Skew bracoids and the Yang-Baxter equation
Ilaria Colazzo Alan Koch Isabel Martin-Lyons Paul J. Truman


Skew bracoids and the Yang-Baxter equation]Skew bracoids and the Yang-Baxter equation


University of Exeter 
 Department of Mathematics and Statistics 
 Exeter 
 EX4 4QF 
 UK
University of Leeds 
 School of Mathematics 
 Leeds 
 LS2 9JT 
 UK
I.Colazzo@leeds.ac.uk


Department of Mathematics, Agnes Scott College, 141 E. College Ave., Decatur, GA 30030 USA
AKoch@agnesscott.edu


School of Computing and Mathematics 
 Keele University 
 Staffordshire 
 ST5 5BG 
 UK
I.D.Martin-Lyons@keele.ac.uk


School of Computing and Mathematics 
 Keele University 
 Staffordshire 
 ST5 5BG 
 UK
P.J.Truman@keele.ac.uk

[2020]Primary 16T25; Secondary 20N99




Skew braces provide an algebraic framework for studying bijective nondegenerate solutions of the set-theoretic Yang-Baxter equation. We show that left skew bracoids, recently introduced by two of the authors, can be used to obtain right nondegenerate solutions, and give examples arising from various methods of constructing left skew bracoids. We compare the solutions we obtain with the left nondegenerate solutions obtained via (left cancellative) left semibraces, and establish a correspondence between left semibraces and a class of left skew bracoids. 


[
    Paul J. Truman
    April 29, 2024
==================





§ INTRODUCTION


The Yang-Baxter equation is a fundamental equation of theoretical physics and statistical mechanics <cit.>, <cit.>. We will be concerned with its set theoretic analogue, introduced by Drinfeld <cit.>. A solution of the set theoretic Yang-Baxter equation on a nonempty set G (hereafter, simply a solution on G) is a map r : G × G → G × G such that

    (r ×𝕀)(𝕀× r)(r ×𝕀) = (𝕀× r)(r ×𝕀)(𝕀× r)

as functions on G × G × G. We say that solution is bijective if r is a bijective function, and involutive if r^2 = 𝕀. For x,y ∈ G we write

    r(x,y) = (λ_x(y), ρ_y(x));

we say that a solution is left nondegenerate if λ_x is bijective for each x ∈ G, right nondegenerate if ρ_y is bijective for each y ∈ G, and nondegenerate if it is both left and right nondegenerate. 

A wide variety of techniques have been brought to bear on the problems of determining and studying solutions with various properties. A particularly fruitful algebraic framework is provided by the theory of skew braces; these objects were introduced by Guanieri and Vendramin <cit.>, generalizing Rump's earlier definition of braces <cit.>. A skew brace is a set G together with two binary operations, each making G into a group, that satisfy a certain compatibility relation (see Section <ref> for more details). Skew braces yield bijective nondegenerate solutions, and have also been found to have connections with numerous other topics including braids, knots, regular subgroups of permutation groups, and Hopf-Galois theory. In <cit.> two of the authors introduce skew bracoids; these consist of two groups G and N, together with a transitive action of the former on the latter that interacts with the binary operation on N in a manner analogous to the compatibility relation for skew braces. Skew bracoids can be used to broaden the connection between skew braces and Hopf-Galois theory <cit.>. It is also natural to ask what connections they might have with the set theoretic Yang-Baxter equation; the purpose of this paper is to address this question. 

Our main result (Theorem <ref>) is that left skew bracoids (G,N) satisfying a mild technical hypothesis yield right nondegenerate solutions on G. Our approach employs a result of Lu, Yan, and Zhu for constructing solutions from groups with certain additional data; we summarize this result, along with some of the theory of skew braces, in Section <ref>. In Section <ref> we recall some fundamental definitions and results concerning skew bracoids, specify the family of skew bracoids we will study (Hypothesis <ref>), and state and prove Theorem <ref>. We construct examples of skew bracoids satisfying Hypothesis <ref> in Section <ref>, drawing upon a variety of constructions in the literature; we also exhibit a family of skew bracoids that do not satisfy this Hypothesis. Finally, in Section <ref> we compare the right nondegenerate solutions arising from our left skew bracoids with the left nondegenerate solutions arising from left semibraces found in <cit.>.



 §.§ Acknowledgements

We gratefully acknowledge the support of the Engineering and Physical Sciences Research Council, project references EP/V005995/1 (Ilaria Colazzo) and EP/W012154/1 (Paul J. Truman), and the London Mathematical Society, joint research group 32312, which enabled parts of this collaboration to take place.



§ PRELIMINARIES: SOLUTIONS ARISING FROM GROUPS, AND SKEW BRACES
 

In <cit.> Lu, Yan, and Zhu prove that given a group G, a homomorphism λ : G →(G), and an antihomomorphism ρ : G →(G) such that λ_x(y)ρ_y(x) = xy for all x,y ∈ G, the map r : G × G → G × G defined by r(x,y) = (λ_x(y), ρ_y(x)) for all x,y ∈ G is a bijective nondegenerate solution on G. We shall require the following mild generalization of their result; we sketch a proof for the convenience of the reader.

