Constructing finite simple solutions of the Yang-Baxter equation
aa r X i v : . [ m a t h . QA ] D ec Constructing finite simple solutions of theYang-Baxter equation ∗ F. Ced´o J. Okni´nski
Abstract
We study involutive non-degenerate set-theoretic solutions (
X, r ) ofthe Yang-Baxter equation on a finite set X . The emphasis is on thecase where ( X, r ) is indecomposable, so the associated permutation group G ( X, r ) acts transitively on X . One of the major problems is to determinehow such solutions are built from the imprimitivity blocks; and also howto characterize these blocks. We focus on the case of so called simplesolutions, which are of key importance. Several infinite families of suchsolutions are constructed for the first time. In particular, a broad class ofsimple solutions of order p , for any prime p , is completely characterized. The Yang-Baxter equation appeared independently in a paper of Yang [36] anin a paper of Baxter [4] and then it became one of the important equations ofmathematical physics. It also appears in the foundations of quantum groupsand Hopf algebras (see for example [5, 26]). Recall that a solution of the Yang-Baxter equation is an automorphism R of a vector space V ⊗ V , such that R R R = R R R , where R ij denotes the map V ⊗ V ⊗ V → V ⊗ V ⊗ V acting as R on the( i, j ) tensor factors and as the identity on the remaining factor. A difficultand important open problem is to find all the solutions of the Yang-Baxterequation. Drinfeld in [17] suggested the study of the set-theoretic solutions ofthe Yang-Baxter equation, these are pairs ( X, r ), where X is a non-empty setand r : X × X → X × X is a bijective map such that r r r = r r r , where r ij denotes the map X × X × X → X × X × X acting as r on the ( i, j )components and as the identity on the remaining component. ∗ The first author was partially supported by the grants MINECO-FEDER MTM2017-83487-P and AGAUR 2017SGR1725 (Spain). The second author is supported by the NationalScience Centre grant. 2016/23/B/ST1/01045 (Poland). 2010 MSC: Primary 16T25, 20B15,20F16. Keywords: Yang-Baxter equation, set-theoretic solution, primitive group, brace. r : X × X → X × X of the Yang-Baxter equation, written in the form r ( x, y ) =( σ x ( y ) , γ y ( x )), for x, y ∈ X , is involutive if r = id, and it is non-degenerate if σ x and γ x are bijective maps from X to X , for all x ∈ X .In [18] and [21] a number of very fruitful algebraic structures were introducedto study this class of solutions: the structure group, the structure monoid andthe structure algebra associated to a solution. More recently, in [29], Rumpintroduced a new algebraic structure, called (left) brace, in this context. Thisattracted a lot of attention (see for example [1, 6, 8, 16, 19, 33] and the referencesin these papers). And this allowed to construct several new families of solutionsand led to a discovery of several unexpected connections to a variety of otherareas [10, 30].In the study of involutive non-degenerate set-theoretic solutions of the Yang-Baxter equation, a very fruitful approach is based on the notions of indecompos-able solutions and irretractable solutions [18]. Roughly speaking, the underlyingidea is to show that several classes of solutions come from solutions of smallercardinality.Every (involutive non-degenerate set-theoretic) solution ( X, r ) of the YBEis equipped with a permutation group G ( X, r ) acting on the set X . It has beenexpected for a long time that a possible classification of solutions ( X, r ) of theYBE would have to be based on the associated groups G ( X, r ) ⊆ Sym X . There-fore, it is natural to anticipate that some aspects of the theory of permutationgroups will be crucial in this context. However, so far, mainly transitivity of thegroup G ( X, r ) (equivalent to indecomposability of the solution (
X, r )) has beenexplored. This has led to important results on decomposability of solutions, buton the other hand it turned out that in general indecomposable solutions aredifficult to construct and classify [7, 9, 22, 31, 34]. In particular, a fundamentalresult of Rump [28] shows that all finite square-free involutive non-degenerateset-theoretic solutions (
X, r ), with | X | >
1, are decomposable. However, this isno longer true in full generality.The second approach, based on the retract relation, allowed to introduce theclass of multipermutation solutions and to define the multipermutation level,which is a measure of their complexity. It is related to the so called structuregroup G ( X, r ) (which is a Bieberbach group if X is finite) associated to eachsolution ( X, r ). Certain positive results in this direction were obtained in [11,12, 20] and later in [2] it was shown that finite multipermutation solutionscoincide with solutions whose structure groups are poly- Z -groups. However,this approach also fails in full generality because there exist solutions that arenot retractable.In a talk [3] during a workshop in Oberwolfach [23], Ballester-Bolinches askedfor a description of all finite primitive solutions, i.e. solutions such that G ( X, r )acts on X as a primitive permutation group. It was recently shown in [14] that2here is only one class of such solutions of cardinality >
1. Namely, every finiteprimitive solution (
X, r ) is of prime order, i.e. | X | = p is a prime number,and it is a so called permutation solution determined by a cyclic permutationof length p . This opens a new perspective on the classification problem of allfinite solutions because, roughly speaking, this shows that every finite solution( X, r ) which is not of the above form is built on an information coming fromits imprimitivity blocks, which are sets of smaller cardinality. And now thechallenge is to understand the structure of possible imprimitivity blocks of anarbitrary solution and to see how it can be built from its imprimitivity blocks.On the other hand, recent results from [7] show that every finite indecomposablesolution of the YBE is a so called dynamical extension (introduced by Vendraminin [35]) of a simple solution. The latter are defined as solutions (
X, r ) that donot admit any nontrivial epimorphism of solutions (
X, r ) → ( Y, r ′ ) (meaningthat 1 < | Y | < | X | ). For example, all finite primitive solutions are simple. Thishas been, till now, the only known infinite family of finite simple solutions. Sothe challenge is to construct and classify all simple solutions. This is our mainmotivation in this paper.The paper is organized as follows. In Section 2 we introduce the necessarybackground on involutive non-degenerate set-theoretic solutions of the Yang-Baxter equation, left braces and on cycle-sets, another structure introduced byRump in [28] to study this class of solutions. In Section 3, we expose the recentadvances in the study of indecomposable solutions and we introduce simple solu-tions. In Section 4 some properties of simple solutions are studied. For example,we prove that finite simple solutions are indecomposable and irretractable (iftheir order is not a prime). Then we focus on solutions of a special but quitegeneral type (Proposition 4.6), which allows us to construct infinite families offinite simple solutions of the Yang-Baxter equation, in Theorems 4.7, 4.9 and4.12. In particular, we show that for every positive integer n , distinct primes p , . . . , p n and positive integers m , . . . , m n , such that P ni =1 m i > n , there ex-ists a simple solution of the Yang-Baxter equation of cardinality p m · · · p m n n . InSection 5 a wide class of simple solutions of cardinality p , for every prime p ,is characterized; see Theorems 5.1 and 5.3. In Section 6 we give an alternativeconstruction of a family of simple solutions of square cardinality introduced inSection 4, in terms of the asymmetric product of left braces. Finally, in Section7 some open questions related to simple solutions are proposed. Let X be a non-empty set and let r : X × X → X × X be a map. For x, y ∈ X weput r ( x, y ) = ( σ x ( y ) , γ y ( x )). Recall that ( X, r ) is an involutive, non-degenerate,set-theoretic solution of the Yang-Baxter equation if r = id, all the maps σ x and γ y are bijective maps from X to itself and r r r = r r r , r = r × id X and r = id X × r are maps from X to itself. Because r = id, one easily verifies that γ y ( x ) = σ − σ x ( y ) ( x ), for all x, y ∈ X (see forexample [18, Proposition 1.6]). Convention.
Throughout the paper a solution of the YBE will mean an invo-lutive, non-degenerate, set-theoretic solution of the Yang-Baxter equation.To study this class of solutions of the YBE, Rump [29] introduced the al-gebraic structure called a left brace. We recall some essential background (see[10] for details). A left brace is a set B with two binary operations, + and ◦ ,such that ( B, +) is an abelian group (the additive group of B ), ( B, ◦ ) is a group(the multiplicative group of B ), and for every a, b, c ∈ B , a ◦ ( b + c ) + a = a ◦ b + a ◦ c. (1)Note that if we denote by 0 the neutral element of ( B, +) and by 1 the neutralelement of ( B, ◦ ), then1 = 1 ◦ (0 + 0) + 1 = 1 ◦ ◦ . In any left brace B there is an action λ : ( B, ◦ ) → Aut( B, +), called the lambdamap of B , defined by λ ( a ) = λ a and λ a ( b ) = − a + a ◦ b , for a, b ∈ B . We shallwrite a ◦ b = ab , for all a, b ∈ B . A trivial brace is a left brace B such that ab = a + b , for all a, b ∈ B , i.e. all λ a = id. The socle of a left brace B isSoc( B ) = { a ∈ B | ab = a + b, for all b ∈ B } . Note that Soc( B ) = Ker( λ ), and thus it is a normal subgroup of the multiplica-tive group of B . The solution of the YBE associated to a left brace B is ( B, r B ),where r B ( a, b ) = ( λ a ( b ) , λ − λ a ( b ) ( a )), for all a, b ∈ B (see [12, Lemma 2]).A left ideal of a left brace B is a subgroup L of the additive group of B suchthat λ a ( b ) ∈ L , for all b ∈ L and all a ∈ B . An ideal of a left brace B is anormal subgroup I of the multiplicative group of B such that λ a ( b ) ∈ I , for all b ∈ I and all a ∈ B . Note that ab − = a − λ ab − ( b ) (2)for all a, b ∈ B , and a − b = a + λ b ( b − ) = aλ a − ( λ b ( b − )) = aλ a − b ( b − ) , (3)for all a, b ∈ B . Hence, every left ideal L of B also is a subgroup of themultiplicative group of B , and every ideal I of a left brace B also is a subgroupof the additive group of B . For example, it is known that Soc( B ) is an ideal ofthe left brace B (see [29, Proposition 7]). Note that B/I inherits a natural leftbrace structure.Let B be a left brace. We define another binary operation ∗ on B by a ∗ b : = − a + ab − b = λ a ( b ) − b, a, b ∈ B . By [29, Corollary of Proposition 6], the subgroup B = B ∗ B =gr( a ∗ b | a, b ∈ B ) + of ( B, +) generated by all the elements of the form a ∗ b isan ideal of B .Recall that if ( X, r ) is a solution of the YBE, with r ( x, y ) = ( σ x ( y ) , γ y ( x )),then its structure group G ( X, r ) = gr( x ∈ X | xy = σ x ( y ) γ y ( x ) , for all x, y ∈ X ) has a natural structure of a left brace such that λ x ( y ) = σ x ( y ), for all x, y ∈ X . The additive group of G ( X, r ) is the free abelian group with basis X .The permutation group G ( X, r ) = gr( σ x | x ∈ X ) of ( X, r ) is a subgroup of thesymmetric group Sym X on X . The map x σ x , from X to G ( X, r ) extendsto a group homomorphism φ : G ( X, r ) −→ G ( X, r ) and Ker( φ ) = Soc( G ( X, r )).Hence there is a unique structure of a left brace on G ( X, r ) such that φ is ahomomorphism of left braces, this is the natural structure of a left brace on G ( X, r ). Lemma 2.1 [14, Lemma 2.1] Let ( X, r ) be a solution of the YBE. Then λ g ( σ x ) = σ g ( x ) , for all g ∈ G ( X, r ) and all x ∈ X . The following result is well known (see [29, Proposition 7]), but we give anew proof for the convenience of the reader.
Lemma 2.2
Let B be a left brace. Then B/ Soc( B ) ∼ = G ( B, r B ) as left braces. Proof.
