PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 99(113) (2016), 15–30 DOI: 10.2298/PIM1613015D

RESIDUAL CHARACTER OF QUASILINEAR VARIETIES OF GROUPOIDS

Petar Ðapić and Petar Marković

This paper is dedicated to Professor Jaroslav “Jarda” Ježek, coauthor, mentor and friend, who passed away five years ago.

Abstract. We consider the quasilinear varieties of groupoids which were characterized in [4] and find the residual character for all of them. Those varieties which are residually small turn out to be residually finite. We compute the residual bounds and find all subdirectly irreducible algebras in them.

1. Introduction

In this paper we complete the investigation of quasilinear varieties of groupoids started in [3] and [4] by investigating the residual character of those varieties. We have already determined the residual character of linear varieties in [3], so here we deal with quasilinear varieties which are not linear.

Finding the residual bound of finitely generated varieties is related to the Restricted Quackenbush Conjecture. The Restricted Quackenbush Conjecture claims that no finitely generated variety in a finite language has residual bound exactly ω (i.e. it is impossible for such varieties to contain only finite subdirectly irreducible algebras, but of unbounded size). All quasilinear varieties were shown to be finitely generated in [4]. Here we will prove that all quasilinear varieties have either a finite residual bound or are residually large. Thus, we establish that there are no counterexamples to the Restricted Quackenbush Conjecture among quasilinear varieties of groupoids. For those quasilinear varieties which are residually finite we find all subdirectly irreducible algebras that contain them, thus semantically describing them up to Horn formulae (rather than just identities). For the residually large quasilinear varieties, we had to invent new constructions to prove residual largeness, which may be useful in future research.

2010 Mathematics Subject Classification: Primary 08B26; Secondary 08B15, 08A45, 08A30, 08A62.

Key words and phrases: variety, equational theory, linear term, subdirectly irreducible, residually large, residually finite.

Supported by the grant 174018 of the Ministry of Education and Science of Serbia.

Communicated by Žarko Mijajlović.

We introduce some of the definitions, notation and basic facts about universal algebra which we will need in this paper. This is not comprehensive, the reader is expected to know standard universal algebra (e.g. free algebras, varieties etc.) in order to follow the arguments in our paper. Good textbooks are [2, 7, 1].

Let F be an algebraic language. We say that an F-term is linear if it is a variable, or equal to f(x_{1}, …, x_{n}), where f is an operation symbol of arity n and x_{1}, …, x_{n} are variables (thus constant symbols are also linear terms, being nullary operation symbols).

In the previous papers [4] and [3] we introduced equational theories which are represented on linear terms, as defined below.

Definition 1.1. An equational theory E is linear if any term is E-equivalent to a unique linear term. E is quasilinear if it is idempotent and each term is Eequivalent to a linear term (not necessarily unique). The variety of all models of a linear (quasilinear) equational theory is called a linear (quasilinear) variety.

Note that we called these properties ∗-linear and ∗-quasilinear in [3] and [4]. This was because we needed to distinguish n-linear and n-quasilinear for the equational theories which have the above properties just for terms with up to n variables. Now we don’t need these weaker properties and thus we simplify our terminology.

We recall some basic facts about simple and subdirectly irreducible algebras. An algebra is simple if it has more than one element and only the two trivial congruences, the equality and full relation. An algebra is subdirectly irreducible if it has more than one element and the least congruence which is not the equality relation. The least congruence different from the equality relation is called the monolith of a subdirectly irreducible algebra. A class of algebras is residually small if it has a cardinal bound on the size of subdirectly irreducible algebras in the class. The class of algebras is residually large otherwise. The property of residual smallness/largeness is called the residual character of the class of algebras in question. The residual bound of a class of algebras K, denoted by resb(K) is the least cardinal κ such that all subdirectly irreducible algebras in K are of size strictly smaller than κ, if K is residually small, or resb(K) = ∞ if K is residually large.

Let A be an algebra. We use the standard notation ConA for the set of all congruences and Con A for the lattice of all congruences of an algebra A. For X ⊆ A × A, Cg^{A} (X) is the congruence of A generated by X. When X = {⟨a, b⟩}, we use Cg^{A} (a, b) as short notation for the principal congruence Cg^{A} (X).

Let A be an algebra. A basic translation is either the identity map id_{A} or the operation p(x) = f^{A} (a_{1}, …, a_{i−1} , x, a_{i+1} , …, a_{n}), where f ∈ F is an operation symbol of arity n ⩾ 1, 1 ⩽ i ⩽ n and a_{j} ∈ A for all j. A translation is a composition of basic translations. The set of all translations of A is denoted by TrA. In the case of a groupoid G = (G; ·), the basic translations are id_{G}, the maps λ_{a}(x) = ax and ρ_{a}(x) = xa for all a ∈ G.

In general, Cg^{A} (X) is the transitive closure of the set

{⟨u, v⟩ ∈ A^{2} :(∃⟨x, y⟩ ∈ X)(∃p(x) ∈ TrA)({u, v} = {p(x), p(y)})}.

Some authors call this a Mal’cev chain, each pair of the form {p(x), p(y)} forming a link in the chain.

For shorter notation, we will write x_{1}x_{2}x_{3} …x_{n} instead of (((x_{1}x_{2})x_{3}) …)x_{n} (the parentheses are grouped to the left), x · yz instead of x(yz), etc. By a leftassociated term we mean a term x_{1}x_{2} …x_{n} where n ⩾ 1 and x_{1}, …, x_{n} are variables. We finish the section by introducing some more notation.

Let G be a nonvoid set, and κ a cardinal. For any a, b ∈ G and j ∈ κ, the element a_{j↦b}^{κ} of G^{κ} is defined by

a_{j↦b}^{κ} (i) = \{ \begin{matrix}a, & for all i ∈ κ ∖ \{j\}; \\ b, & for i = j. \end{matrix}

Let α ∈ G^{κ} and e, f ∈ G. Then the element α_{e↦f} in G^{κ} is defined by

α_{e↦f} (i) = \{ \begin{matrix}α(i), & if α(i) ≠ e; \\ f, & if α(i) = e. \end{matrix}

2. Our result, statement and easy cases

We recall the poset of all quasilinear varieties of groupoids, which was discovered in [4].

