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Several types of groupoids induced by twovariable functions
SpringerPlus volume 5, Article number: 1715 (2016)
Abstract
In this paper, we introduce the concept of several types of groupoids related to semigroups, viz., twisted semigroups for which twisted versions of the associative law hold. Thus, if \((X,*)\) is a groupoid and if \(\varphi : X^2\rightarrow X^2\) is a function \(\varphi (a, b) = (u, v)\), then \((X,*)\) is a lefttwisted semigroup with respect to \(\varphi \) if for all \(a, b, c\in X\), \(a*(b*c) = (u*v)*c\). Other types are righttwisted, middletwisted and their duals, a dual lefttwisted semigroup obeying the rule \((a*b)*c = u*(v*c)\) for all \(a, b, c\in X\). Besides a number of examples and a discussion of homomorphisms, a class of groupoids of interest is the class of groupoids defined over a field \((X,+, \cdot )\) via a formula \(x*y=\lambda x + \mu y\), with \(\lambda , \mu \in X\), fixed structure constants. Properties of these groupoids as twisted semigroups are discussed with several results of interest obtained, e.g., that in this setting simultaneous lefttwistedness and righttwistedness of \((X,*)\) implies the fact that \((X,*)\) is a semigroup.
Introduction and preliminaries
Suppose that \(X=\mathbf{R}\) is the set of all real numbers and that we consider the binary operation \((\mathbf{R}, )\) where “−” is the usual subtraction. Then \((xy)z\not = x(yz) = xy + z\) in general, i.e., \((\mathbf{R}, )\) is not a semigroup. Since \((xy)z = x(y(z))\), if we define \(u:=x, v:=z\), then we have \((xy)z = u(yv)\), which looks like that “−” satisfies a version of the associative law in \(\mathbf{R}\), i.e., there exists a map \(\varphi : \mathbf{R}^2 \rightarrow \mathbf{R}^2 \) such that \(\varphi (x, z)=(x, z)=(u, v)\) and \((xy)z = u(yv)\). Thus, we obtain a “twisted” associated law for \((\mathbf{R}, )\), with the function \(\varphi \) defining the “nature” of the “twisted semigroup” of a particular type. Obviously, a twisted semigroup need not be a semigroup. However, semigroups are twisted semigroups where the twist is \(\varphi (x, y) =(x, y)\). Twisted semigroups of several types will be the topic of investigation in the following. As algebraic objects they include many familiar examples of groupoids which are definitely not semigroups but whose study benefits from the approach taken in what follows.
In particular, if \((X,+, \cdot )\) is a field and if \(x*y=\lambda x + \mu y, \lambda , \mu \in X\), defines a (linear) product, the resulting class of groupoids has a membership whose structure depends both on the nature of the formula as well as on properties of the field \((X,+,\cdot )\) itself. Several conclusions are obtained. A main result is the conclusion that if \((X,*)\) is a lefttwisted semigroup with respect to a map \(\varphi _1\) and if \((X,*)\) is a righttwisted semigroup with respect to a map \(\varphi _2\), then \((X,*)\) is already a semigroup. Homomorphisms of twisted semigroups of the various types are also discussed, and from a counterexample it follows that the class of dual lefttwisted semigroups is not a variety, even though direct products and (groupoid) epimorphic images of any of the types of twisted semigroups are also of at least the same type. Although examples of certain twisted semigroups have long been studied in various settings, the notion of a groupoid \((X,*)\) equipped with a twisting mapping \(\varphi : X^2\rightarrow X^2\) to produce a twisted semigroup \((X,*)\) of a certain type, appears to be new. For general references on semigroups we refer to Clifford et al. (1961), Howie (1995).
Kim and Neggers (2008) introduced the notion of Bin(X), the collection of all groupoids defined on a nonempty set X. Given arbitrary groupoids \((X, *)\) and \((X, \bullet )\), we define a product \((X, \Box ):= (X, *)\Box (X, \bullet )\) where \(x\Box y := (x*y)\Box (y*x)\) for all \(x, y\in X\). They showed that \((Bin(X), \Box )\) is a semigroup and the left zero semigroup on X acts as an identity in \((Bin(X), \Box )\). Let \((R, +, \cdot )\) be a commutative ring with identity and let L(R) denote the collection of all groupoids \((R, *)\) such that, for all \(x, y\in R\), \(x*y:= ax + by + c\), where \(a, b, c\, (\in R)\) are fixed constants. Such a groupoid \((R, *)\) is said to be a linear groupoid. They showed that \((L(R), \Box )\) is a semigroup with identity.
Some researchers studied on linear groupoids and quadratic groupoids in several algebras. Neggers et al. (2001) introduced the notion of a Qalgebra, and showed that every quadratic Qalgebra \((X,*, e), e\in X\), has of the form \(x*y= xy +e\) when X is a field with \(X\ge 3\). Moreover, Kim and So (2012) investigated some properties of \(\beta \)algebras and they obtained linear \(\beta \)algebras.
