Y .2) Let p be a prime number and X = Y := Z /p Z and consider X as thecycle set given by x · y := y + 1, for all x, y ∈ X . Moreover, let α be thefunction from X × X × Y to Sym( Y ) given by α ( x,y ) ( s, t ) := t + x , for all( x, y ) , ( s, t ) ∈ X × Y . Then X × α Y is an orthogonal dynamical extensionof X by α .However, several examples of dynamical extension that are not orthogonaloccur in literature. 12 xamples 3.
1) Every indecomposable dynamical extension having prime-power size andcyclic permutation group can not be obtained as an orthogonal dynamicalextension.2) Let k be a natural number, X := Z /k Z the cycle set given by x · y := y + 1,for all x, y ∈ X , A = B := Z / Z , and S := A × B . Moreover, define β : A × A × X → Sym( B ) the function given by β ( a,b ) ( x, − ) := id B , if a = b ,and β ( a,b ) ( x, − ) := (0 1), otherwise, define γ : B → Sym( A ) the functiongiven by γ b ( a ) := a − b −
1, for every b ∈ B , and let α : X × X × S → Sym( S )be the function given by α ( x,y ) (( a, b ) , ( c, d )) := ( ( c, β ( a,c ) ( x, d )) , if x = y ( γ b ( c ) , d ) , if x = y , for all x, y ∈ X , ( a, b ) , ( c, d ) ∈ S . Then, X × α S is a dynamical extensionof X by α that is not orthogonal. Indeed, if A = { A , . . . , A k } is asystem of blocks orthogonal to B X , then every block A i is of the form A i := { (1 , a , b ) , . . . , ( k, a k , b k ) } . But one can show that there exists g ∈G ( X × α S ) such that g (1 , a , b ) = (1 , a , b ) and g ( k, a k , b k ) = ( k, a g , b k )for an element a g different from a k , a contradiction.From now on, in the rest of this section we focus on constant orthogonaldynamical extensions. As noted in [34, Section 3], to obtain constant dynamicalextensions, the cocycle-condition (4) can be simplified. Now, we show that forconstant orthogonal dynamical extensions the condition (4) can be even moresimplified. Proposition 16.
Let X × α S be a constant orthogonal dynamical extension of X by α . Then, after renaming the elements of the set S , the function α dependsonly on the first variable, i.e., α ( x,y ) ( − , t ) = α ( x,y ′ ) ( − , t ) , for all x, y, y ′ ∈ X and t ∈ S .Proof. Let q be a natural number such that A := { A , ..., A q } is a system ofblocks orthogonal to B X . Thus, we necessarily have that A i = { ( x i , s i ) , . . . , ( x i | X | , s i | X | ) } where x i k = x i j if k = j , for every i ∈ { , ..., q } . Therefore, without loss ofgenerality, after renaming the elements of S , we can suppose that s i = · · · = s i | X | , for every i ∈ { , ..., q } . In this way, for every A ∈ A , there exists anelement s ∈ S such that A = X × { s } , hence we can identify B X × A with X × S . By Lemma 15, we have( x, s ) · ( y, t ) = ( δ ( x,s ) ( y ) , δ ( x,s ) ( t )) , x, y ∈ X and s, t ∈ S , where δ ( x,s ) is an element of G ( X × α S ) dependingon x and s . It follows that( x · y, α ( x,y ) ( − , t )) = ( δ ( x,s ) ( y ) , δ ( x,s ) ( t )) , for all x, y ∈ X and s, t ∈ S , therefore α ( x,y ) ( − , t ) = δ ( x,s ) ( t ), i.e., α onlydepends on x .Therefore, a constant orthogonal dynamical extension X × α S can be thoughtas a cycle set on X × S obtained by a map β : X −→ Sym( S ) , x β x from X to the symmetric group of S such that β x · y β x = β y · x β y and α ( x,y ) ( s, − ) = β x ,for all x, y ∈ X , s ∈ S .As a consequence of the previous proposition, we give a characterization forconstant orthogonal dynamical extensions X × α S . Theorem 17.
Let X be a finite indecomposable cycle set. Moreover, let S bea finite set and α : X −→ Sym( S ) a map from X to the symmetric group of S such that α x · y α x = α y · x α y , for all x, y ∈ X . Then, the binary operation on X × S given by ( x, s ) · ( y, t ) := ( x · y, α x ( t )) , makes X × S into a constant orthogonal dynamical extension of X by α if andonly if the subgroup H y of Sym( X × S ) given by H y := { ( σ ǫ x · · · σ ǫ n x n , α ǫ x · · · α ǫ n x n ) | n ∈ N , ǫ , ..., ǫ n ∈ {− , } σ ǫ x · · · σ ǫ n x n ( y ) = y } acts transitively on { y } × S , for some y ∈ X . In this case, the set { X × { s }} s ∈ S is a system of blocks orthogonal to B X .Moreover, every constant orthogonal dynamical extension can be constructed inthis way.Proof. By a long but simple calculation, one can verify that ( X × S, · ) is adynamical extension of X by α . Moreover, by [5, Theorem 7] we have that it isindecomposable if and only if H y acts transitively on { y } × S , for some y ∈ X .Now, since σ ( x,s ) ( y, t ) ∈ X × { α x ( t ) } , for all x, y ∈ X, s, t ∈ S , it follows that { X × { s }} s ∈ S is a system of blocks.Obviously, it is orthogonal to B X .The rest of the assertion follows by Proposition 16 and [5, Theorem 7]. Remark 1.
