PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 81(95) (2007), 103–109 DOI 102298/PIM0795103J

CANONICAL BIASSOCIATIVE GROUPOIDS

Biljana Janeva, Snežana Ilić, and Vesna Celakoska-Jordanova

Abstract. In the paper Free biassociative groupoids, the variety of biassociative groupoids (i.e., groupoids satisfying the condition: every subgroupoid generated by at most two elements is a subsemigroup) is considered and free objects are constructed using a chain of partial biassociative groupoids that satisfy certain properties. The obtained free objects in this variety are not canonical. By a canonical groupoid in a variety V of groupoids we mean a free groupoid (R, ∗) in V with a free basis B such that the carrier R is a subset of the absolutely free groupoid (T_{B}, ·) with the free basis B and (tu ∈ R ⇒ t, u ∈ R & t∗u = tu). In the present paper, a canonical description of free objects in the variety of biassociative groupoids is obtained.

1. Preliminaries

Let G = (G, ·) be a groupoid and a, b ∈ G. We denote by ⟨a, b⟩ the subgroupoid of G generated by a, b and by ⟨a⟩ the subgroupoid generated by a. Clearly, ⟨a⟩ ⊆ ⟨a, b⟩ and if b ∈ ⟨a⟩, then ⟨a, b⟩ = ⟨a⟩; specially, ⟨a, a⟩ = ⟨a⟩. The subgroupoids ⟨a, b⟩ and ⟨b, a⟩ are equal.

Let a_{1}, a_{2}, …, a_{n} be a finite sequence of elements in a groupoid G. We denote by a_{1}a_{2} ⋯ a_{n} the product of the sequence a_{1}, a_{2}, …, a_{n} in G defined as follows:

i) if n = 3, then a_{1}a_{2}a_{3} =^{def} a_{1}(a_{2}a_{3}) and

ii) if n ⩾ 3, then a_{1}a_{2} ⋯ a_{n} =^{def} a_{1}(a_{2} ⋯ a_{n}).

We call a_{1}a_{2} ⋯ a_{n} the main product of the sequence a_{1}, a_{2}, …, a_{n}. If n = 1 and n = 2, then a_{1} and a_{1}a_{2} will also be called the main products of the sequences a_{1} and a_{1}, a_{2} respectively. If c = a_{1}a_{2} ⋯ a_{n}, then we say that c is presented as a main product of the sequence a_{1}, a_{2}, …, a_{n}.

Let G be a groupoid and A ⊆ G. If Q is the subgroupoid of G generated by A, i.e., Q = ⟨A⟩, then Q = ⋃{A_{k} : k ⩾ 0}, where A_{0} = A, A_{k+1} = A_{k} ∪ A_{k}A_{k}.

If x ∈ Q, then a hierarchy of x in Q is the nonnegative integer χ_{Q}(x), defined by χ_{Q}(x) = min{k ∈ ℕ_{0} : x ∈ A_{k}}, where ℕ_{0} is the set of nonnegative integers.

2000 Mathematics Subject Classification: Primary 08B20; Secondary 03C05.

Key words and phrases: Groupoid, subgroupoid generated by two elements, subsemigroup, free groupoid, canonical groupoid.

In the sequel B will be an arbitrary nonempty set whose elements are called variables. By T_{B} we will denote the set of all groupoid terms over B in the signature ·. The terms are denoted by t, u, v, …, x, y, ⋯ T_{B} = (T_{B}, ·) is the absolutely free groupoid with the free basis B, where the operation is defined by (u, v) ↦ uv. The groupoid T_{B} is injective, i.e., if x, y, v, w ∈ T_{B}, then xy = vw ⇒ x = v, y = w; in other words the operation · is an injective mapping.

Note that T_{B} = ⋃{B_{k} : k ⩾ 0}, where B_{0} = B, B_{k+1} = B_{k} ∪ B_{k}B_{k}. The hierarchy χ : T_{B} → ℕ_{0}, defined by χ(t) = min{k ∈ ℕ_{0} : t ∈ B_{k}}, for any t ∈ T_{B}, has the property:

χ(tu) = 1 + max{χ(t), χ(u)},

for all t, u ∈ T_{B}.

For any term v ∈ T_{B} we define the length |v| of v and the set of subterms P (v) of v in the following way:

|b| = 1, |tu| = |t| + |u|; P (b) = {b}, P (tu) = {tu} ∪ P (t) ∪ P (u),

for any b ∈ B and t, u ∈ T_{B}.