Twisted semigroups
Let \((X,*)\) be a groupoid for which there exists a function \(\varphi : X^2\rightarrow X^2\) such that, for all \(a, b, c\in X\),
where \(\varphi (a,b) =(u, v)\), i.e., \(u=u(a, b), v=v(a, b)\) are functions of two variables. Then \((X,*)\) is said to be a lefttwisted semigroup with respect to the map \(\varphi \). Such a map \(\varphi \) is called an associator function of the groupoid \((X,*)\).
We may think of a dual equation of (1) as follows:
where \(\varphi (a,b) =(u, v)\), i.e., \(u=u(a, b), v=v(a, b)\) are functions of two variables. Then \((X,*)\) is said to be a dual lefttwisted semigroup with respect to the map \(\varphi \). The function \(\varphi \) is not necessarily unique.
Suppose we replace the Eq. (1) by
where \(\varphi (b,c) =(u, v)\), i.e., \(u=u(b, c), v=v(b, c)\) are functions of two variables. Then \((X,*)\) is said to be a righttwisted semigroup with respect to the map \(\varphi \).
We may think of a dual equation of (3) as follows:
where \(\varphi (b,c) =(u, v)\), i.e., \(u=u(b, c), v=v(b, c)\) are functions of two variables. Then \((X,*)\) is said to be a dual righttwisted semigroup with respect to the map \(\varphi \).
If we replace the Eq. (1) by
where \(\varphi (a,c) =(u, v)\), i.e., \(u=u(a, c), v=v(a, c)\) are functions of two variables. Then \((X,*)\) is said to be a middletwisted semigroup with respect to the map \(\varphi \).
We may think of a dual equation of (5) as follows:
where \(\varphi (a,c) =(u, v)\), i.e., \(u=u(a, c), v=v(a, c)\) are functions of two variables. Then \((X,*)\) is said to be a dual middletwisted semigroup with respect to the map \(\varphi \).
If \(\varphi : X^2\rightarrow X^2\) is the identity map \(\varphi (a, b)= (a, b)\), then the Eqs. (1–6) reduce to the associative law and thus:
Proposition 1
If \((X,*)\) is a semigroup, then it is a (dual) left(right, middle)twisted semigroup.
Example 1
Consider \((\mathbf{R}, )\), the real numbers \(\mathbf R\) with the subtraction operation “−”. Since \(a(bc)\not = (ab)c\), the groupoid \((\mathbf{R}, )\) is not a semigroup. Consider the expression \(((a+1)b)(1c) = (ab) +c= a(bc)\). Thus \(\varphi (a,c)= (a+1, 1c)\) produces \((\mathbf{R}, )\) as a middletwisted semigroup which is not a semigroup.
Example 2
Consider \(X:=2^A\) where \(A\not = \emptyset \). If we define \(a*b:= ab\) for any \(a, b\in X\), then \((a*b)*c \not = a*(b*c)\). On the other hand, if we let \(\varphi (b,c):=(b\cup c, \emptyset )\), then \((a*b)*c = (ab)c = a(b\cup c)\), and \(a*(u*v) = a(b\cup c  \emptyset )=a(b\cup c)\), proving that \((X,*)\) is a righttwisted semigroup w.r.t. \(\varphi \).
Note that Example 2 is a typical example of a BCKalgebra which is also a righttwisted semigroup.
Example 3
In Example 1 if we define \(\psi (b, c):=(b+c, 0)\), then \(a((b+c)0) = (ab)c\) and hence it is a righttwisted semigroup with respect to \(\psi \).
Example 4
Let \((\mathbf{R}, +, \cdot )\) be a real field. Define a binary operation “\(*\)” on \(\mathbf R\) by
Given elements \(a, b, c\in \mathbf{R}\), define a map \(\varphi : \mathbf{R}^2 \rightarrow \mathbf{R}^2\) by \(\varphi (a, b) := (u, v)\) where
with \(v^3 = a(ab)(v^2  1)\). Then it is easy to show that \((a*b)*c = u*(v*c)\). This shows that \((\mathbf{R}, *)\) is a dual lefttwisted semigroup with respect to \(\varphi \). Moreover, it can be shown that the function \(\varphi \) is the unique function for making \((\mathbf{R}, *)\) a dual lefttwisted semigroup.
Proposition 2
There is no map \(\varphi : \mathbf{R}^2\rightarrow \mathbf{R}^2\) such that \((\mathbf{R}, )\) is a lefttwisted semigroup.
Proof
If we assume that there is a map \(\varphi : \mathbf{R}^2\rightarrow \mathbf{R}^2\) such that \((\mathbf{R}, )\) is a lefttwisted semigroup,
for the map \(\varphi (a,b)=(u,v)\). If we let \(c:=0\), then \(ab= uv\). Hence, by (7) we obtain \(c=c\), for all \(c\in \mathbf{R}\), a contradiction. \(\square \)
Note that Proposition 2 shows that \((\mathbf{R}, )\) is a groupoid which can not be a lefttwisted semigroup. Moreover, Proposition 2 shows that not every right(middle)twisted semigroup is a lefttwisted semigroup. Examples 1 and 3 together show that a groupoid can fail to be a semigroup and yet be a middletwisted as well as a righttwisted semigroup.