By the definition of the lambda map of B , we have λ a ( b ) λ − λ a ( b ) ( a ) = λ a ( b ) + λ λ a ( b ) ( λ − λ a ( b ) ( a )) = λ a ( b ) + a = ab, for all a, b ∈ B . Hence, by the definition of the group G ( B, r B ), there is ahomomorphism of left braces ϕ : G ( B, r B ) −→ B , such that ϕ ( a ) = a , for all a ∈ B . Let g ∈ ϕ − (Soc( B )). Then λ ϕ ( g ) ( b ) = b for b ∈ B . In G ( B, r B ) wehave that λ g ( b ) ∈ B . Hence b = λ ϕ ( g ) ( b ) = ϕ ( g ) b − ϕ ( g ) = ϕ ( g ) ϕ ( b ) − ϕ ( g ) = ϕ ( gb − g ) = ϕ ( λ g ( b )) = λ g ( b ) , for all b ∈ B . Since the additive group of G ( B, r B ) is the free abelian group withbasis B and λ g ∈ Aut( G ( B, r B ) , +), we get that λ g ( h ) = h , for all h ∈ G ( B, r B ).Therefore g ∈ Soc( G ( B, r B )) and hence ϕ − (Soc( B )) = Soc( G ( B, r B )) follows.Thus G ( B, r B ) ∼ = G ( B, r B ) / Soc( G ( B, r B )) ∼ = B/ Soc( B ) , as left braces.Let ( X, r ) and (
Y, s ) be solutions of the YBE. We write r ( x, y ) = ( σ x ( y ) , γ y ( x ))and s ( t, z ) = ( σ ′ t ( z ) , γ ′ z ( t )), for all x, y ∈ X and t, z ∈ Y . A homomorphismof solutions f : ( X, r ) −→ ( Y, s ) is a map f : X −→ Y such that f ( σ x ( y )) = σ ′ f ( x ) ( f ( y )) and f ( γ y ( x )) = γ ′ f ( y ) ( f ( x )), for all x, y ∈ X . Since γ y ( x ) = σ − σ x ( y ) ( x )and γ ′ z ( t ) = ( σ ′ ) − σ ′ t ( z ) ( t ), it is clear that f is a homomorphism of solutions if andonly if f ( σ x ( y )) = σ ′ f ( x ) ( f ( y )), for all x, y ∈ X .5ote that every homomorphism of solutions f : ( X, r ) −→ ( Y, s ) extends toa homomorphism of left braces f : G ( X, r ) −→ G ( Y, s ) that we also denote by f , and induces a homomorphism of left braces ¯ f : G ( X, r ) −→ G ( Y, s ).In [18], Etingof, Schedler and Soloviev introduced the retract relation onsolutions (
X, r ) of the YBE. This is the binary relation ∼ on X defined by x ∼ y if and only if σ x = σ y . Then, ∼ is an equivalence relation and r induces asolution r on X = X/ ∼ . The retract of the solution ( X, r ) is Ret(
X, r ) = (
X, r ).Note that the natural map f : X −→ X : x ¯ x is an epimorphism of solutionsfrom ( X, r ) onto Ret(
X, r ).Define Ret ( X, r ) = Ret(
X, r ) and for every positive integer n , Ret n +1 ( X, r ) =Ret(Ret n ( X, r )). The solution (
X, r ) is said to be a multipermutation solutionof level n if n is the smaller positive integer such that Ret n ( X, r ) has cardinality1. Recall that a solution (
X, r ) is said to be irretractable if σ x = σ y for alldistinct elements x, y ∈ X , that is ( X, r ) = Ret(
X, r ), otherwise the solution(
X, r ) is retractable.Another algebraic structure introduced by Rump [28] to study solutions ofthe YBE is the structure of a left cycle set. Recall that a left cycle set is a pair( X, · ) of a non-empty set X and a binary operation · such that( x · y ) · ( x · z ) = ( y · x ) · ( y · z ) , for all x, y, z ∈ X , and the map y x · y is a bijective map from X to itself forevery x ∈ X . A left cycle set ( X, · ) is non-degenerate if the map x x · x is abijective map from X to itself. In [28, Propositions 1 and 2] it is proven thatthere is a bijective correspondence between non-degenerate left cycle sets andsolutions of the YBE. That is, if ( X, · ) is a non-degenerate left cycle set, and σ x is the inverse of the map y x · y , then ( X, r ) is the corresponding solution ofthe YBE, where r ( x, y ) = ( σ x ( y ) , σ x ( y ) · x ), for all x, y ∈ X .In [35] Vendramin introduced the dynamical extension of a cycle set. Let I be a left cycle set and let S be a non-empty set. A map α : I × I × S −→ Sym S is a dynamical cocycle of I with values in S if, for all i, j, k ∈ I and r, s, t ∈ S , α ( i · j,i · k ) ( α ( i,j ) ( r, s ) , α ( i,k ) ( r, t )) = α ( j · i,j · k ) ( α ( j,i ) ( s, r ) , α ( j,k ) ( s, t )) , where α ( i,j ) ( r, s ) : = α ( i, j, r )( s ). The dynamical extension of I by α is the leftcycle set S × α I : = ( S × I, · ), where( s, i ) · ( t, j ) : = ( α ( i,j ) ( s, t ) , i · j ) , for all i, j ∈ I and s, t ∈ S . Let (
X, r ) be a solution of the YBE. We say that (
X, r ) is indecomposable if G ( X, r ) acts transitively on X .The following definition is due to Adolfo Ballester-Bolinches [3].6 efinition 3.1 A finite solution ( X, r ) of the YBE is said to be primitive if itspermutation group G ( X, r ) acts primitively on X . By [18, Theorem 2.13], for each prime p , there is, up to isomorphism, aunique indecomposable solution ( X, r ) of the YBE of cardinality p . In this case, X = Z / ( p ) and σ i ( j ) = j + 1, for all i, j ∈ Z / ( p ). Thus G ( X, r ) ∼ = Z / ( p ), and( X, r ) is primitive and it is a multipermutation solution of level 1. In [14] it isproven that these are all the finite primitive solutions of the YBE of cardinality > X, r ) of the YBE has primitivelevel k if k is the biggest positive integer such that there exist solutions ( X , r ) =( X, r ) , ( X , r ) , . . . , ( X k , r k ) and epimorphisms of solutions p i +1 : ( X i , r i ) −→ ( X i +1 , r i +1 ), with | X i | > | X i +1 | >
1, for 1 ≤ i ≤ k −
1, and ( X k , r k ) is primitive. Question 3.2
Describe solutions of primitive level . As said above, solutionsof primitive level (in other words, primitive solutions) admit a very simpledescription. In [34, Theorems 5.3 and 5.4] Agata Smoktunowicz and Alicja Smoktunowiczgave a method to construct any finite indecomposable solution of the YBE,assuming that one is able to construct all finite left braces. Indeed, let B be afinite left brace and let x ∈ B . Consider the left subbrace B ( x ) generated by x . Let X = { λ a ( x ) | a ∈ B ( x ) } . Then ( X, r ) is indecomposable, where r is therestriction of r B to X × X , and ( B, r B ) is the solution associated to the left brace B . Furthermore, every finite indecomposable solution of the YBE is constructedin this way. However, it is difficult to describe the structure of B ( x ) and theproperties of the indecomposable solution ( X, r ). For example, it is unclear howone can choose B and x ∈ B such that ( X, r ) is irretractable. In [15, Section 6]it is proven that this method also is valid for infinite indecomposable solutions.In [31] Rump gave another method to construct indecomposable solutionsof the YBE. Let (
X, r ) be an indecomposable finite solution of the YBE. Let G = G ( X, r ) and let x ∈ X . Let c x, G : G × G −→ G × G be the map defined by c x, G ( g, h ) = ( hσ − g − ( x ) , gσ σ g − x ) h − ( x ) ) , for all g, h ∈ G . Then ( G , c x, G ) is an indecomposable finite solution of the YBE.Furthermore, the map p : G −→ X defined by p ( g ) = g − ( x ) is an epimorphismof solutions, G ( G , c x, G ) ∼ = G and for every indecomposable finite solution ( Y, s ) ofthe YBE and every epimorphism q : Y −→ X of solutions such that the inducedhomomorphism of groups G ( Y, s ) −→ G is an isomorphism, there exists a uniqueepimorphism of solutions q ′ : G −→ Y such that p = qq ′ . Rump uses cycle setsand p is called the universal covering. The fundamental group of the cycle setcorresponding to the solution ( X, r ) is the stabilizer of x , i.e. π ( X ) = π ( X, x ) = { g ∈ G | g ( x ) = x } . Then Rump presents a method to construct all the finite indecomposable solu-tions (
Z, t ) of the YBE such that the natural structure of left brace of G ( Z, t )7s isomorphic to the left brace G . Thus this yields another method to constructall the finite indecomposable solutions, but again assuming that one is able toconstruct all finite left braces.We now introduce the key notion studied in this paper. Definition 3.3
A solution ( X, r ) of the YBE is simple if | X | > and for everyepimorphism of f : ( X, r ) → ( Y, s ) of solutions either f is an isomorphism or | Y | = 1 . Remark 3.4
In [35], Vendramin introduced finite simple cycle sets. His def-inition does not coincide with the above definition of simplicity, but for finiteindecomposable cycle sets both definitions coincide by [7, Proposition 2].
Suppose that (
X, r ) is an indecomposable finite solution of the YBE which isnot simple. Thus there exists a solution (
Y, s ) of the YBE and an epimorphism p : X −→ Y of solutions such that 1 < | Y | < | X | . Note that ( Y, s ) also isindecomposable. In [7, Proposition 2] Castelli, Catino and Pinto proved thatthe cycle set corresponding to (
X, r ) is a dynamical extension of the cycle set( Y, · ) corresponding to ( Y, s ) by a dynamical cocycle α of the cycle set ( Y, · )with values in p − ( y ), for some y ∈ Y . We also say that the solution ( X, r ) is adynamical extension of the solution (
Y, s ). By [7, Lemma 1], | p − ( y ) | = | p − ( y ′ ) | for all y, y ′ ∈ Y , i.e. p is a covering of the corresponding cycle sets. Furthermore,in the proof of [7, Lemma 1] it is proved that g ( p − ( p ( x ))) = p − ( p ( g ( x ))), forall x ∈ X and g ∈ G ( X, r ). Hence { p − ( y ) | y ∈ Y } is a set of imprimitivityblocks of X under the action of G ( X, r ). For every y ∈ Y , let H y be the followingsubgroup of G ( X, r ) H y = { g ∈ G ( X, r ) | g ( p − ( y )) = p − ( y ) } . Since (
X, r ) is indecomposable, H y acts transitively on p − ( y ) (see [7, Theorem7]). In [7] there are some concrete examples of indecomposable solutions con-structed in this way (using dynamical extensions of indecomposable cycle sets).[7, Proposition 10] gives a criterion to check whether a dynamical extension ofa cycle set is irretractable.Since every finite indecomposable solution of the YBE is a dynamical ex-tension of a simple solution of the YBE, another strategy to construct all thefinite indecomposable solutions is to construct all the finite simple solutions ofthe YBE and to determine all the dynamical cocycles on every cycle set corre-sponding to a finite simple solution of the YBE.We define the composition length of a finite indecomposable solution ( X, r )as the biggest positive integer k such that there exist solutions ( X , r ) =( X, r ) , ( X , r ) , . . . , ( X k , r k ) and epimorphisms of solutions p i +1 : ( X i , r i ) −→ ( X i +1 , r i +1 ), with | X i | > | X i +1 | , for 1 ≤ i ≤ k −
1, and ( X k , r k ) is simple.From a computer calculation one obtains that the following two examplesare the only solutions on a set of cardinality four that are indecomposable andirretractable. 8 xample 3.5 Let X = { , , , } . Define permutations σ = (2 , , σ = (1 , , σ = (1 , , , , σ = (1 , , , ∈ Sym X . Then ( X, r ) is a solution of the YBE, with r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )) , for all x, y ∈ X . Notice that G ( X, r ) = h σ , σ , σ , σ i is isomorphic to the dihedralgroup of order . This is an example of an indecomposable and irretractablesolution, see [24] or Example 8.2.14 in [25]; actually the first known example ofa solution whose structure group is not a poly- Z group. It is clear that G ( X, r ) acts transitively on X and X = { , } and X = { , } form imprimitivityblocks for the action of the group G ( X, r ) on X . The second example is as follows.
Example 3.6
Let X = { , , , } . Let σ = (1 , , σ = (3 , , , , σ = (2 , , , , σ = (3 , . It is easy to check that ( X, r ) is an indecomposable and irretractable solutionof the YBE, with r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )) , for all x, y ∈ X . Also in thiscase G ( X, r ) ∼ = D . The imprimitivity blocks for the action of G ( X, r ) on X are { , } and { , } . This solution is not isomorphic to the first one because wehave σ i ( i ) = i , for all i ∈ { , , , } , while this is not true in Example 3.5. In fact, we will see that these two solutions of cardinality four are the onlysimple solutions of the YBE of this cardinality (see Remark 4.8).
Note that every indecomposable solution of the YBE of prime cardinality issimple. And these are the finite primitive solutions of cardinality bigger than 1.In this section we construct new examples of finite simple solutions of theYBE. First, we study some properties of this class of solutions.
Lemma 4.1
Let ( X, r ) be a simple solution of the YBE. If | X | > then ( X, r ) is indecomposable. Proof.
Suppose that (
X, r ) is decomposable. Then X is a disjoint union X = X ∪ X of two non-empty subsets X and X of X such that r ( X i × X i ) = X i × X i , for i = 1 ,
2. Note that then r ( X × X ) = X × X and r ( X × X ) = X × X .Let Y = { , } and s : Y × Y → Y × Y be defined by s ( i, j ) = ( j, i ), for all i, j ∈ Y . Thus ( Y, s ) is the trivial solution. Let f : X → Y be the map definedby f ( x i ) = i if x i ∈ X i . Clearly f is an epimorphism of solutions and it isnot an isomorphism because | X | >
2. But, by assumption, (
X, r ) is simple, acontradiction. Therefore (
X, r ) is indecomposable.As mentioned earlier (see [18]), if (
X, r ) is an indecomposable solution of theYBE and | X | is a prime, then it is a multipermutation solution of level 1 (inparticular it is retractable). 9 roposition 4.2 Let ( X, r ) be a finite simple solution of the YBE. If | X | isnot prime, then ( X, r ) is irretractable. Proof.