2. Main biproducts

Let t, u ∈ T_{B} and ⟨t, u⟩ be the subgroupoid of T_{B} generated by t, u:
⟨t, u⟩ = {t, u, tt, tu, ut, uu, t(tt), t(tu), t(ut), t(uu), (tt)t, (tu)t, ⋯ }.

Each element x of ⟨t, u⟩ is a product of a finite sequence of elements x_{1}, …, x_{n} (n ⩾ 1), where each x_{i} is either t or u, i.e., {x_{1}, x_{2}, …, x_{n}} ⊆ {t, u}. Any such product is constructed by the two generators t, u and therefore we call it a binary product or shortly biproduct.

Thus, if a term x ∈ T_{B} is an element of ⟨t, u⟩, then we say that x has a representation as a biproduct (or shortly, x is a biproduct) with the generating pair {t, u} and denote it by x_{⟨t,u⟩}. (In this case we also say that x is the carrier of the biproduct x_{⟨t,u⟩}.)

If u = t or u ∈ ⟨t⟩, then ⟨t, u⟩ = ⟨t⟩. In that case if x ∈ ⟨t⟩, we say again that x is a biproduct with the generator t and denote it by x_{⟨t⟩}. Specially, t ∈ ⟨t⟩ and t has a representation as a biproduct with the generator t: t_{⟨t⟩} = t. We say that t_{⟨t⟩} is a trivial biproduct of t. Since t ∈ ⟨t, u⟩ we have t_{⟨t,u⟩} = t and we say also that t_{⟨t,u⟩} is the trivial biproduct of t in ⟨t, u⟩.

If t ∉ ⟨u⟩ and u ∉ ⟨t⟩, then no two elements of the subgroupoid ⟨t, u⟩ are equal, since the groupoid T_{B} is injective. Therefore:

Proposition 2.1. If t, u, x are terms of T_{B} and x is such that x ∈ ⟨t, u⟩, t ∉ ⟨u⟩ and u ∉ ⟨t⟩, then x has a unique representation as a biproduct with the generating pair {t, u}.

Note that a term of T_{B} may have representations as biproducts with different pairs of generators.

Example 2.1. Let a, b be two distinct variables and x the term ((ab)b)(ab).

1) x ∈ ⟨x⟩, and thus x_{⟨x⟩} = x is the biproduct of x with the generator x.

2) Put t = (ab)b and u = ab. Then x ∈ ⟨t, u⟩ and x_{⟨t,u⟩} = tu is the biproduct of x with the generating pair {t, u}.

3) If u = ab and v = b, then x ∈ ⟨u, v⟩ and x_{⟨u,v⟩} = (uv)u is the biproduct of x with the generating pair {u, v}.

4) x ∈ ⟨a, b⟩ and thus x_{⟨a,b⟩} = ((ab)b)(ab) is the biproduct of x with the generating pair {a, b}.

(Note that there is no biproduct of x other than those enumerated above.)

A biproduct x_{⟨t,u⟩} of a term x is said to be maximal in T_{B} if and only if for any biproduct x_{⟨α,β⟩} of x, the hierarchy χ_{⟨α,β⟩}(x) does not exceed the hierarchy χ_{⟨t,u⟩}(x), i.e., χ_{⟨α,β⟩}(x) ⩽ χ_{⟨t,u⟩}(x).

Proposition 2.2. Any term x of T_{B} has a finite number of representations as a biproduct in T_{B}, i.e., x ∈ T_{B} is the carrier of a finite number of biproducts in T_{B}. Any term x of T_{B} is the carrier of maximal biproducts in T_{B}.

Proof. The length |x| of any x ∈ T_{B} is finite, and thus the set P (x) of subterms of x is finite. As the generators of any biproduct of x are subterms of x, and the set of subterms P (x) of x is finite, it follows that x has a finite number of biproducts. The set of nonnegative integers that are hierarchies of x (with respect to the pair of generators of all biproducts of x, including the pairs {t, t} = {t}) is finite, and thus it has the largest element. Therefore, there is the largest hierarchy of x, i.e., a maximal biproduct of x. □

Note that a given term x of T_{B} may have more than one maximal biproducts.