[figure: S_{5}; S_{4}; S_{3}; S_{2}; S_{1}; N_{2}; N_{1}; L_{1}; L_{2}; L_{3}; V_{C}; V_{A}; V_{B}; S_{3}^{∂}; S_{2}^{∂}; N_{2}^{∂}; S_{1}^{∂}; N_{1}^{∂}; L_{1}^{∂}; L_{2}^{∂}; L_{3}^{∂}; V_{C}^{∂}; V_{A}^{∂}; V_{B}^{∂}; W; W^{∂}; D; R]

Figure 1. The poset of all ∗-quasilinear varieties.

In this paper we prove that these varieties have the residual characters as depicted in Table 1. First we get the easy cases out of the way.

Table 1. Residual bounds of idempotent ∗-quasilinear varieties.

Variety

Residual bound

S_{5}  2
S_{3}, S_{3}^{∂}, S_{4}, R,  3
S_{2}, S_{2}^{∂}, V_{C}, V_{C}^{∂}, D,  4
W, W^{∂}  5

L_{1}, L_{2}, L_{3}, N_{1}, N_{2}, S_{1}, V_{A}, V_{B}, residually large L_{1}^{∂}, L_{2}^{∂}, L_{3}^{∂}, N_{1}^{∂}, N_{2}^{∂}, S_{1}^{∂}, V_{A}^{∂}, V_{B}^{∂},

If A = ⟨A; ·⟩ is a groupoid, then A^{∂} = ⟨A; ∗⟩ such that x ∗ y = y · x for all x, y ∈ A, while V^{∂} stands for the variety {A^{∂} : A ∈ V}. Clearly, V and V^{∂} have the same residual bounds. Recall that all subdirectly irreducible groupoids in the variety S_{2} (the variety of idempotent semigroups which satisfy xyz ≈ xzy) were described in [6]. Thus we know that resb(S_{2}) = 4, and that its subvarieties S_{3} (the left-zero semigroups, i.e. all models of xy ≈ x) and S_{4} (semilattices) have only the two-element subdirectly irreducible algebras. Also, it is well known (and easy to prove) that the variety of rectangular bands R has only the two-element leftzero semigroup and the two-element right-zero semigroup as nontrivial subdirectly irreducibles. Therefore, resb(S_{4}) = resb(S_{3}) = resb(R) = 3. The trivial variety S_{5} of one-element groupoids is conventionally given the residual bound 2.

We also know that the variety S_{1} of idempotent semigroups, which satisfy xyx ≈ xy, contains subdirectly irreducible algebras of size at least κ where κ is any cardinal (cf. Section 3 of [6], and replace the set of natural numbers with κ). Thus all varieties which contain S_{1} are residually large, and those are N_{1}, L_{1}, L_{2} and L_{3}.

The variety D is the class of all models of the identities

(D1) xx ≈ x  (D2) xy ≈ yx
(D3) x · xy ≈ y  (D4) xy · zt ≈ xz · yt

It is proved in [4], Theorem 9.7, that the variety D is ∗-quasilinear and generated by the three-element groupoid G_{7}.

\begin{matrix}G_{7} & a & b & c \\ a & a & c & b \\ b & c & b & a \\ c & b & a & c \end{matrix}

Since the equational base of the variety D contains the identities (D1) xx ≈ x, (D2) xy ≈ yx, (D3) x · xy ≈ y, it follows that D is a variety of Steiner quasigroups. Actually it is the least such nontrivial variety. The subdirectly irreducible Steiner quasigroups were investigated in the influential paper [9], but since the variety D is not generated by the so-called planar ones, we would have to work through the proofs given in [9] to realize that resb(D) = 4.

The easier way is to realize that (D1) − (D4) are alternative axioms for the subvariety denoted by V_{3} of the variety of SIE groupoids in the terminology of B. Roszkowska-Lech. This V_{3} is a variety which is axiomatized by (D1), (D4) and

(D2′) x ≈ y · xy (D3′) xyy ≈ x

in place of (D2) and (D3). We only need to prove (D2′) and (D3′) in D, though it can be proved that (D2) and (D3) hold in V_{3}, too. (D3′) follows trivially from commutativity and (D3), while x ≈_{(D3)} y · yx ≈_{(D2)} y · xy. Then we invoke [10], where it is proved that all algebras in V_{k} for odd k are polynomially equivalent to Abelian groups which additionally satisfy kx = 0. Thus the only subdirectly irreducible algebra in D = V_{3} is the one polynomially equivalent to the threeelement Abelian group, and resb(D) = 4. The reader is referred to papers [11] and [12] for more details, and to [5] for yet another approach to the same variety.

It remains to find the residual bounds of N_{2}, V_{A}, V_{B}, V_{C} and W, which we will do in the following sections.

3. The varieties N _{2}, V_{A} and V_{B} are residually large

We first consider the variety N_{2}, given by its equational base

(N1) xx ≈ x  (N2) x · yx ≈ xy
(N3) x · xyz ≈ x · yz  (N4) x · yz ≈ x · zy

Lemma 3.1. The variety N_{2} is generated by the subdirectly irreducible groupoid G_{N_{2}}

\begin{matrix}G_{N_{2}} & a & b & c & d \\ a & a & d & c & d \\ b & b & b & b & b \\ c & c & b & c & b \\ d & d & d & d & d \end{matrix}

Proof. One can directly verify that the groupoid G_{N_{2}} satisfies the identities (1)–(4). The groupoid G_{N_{2}} is subdirectly irreducible with the monolith equal to the congruence Cg^{G_{N_{2}}} (b, d). The groupoid G_{N_{2}} is not a semigroup, since acb ≠ a · cb. Therefore, V(G_{N_{2}} ) ⊈ S_{2}, but V(G_{N_{2}} ) ⊆ N_{2}. From the fact that the varieties depicted in Figure 1 are a down-set (order ideal) in the lattice of all groupoid varieties, we deduce that V(G_{N_{2}} ) can be only N_{2}. □

Theorem 3.1. The variety N_{2} is residually large.

Proof. The groupoid G_{N_{2}} generates N_{2}, according to Lemma 3.1. Let κ be a cardinal. Consider the groupoid G := (G_{N_{2}} )^{κ} and denote by b^{κ} and d^{κ} the elements of the groupoid G such that b^{κ}(i) = b and d^{κ}(i) = d for all i ∈ κ. Define the relation ρ on the groupoid G_{N_{2}}^{κ} by: x ρ y iff one of the following three conditions hold:

(1) x = y,
(2) {x, y} = {b^{κ}, d^{κ}},

(3) {x, y} ∩ {b^{κ}, d^{κ}} = ∅, d ∈ x(κ), d ∈ y(κ), x(κ) ⊈ {a, d}, y(κ) ⊈ {a, d}
and for each i ∈ κ, either {x(i), y(i)} = {b, d}, or x(i) = y(i).