Theorem 1
If a lefttwisted semigroup has a right identity element, then it is a semigroup.
Proof
Let \(c:=e\) be the right identity element. Then by (1) we have \(a*b = a*(b*e) = (u*v)*e = u*v\) and thus \((u*v)*c = (a*b)*c\), so that \(a*(b*c) = (a*b)*c\), i.e., the groupoid is a semigroup as claimed. \(\square \)
Dually, if a righttwisted semigroup has a left identity element, then it is a semigroup.
Notice that \((\mathbf{R}, )\) has a right identity element 0, but it is not a semigroup. By Theorem 1 it cannot be lefttwisted, as shown in Proposition 2.
Proposition 3
The groupoid \((\mathbf{R}, )\) is a middletwisted semigroup with respect to \(\varphi \) if and only if \(u(a, c) + v(a, c) = a c\) for any \(a, c\in \mathbf{R}\).
Proof
Given \(a,b, c\in \mathbf{R}\), \((ab)c= u(bv)\), where \(u=u(a, b), v=v(a, b)\), means that \(abc = ub +v\), and hence \(u(a, c) + v(a, c) = a c\). The converse is straightforward. \(\square \)
If we let \(u:=a\alpha , v:=c +\alpha \) where \(\alpha =\alpha (a, c)\), then \((\mathbf{R}, )\) is a middletwisted semigroup.
Twisted semigroups in a field
Let \(X=(X,+,\cdot )\) be a field and \(\lambda , \mu \in X\) (not all zero). If we define a binary operation “\(*\)” on X as follows:
for any \(x, y\in X\), we call such a groupoid \((X,*)\) a linear groupoid over a field X. We define its associator function \(\varphi (a,b) :=(u, v)\), i.e., \(u=u(a, b), v=v(a, b)\) are functions of two variables \(a, b\in X\).
Example 5
Let \(\mathbf{R}=(\mathbf{R},+,\cdot )\) be a real field and \(\lambda \not = 0, \mu \in \mathbf{R}\). We define a binary operation “\(*\)” on \(\mathbf{R}\) as follows: \(x*y:= \lambda x + \mu y\) for any \(x, y\in \mathbf{R}\). If we define a map \(\varphi (a,b) :=(\frac{a}{\lambda }, b)\) and \(\mu ^2 =\mu \), then \((\mathbf{R},*)\) is a lefttwisted semigroup with respect to \(\varphi \).
Proposition 4
Let \((X,*)\) be a linear groupoid over a field X and its associator function defined by \(\varphi (a, b):=(\lambda a, b)\). If \((X,*)\) is a dual lefttwisted semigroup with respect to \(\varphi \), then it has the form \(x*y=\lambda x \) or \(x*y=\lambda x + y\).
Proof
Since \((X,*)\) is a dual lefttwisted semigroup with respect to \(\varphi \), \((a*b)*c = \lambda ^2 a + \lambda \mu b + \mu c\) and \(u*(v*c) = \lambda ^2 a + \mu \lambda b + \mu ^2 c\) for any \(a, b, c\in X\). It follows that \(\mu =\mu ^2\), proving the proposition. \(\square \)
In Proposition 4, if we let \(\lambda := 2\) and \(\mu :=1\), and we define \(\varphi (a,b):=(2a, b)\), then \((X,*)\) is certainly a dual lefttwisted semigroup with respect to \(\varphi \), but it is not a lefttwisted semigroup with respect to \(\varphi \), since \(a*(b*c)= 2a + 2b + c\) and \((u*v)*c = 8a + 2b + c\).
Example 6
Let \((\mathbf{R}, +, \cdot )\) be a real field. Define a binary operation “\(*\)” on \(\mathbf{R}\) as follows: \(x*y:= x + 2y, \forall x, y\in \mathbf{R}\). If we define a map \(\varphi (b, c):=(b, \frac{c}{2}), \forall a, b\in \mathbf{R}\), then \((\mathbf{R}, *)\) is a righttwisted semigroup with respect to \(\varphi \), but not a (dual) lefttwisted semigroup with respect to \(\psi (a, b):=(a, \frac{b}{2})\), since \((a*b)*c = a + 2b + 2c\), while \(u*(v*c) = a + b + 4c\) and \((u*v)*c = a + b + 2c\).
Note that Examples 5 and 6 show that twistedsemigroups may have role to play in the theory of linear groupoids in various algebraic structures (see Kim and Neggers 2008; Kim and So 2012; Neggers et al. 2001).