Suppose that (
X, r ) is retractable. Consider the natural projection π : ( X, r ) → Ret(
X, r ). Since, π is an epimorphism of solutions and becauseit is not an isomorphism and ( X, r ) is simple, we have that the cardinality ofRet(
X, r ) is 1. Again write r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )) for x, y ∈ X . Then wehave that σ x = σ y for all x, y ∈ X . Thus r ( x, y ) = ( σ ( y ) , σ − ( x )), for some permutation σ ∈ Sym X , that is ( X, r ) is a permutation solution, introduced by Lyubashenko; see[17]. By Lemma 4.1, since | X | >
2, (
X, r ) is indecomposable. By [18, p. 184], σ is a cycle of length | X | , that is σ = ( x , . . . , x n ) with n = | X | . Since n isnot prime, there exist integers 1 < d, m < n such that n = dm . Let Y = Z / ( d )and let s : Y × Y → Y × Y be the map defined by s ( i, j ) = ( j + 1 , i −
1) forall i, j ∈ Y . Then ( Y, s ) is a solution of the YBE (a permutation solution). Let f : X → Y be the map defined by f ( x i ) = i ( ∈ Z / ( d )), for all i = 1 , . . . , n . Notethat f ( σ x i ( x j )) = (cid:26) f ( x j +1 ) if j < n j = n . Hence f is an epimorphism of solutions and 1 < | Y | < | X | , a contradiction.Therefore ( X, r ) is irretractable.Let (
X, r ) be a finite simple solution such that | X | is not prime. By Lemma4.1 and Proposition 4.2, we know that ( X, r ) is indecomposable and irretractable.Consider the permutation group G = G ( X, r ). Then the map x σ x is an in-jective morphism of solutions from ( X, r ) to ( G , r G ). Furthermore, by Lemma2.1, σ ( X ) = { σ x | x ∈ X } is an orbit by the action of the lambda map in theleft brace G that generates the multiplicative (and the additive) group of theleft brace G . Proposition 4.3
With the above conditions, if I is a minimal nonzero ideal of G , then G /I is a trivial cyclic brace, and I = G = G ∗ G = h σ x − σ y | x, y ∈ X i + .Furthermore Soc( G ) = { } . Proof.
Let I be a minimal nonzero ideal of G . Let π : G −→ G /I be the naturalmap. Let g ∈ I be a nonzero element. By Lemma 2.1, λ g ( σ x ) = σ g ( x ) , for all x ∈ X . Hence π ( σ g ( x ) ) = π ( λ g ( σ x )) = λ π ( g ) ( π ( σ x )) = λ ( π ( σ x )) = π ( σ x ) . Thus the restriction π | σ ( X ) of π induces an epimorphism x π ( σ x ) of solutionsfrom ( X, r ) to ( π ( σ ( X )) , s ), where s ( π ( σ x ) , π ( σ y )) = ( π ( σ σ x ( y ) ) , π ( σ σ − σx ( y ) ( x ) )) . g = 0, there exists x ∈ X such that x = g ( x ). Furthermore, since π ( σ x ) = π ( σ g ( x ) ) and ( X, r ) is simple, we have that π ( σ x ) = π ( σ y ), for all x, y ∈ X . Hence h σ x − σ y | x, y ∈ X i + ⊆ I. Note that G = h g ∗ h | g, h ∈ Gi + = h λ g ( h ) − h | g, h ∈ Gi + = h λ g ( σ x ) − σ x | g ∈ G , x ∈ X i + (since G = h σ x | x ∈ X i + )= h σ g ( x ) − σ x | g ∈ G , x ∈ X i + (by the above)= h σ y − σ x | x, y ∈ X i + , (since ( X, r ) is indecomposable)By the minimality of I we get I = G = h σ x − σ y | x, y ∈ X i + . Now it is easyto see that if x ∈ X , G /I = h π ( σ x ) i + = h π ( σ x ) i is the trivial brace of the cyclicgroup h π ( σ x ) i . Finally let g ∈ Soc( G ). By Lemma 2.1, we have that σ g ( x ) = λ g ( σ x ) = σ x , for all x ∈ X . Since ( X, r ) is irretractable, we get that g ( x ) = x for all x ∈ X .This proves that Soc( G ) = { } , and the result follows.Of course, nontrivial finite simple left braces B satisfy B = B and Soc( B ) = { } . There are several constructions of finite simple left braces, see for example[13] and the references therein, but we do not know the answer to the followingquestions. Question 4.4
Is there a nontrivial finite simple left brace B with an orbit X by the action of the lambda map such that B = h X i + ? Question 4.5
Is there a finite simple solution ( X, r ) of the YBE such that G ( X, r ) is a nontrivial simple left brace? Note that an affirmative answer to Question 4.5 implies an affirmative answerto Question 4.4.We will study the simple solutions (
X, r ) of the YBE, where X = Y × Z and r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )), for all x, y ∈ X and σ ( i,j ) ( k, l ) = ( σ j ( k ) , d i,σ j ( k ) ( l )) , for all i, k ∈ Y and j, l ∈ Z . Note that in this case the sets X i = { ( i, j ) | j ∈ Z } ,for i ∈ Y , are imprimitivity blocks for the action of G ( X, r ) on X , and we areassuming that σ i,j X k = X σ j ( k ) is independent of i . We also simplify the secondcomponent of σ ( i,j ) ( k, l ), imposing a link between j and k , which is σ j ( k ). Proposition 4.6
Let
Y, Z be finite non-empty sets such that | Y | , | Z | > . Let X = Y × Z . Let r : X × X −→ X × X be a map and write r ( x, y ) = ( σ x ( y ) , γ y ( x )) . ssume that σ ( i,j ) ( k, l ) = ( σ j ( k ) , d i,σ j ( k ) ( l )) , for all i, k ∈ Y and all j, l ∈ Z .Then ( X, r ) is an indecomposable and irretractable solution of the YBE if andonly if σ j ∈ Sym Y , d i,k ∈ Sym Z , the permutation subgroups F = h σ j | j ∈ Z i ⊆ Sym Y and W = h d i,k | i, k ∈ Y i ⊆ Sym Z are transitive,1. r (( i, j ) , ( k, l )) = (( σ j ( k ) , d i,σ j ( k ) ( l )) , ( σ − d i,σj ( k ) ( l ) ( i ) , d − σ j ( k ) ,i ( j ))) ,2. σ j ◦ σ d − i,k ( l ) = σ l ◦ σ d − k,i ( j ) ,3. d i,k = d k,i ,4. d i,w ◦ d σ − j ( k ) ,σ − j ( w ) = d k,w ◦ d σ − l ( i ) ,σ − l ( w ) , for all i, k, w ∈ Y and j, l ∈ Z , and(i) σ j = σ l for j = l ,(ii) if d i,k = d i ′ ,k for every k , then i = i ′ . Proof.
First, we shall see that (
X, r ) is a solution of the YBE if and only if σ j ∈ Sym Y , d i,k ∈ Sym Z , conditions 1. and 2. are satisfied and(4’) d i,σ j σ d − i,k ( l ) ( u ) ◦ d σ − j ( k ) ,σ d − i,k ( l ) ( u ) = d k,σ l σ d − k,i ( j ) ( u ) ◦ d σ − l ( i ) ,σ d − k,i ( l ) ( u ) , for all i, k, u ∈ Y and j, l ∈ Z .We know from [12] that ( X, r ) is a solution of the YBE if and only if σ ( i,j ) is bijective for all ( i, j ) ∈ X and the following conditions hold:(a) γ y ( x ) = σ − σ x ( y ) ( x ) for all x, y ∈ X .(b) σ x σ σ − x ( y ) = σ y σ σ − y ( x ) , for all x, y ∈ X .Clearly σ ( i,j ) is bijective if and only if σ j and d i,σ j ( k ) are bijective for all k ∈ Y .Condition (a) is equivalent to r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )), andby the definition of the maps σ j and d i,k , we have r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j ))= (( σ j ( k ) , d i,σ j ( k ) ( l )) , ( σ − σ j ( k ) ,d i,σj ( k ) ( l )) ( i, j ))= (( σ j ( k ) , d i,σ j ( k ) ( l )) , ( σ − d i,σj ( k ) ( l ) ( i ) , d − σ j ( k ) ,i ( j ))) . Hence, condition (a) is equivalent to condition 1.Note that σ ( i,j ) σ σ − i,j ) ( k,l ) ( u, v ) = σ ( i,j ) σ ( σ − j ( k ) ,d − i,k ( l )) ( u, v )= σ ( i,j ) (cid:18) σ d − i,k ( l ) ( u ) , d σ − j ( k ) , σ d − i,k ( l ) ( u ) ( v ) (cid:19) = ( σ j ( σ d − i,k ( l ) ( u )) , d i,σ j σ d − i,k ( l ) ( u ) ( d σ − j ( k ) σ d − i,k ( l ) ( u ) ( v ))) . σ j ◦ σ d − i,k ( l ) = σ l ◦ σ d − k,i ( j ) and d i σ j σ d − i,k ( l ) ( u ) ◦ d σ − j ( k ) , σ d − i,k ( l ) ( u ) = d k σ l σ d − k,i ( j ) ( u ) ◦ d σ − l ( i ) , σ d − k,i ( j ) ( u ) , for all i, k, u ∈ Y and j, l ∈ Z . Hence condition (b) holds if and only if conditions2. and (4’) hold. Therefore the claim follows.Suppose that ( X, r ) is a solution of the YBE. It is clear that (
X, r ) is inde-composable if and only if the permutation groups F and W are transitive.Let i, i ′ ∈ Y and j, j ′ ∈ Z be such that σ ( i,j ) = σ ( i ′ ,j ′ ) . This is equivalent to( σ j ( k ) , d i,σ j ( k ) ( l )) = ( σ j ′ ( k ) , d i ′ ,σ j ′ ( k ) ( l )) , for all k ∈ Y and l ∈ Z . Hence ( X, r ) is irretractable if and only if conditions(i) and (ii) are satisfied.Suppose that (
X, r ) is an irretractable solution of the YBE. By condition 2.with j = l , we have that σ d − i,k ( j ) = σ d − k,i ( j ) , for all j ∈ Z . Therefore σ ( u,d − i,k ( j )) = σ ( u,d − k,i ( j )) , for all i, k, u ∈ Y and all j ∈ Z . Since ( X, r ) is irretractable, we havethat d i,k = d k,i for all i, k ∈ Y , which is condition 3. Then, by condition (4’),for w = σ j σ d − i,k ( l ) ( u ), we get d i,w ◦ d σ − j ( k ) ,σ − j ( w ) = d k,w ◦ d σ − l ( i ) ,σ − l ( w ) . Thus condition 4. is satisfied.Hence, if (
X, r ) is an irretractable solution of the YBE, conditions 2. and(4’) are equivalent to conditions 2. and 4.Therefore the result follows.Now, we shall construct concrete examples of indecomposable and irre-tractable solutions (
X, r ), of the form described in Proposition 4.6, with | X | = n for every integer n > Theorem 4.7
Let n > be an integer. Let t ∈ Z / ( n ) be an invertible element.Let j , . . . , j n − ∈ Z / ( n ) be elements such that j i = j − i and j t s i = t s j i − ( t s − j , (4) for all i ∈ Z / ( n ) and all s ∈ Z . Suppose that for every nonzero i ∈ Z / ( n ) thereexists k ∈ Z / ( n ) such that j i + k − j k is invertible. Let r : ( Z / ( n )) × ( Z / ( n )) −→ ( Z / ( n )) × ( Z / ( n )) be the map defined by r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) , where σ ( i,j ) ( k, l ) = ( tk + j, t ( l − j tk + j − i )) , for all i, j, k, l ∈ Z / ( n ) . Then (( Z / ( n )) , r ) is an indecomposable and irretractable solution of the YBE.Assume, moreover, that i) j − j i is invertible for every nonzero i ∈ Z / ( n ) , and(ii) j i − j k is invertible for all j i = j k .Then (( Z / ( n )) , r ) is a simple solution of the YBE. Proof.