 
Suppose that λ, ρ : G →Map(G,G) satisfy the following, for all x,y ∈ G:

    λ_xy    =    λ_xλ_y; 
    ρ_xy    =    ρ_yρ_x; 
    
    xy     =    λ_x(y)ρ_y(x) .

Then the map r : G × G → G × G defined by r(x,y) = (λ_x(y), ρ_y(x)) for all x,y ∈ G is a solution on G. 


Let x,y,z ∈ G. By straightforward computation we have

    (r ×𝕀)(𝕀× r)(r ×𝕀)(x,y,z) = 
    (λ_λ_x(y)(λ_ρ_y(x)(z)), ρ_λ_ρ_y(x)(z)(λ_x(y)), ρ_z(ρ_y(z)))

and

    (𝕀× r)(r ×𝕀)(𝕀× r)(x,y,z) = 
    (λ_x(λ_y(z)), λ_ρ_λ_y(z)(ρ_z(y)), ρ_ρ_z(y)(ρ_λ_y(z)(x))).

Hypotheses (<ref>) and (<ref>) imply that the first components of the right hand sides of (<ref>) and (<ref>) agree, and hypotheses (<ref>) and (<ref>) imply that the third components agree. Hypothesis (<ref>) also implies that the products of the three components agree, and so we conclude that the second components also agree. 


Following <cit.> we shall call a pair of maps λ, ρ as in Proposition <ref> a Lu-Yan-Zhu pair on G. 

We record for future use that if λ, ρ form a Lu-Yan-Zhu pair on G then by repeated application of (<ref>) for all x,y,z ∈ G we have

    λ_x(yz)ρ_yz(x) =      xyz  
    
     =     λ_x(y)ρ_y(x)z  
    
     =     λ_x(y)λ_ρ_y(x)(z) ρ_z(ρ_y(x))  
    
     =     λ_x(y)λ_ρ_y(x)(z) ρ_yz(x) by (<ref>)

which yields the useful formula

    λ_x(yz) = λ_x(y)λ_ρ_y(x)(z).
 

Lu-Yan-Zhu pairs leading to bijective nondegenerate solutions can be obtained from skew braces (in fact, these notions are equivalent: see <cit.> and <cit.>). A skew (left) brace is a triple (G,⋆,·) in which (G,⋆) and (G,·) are groups (called the additive and multiplicative groups of the skew brace) and the equation

    x · (y ⋆ z) = (x · y) ⋆x̅⋆ (x ⋆ z)

holds for all x,y,z ∈ G, where x̅ denotes the inverse of x with respect to ⋆ <cit.>. For brevity, we shall refer to skew braces simply as braces; we note, however, that elsewhere in the literature this term is often used to refer to skew braces with abelian additive group, as originally introduced by Rump in <cit.>.

Equation (<ref>) implies that for each x ∈ G the function γ_x : G → G defined by γ_x(y) = x̅⋆ (x · y) is an automorphism of (G,⋆), and that the resulting function γ : G →(G) is a homomorphism (sometimes called the γ-function of the brace) <cit.>; it can be shown (<cit.> or <cit.>) that if we set λ_x(y) = γ_x(y) and define ρ_y(x) by λ_x(y)^-1· x · y then λ, ρ form a Yu-Lan-Zhu pair and hence, via Proposition <ref>, a solution on G. This solution is always bijective and nondegenerate. It is involutive if and only if the group (G,⋆) is abelian; if (G,⋆) is nonabelian then the inverse of the solution arising from the brace (G,⋆,·) is the solution arising from the opposite brace (G,⋆^op,·): see <cit.>. 



§ SKEW BRACOIDS CONTAINING A BRACE
 

A (left) skew bracoid is a quintuple (G,·,N,⋆,⊙) in which (G,·) and (N,⋆) are groups and ⊙ is a transitive (left) action of (G,·) on N such that the equation

    x ⊙ (η⋆μ) = (x ⊙η) ⋆ (x ⊙ e_N)^-1⋆ (x ⊙μ)

holds for all x ∈ G and η, μ∈ N (see <cit.>). In analogy with braces, we call (G,·) the multiplicative group, and (N,⋆) the additive group, of the skew bracoid (G,·,N,⋆,⊙). We suppress the notation for the binary operations in G and N whenever possible. We retain the symbol ⊙ for the action of G on N in our arguments, but often suppress it when specifying skew bracoids, viz (G,N). For brevity, we shall refer to skew bracoids simply as bracoids. 

In this section we show that a large family of bracoids with multiplicative group G yield Lu-Yan-Zhu pairs λ, ρ on G in which ρ_y is bijective for all y ∈ G, and so, via Proposition <ref>, right nondegenerate solutions on G. 