Given two cycle sets X and S and α : X → Aut( S ) a function suchthat α x · y α x = α y · x α y , for all x, y ∈ X , then the binary operation · on X × S given by ( x, s ) · ( y, t ) := ( x · y, α x ( s · t )) makes X × S into a cycle set. Moreover,14t is easy to show that it is isomorphic to the Rump’s semidirect product of thecycle sets X and S via α , a construction of cycle sets introduced by Rump in[27]. Since a set S can be endowed with a cycle set structure by x · y := y , for all x, y ∈ S , the previous results ensure that every constant orthogonal dynamicalextension can be obtained as particular semidirect product of cycle sets. On theother hand, one can show that every semidirect product X ⋉ α S of two finitecycle sets X and S via α gives rise to orthogonal dynamical extensions whenever X ⋉ α S is indecomposable.Theorem 17 turns out to be useful in the investigation of indecomposable cyclesets for which the retraction has an orthogonal system of blocks. Indeed, if X is a finite indecomposable cycle set, by [5, Proposition 8], X is isomorphic to aconstant dynamical extension of Ret( X ), hence we have the following corollary. Corollary 18.
Let X be a finite indecomposable cycle set such that B Ret( X ) has an orthogonal system of blocks. Then, there exist a set S and a function α : Ret( X ) −→ Sym( S ) such that the operation · given by ( x, s ) · ( y, t ) := ( x · y, α x ( t )) , for all x, y ∈ Ret( X ) and s, t ∈ S makes Ret( X ) × S into a cycle set isomorphicto I .Proof. It follows by [5, Proposition 8] and Theorem 17.In other words, the previous corollary states that to construct all the finiteindecomposable cycle sets having X as retraction and such that there exists asystem of blocks orthogonal to B X one has to classify all the pair ( S, α ) where S is a set, α is a function from X to Sym( S ) such that α x · y α x = α y · x α y , for all x, y ∈ X , and the subgroup H y of Theorem 17 acts transitively on { y } × S .Note that if X × α S is a constant indecomposable dynamical extension, thesystem of blocks B Ret( X × α S ) induced by the retraction is in general differentfrom the system of blocks B X induced by X (see, for instance, Example 6).However, B Ret( X × α S ) and B X are closely related, as one can see in the simplebut useful Proposition 19. Before introducing such a proposition, let us recallthat if G is an imprimitive group acting on a finite set X and A , B are systemof blocks, then A is said to be a refinement of B if there exist A ∈ A , B ∈ B such that A ⊆ B . Proposition 19.
Let X × α S be a constant indecomposable dynamical extension.Then, B X is a refinement of B Ret( X × α S ) .Proof. By [5, Propositon 2], we have that( x, s ) · ( y, t ) := ( x · y, α x,y ( t )) , x, y ∈ X and s, t ∈ S . Therefore, if B ∈ B Ret( X × α S ) and ( x, s ) ∈ B , then { x } × S ⊆ B , hence the claim follows.
4. Description of constant orthogonal dynamical extensions by leftbraces
The goal of this section is to give a relationship between constant orthogonaldynamical extensions and semidirect product of left braces. This link naturallyleads to a description of constant orthogonal dynamical extensions and providesa way to investigate these extensions by left braces.For our purpose, it is useful to recall the semidirect product of left braces inthe same terms used in [7]. Given two left braces A and H and a homomorphism α : A → Aut( H ) from the group ( A, ◦ ) to the automorphisms of the left brace H , the semidirect product of A and H via α is the left brace ( A × H, + , ◦ ), wherethe sum is the direct sum of the groups ( A, +) and ( H, +) and the multiplicationis the semidirect product of the multiplicative groups ( A, ◦ ) and ( H, ◦ ) via α .Let A be a left brace, H a trivial left brace, and X (resp. S ) a cycle base of A (resp. H ). By the results provided in [27, 25], one can easily show that semidi-rect products of A and H and semidirect products of X and S are “compatible”in the following sense: if A ⋉ α H is a semidirect product such that α a ( S ) = S ,for every a ∈ A , then α induces a semidirect product of the cycle sets X and S ;on the other side, we have that if X is a cycle set and S a trivial cycle set, thena semidirect product of the cycle sets X and S induces a semidirect productof the left braces G X and Z S , where G X is the left brace having as additivegroup the free abelian group Z X and as multiplicative group the one generatedby the elements of X subject to the relations x ∗ y = σ − x ( y ) ∗ ( σ − x ( y ) · x ), forall x, y ∈ X , and Z S is the trivial left brace on the free abelian group Z S .In the previous section, we showed that to obtain a constant orthogonal dy-namical extension X × α S , it is required the transitivity of a subgroup H y of G ( X × α S ) on the set { y } × S . Now, using the embedding of X into a left brace A as transitive cycle base, we show that the transitivity of H y on { y } × S canbe carried into the richer structure of the left brace A . Theorem 20.