Consider the permutations σ j , d i,k ∈ Sym Z / ( n ) defined by σ j ( k ) = tk + j and d i,k ( l ) = t ( l − j k − i ) , for all i, j, k, l ∈ Z / ( n ). Then σ ( i,j ) ( k, l ) = ( σ j ( k ) , d i,σ j ( k ) ( l )) , for all i, j, k, l ∈ Z / ( n ). Note that σ σ − ( k ) = σ ( t − k ) = k + 1, for all k ∈ Z / ( n ). Therefore the group F = h σ j | j ∈ Z / ( n ) i is transitive on Z / ( n ). Notealso that d − , k d ,k ( l ) = d − , k ( t ( l − j k )) = l − j k + j k for all k, l ∈ Z / ( n ).Since there exists k ∈ Z / ( n ) such that j k − j k is invertible, the group W = h d i,j | i, j ∈ Z / ( n ) i is transitive on Z / ( n ). It is clear that condition (i) ofProposition 4.6 is satisfied. Suppose that d i,u = d i ′ ,u , for all u ∈ Z / ( n ). Then d i,u ( l ) = t ( l − j u − i ) = d i ′ ,u ( l ) = t ( l − j u − i ′ ), for all u, l ∈ Z / ( n ). Therefore i = i ′ and condition (ii) of Proposition 4.6 is satisfied. Obviously condition 1.of Proposition 4.6 is satisfied. Since j i = j − i , we have that d i,k = d k,i and thuscondition 3. of Proposition 4.6 is satisfied. We shall check conditions 2. and 4.Note that σ j σ d − i,k ( l ) ( u ) = σ j ( tu + t − l + j i − k )= t ( tu + t − l + j i − k ) + j = t u + l + tj i − k + j and σ l σ d − k,i ( j ) ( u ) = σ l ( tu + t − j + j k − i )= t ( tu + t − j + j k − i ) + l = t u + j + tj k − i + l. Since j i = j − i , we have that σ j ◦ σ d − i,k ( l ) = σ l ◦ σ d − k,i ( j ) , thus condition 2. issatisfied. Now, (using assumption (4) in the fifth equality below), we have d i,w d σ − j ( k ) ,σ − j ( w ) ( u ) = d i,w ( t ( u − j σ − j ( k ) − σ − j ( w ) ))= d i,w ( t ( u − j t − ( k − w ) ))= t ( t ( u − j t − ( k − w ) ) − j i − w )= t u − t j t − ( k − w ) − tj i − w = t u − t ( t − j k − w − ( t − − j ) − tj i − w = t u − tj k − w + t ( t − − j − tj i − w and thus d k,w d σ − l ( i ) ,σ − l ( w ) ( u ) = t u − tj i − w + t ( t − − j − tj k − w . d i,w ◦ d σ − j ( k ) ,σ − j ( w ) = d k,w ◦ d σ − l ( i ) ,σ − l ( w ) , for all i, j, k, l, w ∈ Z / ( n ),and thus condition 4. of Proposition 4.6 is satisfied. Therefore, by Proposition4.6 (( Z / ( n )) , r ) is an indecomposable and irretractable solution of the YBE.Assume now that conditions ( i ) and ( ii ) are satisfied.Let f : (( Z / ( n )) , r ) → ( Y, s ) be an epimorphism of solutions. Suppose that f is not an isomorphism, so that | Y | < n . Since (( Z / ( n )) , r ) is indecompos-able, ( Y, s ) also is indecomposable, and by [7, Lemma 1], | f − ( y ) | = | f − ( y ′ ) | ,for all y, y ′ ∈ Y . We write s ( y, z ) = ( σ ′ y ( z ) , γ ′ z ( y )).Suppose first that there exists j ∈ Z / ( n ) such that f ( i, j ) = f ( k, j ) for some i = k . Then f ( σ ( i,j ) ( u, v )) = f ( σ ( k,j ) ( u, v )) , that is f ( tu + j, t ( v − j tu + j − i )) = f ( tu + j, t ( v − j tu + j − k )) , for all u, v ∈ Z / ( n ). In particular, for u = t − ( i − j ) and v = w + j , we get f ( i, tw ) = f ( i, t ( w + j − j i − k )) , for all w ∈ Z / ( n ). Since t and j − j i − k are invertible, repeating this argumentseveral times, we obtain that f ( i,
0) = f ( i, w ) , for all w ∈ Z / ( n ). This implies that f ( σ ( i, ( i, v + j ti − i )) = σ ′ f ( i, ( f ( i, v + j ti − i ))= σ ′ f ( i,u ) ( f ( i, v + j ti + u − i ))= f ( σ ( i,u ) ( i, v + j ti + u − i )) , that is f ( ti, tv ) = f ( ti + u, tv ) , for all u ∈ Z / ( n ), and by the above argument, we get that f ( x i,w ) = f ( x u,v ) , for all u, v, w ∈ Z / ( n ). Hence | Y | = 1, as desired.Suppose that | Y | >
1. We have seen above that then for every y ∈ Y , f − ( y ) = { ( i , k ) , . . . , ( i m , k m ) } with |{ k , . . . , k m }| = m . Since | f − ( y ) | = | f − ( y ′ ) | = m , for all y, y ′ ∈ Y , we may assume, for a particular choice of y ,that ( i , k ) = (0 , z = f ( σ (0 , (0 , ∈ Y . Now we have that f ( σ ( i l ,k l ) (0 , f ( σ (0 , (0 , z, for all l = 1 , . . . , m . That is f − ( z ) ⊇ { ( k l , − tj k l − i l ) | l ∈ { , . . . , m }} . |{ k l | l ∈ { , . . . , m }}| = m = | f − ( z ) | , we have that f − ( z ) = { ( k l , − tj k l − i l ) | l ∈ { , . . . , m }} . (5)We also have that, for every u ∈ { , . . . , m } , f ( σ ( i u ,k u ) ( i l , k l )) = f ( σ ( i u ,k u ) (0 , z, for all l = 1 , . . . , m . That is, for every u ∈ { , . . . , m } , f − ( z ) = { ( ti l + k u , t ( k l − j ti l + k u − i l )) | l ∈ { , . . . , m }} . (6)Hence { k l | l ∈ { , . . . , m }} = { ti l + k u | l ∈ { , . . . , m }} , (7)for all u ∈ { , . . . , m } . In particular, for u = 1, we have k = 0 and { k l | l ∈ { , . . . , m }} = { ti l | l ∈ { , . . . , m }} . Thus there exists a permutation τ ∈ Sym m such that ti l = k τ ( l ) , for all l ∈{ , . . . , m } . Now it is clear from (7) that H = { k l | l ∈ { , . . . , m }} is asubgroup of Z / ( n ). Furthermore the map η : { i l | l ∈ { , . . . , m }} −→ { k l | l ∈ { , . . . , m }} , defined by η ( i l ) = ti l = k τ ( l ) is an isomorphism, thus in fact { i l | l ∈ { , . . . , m }} = { k l | l ∈ { , . . . , m }} is the unique subgroup of Z / ( n ) of order m . Moreover, by (5) and (6) for u = 1,we have that ( ti l , t ( k l − j ti l − i l )) = ( k τ ( l ) , − tj k τ ( l ) − i τ ( l ) ) . Hence, since t is invertible, k l − j ti l − i l = − j k τ ( l ) − i τ ( l ) . In particular0 = k = k = j ti − i − j k τ (2) − i τ (2) . By condition ( ii ), k is invertible in Z / ( n ), thus its additive order is n . Since { k l | l ∈ { , . . . , m }} is a subgroup of order m of Z / ( n ), we have that n = m .By (5), f − ( z ) = { ( k l , − tj k l − i l ) | l ∈ { , . . . , n }} , and by the description of f − ( z ) obtained in the first part of the proof n = |{− tj k l − i l | l ∈ { , . . . , n }}| = |{ j k l − i l | l ∈ { , . . . , n }}| . Since j i = j − i , this implies that n = 2. In this case, t = 1, j = j and f (0 ,
0) = f (1 ,
1) and f (0 ,
1) = f (1 , . Furthermore, | Y | = 2. Thus Y = { y , y } . Since ( Y, s ) is indecomposable, it isknown that σ ′ y = σ ′ y . Hence f (1 , j ) = f ( σ (0 , (1 , f ( σ (1 , (1 , f (1 , j ) . Since j = j , we get f (1 ,
0) = f (1 , | Y | = 2. Thiscontradicts the assumption that | Y | >
1. Therefore | Y | = 1 and the resultfollows. 16 emark 4.8 Note that for n = 2 there are two indecomposable and irretractablesolutions of the YBE constructed as in Theorem 4.7, with t = 1 , and correspond-ing to either ( j , j ) = (1 , or ( j , j ) = (0 , . In this case, also conditions (i)and (ii) are satisfied. Hence these two solutions are in fact simple. The solutioncorresponding to ( j , j ) = (0 , is isomorphic to the solution of Example 3.5.The solution corresponding to ( j , j ) = (1 , is isomorphic to the solution ofExample 3.6. In Theorem 4.7, the assumptions on the elements j i are very strong. Notethat for t = 1, condition (4) is empty. In the following result we see that for t = 1 the other assumptions on the j i in Theorem 4.7 can be relaxed to obtainindecomposable and irretractable solutions and condition ( ii ) is not needed toobtain simple solutions. Theorem 4.9
Let n > be an integer. Let j , . . . , j n − ∈ Z / ( n ) be elementssuch that j i = j − i , for all i ∈ Z / ( n ) . Suppose that for every nonzero i ∈ Z / ( n ) there exists k ∈ Z / ( n ) such that j i + k = j k . Assume also that h j i | i ∈ Z / ( n ) i = Z / ( n ) . Let r : ( Z / ( n )) × ( Z / ( n )) −→ ( Z / ( n )) × ( Z / ( n )) be the map definedby r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) , where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / ( n ) . Then (( Z / ( n )) , r ) is an indecomposable andirretractable solution of the YBE.Assume, moreover, that j − j i is invertible for every nonzero i ∈ Z / ( n ) .Then (( Z / ( n )) , r ) is a simple solution of the YBE. Proof.
As in the proof of Theorem 4.7, for t = 1, σ j ( l ) = l + j and d i,k ( l ) = l − j k − i , for all i, j, k, l ∈ Z / ( n ), one can see that (( Z / ( n )) , r ) is a solution of theYBE. Note that h σ j | j ∈ Z / ( n ) i is a transitive subgroup of Sym Z / ( n ) and, since h j i | i ∈ Z / ( n ) i = Z / ( n ), it is clear that h d i,k | i, k ∈ Z / ( n ) i also is a transitivesubgroup of Sym Z / ( n ) . Hence the solution (( Z / ( n )) , r ) is indecomposable. Notethat σ ( i,j ) ( u, v ) = σ ( i ′ ,j ′ ) ( u, v ) , for all u, v ∈ Z / ( n ), if and only if ( u + j, v − j u + j − i ) = ( u + j ′ , v − j u + j ′ − i ′ ),for all u, v ∈ Z / ( n ). This is equivalent to j = j ′ and j u + j − i = j u + j − i ′ , for all u ∈ Z / ( n ). Since for every nonzero i ∈ Z / ( n ) there exists k ∈ Z / ( n ) such that j i + k = j k , choosing u = i ′ − j + k we get that j u + j − i = j u + j − i ′ , for all u ∈ Z / ( n )if and only if i = i ′ . Therefore the solution (( Z / ( n )) , r ) is irretractable.Assume now that j − j i is invertible for every nonzero i ∈ Z / ( n ). Let f : (( Z / ( n )) , r ) → ( Y, s ) be an epimorphism of solutions. Suppose that f isnot an isomorphism, so that | Y | < n . Since (( Z / ( n )) , r ) is indecomposable,( Y, s ) also is indecomposable, and by [7, Lemma 1], | f − ( y ) | = | f − ( y ′ ) | , for all y, y ′ ∈ Y . We write s ( y, z ) = ( σ ′ y ( z ) , γ ′ z ( y )).Suppose first that there exists j ∈ Z / ( n ) such that f ( i, j ) = f ( k, j ) for some i = k . As in the proof of Theorem 4.7, one prove that | Y | = 1.Suppose that | Y | >
1. Hence for every y ∈ Y , f − ( y ) = { ( i , k ) , . . . , ( i m , k m ) } with |{ k , . . . , k m }| = m . Since | f − ( y ) | = | f − ( y ′ ) | = m , for all y, y ′ ∈ Y ,17e may assume, for a particular choice of y , that ( i , k ) = (0 , z = f ( σ (0 , (0 , ∈ Y . Now we have that f ( σ ( i l ,k l ) (0 , f ( σ (0 , (0 , z, for all l = 1 , . . . , m . That is f − ( z ) ⊇ { ( k l , − j k l − i l ) | l ∈ { , . . . , m }} . Since |{ k l | l ∈ { , . . . , m }}| = m = | f − ( z ) | , we have that f − ( z ) = { ( k l , − j k l − i l ) | l ∈ { , . . . , m }} . (8)In particular, by the form of f − ( z ) explained before, we get that j = j k − i = j k − i . Since (( Z / ( n )) , r ) is indecomposable, there exist ( u , v ) , . . . , ( u l , v l ) ∈ ( Z / ( n )) such that σ ( u ,v ) · · · σ ( u l ,v l ) (0 , − j ) = ( k , − j k − i ) . (9)Thus k = v + · · · + v l and − j k − i = − j − j v l − u l − j v l + v l − − u l − −· · ·− j v l + ··· + v − u . (10)Note that by (9) and (10), σ ( u ,v ) · · · σ ( u l ,v l ) ( v, w )= ( v + v + · · · + v l , w − j v l − u l − j v l + v l − − u l − − · · · − j v l + ··· + v − u )= ( v + k , w + j − j k − i ) (11)for all v, w ∈ Z / ( n ). Since, by (8), f ( k , − j k − i ) = f ( k , − j k − i ) = f (0 , − j ) = z , we have by (11), z = f (( σ ( u ,v ) · · · σ ( u l ,v l ) ) h (0 , − j ))= f ( hk , − j + h ( j − j k − i ))for all positive integers h . Since 0 = k − i = k − i , by our assuption j − j k − i is invertible and therefore |{− j + h ( j − j k − i ) | h ∈ Z / ( n ) }| = n. Hence, again using the properties of f − ( y ′ ), we have that m = | f − ( y ′ ) | ≥ n ,for every y ′ ∈ Y . In particular, by (8) |{ ( k l , − j k l − i l ) | l ∈ { , . . . , n }}| = n = |{ j k l − i l | l ∈ { , . . . , n }}| . Since j i = j − i , this implies that n = 2. Now the result follows as in the proofof Theorem 4.7. 18 emark 4.10 Let n be an integer greater than . Let j , . . . , j n − ∈ Z / ( n ) be elements such that j i = j − i for all i ∈ Z / ( n ) . Suppose that the solution ( X, r ) , with X = ( Z / ( n )) and r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) ,where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / ( n ) , is indecom-posable and irretractable. Is ( X, r ) a simple solution? The answer is negative.For example, take n = 6 , j = 1 and j i = 3 for i ∈ Z / (6) \ { } . In this case σ ( i,j ) ( k, l ) = ( k + j, l − δ k + j − i, ) . By Theorem 4.9, (( Z / (6)) , r ) , with r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) ,where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / (6) , is an indecompos-able and irretractable solution of the YBE. Let Y = { , } and s : Y −→ Y bethe map s ( y , y ) = ( σ ( y ) , σ ( y )) , for all y , y ∈ Y , where σ = (1 , ∈ Sym Y .Thus ( Y, s ) is a solution of the YBE. Let f : X −→ Y be the map defined by f ( x i,k ) = (cid:26) if k ∈ (2 Z ) / (6) , otherwise . It is easy to see that f is an epimorphism of solutions. Hence (( Z / (6)) , r ) isnot simple. Leandro Vendramin has calculated all 685 irretractable solutions of the YBEof cardinality 9 and up to isomorphism there are only 3 indecomposable irre-tractable solutions, (
X, r ) ( X, r ) and ( X, r ), where X = { , , , , , , , , } and r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )), with σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , σ = (1 , , , , σ = (2 , , , , r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )), with σ = (1 , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (4 , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (2 , , r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )), with σ = (1 , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , , , , , σ = (1 , , , , , , , , σ = (1 , , , , , , Remark 4.11
One can check that the three indecomposable and irretractablesolutions of the YBE of cardinality are simple solutions. The solution ( X, r ) is somorphic to (( Z / (3)) , r ′ ) with r ′ (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) ,where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / (3) , and j = 0 , j = j = 1 .The solution ( X, r ) is isomorphic to (( Z / (3)) , r ′ ) with r ′ (( i, j ) , ( k, l )) =( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) , where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / (3) , and j = 1 , j = j = 0 .The solution ( X, r ) is isomorphic to (( Z / (3)) , r ′ ) with r ′ (( i, j ) , ( k, l )) =( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) , where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / (3) , and j = 1 , j = j = 2 . Note that all the finite simple solutions of the YBE constructed so far havecardinality n for some integer n >