In order to specify the family of bracoids we shall study, we recall some definitions from <cit.>. First, we say that a bracoid (G,N) in which the subgroup S = (e_N) of G is trivial is essentially a brace (see <cit.>), since in this case we may define a second group operation on G via the rule

    (x ⋆_G y) ⊙ e_N = (x ⊙ e_N) ⋆ (y ⊙ e_N)

such that (G,⋆_G) ≅ (N,⋆) and (G,⋆_G,·) is a brace. Second, a subbracoid of a bracoid (G,N,⊙) consists of a subgroup H of G and a subgroup M of N such that (H,M,⊙) is a bracoid <cit.>. The family of bracoids we shall study are those satisfying

 
For (G,N) a bracoid, there is a subgroup H of G such that the subbracoid (H,N) is essentially a brace. Abusing terminology, we shall say that the bracoid (G,N) contains a brace. 
 



We shall construct examples of bracoids containing a brace in Section <ref>. If (G,N) is a such a bracoid then the map a : H → N given by a(h) = h ⊙ e_N is a bijection; we denote its inverse by b : N → H, so that b(η) ⊙ e_N = η for all η∈ N. Using the bijection b we can reinterpret Equation (<ref>) as

    x ⋆_H y = b( a(x) ⋆ a(y) )  x,y ∈ H.


Finally, recall that the γ-function of a bracoid (G,N) is the homomorphism γ : G →(N) defined by 

    γ_x(η) = (x ⊙ e_N)^-1(x ⊙η)  x ∈ G η∈ N.

(see <cit.>). 

We can now state our main result:

 
Suppose that (G,N) is a bracoid containing a brace (H,N). For x,y ∈ G define 

    λ_x(y) = b(γ_x(y ⊙ e_N))

and

    ρ_y(x) = λ_x(y)^-1 xy.

Then λ, ρ form a Lu-Yan-Zhu pair in which λ_x(G)=H for all x ∈ G and ρ_y is bijective for each y ∈ G. 


Applying Proposition <ref> we obtain immediately


The map r : G × G → G × G defined by r(x,y) = (λ_x(y), ρ_y(x)) for all x,y ∈ G is a right nondegenerate, non-bijective solution. 



Let x,y,z ∈ G. First we show that 

    λ_xy = λ_xλ_y.

We have 

    λ_xλ_y(z) =     λ_x ( b (γ_y(z ⊙ e_N) ) )
    
    =      b (γ_x ( b (γ_y(z ⊙ e_N)) ⊙ e_N ) ) 
    
    =      b (γ_xγ_y(z ⊙ e_N)) definition of b
    
    =      b (γ_xy (z ⊙ e_N)) γ is a homomorphism
    
    =     λ_xy(z).

In particular, Equation (<ref>) implies that λ_x(z) = λ_e(λ_x(z)), so that λ_x(G) ⊆λ_e(G), and that λ_e(z) = λ_x(λ_x^-1(z)), so that λ_e(G) ⊆λ_x(G). Thus λ_x(G) = λ_e(G), which is easily seen to be equal to H. 

Next we show that 

    ρ_xy = ρ_yρ_x.

We have

    ρ_xy(z) = λ_z(xy)^-1 zxy,
 
whereas

    ρ_yρ_x(z)     =    ρ_y( λ_z(x)^-1 zx ) 
        =    λ_λ_z(x)^-1zx(y)^-1λ_z(x)^-1 zxy.

Using (<ref>), we see that (<ref>) and (<ref>) agree if and only if

    λ_λ_z(x)^-1λ_zx(y) = λ_z(x)^-1λ_z(xy).

To show that (<ref>) holds we note that each side lies in H, so it is sufficient to show that their images under the bijection a : H → N agree. 

To simplify notation, write h = λ_z(x) ∈ H. Then the image of the right hand side of (<ref>) under a is 

    (h^-1λ_z(xy)) ⊙ e_N = h^-1⊙ (γ_z(xy ⊙ e_N))

On the other hand, the image of the left hand side of (<ref>) under a is 

    λ_h^-1 (λ_zx(y)) ⊙ e_N
    
    =     γ_h^-1(γ_zx(y ⊙ e_N)) definition of λ and b
    
    =       (h^-1⊙ e_N)^-1 (h^-1⊙γ_zx(y ⊙ e_N)) definition of γ_h^-1
    
    =       (h^-1⊙ e_N)^-1 (h^-1⊙γ_zγ_x(y ⊙ e_N)) γ is a homomorphism
    
    =       (h^-1⊙ e_N)^-1 (h^-1⊙γ_z ( (x ⊙ e_N)^-1(xy ⊙ e_N))) definition of γ_x
    
    =       (h^-1⊙ e_N)^-1 (h^-1⊙γ_z(x ⊙ e_N)^-1γ_z(xy ⊙ e_N)) γ_z∈(N)
    
    =       (h^-1⊙ e_N)^-1 (h^-1⊙γ_z(x ⊙ e_N)^-1) (h^-1⊙ e_N)^-1 (h^-1⊙γ_z(xy ⊙ e_N)) by (<ref>)
    