Example 2.2. Let x = ((ab)b)(b^{2}(ab)) (where a, b are variables). Put t = ab and u = b. Then x_{⟨t,u⟩} = (tu)(u^{2}t) and χ_{⟨t,u⟩}(x) = 3. If we take {a, b} as the generating pair, then x_{⟨a,b⟩} = ((ab)b)(b^{2}(ab)) is a biproduct of x and χ_{⟨a,b⟩}(x) = 3. For all other biproducts x_{⟨α,β⟩} one obtains that χ_{⟨α,β⟩}(x) ⩽ 3. Thus, x_{⟨t,u⟩} and x_{⟨a,b⟩} are maximal biproducts of x.

Let x = x_{1}x_{2} ⋯ x_{m} be the main product of x_{1}, x_{2}, …, x_{m} in T_{B}. If
{x_{1}, x_{2}, …, x_{m}} ⊆ {t, u},

for some terms t, u of T_{B}, then we call x_{1}x_{2} ⋯ x_{m} the main biproduct of x in T_{B} with the generating pair {t, u} and denote it by x_{t,u}. (If u = t, i.e., the generating “pair” is {t, t}, we write x_{t} instead of x_{t,t}.)

Below we will state some properties about main biproducts.

(1) Note that any term x of T_{B} has at least one main biproduct – the trivial one, x_{x}. If x ∈ T_{B} ∖ B, then x = αβ for some α, β ∈ T_{B}, and x_{α,β} = αβ is another main biproduct of x in T_{B}.

(2) The hierarchy of a main biproduct x_{1}x_{2} ⋯ x_{m}, with a generating pair {t, u}, equals m − 1. Therefore, if two main biproducts x_{1}x_{2} ⋯ x_{m} and y_{1}y_{2} ⋯ y_{m+k} are maximal biproducts of x in T_{B}, then they have to satisfy k = 0 (or the hierarchies would differ) and x_{i} = y_{i}, for 1 ⩽ i ⩽ m.

Proposition 2.3. If x ∈ T_{B} has two nontrivial main biproducts x_{t,u} and x_{v,w} in T_{B}, then one generator of the one generating pair coincides with a generator of the other generating pair.

Proof. Let x_{t,u} = x_{1}x_{2} ⋯ x_{m} and x_{v,w} = y_{1}y_{2} ⋯ y_{n} be two main biproducts of x in T_{B}. Then x_{1}x_{2} ⋯ x_{m} = y_{1}y_{2} ⋯ y_{n} implies x_{1} = y_{1}. Since x_{ν} ∈ {t, u} and y_{λ} ∈ {v, w} it follows that x_{1} is either t or u, and y_{1} is either v or w. If, for example, x_{1} = t and y_{1} = v, then v = t (and in that case x_{t,u} = x_{t,w}). □

Using the property (2) stated above, we obtain the following:

Theorem 2.1. If x = x_{1}x_{2} ⋯ x_{m} and x = x′_{1}x′_{2} ⋯ x′_{n} are main biproducts of x in T_{B} with the same generating pair {t, u}, then m = n and x_{i}=x′_{i}, for i = 1, 2, …, m. Specially, any maximal biproduct of x ∈ T_{B}, that is a main biproduct, is uniquely determined.

3. A construction of canonical biassociative groupoids

A groupoid G = (G, ·) is said to be biassociative [1] if and only if for any a, b ∈ G the subgroupoid S of G generated by {a, b}, i.e., S = ⟨a, b⟩, is a subsemigroup of G. The class of all biassociative groupoids will be denoted by Bass. This class is hereditary and closed under the formation of homomorphic images and direct products, i.e., Bass is a variety of groupoids.

Assuming that B is a nonempty set and T_{B} = (T_{B}, ·) the absolutely free groupoid with the free basis B, we are looking for a canonical groupoid in Bass, i.e., a groupoid R = (R, ∗) with the following properties:

i) B ⊂ R ⊂ T_{B}; ii) tu ∈ R ⇒ t, u ∈ R; iii) tu ∈ R ⇒ t ∗ u = tu iv) R is a free groupoid in Bass with the free basis B.

A “candidate” for the carrier R of the desired groupoid R is the set defined by:

(3.1) R = {x ∈ T_{B} : every biproduct of any subterm of x is a main biproduct}.

The following properties of R are obvious corollaries of (3.1).