The relation ρ defined above is an equivalence relation and has at least 2^{κ} equivalence classes. Next we are showing that ρ is a congruence of G. It suffices to show that the following holds for all x, y, z ∈ G: If xρy, then xzρyz and zxρzy.

Let xρy and let z ∈ G be arbitrary. By the definition of the relation ρ, it follows that {i ∈ κ : x(i) ≠ y(i)} = {i ∈ κ : {x(i), y(i)} = {b, d}}. The fact that ub = ud in G_{N_{2}} for all u ∈ {a, b, c, d} implies that zx = zy.

Now we show ⟨xz, yz⟩ ∈ ρ. If x = b^{κ} and y = d^{κ}, then xz = x and yz = y, as b and d are left zeros of the groupoid G_{N_{2}} . Let x and y be distinct elements of G and {x, y} ∩ {b^{κ}, d^{κ}} = ∅. Then d ∈ x(κ) ∩ y(κ) and since d is a left zero in G_{N_{2}} , thus d ∈ xz(κ) ∩ yz(κ). Since xρy, it follows that {b, c} ∩ x(κ) ≠ ∅, but since {b, c} · G = {b, c}, it follows that {b, c} ∩ xz(κ) ≠ ∅, i.e. xz(κ) ⊈ {a, d}. Analogously we show that yz(κ) ⊈ {a, d}. If x(i) ≠ y(i), then {x(i), y(i)} = {b, d}, from which it follows that {xz(i), yz(i)} = {b, d}. Therefore, {xz(i), yz(i)} ∈ {{a}, {b}, {c}, {d}, {b, d}}, for all i ∈ κ. This proves that ρ is a congruence of G.

Note: the set {b, d}^{κ} is the union of two ρ-classes. One is {b^{κ}, d^{κ}}, and the other one is all other elements of {b, d}^{κ}. In other words, for all x, y ∈ G, if x(κ) = y(κ) = {b, d}, then xρy.

Next we show that G⧸ρ is subdirectly irreducible. Let the congruence θ ∈ ConG be such that ρ ⊊ θ.

We want to show that there exists p ∈ G which satisfies p(κ) = {b, d} such antecedent that ⟨b^{κ}, p⟩ ∈ θ. Considering the discussion on ρ restricted to {b, d}^{κ}, this would imply that every such θ has {b, d}^{κ} in a single θ-class, and the congruence (ρ ∨ Cg^{G} (b^{κ}, p))⧸ρ (which is the same no matter which p is chosen) would be the monolith of G⧸ρ.

Let x, y ∈ G be such that ⟨x, y⟩ ∈ θ ∖ ρ. We consider all cases:

Case 1: There exists j ∈ κ such that {x(j), y(j)} ∈ P_{2}({a, b, c, d}) ∖ {b, d}, where P_{2}(X) is the set of all subsets of X with exactly two elements. Then b_{j↦a}^{κ} x(i) = b_{j↦a}^{κ} y(i) = b for all i ∈ κ∖{j} and {b_{j↦a}^{κ} x(j), b_{j↦a}^{κ} y(j)} ∈ P_{2}({a, c, d}). If c ∈ {b_{j↦a}^{κ} x(j), b_{j↦a}^{κ} y(j)}, then

{b_{j↦a}^{κ} xd^{κ}, b_{j↦a}^{κ} yd^{κ}} = {b^{κ}, b_{j↦d}^{κ} },

therefore, ⟨b^{κ}, b_{j↦d}^{κ} ⟩ ∈ θ. If {b_{j↦a}^{κ} x(j), b_{j↦a}^{κ} y(j)} = {a, d}, then

{b_{j↦a}^{κ} xc^{κ}d^{κ}, b_{j↦a}^{κ} yc^{κ}d^{κ}} = {b^{κ}, b_{j↦d}^{κ} },

and hence ⟨b^{κ}, b_{j↦d}^{κ} ⟩ ∈ θ. Thus, b_{j↦d}^{κ} can be chosen for p.

Case 2: For each i ∈ κ,

{x(i), y(i)} ∈ {{a}, {b}, {c}, {d}, {b, d}}.

Our assumptions about x and y imply that x ≠ y, {x, y} ≠ {b^{κ}, d^{κ}}.

If x = b^{κ}, then y(κ) = {b, d}, so b^{κ} = x θ y and y can be chosen for the element p which we need. If x = d^{κ}, then b^{κ}ρ d^{κ} = x θ y and y can be chosen for p, as ρ ⊆ θ.

Let {x, y} ∩ {b^{κ}, d^{κ}} = ∅. Suppose that d ∉ x(κ) and d ∉ y(κ). Assumptions of Case 2 imply that x = y, a contradiction with ⟨x, y⟩ ∉ ρ. Let d ∈ x(κ) and d ∉ y(κ). Then there exists some j ∈ κ such that x(j) = d and y(j) = b. Thus b^{κ} = yy_{a↦c} b^{κ}θ xy_{a↦c} b^{κ} and since xy_{a↦c} b^{κ}(κ) = {b, d}, we can take xy_{a↦c} b^{κ} for p. The case d ∉ x(κ) and d ∈ y(κ) is analogous.

Now let {x, y} ∩ {b^{κ}, d^{κ}} = ∅ and d ∈ x(κ), d ∈ y(κ). Assume that x(κ) ⊆ {a, d}. Then b^{κ}ρ d^{κ} = xb^{κ}θ yb^{κ}, and since yb^{κ}(κ) = {b, d}, it follows that yb^{κ} can be taken for p. The case when y(κ) ⊆ {a, d} is analogous. All remaining cases for ⟨x, y⟩ are already in ρ.

Therefore, the groupoid G⧸ρ is subdirectly irreducible, and the congruence Cg^{G⧸ρ} (b^{κ}⧸ρ, b_{j↦d}^{κ} ⧸ρ) is its monolith. □

We turn to the variety V_{A} = HSP (A). According to [4], more precisely the definition of V_{A} and Theorem 7.15, it is the variety generated by

\begin{matrix}A & a & b & c & d & e \\ a & a & a & a & a & a \\ b & b & b & b & b & b \\ c & d & e & c & c & c \\ d & d & d & d & d & d \\ e & e & e & e & e & e \end{matrix}

whose equational base is

(A1) xx ≈ x  (A2) xyy ≈ xy
(A3) x · yz ≈ xy  (A4) xyzy ≈ xyz

Theorem 3.2. The variety V_{A} = HSP (A) is residually large.