    =        (h^-1⊙γ_z(x ⊙ e_N))^-1 (h^-1⊙γ_z(xy ⊙ e_N)) <cit.>
    
    =        (h^-1⊙ (h ⊙ e_N))^-1 (h^-1⊙γ_z(xy ⊙ e_N)) definitions of λ and h
    
    =       (h^-1⊙γ_z(xy ⊙ e_N))

Since (<ref>) and (<ref>) agree, (<ref>) holds, and so
(<ref>) holds. In particular, this implies that ρ_yρ_y^-1 = ρ_y^-1ρ_y = ρ_e. Since ρ_e(z) = λ_z(e)^-1 ze = z, we see that ρ_y is bijective (with inverse ρ_y^-1), as claimed. 

Finally, it is clear from the definition of ρ_y that we have λ_x(y)ρ_y(x)=xy. This completes the proof.


Next we study some of the properties of solutions arising via Theorem <ref>. The following alternative characterization of Hypothesis <ref> is useful:

 
A bracoid (G,N) contains a brace if and only if the subgroup S = (e_N) has a complement H in G. 


If (G,N) contains a brace then there is a subgroup H of G such that (H,N) is essentially a brace. In particular, the stabilizer of e_N in H is trivial, and so H ∩ S is trivial. Given x ∈ G there exists a unique element h ∈ H such that x ⊙ e_N = h ⊙ e_N, so s = h^-1x ∈ S, and so x = hs. Hence H is a complement to S in G. 

Conversely, if S has a complement H in G then given x ∈ G we may write x=hs with h ∈ H and s ∈ S, and then we have x ⊙ e_N = hs ⊙ e_N = h ⊙ e_N; thus H acts transitively on N. Since H ∩ S is trivial, the stabilizer of e_N in H is trivial, and so (H,N) is essentially a brace. Hence (G,N) contains a brace. 



If (G,N) contains a brace (H,N) then we also have an exact factorization G=SH. The functions λ, ρ defined in Theorem <ref> allow us to translate between these two exact factorizations: if h ∈ H and s ∈ S then 

    sh = λ_s(h)ρ_h(s);

we have λ_s(h) ∈ H by definition, and 

    λ_s(h) ⊙ e_N   = (s ⊙ e_N)(sh ⊙ e_N) 
       = sh ⊙ e_Nsince s ∈ S
       = λ_s(h)ρ_h(s) ⊙ e_Nusing Equation (<ref>)

Thus ρ_h(s) ∈ S, and so λ_s(h)ρ_h(s) is the unique representation of the element sh using the exact factorization HS. 




We recall that if (G,N) is a bracoid containing a brace (H,N) then the bijection b : N → H can be used to define a new binary operation ⋆_H on H such that (H,⋆_H,·) is a brace. We can identify the solution arising from this brace as a subsolution of the one arising from the bracoid.  

 
Suppose that (G,N) is a bracoid containing a brace (H,N) and let r(x,y) = (λ_x(y), ρ_y(x)) be the corresponding solution on G. Then r restricts to a solution r_H on H, which coincides with the solution arising from the brace (H,⋆_H,·), and r restricts to the solution r_S on S given by r_S(x,y) = (e_G,xy) for all x,y ∈ S. 


First suppose that x,y ∈ H. We have λ_x(y) ∈ H by definition, and ρ_y(x) = λ_x(y)^-1xy ∈ H; hence r(x,y) ∈ H × H. A short computation shows that the γ-function of the brace (H,⋆_H,·) is precisely λ_x(y), so the restriction of r to H coincides with the solution arising from this brace. 

Now suppose that x,y ∈ S. Then we have

    λ_x(y)     =     b( (x ⊙ e_N)(xy ⊙ e_N) ) 
        =     b( e_N e_N ) 
        =     e_G,

and ρ_y(x) = xy ∈ S; hence r(x,y) = (e_G, xy) ∈ S × S. 


It is natural to ask whether the original solution r on G can be reconstructed from r_H and r_S. In <cit.> the notion of the matched product of solutions is introduced: given sets X, Y, solutions r_X, r_Y on these sets, and maps α : Y →(X) and β : X →(Y), all satisfying various compatibility conditions, we can construct a solution 

    r_X⋈ r_Y : (X × Y) × (X × Y) → (X × Y) × (X × Y),

called the matched product of the solutions r_X and r_Y (via α and β). 

We claim that the solution r(x,y)=(λ_x(y), ρ_y(x)) on G obtained via Theorem <ref> is isomorphic to a matched product of the solutions r_S and r_H defined in Proposition <ref>. Define α : H →(S) by α_h(s)=(ρ_h^-1(s^-1))^-1 and β:S →(H) by  β_s(h)=λ_s(h); lengthy but routine calculations show that the conditions of <cit.> are satisfied, and that the resulting matched product of solutions is given by

    (r_S⋈ r_H)((a, u), (b, v))    = ((e, β_aλ_u(v)), (α^-1_Uρ_α_u(b)(a), ρ_β_a(v)(u))) 
       =( λ_(a,u)(b,v), ρ_(b,v)(a,u) )

say, where a=α^-1_u(a), u=β^-1_a(u), A= α_uλ_a(b), U=β_aλ_u(v), A=α^-1_U(A) and  U=β^-1_A(U).

