Left non-degenerate set-theoretic solutions of the Yang-Baxter equation and dynamical extensions of q-cycle sets
aa r X i v : . [ m a t h . QA ] J a n Left non-degenerate set-theoretic solutions of theYang-Baxter equation and dynamical extensions ofq-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
A first aim of this paper is to give sufficient conditions on left non-degeneratebijective set-theoretic solutions of the Yang-Baxter equation so that they arenon-degenerate. In particular, we extend the results on involutive solutionsobtained by Rump in [36] and answer in a positive way to a question posed byCed´o, Jespers, and Verwimp [18, Question 4.2]. Moreover, we develop a theoryof extensions for left non-degenerate set-theoretic solutions of the Yang-Baxterequation that allows one to construct new families of set-theoretic solutions.
Keywords: q-cycle set, cycle set, set-theoretic solution,Yang-Baxter equation, brace, skew brace
1. Introduction A set-theoretic solution of the Yang-Baxter equation , or shortly a solution , isa pair ( X, r ) where X is a non-empty set and r is a map from X × X into itselfsuch that r r r = r r r , where r := r × id X and r := id X × r . Let λ x : X → X and ρ y : X → X bemaps such that r ( x, y ) = ( λ x ( y ) , ρ y ( x )) ✩ This work was partially supported by the Dipartimento di Matematica e Fisica “EnnioDe Giorgi” - Universit 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)
January 30, 2020 or all x, y ∈ X . A set-theoretic solution of the Yang-Baxter equation ( X, r ) issaid to be a left [ right ] non-degenerate if λ x ∈ Sym( X ) [ ρ x ∈ Sym( X ) ], forevery x ∈ X , and non-degenerate if it is left and right non-degenerate.Drinfeld’s paper [21] stimulated much interest in this subject. In recent years,after the seminal papers by Gateva-Ivanova and Van den Bergh [25] and Etingof,Schedler and Soloviev [22] the involutive solutions r , i.e., r = id X × X , havereceived a lot of attention.To study involutive solutions, many theory involving a lot of algebraic struc-tures has developed. Several examples of involutive solutions have obtained byusing groups, racks, and quandles [23, 24, 15, 2, 17]. In 2005, Rump introduced cycle sets , non-associative algebraic structures that allow one to find involutiveleft non-degenerate solutions. A set X endowed of an operation · is said to be acycle set if the left multiplication σ x : X −→ X, y x · y is invertible, for every x ∈ X , and the relation ( x · y ) · ( x · z ) = ( y · x ) · ( y · z )is satisfied, for all x, y, z ∈ X . By cycle sets several families of involutive so-lutions were determined and several interesting results were obtained (see, forexample, [20, 6, 8, 10, 42, 9, 5]).In 2000, Lu, Yan, and Zhu [34] and Soloviev [41] started the study of non-degenerate solutions that are not necessarily involutive. To obtain new familiesof bijective non-degenerate solutions, in 2015 Guarnieri and Vendramin [26] in-troduced the algebraic structure of skew brace , a generalisation of the bracesintroduced by Rump in [37]. Some works where this structure is studied are[11, 13, 16, 40, 19, 29], just to name a few. As skew braces are the analogueversion of braces for non-involutive non-degenerate solutions, q-cycle sets , in-troduced recently by Rump [39], are the analogue version of cycle sets for leftnon-degenerate solutions that are not necessarily involutive. Recall that a set X with two binary operations · and : is a q-cycle set if σ x : X −→ X, y x · y is invertible, for every x ∈ X , and( x · y ) · ( x · z ) = ( y : x ) · ( y · z )( x : y ) : ( x : z ) = ( y · x ) : ( y : z )( x · y ) : ( x · z ) = ( y : x ) · ( y : z )hold, for all x, y, z ∈ X . If ( X, · , :) is a q-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 left non-degenerate solution.Conversely, if ( X, r ) is a left non-degenerate solution, then the operations · and :given by x · y := λ − x ( y ) and x : y := ρ λ − y ( x ) ( y ), for all x, y ∈ X , give rise to a q-cycle set. Thanks to this correspondence, one can move from left non-degeneratesolutions to q-cycle sets. 2n the first part of the paper, we focus on non-degeneracy of bijective solutions.In particular, using q-cycle sets, we show that any finite left non-degenerate bi-jective solutions is right non-degenerate, giving a positive answer to [18, Ques-tion 4.2]. In this way, we also extend the corresponding result for finite involutivesolutions provided by Rump in [36, Theorem 2] in terms of cycle sets, and byJespers and Okni´nski in [30, Corollary 2.3] in terms of monoids of I -type. Inaddition, using the result on finite left non-degenerate solutions, we give suffi-cient conditions to find a class non-degenerate solutions that include properlythe finite left non-degenerate ones.In Section 4, we introduce an equivalence relation for a q-cycle set which wecall retraction , in analogy to the retraction of cycle sets, that is compatible withrespect to the two operations. Since the retraction of a non-degenerate q-cycleset is again a non-degenerate q-cycle set, we obtain an alternative proof of [28,Theorem 3.3] for non-degenerate solutions ( X, r ) with the additional propertyof r bijective. Clearly, we include the result showed in [33, Lemma 8.4].The final goal of this paper is to develop a theory of extensions of q-cycle sets.Following the ideas of Vendramin for cycle sets [42] and of Nelson and Watter-berg for biracks [35], we introduce a suitable notion of dynamical extension ofq-cycle sets. Even if dynamical extensions are often hard to find, we introduceseveral examples of dynamical extensions that are relatively easy to compute.Moreover, we introduce several families of dynamical extensions that providenon-degenerate solutions that are different from those obtained by skew braces.As an application, we construct a semidirect product of q-cycle sets, which is ageneralization of the semidirect product of cycle sets introduced by Rump [38]and, referring to [4, Problem 4.15], gives rise to a general definition of semidirectproduct of biquandles.