Proposition 3.1. a) R satisfies i) and ii).

b) x, y ∈ R ⇒ {xy ∉ R ⇔ xy has a biproduct that is not a main biproduct in T_{B}}.

c) x, y ∈ T_{B} ⇒ {xy ∈ R ⇔ x, y ∈ R & every biproduct of any subterm of xy in T_{B} is a main biproduct}.

Lemma 3.1. For any x ∈ R there is a unique maximal biproduct of x in T_{B} that is a main biproduct.

Proof. Existence. By Proposition 2.2, any x ∈ T_{B} has maximal biproducts in T_{B} and thus any x ∈ R has maximal biproducts in T_{B}. By the definition of R, every biproduct of any subterm of x is a main biproduct and therefore the maximal biproducts of x are main biproducts, too.

Uniqueness. Let x ∈ R and x_{⟨t,u⟩}, x_{⟨v,w⟩} be maximal biproducts of x in T_{B}. Since x ∈ R, both maximal biproducts x_{⟨t,u⟩}, x_{⟨v,w⟩} are main biproducts and we will denote them by x_{t,u}, x_{v,w}. Let x = x_{1}x_{2} ⋯ x_{m} and x = x′_{1}x′_{2} ⋯ x′_{m}x′_{m+1} ⋯ x′_{m+k}, k ⩾ 0, be the representations of x as main biproducts in ⟨t, u⟩ and ⟨v, w⟩, respectively. By the property (2) we have that

m − 1 = χ_{⟨t,u⟩}(x_{1}x_{2} ⋯ x_{m}) = χ_{⟨v,w⟩}(x′_{1}x′_{2} ⋯ x′_{m+k}) = m + k − 1,

which implies that k = 0 and that x_{i} = x′_{i}, for 1 ⩽ i ⩽ m. Therefore, the maximal biproducts x_{t,u} and x_{v,w} are in fact the same biproduct. □

Bellow, for x ∈ R, we will denote by x = x_{1}x_{2} ⋯ x_{m} the maximal main biproduct of x in T_{B} (if it is not stated otherwise).

Lemma 3.2. Let x ∈ R, let the maximal biproduct of x be generated by {t, u}, and let another biproduct of x be generated by {v, w}. Then v, w ∈ ⟨t, u⟩.

Proof. Let x = x_{1} ⋯ x_{m} be the maximal biproduct of x generated by {t, u} and let x = x′_{1} ⋯ x′_{n} be another biproduct of x generated by {v, w}. By Proposition 2.3 we may put t = v. Both biproducts are equal and since x ∈ R, they are main biproducts. By Lemma 3.1, n < m, i.e., m = n + k, k ⩾ 1, so

x′_{1} ⋯ x′_{n} = x_{1} ⋯ x_{n}x_{n+1} ⋯ x_{n+k}.

Using this facts, we obtain that x′_{i} = x_{i} = t, for i ∈ {1, …, n − 1}. Clearly, x′_{n} = w and x_{i} ∈ {t, u}, for i ∈ {n, …, n + k}. Therefore, v, w ∈ ⟨t, u⟩. □

Proposition 3.2. Let x, y ∈ R and the maximal biproducts x = x_{1}x_{2} ⋯ x_{m}, y = y_{1}y_{2} ⋯ y_{n} have generating pairs {t, u}, {v, w}, respectively. Then xy ∈ R if and only if (a) or (b), where

(a) y ∉ ⟨t, u⟩, and for any biproduct of x with a generating pair {t_{1}, u_{1} }, if t_{1}, u_{1} ∈ ⟨v, w⟩, then t_{1} = u_{1} = x

(b) y ∈ ⟨t, u⟩ and t = u = x ∈ B.

Proof. Let xy ∈ R. There are two possible cases for y: 1) y ∉ ⟨t, u⟩ and 2) y ∈ ⟨t, u⟩.

Case 1). Since {v, w} is the generating pair for the biproduct y = y_{1}y_{2} ⋯ y_{n}, and y ∉ ⟨t, u⟩, we should consider the cases when some of the biproducts of x has a generating pair {t_{1}, u_{1}}, such that t_{1}, u_{1} ∈ ⟨v, w⟩. Let x = z_{1}z_{2} ⋯ z_{k} be such a biproduct of x. Then z_{i} ∈ {t_{1}, u_{1}} ⊆ ⟨v, w⟩. The product xy = (z_{1}z_{2} ⋯ z_{k})(y_{1}y_{2} ⋯ y_{n}) will be a main biproduct only if k = 1, i.e., x = z_{1}, and z_{1} = v or z_{1} = w. Since x = z_{1} is generated by {t_{1}, u_{1}}, it follows that t_{1} = u_{1} = x.