Now define f : S× H → G by f(a,u)=u(ρ_u(a^-1))^-1; this is a bijection since G=HS exactly and ρ_u is bijective for all u. Then employing the useful fact that λ_x(hs) = λ_x(h) for any x ∈ G, h ∈ H, and s∈ S, we have


    λ_f(a,u)f(b,v)    = λ_u(ρ_u(a^-1))^-1(v(ρ_v(b^-1))^-1) 
       = λ_aλ_a^-1(u)(v) ρ_v(b^-1) ∈ S by Remark <ref>
       =λ_aλ_λ_a^-1(u)(v) 
       = β_aλ_β_a^-1(u)(v) definition of β
       = f(λ_(a,u)(b,v)).


A similar but longer calculation establishes that f(ρ_(b,v)(a,u))=ρ_f(b,v)f(a,u). Therefore the solution of G arising via Theorem <ref> is isomorphic to the matched product of solutions r_S⋈ r_H. 



§ EXAMPLES OF BRACOIDS CONTAINING A BRACE
 

In this section we construct examples of bracoids satisfying Hypothesis <ref>, often by employing Lemma <ref>. 



A large family of bracoids arises from a certain quotienting procedure on braces. Recall that a strong left ideal of a brace (G,⋆,·) is a subset S of G such that (S,⋆)  (G,⋆) and γ_x(s) ∈ S for all x ∈ G and s ∈ S (these conditions imply that (S,·) ≤ (G,·)). If S is a strong left ideal of G then (G,·) acts transitively on the group (G/S, ⋆) by left translation of cosets and (G,·,G/S,⋆,⊙) is a bracoid in which _G(eS)=S <cit.>. If in addition S has a complement H in G then we can apply Theorem <ref> to obtain a right nondegenerate solution on G. 

 
Suppose that H, S are groups and that there is a homomorphism α : S →(H). Then the set H × S, together with the operations

    (h,s) ⋆ (h',s') = (hh', ss'),  (h,s)·(h',s') = (h α_s(h'), ss')

forms a brace (see <cit.>). It is straightforward to check that the γ-function of this brace is γ_(h,s)(h',s') = (α_s(h'),s'), and it follows quickly that { e }× S is a strong left ideal. Since { e }× S has a complement H ×{ e } in (H × S,·), Theorem <ref> applies. 


We note that in this example H ×{ e } is a normal complement to { e }× S in (H × S,·). Bracoids (G,N) such that (e_N) has a normal complement in G have important applications in Hopf-Galois theory;  in the framework of <cit.> they correspond to Hopf-Galois structures on so-called almost classically Galois field extensions, which occupy a distinguished place in the theory: see <cit.>, <cit.>, <cit.>, for example. It is therefore interesting to note that Lemma <ref> and Theorem <ref> imply that bracoids (G,N) such that (e_N) has a normal complement in G produce solutions on G. See <cit.> for further explorations of these ideas. 

In <cit.> it is shown that given a group (G,·) and an endomorphism ψ of (G,·) with abelian image (an abelian map) we may define a new binary operation ⋆ = ⋆_ψ on G such that (G,⋆,·) is a brace. The strong left ideals in braces of this form are characterized in <cit.>, as follows: writing ϕ(x)=x ψ(x)^-1 for all x ∈ G, a subgroup (S,·) of (G,·) is a strong left ideal if and only if [G,ϕ(S)] ≤ S (where the commutators are computed with respect to ·. If this is the case then we may apply  <cit.> and form the bracoid (G,G/S), as described above. It is possible for bracoids of this form to contain a brace:


Let p and q be distinct odd prime numbers, and let 

    G = ⟨ x,y,z | x^pq = y^2 = z^2 = e,   yxy^-1 = zxz^-1 = x^-1,   yz=zy ⟩≅ C_pq⋊ (C_2× C_2).

The map ψ: G → G defined by ψ(x^iy^jz^k) = y^jz^k is an abelian map, and we have ϕ(x^iy^jz^k) = x^i. Now let S = ⟨ x^q, z ⟩; then ψ(S) = ⟨ x^q⟩ and [G,ϕ(S)] ⊆⟨ x^q⟩⊆ S, so we may form the bracoid (G,G/S). Finally, note that the subgroup S = ⟨ x^q, z ⟩ has a complement H = ⟨ x^p, y ⟩ in G (neither of H,S is normal in G). Therefore by Lemma <ref> the bracoid (G,G/S) contains a brace, and Theorem <ref> applies. 