Let A be a left brace, S a finite set and ¯ α : A −→ Sym( S ) anhomomorphism from ( A, ◦ ) to Sym( S ) . Moreover, let X be a finite transitivecycle base of A and e an element of X and suppose that the stabilizer A e of e in A (by the map λ ) acts transitively on S by ¯ α . Then, the binary operation · on X × S given by ( x, s ) · ( y, t ) := ( λ − x ( y ) , ¯ α − x ( t ))16 akes X × S into a constant orthogonal dynamical extension of X by ¯ α − | X .Conversely, any constant orthogonal dynamical extension can be constructed inthis way.Proof. Clearly, the binary operation given by x · y := λ − x ( y ) makes X into acycle set. Moreover, we have that¯ α x + y = ¯ α y + x ⇒ ¯ α x ◦ λ − x ( y ) = ¯ α y ◦ λ − y ( x ) ⇒ ¯ α − λ − x ( y ) ¯ α − x = ¯ α − λ − y ( x ) ¯ α − y for all x, y ∈ X , hence ( X × S, · ) is a cycle set. Clearly, it is a constant dynamicalextension of X by S having dynamical cocycle α := ¯ α − | X .Since X is a transitive cycle base of A , by [31, Proposition 9] X is an indecom-posable cycle set. Therefore, it remains to show that the subgroup H e defined asin Theorem 17 acts transitively on { e }× S . Let s and s be elements of S . Since A e acts transitively on S by ¯ α , there exists a e ∈ A e such that ¯ α ( a e )( s ) = s .Moreover, by [31, Proposition 8], there exist g , . . . , g n ∈ X , ǫ , . . . , ǫ n ∈ {− , } such that b e = g ǫ ◦ · · · ◦ g ǫ n n . Then, it follows that¯ α ( a e )( s ) = s ⇒ ¯ α ( g ǫ ◦ · · · ◦ g ǫ n n )( s ) = s ⇒ ¯ α g ǫ · · · ¯ α g ǫnn ( s ) = s ⇒ ¯ α ǫ g · · · ¯ α ǫ n g n ( s ) = s and since e = λ a e ( e ) = λ ǫ g · · · λ ǫ n g n ( e ) it follows that H e acts transitively on { e } × S , hence X × S is indecomposable. Clearly, { X × { s }} s ∈ S is a system ofblocks orthogonal to B X .Conversely, suppose that X × α S is a constant orthogonal dynamical extension.By [31, Proposition 9], X is a transitive cycle base of the left brace G X having asadditive group the free abelian group Z X and as multiplicative group the groupgenerated by the elements of X subject to the relation x ∗ y = σ − x ( y ) ∗ ( σ − x ( y ) · x ),for all x, y ∈ X . Then, since α x · y α x = α y · x α y , for all x, y ∈ X , replacing y by σ − x ( y ) we obtain that α y α x = α σ − x ( y ) · x α σ − x ( y ) and hence α − x α − y = α − σ − x ( y ) α − σ − x ( y ) · x . Therefore, the assignment x α − x can be extended to anaction ¯ α from the multiplicative group of G X to Sym( S ). Using Theorem 17,by similar computations seen in the previous implication, we obtain that thestabilizer ( G X ) e of an element e ∈ X acts transitively on S by ¯ α . Moreover,the indecomposable cycle set on X × S constructed starting from G X and ¯ α asin the previous implication is exactly the dynamical extension X × α S . Hencethe proof is complete. Remarks 1.
1. By Theorem 20, the classification of constant orthogonal dynamical exten-sions of a finite indecomposable cycle set X reduces to the classification17f the actions ¯ α : A −→ Sym( S ), where S is a finite set and A is a leftbrace having X as transitive cycle base and such that the stabilizer of anelement e ∈ X (by maps λ ) acts transitively on S by ¯ α .2. An action of the multiplicative group ( A, ◦ ) of a left brace A on a set S isa particular case of an action of a skew left brace. This algebraic objecthas been recently introduced and studied by De Commer in [11].In the following result, we obtain that left braces are a powerful tools to de-scribe constant orthogonal dynamical extensions and to check them by semidi-rect products. Corollary 21.