Proposition 5
Let \((X,*)\) be a linear groupoid over a field \((X, +, \cdot )\), i.e., \(x*y:= \lambda x + \mu y\) for all \(x, y\in X\) where \(\lambda , \mu \) are not all zero in X. If \((X,*)\) is a righttwisted semigroup with respect to \(\varphi (b, c) = (u, v)\), then it has one of the forms: (i) \(x*y= x \); (ii) \(x*y=\mu y\), \(\mu \not = 0\) and \(u=b, v =\frac{c}{\mu }\); (iii) \(x*y= x + \mu y\), \(\mu \not = 0\) and \(u=b, v= \frac{c}{\mu }\).
Proof
Since \((X,*)\) is a linear groupoid over X, there exist \(\lambda , \mu \in X\) (not all zero) such that \(x*y = \lambda x + \mu y\) for all \(x, y\in X\). Let \((X,*)\) be a righttwisted semigroup with respect to \(\varphi (b, c) = (u, v)\). Then \((a*b)*c = (\lambda a + \mu b)*c = \lambda (\lambda a + \mu b) + \mu c = \lambda ^2 a + \lambda \mu b + \mu c\) and \(a*(u*v) = a*(\lambda u + \mu v) = \lambda a + \mu (\lambda u + \mu v) = \lambda a + \lambda \mu u + \mu ^2 v\). It follows that
Assume that \(\mu \not = 0\). If we let \(u:= b, v:= \frac{c}{\mu }\), then
and hence \(\lambda ^2 a =\lambda a\), i.e., \(\lambda = 0\) or \(\lambda = 1\). If \(\lambda = 0\), then \(x*y = \mu y\). If \(\lambda \not = 0\), then \(x*y = x + \mu y\). Assume \(\mu = 0\). Then \(x*y = \lambda x\) and hence
It follows that \(\lambda = 0\) or \(\lambda = 1\), which shows that \(x*y = x\), since \((X,*)\) is a linear groupoid. \(\square \)
Proposition 6
Let \((X,*)\) be a linear groupoid over a field X. If we define a map \(\varphi (b, c):=(b, \mu c), \forall b, c\in X\), then \((X,*)\) is a dual righttwisted semigroup with respect to \(\varphi \) when \(\lambda ^2 = \lambda \).
Proof
Straightforward. \(\square \)
Theorem 2
Let \((X,*)\) be a linear groupoid over a field X, i.e., \(x*y:=\lambda x + \mu y\) for all \(x, y\in X\) and let \(\lambda \mu \not = 0\). If \((X,*)\) is both a lefttwisted semigroup with respect to a map \(\varphi (a, b) = (\frac{a}{\lambda }, b)\) and a righttwisted semigroup with respect to a map \(\psi (b, c) = (b, \frac{c}{\mu })\), then \((X,*)\) is an additive group of the field X.
Proof
Since it is a lefttwisted semigroup, given \(a, b, c\in X\), \(a*(b*c) = \lambda a + \lambda \mu b + \mu ^2 c\) and \((u*v)*c = \lambda ^2 u + \lambda \mu v + \mu c\) for some \(u= u(a, b), v=v(a, b)\). Hence we have
Similarly, since it is a righttwisted semigroup, we have
for some \(u^{\prime } = u^{\prime }(a, b), v^{\prime } = v^{\prime }(a, b)\). If we put \(u:= \frac{a}{\lambda }, v:= b\) in (11), then \(\mu ^2 c = \mu c\). It follows that \(\mu =1\). If we put \(u^{\prime }:= b, v^{\prime }:= \frac{c}{\mu }\) in (12), then \(\lambda ^2 a =\lambda a\), proving that \(\lambda = 1\). Hence \(x*y = x + y\), i.e., \((X,*)\) is an additive group of the field X. \(\square \)
Proposition 7
Let \((X,*)\) be a linear groupoid over a field X, i.e., \(x*y =\lambda x + \mu y\) for all \(x, y\in X\) where \(\lambda , \mu \) are not all zero. If we define a map \(\varphi (a, c) :=(\lambda a, \frac{1}{\mu }c), \forall a,c \in X\) where \(\mu \not = 0\), then \((X, *)\) is a middletwisted semigroup with respect to \(\varphi \).
Proof
Straightforward. \(\square \)
Example 7
Let \((\mathbf{R}, +, \cdot )\) be a real field. Define a binary operation “\(*\)” on \(\mathbf{R}\) by \(x*y := xy, \forall x, y\in \mathbf{R}\) and define a map \(\varphi (a, c) :=(a, c), \forall a, c \in \mathbf{R}\). Then \((\mathbf{R}, *)\) is a dual middletwisted semigroup with respect to \(\varphi \).
Twisted semigroups on groups
Proposition 8
Let \((G, \cdot )\) be a group and \(n\in \mathbf{N}\) (fixed). Define a binary operation “\(*\)” on G by \(a*b:= a^nb, \forall a, b\in G\). If we define a map \(\varphi (a, b):= (a^nb, e), \forall a, b\in G\) where e is the identity of G, then \((G,*)\) is a dual lefttwisted semigroup with respect to \(\varphi \).