Proof. Let κ be any cardinal. Consider the groupoid A^{κ} and the set G = {d_{i↦u}^{κ} | u ∈ {a, b, c}, i ∈ κ} ∪ {d, e}^{κ}. Then G is a subgroupoid of A^{κ}. Define the relation ρ on the groupoid G by: x ρ y iff one of the conditions

(1) x = y, or

(2) {x, y} ∈ {d, e}^{κ} ∖ {d^{κ}}.

is satisfied. The equivalence relation ρ has κ equivalence classes. Let us show that ρ is a congruence of G. It suffices to prove for all x, y, z ∈ G, if xρy, then xzρyz and zxρzy. Since d and e are left zeros and right neutral elements of A, it follows that ρ is a congruence of G.

Next we show that G⧸ρ is subdirectly irreducible. Let θ ∈ ConG satisfy ρ ⊊ θ. We aim to show that there exists p ∈ {d, e}^{κ} ∖ {d^{κ}} such that ⟨d^{κ}, p⟩ ∈ θ. Note that {d^{κ}} and {d, e}^{κ} ∖ {d^{κ}} are ρ-classes. If we show the claim, then it will follow that the pair ⟨{d, e}^{κ} ∖ {d^{κ}}, {d^{κ}}⟩ is contained in any congruence of G⧸ρ which is not the equality relation, i.e. it will follow that G⧸ρ is subdirectly irreducible.

Let ⟨x, y⟩ ∈ θ ∖ ρ. We consider the following cases for ⟨x, y⟩:

Case 1: There exists j ∈ κ such that ⟨x(j), y(j)⟩ ∈ {a}×{b, c, d, e}∪{b, c, d, e}× {a}. Then {d_{j→c}^{κ} x, d_{j→c}^{κ} y} = {d^{κ}, d_{j→c}^{κ} } or {d_{j→c}^{κ} x, d_{j→c}^{κ} y} = {d^{κ}, d_{j→e}^{κ} }. Hence,

⟨(d_{j↦c}^{κ} x)d_{j↦b}^{κ} , (d_{j↦c}^{κ} y)d_{j↦b}^{κ} ⟩ ∈ {⟨d^{κ}, d_{j↦e}^{κ} ⟩, ⟨d_{j↦e}^{κ} , d^{κ}⟩},

and thus ⟨d^{κ}, d_{j↦e}^{κ} ⟩ ∈ Cg^{G} (x, y) ⊆ θ, so we take p = d_{j↦e}^{κ} .

Case 2: There exists j ∈ κ such that ⟨x(j), y(j)⟩ ∈ {b}×{c, d, e}∪{c, d, e}×{b}. Then {d_{j→c}^{κ} x, d_{j→c}^{κ} y} = {d_{j→e}^{κ} , d_{j→c}^{κ} }. Hence,

⟨(d_{j↦c}^{κ} x)d_{j↦a}^{κ} , (d_{j↦c}^{κ} y)d_{j↦a}^{κ} ⟩ ∈ {⟨d^{κ}, d_{j↦e}^{κ} ⟩, ⟨d_{j↦e}^{κ} , d^{κ}⟩}.

Therefore, ⟨d^{κ}, d_{j↦e}^{κ} ⟩ ∈ Cg^{G} (x, y) ⊆ θ, so we take p = d_{j↦e}^{κ} .

Case 3: There exists j ∈ κ such that {x, y} = {d_{j↦c}^{κ} , p} for some p ∈ {d, e}^{κ} ∖ {d^{κ}}. Without loss of generality, let y = p. Then ⟨d^{κ}, p⟩ = ⟨xd_{j↦a}^{κ} , yd_{j↦a}^{κ} ⟩ ∈ Cg^{G} (x, y) ⊆ θ, and since p ∈ {d, e}^{κ} ∖ {d^{κ}}, the case is done.

Case 4: There exist j, l ∈ κ so that {x, y} = {d_{j→c}^{κ} , d_{l→c}^{κ} }. Then {xd_{l→b}^{κ} , yd_{l→b}^{κ} } = {d_{j→c}^{κ} , d_{l→e}^{κ} }, and since ⟨xd_{l→b}^{κ} , yd_{l→b}^{κ} ⟩ ∈ Cg^{G} (x, y) ⊆ θ, θ is also in Case 3.

Case 5: There exists j ∈ κ such that x = d_{j↦c}^{κ} and y = d^{κ} (or vice versa, but it is analogous). Then ⟨d_{j↦e}^{κ} , d^{κ}⟩ = ⟨xd_{j↦b}^{κ} , yd_{j↦b}^{κ} ⟩ ∈ Cg^{G} (x, y) ⊆ θ, so we take p = d_{j↦e}^{κ} . □

We turn our attention to the variety V_{B} = HSP (B). According to [4], more precisely the definition of V_{B} and Theorem 7.15, it is the variety generated by

\begin{matrix}B & a & b & c & d & e \\ a & a & a & a & a & a \\ b & b & b & b & b & b \\ c & d & c & c & c & c \\ d & d & e & d & d & d \\ e & d & e & e & e & e \end{matrix}

(actually, it is also generated by the subalgebra with universe {a, b, d, e}). The equational base of V_{B} is

(B1) xx ≈ x  (B2) xyy ≈ xy
(B3) x · yz ≈ xy  (B4) xyzy ≈ xzy

Theorem 3.3. The variety V_{B} is residually large.

Proof. Let κ be any cardinal. For any x ∈ κ, x′ and x′′are short notation for ⟨0, x⟩ and ⟨1, x⟩, respectively. Consider the groupoid G = ({0, 1} × κ, ·) where the operation · is defined like this:

x · y = \{ \begin{matrix}1′′, & if x = 0′′, y = 0′; \\ 0′′, & if x = k′′, y = l′, l ≠ 0 and k ⩽ l; \\ x, & else. \end{matrix}

The operation is given in Table 2. Notice that the elements of {0}×κ are left zeros, while the elements of {1} × κ are right neutral. Besides, we note that 1′′y = 0′′y for all y ∈ {0} × κ.

We show that G ∈ V_{B}, by showing G satisfies (B1)–(B4).

Identity (B1): G is idempotent since it consists of left zeros and right units.