Let A be a left brace having X as a finite transitive cycle base, H a trivial left brace having S as a finite cycle base. Assume that α : A → Aut( H ) is a homomorphism from the group ( A, ◦ ) to the automorphisms of the left brace H such that α ( a )( S ) = S , for every a ∈ A , and there exists e ∈ X such that thestabilizer A e acts transitively on S by α . Then, the semidirect product of theleft brace A ⋉ α H has the constant dynamical extension X × ¯ α S as transitivecycle base, where ¯ α : X → Sym( S ) is given by ¯ α ( x ) := α − ( x ) , for every x ∈ X .Moreover, X × ¯ α S is a constant orthogonal dynamical extension.Conversely, every constant orthogonal dynamical extension can be constructedin this way.Proof. Note that the homomorphism α : A → Aut( H, +) induces a homomor-phism α ∗ : A → Sym( S ) such that A e acts transitively on S by α ∗ . Then, byTheorem 20, X × ¯ α S is a constant orthogonal dynamical extension. Since theadditive group of the left brace A ⋉ α H is the direct sum of ( A, +) and ( H, +),it follows that X × ¯ α S is a transitive cycle base.For the converse part, given a constant orthogonal dynamical extension X × α S ,it is sufficient to consider X as a transitive cycle base of the left brace G X and S as the cycle base of the trivial left brace over the free abelian group Z S .In this way, by Theorem 20, the function α induces an action ¯ α of G X on S .Again, ¯ α induces an action ¯¯ α by automorphisms of G X on the free abelian group Z S . Finally, the semidirect product of left braces G X ⋉ ¯¯ α Z S has the dynamicalextension X × α S as a transitive cycle base.We conclude this section by giving an example of constant dynamical exten-sions obtained by using Corollary 21. Example 4.
Let A be the left brace having ( Z / Z , +) as additive group andwith multiplication given by a ◦ b := a + b + 2 ab , for all a, b ∈ A , and let H be thetrivial left brace on ( Z / Z , +). Consider G := A ⋉ α H the semidirect productof the left braces A and H where α : A −→ Aut( H ) is the homomorphism from( A, ◦ ) to the automorphism group of the left brace H given by α (0) = α (3) :=18d H and α (1) = α (2) := (1 2). Then, X := { , } is the unique transitivecycle base of A and S := { , } is a cycle base of H . Moreover, the stabilizer A := { , } of 1 is a subgroup of ( A, ◦ ) that acts transitively on S . Hence, if ¯ α is the function defined as in Corollary 21, it follows that X × ¯ α S is a constantorthogonal dynamical extension. In particular, it is isomorphic to the that on Z / Z × Z / Z given by ( x, s ) · ( y, t ) := ( y + 1 , t + x ) , for all ( x, s ) , ( y, t ) ∈ Z / Z × Z / Z .
5. Orthogonal dynamical extensions: examples and constructions
In this section, we specialize Theorem 17 to provide examples and constructionsof indecomposable cycle sets, giving special attention to the ones having abelianpermutation group. Finally, we show that, by semidirect products of cycle sets,one can construct examples of orthogonal dynamical extensions that are notconstant.Let us recall that, if X is a finite indecomposable cycle set with abelian permu-tation group and A is a finite abelian group, the constant orthogonal dynamicalextensions having permutation group isomorphic to G ( X ) × A simply correspondto the maps α : X −→ A from X to the translations group of A (which we iden-tify with A ) such that α ( x · y ) + α ( x ) = α ( y · x ) + α ( y ), for all x, y ∈ X , andthe group H y := { ǫ α ( x )+ · · · + ǫ n α ( x n ) | n ∈ N , ǫ , ..., ǫ n ∈ {− , } σ ǫ x · · · σ ǫ n x n ( y ) = y } is equal to A , for some y ∈ X .We underline that the indecomposable cycle sets constructed in this section havenot cyclic permutation group, hence they are different from those obtained in[6, 30].At first, we introduce the following example that, in particular, coincides with[5, Example 10]. Example 5.
Let X := Z /k Z be the cycle set given by x · y := y + 1, for all x, y ∈ X . Moreover, let S := Z /k Z and α : X −→ Sym( S ) the function givenby α x := t x , for every x ∈ X , where t x is the translation by x . Then,( x, s ) · ( y, t ) = ( y + 1 , α x ( t )) , for all x, y ∈ X, s, t ∈ S and X × α S is an orthogonal dynamical extension of X by α having multipermutation level 2 and permutation group isomorphic to Z /k Z × Z /k Z . 19he previous example is an orthogonal dynamical extension X × α S for whichRet( X × α S ) = X . However, there exist orthogonal dynamical extensions X × α S for which Ret( X × α S ) is not isomorphic to X , as we can see in the followingexample. Example 6.