Proof
Given \(a, b, c\in G\), we have \((a*b)*c = (a^nb)^nc\). Since \(\varphi (a, b):= (a^nb, e)\), if we let \(u:= a^nb, v:=e\), then \(u*(v*c)= u^n(v^nc)= (a^nb)^n(e^nc) = (a^nb)^nc\), proving the proposition. \(\square \)
Let n be a natural number. A group \((G, \cdot )\) is said to have the \(n{\rm th}\) power property if there are \(a, b\in G\) such that \(a^nb^n = x^n\) has no solution x in G. For example, consider the dihedral group \(D_4=\{r_0, r_1, r_2, r_3, h, v, d, t\} \) in (Hungerford 1990, p.158). It is easy to show that \(r_3^3\cdot v^3 = d\), but there is no element \(x\in D_4\) such that \(x^3= d\).
Proposition 9
Let \((G, \cdot )\) be a group having the \(n^{th}\) power property. Define a binary operation “\(*\)” on G by \(a*b:= a^nb, \forall a, b\in G\). Then \((G,*)\) is not a lefttwisted semigroup with respect to any mapping \(\varphi \).
Proof
Assume that \((G,*)\) is a lefttwisted semigroup with respect to some mapping \(\varphi \). Then for any \(a, b, c\in G\), there exist \(u= u(a, b), v = v(a, b)\) in G such that \(\varphi (a, b) = (u, v)\) and \(a*(b*c) = (u*v)*c\). This means that \(a^nb^nc = (u^nc)^nc\). Since \((G, \cdot )\) is a group, we obtain \(a^nb^n = (u^nv)^n\). If we let \(x:= u^nv\), then \(a^nb^n = x^n\) has a solution \(x=u^nv\), a contradiction. \(\square \)
Proposition 10
Let \((G, \cdot )\) be a group having the \(n^{th}\) power property. Define a binary operation “\(*\)” on G by \(a*b:= ab^n, \forall a, b\in G\). Then \((G,*)\) can not be a righttwisted semigroup with respect to any mapping \(\varphi \).
Proof
Assume that \((G,*)\) is a righttwisted semigroup with respect to \(\varphi \). Since \((G, \cdot )\) has the \(n^{th}\) power property, there are \(b, c\in G\) such that \(b^nc^n = x^n\) has no solution in G. Since \((G, *)\) is a righttwisted semigroup, for any \(a\in G\), there exist \(u=u(b, c), v=v(b, c)\) in G such that \((a*b)*c = a*(u*v)\). Hence \((ab^n)c^n = a(uv^n)^n\), i.e., \(b^nc^n = (uv^n)^n\). If we let \(x:= uv^n\), then \(b^nc^n = x^n\) has a solution, a contradiction. \(\square \)
Homomorphisms of twisted semigroups
Theorem 3
Let \((X, *)\) be a lefttwisted semigroup with respect to a map \(\varphi : X^2\rightarrow X^2\). If \(f: (X,*)\rightarrow (Y, \bullet )\) is an epimorphism of groupoids, i.e., \(f(x*y)= f(x)\bullet f(y), \forall x, y\in X\), then there exists a map \(\psi : Y^2\rightarrow Y^2\) such that \((Y, \bullet )\) is a lefttwisted semigroup with respect to the map \(\psi \).
Proof
Given \(\alpha , \beta \in Y\), since \(f: X\rightarrow Y\) is onto, there exist \(a, b\in X\) such that \(\alpha = f(a), \beta = f(b)\). Since \((X, *)\) is a lefttwisted semigroup with respect to a map \(\varphi \), for all \(c\in X\), there exist \(u_0= u_0(a, b), v_0 = v_0(a, b)\in X\) such that \(a*(b*c) = (u_0 * v_0)*c\) and \(\varphi (a, b) = (u_0, v_0)\). We define a set \(\Gamma _{(\alpha , \beta )}\) as follows:
If we let \(v:= f(u_0), w:= f(v_0)\), then \((v, w) \in \Gamma _{(\alpha , \beta )}\). We define \(\psi : Y^2 \rightarrow Y^2\) by \(\psi (\alpha , \beta ) := (v, w)\). Then it is welldefined. In fact, if we assume \(\psi (\alpha , \beta ) = (v, w)\) and \(\psi (\alpha , \beta )= (p, q)\), then there exist \(u_1, v_1 \in X\) such that \(p= f(u_1), q= f(v_1)\) and \((p, q)\in \Gamma _{(\alpha , \beta )}\). It follows that \(\varphi (a, b) = (u_1, v_1)\) and \(\varphi (a, b) = (u_0, v_0)\). Since \(\varphi \) is a mapping, we obtain \(p=v, q= w\), which proves that \(\psi : Y^2 \rightarrow Y^2\) is a mapping.