Case 2). In this case xy has a biproduct with a generating pair {t, u}. xy is a main biproduct, since xy ∈ R and, therefore m = 1, i.e., x = x_{1}. The maximal biproduct of x is generated by {t, u}, so t = u = x. Moreover, x ∈ B, because if x ∉ B (for example x = ab, i.e., t = u = ab), then the biproduct of xy generated by {a, b} can not be a main biproduct, that contradicts the assumption that xy ∈ R.

For the converse, let (a) or (b) hold. If (b) holds, then it is clear that xy ∈ R. Let (a) holds and suppose xy ∉ R. From 1) we obtain that x ∈ ⟨v, w⟩. There-

fore, there is a biproduct of x with a generating pair {v, w}. By Lemma 3.2 it follows that v, w ∈ ⟨t, u⟩, that contradicts the assumption that y ∉ ⟨t, u⟩. □

Now we define an operation ∗ on R as follows. Let x, y ∈ R, x = x_{1}x_{2} ⋯ x_{m}, y = y_{1}y_{2} ⋯ y_{n} and put

(3.2)  x ∗ y = \begin{cases}xy, & if xy ∈ R \\ x_{1}x_{2} ⋯ x_{m}y_{1}y_{2} ⋯ y_{n}, & if xy ∉ R. \end{cases}

The operation ∗ is well-defined, i.e., R = (R, ∗) is a groupoid. Namely, let x, y ∈ R. If xy ∈ R, then x ∗ y is a uniquelly determined element of R. If xy ∉ R, then z = x_{1}x_{2} ⋯ x_{m}y_{1}y_{2} ⋯ y_{n} is a term of T_{B} that is a main biproduct. Clearly, every biproduct of any subterm of x_{1}x_{2} ⋯ x_{m}y_{1}y_{2} ⋯ y_{n} is a main biproduct. Therefore, by (3.1), z ∈ R. Since x_{1}x_{2} ⋯ x_{m}y_{1}y_{2} ⋯ y_{n} as a maximal biproduct in T_{B} is unique (by Lemma 3.1), it follows that x ∗ y is uniquely determined element of R in the case xy ∉ R. Thus, R = (R, ∗) is a groupoid.

By (3.2) it follows directly that:

1^{∘}. If xy ∈ R, then x, y ∈ R & x ∗ y = xy (i.e., R satisfies ii) and iii)).

2^{∘}. (∀x, y ∈ R) |x ∗ y| = |x| + |y|.

The following three properties of R (3^{∘}–5^{∘}) show that the groupoid R = (R, ∗) is free in Bass with the free basis B.

3^{∘}. R ∈ Bass.

Proof of 3^{∘}. We have to show that every subgroupoid of R generated by two elements is a subsemigroup of R.

For this purpose, let t, u ∈ R and ⟨t, u⟩_{∗} be the subgroupoid of R generated by {t, u}. According to the definition of ∗, any x ∈ ⟨t, u⟩_{∗} is a maximal biproduct with the generating pair {t, u}. Therefore, if x, y, z ∈ ⟨t, u⟩_{∗}, then x = x_{1}x_{2} ⋯ x_{m}, y = y_{1}y_{2} ⋯ y_{n}, z = z_{1}z_{2} ⋯ z_{p} (x_{i}, y_{j}, z_{k} ∈ {t, u}) and by (3.2):

(x ∗ y) ∗ z = x_{1}x_{2} ⋯ x_{m}y_{1}y_{2} ⋯ y_{n}z_{1}z_{2} ⋯ z_{p} = x ∗ (y ∗ z),

i.e., the subgroupoid ⟨t, u⟩_{∗} is a subsemigroup of R. Hence, R ∈ Bass. 4^{∘}. The set of primes in R coincides with B and generates R. (An element a in a groupoid G = (G, ·) is said to be prime in G if and only if

a ≠ xy, for any x, y ∈ G.)