2. Basic results and examples
To study non-degenerate solutions, Rump recently introduced in [39] the notionof q-cycle set. Recall that a set X together with two binary operations · and: is a q-cycle set if the function σ x : X −→ X, y x · y is bijective, for every x ∈ X , and ( x · y ) · ( x · z ) = ( y : x ) · ( y · z ) (1)( x : y ) : ( x : z ) = ( y · x ) : ( y : z ) (2)( x · y ) : ( x · z ) = ( y : x ) · ( y : z ) (3)hold for all x, y, z ∈ X . Hereinafter, we denote by q and q ′ the squaring mapsrelated to · and :, i.e., the maps given by q ( x ) := x · x and q ′ ( x ) := x : x, x ∈ X .A q-cycle set ( X, · , :) is said to be regular if the function δ x : X −→ X, y x : y is bijective, for every x ∈ X , and non-degenerate if it is regular and q and q ′ arebijective. At first sight, the notion of non-degeneracy introduced by Rump seemsdifferent, but [39, Corollary 2] ensures that the two definitions are equivalent.The left non-degenerate solution ( X, r ) provided by a q-cycle set X is given by r ( x, y ) := ( σ − x ( y ) , σ − x ( y ) : x ) , for all x, y ∈ X , is a left-non degenerate solution. Conversely, if ( X, r ) is a leftnon-degenerate solution and r ( x, y ) = ( λ x ( y ) , ρ y ( x )), for all x, y ∈ X , then theoperations · and : given by x · y := λ − x ( y ) x : y := ρ λ − y ( x ) ( y )for all x, y ∈ X , give rise to a q-cycle set, see [39, Proposition 1]. As one wouldexpect, non-degenerate q-cycle sets corresponds to non-degenerate bijective so-lutions.Q-cycle sets allow us to construct a lot of families of solutions obtained inseveral recent papers. To show this, in the rest of this section we collect severalexamples of q-cycle sets and highlight some connections with other algebraicstructures.At first, note that q-cycle sets with x · y = x : y , for all x, y ∈ X , correspond tocycle sets. Clearly, in this case, (1), (2), and (3) are reduced to( x · y ) · ( x · z ) = ( y · x ) · ( y · z ) . (4)These examples of q-cycle sets determine the “celebrated” left non-degeneratesolutions that are also involutive, that in particular are maps r : X × X → X × X of the form r ( x, y ) = (cid:0) σ − x ( y ) , σ − x ( y ) · x (cid:1) , for all x, y ∈ X .In [39] Rump showed some connections between q-cycle sets and other algebraicstructures. In particular, he observed that skew braces correspond to particularq-cycle sets for which the squaring maps q and q ′ coincide (see [39, Corollary2]). In this context, we note that an analogue result follows for left semi-braces,algebraic structures that are a generalisation of skew braces. We specify that,in this paper, for a left semi-brace we mean the structure introduced in [12],named left cancellative left semi-brace in [31]. Specifically, we say that a set B with two operations + and ◦ is a left semi-brace if ( B, +) is a left cancellativesemigroup, ( B, ◦ ) is a group, and a ◦ ( b + c ) = a ◦ b + a ◦ (cid:0) a − + c (cid:1) a, b, c ∈ B , where a − is the inverse of a with respect to ◦ .If ( B, + , ◦ ) is a left semi-brace, by [12, Theorem 9] we have that the map r B : B × B → B × B defined by r B ( a, b ) = (cid:16) a ◦ (cid:0) a − + b (cid:1) , (cid:0) a − + b (cid:1) − ◦ b (cid:17) , for all a, b ∈ B , is a left non-degenerate solution. Hence, the q-cycle set associ-ated to B is the structure ( B, · , :) where · and : are given by a · b = λ a − ( b ) = a − ◦ ( a + b ) a : b = (cid:0) ρ a (cid:0) b − (cid:1)(cid:1) − = a − ◦ ( b + a ) , for all a, b ∈ B . Therefore, it is now easy to see that the squaring maps q and q ′ coincide also for q-cycle sets obtained by left semi-braces that are notskew-braces. In the following example, we show a family of concrete examplesof q-cycle sets determined by left semi-braces. Example 1.
Let ( B, ◦ ) be a group, f an idempotent endomorphism of ( B, ◦ ),and ( B, + , ◦ ) the left semi-brace in [12, Example 10] where the sum is given by a + b := b ◦ f ( a ) , for all a, b ∈ B . Since the left non-degenerate solution associated to ( B, + , ◦ ) isthe map r B : B : B × B → B × B given by r ( a, b ) = (cid:16) a ◦ b ◦ f ( a ) − , f ( a ) (cid:17) , we obtain that ( B, · , :) is a q-cycle set with · and : defined by a · b := a − ◦ b ◦ f ( a ) a : b := f ( b ) , for all a, b ∈ B . Clearly, if f = id B , then ( B, · , :) is not regular.More in general, let us observe that the hypothesis of idempotency on f isunnecessary to have a structure of q-cycle set on the group ( B, ◦ ). Obviously,( B, · , :) is regular if and only if the map f is bijective.However, in general, the squaring maps q and q ′ do not coincide, as one cansee in Examples 2 and Examples 3 below. Examples 2.
1) Let k ∈ N , B := Z /m Z , · and : the binary operations on B given by x · y := y + k and x : y = y , for all x, y ∈ B . Then, ( B, · , :) is a regular q-cycleset and the solution associated to ( B, · , :) is the map r : B × B → B × B given by r ( x, y ) = ( y − k, x ) , for all x, y ∈ B . 5) Let B := Z / Z and α the element of Sym( B ) given by α := (0 1).Moreover, let · and : the binary operations on B given by x · y := α ( y )if x ∈ { , } and x · y := y otherwise and x : y := y if x ∈ { , } and x : y := α ( y ) otherwise. Then, ( B, · , :) is a regular q-cycle set and thesolution r associated to ( B, · , :) is defined by r ( x, y ) = ( ( α ( y ) , x ) if x ∈ { , } ( y,
2) if x = 2 , for all x, y ∈ B . It is a routine computation to verify that r satisfies r = id B × B .The previous examples are all regular q-cycle sets. We conclude the section byshowing examples of q-cycle sets that are not regular. Examples 3.
1) Let X be a set, · and : two operations on X given by x · y := y x : y := k for all x, y ∈ X , where k ∈ X . Then, it is easy to check that the structure( X, · , :) is a q-cycle set. Clearly, if | X | >
1, then ( X, · , :) is not regular.Moreover, the left non-degenerate solution associated to ( X, · , :) is themap r : X × X → X × X given by r ( x, y ) = ( y, k ) , for all x, y ∈ X . In particular, let us observe that r satisfies the relation r = r .2) Let X be a left quasi-normal semigroup, i.e., X is a semigroup such that xyz = xzyz , for all x, y, z ∈ X . Then, it is straightforward to check that X endowed by the operations · and : defined by x · y := y x : y := yx, for all x, y ∈ X , is a q-cycle set. Clearly, ( X, · , :) in general is not regular.Note that the solution associated to X is given by r ( x, y ) = ( y, xy ) , for all x, y ∈ X , and it coincides with that provided in [14, Examples 6.2].In particular, we have that r satisfies the property r = r .6 . Non-degeneracy of q-cycle sets This section focuses on non-degeneracy of q-cycle sets. Rump in [36, The-orem 2] and Jespers and Okni´nski in [30, Corollary 2.3], using different lan-guages, showed that every finite involutive left non-degenerate solution is non-degenerate. Recently, Ced´o, Jespers and Verwimp asked if the same result istrue for every finite bijective left non-degenerate solution.
Question. [18, Question 4.2] Is every finite left non-degenerate solution non-degenerate?A first aim is to use the theory of q-cycle sets to answer in the positive sense.Moreover, we give sufficient conditions to find a class of non-degenerate solutionsthat include properly the finite left non-degenerate ones.Initially, we provide the “q-version” of the proof of [36, Theorem 2] providedrecently by Jedlicka, Pilitowska, and Zamojska-Dzienio in [28, Proposition 4.7].Hereinafter, for any regular q-cycle set X , we denote by G ( X ) the subgroup ofSym( X ) given by G ( X ) := < { σ x | x ∈ X } ∪ { δ x | x ∈ X } > and we call it the permutation group associated to X . Lemma 4.
Let ( X, · , :) be a regular q-cycle set such that the associated permu-tation group G ( X ) is finite. Then the squaring maps q and q ′ are surjective.Proof. At first, note that thanks to (2), we have( σ − x ( y ) : x ) : ( σ − x ( y ) : x ) = y : ( x : x ) (5)for every x, y ∈ X . Now, since G ( X ) is finite, there exists a natural number m such that δ mz ( z ) = z , for every z ∈ X . If m ∈ { , } then q ′ is surjective by (5).If m > z ∈ X , then z = δ mz ( z )= δ m − z ( z : ( z : z ))= δ m − z ( q ′ ( σ − z ( z ) : z )))= δ m − z ( z : q ′ ( σ − z ( z ) : z )))and applying repeatedly (5), we obtain that q ′ ( v ) = z , for some v ∈ X , hence q ′ is surjective. In a similar way, one can show that q is surjective, therefore thethesis follows. Theorem 5.
Let ( X, · , :) be a finite regular q-cycle set. Then X is non-degenerate. roof. If X is finite then so G ( X ) is, hence by the previous lemma q and q ′ are surjective. Again, since X is finite, q and q ′ are also injective, hence thethesis.As a consequence of the previous theorem, we obtain the following result thatextends the analogous one for finite involutive solutions showed by Rump in [36,Theorem 2] and by Jespers and Okni´nski in [30, Corollary 2.3]. Corollary 6.