1. We shall construct now finite simplesolutions of the YBE of non-square cardinality.
Theorem 4.12
Let n, m > be integers. Let X = Z / ( mn ) × ( n Z ) / ( mn ) ∼ = Z / ( mn ) × Z / ( m ) . Consider σ ( i,j ) ∈ Sym X defined by σ ( i,j ) ( k, l ) = (cid:26) ( k + j/n, l ) if i = k + j/n ( k + j/n, l + n ) if i = k + j/n , for all ( i, j ) , ( k, l ) ∈ X , where j/n means the unique element of the subset { , , . . . , m − } of Z / ( mn ) such that n · ( j/n ) = j ∈ Z / ( mn ) . Let r : X × X −→ X × X be the map defined by r (( i, j ) , ( k, l )) = ( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) , forall ( i, j ) , ( k, l ) ∈ X . Then ( X, r ) is a simple solution of the YBE. Proof.
For every j ∈ ( n Z ) / ( mn ), let σ j ∈ Sym Z / ( mn ) be the permutationdefined by σ j ( k ) = k + j/n , for all k ∈ Z / ( mn ). For i, k ∈ Z / ( mn ), we define d i,k ∈ Sym ( n Z ) / ( mn ) by d i,k ( l ) = l + nδ i,k , for all l ∈ ( n Z ) / ( mn ). Note that h σ n i is a transitive subgroup of Sym Z / ( mn ) ,and h d , i is a transitive subgroup of Sym ( n Z ) / ( mn ) . Also it is easy to check thatconditions of Proposition 4.6 are satisfied. Hence ( X, r ) is an indecomposableand irretractable solution of the YBE.Note that σ (1 ,n ) is a cycle of length nm of the form:(0 , (1 , n ) (2 , n )
7→ · · · 7→ ( mn − , n ) (0 , n ) (1 , n ) (2 , n )
7→ · · · 7→ ( mn − , n ) (0 , ( m − n ) (1 , (2 ,
7→ · · · 7→ ( mn − , (0 ,
0) (12)We know that, if f : ( X, r ) → ( T, s ) is an epimorphism of solutions, and 1 < | T | < nm , then f − ( t ) , t ∈ T , form a system of imprimitivity blocks in X (with20espect to G ( X, r )). By the comment after Remark 3.4, (
X, r ) is a dynamicalextension of the solution (
T, s ). So, elements from the same block permute theset of blocks in the same way. And there exists an integer q > q | nm and each block consists of all elements in the full cycle (12) which are in distancebeing a multiple of q . If q | mn , then mn = dq for some positive integer d and theblocks are of the form B i = { ( i + qk, j ) | k = 0 , , . . . , d − j ∈ ( n Z ) / ( mn ) } , for i ∈ { , , . . . , q − } ⊆ Z / ( mn ). But then σ (1 , ( B i ) ⊆ B i for every block B i and σ (1 ,n ) ( B i ) ⊆ B i +1 , thus σ (1 , and σ (1 ,n ) determine different permutations ofthe set of blocks, while (1 , , (1 , n ) ∈ B . So X cannot come from a dynamicalcocycle, a contradiction. Hence q is not a divisor of mn . Suppose that q < mn .In this case, mn = dq + t for some positive integers d, t with t < q . Then theblock containing (0 ,
0) is B (0 , = { (0 , , ( q, n ) , . . . , ( dq, n ) , (( d + 1) q, n ) , . . . } . Now σ (0 , (0 ,
0) = (0 , n ) / ∈ B (0 , , but σ (0 , ( q, n ) = ( q, n ) ∈ B (0 , , a contradic-tion. Therefore q > mn . Hence q = dmn + t for some positive integers d, t with t < mn . Then B (0 , = { (0 , , ( t, ( d + 1) n ) , . . . } . Now σ (0 , (0 ,
0) = (0 , n ) / ∈ B (0 , , but σ (0 , ( t, ( d + 1) n ) = ( t, ( d + 1) n ) ∈ B (0 , ,a contradiction. It follows that ( X, r ) is a simple solution.
Remark 4.13
Note that if p , . . . p k are distinct primes and m , . . . , m k arepositive integers and m > , then we can take m = p and n = p m − p m · · · p m k k in Theorem 4.12 to obtain a simple solution ( X, r ) of the YBE of cardinality m n = p m · · · p m k k . p p be a prime number. Let X = Z / ( p ) × Z / ( p ). We will study the simplesolutions ( X, r ) of the YBE, where r ( x, y ) = ( σ x ( y ) , σ − σ x ( y ) ( x )), for all x, y ∈ X and σ ( i,j ) ( k, l ) = ( σ j ( k ) , d i,σ j ( k ) ( l )) , for all i, j, k, l ∈ Z / ( p ). The main result in this section shows that all the simplesolutions of this form are isomorphic to the simple solutions of cardinality p constructed as in Theorem 4.7, but with weaker conditions on the parameters j , . . . , j p − . Theorem 5.1
Let p be a prime number. Let t ∈ Z / ( p ) be a nonzero element.Let j , . . . , j p − ∈ Z / ( p ) be elements such that j i = j − i and j t s i = t s j i − ( t s − j , for all i ∈ Z / ( p ) and all s ∈ Z . Suppose that j i = j k for some i = k . Let r : ( Z / ( p )) × ( Z / ( p )) −→ ( Z / ( p )) × ( Z / ( p )) be the map defined by r (( i, j ) , ( k, l )) =21 σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) , where σ ( i,j ) ( k, l ) = ( tk + j, t ( l − j tk + j − i )) , for all i, j, k, l ∈ Z / ( p ) . Then (( Z / ( p )) , r ) is a simple solution of the YBE. Proof.
It is easy to see that since p is prime and there exist i = k such that j i = j k , we have that for every nonzero u ∈ Z / ( p ) there exists l ∈ Z / ( p ) suchthat j u + l − j l is invertible. Hence, by the first part of Theorem 4.7, (( Z / ( p )) , r )is an indecomposable and irretractable solution of the YBE.Let f : (( Z / ( p )) , r ) → ( Y, s ) be an epimorphism of solutions. Suppose that f is not an isomorphism and that | Y | >
1. Since (( Z / ( p )) , r ) is indecomposable,( Y, s ) also is indecomposable, and by [7, Lemma 1], | Y | = p and | f − ( y ) | = p ,for all y ∈ Y . We write s ( y, z ) = ( σ y ( z ) , γ z ( y )), for all y, z ∈ Y . It is known that σ y = σ z = σ for all y, z ∈ Y , where σ ∈ Sym Y is a cycle of length p (see [18]).The epimorphism f induces an epimorphism of groups ˜ f : G (( Z / ( p )) , r ) −→G ( Y, s ), such that ˜ f ( σ ( i,j ) ) = σ f ( i,j ) = σ , for all i, j ∈ Z / ( p ). Furthermore, f ( σ ( i,j ) ( k, l )) = σ ( f ( k, l )), for all i, j, k, l ∈ Z / ( p ). Now σ ( k, (0 ,
0) = (0 , − tj k )and σ ( j − k − ,j ) (0 ,
0) = ( j, − tj k +1 ), for all j, k ∈ Z / ( p ). Therefore f (0 , − tj k ) = f ( σ ( k, (0 , σ ( f (0 , f ( σ ( j − k − ,j ) (0 , f ( j, − tj k +1 ) , for all j, k ∈ Z / ( p ). Since there exists k ∈ Z / ( p ) such that j k = j k +1 , wehave that | f − ( f (0 , − tj k )) | > p , a contradiction. Therefore either f is an iso-morphism or | Y | = 1. Hence, (( Z / ( p )) , r ) is a simple solution of the YBE. Remark 5.2
The hypothesis in the statement of Theorem 5.1 implies that themultiplicative order of t is odd. Indeed if the multiplicative order of t is even,then p should be odd and there exists a positive integer s such that t s = − . Inthis case, the condition j t s i = t s j i − ( t s − j , implies that j − i = − j i + 2 j .Since j − i = j i and p is odd, we get that j i = j for all i , in contradiction withthe hypothesis that j i = j k for some i = k . We are now ready for the main result of this section.
Theorem 5.3
Let p be a prime number. Let X = Z / ( p ) × Z / ( p ) . Let ( X, r ) bean indecomposable and irretractable solution of the YBE such that r (( i, j ) , ( k, l )) =( σ i,j ( k, l ) , σ − σ i,j ( k,l ) ( i, j )) , where σ ij ( k, l ) = ( σ j ( k ) , d i,σ j ( k ) ( l )) , (13) for all i, j, k, l ∈ Z / ( p ) . Then ( X, r ) is isomorphic to one of the simple solutionsconstructed in Theorem 5.1. Proof.