We can also construct examples of bracoids containing a brace that do not arise as the quotient of a brace by a strong left ideal as in Example <ref>. Recall that the holomorph of a group N is the group (N) = N ⋊(N), which acts on N via the formula (η,α)(μ) = ηα(μ) for all η,μ∈ N and α∈(N). If J is a transitive subgroup of (N) then (J,N) is a bracoid in which S = (e_N) = { (η,α) ∈ J |η = e } <cit.>. 


Let N be an elementary abelian group of order 8. By <cit.> the holomorph of N contains a transitive subgroup J isomorphic to the simple group GL_3(𝔽_2) of order 168. The resulting bracoid (J,N) does not arise as the quotient of a brace by a strong left ideal: see <cit.>. However, writing S = (e_N) as usual we have |S|=21 by the Orbit-Stabilizer Theorem, so taking H to be a Sylow 2-subgroup of J we have an exact factorization J = HS, and Lemma <ref> implies that we may apply Theorem <ref>. 


We can also use the holomorph framework to construct examples of bracoids that do not contain a brace. We seek a group N and a transitive subgroup J of (N) such that S = (e_N) does not have a complement in J. Using results of Darlington <cit.> we have

 
Let p and q be prime numbers with p ≡ 1 q^2, and let N be a cyclic group of order pq, presented as

    N = ⟨σ, τ|σ^p=τ^q=1,  στ = τσ⟩.

Let α be an automorphism of N that has order q^2 and fixes τ (for example, let α be the (p-1)/q^2 power of a generator of (⟨σ⟩)), and let 

    J = ⟨ (σ, 𝕀), (τ, α) ⟩≤(N).

Then J is a transitive subgroup of (N) of order pq^2, and (J,N) is a bracoid. 

We claim that S = (e_N) does not have a complement in J. First we determine S. Since α fixes τ we have (τ, α)^i = (τ^i, α^i) for each i; it follows that S = ⟨ (e_N,α^q) ⟩ and that |S| = q. If H is a complement to S in J then in particular there exist h ∈ H and s ∈ S such that hs = (τ, α); hence (τ,α^i) ∈ H for some i ≡ 1 q. An element of this form has order q^2, but H must have order pq, so this is impossible. Hence S does not have a complement in J. 


More generally, given a bracoid (G,N,⊙) the image of the permutation representation λ_⊙ : G →(N) is contained in (N) <cit.>. If S = (e_N) has a complement H in G then the transitive subgroup λ_⊙(G) ≤(N) contains a regular subgroup (namely λ_⊙(H)). If λ_⊙ is injective (that is, if the bracoid is reduced <cit.>) then the converse is true. We expect the situation of Example <ref> to be relatively rare. 



§ CONNECTIONS WITH SOLUTIONS OBTAINED FROM OTHER ALGEBRAIC STRUCTURES
 

In <cit.> it is shown that left nondegenerate solutions may be obtained from left semibraces. In this section we obtain a correspondence between left bracoids containing a brace and left semibraces, and explain how the right nondegenerate solution arising from the bracoid and the left nondegenerate solution arising from the corresponding left semibrace are related to one another. 

We begin by recalling some definitions and properties concerning semibraces. In <cit.> a left semibrace is defined to be a triple (G,+,·) in which (G,·) is a group, (G,+) is a left cancellative semigroup, and the equation 

    x · (y+z) = x · y + x·(x^-1 + y)

holds for all x,y,z ∈ G. 

In <cit.>, Jespers and Van Antwerpen give an extended the definition of semibraces that does not require the additive semigroup to be left cancellative. Left semibraces here defined are what they call left cancellative left semibrace. For simplicity's sake we will use the term left semibrace to refer to left cancellative left semibraces.

As when working with bracoids, we suppress the notation · where possible. 

If (G,+,·) is a left semibrace then for x,y ∈ G we define 

    _x(y) = x(x^-1 + y)

and

    _y(x) = _x(y)^-1xy = (x^-1 + y)^-1y;

the maps , then form a Lu-Yan-Zhu pair in which _x is bijective for each x ∈ G and _y(G) = (G,+) for each y ∈ G, and so by Proposition <ref> the map s : G × G → G × G defined by s(x,y) = (_x(y), _y(x)) for all x,y ∈ G is a left nondegenerate solution on G. 

We note that we may rewrite Equation (<ref>) as

    x · (y+z) = x · y + _x(y).

The left semigroup (G,+) admits a decomposition (G+e,+) ⊕ (E,+) in which (G+e,+) is a group and E denotes the set of idempotents of (G,+). The identity element of e of (G,·) lies in E, and an element x ∈ G lies in E if and only if x+e = e. If x ∈ E then x+y = y for all y ∈ G. 

To connect semibraces with bracoids containing a brace, it is convenient to work with bracoids using a particular presentation. Recall that if (G,N) is a bracoid containing a brace and H is a subgroup of G such that (H,N) is essentially a brace then we may use the bijection b : N → H to define a second group operation on H via the rule

    h ⋆ k = b((h ⊙ e_N)(k ⊙ e_N))  h,k ∈ H.