Let ( X, r ) be a finite bijective left non-degenerate solution. Then ( X, r ) is non-degenerate.Proof. It follows by the previous theorem and [39, Proposition 1].Now, we use the previous result on finite solutions to find a larger class ofnon-degenerate solutions. At first, we have to introduce a preliminary lemma.We recall that a subset Y of a q-cycle set X is said to be G ( X )-invariant if g ( y ) ∈ Y , for all g ∈ G ( X ) and y ∈ Y . Lemma 7.
Let X be a regular q-cycle set and Y a G ( X ) -invariant subset of X . Then, Y is a regular q-cycle set with respect to the operations induced by X .Moreover, if X is non-degenerate then so Y is.Proof. Is is a straightforward calculation.
Theorem 8.
Let X be a regular q-cycle set such that the orbits of X with respectto the action of G ( X ) have finite size. Then, X is non-degenerate.Proof. We have to show that the squaring maps q and q ′ are bijective. Let x be an element of X and Y the orbit of X such that x ∈ Y . By the previouslemma and the hypothesis, Y is a finite regular q-cycle set and, by Theorem 5,it is non-degenerate. Therefore, there exist x , x ∈ Y such that q ( x ) = x and q ′ ( x ) = x . Hence, we have that the squaring maps of X are surjective.Now, suppose that y, z, t are elements of X such that q ( y ) = q ( z ) = t and let T be the orbit of X such that t ∈ T . Then, we have that y = σ − y ( t ) ∈ T and z = σ − z ( t ) ∈ T . Since T is non-degenerate, we obtain that y = z , hencethe squaring map q of X is injective. In a similar way, one can show that thesquaring map q ′ of X is injective.We conclude the section by translating the previous result in terms of solutions.Note that, if X is a regular q-cycle set and ( X, r ) the associated solution, where r ( x, y ) := ( λ x ( y ) , ρ y ( x )), for all x, y ∈ X , then the associated permutation group G ( X ) coincides with the subgroup of Sym( X ) generated by the set { λ x | x ∈ X } ∪ { η x | x ∈ X } , where η x ( y ) = ρ λ − y ( x ) ( y ), for all x, y ∈ X .8 orollary 9. Let ( X, r ) be a bijective left non-degenerate solution such thatthe orbits of X respect to the action of G ( X ) have finite size. Then ( X, r ) isnon-degenerate.Proof. It follows by the previous theorem and [39, Proposition 1].
4. Retraction of regular q-cycle sets
In this section we introduce the “q-version” of the retract relation introducedby Rump in [36] for cycle sets. Namely, we introduce a suitable notion ofretraction that is a congruence of q-cycle sets, i.e., it is a congruence with respectto the two operations · and :. Furthermore, we show that, for the specific classof non-degenerate q-cycle sets, the quotient of a q-cycle set by its retraction isstill a q-cycle set. Definition 1.
Let ( X, · , :) be a regular q-cycle set and ∼ the relation on X given by x ∼ y : ⇐⇒ σ x = σ y and δ x = δ y for every x, y ∈ X . Then ∼ will be called the retract relation .In analogy to [36, Lemma 2], we show that ∼ is always a congruence of aq-cycle set. Proposition 10.
Let ( X, · , :) be a regular q-cycle set. Then, the retraction ∼ is a congruence of ( X, · , :) .Proof. Let x, y, z, t be such that x ∼ y and z ∼ t . Then, using (1),( x · z ) · ( x · k ) = ( y · z ) · ( y · k ) = ( z : y ) · ( z · k ) = ( t : y ) · ( t · k )= ( y · t ) · ( y · k ) , therefore σ x · z = σ y · t . Moreover, using (2),( x : z ) · ( x · k ) = ( y : z ) · ( y · k ) = ( z · y ) · ( z · k ) = ( t · y ) · ( t · k )= ( y : t ) · ( y · k ) , therefore σ x : z = σ y : t . Similarly, one can show that δ x · z = δ y · t and δ x : z = δ y : t ,hence the thesis.To prove the main result of this section we make use of the extension on thefree group F ( X ) of a q-cycle set X . Initially, we recall the following propositioncontained in [39]. Proposition 11. [39, Theorem 1]
Let X be a non-degenerate regular q-cycleset. Then X can be extended to a q-cycle set on the free group ( F ( X ) , ◦ ) suchthat · a = 1 : a = a and ) ( a ◦ b ) · c = a · ( b · c ) ( a ◦ b ) : c = a : ( b : c ) a · ( b ◦ c ) = (( c : a ) · b ) ◦ ( a · c ) a : ( b ◦ c ) = (( c · a ) : b ) ◦ ( a : c ) ,are satisfied, for all a, b, c ∈ F ( X ) . Furthermore, in F ( X ) we have that x · y − = ( δ − y ( x ) · y ) − and x : y − = ( σ − y ( x ) : y ) − hold, for all x, y ∈ X . As a consequence of the previous proposition we have the following result.
Lemma 12.
Let X be a non-degenerate regular q-cycle set. Then, the followinghold:1) a · x ∈ X , for all a ∈ F ( X ) and x ∈ X ;2) a : x ∈ X , for all a ∈ F ( X ) and x ∈ X ;3) for all x, x , . . . , x n ∈ X and ǫ, ǫ , . . . , ǫ n ∈ {− , } , there exist t , ..., t n ∈ X and η, η , ..., η n ∈ { , − } such that σ x ǫ ( x ǫ ◦ · · · ◦ x ǫ n n ) = t η ◦ · · · ◦ t η n n ;
4) for all x, x , . . . , x n ∈ X and ǫ, ǫ , . . . , ǫ n ∈ {− , } , there exist t , ..., t n ∈ X and η, η , ..., η n , ǫ ∈ { , − } such that δ x ǫ ( x ǫ ◦ · · · ◦ x ǫ n n ) = t η ◦ · · · ◦ t η n n . Proof.
1) If x, y ∈ X , by 1) in Proposition 11, it follows that σ x − σ x ( y ) = x − · ( x · y ) = (cid:0) x − ◦ x (cid:1) · y = 1 · y = y and, analogously, σ x σ x − ( y ) = y , i.e., σ x − = σ − x . Thus, if a ∈ F ( X ), thereexist x , . . . , x n ∈ X and ǫ , . . . , ǫ n ∈ {− , } such that a = x ǫ ◦ · · ·◦ x ǫ n n , hence,by 1) in Proposition 11, it holds that a · x = σ x ǫ . . . σ x ǫnn ( x ) = σ ǫ x . . . σ ǫ n x n ( x ) ∈ X.
2) The proof is similar to 1).3) We proceed by induction on n ∈ N . Let n = 1, x, x ∈ X , ǫ, ǫ ∈ {− , } .Let us note that if ǫ = 1, then the thesis follows by 1). If ǫ = −
1, since byProposition 11, a · a ∈ F ( X ), by 3) in Proposition 11 it followsthat 1 = x ǫ · (cid:0) x ◦ x − (cid:1) = (cid:0)(cid:0) x − : x ǫ (cid:1) · x (cid:1) ◦ (cid:0) x ǫ · x − (cid:1) and so σ x ǫ (cid:0) x − (cid:1) = (cid:0)(cid:0) x − : x ǫ (cid:1) · x (cid:1) − (cid:0) x − : x ǫ (cid:1) · x ∈ X by 1). Now, suppose the thesis holds for a naturalnumber n and let x , . . . , x n ∈ X , ǫ , . . . , ǫ n ∈ {− , } . Then, by 3) in Proposi-tion 11, we have that σ x ǫ (cid:0) x ǫ ◦ · · · ◦ x ǫ n +1 n +1 (cid:1) = (cid:0)(cid:0) x ǫ n +1 n +1 : x ǫ (cid:1) · ( x ǫ ◦ · · · ◦ x ǫ n n ) (cid:1) ◦ (cid:0) x ǫ · x ǫ n +1 n +1 (cid:1) , therefore, by induction hypothesis, the thesis follows.4) The proof is similar to 3).Now, let us introduce the following preliminary lemmas. Lemma 13.