The assumptions allow to use Proposition 4.6, in particular σ j ◦ σ d − i,k ( l ) = σ l ◦ σ d − k,i ( j ) , (14) d i,k = d k,i , (15)22 i,w ◦ d σ − j ( k ) ,σ − j ( w ) = d k,w ◦ d σ − l ( i ) ,σ − l ( w ) . (16)If i = k then it follows that d σ − j ( i ) ,σ − j ( w ) = d σ − l ( i ) ,σ − l ( w ) for all i, j, l, w . Thus d u,v = d σ − l σ j ( u ) ,σ − l σ j ( v ) , (17)for all j, l, u, v .By Proposition 4.6 we also know that F = h σ j | j ∈ Z / ( p ) i ⊆ Sym Z / ( p ) and W = h d i,k | i, k ∈ Z / ( p ) i ⊆ Sym Z / ( p ) are transitive. Let F ′ = h σ j σ − l | l, j ∈ Z / ( p ) i ⊆ F . By (14) we know that also σ − j σ l ∈ F ′ , thus F ′ is anormal subgroup of F . If F ′ does not act transitively on Z / ( p ), then by [27,Proposition 4.4], we know that the F ′ -orbits on Z / ( p ) are of the same cardinality.So they are singletons, which means that F ′ acts trivially on Z / ( p ). So F ′ = { id } , in particular σ j = σ k for all j, k ∈ Z/ ( p ), in contradiction with condition ( i )in Proposition 4.6. Thus F ′ acts transitively on Z / ( p ). Let S i be the stabilizerof i ∈ Z / ( p ) in F . Then [ F : S i ] = p and [ F ′ : ( S i ∩ F ′ )] = p . So F ′ contains allcycles of length p in F . Notice that σ l ∈ σ j F ′ , so that F = h σ j i F ′ for every j .Let W ′ = h d i,j d − k,l | i, j, k, l ∈ Z / ( p ) i ⊆ W . By (16) we know that also d − i,j d k,l ∈ W ′ , thus W ′ is a normal subgroup of W . As above one can see that W ′ acts transitively on Z / ( p ), using condition ( ii ) of Proposition 4.6. Noticethat d i,j ∈ d k,l W ′ , so that W = h d i,j i W ′ for every i, j ∈ Z / ( p ).First we shall prove that F and W are abelian-by-cyclic groups.Note that condition (14) can be written in the form σ j σ m = σ d i,k ( m ) σ d − i,k ( j ) for i, k, j, m ∈ Z / ( p ) . So for every i, k ∈ Z / ( p ), we get the permutation solution ( Z / ( p ) , s ) correspond-ing to the permutation d i,k , where s : ( j, m ) ( d i,k ( m ) , d − i,k ( j )). Then we havea natural homomorphism of groups G ( Z / ( p ) , s ) → F such that z σ z , forall z ∈ Z / ( p ). But the structure group G ( Z / ( p ) , s ) embeds into Z p ⋊ h d i,k i (see [18, Propositions 2.3 and 2.4]). So it is abelian-by-cyclic. Hence, so is F = h σ j | j ∈ Z i ⊆ Sym Z / ( p ) . Recall that we know that F is transitive. Now, F has a maximal abelian normal subgroup A such that F/A is cyclic and | F/A | divides the order of d i,k , for all i, k ∈ Z / ( p ).Suppose that A is not transitive on Z / ( p ). From [27, Proposition 4.4], weknow that the A -orbits on Z / ( p ) are of the same cardinality. So A = { id } .Then F is cyclic, so F = { id } by the maximality of A , in contradiction with thetransitivity of F . Therefore A is transitive on Z / ( p ).By [27, Proposition 3.2], we must have A ∼ = Z / ( p ). Since p ∤ | F | , it followsthat p ∤ | F/A | . Hence F = A ⋊ N for a cyclic subgroup N such that the orderof N divides the order of d i,k , for all i, k ∈ Z / ( p ). And | N | is not divisible by p .Recall that p ∤ | F/F ′ | . Hence A ⊆ F ′ . Let A = h a i . By (17), we have that d i,j = d a t ( i ) ,a t ( j ) , and by (16), d i,w ◦ d a t σ − j ( k ) ,a t σ − j ( w ) = d k,w ◦ d a t σ − j ( i ) ,a t σ − j ( w ) , t ∈ Z / ( p ). Since A is transitive, for every w ∈ Z / ( p ) there exists t j ∈ Z / ( p ) such that a t j σ − j ( w ) = w . Hence d i,w ◦ d a tj σ − j ( k ) ,w = d k,w ◦ d a tj σ − j ( i ) ,w , (18)for all i, j, k, w ∈ Z / ( p ). Taking u = a t j σ − j ( k ), we have that d i,w ◦ d u,w = d σ j a − tj ( u ) ,w ◦ d a tj σ − j ( i ) ,w . Hence for every the permutation τ j = σ j a − t j and w , we get a permutationsolution ( Z / ( p ) , s ′ ) corresponding to the permutation τ j , where s ′ : ( i, u ) ( τ j ( u ) , τ − j ( i )). Since d i,j = d a t ( i ) ,a t ( j ) , it follows that W = h d i,w | i ∈ Z / ( p ) i .Then we have the natural homomorphism of groups G ( Z / ( p ) , s ′ ) → W suchthat i d i,w , and it is surjective. But the structure group G ( Z / ( p ) , s ′ ) embedsinto Z p ⋊ h τ j i . So it is abelian-by-cyclic. Hence, so is W . Recall that we knowthat W is transitive. Now, W has a maximal abelian normal subgroup A ′ suchthat W/A ′ is cyclic and | W/A ′ | divides the order of τ j , for all j ∈ Z / ( p ).Since A is a normal subgroup of F , τ mj ∈ σ mj A , for all positive integers m .Hence σ mj / ∈ A if and only if τ mj / ∈ A .Note that if σ j / ∈ A , then the order of σ j is not divisible by p , becausein Sym Z / ( p ) an element has order divisible by p if and only if it has order p ,and A is the Sylow p -subgroup of F . Thus in this case, A ∩ h σ j i = { id } and A ∩ h τ j i = { id } . Since τ mj ∈ σ mj A , for all positive integers m , we have that theorder of τ j is equal to the order of σ j , in this case.On the other hand, if σ j ∈ A , then since a t j σ − j ( w ) = w and A is a transitivesubgroup of order p of Sym Z / ( p ) , we have that a t j = σ j . Thus, by (18), we havethat d i,w ◦ d k,w = d k,w ◦ d i,w , for all i, k, w ∈ Z / ( p ). Since W = h d i,w | i ∈ Z / ( p ) i , we have that W is abelianand thus W = A ′ in this case.By the maximality of A ′ and the transitivity of W , one can see that A ′ istransitive and then A ′ ∼ = Z / ( p ). Hence A ′ is the Sylow p -subgroup of W , andit follows that p ∤ | W/A ′ | . Hence W = A ′ ⋊ N ′ for a cyclic subgroup N ′ suchthat the order of N ′ divides the order of τ j , for all j ∈ Z / ( p ). And | N ′ | is notdivisible by p .Now we shall see that F = A if and only if W = A ′ .Suppose first that F = A ∼ = Z / ( p ). Thus σ j ∈ A and we have seen abovethat, in this case, W = A ′ ∼ = Z / ( p ).Now, suppose that W = A ′ ∼ = Z / ( p ). Since the order of N divides theorder of d i,k and | N | is not divisible by p , it follows that | N | = 1 and therefore F = A ∼ = Z / ( p ).We shall study two cases. 24 ase 1. Suppose that F = A ∼ = Z / ( p ) ∼ = A ′ = W .In this case, F = h σ t i for some t ∈ Z / ( p ). Hence there exists a permutation η ∈ Sym Z / ( p ) such that σ t ( η ( l )) = η ( l + 1) for all l ∈ Z / ( p ) . (19)Also there exists a permutation ν ∈ Sym Z / ( p ) such that σ ν ( l ) = σ lt = σ lν (1) , forall l ∈ Z / ( p ). Then condition (14) implies that σ j − lν (1) = σ − ν ( l ) ◦ σ ν ( j ) = σ ν ( ν − ( d − k,i ( ν ( j )))) ◦ σ − ν ( ν − ( d − i,k ( ν ( l )))) = σ ν − ( d − k,i ( ν ( j ))) − ν − ( d − k,i ( ν ( l ))) ν (1) . Hence ν − d − i,k ν ( j ) − ν − d − i,k ν ( l ) = j − l, (20)for all i, k, j, l ∈ Z / ( p ). In particular, ν − d − i,k ν ( j ) = j + ν − d − i,k ν (0) for all i, k, j ∈ Z / ( p ) . (21)Note that σ ν (0) = id. Hence condition (17) implies, in view of (19), that d η ( l ) ,η ( t ) = d σ ν ( j ) ( η ( l )) ,σ ν ( j ) ( η ( t )) = d η ( l + j ) ,η ( t + j ) , (22)for all l, t, j ∈ Z / ( p ). In particular, for j = − t we have that d η ( l ) ,η ( t ) = d η ( l − t ) ,η (0) , (23)for all l, t ∈ Z / ( p ). We define j i = − ν − d − η ( i ) ,η (0) ν (0)for all i ∈ Z / ( p ). Note that by (23), j i = − ν − d − η ( i ) ,η (0) ν (0) = − ν − d − η (0) ,η ( i ) ν (0) = − ν − d − η ( − i ) ,η (0) ν (0) = j − i , for all i ∈ Z / ( p ). Since W ∼ = Z / ( p ), d η ( i ) ,η (0) = d η ( k ) ,η (0) if and only if d − η ( i ) ,η (0) ν (0) = d − η ( k ) ,η (0) ν (0) , and this is equivalent to j i = j k . By condition (ii) in Proposition 4.6, if i = i ′ ,there exists k ∈ Z / ( p ) such that d η ( i ) ,η ( k ) = d η ( i ′ ) ,η ( k ) . Then, by condition (23), d η ( i − k ) ,η (0) = d η ( i ′ − k ) ,η (0) , and therefore j i − k = j i ′ − k . Hence the elements j i satisfy the hypothesis of Theorem 5.1 for t = 1. Consider the correspondingsimple solution ( X, r ′ ), thus r ′ (( i, j ) , ( k, l )) = ( σ ′ ( i,j ) ( k, l ) , (cid:16) σ ′ σ ′ ( i,j ) ( k,l ) (cid:17) − ( i, j )) , σ ′ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ) , for all i, j, k, l ∈ Z / ( p ). We shall prove that ( X, r ) and (
X, r ′ ) are isomorphic.Let f : X → X be the map defined by f ( i, j ) = ( η ( i ) , ν ( j )) for all i, j ∈ Z / ( p ). Note that f ( σ ′ i,j ( k, l )) = f ( k + j, l − j k + j − i )= ( η ( k + j ) , ν ( l − j k + j − i ))= ( η ( k + j ) , ν ( l + ν − d η ( k + j − i ) ,η (0) ν (0)))= ( η ( k + j ) , ν ( ν − d η ( k + j − i ) ,η (0) ν ( l ))) by (21)= ( η ( k + j ) , d η ( i ) ,η ( k + j ) ν ( l )) by (15) and (22)= ( σ ν ( j ) ( η ( k )) , d η ( i ) ,σ ν ( j ) η ( k ) ( ν ( l ))) by (19)= σ ( η ( i ) ,ν ( j )) ( η ( k ) , ν ( l ))= σ f ( i,j ) f ( k, l ) . Hence f is an isomorphism from ( X, r ′ ) to ( X, r ). Case 2.