Table 2. The Cayley table of the subdirectly irreducible groupoid in the variety V_{B} of size 2κ.

0′ 1′ 2′ 3′ 4′ … 4′′ 3′′ 2′′ 1′′ 0′′
0′ 0′ 0′ 0′ 0′ 0′ … 0′ 0′ 0′ 0′ 0′
1′ 1′ 1′ 1′ 1′ 1′ … 1′ 1′ 1′ 1′ 1′
2′ 2′ 2′ 2′ 2′ 2′ … 2′ 2′ 2′ 2′ 2′
3′ 3′ 3′ 3′ 3′ 3′ … 3′ 3′ 3′ 3′ 3′
4′ 4′ 4′ 4′ 4′ 4′ … 4′ 4′ 4′ 4′ 4′
⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
4′′ 4′′ 4′′ 4′′ 4′′ 4′′ … 4′′ 4′′ 4′′ 4′′ 4′′
3′′ 3′′ 3′′ 3′′ 3′′ 3′′ … 3′′ 3′′ 3′′ 3′′ 3′′
2′′ 2′′ 2′′ 2′′ 2′′ 2′′ … 2′′ 2′′ 2′′ 2′′ 2′′
1′′ 1′′ 1′′ 1′′ 1′′ 1′′ … 1′′ 1′′ 1′′ 1′′ 1′′
0′′ 0′′ 0′′ 0′′ 0′′ 0′′ … 0′′ 0′′ 0′′ 0′′ 0′′

Identity (B2): If x = k′ is a left zero or y a right unit, (B2) holds trivially. Assume that x = k′′ and y = l′. In the case x = 0′′ and y = 0′ we verify (B2) directly, since both sides equal 1′′. If l ≠ 0 and k ⩽ l, then xyy = 0′′y = 0′′= xy. If k > l, then xy = x and xyy = xy (= x).

Identity (B3): If x is a left zero, y a left zero, or z a right unit, the equality trivially holds. Assume that x = k′′, y = l′′and z = m′. Then yz ∈ {1} × κ, say yz = s′′, and hence x · yz = k′′s′′= k′′= k′′l′′= xy.

Identity (B4): If x is a left zero, or y a right unit, (B4) trivially holds. If z is a right unit, (B4) reduces to (B2). So we assume that x = k′′, y = l′ and z = m′.

Case x = 0′′: If y = 0′, then xyzy = 1′′zy = 0′′zy = xzy, and if y ≠ 0′, then xy = x, so xyzy = xzy.

Case k > 0: Then xy, xz ∈ {x, 0′′}. If xy = x, then xyzy = xzy. If xy = 0′′ and xz = x, then xyzy = 0′′zy ∈ {0′′y, 1′′y} = {1′′y} since 0′′y = 1′′y. Thus xyzy = 1′′y = xyy =_{(B2)} xy = xzy. Finally, if xy = xz = 0′′, then xyzy = 0′′zy = xzzy =_{(B2)} xzy.

To complete the proof, we are proving that G is subdirectly irreducible with the monolith Cg^{G} (0′′, 1′′). Let x, y ∈ G be distinct. We want to prove that ⟨0′′, 1′′⟩ ∈ Cg^{G} (x, y).

Case x = k′′, y = l′′: Without loss of generality, assume k < l. Then

⟨0′′, y⟩= ⟨xk′, yk′⟩ ∈ Cg^{G} (x, y),
⟨1′′, y⟩= ⟨0′′0′, y0′⟩ ∈ Cg^{G} (x, y),

so transitivity implies that ⟨0′′, 1′′⟩ ∈ Cg^{G} (x, y).

Case x = k′, y = l′: Without loss of generality, assume k < l. Then

⟨l′′, 0′′⟩ = ⟨l′′x, l′′y⟩ ∈ Cg^{G} (x, y).

From the previous case we obtain ⟨0′′, 1′′⟩ ∈ Cg^{G} (0′′, l′′, y) ⊆ Cg^{G} (x, y).

Case x = k′, y = l′′: If k = 0, then ⟨1′′, 0′′⟩ = ⟨0′′x, 0′′y⟩ ∈ Cg^{G} (x, y). If k ≠ 0, then ⟨0′′, 1′′⟩ = ⟨1′′x, 1′′y⟩ ∈ Cg^{G} (x, y). □

4. Residual bound of V_{C}

We turn to the variety V_{C}. According to [4], more precisely the definition of V_{C} and Lemma 7.12, V_{C} is the variety generated by

\begin{matrix}G_{5} & x & y & xy & yx \\ x & x & xy & x & xy \\ y & yx & y & yx & y \\ xy & xy & xy & xy & xy \\ yx & yx & yx & yx & yx \end{matrix}

which is also its 2-generated free groupoid. The equational base of V_{C} is

(C1) xx ≈ x  (C2) xyy ≈ xy
(C3) x · yz ≈ xy  (C4) xyz ≈ xzy

Lemma 4.1. Let G ∈ V_{C}, and assume that ab = b for some a, b ∈ G. Then a = b.

Proof. a = aa =_{(C3)} a · ab = ab = b. □

Let G be an n-element groupoid in V_{C}, and define for any a ∈ G the set C_{a} as

C_{a} = a · G ∪ a · G · G ∪ ⋯∪ a · \underbrace{G · . . . · G}_{n−1}.

Here, a · \underbrace{G · . . . · G}_{n−1}= \{ag_{1}g_{2} …g_{n−1} : (∀i < n)g_{i} ∈ G\}. Note that a · \underbrace{G · . . . · G}_{k−1}⊆

a · \underbrace{G · . . . · G}_{k}by idempotence. Also, note that the left-associated products need

only go as far as n = |G| terms since any repetition is cancelled by applying (C4) several times, and then (C1) or (C2). It follows that C_{a} · G = C_{a}. The relation ρ_{a} on G is defined by:

b ρ_{a} c iff b = c or b, c ∈ C_{a}.

Lemma 4.2. Let G ∈ V_{C} and |G| = n. Then ρ_{a} is a congruence on G.

Proof. Let b ρ_{a} c for distinct elements b, c and let d ∈ G. Then there exist

i, j ⩽ n such that b ∈ a · \underbrace{G · . . . · G}_{i}and c ∈ a · \underbrace{G · . . . · G}_{j}. According to (C3) and

transitivity of equality, it follows that da = db and da = dc. Therefore, ⟨db, dc⟩ ∈ ρ_{a}. On the other hand, bd, cd ∈ C_{a}, hence ⟨bd, cd⟩ ∈ ρ_{a}. □

Corollary 4.1. Let G ∈ V_{C} be a finite groupoid, and let a ∈ G. Then for all b, c ∈ C_{a}, and all x ∈ G, we have xb = xc.