Given \(\gamma =f(c)\in Y\), since f is an epimorphism, we have
This proves that \((Y, \bullet )\) is a lefttwisted semigroup with respect to a map \(\psi : Y^2\rightarrow Y^2\). \(\square \)
Let \((X,*)\) and \((Y,\bullet )\) be groupoids. A map \(f: (X,*)\rightarrow (Y,\bullet )\) is said to be lefttwistedinjective if \(f(a*(b*c)) = f((u*v)*c)\), then there exist \(u^{\prime }\) and \(v^{\prime }\) in X such that \(a*(b*c) = (u^{\prime }*v^{\prime })*c\) and \(f(u)= f(u^{\prime }), f(v)= f(v^{\prime })\) where \(a, b, c, u, v\in X\). For example, the canonical group homomorphism \(\pi : G\rightarrow G/N\) is lefttwistedinjective.
Proposition 11
Let \((Y, \bullet )\) be a lefttwisted semigroup with respect to a map \(\psi \). If \(f: (X,*)\rightarrow (Y, \bullet )\) is a lefttwistedinjective epimorphism, then there exists a map \(\varphi : X^2\rightarrow X^2\) such that \((X, *)\) is a lefttwisted semigroup with respect to the map \(\varphi \).
Proof
For any \(a, b, c\in X\), since \(f:X\rightarrow Y\) is onto, there exist \(\alpha , \beta , \gamma \in Y\) such that \(f(a) = \alpha , f(b) = \beta , f(c) = \gamma \). Since \((Y,\bullet )\) is a lefttwisted semigroup, there exist \(v, w\in Y\) such that \(\alpha \bullet (\beta \bullet \gamma ) = (v\bullet w)\bullet \gamma \) and \(\psi (\alpha , \beta ) = (v, w)\). Since f is onto, there exist \(p, q\in X\) such that \(f(p)= v, f(q) = w\) and hence \(f(a)\bullet (f(b)\bullet f(c)) = (f(p)\bullet f(q))\bullet f(c)\). Since f is lefttwistedinjective, we have
for some \(p^{\prime }, q^{\prime }\in X\) where \(f(p) = f(p^{\prime }), f(q) = f(q^{\prime })\). Given \(a, b\in X\), we obtain \(p^{\prime }, q^{\prime }\in X\) satisfying (13), which means that \(p^{\prime }, q^{\prime }\in X\) are determined by choosing \(a, b\in X\), i,e., there exists a map \(\varphi : X^2\rightarrow X^2\) such that \(\varphi (a, b) = (p^{\prime }, q^{\prime })\). Hence \((X,*)\) is also a lefttwisted semigroup with respect to \(\varphi \). \(\square \)
In Theorem 3, given \(a, b, c\in X\), there exist \(u, v\in X\) such that \(a*(b*c)= (u*v)*c\) where \(\varphi (a, b) = (u, v)\). Since \(f: (X, *)\rightarrow (Y, \bullet )\) is an epimorphism, we have \(f(a)\bullet (f(b)\bullet f(c))= f(a*(b*c)) = f((u*v)*c) = (f(u)\bullet f(v))\bullet f(c)\). Now, since \((Y,\bullet )\) is a lefttwisted semigroup with respect to \(\psi \), we obtain \(p, q\in Y\) such that \(f(a)\bullet (f(b)\bullet f(c)) = (p\bullet q)\bullet f(c)\) where \(\psi (f(a), f(b))= (p, q)\). It follows that \((p\bullet q)\bullet f(c)=(f(u)\bullet f(v))\bullet f(c)\).
Using the notion of this concept, we introduce an equivalence relation on the Cartesian product \(Y\times Y\) of any groupoid \((Y, \bullet )\) (not necessarily a (lefttwisted) semigroup). Let \((X,*)\) and \((Y,\bullet )\) be groupoids and let \(f:X\rightarrow Y\) be a map. Define a relation “\(\equiv \)” on \(Y^2\) using f by
where \(\alpha , \beta , \gamma , \delta \in Y\). Then it is easy to show that \(\equiv (\hbox {mod}\, f)\) is an equivalence relation on \(Y^2\). Hence \(Y^2\) is partitioned into equivalence classes \([(\alpha , \beta )]:= \{(\gamma , \delta )\,\, (\alpha , \beta ) \equiv (\gamma , \delta ) (\hbox {mod}\, f)\}\).
Example 8
Consider a dalgebra \((X,*, 0)\) and a BCKalgebra \((Y,\bullet , 0)\) as follows:
If we define a map \(f: X\rightarrow Y\) by \(f(0)= 0, f(1)= b, f(2)=a\), then \(Y^2 = \{[(0,0)], [(a,0)], [(b, a)]\}\) where \([(0, 0)] =\{(0, 0), (0, a), (0, b), (a, a),\) \( (b, b)\}\), \([(a, 0)]=\{(a, 0), (a, b)\}, [(b, a)]=\{(b, a), (b, 0)\}\).