Let X be a non-degenerate regular q-cycle set and x, y ∈ X . Then,the following statements are equivalent1) x · z = y · z and x : z = y : z , for every z ∈ X ;2) x · z = y · z and x : z = y : z , for every z ∈ F ( X ) .Proof. Since every element of F ( X ) can be written as x ǫ ◦ ... ◦ x ǫ n n for some x , ..., x n ∈ X and ǫ , ..., ǫ n ∈ { , − } we prove “1) ⇒ n .Hence, suppose that x · t = y · t and x : t = y : t , for every t ∈ X .If n = 1 and z = x ǫ for some x ∈ X, ǫ ∈ { , − } we have two cases: if ǫ = 1, then x : z = y : z by the hypothesis, if ǫ = −
1, by equality (2) and thehypothesis we have that x : ( x : x ) = y : ( x : x ) ⇒⇒ ( x · σ − x ( x )) : ( x : x ) = ( x · σ − x ( y )) : ( x : x ) ⇒ ( σ − x ( x ) : x ) : ( σ − x ( x ) : x ) = ( σ − x ( y ) : x ) : ( σ − x ( y ) : x ) ⇒ q ′ ( σ − x ( x ) : x ) = q ′ ( σ − x ( y ) : x ) ⇒ σ − x ( x ) : x = σ − x ( y ) : x and, since F ( X ) is a group, it follows that ( σ − x ( x ) : x ) − = ( σ − x ( y ) : x ) − and hence, by Proposition 11, x : x − = y : x − . Similarly, one can show that x · x − = y · x − .Now, suppose the thesis for a natural number n and let z := x ǫ ◦ ... ◦ x ǫ n +1 n +1 ∈ F ( X ), where x , ..., x n +1 ∈ X and ǫ , ..., ǫ n +1 ∈ { , − } . Then, by 4) in Propo-sition 11, we have that x : ( x ǫ ◦ ... ◦ x ǫ n +1 n +1 ) = (( x ǫ n +1 n +1 · x ) : x ǫ ◦ ... ◦ x ǫ n n ) ◦ ( x : x ǫ n +1 n +1 )= (( x ǫ n +1 n +1 · x ) : ( x ǫ n +1 n +1 : δ − x ǫn +1 n +1 ( x ǫ ◦ ... ◦ x ǫ n n ))) ◦ ( x : x ǫ n +1 n +1 )= (( x : x ǫ n +1 n +1 ) : ( x : δ − x ǫn +1 n +1 ( x ǫ ◦ ... ◦ x ǫ n n ))) ◦ ( x : x ǫ n +1 n +1 )and similarly y : ( x ǫ ◦ ... ◦ x ǫ n +1 n +1 ) = (( y : x ǫ n +1 n +1 ) : ( y : δ − x ǫn +1 n +1 ( x ǫ ◦ ... ◦ x ǫ n n ))) ◦ ( y : x ǫ n +1 n +1 ) . t , ..., t n ∈ X and η , ..., η n ∈ { , − } suchthat δ − x ǫn +1 n +1 ( x ǫ ◦ ... ◦ x ǫ n n ) = t η ◦ ... ◦ t η n n , and this fact, together with the inductivehypothesis, implies that x : δ − x ǫn +1 n +1 ( x ǫ ◦ ... ◦ x ǫ n n )) = y : δ − x ǫn +1 n +1 ( x ǫ ◦ ... ◦ x ǫ n n )) . Moreover, again by inductive hypothesis, it follows that x : x ǫ n +1 n +1 = y : x ǫ n +1 n +1 .Therefore, we showed that x : z = y : z . In a similar way, one can show that x · z = y · z , hence “1) ⇒ X be a regular non-degenerate q-cycle set and F ( X ) its extension to thefree group on X . In [39, Section 4] Rump define the socle of F ( X ), denoted by Soc ( F ( X )), as the set Soc ( F ( X )) := { a | a ∈ F ( X ) , a · b = a : b = b ∀ b ∈ F ( X ) } and he showed that Soc ( F ( X )) is a subgroup of F ( X ) and that the factor F ( X ) /Soc ( F ( X )) is again a non-degenerate q-cycle set. From this fact thefollowing lemma follows. Lemma 14.
Let X be a non-degenerate q-cycle set and F ( X ) its extension tothe free group on X . Then, Ret( F ( X )) is a regular non-degenerate q-cycle set.Proof. It is easy to see that x ∼ y if and only if x ◦ y − ∈ Soc ( F ( X )) for every x, y ∈ F ( X ), hence the result follows by the previous remark. Theorem 15.
Let X be a regular q-cycle set and Ret( X ) its retraction. Then, X is non-degenerate if and only if Ret( X ) is a non-degenerate q-cycle set.Proof. To avoid confusion, for any element x ∈ X we denote by ¯ x its equivalenceclass with respect to the retract relation on X and by ˜ x its equivalence classwith respect to the retract relation on F ( X ).Suppose that X is non-degenerate. Then, the function G : X −→ Ret( F ( X )) , x ˜ x is a homomorphism of q-cycle sets. Moreover, by Lemma 13, we have that G ( x ) = G ( y ) if and only if ¯ x = ¯ y , hence Ret( X ) ∼ = G ( X ) as algebraic structures.If z ∈ X then ˜ z · ˜ x = g z · x ˜ z : ˜ x = g z : x g z − · ˜ x = ^ z − · x = ^ σ − z ( x )and g z − : ˜ x = ^ z − : x = ^ δ − z ( x )12nd these equalities imply that G ( X ) is a G (Ret( F ( X )))-invariant subset ofRet( F ( X )), hence, by Lemma 7, it is a non-degenerate q-cycle set. Therefore,Ret( X ) is a non-degenerate q-cycle set.Conversely, suppose that Ret( X ) is a non-degenerate q-cycle set. We haveto show that the maps q and q ′ of X are bijective. Let x, y ∈ X such that x · x = y · y . Then, ¯ x · ¯ x = ¯ y · ¯ y and, by the non-degeneracy of Ret( X ), we havethat ¯ x = ¯ y . Hence, y · x = x · x = y · y , therefore x = y and the injectivity of q follows. Moreover, if x ∈ X , there exists a unique ¯ y ∈ Ret( X ) such that¯ y · ¯ y = ¯ x . Since ¯ x = ¯ y · σ − y ( x ), we have that σ − y ( x ) = ¯ y , and hence x = σ y ( σ − y ( x )) = σ σ − y ( x ) ( σ − y ( x )) = q ( σ − y ( x ))therefore q is bijective. In the same way, one can show that q ′ is bijective.If X is a non-degenerate q-cycle set and ( X, r ) is the associated solution, it iseasy to see that x ∼ y if and only if λ x = λ y and ρ x = ρ y . Indeed, if x, y ∈ X , x ∼ y ⇔ σ x = σ y and δ x = δ y ⇔ σ − x = σ − y and δ x q ′ = δ y q ′ ⇔ σ − x = σ − y and q ′ ( σ − z ( x ) : z ) = q ′ ( σ − z ( y ) : z ) ∀ z ∈ X ⇔ λ x = λ y and ρ x = ρ y . Hence, the previous theorem give us an alternative proof of [28, Theorem 3.3],in terms of solutions (
X, r ) having bijective map r . Moreover, it shows a kindof converse for left non-degenerate bijective solutions: indeed, if X retracts toa non-degenerate solution then so X is. Corollary 16.