Proof of 4^{∘}. If b ∈ B, then by (3.2) b ≠ x ∗ y, for all x, y ∈ R. Hence, every b ∈ B is prime in R. To show that no element of R ∖ B is prime in R, let x ∈ T_{B} ∖ B be a term belonging to R. Then by (3.1), every biproduct of any subterm of x is a main biproduct, and thus the maximal biproduct of x in T_{B} is a main biproduct. Therefore, x = x_{1}x_{2} ⋯ x_{m}, where m ⩾ 2 (since x ∈ T_{B} ∖ B). Thus, x = x_{1} ∗ (x_{2} ⋯ x_{m}), i.e., x is not prime in R.

Let Q be the subgroupoid of R generated by B, Q = ⟨B⟩_{∗}. We will show that R = Q. Clearly, Q ⊆ R. To show that R ⊆ Q, let x ∈ R. If x ∈ B, then x ∈ ⟨B⟩_{∗} = Q, i.e., (x ∈ R & | x | = 1 ⇒ x ∈ Q).

Suppose that (x ∈ R & | x | ⩽ k ⇒ x ∈ Q) is true. If x ∈ R is such that | x | = k + 1, then x = x_{1}x_{2} in T_{B} and | x_{1} |, | x_{2} | ⩽ k. By the inductive hypothesis we have x_{1}, x_{2} ∈ Q, and since Q is a groupoid, it follows that x = x_{1}x_{2} = x_{1} ∗ x_{2} ∈ Q. Thus, R ⊆ Q. Therefore, R = Q = ⟨B⟩_{∗}.

5^{∘}. If G ∈ Bass and λ : B → G is a mapping, then there is a homomorphism ψ : R → G that extends λ, i.e., ψ(b) = λ(b), for all b ∈ B.

Proof of 5^{∘}. Let φ : T_{B} → G be the homomorphism that extends λ. Denote by ψ the restriction of φ on R (i.e., ψ = φ|_{R}). It suffices to show that

(∀x, y ∈ R) φ(x ∗ y) = φ(x)φ(y).

Let x, y ∈ R. If xy ∈ R, then φ(x ∗ y) = φ(xy) = φ(x)φ(y). If xy ∉ R (i.e., x = x_{1}x_{2} ⋯ x_{m}, y = y_{1}y_{2} ⋯ y_{n}, where x_{i}, y_{j} ∈ {t, u} and m ⩾ 2) then using the fact: (x_{i}, y_{j} ∈ {t, u} ⇒ φ(x_{i}), φ(y_{j}) ∈ {φ(t), φ(u)}) we have

φ(x ∗ y) = φ(x_{1} ⋯ x_{m}y_{1} ⋯ y_{n}) = φ(x_{1}) ⋯ φ(x_{m})φ(y_{1}) ⋯ φ(y_{n})
= [G ∈ Bass] = φ(x_{1} ⋯ x_{m})φ(y_{1} ⋯ y_{n}) = φ(x)φ(y).

So, the conditions i)–iv) at the beginning of this section are fulfilled and thus we proved the following

Theorem 3.1. The groupoid R = (R, ∗), defined by (3.1) and (3.2) is a canonical biassociative groupoid with a free basis B.

Acknowledgements

This work is partly supported by Macedonian Academy of Sciences and Arts within the project “Algebraic structures”. Authors would like to express their sincere gratitude to Professor G. Čupona, the project manager, and special thanks to the referee for the valuable comments and suggestions.

References

[1] S. Ilić, B. Janeva, N. Celakoski, Free Biassociative groupoids, Novi Sad J. Math. 35:1 (2005), 15–23

[2] R. N. McKenzie, G. F. McNulty, W. F. Taylor, Algebras, Latices, Varieties, Vol. I, Wadsworth and Brooks/Cole, Monterey, 1987

[3] G. Čupona, N. Celakoski, S. Ilić, On Monoassociative groupoids, Mat. Bilten 26 (2002), 5–16 [4] G. Čupona, N. Celakoski, B. Janeva, Canonical groupoids with x^{m} · y^{n} = xy, Mat. Bilten 23

(1999), 11–18

Prirodno-matematički fakultet

Skopje, p.f. 162 Macedonia (B. Janeva, V. Celakoska-Jordanova)

    biljana@ii.edu.mk
    vesnacj@iunona.pmf.ukim.edu.mk

Prirodno-matematički fakultet, 18000 Niš, Serbia

    sneska@pmf.ni.ac.yu