Proof. By a repeated use of the identity (C3), we obtain xa = xd for any d ∈ C_{a} and any x ∈ G. □

Let b ∈ G, G ∈ V_{C}. Denote by F_{b} = {g ∈ G : bg = b}.

Lemma 4.3. Let G ∈ V_{C} be a finite groupoid, and let a ∈ G, b ∈ a · G, b ≠ a. For all c ∈ G we have

(ac = a) ∨ (ac = b) ⇒ bc = b.

Consequently, |F_{b}| > |F_{a}| and C_{b} ⊆ C_{a}.

Proof. From b ∈ a · G we know that there exists d ∈ G such that ad = b. Moreover, bc = adc =_{(C4)} acd = ad = b if ac = a, or bc = acc =_{(C2)} ac = b if ac = b. From the first part we get F_{a} ⊆ F_{b}, but we also know that ad = b ≠ a and

bd = b. Finally, b · \underbrace{G · . . . · G}_{i}⊆ a · \underbrace{G · . . . · G}_{i+1}since b ∈ a · G, hence C_{b} ⊆ C_{a}. □

Lemma 4.4. Let G ∈ V_{C} be a finite groupoid, |G| = n, c ∈ C_{a} so that c = ab_{1} …b_{i} for some i < n and b_{1}, …, b_{i} ∈ G. If c · G = c (c is a left zero), then for every d ∈ C_{a}, db_{1} …b_{i} = c.

Proof. Let d = ab′_{1} …b′_{j} for some j ⩽ n and b′_{1}, …, b′_{j} ∈ G. Then

db_{1} …b_{i} = ab′_{1} …b′_{j}b_{1} …b_{i} =_{(C4)} … =_{(C4)} ab_{1} …b_{i}b′_{1} …b′_{j} = cb′_{1} …b′_{j} = c. □

Lemma 4.5. Let G ∈ V_{C} be a finite groupoid, |G| = n, and let a ∈ G. Then there exists a unique c ∈ C_{a} such that c · G = c.

Proof. Assume the opposite. Let b ∈ C_{a} be such that |F_{b}| is maximal. Since {b} ≠ b · G, then there exists c ∈ b · G, c ≠ b, so by Lemma 4.3 we get |F_{c}| > |F_{b}|. Moreover, the conditions c ∈ b · G and b ∈ C_{a} mean that we can select some i < n and b_{1}, …, b_{i}, d ∈ G such that b = ab_{1}b_{2} …b_{i} and c = bd. Thus c ∈ C_{a}, which contradicts the choice of b. Uniqueness of c in C_{a} follows from Lemma 4.4. □

Lemma 4.6. Let G ∈ V_{C} be a finite groupoid, a ∈ G, and let |C_{a}| = 3. Then G is not subdirectly irreducible.

Proof. Assume that C_{a} = {a, b, c} and according to Lemma 4.5, one of b, c is a left zero, say c · G = c.

First we show that C_{a} = a · G. We know that {a} ⊆ a · G ⊆ a · G · G ⊆ …, and that once this sequence stabilizes, it stays constant and equal to C_{a}. Therefore, either a · G = {a, b, c} = C_{a}, or {a} ⊊ a · G ⊊ a · G · G = {a, b, c}. If a · G = {a, c}, then {a, b, c} = a·G·G = a·G∪c·G = {a, c}∪{c}, a contradiction. If a·G = {a, b}, then ade = c for some d, e ∈ G. We know ad ∈ a · G = {a, b}. If ad = a, then c = ade = ae ∈ a · G. So, we may assume that ad = b. According to (C4), c = ade = aed, so an analogous argument proves that we may assume ae = b. However, from Lemma 4.3 it follows that be = b, and this leads to a contradiction: c = ade = be = b.

Let ad = c. According to Lemma 4.4 we have {a, b, c} · d = c. Lemma 4.3 implies that b · G ⊆ {b, c}, and Lemma 4.4 implies c ∈ b · G, while (C1) implies b ∈ b · G, so b · G = {b, c}. Thus C_{b} = {b, c}, as c is a left zero. Lemma 4.2 implies that α = Δ_{G} ∪ {⟨b, c⟩, ⟨c, b⟩} is a congruence of G. Let us show that β = Δ_{G} ∪ {⟨a, b⟩, ⟨b, a⟩} is a congruence of G. Let ae = b. If g ∈ G, then gb = g·ae =_{(C3)} ga. On the other hand, we know {ag, bg} ⊆ {a, b, c}. If ag ∈ {a, b}, Lemma 4.3 implies bg = b. If ag = c, then Lemma 4.4 implies bg = c. So, α, β ∈ ConG ∖ {Δ_{G}}, but α ∩ β = Δ_{G}, hence G is not subdirectly irreducible. □

Lemma 4.7. Let G ∈ V_{C} be a finite groupoid, a, b ∈ G, and let |C_{a}| = |C_{b}| = 2. Then G is not subdirectly irreducible.

Proof. Lemma 4.2 implies that ρ_{a}, ρ_{b} are atoms in Con G. Assume that they are equal. Then from a ∈ C_{a} and b ∈ C_{b} we get C_{a} = C_{b} = {a, b}. According to Lemma 4.5, one of a, b must be a left zero, but this contradicts the assumption that |C_{a}| = |C_{b}| = 2. Thus ρ_{a} and ρ_{b} are distinct atoms in Con G, so G is not subdirectly irreducible. □

Theorem 4.1. resb(V_{C}) ⩽ 4.

Proof. Let G ∈ V_{C} be a subdirectly irreducible groupoid with at least 4 elements. V_{C} is finitely generated, thus it is locally finite. Recall the Quackenbush Lemma from [8]: if a locally finite variety has an infinite subdirectly irreducible algebra, it has no bound on sizes of finite subdirectly irreducibles. So we may assume that G is finite.