Let ( X, r ) be a bijective left non-degenerate solution of the Yang-Baxter equation and Ret( X ) the retraction of the associated q-cycle set. If ( X, r ) is non-degenerate then, for every ¯ x ∈ Ret( X ) , we can define the function ¯ λ ¯ x by ¯ λ ¯ x : Ret( X ) −→ Ret( X ) , ¯ y λ x ( y ) for every ¯ y ∈ Ret( X ) . Moreover, the pair (Ret( X ) , ¯ r ) , where ¯ r is given by ¯ r (¯ x, ¯ y ) := ( λ x ( y ) , λ x ( y ) : x ) is a bijective non-degenerate solution of the Yang-Baxter equation.Conversely, suppose that the function ¯ λ ¯ x is well defined for every ¯ x ∈ Ret( X ) and that the pair (Ret( X ) , ¯ r ) defined as above is a non-degenerate bijective so-lution. Then, ( X, r ) is non-degenerate. . Dynamical extensions of q-cycle sets Inspired by the extensions of cycle sets and racks, introduced by Vendraminin [42] and Andruskiewitsch and Gra˜na in [1], and the dynamical cocycles ofbiracks [35], introduced by Nelson and Watterberg, in this section we develop atheory of dynamical extensions of q-cycle sets.
Theorem 17.
Let ( X, · , :) be a q-cycle set, S a set, α : X × X × S −→ Sym( S ) and α ′ : X × X × S −→ S S maps such that α ( x · y ) , ( x · z ) ( α ( x,y ) ( s, t ) , α ( x,z ) ( s, u )) = α ( y : x ) , ( y · z ) ( α ′ ( y,x ) ( t, s ) , α ( y,z ) ( t, u )) (6) α ′ ( x : y ) , ( x : z ) ( α ′ ( x,y ) ( s, t ) , α ′ ( x,z ) ( s, u )) = α ′ ( y · x ) , ( y : z ) ( α ( y,x ) ( t, s ) , α ′ ( y,z ) ( t, u )) (7) α ′ ( x · y ) , ( x · z ) ( α ( x,y ) ( s, t ) , α ( x,z ) ( s, u )) = α ( y : x ) , ( y : z ) ( α ′ ( y,x ) ( t, s ) , α ′ ( y,z ) ( t, u )) (8) hold, for all x, y, z ∈ X , s, t, u ∈ S . Then, the triple ( X × S, · , :) where ( x, s ) · ( y, t ) := ( x · y, α ( x,y ) ( s, t ))( x, s ) : ( y, t ) := ( x : y, α ′ ( x,y ) ( s, t )) , for all x, y ∈ X and s, t ∈ S , is a q-cycle set. Moreover, ( X × S, · , :) is regularif and only if α ′ ( x,y ) ( s, − ) ∈ Sym( S ) , for all x, y ∈ X and s ∈ S .Proof. Since(( x, s ) · ( y, t )) · (( x, s ) · ( z, u )) = (( x · y ) · ( x · z ) , α ( x · y ) , ( x · z ) ( α ( x,y ) ( s, t ) , α ( x,z ) ( s, u )))and(( y, t ) : ( x, s )) · (( y, t ) · ( z, u )) = (( y : x ) · ( y · z ) , α ( y : x ) , ( y · z ) ( α ′ ( y,x ) ( t, s ) , α ( y,z ) ( t, u ))) , for all x, y, z ∈ X , s, t, u ∈ S , we have that (6) is equivalent to (1). In the sameway one can show that (7) and (8) are equivalent to (2) and (3). The rest ofthe proof is a straightforward calculation. Definition 2.
We call the pair ( α, α ′ ) dynamical pair and we call the q-cycleset X × α,α ′ S := ( X × S, · , :) dynamical extension of X by S .Clearly, every dynamical extension of a cycle set [42] is a dynamical extensionof a q-cycle set for which α = α ′ .If X and S are q-cycle sets and α ( x,y ) ( s, t ) = α ′ ( x,y ) ( s, t ) := t for all x, y ∈ X , s, t ∈ S , X × α,α ′ S , is a dynamical extension which we call trivial dynamical extension of X by S . The goal of the next section is to providefurther examples of dynamical extensions.14 efinition 3. Let X and Y be q-cycle sets. A homomorphism p : X −→ Y iscalled covering map if it is surjective and all the fibers p − ( y ) := { x | x ∈ X, p ( x ) = y } have the same cardinality. Moreover, a q-cycle set X is simple if, for everycovering map p : X −→ Y , we have that | Y | = 1 or p is an isomorphism. Examples 18.
1) Every q-cycle set having a prime number of elements is simple.2) Let X := { , , , } be the q-cycle set given by σ := (1 3) σ := (0 3) σ := (0 1 3) σ := (0 1) δ = δ = δ := id X δ := (0 1 3) . Suppose that p : X −→ Y is a covering map. Hence, Y has necessarilytwo elements. Moreover, since p (0) · p (0) = p (0) : p (0) = p (0), Y is theq-cycle set given by σ y = δ y = id Y , for every y ∈ Y . This fact impliesthat X has a subset I of 2 elements such that σ x ( I ) = δ x ( I ) = I , for every x ∈ X , but this is a contradiction. Then, X is simple.The following theorem, that is a “q-version” of [42, Theorem 2.12], states thatevery non-simple q-cycle set can be found as a dynamical extension of a smallerq-cycle set. Theorem 19.
Let X and Y be q-cycle sets and suppose that p : Y −→ X is acovering map. Then, there exist a set S and a dynamical pair ( α, α ′ ) such that Y ∼ = X × α,α ′ S .Proof. Since all the fibers p − ( x ) are equipotent, there exist a set S and abijection f x : p − ( x ) −→ S . Let α : X × X × S −→ Sym( S ) and α ′ : X × X × S −→ S S given by α ( x,z ) ( s, t ) := f x · z ( f − x ( s ) · f − z ( t ))and α ′ ( x,z ) ( s, t ) := f x : z ( f − x ( s ) : f − z ( t ))for all x, z ∈ X , s, t ∈ S . Then, α ( x · y,x · z ) ( α ( x,y ) ( r, s ) , α ( x,z ) ( r, t )) == f ( x · y ) · ( x · z ) ( f − x · y ( α ( x,y ) ( r, s )) · f − x · z ( α ( x,z ) ( r, t )))= f ( x · y ) · ( x · z ) ( f − x · y ( f x · y ( f − x ( r ) · f − y ( s )) · f − x · z ( f x · z ( f − x ( r ) · f − z ( t ))= f ( x · y ) · ( x · z ) (( f − x ( r ) · f − y ( s )) · ( f − x ( r ) · f − z ( t ))= f ( y : x ) · ( y · z ) (( f − y ( s ) : f − x ( r )) · ( f − y ( s ) · f − z ( t ))= f ( y : x ) · ( y · z ) ( f − y : x ( f y : x ( f − y ( s ) : f − x ( r )) · f − y · z ( f y · z ( f − y ( s ) · f − z ( t ))= α ( y : x,y · z ) ( α ′ ( y,x ) ( s, r ) , α ( y,z ) ( s, t )) , x, y, z ∈ X and r, s, t ∈ S , hence (6) holds. In the same way, one cancheck that the pair ( α, α ′ ) verifies (7) and (8), hence, by Theorem 17, X × α,α ′ S is a q-cycle set.The map φ : Y −→ X × α,α ′ S , y ( p ( y ) , f p ( y ) ( y )) is a bijective homomorphism.Indeed, φ ( y · z ) = ( p ( y · z ) , f p ( y · z ) ( y · z ))= ( p ( y ) · p ( z ) , f p ( y ) · p ( z ) ( f − p ( y ) ( f p ( y ) ( y )) · f − p ( z ) ( f p ( z ) ( z ))))= ( p ( y ) · p ( z ) , α ( p ( y ) ,p ( z )) ( f p ( y ) ( y ) , f p ( z ) ( z )))= ( p ( y ) , f p ( y ) ( y )) · ( p ( z ) , f p ( z ) ( z ))= φ ( y ) · φ ( z ) , for all y, z ∈ Y . In a similar way one can show that φ ( y : z ) = φ ( y ) : φ ( z ),for all y, z ∈ Y . By the surjectivity of p and the bijectivity of the maps f y thebijectivity of φ also follows, hence the thesis.