Let X := Z /p Z be the cycle set given by x · y := y + 1, for all x, y ∈ X . Moreover, let H := p Z /p Z , S := X/H , and α : X −→ Sym( S ) thefunction given by α x := t [ x ] , for every x ∈ X , where [ x ] is the image of x in X/H under the canonical epimorphism and t [ x ] is the translation by [ x ]. Then,( x, s ) · ( y, t ) = ( y + 1 , t + [ x ]) , for all x, y ∈ X, s, t ∈ S , and X × α S is a constant orthogonal dynamicalextension of X by α having multipermutation level 2: indeed, | Ret( X × S ) | = p and α x · y α x = t [ x + y +1] , which is symmetric with respect to x and y . Moreover, σ p − , σ (1 , ∈ H and < σ p − , σ (1 , > acts transitively on { } × S . Clearly, thepermutation group is isomorphic to Z /p Z × Z /p Z .The indecomposable cycle sets given before are of multipermutation level 2and with abelian permutation group. This class of cycle sets is studied in detailin [18, 6]. Now, we give further examples of indecomposable cycle sets havinggreater multipermutation level. Such examples are also different from the onesprovided in [5]: indeed, the multipermutation cycle sets contained in [5] havemultipermutation level less than 2. Example 7.
Let X := Z / Z be the cycle set given by σ = σ := t and σ = σ := t − , where t and t − are the translations by 1 and −
1, respectively.Moreover, let S be the group Z / Z and α : X −→ Sym( S ) given by α = α := id S α = α := t . By a long but easy calculation, one can show that α x · y α x = α y · x α y , for all x, y ∈ X , and that the subgroup < σ , σ (2 , > of H acts transitively on { } × S . Moreover, the function θ : X × S −→ X , ( x, s ) x induces anisomorphism from Ret( X × S ) to X . Therefore, X × α S is a constant orthogonaldynamical extension of X by S of multipermutation level 3 having permutationgroup isomorphic to Z / Z × Z / Z . In particular, X × α S is isomorphic to thecycle set on Z / Z given by σ = σ := (0 1 2 3)(4 5 6 7) σ = σ := (0 3 2 1)(4 7 6 5) σ = σ := (0 5 2 7)(1 6 3 4) σ = σ := (0 7 2 5)(1 4 3 6) . In the following proposition, we construct a family of indecomposable cyclesets of multipermutation level 3 obtained as constant orthogonal dynamicalextensions of cycle sets having multipermutation level 2.20 roposition 22.
Let p be an odd prime number and X := Z /p s Z the inde-composable cycle set of multipermutation level given by x · y := y + 1 + xp s ,for all x, y ∈ X . Let k be a number coprime with p and f the function from X into itself given by f ( j ) := k ( j + p s j ( j − , for every j ∈ X , where p s j ( j − is the element a ∈ X such that a = p s j ( j − .Let α : X → Z /p s Z be the function given by α x := t f ( x ) , for all x ∈ X .Then, X × α Z /p s Z is a constant orthogonal dynamical extension of X by α having multipermutation level . Moreover, Ret( X × Z /p s Z ) ∼ = X and G ( X × Z /p s Z ) ∼ = Z /p s Z × Z /p s Z .Proof. To show that X × α Z /p s Z is a cycle set, it is sufficient to see that f ( i · j ) + f ( i ) = f ( j · i ) + f ( j ), for all i, j ∈ X . Let i, j be elements of X . Then, f ( i · j ) + f ( i ) = f ( j + 1 + p s i ) + f ( i )= k ( j + 1 + p s i + p s ( j + 1 + p s i )( j + 1 + p s i − k ( i + p s i ( i − k ( j + 1 + p s i + p s ( j + 1) j k ( i + p s i ( i − k ( j + i + 1 + p s ( j + 1) j p s ( i + 1) i i and j it follows that f ( i · j ) + f ( i ) = f ( j · i ) + f ( j ), hence X × α Z /p s Z is a cycle set. Moreover, thesubgroup < σ p s − p s − , σ (1 , > of H acts transitively on { }× Z /p s Z , hence, byTheorem 17 the cycle set X × α Z /p s Z is indecomposable and G ( X × α Z /p s Z )is isomorphic to Z /p s Z × Z /p s Z .Finally, let ρ : X × Z /p s Z −→ X be the function given by ρ ( x, s ) := x , for every( x, s ) ∈ X × Z /p s Z . Clearly, ρ is an epimorphism of cycle sets. Moreover, since f ( j + ap s ) = f ( j ) + kap s , for all j, a ∈ Z /p s Z , it follows that σ ( x,s ) = σ ( y,t ) ifand only if ρ ( x ) = ρ ( y ), therefore Ret( X × α Z /p s ) ∼ = X and X × Z /p s hasmultipermutation level 3.The following proposition shows that other examples of indecomposable cyclesets that are constant dynamical extensions can be easily obtained by thosehaving cyclic permutation group and arbitrary multipermutation level. Proposition 23.