Thus, let \(f: (X,*)\rightarrow (Y,\bullet )\) be an epimorphism of groupoids, where \((X,*)\) and \((Y,\bullet )\) are lefttwisted semigroups with respect to \(\varphi \) and \(\psi \) respectively, and let \(\varphi (a, b) = (u, v)\) for some \(u, v\in X\) and \(\psi (f(a), f(b))= (p, q)\) for some \(p, q\in Y\). Then \(a*(b*c) = (u*v)*c\) and \(f(a)\bullet (f(b)\bullet \gamma ) = (p\bullet q)\bullet \gamma \) for any \(c\in X, \gamma \in Y\). Since f is an epimorphism, we have \(f(a)\bullet (f(b)\bullet f(c)) = (f(u)\bullet f(v))\bullet f(c)\) and hence \((f(u)\bullet f(v))\bullet \gamma = (p\bullet q)\bullet \gamma \) for any \(\gamma \in Y\). It follows that \((f(u), f(v))\equiv (p, q)(\hbox {mod}\, f)\). We summarize:
Proposition 12
Let \(f: (X,*)\rightarrow (Y, \bullet )\) be an epimorphism of groupoids, where \((X,*)\) and \((Y,\bullet )\) are lefttwisted semigroups with respect to \(\varphi \) and \(\psi \) respectively. If \(\varphi (a, b) = (u, v)\), then \(\psi (f(a), f(b)) \equiv (f(u), f(v)) (\hbox {mod}\, f)\).
Given an epimorphism \(f: (X,*)\rightarrow (Y,\bullet )\) as in Proposition 12, we define a new map \(\widetilde{f}: X^2\rightarrow Y^2\) by \(\widetilde{f}(x_1, x_2):=(f(x_1), f(x_2)), \forall x_1, x_2\in X\). Using this map we obtain \((\psi \circ \widetilde{f})(a, b) = \psi ( \widetilde{f}(a, b))=\psi (f(a), f(b)))= (p, q)\) and \((\widetilde{f}\circ \varphi )(a,b) = \widetilde{f}(\varphi (a, b)))= \widetilde{f}(u, v) = (f(u), f(v))\) where “\(\circ \)” is a composition of functions. Hence \(\psi \circ \widetilde{f}(a, b) \equiv \widetilde{f}\circ \varphi (a, b) (\hbox {mod}\, f)\), i.e., the following diagram is a commuting diagram:
For the other types of twisted semigroups we have similar situations. In the case of a dual lefttwisted semigroup, we obtain the following results.
Theorem 3 \(^{\prime }\). Let \((X, *)\) be a dual lefttwisted semigroup with respect to a map \(\varphi : X^2\rightarrow X^2\). If \(f: (X,*)\rightarrow (Y, \bullet )\) is an epimorphism of groupoids, i.e., \(f(x*y)= f(x)\bullet f(y), \forall x, y\in X\), then there exists a map \(\psi : Y^2\rightarrow Y^2\) such that \((Y, \bullet )\) is a dual lefttwisted semigroup with respect to the map \(\psi \).
Let \((X,*)\) and \((Y,\bullet )\) be groupoids. A map \(f: (X,*)\rightarrow (Y,\bullet )\) is said to be dual lefttwistedinjective if \(f((a*b)*c)) = f(u*(v*c))\), then there exist \(u^{\prime }\) and \(v^{\prime }\) in X such that \((a*b)*c = u^{\prime }*(v^{\prime }*c)\) and \(f(u)= f(u^{\prime }), f(v)= f(v^{\prime })\) where \(a, b, c, u, v\in X\).
Proposition 11 \(^{\prime }\). Let \((Y, \bullet )\) be a dual lefttwisted semigroup with respect to \(\psi \). If \(f: (X,*)\rightarrow (Y, \bullet )\) is a dual lefttwistedinjective epimorphism, then there exists a map \(\varphi : X^2\rightarrow X^2\) such that \((X, *)\) is a dual lefttwisted semigroup with respect to the map \(\varphi \).
The corresponding relationship on \(Y^2\) is now the following. Let \((X,*)\) and \((Y,\bullet )\) be groupoids and let \(f:X\rightarrow Y\) be a map. Define a relation “\(\equiv \)” on \(Y^2\) using f by
where \(\alpha , \beta , \gamma , \delta \in Y\). Then it is easy to show that \(\equiv (\hbox {mod}\, f)^{*}\) is an equivalence relation on \(Y^2\). Hence \(Y^2\) is partitioned into equivalence classes \([(\alpha , \beta )]^{*}:= \{(\gamma , \delta )\,\, (\alpha , \beta ) \equiv (\gamma , \delta ) (\hbox {mod}\, f)^{*}\}\).
The other four types go in entirely the same way. We list the relevant information.
 (1):

righttwisted semigroups: \(\psi (\alpha , \beta )\) is defined as previous cases, and the equivalence relation on \(Y^2\) is \((\alpha , \beta ) \equiv (\gamma , \delta ) [\hbox {mod}\, f]\) provided for all \(x\in X\), \(f(x)\bullet (\alpha \bullet \beta ) = f(x)\bullet (\gamma \bullet \delta )\).