6. Examples and constructions
In this section, we collect several both known and new examples of dynamicalextension of q-cycle sets. Finally, we introduce a new construction of q-cyclesets, the semidirect product .Some dynamical cocycles of biracks founded in [4, 35, 27] provide dynamicalextensions of q-cycle sets (one can see this fact by a long but easy calculation).Using non-degenerate cycle sets, in [42, 7, 3] several examples of dynamicalextensions were obtained.We start by giving an example of a dynamical extension of a degenerate cycleset that is similar to those obtained in [7] and [3].
Example 20.
Let X be the cycle set on Z given by x · y := y − min { , x } , forall x, y ∈ X , (see [36, Example 1]), and let S be the Klein group Z / Z × Z / Z .Let α : X × X × S −→ Sym( S ) be the function given by α ( i,j ) (( a, b ) , ( c, d )) := ( ( c, d − ( a − c )) if i = j, ( c − b, d ) if i = j for all i, j ∈ X , ( a, b ) , ( c, d ) ∈ S , and set α ′ := α . Then, X × α,α ′ S is a dynamicalextension of X by S .By an easy calculation one can see that the previous example is an irre-tractable cycle set. Moreover, it is degenerate: indeed, q ( − , ,
0) is equal to q ( − , , Example 21.
Let X be the cycle set given by x · y = y for every x, y ∈ X , and G an abelian group. Let α, α ′ : X × X × ( G × G ) −→ Sym( G × G ) be the mapsgiven by α ( x,y ) (( s , s ) , ( t , t )) := ( ( t + t − s , t ) if x = y ( t , t + s ) if x = y and α ′ ( x,y ) (( s , s ) , ( t , t )) := ( ( t − t + s , t ) if x = y ( t , t + s ) if x = y for all x, y ∈ X , s , s , t , t ∈ G . Then ( α, α ′ ) is a dynamical pair and hence X × α,α ′ ( G × G ) is a q-cycle set. Example 22.
Let B be a group, f an endomorphism of B and ( B, · , :) theq-cycle in Example 1 where x · y = x − yf ( x ) and x : y = f ( y ), for all x, y ∈ B .Let S = B , α : B × B × S → Sym ( S ) and α ′ : B × B × S → S S be the mapsdefined by α ( x,y ) ( s, t ) := x − tα ′ ( x,y ) ( s, t ) := f ( t ) , for all x, y, s, t ∈ B . Then, it is a routine computation to verify that ( α, α ′ ) isa dynamical pair. Thus, the dynamical extension B × α,α ′ S of B by S is suchthat ( x, s ) · ( y, t ) = ( x − yf ( x ) , x − t )( x, s ) : ( y, t ) = ( f ( y ) , f ( t )) , for all x, y, s, t ∈ B . Let us observe that the q-cycle set B × α,α ′ S is regular ifand only if f is bijective. Moreover, the left non-degenerate solution associatedto B × α,α ′ S is given by r (( x, s ) , ( y, t )) = (cid:0)(cid:0) xyf ( x ) − , xt (cid:1) , ( f ( x ) , f ( s )) (cid:1) , for all x, y, s, t ∈ B . Example 23.
Let X be a left quasi-normal semigroup and ( X, · , :) the q-cycleset in 2) of Examples 3 where x · y = y and x : y = yx , for all x, y ∈ X . Let17 = X , α : X × X × S → Sym ( S ) and α ′ : X × X × S → S S be the mapsdefined by α ( x,y ) ( s, t ) = tα ′ ( x,y ) ( s, t ) = tx, for all x, y, s, t ∈ X . Then, ( α, α ′ ) is a dynamical pair. In fact, clearly (6) issatisfied. Moreover, if x, y, s, t ∈ X , we have that α ′ ( x : y ) , ( x : z ) ( α ′ ( x,y ) ( s, t ) , α ′ ( x,z ) ( s, u )) = α ′ ( yx,zx ) ( tx, ux ) = uxyx = uyx = α ′ ( x,zy ) ( s, uy ) = α ′ ( y · x ) , ( y : z ) ( α ( y,x ) ( t, s ) , α ′ ( y,z ) ( t, u ))and α ′ ( x · y ) , ( x · z ) ( α ( x,y ) ( s, t ) , α ( x,z ) ( s, u )) = α ′ ( y,z ) ( t, u ) = uy = α ′ ( y,z ) ( t, u )= α ( y : x ) , ( y : z ) ( α ′ ( y,x ) ( t, s ) , α ′ ( y,z ) ( t, u )) , i.e., (7) and (8) hold. Therefore, the dynamical extension X × α,α ′ S of X by S is such that ( x, s ) · ( y, t ) = ( y, t )( x, s ) : ( y, t ) = ( yx, tx ) , for all x, y, s, t ∈ X . Note that X × α,α ′ S in general is not regular. Moreover,the left non-degenerate solution associated to the q-cycle set X × α,α ′ S is givenby r (( x, s ) , ( y, t )) = (( y, t ) , ( xy, sy )) . Let us observe that this solution satisfies the property r = r , just like for thesolution associated to the q-cycle set ( X, · , :).The following example is the analogue of the abelian extension of cycle setsstudied by Lebed and Vendramin [32]. Example 24.
Let X be a cycle set, A an abelian group, a, b, a ′ , b ′ ∈ A and f, f ′ : X × X −→ A given by f ( x, y ) := ( a if x = yb if x = y and f ′ ( x, y ) := ( a ′ if x = yb ′ if x = y x, y ∈ X . Define on X × A the function α, α ′ : X × X × A −→ Sym( A )given by α ( x,y ) ( s, t ) = t + f ( x, y )and α ′ ( x,y ) ( s, t ) = t + f ′ ( x, y ) , for all x, y ∈ X . Then, ( α, α ′ ) is a dynamical pair and hence X × α,α ′ A is aq-cycle set. In particular, we have that( x, s ) · ( y, t ) = ( x · y, t + f ( x, y ))( x, s ) : ( y, t ) = ( x · y, t + f ′ ( x, y )) , for all x, y ∈ X .A particular family of dynamical extensions allow us to define a kind of semidi-rect product of q-cycle sets. Since this construction is similar to the ones ofother algebraic structures, for the convenience of the reader we write the twooperations · and : of the q-cycle set without using the dynamical pair. Proposition 25.