Let n, k ∈ N , X := { , . . . , p k − } be an indecomposable cycleset of multpermutation level n and with cyclic permutation group constructedas in Theorem 2. Moreover, let j ∈ N such that | Ret n − ( X ) | = p j and s anatural number in { , . . . , j } . Let α : X → Sym( Z /p s Z ) be the function given by α x := t [ x ] , where [ x ] is the class of x module p s . Then, X × α Z /p s Z is a constant rthogonal dynamical extension of X such that G ( X × α Z /p s Z ) ∼ = Z /p k Z × Z /p s Z and Ret( X × α S ) ∼ = Ret( X ) .Proof. Similarly to Proposition 22, one can show that X × α Z /p s Z is an inde-composable cycle set such that G ( X × α Z /p s Z ) ∼ = Z /p k Z × Z /p s Z . Now, weshow that Ret( X × α Z /p s Z ) is isomorphic to Ret( X ). Let ρ : X × S −→ Ret( X )be the function given by ρ ( x, s ) := σ x , for all x ∈ X, s ∈ S . Clearly, ρ is anepimorphism of cycle sets. Moreover, σ x = σ y ⇒ | Ret( X ) | | y − x ⇒ [ x ] = [ y ] , for all x, y ∈ X , therefore the equivalence relation induced by ρ coincides withthe retract relation of X × Z /p s Z , hence the proof is complete. Example 8.
Let X := Z / Z be the cycle set of multipermutation level 3 andcyclic permutation group given by σ x := t if x ≡ mod t if x ≡ mod t if x ≡ mod t if x ≡ mod t if x ≡ mod t if x ≡ mod t if x ≡ mod t if x ≡ mod α : X −→ Z / Z the function given by α x := t [ x ] , where [ x ] is the class of x module 2. By Proposition 23, X × α Z / Z is a constant orthogonal dynamicalextension of X by α having multipermutation level 3 and permutation groupisomorphic to Z / Z × Z / Z .The following example ensures that the class of constant orthogonal dynamicalextensions is not contained in that of cycle sets with abelian permutation group.Furthermore, this example of cycle set is different from the previous ones sinceit is indecomposable and has not finite multipermutation level. Example 9.
Let X := { , , , } be the irretractable indecomposable cycleset given by σ := (1 , σ := (0 , σ := (0 , , , σ := (0 , , , S := Z / Z and α : X −→ S the function given by α = α := t and α = α := t . Then, by Theorem 17, the cycle set X × α S is an indecomposablecycle set of size 16: indeed, α x · y α x is symmetric with respect to x and y , X is indecomposable and the subgroup of H generated by σ (0 , acts transitivelyon { } × S . Moreover, since Ret( X × α S ) is isomorphic to X , X × α S has notfinite multipermutation level. Since G ( X ) is not abelian, G ( X × α S ) is not.22roviding examples of orthogonal dynamical extensions that are not constantis rather difficult. However, the semidirect product of cycle sets belonging tothe class of quasigroups allows for obtaining a family of examples of orthogonaldynamical extensions. We recall that a quasigroup is an algebraic structure forwhich left and right multiplications are bijective, see [3]. Proposition 24.
Let
X, S be finite cycle sets and α : X → Aut( S ) a functionfrom X to Aut( S ) such that α x · y α x = α y · x α y , for all x, y ∈ X . Then thesemidirect product of X and S via α is a quasigroup if and only if X and S are quasigroups. Moreover, if X ⋉ α S is a quasigroup, then it is an orthogonaldynamical extension.Proof. It is a routine computation to check that X ⋉ α S is a quasigroup if andonly if X and S are quasigroups. Clearly, every cycle set that also is a quasigroupis indecomposable. Therefore, if X ⋉ α S is a quasigroup, by Remark 1, it is anorthogonal dynamical extension. Remarks 2.
1. To construct examples of orthogonal dynamical extensions using Propo-sition 24, one has to provide cycle sets that also are quasigroups. Evenif these cycle sets are not easy to compute, several instances have beenprovided by Bonatto, Kinyon, Stanovsk´y, and Vojtˇechovsk´y in [3]. Foran exhaustive study of these structures, we refer the reader to the samepaper [3].2. The indecomposable cycle sets given in [6, 30, 5, 18] are not quasigroups:in this sense, Proposition 24 provides different examples of indecomposablecycle sets.3. Proposition 24 allows for determining “genuine” examples of orthogonaldynamical extension. Indeed, if X ⋉ α S is a quasigroup, then it clearly isan irretractable cycle set, hence it can not be constructed by a constantorthogonal dynamical extension.We close this section by showing a concrete example of orthogonal dynamicalextension which we construct using Proposition 24. Example 10.
Let X := { , , , } be the indecomposable cycle set given by σ := (1 , σ := (0 , σ := (0 , , , σ := (0 , , , S, · ) be a copy of X and α : X → Aut( S ) the function given by α = α :=(0 , ,
3) and α = α := id S . By a long but easy calculation, it follows that X and S are quasigroups and α x · y α x = α y · x α y , for all x, y ∈ X . Therefore, X ⋉ α S is an orthogonal dynamical extension.23 . One-generator left braces of multipermutation level 2 In this section, which is self-contained with respect to the previous ones, westudy the class of one-generator left braces, which is closely related to indecom-posable cycle sets (see [33, 31] for more details). Specifically, we investigateone-generator left braces of multipermutation level at most 2 focusing our at-tention on the properties related to the map λ . In this way, we will prove [31,Theorem 2] by a different proof. At this purpose, let us recall that a left brace A is said to be a one-generator left brace if there exists x ∈ A such that A = A ( x ),where A ( x ) is the smallest left sub-brace of A containing x .At first, for any subset S of a left brace A , let us put − S := {− s | s ∈ S } and S − := { s − | s ∈ S } . Lemma 25.