 (2):

dual righttwisted semigroups: \(\psi (\alpha , \beta )\) is defined as previous cases, and the equivalence relation on \(Y^2\) is \((\alpha , \beta ) \equiv (\gamma , \delta ) [\hbox {mod}\, f]^{*}\) provided for all \(x\in X\), \((f(x)\bullet \alpha )\bullet \beta = (f(x)\bullet \gamma )\bullet \delta \).
 (3):

middletwisted semigroups: \(\psi (\alpha , \beta )\) is defined as previous cases, and the equivalence relation on \(Y^2\) is \((\alpha , \beta ) \equiv (\gamma , \delta ) <\hbox {mod}\, f>\) provided for all \(x\in X\), \(\alpha \bullet (f(x)\bullet \beta ) = \gamma \bullet (f(x)\bullet \delta )\).
 (4):

dual middletwisted semigroups: \(\psi (\alpha , \beta )\) is defined as previous cases, and the equivalence relation on \(Y^2\) is \((\alpha , \beta ) \equiv (\gamma , \delta ) <\hbox {mod}\, f>^{*}\) provided for all \(x\in X\), \((\alpha \bullet f(x))\bullet \beta = (\gamma \bullet f(x))\bullet \delta \).
Given one of the six types of twisted semigroups, let \(\{(X_{\alpha }, *_{\alpha }, \varphi _{\alpha })\}_{\alpha \in I}\) be an indexed family of one of these types of twisted semigroups. Let \(X=\prod _{\alpha \in I}X_{\alpha }\) be the direct product and let \(\pi _{\alpha }: X \rightarrow X_{\alpha }\) be the canonical surjection. Let X be equipped with the product binary operation given by the formula \((x_{\alpha })*(y_{\alpha }) = (x_{\alpha }*_{\alpha } y_{\alpha })\). If we define a map \(\varphi ((a_{\alpha }), (b_{\alpha })) = (\varphi _{\alpha }(a_{\alpha }, b_{\alpha }))= ((u_{\alpha }, v_{\alpha }))\), then it follows that \((X,*)\) is a twisted semigroup of the same type with respect to the map \(\varphi \).
In order to obtain varieties we must be able to claim that if \((X,*)\) is a twisted semigroup of a given (one of the six) type(s) with respect to a map \(\varphi \) and if \((A,*)\) is a subgroupoid of \((X,*)\), then it is also of the same type, i.e., there is a function \(\psi : A^2 \rightarrow A^2\) (rather than \(\psi : A^2 \rightarrow X^2, \psi =\) \({\varphi \mid A^2}\)) which satisfies the required identity belonging to the special type in question. We give a counterexample that a subgroupoid \((A,*)\) of a dual lefttwisted semigroup \((X,*)\) need not be a dual lefttwisted semigroup.
Example 9
In Example 4, let Z be the collection of all integers. Then \((Z,*)\) is a subgroupoid of \((\mathbf{R},*)\) and \((\mathbf{R}, *)\) is a dual lefttwisted semigroup with respect to \(\varphi \). If we let \(a:=13, b:=12\), then \(v^3= 13(1312)(v^21)=13(v^21)\). Assume v is an integer. Then \(v= 13k\) for some integer k. This means that \((13)^2k^2(1k) = 1\), which is impossible for an integer k. Hence there is no mapping \(\varphi : Z^2 \rightarrow Z^2\) such that \((Z,*)\) is a dual lefttwisted semigroup with respect to \(\varphi \).
Conclusions
In this paper, we introduced the concept of several types of groupoids related to semigroups, viz., twisted semigroups for which twisted versions of the associative law hold. Besides a number of examples and a discussion of homomorphisms, a class of groupoids of interest was the class of groupoids defined over a field \((X,+, \cdot )\) via a formula \(x*y=\lambda x + \mu y\), with \(\lambda , \mu \in X\), fixed structure constants. Properties of these groupoids as twisted semigroups were discussed with several results of interest obtained, e.g., that in this setting simultaneous lefttwistedness and righttwistedness of \((X,*)\) implies the fact that \((X,*)\) is a semigroup.
In the investigation of “residual associativity” in groupoids one encounters a number of levels. The strongest version of such residual associativity is “associativity” itself. Making a study of twisted semigroups, besides being of interest in itself, is also relevant in several other ways. One may wish to determine as precisely as possible how different twisted semigroups may differ from semigroups. Accordingly we note that the study of twisted semigroups, commenced in this paper, will prove to be a rich area for research both in itself and as embedded in the area of the study of “general theory of groupoids”.
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Allen, P.J., Kim, H.S. & Neggers, J. Several types of groupoids induced by twovariable functions. SpringerPlus 5, 1715 (2016). https://doi.org/10.1186/s400640163411y
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Keywords
 Groupoid
 (Twisted) semigroup
 Linear groupoid over a field
 \(n\mathrm{th}\) power property
 Homomorphism
Mathematics Subject Classification
 Primary 20N02
 20M99