According to Lemma 4.6, for all a ∈ G, |C_{a}| ≠ 3. Assume that a ∈ G is such that |C_{a}| > 3. Lemma 4.5 implies that there exists a unique c ∈ C_{a} such that c · G = c. Lemma 4.4 implies that for all d ∈ C_{a}, c ∈ C_{d}. Let b ∈ C_{a} ∖ {c} be such that C_{b} is minimal, and such that |F_{b}| is maximal among |F_{d}| which satisfy d ∈ C_{a}∖{c} and C_{d} = C_{b}. If C_{b} ≠ {b, c}, then b·G ≠ {b, c} (otherwise we would get b·G·G = b·G∪c·G = {b, c}). Let d ∈ b·G∖{b, c}. Then d ∈ C_{b} and b ∈ C_{a} imply d ∈ C_{a}. Moreover, Lemma 4.3 implies that C_{d} ⊆ C_{b} and |F_{b}| < |F_{d}|, contradicting the choice of b. Therefore, C_{b} = {c, b}. Next we select d ∈ C_{a}∖{b, c} such that C_{d} is minimal, and such that |F_{d}| is maximal among |F_{e}| which satisfy e ∈ C_{a} ∖{b, c} and C_{e} = C_{d}. The analogous argument proves that either C_{d} = {c, d}, or C_{d} = {b, c, d}. But this is a contradiction with Lemma 4.7, or with Lemma 4.6.

Assume now that |C_{a}| ⩽ 2 for all a ∈ G. Lemma 4.7 implies that at most one a ∈ G is such that |C_{a}| = 2. If |C_{a}| = 1 for all a ∈ G, then G is a left zero semigroup, and the only subdirectly irreducible left zero semigroup has two elements. Let a ∈ G be the only element such that |C_{a}| = 2, say C_{a} = {a, b}, and all other elements of G are left zeros. Since |G| ⩾ 4, then there exist left zeros u, v ∈ G ∖ {a} such that au = av (since G contains at least three left zeros, and a · G = {a, b}). But then Cg^{G} (a, b) = ρ_{a} = {a, b}^{2} ∪ Δ_{G} and Cg^{G} (u, v) = {u, v}^{2} ∪ Δ_{G} are distinct atoms in Con G, which is a contradiction. □

Theorem 4.2. All subdirectly irreducible members of V_{C} are depicted in Table 3. Thus, resb(V_{C}) = 4.

Proof. Let G ∈ V_{C} be subdirectly irreducible. By Theorem 4.1, |G| < 4. emphCase |G| = 3: Let G = {a, b, c}. G is not a left zero semigroup, since the

three-element left zero semigroup is not subdirectly irreducible. From Lemma 4.1 it follows that we may assume ab = c, from Lemmas 4.6 and 4.7 it follows that there is at most one element of G which is not a left zero, and it must be a, so b and c are left zeros. Finally, ac = a since ac ∈ C_{a} = {a, c} and ac = c would imply Table 3. Cayley tables of subdirectly irreducible groupoids in V_{C}.

a b c  a b
a a c a  a a a
b b b b  b b b
c c c c

a = c by Lemma 4.1. This means that G is the left-hand side groupoid depicted in Table 3, which is subdirectly irreducible with the monolith Cg^{g}(a, c).

Case |G| = 2: Let G = {a, b}. We know that at least one of {a, b} is a left zero. If a is not a left zero, then idempotence implies that ab = b, but from Lemma 4.1 it follows that a = b. So G is a left zero semigroup. □

5. Residual bound of W

We turn to the variety W. According to [4], more precisely Lemma 6.1 and Theorem 6.3, W is the variety generated by

\begin{matrix}G_{4} & x & y & xy & yx \\ x & x & xy & x & xy \\ y & yx & y & yx & y \\ xy & xy & xy & xy & xy \\ yx & yx & yx & yx & yx \end{matrix}

which is also its 2-generated free groupoid. The equational base of W is

(W1) xx ≈ x,  (W2) xyy ≈ x,
(W3) x · yz ≈ xy,  (W4) xyz ≈ xzy,

Lemma 5.1. Let G ∈ W. The groupoid G is right cancellative, i.e. for all a, b, c ∈ G, ba = ca ⇒ b = c. If a, b ∈ G are distinct, then ab ≠ b.

Proof. The proof follows from (W 2) and idempotence (W 1). □

Lemma 5.2. Let G ∈ W and assume a, b, c ∈ G are pairwise distinct elements which satisfy ab = c. Then for every pair ⟨d, e⟩ ∈ Cg^{G} (a, c) and all z ∈ G, zd = ze. The congruence α = Cg^{G} (a, c) is an atom in the lattice Con G and all α-classes have at most two elements.

Proof. From (W 1)–(W 4) we get that ac = a · ab = aa = a, ca = aba = aab = ab = c, cb = abb = a, i.e. the following equalities hold in G:

\begin{matrix}G & a & b & c & … \\ a & a & c & a & … \\ b & . & b & . & … \\ c & c & a & c & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix}

Recall that the congruences are generated by Mal’cev chains, so it suffices to prove that for any z ∈ G and translation p(x) ∈ TrG, zp(a) = zp(c). We prove this by an induction on the complexity of the shortest composition of basic translations equal to p(x). Since a = cb, it follows that for all z ∈ G, za = zc (za = z(cb) =_{(W 3)} zc), which proves the inductive base. Assume that the shortest composition of basic translations which equals p(x) uses any left translation λ_{u}, so p(x) = (r ∘λ_{u} ∘q)(x), where r, q ∈ TrG, then (λ_{u} ∘ q)(a) = uq(a) = uq(c) = (λ_{u} ∘ q)(c) by the inductive assumption on q(x). Therefore, p(a) = p(c), and also zp(a) = zp(c). The only case we need to consider is when p is a composition of right translations.

For each z ∈ G we have

z · ad_{1} …d_{n}=_{(W 3)} z · ad_{1} …d_{n−1} =_{(W 3)} ⋯=_{(W 3)} za = zc
=_{(W 3)} z · cd_{1} =_{(W 3)} ⋯=_{(W 3)} z · cd_{1} …d_{n}.

We obtain that all translations we need to consider for generating links in the Mal’cev chains in Cg^{G} (a, c) are of the form p(x) = xd_{1} …d_{n}. Moreover,

ad_{1} …d_{n}b=_{(W 4)} ad_{1} …bd_{n} =_{(W 4)} ⋯=_{(W 4)} abd_{1} …d_{n} = cd_{1} …d_{n},

and equivalently, cd_{1} …d_{n}b = ad_{1} …d_{n}, since cb = a can replace ab = c and then the proof is analogous to above derivation. We have proved for any u, v ∈ G that there exists a translation p(x) such that u = p(a) and v = p(c) iff there exists a translation q(x) such that u = q(c) and v = q(a).