Let A be a left brace and x ∈ A . Define inductively A := { x, − x, x − , } and A i = ( A i − + A i − ) ∪ ( A i − − A i − ) ∪ ( A i − ◦ A i − ) ∪ ( A i − ◦ A − i − ) ∪ ( A − i − ◦ A i − ) , for every i ∈ N . Then, S n ∈ N A n = A ( x ) .Proof. Let U := S n ∈ N A n . At first, U is closed under the operations + and ◦ .Indeed, if a ∈ A s and b ∈ A t , since A k ⊆ A k + j , for all k, j ∈ N , a + b, a ◦ b ∈ A s + t +1 . Moreover, 0 ∈ U , hence if a ∈ A s we have that a − = 0 ◦ a − ∈ A s +1 and, similarly, − a = 0 − a ∈ A s +1 . Therefore, U is a left brace containing x .Now, if B is a left sub-brace of A containing x , by induction on i , it is easy toshow that A i ⊆ B , hence U ⊆ B . Therefore, U is equal to A ( x ).The following result is originally contained in [19, Theorem 9.2]. In partic-ular, it shows that the map λ satisfies an additional property with respect toProposition 3 if the left brace has multipermutation level at most 2. Lemma 26.
Let A be a left brace. Then the map λ : A −→ Aut( A, +) is ahomomorphism from ( A, +) into Aut( A, +) if and only if A is a trivial left braceor A has multipermutation level . Following [2], a left brace A is said to be λ - cyclic if λ ( A ) is a cyclic group. Inthe following proposition, we show that one-generator left braces of multiper-mutation level 2 are always λ -cyclic. Proposition 27.
Let A be a one-generator left brace of multipermutation level , x ∈ A , and suppose that A = A ( x ) . Then, the subgroup generated by λ ( A ) isequal to the one generated by λ x . roof. It is sufficient to show that λ a = λ kx , for some k ∈ Z , for every a ∈ A i ,proceeding by induction on i . Since A = { , x, − x, x − } , by Lemma 26 we havethat id A = λ x − x = λ x λ − x , hence λ − x = λ − x and, by Proposition 3, λ x − = λ − x .Now, suppose that λ a = λ kx , for some k ∈ Z , for every a ∈ A i , and let a , a be elements of A i and k , k ∈ Z such that λ a = λ k x and λ a = λ k x . Then,by Lemma 26, λ a + a = λ k + k x and λ a − a = λ k − k x . By Proposition 3, wehave that λ a ◦ a = λ k + k x and λ a ◦ a − = λ a − ◦ a = λ k − k x , hence the claimfollows.In the following lemma, we provide a necessary and sufficient condition thatcharacterizes the trivial one-generator left braces. This result will be of crucialimportance to give a precise description of one-generator left braces of multi-permutation level 2. Lemma 28.
Let A be a one-generator left brace and x ∈ A such that A = A ( x ) .Then, the following statements are equivalent:1) the map λ x fixes x ;2) A is a trivial left brace with ( A, +) the additive subgroup generated by x .Proof. Suppose that 1) holds. At first, we denote by ka the element of A givenby ka := a + · · · + a | {z } k times , for all a ∈ A , k ∈ N . Similarly, let a ◦ k be the element of A given by a ◦ k := a ◦ · · · ◦ a | {z } k times . Since λ x fixes x , it is easy to show by inductionon k that kx = x ◦ k and k ( − x ) = ( − x ) ◦ k = ( x − ) ◦ k , for every k ∈ N . In thisway, we obtain that kx ◦ hx = kx + λ kx ( hx ) = kx + hx , for all k, h ∈ Z , hencethe additive subgroup generated by x is a trivial left brace containing x . Since A = A ( x ), we obtain that 1) implies 2). The converse implication is trivial.Now, we are able to prove a characterization of one-generator left braces ofmultipermutation level 2 by means of transitive cycle bases. Theorem 29.