We may also transport the transitive action of G of N to a transitive action of G on H via the rule

    x ⊙ h = b(x ⊙ (h ⊙ e_N)) = b( xh ⊙ e_N)  x ∈ G  h ∈ H;
 
we then find that (G,H) is a bracoid, and the pair (𝕀,b) : (G,N) → (G,H) is an isomorphism of bracoids (see <cit.>). 

If (G,H) is a bracoid of this form containing a brace then we have h ⊙ e = h for all h ∈ H, and h ⊙ k = hk for all h,k ∈ H. In particular, we have γ_x(y) ⊙ e = γ_x(y) for all x,y ∈ G, and so we have λ_x(y) = γ_x(y ⊙ e) in Theorem <ref>. 

 
Let G = (G,·) be a group with an exact factorization HS. There is a bijection between


  * binary operations ⋆ on H and transitive actions ⊙ of G on H such that (G,·,H,⋆,⊙) is a left bracoid containing a brace and with _G(e)=S;

  * binary operations + on G such that (G,+,·) is a left semibrace in which G+e = H and E=S.



First suppose that (G,·,H,⋆,⊙) is a left bracoid containing a brace and with (e)=S. For x,y ∈ G let λ_x(y) = γ_x(y ⊙ e) and ρ_y(x) = λ_x(y)^-1xy, and define 

    x + y = y λ_y^-1(x).

We claim that (G,+,·) is a left semibrace. First we show that + is associative on G. For x,y,z ∈ G we have

    x + (y + z) =      x + zλ_z^-1(y) 
    
     =      zλ_z^-1(y)λ_(zλ_z^-1(y))^-1(x) 
    
     =      zλ_z^-1(y)λ_ρ_y(z^-1)y^-1(x) since λ_z^-1(y)ρ_y(z^-1)=z^-1y,

and

    (x+y) + z  =      yλ_y^-1(x) + z 
    
     =      zλ_z^-1(yλ_y^-1(x)) 
    
     =      zλ_z^-1(y)λ_ρ_y(z^-1)(λ_y^-1(x)) by (<ref>)
    
     =      zλ_z^-1(y)λ_ρ_y(z^-1)y^-1(x)  since λ is a homomorphism.

Thus + is associative, and so (G,+) is a semigroup. 

Next we show that (G,+) is left cancellative. For x,y,z ∈ G we have

    x + y = x + z 
    ⇒     yλ_y^-1(x) = z λ_z^-1(x) 
    ⇒     xρ_x(y^-1)^-1 = xρ_x(z^-1)^-1
    ⇒    ρ_x(y^-1) = ρ_x(z^-1) 
    ⇒     y = z  since ρ_x is injective.

Thus (G,+) is left cancellative. 

Finally we show that the left semibrace relation is satisfied. For x,y,z ∈ G we have 

    xy + x(x^-1+z) =      xy + xzλ_z^-1(x^-1) 
    
     =      xzλ_z^-1(x^-1) λ_xzλ_z^-1(x^-1)(xy) 
    
     =      xzλ_z^-1(x^-1)λ_ρ_x^-1(z^-1)(xy) 
    
     =      xzλ_z^-1(x^-1xy) by (<ref>)
    
     =      xzλ_z^-1(y) 
    
     =      x(y+z).

Thus (G,+,·) is indeed a left semibrace. 

For the remaining claims, we note that for x ∈ G we have x+e = eλ_e^-1(x) = λ_e(x); thus G+e = λ_e(G) = H. Finally, we have

    x + x = x 
    ⇔     xλ_x^-1(x) = x 
    ⇔    λ_x^-1(x) = e 
    ⇔    λ_x^-1(x) ⊙ e = e note λ_x^-1(x) ∈ H
    ⇔    γ_x^-1(x ⊙ e) = e relationship between λ and γ
    ⇔     x ⊙ e = e γ_x∈(H,⋆)
    ⇔     x ∈ S.

Thus E=S. 

Conversely, suppose that + is a binary operation on G such that (G,+,·) is a left semibrace in which G+e = H and E=S. For h,k ∈ H define

    h ⋆ k = k + h;

then (H,⋆) is the opposite group to (H,+). In addition, for x ∈ G and h ∈ H define x ⊙ h = xh + e. We claim that (G,·,H,⋆,⊙) is a bracoid containing a brace in which (e)=S. It is straightforward to verify that ⊙ is a transitive left action of (G,·) on H; it remains to show that the bracoid relation (<ref>) is satisfied. By the definition of the binary operation ⋆, it suffices to show that

    x ⊙ (h + k) = (x ⊙ h) - (x ⊙ e) + (x ⊙ k)

for all x ∈ G and h,k ∈ H. Beginning with the right hand side of (<ref>) we have:

    (x ⊙ h) - (x ⊙ e) + (x ⊙ k)      =     (x ⊙ h) - (x ⊙ e) + (xk + e) 
        =     (x ⊙ h) - (x ⊙ e) + (x(e+k) + e) 
        =     (x ⊙ h) - (x ⊙ e) + (xe) + _x(k) + e 
        =     (x ⊙ h) - (x ⊙ e) + (xe + e) + _x(k) + e 
        =     xh + _x(k) + e 
        =     x(h+k)+e 
        =     x ⊙ (h+k).