Let d_{1}, …, d_{n}, e_{1}, …, e_{m} ∈ G be parameters such that ad_{1} …d_{n} ≠ cd_{1} …d_{n} and ae_{1} …e_{m} ≠ ce_{1} …e_{m}. We claim that

|{ad_{1} …d_{n}, cd_{1} …d_{n}} ∩ {ae_{1} …e_{m}, ce_{1} …e_{m}}| ≠ 1.

Assume the opposite. By the above considerations, it suffices to consider the case when ad_{1} …d_{n} = ae_{1} …e_{m} and cd_{1} …d_{n} ≠ ce_{1} …e_{m} and also the case when ad_{1} …d_{n} ≠ ae_{1} …e_{m}, cd_{1} …d_{n} = ce_{1} …e_{m}. The two cases are symmetric, thus we may assume without loss of generality that ad_{1} …d_{n} = ae_{1} …e_{m} and cd_{1} …d_{n} ≠ ce_{1} …e_{m}. We get

cd_{1} …d_{n} = ad_{1} …d_{n}b = ae_{1} …e_{m}b = ce_{1} …e_{m} ≠ cd_{1} …d_{n},

a contradiction (the third equality was proved above). We conclude that all classes of the congruence Cg^{G} (a, c) have at most two elements. Since the identity (W 2) implies that

Cg^{G} (a, c) = Cg^{G} (ad_{1} …d_{n}, cd_{1} …d_{n}),

it follows that Cg^{G} (a, c) is an atom in the lattice Con G. □

Lemma 5.3. Let G ∈ W be subdirectly irreducible and for three distinct elements a, b, c ∈ G let ab = c hold. Then {a, c} · G = {a, c} and all elements in G ∖ {a, c} are left zeros.

Proof. The table in the proof of Lemma 5.2 proves that {a, c} · {a, b, c} = {a, c}.

Assume that there exists an element d ∈ G ∖ {a, b, c} such that ad ∉ {a, c}. Then ad ≠ d by Lemma 5.1, so ad ∉ {a, d} and Lemma 5.2 proves that Cg^{G} (a, ad) is an atom in the lattice Con G, which is incomparable with Cg^{G} (a, c). This contradicts the assumption that G is subdirectly irreducible. The proof that cd ∈ Table 4. Cayley tables of all subdirectly irreducible groupoids in W.

a b c d  a b c  a b
a a c a a  a a c a  a a a
b b b b b  b b b b  b b b
c c a c c  c c a c
d d d d d

{a, c} is analogous, obtained by interchanging a and c above. Therefore, {a, c}·G = {a, c}. Thus Cg^{G} (a, c) = Δ_{G} ∪ {a, c}^{2}, according to Lemma 5.2.

Assume that there exist d ∈ G∖{a, c} and e ∈ G∖{d} such that d ≠ de. Then de ≠ e by Lemma 5.1, so de ∉ {d, e}. The argument from the preceding paragraph proves that Cg^{G} (d, de) = Δ_{G} ∪ {d, de}^{2} and Cg^{G} (a, c) ∩ Cg^{G} (d, de) = Δ_{G}, which is impossible since G is subdirectly irreducible. Thus all d ∈ G ∖ {a, c} are left zeros. □

Theorem 5.1. resb(W) ⩽ 5.

Proof. Assume the opposite, i.e. let G ∈ W be a subdirectly irreducible groupoid with |G| ⩾ 5. G can not be a left zero semigroup, since the only subdirectly irreducible left zero semigroup has two elements. From Lemma 5.1 and idempotence it follows that there exist a, b, c ∈ G such that ab = c. Lemma 5.2 proves that Cg^{G} (a, c) is an atom in the congruence lattice Con G and Cg^{G} (a, c) ≠ ∇_{G} since |G| ⩾ 5 and all equivalence classes of Cg^{G} (a, c) have at most two elements.

Lemma 5.3 implies that {a, c} · G = {a, c} and all elements of G ∖ {a, c} are left zeros. Since |G| ⩾ 5, there exist elements d, e ∈ G ∖ {a, c} such that ad = ae. Thus for each z ∈ G∖{c} we have zd = ze, for z ∈ G∖{a, c} are left zeros. On the other hand, if z = c, we get cd = abd =_{(W 4)} adb = aeb =_{(W 4)} abe = ce. Therefore, Cg^{G} (d, e) = Δ_{G} ∪ {d, e}^{2}, and Cg^{G} (a, c) ∩ Cg^{G} (d, e) = Δ_{G}, a contradiction. □

Theorem 5.2. resb(W) = 5 and all subdirectly irreducible groupoids in W are given in Table 4.

Proof. Let G ∈ W be a subdirectly irreducible groupoid. From Theorem 5.1 we know that |G| < 5.

Case |G| = 4: Let G = {a, b, c, d} and we may assume that ab = c since G can’t be a left zero semigroup and ab ≠ b by Lemma 5.1. In the proof of Lemma 5.2 we showed that ac = cb = a and ab = ca = c in this case. From Lemma 5.3 it follows that b and d are left zeros and that {a, c} · G = {a, c}. If ab = ad, it would follow that cd = ab·d = ad·b = ab·b = cb, and Δ_{G}∪{b, d}^{2} would have been a congruence, a contradiction. Similarly we conclude cb ≠ cd, so ad ≠ ab = c forces ad = a, and we also get cd = c. This is the leftmost groupoid in Table 4 for which we verify that it is subdirectly irreducible and in W by checking directly the identities and the congruence lattice.

Case |G| = 3: Let G = {a, b, c}. We may assume that ab = c like in the previous case, and the proofs that b is a left zero, that ac = cb = a and ab = ca = c are also just as above. Thus we obtain the middle groupoid in 4 which is subdirectly irreducible with the monolith Cg^{g}(a, c).

Case |G| = 2: Let G = {a, b}. By idempotence and Lemma 5.1 we get that G is the left zero semigroup. □

Acknowledgements. We express thanks to Professor David Stanovsky for his explanations regarding Steiner quasigroups, SIE groupoids and quandles. We profoundly thank the referee for numerous useful comments, catching several mistakes in our proofs and improving the presentation of our proofs.

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Department of Mathematics and Informatics (Received 23 12 2015) University of Novi Sad (Revised 29 01 2016) Novi Sad, Serbia

 ptdjapic@dmi.uns.ac.rs
 pera@dmi.uns.ac.rs