Let
X, S be q-cycle sets, θ : X −→ Aut( S ) , x θ x such that θ x · y θ x = θ y : x θ y , for all x, y ∈ X . Define on X × S the operations · and : by ( x, s ) · ( y, t ) := ( x · y, θ x · y ( s ) · θ y : x ( t ))( x, s ) : ( y, t ) := ( x : y, θ x : y ( s ) : θ y · x ( t )) for all x, y ∈ X , s, t ∈ S . Then, ( X × S, · , :) is a q-cycle set which we call thesemidirect product of X and S .Proof. Let ( x, s ) , ( y, t ) , ( z, u ) ∈ X × S . Then(( x, s ) · ( y, t )) · (( x, s ) · ( z, u ))= ( x · y, θ x · y ( s ) · θ y : x ( t )) · ( x · z, θ x · z ( s ) · θ z : x ( u ))= (( x · y ) · ( x · z ) , θ ( x · y ) · ( x · z ) ( θ x · y ( s ) · θ y : x ( t )) · θ ( x · z ):( x · y ) ( θ x · z ( s ) · θ z : x ( u ))and, thanks to the hypothesis, we obtain that θ ( x · y ) · ( x · z ) ( θ x · y ( s ) · θ y : x ( t )) · θ ( x · z ):( x · y ) ( θ x · z ( s ) · θ z : x ( u )) (9)= ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) : θ ( x · y ) · ( x · z ) ( θ x · y ( s ))) · ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) · θ ( z : y ):( z : x ) ( θ z : y ( u ))) and θ ( y : x ) · ( y · z ) ( θ y : x ( t ) : θ x · y ( s ))) · θ ( y · z ):( y : x ) ( θ y · z ( t ) · θ z : y ( u ) (10)= ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) : θ ( x · y ) · ( x · z ) ( θ x · y ( s ))) · ( θ ( y : x ) · ( y · z ) ( θ ( y : x ) ( t )) · θ ( z : y ):( z : x ) ( θ z : y ( u ))) . θ ( x · y ) · ( x · z ) ( θ x · y ( s ) · θ y : x ( t )) · θ ( x · z ):( x · y ) ( θ x · z ( s ) · θ z : x ( u )= ( θ ( x · y ) · ( x · z ) ( θ x · y ( s )) · θ ( x · y ) · ( x · z ) ( θ y : x ( t ))) · ( θ ( x · z ):( x · y ) ( θ x · z ( s )) · θ ( x · z ):( x · y ) ( θ z : x ( u )))= ( θ ( x · y ) · ( x · z ) ( θ x · y ( s )) · θ ( x · y ) · ( x · z ) ( θ y : x ( t ))) · ( θ ( x · y ) · ( x · z ) ( θ x · y ( s )) · θ ( x · z ):( x · y ) ( θ z : x ( u )))= ( θ ( x · y ) · ( x · z ) ( θ y : x ( t )) : θ ( x · y ) · ( x · z ) ( θ x · y ( s ))) · ( θ ( x · y ) · ( x · z ) ( θ y : x ( t )) · θ ( x · z ):( x · y ) ( θ z : x ( u )))= ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) : θ ( x · y ) · ( x · z ) ( θ x · y ( s ))) · ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) · θ ( z : x ) · ( z : y ) ( θ z : x ( u )))= ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) : θ ( x · y ) · ( x · z ) ( θ x · y ( s ))) · ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) · θ ( z : y ):( z : x ) ( θ z : y ( u ))) , i.e., (9) holds. On the other hand, we have that (( y, t ) : ( x, s )) · (( y, t ) · ( z, u ))= ( y : x, θ y : x ( t ) : θ x · y ( s )) · ( y · z, θ y · z ( t ) · θ z : y ( u ))= (( y : x ) · ( y · z ) , θ ( y : x ) · ( y · z ) ( θ y : x ( t ) : θ x · y ( s ))) · θ ( y · z ):( y : x ) ( θ y · z ( t ) · θ z : y ( u )) and, by the hypothesis, it follows that θ ( y : x ) · ( y · z ) ( θ y : x ( t ) : θ x · y ( s ))) · θ ( y · z ):( y : x ) ( θ y · z ( t ) · θ z : y ( u )= ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) : θ ( y : x ) · ( y · z ) ( θ x · y ( s ))) · ( θ ( y · z ):( y : x ) ( θ y · z ( t )) · θ ( y · z ):( y : x ) ( θ z : y ( u )))= ( θ ( y : x ) · ( y · z ) ( θ y : x ( t )) : θ ( x · y ) · ( x · z ) ( θ x · y ( s ))) · ( θ ( y : x ) · ( y · z ) ( θ ( y : x ) ( t )) · θ ( z : y ):( z : x ) ( θ z : y ( u ))) , i.e., (10) holds. Therefore, since we showed that the first members of (9) and(10) coincide and X is a q-cycle set, we have that equality (1) holds. In a similarway, one can check that (2) and (3) hold, hence it remains to show that σ ( x,s ) is bijective for every ( x, s ) ∈ X × S . Suppose that ( x, s ) · ( y, t ) = ( x, s ) · ( z, u ) forsome ( x, s ) , ( y, t ) , ( z, u ) ∈ X × S . Then,( x · y, θ x · y ( s ) · θ y : x ( t )) = ( x · z, θ x · z ( s ) · θ z : x ( u ))and, since X and S are q-cycle sets, it follows that y = z and θ y : x ( t ) = θ z : x ( u ).Therefore, from the bijectivity of θ y : x , we have that t = u , hence σ ( x,s ) is in-jective. Finally, we have that ( x, s ) · ( σ − x ( y ) , θ − σ − x ( y ): x ( σ − θ y ( s ) ( t ))) = ( y, t ), hencethe thesis.By semidirect product of q-cycle sets we are able to construct further dynamicalextensions. Example 26.
Let X := { , , } be the q-cycle set given by x · y := y , for all x, y ∈ X , and x : 0 = x : 1 := 0 x : 2 = 2 , for every x ∈ X . Moreover, let Y := { , , } be the q-cycle set given by x · y = x : y = y, for all x, y ∈ Y , and θ : X −→ Aut( Y ) given by θ (0) = θ (1) := (0 1) and θ (2) := id Y . Then, the semidirect product ( X × Y, · , :) is a q-cycle set of order9. In particular, note that it is not regular.20ote that, if X and S are cycle sets, the semidirect product (as q-cycle sets) of X and S coincides with the semidirect product of cycle sets developed by Rumpin [38]. Moreover, since biquandles in [4] can be viewed as particular q-cyclesets, referring to Problem 4.15 in [4] itself, the previous proposition gives riseto a general definition of semidirect product of biquandles. References [1] N. Andruskiewitsch, M. Gra˜na, From racks to pointed Hopf algebras, Adv.Math. 178 (2) (2003) 177–243.URL https://doi.org/10.1142/S1005386716000183 [2] D. Bachiller, Solutions of the Yang-Baxter equation associated to skew leftbraces, with applications to racks, J. Knot Theory Ramifications 27 (8)(2018) 1850055, 36.URL https://doi.org/10.1142/S0218216518500554 [3] D. Bachiller, F. Ced´o, E. Jespers, J. Okni´nski, A family of irretractablesquare-free solutions of the Yang-Baxter equation, Forum Math. 29 (6)(2017) 1291–1306.URL https://doi.org/10.1515/forum-2015-0240 [4] V. Bardakov, T. Nasybullov, M. Singh, General constructions of biquandlesand their symmetries, Preprint.URL https://arxiv.org/abs/1908.08301 [5] M. Bonatto, M. Kinyon, D. Stanovsk´y, P. Vojtechovsk´y, Involutive latinsolutions of the Yang-Baxter equation, Preprint.URL https://arxiv.org/pdf/1910.02148.pdf [6] M. Castelli, F. Catino, M. M. Miccoli, G. Pinto, Dynamical extensions ofquasi-linear left cycle sets and the Yang-Baxter equation, J. Alg. Appl.18 (11) (2019) 1950220.URL https://doi.org/10.1142/S0219498819502207 [7] M. Castelli, F. Catino, G. Pinto, A new family of set-theoretic solutions ofthe Yang-Baxter equation, Comm. Algebra 46 (4) (2017) 1622–1629.URL http://dx.doi.org/10.1080/00927872.2017.1350700 [8] M. Castelli, F. Catino, G. Pinto, About a question of Gateva-Ivanova andCameron on square-free set-theoretic solutions of the Yang-Baxter equa-tion, Comm. Algebra, In press.URL https://doi.org/10.1080/00927872.2020.1713328 https://doi.org/10.1016/j.jpaa.2019.01.017 [10] M. Castelli, G. Pinto, W. Rump, On the indecomposable involutive set-theoretic solutions of the Yang-Baxter equation of prime-power size, Comm.Algebra, In press.URL https://doi.org/10.1080/00927872.2019.1710163 [11] F. Catino, I. Colazzo, P. Stefanelli, On regular subgroups of the affinegroup, Bull. Aust. Math. Soc. 91 (1) (2015) 76–85.URL http://dx.doi.org/10.1017/S000497271400077X [12] F. Catino, I. Colazzo, P. Stefanelli, Semi-braces and the Yang-Baxter equa-tion, J. Algebra 483 (2017) 163–187.URL https://doi.org/10.1016/j.jalgebra.2017.03.035 [13] F. Catino, I. Colazzo, P. Stefanelli, Skew left braces with non-trivial anni-hilator, J. Algebra Appl. 18 (2) (2019) 1950033, 23.URL https://doi.org/10.1142/S0219498819500336 [14] F. Catino, M. Mazzotta, P. Stefanelli, Set-theoretical solutions of the Yang-Baxter and pentagon equations on semigroups, Accepted on SemigroupForum.URL https://arxiv.org/abs/1910.05393 [15] F. Ced´o, E. Jespers, ´A. Del R´ıo, Involutive Yang-Baxter groups, Transac-tions of the American Mathematical Society 362 (5) (2010) 2541–2558.URL https://doi.org/10.1090/S0002-9947-09-04927-7 [16] F. Ced´o, E. Jespers, J. Okni´nski, Braces and the Yang-Baxter equation,Comm. Math. Phys. 327 (1) (2014) 101–116.URL https://doi.org/10.1007/s00220-014-1935-y [17] F. Ced´o, E. Jespers, J. Okni´nski, Set-theoretic solutions of the Yang-Baxterequation, associated quadratic algebras and the minimality condition, Rev.Mat. Complut. (2020).URL https://doi.org/10.1007/s13163-019-00347-6 [18] F. Ced´o, E. Jespers, C. Verwimp, Structure monoids of set-theoretic solu-tions of the Yang-Baxter equation, Preprint.URL https://arxiv.org/abs/1912.09710 [19] F. Ced´o, A. Smoktunowicz, L. Vendramin, Skew left braces of nilpotenttype, Proc. Lond. Math. Soc. (3) 118 (6) (2019) 1367–1392.URL https://doi.org/10.1112/plms.12209 https://doi.org/10.1142/S0219498815500012 [21] V. G. Drinfel ′ d, On some unsolved problems in quantum group theory, in:Quantum groups (Leningrad, 1990), vol. 1510 of Lecture Notes in Math.,Springer, Berlin, 1992, pp. 1–8.URL https://doi.org/10.1007/BFb0101175 [22] P. Etingof, T. Schedler, A. Soloviev, Set-theoretical solutions to the Quan-tum Yang-Baxter equation, Duke Math. J. 100 (2) (1999) 169–209.URL http://doi.org/10.1215/S0012-7094-99-10007-X [23] T. Gateva-Ivanova, Set-theoretic solutions of the Yang-Baxter equation,braces and symmetric groups, Adv. Math. 338 (2018) 649–701.URL https://doi.org/10.1016/j.aim.2018.09.005 [24] T. Gateva-Ivanova, S. Majid, Matched pairs approach to set theoretic so-lutions of the Yang–Baxter equation, J. Algebra 319 (4) (2008) 1462–1529.URL https://doi.org/10.1016/j.jalgebra.2007.10.035 [25] T. Gateva-Ivanova, M. Van den Bergh, Semigroups of I-Type, J. Algebra206 (1) (1998) 97–112.URL https://doi.org/10.1006/jabr.1997.7399 [26] L. Guarnieri, L. Vendramin, Skew braces and the Yang-Baxter equation,Math. Comp. 86 (307) (2017) 2519–2534.URL https://doi.org/10.1090/mcom/3161 [27] E. Horvat, Constructing biquandles, Preprint.URL https://arxiv.org/abs/1810.03027 [28] P. Jedlicka, A. Pilitowska, A. Zamojska-Dzienio, The retraction relation forbiracks, J. Pure Appl. Algebra 223 (8) (2019) 3594–3610.URL https://doi.org/10.1016/j.jpaa.2018.11.020 [29] E. Jespers, L. Kubat, A. Van Antwerpen, L. Vendramin, Factorizations ofskew braces, Math. Ann. 375 (3-4) (2019) 1649–1663.URL https://doi.org/10.1007/s00208-019-01909-1 [30] E. Jespers, J. Okni´nski, Monoids and groups of I -type, Algebr. Represent.Theory 8 (5) (2005) 709–729.URL https://doi.org/10.1007/s10468-005-0342-7 [31] E. Jespers, A. Van Antwerpen, Left semi-braces and solutions of the Yang-Baxter equation, Forum Math. 31 (1) (2019) 241–263.URL https://doi.org/10.1515/forum-2018-0059 https://doi.org/10.1142/S0218196716500570 [33] V. Lebed, L. Vendramin, On structure groups of set-theoretic solutions tothe Yang-Baxter equation, Proc. Edinb. Math. Soc. (2) 62 (3) (2019) 683–717.URL https://doi.org/10.1017/s0013091518000548 [34] J.-H. Lu, M. Yan, Y.-C. Zhu, On the set-theoretical Yang-Baxter equation,Duke Math. J. 104 (1) (2000) 1–18.URL http://dx.doi.org/10.1215/S0012-7094-00-10411-5 [35] S. Nelson, E. Watterberg, Birack dynamical cocycles and homomorphisminvariants, J. Algebra Appl. 12 (8) (2013) 1350049, 14.URL https://doi.org/10.1142/S0219498813500497 [36] W. Rump, A decomposition theorem for square-free unitary solutions ofthe quantum Yang-Baxter equation, Adv. Math. 193 (2005) 40–55.URL https://doi.org/10.1016/j.aim.2004.03.019 [37] W. Rump, Braces, radical rings, and the quantum Yang-Baxter equation,J. Algebra 307 (1) (2007) 153–170.URL https://doi.org/10.1016/j.jalgebra.2006.03.040 [38] W. Rump, Semidirect products in algebraic logic and solutions of the quan-tum Yang-Baxter equation, J. Algebra Appl. 7 (4) (2008) 471–490.URL https://doi.org/10.1142/S0219498808002904 [39] W. Rump, A covering theory for non-involutive set-theoretic solutions tothe Yang-Baxter equation, J. Algebra 520 (2019) 136–170.URL https://doi.org/10.1016/j.jalgebra.2018.11.007 [40] A. Smoktunowicz, L. Vendramin, On skew braces (with an appendix by N.Byott and L. Vendramin), J. Comb. Algebra 2 (1) (2018) 47–86.URL https://doi.org/10.4171/JCA/2-1-3 [41] A. Soloviev, Non-unitary set-theoretical solutions to the quantum Yang-Baxter equation, Math. Res. Lett. 7 (5-6) (2000) 577–596.URL https://doi.org/10.4310/MRL.2000.v7.n5.a4 [42] L. Vendramin, Extensions of set-theoretic solutions of the Yang-Baxterequation and a conjecture of Gateva-Ivanova, J. Pure Appl. Algebra 220(2016) 2064–2076.URL https://doi.org/10.1142/S1005386716000183https://doi.org/10.1142/S1005386716000183