Suppose that F = A and W = A ′ .Recall that in this case F = A ⋊ N , where N is a cyclic group such that | N | > d i,k , for all i, k , and p is not a divisor of | N | .Also W = A ′ ⋊ N ′ , where N ′ is a cyclic group such that | N ′ | > τ j = σ j a − t j , for every j ∈ Z / ( p ), where A = h a i . In particular, τ j / ∈ A and we have seen that in this case the order of σ j is equal to the orderof τ j , for all j ∈ Z / ( p ), and p is not a divisor of | N ′ | . Hence- every d i,k = id and has order not divisible by p ,- also every σ j = id and has order prime to p .In this case, applying condition (14) twice, σ j ◦ σ l = σ d i,k ( l ) ◦ σ d − i,k ( j ) = σ d u,v d − i,k ( j ) ◦ σ d − u,v d i,k ( l ) , (24)for all j, l ∈ Z and i, k, u, v ∈ Z / ( p ). Since W ′ acts transitively on Z / ( p ), forevery j, l ∈ Z / ( p ) there exists j ′ ∈ Z / ( p ) such that σ j ◦ σ l = σ l ◦ σ j ′ . In particular A acts by conjugation on the set { σ j | j ∈ Z / ( p ) } . Recall that A = h a i ∼ = Z / ( p ). Let j ∈ Z / ( p ). We shall see that { a t σ j a − t | t ∈ Z / ( p ) } = { σ j | j ∈ Z / ( p ) } . σ j = a t σ j a − t for some t = 0, then h a, σ j i is abelian and transitiveon Z / ( p ), and therefore σ j ∈ A , a contradiction because σ j = id and has orderprime to p .Thus indeed { a t σ j a − t | t ∈ Z / ( p ) } = { σ j | j ∈ Z / ( p ) } . Therefore σ − j σ l = a t σ − j a − t a t σ j a − t for some t , t ∈ Z / ( p ). Since A is normal in F , we havethat F ′ = h σ − j σ l | j, l ∈ Z i = A .Since F = F ′ h σ j i = A ⋊ N , for all j ∈ Z / ( p ), and F ′ = A ∼ = Z / ( p ), we havethat the order of σ j is equal to the order of N , for all j ∈ Z / ( p ), because in thiscase the order of σ j is relatively prime to p .Recall that W = A ′ ⋊ N ′ , with A ′ ∼ = Z / ( p ) and | N ′ | > τ j which is equal to the order of σ j , and p ∤ | N ′ | . Since the order of σ j is equal to de order of N , we have that | N ′ | divides | N | . Recall that | N | dividesthe order of any d i,k . Since the order of d i,k is not divisible by p , clearly theorder of d i,k divides | N ′ | . Therefore | N ′ | = | N | . Let b ∈ A ′ be an element oforder p . Thus A ′ = h b i .Since A ∼ = A ′ ∼ = Z / ( p ), and N and N ′ are isomorphic to subgroups ofAut( Z / ( p )) of the same order and Aut( Z / ( p )) is cyclic, we have that F ∼ = W ∼ = A ⋊ N .Consider the map ϕ : Z / ( p ) ⋊ Aut( Z / ( p )) −→ Sym Z / ( p ) defined by ϕ ( j, g )( l ) = j + g ( l ) , for all l ∈ Z / ( p ). It is clear the ϕ is a monomorphism of groups. Since F is atransitive subgroup of Sym Z / ( p ) , it is primitive. By [32, Theorem 2.1.6], thereexist a subgroup H of Aut( Z / ( p )) and a permutation η ∈ Sym Z / ( p ) such that ϕ ( Z / ( p ) ⋊ H ) = η − F η . Since Aut( Z / ( p )) is cyclic, there exists h ∈ H such that H = h h i . Note that ϕ (1 , id) has order p , thus ηϕ (1 , id) η − is an element of order p in F . Hence we may assume that a = ηϕ (1 , id) η − . For every j ∈ Z / ( p ), since σ j / ∈ A , there exist integers m j , k j such that η − σ j η = ϕ ( m j , h k j ) and h k j = id.Since, for every l ∈ Z / ( p ), { a k σ l a − k | k ∈ Z / ( p ) } = { σ j | j ∈ Z / ( p ) } , we havethat for every j ∈ Z / ( p ) there exists m ∈ Z / ( p ) such that σ j = a m σ a − m .Hence ϕ ( m j , h k j ) = η − σ j η = η − a m σ a − m η = ϕ ((1 , id) m ( m , h k )(1 , id) − m ) = ϕ (( m + m , h k )( − m, id))= ϕ ( m + m − h k ( m ) , h k ) . It follows that h k j = h k , for all j ∈ Z / ( p ). Hence { η − σ j η | j ∈ Z / ( p ) } = { ϕ ( l, h k ) | l ∈ Z / ( p ) } . Since F = h σ j | j ∈ Z / ( p ) i we have ϕ ( Z / ( p ) ⋊ h h i ) = η − F η = h η − σ j η | j ∈ Z / ( p ) i = ϕ ( h ( l, h k ) | l ∈ Z / ( p ) i )= ϕ ( Z / ( p ) ⋊ h h k i ) , h h i = h h k i . Thus we may assume that h = h k . So there exists j ∈ Z / ( p ) such that η − σ j η = ϕ (0 , h ). Since h ∈ Aut( Z / ( p )) \ { id } , there exists t ∈ Z / ( p ) \ { , } such that h ( k ) = tk for all k ∈ Z / ( p ).Now there exists a permutation ν such that σ ν ( k ) = a (1 − t ) − k σ j a − (1 − t ) − k .Since η − aη = ϕ (1 , id), it follows that η − a k η = ϕ ( k, id). Therefore η − σ ν ( k ) η = η − a (1 − t ) − k σ j a − (1 − t ) − k η = ϕ (((1 − t ) − k, id)(0 , h )( − (1 − t ) − k, id))= ϕ ((1 − t ) − k − h ((1 − t ) − k ) , h )= ϕ ((1 − t ) − k − t (1 − t ) − k, h )= ϕ ( k, h ) . Hence η − σ ν ( k ) η ( l ) = ϕ ( k, h )( l ) = k + h ( l ) = k + tl , and thus σ ν ( k ) ( η ( l )) = η ( tl + k ) . (25)By (14), σ u σ v ( l ) = σ d k, ( v ) σ d − k, ( u ) ( l ) , for all u, v, k, l ∈ Z / ( p ). In particular, σ ν ( u ) σ ν ( v ) ( η ( l )) = σ d k, ( ν ( v )) σ d − k, ( ν ( u )) ( η ( l )) . Hence, applying this and condition (25) several times, we get η ( t ( tl + v ) + u ) = σ ν ( u ) ( η ( tl + v ))= σ ν ( u ) ( σ ν ( v ) ( η ( l )))= σ d k, ( ν ( v )) σ d − k, ( ν ( u )) ( η ( l ))= σ d k, ( ν ( v )) ( η ( tl + ν − ( d − k, ( ν ( u )))))= η ( ν − ( d k, ( ν ( v ))) + t ( tl + ν − ( d − k, ( ν ( u ))))) , and thus tv + u = ν − ( d k, ( ν ( v ))) + tν − ( d − k, ( ν ( u ))) , for all u, v, k ∈ Z / ( p ). In particular, for u = 0 we get ν − ( d k, ( ν ( v ))) = t ( v − ν − ( d − k, ( ν (0)))) . Let j k = ν − ( d − η ( k ) ,η (0) ( ν (0))). Hence d η ( k ) ,η (0) ( ν ( v )) = ν ( t ( v − j k )) . (26)As η − a k η ( u ) = ϕ ( k, id)( u ) = u + k , by (17) we have d η ( u ) ,η ( v ) = d a k ( η ( u )) ,a k ( η ( v )) = d η ( u + k ) ,η ( v + k ) , (27)28or all k, u, v ∈ Z / ( p ). Hence j k = ν − ( d − η ( k ) ,η (0) ( ν (0))) = ν − ( d − η (0) ,η ( − k ) ( ν (0))) = j − k , where the last equality follows by (15).By (16), d η ( i ) ,η (0) d σ − ν ( j ) ( η ( k )) ,σ − ν ( j ) ( η (0)) ( ν ( l )) = d η ( k ) ,η (0) d σ − ν ( j ′ ) ( η ( i )) ,σ − ν ( j ′ ) ( η (0)) ( ν ( l )) , for all i, j, j ′ , k, l ∈ Z / ( p ). Now, using (25), we get d η ( i ) ,η (0) d σ − ν ( j ) ( η ( k )) ,σ − ν ( j ) ( η (0)) ( ν ( l )) = d η ( i ) ,η (0) d η ( t − ( k − j )) ,η ( − t − j ) ( ν ( l )) by (27)= d η ( i ) ,η (0) d η ( t − k ) ,η (0) ( ν ( l )) by (26)= d η ( i ) ,η (0) ( ν ( t ( l − j t − k ))) by (26)= ν ( t ( t ( l − j t − k ) − j i )) and similarly d η ( k ) ,η (0) d σ − ν ( j ′ ) ( η ( i )) ,σ − ν ( j ′ ) ( η (0)) ( ν ( l )) = ν ( t ( t ( l − j t − i ) − j k )) . Hence tj t − i + j k = tj t − k + j i . In particular, for i = 0 and t − k = u , we getthat tj + j tu = tj u + j . Thus j tu = tj u − ( t − j . By induction on s one can see that j t s u = t s j u − ( t s − j . Finally, suppose that i is a nonzero element of Z / ( p ). Suppose that j i + k = j k for all k ∈ Z / ( p ). Then, using (27) and the definition of j k we get d η ( i ) ,η ( − k ) = d η ( i + k ) ,η (0) = d η ( k ) ,η (0) = d η (0) ,η ( − k ) , for all k ∈ Z / ( p ). But this is not possible because ( X, r ) is irretractable andshould satisfy condition (ii) of Proposition 4.6. Therefore, for every nonzeroelement i ∈ Z / ( p ), there exists k ∈ Z / ( p ) such that j i + k = j k . Hence t andthe j i satisfy the hypothesis of Theorem 5.1. Consider the corresponding simplesolution ( X, r ′ ), with r (( i, j ) , ( k, l )) = ( σ ′ ( i,j ) ( k, l ) , ( σ ′ σ ′ ( i,j ) ( k,l ) ) − ( i, j )), where σ ′ ( i,j ) ( k, l ) = ( tk + j, t ( l − j tk + j − i )) , for all i, j, k, l ∈ Z / ( p ). We shall prove that the solutions ( X, r ) and (
X, r ′ ) areisomorphic. Let f : X −→ X be the map defined by f ( i, j ) = ( η ( i ) , ν ( j )) for all i, j ∈ Z / ( p ). Note that f ( σ ′ ( i,j ) ( k, l )) = f ( tk + j, t ( l − j k + j − i ))= ( η ( tk + j ) , ν ( t ( l − j tk + j − i ))) by (26)= ( η ( tk + j ) , ν ( ν − d η ( tk + j − i ) ,η (0) ν ( l ))) by (27) and (15)= ( η ( tk + j ) , d η ( i ) ,η ( tk + j ) ν ( l )) by (25)= ( σ ν ( j ) ( η ( k )) , d η ( i ) ,σ ν ( j ) η ( k ) ( ν ( l )))= σ ( η ( i ) ,ν ( j )) ( η ( k ) , ν ( l ))= σ f ( i,j ) f ( k, l ) . f is an isomorphism from ( X, r ′ ) to ( X, r ). Therefore the result follows.
Remark 5.4
Clearly, if t = 1 in Theorem 5.1, then one gets many examples asthe hypothesis on the relations between j i are satisfied trivially. The followingexample shows that there are simple solutions that satisfy the hypothesis of The-orem 5.1 with t = 1 . Indeed, for p = 7 , we have that { j | j ∈ Z } = { , , } .If we take σ j ( l ) = 2 l + j and d , ( l ) = 2 l , d , ( l ) = 2( l − , d , ( l ) = 2( l − , d , ( l ) = 2( l − and d k,k = d , , d k +1 ,k = d k,k +1 = d , ,d k +2 ,k = d k,k +2 = d , and d k +4 ,k = d k,k +4 = d , , for all k ∈ Z / (7) , one can check that ( Z / (7) × Z / (7) , r ) , where r (( i, j ) , ( k, l )) =( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ( i, j )) and σ ( i,j ) ( k, l ) = ( σ j ( k ) , d i,σ j ( k ) ( l )) , satisfies the hy-pothesis of Theorem 5.1, and thus it is a simple solution of the YBE. n . In this section we shall see an alternative construction of some of the simplesolutions constructed in Theorems 5.1 and 4.9.Consider the standard basis B = ( e , . . . , e n ), e = (1 , , . . . , , e = (0 , , , . . . , , . . . , e n = (0 , . . . , , Z / ( n )-module ( Z / ( n )) n . The indices are viewed as elements of Z / ( n ).Consider the wreath product Z / ( n ) ≀ S = ( Z / ( n )) n ⋊ α Z / ( n ) , where α ( i )( e j ) = e i + j , for all i, j ∈ Z / ( n ). For j , j , . . . , j n − ∈ Z / ( n ), such that j i = j − i for all i ∈ Z / ( n ) (interpreting the indices as elements of Z / ( n )), consider the bilinearform b j ,...,j n − : ( Z / ( n )) n × ( Z / ( n )) n → Z / ( n ) , with matrix with respect to B M ( b j ,...,j n − , B ) = j j . . . . . . j n − j n − j . . . ...... . . . . . . . . . ...... . . . . . . j j j . . . j n − j . α ( i ) is in the orthogonal group of (( Z / ( n )) n , b j ,...,j n − ) for all i ∈ Z / ( n ). Consider the asymmetric product ([8]) of the trivial braces ( Z / ( n )) n and Z / ( n ) defined via α and b j ,...,j n − : B j ,...,j n − = ( Z / ( n )) n ⋊ ◦ Z / ( n ) . Hence, ( B j ,...,j n − , ◦ ) ∼ = ( Z / ( n )) n ⋊ α Z / ( n ) and the addition is defined by( u, i ) + ( v, j ) = ( u + v, i + j + b j ,...,j n − ( u, v )) , for all u, v ∈ ( Z / ( n )) n and i, j ∈ Z / ( n ). We know that ( B j ,...,j n − , + , ◦ ) is aleft brace. We denote the lambda map of B j ,...,j n − by λ ( j ,...,j n − ) . We have λ ( j ,...,j n − )( u,i ) ( v, j ) = ( u, i )( v, j ) − ( u, i )= ( u + α ( i )( v ) , i + j ) − ( u, i )= ( α ( i )( v ) , j − b j ,...,j n − ( u, α ( i )( v ))) . (28)Note that λ ( j ,...,j n − )( u,i ) is determined by the action on X = { ( e i , j ) | i, j ∈ Z / ( n ) } .Put x ij = ( e i , j ) . Note that λ ( j ,...,j n − ) x ij ( x kl ) = λ ( e i ,j ) ( e k , l ) = ( e k + j , l − a k + j − i ) = x k + j,l − j k + j − i . (29)Let r j ,...,j n − : X × X → X × X be the map defined by r j ,...,j n − ( x ij , x kl ) = ( λ ( j ,...,j n − ) x ij ( x kl ) , ( λ ( j ,...,j n − ) λ ( j ,...,jn − xij ( x kl ) ) − ( x ij )) . Note that r j ,...,j n − is the restriction to X of the solution associated to the leftbrace B j ,...,j n − (see [12, Lemma 2]). Thus ( X, r j ,...,j n − ) is a solution of theYBE.Consider the map f : ( Z ( n )) → X defined by f ( i, j ) = x ij , for all i, j ∈ Z / ( n ). Then, by (29), f is an isomorphism of the solution (( Z ( n )) , r ) to( X, r j ,...,j n − ), where r is defined as in Theorem 4.9, that is r ( i,j ) ( k, l ) =( σ ( i,j ) ( k, l ) , σ − σ ( i,j ) ( k,l ) ), where σ ( i,j ) ( k, l ) = ( k + j, l − j k + j − i ), for all i, j, k, l ∈ Z / ( n ). Thus the following result is an easy consequence. Proposition 6.1
The solution ( X, r j ,...,j n − ) is indecomposable if and only if Z / ( n ) = h j , . . . , j n − i .The solution ( X, r j ,...,j n − ) is irretractable if and only if the n rows of M ( b j ,...,j n − , B ) are distinct (i.e. if j i + k = j i ′ + k for every k then i = i ′ ). roposition 6.2 The left braces B j ,...,j n − and G ( X, r j ,...,j n − ) are isomor-phic if and only if b j ,...,j n − is non-singular. Proof.