Let A be a left brace. Then, the following conditions are equiv-alent:1) A is a one-generator left brace of multipermutation level ;2) A has a transitive cycle base X , with | X | > , that is an indecomposablecycle set having multipermutation level .Proof. Suppose that A is a one-generator left brace of multipermutation level2 and let x be such that A = A ( x ). By Proposition 27, the orbit of x underthe action of A by λ is equal to X := { λ kx ( x ) | k ∈ Z } . Moreover, since A has multipermutation level 2, by Lemma 28 it follows that | X | >
1. By [9,25roposition 6.4], the additive subgroup generated by X is equal to A , hence X is a transitive cycle base. Moreover, X is an indecomposable cycle set havingpermutation group equal to the cyclic subgroup generated by λ x | X . Since X isa cycle base, we have that λ x = λ y if and only if λ x | X = λ y | X , hence we canidentify Ret( X ) with { λ kx ( x ) + Soc( A ) | k ∈ Z } . Since A has multipermutationlevel 2, we have that λ λ kx ( x )+Soc( A ) = λ x +Soc( A ) = id A/ Soc( A ) , for every k ∈ N , and since Ret( X ) is again indecomposable it follows that | Ret( X ) | = 1,therefore 2) follows.Now, suppose that 2) holds and let x be an element of X . At first, we showthat λ ǫ x + ··· + ǫ n x n = λ P ǫ i x , for all n ∈ N , ǫ , . . . , ǫ n ∈ { , − } , and x , . . . , x n ∈ X .For n = 1, if ǫ = 1, the claim follows because X has multipermutation level 1,if ǫ = − x ∈ X , we obtain that λ = λ − x + x = λ ( − x ) ◦ λ ( − x − ( x ) = λ − x λ λ ( − x − ( x ) and, since λ ( − x ) − ( x ) ∈ X , we have that λ λ ( − x − ( x ) = λ x , hence λ − x = λ − x .Now, suppose that n is a natural number greater than 1. Therefore λ ǫ x + ··· + ǫ n x n = λ ǫ x ◦ ( λ ( ǫ x − ( ǫ x )+ ··· + λ ( ǫ x − ( ǫ n x n )) = λ ǫ x λ λ ( ǫ x − ( ǫ x )+ ··· + λ ( ǫ x − ( ǫ n x n ) = λ P ǫ i x where λ λ ( ǫ x − ( ǫ x )+ ··· + λ ( ǫ x − ( ǫ n x n ) = λ n − P i =1 ǫ i x and λ ǫ x = λ ǫ x , by inductivehypothesis. Then, we have that X = { λ kx ( x ) | k ∈ Z } . By induction on k , itis easy to show that λ kx ( x ) and λ − kx ( x ) belong to A ( x ), for every k ∈ N , hence A = X + ⊆ A ( x ). To show 1), since | X | >
1, by Lemma 26 it is sufficient to seethat λ a + b = λ a ◦ b , for all a, b ∈ A .Finally, let a, b ∈ A , m, n ∈ N ǫ , . . . , ǫ m , η , . . . , η n ∈ {− , } and x , . . . , x m , y , . . . , y n ∈ X such that a = ǫ x + · · · + ǫ m x m and b = η y + · · · + η n y n .Then, λ a + b = λ ǫ x + ··· + ǫ m x m + η y + ··· + η n y n = λ P ǫ i + P η j x = λ P ǫ i x λ P η j x = λ a λ b = λ a ◦ b , hence the claim follows. 26s a final result of this paper, we give a precise description of one-generatorleft braces of multipermutation level 2 answering [33, Question 5.5]. We pointout that such a description has already contained in [31, Theorem 2]; however,here we provide a different proof about it. Corollary 30.
A one-generator left brace of multipermutation level is equiv-alent to an abelian group ( A, +) and an automorphism ψ of ( A, +) such that1) there exists an element x of A whose orbit { ψ z ( x ) | z ∈ Z } has size greaterthan and generates ( A, +) ;2) if k P i =1 a i ψ z i ( x ) = 0 implies ψ k P i =1 a i = id A , for all k ∈ N and a , . . . , a k , z , . . . , z k ∈ Z .Proof. Suppose that A is a one-generator left brace of multipermutation level 2.By Theorem 29, there exists a transitive cycle base X with | X | >
1. Let x ∈ X and ψ := λ x . Then, ψ is an automorphism of ( A, +) and the additive groupgenerated by X = { ψ z ( x ) | z ∈ Z } is equal to A . Now, if k P i =1 a i ψ z i ( x ) = 0, byTheorem 29-2) and Lemma 26, it follows that ψ k P i =1 a i = λ ( k P i =1 a i ψ z i ( x )) = id A .Conversely, suppose that ( A, +) is an abelian group and ψ an automorphismof A such that 1) and 2) hold and let x be the element of A such that itsorbit X := { ψ z ( x ) | z ∈ Z } generates ( A, +). Now, by 1) and 2), the map λ ′ : X → Aut( A, +) given by λ ′ ( t ) := ψ , for every t ∈ X , can be extended to agroup homomorphism λ from ( A, +) into Aut( A, +). Now, let ◦ be the binaryoperation on A given by a ◦ b = a + λ a ( b ), for all a, b ∈ A . To prove that ( A, + , ◦ )is a left brace, by [26, Proposition 2] it remains to show that λ ( a ◦ b ) = λ a λ b ,for all a, b ∈ A . Therefore, let a, b ∈ A and suppose that a = P a i ψ z i ( x ) and b = P b i ψ z i ( x ). 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