Hence (<ref>) holds, and so (G,·,H,⋆,⊙) is indeed a left bracoid.

For the remaining claims: first we note that x ∈ S if and only if x + e = e, which occurs if and only if x ∈ E; second, for h ∈ H we have h ⊙ e = h + e = h, so the action of H on itself is transitive, and so (H,+,·) is a brace. 

The constructions described above are mutually inverse: if (G,·,H,⋆,⊙) is a bracoid as in the statement of the theorem and (G,+,·) is the corresponding semibrace, then the bracoid corresponding to (G,+,·) is (G,·,H,⋆̂, ⊙̂), where for all h,k ∈ H and x ∈ G we have h ⋆̂  k = k + h and x  ⊙̂  h = xh + e. Since k + h = h ⋆ k, we quickly see that ⋆̂ conincides with ⋆. Since ⊙̂ and ⊙ are both transitive actions of (G,·) on H, they coincide if and only if x  ⊙̂  e = x ⊙ e for all x ∈ G. Writing x=hs, with h ∈ H and s ∈ S, we have: 

    x  ⊙̂  e =      hs  ⊙̂  e 
    
    =      h  ⊙̂  e  since s is idempotent in (G,+)
    
    =      h + e 
    
    =      h 
    
    =      h ⊙ e 
    
    =      hs ⊙ e 
    
    =      x ⊙ e.

Thus the bracoid corresponding to (G,+,·) is the original bracoid (G,·,H,⋆,⊙).

Conversely, if (G,+,·) is a semibrace as in the statement of the theorem and (G,·,H,⋆,⊙) is the corresponding bracoid, then the semibrace corresponding to (G,·,H,⋆,⊙) is (G,+̂,·), where x  +̂  y = yγ_y^-1(x). By the bracoid relation (<ref>) we may write

    γ_y^-1(x) = y^-1⊙ ((y ⊙ e)⋆(x ⊙ e)),

and so

    x  +̂  y =      yλ_y^-1(x) 
    
     =      y(y^-1⊙ ((y ⊙ e)⋆(x ⊙ e))) 
    
     =      y(y^-1⊙ ((y + e)⋆(x + e))) 
    
     =      y(y^-1⊙ (x+e+y+e)) 
    
     =      y(y^-1(x+y+e) + e) 
    
     =      x+y+e + _y(e) 
    
     =      x+y

Thus the semibrace corresponding to (G,·,H,⋆,⊙) is the original semibrace (G,+,·). 

Hence we obtain the bijection in the statement of the theorem. 


 
Suppose that the left bracoid (G,·,H,⋆,⊙) and the left semibrace (G,+,·) correspond as in Theorem <ref>. Let r be the right nondegenerate solution arising from (G,·,H,⋆,⊙), let s be the left nondegenerate solution arising from (G,+,·), and define ι, τ : G × G → G × G by 

    ι(x,y) = (x^-1,y^-1) τ(x,y) = (y,x).

Then 

    s(x,y) = ιτ r τι (x,y).



Let x,y ∈ G and recall that r(x,y) = (λ_x(y), ρ_y(x)) and s(x,y) = (_x(y), _y(x)). From the first of these we obtain quickly that

    ιτ r τι (x,y) = (ρ_x^-1(y^-1)^-1, λ_y^-1(x^-1)^-1).

The relationship between + and λ implies that x^-1 + y = y λ_y^-1(x^-1), so 

    λ_y^-1(x^-1)^-1 = (x^-1+y)^-1y = _y(x).

Now we recall that

    λ_y^-1(x^-1) ρ_x^-1(y^-1) = y^-1x^-1 = (xy)^-1,

so 

    ρ_x^-1(y^-1)^-1λ_y^-1(x^-1)^-1 = xy = _x(y)_y(x).

We have already shown that λ_y^-1(x^-1)^-1 = _y(x), so we must also have ρ_x^-1(y^-1)^-1 = _x(y). This  completes the proof. 



If G=H then bracoids and the semibraces considered in Theorem <ref> are braces, and the corresponding solutions are bijective and nondegenerate. In this case our constructions pair a brace with its opposite, and a solution with its inverse. If we adjust our method of constructing a semibrace from a bracoid by defining λ̂_x(y) = x ⊙ (y ⊙ e) ⋆ (x^-1⊙ e) (this is the γ-function of the opposite bracoid) and x + y = y λ̂_y^-1(x) then we obtain a correspondence such that in the case G=H the braces (and solutions) coincide. This may be more satisfying from an algebraic perspective, but the connection between the left and right degenerate solutions appears to be much less transparent under this formulation. 




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