First we will show that Soc( B j ,...,j n − ) = { (0 , } if and only if b j ,...,j n − is non-singular. Let ( u, i ) ∈ Soc( B j ,...,j n − ). By (28) and the definition of α , we have that i = 0 and b j ,...,j n − ( u, v ) = 0, for all v ∈ ( Z / ( n )) n . ThusSoc( B j ,...,j n − ) = { (0 , } if and only if b j ,...,j n − is non-singular.Let ( B j ,...,j n − , r B j ,...,jn − ) be the solution associated to the left brace B j ,...,j n − ,that is, r B j ,...,jn − : B j ,...,j n − × B j ,...,j n − −→ B j ,...,j n − × B j ,...,j n − is definedby r B j ,...,jn − (( u, i ) , ( v, j )) = ( λ ( j ,...,j n − )( u,i ) ( v, j ) , ( λ ( j ,...,j n − ) λ ( j ,...,jn − u,i ) ( v,j ) ) − ( u, i )) , for all u, v ∈ ( Z / ( n )) n and i, j ∈ Z / ( n ). By Lemma 2.2, B j ,...,j n − / Soc( B j ,...,j n − ) ∼ = G ( B j ,...,j n − , r B j ,...,jn − )as left braces. Note that the map ϕ : G ( B j ,...,j n − , r B j ,...,jn − ) −→ G ( X, r j ,...,j n − )defined by ϕ ( λ ( j ,...,j n − )( v,j ) ) = λ ( j ,...,j n − )( v,j ) | X , for all ( v, j ) ∈ B j ,...,j n − , is anisomorphism of the multiplicative groups.Furthermore, the inclusion map X −→ B j ,...,j n − is a homomorphism ofsolutions from ( X, r j ,...,j n − ) to ( B j ,...,j n − , r B j ,...,jn − ). Hence it induces ahomomorphism of left braces G ( X, r j ,...,j n − ) −→ G ( B j ,...,j n − , r B j ,...,jn − ) , which is exactly ϕ − . Therefore the result follows. Proposition 6.3
Let a , . . . , a n − , c , . . . , c n − ∈ Z / ( n ) be elements such that a i = a − i and c i = c − i for all i ∈ Z / ( n ) . If there exists an invertible element a ∈ Z / ( n ) such that aa i = c ai for all i ∈ Z / ( n ) , then the solutions ( X, r a ,...,a n − ) and ( X, r c ,...,c n − ) are isomorphic. Furthermore, if b a ,...,a n − and b c ,...,c n − are non-singular, thenthe converse holds. Proof.
Suppose that there exists an invertible element a ∈ Z / ( n ) such that aa i = c ai , for all i ∈ Z / ( n ). We define h : X −→ X by h ( x ij ) = x ai,aj , for all i, j ∈ Z / ( n ). We have h ( λ ( a ,...,a n − ) x i,j ( x k,l )) = h ( x k + j,l − a k + j − i ) = x a ( k + j ) ,a ( l − a k + j − i ) λ ( c ,...,c n − ) h ( x i,j ) ( h ( x k,l )) = λ ( c ,...,c n − ) x ai,aj ( x ak,al )= x ak + aj,al − c ak + aj − ai = x a ( k + j ) ,a ( l − a k + j − i ) , for all i, j, k, l ∈ Z / ( n ). Therefore h is an isomorphism of solutions from( X, r a ,...,a n − ) to ( X, r c ,...,c n − ).Suppose that b a ,...,a n − and b c ,...,c n − are non-singular, and ( X, r a ,...,a n − )and ( X, r c ,...,c n − ) are isomorphic. Let g : X −→ X be an isomorphism ofsolutions from ( X, r a ,...,a n − ) to ( X, r c ,...,c n − ). Then g induces an isomorphismof left braces ˜ g : G ( X, r a ,...,a n − ) −→ G ( X, r c ,...,c n − ) , such that ˜ g ( λ ( a ,...,a n − ) x i,j ) = λ ( c ,...,c n − ) g ( x i,j ) , for all x i,j ∈ X . Since b a ,...,a n − and b c ,...,c n − are non-singular, by Proposition 6.2, and its proof g induces anisomorphism of left braces from B a ,...,a n − to B c ,...,c n − that we also denote by g . Note the multiplicative groups of B a ,...,a n − and B c ,...,c n − are the same,that is ( Z / ( n )) n ⋊ α Z / ( n ). In particular, g is an automorphism of this group.In this group we have( u, k ) x i, ( u, k ) − = ( u, k )( e i , − α ( − k )( u ) , − k ) = ( e i + k ,
0) = x i + k, , for all u ∈ ( Z / ( n )) n and i, k ∈ Z / ( n ). Hence { x i, | i ∈ Z / ( n ) } is a conjugacyclass contained in X . We shall see that if n >
2, then this is the only conjugacyclass of ( Z / ( n )) n ⋊ α Z / ( n ) contained in X . So, assume that n >
2. Let j ∈ Z / ( n )be a nonzero element. We have x i + j,j x i,j x − i + j,j = ( e i + j , j )( e i , j )( − e i , − j )= (2 e i + j , j )( − e i , − j )= (2 e i + j − e i +2 j , j ) / ∈ X, for all i ∈ Z / ( n ), because 2 = 0 in Z / ( n ). Hence { x i, | i ∈ Z / ( n ) } indeed is theonly conjugacy class of ( Z / ( n )) n ⋊ α Z / ( n ) contained in X . Since g ( X ) = X and g ( C ) is a conjugacy class of ( Z / ( n )) n ⋊ α Z / ( n ), for every conjugacy class C , wehave that g ( { x i, | i ∈ Z / ( n ) } ) ⊆ X is a conjugacy class of ( Z / ( n )) n ⋊ α Z / ( n ).Thus, we have that g ( { x i, | i ∈ Z / ( n ) } ) = { x i, | i ∈ Z / ( n ) } . Hence there exists ν ∈ Sym Z / ( n ) such that g ( x i, ) = x ν ( i ) , , for all i ∈ Z / ( n ).Let g (0 ,
1) = ( u, a ), for some u ∈ ( Z / ( n )) n and a ∈ Z / ( n ). Let g ( x , ) = x k,l .Since x − , x , = ( − e , e ,
1) = (0 ,
1) in B a ,...,a n − , we have that ( u, a ) = g ( x , ) − g ( x , ) = x − ν (1) , x k,l = ( e k − e ν (1) , l ) in B c ,...,c n − . Hence u = e k − e ν (1) and a = l . Note that g ( x , ) = g (( e ,
0) + (0 , g ( e ,
0) + g (0 , g ( x , ) + g (0 ,
1) = x ν (2) , + ( e k − e ν (1) , a )= ( e ν (2) + e k − e ν (1) , a + b c ,...,c n − ( e ν (2) , e k − e ν (1) )) ∈ X. k = ν (1), and thus g (0 ,
1) = (0 , a ) and g ( x i,j ) = g (( e i ,
0) + j (0 , g ( e i ,
0) + jg (0 ,
1) = g ( x i, ) + j (0 , a )= x ν ( i ) , + (0 , ja ) = x ν ( i ) ,ja , for all i, j ∈ Z / ( n ). Since the additive order of (0 ,
1) in B a ,...,a n − is n , theadditive order of (0 , a ) = g (0 ,
1) in B c ,...,c n − is n . Thus a ∈ Z / ( n ) is invertible.Now, by (29), g ( λ ( a ,...,a n − ) x i,j ( x k,l )) = g ( x k + j,l − a k + j − i ) = x ν ( k + j ) ,a ( l − a k + j − i ) and λ ( c ,...,c n − ) g ( x i,j ) ( g ( x k,l )) = λ ( c ,...,c n − ) x ν ( i ) ,aj ( x ν ( k ) ,al ) = x ν ( k )+ aj,al − c ν ( k )+ aj − ν ( i ) . Since g ( λ ( a ,...,a n − ) x i,j ( x k,l )) = λ ( c ,...,c n − ) g ( x i,j ) ( g ( x k,l )), we have that ν ( k + j ) = ν ( k )+ aj , for all k, j ∈ Z / ( n ) and aa k + j − i = c ν ( k )+ aj − ν ( i ) = c ν ( i + k − i )+ aj − ν ( i ) = c a ( k − i )+ aj , as desired.Finally, for n = 2, there are only two solutions ( X, r , ) and ( X, r , ) with b , and b , non-singular. The solution ( X, r , ) is isomorphic to the solution ofExample 3.5, and the solution ( X, r , ) is isomorphic to the solution of Example3.6. Hence ( X, r , ) and ( X, r , ) are not isomorphic.Thus the result follows. To conclude, we propose some open questions related to the results obtained inthe previous sections.In Sections 4, 5 and 6 we study finite indecomposable and irretractable solu-tions (
X, r ) of the YBE, where X = Y × Z , the X y = { ( y, z ) | z ∈ Z } , for y ∈ Y ,are blocks of imprimitivity for the action of G ( X, r ) on X , and r (( y, z ) , ( y ′ , z ′ )) =( σ ( y,z ) ( y ′ , z ′ ) , σ − σ ( y,z ) ( y ′ ,z ′ ) ( y, z )), where σ ( y,z ) ( y ′ , z ′ ) = ( σ z ( y ′ ) , d y,σ z ( y ′ ) ( z ′ )), forsome permutations σ z ∈ Sym Y and d y,y ′ ∈ Sym Z , for all y, y ′ ∈ Y and z, z ′ ∈ Z . Question 7.1
Let
Y, Z be finite sets of cardinality > . Let X = Y × Z .Study the indecomposable and irretractable solutions ( X, r ) , where the sets X y = { ( y, z ) | z ∈ Z } , for y ∈ Y , are blocks of imprimitivity for the action of G ( X, r ) on X , and r (( y, z ) , ( y ′ , z ′ )) = ( σ ( y,z ) ( y ′ , z ′ ) , σ − σ ( y,z ) ( y ′ ,z ′ ) ( y, z )) , where σ ( y,z ) ( y ′ , z ′ ) = ( β y,z ( y ′ ) , α y,z,y ′ ( z ′ )) , for some permutations β y,z ∈ Sym Y and α y,z,y ′ ∈ Sym Z , for all y, y ′ ∈ Y and z, z ′ ∈ Z . uestion 7.2 With the same notation as in Question 7.1, assume that | Y | = | Z | = p is prime. Suppose that ( X, r ) is an indecomposable and irretractablesolution of the YBE. Is it true that β y,z = β y ′ ,z , for all y, y ′ ∈ Y and all z ∈ Z ?Is it true that α y,z,y ′ = α y,z ′ ,y ′′ if and only if β y,z ( y ′ ) = β y,z ′ ( y ′′ ) ? Question 7.3
Let p be a prime. Let ( X, r ) be an indecomposable and irre-tractable solution of the YBE of cardinality p . Is ( X, r ) simple? Is it isomor-phic to one of the simple solutions constructed in Theorem 5.1? Question 7.4
Consider the indecomposable and irretractable solutions constructedin Theorem 4.9. Can they be constructed using the asymmetric product of leftbraces?
Note that for t = 1 we give such a construction in Section 6. Question 7.5
Does there exist a simple solution ( X, r ) of the YBE such that | X | = p p · · · p n , for n > distinct primes p , p , . . . , p n ? Note that if p , p , . . . , p n are n distinct primes, k , . . . , k n are positive integersand n >
1, then by Remark 4.13, there exist simple solutions of the YBE ofcardinality p k p k · · · p k n n if P ni =1 k i > n .Note also that the finite simple solutions ( X, r ) of the YBE, with X = Y × Z ,constructed in Sections 4, 5 and 6 satisfy that 1 < | Y | , | Z | and | Z | is a divisorof | Y | . Thus it seems natural to ask the following question. Question 7.6
Does there exist a finite simple solution ( X, r ) of the YBE suchthat X = Y × Z , the sets X y = { ( y, z ) | z ∈ Z } , for y ∈ Y , are blocks ofimprimitivity for the action of G ( X, r ) on X , and | Z | is not a divisor of | Y | ? Acknowledgements
The authors thank Eric Jespers for his comments and suggestions, and Lean-dro Vendramin for the information about the indecomposable and irretractablesolutions of cardinality 9.